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diff --git a/old/38769-t.tex b/old/38769-t.tex new file mode 100644 index 0000000..eda422c --- /dev/null +++ b/old/38769-t.tex @@ -0,0 +1,29060 @@ +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% The Project Gutenberg EBook of A Course of Pure Mathematics, by % +% G. H. (Godfrey Harold) Hardy % +% % +% This eBook is for the use of anyone anywhere at no cost and with % +% almost no restrictions whatsoever. You may copy it, give it away or % +% re-use it under the terms of the Project Gutenberg License included % +% with this eBook or online at www.gutenberg.net % +% % +% % +% Title: A Course of Pure Mathematics % +% Third Edition % +% % +% Author: G. H. (Godfrey Harold) Hardy % +% % +% Release Date: February 5, 2012 [EBook #38769] % +% % +% Language: English % +% % +% Character set encoding: ISO-8859-1 % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK A COURSE OF PURE MATHEMATICS *** +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{38769} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% inputenc: Standard DP encoding. Required. %% +%% %% +%% ifthen: Logical conditionals. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. Required. %% +%% amssymb: Additional mathematical symbols. Required. %% +%% %% +%% alltt: Fixed-width font environment. Required. %% +%% %% +%% footmisc: Extended footnote capabilities. Required. %% +%% %% +%% indentfirst: Indent first word of each sectional unit. Required. %% +%% icomma: Make the comma a decimal separator in math. Required. %% +%% %% +%% calc: Length calculations. Required. %% +%% %% +%% fancyhdr: Enhanced running headers and footers. Required. %% +%% %% +%% graphicx: Standard interface for graphics inclusion. Required. %% +%% caption: Caption customization. Required. %% +%% %% +%% geometry: Enhanced page layout package. Required. %% +%% hyperref: Hypertext embellishments for pdf output. Required. %% +%% %% +%% %% +%% Producer's Comments: %% +%% %% +%% Changes are noted in this file in multiple ways. %% +%% 1. \DPnote{} for in-line `placeholder' notes. %% +%% 2. \DPtypo{}{} for typographical corrections, showing original %% +%% and replacement text side-by-side. %% +%% 3. \DPchg (stylistic uniformity) and \DPmod (modernization). %% +%% 4. [** TN: Note]s for lengthier or stylistic comments. %% +%% %% +%% %% +%% Compilation Flags: %% +%% %% +%% The following behavior may be controlled by boolean flags. %% +%% %% +%% ForPrinting (false by default): %% +%% Compile a screen-optimized PDF file. Set to true for print- %% +%% optimized file (two-sided layout, black hyperlinks). %% +%% %% +%% Modernize (true by default): %% +%% Modernize the mathematical notation (see below for details). %% +%% %% +%% %% +%% PDF pages: 587 (if ForPrinting set to false) %% +%% PDF page size: 5.5 x 8in (non-standard) %% +%% %% +%% Images: 68 pdf diagrams, 1 png image (CUP device) %% +%% %% +%% Summary of log file: %% +%% * One overfull hbox (7.3pt too wide). %% +%% * Three underfull hboxes, four underfull vboxes. %% +%% %% +%% %% +%% Compile History: %% +%% %% +%% January, 2012: (Andrew D. Hwang) %% +%% texlive2007, GNU/Linux %% +%% %% +%% Command block: %% +%% %% +%% pdflatex x3 %% +%% %% +%% %% +%% February 2012: pglatex. %% +%% Compile this project with: %% +%% pdflatex 38769-t.tex ..... THREE times %% +%% %% +%% pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian) %% +%% %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\listfiles +\documentclass[12pt]{book}[2005/09/16] + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\usepackage[latin1]{inputenc}[2006/05/05] + +\usepackage{ifthen}[2001/05/26] %% Logical conditionals + +\usepackage{amsmath}[2000/07/18] %% Displayed equations +\usepackage{amssymb}[2002/01/22] %% and additional symbols + +\usepackage{alltt}[1997/06/16] %% boilerplate, credits, license + + %% extended footnote capabilities +\usepackage[symbol,perpage]{footmisc}[2005/03/17] + +\usepackage{indentfirst}[1995/11/23] +\usepackage{icomma}[2002/03/10] + +\usepackage{calc}[2005/08/06] + +\usepackage{fancyhdr} + +\usepackage{graphicx}[1999/02/16]%% For diagrams +\usepackage[labelformat=empty,textfont=small]{caption}[2007/01/07] + +% Modernize notation: Use square root signs instead of surds, square +% brackets for closed intervals, reverse roles of delta and epsilon. +\newboolean{Modernize} +%% COMMENT the line below to revert to the original notation. +%% Figures 27 (p117.pdf) and 30 (p176.pdf) will automatically work +%% if all the Project Gutenberg diagram files are present in images/. +%% (This switch does not affect typographical corrections.) +\setboolean{Modernize}{true} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%% Interlude: Set up PRINTING (default) or SCREEN VIEWING %%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +% ForPrinting=true (default) false +% Asymmetric margins Symmetric margins +% Black hyperlinks Blue hyperlinks +% Start Preface, ToC, etc. recto No blank verso pages +% +\newboolean{ForPrinting} +%% UNCOMMENT the next line for a PRINT-OPTIMIZED VERSION of the text %% +%\setboolean{ForPrinting}{true} + +%% Initialize values to ForPrinting=false +\newcommand{\ChapterSpace}{} +\newcommand{\Margins}{hmarginratio=1:1} % Symmetric margins +\newcommand{\HLinkColor}{blue} % Hyperlink color +\newcommand{\PDFPageLayout}{SinglePage} +\newcommand{\TransNote}{Transcriber's Note} +%% TN at the beginning +\newcommand{\TransNoteCommon}{% + Minor typographical corrections and presentational changes have been + made without comment. \ifthenelse{\boolean{Modernize}}{Notational + modernizations are listed in the transcriber's note at the end of the + book.}{} All changes are detailed + in the \LaTeX\ source file, which may be downloaded from + \begin{center} + \texttt{www.gutenberg.org/ebooks/\ebook}. + \end{center} + \bigskip +} +%% TN at the end, regarding modernization +\newcommand{\ModernizationNote}{% + The notational modernizations listed below have been made. These + changes may be reverted by commenting out one line in the \LaTeX\ + source file and recompiling the book. + \begin{itemize} + \item Closed intervals are denoted with square brackets, \eg,~$[a, b]$, + instead of round parentheses,~$(a, b)$. + + \item Repeating decimals are denoted with an overline, + \eg,~$.217\Repeat{13}$, instead of with dot + accents,~$.217\dot{1}\dot{3}$. + + \item The roles of $\delta$~and~$\epsilon$ in the definition of limits, + \PageRef{p.}{113}~\textit{ff.}, have been interchanged in accordance + with modern convention: ``For every $\epsilon > 0$, there exists a + $\delta > 0$ such that~\dots''. + \end{itemize} +} +%% TN at the end, regarding change of formula in Figure 16 +\newcommand{\ChangeNote}{% + In Example~11, \PageRef{p.}{57}~\textit{ff.}, the text refers to the + formula + \[ + y = \left\{ + \begin{array}{@{}cl} + \sqrt{(1 + p^{2})(1 + q^{2})} & \text{if $x = p/q$ in lowest terms,} \\ + x & \text{if $x$~is irrational.} + \end{array} + \right. + \] + The computer-generated \Fig{16} instead depicts the formula + \[ + y = \left\{ + \begin{array}{@{}cl} + \sqrt{(10 + p^{2})(10 + q^{2})} & \text{if $x = p/q$ in lowest terms,} \\ + x & \text{if $x$~is irrational,} + \end{array} + \right. + \] + which exhibits the same mathematical behavior, but better matches + the hand-drawn diagram in the original. +} + +\newcommand{\TransNoteText}{% + \TransNoteCommon + + This PDF file is optimized for screen viewing, but may easily be + recompiled for printing. Please consult the preamble of the \LaTeX\ + source file for instructions. +} +%% Re-set if ForPrinting=true +\ifthenelse{\boolean{ForPrinting}}{% + \renewcommand{\ChapterSpace}{\vspace*{1in}} + \renewcommand{\Margins}{hmarginratio=2:3} % Asymmetric margins + \renewcommand{\HLinkColor}{black} % Hyperlink color + \renewcommand{\PDFPageLayout}{TwoPageRight} + \renewcommand{\TransNoteText}{% + \TransNoteCommon + + This PDF file is optimized for printing, but may easily be + recompiled for screen viewing. Please consult the preamble of the + \LaTeX\ source file for instructions. + } + \newcommand{\longpage}{} +}{% If ForPrinting=false, don't skip to recto + \renewcommand{\cleardoublepage}{\clearpage} + \newcommand{\longpage}{\enlargethispage{\baselineskip}} +} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%%%% End of PRINTING/SCREEN VIEWING code; back to packages %%%% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +\ifthenelse{\boolean{ForPrinting}}{% + \setlength{\paperwidth}{8.5in}% + \setlength{\paperheight}{11in}% + \usepackage[body={5.25in,8.5in},\Margins]{geometry}[2002/07/08] +}{% + \setlength{\paperwidth}{5.5in}% + \setlength{\paperheight}{8in}% + \usepackage[body={5.25in,6.9in},\Margins,includeheadfoot]{geometry}[2002/07/08] +} + +\providecommand{\ebook}{00000} % Overridden during white-washing +\usepackage[pdftex, + hyperfootnotes=false, + pdftitle={The Project Gutenberg eBook \#\ebook: A Course of Pure Mathematics}, + pdfauthor={Godfrey Harold Hardy}, + pdfkeywords={Andrew D. Hwang, Brenda Lewis, + Project Gutenberg Online Distributed Proofreading Team, + Internet Archive/American Libraries}, + pdfstartview=Fit, % default value + pdfstartpage=1, % default value + pdfpagemode=UseNone, % default value + bookmarks=true, % default value + linktocpage=false, % default value + pdfpagelayout=\PDFPageLayout, + pdfdisplaydoctitle, + pdfpagelabels=true, + bookmarksopen=true, + bookmarksopenlevel=1, + colorlinks=true, + linkcolor=\HLinkColor]{hyperref}[2007/02/07] + +%%%% Fixed-width environment to format PG boilerplate %%%% +\newenvironment{PGtext}{% +\begin{alltt} +\fontsize{9.2}{10.5}\ttfamily\selectfont}% +{\end{alltt}} + +%%%% Global style parameters %%%% +% No hrule in page header +\renewcommand{\headrulewidth}{0pt} +\setlength{\headheight}{15pt} + +% Loosen horizontal spacing +\setlength{\emergencystretch}{1.5em} + +% Local spacing coercion +\newcommand{\Loosen}{\spaceskip 0.375em plus 0.75em minus 0.25em} +% Used only once, to coax a wide display into the text block +\newcommand{\Squeeze}[2][0.98]{\scalebox{#1}[1]{#2}} + +% Allow \quad to compress a bit +\let\oldquad=\quad +\renewcommand{\quad}{\oldquad\hspace{0pt minus 3pt}} + +% Misc spacing parameters +\setlength{\multlinegap}{2\parindent} +\newcommand{\Medskip}{\vspace{0pt plus 0.5\baselineskip}} +% "Scratch pad" for length calculations +\newlength{\TmpLen} + +%% Parametrized vertical space %% +\newcommand{\Strut}[1][12pt]{\rule{0pt}{#1}} + +%%%% Corrections and in-line transcriber's notes %%%% +% In-line notes +\newcommand{\DPnote}[1]{} +% Errors +\newcommand{\DPtypo}[2]{#2} + +%%%% Notational modernizations %%%% +% ** If \epsilon -> \varepsilon, figures p117 and p176 must be recompiled +\ifthenelse{\boolean{Modernize}}{% +% Stylistic changes made for clarity or consistency + \newcommand{\DPchg}[2]{#2} + \newcommand{\Add}[1]{#1} + +% Modernize notation + \newcommand{\DPmod}[2]{#2} + +% ** Incarnations of \sqrt; see below for significance + \newcommand{\sqrtp}[2][\ ]{\sqrt[#1]{#2}} + \newcommand{\sqrtb}[2][\ ]{\sqrt[#1]{#2}} + \newcommand{\sqrtbr}[2][\ ]{\sqrt[#1]{#2}} + \newcommand{\bigsqrt}[2][\ ]{\sqrt[#1]{#2}} + \newcommand{\bigsqrtb}[2][\ ]{\sqrt[#1]{#2}} + \newcommand{\bigsqrtp}[2][\ ]{\sqrt[#1]{#2}} + +% Exchange delta, epsilon in the definition of limits + \newcommand{\DELTA}{\epsilon} + \newcommand{\EPSILON}{\delta} + +% Add visual delimiters to large integers/long decimals + \newcommand{\MC}{,}% "Math comma" + \newcommand{\MS}{\,}% "Math space" +}{% Modernize = false + \newcommand{\DPchg}[2]{#1} + \newcommand{\Add}[1]{} + \newcommand{\DPmod}[2]{#1} + % Use surd sign... + \let\oldsqrt=\sqrt% + \renewcommand*{\sqrt}[2][\ ]{\oldsqrt[#1]{\vphantom{|}}#2} + % ... with parentheses or curly braces around radicand + \newcommand{\sqrtp}[2][\ ]{\sqrt[#1]{(#2)}} + \newcommand{\sqrtb}[2][\ ]{\sqrt[#1]{\{#2\}}} + \newcommand{\sqrtbr}[2][\ ]{\sqrt[#1]{\,[#2]}} + \newcommand{\bigsqrt}[2][\ ]{\oldsqrt[#1]{\vphantom{#2}}#2} + \newcommand{\bigsqrtb}[2][\ ]{\oldsqrt[#1]{\vphantom{\bigg|}}\left\{#2\right\}} + \newcommand{\bigsqrtp}[2][\ ]{\oldsqrt[#1]{\vphantom{#2}}\!\!\left(#2\right)} + + \newcommand{\DELTA}{\delta} + \newcommand{\EPSILON}{\epsilon} +% Don't add visual delimiters to large integers/long decimals + \newcommand{\MC}{} + \newcommand{\MS}{} +} +%% End of modernization code %% + +%%%% Running heads %%%% +\newcommand{\FlushRunningHeads}{% + \clearpage + \pagestyle{fancy} + \fancyhf{} + \cleardoublepage + \thispagestyle{empty} + + \ifthenelse{\boolean{ForPrinting}} + {\fancyhead[RO,LE]{\thepage}} + {\fancyhead[R]{\thepage}} +} + +% ** \Chapter{X} uses optional argument to set separate running heads +\newcommand{\SetCenterHeads}[2][]{% + \ifthenelse{\equal{#1}{}}{% + \fancyhead[C]{{\footnotesize #2}}% + }{% + \fancyhead[CE]{{\footnotesize #1}}% + \fancyhead[CO]{{\footnotesize #2}}% + }% +} + +\newcommand{\SetCornerHeads}[1]{% + \ifthenelse{\boolean{ForPrinting}}{% + \fancyhead[RE]{[\ChapNo}% + \fancyhead[LO]{#1]}% + }{% + \fancyhead[L]{[\ChapNo\,:\,#1]}% + }% +} + +\newcommand{\BookMark}[3][]{% + \phantomsection% + \ifthenelse{\equal{#1}{}}{% + \pdfbookmark[#2]{#3}{#3}% + }{% + \pdfbookmark[#2]{#3}{#1}% + }% +} + +%%%% Major document divisions %%%% +\newcommand{\FrontMatter}{% + \cleardoublepage + \frontmatter + \BookMark{-1}{Front Matter} +} +\newcommand{\PGBoilerPlate}{% + \pagenumbering{Alph} + \pagestyle{empty} + \BookMark{0}{PG Boilerplate} +} +\newcommand{\MainMatter}{% + \FlushRunningHeads + \mainmatter + \BookMark{-1}{Main Matter} +} +\newcommand{\BackMatter}{% + \FlushRunningHeads + \backmatter + \BookMark{-1}{Back Matter} +} +\newcommand{\PGLicense}{% + \FlushRunningHeads + \pagenumbering{Roman} + \BookMark{-1}{PG License} + \SetCenterHeads{License} +} + +\newcommand{\TranscribersNote}[2][]{% + \begin{minipage}{0.85\textwidth} + \small + \BookMark[#1]{0}{Transcriber's Note} + \subsection*{\centering\normalfont\scshape\normalsize\TransNote} + #2 + \end{minipage} +} + +%%%% Table of Contents %%%% +% Misc. macros for internal use +\newcounter{tocentry} +\setcounter{tocentry}{0} +\newcommand{\ToCAnchor}{} +\newcommand{\SectPageLine}{% + \parbox{\textwidth}{\scriptsize SECT.\hfill PAGE}\\% +} + +% Contents heading +\newcommand{\Contents}{% + \FlushRunningHeads + \setlength{\headheight}{15pt} + \SetCenterHeads{CONTENTS} + \BookMark{0}{Contents} + \Section{CONTENTS} +} + +% Chapter entries +\newcommand{\ToCChap}[2]{% + \subsection*{\centering\normalfont\small #1} + \subsubsection*{\centering\normalfont\footnotesize #2} +} + +% Section(s) entries +% ** Macro discards third argument (original page number) +\newcommand{\ToCSect}[4]{% + \noindent\Strut% Issue vertical space to see if we'll be set on a new page +% If #1 is empty, generate our own label, and update \ToCAnchor + \ifthenelse{\equal{#1}{}}{% "Miscellaneous examples" line + \stepcounter{tocentry}\label{toc:special\thetocentry}% + \ifthenelse{\not\equal{\pageref{toc:special\thetocentry}}{\ToCAnchor}}{% + \renewcommand{\ToCAnchor}{\pageref{toc:special\thetocentry}}% + \SectPageLine% + }{}% + }{% else use #1 to generate label, and update \ToCAnchor + \label{toc:#1}% + \ifthenelse{\not\equal{\pageref{toc:#1}}{\ToCAnchor}}{% + \renewcommand{\ToCAnchor}{\pageref{toc:#1}}% + \SectPageLine% + }{}% + }% + \settowidth{\TmpLen}{999--999. }% Maximum heading width + % ** Width (2em) must match \ToCPage width below + \parbox[b]{\textwidth-2em}{\Strut\small\hangindent1.5\TmpLen% + \makebox[\TmpLen][l]{#1}#2\ \dotfill}\ToCPage{#4}% +} + +% Appendix entries +% ** Macro discards third argument (original page number) +\newcommand{\ToCApp}[3]{% +\noindent\parbox[b]{\textwidth-2em}{% + \small\textsc{Appendix}~\makebox[2em][l]{#1.} #2\ \dotfill}\ToCPage{appendix:#1}% +} + +% Page numbers +\newcommand{\ToCPage}[1]{\makebox[2em][r]{\small\pageref{#1}}} +% ** Approximate; refers to original page separators, no hyperlink +\newcommand{\PgNo}[2][]{% + \ifthenelse{\equal{#1}{}}{% + \pageref*{pg:#2}% + }{% + \pageref{page:#2}% + }% +} + +\newenvironment{ToCPar}{\begin{quote}\footnotesize}{\end{quote}} + +%%%% Document Sectioning %%%% +\newcommand{\RunInHead}[1]{\paragraph*{\indent #1}} +\newcommand{\ChapNo}{} + +\newcommand{\Preface}[1]{% + \section*{\centering\normalfont#1} +} + +% \Chapter[Running head]{Number}{Heading title} +\newcommand{\Chapter}[3][]{% + \FlushRunningHeads + \phantomsection + \label{chapter:#2} + \BookMark{0}{Chapter #2}% + \renewcommand{\ChapNo}{#2} + \thispagestyle{plain} +% + \ifthenelse{\equal{#1}{}}{% + \ifthenelse{\equal{#2}{IV}}{% Chapters IV, X get asymmetric heading + \SetCenterHeads[LIMITS OF FUNCTIONS OF A] + {POSITIVE INTEGRAL VARIABLE}% + }{% Not Chapter IV + \ifthenelse{\equal{#2}{X}}{% + \SetCenterHeads[THE GENERAL THEORY OF THE LOGARITHMIC,] + {EXPONENTIAL, AND CIRCULAR FUNCTIONS}% + }{% Not Chapter X, but running head manually specified + \SetCenterHeads{#3}% + }% + }% + }{% Otherwise, use chapter title as running head + \SetCenterHeads{#1}% + } + \ChapterSpace + \section*{\centering CHAPTER #2} + \subsection*{\centering\normalfont\small #3} +} + +\newcommand{\Section}[1]{\subsection*{\centering\normalfont #1}} + +\newcommand{\Appendix}[3]{% + \FlushRunningHeads + \label{appendix:#1} + \BookMark{0}{Appendix #1} + \renewcommand{\ChapNo}{A.#2} + \thispagestyle{plain} +% + \SetCenterHeads{APPENDIX #1} + \ChapterSpace + \section*{\centering\normalfont APPENDIX #1} + \subsection*{\centering\normalfont\scshape #2} + \subsubsection*{\centering\normalfont\itshape #3} +} + +% Numbered sections; use dedicated counter (not macro arg.) to create labels +\newcounter{ParNo} +\newcommand{\Paragraph}[1]{% + \RunInHead{#1}% + \stepcounter{ParNo}\phantomsection\label{par:\theParNo}% + \SetCornerHeads{\theParNo}% +} + +\newcommand{\Par}[1]{\RunInHead{\normalfont\itshape #1}} + +%%%% Other semantic units %%%% +% Numbered item +\newcommand{\Item}[1]{\makebox[1.5em][l]{\normalfont\upshape#1\Strut}} +% Parenthesized item +\newcommand{\Itemp}[1]{% + \ifmmode\makebox[2.25em][l]{\normalfont\upshape#1}% + \else\makebox[2.25em][l]{\normalfont\upshape#1\Strut}% + \fi% +} +\newcommand{\SubItem}[1]{\quad\Itemp{#1}} +\newcommand{\Hang}[1][6em]{\hangindent#1} + +% Template for definitions, theorems, corollaries +\newenvironment{MyEnvt}[2]{% + \Medskip\par% + \ifthenelse{\equal{#1}{}}{% + \textsc{#2.} + }{% + \textsc{#2 #1} + }% + \itshape\ignorespaces +}{\normalfont\Medskip} + +% Document-level environments +\newenvironment{Theorem}[1][]{\begin{MyEnvt}{#1}{Theorem}}{\end{MyEnvt}} +\newenvironment{Corollary}[1][]{\begin{MyEnvt}{#1}{Corollary}}{\end{MyEnvt}} +\newenvironment{Cor}[1][]{\begin{MyEnvt}{#1}{Cor}}{\end{MyEnvt}} +\newenvironment{Definition}[1][]{\begin{MyEnvt}{#1}{Definition}}{\end{MyEnvt}} +\newenvironment{Definitions}[1][]{\begin{MyEnvt}{#1}{Definitions}}{\end{MyEnvt}} + +% Miscellaneous italicized constructs +\newenvironment{Construction}[1][\Medskip]{#1\itshape}{\normalfont\Medskip} +\newenvironment{Defn}{\itshape\ignorespaces}{\normalfont} +\newenvironment{Result}{\itshape\ignorespaces}{\par\Medskip\normalfont} + +\newenvironment{ParTheorem}[1]{\RunInHead{#1}\itshape}{\normalfont\Medskip} + +% "Examples" sections; auto-number using dedicated counter +\newcounter{ExNo} +\newenvironment{Examples}[1]{% + \small\ifthenelse{\not\equal{#1}{}}{% + \RunInHead{Examples #1}% + \stepcounter{ExNo}\phantomsection\label{examples:\roman{ExNo}}% + }{ + \phantomsection\label{misc:\ChapNo}% + }% +}{\par\Medskip\normalsize} + +% Passages of small text having no special run-in heading +\newenvironment{Remark}{\Medskip\par\small}{\normalsize\Medskip} + +\newcommand{\Signature}[2]{% + \null\hfill#1\hspace*{\parindent}\\ + \hspace*{\parindent}#2% +} + +% Equation-like entities with \Item-like numbering +\newcommand{\CenterLine}[3][\qquad]{% + \[ + \makebox[\textwidth]{\indent#2 \hfill #3 #1 \hfill} + \] +} + +% Same, but with "Df." tag at right margin +\newcommand{\CenterDef}[3][]{% + \[ + \makebox[\textwidth]{\indent#2 \hfill #3\qquad \hfill\text{Df.}\rlap{#1}\quad} + \] +} + +\newcommand{\MathTrip}[1]{% + \pagebreak[0]% + \hfil\allowbreak\null\nobreak\hfill\nobreak\mbox{(\textit{Math.\ Trip.}\ #1)}% + \pagebreak[1]% +} + +%%%% Misc. textual macros %%%% +\newcommand{\First}[1]{\textsc{#1}} +\newcommand{\continued}{{\normalfont\textit{continued}}} +\newcommand{\Emph}[1]{{\textbf{\upshape#1}}} +\newcommand{\Topic}[1]{\textbf{\upshape#1}} + +\newcommand{\eg}{\textit{e.g.}} +\newcommand{\ie}{\textit{i.e.}} +\newcommand{\Ie}{\textit{I.e.}} + +\renewcommand{\(}{{\upshape(}} +\renewcommand{\)}{{\upshape)}} + +% Fixed-width lines on the copyright page +\newcommand{\SetLine}[2][\TmpLen]{\makebox[#1][s]{#2}} + +%%%% Illustrations %%%% +% Inclusion wrapper +\newcommand{\Graphic}[3][pdf]{\includegraphics[width=#2]{./images/#3.#1}} + +%\Figure[width]{Figure number}{File name} +\newcommand{\Figure}[3][0.8\textwidth]{% + \begin{figure}[hbt!] + \centering + \Graphic{#1}{#3} + \caption{Fig.~#2.} + \label{fig:#2} + \end{figure}\ignorespaces% +} + +% \Figures{width1}{fig1}{graphic1}{width2}{fig2}{graphic2} +\newcommand{\Figures}[6]{% + \begin{figure}[hbt!] + \begin{minipage}{0.5\textwidth} + \centering + \Graphic{#1}{#3} + \caption{Fig.~#2.} + \label{fig:#2} + \end{minipage}% + \begin{minipage}{0.5\textwidth} + \centering + \Graphic{#4}{#6} + \caption{Fig.~#5.} + \label{fig:#5} + \end{minipage}% + \end{figure}\ignorespaces% +} + +%%%% Cross-referencing %%%% + +% Original page separators; generated numbers used in the ToC +\newcommand{\PageSep}[1]{\PageLabel[pg]{#1}\ignorespaces} + +%% Anchors +\newcommand{\PageLabel}[2][page]{\phantomsection\label{#1:#2}} + +% Code stub; cross-referencing eqn numbers not feasible +\newcommand{\Tag}[1]{\tag*{#1}} + +%% Links +\newcommand{\PageRef}[2]{\hyperref[page:#2]{#1~\pageref*{page:#2}}} + +\newcommand{\Fig}[1]{\hyperref[fig:#1]{Fig.~#1}} + +% Chapter/appendix reference +\newcommand{\Ref}[2]{% + \ifthenelse{\equal{#1}{Appendix}}{% + \hyperref[appendix:#2]{#1~#2}% + }{% + \ifthenelse{\not\equal{#1}{}}{% + \hyperref[chapter:#2]{#1~#2}% + }{% + \hyperref[chapter:#2]{#2}% + }% + }% +} + +% Paragraph reference +\newcommand{\SecNo}[2][]{% + \ifthenelse{\equal{#1}{}}{% + \hyperref[par:#2]{{\normalfont\upshape #2}}% + }{% + \hyperref[par:#2]{#1\:{\normalfont\upshape #2}}% + }% +} + +% "Examples" section reference +\newcommand{\Ex}[1]{% + \hyperref[examples:#1]{Ex.~\textsc{#1}}% +} +\newcommand{\Exs}[2][Exs.~]{% + \hyperref[examples:#2]{#1\textsc{#2}}% +} +\newcommand{\MiscEx}[1]{\hyperref[misc:#1]{Misc.~Ex.}} +\newcommand{\MiscExs}[1]{\hyperref[misc:#1]{Misc.~Exs.}} + +% Code stub; no hyperlinking +\newcommand{\Eq}[1]{{\upshape#1}} + +%%%% Typographical conveniences %%%% +\newcommand{\Inum}[1]{{\upshape#1}} + +\newcommand{\ia}{\textit{a}} +\newcommand{\ib}{\textit{b}} +\newcommand{\ic}{\textit{c}} +\newcommand{\id}{\textit{d}} + +\newcommand{\TEntry}[1]{\multicolumn{1}{c}{#1}} + +%%%% Misc. mathematical macros %%%% +\newcommand{\ds}{\displaystyle} +\renewcommand{\leq}{\leqq} +\renewcommand{\geq}{\geqq} + +\newcommand{\bigint}[1][1.3]{\scalebox{#1}{$\ds\int$}} + +\newcommand{\btw}{\mathbin{)\kern-5pt(}} +\newcommand{\dd}{\partial} +\newcommand{\tsum}{{\textstyle\sum}} + +\newcommand{\Mu}{\mathrm{M}} + +% Duplicate Hardy's notation +\renewcommand{\limsup}{\varlimsup} +\renewcommand{\liminf}{\varliminf} + +%% Named operators +\DeclareMathOperator{\ArcCos}{arc\,cos} +\DeclareMathOperator{\ArcCosec}{arc\,cosec} +\DeclareMathOperator{\ArcCot}{arc\,cot} +\DeclareMathOperator{\ArcSec}{arc\,sec} +\DeclareMathOperator{\ArcSin}{arc\,sin} +\DeclareMathOperator{\ArcTan}{arc\,tan} +\DeclareMathOperator{\cosec}{cosec} +\DeclareMathOperator{\sech}{sech} +\DeclareMathOperator{\cosech}{cosech} + +\DeclareMathOperator{\argcosh}{arg\,cosh} +\DeclareMathOperator{\argcoth}{arg\,coth} +\DeclareMathOperator{\argsinh}{arg\,sinh} +\DeclareMathOperator{\argtanh}{arg\,tanh} + +\newcommand{\arccosec}{\ArcCosec} +\newcommand{\arccot}{\ArcCot} +\newcommand{\arcsec}{\ArcSec} + +\renewcommand{\arccos}{\ArcCos} +\renewcommand{\arcsin}{\ArcSin} +\renewcommand{\arctan}{\ArcTan} + +\DeclareMathOperator{\Cis}{Cis} +\DeclareMathOperator{\Log}{Log} +\DeclareMathOperator{\sgn}{\textit{sgn}\,} +\DeclareMathOperator{\am}{am} + +\DeclareMathOperator{\Real}{\mathbf{R}} +\DeclareMathOperator{\Imag}{\mathbf{I}} +\renewcommand{\Re}{\Real} +\renewcommand{\Im}{\Imag} + +% Handle degree symbols and centered dots as Latin-1 characters +\DeclareInputText{176}{\ifmmode{{}^\circ}\else\textdegree\fi} +\DeclareInputText{183}{\ifmmode\cdot\else\textperiodcentered\fi} + +% Repeating decimals +\newcommand{\Repeat}[1]{\overline{#1\vphantom{|}}} +% Line segments +\newcommand{\Seg}[1]{\overline{#1\vphantom{P'}}} +% Wide ellipsis (used only once) +\newcommand{\DotRow}[1]{\makebox[#1][c]{\dotfill}} + +%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%% +\begin{document} +\FrontMatter +%%%% PG BOILERPLATE %%%% +\PGBoilerPlate +\begin{center} +\begin{minipage}{\textwidth} +\small +\begin{PGtext} +The Project Gutenberg EBook of A Course of Pure Mathematics, by +G. H. (Godfrey Harold) Hardy + +This eBook is for the use of anyone anywhere at no cost and with +almost no restrictions whatsoever. You may copy it, give it away or +re-use it under the terms of the Project Gutenberg License included +with this eBook or online at www.gutenberg.net + + +Title: A Course of Pure Mathematics + Third Edition + +Author: G. H. (Godfrey Harold) Hardy + +Release Date: February 5, 2012 [EBook #38769] + +Language: English + +Character set encoding: ISO-8859-1 + +*** START OF THIS PROJECT GUTENBERG EBOOK A COURSE OF PURE MATHEMATICS *** +\end{PGtext} +\end{minipage} +\end{center} +\clearpage + +%%%% Credits and transcriber's note %%%% +\begin{center} +\begin{minipage}{\textwidth} +\begin{PGtext} +Produced by Andrew D. Hwang, Brenda Lewis, and the Online +Distributed Proofreading Team at http://www.pgdp.net (This +file was produced from images generously made available +by The Internet Archive/American Libraries.) +\end{PGtext} +\end{minipage} +\vfill +\TranscribersNote{\TransNoteText} +\end{center} +%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%% +\PageSep{i} +\cleardoublepage +\pagenumbering{roman} +\null\vfill +\begin{center} +\bfseries +\LARGE A COURSE \\[\baselineskip] +OF \\[\baselineskip] +\Huge PURE MATHEMATICS +\end{center} +\vfill +\clearpage +\PageSep{ii} +\null\vfill +\begin{center} +CAMBRIDGE UNIVERSITY PRESS \\ +C. F. CLAY, \textsc{Manager} \\ +LONDON: FETTER LANE, E.C. 4 \\ +% [Publisher's device] +\Graphic[png]{1in}{device} + +\small +\settowidth{\TmpLen}{TOKYO: MARUZEN-KABUSHIKI-KAISHA\quad}% +\SetLine{NEW YORK : THE MACMILLAN CO.} \\ +\SetLine{$\left.\kern -1pt%\setlength{\arraycolsep}{0pt} +\begin{array}{@{}l@{}} +\text{BOMBAY} \\ +\text{CALCUTTA} \\ +\text{MADRAS} +\end{array} +\right\}$ +MACMILLAN AND CO., \textsc{Ltd.}} \\ +\SetLine{TORONTO : THE MACMILLAN CO. OF} \\ +\SetLine{\hfil CANADA, \textsc{Ltd.}\hfil} \\ +\SetLine{TOKYO : MARUZEN-KABUSHIKI-KAISHA} \\[24pt] + +\footnotesize +ALL RIGHTS RESERVED +\end{center} +\vfill +\clearpage +\PageSep{iii} +\begin{center} +\bfseries +\LARGE A COURSE \\[12pt] +\large OF \\[12pt] +\Huge PURE MATHEMATICS +\par +\vfil + +\normalfont +\normalsize BY \\ +\Large G.~H. HARDY, M.A., F.R.S. +\bigskip + +\footnotesize +FELLOW OF NEW COLLEGE \\[4pt] +SAVILIAN PROFESSOR OF GEOMETRY IN THE UNIVERSITY \\[4pt] +OF OXFORD \\[4pt] +LATE FELLOW OF TRINITY COLLEGE, CAMBRIDGE +\vfil\vfil +\textsf{THIRD EDITION} +\vfil\vfil\vfil +\Large Cambridge \\ +at the University Press \\ +1921 +\end{center} +\clearpage +\PageSep{iv} +\null\vfill +\begin{center} +\textit{First Edition} 1908 \\ +\textit{Second Edition} 1914 \\ +\textit{Third Edition} 1921 +\end{center} +\vfill\clearpage +\PageSep{v} + + +\Preface{PREFACE TO THE THIRD EDITION} + +\First{No} extensive changes have been made in this edition. The most +important are in \SecNo[§§]{80}--\SecNo{82}, which I have rewritten in accordance +with suggestions made by Mr~S.~Pollard. + +The earlier editions contained no satisfactory account of the +genesis of the circular functions. I have made some attempt to +meet this objection in \SecNo[§]{158} and \Ref{Appendix}{III}\@. \Ref{Appendix}{IV} is also +an addition. + +It is curious to note how the character of the criticisms I have +had to meet has changed. I was too meticulous and pedantic for +my pupils of fifteen years ago: I am altogether too popular for the +Trinity scholar of to-day. I need hardly say that I find such +criticisms very gratifying, as the best evidence that the book has +to some extent fulfilled the purpose with which it was written. + +\Signature{G.~H.~H.}{\textit{August} 1921} + + +\Preface{EXTRACT FROM THE PREFACE TO +THE SECOND EDITION} + +\First{The} principal changes made in this edition are as follows. +I have inserted in \Ref{Chapter}{I} a sketch of Dedekind's theory +of real numbers, and a proof of Weierstrass's theorem concerning +points of condensation; in \Ref{Chapter}{IV} an account of `limits of +indetermination' and the `general principle of convergence'; in +\Ref{Chapter}{V} a proof of the `Heine-Borel Theorem', Heine's theorem +concerning uniform continuity, and the fundamental theorem +concerning implicit functions; in \Ref{Chapter}{VI} some additional +matter concerning the integration of algebraical functions; and +in \Ref{Chapter}{VII} a section on differentials. I have also rewritten +in a more general form the sections which deal with the definition +of the definite integral. In order to find space for these +insertions I have deleted a good deal of the analytical geometry +and formal trigonometry contained in Chapters II~and~III of +the first edition. These changes have naturally involved a +large number of minor alterations. + +\Signature{G.~H.~H.}{\textit{October} 1914} +\PageSep{vi} + +\Preface{EXTRACT FROM THE PREFACE TO THE +FIRST EDITION} + +\First{This} book has been designed primarily for the use of first +year students at the Universities whose abilities reach or +approach something like what is usually described as `scholarship +standard'. I hope that it may be useful to other classes of +readers, but it is this class whose wants I have considered first. +It is in any case a book for mathematicians: I have nowhere +made any attempt to meet the needs of students of engineering +or indeed any class of students whose interests are not primarily +mathematical. + +I regard the book as being really elementary. There are +plenty of hard examples (mainly at the ends of the chapters): to +these I have added, wherever space permitted, an outline of the +solution. But I have done my best to avoid the inclusion of +anything that involves really difficult ideas. For instance, I make +no use of the `principle of convergence': uniform convergence, +double series, infinite products, are never alluded to: and I prove +no general theorems whatever concerning the inversion of limit-operations---I +never even define $\dfrac{\dd^{2} f}{\dd x\, \dd y}$ and $\dfrac{\dd^{2} f}{\dd y\, \dd x}$. In the last two +chapters I have occasion once or twice to integrate a power-series, +but I have confined myself to the very simplest cases and given +a special discussion in each instance. Anyone who has read this +book will be in a position to read with profit Dr~Bromwich's +\textit{Infinite Series}, where a full and adequate discussion of all these +points will be found. + +\Signature{}{\textit{September} 1908} +\PageSep{vii} + +\Contents + +\ToCChap{CHAPTER I}{REAL VARIABLES} + +% SECT. PAGE + +\ToCSect{1--2.}{Rational numbers}{1}{par:1} + +\ToCSect{3--7.}{Irrational numbers}{3}{par:3} + +\ToCSect{8.}{Real numbers}{13}{par:8} + +\ToCSect{9.}{Relations of magnitude between real numbers}{15}{par:9} + +\ToCSect{10--11.}{Algebraical operations with real numbers}{17}{par:10} + +\ToCSect{12.}{The number~$\sqrt{2}$}{19}{par:12} + +\ToCSect{13--14.}{Quadratic surds}{19}{par:13} + +\ToCSect{15.}{The continuum}{23}{par:15} + +\ToCSect{16.}{The continuous real variable}{26}{par:16} + +\ToCSect{17.}{Sections of the real numbers. Dedekind's Theorem}{27}{par:17} + +\ToCSect{18.}{Points of condensation}{29}{par:18} + +\ToCSect{19.}{Weierstrass's Theorem}{30}{par:19} + +\ToCSect{}{Miscellaneous Examples}{31}{misc:I} + +\begin{ToCPar} +Decimals,~\PgNo{1}. Gauss's Theorem,~\PgNo{6}. Graphical solution of quadratic +equations,~\PgNo{20}. Important inequalities,~\PgNo{32}. Arithmetical and geometrical +means,~\PgNo{32}. Schwarz's Inequality,~\PgNo{33}. Cubic and other surds,~\PgNo{34}. +Algebraical numbers,~\PgNo{36}. +\end{ToCPar} + + +\ToCChap{CHAPTER II}{FUNCTIONS OF REAL VARIABLES} + +\ToCSect{20.}{The idea of a function}{38}{par:20} + +\ToCSect{21.}{The graphical representation of functions. Coordinates}{41}{par:21} + +\ToCSect{22.}{Polar coordinates}{43}{par:22} + +\ToCSect{23.}{Polynomials}{44}{par:23} + +\ToCSect{24--25.}{Rational functions}{47}{par:24} + +\ToCSect{26--27.}{Algebraical functions}{49}{par:26} + +\ToCSect{28--29.}{Transcendental functions}{52}{par:28} + +\ToCSect{30.}{Graphical solution of equations}{58}{par:30} + +\ToCSect{31.}{Functions of two variables and their graphical representation}{59}{par:31} +\PageSep{viii} +\ToCSect{32.}{Curves in a plane}{60}{par:32} + +\ToCSect{33.}{Loci in space}{61}{par:33} + +\ToCSect{}{Miscellaneous Examples}{65}{misc:II} + +\begin{ToCPar} +Trigonometrical functions,~\PgNo{53}. Arithmetical functions,~\PgNo{55}. Cylinders,~\PgNo{62}. +Contour maps,~\PgNo{62}. Cones,~\PgNo{63}. Surfaces of revolution,~\PgNo{63}. Ruled +surfaces,~\PgNo{64}. Geometrical constructions for irrational numbers,~\PgNo{66}. +Quadrature of the circle,~\PgNo{68}. +\end{ToCPar} + + +\ToCChap{CHAPTER III}{COMPLEX NUMBERS} + +\ToCSect{34--38.}{Displacements}{69}{par:34} + +\ToCSect{39--42.}{Complex numbers}{78}{par:39} + +\ToCSect{43.}{The quadratic equation with real coefficients}{81}{par:43} + +\ToCSect{44.}{Argand's diagram}{84}{par:44} + +\ToCSect{45.}{\DPchg{de~Moivre's}{De~Moivre's} Theorem}{86}{par:45} + +\ToCSect{46.}{Rational functions of a complex variable}{88}{par:46} + +\ToCSect{47--49.}{Roots of complex numbers}{98}{par:47} + +\ToCSect{}{Miscellaneous Examples}{101}{misc:III} + +\begin{ToCPar} +Properties of a triangle,~\PgNo{90},~\PgNo{101}. Equations with complex coefficients,~\PgNo{91}. +Coaxal circles,~\PgNo{93}. Bilinear and other transformations,~\PgNo{94},~\PgNo{97},~\PgNo{104}. +Cross ratios,~\PgNo{96}. Condition that four points should be concyclic,~\PgNo{97}. +Complex functions of a real variable,~\PgNo{97}. Construction of regular polygons +by Euclidean methods,~\PgNo{100}. Imaginary points and lines,~\PgNo{103}. +\end{ToCPar} + + +\ToCChap{CHAPTER IV}{LIMITS OF FUNCTIONS OF A POSITIVE INTEGRAL VARIABLE} + +\ToCSect{50.}{Functions of a positive integral variable}{106}{par:50} + +\ToCSect{51.}{Interpolation}{107}{par:51} + +\ToCSect{52.}{Finite and infinite classes}{108}{par:52} + +\ToCSect{53--57.}{Properties possessed by a function of~$n$ for large values of~$n$}{109}{par:53} + +\ToCSect{58--61.}{Definition of a limit and other definitions}{116}{par:58} + +\ToCSect{62.}{Oscillating functions}{121}{par:62} + +\ToCSect{63--68.}{General theorems concerning limits}{125}{par:63} + +\ToCSect{69--70.}{Steadily increasing or decreasing functions}{131}{par:69} + +\ToCSect{71.}{Alternative proof of Weierstrass's Theorem}{134}{par:71} + +\ToCSect{72.}{The limit of~$x^{n}$}{134}{par:72} + +\ToCSect{73.}{The limit of $\left(1 + \dfrac{1}{n}\right)^{n}$}{137}{par:73} + +\ToCSect{74.}{Some algebraical lemmas}{138}{par:74} + +\ToCSect{75.}{The limit of $n(\sqrt[n]{x} - 1)$}{139}{par:75} + +\ToCSect{76--77.}{Infinite series}{140}{par:76} + +\ToCSect{78.}{The infinite geometrical series}{143}{par:78} +\PageSep{ix} +\ToCSect{79.}{The representation of functions of a continuous real variable +by means of limits}{147}{par:79} + +\ToCSect{80.}{The bounds of a bounded aggregate}{149}{par:80} + +\ToCSect{81.}{The bounds of a bounded function}{149}{par:81} + +\ToCSect{82.}{The limits of indetermination of a bounded function}{150}{par:82} + +\ToCSect{83--84.}{The general principle of convergence}{151}{par:83} + +\ToCSect{85--86.}{Limits of complex functions and series of complex terms}{153}{par:85} + +\ToCSect{87--88.}{Applications to~$z^{n}$ and the geometrical series}{156}{par:87} + +\ToCSect{}{Miscellaneous Examples}{157}{misc:IV} + +\begin{ToCPar} +Oscillation of $\sin n\theta\pi$,~\PgNo{121},~\PgNo{123},~\PgNo{151}. Limits of $n^{k} x^{n}$, $\sqrt[n]{x}$, $\sqrt[n]{n}$, $\sqrtp[n]{n!}$, +$\dfrac{x^{n}}{n!}$, $\dbinom{m}{n} x^{n}$,~\PgNo{136},~\PgNo{139}. Decimals,~\PgNo{143}. Arithmetical series,~\PgNo{146}. Harmonical +series,~\PgNo{147}. Equation $x_{n+1} = f(x_{n})$,~\PgNo{158}. Expansions of rational functions,~\PgNo{159}. +Limit of a mean value,~\PgNo{160}. +\end{ToCPar} + + +\ToCChap{CHAPTER V}{LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE\@. CONTINUOUS +AND DISCONTINUOUS FUNCTIONS} + +\ToCSect{89--92.}{Limits as $x\to\infty$ or $x\to-\infty$}{162}{par:89} + +\ToCSect{93--97.}{Limits as $x\to a$}{165}{par:93} + +\ToCSect{98--99.}{Continuous functions of a real variable}{174}{par:98} + +\ToCSect{100--104.}{Properties of continuous functions. Bounded functions. +The oscillation of a function in an interval}{179}{par:100} + +\ToCSect{105--106.}{Sets of intervals on a line. The Heine-Borel Theorem}{185}{par:105} + +\ToCSect{107.}{Continuous functions of several variables}{190}{par:107} + +\ToCSect{108--109.}{Implicit and inverse functions}{191}{par:108} + +\ToCSect{}{Miscellaneous Examples}{194}{misc:V} + +\begin{ToCPar} +Limits and continuity of polynomials and rational functions,~\PgNo{169},~\PgNo{176}. +Limit of $\dfrac{x^{m} - a^{m}}{x - a}$,~\PgNo{171}. Orders of smallness and greatness,~\PgNo{172}. Limit of +$\dfrac{\sin{x}}{x}$,~\PgNo{173}. Infinity of a function,~\PgNo{177}. Continuity of $\cos x$ and $\sin x$,~\PgNo{177}. +Classification of discontinuities,~\PgNo{178}. +\end{ToCPar} + + +\ToCChap{CHAPTER VI}{DERIVATIVES AND INTEGRALS} + +\ToCSect{110--112.}{Derivatives}{197}{par:110} + +\ToCSect{113.}{General rules for differentiation}{203}{par:113} + +\ToCSect{114.}{Derivatives of complex functions}{205}{par:114} + +\ToCSect{115.}{The notation of the differential calculus}{205}{par:115} + +\ToCSect{116.}{Differentiation of polynomials}{207}{par:116} + +\ToCSect{117.}{Differentiation of rational functions}{209}{par:117} + +\ToCSect{118.}{Differentiation of algebraical functions}{210}{par:118} +\PageSep{x} +\ToCSect{119.}{Differentiation of transcendental functions}{212}{par:119} + +\ToCSect{120.}{Repeated differentiation}{214}{par:120} + +\ToCSect{121.}{General theorems concerning derivatives. Rolle's Theorem}{217}{par:121} + +\ToCSect{122--124.}{Maxima and minima}{219}{par:122} + +\ToCSect{125--126.}{The Mean Value Theorem}{226}{par:125} + +\ToCSect{127--128.}{Integration. The logarithmic function}{228}{par:127} + +\ToCSect{129.}{Integration of polynomials}{232}{par:129} + +\ToCSect{130--131.}{Integration of rational functions}{233}{par:130} + +\ToCSect{132--139.}{Integration of algebraical functions. Integration by +rationalisation. Integration by parts}{236}{par:132} + +\ToCSect{140--144.}{Integration of transcendental functions}{245}{par:140} + +\ToCSect{145.}{Areas of plane curves}{249}{par:145} + +\ToCSect{146.}{Lengths of plane curves}{251}{par:146} + +\ToCSect{}{Miscellaneous Examples}{253}{misc:VI} + +\begin{ToCPar} +Derivative of $x^{m}$,~\PgNo{201}. Derivatives of $\cos{x}$ and $\sin{x}$,~\PgNo{201}. Tangent +and normal to a curve,~\PgNo{201},~\PgNo{214}. Multiple roots of equations,~\PgNo{208},~\PgNo{255}. +Rolle's Theorem for polynomials,~\PgNo{209}. Leibniz' Theorem,~\PgNo{215}. Maxima +and minima of the quotient of two quadratics,~\PgNo{223},~\PgNo{256}. Axes of a conic,~\PgNo{226}. +Lengths and areas in polar coordinates,~\PgNo{253}. Differentiation of a +determinant,~\PgNo{254}. Extensions of the Mean Value Theorem,~\PgNo{258}. Formulae +of reduction,~\PgNo{259}. +\end{ToCPar} + + +\ToCChap{CHAPTER VII}{ADDITIONAL THEOREMS IN THE DIFFERENTIAL AND INTEGRAL CALCULUS} + +\ToCSect{147.}{Taylor's Theorem}{262}{par:147} + +\ToCSect{148.}{Taylor's Series}{266}{par:148} + +\ToCSect{149.}{Applications of Taylor's Theorem to maxima and minima}{268}{par:149} + +\ToCSect{150.}{Applications of Taylor's Theorem to the calculation of limits}{268}{par:150} + +\ToCSect{151.}{The contact of plane curves}{270}{par:151} + +\ToCSect{152--154.}{Differentiation of functions of several variables}{274}{par:152} + +\ToCSect{155.}{Differentials}{280}{par:155} + +\ToCSect{156--161.}{Definite Integrals. Areas of curves}{283}{par:156} + +\ToCSect{162.}{Alternative proof of Taylor's Theorem}{298}{par:162} + +\ToCSect{163.}{Application to the binomial series}{299}{par:163} + +\ToCSect{164.}{Integrals of complex functions}{299}{par:164} + +\ToCSect{}{Miscellaneous Examples}{300}{misc:VII} + +\begin{ToCPar} +Newton's method of approximation to the roots of equations,~\PgNo{265}. +Series for $\cos{x}$ and $\sin{x}$,~\PgNo{267}. Binomial series,~\PgNo{267}. Tangent to a curve, +\PgNo{272},~\PgNo{283},~\PgNo{303}. Points of inflexion,~\PgNo{272}. Curvature,~\PgNo{273},~\PgNo{302}. Osculating +conics,~\PgNo{274},~\PgNo{302}. Differentiation of implicit functions,~\PgNo{283}. Fourier's +integrals,~\PgNo{290},~\PgNo{294}. The second mean value theorem,~\PgNo{296}. Homogeneous +functions,~\PgNo{302}. Euler's Theorem,~\PgNo{302}. Jacobians,~\PgNo{303}. Schwarz's inequality +for integrals,~\PgNo{306}. Approximate values of definite integrals,~\PgNo{307}. +Simpson's Rule,~\PgNo{307}. +\end{ToCPar} +\PageSep{xi} + + +\ToCChap{CHAPTER VIII}{THE CONVERGENCE OF INFINITE SERIES AND INFINITE INTEGRALS} + +\ToCSect{165--168.}{Series of positive terms. Cauchy's and d'Alembert's tests +of convergence}{308}{par:165} + +\ToCSect{169.}{Dirichlet's Theorem}{313}{par:169} + +\ToCSect{170.}{Multiplication of series of positive terms}{313}{par:170} + +\ToCSect{171--174.}{Further tests of convergence. Abel's Theorem. Maclaurin's +integral test}{315}{par:171} + +\ToCSect{175.}{The series $\sum n^{-s}$}{319}{par:175} + +\ToCSect{176.}{Cauchy's condensation test}{320}{par:176} + +\ToCSect{177--182.}{Infinite integrals}{321}{par:177} + +\ToCSect{183.}{Series of positive and negative terms}{335}{par:183} + +\ToCSect{184--185.}{Absolutely convergent series}{336}{par:184} + +\ToCSect{186--187.}{Conditionally convergent series}{338}{par:186} + +\ToCSect{188.}{Alternating series}{340}{par:188} + +\ToCSect{189.}{Abel's and Dirichlet's tests of convergence}{342}{par:189} + +\ToCSect{190.}{Series of complex terms}{344}{par:190} + +\ToCSect{191--194.}{Power series}{345}{par:191} + +\ToCSect{195.}{Multiplication of series in general}{349}{par:195} + +\ToCSect{}{Miscellaneous Examples}{350}{misc:VIII} + +\begin{ToCPar} +The series $\sum n^{k}r^{n}$ and allied series,~\PgNo{311}. Transformation of infinite +integrals by substitution and integration by parts,~\PgNo{327},~\PgNo{328},~\PgNo{333}. The +series $\sum a_{n} \cos n\theta$, $\sum a_{n} \sin n\theta$,~\PgNo{338},~\PgNo{343},~\PgNo{344}. Alteration of the sum of a +series by rearrangement,~\PgNo{341}. Logarithmic series,~\PgNo{348}. Binomial series, +\PgNo{348},~\PgNo{349}. Multiplication of conditionally convergent series,~\PgNo{350},~\PgNo{354}. +Recurring series,~\PgNo{352}. Difference equations,~\PgNo{353}. Definite integrals,~\PgNo{355}. +Schwarz's inequality for infinite integrals,~\PgNo{356}. +\end{ToCPar} + + +\ToCChap{CHAPTER IX}{THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS OF A REAL VARIABLE} + +\ToCSect{196--197.}{The logarithmic function}{357}{par:196} + +\ToCSect{198.}{The functional equation satisfied by $\log x$}{360}{par:198} + +\ToCSect{199--201.}{The behaviour of $\log x$ as $x$~tends to infinity or to zero}{360}{par:199} + +\ToCSect{202.}{The logarithmic scale of infinity}{362}{par:202} + +\ToCSect{203.}{The number~$e$}{363}{par:203} + +\ToCSect{204--206.}{The exponential function}{364}{par:204} + +\ToCSect{207.}{The general power~$a^{x}$}{366}{par:207} + +\ToCSect{208.}{The exponential limit}{368}{par:208} + +\ToCSect{209.}{The logarithmic limit}{369}{par:209} + +\ToCSect{210.}{Common logarithms}{369}{par:210} + +\ToCSect{211.}{Logarithmic tests of convergence}{374}{par:211} +\PageSep{xii} +\ToCSect{212.}{The exponential series}{378}{par:212} + +\ToCSect{213.}{The logarithmic series}{381}{par:213} + +\ToCSect{214.}{The series for $\arctan x$}{382}{par:214} + +\ToCSect{215.}{The binomial series}{384}{par:215} + +\ToCSect{216.}{Alternative development of the theory}{386}{par:216} + +\ToCSect{}{Miscellaneous Examples}{387}{misc:IX} + +\begin{ToCPar} +Integrals containing the exponential function,~\PgNo{370}. The hyperbolic +functions,~\PgNo{372}. Integrals of certain algebraical functions,~\PgNo{373}. Euler's +constant,~\PgNo{377},~\PgNo{389}. Irrationality of~$e$,~\PgNo{380}. Approximation to surds by the +binomial theorem,~\PgNo{385}. Irrationality of~$\log_{10} n$,~\PgNo{387}. Definite integrals,~\PgNo{393}. +\end{ToCPar} + + +\ToCChap{CHAPTER X}{THE GENERAL THEORY OF THE LOGARITHMIC, EXPONENTIAL, +AND CIRCULAR FUNCTIONS} + +\ToCSect{217--218.}{Functions of a complex variable}{395}{par:217} + +\ToCSect{219.}{Curvilinear integrals}{396}{par:219} + +\ToCSect{220.}{Definition of the logarithmic function}{397}{par:220} + +\ToCSect{221.}{The values of the logarithmic function}{399}{par:221} + +\ToCSect{222--224.}{The exponential function}{403}{par:222} + +\ToCSect{225--226.}{The general power~$a^{z}$}{404}{par:225} + +\ToCSect{227--230.}{The trigonometrical and hyperbolic functions}{409}{par:227} + +\ToCSect{231.}{The connection between the logarithmic and inverse +trigonometrical functions}{413}{par:231} + +\ToCSect{232.}{The exponential series}{414}{par:232} + +\ToCSect{233.}{The series for $\cos z$ and $\sin z$}{416}{par:233} + +\ToCSect{234--235.}{The logarithmic series}{417}{par:234} + +\ToCSect{236.}{The exponential limit}{421}{par:236} + +\ToCSect{237.}{The binomial series}{422}{par:237} + +\ToCSect{}{Miscellaneous Examples}{425}{misc:X} + +\begin{ToCPar} +The functional equation satisfied by $\Log z$,~\PgNo{402}. The function~$e^{z}$,~\PgNo{407}. +Logarithms to any base,~\PgNo{408}. The inverse cosine, sine, and tangent of a +complex number,~\PgNo{412}. Trigonometrical series,~\PgNo{417},~\PgNo{420},~\PgNo{431}. Roots of +transcendental equations,~\PgNo{425}. Transformations,~\PgNo{426},~\PgNo{428}. Stereographic +projection,~\PgNo{427}. Mercator's projection,~\PgNo{428}. Level curves,~\PgNo{429}. Definite +integrals,~\PgNo{432}. +\end{ToCPar} + +\ToCApp{I}{The proof that every equation has a root}{433} + +\ToCApp{II}{A note on double limit problems}{439} + +\ToCApp{III}{The circular functions}{443} + +\ToCApp{IV}{The infinite in analysis and geometry}{445} +\MainMatter +\PageSep{1} + + +\Chapter{I}{REAL VARIABLES} + +\Paragraph{1. Rational numbers.} A fraction $r = p/q$, where $p$~and~$q$ +are positive or negative integers, is called a \emph{rational number}. We +can suppose (i)~that $p$~and~$q$ have no common factor, as if they +have a common factor we can divide each of them by it, and +(ii)~that $q$~is positive, since +\[ +p/(-q) = (-p)/q,\quad (-p)/(-q) = p/q. +\] +To the rational numbers thus defined we may add the `rational +number~$0$' obtained by taking $p = 0$. + +We assume that the reader is familiar with the ordinary +arithmetical rules for the manipulation of rational numbers. The +examples which follow demand no knowledge beyond this. + +\begin{Examples}{I.} +\Item{1.} If $r$~and~$s$ are rational numbers, then $r + s$, $r - s$, $rs$, and +$r/s$ are rational numbers, unless in the last case $s = 0$ (when $r/s$~is of course +meaningless). + +\Item{2.} {\Loosen If $\lambda$,~$m$, and~$n$ are positive rational numbers, and $m > n$, then +$\lambda(m^{2} - n^{2})$, $2\lambda mn$, and $\lambda(m^{2} + n^{2})$ are positive rational numbers. Hence show +how to determine any number of right-angled triangles the lengths of all of +whose sides are rational.} + +\Item{3.} Any terminated decimal represents a rational number whose denominator +contains no factors other than $2$~or~$5$. Conversely, any such rational +number can be expressed, and in one way only, as a terminated decimal. + +[The general theory of decimals will be considered in \Ref{Ch.}{IV}.] + +\Item{4.} The positive rational numbers may be arranged in the form of a simple +series as follows: +\[ +\tfrac{1}{1},\quad +\tfrac{2}{1},\quad +\tfrac{1}{2},\quad +\tfrac{3}{1},\quad +\tfrac{2}{2},\quad +\tfrac{1}{3},\quad +\tfrac{4}{1},\quad +\tfrac{3}{2},\quad +\tfrac{2}{3},\quad +\tfrac{1}{4},\ \dots. +\] + +Show that $p/q$ is the $[\frac{1}{2}(p + q - 1)(p + q - 2) + q]$th term of the series. + +[In this series every rational number is repeated indefinitely. Thus $1$ +occurs as $\frac{1}{1}$,~$\frac{2}{2}$, $\frac{3}{3}, \dots$. We can of course avoid this by omitting every number +\PageSep{2} +which has already occurred in a simpler form, but then the problem of determining +the precise position of~$p/q$ becomes more complicated.] +\end{Examples} + +\Paragraph{2. The representation of rational numbers by points +on a line.} It is convenient, in many branches of mathematical +analysis, to make a good deal of use of geometrical illustrations. + +The use of geometrical illustrations in this way does not, of +course, imply that analysis has any sort of dependence upon +geometry: they are illustrations and nothing more, and are employed +merely for the sake of clearness of exposition. This being +so, it is not necessary that we should attempt any logical analysis +of the ordinary notions of elementary geometry; we may be content +to suppose, however far it may be from the truth, that we know +what they mean. + +Assuming, then, that we know what is meant by a \emph{straight +line}, a \emph{segment} of a line, and the \emph{length} of a segment, let us take +a straight line~$\Lambda$, produced indefinitely in both directions, and a +segment~$A_{0}A_{1}$ of any length. We call $A_{0}$ the \emph{origin}, or \emph{the point~$0$}, +and $A_{1}$ \emph{the point~$1$}, and we regard these points as representing +the numbers $0$~and~$1$. + +In order to obtain a point which shall represent a positive +rational number $r = p/q$, we choose the point~$A_{r}$ such that +\[ +A_{0}A_{r}/A_{0}A_{1} = r, +\] +$A_{0}A_{r}$ being a stretch of the line extending in the same direction +along the line as~$A_{0}A_{1}$, a direction which we shall suppose to be +from left to right when, as in \Fig{1}, the line is drawn horizontally +across the paper. In order to obtain a point to represent a +%[Illustration: Fig. 1.] +\Figure[0.9\textwidth]{1}{p002} +negative rational number $r = -s$, it is natural to regard length as +a magnitude capable of sign, positive if the length is measured in +one direction (that of~$A_{0}A_{1}$), and negative if measured in the +other, so that $AB = -BA$; and to take as the point representing +$r$ the point~$A_{-s}$ such that +\[ +A_{0}A_{-s} = -A_{-s}A_{0} = -A_{0}A_{s}. +\] +\PageSep{3} + +We thus obtain a point~$A_{r}$ on the line corresponding to every +rational value of~$r$, positive or negative, and such that +\[ +A_{0}A_{r} = r · A_{0}A_{1}; +\] +{\Loosen and if, as is natural, we take $A_{0}A_{1}$ as our unit of length, and write +$A_{0}A_{1} = 1$, then we have} +\[ +A_{0}A_{r} = r. +\] +We shall call the points~$A_{r}$ the \emph{rational points} of the line. + +\Paragraph{3. Irrational numbers.} If the reader will mark off on the +line all the points corresponding to the rational numbers whose +denominators are $1$,~$2$, $3, \dots$ in succession, he will readily convince +himself that he can cover the line with rational points as closely +as he likes. We can state this more precisely as follows: \emph{if we +take any segment~$BC$ on~$\Lambda$, we can find as many rational points as +we please on~$BC$}. + +Suppose, for example, that $BC$~falls within the segment~$A_{1}A_{2}$. +It is evident that if we choose a positive integer~$k$ so that +\[ +k · BC > 1,\footnotemark +\Tag{(1)} +\] +\footnotetext{The assumption that this is possible is equivalent to the assumption of what + is known as the Axiom of Archimedes.}% +and divide~$A_{1}A_{2}$ into $k$~equal parts, then at least one of the points +of division (say~$P$) must fall inside~$BC$, without coinciding with +either $B$~or~$C$. For if this were not so, $BC$~would be entirely +included in one of the $k$~parts into which~$A_{1}A_{2}$ has been divided, +which contradicts the supposition~\Eq{(1)}. But $P$~obviously corresponds +to a rational number whose denominator is~$k$. Thus at +least one rational point~$P$ lies between $B$~and~$C$. But then we +can find another such point~$Q$ between $B$~and~$P$, another between +$B$~and~$Q$, and so on indefinitely; \ie, as we asserted above, we can +find as many as we please. We may express this by saying that +$BC$~includes \emph{infinitely many} rational points. + +\begin{Remark} +The meaning of such phrases as `\emph{infinitely many}' or `\emph{an infinity of}', in +such sentences as `$BC$~includes infinitely many rational points' or `there are +an infinity of rational points on~$BC$' or `there are an infinity of positive +integers', will be considered more closely in \Ref{Ch.}{IV}\@. The assertion `there are +an infinity of positive integers' means `given any positive integer~$n$, however +large, we can find more than~$n$ positive integers'. This is plainly true +\PageSep{4} +whatever $n$~may be, \eg\ for $n = 100,000$ or $100,000,000$. The assertion means +exactly the same as `we can find \emph{as many positive integers as we please}'. + +The reader will easily convince himself of the truth of the following +assertion, which is substantially equivalent to what was proved in the second +paragraph of this section: given any rational number~$r$, and any positive +integer~$n$, we can find another rational number lying on either side of~$r$ and +differing from~$r$ by less than~$1/n$. It is merely to express this differently to +say that we can find a rational number lying on either side of~$r$ and differing +from~$r$ \emph{by as little as we please}. Again, given any two rational numbers +$r$~and~$s$, we can interpolate between them a chain of rational numbers in +which any two consecutive terms differ by as little as we please, that is to +say by less than~$1/n$, where $n$~is any positive integer assigned beforehand. +\end{Remark} + +From these considerations the reader might be tempted to +infer that an adequate view of the nature of the line could be +obtained by imagining it to be formed simply by the rational +points which lie on it. And it is certainly the case that if we +imagine the line to be made up solely of the rational points, +and all other points (if there are any such) to be eliminated, +the figure which remained would possess most of the properties +which common sense attributes to the straight line, and would, +to put the matter roughly, look and behave very much like +a line. + +A little further consideration, however, shows that this view +would involve us in serious difficulties. + +Let us look at the matter for a moment with the eye of +common sense, and consider some of the properties which we may +reasonably expect a straight line to possess if it is to satisfy the +idea which we have formed of it in elementary geometry. + +The straight line must be composed of points, and any segment +of it by all the points which lie between its end points. With +any such segment must be associated a certain entity called its +\emph{length}, which must be a \emph{quantity} capable of \emph{numerical measurement} +in terms of any standard or unit length, and these lengths +must be capable of combination with one another, according to +the ordinary rules of algebra, by means of addition or multiplication. +Again, it must be possible to construct a line whose +length is the sum or product of any two given lengths. If the +length~$PQ$, along a given line, is~$a$, and the length~$QR$, along +the same straight line, is~$b$, the length~$PR$ must be~$a + b$. +\PageSep{5} +Moreover, if the lengths $OP$,~$OQ$, along one straight line, are +$1$~and~$a$, and the length~$OR$ along another straight line is~$b$, +and if we determine the length~$OS$ by Euclid's construction (Euc.~\textsc{vi}.~12) +for a fourth proportional to the lines $OP$,~$OQ$,~$OR$, this +length must be~$ab$, the algebraical fourth proportional to $1$,~$a$,~$b$. +And it is hardly necessary to remark that the sums and products +thus defined must obey the ordinary `laws of algebra'; viz. +\begin{gather*} +a + b = b + a,\quad a + (b + c) = (a + b) + c,\\ +ab = ba,\quad a(bc) = (ab)c,\quad a(b + c) = ab + ac. +\end{gather*} +The lengths of our lines must also obey a number of obvious +laws concerning inequalities as well as equalities: thus if +$A$,~$B$,~$C$ are three points lying along~$\Lambda$ from left to right, we must +have $AB < AC$, and so on. Moreover it must be possible, on our +fundamental line~$\Lambda$, to find a point~$P$ such that~$A_{0}P$ is equal to +any segment whatever taken along~$\Lambda$ or along any other straight +line. All these properties of a line, and more, are involved in the +presuppositions of our elementary geometry. + +Now it is very easy to see that the idea of a straight line as +composed of a series of points, each corresponding to a rational +number, cannot possibly satisfy all these requirements. There are +various elementary geometrical constructions, for example, which +purport to construct a length~$x$ such that $x^{2} = 2$. For instance, we +%[Illustration: Fig. 2.] +\Figure{2}{p005} +may construct an isosceles right-angled triangle~$ABC$ such that +$AB = AC = 1$. Then if $BC = x$, $x^{2} = 2$. Or we may determine +the length~$x$ by means of Euclid's construction (Euc.~\textsc{vi}.~13) for +a mean proportional to $1$~and~$2$, as indicated in the figure. Our +requirements therefore involve the existence of a length measured +by a number~$x$, and a point~$P$ on~$\Lambda$ such that +\[ +A_{0}P = x,\quad x^{2} = 2. +\] +\PageSep{6} +But it is easy to see that \emph{there is no rational number such that +its square is~$2$}. In fact we may go further and say that there +is no rational number whose square is~$m/n$, where $m/n$~is any +positive fraction in its lowest terms, unless $m$~and~$n$ are both +perfect squares. + +For suppose, if possible, that +\[ +p^{2}/q^{2} = m/n\DPtypo{.}{,} +\] +$p$~having no factor in common with~$q$, and $m$~no factor in common +with~$n$. Then $np^{2} = mq^{2}$. Every factor of~$q^{2}$ must divide~$np^{2}$, and +as $p$~and~$q$ have no common factor, every factor of~$q^{2}$ must divide~$n$. +Hence $n = \lambda q^{2}$, where $\lambda$~is an integer. But this involves +$m = \lambda p^{2}$: and as $m$~and~$n$ have no common factor, $\lambda$~must be unity. +Thus $m = p^{2}$, $n = q^{2}$, as was to be proved. In particular it follows, +by taking $n = 1$, that an integer cannot be the square of a rational +number, unless that rational number is itself integral. + +It appears then that our requirements involve the existence of +a number~$x$ and a point~$P$, not one of the rational points already +constructed, such that $A_{0}P = x$, $x^{2} = 2$; and (as the reader will +remember from elementary algebra) we write $x = \sqrt{2}$. + +\begin{Remark} +The following alternative proof that no rational number can have its +square equal to~$2$ is interesting. + +Suppose, if possible, that $p/q$~is a positive fraction, in its lowest terms, +such that $(p/q)^{2} = 2$ or $p^{2} = 2q^{2}$. It is easy to see that this involves +$(2q - p)^{2} = 2(p - q)^{2}$; and so $(2q - p)/(p - q)$ is another fraction having the +same property. But clearly $q < p < 2q$, and so $p - q < q$. Hence there is +another fraction equal to~$p/q$ and having a smaller denominator, which +contradicts the assumption that $p/q$~is in its lowest terms. +\end{Remark} + +\begin{Examples}{II.} +\Item{1.} Show that no rational number can have its cube equal +to~$2$. + +\Item{2.} Prove generally that a rational fraction~$p/q$ in its lowest terms cannot +be the cube of a rational number unless $p$~and~$q$ are both perfect cubes. + +\Item{3.} A more general proposition, which is due to Gauss and includes those +which precede as particular cases, is the following: \emph{an algebraical equation +\[ +x^{n} + p_{1}x^{n-1} + p_{2}x^{n-2} + \dots + p_{n} = 0, +\] +with integral coefficients, cannot have a rational but non-integral root}. + +[For suppose that the equation has a root~$a/b$, where $a$~and~$b$ are integers +\PageSep{7} +without a common factor, and $b$~is positive. Writing~$a/b$ for~$x$, and multiplying +by~$b^{n-1}$, we obtain +\[ +-\frac{a^{n}}{b} = p_{1}a^{n-1} + p_{2}a^{n-2}b + \dots + p_{n}b^{n-1}, +\] +a fraction in its lowest terms equal to an integer, which is absurd. Thus $b = 1$, +and the root is~$a$. It is evident that $a$~must be a divisor of~$p_{n}$.] + +\Item{4.} Show that if $p_{n} = 1$ and neither of +\[ +1 + p_{1} + p_{2} + p_{3} + \dots,\quad +1 - p_{1} + p_{2} - p_{3} + \dots +\] +is zero, then the equation cannot have a rational root. + +\Item{5.} Find the rational roots (if any) of +\[ +x^{4} - 4x^{3} - 8x^{2} + 13x + 10 = 0. +\] + +[The roots can only be integral, and so $±1$, $±2$, $±5$, $±10$ are the only +possibilities: whether these are roots can be determined by trial. It is clear +that we can in this way determine the rational roots of any such equation.] +\end{Examples} + +\Paragraph{4. Irrational numbers (\continued).} The result of our +geometrical representation of the rational numbers is therefore to +suggest the desirability of enlarging our conception of `number' +by the introduction of further numbers of a new kind. + +The same conclusion might have been reached without the use +of geometrical language. One of the central problems of algebra +is that of the solution of equations, such as +\[ +x^{2} = 1,\quad x^{2} = 2. +\] +The first equation has the two rational roots $1$~and~$-1$. But, +if our conception of number is to be limited to the rational +numbers, we can only say that the second equation has no roots; +and the same is the case with such equations as $x^{3} = 2$, $x^{4} = 7$. +These facts are plainly sufficient to make some generalisation of +our idea of number desirable, if it should prove to be possible. + +Let us consider more closely the equation $x^{2} = 2$. + +We have already seen that there is no rational number~$x$ which +satisfies this equation. The square of any rational number is +either less than or greater than~$2$. We can therefore divide the +rational numbers into two classes, one containing the numbers +whose squares are less than~$2$, and the other those whose squares +are greater than~$2$. We shall confine our attention to the \emph{positive} +rational numbers, and we shall call these two classes \emph{the class~$L$}, or +\emph{the lower class}, or \emph{the left-hand class}, and \emph{the class~$R$}, or \emph{the upper +\PageSep{8} +class}, or \emph{the right-hand class}. It is obvious that every member of~$R$ +is greater than all the members of~$L$. Moreover it is easy to +convince ourselves that we can find a member of the class~$L$ whose +square, though less than~$2$, differs from~$2$ by as little as we please, +and a member of~$R$ whose square, though greater than~$2$, also +differs from~$2$ by as little as we please. In fact, if we carry out +the ordinary arithmetical process for the extraction of the square +root of~$2$, we obtain a series of rational numbers, viz. +\[ +1,\quad 1.4,\quad 1.41,\quad 1.414,\quad 1.4142,\ \dots +\] +whose squares +\[ +1,\quad 1.96,\quad 1.9881,\quad 1.999\MS396,\quad 1.999\MS961\MS64,\ \dots +\] +are all less than~$2$, but approach nearer and nearer to it; and by +taking a sufficient number of the figures given by the process we +can obtain as close an approximation as we want. And if we +increase the last figure, in each of the approximations given above, +by unity, we obtain a series of rational numbers +\[ +2,\quad 1.5,\quad 1.42,\quad 1.415,\quad 1.4143,\ \dots +\] +whose squares +\[ +4,\quad 2.25,\quad 2.0164,\quad 2.002\MS225,\quad 2.000\MS244\MS49,\ \dots +\] +are all greater than~$2$ but approximate to~$2$ as closely as we please. + +\begin{Remark} +The reasoning which precedes, although it will probably convince the +reader, is hardly of the precise character required by modern mathematics. +We can supply a formal proof as follows. In the first place, we can find +a member of~$L$ and a member of~$R$, differing by as little as we please. For +we saw in~\SecNo[§]{3} that, given any two rational numbers $a$~and~$b$, we can construct +a chain of rational numbers, of which $a$~and~$b$ are the first and last, and in +which any two consecutive numbers differ by as little as we please. Let us +then take a member~$x$ of~$L$ and a member~$y$ of~$R$, and interpolate between +them a chain of rational numbers of which $x$~is the first and $y$~the last, and +in which any two consecutive numbers differ by less than~$\delta$, $\delta$~being any +positive rational number as small as we please, such as $.01$ or $.0001$ or $.000\MS001$. +In this chain there must be a last which belongs to~$L$ and a first which belongs +to~$R$, and these two numbers differ by less than~$\delta$. + +We can now prove that \emph{an~$x$ can be found in~$L$ and a~$y$ in~$R$ such that +$2 - x^{2}$ and $y^{2} - 2$ are as small as we please}, say less than~$\delta$. Substituting $\frac{1}{4}\delta$ +for~$\delta$ in the argument which precedes, we see that we can choose $x$~and~$y$ so +that $y - x < \frac{1}{4}\delta$; and we may plainly suppose that both $x$~and~$y$ are less +than~$2$. Thus +\[ +y + x < 4,\quad +y^{2} - x^{2} = (y - x)(y + x) < 4(y - x) < \delta; +\] +\PageSep{9} +and since $x^{2} < 2$ and $y^{2} > 2$ it follows \textit{a~fortiori} that $2 - x^{2}$ and $y^{2} - 2$ are each +less than~$\delta$. +\end{Remark} + +It follows also that \emph{there can be no largest member of~$L$ or +smallest member of~$R$}. For if $x$~is any member of~$L$, then $x^{2} < 2$. +Suppose that $x^{2} = 2 - \delta$. Then we can find a member~$x_{1}$ of~$L$ +such that $x_{1}^{2}$~differs from~$2$ by less than~$\delta$, and so $x_{1}^{2} > x^{2}$ or $x_{1} > x$. +Thus there are larger members of~$L$ than~$x$; and as $x$~is \emph{any} +member of~$L$, it follows that no member of~$L$ can be larger than +all the rest. Hence $L$~has no largest member, and similarly $R$~has +no smallest. + +\Paragraph{5. Irrational numbers (\continued).} We have thus divided +the positive rational numbers into two classes, $L$~and~$R$, such that +(i)~every member of~$R$ is greater than every member of~$L$, (ii)~we +can find a member of~$L$ and a member of~$R$ whose difference is as +small as we please, (iii)~$L$~has no greatest and $R$~no least member. +Our common-sense notion of the attributes of a straight line, the +requirements of our elementary geometry and our elementary +algebra, alike demand \emph{the existence of a number~$x$ greater than all +the members of~$L$ and less than all the members of~$R$, and of +a corresponding point~$P$ on~$\Lambda$ such that $P$~divides the points which +correspond to members of~$L$ from those which correspond to members +of~$R$}. +%[Illustration: Fig. 3.] +\Figure[0.9\textwidth]{3}{p009} + +Let us suppose for a moment that there is such a number~$x$, +and that it may be operated upon in accordance with the laws of +algebra, so that, for example, $x^{2}$~has a definite meaning. Then $x^{2}$ +cannot be either less than or greater than~$2$. For suppose, for +example, that $x^{2}$~is less than~$2$. Then it follows from what precedes +that we can find a positive rational number~$\xi$ such that $\xi^{2}$~lies +\PageSep{10} +between $x^{2}$~and~$2$. That is to say, we can find a member of~$L$ +greater than~$x$; and this contradicts the supposition that $x$~divides +the members of~$L$ from those of~$R$. Thus $x^{2}$~cannot be less than~$2$, +and similarly it cannot be greater than~$2$. We are therefore +driven to the conclusion that $x^{2} = 2$, and that $x$~is the number +which in algebra we denote by~$\sqrt{2}$. And of course this number +$\sqrt{2}$~is not rational, for no rational number has its square equal to~$2$. +It is the simplest example of what is called an \Emph{irrational} +number. + +But the preceding argument may be applied to equations +other than $x^{2} = 2$, almost word for word; for example to $x^{2} = N$, +where $N$~is any integer which is not a perfect square, or to +\[ +x^{3} = 3,\quad +x^{3} = 7,\quad +x^{4} = 23, +\] +or, as we shall see later on, to $x^{3} = 3x + 8$. We are thus led to +believe in the existence of irrational numbers~$x$ and points~$P$ on~$\Lambda$ +such that $x$~satisfies equations such as these, even when these +lengths cannot (as $\sqrt{2}$~can) be constructed by means of elementary +geometrical methods. + +\begin{Remark} +The reader will no doubt remember that in treatises on elementary algebra +the root of such an equation as $x^{q} = n$ is denoted by $\sqrt[q]n$~or~$n^{1/q}$, and that a +meaning is attached to such symbols as +\[ +n^{p/q},\quad +n^{-p/q} +\] +by means of the equations +\[ +n^{p/q} = (n^{1/q})^{p},\quad +n^{p/q} n^{-p/q} = 1. +\] +And he will remember how, in virtue of these definitions, the `laws of indices' +such as +\[ +n^{r} × n^{s} = n^{r+s},\quad +(n^{r})^{s} = n^{rs} +\] +are extended so as to cover the case in which $r$~and~$s$ are any rational numbers +whatever. +\end{Remark} + +The reader may now follow one or other of two alternative +courses. He may, if he pleases, be content to assume that +`irrational numbers' such as $\sqrt{2}$,~$\sqrt[3]{3}, \dots$ exist and are amenable to +the algebraical laws with which he is familiar.\footnote + {This is the point of view which was adopted in the first edition of this book.} +If he does this +he will be able to avoid the more abstract discussions of the next +few sections, and may pass on at once to \SecNo[§§]{13}~\textit{et~seq.} + +If, on the other hand, he is not disposed to adopt so \textit{naive} an +\PageSep{11} +attitude, he will be well advised to pay careful attention to the +sections which follow, in which these questions receive fuller +consideration.\footnote + {In these sections I have borrowed freely from Appendix~I of Bromwich's + \textit{Infinite Series}.} + +\begin{Examples}{III.} +\Item{1.} Find the difference between~$2$ and the squares of the +decimals given in \SecNo[§]{4} as approximations to~$\sqrt{2}$. + +\Item{2.} Find the differences between~$2$ and the squares of +\[ +\tfrac{1}{1},\quad +\tfrac{3}{2},\quad +\tfrac{7}{5},\quad +\tfrac{17}{12},\quad +\tfrac{41}{29},\quad +\tfrac{99}{70}. +\] + +\Item{3.} Show that if $m/n$ is a good approximation to~$\sqrt{2}$, then~$(m + 2n)/(m + n)$ +is a better one, and that the errors in the two cases are in opposite directions. +Apply this result to continue the series of approximations in the last +example. + +\Item{4.} If $x$~and~$y$ are approximations to~$\sqrt{2}$, by defect and by excess respectively, +and $2 - x^{2} < \delta$, $y^{2} - 2 < \delta$, then $y - x < \delta$. + +\Item{5.} The equation $x^{2} = 4$ is satisfied by $x = 2$. Examine how far the argument +of the preceding sections applies to this equation (writing~$4$ for~$2$ +throughout). [If we define the classes $L$,~$R$ as before, they do not include \emph{all} +rational numbers. The rational number~$2$ is an exception, since~$2^{2}$ is neither +less than or greater than~$4$.] +\end{Examples} + +\Paragraph{6. Irrational numbers (\continued).} In \SecNo[§]{4} we discussed +a special mode of division of the positive rational numbers~$x$ into +two classes, such that $x^{2} < 2$ for the members of one class and +$x^{2} > 2$ for those of the others. Such a mode of division is called a +\Emph{section} of the numbers in question. It is plain that we could +equally well construct a section in which the numbers of the two +classes were characterised by the inequalities $x^{3} < 2$ and $x^{3} > 2$, or +$x^{4} < 7$ and $x^{4} > 7$. Let us now attempt to state the principles +of the construction of such `sections' of the positive rational +numbers in quite general terms. + +Suppose that $P$~and~$Q$ stand for two properties which are +mutually exclusive and one of which must be possessed by every +positive rational number. Further, suppose that every such +number which possesses~$P$ is less than any such number which +possesses~$Q$. Thus $P$~might be the property `$x^{2} < 2$' and $Q$~the +property `$x^{2} > 2$.' Then we call the numbers which possess~$P$ the +lower or left-hand class~$L$ and those which possess~$Q$ the upper or +\PageSep{12} +right-hand class~$R$. In general both classes will exist; but it may +happen in special cases that one is non-existent and that every +number belongs to the other. This would obviously happen, for +example, if $P$ (or~$Q$) were the property of being rational, or of +being positive. For the present, however, we shall confine +ourselves to cases in which both classes do exist; and then it +follows, as in \SecNo[§]{4}, that we can find a member of~$L$ and a member +of~$R$ whose difference is as small as we please. + +In the particular case which we considered in \SecNo[§]{4}, $L$~had no +greatest member and $R$~no least. This question of the existence +of greatest or least members of the classes is of the utmost importance. +We observe first that it is impossible in any case that +$L$~should have a greatest member \emph{and} $R$~a least. For if $l$ were +the greatest member of~$L$, and $r$~the least of~$R$, so that $l < r$, then +$\frac{1}{2}(l + r)$ would be a positive rational number lying between $l$~and~$r$, +and so could belong neither to~$L$ nor to~$R$; and this contradicts +our assumption that every such number belongs to one class or to +the other. This being so, there are but three possibilities, which +are mutually exclusive. Either (i)~$L$~has a greatest member~$l$, or +(ii)~$R$~has a least member~$r$, or (iii)~$L$~has no greatest member and +$R$~no least. + +\begin{Remark} +The section of \SecNo[§]{4} gives an example of the last possibility. An example +of the first is obtained by taking~$P$ to be `$x^{2} \leq 1$' and $Q$~to be `$x^{2} > 1$'; +here $l = 1$. If $P$~is `$x^{2} < 1$' and $Q$~is `$x^{2} \geq 1$', we have an example of the +second possibility, with $r = 1$. It should be observed that we do not obtain +a section at all by taking $P$ to be `$x^{2} < 1$' and $Q$~to be `$x^{2} > 1$'; for the special +number~$1$ escapes classification (cf.\ \Ex{iii}.~5). +\end{Remark} + +\Paragraph{7. Irrational numbers (\continued).} In the first two cases +we say that the section \emph{corresponds} to a positive rational number~$a$, +which is~$l$ in the one case and $r$~in the other. Conversely, it is +clear that to any such number~$a$ corresponds a section which +we shall denote by~$\alpha$.\footnote + {It will be convenient to denote a section, corresponding to a rational number + denoted by an English letter, by the corresponding Greek letter.} +For we might take $P$~and~$Q$ to be the +properties expressed by +\[ +x \leq a,\quad x > a +\] +respectively, or by $x < a$ and $x \geq a$. In the first case $a$~would be +the greatest member of~$L$, and in the second case the least member +\PageSep{13} +of~$R$. There are in fact just two sections corresponding to any +positive rational number. In order to avoid ambiguity we select +one of them; let us select that in which the number itself belongs +to the \emph{upper} class. In other words, let us agree that we will consider +only sections in which the lower class~$L$ has no greatest number. + +There being this correspondence between the positive rational +numbers and the sections defined by means of them, it would be +perfectly legitimate, for mathematical purposes, to replace the +numbers by the sections, and to regard the symbols which occur +in our formulae as standing for the sections instead of for the +numbers. Thus, for example, $\alpha > \alpha'$ would mean the same as +$a > a'$, if $\alpha$~and~$\alpha'$ are the sections which correspond to $a$~and~$a'$. + +But when we have in this way substituted sections of rational +numbers for the rational numbers themselves, we are almost forced +to a generalisation of our number system. For there are sections +(such as that of \SecNo[§]{4}) which do \emph{not} correspond to any rational +number. The aggregate of sections is a larger aggregate than that +of the positive rational numbers; it includes sections corresponding +to all these numbers, and more besides. It is this fact which we +make the basis of our generalisation of the idea of number. We +accordingly frame the following definitions, which will however be +modified in the next section, and must therefore be regarded as +temporary and provisional. + +\begin{Defn} +A section of the positive rational numbers, in which both classes +exist and the lower class has no greatest member, is called a +\Emph{positive real number}. +\end{Defn} + +\begin{Defn} +A positive real number which does not correspond to a positive +rational number is called a positive \Emph{irrational} number. +\end{Defn} + +\Paragraph{8. Real numbers.} We have confined ourselves so far to +certain sections of the positive rational numbers, which we have +agreed provisionally to call `positive real numbers.' Before we +frame our final definitions, we must alter our point of view a +little. We shall consider sections, or divisions into two classes, +not merely of the positive rational numbers, but of all rational +numbers, including zero. We may then repeat all that we have +said about sections of the positive rational numbers in \SecNo[§§]{6},~\SecNo{7}, +merely omitting the word positive occasionally. +\PageSep{14} + +\begin{Definitions} +A section of the rational numbers, in which both +classes exist and the lower class has no greatest member, is called +a \Emph{real number}, or simply a \Emph{number}. + +A real number which does not correspond to a rational number +is called an \Emph{irrational} number. +\end{Definitions} + +If the real number does correspond to a rational number, we +shall use the term `rational' as applying to the real number also. + +\begin{Remark} +The term `rational number' will, as a result of our definitions, be +ambiguous; it may mean the rational number of \SecNo[§]{1}, or the corresponding +real number. If we say that $\frac{1}{2} > \frac{1}{3}$, we may be asserting either of two different +propositions, one a proposition of elementary arithmetic, the other a proposition +concerning sections of the rational numbers. Ambiguities of this kind are +common in mathematics, and are perfectly harmless, since the relations +between different propositions are exactly the same whichever interpretation +is attached to the propositions themselves. From $\frac{1}{2} > \frac{1}{3}$ and $\frac{1}{3} > \frac{1}{4}$ we can +infer $\frac{1}{2} > \frac{1}{4}$; the inference is in no way affected by any doubt as to whether +$\frac{1}{2}$,~$\frac{1}{3}$, and~$\frac{1}{4}$ are arithmetical fractions or real numbers. Sometimes, of course, +the context in which (\eg)~`$\frac{1}{2}$' occurs is sufficient to fix its interpretation. +When we say (see \SecNo[§]{9}) that $\frac{1}{2} < \sqrt{\frac{1}{3}}$, we \emph{must} mean by~`$\frac{1}{2}$' the real number~$\frac{1}{2}$. + +The reader should observe, moreover, that no particular logical importance +is to be attached to the precise form of definition of a `real number' that we +have adopted. We defined a `real number' as being a section, \ie\ a pair of +classes. We might equally well have defined it as being the lower, or the +upper, class; indeed it would be easy to define an infinity of classes of +entities each of which would possess the properties of the class of real +numbers. What is essential in mathematics is that its symbols should be +capable of \emph{some} interpretation; generally they are capable of \emph{many}, and +then, so far as mathematics is concerned, it does not matter which we adopt. +Mr~Bertrand Russell has said that `mathematics is the science in which +we do not know what we are talking about, and do not care whether what +we say about it is true', a remark which is expressed in the form of a +paradox but which in reality embodies a number of important truths. It +would take too long to analyse the meaning of Mr~Russell's epigram in detail, +but one at any rate of its implications is this, that the symbols of mathematics +are capable of varying interpretations, and that we are in general at +liberty to adopt whichever we prefer. +\end{Remark} + +There are now three cases to distinguish. It may happen that +all negative rational numbers belong to the lower class and zero +and all positive rational numbers to the upper. We describe +this section as the \Emph{real number zero}. Or again it may happen +that the lower class includes some positive numbers. Such a section +\PageSep{15} +we describe as a \Emph{positive real number}. Finally it may happen +that some negative numbers belong to the upper class. Such +a section we describe as a \Emph{negative real number}.\footnote + {There are also sections in which every number belongs to the lower or to + the upper class. The reader may be tempted to ask why we do not regard these + sections also as defining numbers, which we might call the \emph{real numbers positive + and negative infinity}. + + There is no logical objection to such a procedure, but it proves to be inconvenient + in practice. The most natural definitions of addition and multiplication do + not work in a satisfactory way. Moreover, for a beginner, the chief difficulty in the + elements of analysis is that of learning to attach precise senses to phrases containing + the word `infinity'; and experience seems to show that he is likely to be confused by + any addition to their number.} + +\begin{Remark} +The difference between our present definition of a positive real number~$a$ +and that of \SecNo[§]{7} amounts to the addition to the lower class of zero and all the +negative rational numbers. An example of a negative real number is given +by taking the property~$P$ of \SecNo[§]{6} to be $x + 1 < 0$ and $Q$~to be $x + 1 \geq 0$. +This section plainly corresponds to the negative rational number~$-1$. If we +took $P$~to be $x^{3} < -2$ and $Q$~to be $x^{3} > -2$, we should obtain a negative real +number which is not rational. +\end{Remark} + +\Paragraph{9. Relations of magnitude between real numbers.} It +is plain that, now that we have extended our conception of +number, we are bound to make corresponding extensions of our +conceptions of equality, inequality, addition, multiplication, and so +on. We have to show that these ideas can be applied to the new +numbers, and that, when this extension of them is made, all the +ordinary laws of algebra retain their validity, so that we can +operate with real numbers in general in exactly the same way +as with the rational numbers of \SecNo[§]{1}. To do all this systematically +would occupy a considerable space, and we shall be content to +indicate summarily how a more systematic discussion would +proceed. + +We denote a real number by a Greek letter such as $\alpha$, $\beta$, $\gamma, \dots$; +the rational numbers of its lower and upper classes by the corresponding +English letters $a$,~$A$; $b$,~$B$; $c$,~$C$;~\dots. The classes themselves +we denote by $(a)$,~$(A), \dots$. + +If $\alpha$~and~$\beta$ are two real numbers, there are three possibilities: + +\Itemp{(i)} every~$a$ is a~$b$ and every~$A$ a~$B$; in this case $(a)$~is identical +with~$(b)$ and $(A)$~with~$(B)$; +\PageSep{16} + +\Itemp{(ii)} every~$a$ is a~$b$, but not all~$A$'s are~$B$'s; in this case $(a)$~is +a proper part of~$(b)$,\footnote + {\Ie\ is included in but not identical with~$(b)$.} +and $(B)$~a proper part of~$(A)$; + +\Itemp{(iii)} every~$A$ is a~$B$, but not all~$a$'s are~$b$'s. + +These three cases may be indicated graphically as in \Fig{4}. + +In case~(i) we write $\alpha = \beta$, in case~(ii) $\alpha < \beta$, and in case~(iii) +$\alpha > \beta$. It is clear that, when +$\alpha$~and~$\beta$ are both rational, these +%[Illustration: Fig. 4.] +\Figure[0.4\textwidth]{4}{p016} +definitions agree with the ideas of +equality and inequality between +rational numbers which we began +by taking for granted; and that +any positive number is greater +than any negative number. + +It will be convenient to define at this stage the negative~$-\alpha$ +of a positive number~$\alpha$. If $(a)$,~$(A)$ are the classes which constitute~$\alpha$, +we can define another section of the rational numbers by +putting all numbers~$-A$ in the lower class and all numbers~$-a$ +in the upper. The real number thus defined, which is clearly +negative, we denote by~$-\alpha$. Similarly we can define~$-\alpha$ when $\alpha$~is +negative or zero; if $\alpha$~is negative, $-\alpha$~is positive. It is plain +also that $-(-\alpha) = \alpha$. Of the two numbers $\alpha$~and~$-\alpha$ one is always +positive (unless $\alpha = 0$). The one which is positive we denote by~$|\alpha|$ +and call the \emph{modulus} of~$\alpha$. + +\begin{Examples}{IV.} +\Item{1.} Prove that $0 = -0$. + +\Item{2.} Prove that $\beta = \alpha$, $\beta < \alpha$, or $\beta > \alpha$ according as $\alpha = \beta$, $\alpha > \beta$, or $\alpha < \beta$. + +\Item{3.} If $\alpha = \beta$ and $\beta = \gamma$, then $\alpha = \gamma$. + +\Item{4.} If $\alpha \leq \beta$, $\beta < \gamma$, or $\alpha < \beta$, $\beta \leq \gamma$, then $\alpha < \gamma$. + +\Item{5.} Prove that $-\beta = -\alpha$, $-\beta < -\alpha$, or $-\beta > -\alpha$, according as $\alpha = \beta$, $\alpha < \beta$, +or $\alpha > \beta$. + +\Item{6.} Prove that $\alpha > 0$ if $\alpha$~is positive, and $\alpha < 0$ if $\alpha$~is negative. + +\Item{7.} Prove that $\alpha \leq |\alpha|$. + +\Item{8.} Prove that $1 < \sqrt{2} < \sqrt{3} < 2$. + +\Item{9.} Prove that, if $\alpha$~and~$\beta$ are two different real numbers, we can always +find an infinity of rational numbers lying between $\alpha$~and~$\beta$. + +[All these results are immediate consequences of our definitions.] +\end{Examples} +\PageSep{17} + +\Paragraph{10. Algebraical operations with real numbers.} We now +proceed to define the meaning of the elementary algebraical operations +such as addition, as applied to real numbers in general. + +\Par{\Itemp{(i)} Addition.} In order to define the sum of two numbers +$\alpha$~and~$\beta$, we consider the following two classes: (i)~the class~$(c)$ +formed by all sums $c = a + b$, (ii)~the class~$(C)$ formed by all sums +$C = A + B$. Plainly $c < C$ in all cases. + +Again, there cannot be more than one rational number which +does not belong either to~$(c)$ or to~$(C)$. For suppose there were +two, say $r$~and~$s$, and let $s$~be the greater. Then both $r$~and~$s$ +must be greater than every~$c$ and less than every~$C$; and so $C - c$ +cannot be less than $s - r$. But +\[ +C - c = (A - a) + (B - b); +\] +and we can choose $a$, $b$, $A$, $B$ so that both $A - a$ and $B - b$ +are as small as we like; and this plainly contradicts our +hypothesis. + +If every rational number belongs to~$(c)$ or to~$(C)$, the classes $(c)$,~$(C)$ +form a section of the rational numbers, that is to say, a number~$\gamma$. +If there is one which does not, we add it to~$(C)$. We have +now a section or real number~$\gamma$, which must clearly be rational, +since it corresponds to the least member of~$(C)$. \emph{In any case +we call~$\gamma$ the sum of $\alpha$~and~$\beta$, and write} +\[ +\gamma = \alpha + \beta. +\] + +\begin{Remark} +If both $\alpha$~and~$\beta$ are rational, they are the least members of the upper +classes $(A)$~and~$(B)$. In this case it is clear that $\alpha + \beta$ is the least member +of~$(C)$, so that our definition agrees with our previous ideas of addition. +\end{Remark} + +\Par{\Itemp{(ii)} Subtraction.} We define $\alpha - \beta$ by the equation +\[ +\alpha - \beta = \alpha + (-\beta). +\] +The idea of subtraction accordingly presents no fresh difficulties. + +\begin{Examples}{V.} +\Item{1.} Prove that $\alpha + (-\alpha) = 0$. + +\Item{2.} Prove that $\alpha + 0 = 0 + \alpha = \alpha$. + +\Item{3.} Prove that $\alpha + \beta = \beta + \alpha$. [This follows at once from the fact that the +classes $(a + b)$~and~$(b + a)$, or $(A + B)$~and~$(B + A)$, are the same, since, \eg, +$a + b = b + a$ when $a$~and~$b$ are rational.] + +\Item{4.} Prove that $\alpha + (\beta + \gamma) = (\alpha + \beta) + \gamma$. +\PageSep{18} + +\Item{5.} Prove that $\alpha - \alpha = 0$. + +\Item{6.} Prove that $\alpha - \beta = -(\beta - \alpha)$. + +\Item{7.} From the definition of subtraction, and Exs.\ 4,~1, and~2 above, it +follows that +\[ +(\alpha - \beta) + \beta + = \{\alpha + (-\beta)\} + \beta + = \alpha + \{(-\beta) + \beta\} + = \alpha + 0 = \alpha. +\] +We might therefore define the difference $\alpha - \beta = \gamma$ by the equation $\gamma + \beta = \alpha$. + +\Item{8.} Prove that $\alpha - (\beta - \gamma) = \alpha - \beta + \gamma$. + +\Item{9.} Give a definition of subtraction which does not depend upon a previous +definition of addition. [To define $\gamma = \alpha - \beta$, form the classes $(c)$,~$(C)$ for which +$c = a - B$, $C = A - b$. It is easy to show that this definition is equivalent to +that which we adopted in the text.] + +\Item{10.} Prove that +\[ +\big||\alpha| - |\beta|\big| \leq |\alpha ± \beta| \leq |\alpha| + |\beta|. +\] +\end{Examples} + +\Paragraph{11. Algebraical operations with real numbers (\continued).} +\Itemp{(iii)}~\emph{Multiplication}. When we come to multiplication, +it is most convenient to confine ourselves to \emph{positive} numbers +(among which we may include~$0$) in the first instance, and to go +back for a moment to the sections of positive rational numbers +only which we considered in \SecNo[§§]{4}--\SecNo{7}. We may then follow practically +the same road as in the case of addition, taking~$(c)$ to be~$(ab)$ +and $(C)$ to be~$(AB)$. The argument is the same, except when we +are proving that all rational numbers with at most one exception +must belong to $(c)$~or~$(C)$. This depends, as in the case of addition, +on showing that we can choose $a$,~$A$, $b$, and~$B$ so that $C - c$ is +as small as we please. Here we use the identity +\[ +C - c = AB - ab = (A - a)B + a(B - b). +\] + +Finally we include negative numbers within the scope of our +definition by agreeing that, if $\alpha$~and~$\beta$ are positive, then +\[ +(-\alpha)\beta = -\alpha\beta,\quad +\alpha(-\beta) = -\alpha\beta,\quad +(-\alpha)(-\beta) = \alpha\beta. +\] + +\Par{\Itemp{(iv)} Division.} In order to define division, we begin by defining +the reciprocal~$1/\alpha$ of a number~$\alpha$ (other than zero). Confining +ourselves in the first instance to positive numbers and +sections of positive rational numbers, we define the reciprocal of a +positive number~$\alpha$ by means of the lower class~$(1/A)$ and the upper +class~$(1/a)$. We then define the reciprocal of a negative number~$-\alpha$ +by the equation $1/(-\alpha) = -(1/\alpha)$. Finally we define $\alpha/\beta$ by +the equation +\[ +\alpha/\beta = \alpha × (1/\beta). +\] +\PageSep{19} + +We are then in a position to apply to all real numbers, rational +or irrational, the whole of the ideas and methods of elementary +algebra. Naturally we do not propose to carry out this task in +detail. It will be more profitable and more interesting to turn +our attention to some special, but particularly important, classes +of irrational numbers. + +\begin{Examples}{VI.} +Prove the theorems expressed by the following +formulae: + +%[** TN: One-off two-column layout] +\begin{minipage}{0.5\textwidth-\parindent} +\Item{1.} $\alpha × 0 = 0 × \alpha = 0$. + +\Item{2.} $\alpha × 1 = 1 × \alpha = \alpha$. + +\Item{3.} $\alpha × (1/\alpha) = 1$. + +\Item{4.} $\alpha\beta = \beta\alpha$. +\end{minipage}% +\begin{minipage}{0.5\textwidth} +\Item{5.} $\alpha(\beta\gamma) = (\alpha\beta)\gamma$. + +\Item{6.} $\alpha(\beta + \gamma) = \alpha\beta + \alpha\gamma$. + +\Item{7.} $(\alpha + \beta)\gamma = \alpha\gamma + \beta\gamma$. + +\Item{8.} $|\alpha\beta| = |\alpha|\, |\beta|$. +\end{minipage} +\end{Examples} + +\Paragraph{12. The number $\sqrt{2}$.} Let us now return for a moment to +the particular irrational number which we discussed in \SecNo[§§]{4}--\SecNo{5}. +We there constructed a section by means of the inequalities +$x^{2} < 2$, $x^{2} > 2$. This was a section of the positive rational numbers +only; but we replace it (as was explained in \SecNo[§]{8}) by a section of +all the rational numbers. We denote the section or number thus +defined by the symbol~$\sqrt{2}$. + +The classes by means of which the product of $\sqrt{2}$ by itself is +defined are (i)~$(aa')$, where $a$~and~$a'$ are positive rational numbers +whose squares are less than~$2$, (ii)~$(AA')$, where $A$~and~$A'$ are +positive rational numbers whose squares are greater than~$2$. These +classes exhaust all positive rational numbers save one, which can +only be~$2$ itself. Thus +\[ +(\sqrt{2})^{2} = \sqrt{2}\sqrt{2} = 2. +\] + +Again +\[ +(-\sqrt{2})^{2} + = (-\sqrt{2})(-\sqrt{2}) + = \sqrt{2}\sqrt{2} + = (\sqrt{2})^{2} = 2. +\] +Thus \emph{the equation $x^{2} = 2$ has the two roots $\sqrt{2}$~and~$-\sqrt{2}$}. Similarly +we could discuss the equations $x^{2} = 3$, $x^{3} = 7, \dots$ and the corresponding +irrational numbers $\sqrt{3}$,~$-\sqrt{3}$, $\sqrt[3]{7}, \dots$. + +\Paragraph{13. Quadratic surds.} A number of the form~$±\sqrt{a}$, where +$a$~is a positive rational number which is not the square of another +rational number, is called a \emph{pure quadratic surd}. A number of +the form $a ± \sqrt{b}$, where $a$~is rational, and $\sqrt{b}$~is a pure quadratic +surd, is sometimes called a mixed quadratic surd. +\PageSep{20} + +\begin{Remark} +The two numbers $a ± \sqrt{b}$ are the roots of the quadratic equation +\[ +x^{2} - 2ax + a^{2} - b = 0. +\] +Conversely, the equation $x^{2} + 2px + q = 0$, where $p$~and~$q$ are rational, and +$p^{2} - q > 0$, has as its roots the two quadratic surds $-p ± \sqrtp{p^{2} - q}$. +\end{Remark} + +The only kind of irrational numbers whose existence was +suggested by the geometrical considerations of \SecNo[§]{3} are these +quadratic surds, pure and mixed, and the more complicated +irrationals which may be expressed in a form involving the +repeated extraction of square roots, such as +\[ +\sqrt{2} + \sqrtp{2 + \sqrt{2}} + \sqrtb{2 + \sqrtp{2 + \sqrt{2}}}. +\] + +It is easy to construct geometrically a line whose length is +equal to any number of this form, as the reader will easily see for +himself. That irrational numbers of these kinds \emph{only} can be constructed +by Euclidean methods (\ie~by geometrical constructions +with ruler and compasses) is a point the proof of which must +be deferred for the present.\footnote + {See \Ref{Ch.}{II}, \MiscExs{II}~22.} +This property of quadratic surds +makes them especially interesting. + +\begin{Examples}{VII.} +\Item{1.} Give geometrical constructions for +\[ +\sqrt{2},\quad +\sqrtp{2 + \sqrt{2}},\quad +\sqrtb{2 + \sqrtp{2 + \sqrt{2}}}. +\] + +\Item{2.} The quadratic equation $ax^{2} + 2bx + c = 0$ has two real roots\footnote + {\Ie\ there are two values of~$x$ for which $ax^{2} + 2bx + c = 0$. If $b^{2} - ac < 0$ there + are no such values of~$x$. The reader will remember that in books on elementary + algebra the equation is said to have two `complex' roots. The meaning to be + attached to this statement will be explained in \Ref{Ch.}{III}\@. + + When $b^{2} = ac$ the equation has only one root. For the sake of uniformity + it is generally said in this case to have `two equal' roots, but this is a mere + convention.} +if +$b^{2} - ac > 0$. Suppose $a$,~$b$,~$c$ rational. Nothing is lost by taking all three +to be integers, for we can multiply the equation by the least common +multiple of their denominators. + +The reader will remember that the roots are $\{-b ± \sqrtp{b^{2} - ac}\}/a$. It is +easy to construct these lengths geometrically, first constructing $\sqrtp{b^{2} - ac}$. +A much more elegant, though less straightforward, construction is the +following. +\PageSep{21} + +\begin{Construction} +Draw a circle of unit radius, a diameter~$PQ$, and the tangents at the ends +of the diameters. +%[Illustration: Fig. 5.] +\Figure[0.7\textwidth]{5}{p021} + +Take $PP' = -2a/b$ and $QQ' = -c/2b$, having regard to sign.\footnote + {The figure is drawn to suit the case in which $b$~and~$c$ have the same and $a$ + the opposite sign. The reader should draw figures for other cases.} +Join $P'Q'$, +cutting the circle in $M$~and~$N$. Draw $PM$~and~$PN$, cutting~$QQ'$ in $X$~and~$Y$. +Then $QX$~and~$QY$ are the roots of the equation with their proper signs.\footnote + {I have taken this construction from Klein's \textit{Leçons sur certaines questions de + géométrie élémentaire} (French translation by J.~Griess, Paris, 1896).} +\end{Construction} + +The proof is simple and we leave it as an exercise to the reader. +Another, perhaps even simpler, construction is the following. \begin{Construction}[]Take a line +$AB$ of unit length. Draw $BC = -2b/a$ perpendicular to~$AB$, and $CD = c/a$ + perpendicular to~$BC$ and in the same direction as~$BA$. On~$AD$ as diameter +describe a circle cutting~$BC$ in $X$~and~$Y$. Then $BX$~and~$BY$ are the roots. +\end{Construction} + +\Item{3.} If $ac$ is positive $PP'$~and~$QQ'$ will be drawn in the same direction. +Verify that $P'Q'$~will not meet the circle if $b^{2} < ac$, while if $b^{2} = ac$ it will be +a tangent. Verify also that if $b^{2} = ac$ the circle in the second construction +will touch~$BC$. + +\Item{4.} Prove that +\[ +\sqrtp{pq} = \sqrt{p} × \sqrt{q},\quad +\sqrtp{p^{2}q} = p\sqrt{q}. +\] +\end{Examples} + +\Paragraph{14. Some theorems concerning quadratic surds.} Two +pure quadratic surds are said to be \emph{similar} if they can be expressed +as rational multiples of the same surd, and otherwise to be +\emph{dissimilar}. Thus +\[ +\sqrt{8} = 2\sqrt{2},\quad +\sqrt{\tfrac{25}{2}} = \tfrac{5}{2}\sqrt{2}, +\] +and so $\sqrt{8}$,~$\sqrt{\frac{25}{2}}$ are similar surds. On the other hand, if $M$~and~$N$ +are integers which have no common factor, and neither of which +is a perfect square, $\sqrt{M}$~and~$\sqrt{N}$ are dissimilar surds. For suppose, +if possible, +\[ +\sqrt{M} = \frac{p}{q}\bigsqrt{\frac{t}{u}},\quad +\sqrt{N} = \frac{r}{s}\bigsqrt{\frac{t}{u}}, +\] +where all the letters denote integers. +\PageSep{22} + +Then $\DPtypo{\sqrt{MN}}{\sqrtp{MN}}$ is evidently rational, and therefore (\Ex{ii}.~3) +integral. Thus $MN = P^{2}$, where $P$~is an integer. Let $a$,~$b$, $c, \dots$ +be the prime factors of~$P$, so that +\[ +MN = a^{2\alpha} b^{2\beta} c^{2\gamma}\ \dots, +\] +where $\alpha$,~$\beta$, $\gamma, \dots$ are positive integers. Then $MN$~is divisible by~$a^{2\alpha}$, +and therefore either (1)~$M$~is divisible by~$a^{2\alpha}$, or (2)~$N$~is +divisible by~$a^{2\alpha}$, or (3)~$M$ and~$N$ are both divisible by~$a$. The last +case may be ruled out, since $M$~and~$N$ have no common factor. +This argument may be applied to each of the factors $a^{2\alpha}$,~$b^{2\beta}$, $c^{2\gamma}, \dots$, so that $M$~must be divisible by some of these factors and $N$~by +the remainder. Thus +\[ +M = P_{1}^{2},\quad +N = P_{2}^{2}, +\] +where $P_{1}^{2}$~denotes the product of some of the factors $a^{2\alpha}$,~$b^{2\beta}$, $c^{2\gamma}, \dots$ +and $P_{2}^{2}$~the product of the rest. Hence $M$~and~$N$ are both perfect +squares, which is contrary to our hypothesis. + +\begin{Theorem} +If $A$, $B$, $C$, $D$ are rational and +\[ +A + \sqrt{B} = C + \sqrt{D}, +\] +then either \Inum{(i)}~$A = C$, $B = D$ or \Inum{(ii)}~$B$ and~$D$ are both squares of +rational numbers. +\end{Theorem} + +For $B - D$ is rational, and so is +\[ +\sqrt{B} - \sqrt{D} = C - A. +\] +If $B$~is not equal to~$D$ (in which case it is obvious that $A$~is also +equal to~$C$), it follows that +\[ +\sqrt{B} + \sqrt{D} = (B - D)/(\sqrt{B}- \sqrt{D}) +\] +is also rational. Hence $\sqrt{B}$~and~$\sqrt{D}$ are rational. + +\begin{Corollary} +If $A + \sqrt{B} = C + \sqrt{D}$, then $A - \sqrt{B} = C - \sqrt{D}$ +\(unless $\sqrt{B}$~and~$\sqrt{D}$ are both rational\). +\end{Corollary} + +\begin{Examples}{VIII.} +\Item{1.} Prove \textit{ab initio} that $\sqrt{2}$~and~$\sqrt{3}$ are not similar +surds. + +\Item{2.} Prove that $\sqrt{a}$~and~$\sqrtp{1/a}$, where $a$~is rational, are similar surds +(unless both are rational). + +\Item{3.} If $a$~and~$b$ are rational, then $\sqrt{a} + \sqrt{b}$ cannot be rational unless $\sqrt{a}$~and~$\sqrt{b}$ +are rational. The same is true of $\sqrt{a}- \sqrt{b}$, unless $a = b$. +\PageSep{23} + +\Item{4.} If +\[ +\sqrt{A} + \sqrt{B} = \sqrt{C} + \sqrt{D}, +\] +then either (\ia)~$A = C$ and~$B = D$, or (\ib)~$A = D$ and~$B = C$, or (\ic)~$\sqrt{A}$, $\sqrt{B}$, $\sqrt{C}$, +$\sqrt{D}$ are all rational or all similar surds. [Square the given equation and +apply the theorem above.] + +\Item{5.} Neither $(a + \sqrt{b})^{3}$ nor $(a - \sqrt{b})^{3}$ can be rational unless $\sqrt{b}$~is rational. + +\Item{6.} Prove that if $x = p + \sqrt{q}$, where $p$~and~$q$ are rational, then $x^{m}$, where +$m$~is any integer, can be expressed in the form $P + Q \sqrt{q}$, where $P$~and~$Q$ +are rational. For example, +\[ +(p + \sqrt{q})^{2} = p^{2} + q + 2p\sqrt{q},\quad +(p + \sqrt{q})^{3} = p^{3} + 3pq + (3p^{2} + q)\sqrt{q}. +\] +Deduce that any polynomial in~$x$ with rational coefficients (\ie~any expression +of the form +\[ +a_{0}x^{n} + a_{1}x^{n-1} + \dots + a_{n}, +\] +where $a_{0}$,~\dots\Add{,} $a_{n}$ are rational numbers) can be expressed in the form $P + Q\sqrt{q}$. + +\Item{7.} If $a + \sqrt{b}$, where $b$~is not a perfect square, is the root of an algebraical +equation with rational coefficients, then $a - \sqrt{b}$ is another root of the same +equation. + +\Item{8.} Express $1/(p + \sqrt{q})$ in the form prescribed in Ex.~6. [Multiply +numerator and denominator by~$p - \sqrt{q}$.] + +\Item{9.} Deduce from Exs.\ 6~and~8 that any expression of the form $G(x)/H(x)$, +where $G(x)$~and~$H(x)$ are polynomials in~$x$ with rational coefficients, can be +expressed in the form $P + Q\sqrt{q}$, where $P$~and~$Q$ are rational. + +\Item{10.} If $p$,~$q$, and $p^{2} - q$ are positive, we can express $\sqrtp{p + \sqrt{q}}$ in the form +$\sqrt{x} + \sqrt{y}$, where +\[ +x = \tfrac{1}{2}\{p + \sqrtp{p^{2} - q}\},\quad +y = \tfrac{1}{2}\{p - \sqrtp{p^{2} - q}\}. +\] + +\Item{11.} Determine the conditions that it may be possible to express $\sqrtp{p + \sqrt{q}}$, +where $p$~and~$q$ are rational, in the form $\sqrt{x} + \sqrt{y}$, where $x$~and~$y$ are rational. + +\Item{12.} If $a^{2} - b$ is positive, the necessary and sufficient conditions that +\[ +\sqrtp{a + \sqrt{b}} + \sqrtp{a - \sqrt{b}} +\] +should be rational are that $a^{2} - b$ and $\frac{1}{2}\{a + \sqrtp{a^{2} - b}\}$ should both be squares +of rational numbers. +\end{Examples} + +\Paragraph{15. The continuum.} The aggregate of all real numbers, +rational and irrational, is called the \Emph{arithmetical continuum}. + +It is convenient to suppose that the straight line~$\Lambda$ of \SecNo[§]{2} +is composed of points corresponding to all the numbers of the +arithmetical continuum, and of no others.\footnote + {This supposition is merely a hypothesis adopted (i)~because it suffices for the + purposes of our geometry and (ii)~because it provides us with convenient geometrical + illustrations of analytical processes. As we use geometrical language only for + purposes of illustration, it is not part of our business to study the foundations + of geometry.} +The points of the +\PageSep{24} +line, the aggregate of which may be said to constitute the \Emph{linear +continuum}, then supply us with a convenient image of the +arithmetical continuum. + +We have considered in some detail the chief properties of a +few classes of real numbers, such, for example, as rational numbers +or quadratic surds. We add a few further examples to show how +very special these particular classes of numbers are, and how, to +put it roughly, they comprise only a minute fraction of the infinite +variety of numbers which constitute the continuum. + +\begin{Remark} +\Itemp{(i)} Let us consider a more complicated surd expression such as +\[ +z = \sqrtp[3]{4 + \sqrt{15}} + \sqrtp[3]{4 - \sqrt{15}}. +\] +Our argument for supposing that the expression for~$z$ has a meaning might be +as follows. We first show, as in \SecNo[§]{12}, that there is a number $y = \sqrt{15}$ such that +$y^{2} = 15$, and we can then, as in \SecNo[§]{10}, define the numbers $4 + \sqrt{15}$, $4 - \sqrt{15}$. +Now consider the equation in~$z_{1}$, +\[ +z_{1}^{3} = 4 + \sqrt{15}. +\] +The right-hand side of this equation is not rational: but exactly the same +reasoning which leads us to suppose that there is a real number~$x$ such that +$x^{3} = 2$ (or any other rational number) also leads us to the conclusion that there +is a number~$z_{1}$ such that $z_{1}^{3} = 4 + \sqrt{15}$. We thus define $z_{1} = \sqrtp[3]{4 + \sqrt{15}}$, and +similarly we can define $z_{2} = \sqrtp[3]{4 - \sqrt{15}}$; and then, as in \SecNo[§]{10}, we define $z = z_{1} + z_{2}$. + +Now it is easy to verify that +\[ +z^{3} = 3z + 8. +\] +And we might have given a direct proof of the existence of a unique number~$z$ +such that $z^{3} = 3z + 8$. It is easy to see that there cannot be two such +numbers. For if $z_{1}^{3} = 3z_{1} + 8$ and $z_{2}^{3} = 3z_{2} + 8$, we find on subtracting and +dividing by $z_{1} - z_{2}$ that $z_{1}^{2} + z_{1}z_{2} + z_{2}^{2} = 3$. But if $z_{1}$~and~$z_{2}$ are positive $z_{1}^{3}>8$, +$z_{2}^{3}>8$ and therefore $z_{1} > 2$, $z_{2} > 2$, $z_{1}^{2} + z_{1}z_{2} + z_{2}^{2} > 12$, and so the equation +just found is impossible. And it is easy to see that neither $z_{1}$ nor~$z_{2}$ can +be negative. For if $z_{1}$~is negative and equal to~$-\zeta$, $\zeta$~is positive and +$\zeta^{3} - 3\zeta + 8 = 0$, or $3 - \zeta^{2} = 8/\zeta$. Hence $3 - \zeta^{2} > 0$, and so $\zeta < 2$. But then +$8/\zeta > 4$, and so $8/\zeta$ cannot be equal to~$3 - \zeta^{2}$, which is less~than~$3$. + +Hence there is at most one~$z$ such that $z^{3} = 3z + 8$. And it cannot be +rational. For any rational root of this equation must be integral and a +factor of~$8$ (\Ex{ii}.~3), and it is easy to verify that no one of $1$, $2$, $4$,~$8$ is a root. + +Thus $z^{3} = 3z + 8$ has at most one root and that root, if it exists, is positive +and not rational. We can now divide the positive rational numbers~$x$ into +two classes $L$,~$R$ according as $x^{3} < 3x + 8$ or $x^{3} > 3x + 8$. It is easy to see that +if $x^{3} > 3x + 8$ and $y$~is any number greater than~$x$, then also $y^{3} > 3y + 8$. For +suppose if possible $y^{3} \leq 3y + 8$. Then since $x^{3} > 3x + 8$ we obtain on subtracting +$y^{3} - x^{3} < 3(y - x)$, or $y^{2} + xy + x^{2} < 3$, which is impossible; for $y$~is +\PageSep{25} +positive and $x > 2$ (since $x^{3} > 8$). Similarly we can show that if $x^{3} < 3x + 8$ +and $y < x$ then also $y^{3} < 3y + 8$. + +Finally, it is evident that the classes $L$~and~$R$ both exist; and they form +a section of the positive rational numbers or positive real number~$z$ which +satisfies the equation $z^{3} = 3z + 8$. The reader who knows how to solve cubic +equations by Cardan's method will be able to obtain the explicit expression of~$z$ +directly from the equation. +\end{Remark} + +\Itemp{(ii)} The direct argument applied above to the equation +$x^{3} = 3x + 8$ could be applied (though the application would be +a little more difficult) to the equation +\[ +x^{5} = x + 16\DPtypo{.}{,} +\] +and would lead us to the conclusion that a unique positive real +number exists which satisfies this equation. In this case, however, +it is not possible to obtain a simple explicit expression +for~$x$ composed of any combination of surds. It can in fact +be proved (though the proof is difficult) that it is \emph{generally} +impossible to find such an expression for the root of an equation +of higher degree than~$4$. Thus, besides irrational numbers which +can be expressed as pure or mixed quadratic or other surds, or +combinations of such surds, there are others which are roots of +algebraical equations but cannot be so expressed. It is only in +very special cases that such expressions can be found. + +\Itemp{(iii)} But even when we have added to our list of irrational +numbers roots of equations (such as $x^{5} = x + 16$) which cannot be +explicitly expressed as surds, we have not exhausted the different +kinds of irrational numbers contained in the continuum. Let us +draw a circle whose diameter is equal to~$A_{0}A_{1}$, \ie~to unity. It is +natural to suppose\footnote + {A proof will be found in \Ref{Ch.}{VII}\@.} +that the circumference of such a circle has a +length capable of numerical measurement. This length is usually +denoted by~$\pi$. And it has been shown\footnote + {See Hobson's \textit{Trigonometry} (3rd~edition), pp.~305 \textit{et~seq.}, or the same writer's + \textit{Squaring the Circle} (Cambridge,~1913).} +(though the proof is unfortunately +long and difficult) that this number~$\pi$ is not the +root of any algebraical equation with integral coefficients, such, +for example, as +\[ +\pi^{2} = n,\quad +\pi^{3} = n,\quad +\pi^{5} = \pi + n, +\] +\PageSep{26} +where $n$~is an~integer. In this way it is possible to define a +number which is not rational nor yet belongs to any of the classes +of irrational numbers which we have so far considered. And this +number~$\pi$ is no isolated or exceptional case. Any number of other +examples can be constructed. In fact it is only special classes of +irrational numbers which are roots of equations of this kind, just +as it is only a still smaller class which can be expressed by means +of surds. + +\Paragraph{16. The continuous real variable.} The `real numbers' +may be regarded from two points of view. We may think of +them \emph{as an aggregate}, the `arithmetical continuum' defined in +the preceding section, or \emph{individually}. And when we think of +them individually, we may think either of a particular \emph{specified} +number (such as $1$, $-\frac{1}{2}$, $\sqrt{2}$, or~$\pi$) or we may think of \emph{any} number, +\emph{an unspecified} number, \emph{the number~$x$}. This last is our point of +view when we make such assertions as `$x$~is a~number', `$x$~is the +measure of a length', `$x$~may be rational or irrational'\DPtypo{,}{.} The~$x$ +which occurs in propositions such as these is called \emph{the continuous +real variable}: and the individual numbers are called the \emph{values} of +the variable. + +A `variable', however, need not necessarily be continuous. +Instead of considering the aggregate of \emph{all} real numbers, we +might consider some partial aggregate contained in the former +aggregate, such as the aggregate of rational numbers, or the +aggregate of positive integers. Let us take the last case. Then +in statements about \emph{any} positive integer, or \emph{an unspecified} positive +integer, such as `$n$~is either odd or~even', $n$~is called the variable, +a \emph{positive integral variable}, and the individual positive integers +are its values. + +Naturally `$x$'~and~`$n$' are only examples of variables, the +variable whose `field of variation' is formed by all the real +numbers, and that whose field is formed by the positive integers. +These are the most important examples, but we have often to +consider other cases. In the theory of decimals, for instance, we +may denote by~$x$ any figure in the expression of any number as a +decimal. Then $x$~is a variable, but a variable which has only ten +different values, viz.\ $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$,~$9$. The reader should +\PageSep{27} +think of other examples of variables with different fields of variation. +He will find interesting examples in ordinary life: policeman~$x$, +the driver of cab~$x$, the year~$x$, the $x$th~day of the week. The +values of these variables are naturally not numbers. + +\Paragraph{17. Sections of the real numbers.} In \SecNo[§§]{4}--\SecNo{7} we considered +`sections' of the rational numbers, \ie\ modes of division of +the rational numbers (or of the positive rational numbers only) +into two classes $L$~and~$R$ possessing the following characteristic +properties: + +\Itemp{(i)} that every number of the type considered belongs to one +and only one of the two classes; + +\Itemp{(ii)} that both classes exist; + +\Itemp{(iii)} that any member of~$L$ is less than any member of~$R$. + +It is plainly possible to apply the same idea to the aggregate +of all real numbers, and the process is, as the reader will find in +later chapters, of very great importance. + +Let us then suppose\footnote + {The discussion which follows is in many ways similar to that of \SecNo[§]{6}. We + have not attempted to avoid a certain amount of repetition. The idea of a `section,' + first brought into prominence in Dedekind's famous pamphlet \textit{Stetigkeit und + irrationale Zahlen}, is one which can, and indeed must, be grasped by every reader + of this book, even if he be one of those who prefer to omit the discussion of the + notion of an irrational number contained in \SecNo[§§]{6}--\SecNo{12}.} +that $P$~and~$Q$ are two properties which +are mutually exclusive, and one of which is possessed by every +real number. Further let us suppose that any number which +possesses~$P$ is less than any which possesses~$Q$. We call the +numbers which possess~$P$ the \emph{lower} or \emph{left-hand class}~$L$, and +those which possess~$Q$ the \emph{upper} or \emph{right-hand class}~$R$. + +\begin{Remark} +Thus~$P$ might be $x \leq \sqrt{2}$ and $Q$~be $x > \sqrt{2}$. It is important to observe +that a pair of properties which suffice to define a section of the rational +numbers may not suffice to define one of the real numbers. This is so, for +example, with the pair `$x < \sqrt{2}$' and `$x > \sqrt{2}$' or (if we confine ourselves +to positive numbers) with `$x^{2} < 2$' and `$x^{2} > 2$'. Every rational number +possesses one or other of the properties, but not every real number, since in +either case $\sqrt{2}$~escapes classification. +\end{Remark} + +There are now two possibilities.\footnote + {There were three in \SecNo[§]{6}.} +Either $L$~has a greatest +member~$l$, or $R$~has a least member~$r$. \emph{Both} of these events +\PageSep{28} +cannot occur. For if $L$~had a greatest member~$l$, and $R$~a least +member~$r$, the number $\frac{1}{2}(l + r)$ would be greater than all members +of~$L$ and less than all members of~$R$, and so could not belong to +either class. On the other hand \emph{one} event must occur.\footnote + {This was not the case in \SecNo[§]{6}.} + +For let $L_{1}$~and~$R_{1}$ denote the classes formed from $L$~and~$R$ by +taking only the rational members of $L$~and~$R$. Then the classes +$L_{1}$~and~$R_{1}$ form a section of the rational numbers. There are now +two cases to distinguish. + +It may happen that $L_{1}$~has a greatest member~$\alpha$. In this case +$\alpha$~must be also the greatest member of~$L$. For if not, we could find +a greater, say~$\beta$. There are rational numbers lying between $\alpha$~and~$\beta$, +and these, being less than~$\beta$, belong to~$L$, and therefore to~$L_{1}$; +and this is plainly a contradiction. Hence $\alpha$~is the greatest +member of~$L$. + +On the other hand it may happen that $L_{1}$~has no greatest +member. In this case the section of the rational numbers formed +by $L_{1}$~and~$R_{1}$ is a real number~$\alpha$. This number~$\alpha$ must belong +to~$L$ or to~$R$. If it belongs to~$L$ we can \DPtypo{shew}{show}, precisely as before, +that it is the greatest member of~$L$, and similarly, if it belongs +to~$R$, it is the least member of~$R$. + +Thus in any case either $L$~has a greatest member or $R$~a +least. Any section of the real numbers therefore `corresponds' to +a real number in the sense in which a section of the rational +numbers sometimes, but not always, corresponds to a rational +number. This conclusion is of very great importance; for it shows +that the consideration of sections of all the real numbers does not +lead to any further generalisation of our idea of number. Starting +from the rational numbers, we found that the idea of a section of +the rational numbers led us to a new conception of a number, that +of a real number, more general than that of a rational number; +and it might have been expected that the idea of a section of the +real numbers would have led us to a conception more general still. +The discussion which precedes shows that this is not the case, and +that the aggregate of real numbers, or the continuum, has a kind +of completeness which the aggregate of the rational numbers +lacked, a completeness which is expressed in technical language +by saying that the continuum is closed. +\PageSep{29} + +The result which we have just proved may be stated as follows: + +\begin{ParTheorem}{Dedekind's Theorem.} +If the real numbers are divided into +two classes $L$~and~$R$ in such a way that + +\Itemp{(i)} every number belongs to one or other of the two classes, + +\Itemp{(ii)} each class contains at least one number, + +\Itemp{(iii)} any member of~$L$ is less than any member of~$R$, \\ +then there is a number~$\alpha$, which has the property that all the numbers +less than it belong to~$L$ and all the numbers greater than it to~$R$. +The number~$\alpha$ itself may belong to either class. +\end{ParTheorem} + +\begin{Remark} +In applications we have often to consider sections not of \emph{all} numbers but +of all those contained in an \emph{interval} $\DPmod{(\beta, \gamma)}{[\beta, \gamma]}$, that is to say of all numbers~$x$ +such that $\beta \leq x \leq \gamma$. A `section' of such numbers is of course a division of +them into two classes possessing the properties (i),~(ii), and~(iii). Such +a section may be converted into a section of \emph{all} numbers by adding to~$L$ all +numbers less than~$\beta$ and to~$R$ all numbers greater than~$\gamma$. It is clear that +the conclusion stated in Dedekind's Theorem still holds if we substitute `the +real numbers of the interval $\DPmod{(\beta, \gamma)}{[\beta, \gamma]}$' for `the real numbers', and that the +number~$\alpha$ in this case satisfies the inequalities $\beta \leq \alpha \leq \gamma$. +\end{Remark} + +\Paragraph{18. Points of accumulation.} A system of real numbers, or +of the points on a straight line corresponding to them, defined in +any way whatever, is called an \Emph{aggregate} or \Emph{set} of numbers or +points. The set might consist, for example, of all the positive +integers, or of all the rational points. + +It is most convenient here to use the language of geometry.\footnote + {The reader will hardly require to be reminded that this course is adopted + solely for reasons of linguistic convenience.} +Suppose then that we are given a set of points, which we will +denote by~$S$. Take any point~$\xi$, which may or may not belong to $S$. +Then there are two possibilities. Either (i)~it is possible to choose +a positive number~$\delta$ so that the interval $\DPmod{(\xi - \delta, \xi + \delta)}{[\xi - \delta, \xi + \delta]}$ does not contain +any point of~$S$, other than $\xi$~itself,\footnote + {This clause is of course unnecessary if $\xi$~does not itself belong to~$S$.} +or (ii)~this is not possible. + +\begin{Remark} +Suppose, for example, that $S$~consists of the points corresponding to all +the positive integers. If $\xi$~is itself a positive integer, we can take $\delta$ to be any +number less than~$1$, and (i)~will be true; or, if $\xi$~is halfway between two +positive integers, we can take $\delta$ to be any number less than~$\frac{1}{2}$. On the other +hand, if $S$~consists of all the rational points, then, whatever the value of~$\xi$, +(ii)~is true; for any interval whatever contains an infinity of rational points. +\end{Remark} +\PageSep{30} + +Let us suppose that (ii)~is true. Then any interval $\DPmod{(\xi - \delta, \xi + \delta)}{[\xi - \delta, \xi + \delta]}$, +however small its length, contains at least one point~$\xi_{1}$ which +belongs to~$S$ and does not coincide with~$\xi$; and this whether $\xi$~itself +be a member of~$S$ or not. In this case we shall say that $\xi$~is +a \Emph{point of accumulation} of~$S$. It is easy to see that the interval +$\DPmod{(\xi - \delta, \xi + \delta)}{[\xi - \delta, \xi + \delta]}$ must contain, not merely one, but infinitely many +points of~$S$. For, when we have determined~$\xi_{1}$, we can take an +interval $\DPmod{(\xi - \delta_{1}, \xi + \delta_{1})}{[\xi - \delta_{1}, \xi + \delta_{1}]}$ surrounding~$\xi$ but not reaching as far as~$\xi_{1}$. +But this interval also must contain a point, say~$\xi_{2}$, which is a +member of~$S$ and does not coincide with~$\xi$. Obviously we may +repeat this argument, with $\xi_{2}$~in the place of~$\xi_{1}$; and so on +indefinitely. In this way we can determine as many points +\[ +\xi_{1},\quad \xi_{2},\quad \xi_{3},\ \dots +\] +as we please, all belonging to~$S$, and all lying inside the interval +$\DPmod{(\xi - \delta, \xi + \delta)}{[\xi - \delta, \xi + \delta]}$. + +A point of accumulation of~$S$ may or may not be itself a point +of~$S$. The examples which follow illustrate the various possibilities. + +\begin{Examples}{IX.} +\Item{1.} If $S$~consists of the points corresponding to the +positive integers, or all the integers, there are no points of accumulation. + +\Item{2.} If $S$~consists of all the rational points, every point of the line is a +point of accumulation. + +\Item{3.} If $S$~consists of the points $1$, $\frac{1}{2}$, $\frac{1}{3}, \dots$, there is one point of accumulation, +viz.\ the origin. + +\Item{4.} If $S$~consists of all the positive rational points, the points of accumulation +are the origin and all positive points of the line. +\end{Examples} + +\Paragraph{19. Weierstrass's Theorem.} The general theory of sets +of points is of the utmost interest and importance in the higher +branches of analysis; but it is for the most part too difficult to be +included in a book such as this. There is however one fundamental +theorem which is easily deduced from Dedekind's Theorem +and which we shall require later. + +\begin{Theorem} +If a set~$S$ contains infinitely many points, and is +entirely situated in an interval $\DPmod{(\alpha, \beta)}{[\alpha, \beta]}$, then at least one point of the +interval is a point of accumulation of~$S$. +\end{Theorem} + +We divide the points of the line~$\Lambda$ into two classes in the +following manner. The point~$P$ belongs to~$L$ if there are an +\PageSep{31} +infinity of points of~$S$ to the right of~$P$, and to~$R$ in the contrary +case. Then it is evident that conditions (i)~and~(iii) of Dedekind's +Theorem are satisfied; and since $\alpha$~belongs to~$L$ and $\beta$~to~$R$, +condition~(ii) is satisfied also. + +Hence there is a point~$\xi$ such that, however small be~$\delta$, $\xi - \delta$ +belongs to~$L$ and $\xi + \delta$ to~$R$, so that the interval $\DPmod{(\xi-\delta, \xi+\delta)}{[\xi - \delta, \xi + \delta]}$ +contains an infinity of points of~$S$. Hence $\xi$~is a~point of accumulation +of~$S$. + +\begin{Remark} +This point may of course coincide with $\alpha$~or~$\beta$, as for instance when $\alpha = 0$, +$\beta = 1$, and $S$~consists of the points $1$, $\frac{1}{2}$, $\frac{1}{3}, \dots$. In this case $0$~is the sole +point of accumulation. +\end{Remark} + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER I.} + +\begin{Examples}{} +\Item{1.} What are the conditions that $ax + by + cz = 0$, (1)~for all values of +$x$, $y$,~$z$; (2)~for all values of $x$, $y$,~$z$ subject to $\alpha x + \beta y + \gamma z=0$; (3)~for all +values of $x$, $y$,~$z$ subject to both $\alpha x + \beta y + \gamma z = 0$ and $Ax + By + Cz = 0$? + +\Item{2.} Any positive rational number can be expressed in one and only one +way in the form +\[ +a_{1} + \frac{a_{2}}{1·2} + \frac{a_{3}}{1·2·3} + \dots + + \frac{a_{k}}{1·2·3\dots k}, +\] +where $a_{1}$, $a_{2}$, \dots,~$a_{k}$ are integers, and +\[ +0 \leq a_{1},\quad +0 \leq a_{2} < 2,\quad +0 \leq a_{3} < 3,\ \dots\quad +0 < a_{k} < k. +\] + +\Item{3.} Any positive rational number can be expressed in one and one way +only as a simple continued fraction +\[ +%[** TN: Modernized notation, added second-to-last numerator] +a_{1} + \cfrac{1}{a_{2} + \cfrac{1}{a_{3} + \cfrac{1}{\dots + \cfrac{1}{a_{n}}}}}\;, +\] +where $a_{1}$, $a_{2}, \dots$ are positive integers, of which the first only may be zero. + +[Accounts of the theory of such continued fractions will be found in text-books +of algebra. For further information as to modes of representation of +rational and irrational numbers, see Hobson, \textit{Theory of Functions of a Real +Variable}, pp.~45--49.] + +\Item{4.} Find the rational roots (if any) of $9x^{3} - 6x^{2} + 15x - 10 = 0$. + +\Item{5.} A line~$AB$ is divided at~$C$ \textit{in aurea sectione} (Euc.~\textsc{ii}.~11)---\ie\ so that +$AB·AC = BC^{2}$. Show that the ratio~$AC/AB$ is irrational. + +[A direct geometrical proof will be found in Bromwich's \textit{Infinite Series}, +§\:143, p.~363.] + +\Item{6.} $A$~is irrational. In what circumstances can $\smash{\dfrac{aA + b}{cA + d}}$, where $a$, $b$, $c$,~$d$ +are rational, be rational? +\PageSep{32} + +\Item{7.} \Topic{Some elementary inequalities.} In what follows $a_{1}$, $a_{2}, \dots$ denote +positive numbers (including zero) and $p$, $q, \dots$ positive integers. Since +$a_{1}^{p} - a_{2}^{p}$ and $a_{1}^{q} - a_{2}^{q}$ have the same sign, we have $(a_{1}^{p} - a_{2}^{p}) (a_{1}^{q} - a_{2}^{q}) \geq 0$, or +\[ +a_{1}^{p+q} + a_{2}^{p+q} \geq a_{1}^{p} a_{2}^{q} + a_{1}^{q} a_{2}^{p}, +\Tag{(1)} +\] +an inequality which may also be written in the form +\[ +\frac{a_{1}^{p+q} + a_{2}^{p+q}}{2} + \geq \left(\frac{a_{1}^{p} + a_{2}^{p}}{2}\right) + \left(\frac{a_{1}^{q} + a_{2}^{q}}{2}\right). +\Tag{(2)} +\] +By repeated application of this formula we obtain +\[ +\frac{a_{1}^{p+q+r+\dots} + a_{2}^{p+q+r+\dots}}{2} + \geq \left(\frac{a_{1}^{p} + a_{2}^{p}}{2}\right) + \left(\frac{a_{1}^{q} + a_{2}^{q}}{2}\right) + \left(\frac{a_{1}^{r} + a_{2}^{r}}{2}\right) \dots, +\Tag{(3)} +\] +and in particular +\[ +\frac{a_{1}^{p} + a_{2}^{p}}{2} \geq \left(\frac{a_{1} + a_{2}}{2}\right)^{p}. +\Tag{(4)} +\] +When $p = q = 1$ in~\Eq{(1)}, or $p = 2$ in~\Eq{(4)}, the inequalities are merely different +forms of the inequality $a_{1}^{2} + a_{2}^{2} \geq 2a_{1} a_{2}$, which expresses the fact that the +arithmetic mean of two positive numbers is not less than their geometric +mean. + +\Item{8.} \Topic{Generalisations for $n$~numbers.} If we write down the $\frac{1}{2} n(n - 1)$ +inequalities of the type~\Eq{(1)} which can be formed with $n$~numbers $a_{1}$, $a_{2}$, \dots,~$a_{n}$, +and add the results, we obtain the inequality +\[ +n \tsum{a^{p+q}} \geq \tsum a^{p} \tsum a^{q}, +\Tag{(5)} +\] +or +\[ +\left(\tsum a^{p+q}\right)/n + \geq \left\{\left(\tsum a^{p}\right)/n\right\} + \left\{\left(\tsum a^{q}\right)/n\right\}. +\Tag{(6)} +\] +Hence we can deduce an obvious extension of~\Eq{(3)} which the reader may +formulate for himself, and in particular the inequality +\[ +\left(\tsum a^{p}\right)/n \geq \left\{\left(\tsum a\right)/n\right\}^{p}. +\Tag{(7)} +\] + +\Item{9.} \Topic{The general form of the theorem concerning the arithmetic and +geometric means.} An inequality of a slightly different character is +that which asserts that the arithmetic mean of $a_{1}$, $a_{2}$, \dots,~$a_{n}$ is not less +than their geometric mean. Suppose that $a_{r}$~and~$a_{s}$ are the greatest and +least of the~$a$'s (if there are several greatest or least~$a$'s we may choose any +of them indifferently), and let $G$ be their geometric mean. We may suppose +$G > 0$, as the truth of the proposition is obvious when $G = 0$. If now we replace +$a_{r}$~and~$a_{s}$ by +\[ +a_{r}' = G,\quad +a_{s}' = a_{r}a_{s}/G, +\] +we do not alter the value of the geometric mean; and, since +\[ +a_{r}' + a_{s}' - a_{r} - a_{s} = (a_{r} - G)(a_{s} - G)/G \leq 0, +\] +we certainly do not increase the arithmetic mean. + +It is clear that we may repeat this argument until we have replaced each +of $a_{1}$, $a_{2}$, \dots,~$a_{n}$ by~$G$; at most $n$~repetitions will be necessary. As the final +value of the arithmetic mean is~$G$, the initial value cannot have been less. +\PageSep{33} + +\Item{10.} \Topic{Schwarz's inequality.} Suppose that $a_{1}$, $a_{2}$, \dots,~$a_{n}$ and $b_{1}$, $b_{2}$, \dots,~$b_{n}$ +are any two sets of numbers positive or negative. It is easy to verify the +identity +\[ +\left(\tsum a_{r} b_{r}\right)^{2} + = \tsum a_{r}^{2} \tsum a_{s}^{2} + - \tsum (a_{r} b_{s} - a_{s} b_{r})^{2}, +\] +where $r$~and~$s$ assume the values $1$, $2$, \dots,~$n$. It follows that +\[ +\left(\tsum a_{r} b_{r}\right)^{2} \leq \tsum a_{r}^{2} \tsum b_{r}^{2}, +\] +an inequality usually known as Schwarz's (though due originally to Cauchy). + +\Item{11.} If $a_{1}$, $a_{2}$, \dots,~$a_{n}$ are all positive, and $s_{n} = a_{1} + a_{2} + \dots + a_{n}$, then +\[ +(1 + a_{1})(1 + a_{2}) \dots (1 + a_{n}) + \leq 1 + s_{n} + \frac{s_{n}^{2}}{2!} + \dots + \frac{s_{n}^{n}}{n!}. +\] +\MathTrip{1909.} + +\Item{12.} If $a_{1}$, $a_{2}$, \dots,~$a_{n}$ and $b_{1}$, $b_{2}$, \dots,~$b_{n}$ are two sets of positive numbers, +arranged in descending order of magnitude, then +\[ +(a_{1} + a_{2} + \dots + a_{n}) +(b_{1} + b_{2} + \dots + b_{n}) + \leq n(a_{1}b_{1} + a_{2}b_{2} + \dots + a_{n}b_{n}). +\] + +\Item{13.} If $a$, $b$, $c$, \dots~$k$ and $A$, $B$, $C$, \dots~$K$ are two sets of numbers, and all of +the first set are positive, then +\[ +\frac{aA + bB + \dots + kK}{a + b + \dots + k} +\] +lies between the algebraically least and greatest of $A$, $B$, \dots,~$K$. + +\Item{14.} If $\sqrt{p}$,~$\sqrt{q}$ are dissimilar surds, and $a + b\sqrt{p} + c\sqrt{q} + d\sqrtp{pq} = 0$, +where $a$, $b$, $c$,~$d$ are rational, then $a = 0$, $b = 0$, $c = 0$, $d = 0$. + +[Express $\sqrt{p}$ in the form $M + N \sqrt{q}$, where $M$~and~$N$ are rational, and apply +the theorem of \SecNo[§]{14}.] + +\Item{15.} Show that if $a\sqrt{2} + b\sqrt{3} + c\sqrt{5} = 0$, where $a$,~$b$,~$c$ are rational numbers, +then $a = 0$, $b = 0$, $c = 0$. + +\Item{16.} Any polynomial in $\sqrt{p}$~and~$\sqrt{q}$, with rational coefficients (\ie\ any +sum of a finite number of terms of the form $A(\sqrt{p})^{m}(\sqrt{q})^{n}$, where $m$~and~$n$ +are integers, and $A$~rational), can be expressed in the form +\[ +a + b\sqrt{p} + c\sqrt{q} + d\DPtypo{\sqrt{pq}}{\sqrtp{pq}}, +\] +where $a$, $b$, $c$,~$d$ are rational. + +\Item{17.} Express $\dfrac{a + b\sqrt{p} + c\sqrt{q}}{d + e\sqrt{p} + f\sqrt{q}}$, where $a$,~$b$,~etc.\ are rational, in the form +\[ +A + B\sqrt{p} + C\sqrt{q} + D\DPtypo{\sqrt{pq}}{\sqrtp{pq}}, +\] +where $A$, $B$, $C$,~$D$ are rational. + +[Evidently +%[** TN: Set on one line in the original] +\begin{align*} +\frac{a + b\sqrt{p} + c\sqrt{q}}{d + e\sqrt{p} + f\sqrt{q}} + &= \frac{(a + b\sqrt{p} + c\sqrt{q})(d + e\sqrt{p} - f\sqrt{q})} + {(d + e\sqrt{p})^{2} - f^{2}q} \\ + &= \frac{\alpha + \beta\sqrt{p} + \gamma\sqrt{q} + \delta\DPtypo{\sqrt{pq}}{\sqrtp{pq}}} + {\epsilon + \zeta\sqrt{p}}, +\end{align*} +where $\alpha$, $\beta$,~etc.\ are rational numbers which can easily be found. The required +\PageSep{34} +reduction may now be easily completed by multiplication of numerator and +denominator by $\epsilon - \zeta\sqrt{p}$. For example, prove that +\[ +\frac{1}{1 + \sqrt{2} + \sqrt{3}} + = \frac{1}{2} + \frac{1}{4}\sqrt{2} - \frac{1}{4}\sqrt{6}.] +\] + +\Item{18.} If $a$, $b$, $x$, $y$ are rational numbers such that +\[ +(ay - bx)^{2} + 4(a - x)(b - y) = 0, +\] +then either (i)~$x = a$, $y = b$ or (ii)~$1 - ab$ and~$1 - xy$ are squares of rational +numbers. \MathTrip{1903.} + +\Item{19.} If all the values of $x$~and~$y$ given by +\[ +ax^{2} + 2hxy + by^{2} = 1,\quad +a'x^{2} + 2h'xy + b'y^{2} = 1 +\] +(where $a$,~$h$,~$b$, $a'$,~$h'$,~$b'$ are rational) are rational, then +\[ +(h - h')^{2} - (a - a')(b - b'),\quad +(ab' - a'b)^{2} + 4(ah' - a'h)(bh' - b'h) +\] +are both squares of rational numbers. \MathTrip{1899.} + +\Item{20.} Show that $\sqrt{2}$~and~$\sqrt{3}$ are cubic functions of $\sqrt{2} + \sqrt{3}$, with rational +coefficients, and that $\sqrt{2} - \sqrt{6} + 3$ is the ratio of two linear functions of +$\sqrt{2} + \sqrt{3}$. \MathTrip{1905.} + +\Item{21.} The expression +\[ +\sqrtb{a + 2m\sqrtp{a - m^{2}}} + \sqrtb{a - 2m\sqrtp{a - m^{2}}} +\] +is equal to~$2m$ if $2m^{2} > a > m^{2}$, and to $2\sqrtp{a - m^{2}}$ if $a > 2m^{2}$. + +\Item{22.} Show that any polynomial in~$\sqrt[3]{2}$, with rational coefficients, can be +expressed in the form +\[ +a + b\sqrt[3]{2} + c\sqrt[3]{4}, +\] +where $a$,~$b$,~$c$ are rational. + +More generally, if $p$~is any rational number, any polynomial in~$\sqrt[m]{p}$ with +rational coefficients can be expressed in the form +\[ +a_{0} + a_{1}\alpha + a_{2}\alpha^{2} + \dots + a_{m-1}\alpha^{m-1}, +\] +where $a_{0}$, $a_{1}, \dots$ are rational and $\alpha = \sqrt[m]{p}$. For any such polynomial is of the +form +\[ +b_{0} + b_{1}\alpha + b_{2}\alpha^{2} + \dots + b_{k}\alpha^{k}, +\] +where the~$b$'s are rational. If $k \leq m - 1$, this is already of the form required. If +$k > m - 1$, let $\alpha^{r}$~be any power of~$\alpha$ higher than the~$(m - 1)$th. Then $r = \lambda m + s$, +where $\lambda$~is an integer and $0 \leq s \leq m - 1$; and $\alpha^{r} = \alpha^{\lambda m + s} = p^{\lambda}\alpha^{s}$. Hence we can +get rid of all powers of~$\alpha$ higher than the~$(m - 1)$th. + +\Item{23.} Express $(\sqrt[3]{2} - 1)^{5}$ and $(\sqrt[3]{2} - 1)/(\sqrt[3]{2} + 1)$ in the form $a + b\sqrt[3]{2} + c\sqrt[3]{4}$, +where $a$,~$b$,~$c$ are rational. [Multiply numerator and denominator of the +second expression by $\sqrt[3]{4} - \sqrt[3]{2} + 1$.] + +\Item{24.} If +\[ +a + b\sqrt[3]{2} + c\sqrt[3]{4} = 0, +\] +where $a$,~$b$,~$c$ are rational, then $a = 0$, $b = 0$, $c = 0$. +\PageSep{35} + +[Let $y = \sqrt[3]{2}$. Then $y^{3} = 2$ and +\[ +cy^{2} + by + a = 0. +\] +Hence $2cy^{2} + 2by + ay^{3} = 0$ or +\[ +ay^{2} + 2cy + 2b = 0. +\] + +Multiplying these two quadratic equations by $a$~and~$c$ and subtracting, +we obtain $(ab - 2c^{2})y + a^{2} - 2bc = 0$, or $y = -(a^{2} - 2bc)/(ab - 2c^{2})$, a rational +number, which is impossible. The only alternative is that $ab - 2c^{2} = 0$, +$a^{2} - 2bc = 0$. + +Hence $ab = 2c^{2}$, $a^{4} = 4b^{2}c^{2}$. If neither $a$~nor~$b$ is zero, we can divide the +second equation by the first, which gives $a^{3} = 2b^{3}$: and this is impossible, +since $\sqrt[3]{2}$~cannot be equal to the rational number~$a/b$. Hence $ab = 0$, $c = 0$, +and it follows from the original equation that $a$,~$b$, and~$c$ are all zero. + +As a corollary, if $a + b\sqrt[3]{2} + c\sqrt[3]{4} = d + e\sqrt[3]{2} + f\sqrt[3]{4}$, then $a = d$, $b = e$, $c = f$. + +It may be proved, more generally, that if +\[ +a_{0} + a_{1}p^{1/m} + \dots + a_{m-1}p^{(m-1)/m} = 0, +\] +$p$ not being a perfect $m$th~power, then $a_{0} = a_{1} = \dots = a_{m-1} = 0$; but the proof is +less simple.] + +\Item{25.} If $A + \sqrt[3]{B} = C + \sqrt[3]{D}$, then either $A = C$, $B = D$, or $B$~and~$D$ are both +cubes of rational numbers. + +\Item{26.} If $\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C} = 0$, then either one of $A$,~$B$,~$C$ is zero, and the other +two equal and opposite, or $\sqrt[3]{A}$,~$\sqrt[3]{B}$,~$\sqrt[3]{C}$ are rational multiples of the same +surd~$\sqrt[3]{X}$. + +\Item{27.} Find rational numbers $\alpha$,~$\beta$ such that +\[ +\sqrtp[3]{7 + 5\sqrt{2}} = \alpha + \beta\sqrt{2}. +\] + +\Item{28.} If $(a - b^{3})b > 0$, then +\[ +\bigsqrtb[3]{a + \frac{9b^{3} + a}{3b}\bigsqrtp{\frac{a - b^{3}}{3b}}} + +\bigsqrtb[3]{a - \frac{9b^{3} + a}{3b}\bigsqrtp{\frac{a - b^{3}}{3b}}} +\] +is rational. [Each of the numbers under a cube root is of the form +\[ +\left\{\alpha + \beta\bigsqrtp{\frac{a - b^{3}}{3b}}\right\}^{3} +\] +where $\alpha$~and~$\beta$ are rational.] + +\Item{29.} If $\alpha = \sqrt[n]{p}$, any polynomial in~$\alpha$ is the root of an equation of degree~$n$, +with rational coefficients. + +[We can express the polynomial ($x$~say) in the form +\[ +x = l_{1} + m_{1}\alpha + \dots + r_{1}\alpha^{(n-1)}, +\] +where $l_{1}$,~$m_{1}, \dots$ are rational, as in Ex.~22. +\PageSep{36} + +Similarly +\begin{alignat*}{4} +x^{2} &= l_{2} &&+ m_{2}a &&+ \dots &&+ r_{2}a^{(n-1)}, \\ +\multispan{9}{\dotfill} \\ +x^{n} &= l_{n} &&+ m_{n}a &&+ \dots &&+ r_{n}a^{(n-1)}. +\end{alignat*} + +Hence +\[ +L_{1}x + L_{2}x^{2} + \dots + L_{n}x^{n} = \Delta, +\] +where $\Delta$~is the determinant +\[ +\left| \begin{array}{cccc} +l_{1} & m_{1} & \dots & r_{1} \\ +l_{2} & m_{2} & \dots & r_{2} \\ +\hdotsfor{4} \\ +l_{n} & m_{n} & \dots & r_{n} \\ +\end{array} \right| +\] +and $L_{1}$, $L_{2}, \dots$ the minors of $l_{1}$, $l_{2}, \dots$.] + +\Item{30.} Apply this process to $x = p + \sqrt{q}$, and deduce the theorem of \SecNo[§]{14}. + +\Item{31.} Show that $y = a + bp^{1/3} + cp^{2/3}$ satisfies the equation +\[ +y^{3} - 3ay^{2} + 3y(a^{2} - bcp) - a^{3} - b^{3}p - c^{3}p^{2} + 3abcp = 0. +\] + +\Item{32.} \Topic{Algebraical numbers.} We have seen that some irrational numbers +(such as~$\sqrt{2}$) are roots of equations of the type +\[ +a_{0}x^{n} + a_{1}x^{n-1} + \dots + a_{n} = 0, +\] +where $a_{0}$, $a_{1}$, \dots,~$a_{n}$ are integers. Such irrational numbers are called \emph{algebraical} +numbers: all other irrational numbers, such as~$\pi$ (\SecNo[§]{15}), are called +\emph{transcendental} numbers. Show that if $x$~is an algebraical number, then so are~$kx$, +where $k$~is any rational number, and~$x^{m/n}$, where $m$~and~$n$ are any integers. + +\Item{33.} If $x$~and~$y$ are algebraical numbers, then so are $x + y$, $x - y$, $xy$ and~$x/y$. + +[We have equations +\begin{alignat*}{4} +a_{0}x^{m} &+ a_{1}x^{m-1} &&+ \dots &&+ a_{m} &&= 0, \\ +b_{0}y^{n} &+ b_{1}y^{n-1} &&+ \dots &&+ b_{n} &&= 0, +\end{alignat*} +where the $a$'s~and~$b$'s are integers. Write $x + y = z$, $y = z - x$ in the second, +and eliminate~$x$. We thus get an equation of similar form +\[ +c_{0}z^{p} + c_{1}z^{p-1} + \dots + c_{p} = 0, +\] +satisfied by~$z$. Similarly for the other cases.] + +\Item{34.} If +\[ +a_{0}x^{n} + a_{1}x^{n-1} + \dots + a_{n} = 0, +\] +where $a_{0}$, $a_{1}$, \dots,~$a_{n}$ are any algebraical numbers, then $x$~is an algebraical +number. [We have $n + 1$~equations of the type +\[ +a_{0, r}a_{r}^{m_{r}} + a_{1, r}a_{r}^{m_{r}-1} + \dots + a_{m_{r}, r} = 0\quad +(r = 0,\ 1,\ \dots,\ n), +\] +in which the coefficients $a_{0, r}$, $a_{1, r}, \dots$ are integers\Add{.} Eliminate $a_{0}$\DPtypo{.}{,} $a_{1}$, \dots,~$a_{n}$ +between these and the original equation for~$x$.] + +\Item{35.} Apply this process to the equation $x^{2} - 2x\sqrt{2} + \sqrt{3} = 0$. + +[The result is $x^{8} - 16x^{6} + 58x^{4} - 48x^{2} + 9 = 0$.] +\PageSep{37} + +\Item{36.} Find equations, with rational coefficients, satisfied by +\[ +1 + \sqrt{2} + \sqrt{3},\quad +\frac{\sqrt{3} + \sqrt{2}} + {\sqrt{3} - \sqrt{2}},\quad +\sqrtb{\sqrt{3}+ \sqrt{2}} + \sqrtb{\sqrt{3} - \sqrt{2}},\quad +\sqrt[3]{2} + \sqrt[3]{3}. +\] + +\Item{37.} If $x^{3} = x + 1$, then $x^{3n} = a_{n}x + b_{n} + c_{n}/x$, where +\[ +a_{n+1} = a_{n} + b_{n},\quad +b_{n+1} = a_{n} + b_{n} + c_{n},\quad +c_{n+1} = a_{n} + c_{n}. +\] + +\Item{38.} If $x^{6} + x^{5} - 2x^{4} - x^{3} + x^{2} + 1 = 0$ and $y = x^{4} - x^{2} + x - 1$, then $y$~satisfies +a quadratic equation with rational coefficients. \MathTrip{1903.} + +[It will be found that $y^{2} + y + 1 = 0$.] +\end{Examples} +\PageSep{38} + + +\Chapter{II}{FUNCTIONS OF REAL VARIABLES} + +\Paragraph{20. The idea of a function.} Suppose that $x$~and~$y$ are +two continuous real variables, which we may suppose to be represented +geometrically by distances $A_{0}P = x$, $B_{0}Q = y$ measured +from fixed points $A_{0}$,~$B_{0}$ along two straight lines $\Lambda$,~$\Mu$. And +let us suppose that the positions of the points $P$~and~$Q$ are not +independent, but connected by a relation which we can imagine +to be expressed as a relation between $x$~and~$y$: so that, when +$P$~and~$x$ are known, $Q$~and~$y$ are also known. We might, +for example, suppose that $y = x$, or $y = 2x$, or~$\frac{1}{2}x$, or~$x^{2} + 1$. In +all of these cases the value of~$x$ determines that of~$y$. Or +again, we might suppose that the relation between $x$~and~$y$ is +given, not by means of an explicit formula for~$y$ in terms of~$x$, +but by means of a geometrical construction which enables us to +determine~$Q$ when~$P$ is known. + +In these circumstances $y$~is said to be a \emph{function} of~$x$. This +notion of functional dependence of one variable upon another is +perhaps the most important in the whole range of higher mathematics. +In order to enable the reader to be certain that he +understands it clearly, we shall, in this chapter, illustrate it by +means of a large number of examples. + +But before we proceed to do this, we must point out that +the simple examples of functions mentioned above possess three +characteristics which are by no means involved in the general +idea of a function, viz.: + +\Item{(1)} $y$~is determined \emph{for every value of~$x$}; + +\Item{(2)} to each value of~$x$ for which $y$~is given corresponds \emph{one +and only one value of~$y$}; + +\Item{(3)} the relation between $x$~and~$y$ is expressed by means of +\emph{an analytical formula}, from which the value of~$y$ corresponding to +a given value of~$x$ can be calculated by direct substitution of the +latter. +\PageSep{39} + +It is indeed the case that these particular characteristics are +possessed by many of the most important functions. But the consideration +of the following examples will make it clear that they +are by no means essential to a function. All that is essential is +that there should be some relation between $x$~and~$y$ such that to +some values of~$x$ at any rate correspond values of~$y$. + +\begin{Examples}{X.} +\Item{1.} Let $y = x$ or~$2x$ or~$\frac{1}{2}x$ or $x^{2} +1$. Nothing further need +be said at present about cases such as these. + +\Item{2.} Let $y = 0$ whatever be the value of~$x$. Then $y$~is a function of~$x$, for we +can give~$x$ any value, and the corresponding value of~$y$ (viz.~$0$) is known. In +this case the functional relation makes the same value of~$y$ correspond to all +values of~$x$. The same would be true were $y$~equal to~$1$ or~$-\frac{1}{2}$ or~$\sqrt{2}$ instead +of~$0$. Such a function of~$x$ is called \emph{a constant}. + +\Item{3.} Let $y^{2} = x$. Then if $x$~is positive this equation defines \emph{two} values of~$y$ +corresponding to each value of~$x$, viz.~$±\sqrt{x}$. If $x = 0$, $y = 0$. Hence to the +particular value~$0$ of~$x$ corresponds \emph{one} and only one value of~$y$. But if $x$~is +negative there is \emph{no} value of~$y$ which satisfies the equation. That is to say, +the function~$y$ is not defined for negative values of~$x$. This function therefore +possesses the characteristic~(3), but neither (1)~nor~(2). + +\Item{4.} Consider a volume of gas maintained at a constant temperature and +contained in a cylinder closed by a sliding piston.\footnote + {I borrow this instructive example from Prof.\ H.~S. Carslaw's \textit{Introduction to + the Calculus.}} + +Let $A$ be the area of the cross section of the piston and $W$~its weight. +The gas, held in a state of compression by the piston, exerts a certain pressure +$p_{0}$~per unit of area on the piston, which balances the weight~$W$, so that +\[ +W = Ap_{0}. +\] + +Let $v_{0}$ be the volume of the gas when the system is thus in equilibrium. +If additional weight is placed upon the piston the latter is forced downwards. +The volume~($v$) of the gas diminishes; the pressure~($p$) which it exerts +upon unit area of the piston increases. Boyle's experimental law asserts that +the product of $p$~and~$v$ is very nearly constant, a correspondence which, if +exact, would be represented by an equation of the type +\[ +pv = a, +\Tag{(i)} +\] +where $a$~is a number which can be determined approximately by experiment. + +Boyle's law, however, only gives a reasonable approximation to the facts +provided the gas is not compressed too much. When $v$~is decreased and $p$~increased +beyond a certain point, the relation between them is no longer +expressed with tolerable exactness by the equation~\Eq{(i)}. It is known that a +\PageSep{40} +much better approximation to the true relation can then be found by means +of what is known as `van~der Waals' law', expressed by the equation +\[ +\left(p + \frac{\alpha}{v^{2}}\right)(v - \beta) = \gamma, +\Tag{(ii)} +\] +where $\alpha$, $\beta$, $\gamma$ are numbers which can also be determined approximately by +experiment. + +Of course the two equations, even taken together, do not give anything +like a complete account of the relation between $p$~and~$v$. This relation is no +doubt in reality much more complicated, and its form changes, as $v$~varies, +from a form nearly equivalent to~\Eq{(i)} to a form nearly equivalent to~\Eq{(ii)}. But, +from a mathematical point of view, there is nothing to prevent us from contemplating +an ideal state of things in which, for all values of~$v$ not less than +a certain value~$V$, \Eq{(i)}~would be exactly true, and \Eq{(ii)}~exactly true for all +values of~$v$ less than~$V$. And then we might regard the two equations as +together defining~$p$ as a function of~$v$. It is an example of a function which +for some values of~$v$ is defined by one formula and for other values of~$v$ is +defined by another. + +This function possesses the characteristic~(2)\DPtypo{.}{;} to any value of~$v$ only one +value of~$p$ corresponds: but it does not possess~(1). For $p$~is not defined as +a function of~$v$ for negative values of~$v$; a `negative volume' means +nothing, and so negative values of~$v$ do not present themselves for consideration +at all. + +\Item{5.} Suppose that a perfectly elastic ball is dropped (without rotation) +from a height~$\frac{1}{2}g\tau^{2}$ on to a fixed horizontal plane, and rebounds continually. + +The ordinary formulae of elementary dynamics, with which the reader is +probably familiar, show that $h = \frac{1}{2}gt^{2}$ if $0 \leq t \leq \tau$, $h = \frac{1}{2}g(2\tau - t)^{2}$ if $\tau \leq t \leq 3\tau$, and +generally +\[ +h = \tfrac{1}{2}g(2n\tau - t)^{2} +\] +if $(2n - 1)\tau \leq t \leq (2n + 1)\tau$, $h$~being the depth of the ball, at time~$t$, below its +original position. Obviously $h$~is a function of~$t$ which is only defined for +positive values of~$t$. + +\Item{6.} Suppose that $y$~is defined as being \emph{the largest prime factor of~$x$}. This +is an instance of a definition which only applies to a particular class of values +of~$x$, viz.\ \emph{integral} values. `The largest prime factor of~$\frac{11}{3}$ or of~$\sqrt{2}$ or of~$\pi$' +means nothing, and so our defining relation fails to define for such values of~$x$ +as these. Thus this function does not possess the characteristic~(1). It does +possess~(2), but not~(3), as there is no simple formula which expresses~$y$ in +terms of~$x$. + +\Item{7.} Let $y$~be defined as \emph{the denominator of~$x$ when $x$~is expressed in its +lowest terms}. This is an example of a function which is defined if and only +if $x$~is \emph{rational}. Thus $y = 7$ if $x = -11/7$: but $y$~is not defined for $x = \sqrt{2}$, `the +denominator of~$\sqrt{2}$' being a meaningless form of words. +\PageSep{41} + +\Item{8.} Let $y$~be defined as \emph{the height in inches of policeman~$Cx$, in the +Metropolitan Police, at {\upshape5.30}~\DPchg{p.m.}{\textsc{p.m.}}\ on {\upshape8}~Aug.~{\upshape1907}}. Then $y$~is defined for a +certain number of integral values of~$x$, viz.\ $1$, $2$, \dots,~$N$, where $N$~is the total +number of policemen in division~$C$ at that particular moment of time. +\end{Examples} + +\Paragraph{21. The graphical representation of functions.} Suppose +that the variable~$y$ is a function of the variable~$x$. It will +generally be open to us also to regard~$x$ as a function of~$y$, in virtue +of the functional relation between $x$~and~$y$. But for the present we +shall look at this relation from the first point of view. We shall +then call~$x$ the \emph{independent variable} and~$y$ the \emph{dependent variable}; +and, when the particular form of the functional relation is not +specified, we shall express it by writing +\[ +y = f(x) +\] +(or $F(x)$, $\phi(x)$, $\psi(x), \dots$, as the case may be). + +The nature of particular functions may, in very many cases, be +illustrated and made easily intelligible as follows. Draw two lines +$OX$,~$OY$ at right angles to one another +and produced indefinitely in both directions. +We can represent values of $x$~and~$y$ by distances measured from~$O$ +along the lines $OX$,~$OY$ respectively, +regard being paid, of course, to sign, +and the positive directions of measurement +being those indicated by arrows +in \Fig{6}. +%[Illustration: Fig. 6.] +\Figure[1.75in]{6}{p041} + +Let $a$ be any value of~$x$ for which +$y$~is defined and has (let us suppose) +the single value~$b$. Take $OA = a$, +$OB = b$, and complete the rectangle~$OAPB$. +Imagine the point~$P$ marked on the diagram. This +marking of the point~$P$ may be regarded as showing that the +value of~$y$ for $x = a$ is~$b$. + +If to the value~$a$ of~$x$ correspond several values of~$y$ (say +$b$,~$b'$,~$b''$), we have, instead of the single point~$P$, a number of +points $P$,~$P'$,~$P''$. + +We shall call~$P$ the \emph{point} $(a, b)$; $a$~and~$b$ the \emph{coordinates of~$P$ +referred to the axes $OX$,~$OY$}; $a$~the \emph{abscissa}, $b$~the \emph{ordinate} of~$P$; +$OX$~and~$OY$ the \emph{axis of~$x$} and the \emph{axis of~$y$}, or together the +\PageSep{42} +\emph{axes of coordinates}, and $O$~the \emph{origin of coordinates}, or simply +the \emph{origin}. + +Let us now suppose that for all values~$a$ of~$x$ for which $y$~is +defined, the value~$b$ (or values $b$,~$b'$,~$b'', \dots$) of~$y$, and the corresponding +point~$P$ (or points $P$,~$P'$,~$P'', \dots$), have been determined. +We call the aggregate of all these points the \Emph{graph} of the +function~$y$. + +To take a very simple example, suppose that $y$~is defined as +a function of~$x$ by the equation +\[ +Ax + By + C = 0, +\Tag{(1)} +\] +where $A$, $B$, $C$ are any fixed numbers.\footnote + {If $B = 0$, $y$~does not occur in the equation. We must then regard~$y$ as a + function of~$x$ defined for one value only of~$x$, viz.\ $x = -C/A$, and then having \emph{all} + values.} +Then $y$~is a function of~$x$ +which possesses all the characteristics (1),~(2),~(3) of \SecNo[§]{20}. It is +easy to show that \emph{the graph of~$y$ is a straight line}. The reader is +in all probability familiar with one or other of the various proofs +of this proposition which are given in text-books of Analytical +Geometry. + +We shall sometimes use another mode of expression. We +shall say that when $x$~and~$y$ vary in such a way that equation~\Eq{(1)} +is always true, \emph{the locus of the point~$(x, y)$ is a straight line}, and +we shall call~\Eq{(1)} \emph{the equation of the locus}, and say that the equation +\emph{represents} the locus. This use of the terms `locus', `equation of +the locus' is quite general, and may be applied whenever the +relation between $x$~and~$y$ is capable of being represented by an +analytical formula. + +The equation $Ax + By + C = 0$ is \emph{the general equation of the first +degree}, for $Ax + By + C$ is the most general polynomial in $x$~and~$y$ +which does not involve any terms of degree higher than the first +in $x$~and~$y$. Hence \emph{the general equation of the first degree represents +a straight line}. It is equally easy to prove the converse +proposition that \emph{the equation of any straight line is of the first +degree}. + +We may mention a few further examples of interesting geometrical +loci defined by equations. An equation of the form +\[ +(x - \alpha)^{2} + (y - \beta)^{2} = \rho^{2}, +\] +\PageSep{43} +or +\[ +x^{2} + y^{2} + 2Gx + 2Fy + C = 0, +\] +where $G^{2} + F^{2} - C > 0$, represents a circle. The equation +\[ +Ax^{2} + 2Hxy + By^{2} + 2Gx + 2Fy + C = 0 +\] +(\emph{the general equation of the second degree}) represents, assuming +that the coefficients satisfy certain inequalities, a conic section, +\ie\ an ellipse, parabola, or hyperbola. For further discussion of +these loci we must refer to books on Analytical Geometry. + +\Paragraph{22. Polar coordinates.} In what precedes we have determined +the position of~$P$ by the lengths of its coordinates $OM = x$, $MP = y$. +If $OP = r$ and $MOP = \theta$, $\theta$~being an +angle between $0$~and~$2\pi$ (measured in +the positive direction), it is evident that +\begin{gather*} +x = r\cos\theta,\qquad +y = r\sin\theta, \\ +r = \sqrtp{x^{2} + y^{2}},\quad +\cos\theta : \sin\theta : 1 :: x : y : r, +\end{gather*} +and that the position of~$P$ is equally well +determined by a knowledge of $r$~and~$\theta$. +We call $r$~and~$\theta$ the \emph{polar coordinates} +of~$P$. The former, it should be observed, is essentially positive.\footnote + {Polar coordinates are sometimes defined so that $r$~may be positive or negative. + In this case two pairs of coordinates---\eg\ $(1, 0)$ and $(-1, \pi)$---correspond to the + same point. The distinction between the two systems may be illustrated by means + of the equation $l/r = 1 - e\cos\theta$, where $l > 0$, $e > 1$. According to our definitions $r$~must + be positive and therefore $\cos\theta < 1/e$: the equation represents one branch only + of a hyperbola, the other having the equation $-l/r = 1 - e\cos\theta$. With the system + of coordinates which admits negative values of~$r$, the equation represents the whole + hyperbola.} +%[Illustration: Fig. 7.] +\Figure[1.75in]{7}{p043} + +If $P$~moves on a locus there will be some relation between $r$~and~$\theta$, +say $r = f(\theta)$ or $\theta = F(r)$. This we call the \emph{polar equation} +of the locus. The polar equation may be deduced from the $(x, y)$ +equation (or \textit{vice versa}) by means of the formulae above. + +Thus the polar equation of a straight line is of the form +\[ +r\cos(\theta - \alpha) = p, +\] +where $p$~and~$\alpha$ are constants. The equation $r = 2a\cos\theta$ represents +a circle passing through the origin; and the general equation of +a circle is of the form +\[ +r^{2} + c^{2} - 2rc\cos(\theta - \alpha) = A^{2}, +\] +where $A$, $c$, and~$\alpha$ are constants. +\PageSep{44} + +\Paragraph{23. Further examples of functions and their graphical +representation.} The examples which follow will give the +reader a better notion of the infinite variety of possible types of +functions. + +\Topic{\Item{A.} Polynomials.} A \emph{polynomial} in~$x$ is a function of the +form +\[ +a_{0}x^{m} + a_{1}x^{m-1} + \dots + a_{m}, +\] +where $a_{0}$, $a_{1}$, \dots,~$a_{m}$ are constants. The simplest polynomials are +the simple powers $y = x$, $x^{2}$, $x^{3}$,~\dots, $x^{m}, \dots$. The graph of the function~$x^{m}$ +is of two distinct types, according as $m$~is even or odd. + +First let $m = 2$. Then three points on the graph are $(0, 0)$, +$(1, 1)$, $(-1, 1)$. Any number of additional points on the graph +may be found by assigning other special values to~$x$: thus the +values +\begin{alignat*}{6} +x &= \tfrac{1}{2},\quad &&2,\quad &&3,\quad -&&\tfrac{1}{2},\quad -&&2,\quad &&3 \\ +\intertext{give} +y &= \tfrac{1}{4},\quad &&4,\quad &&9,\quad &&\tfrac{1}{4},\quad &&4,\quad &&9. +\end{alignat*} +If the reader will plot off a fair number of points on the graph, he +will be led to conjecture that the +form of the graph is something +like that shown in \Fig{8}. If +he draws a curve through the +special points which he has proved +to lie on the graph and then tests +its accuracy by giving~$x$ new +values, and calculating the corresponding +values of~$y$, he will +find that they lie as near to the curve as it is reasonable to expect, +when the inevitable inaccuracies of drawing are considered. The +curve is of course a parabola. +%[Illustration: Fig. 8.] +\Figure[2.25in]{8}{p044} + +There is, however, one fundamental question which we cannot +answer adequately at present. The reader has no doubt some +notion as to what is meant by a \emph{continuous} curve, a curve without +breaks or jumps; such a curve, in fact, as is roughly represented +in \Fig{8}. The question is whether the graph of the function +$y = x^{2}$ is in fact such a curve. This cannot be \emph{proved} by merely +\PageSep{45} +constructing any number of isolated points on the curve, although +the more such points we construct the more probable it will +appear. + +This question cannot be discussed properly until \Ref{Ch.}{V}\@. In +that chapter we shall consider in detail what our common sense +idea of continuity really means, and how we can prove that such +graphs as the one now considered, and others which we shall +consider later on in this chapter, are really continuous curves. +For the present the reader may be content to draw his curves as +common sense dictates. + +\begin{Remark} +It is easy to see that the curve $y = x^{2}$ is everywhere convex to the axis of~$x$. +Let $P_{0}$,~$P_{1}$ (\Fig{8}) be the points $(x_{0}, x_{0}^{2})$, $(x_{1}, x_{1}^{2})$. Then the coordinates of +a point on the chord~$P_{0}P_{1}$ are $x = \lambda x_{0} + \mu x_{1}$, $y = \lambda x_{0}^{2} + \mu x_{1}^{2}$, where $\lambda$~and~$\mu$ are +positive numbers whose sum is~$1$. And +\[ +y - x^{2} + = (\lambda + \mu)(\lambda x_{0}^{2} + \mu x_{1}^{2}) + - (\lambda x_{0} + \mu x_{1} )^{2} + = \lambda\mu(x_{1} - x_{0})^{2} \geq 0, +\] +so that the chord lies entirely above the curve. +\end{Remark} + +The curve $y = x^{4}$ is similar to $y = x^{2}$ in general appearance, but +flatter near~$O$, and steeper beyond the points $A$,~$A'$ (\Fig{9}), +and $y = x^{m}$, where $m$~is even and greater than~$4$, is still more so. +As $m$~gets larger and larger the flatness and steepness grow +more and more pronounced, until the curve is practically indistinguishable +from the thick line in the figure. +%[Illustration: Fig. 9.] +%[Illustration: Fig. 10.] +%[** TN: Captions vertically aligned in the original] +\Figures{2.25in}{9}{p045a}{2.25in}{10}{p045b}\PageLabel{45} + +The reader should next consider the curves given by $y = x^{m}$, +when $m$~is odd. The fundamental difference between the two +cases is that whereas when $m$~is even $(-x)^{m} = x^{m}$, so that the +curve is symmetrical about~$OY$, when $m$~is odd $(-x)^{m} = -x^{m}$, so +\PageSep{46} +that $y$~is negative when $x$~is negative. \Fig{10} shows the curves +$y = x$, $y = x^{3}$, and the form to which $y = x^{m}$ approximates for +larger odd values of~$m$. + +It is now easy to see how (theoretically at any rate) the graph +of any polynomial may be constructed. In the first place, from +the graph of $y = x^{m}$ we can at once derive that of~$Cx^{m}$, where $C$~is +a constant, by multiplying the ordinate of every point of the +curve by~$C$. And if we know the graphs of $f(x)$~and~$F(x)$, we +can find that of $f(x) + F(x)$ by taking the ordinate of every point +to be the sum of the ordinates of the corresponding points on the +two original curves. + +The drawing of graphs of polynomials is however so much +facilitated by the use of more advanced methods, which will be +explained later on, that we shall not pursue the subject further +here. + +\begin{Examples}{XI.} +\Item{1.} Trace the curves $y = 7x^{4}$, $y = 3x^{5}$, $y = x^{10}$. + +[The reader should draw the curves carefully, and all three should be +drawn in one figure.\footnote + {It will be found convenient to take the scale of measurement along the axis + of~$y$ a good deal smaller than that along the axis of~$x$, in order to prevent the + figure becoming of an awkward size.} +He will then realise how rapidly the higher powers +of~$x$ increase, as $x$~gets larger and larger, and will see that, in such a +polynomial as +\[ +x^{10} + 3x^{5} + 7x^{4} +\] +(or even $x^{10} + 30x^{5} + 700x^{4}$), it is the \emph{first} term which is of really preponderant +importance when $x$~is fairly large. Thus even when $x = 4$, $x^{10} > 1,000,000$, +while $30x^{5} < 35,000$ and $700x^{4} < 180,000$; while if $x = 10$ the preponderance +of the first term is still more marked.] + +\Item{2.} Compare the relative magnitudes of $x^{12}$, $1,000,000x^{6}$, $1,000,000,000,000x$ +when $x = 1$, $10$, $100$,~etc. + +[The reader should make up a number of examples of this type for himself. +This idea of the \emph{relative rate of growth} of different functions of~$x$ is one with +which we shall often be concerned in the following chapters.] + +\Item{3.} Draw the graph of $ax^{2} + 2bx + c$. + +[Here $y - \{(ac - b^{2})/a\} = a\{x + (b/a)\}^{2}$. If we take new axes parallel to the +old and passing through the point $x = -b/a$, $y = (ac - b^{2})/a$, the new equation +is $y' = ax'^{2}$. The curve is a parabola.] + +\Item{4.} Trace the curves $y = x^{3} - 3x + 1$, $y = x^{2}(x - 1)$, $y = x(x - 1)^{2}$. +\end{Examples} +\PageSep{47} + +\Paragraph{24.} \Topic{\Item{B.} Rational Functions.} The class of functions which +ranks next to that of polynomials in simplicity and importance +is that of \emph{rational functions}. A rational function is the quotient +of one polynomial by another: thus if $P(x)$,~$Q(x)$ are polynomials, +we may denote the general rational function by +\[ +R(x) = \frac{P(x)}{Q(x)}. +\] + +In the particular case when $Q(x)$~reduces to unity or any other +constant (\ie\ does not involve~$x$), $R(x)$~reduces to a polynomial: +thus the class of rational functions includes that of polynomials +as a sub-class. The following points concerning the definition +should be noticed. + +\begin{Remark} +\Item{(1)} We usually suppose that $P(x)$~and~$Q(x)$ have no common factor $x + a$ +or $x^{p} + ax^{p-1} + bx^{p-2} + \dots + k$, all such factors being removed by division. + +\Item{(2)} It should however be observed that this removal of common factors +\emph{does as a rule change the function}. Consider for example the function~$x/x$, +which is a rational function. On removing the common factor~$x$ we obtain +$1/1 = 1$. But the original function is not \emph{always} equal to~$1$: it is equal to~$1$ +only so long as $x\neq 0$. If $x = 0$ it takes the form~$0/0$, which is meaningless. +Thus the function~$x/x$ is equal to~$1$ if $x\neq 0$ and is undefined when $x = 0$. It +therefore differs from the function~$1$, which is \emph{always} equal to~$1$. + +\Item{(3)} Such a function as +\[ +\left(\frac{1}{x + 1} + \frac{1}{x - 1}\right) \bigg/ +\left(\frac{1}{x} + \frac{1}{x - 2}\right) +\] +may be reduced, by the ordinary rules of algebra, to the form +\[ +\frac{x^{2}(x - 2)}{(x - 1)^{2} (x + 1)}, +\] +which is a rational function of the standard form. But here again it must be +noticed that the reduction is not \emph{always} legitimate. In order to calculate the +value of a function for a given value of~$x$ we must substitute the value for~$x$ +in the function \emph{in the form in which it is given}. In the case of this function +the values $x = -1$, $1$,~$0$,~$2$ all lead to a meaningless expression, and so the +function is not defined for these values. The same is true of the reduced +form, so far as the values $-1$~and~$1$ are concerned. But $x = 0$ and $x = 2$ give +the value~$0$. Thus once more the two functions are not the same. + +\Item{(4)} But, as appears from the particular example considered under~(3), +there will generally be a certain number of values of~$x$ for which the function +is not defined even when it has been reduced to a rational function of the +standard form. These are the values of~$x$ (if any) for which the denominator +vanishes. Thus $(x^{2} - 7)/(x^{2} - 3x + 2)$ is not defined when $x = 1$ or~$2$. +\PageSep{48} + +\Item{(5)} Generally we agree, in dealing with expressions such as those considered +in (2)~and~(3), to disregard the exceptional values of~$x$ for which such +processes of simplification as were used there are illegitimate, and to reduce +our function to the standard form of rational function. The reader will +easily verify that (on this understanding) the sum, product, or quotient of +two rational functions may themselves be reduced to rational functions of +the standard type. And generally \emph{a rational function of a rational function +is itself a rational function}: \ie\ if in $z = P(y)/Q(y)$, where $P$~and~$Q$ are +polynomials, we substitute $y = P_{1}(x)/Q_{1}(x)$, we obtain on simplification an +equation of the form $z = P_{2}(x)/Q_{2}(x)$. + +\Item{(6)} It is in no way presupposed in the definition of a rational function +that the constants which occur as coefficients should be rational \emph{numbers}. +The word rational has reference solely to the way in which the variable~$x$ +appears in the function. Thus +\[ +\frac{x^{2} + x + \sqrt{3}}{x\sqrt[3]{2} - \pi} +\] +is a rational function. + +The use of the word rational arises as follows. The rational function +$P(x)/Q(x)$ may be generated from~$x$ by a finite number of operations upon~$x$, +including only multiplication of $x$ by itself or a constant, addition of terms +thus obtained and division of one function, obtained by such multiplications +and additions, by another. In so far as the variable~$x$ is concerned, this procedure +is very much like that by which all rational numbers can be obtained +from unity, a procedure exemplified in the equation +\[ +\frac{5}{3} = \frac{1 + 1 + 1 + 1 + 1}{1 + 1 + 1}. +\] + +Again, \emph{any} function which can be deduced from~$x$ by the elementary +operations mentioned above using at each stage of the process functions +which have already been obtained from~$x$ in the same way, can be reduced to +the standard type of rational function. The most general kind of function +which can be obtained in this way is sufficiently illustrated by the example +\[ +\Biggl(\frac{x}{x^{2} + 1} + + \frac{2x + 7}{x^{2} + \dfrac{11x - 3\sqrt{2}}{9x + 1}}\Biggr) \Bigg/ +\left(17 + \frac{2}{x^{3}}\right), +\] +which can obviously be reduced to the standard type of rational function. +\end{Remark} + +\Paragraph{25.} The drawing of graphs of rational functions, even more +than that of polynomials, is immensely facilitated by the use of +methods depending upon the differential calculus. We shall +therefore content ourselves at present with a very few examples. + +\begin{Examples}{XII.} +\Item{1.} Draw the graphs of $y = 1/x$, $y = 1/x^{2}$, $y = 1/x^{3}$,~\dots. + +[The figures show the graphs of the first two curves. It should be +observed that since $1/0$,~$1/0^{2}$,~\dots\ are meaningless expressions, these functions +are not defined for $x = 0$.] +\PageSep{49} +%[Illustration: Fig. 11.] +%[Illustration: Fig. 12.] +%[** TN: Moved up three paragraphs] +\Figures{2.25in}{11}{p049a}{2.25in}{12}{p049b} + +\Item{2.} Trace $y = x + (1/x)$, $x - (1/x)$, $x^{2} + (1/x^{2})$, $x^{2} - (1/x^{2})$ and $ax + (b/x)$ +taking various values, positive and negative, for $a$~and~$b$. + +\Item{3.} Trace +\[ +y = \frac{x + 1}{x - 1},\quad +\left(\frac{x + 1}{x - 1}\right)^{2},\quad +\frac{1}{(x - 1)^{2}},\quad +\frac{x^{2} + 1}{x^{2} - 1}. +\] + +\Item{4.} Trace $y = 1/(x - a)(x - b)$, $1/(x - a)(x - b)(x - c)$, where $a < b < c$. + +\Item{5.} Sketch the general form assumed by the curves $y = 1/x^{m}$ as $m$~becomes +larger and larger, considering separately the cases in which $m$~is +odd or even. +\end{Examples} + +\Paragraph{26.} \Topic{\Item{C.} Explicit Algebraical Functions.} The next important +class of functions is that of \emph{explicit algebraical functions}. +These are functions which can be generated from~$x$ by a finite +number of operations such as those used in generating rational +functions, together with a finite number of operations of root +extraction. Thus +\[ +%[** TN: On two lines in the original] +\frac{\sqrtp{1 + x} - \sqrtp[3]{1 - x}} + {\sqrtp{1 + x} + \sqrtp[3]{1 - x}},\quad +\sqrt{x} + \sqrtp{x +\sqrt{x}},\quad +\left(\frac{x^{2} + x + \sqrt{3}}{x\sqrt[3]{2} - \pi}\right)^{\frac{2}{3}} +\] +are explicit algebraical functions, and so is $x^{m/n}$ (\ie~$\sqrt[n]{x^{m}}$), where $m$~and~$n$ +are any integers. + +It should be noticed that there is an ambiguity of notation +involved in such an equation as $y = \sqrt{x}$. We have, up to the +present, regarded (\eg)~$\sqrt{2}$ as denoting the \emph{positive} square root +of~$2$, and it would be natural to denote by~$\sqrt{x}$, where $x$~is any +\PageSep{50} +positive number, the positive square root of~$x$, in which case +$y = \sqrt{x}$ would be a one-valued function of~$x$. It is however +often more convenient to regard~$\sqrt{x}$ as standing for the two-valued +function whose two values are the positive and negative square +roots of~$x$. + +The reader will observe that, when this course is adopted, the +function~$\sqrt{x}$ differs fundamentally from rational functions in two +respects. In the first place a rational function is always defined +for all values of~$x$ with a certain number of isolated exceptions. +But $\sqrt{x}$~is undefined for a \emph{whole range} of values of~$x$ (\ie\ all +negative values). Secondly the function, when $x$~has a value +for which it is defined, has generally two values of opposite signs. + +The function~$\sqrt[3]{x}$, on the other hand, is one-valued and defined +for all values of~$x$. + +\begin{Examples}{XIII.} +\Item{1.} $\sqrtb{(x - a)(b - x)}$, where $a < b$, is defined only for +$a \leq x \leq b$. If $a < x < b$ it has two values: if $x = a$ or $b$ only one, viz.~$0$. + +\Item{2.} Consider similarly +\begin{gather*} +\sqrtb{(x - a)(x - b)(x - c)} \quad (a < b < c), \\ +\sqrtb{x(x^{2} - a^{2})},\quad +\sqrtb[3]{(x - a)^{2}(b - x)}\quad (a < b), \\ +\frac{\sqrtp{1 + x} - \sqrtp{1 - x}} + {\sqrtp{1 + x} + \sqrtp{1 - x}},\quad +\sqrtb{x + \sqrt{x}}. +\end{gather*} + +\Item{3.} Trace the curves $y^{2} = x$, $y^{3} = x$, $y^{2} = x^{3}$. + +\Item{4.} Draw the graphs of the functions +%[** TN: Not displayed in the original] +\[ +y = \sqrtp{a^{2} - x^{2}},\quad +y = b\sqrtb{1 - (x^{2}/a^{2})}. +\] +\end{Examples} + +\Paragraph{27.} \Topic{\Item{D.} Implicit Algebraical Functions.} It is easy to +verify that if +\[ +y = \frac{\sqrtp{1 + x} - \sqrtp[3]{1 - x}} + {\sqrtp{1 + x} + \sqrtp[3]{1 - x}}, +\] +then +\[ +\left(\frac{1 + y}{1 - y}\right)^{6} = \frac{(1 + x)^{3}}{(1 - x)^{2}}; +\] +or if +\[ +y = \sqrt{x} + \sqrtp{x + \sqrt{x}}, +\] +then +\[ +y^{4} - (4y^{2} + 4y + 1)x = 0. +\] +Each of these equations may be expressed in the form +\[ +y^{m} + R_{1}y^{m-1} + \dots + R_{m} = 0, +\Tag{(1)} +\] +where $R_{1}$, $R_{2}$, \dots,~$R_{m}$ are rational functions of~$x$: and the reader +will easily verify that, if $y$~is any one of the functions considered +in the last set of examples, $y$~satisfies an equation of this form. +\PageSep{51} +It is naturally suggested that the same is true of any explicit +algebraic function. And this is in fact true, and indeed not +difficult to prove, though we shall not delay to write out a formal +proof here. An example should make clear to the reader the lines +on which such a proof would proceed. Let +\[ +y = \frac{x + \sqrt{x} + \sqrtb{x + \sqrt{x}} + \sqrtp[3]{1 + x}} + {x - \sqrt{x} + \sqrtb{x + \sqrt{x}} - \sqrtp[3]{1 + x}}. +\] +Then we have the equations +\begin{gather*} +y = \frac{x + u + v + w} + {x - u + v - w}, \\ +u^{2} = x,\quad +v^{2} = x + u,\quad +w^{3} = 1 + x, +\end{gather*} +and we have only to eliminate $u$,~$v$,~$w$ between these equations in +order to obtain an equation of the form desired. + +We are therefore led to give the following definition: \emph{a function +$y = f(x)$ will be said to be an algebraical function of~$x$ if it is the +root of an equation such as~\Eq{(1)}, \ie~the root of an equation of the +$m$\textsuperscript{th}~degree in~$y$, whose coefficients are rational functions of~$x$}. There +is plainly no loss of generality in supposing the first coefficient to +be unity. + +This class of functions includes all the explicit algebraical +functions considered in \SecNo[§]{26}. But it also includes other functions +which cannot be expressed as explicit algebraical functions. For +it is known that in general such an equation as~\Eq{(1)} cannot be +solved explicitly for~$y$ in terms of~$x$, when $m$~is greater than~$4$, +though such a solution is always possible if $m = 1$, $2$,~$3$, or~$4$ and +in special cases for higher values of~$m$. + +The definition of an algebraical function should be compared +with that of an algebraical number given in the last chapter +(\MiscExs{I}~32). + +\begin{Examples}{XIV.} +\Item{1.} If $m = 1$, $y$~is a rational function. + +\Item{2.} If $m = 2$, the equation is $y^{2} + R_{1}y + R_{2} = 0$, so that +\[ +y = \tfrac{1}{2}\{-R_{1} ± \sqrtp{R_{1}^{2} - 4R_{2}}\}. +\] +This function is defined for all values of~$x$ for which $R_{1}^{2} \geq 4R_{2}$. It has two +values if $R_{1}^{2} > 4R_{2}$ and one if $R_{1}^{2} = 4R_{2}$. + +If $m = 3$ or~$4$, we can use the methods explained in treatises on Algebra for +the solution of cubic and biquadratic equations. But as a rule the process is +complicated and the results inconvenient in form, and we can generally study +the properties of the function better by means of the original equation. +\PageSep{52} + +\Item{3.} Consider the functions defined by the equations +\[ +y^{2} - 2y - x^{2} = 0,\quad +y^{2} - 2y + x^{2} = 0,\quad +y^{4} - 2y^{2} + x^{2} = 0, +\] +in each case obtaining~$y$ as an explicit function of~$x$, and stating for what +values of~$x$ it is defined. + +\Item{4.} Find algebraical equations, with coefficients rational in~$x$, satisfied by +each of the functions +\[ +\sqrt{x} + \sqrtp{1/x},\quad +\sqrt[3]{x} + \sqrtp[3]{1/x},\quad +\sqrtp{x + \sqrt{x}},\quad +\sqrt{x + \sqrtp{x + \sqrt{x}}}. +\] + +\Item{5.} Consider the equation $y^{4} = x^{2}$. + +[Here $y^{2} = ±x$. If $x$~is positive, $y = \sqrt{x}$: if negative, $y = \sqrtp{-x}$. Thus the +function has two values for all values of~$x$ save $x = 0$.] + +\Item{6.} An algebraical function of an algebraical function of~$x$ is itself an +algebraical function of~$x$. + +[For we have +\begin{alignat*}{4} +y^{m} &+ R_{1}(z)y^{m-1} &&+ \dots &&+ R_{m}(z) &&= 0, +\intertext{where} +z^{n} &+ S_{1}(x)z^{n-1} &&+ \dots &&+ S_{n}(x) &&= 0. +\intertext{Eliminating~$z$ we find an equation of the form} +y^{p} &+ T_{1}(x)y^{p-1} &&+ \dots &&+ T_{p}(x) &&= 0. +\end{alignat*} +Here all the capital letters denote rational functions.] + +\Item{7.} An example should perhaps be given of an algebraical function which +cannot be expressed in an explicit algebraical form. Such an example is the +function~$y$ defined by the equation +\[ +y^{5} - y - x = 0. +\] +But the proof that we cannot find an explicit algebraical expression for~$y$ in +terms of~$x$ is difficult, and cannot be attempted here. +\end{Examples} + +\Paragraph{28. Transcendental functions.} All functions of~$x$ which +are not rational or even algebraical are called \emph{transcendental} +functions. This class of functions, being defined in so purely +negative a manner, naturally includes an infinite variety of whole +kinds of functions of varying degrees of simplicity and importance. +Among these we can at present distinguish two kinds which are +particularly interesting. + +\Topic{\Item{E.} The direct and inverse trigonometrical or circular +functions.} These are the sine and cosine functions of elementary +trigonometry, and their inverses, and the functions derived from +them. We may assume provisionally that the reader is familiar +with their most important properties.\footnote + {The definitions of the circular functions given in elementary trigonometry presuppose + that any sector of a circle has associated with it a definite number called its + \emph{area}. How this assumption is justified will appear in \Ref{Ch.}{VII}\@.} +\PageSep{53} + +\begin{Examples}{XV.} +\Item{1.} Draw the graphs of $\cos x$, $\sin x$, and $a\cos x + b\sin x$. + +[Since $a\cos x + b\sin x = \beta\cos(x - \alpha)$, where $\beta = \sqrtp{a^{2} + b^{2}}$, and $\alpha$~is an angle +whose cosine and sine are $a/\sqrtp{a^{2} + b^{2}}$ and $b/\sqrtp{a^{2} + b^{2}}$, the graphs of these +three functions are similar in character.] + +\Item{2.} Draw the graphs of $\cos^{2} x$, $\sin^{2} x$, $a\cos^{2} x + b\sin^{2} x$. + +\Item{3.} Suppose the graphs of $f(x)$~and~$F(x)$ drawn. Then the graph of +\[ +f(x)\cos^{2} x + F(x)\sin^{2} x +\] +is a wavy curve which oscillates between the curves $y = f(x)$, $y = F(x)$. Draw +the graph when $f(x) = x$, $F(x) = x^{2}$. + +\Item{4.} Show that the graph of $\cos px + \cos qx$ lies between those of +$2\cos\frac{1}{2}(p - q)x$ and $-2\cos\frac{1}{2}(p + q)x$, touching each in turn. Sketch the +graph when $(p - q)/(p + q)$ is small. \MathTrip{1908.} + +\Item{5.} Draw the graphs of $x + \sin x$, $(1/x) + \sin x$, $x\sin x$, $(\sin x)/x$. + +\Item{6.} Draw the graph of~$\sin(1/x)$. + +[If $y = \sin(1/x)$, then $y = 0$ when $x = 1/m\pi$, where $m$~is any integer. Similarly +$y = 1$ when $x = 1/(2m + \frac{1}{2})\pi$ and $y = -1$ when $x = 1/(2m - \frac{1}{2})\pi$. The curve is +entirely comprised between the lines $y = -1$ and $y = 1$ (\Fig{13}). It oscillates +up and down, the rapidity of the oscillations becoming greater and greater as +$x$~approaches~$0$. For $x = 0$ the function is undefined. When $x$~is large $y$~is +small.\footnote + {See \Ref{Chs.}{IV}~and~\Ref{}{V} for explanations as to the precise meaning of this phrase.} +The negative half of the curve is similar in character to the positive +half.] + +\Item{7.} Draw the graph of $x\sin(1/x)$. + +[This curve is comprised between the lines $y = -x$ and $y = x$ just as the +last curve is comprised between the lines $y = -1$ and $y = 1$ (\Fig{14}).] +%[Illustration: Fig. 13.] +%[Illustration: Fig. 14.] +\Figures{2.25in}{13}{p053a}{2.25in}{14}{p053b}\PageLabel{53} +\PageSep{54} + +\Item{8.} Draw the graphs of $x^{2}\sin(1/x)$, $(1/x)\sin(1/x)$, $\sin^{2}(1/x)$, $\{x\sin(1/x)\}^{2}$, +$a\cos^{2}(1/x) + b\sin^{2}(1/x)$, $\sin x + \sin(1/x)$, $\sin x\sin(1/x)$. + +\Item{9.} Draw the graphs of $\cos x^{2}$, $\sin x^{2}$, $a\cos x^{2} + b\sin x^{2}$. + +\Item{10.} Draw the graphs of $\arccos x$ and $\arcsin x$. + +[If $y = \arccos x$, $x = \cos y$. This enables us to draw the graph of~$x$, considered +as a function of~$y$, and the same curve shows $y$~as a function of~$x$. +It is clear that $y$~is only defined for $-1 \leq x \leq 1$, and is infinitely many-valued +for these values of~$x$. As the reader no doubt remembers, there is, +when $-1 < x < 1$, a value of~$y$ between $0$~and~$\pi$, say~$\alpha$, and the other values +of~$y$ are given by the formula~$2n\pi ± \alpha$, where $n$~is any integer, positive or +negative.] + +\Item{11.} Draw the graphs of +\[ +\tan x,\quad +\cot x,\quad +\sec x,\quad +\cosec x,\quad +\tan^{2} x,\quad +\cot^{2} x,\quad +\sec^{2} x,\quad +\cosec^{2} x. +\] + +\Item{12.} Draw the graphs of $\arctan x$, $\arccot x$, $\arcsec x$, $\arccosec x$. Give +formulae (as in Ex.~10) expressing all the values of each of these functions +in terms of any particular value. + +\Item{13.} Draw the graphs of $\tan(1/x)$, $\cot(1/x)$, $\sec(1/x)$, $\cosec(1/x)$. + +\Item{14.} Show that $\cos x$ and $\sin x$ are not rational functions of~$x$. + +[A function is said to be \emph{periodic}, with period~$a$, if $f(x) = f(x + a)$ for all +values of~$x$ for which $f(x)$~is defined. Thus $\cos x$ and $\sin x$ have the period~$2\pi$. +It is easy to see that no periodic function can be a rational function, +unless it is a constant. For suppose that +\[ +f(x) = P(x)/Q(x), +\] +where $P$~and~$Q$ are polynomials, and that $f(x) = f(x + a)$, each of these equations +holding for all values of~$x$. Let $f(0) = k$. Then the equation $P(x) - kQ(x) = 0$ +is satisfied by an infinite number of values of~$x$, viz.\ $x = 0$, $a$,~$2a$,~etc., and +therefore for all values of~$x$. Thus $f(x) = k$ for all values of~$x$, \ie\ $f(x)$~is a +constant.] + +\Item{15.} Show, more generally, that no function with a period can be an +algebraical function of~$x$. + +[Let the equation which defines the algebraical function be +\[ +y^{m} + R_{1}y^{m-1} + \dots + R_{m} = 0 +\Tag{(1)} +\] +where $R_{1}$,~\dots\ are rational functions of~$x$. This may be put in the form +\[ +P_{0}y^{m} + P_{1}y^{m-1} + \dots + P_{m} = 0, +\] +where $P_{0}$, $P_{1}$,~\dots\ are polynomials in~$x$. Arguing as above, we see that +\[ +P_{0}k^{m} + P_{1}k^{m-1} + \dots + P_{m} = 0 +\] +\PageSep{55} +for all values of~$x$. Hence $y = k$ satisfies the equation~\Eq{(1)} for all values of~$x$, +and one set of values of our algebraical function reduces to a constant. + +Now divide~\Eq{(1)} by $y - k$ and repeat the argument. Our final conclusion is +that our algebraical function has, for any value of~$x$, the same set of values +$k$,~$k'$,~\dots; \ie\ it is composed of a certain number of constants.] + +\Item{16.} The inverse sine and inverse cosine are not rational or algebraical +functions. [This follows from the fact that, for any value of~$x$ between $-1$ +and~$+1$, $\arcsin x$ and $\arccos x$ have infinitely many values.] +\end{Examples} + +\Paragraph{29.} \Topic{\Item{F.} Other classes of transcendental functions.} Next +in importance to the trigonometrical functions come the exponential +and logarithmic functions, which will be discussed in +\Ref{Chs.}{IX}~and~\Ref{}{X}\@. But these functions are beyond our range at +present. And most of the other classes of transcendental functions +whose properties have been studied, such as the elliptic +functions, Bessel's and Legendre's functions, Gamma-functions, +and so forth, lie altogether beyond the scope of this book. +There are however some elementary types of functions which, +though of much less importance theoretically than the rational, +algebraical, or trigonometrical functions, are particularly instructive +as illustrations of the possible varieties of the functional +relation. + +\begin{Examples}{XVI.} +\Item{1.} Let $y = [x]$, where $[x]$~denotes the greatest integer +not greater than~$x$. The graph is shown in \Fig{15a}. The left-hand end +points of the thick lines, but not the right-hand ones, belong to the graph. + +\Item{2.} $y = x - [x]$. (\Fig{15b}.) +%[Illustration: Fig. 15a.] +%[Illustration: Fig. 15b.] +\Figures{2.25in}{15a}{p055a}{2.25in}{15b}{p055b} +\PageSep{56} + +\Item{3.} $y = \sqrtb{x - [x]}$. (\Fig{15c}.) + +\Item{4.} $y = [x] + \sqrtb{x - [x]}$. (\Fig{15d}.) + +\Item{5.} $y = (x - [x])^{2}$, $[x] + (x - [x])^{2}$. + +\Item{6.} $y = [\sqrt{x}]$, $[x^{2}]$, $\sqrt{x} - [\sqrt{x}]$, $x^{2} - [x^{2}]$, $[1 - x^{2}]$. +%[Illustration: Fig. 15c.] +%[Illustration: Fig. 15d.] +\Figures{2.25in}{15c}{p056a}{2.25in}{15d}{p056b} + +\Item{7.} Let $y$~be defined as \emph{the largest prime factor of~$x$} (cf.\ \Exs{x}.~6). +Then $y$~is defined only for integral values of~$x$. If +\begin{alignat*}{3} +x &= 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ &10,&\ 11,\ &12,&\ 13,\ \dots, \\ +\intertext{then} +y &= 1,\ 2,\ 3,\ 2,\ 5,\ 3,\ 7,\ 2,\ 3,\ & 5,&\ 11,\ & 3,&\ 13,\ \dots. +\end{alignat*} +The graph consists of a number of isolated points. + +\Item{8.} Let $y$~be \emph{the denominator of~$x$} (\Exs{x}.~7). In this case $y$~is defined +only for rational values of~$x$. We can mark off as many points on the graph +as we please, but the result is not in any ordinary sense of the word a curve, +and there are no points corresponding to any irrational values of~$x$. + +Draw the straight line joining the points $(N - 1, N)$,~$(N, N)$, where $N$~is a +positive integer. Show that the number of points of the locus which lie on +this line is equal to the number of positive integers less than and prime to~$N$. + +\Item{9.} Let $y = 0$ when $x$~is an integer, $y = x$ when $x$~is not an integer. The +graph is derived from the straight line $y = x$ by taking out the points +\[ +\dots\ (-1, -1),\quad (0, 0),\quad (1, 1),\quad (2, 2),\ \dots +\] +and adding the points $(-1, 0)$, $(0, 0)$, $(1, 0)$,~\dots\ on the axis of~$x$. + +The reader may possibly regard this as an unreasonable function. \emph{Why}, +he may ask, if $y$~is equal to~$x$ for all values of~$x$ save integral values, should it +not be equal to~$x$ for integral values too? The answer is simply, \emph{why should +it}? The function~$y$ does in point of fact answer to the definition of a +function: there is a relation between $x$~and~$y$ such that when $x$~is known $y$~is +known. We are perfectly at liberty to take this relation to be what we please, +however arbitrary and apparently futile. This function~$y$ is, of course, a quite +different function from that one which is \emph{always} equal to~$x$, whatever value, +integral or otherwise, $x$~may have. +\PageSep{57} + +\Item{10.} Let $y = 1$ when $x$~is rational, but $y = 0$ when $x$~is irrational. The graph +consists of two series of points arranged upon the lines $y = 1$ and $y = 0$. To +the eye it is not distinguishable from two continuous straight lines, but in +reality an infinite number of points are missing from each line. + +\Item{11.} Let $y = x$ when $x$~is irrational and $y = \sqrtb{(1 + p^{2})/(1 + q^{2})}$ when $x$~is a\PageLabel{57} +rational fraction~$p/q$. +%[Illustration: Fig. 16.] +%[** TN: The formula depicted for x = p/q, chosen to match the book visually, +% is sqrt{(10 + p^2)/(10 + q^2)}. See Transcriber's Note at end for details.] +\Figure{16}{p057} + +The irrational values of~$x$ contribute to the graph a curve in reality discontinuous, +but apparently not to be distinguished from the straight line $y = x$. + +Now consider the rational values of~$x$. First let $x$~be positive. Then +$\sqrtb{(1 + p^{2})/(1 + q^{2})}$ cannot be equal to~$p/q$ unless $p = q$, \ie\ $x = 1$. Thus all +the points which correspond to rational values of~$x$ lie off the line, except +the one point~$(1, 1)$. Again, if $p < q$, $\sqrtb{(1 + p^{2})/(1 + q^{2})} > p/q$; if $p > q$, +$\sqrtb{(1 + p^{2})/(1 + q^{2})} < p/q$. Thus the points lie above the line $y = x$ if $0 < x < 1$, +below if $x > 1$. If $p$~and~$q$ are large, $\sqrtb{(1 + p^{2})/(1 + q^{2})}$ is nearly equal to~$p/q$. +Near any value of~$x$ we can find any number of rational fractions with large +numerators and denominators. Hence the graph contains a large number of +points which crowd round the line $y = x$. Its general appearance (for positive +values of~$x$) is that of a line surrounded by a swarm of isolated points which +gets denser and denser as the points approach the line. + +The part of the graph which corresponds to negative values of~$x$ consists +of the rest of the discontinuous line together with the reflections of all these +isolated points in the axis of~$y$. Thus to the left of the axis of~$y$ the swarm +of points is not round $y = x$ but round $y = -x$, which is not itself part of the +graph. See \Fig{16}. +\end{Examples} +\PageSep{58} + +\Paragraph{30. Graphical solution of equations containing a single +unknown number.} Many equations can be expressed in the +form +\[ +f(x) = \phi(x), +\Tag{(1)} +\] +where $f(x)$~and~$\phi(x)$ are functions whose graphs are easy to draw. +And if the curves +\[ +y = f(x),\quad +y = \phi(x) +\] +intersect in a point~$P$ whose abscissa is~$\xi$, then $\xi$~is a root of the +equation~\Eq{(1)}. + +\begin{Examples}{XVII.} +\Item{1.} \Topic{The quadratic equation $ax^{2} + 2bx + c = 0$.} This +may be solved graphically in a variety of ways. For instance we may draw +the graphs of +\[ +y = ax + 2b,\quad +y = -c/x, +\] +whose intersections, if any, give the roots. Or we may take +\[ +y = x^{2},\quad +y = -(2bx + c)/a. +\] +But the most elementary method is probably to draw the circle +\[ +a(x^{2} + y^{2}) + 2bx + c = 0, +\] +whose centre is~$(-b/a, 0)$ and radius $\{\sqrtp{b^{2} - ac}\}/a$. The abscissae of its +intersections with the axis of~$x$ are the roots of the equation. + +\Item{2.} Solve by any of these methods +\[ +x^{2} + 2x - 3 = 0,\quad +x^{2} - 7x + 4 = 0,\quad +3x^{2} + 2x - 2 = 0. +\] + +\Item{3.} \Topic{The equation $x^{m} + ax + b = 0$.} This may be solved by constructing +the curves $y = x^{m}$, $y = -ax - b$. Verify the following table for the number of +roots of +\begin{gather*} +x^{m} + ax + b = 0: \\ +\begin{alignedat}{3} +&\Item{(\ia)} &&m~\emph{even} &&\left\{ + \begin{aligned} + &\text{$b$~positive, \emph{two or none},}\\ + &\text{$b$~negative, \emph{two}\Add{;}} + \end{aligned} +\right. \\ +&\Item{(\ib)} &&m~\emph{odd} &&\left\{ + \begin{aligned} + &\text{$a$~positive, \emph{one},}\\ + &\text{$a$~negative, \emph{three or one}.\qquad\qquad\qquad\qquad\qquad} + \end{aligned} +\right. +\end{alignedat} +\end{gather*} +Construct numerical examples to illustrate all possible cases. + +\Item{4.} Show that the equation $\tan x = ax + b$ has always an infinite number +of roots. + +\Item{5.} Determine the number of roots of +\[ +\sin x = x,\quad +\sin x = \tfrac{1}{3} x,\quad +\sin x = \tfrac{1}{8} x,\quad +\sin x = \tfrac{1}{120} x. +\] + +\Item{6.} Show that if $a$~is small and positive (\eg\ $a = .01$), the equation +\[ +x - a = \tfrac{1}{2}\pi\sin^{2} x +\] +has three roots. Consider also the case in which $a$~is small and negative. +Explain how the number of roots varies as $a$~varies. +\end{Examples} +\PageSep{59} + +\Paragraph{31. Functions of two variables and their graphical +representation.} In \SecNo[§]{20} we considered two variables connected +by a relation. We may similarly consider \emph{three} variables ($x$,~$y$, +and~$z$) connected by a relation such that when the values of $x$~and~$y$ +are both given, the value or values of~$z$ are known. In this case +we call~$z$ a \emph{function of the two variables} $x$~and~$y$; $x$~and~$y$ the +\emph{independent} variables, $z$~the \emph{dependent} variable; and we express +this dependence of~$z$ upon $x$~and~$y$ by writing +\[ +z = f(x, y). +\] +The remarks of \SecNo[§]{20} may all be applied, \textit{mutatis mutandis}, to this +more complicated case. + +The method of representing such functions of two variables +graphically is exactly the same in principle as in the case of +functions of a single variable. We must take three axes, $OX$, $OY$, +$OZ$ in space of three dimensions, each axis being perpendicular +to the other two. The point~$(a, b, c)$ is the point whose distances +from the planes $YOZ$, $ZOX$, $XOY$, measured parallel to $OX$, $OY$, +$OZ$, are $a$,~$b$, and~$c$. Regard must of course be paid to sign, +lengths measured in the directions $OX$, $OY$, $OZ$ being regarded +as positive. The definitions of \emph{coordinates}, \emph{axes}, \emph{origin} are the +same as before. + +Now let +\[ +z = f(x, y). +\] +As $x$~and~$y$ vary, the point~$(x, y, z)$ will move in space. The +aggregate of all the positions it assumes is called the \emph{locus} of the +point $(x, y, z)$ or the \emph{graph} of the function $z = f(x, y)$. When the +relation between $x$,~$y$, and~$z$ which defines~$z$ can be expressed in an +analytical formula, this formula is called the \emph{equation} of the locus. +It is easy to show, for example, that the equation +\[ +Ax + By + Cz + D = 0 +\] +(\emph{the general equation of the first degree}) represents a \emph{plane}, and +that the equation of any plane is of this form. The equation +\[ +(x - \alpha)^{2} + (y - \beta)^{2} + (z - \gamma)^{2} = \rho^{2}, +\] +or +\[ +x^{2} + y^{2} + z^{2} + 2Fx + 2Gy + 2Hz + C = 0, +\] +where $F^{2} + G^{2} + H^{2} - C > 0$, represents a \emph{sphere}; and so on. For +proofs of these propositions we must again refer to text-books of +Analytical Geometry. +\PageSep{60} + +\Paragraph{32. Curves in a plane.} We have hitherto used the notation +\[ +y = f(x) +\Tag{(1)} +\] +to express functional dependence of~$y$ upon~$x$. It is evident that +this notation is most appropriate in the case in which $y$~is expressed +explicitly in terms of~$x$ by means of a formula, as when +for example +\[ +y = x^{2},\quad +\sin x,\quad +a\cos^{2}x + b\sin^{2}x. +\] + +We have however very often to deal with functional relations +which it is impossible or inconvenient to express in this form. +If, for example, $y^{5} - y - x = 0$ or $x^{5} + y^{5} - ay = 0$, it is known +to be impossible to express~$y$ explicitly as an algebraical function +of~$x$. If +\[ +x^{2} + y^{2} + 2Gx + 2Fy+ C = 0, +\] +$y$~can indeed be so expressed, viz.\ by the formula +\[ +y = -F + \sqrtp{F^{2} - x^{2} - 2Gx - C}; +\] +but the functional dependence of~$y$ upon~$x$ is better and more +simply expressed by the original equation. + +It will be observed that in these two cases the functional +relation is fully expressed \emph{by equating a function of the two +variables $x$~and~$y$ to zero}, \ie\ by means of an equation +\[ +f(x, y) = 0. +\Tag{(2)} +\] + +We shall adopt this equation as the standard method of +expressing the functional relation. It includes the equation~\Eq{(1)} +as a special case, since $y - f(x)$ is a special form of a function of $x$~and~$y$. +We can then speak of the locus of the point $(x, y)$ subject +to $f(x, y) = 0$, the graph of the function~$y$ defined by $f(x, y) = 0$, +the curve or locus $f(x, y) = 0$, and the equation of this curve or +locus. + +There is another method of representing curves which is often +useful. Suppose that $x$~and~$y$ are both functions of a third +variable~$t$, which is to be regarded as essentially auxiliary and +devoid of any particular geometrical significance. We may write +\[ +x = f(t),\quad +y = F(t). +\Tag{(3)} +\] +If a particular value is assigned to~$t$, the corresponding values of +$x$ and of~$y$ are known. Each pair of such values defines a point~$(x, y)$. +\PageSep{61} +If we construct all the points which correspond in this +way to different values of~$t$, we obtain \emph{the graph of the locus +defined by the equations}~\Eq{(3)}. Suppose for example +\[ +x = a\cos t,\quad +y = a\sin t. +\] +Let $t$~vary from~$0$ to~$2\pi$. Then it is easy to see that the point +$(x, y)$ describes the circle whose centre is the origin and whose +radius is~$a$. If $t$~varies beyond these limits, $(x, y)$ describes the +circle over and over again. We can in this case at once obtain +a direct relation between $x$~and~$y$ by squaring and adding: we +find that $x^{2} + y^{2} = a^{2}$, $t$~being now eliminated. + +\begin{Examples}{XVIII.} +\Item{1.} The points of intersection of the two curves whose +equations are $f(x, y) = 0$, $\phi(x, y) = 0$, where $f$~and~$\phi$ are polynomials, can be +determined if these equations can be solved as a pair of simultaneous equations +in $x$~and~$y$. The solution generally consists of a finite number of pairs of +values of $x$~and~$y$. The two equations therefore generally represent a finite +number of isolated points. + +\Item{2.} Trace the curves $(x + y)^{2} = 1$, $xy = 1$, $x^{2} - y^{2} = 1$. + +\Item{3.} The curve $f(x, y) + \lambda\phi(x, y) = 0$ represents a curve passing through +the points of intersection of $f = 0$ and $\phi = 0$. + +\Item{4.} What loci are represented by +\[ +\Item{$(\alpha)$}\ x = at + b,\quad y = ct + d,\qquad +\Item{$(\beta)$}\ x/a = 2t/(1 + t^{2}),\quad y/a = (1 - t^{2})/(1 + t^{2}), +\] +when $t$~varies through all real values? +\end{Examples} + +\Paragraph{33. Loci in space.} In space of three dimensions there are +two fundamentally different kinds of loci, of which the simplest +examples are the plane and the straight line. + +A particle which moves along a straight line has only \emph{one +degree of freedom}. Its direction of motion is fixed; its position +can be completely fixed by one measurement of position, \eg\ by +its distance from a fixed point on the line. If we take the line as +our fundamental line~$\Lambda$ of \Ref{Chap.}{I}, the position of any of its points +is determined by a single coordinate~$x$. A particle which moves +in a plane, on the other hand, has \emph{two} degrees of freedom; its +position can only be fixed by the determination of two coordinates. + +A locus represented by a single equation +\[ +z = f(x, y) +\] +plainly belongs to the second of these two classes of loci, and is +called a \emph{surface}. It may or may not (in the obvious simple cases +\PageSep{62} +it will) satisfy our common-sense notion of what a surface +should be. + +The considerations of \SecNo[§]{31} may evidently be generalised so +as to give definitions of a function $f(x, y, z)$ of \emph{three} variables (or +of functions of any number of variables). And as in \SecNo[§]{32} we +agreed to adopt $f(x, y) = 0$ as the standard form of the equation +of a plane curve, so now we shall agree to adopt +\[ +f(x, y, z) = 0 +\] +as the standard form of equation of a surface. + +{\Loosen The locus represented by \emph{two} equations of the form $z = f(x, y)$ +or $f(x, y, z) = 0$ belongs to the first class of loci, and is called +a \emph{curve}. Thus a \emph{straight line} may be represented by two equations +of the type $Ax + By + Cz + D = 0$. A \emph{circle} in space may be +regarded as the intersection of a sphere and a plane; it may +therefore be represented by two equations of the forms} +\[ +(x - \alpha)^{2} + (y - \beta)^{2} + (z - \gamma)^{2} = \rho^{2},\quad +Ax + By + Cz + D = 0. +\] + +\begin{Examples}{XIX.} +\Item{1.} What is represented by \emph{three} equations of the type +$f(x, y, z) = 0$? + +\Item{2.} Three linear equations in general represent a single point. What are +the exceptional cases? + +\Item{3.} What are the equations of a plane curve $f(x, y) = 0$ in the plane~$XOY$, +when regarded as a curve in space? [$f(x, y) = 0$, $z = 0$.] + +\Item{4.} \Topic{Cylinders.} What is the meaning of a single equation $f(x, y) = 0$, +considered as a locus in space of three dimensions? + +[All points on the surface satisfy $f(x, y) = 0$, whatever be the value of~$z$. The +curve $f(x, y) = 0$, $z = 0$ is the curve in which the locus cuts the plane~$XOY$. +The locus is the surface formed by drawing lines parallel to~$OZ$ through all +points of this curve. Such a surface is called a \emph{cylinder}.] + +\Item{5.} \Topic{Graphical representation of a surface on a plane. Contour Maps.} +It might seem to be impossible to represent a surface adequately by a +drawing on a plane; and so indeed it is: but a very fair notion of the +nature of the surface may often be obtained as follows. Let the equation of +the surface be $z = f(x, y)$. + +If we give~$z$ a particular value~$a$, we have an equation $f(x, y) = a$, which +we may regard as determining a plane curve on the paper. We trace this +curve and mark it~$(a)$. Actually the curve~$(a)$ is the projection on the plane~$XOY$ +\PageSep{63} +of the section of the surface by the plane $z = a$. We do this for all +values of~$a$ (practically, of course, for a selection of values of~$a$). We obtain +some such figure as is shown in \Fig{17}. It will at once suggest a contoured +Ordnance Survey map: and in fact this is the principle on which such maps +are constructed. The contour line~$1000$ is the projection, on the plane of the +sea level, of the section of the surface of the land by the plane parallel to the +plane of the sea level and $1000$~ft.\ above it.\footnote + {We assume that the effects of the earth's curvature may be neglected.} +%[Illustration: Fig. 17.] +\Figure{17}{p063} + +\Item{6.} {\Loosen Draw a series of contour lines to illustrate the form of the surface +$2z = 3xy$.} + +\Item{7.} \Topic{Right circular cones.} Take the origin of coordinates at the +vertex of the cone and the axis of~$z$ along the axis of the cone; and let~$\alpha$ be +the semi-vertical angle of the cone. The equation of the cone (which must +be regarded as extending both ways from its vertex) is $x^{2} + y^{2} - z^{2}\tan^{2} \alpha = 0$. + +\Item{8.} \Topic{Surfaces of revolution in general.} The cone of Ex.~7 cuts~$ZOX$ in +two lines whose equations may be combined in the equation $x^{2} = z^{2}\tan^{2}\alpha$. +That is to say, the equation of the surface generated by the revolution of +the curve $y = 0$, $x^{2} = z^{2}\tan^{2}\alpha$ round the axis of~$z$ is derived from the second of +these equations by changing~$x^{2}$ into~$x^{2} + y^{2}$. Show generally that the equation +of the surface generated by the revolution of the curve $y = 0$, $x = f(z)$, round +the axis of~$z$, is +\[ +\sqrtp{x^{2} + y^{2}} = f(z). +\] + +\Item{9.} \Topic{Cones in general.} A surface formed by straight lines passing +through a fixed point is called a \emph{cone}: the point is called the \emph{vertex}. A +particular case is given by the right circular cone of Ex.~7. Show that the +equation of a cone whose vertex is~$O$ is of the form $f(z/x, z/y) = 0$, and that any +equation of this form represents a cone. [If $(x, y, z)$ lies on the cone, so must +$(\lambda x, \lambda y, \lambda z)$, for any value of~$\lambda$.] +\PageSep{64} + +\Item{10.} \Topic{Ruled surfaces.} Cylinders and cones are special cases of \emph{surfaces +composed of straight lines}. Such surfaces are called \emph{ruled surfaces}. + +The two equations +\[ +x = az + b,\quad +y = cz + d, +\Tag{(1)} +\] +represent the intersection of two planes, \ie\ a straight line. Now suppose +that $a$, $b$, $c$, $d$ instead of being fixed are \emph{functions of an auxiliary variable~$t$}. +For any particular value of~$t$ the equations~\Eq{(1)} give a line. As $t$~varies, +this line moves and generates a surface, whose equation may be found by +eliminating~$t$ between the two equations~\Eq{(1)}. For instance, in Ex.~7 the +equations of the line which generates the cone are +\[ +x = z\tan \alpha\cos t,\quad +y = z\tan \alpha\sin t, +\] +where $t$~is the angle between the plane~$XOZ$ and a plane through the line and +the axis of~$z$. + +Another simple example of a ruled surface may be constructed as follows. +Take two sections of a right circular cylinder perpendicular to the axis and +at a distance~$l$ apart (\Fig{18a}). We can imagine the surface of the cylinder +to be made up of a number of thin parallel rigid rods of length~$l$, such as~$PQ$, +the ends of the rods being fastened to two circular rods of radius~$a$. + +Now let us take a third circular rod of the same radius and place it +round the surface of the cylinder at a distance~$h$ from one of the first two +rods (see \Fig{18a}, where $Pq = h$). Unfasten the end~$Q$ of the rod~$PQ$ and +turn~$PQ$ about~$P$ until $Q$~can be fastened to the third circular rod in the +position~$Q'$. The angle $qOQ' = \alpha$ in the figure is evidently given by +\[ +l^{2} - h^{2} = qQ'^{2} = \left (2a\sin\tfrac{1}{2} \alpha\right)^{2}. +\] +Let all the other rods of which the cylinder was composed be treated in the +same way. We obtain a ruled surface whose form is indicated in \Fig{18b}. +It is entirely built up of straight lines; but the surface is curved everywhere, +and is in general shape not unlike certain forms of table-napkin rings (\Fig{18c}). +%[Illustration: Fig. 18a.] +%[Illustration: Fig. 18b.] +%[Illustration: Fig. 18c.] +\begin{figure}[hbt!] + \begin{minipage}{0.3\textwidth} + \centering + \Graphic{1.5in}{p064a} + \caption{Fig.~18a.} + \label{fig:18a} + \end{minipage}\hfill + \begin{minipage}{0.3\textwidth} + \centering + \Graphic{1.5in}{p064b} + \caption{Fig.~18b.} + \label{fig:18b} + \end{minipage}\hfill + \begin{minipage}{0.3\textwidth} + \centering + \Graphic{1.5in}{p064c} + \caption{Fig.~18c.} + \label{fig:18c} + \end{minipage} +\end{figure} +\end{Examples} +\PageSep{65} + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER II.} + +\begin{Examples}{} +\Item{1.} Show that if $y = f(x) = (ax + b)/(cx - a)$ then $x = f(y)$. + +\Item{2.} If $f(x) = f(-x)$ for all values of~$x$, $f(x)$~is called an \emph{even} function. +If $f(x) = -f(-x)$, it is called an \emph{odd} function. Show that any function of~$x$, +defined for all values of~$x$, is the sum of an even and an odd function of~$x$. + +[Use the identity $f(x) = \frac{1}{2}\{f(x) + f(-x)\} + \frac{1}{2}\{f(x) - f(-x)\}$.] + +\Item{3.} Draw the graphs of the functions +\[ +3\sin x + 4\cos x,\quad +\sin\left(\frac{\pi}{\sqrt{2}} \sin x\right). +\] +\MathTrip{1896.} + +\Item{4.} Draw the graphs of the functions +\[ +\sin x(a\cos^{2} x + b\sin^{2} x),\quad +\frac{\sin x}{x}(a\cos^{2} x + b\sin^{2} x),\quad +\left(\frac{\sin x}{x}\right)^{2}. +\] + +\Item{5.} Draw the graphs of the functions $x[1/x]$, $[x]/x$. + +\Item{6.} Draw the graphs of the functions +\begin{align*} +\Itemp{(i)} & \arccos(2x^{2} - 1) - 2 \arccos{x}, \\ +\Itemp{(ii)} & \arctan \frac{a + x}{1 - ax} - \arctan{a} - \arctan{x}, +\end{align*} +where the symbols $\arccos a$, $\arctan a$ denote, for any value of~$a$, the least +positive (or zero) angle, whose cosine or tangent is~$a$. + +\Item{7.} Verify the following method of constructing the graph of $f\{\phi(x)\}$ by +means of the line $y = x$ and the graphs of $f(x)$~and~$\phi(x)$: take $OA = x$ along~$OX$, +draw $AB$ parallel to~$OY$ to meet $y = \phi(x)$ in~$B$, $BC$~parallel to~$OX$ to +meet $y = x$ in~$C$, $CD$~parallel to~$OY$ to meet $y = f(x)$ in~$D$, and $DP$~parallel to~$OX$ +to meet~$AB$ in~$P$; then $P$~is a point on the graph required. + +\Item{8.} Show that the roots of $x^{3} + px + q = 0$ are the abscissae of the points of +intersection (other than the origin) of the parabola $y = x^{2}$ and the circle +\[ +x^{2} + y^{2} + (p - 1)y + qx = 0. +\] + +\Item{9.} The roots of $x^{4} + nx^{3} + px^{2} + qx + r = 0$ are the abscissae of the points of +intersection of the parabola $x^{2} = y - \frac{1}{2}nx$ and the circle +\[ +x^{2} + y^{2} + + (\tfrac{1}{8}n^{2} - \tfrac{1}{2}pn + \tfrac{1}{2}n + q)x + + (p - 1 - \tfrac{1}{4}n^{2})y + r = 0. +\] + +\Item{10.} Discuss the graphical solution of the equation +\[ +x^{m} + ax^{2} + bx + c = 0 +\] +by means of the curves $y = x^{m}$, $y = -ax^{2} - bx - c$. Draw up a table of the +various possible numbers of roots. + +\Item{11.} Solve the equation $\sec\theta + \cosec\theta = 2\sqrt{2}$; and show that the equation +$\sec\theta + \cosec\theta = c$ has two roots between $0$~and~$2\pi$ if $c^{2} < 8$ and four if $c^{2} > 8$. +\PageSep{66} + +\Item{12.} Show that the equation +\[ +2x = (2n + 1)\pi(1 - \cos x), +\] +where $n$~is a positive integer, has $2n + 3$ roots and no more, indicating +their localities roughly. \MathTrip{1896.} + +\Item{13.} Show that the equation $\frac{2}{3}x\sin x = 1$ has four roots between $-\pi$~and~$\pi$. + +\Item{14.} Discuss the number and values of the roots of the equations + +%[** TN: Items in multiple columns in the original] +\SubItem{(1)} $\cot x + x - \frac{3}{2}\pi = 0$, + +\SubItem{(2)} $x^{2} + \sin^{2} x = 1$, + +\SubItem{(3)} $\tan x = 2x/(1 + x^{2})$, + +\SubItem{(4)} $\sin x - x + \frac{1}{6}x^{3} = 0$, + +\SubItem{(5)} $(1 - \cos x)\tan\alpha - x + \sin x = 0$. + +\Item{15.} The polynomial of the second degree which assumes, when $x = a$, $b$,~$c$ +the values $\alpha$,~$\beta$,~$\gamma$ is +\[ +\alpha\frac{(x - b)(x - c)}{(a - b)(a - c)} + +\beta \frac{(x - c)(x - a)}{(b - c)(b - a)} + +\gamma\frac{(x - a)(x - b)}{(c - a)(c - b)}. +\] +Give a similar formula for the polynomial of the $(n - 1)$th~degree which +assumes, when $x = a_{1}$, $a_{2}$, \dots~$a_{n}$, the values $\alpha_{1}$, $\alpha_{2}$, \dots~$\alpha_{n}$. + +\Item{16.} Find a polynomial in~$x$ of the second degree which for the values +$0$,~$1$,~$2$ of~$x$ takes the values $1/c$, $1/(c + 1)$, $1/(c + 2)$; and show that when +$x = c + 2$ its value is~$1/(c + 1)$. \MathTrip{1911.} + +\Item{17.} Show that if $x$~is a rational function of~$y$, and $y$~is a rational function +of~$x$, then $Axy + Bx + Cy + D = 0$. + +\Item{18.} If $y$~is an algebraical function of~$x$, then $x$~is an algebraical function +of~$y$. + +\Item{19.} Verify that the equation +\[ +\cos\tfrac{1}{2}\pi x + = 1 - \frac{x^{2}}{x + (x - 1)\bigsqrtp{\dfrac{2 - x}{3}}} +\] +is approximately true for all values of~$x$ between $0$~and~$1$. [Take $x = 0$, $\frac{1}{6}$, $\frac{1}{3}$, +$\tfrac{1}{2}$, $\frac{2}{3}$, $\frac{5}{6}$,~$1$, and use tables. For which of these values is the formula exact?] + +\Item{20.} What is the form of the graph of the functions +\[ +z = [x] + [y],\quad +z = x + y - [x] - [y]? +\] + +\Item{21.} What is the form of the graph of the functions $z = \sin x + \sin y$, +$z = \sin x\sin y$, $z = \sin xy$, $z = \sin(x^{2} + y^{2})$? + +\Item{22.} \Topic{Geometrical constructions for irrational numbers.} In \Ref{Chapter}{I} +we indicated one or two simple geometrical constructions for a length equal to~$\sqrt{2}$, +starting from a given unit length. We also showed how to construct +the roots of any quadratic equation $ax^{2} + 2bx + c = 0$, it being supposed that +we can construct lines whose lengths are equal to any of the ratios of the +coefficients $a$,~$b$,~$c$, as is certainly the case if $a$,~$b$,~$c$ are rational. All these constructions +were what may be called Euclidean constructions; they depended +on the ruler and compasses only. +\PageSep{67} + +It is fairly obvious that we can construct by these methods the length +measured by any irrational number which is defined by any combination of +square roots, however complicated. Thus +\[ +\bigsqrtb[4]{\bigsqrtp{\frac{17 + 3\sqrt{11}}{17 - 3\sqrt{11}}} + - \bigsqrtp{\frac{17 - 3\sqrt{11}}{17 + 3\sqrt{11}}}} +\] +is a case in point. This expression contains a fourth root, but this is of +course the square root of a square root. We should begin by constructing~$\sqrt{11}$, +\eg\ as the mean between $1$~and~$11$: then $17 + 3\sqrt{11}$ and $17 - 3\sqrt{11}$, and +so on. Or these two mixed surds might be constructed directly as the roots of +$x^{2} - 34x + 190 = 0$. + +Conversely, \emph{only} irrationals of this kind can be constructed by Euclidean +methods. Starting from a unit length we can construct any \emph{rational} length. +And hence we can construct the line $Ax + By + C = 0$, provided that the ratios +of $A$,~$B$,~$C$ are rational, and the circle +\[ +(x - \alpha)^{2} + (y - \beta)^{2} = \rho ^{2} +\] +(or $x^{2} + y^{2} + 2gx + 2fy + c = 0$), provided that $\alpha$,~$\beta$,~$\rho$ are rational, a condition +which implies that $g$,~$f$,~$c$ are rational. + +Now in any Euclidean construction each new point introduced into the +figure is determined as the intersection of two lines or circles, or a line and +a circle. But if the coefficients are rational, such a pair of equations as +\[ +Ax + By + C = 0,\quad +x^{2} + y^{2} + 2gx + 2fy + c = 0 +\] +give, on solution, values of $x$~and~$y$ of the form $m + n\sqrt{p}$, where $m$,~$n$,~$p$ are +rational: for if we substitute for~$x$ in terms of~$y$ in the second equation we +obtain a quadratic in~$y$ with rational coefficients. Hence the coordinates of +all points obtained by means of lines and circles with rational coefficients +are expressible by rational numbers and quadratic surds. And so the same +is true of the distance $\sqrtb{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}$ between any two points so +obtained. + +With the irrational distances thus constructed we may proceed to construct +a number of lines and circles whose coefficients may now themselves involve +quadratic surds. It is evident, however, that all the lengths which we can +construct by the use of such lines and circles are still expressible by square +roots only, though our surd expressions may now be of a more complicated +form. And this remains true however often our constructions are repeated. +Hence \emph{Euclidean methods will construct any surd expression involving square +roots only, and no others}. + +One of the famous problems of antiquity was that of the duplication of +the cube, that is to say of the construction by Euclidean methods of a +length measured by~$\sqrt[3]{2}$. It can be shown that $\sqrt[3]{2}$~cannot be expressed by +means of any finite combination of rational numbers and square roots, and so +that the problem is an impossible one. See Hobson, \textit{Squaring the Circle}, +pp.~47~\textit{et~seq.}; the first stage of the proof, viz.\ the proof that $\sqrt[3]{2}$~cannot be a +root of a quadratic equation $ax^{2} + 2bx + c = 0$ with rational coefficients, was +given in \Ref{Ch.}{I} (\MiscExs{I}~24). +\PageSep{68} + +\Item{23.} \Topic{Approximate quadrature of the circle.} Let $O$~be the centre of +a circle of radius~$R$. On the tangent at~$A$ take $AP = \frac{11}{5}R$ and $AQ = \frac{13}{5}R$, +in the same direction. On~$AO$ take $AN = OP$ and draw~$NM$ parallel to~$OQ$ +and cutting~$AP$ in~$M$. Show that +\[ +AM/R = \tfrac{13}{25}\sqrt{146}, +\] +and that to take~$AM$ as being equal to the circumference of the circle would +lead to a value of~$\pi$ correct to five places of decimals. If $R$~is the earth's +radius, the error in supposing $AM$ to be its circumference is less than $11$~yards. + +\Item{24.} Show that the only lengths which can be constructed with the ruler +only, starting from a given unit length, are rational lengths. + +\Item{25.} \Topic{Constructions for $\sqrt[3]{2}$.} $O$~is the vertex and $S$~the focus of the +parabola $y^{2} = 4x$, and $P$~is one of its points of intersection with the parabola +$x^{2} = 2y$. Show that $OP$~meets the latus rectum of the first parabola in a point~$Q$ +such that $SQ = \sqrt[3]{2}$. + +\Item{26.} Take a circle of unit diameter, a diameter~$OA$ and the tangent at~$A$. +Draw a chord~$OBC$ cutting the circle at~$B$ and the tangent at~$C$. On this +line take $OM = BC$. Taking $O$~as origin and $OA$~as axis of~$x$, show that the +locus of~$M$ is the curve +\[ +(x^{2} + y^{2})x - y^{2} = 0 +\] +(the \emph{Cissoid of Diocles}). Sketch the curve. Take along the axis of~$y$ a length +$OD = 2$. Let $AD$~cut the curve in~$P$ and $OP$~cut the tangent to the circle +at~$A$ in~$Q$. Show that $AQ = \sqrt[3]{2}$. +\end{Examples} +\PageSep{69} + + +\Chapter{III}{COMPLEX NUMBERS} + +\Paragraph{34. Displacements along a line and in a plane.} The +`real number'~$x$, with which we have been concerned in the two +preceding chapters, may be regarded from many different points +of view. It may be regarded as a pure number, destitute of +geometrical significance, or a geometrical significance may be +attached to it in at least three different ways. It may be regarded +as \emph{the measure of a length}, viz.~the length~$A_{0}P$ along the +line~$\Lambda$ of \Ref{Chap.}{I}\@. It may be regarded as \emph{the mark of a point}, +viz.~the point~$P$ whose distance from~$A_{0}$ is~$x$. Or it may be +regarded as \emph{the measure of a displacement} or \emph{change of position} +on the line~$\Lambda$. It is on this last point of view that we shall now +concentrate our attention. + +Imagine a small particle placed at~$P$ on the line~$\Lambda$ and then +displaced to~$Q$. We shall call the displacement or change of +position which is needed to transfer the particle from $P$ to~$Q$ \emph{the +displacement~$\Seg{PQ}$}. To specify a displacement completely three +things are needed, its \emph{magnitude}, its \emph{sense} forwards or backwards +along the line, and what may be called its \emph{point of application}, +\ie\ the original position~$P$ of the particle. But, when we are +thinking merely of the change of position produced by the displacement, +it is natural to disregard the point of application and +to consider all displacements as equivalent whose lengths and +senses are the same. Then the displacement is completely specified +by the length $PQ = x$, the sense of the displacement being +fixed by the sign of~$x$. We may therefore, without ambiguity, +speak of \emph{the displacement~$[x]$},\footnote + {It is hardly necessary to caution the reader against confusing this use of the + symbol~$[x]$ and that of \Ref{Chap.}{II} (\Exs{xvi}.\ and \MiscExs{II}).} +and we may write $\Seg{PQ} = [x]$. +\PageSep{70} + +We use the square bracket to distinguish the displacement~$[x]$ +from the length or number~$x$.\footnote + {Strictly speaking we ought, by some similar difference of notation, to distinguish + the actual length~$x$ from the number~$x$ which measures it. The reader + will perhaps be inclined to consider such distinctions futile and pedantic. But + increasing experience of mathematics will reveal to him the great importance of + distinguishing clearly between things which, however intimately connected, are not + the same. If cricket were a mathematical science, it would be very important to + distinguish between the \emph{motion} of the batsman between the wickets, the \emph{run} which + he scores, and the \emph{mark} which is put down in the score-book.} +If the coordinate of~$P$ is~$a$, that +of~$Q$ will be~$a + x$; the displacement~$[x]$ therefore transfers a +particle from the point~$a$ to the point~$a + x$. + +We come now to consider \emph{displacements in a plane}. We may +define the displacement~$\Seg{PQ}$ as before. But now more data are +required in order to specify it completely. We require to know: +(i)~the \emph{magnitude} of the displacement, \ie\ the length of the +straight line~$PQ$; (ii)~the \emph{direction} of the displacement, which is +determined by the angle which $PQ$~makes with some fixed line in +the plane; (iii)~the \emph{sense} of the displacement; and (iv)~its \emph{point +of application}. Of these requirements we may disregard the +fourth, if we consider two displacements as equivalent if they are +%[Illustration: Fig. 19.] +\Figure[2in]{19}{p070} +the same in magnitude, direction, and sense. In other words, if +$PQ$~and~$RS$ are equal and parallel, and the sense of motion from +$P$~to~$Q$ is the same as that of +motion from $R$~to~$S$, we regard +the displacements $\Seg{PQ}$ and~$\Seg{RS}$ as +equivalent, and write +\[ +\Seg{PQ} = \Seg{RS}. +\] + +Now let us take any pair of +coordinate axes in the plane (such +as $OX$,~$OY$ in \Fig{19}). Draw a +line~$OA$ equal and parallel to~$PQ$, the sense of motion from $O$ +to~$A$ being the same as that from $P$ to~$Q$. Then $\Seg{PQ}$~and~$\Seg{OA}$ +are equivalent displacements. Let $x$~and~$y$ be the coordinates +of~$A$. Then it is evident that $\Seg{OA}$~is completely specified +if $x$~and~$y$ are given. We call $\Seg{OA}$ \emph{the displacement $[x, y]$} and +write +\[ +\Seg{OA} = \Seg{PQ} = \Seg{RS} = [x, y]. +\] +\PageSep{71} + +\Paragraph{35. Equivalence of displacements. Multiplication of +displacements by numbers.} If $\xi$~and~$\eta$ are the coordinates +of~$P$, and $\xi'$~and~$\eta'$ those of~$Q$, it is evident that +\[ +x = \xi' - \xi,\quad +y = \eta' - \eta. +\] +The displacement from $(\xi, \eta)$ to $(\xi', \eta')$ is therefore +\[ +[\xi' - \xi, \eta' - \eta]. +\] + +It is clear that two displacements $[x, y]$, $[x', y']$ are equivalent +if, and only if, $x = x'$, $y = y'$. Thus $[x, y] = [x', y']$ if and only if +\[ +x = x',\quad +y = y'. +\Tag{(1)} +\] + +The reverse displacement $\Seg{QP}$ would be $[\xi - \xi', \eta - \eta']$, and it +is natural to agree that +\begin{align*} +[\xi - \xi', \eta - \eta'] &= -[\xi' - \xi, \eta' - \eta],\\ +\Seg{QP} &= -\Seg{PQ}, +\end{align*} +{\Loosen these equations being really definitions of the meaning of the +symbols $-[\xi' - \xi, \eta' - \eta]$, $-\Seg{PQ}$. Having thus agreed that} +\[ +-[x, y] = [-x, -y], +\] +it is natural to agree further that +\[ +\alpha[x, y] = [\alpha x, \alpha y], +\Tag{(2)} +\] +{\Loosen where $\alpha$~is any real number, positive or negative. Thus (\Fig{19}) +if $OB = -\frac{1}{2}OA$ then} +\[ +\Seg{OB} = -\tfrac{1}{2}\Seg{OA} = -\tfrac{1}{2}[x, y] + = [-\tfrac{1}{2}x, -\tfrac{1}{2}y]. +\] + +The equations \Eq{(1)}~and~\Eq{(2)} define the first two important ideas +connected with displacements, viz.\ \emph{equivalence} of displacements, +and \emph{multiplication of displacements by numbers}. + +\Paragraph{36. Addition of displacements.} We have not yet given +any definition which enables us to attach any meaning to the +expressions +\[ +\Seg{PQ} + \Seg{P'Q'},\quad +[x, y] + [x', y']. +\] +Common sense at once suggests that we should define the sum +of two displacements as the displacement which is the result +of the successive application of the two given displacements. In +\PageSep{72} +other words, it suggests that if $QQ_{1}$~be drawn equal and parallel +to~$P'Q'$, so that the result of successive displacements $\Seg{PQ}$,~$\Seg{P'Q'}$ on +a particle at~$P$ is to transfer it first to~$Q$ and then to~$Q_{1}$ then we +should define the sum of $\Seg{PQ}$~and~$\Seg{P'Q'}$ as being~$\Seg{PQ_{1}}$. If then we +draw $OA$~equal and parallel to~$PQ$, and $OB$~equal and parallel to~$P'Q'$, +and complete the parallelogram~$OACB$, we have +%[Illustration: Fig. 20.] +\Figure[3.5in]{20}{p072} +\[ +\Seg{PQ} + \Seg{P'Q'} = \Seg{PQ_{1}} = \Seg{OA} + \Seg{OB} = \Seg{OC}. +\] + +Let us consider the consequences of adopting this definition. +If the coordinates of~$B$ are $x'$,~$y'$, then those of the middle point of~$AB$ +are $\frac{1}{2}(x + x')$, $\frac{1}{2} (y + y')$, and those of~$C$ are $x + x'$, $y + y'$. Hence +\[ +[x, y] + [x', y'] = [x + x', y + y'], +\Tag{(3)} +\] +which may be regarded as the symbolic definition of addition of +displacements. We observe that +\begin{align*} +[x', y'] + [x, y] + &= [x' + x, y' + y]\\ + &= [x + x', y + y'] = [x, y] + [x', y'] +\end{align*} +In other words, \emph{addition of displacements obeys the commutative +law} expressed in ordinary algebra by the equation $a + b = b + a$. +This law expresses the obvious geometrical fact that if we move +from~$P$ first through a distance~$PQ_{2}$ equal and parallel to~$P'Q'$, +and then through a distance equal and parallel to~$PQ$, we shall +arrive at the same point~$Q_{1}$ as before. +\PageSep{73} + +In particular +\[ +[x, y] = [x, 0] + [0, y]. +\Tag{(4)} +\] +Here $[x, 0]$ denotes a displacement through a distance~$x$ in +a direction parallel to~$OX$. It is in fact what we previously +denoted by~$[x]$, when we were considering only displacements +along a line. We call $[x, 0]$~and~$[0, y]$ the \emph{components} of~$[x, y]$, +and $[x, y]$ their \emph{resultant}. + +When we have once defined addition of two displacements, +there is no further difficulty in the way of defining addition of +any number. Thus, by definition, +\begin{gather*} +[x, y] + [x', y'] + [x'', y''] + = ([x, y] + [x', y']) + [x'', y'']\\ + = [x + x', y + y'] + [x'', y''] + = [x + x' + x'', y + y' + y'']. +\end{gather*} + +We define \emph{subtraction} of displacements by the equation +\[ +[x, y] - [x', y'] = [x, y] + (-[x', y']), +\Tag{(5)} +\] +which is the same thing as $[x, y] + [-x', -y']$ or as $[x - x', y - y']$. +In particular +\[ +[x, y] - [x, y] = [0, 0]. +\] + +The displacement~$[0, 0]$ leaves the particle where it was; it is +the \emph{zero displacement}, and we agree to write $[0, 0] = 0$. + +\begin{Examples}{XX.} +\Item{1.} Prove that + +\SubItem{(i)} $\alpha [\beta x, \beta y] = \beta [\alpha x, \alpha y] = [\alpha \beta x, \alpha \beta y]$, + +\SubItem{(ii)} $([x, y] + [x', y']) + [x'', y''] = [x, y] + ([x', y'] + [x'', y''])$, + +\SubItem{(iii)} $[x, y] + [x', y'] = [x', y'] + [x, y]$, + +\SubItem{(iv)} $(\alpha + \beta) [x, y] = \alpha [x, y] + \beta [x, y]$, + +\SubItem{(v)} $\alpha \{[x, y] + [x', y']\} = \alpha [x, y] + \alpha [x', y']$. + +[We have already proved~(iii). The remaining equations follow with equal +ease from the definitions. The reader should in each case consider the +geometrical significance of the equation, as we did above in the case of~(iii).] + +\Item{2.} If $M$~is the middle point of~$PQ$, then $\Seg{OM} = \frac{1}{2}(\Seg{OP} + \Seg{OQ})$. More generally, +if $M$~divides~$PQ$ in the ratio~$\mu : \lambda$, then +\[ +\Seg{OM} + = \frac{\lambda}{\lambda + \mu}\, \Seg{OP} + + \frac{\mu}{\lambda + \mu}\, \Seg{OQ}. +\] + +\Item{3.} If $G$~is the centre of mass of equal particles at $P_{1}$, $P_{2}$, \dots,~$P_{n}$, then +\[ +\Seg{OG} = (\Seg{OP_{1}} + \Seg{OP_{2}} + \dots + \Seg{OP_{n}})/n. +\] +\PageSep{74} + +\Item{4.} If $P$,~$Q$,~$R$ are collinear points in the plane, then it is possible to find +real numbers $\alpha$,~$\beta$,~$\gamma$, not all zero, and such that +\[ +\alpha · \Seg{OP} + \beta · \Seg{OQ} + \gamma · \Seg{OR} = 0; +\] +and conversely. [This is really only another way of stating Ex.~2.] + +\Item{5.} If $\Seg{AB}$~and~$\Seg{AC}$ are two displacements not in the same straight line, +and +\[ +\alpha · \Seg{AB} + \beta · \Seg{AC} = \gamma · \Seg{AB} + \delta · \Seg{AC}, +\] +then $\alpha = \gamma$ and $\beta = \delta$. + +[Take $AB_{1} = \alpha · AB$, $AC_{1} = \beta · AC$. Complete the parallelogram $AB_{1}P_{1}C_{1}$. +Then $\Seg{AP_{1}} = \alpha · \Seg{AB} + \beta · \Seg{AC}$. It is evident that $\Seg{AP_{1}}$~can only be expressed +in this form in one way, whence the theorem follows.] + +\Item{6.} $ABCD$~is a parallelogram. Through~$Q$, a point inside the parallelogram, +$RQS$~and~$TQU$ are drawn +parallel to the sides. Show that +$RU$,~$TS$ intersect on~$AC$. +%[Illustration: Fig. 21.] +\Figure[2.75in]{21}{p074} + +[Let the ratios $AT:AB$, $AR:AD$ +be denoted by $\alpha$,~$\beta$. Then +\begin{gather*} +\Seg{AT} = \alpha · \Seg{AB},\quad +\Seg{AR} = \beta · \Seg{AD}, \\ +\Seg{AU} = \alpha · \Seg{AB} + \Seg{AD},\quad +\Seg{AS} = \Seg{AB} + \beta · \Seg{AD}. +\end{gather*} + +Let $RU$~meet $AC$ in~$P$. Then, +since $R$,~$U$,~$P$ are collinear, +\[ +\Seg{AP} + = \frac{\lambda}{\lambda + \mu}\, \Seg{AR} + + \frac{\mu}{\lambda + \mu}\, \Seg{AU}, +\] +where $\mu/\lambda$ is the ratio in which $P$~divides~$RU$. That is to say +\[ +\Seg{AP} + = \frac{\alpha\mu}{\lambda + \mu}\, \Seg{AB} + + \frac{\beta\lambda + \mu}{\lambda + \mu}\, \Seg{AD}. +\] + +But since $P$~lies on~$AC$, $\Seg{AP}$~is a numerical multiple of~$\Seg{AC}$; say +\[ +\Seg{AP} = k · \Seg{AC} = k · \Seg{AB} + k · \Seg{AD}. +\] +Hence (Ex.~5) $\alpha\mu = \beta\lambda + \mu = (\lambda + \mu)k$, from which we deduce +\[ +k = \frac{\alpha\beta}{\alpha + \beta - 1}. +\] +The symmetry of this result shows that a similar argument would also give +\[ +\Seg{AP'} = \frac{\alpha\beta}{\alpha + \beta - 1}\, \Seg{AC}, +\] +if $P'$~is the point where $TS$~meets~$AC$. Hence $P$~and~$P'$ are the same point.] + +\Item{7.} $ABCD$~is a parallelogram, and $M$~the middle point of~$AB$. Show that +$DM$~trisects and is trisected by~$AC$.\footnote + {The two preceding examples are taken from Willard Gibbs' \textit{Vector Analysis}.} +\end{Examples} +\PageSep{75} + +\Paragraph{37. Multiplication of displacements.} So far we have +made no attempt to attach any meaning whatever to the notion +of the \emph{product} of two displacements. The only kind of multiplication +which we have considered is that in which a displacement +is multiplied by a number. The expression +\[ +[x, y] × [x', y'] +\] +so far means nothing, and we are at liberty to define it to mean +anything we like. It is, however, fairly clear that if any definition +of such a product is to be of any use, the product of two displacements +must itself be a displacement. + +We might, for example, define it as being equal to +\[ +[x + x', y + y']; +\] +in other words, we might agree that the product of two displacements +was to be always equal to their sum. But there would be +two serious objections to such a definition. In the first place our +definition would be futile. We should only be introducing a new +method of expressing something which we can perfectly well +express without it. In the second place our definition would be +inconvenient and misleading for the following reasons. If $\alpha$~is +a real number, we have already defined $\alpha [x, y]$ as~$[\alpha x, \alpha y]$. Now, +as we saw in \SecNo[§]{34}, the real number~$\alpha$ may itself from one point of +view be regarded as a displacement, viz.\ the displacement~$[\alpha]$ +along the axis~$OX$, or, in our later notation, the displacement +$[\alpha, 0]$. It is therefore, if not absolutely necessary, at any rate +most desirable, that our definition should be such that +\[ +[\alpha, 0] [x, y] = [\alpha x, \alpha y], +\] +and the suggested definition does not give this result. + +A more reasonable definition might appear to be +\[ +[x, y] [x', y'] = [xx', yy']. +\] +But this would give +\[ +[\alpha, 0] [x, y] = [\alpha x, 0]; +\] +and so this definition also would be open to the second objection. + +In fact, it is by no means obvious what is the best meaning +to attach to the product $[x, y] [x', y']$. All that is clear is (1)~that, +if our definition is to be of any use, this product must itself be +\PageSep{76} +a displacement whose coordinates depend on $x$~and~$y$, or in other +words that we must have +\[ +[x, y] [x', y'] = [X, Y], +\] +where $X$~and~$Y$ are functions of $x$,~$y$,~$x'$, and~$y'$; (2)~that the +definition must be such as to agree with the equation +\[ +[x, 0] [x', y'] = [xx', xy']; +\] +and (3)~that the definition must obey the ordinary commutative, +distributive, and associative laws of multiplication, so that +\begin{align*} +[x, y] [x', y'] &= [x', y'] [x, y],\\ +([x, y] + [x', y']) [x'', y''] &= [x, y] [x'', y''] + [x', y'] [x'', y''],\\ +[x, y] ([x', y'] + [x'', y'']) &= [x, y] [x', y'] + [x, y] [x'', y''],\\ +\intertext{and} +[x, y] ([x', y'] [x'', y'']) &= ([x, y] [x', y']) [x'', y'']. +\end{align*} + +\Paragraph{38.} The right definition to take is suggested as follows. We +know that, if $OAB$,~$OCD$ are two similar triangles, the angles +corresponding in the order in which they are written, then +\[ +OB/OA = OD/OC, +\] +or $OB · OC = OA · OD$. This suggests that we should try to define +multiplication and division of displacements in such a way that +\[ +\Seg{OB}/\Seg{OA} = \Seg{OD}/\Seg{OC},\quad +\Seg{OB} · \Seg{OC} = \Seg{OA} · \Seg{OD}. +\] +%[Illustration: Fig. 22.] +\Figure[3.5in]{22}{p076} + +Now let +\[ +\Seg{OB} = [x, y],\quad +\Seg{OC} = [x', y'],\quad +\Seg{OD} = [X, Y], +\] +\PageSep{77} +and suppose that $A$~is the point~$(1, 0)$, so that $\Seg{OA} = [1, 0]$. Then +\[ +\Seg{OA} · \Seg{OD} = [1, 0] [X, Y] = [X, Y], +\] +and so +\[ +[x, y] [x', y'] = [X, Y]. +\] +The product $\Seg{OB} · \Seg{OC}$ is therefore to be defined as $\Seg{OD}$, $D$~being +obtained by constructing on~$OC$ a triangle similar to~$OAB$. In +order to free this definition from ambiguity, it should be observed +that on~$OC$ we can describe \emph{two} such triangles, $OCD$~and~$OCD'$. +We choose that for which the angle~$COD$ is equal to~$AOB$ in sign +as well as in magnitude. We say that the two triangles are then +\emph{similar in the same sense}. + +If the polar coordinates of $B$~and~$C$ are $(\rho, \theta)$ and $(\sigma, \phi)$, so +that +\[ +x = \rho\cos\theta,\quad +y = \rho\sin\theta,\quad +x' = \sigma\cos\phi,\quad +y' = \sigma\sin\phi, +\] +then the polar coordinates of~$D$ are evidently $\rho\sigma$ and $\theta + \phi$. Hence +\begin{alignat*}{2} +X &= \rho\sigma\cos(\theta + \phi) &&= xx' - yy',\\ +Y &= \rho\sigma\sin(\theta + \phi) &&= xy' + yx'. +\end{alignat*} +The required definition is therefore +\[ +[x, y] [x', y'] = [xx' - yy', xy' + yx']. +\Tag{(6)} +\] + +We observe (1)~that if $y = 0$, then $X = xx'$, $Y = xy'$, as we +desired; (2)~that the right-hand side is not altered if we interchange +$x$~and~$x'$, and $y$~and~$y'$, so that +\[ +[x, y] [x', y'] = [x', y'] [x, y]; +\] +and (3)~that +\begin{multline*} +\{[x, y] + [x', y']\} [x'', y''] = [x + x', y + y'] [x'', y'']\\ +\begin{aligned}[t] + &= [(x + x') x'' - (y + y') y'', (x + x') y'' + (y + y') x'']\\ + &= [xx'' - yy'', xy'' + yx''] + [x'x'' - y'y'', x'y'' + y'x'']\\ + &= [x, y] [x'', y''] + [x', y'] [x'', y'']. +\end{aligned} +\end{multline*} + +Similarly we can verify that all the equations at the end of \SecNo[§]{37} +are satisfied. Thus the definition~\Eq{(6)} fulfils all the requirements +which we made of it in \SecNo[§]{37}. + +\begin{Remark} +\Par{Example.} Show directly from the geometrical definition given above +that multiplication of displacements obeys the commutative and distributive +laws. [Take the commutative law for example. The product $\Seg{OB} · \Seg{OC}$ is~$\Seg{OD}$ +(\Fig{22}), $COD$~being similar to~$AOB$. To construct the product $\Seg{OC} · \Seg{OB}$ we +\PageSep{78} +should have to construct on~$OB$ a triangle~$BOD_{1}$ similar to~$AOC$; and so what +we want to prove is that $D$~and~$D_{1}$ coincide, or that $BOD$~is similar to~$AOC$. +This is an easy piece of elementary geometry.] +\end{Remark} + +\Paragraph{39. Complex numbers.} Just as to a displacement~$[x]$ along~$OX$ +correspond a point~$(x)$ and a real number~$x$, so to a displacement~$[x, y]$ +in the plane correspond a point~$(x, y)$ and a \emph{pair +of real numbers $x$,~$y$}. + +We shall find it convenient to denote this pair of real numbers +$x$,~$y$ by the symbol +\[ +x + yi. +\] +The reason for the choice of this notation will appear later. +For the present the reader must regard $x + yi$ as \emph{simply another +way of writing $[x, y]$}. The expression $x + yi$ is called a \emph{complex +number}. + +We proceed next to define \emph{equivalence}, \emph{addition}, and \emph{multiplication} +of complex numbers. To every complex number corresponds +a displacement. Two complex numbers are equivalent if the +corresponding displacements are equivalent. The sum or product +of two complex numbers is the complex number which corresponds +to the sum or product of the two corresponding displacements. +Thus +\[ +x + yi = x' + y'i, +\Tag{(1)} +\] +if and only if $x = x'$, $y = y'$; +\begin{gather*} +(x + yi) + (x' + y'i) = (x + x') + (y + y')i; +\Tag{(2)}\\ +(x + yi) (x' + y'i) = xx' - yy' + (xy' + yx')i. +\Tag{(3)} +\end{gather*} + +In particular we have, as special cases of \Eq{(2)}~and~\Eq{(3)}, +\begin{gather*} +x + yi = (x + 0i) + (0 + yi),\\ +(x + 0i) (x' + y'i) = xx' + xy'i; +\end{gather*} +and these equations suggest that there will be no danger of +confusion if, when dealing with complex numbers, we write $x$~for +$x + 0i$ and $yi$~for $0 + yi$, as we shall henceforth. + +Positive integral powers and polynomials of complex numbers +are then defined as in ordinary algebra. Thus, by putting $x = x'$, +$y = y'$ in~\Eq{(3)}, we obtain +\[ +(x + yi)^{2} = (x + yi) (x + yi) = x^{2} - y^{2} + 2xyi. +\] +\PageSep{79} + +The reader will easily verify for himself that addition and +multiplication of complex numbers obey the laws of algebra +expressed by the equations +\begin{gather*} +\DPchg{x + yi}{(x + yi)} + (x' + y'i) = (x' + y'i) + (x + yi),\\ +\{(x + yi) + (x' + y'i)\} + (x'' + y''i) + = (x + yi) + \{(x' + y'i) + (x'' + y''i)\},\\ +(x + yi) (x' + y'i) = (x' + y'i) (x + yi),\\ +(x + yi)\{(x' + y'i) + (x'' + y''i)\} + = (x + yi)(x' + y'i) + (x + yi)(x'' + y''i),\\ +\Squeeze{$\{(x + yi) + (x' + y'i)\}(x'' + y''i) + = (x + yi)(x'' + y''i) + (x' + y'i)(x'' + y''i)$,}\\ +(x + yi) \{(x' + y'i) (x'' + y''i)\} = \{(x + yi) (x' + y'i)\} (x'' + y''i), +\end{gather*} +the proofs of these equations being practically the same as those +of the corresponding equations for the corresponding displacements. + +Subtraction and division of complex numbers are defined as +in ordinary algebra. Thus we may define $(x + yi) - (x' + y'i)$ as +\[ +(x + yi) + \{- (x' + y'i)\} = x + yi + (-x' - y'i) = (x - x') + (y - y')i; +\] +or again, as the number $\xi + \eta i$ such that +\[ +(x' + y'i) + (\xi + \eta i) = x + yi, +\] +which leads to the same result. And $(x + yi)/(x' + y'i)$ is defined +as being the complex number $\xi + \eta i$ such that +\[ +(x' + y'i) (\xi + \eta i) = x + yi, +\] +or +\[ +x' \xi - y' \eta + (x' \eta + y' \xi)i = x + yi, +\] +or +\[ +x' \xi - y' \eta = x,\quad +x' \eta + y' \xi = y. +\Tag{(4)} +\] + +Solving these equations for $\xi$~and~$\eta$, we obtain +\[ +\xi = \frac{xx' + yy'}{x'^{2} + y'^{2}},\quad +\eta = \frac{yx' - xy'}{x'^{2} + y'^{2}}. +\] +This solution fails if $x'$~and~$y'$ are both zero, \ie\ if $x' + y'i = 0$. +Thus subtraction is always possible; division is always possible +unless the divisor is zero. +\PageSep{80} + +\begin{Remark} +\Par{Examples.} \Item{(1)} From a geometrical point of view, the problem of the +division of the displacement~$\Seg{OB}$ by~$\Seg{OC}$ is that of finding~$D$ so that +the triangles $COB$,~$AOD$ are similar, and this is evidently possible (and +the solution unique) unless $C$~coincides with~$0$, or +$\Seg{OC} = 0$. +%[Illustration: Fig. 23.] +\Figure[2.5in]{23}{p080} + +\Item{(2)} The numbers $x + yi$, $x - yi$ are +said to be \emph{conjugate}. Verify that +\[ +(x + yi)(x - yi) = x^{2} + y^{2}, +\] +so that the product of two conjugate +numbers is real, and that +%[** TN: Set on two lines in the original] +\[ +\frac{x + yi}{x' + y'i} + = \frac{(x + yi)(x' - y'i)}{(x' + y'i)(x' - y'i)}\\ + = \frac{xx' + yy' + (x'y - xy')i}{x'^{2} + y'^{2}}. +\] +\end{Remark} + +\Paragraph{40.} One most important property of real numbers is that +known as \emph{the factor theorem}, which asserts that \emph{the product of two +numbers cannot be zero unless one of the two is itself zero}. To +prove that this is also true of complex numbers we put $x = 0$, +$y = 0$ in the equations~\Eq{(4)} of the preceding section. Then +\[ +x'\xi - y'\eta = 0,\quad +x'\eta + y'\xi = 0. +\] +These equations give $\xi = 0$, $\eta = 0$, \ie +\[ +\xi + \eta i = 0, +\] +unless $x' = 0$ and $y' = 0$, or $x' + y'i = 0$. Thus $x + yi$ cannot vanish +unless either $x' + y'i$ or $\xi + \eta i$ vanishes. + +\Paragraph{41. The equation $i^{2} = -1$.} We agreed to simplify our +notation by writing~$x$ instead of $x + 0i$ and $yi$~instead of $0 + yi$. +The particular complex number~$1i$ we shall denote simply by~$i$. +It is the number which corresponds to a unit displacement along~$OY$. +Also +\[ +i^{2} = ii = (0 + 1i) (0 + 1i) = (0 · 0 - 1 · 1) + (0 · 1 + 1 · 0)i = -1. +\] +Similarly $(-i)^{2} = -1$. Thus the complex numbers $i$~and~$-i$ +satisfy the equation $x^{2} = -1$. + +The reader will now easily satisfy himself that the upshot of +the rules for addition and multiplication of complex numbers is +this, that \emph{we operate with complex numbers in exactly the same +way as with real numbers, treating the symbol~$i$ as itself a number, +\PageSep{81} +but replacing the product $ii = i^{2}$ by~$-1$ whenever it occurs}. Thus, +for example, +\begin{align*} +(x + yi) (x' + y'i) &= xx' + xy'i + yx'i + yy'i^{2}\\ + &= (xx' - yy') + (xy'+ yx')i. +\end{align*} + +\Paragraph{42. The geometrical interpretation of multiplication +by~$i$.} Since +\[ +(x + yi)i = -y + xi, +\] +it follows that if $x + yi$ corresponds to~$\Seg{OP}$, and $OQ$~is drawn equal +to~$OP$ and so that $POQ$~is a positive right angle, then $(x + yi)i$ +corresponds to~$\Seg{OQ}$. In other words, \emph{multiplication of a complex +number by~$i$ turns the corresponding displacement through a right +angle}. + +We might have developed the whole theory of complex +numbers from this point of view. Starting with the ideas of +$x$~as representing a displacement along~$OX$, and of $i$~as a symbol +of operation equivalent to turning~$x$ through a right angle, we +should have been led to regard~$yi$ as a displacement of magnitude~$y$ +along~$OY$. It would then have been natural to define $x + yi$ as +in \SecNo[§§]{37}~and~\SecNo{40}, and $(x + yi)i$ would have represented the displacement +obtained by turning $x + yi$ through a right angle, +\ie\ $-y + xi$. Finally, we should naturally have defined $(x + yi)x'$ +as $xx' + yx'i$, $(x + yi)y'i$ as $-yy' + xy'i$, and $(x + yi) (x' + y'i)$ as the +sum of these displacements, \ie\ as +\[ +xx' - yy' + (xy' + yx')i. +\] + +\Paragraph{43. The equations $z^{2} + 1 = 0$, $az^{2} + 2bz + c = 0$.} There is no +real number~$z$ such that $z^{2} + 1 = 0$; this is expressed by saying +that the equation has \emph{no real roots}. But, as we have just seen, +the two complex numbers $i$~and~$-i$ satisfy this equation. We +express this by saying that the equation has \emph{the two complex roots} +$i$~and~$-i$. Since $i$~satisfies $z^{2} = -1$, it is sometimes written in the +form~$\sqrtp{-1}$. + +Complex numbers are sometimes called \emph{imaginary}.\footnote + {The phrase `real number' was introduced as an antithesis to `imaginary + number'.} +The +expression is by no means a happily chosen one, but it is firmly +\PageSep{82} +established and has to be accepted. It cannot, however, be too +strongly impressed upon the reader that an `imaginary number' +is no more `imaginary', in any ordinary sense of the word, than a +`real' number; and that it is not a number at all, in the sense in +which the `real' numbers are numbers, but, as should be clear from +the preceding discussion, \emph{a pair of numbers $(x, y)$}, united symbolically, +for purposes of technical convenience, in the form $x + yi$. Such +a pair of numbers is no less `real' than any ordinary number +such as~$\frac{1}{2}$, or than the paper on which this is printed, or than +the Solar System. Thus +\[ +i = 0 + 1i +\] +stands for the pair of numbers $(0, 1)$, and may be represented +geometrically by a point or by the displacement $[0, 1]$. And +when we say that $i$~is a root of the equation $z^{2} + 1 = 0$, what we +mean is simply that we have defined a method of combining such +pairs of numbers (or displacements) which we call `multiplication', +and which, when we so combine $(0, 1)$ with itself, gives the +result~$(-1, 0)$. + +Now let us consider the more general equation +\[ +az^{2} + 2bz + c = 0, +\] +where $a$,~$b$,~$c$ are real numbers. If $b^{2} > ac$, the ordinary method of +solution gives two real roots +\[ +\{-b ± \sqrtp{b^{2} - ac}\}/a. +\] +If $b^{2} < ac$, the equation has no real roots. It may be written in +the form +\[ +\{z + (b/a)\}^{2} = -(ac - b^{2})/a^{2}, +\] +an equation which is evidently satisfied if we substitute for +$z + (b/a)$ either of the complex numbers $±i\sqrtp{ac - b^{2}}/a$.\footnote + {We shall sometimes write $x + iy$ instead of $x + yi$ for convenience in printing.} +We +express this by saying that the equation has \emph{the two complex roots} +\[ +\{-b ± i\sqrtp{ac - b^{2}}\}/a. +\] + +If we agree as a matter of convention to say that when $b^{2} = ac$ +(in which case the equation is satisfied by \emph{one} value of~$x$ only, +viz.~$-b/a$), the equation has \emph{two equal roots}, we can say that +\emph{a quadratic equation with real coefficients has two roots in all +cases, either two distinct real roots, or two equal real roots, or two +distinct complex roots}. +\PageSep{83} + +The question is naturally suggested whether a quadratic +equation may not, when complex roots are once admitted, have +more than two roots. It is easy to see that this is not possible. +Its impossibility may in fact be proved by precisely the same +chain of reasoning as is used in elementary algebra to prove that +an equation of the $n$th~degree cannot have more than $n$ real +roots. Let us denote the complex number $x + yi$ by the single +letter~$z$, a convention which we may express by writing +$z = x + yi$. Let $f(z)$~denote any polynomial in~$z$, with real or +complex coefficients. Then we prove in succession: + +\Item{(1)} that the remainder, when $f(z)$~is divided by~$z - a$, $a$~being +any real or complex number, is~$f(a)$; + +\Item{(2)} that if $a$~is a root of the equation $f(z) = 0$, then $f(z)$~is +divisible by~$z - a$; + +\Item{(3)} that if $f(z)$~is of the $n$th~degree, and $f(z) = 0$ has the +$n$~roots $a_{1}$, $a_{2}$, \dots,~$a_{n}$, then +\[ +f(z) = A(z - a_{1}) (z - a_{2}) \dots (z - a_{n}), +\] +where $A$~is a constant, real or complex, in fact the coefficient +of~$z^{n}$ in~$f(z)$. From the last result, and the theorem of \SecNo[§]{40}, +it follows that $f(z)$~cannot have more than $n$ roots. + +We conclude that a quadratic equation with real coefficients has +exactly two roots. We shall see later on that a similar theorem is +true for an equation of any degree and with either real or complex +coefficients: \emph{an equation of the $n$th~degree has exactly $n$~roots}. +The only point in the proof which presents any difficulty is the +first, viz.\ the proof that any equation must have \emph{at least one} +root. This we must postpone for the present.\footnote + {See \Ref{Appendix}{I}.} +We may, however, +at once call attention to one very interesting result of this theorem. +In the theory of number we start from the positive integers and +from the ideas of addition and multiplication and the converse +operations of subtraction and division. We find that these +operations are not always possible unless we admit new kinds of +numbers. We can only attach a meaning to~$3 - 7$ if we admit +\emph{negative} numbers, or to~$\frac{3}{7}$ if we admit \emph{rational fractions}. When +we extend our list of arithmetical operations so as to include root +extraction and the solution of equations, we find that some of +\PageSep{84} +them, such as that of the extraction of the square root of a number +which (like~$2$) is not a perfect square, are not possible unless we +widen our conception of a number, and admit the \emph{irrational} +numbers of \Ref{Chap.}{I}. + +Others, such as the extraction of the square root of~$-1$, are +not possible unless we go still further, and admit the \emph{complex} +numbers of this chapter. And it would not be unnatural to +suppose that, when we come to consider equations of higher +degree, some might prove to be insoluble even by the aid of +complex numbers, and that thus we might be led to the considerations +of higher and higher types of, so to say, \emph{hyper-complex} +numbers. The fact that the roots of any algebraical equation +whatever are ordinary complex numbers shows that this is not the +case. The application of any of the ordinary algebraical operations +to complex numbers will yield only complex numbers. In technical +language `the field of the complex numbers is closed for algebraical +operations'. + +Before we pass on to other matters, let us add that all +theorems of elementary algebra which are proved merely by +the application of the rules of addition and multiplication are +true \emph{whether the numbers which occur in them are real or complex}, +since the rules referred to apply to complex as well as +real numbers. For example, we know that, if $\alpha$~and~$\beta$ are the +roots of +\[ +az^{2} + 2bz + c = 0, +\] +then +\[ +\alpha + \beta = -(2b/a),\quad +\alpha\beta = (c/a). +\] + +Similarly, if $\alpha$,~$\beta$,~$\gamma$ are the roots of +\[ +az^{3} + 3bz^{2} + 3cz + d = 0, +\] +then +\[ +\alpha + \beta + \gamma = -(3b/a),\quad +\beta\gamma + \gamma\alpha + \alpha\beta = (3c/a),\quad +\alpha\beta\gamma = -(d/a). +\] +All such theorems as these are true whether $a$,~$b$,~\dots\Add{,} $\alpha$,~$\beta$,~\dots\ are +real or complex. + +\Paragraph{44. Argand's diagram.} Let $P$ (\Fig{24}) be the point $(x, y)$, +$r$~the length~$OP$, and $\theta$~the angle~$XOP$, so that +\[ +x = r\cos\theta,\quad +y = r\sin\theta,\quad +r = \sqrtp{x^{2} + y^{2}},\quad +\cos\theta : \sin\theta : 1 :: x : y : r. +\] +\PageSep{85} + +We denote the complex number $x + yi$ by~$z$, as in \SecNo[§]{43}, and +we call~$z$ the \emph{complex variable}. +%[Illustration: Fig. 24.] +\Figure[2.5in]{24}{p085} +We call~$P$ \emph{the point}~$z$, or +the point corresponding to~$z$; +$z$~the \emph{argument} of~$P$, $x$~the +\emph{real part}, $y$~the \emph{imaginary +part}, $r$~the \emph{modulus}, and +$\theta$~the \emph{amplitude} of~$z$; and we +write +\[ +%[** TN: Set on two lines in the original] +x = \Real(z),\quad y = \Imag(z),\quad +r = |z|,\quad \theta = \am z. +\] + +When $y = 0$ we say that \emph{$z$~is real}, when $x = 0$ that \emph{$z$~is purely +imaginary}. Two numbers $x + yi$, $x - yi$ which differ only in +the signs of their imaginary parts, we call \emph{conjugate}. It will be +observed that the sum~$2x$ of two conjugate numbers and their +product $x^{2} + y^{2}$ are both real, that they have the same modulus +$\sqrtp{x^{2} + y^{2}}$ and that their product is equal to the square of the +modulus of either. The roots of a quadratic with real coefficients, +for example, are conjugate, when not real. + +It must be observed that $\theta$ or $\am z$ is a many-valued function of +$x$~and~$y$, having an infinity of values, which are angles differing by +multiples of~$2\pi$.\footnote + {It is evident that $|z|$~is identical with the polar coordinate~$r$ of~$P$, and that + the other polar coordinate~$\theta$ is one value of~$\am z$. This value is not necessarily + the \emph{principal} value, as defined below, for the polar coordinate of \SecNo[§]{22} lies between + $0$~and~$2\pi$, and the principal value between $-\pi$~and~$\pi$.} +A line originally lying along~$OX$ will, if turned +through any of these angles, come to lie along~$OP$. We shall +describe that one of these angles which lies between $-\pi$~and~$\pi$ +as the \emph{principal value} of the amplitude of~$z$. This definition +is unambiguous except when one of the values is~$\pi$, +in which case $-\pi$~is also a value. In this case we must make +some special provision as to which value is to be regarded as +the principal value. In general, when we speak of the amplitude +of~$z$ we shall, unless the contrary is stated, mean the principal +value of the amplitude. + +\Fig{24} is usually known as Argand's diagram. +\PageSep{86} + +\Paragraph{45. De~Moivre's Theorem.} The following statements +follow immediately from the definitions of addition and multiplication. + +\Item{(1)} The real (or imaginary) part of the sum of two complex +numbers is equal to the sum of their real (or imaginary) parts. + +\Item{(2)} The modulus of the product of two complex numbers is +equal to the product of their moduli. + +\Item{(3)} The amplitude of the product of two complex numbers is +either equal to the sum of their amplitudes, or differs from it by~$2\pi$. + +\begin{Remark} +It should be observed that it is not always true that the principal value of~$\am(zz')$ +is the sum of the principal values of $\am z$ and~$\am z'$. For example, if +$z = z' = -1 + i$, then the principal values of the amplitudes of $z$~and~$z'$ are each~$\frac{3}{4}\pi$. +But $zz' = -2i$, and the principal value of~$\am(zz')$ is~$-\frac{1}{2}\pi$ and not~$\frac{3}{2}\pi$. +\end{Remark} + +The two last theorems may be expressed in the equation +\[ +%[** TN: Set on two lines in the original] +r(\cos\theta + i\sin\theta) × \rho(\cos\phi + i\sin\phi) + = r\rho\{\cos(\theta + \phi) + i\sin(\theta + \phi)\}, +\] +which may be proved at once by multiplying out and using the +ordinary trigonometrical formulae for $\cos(\theta + \phi)$ and $\sin(\theta + \phi)$. +More generally +\begin{gather*} +r_{1}(\cos\theta_{1} + i\sin\theta_{1}) + × r_{2}(\cos\theta_{2} + i\sin\theta_{2}) × \dots + × r_{n}(\cos\theta_{n} + i\sin\theta_{n})\\ + = r_{1}r_{2} \dots r_{n} \{\cos(\theta_{1} + \theta_{2} + \dots + \theta_{n}) + + i \sin(\theta_{1} + \theta_{2} + \dots + \theta_{n})\}. +\end{gather*} + +A particularly interesting case is that in which +\[ +r_{1} = r_{2} = \dots = r_{n} = 1, \quad +\theta_{1} = \theta_{2} = \dots = \theta_{n} = \theta\Add{.} +\] + +We then obtain the equation +\[ +(\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta, +\] +where $n$~is any positive integer: a result known as \emph{De~Moivre's +Theorem}.\footnote + {It will sometimes be convenient, for the sake of brevity, to denote $\cos\theta + i\sin\theta$ + by~$\Cis\theta$: in this notation, suggested by Profs.\ Harkness and Morley, De~Moivre's + theorem is expressed by the equation $(\Cis\theta)^{n} = \Cis n\theta$.} + +Again, if +\[ +z = r(\cos\theta + i\sin\theta) +\] +then +\[ +1/z = (\cos\theta - i\sin\theta)/r. +\] +Thus the modulus of the reciprocal of~$z$ is the reciprocal of the +modulus of~$z$, and the amplitude of the reciprocal is the negative of +the amplitude of~$z$. We can now state the theorems for quotients +which correspond to \Eq{(2)}~and~\Eq{(3)}. +\PageSep{87} + +\Item{(4)} The modulus of the quotient of two complex numbers is +equal to the quotient of their moduli. + +\Item{(5)} The amplitude of the quotient of two complex numbers +either is equal to the difference of their amplitudes, or differs from +it by~$2\pi$. + +Again +\begin{align*} +(\cos\theta + i\sin\theta)^{-n} + &= (\cos\theta - i\sin\theta)^{n}\\ + &= \{\cos(-\theta) + i\sin(-\theta)\}^{n}\\ + &= \cos(-n\theta) + i\sin(-n\theta). +\end{align*} +Hence \emph{De Moivre's Theorem holds for all integral values of~$n$, +positive or negative}. + +To the theorems (1)--(5) we may add the following theorem, +which is also of very great importance. + +\Item{(6)} The modulus of the sum of any number of complex +numbers is not greater than the sum of their moduli. +%[Illustration: Fig. 25.] +\Figure{25}{p087} + +Let $\Seg{OP}$, $\Seg{OP'}$,~\dots\ be the displacements corresponding to the +various complex numbers. Draw $PQ$ equal and parallel to~$OP'$, +$QR$~equal and parallel to~$OP''$, and so on. Finally we reach a +point~$U$, such that +\[ +\Seg{OU} = \Seg{OP} + \Seg{OP'} + \Seg{OP''} + \dots. +\] +The length~$OU$ is the modulus of the sum of the complex +numbers, whereas the sum of their moduli is the total length +of the broken line $OPQR\dots U$, which is not less than~$OU$. + +A purely arithmetical proof of this theorem is outlined in +\Exs{xxi}.~1. +\PageSep{88} + +\Paragraph{46.} We add some theorems concerning rational functions of +complex numbers. A \emph{rational function} of the complex variable~$z$ +is defined exactly as is a rational function of a real variable~$x$, +viz.\ as the quotient of two polynomials in~$z$. + +\begin{Theorem}[1.] +Any rational function~$R(z)$ can be reduced to +the form $X + Yi$, where $X$~and~$Y$ are rational functions of $x$~and~$y$ +with real coefficients. +\end{Theorem} + +In the first place it is evident that any polynomial $P(x + yi)$ +can be reduced, in virtue of the definitions of addition and multiplication, +to the form $A + Bi$, where $A$~and~$B$ are polynomials +in $x$~and~$y$ with real coefficients. Similarly $Q(x + yi)$ can be +reduced to the form $C + Di$. Hence +\[ +R(x + yi) = P(x + yi)/Q(x + yi) +\] +can be expressed in the form +\begin{align*} +(A + Bi)/(C + Di) + &= (A + Bi) (C - Di)/(C + Di) (C - Di)\\ + &= \frac{AC + BD}{C^{2} + D^{2}} + \frac{BC - AD}{C^{2} + D^{2}} i, +\end{align*} +which proves the theorem. + +\begin{Theorem}[2.] +If $R(x + yi) = X + Yi$, $R$~denoting a rational +function as before, but with \Emph{real} coefficients, then $R(x - yi) = X - Yi$. +\end{Theorem} + +In the first place this is easily verified for a power $(x + yi)^{n}$ +by actual expansion. It follows by addition that the theorem is +true for any polynomial with real coefficients. Hence, in the +notation used above, +\[ +R(x - yi) + = \frac{A - Bi}{C - Di} + = \frac{AC + BD}{C^{2} + D^{2}} - \frac{BC - AD}{C^{2} + D^{2}}i, +\] +the reduction being the same as before except that the sign of~$i$ +is changed throughout. It is evident that results similar to those +of Theorems 1~and~2 hold for functions of any number of complex +variables. + +\begin{Theorem}[3.] +The roots of an equation +\[ +a_{0}z^{n} + a_{1}z^{n-1} + \dots + a_{n} = 0, +\] +whose coefficients are real, may, in so far as they are not themselves +real, be arranged in conjugate pairs. +\end{Theorem} +\PageSep{89} + +For it follows from Theorem~2 that if $x + yi$~is a root then so is~$x - yi$. +A particular case of this theorem is the result (\SecNo[§]{43}) that +the roots of a quadratic equation with real coefficients are either +real or conjugate. + +This theorem is sometimes stated as follows: \emph{in an equation +with real coefficients complex roots occur in conjugate pairs}. It +should be compared with the result of \Exs{viii}.~7, which may be +stated as follows: \emph{in an equation with rational coefficients irrational +roots occur in conjugate pairs}.\footnote + {The numbers $a + \sqrt{b}$, $a - \sqrt{b}$, where $a$,~$b$ are rational, are sometimes said to be + `conjugate'.} + +\begin{Examples}{XXI.} +\Item{1.} Prove theorem~(6) of \SecNo[§]{45} directly from the +definitions and without the aid of geometrical considerations. + +[First, to prove that $|z + z'| \leq |z| + |z'|$ is to prove that +\[ +(x + x')^{2} + (y + y')^{2} + \leq \{\sqrtp{x^{2} + y^{2}} + \sqrtp{x'^{2} + y'^{2}}\}^{2}. +\] +The theorem is then easily extended to the general case.] + +\Item{2.} The one and only case in which +\[ +|z| + |z'| + \dots = |z + z' + \dots|, +\] +is that in which the numbers $z$, $z'$,~\dots\ have all the same amplitude. Prove +this both geometrically and analytically. + +\Item{3.} The modulus of the sum of any number of complex numbers is not +less than the sum of their real (or imaginary) parts. + +\Item{4.} If the sum and product of two complex numbers are both real, then +the two numbers must either be real or conjugate. + +\Item{5.} If +\[ +a + b\sqrt{2} + (c + d \sqrt{2})i = A + B\sqrt{2} + (C + D\sqrt{2})i, +\] +where $a$,~$b$,~$c$,~$d$, $A$,~$B$,~$C$,~$D$ are real rational numbers, then +\[ +a = A,\quad +b = B,\quad +c = C,\quad +d = D. +\] + +\Item{6.} Express the following numbers in the form $A + Bi$, where $A$~and~$B$ are +real numbers: +\[ +(1 + i)^{2},\quad +\left(\frac{1 + i}{1 - i}\right)^{2},\quad +\left(\frac{1 - i}{1 + i}\right)^{2},\quad +\frac{\lambda + \mu i}{\lambda - \mu i},\quad + \left(\frac{\lambda + \mu i}{\lambda - \mu i}\right)^{2} +- \left(\frac{\lambda - \mu i}{\lambda + \mu i}\right)^{2}, +\] +where $\lambda$~and~$\mu$ are real numbers. + +\Item{7.} Express the following functions of $z = x + yi$ in the form $X + Yi$, where +$X$~and~$Y$ are real functions of $x$~and~$y$: $z^{2}$,~$z^{3}$,~$z^{n}$, $1/z$, $z + (1/z)$, $(\alpha + \beta z)/(\gamma + \delta z)$, +where $\alpha$,~$\beta$,~$\gamma$,~$\delta$ are real numbers. + +\Item{8.} Find the moduli of the numbers and functions in the two preceding +examples. +\PageSep{90} + +\Item{9.} The two lines joining the points $z = a$, $z = b$ and $z = c$, $z = d$ will be +perpendicular if +\[ +\am\left(\frac{a - b}{c - d}\right) = ±\tfrac{1}{2} \pi, +\] +\ie\ if $(a - b)/(c - d)$ is purely imaginary. What is the condition that the lines +should be parallel? + +\Item{10.} The three angular points of a triangle are given by $z = \alpha$, $z = \beta$, $z = \gamma$, +where $\alpha$,~$\beta$,~$\gamma$ are complex numbers. Establish the following propositions: + +\SubItem{(i)} \emph{the centre of gravity is given by $z = \frac{1}{3}(\alpha + \beta + \gamma)$}; + +%[** TN: Sole instance of circum-centre, keeping hyphenation] +\SubItem{(ii)} \emph{the circum-centre is given by $|z - \alpha| = |z - \beta| = |z - \gamma|$}; + +\SubItem{(iii)} \emph{the three perpendiculars from the angular points on the opposite +sides meet in a point given by} +\[ +\Re\left(\frac{z - \alpha}{\beta - \gamma}\right) + = \Re\left(\frac{z - \beta}{\gamma - \alpha}\right) + = \Re\left(\frac{z - \gamma}{\alpha - \beta}\right) + = 0; +\] + +\SubItem{(iv)} \emph{there is a point~$P$ inside the triangle such that +\[ +CBP = ACP = BAP = \omega, +\] +and} +\[ +\cot\omega = \cot A + \cot B + \cot C. +\] + +[To prove~(iii) we observe that if $A$,~$B$,~$C$ are the vertices, and $P$~any +point~$z$, then the condition that $AP$~should be perpendicular to~$BC$ is (Ex.~9) +that $(z - \alpha)/(\beta - \gamma)$ should be purely imaginary, or that +\[ +\Re(z - \alpha) \Re(\beta - \gamma) + \Im(z - \alpha) \Im(\beta - \gamma) = 0. +\] +This equation, and the two similar equations obtained by permuting $\alpha$,~$\beta$,~$\gamma$ +cyclically, are satisfied by the same value of~$z$, as appears from the fact that +the sum of the three left-hand sides is zero. + +To prove~(iv), take $BC$~parallel to the positive direction of the axis of~$x$. +Then\footnote + {We suppose that as we go round the triangle in the direction~$ABC$ we leave + it on our left.} +\[ +\gamma - \beta = a,\quad +\alpha - \gamma = - b\Cis(-C),\quad +\beta - \alpha = - c\Cis B. +\] + +We have to determine $z$~and~$\omega$ from the equations +\[ +\frac{(z - \alpha)(\beta_{0} - \alpha_{0})} + {(z_{0} - \alpha_{0})(\beta - \alpha)} + = \frac{(z - \beta)(\gamma_{0} - \beta_{0})} + {(z_{0} - \beta_{0})(\gamma - \beta)} + = \frac{(z - \gamma)(\alpha_{0} - \gamma_{0})} + {(z_{0} - \gamma_{0})(\alpha - \gamma)} + = \Cis 2\omega, +\] +where $z_{0}$, $\alpha_{0}$,~$\beta_{0}$,~$\gamma_{0}$ denote the conjugates of $z$, $\alpha$,~$\beta$,~$\gamma$. + +Adding the numerators and denominators of the three equal fractions, +and using the equation +\[ +i\cot\omega = (1 + \Cis 2\omega)/(1 - \Cis 2\omega), +\] +we find that +\[ +i\cot\omega + = \frac{(\beta - \gamma)(\beta_{0} - \gamma_{0}) + + (\gamma - \alpha)(\gamma_{0} - \alpha_{0}) + + (\alpha - \beta)(\alpha_{0} - \beta_{0})} + {\beta\gamma_{0} - \beta_{0}\gamma + + \gamma\alpha_{0} - \gamma_{0}\alpha + + \alpha\beta_{0} - \alpha_{0}\beta}. +\] +From this it is easily deduced that the value of~$\cot\omega$ is $(a^{2} + b^{2} +c^{2})/4\Delta$, +where $\Delta$~is the area of the triangle; and this is equivalent to the result given. +\PageSep{91} + +To determine~$z$, we multiply the numerators and denominators of the +equal fractions by +$(\gamma_{0} - \beta_{0})/(\beta - \alpha)$, +$(\alpha_{0} - \gamma_{0})/(\gamma - \beta)$, +$(\beta_{0} - \alpha_{0})/(\alpha - \gamma)$, and add +to form a new fraction. It will be found that +\[ +z = \frac{a\alpha \Cis A + b\beta \Cis B + c\gamma \Cis C} + {a\Cis A + b\Cis B + c\Cis C}.] +\] + +\Item{11.} The two triangles whose vertices are the points $a$,~$b$,~$c$ and $x$,~$y$,~$z$ +respectively will be similar if +\[ +\begin{vmatrix} +1 & 1 & 1\\ +a & b & c \\ +x & y & z +\end{vmatrix} += 0 +\] + +{\Loosen[The condition required is that $\Seg{AB}/\Seg{AC} = \Seg{XY}/\Seg{XZ}$ (large letters denoting +the points whose arguments are the corresponding small letters), or +$(b - a)/(c - a) = (y - x)/(z - x)$, which is the same as the condition given.]} + +\Item{12.} Deduce from the last example that if the points $x$,~$y$,~$z$ are collinear +then we can find \emph{real} numbers $\alpha$,~$\beta$,~$\gamma$ such that $\alpha + \beta + \gamma = 0$ and $\alpha x + \beta y + \gamma z = 0$, +and conversely (cf.\ \Exs{xx}.~4). [Use the fact that in this case the triangle +formed by $x$,~$y$,~$z$ is similar to a certain line-triangle on the axis~$OX$, and +apply the result of the last example.] + +\Item{13.} \Topic{The general linear equation with complex coefficients.} The +equation $\alpha z + \beta = 0$ has the one solution $z = -(\beta/\alpha)$, unless $\alpha = 0$. If we put +\[ +\alpha = a + Ai,\quad +\beta = b + Bi,\quad +z = x + yi, +\] +and equate real and imaginary parts, we obtain two equations to determine +the two real numbers $x$~and~$y$. The equation will have a real root if $y = 0$, +which gives $ax + b = 0$, $Ax + B = 0$, and the condition that these equations +should be consistent is~$aB - bA = 0$. + +\Item{14.} \Topic{The general quadratic equation with complex coefficients.} This +equation is +\[ +(a + Ai)z^{2} + 2(b + Bi)z + (c + Ci) = 0. +\] + +Unless $a$~and~$A$ are both zero we can divide through by~$a + iA$. Hence +we may consider +\[ +z^{2} + 2(b + Bi)z + (c + Ci) = 0 +\Tag{(1)} +\] +as the standard form of our equation. Putting $z = x + yi$ and equating real +and imaginary parts, we obtain a pair of simultaneous equations for $x$~and~$y$, +viz. +\[ +x^{2} - y^{2} + 2(bx - By) + c = 0,\quad +2xy + 2(by + Bx) + C = 0. +\] + +If we put +\[ +x + b = \xi,\quad +y + B = \eta,\quad +b^{2} - B^{2} - c = h,\quad +2bB - C = k, +\] +these equations become +\[ +\xi^{2} - \eta^{2} = h,\quad +2\xi\eta = k. +\] +\PageSep{92} + +Squaring and adding we obtain +\[ +\xi^{2} + \eta^{2} = \sqrtp{h^{2} + k^{2}},\quad +\xi = ±\sqrtbr{\tfrac{1}{2}\{\sqrtp{h^{2} + k^{2}} + h\}},\quad +\eta = ±\sqrtbr{\tfrac{1}{2}\{\sqrtp{h^{2} + k^{2}} - h\}}. +\] +We must choose the signs so that $\xi\eta$~has the sign of~$k$: \ie\ if $k$~is positive +we must take like signs, if $k$~is negative unlike signs. + +\Par{Conditions for equal roots.} The two roots can only be equal if both the +square roots above vanish, \ie\ if $h = 0$, $k = 0$, or if $c = b^{2} - B^{2}$, $C = 2bB$. These +conditions are equivalent to the single condition $c + Ci = (b + Bi)^{2}$, which +obviously expresses the fact that the left-hand side of~\Eq{(1)} is a perfect square. + +\Par{Condition for a real root.} If $x^{2} + 2(b + Bi) x + (c + Ci) = 0$, where $x$~is +real, then $x^{2} + 2bx + c = 0$, $2Bx + C = 0$. Eliminating~$x$ we find that the +required condition is +\[ +C^{2} - 4bBC + 4cB^{2} = 0. +\] + +\Par{Condition for a purely imaginary root.} This is easily found to be +\[ +C^{2} - 4bBC - 4b^{2}c = 0. +\] + +\Par{Conditions for a pair of conjugate complex roots.} Since the sum and the +product of two conjugate complex numbers are both real, $b + Bi$ and $c + Ci$ +must both be real, \ie\ $B = 0$, $C = 0$. Thus the equation~\Eq{(1)} can have a pair of +conjugate complex roots only if its coefficients are real. The reader should +verify this conclusion by means of the explicit expressions of the roots. +Moreover, if $b^{2}\geq c$, the roots will be real even in this case. Hence for a pair +of conjugate roots we must have $B = 0$, $C = 0$, $b^{2} < c$. + +\Item{15.} \Topic{The Cubic equation.} Consider the cubic equation +\[ +z^{3} + 3Hz + G = 0, +\] +where $G$~and~$H$ are complex numbers, it being given that the equation has +(\ia)~a real root, (\ib)~a purely imaginary root, (\ic)~a pair of conjugate roots. If +$H = \lambda + \mu i$, $G = \rho + \sigma i$, we arrive at the following conclusions. + +\Par{\Item{(\ia)} Conditions for a real root.} If $\mu$~is not zero, then the real root is~$-\sigma/3\mu$, +and $\sigma^{3} + 27\lambda\mu^{2}\sigma - 27\mu^{3}\rho = 0$. On the other hand, if $\mu = 0$ then we must also +have $\sigma = 0$, so that the coefficients of the equation are real. In this case there +may be three real roots. + +\Par{\Item{(\ib)} Conditions for a purely imaginary root.} If $\mu$~is not zero then the purely +imaginary root is~$(\rho/3\mu)i$, and $\rho^{3} - 27\lambda\mu^{2}\rho - 27\mu^{3}\sigma = 0$. If $\mu = 0$ then also $\rho = 0$, +and the root is~$yi$, where $y$~is given by the equation $y^{3} - 3\lambda y - \sigma = 0$, which has +real coefficients. In this case there may be three purely imaginary roots. + +\Par{\Item{(\ic)} Conditions for a pair of conjugate complex roots.} Let these be $x + yi$ +and $x - yi$. Then since the sum of the three roots is zero the third root +must be~$-2x$. From the relations between the coefficients and the roots of +an equation we deduce +\[ +y^{2} - 3x^{2} = 3H,\quad +2x(x^{2} + y^{2}) = G. +\] +Hence $G$~and~$H$ must both be real. + +In each case we can either find a root (in which case the equation can +be reduced to a quadratic by dividing by a known factor) or we can reduce +the solution of the equation to the solution of a cubic equation with real +coefficients. +\PageSep{93} + +\Item{16.} The cubic equation $x^{3} + a_{1}x^{2} + a_{2}x + a_{3} = 0$, where $a_{1} = A_{1} + A_{1}'i$,~\dots, has +a pair of conjugate complex roots. Prove that the remaining root is +$-A_{1}'a_{3}/A_{3}'$, unless $A_{3}' = 0$. Examine the case in which $A_{3}' = 0$. + +\Item{17.} Prove that if $z^{3} + 3Hz + G = 0$ has two complex roots then the equation +\[ +8\alpha^{3} + 6\alpha H - G = 0 +\] +has one real root which is the real part~$\alpha$ of the complex roots of the +original equation; and show that $\alpha$~has the same sign as~$G$. + +\Item{18.} An equation of any order with complex coefficients will in general +have no real roots nor pairs of conjugate complex roots. How many conditions +must be satisfied by the coefficients in order that the equation should +have (\ia)~a real root, (\ib)~a pair of conjugate roots? + +\Item{19.} \Topic{Coaxal circles.} In \Fig{26}, let $a$,~$b$,~$z$ be the arguments of $A$,~$B$,~$P$. +Then +\[ +\am\frac{z - b}{z - a} = APB, +\] +if the principal value of the amplitude is chosen. If the two circles shown +in the figure are equal, and $z'$,~$z_{1}$,~$z_{1}'$ are the arguments of $P'$,~$P_{1}$,~$P_{1}'$, +and $APB = \theta$, it is easy to see that +\[ +\am\frac{z' - b}{z' - a} = \pi - \theta,\quad +\am\frac{z_{1} - b}{z_{1} - a} = -\theta, +\] +and +\[ +\am\frac{z_{1}' - b}{z_{1}' - a} = -\pi + \theta. +\] + +The locus defined by the equation +\[ +\am\frac{z - b}{z - a} = \theta, +\] +where $\theta$~is constant, is the arc~$APB$. By +writing $\pi - \theta$,~$-\theta$,~$-\pi + \theta$ for~$\theta$, we obtain +the other three arcs shown. +%[Illustration: Fig. 26.] +\Figure[2.25in]{26}{p093} + +The system of equations obtained by +supposing that $\theta$~is a parameter, varying +from~$-\pi$ to~$+\pi$, represents \emph{the system of +circles which can be drawn through the +points $A$,~$B$}. It should however be observed +that each circle has to be divided +into two parts to which correspond different +values of~$\theta$. + +\Item{20.} Now let us consider the equation +\[ +\left|\frac{z - b}{z - a}\right| = \lambda, +\Tag{(1)} +\] +where $\lambda$~is a constant. + +Let $K$ be the point in which the tangent to the circle~$ABP$ at~$P$ meets~$AB$. +Then the triangles $KPA$,~$KBP$ are similar, and so +\[ +AP/PB = PK/BK = KA/KP = \lambda. +\] +\PageSep{94} +Hence $KA/KB = \lambda^{2}$, and therefore $K$~is a fixed point for all positions of~$P$ +which satisfy the equation~\Eq{(1)}. Also $KP^{2} = KA · KB$, and so is constant. +Hence \emph{the locus of~$P$ is a circle whose centre is~$K$}. + +The system of equations obtained by varying~$\lambda$ represents a system of +circles, and every circle of this system cuts at right angles every circle of the +system of Ex.~19. + +The system of Ex.~19 is called \emph{a system of coaxal circles of the common +point kind}. The system of Ex.~20 is called \emph{a system of coaxal circles of the +limiting point kind}, $A$~and~$B$ being the \emph{limiting points} of the system. If $\lambda$~is +very large or very small then the circle is a very small circle containing $A$~or~$B$ +in its interior. + +\Item{21.} \Topic{Bilinear Transformations.} Consider the equation +\[ +z = Z + a, +\Tag{(1)} +\] +where $z = x + yi$ and $Z = X + Yi$ are two complex variables which we may +suppose to be represented in two planes $xoy$,~$XOY$. To every value of~$z$ +corresponds one of~$Z$, and conversely. If $a = \alpha + \beta i$ then +\[ +x = X + \alpha,\quad +y = Y + \beta, +\] +and to the point $(x, y)$ corresponds the point $(X, Y)$. If $(x, y)$ describes a +curve of any kind in its plane, $(X, Y)$ describes a curve in its plane. Thus +to any figure in one plane corresponds a figure in the other. A passage of +this kind from a figure in the plane~$xoy$ to a figure in the plane~$XOY$ by +means of a relation such as~\Eq{(1)} between $z$~and~$Z$ is called a \emph{transformation}. +In this particular case the relation between corresponding figures is very +easily defined. The $(X, Y)$ figure is the same in size, shape, and orientation +as the $(x, y)$ figure, but is shifted a distance~$\alpha$ to the left, and a distance~$\beta$ +downwards. Such a transformation is called a \emph{translation}. + +Now consider the equation +\[ +z = \rho Z, +\Tag{(2)} +\] +where $\rho$~is real. This gives $x = \rho X$, $y = \rho Y$. The two figures are similar and +similarly situated about their respective origins, but the scale of the $(x, y)$ +figure is $\rho$~times that of the $(X, Y)$ figure. Such a transformation is called +a \emph{magnification}. + +Finally consider the equation +\[ +z = (\cos\phi + i \sin\phi)Z. +\Tag{(3)} +\] +It is clear that $|z| = |Z|$ and that one value of $\am z$ is $\am Z + \phi$, and that the +two figures differ only in that the $(x, y)$ figure is the $(X, Y)$ figure turned +about the origin through an angle~$\phi$ in the positive direction. Such a transformation +is called a \emph{rotation}. + +The general linear transformation +\[ +z = aZ + b +\Tag{(4)} +\] +\PageSep{95} +is a combination of the three transformations \Eq{(1)},~\Eq{(2)},~\Eq{(3)}. For, if $|a| = \rho$ and +$\am a = \phi$, we can replace~\Eq{(4)} by the three equations +\[ +z = z' + b,\quad +z' = \rho Z',\quad +Z' = (\cos\phi + i\sin\phi)Z. +\] +Thus \emph{the general linear transformation is equivalent to the combination of a +translation, a magnification, and a rotation}. + +Next let us consider the transformation +\[ +z = 1/Z. +\Tag{(5)} +\] +If $|Z| = R$ and $\am Z = \Theta$, then $|z| = 1/R$ and $\am z = -\Theta$, and to pass from +the $(x, y)$ figure to the $(X, Y)$ figure we invert the former with respect to~$o$, +with unit radius of inversion, and then construct the image of the new figure +in the axis~$ox$ (\ie\ the symmetrical figure on the other side of~$ox$). + +Finally consider the transformation +\[ +z = \frac{aZ + b}{cZ + d}. +\Tag{(6)} +\] +This is equivalent to the combination of the transformations +\[ +z = (a/c) + (bc - ad)(z'/c),\quad +z' = 1/Z',\quad +Z' = cZ + d, +\] +\ie\ to a certain combination of transformations of the types already considered. + +The transformation~\Eq{(6)} is called the \emph{general bilinear transformation}. +Solving for~$Z$ we obtain +\[ +Z = \frac{dz - b}{cz - a}. +\] + +The general bilinear transformation is the most general type of transformation +for which one and only one value of~$z$ corresponds to each value of~$Z$, +and conversely. + +\Par{\Item{22.} The general bilinear transformation transforms circles into circles.} +This may be proved in a variety of ways. We may assume the well-known +theorem in pure geometry, that inversion transforms circles into circles +(which may of course in particular cases be straight lines). Or we may +use the results of Exs.\ 19~and~20. If, \eg, the $(x, y)$ circle is +\[ +|(z - \sigma)/(z - \rho)| = \lambda, +\] +and we substitute for~$z$ in terms of~$Z$, we obtain +\[ +|(Z - \sigma')/(Z - \rho')| = \lambda', +\] +where +\[ +\sigma' = -\frac{b - \sigma d}{a - \sigma c},\quad +\rho' = -\frac{b - \rho d}{a - \rho c},\quad +\lambda' = \left|\frac{a - \rho c}{a - \sigma c}\right|\lambda. +\] + +\Item{23.} Consider the transformations $z = 1/Z$, $z = (1 + Z)/(1 - Z)$, and draw +the $(X, Y)$ curves which correspond to (1)~circles whose centre is the origin, +(2)~straight lines through the origin. +\PageSep{96} + +\Item{24.} The condition that the transformation $z = (aZ + b)/(cZ + d)$ should +make the circle $x^{2} + y^{2} = 1$ correspond to a straight line in the $(X, Y)$ plane +is $|a| = |c|$. + +\Item{25.} \Topic{Cross ratios.} The cross ratio $\DPmod{(z_{1}z_{2}, z_{3}z_{4})}{(z_{1}, z_{2}; z_{3}, z_{4})}$ is defined to be +\[ +\frac{(z_{1} - z_{3}) (z_{2} - z_{4})}{(z_{1} - z_{4}) (z_{2} - z_{3})}. +\] + +If the four points $z_{1}$,~$z_{2}$,~$z_{3}$,~$z_{4}$ are on the same line, this definition agrees +with that adopted in elementary geometry. There are $24$~cross ratios which +can be formed from $z_{1}$,~$z_{2}$,~$z_{3}$,~$z_{4}$ by permuting the suffixes. These consist of +six groups of four equal cross ratios. If one ratio is~$\lambda$, then the six distinct +cross ratios are $\lambda$, $1 - \lambda$, $1/\lambda$, $1/(1 - \lambda)$, $(\lambda - 1)/\lambda$, $\lambda/(\lambda - 1)$. The four points are +said to be \emph{harmonic} or \emph{harmonically related} if any one of these is equal to~$-1$. +In this case the six ratios are $-1$, $2$, $-1$, $\frac{1}{2}$, $2$,~$\frac{1}{2}$. + +\emph{If any cross ratio is real then all are real and the four points lie on a +circle}. For in this case +\[ +\am\frac{(z_{1} - z_{3}) (z_{2} - z_{4})}{(z_{1} - z_{4}) (z_{2} - z_{3})} +\] +must have one of the three values $-\pi$,~$0$,~$\pi$, so that $\am\{(z_{1} - z_{3})/(z_{1} - z_{4})\}$ and +$\am\{(z_{2} - z_{3})/(z_{2} - z_{4})\}$ must either be equal or differ by~$\pi$ (cf.~Ex.~19). + +If $\DPmod{(z_{1}z_{2}, z_{3}z_{4})}{(z_{1}, z_{2}; z_{3}, z_{4})} = - 1$, we have the two equations +\[ +\am\frac{z_{1} - z_{3}}{z_{1} - z_{4}} + = ±\pi + \am\frac{z_{2} - z_{3}}{z_{2} - z_{4}},\quad +\left|\frac{z_{1} - z_{3}}{z_{1} - z_{4}}\right| + = \left|\frac{z_{2} - z_{3}}{z_{2} - z_{4}}\right|. +\] +The four points $A_{1}$, $A_{2}$, $A_{3}$,~$A_{4}$ lie on a circle, $A_{1}$~and~$A_{2}$ being separated +by $A_{3}$~and~$A_{4}$. Also $A_{1}A_{3}/A_{1}A_{4} = A_{2}A_{3}/A_{2}A_{4}$. Let $O$~be the middle point of~$A_{3}A_{4}$. +The equation +\[ +\frac{(z_{1} - z_{3}) (z_{2} - z_{4})}{(z_{1} - z_{4}) (z_{2} - z_{3})} = -1 +\] +may be put in the form +\[ +(z_{1} + z_{2}) (z_{3} + z_{4}) = 2(z_{1}z_{2} + z_{3}z_{4}), +\] +or, what is the same thing, +\[ +\{z_{1} - \tfrac{1}{2}(z_{3} + z_{4})\} +\{z_{2} - \tfrac{1}{2}(z_{3} + z_{4})\} + = \{\tfrac{1}{2}(z_{3} - z_{4})\}^{2}. +\] +But this is equivalent to $\Seg{OA_{1}} · \Seg{OA_{2}} = \Seg{OA_{3}}^{2} = \Seg{OA_{4}}^{2}$. Hence $OA_{1}$ and $OA_{2}$ +make equal angles with~$A_{3}A_{4}$, and $OA_{1} · OA_{2} = OA_{3}^{2} = OA_{4}^{2}$. It will be observed +that the relation between the pairs $A_{1}$,~$A_{2}$ and $A_{3}$,~$A_{4}$ is symmetrical. +Hence, if $O'$~is the middle point of~$A_{1}A_{2}$, $O'A_{3}$~and~$O'A_{4}$ are equally inclined +to~$A_{1}A_{2}$, and $O'A_{3} · O'A_{4} = O'A_{1}^{2} = O'A_{2}^{2}$. + +\Item{26.} If the points $A_{1}$,~$A_{2}$ are given by $az^{2} + 2bz + c = 0$, and the points +$A_{3}$,~$A_{4}$ by $a'z^{2} + 2b'z + c' = 0$, and $O$~is the middle point of~$A_{3}A_{4}$, and +$ac' + a'c - 2bb' = 0$, then $OA_{1}$,~$OA_{2}$ are equally inclined to~$A_{3}A_{4}$ and +$OA_{1} · OA_{2} = OA_{3}^{2} = OA_{4}^{2}$. \MathTrip{1901.} +\PageSep{97} + +\Item{27.} $AB$,~$CD$ are two intersecting lines in Argand's diagram, and $P$,~$Q$ +their middle points. Prove that, if $AB$~bisects the angle~$CPD$ and +$PA^{2} = PB^{2} = PC · PD$, then $CD$~bisects the angle~$AQB$ and $QC^{2} = QD^{2} = QA · QB$. +\MathTrip{1909.} + +\Item{28.} \Topic{The condition that four points should lie on a circle.} A +sufficient condition is that one (and therefore all) of the cross ratios +should be real (Ex.~25); this condition is also necessary. Another form +of the condition is that it should be possible to choose real numbers +$\alpha$,~$\beta$,~$\gamma$ such that +\[ +\begin{vmatrix} +1 & 1 & 1\\ +\alpha & \beta & \gamma\\ +z_{1}z_{4} + z_{2}z_{3} & z_{2}z_{4} + z_{3}z_{1} & z_{3}z_{4} + z_{1}z_{2} +\end{vmatrix} += 0. +\] + +{\Loosen[To prove this we observe that the transformation $Z = 1/(z - z_{4})$ is equivalent +to an inversion with respect to the point~$z_{4}$, coupled with a certain reflexion +(Ex.~21). If $z_{1}$,~$z_{2}$,~$z_{3}$ lie on a circle through~$z_{4}$, the corresponding points +$Z_{1} = 1/(z_{1} - z_{4})$, $Z_{2} = 1/(z_{2} - z_{4})$, $Z_{3} = 1/(z_{3} - z_{4})$ lie on a straight line. Hence +(Ex.~12) we can find real numbers $\alpha'$,~$\beta'$,~$\gamma'$ such that $\alpha' + \beta' + \gamma' = 0$ and +$\alpha'/(z_{1} - z_{4}) + \beta'/(z_{2} - z_{4}) + \gamma'/(z_{3} - z_{4}) = 0$, and it is easy to prove that this is +equivalent to the given condition.]} + +\Item{29.} Prove the following analogue of De~Moivre's Theorem for real +numbers: if $\phi_{1}$,~$\phi_{2}$, $\phi_{3}$,~\dots\ is a series of positive acute angles such that +\begin{alignat*}{2} +\tan\phi_{m+1} &= \tan\phi_{m} \sec\phi_{1} &&+ \sec\phi_{m} \tan\phi_{1},\\ +\intertext{then} +\tan\phi_{m+n} &= \tan\phi_{m} \sec\phi_{n} &&+ \sec\phi_{m} \tan\phi_{n},\\ +\sec\phi_{m+n} &= \sec\phi_{m} \sec\phi_{n} &&+ \tan\phi_{m} \tan\phi_{n}, +\end{alignat*} +and +\[ +\tan\phi_{m} + \sec\phi_{m} = (\tan\phi_{1} + \sec\phi_{1})^{m}. +\] + +[Use the method of mathematical induction.] + +\Item{30.} \Topic{The transformation $z = Z^{m}$.} In this case $r = R^{m}$, and $\theta$~and~$m\Theta$ +differ by a multiple of~$2\pi$. If $Z$~describes a circle round the origin then $z$~describes +a circle round the origin $m$~times. + +The whole $(x, y)$ plane corresponds to any one of $m$~sectors in the $(X, Y)$ +plane, each of angle~$2\pi/m$. To each point in the $(x, y)$ plane correspond +$m$~points in the $(X, Y)$ plane. + +\Item{31.} \Topic{Complex functions of a real variable.} If $f(t)$,~$\phi(t)$ are two real +functions of a real variable~$t$ defined for a certain range of values of~$t$, +we call +\[ +z = f(t) + i\phi(t) +\Tag{(1)} +\] +a complex function of~$t$. We can represent it graphically by drawing the +curve +\[ +x = f(t),\quad +y = \phi(t); +\] +\PageSep{98} +the equation of the curve may be obtained by eliminating~$t$ between these +equations. If $z$~is a polynomial in~$t$, or rational function of~$t$, with complex +coefficients, we can express it in the form~\Eq{(1)} and so determine the curve +represented by the function. + +\SubItem{(i)} Let +\[ +z = a + (b - a)t, +\] +where $a$~and~$b$ are complex numbers. If $a = \alpha + \alpha' i$, $b = \beta + \beta' i$, then +\[ +x = \alpha + (\beta - \alpha)t,\quad +y = \alpha' + (\beta' - \alpha')t. +\] +The curve is the straight line joining the points $z = a$ and $z = b$. The segment +between the points corresponds to the range of values of~$t$ from~$0$ +to~$1$. Find the values of~$t$ which correspond to the two produced segments +of the line. + +\SubItem{(ii)} If +\[ +z = c + \rho\left(\frac{1 + ti}{1 - ti}\right), +\] +where $\rho$~is positive, then the curve is the circle of centre~$c$ and radius~$\rho$. As +$t$~varies through all real values\DPnote{** [sic], no comma} $z$~describes the circle once. + +\SubItem{(iii)} In general the equation $z = (a + bt)/(c + dt)$ represents a circle. +This can be proved by calculating $x$~and~$y$ and eliminating: but this process +is rather cumbrous. A simpler method is obtained by using the result of +Ex.~22. Let $z = (a + bZ)/(c + dZ)$, $Z = t$. As $t$~varies\DPnote{** [sic], no comma} $Z$~describes a straight +line, viz.\ the axis of~$X$. Hence $z$~describes a circle. + +\SubItem{(iv)} The equation +\[ +z = a + 2bt + ct^{2} +\] +represents a parabola generally, a straight line if $b/c$~is real. + +\SubItem{(v)} The equation $z = (a + 2bt + ct^{2})/(\alpha + 2\beta t + \gamma t^{2})$, where $\alpha$,~$\beta$,~$\gamma$ are real, +represents a conic section. + +[Eliminate~$t$ from +\[ +x = (A + 2Bt + Ct^{2})/(\alpha + 2\beta t + \gamma t^{2}),\quad +y = (A' + 2B't + C't^{2})/(\alpha + 2\beta t + \gamma t^{2}), +\] +where $A + A'i = a$, $B + B'i = b$, $C + C'i = c$.] +\end{Examples} + +\Paragraph{47. Roots of complex numbers.} We have not, up to the +present, attributed any meaning to symbols such as $\sqrt[n]{a}$,~$a^{m/n}$, +when $a$~is a complex number, and $m$~and~$n$ integers. It is, +however, natural to adopt the definitions which are given in +elementary algebra for real values of~$a$. Thus we define~$\sqrt[n]{a}$ or~$a^{1/n}$, +where $n$~is a positive integer, as a number~$z$ which satisfies +the equation $z^{n} = a$; and $a^{m/n}$, where $m$~is an integer, as~$(a^{1/n})^{m}$. +These definitions do not prejudge the question as to whether +there are or are not more than one (or any) roots of the equation. + +\Paragraph{48. Solution of the equation $z^{n} = a$.} Let +\[ +a = \rho(\cos\phi + i\sin\phi), +\] +where $\rho$~is positive and $\phi$~is an angle such that $-\pi < \phi \leq \pi$. If +\PageSep{99} +we put $z = r(\cos\theta + i\sin\theta)$, the equation takes the form +\[ +r^{n}(\cos n\theta + i\sin n\theta) = \rho(\cos\phi + i \sin\phi); +\] +so that +\[ +r^{n} = \rho,\quad +\cos n\theta = \cos\phi,\quad +\sin n\theta = \sin\phi. +\Tag{(1)} +\] + +The only possible value of~$r$ is~$\sqrt[n]{\rho}$, the ordinary arithmetical +$n$th~root of~$\rho$; and in order that the last two equations should be +satisfied it is necessary and sufficient that $n\theta = \phi + 2k\pi$, where $k$~is +an integer, or +\[ +\theta = (\phi + 2k\pi)/n. +\] +If $k = pn + q$, where $p$~and~$q$ are integers, and $0 \leq q < n$, the +value of~$\theta$ is~$2p\pi + (\phi + 2q\pi)/n$, and in this the value of~$p$ is a +matter of indifference. Hence \emph{the equation +\[ +z^{n} = a = \rho(\cos\phi + i\sin\phi) +\] +has $n$~roots and $n$~only, given by $z = r(\cos\theta + i\sin\theta)$, where} +\[ +r = \sqrt[n]{\rho},\quad +\theta = (\phi + 2q\pi)/n,\quad +(q = 0,\ 1,\ 2,\ \dots\Add{,} n - 1). +\] + +That these $n$~roots are in reality all distinct is easily seen +by plotting them on Argand's diagram. The particular root +\[ +\sqrt[n]{\rho}\{\cos(\phi/n) + i\sin(\phi/n)\} +\] +is called the \emph{principal value} of~$\sqrt[n]{a}$. + +The case in which $a = 1$, $\rho = 1$, $\phi = 0$ is of particular interest. +The $n$~roots of the equation $x^{n} = 1$ are +\[ +\cos(2q\pi/n) + i\sin(2q\pi/n),\quad +(q = 0,\ 1,\ \dots\Add{,} n - 1). +\] +These numbers are called the $n$th~roots of unity; the principal +value is unity itself. If we write $\omega_{n}$ for $\cos(2\pi/n) + i\sin(2\pi/n)$, +we see that the $n$th~roots of unity are +\[ +1,\quad +\omega_{n},\quad +\omega_{n}^{2},\ \dots,\quad +\omega_{n}^{n-1}. +\] + +\begin{Examples}{XXII.} +\Item{1.} The two square roots of~$1$ are $1$,~$-1$; the three +cube roots are $1$, $\frac{1}{2}(-1 + i\sqrt{3})$, $\frac{1}{2}(-1 - i\sqrt{3})$; the four fourth roots are $1$, +$i$, $-1$, $-i$; and the five fifth roots are +\begin{alignat*}{4} +1,\quad &\tfrac{1}{4} \Bigl[ &&\sqrt{5} - 1 + i\sqrtb{10 + 2\sqrt{5}}\Bigr],\quad + && \tfrac{1}{4} \Bigl[-&&\sqrt{5} - 1 + i\sqrtb{10 - 2\sqrt{5}}\Bigr],\\ + &\tfrac{1}{4} \Bigl[-&&\sqrt{5} - 1 - i\sqrtb{10 - 2\sqrt{5}}\Bigr],\quad + && \tfrac{1}{4} \Bigl[ &&\sqrt{5} - 1 - i\sqrtb{10 + 2\sqrt{5}}\Bigr]. +\end{alignat*} + +\Item{2.} Prove that +\[ +1 + \omega_{n} + \omega_{n}^{2} + \dots + \omega_{n}^{n-1} = 0. +\] + +\Item{3.} Prove that +\[ +(x + y\omega_{3} + z\omega_{3}^{2}) +(x + y\omega_{3}^{2} + z\omega_{3}) + = x^{2} + y^{2} + z^{2} - yz - zx - xy. +\] + +\Item{4.} The $n$th~roots of~$a$ are the products of the $n$th~roots of unity by the +principal value of~$\sqrt[n]{a}$. +\PageSep{100} + +\Item{5.} It follows from \Exs{xxi}.~14 that the roots of +\[ +z^{2} = \alpha + \beta i +\] +are +\[ +± \sqrtbr{\tfrac{1}{2} \{\sqrtp{\alpha^{2} + \beta^{2}} + \alpha\}} +± i\sqrtbr{\tfrac{1}{2} \{\sqrtp{\alpha^{2} + \beta^{2}} - \alpha\}}, +\] +like or unlike signs being chosen according as $\beta$~is positive or negative. Show +that this result agrees with the result of \SecNo[§]{48}. + +\Item{6.} Show that $(x^{2m} - a^{2m})/(x^{2} - a^{2})$ is equal to +\[ +\Bigl(x^{2} - 2ax\cos\frac{\pi}{m} + a^{2}\Bigr) +\Bigl(x^{2} - 2ax\cos\frac{2\pi}{m} + a^{2}\Bigr) \dots +\Bigl(x^{2} - 2ax\cos\frac{(m - 1)\pi}{m} + a^{2}\Bigr). +\] + +[The factors of $x^{2m} - a^{2m}$ are +\[ +(x - a),\quad +(x - a\omega_{2m}),\quad +(x - a\omega_{2m}^{2}),\ \dots\quad +(x - a\omega_{2m}^{2m-1}). +\] +The factor $x - a\omega_{2m}^{m}$ is $x + a$. The factors $(x - a\omega_{2m}^{s})$, $(x - a\omega_{2m}^{2m-s})$ taken together +give a factor $x^{2} - 2ax \cos(s\pi/m) + a^{2}$.] + +\Item{7.} Resolve $x^{2m+1} - a^{2m+1}$, $x^{2m} + a^{2m}$, and $x^{2m+1} + a^{2m+1}$ into factors in a +similar way. + +\Item{8.} Show that $x^{2n} - 2x^{n}a^{n} \cos\theta + a^{2n}$ is equal to +\begin{multline*} +\left(x^{2} - 2xa\cos\frac{\theta}{n} + a^{2}\right) +\left(x^{2} - 2xa\cos\frac{\theta + 2\pi}{n} + a^{2}\right) \dots \\ +\dots\left(x^{2} - 2xa\cos\frac{\theta + 2(n - 1)\pi}{n} + a^{2}\right). +\end{multline*} + +[Use the formula +\[ +x^{2n} - 2x^{n}a^{n} \cos\theta + a^{2n} + = \{x^{n} - a^{n}(\cos\theta + i\sin\theta)\} + \{x^{n} - a^{n}(\cos\theta - i\sin\theta)\}, +\] +and split up each of the last two expressions into $n$~factors.] + +\Item{9.} Find all the roots of the equation $x^{6} - 2x^{3} + 2 = 0$. \MathTrip{1910.} + +\Item{10.} The problem of finding the accurate value of~$\omega_{n}$ in a numerical form +involving square roots only, as in the formula $\omega_{3} = \frac{1}{2}(-1 + i\sqrt{3})$, is the +algebraical equivalent of the geometrical problem of inscribing a regular +polygon of $n$~sides in a circle of unit radius by Euclidean methods, \ie\ by ruler +and compasses. For this construction will be possible if and only if we can +construct lengths measured by $\cos(2\pi/n)$ and $\sin(2\pi/n)$; and this is possible +(\Ref{Ch.}{II}, \MiscExs{II}~22) if and only if these numbers are expressible in a form +involving square roots only. + +Euclid gives constructions for $n = 3$, $4$, $5$, $6$, $8$, $10$, $12$, and~$15$. It is +evident that the construction is possible for any value of~$n$ which can be +found from these by multiplication by any power of~$2$. There are other +special values of~$n$ for which such constructions are possible, the most interesting +being~$n = 17$. +\end{Examples} +\PageSep{101} + +\Paragraph{49. The general form of De~Moivre's Theorem.} It +follows from the results of the last section that if $q$~is a positive +integer then one of the values of $(\cos\theta + i\sin\theta)^{1/q}$ is +\[ +\cos(\theta/q) + i\sin(\theta/q). +\] +Raising each of these expressions to the power~$p$ (where $p$~is any +integer positive or negative), we obtain the theorem that one of +the values of $(\cos\theta + i\sin\theta)^{p/q}$ is $\cos(p\theta/q) + i\sin(p\theta/q)$, or that \emph{if +$\alpha$~is any rational number then one of the values of $(\cos\theta + i\sin\theta)^{\alpha}$ is} +\[ +\cos\alpha\theta + i\sin\alpha\theta. +\] +This is a generalised form of De~Moivre's Theorem (\SecNo[§]{45}). + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER III.} + +\begin{Examples}{} +\Item{1.} The condition that a triangle~$(xyz)$ should be equilateral is that +\[ +x^{2} + y^{2} + z^{2} - yz - zx - xy = 0. +\] + +{\Loosen[Let $XYZ$~be the triangle. The displacement $\Seg{ZX}$ is $\Seg{YZ}$ turned through +an angle~$\frac{2}{3}\pi$ in the positive or negative direction. Since $\Cis\frac{2}{3}\pi = \omega_{3}$, +$\Cis(-\frac{2}{3}\pi) = 1/\omega_{3} = \omega_{3}^{2}$, we have $x - z = (z - y)\omega_{3}$ or $x - z = (z - y)\omega_{3}^{2}$. Hence +$x + y\omega_{3} + z\omega_{3}^{2} = 0$ or $x + y\omega_{3}^{2} + z\omega_{3} = 0$. The result follows from \Exs{xxii}.~3.]} + +\Item{2.} If $XYZ$, $X'Y'Z'$ are two triangles, and +\[ +\Seg{YZ} · \Seg{Y'Z'} = \Seg{ZX} · \Seg{Z'X'} = \Seg{XY} · \Seg{X'Y'}, +\] +then both triangles are equilateral. [From the equations +\[ +(y - z)(y' - z') = (z - x)(z' - x') = (x - y)(x' - y') = \kappa^{2}, +\] +say, we deduce $\sum 1/(y' - z') = 0$, or $\sum x'^{2} - \sum y'z' = 0$. Now apply the result of the +last example.] + +\Item{3.} Similar triangles $BCX$, $CAY$, $ABZ$ are described on the sides of a +triangle~$ABC$. Show that the centres of gravity of $ABC$,~$XYZ$ are coincident. + +[We have $(x - c)/(b - c) = (y - a)/(c - a) = (z - b)/(a - b) = \lambda$, say. Express +$\frac{1}{3}(x + y + z)$ in terms of $a$,~$b$,~$c$.] + +\Item{4.} If $X$,~$Y$,~$Z$ are points on the sides of the triangle $ABC$, such that +\[ +BX/XC = CY/YA = AZ/ZB = r, +\] +and if $ABC$, $XYZ$ are similar, then either $r = 1$ or both triangles are +equilateral. + +\Item{5.} If $A$,~$B$,~$C$,~$D$ are four points in a plane, then +\[ +AD · BC \leq BD · CA + CD · AB. +\] +\PageSep{102} + +[Let $z_{1}$,~$z_{2}$,~$z_{3}$,~$z_{4}$ be the complex numbers corresponding to $A$,~$B$,~$C$,~$D$. +Then we have identically +\[ +(x_{1} - x_{4})(x_{2} - x_{3}) + +(x_{2} - x_{4})(x_{3} - x_{1}) + +(x_{3} - x_{4})(x_{1} - x_{2}) = 0. +\] +Hence +\begin{align*} +|(x_{1} - x_{4})(x_{2} - x_{3})| + &= |(x_{2} - x_{4})(x_{3} - x_{1}) + (x_{3} - x_{4})(x_{1} - x_{2})|\\ + &\leq |(x_{2} - x_{4})(x_{3} - x_{1})| + |(x_{3} - x_{4})(x_{1} - x_{2})|.] +\end{align*} + +\Item{6.} Deduce Ptolemy's Theorem concerning cyclic quadrilaterals from the +fact that the cross ratios of four concyclic points are real. [Use the same +identity as in the last example.] + +\Item{7.} If $z^{2} + z'^{2} = 1$, then the points $z$,~$z'$ are ends of conjugate diameters of an +ellipse whose foci are the points $1$,~$-1$. [If $CP$,~$CD$ are conjugate semi-diameters +of an ellipse and $S$,~$H$ its foci, then $CD$~is parallel to the external +bisector of the angle~$SPH$, and $SP · HP = CD^{2}$.] + +\Item{8.} Prove that $|a + b|^{2} + |a - b|^{2} = 2\{|a|^{2} + |b|^{2}\}$. [This is the analytical +equivalent of the geometrical theorem that, if $M$~is the middle point of~$PQ$, +then $OP^{2} + OQ^{2} = 2OM^{2} + 2MP^{2}$.] + +\Item{9.} Deduce from Ex.~8 that +\[ +|a + \sqrtp{a^{2} - b^{2}}| + |a - \sqrtp{a^{2} - b^{2}}| = |a + b| + |a - b|. +\] + +[If $a + \sqrtp{a^{2} - b^{2}} = z_{1}$, $a - \sqrtp{a^{2} - b^{2}} = z_{2}$, we have +\[ +|z_{1}|^{2} + |z_{2}|^{2} + = \tfrac{1}{2}|z_{1} + z_{2}|^{2} + + \tfrac{1}{2}|z_{1} - z_{2}|^{2} + = 2|a|^{2} + 2|a^{2} - b^{2}|, +\] +and so +\[ +(|z_{1}| + |z_{2}|)^{2} + = 2\{|a|^{2} + |a^{2} - b^{2}| + |b|^{2}\} + = |a + b|^{2} + |a - b|^{2} + 2|a^{2} - b^{2}|. +\] + +{\Loosen Another way of stating the result is: if $z_{1}$~and~$z_{2}$ are the roots of +$\alpha z^{2} + 2\beta z + \gamma = 0$, then} +\[ +|z_{1}| + |z_{2}| = (1/|\alpha|) + \{(|-\beta + \DPtypo{\sqrt{\alpha\gamma}}{\sqrtp{\alpha\gamma}}|) + + (|-\beta - \DPtypo{\sqrt{\alpha\gamma}}{\sqrtp{\alpha\gamma}}|)\}.] +\] + +\Item{10.} Show that the necessary and sufficient conditions that both the roots +of the equation $z^{2} + az + b = 0$ should be of unit modulus are +\[ +|a| \leq 2,\quad +|b| = 1,\quad +\am b = 2\am a. +\] + +[The amplitudes have not necessarily their principal values.] + +\Item{11.} If $x^{4} + 4a_{1}x^{3} + 6a_{2}x^{2} + 4a_{3}x + a_{4} = 0$ is an equation with real coefficients +and has two real and two complex roots, concyclic in the Argand diagram, then +\[ +a_{3}^{2} + a_{1}^{2}a_{4} + a_{2}^{3} - a_{2}a_{4} - 2a_{1}a_{2}a_{3} = 0. +\] + +\Item{12.} The four roots of $a_{0}x^{4} + 4a_{1}x^{3} + 6a_{2}x^{2} + 4a_{3}x + a_{4} = 0$ will be harmonically +related if +\[ +a_{0}a_{3}^{2} + a_{1}^{2}a_{4} + a_{2}^{3} + - a_{0}a_{2}a_{4} - 2a_{1}a_{2}a_{3} = 0. +\] + +[Express $Z_{23, 14} Z_{31, 24} Z_{12, 34}$, where $Z_{23, 14} = (z_{1} - z_{2}) (z_{3} - z_{4}) + (z_{1} -z_{3}) (z_{2} - z_{4})$ +and $z_{1}$,~$z_{2}$, $z_{3}$,~$z_{4}$ are the roots of the equation, in terms of the coefficients.] +\PageSep{103} + +\Item{13.} \Topic{Imaginary points and straight lines.} Let $ax + by + c = 0$ be +an equation with complex coefficients (which of course may be real in special +cases). + +If we give $x$ any particular real or complex value, we can find the corresponding +value of~$y$. The aggregate of pairs of real or complex values of $x$~and~$y$ +which satisfy the equation is called an \emph{imaginary straight line}; the +pairs of values are called \emph{imaginary points}, and are said \emph{to lie on the line}. +The values of $x$~and~$y$ are called the \emph{coordinates} of the point $(x, y)$. When +$x$~and~$y$ are real, the point is called a \emph{real point}: when $a$,~$b$,~$c$ are all real (or +can be made all real by division by a common factor), the line is called a \emph{real +line}. The points $x = \alpha + \beta i$, $y = \gamma + \delta i$ and $x = \alpha - \beta i$, $y = \gamma - \delta i$ are said to be +\emph{conjugate}; and so are the lines +\[ +(A + A'i)x + (B + B'i)y + C + C'i = 0,\quad +(A - A'i)x + (B - B'i)y + C - C'i = 0. +\] + +Verify the following assertions:---every real line contains infinitely many +pairs of conjugate imaginary points; an imaginary line in general contains +one and only one real point; an imaginary line cannot contain a pair of +conjugate imaginary points:---and find the conditions (\ia)~that the line +joining two given imaginary points should be real, and (\ib)~that the point +of intersection of two imaginary lines should be real. + +\Item{14.} Prove the identities +\begin{gather*} +(x + y + z) (x + y\omega_{3} + z\omega_{3}^{2}) + (x + y\omega_{3}^{2} + z\omega_{3}) + = x^{3} + y^{3} + z^{3} - 3xyz,\\ +(x + y + z) (x + y\omega_{5} + z\omega_{5}^{4}) + (x + y\omega_{5}^{2} + z\omega_{5}^{3}) + (x + y\omega_{5}^{3} + z\omega_{5}^{2}) + (x + y\omega_{5}^{4} + z\omega_{5})\\ += x^{5} + y^{5} + z^{5} - 5x^{3}yz + 5xy^{2}z^{2}. +\end{gather*} + +\Item{15.} Solve the equations +\[ +x^{3} - 3ax + (a^{3} + 1) = 0,\quad +x^{5} - 5ax^{3} + 5a^{2}x + (a^{5} + 1) = 0. +\] + +\Item{16.} If $f(x) = a_{0} + a_{1}x + \dots + a_{k}x^{k}$, then +\[ +\{f(x) + f(\omega x) + \dots + f(\omega^{n-1}x)\}/n + = a_{0} + a_{n}x^{n} + a_{2n}x^{2n} + \dots + a_{\lambda n}x^{\lambda n}, +\] +$\omega$~being any root of $x^{n} = 1$ (except $x = 1$), and $\lambda n$~the greatest multiple of~$n$ +contained in~$k$. Find a similar formula for $a_{\mu} + a_{\mu+n}x^{n} + a_{\mu+2n}x^{2n} + \dots$. + +\Item{17.} If +\[ +(1 + x)^{n} = p_{0} + p_{1}x + p_{2}x^{2} + \dots, +\] +$n$~being a positive integer, then +\[ +p_{0} - p_{2} + p_{4} - \dots = 2^{\frac{1}{2} n} \cos\tfrac{1}{4}n\pi,\quad +p_{1} - p_{3} + p_{5} - \dots = 2^{\frac{1}{2} n} \sin\tfrac{1}{4}n\pi. +\] + +\Item{18.} Sum the series +\[ +\frac{x}{2! \DPchg{n - 2!}{(n - 2)!}} + + \frac{x^{2}}{5! \DPchg{n - 5!}{(n - 5)!}} + + \frac{x^{3}}{8! \DPchg{n - 8!}{(n - 8)!}} + \dots + + \frac{x^{n/3}}{\DPchg{n - 1!}{(n - 1)!}}, +\] +$n$~being a multiple of~$3$. \MathTrip{1899.} + +\Item{19.} {\Loosen If $t$~is a complex number such that $|t| = 1$, then the point +$x = (at + b)/(t - c)$ describes a circle as $t$~varies, unless $|c| = 1$, when it +describes a straight line.} +\PageSep{104} + +\Item{20.} If~$t$ varies as in the last example then the point $x = \frac{1}{2}\{at + (b/t)\}$ in +general describes an ellipse whose foci are given by $x^{2} = ab$, and whose axes +are $|a| + |b|$ and $|a| - |b|$. But if $|a| = |b|$ then $x$~describes the finite straight +line joining the points $-\sqrtp{ab}$, $\sqrtp{ab}$. + +\Item{21.} Prove that if $t$~is real and $z = t^{2} - 1 + \sqrtp{t^{4} - t^{2}}$, then, when $t^{2} < 1$, $z$~is +represented by a point which lies on the circle $x^{2} + y^{2} + x = 0$. Assuming that, +when $t^{2} > 1$, $\sqrtp{t^{4} - t^{2}}$ denotes the positive square root of $t^{4} - t^{2}$, discuss the +motion of the point which represents~$z$, as $t$~diminishes from a large positive +value to a large negative value. \MathTrip{1912.} + +\Item{22.} The coefficients of the transformation $z = (aZ + b)/(cZ + d)$ are subject +to the condition $ad - bc = 1$. Show that, if $c \neq 0$, there are two \emph{fixed points} +$\alpha$,~$\beta$, \ie\ points unaltered by the transformation, except when $(a + d)^{2} = 4$, when +there is only one fixed point~$\alpha$; and that in these two cases the transformation +may be expressed in the forms +\[ +\frac{z - \alpha}{z - \beta} = K\frac{Z - \alpha}{Z - \beta},\quad +\frac{1}{z - \alpha} = \frac{1}{Z - \alpha} + K. +\] + +Show further that, if $c = 0$, there will be one fixed point~$\alpha$ unless $a = d$, +and that in these two cases the transformation may be expressed in the +forms +\[ +z - \alpha = K(Z - \alpha),\quad +z = Z + K. +\] + +Finally, if $a$,~$b$,~$c$,~$d$ are further restricted to positive integral values (including +zero), show that the only transformations with less than two fixed +points are of the forms $(1/z) = (1/Z) + K$, $z = Z + K$. \MathTrip{1911.} + +\Item{23.} Prove that the relation $z = (1 + Zi)/(Z + i)$ transforms the part of the +axis of~$x$ between the points $z = 1$ and $z = -1$ into a semicircle passing +through the points $Z = 1$ and $Z = -1$. Find all the figures that can be obtained +from the originally selected part of the axis of~$x$ by successive applications of +the transformation. \MathTrip{1912.} + +\Item{24.} If $z = 2Z + Z^{2}$ then the circle $|Z| = 1$ corresponds to a cardioid in the +plane of~$z$. + +\Item{25.} Discuss the transformation $z = \frac{1}{2}\{Z + (1/Z)\}$, showing in particular +that to the circles $X^{2} + Y^{2} = \alpha^{2}$ correspond the confocal ellipses +\[ +\frac{x^{2}}{\left\{\dfrac{1}{2}\left(\alpha + \dfrac{1}{\alpha}\right)\right\}^{2}} + + \frac{y^{2}}{\left\{\dfrac{1}{2}\left(\alpha - \dfrac{1}{\alpha}\right)\right\}^{2}} + = 1. +\] + +\Item{26.} If $(z + 1)^{2} = 4/Z$ then the unit circle in the $z$-plane corresponds to the +parabola $R\cos^{2} \frac{1}{2}\Theta = 1$ in the $Z$-plane, and the inside of the circle to the +outside of the parabola. + +\Item{27.} Show that, by means of the transformation $z = \{(Z - ci)/(Z + ci)\}^{2}$, +the upper half of the $z$-plane may be made to correspond to the interior of +a certain semicircle in the $Z$-plane. +\PageSep{105} + +\Item{28.} If $z = Z^{2} - 1$, then as $z$~describes the circle $|z| = \kappa$, the two corresponding +positions of~$Z$ each describe the Cassinian oval $\rho_{1}\rho_{2} = \kappa$, where +$\rho_{1}$,~$\rho_{2}$ are the distances of~$Z$ from the points $-1$,~$1$. Trace the ovals for +different values of~$\kappa$. + +\Item{29.} Consider the relation $az^{2} + 2hzZ + bZ^{2} + 2gz + 2fZ + c = 0$. Show that +there are two values of~$Z$ for which the corresponding values of~$z$ are equal, +and \textit{vice versa}. We call these the \emph{branch points} in the $Z$ and $z$-planes respectively. +Show that, if $z$~describes an ellipse whose foci are the branch +points, then so does~$Z$. + +[We can, without loss of generality, take the given relation in the form +\[ +%[** TN: [sic] colon] +z^{2} + 2zZ\cos\omega + Z^{2} = 1: +\] +the reader should satisfy himself that this is the case. The branch points in +either plane are $\cosec\omega$ and $-\cosec\omega$. An ellipse of the form specified is +given by +\[ +|z + \cosec\omega| + |z - \cosec\omega| = C, +\] +where $C$~is a constant. This is equivalent (Ex.~9) to +\[ +|z + \sqrtp{z^{2} - \cosec^{2}\omega}| + +|z - \sqrtp{z^{2} - \cosec^{2}\omega}| = C. +\] +Express this in terms of~$Z$.] + +\Item{30\Add{.}} If $z = aZ^{m} + bZ^{n}$, where $m$,~$n$ are positive integers and $a$,~$b$ real, then +as $Z$~describes the unit circle, $z$~describes a hypo- or epi-cycloid. + +\Item{31.} Show that the transformation +\[ +z = \frac{(a + di)Z_{0} + b}{cZ_{0} - (a - di)}, +\] +where $a$,~$b$,~$c$,~$d$ are real and $a^{2} + d^{2} + bc > 0$, and $Z_{0}$~denotes the conjugate of~$Z$, +is equivalent to an inversion with respect to the circle +\[ +c(x^{2} + y^{2}) - 2ax - 2dy - b = 0. +\] +What is the geometrical interpretation of the transformation when +\[ +a^{2} + d^{2} + bc < 0? +\] + +\Item{32.} The transformation +\[ +\frac{1 - z}{1 + z} = \left(\frac{1 - Z}{1 + Z}\right)^{c}, +\] +where $c$~is rational and $0 < c < 1$, transforms the circle $|z| = 1$ into the boundary +of a circular lune of angle~$\pi/c$. +\end{Examples} +\PageSep{106} + + +\Chapter{IV}{LIMITS OF FUNCTIONS OF A POSITIVE INTEGRAL VARIABLE} + +\Paragraph{50. Functions of a positive integral variable.} In +\Ref{Chapter}{II} we discussed the notion of a function of a real +variable~$x$, and illustrated the discussion by a large number of +examples of such functions. And the reader will remember that +there was one important particular with regard to which the +functions which we took as illustrations differed very widely. +Some were defined for \emph{all} values of~$x$, some for \emph{rational} values +only, some for \emph{integral} values only, and so on. + +\begin{Remark} +Consider, for example, the following functions: (i)~$x$, (ii)~$\sqrt{x}$, (iii)~the +denominator of~$x$, (iv)~the square root of the product of the numerator and +the denominator of~$x$, (v)~the largest prime factor of~$x$, (vi)~the product of +$\sqrt{x}$ and the largest prime factor of~$x$, (vii)~the $x$th~prime number, (viii)~the +height measured in inches of convict~$x$ in Dartmoor prison. + +Then the aggregates of values of~$x$ for which these functions are defined +or, as we may say, the \emph{fields of definition} of the functions, consist of (i)~\emph{all} +values of~$x$, (ii)~\emph{all positive} values of~$x$, (iii)~\emph{all rational} values of~$x$, (iv)~\emph{all +positive rational} values of~$x$, (v)~\emph{all integral} values of~$x$, (vi),~(vii)~\emph{all positive +integral} values of~$x$, (viii)~a certain number of positive integral values of~$x$, +viz., $1$,~$2$, \dots,~$N$, where $N$~is the total number of convicts at Dartmoor at a +given moment of time.\footnote + {In the last case $N$~depends on the time, and convict~$x$, where $x$~has a definite + value, is a different individual at different moments of time. Thus if we take + different moments of time into consideration we have a simple example of a + function $y = F(x, t)$ of two variables, defined for a certain range of values of~$t$, viz.\ + from the time of the establishment of Dartmoor prison to the time of its abandonment, + and for a certain number of positive integral values of~$x$, this number + varying with~$t$.} +\end{Remark} + +Now let us consider a function, such as (vii) above, which is +defined for all positive integral values of~$x$ and no others. This +\PageSep{107} +function may be regarded from two slightly different points of +view. We may consider it, as has so far been our custom, as a +function of the real variable~$x$ defined for some only of the values +of~$x$, viz.\ positive integral values, and say that for all other values +of~$x$ the definition fails. Or we may leave values of~$x$ other +than positive integral values entirely out of account, and regard +our function as a function of the \emph{positive integral variable~$n$}, +whose values are the positive integers +\[ +1,\ 2,\ 3,\ 4,\ \dots. +\] +In this case we may write +\[ +y = \phi(n) +\] +and regard $y$~now as a function of~$n$ defined for all values of~$n$. + +It is obvious that any function of~$x$ defined for all values of~$x$ +gives rise to a function of~$n$ defined for all values of~$n$. Thus from +the function $y = x^{2}$ we deduce the function $y = n^{2}$ by merely +omitting from consideration all values of~$x$ other than positive +integers, and the corresponding values of~$y$. On the other hand +from any function of~$n$ we can deduce any number of functions +of~$x$ by merely assigning values to~$y$, corresponding to values of~$x$ +other than positive integral values, in any way we please. + +\begin{Remark} +\Paragraph{51. Interpolation.} The problem of determining a function of~$x$ which +shall assume, for all positive integral values of~$x$, values agreeing with those +of a given function of~$n$, is of extreme importance in higher mathematics. +It is called the \emph{problem of functional interpolation}. + +Were the problem however merely that of finding \emph{some} function of~$x$ to +fulfil the condition stated, it would of course present no difficulty whatever. +We could, as explained above, simply fill in the missing values as we pleased: +we might indeed simply regard the given values of the function of~$n$ as \emph{all} +the values of the function of~$x$ and say that the definition of the latter +function failed for all other values of~$x$. But such purely theoretical solutions +are obviously not what is usually wanted. What is usually wanted is some +\emph{formula} involving~$x$ (of as simple a kind as possible) which assumes the given +values for $x = 1$, $2$,~\dots. + +In some cases, especially when the function of~$n$ is itself defined by a +formula, there is an obvious solution. If for example $y = \phi(n)$, where $\phi(n)$ +is a function of~$n$, such as $n^{2}$ or $\cos n\pi$, which would have a meaning even +were $n$ not a positive integer, we naturally take our function of~$x$ to be +$y = \phi(x)$. But even in this very simple case it is easy to write down other +almost equally obvious solutions of the problem. For example +\[ +y = \phi(x) + \sin x\pi +\] +assumes the value $\phi(n)$ for $x = n$, since $\sin n\pi = 0$. +\PageSep{108} + +In other cases $\phi(n)$ may be defined by a formula, such as~$(-1)^{n}$, which +ceases to define for some values of~$x$ (as here in the case of fractional values +of~$x$ with even denominators, or irrational values). But it may be possible +to transform the formula in such a way that it does define for all values of~$x$. +In this case, for example, +\[ +(-1)^{n} = \cos n\pi, +\] +if $n$~is an integer, and the problem of interpolation is solved by the +function~$\cos x\pi$. + +In other cases $\phi(x)$ may be defined for some values of~$x$ other than +positive integers, but not for all. Thus from $y = n^{n}$ we are led to $y = x^{x}$. +This expression has a meaning for some only of the remaining values of~$x$. +If for simplicity we confine ourselves to positive values of~$x$, then $x^{x}$~has +a meaning for all rational values of~$x$, in virtue of the definitions of +fractional powers adopted in elementary algebra. But when $x$~is \emph{irrational} +$x^{x}$~has (so far as we are in a position to say at the present moment) no +meaning at all. Thus in this case the problem of interpolation at once +leads us to consider the question of extending our definitions in such a +way that $x^{x}$~shall have a meaning even when $x$~is irrational. We shall see +later on how the desired extension may be effected. + +Again, consider the case in which +\[ +y = 1 · 2 \dots n = n!. +\] +In this case there is no obvious formula in~$x$ which reduces to~$n!$ for $x = n$, +as $x!$~means nothing for values of~$x$ other than the positive integers. This +is a case in which attempts to solve the problem of interpolation have led to +important advances in mathematics. For mathematicians have succeeded +in discovering a function (the Gamma-function) which possesses the desired +property and many other interesting and important properties besides. +\end{Remark} + +\Paragraph{52. Finite and infinite classes.} Before we proceed further +it is necessary to make a few remarks about certain ideas of an +abstract and logical nature which are of constant occurrence in +Pure Mathematics. + +In the first place, the reader is probably familiar with the +notion of \Emph{a class}. It is unnecessary to discuss here any logical +difficulties which may be involved in the notion of a `class': +roughly speaking we may say that a class is the aggregate or +collection of all the entities or objects which possess a certain +property, simple or complex. Thus we have the class of British +subjects, or members of Parliament, or positive integers, or real +numbers. +\PageSep{109} + +Moreover, the reader has probably an idea of what is meant +by a \Emph{finite} or \Emph{infinite} class. Thus the class of \emph{British subjects} +is a finite class: the aggregate of all British subjects, past, +present, and future, has a finite number~$n$, though of course we +cannot tell at present the actual value of~$n$. The class of \emph{present +British subjects}, on the other hand, has a number~$n$ which could +be ascertained by counting, were the methods of the census +effective enough. + +On the other hand the class of positive integers is not finite +but infinite. This may be expressed more precisely as follows. +If $n$~is any positive integer, such as $1000$, $1,000,000$ or any number +we like to think of, then there are more than $n$ positive integers. +Thus, if the number we think of is $1,000,000$, there are obviously +at least $1,000,001$ positive integers. Similarly the class of rational +numbers, or of real numbers, is infinite. It is convenient to +express this by saying that there are \Emph{an infinite number} of +positive integers, or rational numbers, or real numbers. But the +reader must be careful always to remember that by saying this +we mean \emph{simply} that the class in question has not a finite number +of members such as $1000$ or $1,000,000$. + +\Paragraph{53. Properties possessed by a function of~$n$ for large +values of~$n$.} We may now return to the `functions of~$n$' which we +were discussing in \SecNo[§§]{50}--\SecNo{51}. They have many points of difference +from the functions of~$x$ which we discussed in \Ref{Chap.}{II}\@. But there +is one fundamental characteristic which the two classes of functions +have in common: \emph{the values of the variable for which they +are defined form an infinite class}. It is this fact which forms the +basis of all the considerations which follow and which, as we shall +see in the next chapter, apply, \textit{mutatis mutandis}, to functions of~$x$ +as well. + +Suppose that $\phi(n)$ is any function of~$n$, and that $P$~is any +property which $\phi(n)$ may or may not have, such as that of being +a positive integer or of being greater than~$1$. Consider, for each +of the values $n = 1$, $2$, $3$,~\dots, whether $\phi(n)$ has the property~$P$ or +not. Then there are three possibilities:--- + +\Item{(\ia)} $\phi(n)$ may have the property~$P$ for \emph{all} values of~$n$, or for +all values of~$n$ except a finite number~$N$ of such values: +\PageSep{110} + +\Item{(\ib)} $\phi(n)$ may have the property for \emph{no} values of~$n$, or only for +a finite number~$N$ of such values: + +\Item{(\ic)} neither~(\ia) nor~(\ib) may be true. + +If (\ib)~is true, the values of~$n$ for which $\phi(n)$ has the property +form a finite class. If (\ia)~is true, the values of~$n$ for which $\phi(n)$ +has not the property form a finite class. In the third case neither +class is finite. Let us consider some particular cases. + +\begin{Remark} +\Item{(1)} Let $\phi(n) = n$, and let $P$~be the property of being a positive integer. +Then $\phi(n)$ has the property~$P$ for all values of~$n$. + +If on the other hand $P$~denotes the property of being a positive integer +greater than or equal to~$1000$, then $\phi(n)$ has the property for all values of~$n$ +except a finite number of values of~$n$, viz.\ $1$,~$2$, $3$, \dots,~$999$. In either of +these cases (\ia)~is true. + +\Item{(2)} If $\phi(n) = n$, and $P$~is the property of being less than~$1000$, then (\ib)~is +true. + +\Item{(3)} If $\phi(n) = n$, and $P$~is the property of being odd, then~(\ic) is true. For +$\phi(n)$~is odd if $n$~is odd and even if $n$~is even, and both the odd and the even +values of~$n$ form an infinite class. + +\Par{Example.} Consider, in each of the following cases, whether (\ia),~(\ib), or +(\ic) is true: + +\Itemp{(i)} $\phi(n) = n$, $P$~being the property of being a perfect square, + +\Itemp{(ii)} \Hang $\phi(n) = p_{n}$, where $p_{n}$~denotes the $n$th~prime number, $P$~being the \\ %[** TN: Explicit line break avoids minor glue problem] +property of being odd, + +\Itemp{(iii)} $\phi(n) = p_{n}$, $P$~being the property of being even, + +\Itemp{(iv)} $\phi(n) = p_{n}$, $P$~being the property $\phi(n) > n$, + +\Itemp{(v)} $\phi(n) = 1 - (-1)^{n}(1/n)$, $P$~being the property $\phi(n) < 1$, + +\Itemp{(vi)} $\phi(n) = 1 - (-1)^{n}(1/n)$, $P$~being the property $\phi(n) < 2$, + +\Itemp{(vii)} $\phi(n) = 1000\{1 + (-1)^{n}\}/n$, $P$~being the property $\phi(n) < 1$, + +\Itemp{(viii)} $\phi(n) = 1/n$, $P$~being the property $\phi(n) < .001$, + +\Itemp{(ix)} $\phi(n) = (-1)^{n}/n$, $P$~being the property $|\phi(n)| < .001$, + +\Itemp{(x)} \Hang $\phi(n) = 10\MC000/n$, or $(-1)^{n}10\MC000/n$, $P$~being either of the properties +$\phi(n) < .001$ or $|\phi(n)| < .001$, + +\Itemp{(xi)} $\phi(n) = (n - 1)/(n + 1)$, $P$~being the property $1 - \phi(n) < .0001$. +\end{Remark} + +\Paragraph{54.} Let us now suppose that $\phi(n)$~and~$P$ are such that the +assertion~(\ia) is true, \ie\ that $\phi(n)$~has the property~$P$, if not for +all values of~$n$, at any rate for all values of~$n$ except a finite +number~$N$ of such values. We may denote these exceptional +values by +\[ +n_{1},\ n_{2},\ \dots,\ n_{N}. +\] +\PageSep{111} +There is of course no reason why these $N$~values should be the +\emph{first} $N$~values $1$,~$2$, \dots,~$N$, though, as the preceding examples +show, this is frequently the case in practice. But whether this +is so or not we know that $\phi(n)$ has the property~$P$ if $n > n_{N}$. +Thus the $n$th~prime is odd if $n > 2$, $n = 2$ being the only exception +to the statement; and $1/n < .001$ if $n > 1000$, the first $1000$~values +of~$n$ being the exceptions; and +\[ +1000\{1 + (-1)^{n}\}/n < 1 +\] +if $n > 2000$, the exceptional values being $2$,~$4$,~$6$, \dots,~$2000$. That +is to say, in each of these cases the property is possessed \emph{for all +values of~$n$ from a definite value onwards}. + +We shall frequently express this by saying that $\phi(n)$ has the +property for \Emph{large}, or \emph{very large}, or \emph{all sufficiently large} values of~$n$. +Thus when we say that \emph{$\phi(n)$~has the property~$P$} (which will as a +rule be a property expressed by some relation of inequality) \emph{for +large values of~$n$}, what we mean is that we can determine some +definite number, $n_{0}$~say, such that $\phi(n)$ has the property for all +values of~$n$ greater than or equal to~$n_{0}$. This number~$n_{0}$, in the +examples considered above, may be taken to be any number +greater than~$n_{N}$, the greatest of the exceptional numbers: it is +most natural to take it to be~$n_{N} + 1$. + +Thus we may say that `all large primes are odd', or that `$1/n$~is +less than~$.001$ for large values of~$n$'. And the reader must make +himself familiar with the use of the word \emph{large} in statements of +this kind. \emph{Large} is in fact a word which, standing by itself, has +no more absolute meaning in mathematics than in the language +of common life. It is a truism that in common life a number +which is large in one connection is small in another; $6$~goals is a +large score in a football match, but $6$~runs is not a large score in a +cricket match; and $400$~runs is a large score, but £$400$~is not +a large income: and so of course in mathematics \emph{large} generally +means \emph{large enough}, and what is large enough for one purpose +may not be large enough for another. + +We know now what is meant by the assertion `$\phi(n)$~has the +property~$P$ for large values of~$n$'. It is with assertions of this +kind that we shall be concerned throughout this chapter. +\PageSep{112} + +\Paragraph{55. The phrase `$n$~tends to infinity'.} There is a somewhat +different way of looking at the matter which it is natural to +adopt. Suppose that $n$~assumes successively the values $1$,~$2$, $3$,~\dots. +The word `successively' naturally suggests succession in time, and +we may suppose~$n$, if we like, to assume these values at successive +moments of time (\eg\ at the beginnings of successive seconds). +Then as the seconds pass $n$~gets larger and larger and there is +no limit to the extent of its increase. However large a number +we may think of (\eg\ $2\MC147\MC483\MC647$), a time will come when $n$~has +become larger than this number. + +It is convenient to have a short phrase to express this unending +growth of~$n$, and we shall say that \Emph{$n$~tends to infinity}, or $n \to \infty$, +this last symbol being usually employed as an abbreviation for +`infinity'. The phrase `tends to' like the word `successively' +naturally suggests the idea of change in time, and it is convenient +to think of the variation of~$n$ as accomplished in time in the +manner described above. This however is a mere matter of convenience. +The variable~$n$ is a purely logical entity which has in +itself nothing to do with time. + +The reader cannot too strongly impress upon himself that +when we say that $n$~`tends to~$\infty$' we mean simply that $n$~is +supposed to assume a series of values which increase continually +and without limit. \Emph{There is no number `infinity'}: such an +equation as +\[ +n = \infty +\] +is as it stands \emph{absolutely meaningless}: $n$~cannot be equal to~$\infty$, +because `equal to~$\infty$' means nothing. So far in fact the symbol~$\infty$ +means nothing at all except in the one phrase `tends to~$\infty$', +the meaning of which we have explained above. Later on we +shall learn how to attach a meaning to other phrases involving +the symbol~$\infty$, but the reader will always have to bear in mind + +\Item{(1)} that \emph{$\infty$~by itself} means nothing, although \emph{phrases containing +it} sometimes mean something, + +\Item{(2)} that in every case in which a phrase containing the +symbol~$\infty$ means something it will do so simply because we have +previously attached a meaning to this particular phrase by means +of a special definition. +\PageSep{113} + +Now it is clear that if $\phi(n)$~has the property~$P$ for large values +of~$n$, and if $n$~`tends to~$\infty$', in the sense which we have just +explained, then $n$~will ultimately assume values large enough to +ensure that $\phi(n)$ has the property~$P$. And so another way of +putting the question `what properties has $\phi(n)$ for sufficiently +large values of~$n$?'\ is `how does $\phi(n)$ behave as $n$~tends to~$\infty$?' + +\Paragraph{56. The behaviour of a function of~$n$ as $n$~tends to +infinity.} {\Loosen We shall now proceed, in the light of the remarks +made in the preceding sections, to consider the meaning of some +kinds of statements which are perpetually occurring in higher +mathematics. Let us consider, for example, the two following +statements: (\ia)~\emph{$1/n$ is small for large values of~$n$}, (\ib)~\emph{$1 - (1/n)$ is +nearly equal to~$1$ for large values of~$n$}. Obvious as they may +seem, there is a good deal in them which will repay the reader's +attention. Let us take (\ia) first, as being slightly the simpler.} + +We have already considered the statement `\emph{$1/n$~is less than~$.01$ +for large values of~$n$}'. This, we saw, means that the inequality +$1/n < .01$ is true for all values of~$n$ greater than some definite +value, in fact greater than~$100$. Similarly it is true that `\emph{$1/n$~is +less than $.0001$ for large values of~$n$}': in fact $1/n < .0001$ if +$n > 10\MC000$. And instead of $.01$ or $.0001$ we might take $.000\MS001$ or +$.000\MS000\MS01$, or indeed any positive number we like. + +It is obviously convenient to have some way of expressing the\PageLabel{113} +fact that \emph{any} such statement as `\emph{$1/n$~is less than~$.01$ for large +values of~$n$}' is true, when we substitute for~$.01$ any smaller +number, such as $.0001$ or $.000\MS001$ or any other number we care +to choose. And clearly we can do this by saying that `\emph{however +small $\DELTA$ may be} (provided of course it is positive), \emph{then $1/n < \DELTA$ for +sufficiently large values of~$n$}'. That this is true is obvious. For +$1/n < \DELTA$ if $n > 1/\DELTA$, so that our `sufficiently large' values of~$n$ need +only all be greater than~$1/\DELTA$. The assertion is however a complex +one, in that it really stands for the whole class of assertions which +we obtain by giving to~$\DELTA$ special values such as~$.01$. And of course +the smaller $\DELTA$~is, and the larger~$1/\DELTA$, the larger must be the least of +the `sufficiently large' values of~$n$: values which are sufficiently +large when $\DELTA$~has one value are inadequate when it has a smaller. + +The last statement italicised is what is really meant by the +statement~(\ia), that $1/n$~is small when $n$~is large. Similarly +\PageSep{114} +(\ib)~really means ``\emph{if $\phi(n) = 1 - (1/n)$, then the statement `$1 - \phi(n) < \DELTA$ +for sufficiently large values of~$n$' is true whatever positive value +\(such as $.01$ or $.0001$\) we attribute to~$\DELTA$}''. That the statement~(\ib) +is true is obvious from the fact that $1 - \phi(n) = 1/n$. + +There is another way in which it is common to state the facts +expressed by the assertions (\ia)~and~(\ib). This is suggested at once +by \SecNo[§]{55}. Instead of saying `$1/n$~is small for large values of~$n$' we +say `$1/n$~tends to~$0$ as $n$~tends to~$\infty$'. Similarly we say that +`$1 - (1/n)$ tends to~$1$ as $n$~tends to~$\infty$': and these statements are +to be regarded as precisely equivalent to (\ia)~and~(\ib). Thus the +statements +\begin{align*} +&\text{`$1/n$~is small when $n$~is large',} \\ +&\text{`$1/n$~tends to~$0$ as $n$~tends to~$\infty$',} +\end{align*} +are equivalent to one another and to the more formal statement +\begin{quotation} +`if $\DELTA$ is any positive number, however small, then $1/n < \DELTA$ +for sufficiently large values of~$n$', +\end{quotation} +or to the still more formal statement +\begin{quotation} +`if $\DELTA$ is any positive number, however small, then we can +find a number~$n_{0}$ such that $1/n < \DELTA$ for all values of~$n$ greater +than or equal to~$n_{0}$'. +\end{quotation} + +The number~$n_{0}$ which occurs in the last statement is of course +a function of~$\DELTA$. We shall sometimes emphasize this fact by +writing $n_{0}$ in the form~$n_{0}(\DELTA)$. + +\begin{Remark} +The reader should imagine himself confronted by an opponent who +questions the truth of the statement. He would name a series of numbers +growing smaller and smaller. He might begin with~$.001$. The reader would +reply that $1/n < .001$ as soon as $n > 1000$. The opponent would be bound to +admit this, but would try again with some smaller number, such as~$.000\MS000\MS1$. +The reader would reply that $1/n < .000\MS000\MS1$ as soon as $n > 10\MC000\MC000$: and so +on. In this simple case it is evident that the reader would always have the +better of the argument. +\end{Remark} + +We shall now introduce yet another way of expressing this +property of the function~$1/n$. We shall say that `\emph{the \Emph{limit} of~$1/n$ +as $n$~tends to~$\infty$ is~$0$}', a statement which we may express symbolically +in the form +\[ +\lim_{n\to\infty} \frac{1}{n} = 0, +\] +\PageSep{115} +or simply $\lim(1/n) = 0$. We shall also sometimes write +%[** TN: Next expression displayed in the original] +`$1/n \to 0$ +as $n \to \infty$', which may be read `$1/n$~tends to~$0$ as $n$~tends to~$\infty$'; or +simply `$1/n \to 0$'. In the same way we shall write +\[ +\lim_{n\to\infty} \left(1 - \frac{1}{n}\right) = 1,\quad +\lim \left(1 - \frac{1}{n}\right) = 1, +\] +or $1 - (1/n) \to 1$. + +\Paragraph{57.} Now let us consider a different example: let $\phi(n) = n^{2}$. +Then `\emph{$n^{2}$~is large when $n$~is large}'. This statement is equivalent +to the more formal statements +\begin{quotation} +`if $\Delta$ is any positive number, however large, then $n^{2} > \Delta$ +for sufficiently large values of~$n$', +\bigskip + +`we can find a number $n_{0}(\Delta)$ such that $n^{2} > \Delta$ for all values +of~$n$ greater than or equal to~$n_{0}(\Delta)$'. +\end{quotation} +And it is natural in this case to say that `$n^{2}$~tends to~$\infty$ as $n$~tends +to~$\infty$', or `$n^{2}$~tends to~$\infty$ with~$n$', and to write +\[ +n^2 \to \infty. +\] + +Finally consider the function $\phi(n) = -n^{2}$. In this case $\phi(n)$ +is large, but negative, when $n$~is large, and we naturally say that +`$-n^{2}$~tends to~$-\infty$ as $n$~tends to~$\infty$' and write +\[ +-n^{2} \to -\infty. +\] +And the use of the symbol~$-\infty$ in this sense suggests that it +will sometimes be convenient to write $n^{2} \to +\infty$ for $n^{2} \to \infty$ and +generally to use~$+\infty$ instead of~$\infty$, in order to secure greater +uniformity of notation. + +But we must once more repeat that in all these statements +the symbols $\infty$,~$+\infty$,~$-\infty$ mean nothing whatever by themselves, +and only acquire a meaning when they occur in certain special +connections in virtue of the explanations which we have just +given. +\PageSep{116} + +\Paragraph{58. Definition of a limit.} After the discussion which +precedes the reader should be in a position to appreciate the +general notion of a \emph{limit}. Roughly we may say that \emph{$\phi(n)$~tends +to a limit~$l$ as $n$~tends to~$\infty$ if $\phi(n)$~is nearly equal to~$l$ when $n$~is +large}. But although the meaning of this statement should be +clear enough after the preceding explanations, it is not, as it +stands, precise enough to serve as a strict mathematical definition. +It is, in fact, equivalent to a whole class of statements of the +type `\emph{for sufficiently large values of~$n$, $\phi(n)$~differs from~$l$ by less +than~$\DELTA$}'. This statement has to be true for $\DELTA = .01$ or $.0001$ or \emph{any} +positive number; and for any such value of~$\DELTA$ it has to be true for +\emph{any} value of~$n$ after a certain definite value~$n_{0}(\DELTA)$, though the +smaller~$\DELTA$ is the larger, as a rule, will be this value~$n_{0}(\DELTA)$. + +We accordingly frame the following formal definition: + +\begin{Definition}[I.] +The function $\phi(n)$ is said to tend to the limit~$l$ +as $n$~tends to~$\infty$, if, however small be the positive number~$\DELTA$, +$\phi(n)$~differs from~$l$ by less than~$\DELTA$ for sufficiently large values of~$n$; +that is to say if, however small be the positive number~$\DELTA$, we can +determine a number~$n_{0}(\DELTA)$ corresponding to~$\DELTA$, such that $\phi(n)$~differs +from~$l$ by less than~$\DELTA$ for all values of~$n$ greater than or equal to~$n_{0}(\DELTA)$. +\end{Definition} + +It is usual to denote the difference between $\phi(n)$~and~$l$, taken +positively, by $|\phi(n) - l|$. It is equal to $\phi(n) - l$ or to $l - \phi(n)$, +whichever is positive, and agrees with the definition of the +\emph{modulus} of~$\phi(n) - l$, as given in \Ref{Chap.}{III}, though at present +we are only considering real values, positive or negative. + +With this notation the definition may be stated more shortly +as follows: `\emph{if, given any positive number,~$\DELTA$, however small, we +can find~$n_{0}(\DELTA)$ so that $|\phi(n) - l| < \DELTA$ when $n \geq n_{0}(\DELTA)$, then we say +that $\phi(n)$~tends to the limit~$l$ as $n$~tends to~$\infty$, and write} +\[ +\lim_{n \to \infty} \phi(n) = l\text{'.} +\] + +\begin{Remark} +Sometimes we may omit the `$n \to \infty$'; and sometimes it is convenient, for +brevity, to write $\phi(n) \to l$. + +The reader will find it instructive to work out, in a few simple cases, the +explicit expression of~$n_{0}$ as a function of~$\DELTA$. Thus if $\phi(\DPtypo{x}{n}) = 1/n$ then $l = 0$, and +the condition reduces to $1/n < \DELTA$ for $n \geq n_{0}$, which is satisfied if $n_{0} = 1 + [1/\DELTA]$.\footnote + {Here and henceforward we shall use $[x]$ in the sense of \Ref{Chap.}{II}, \ie\ as the + greatest integer not greater than~$x$.} +There is one and only one case in which \emph{the same~$n_{0}$} will do for \emph{all} values of~$\DELTA$. +\PageSep{117} +If, from a certain value~$N$ of~$n$ onwards, $\phi(n)$~is constant, say equal to~$C$, then +it is evident that $\phi(n) - C = 0$ for $n \geq N$, so that the inequality $|\phi(n) - C| < \DELTA$ +is satisfied for $n \geq N$ and all positive values of~$\DELTA$. And if $|\phi(n) - l| < \DELTA$ for +$n \geq N$ and all positive values of~$\DELTA$, then it is evident that $\phi(n) = l$ when $n \geq N$, +so that $\phi(n)$~is constant for all such values of~$n$. +\end{Remark} + +\Paragraph{59.} The definition of a limit may be illustrated geometrically +as follows. The graph of~$\phi(n)$ consists of a number of points +corresponding to the values $n = 1$, $2$,~$3$,~\dots. + +Draw the line $y = l$, and the parallel lines $y = l - \DELTA$, $y = l + \DELTA$ +at distance~$\DELTA$ from it. Then +\[ +\lim_{n \to \infty} \phi(n) = l, +\] +%[Illustration: Fig. 27.] +\ifthenelse{\boolean{Modernize}}{% +\Figure{27}{p117} +}{% +\Figure{27}{p117_orig_notation}% +} +if, when once these lines have been drawn, no matter how close +they may be together, we can always draw a line $x = n_{0}$, as in the +figure, in such a way that the point of the graph on this line, and +all points to the right of it, lie between them. We shall find +this geometrical way of looking at our definition particularly +useful when we come to deal with functions defined for all values +of a real variable and not merely for positive integral values. + +\Paragraph{60.} So much for functions of~$n$ which tend to a limit as~$n$ +tends to~$\infty$. We must now frame corresponding definitions for +functions which, like the functions $n^{2}$~or~$-n^{2}$, tend to positive or +negative infinity. The reader should by now find no difficulty in +appreciating the point of +\begin{Definition}[II.] +The function~$\phi(n)$ is said to tend to~$+\infty$ +(positive infinity) with~$n$, if, when any number~$\Delta$, however large, +is assigned, we can determine~$n_{0}(\Delta)$ so that $\phi(n) > \Delta$ when $n \geq n_{0}(\Delta)$; +\PageSep{118} +that is to say if, however large~$\Delta$ may be, $\phi(n) > \Delta$ for sufficiently +large values of~$n$. +\end{Definition} + +Another, less precise, form of statement is `\emph{if we can make +$\phi(n)$~as large as we please by sufficiently increasing~$n$}'. This is +open to the objection that it obscures a fundamental point, viz.\ +that $\phi(n)$~must be greater than~$\Delta$ for \emph{all} values of~$n$ such that +$n \geq n_{0}(\Delta)$, and not merely for \emph{some} such values. But there is no +harm in using this form of expression if we are clear what it +means. + +When $\phi(n)$ tends to~$+\infty$ we write +\[ +\phi(n) \to +\infty. +\] +We may leave it to the reader to frame the corresponding +definition for functions which tend to negative infinity. + +\Paragraph{61. Some points concerning the definitions.} The reader +should be careful to observe the following points. + +\Item{(1)} We may obviously alter the values of~$\phi(n)$ for any +finite number of values of~$n$, in any way we please, without in +the least affecting the behaviour of~$\phi(n)$ as $n$~tends to~$\infty$. For +example $1/n$~tends to~$0$ as $n$~tends to~$\infty$. We may deduce any +number of new functions from~$1/n$ by altering a finite number of +its values. For instance we may consider the function~$\phi(n)$ which +is equal to~$3$ for $n = 1$,~$2$, $7$, $11$, $101$, $107$, $109$,~$237$ and equal to~$1/n$ +for all other values of~$n$. For this function, just as for the +original function~$1/n$, $\lim\phi(n) = 0$. Similarly, for the function~$\phi(n)$ +which is equal to~$3$ if $n = 1$,~$2$, $7$, $11$, $101$, $107$, $109$,~$237$, and +to~$n^{2}$ otherwise, it is true that $\phi(n) \to +\infty$. + +\Item{(2)} On the other hand we cannot as a rule alter an \emph{infinite} +number of the values of~$\phi(n)$ without affecting fundamentally its +behaviour as $n$~tends to~$\infty$. If for example we altered the function~$1/n$ +by changing its value to~$1$ whenever $n$~is a multiple of~$100$, +it would no longer be true that $\lim\phi(n) = 0$. So long as a finite +number of values only were affected we could always choose the +number~$n_{0}$ of the definition so as to be greater than the greatest +of the values of~$n$ for which $\phi(n)$ was altered. In the examples +above, for instance, we could always take $n_{0} > 237$, and indeed we +should be compelled to do so as soon as our imaginary opponent +\PageSep{119} +of \SecNo[§]{56} had assigned a value of~$\DELTA$ as small as~$3$ (in the first +example) or a value of~$\Delta$ as great as~$3$ (in the second). But +now \emph{however} large $n_{0}$ may be there will be greater values of~$n$ for +which $\phi(n)$~has been altered. + +\Item{(3)} In applying the test of Definition~I it is of course %[xref] +absolutely essential that we should have $|\phi(n) - l| < \DELTA$ not merely +when $n = n_{0}$ but when $n \geq n_{0}$, \ie\ \emph{for $n_{0}$ and for all larger values +of~$n$}. It is obvious, for example, that, if $\phi(n)$~is the function last +considered, then given~$\DELTA$ we can choose~$n_{0}$ so that $|\phi(n)| < \DELTA$ when +$n = n_{0}$: we have only to choose a sufficiently large value of~$n$ +which is not a multiple of~$100$. But, when $n_{0}$ is thus chosen, it +is not true that $|\phi(n)| < \DELTA$ when $n \geq n_{0}$: all the multiples of~$100$ +which are greater than~$n_{0}$ are exceptions to this statement. + +\Item{(4)} If $\phi(n)$ is always greater than~$l$, we can replace +$|\phi(n) - l|$ by $\phi(n) - l$. Thus the test whether $1/n$~tends to the +limit~$0$ as $n$~tends to~$\infty$ is simply whether $1/n < \DELTA$ when $n \geq n_{0}$. +If however $\phi(n) = (-1)^{n}/n$, then $l$~is again~$0$, but $\phi(n) - l$ is sometimes +positive and sometimes negative. In such a case we must +state the condition in the form $|\phi(n) - l| < \DELTA$, for example, in +this particular case, in the form $|\phi(n)| < \DELTA$. + +\Item{(5)} \emph{The limit~$l$ may itself be one of the actual values of +$\phi(n)$.} Thus if $\phi(n) = 0$ for all values of~$n$, it is obvious that +$\lim\phi(n) = 0$. Again, if we had, in (2)~and~(3) above, altered +the value of the function, when $n$~is a multiple of~$100$, to~$0$ +instead of to~$1$, we should have obtained a function $\phi(n)$ which +is equal to~$0$ when $n$~is a multiple of~$100$ and to~$1/n$ otherwise. +The limit of this function as $n$~tends to~$\infty$ is still obviously zero. +This limit is itself the value of the function for an infinite number +of values of~$n$, viz.\ all multiples of~$100$. + +On the other hand \emph{the limit itself need not \(and in general will +not\) be the value of the function for any value of~$n$}. This is +sufficiently obvious in the case of $\phi(n) = 1/n$. The limit is zero; +but the function is never equal to zero for any value of~$n$. + +The reader cannot impress these facts too strongly on his +mind. \Emph{A limit is not a value of the function}: it is something +quite distinct from these values, though it is defined by its relations +\PageSep{120} +to them and may possibly be equal to some of them. For the +functions +\[ +\phi(n) = 0,\ 1, +\] +the limit is equal to \emph{all} the values of~$\phi(n)$: for +\[ +\phi(n) = 1/n,\quad +(-1)^{n}/n,\quad +1 + (1/n),\quad +1 + \{(-1)^{n}/n\} +\] +it is not equal to \emph{any} value of~$\phi(n)$: for +\[ +\phi(n) = (\sin\tfrac{1}{2}n\pi)/n,\quad +1 + \{(\sin\tfrac{1}{2}n\pi)/n\} +\] +(whose limits as $n$~tends to~$\infty$ are easily seen to be $0$~and~$1$, since +$\sin\frac{1}{2}n\pi$ is never numerically greater than~$1$) the limit is equal to +the value which $\phi(n)$ assumes for all even values of~$n$, but the +values assumed for odd values of~$n$ are all different from the limit +and from one another. + +\Item{(6)} A function may be always numerically very large when +$n$~is very large without tending either to~$+\infty$ or to~$-\infty$. A +sufficient illustration of this is given by $\phi(n) = (-1)^{n} n$. A function +can only tend to~$+\infty$ or to~$-\infty$ if, after a certain value of~$n$, +it maintains a constant sign. + +\begin{Examples}{XXIII.} +Consider the behaviour of the following functions +of~$\DPtypo{x}{n}$ as $n$~tends to~$\infty$: + +\Item{1.} $\phi(n) = n^{k}$, where $k$~is a positive or negative integer or rational fraction. +If $k$~is positive, then $n^{k}$~tends to~$+\infty$ with~$n$. If $k$~is negative, then $\lim n^{k} = 0$. +If $k = 0$, then $n^{k} = 1$ for all values of~$n$. Hence $\lim n^{k} = 1$. + +The reader will find it instructive, even in so simple a case as this, to +write down a formal proof that the conditions of our definitions are satisfied. +Take for instance the case of $k > 0$. Let $\Delta$ be any assigned number, however +large. We wish to choose~$n_{0}$ so that $n^{k} > \Delta$ when $n \geq n_{0}$. We have in fact only +to take for~$n_{0}$ any number greater than~$\sqrt[k]{\Delta}$. If \eg\ $k = 4$, then $n^{4} > 10\MC000$ when +$n \geq 11$, $n^{4}> 100\MC000\MC000$ when $n \geq 101$, and so on. + +\Item{2.} $\phi(n) = p_{n}$, where $p_{n}$~is the $n$th~prime number. If there were only +a finite number of primes then $\phi(n)$ would be defined only for a finite number +of values of~$n$. There are however, as was first shown by Euclid, infinitely +many primes. Euclid's proof is as follows. If there are only a finite +number of primes, let them be $1$,~$2$, $3$, $5$, $7$, $11$,~\dots~$N$. Consider the number +$1 + (1 · 2 · 3 · 5 · 7 · 11 \dots N)$. This number is evidently not divisible by +any of $2$,~$3$, $5$,~\dots~$N$, since the remainder when it is divided by any of +these numbers is~$1$. It is therefore not divisible by any prime save~$1$, and +is therefore itself prime, which is contrary to our hypothesis. + +It is moreover obvious that $\phi(n) > n$ for all values of~$n$ (save $n = 1$, $2$,~$3$). +Hence $\phi(n) \to +\infty$. +\PageSep{121} + +\Item{3.} Let $\phi(n)$~be the number of primes less than~$n$. Here again $\phi(n) \to +\infty$. + +\Item{4.} $\phi(n) = [\alpha n]$, where $\alpha$~is any positive number. Here +\[ +\phi(n) = 0\quad (0 \leq n < 1 / \alpha),\qquad +\phi(n) = 1\quad (1/\alpha \leq n < 2/\alpha), +\] +and so on; and $\phi(n) \to +\infty$. + +\Item{5.} If $\phi(n) = 1\MC000\MC000/n$, then $\lim\phi(n) = 0$: and if $\psi(n) = n/1\MC000\MC000$, then +$\psi(n) \to +\infty$. These conclusions are in no way affected by the fact that at first +$\phi(n)$~is much larger than~$\psi(n)$, being in fact larger until $n = 1\MC000\MC000$. + +\Item{6.} $\phi(n) = 1/\{n - (-1)^{n}\}$, $n - (-1)^{n}$, $n\{1 - (-1)^{n}\}$. The first function tends +to~$0$, the second to~$+\infty$, the third does not tend either to a limit or to~$+\infty$. + +\Item{7.} $\phi(n) = (\sin n\theta\pi)/n$, where $\theta$~is any real number. Here $|\phi(n)| < 1/n$, +since $|\sin n\theta\pi| \leq 1$, and $\lim\phi(n) = 0$. + +\Item{8.} $\phi(n) = (\sin n\theta\pi)/\sqrt{n}$, $(a\cos^{2} n\theta + b\sin^{2}n\theta)/n$, where $a$~and~$b$ are any real +numbers. + +\Item{9.} $\phi(n) = \sin n\theta\pi$. If $\theta$~is integral then $\phi(n) = 0$ for all values of~$n$, and +therefore $\lim\phi(n) = 0$. + +Next let $\theta$~be rational, \eg\ $\theta = p/q$, where $p$~and~$q$ are positive integers. +Let $n = aq + b$ where $a$~is the quotient and $b$~the remainder when $n$~is divided +by~$q$. Then $\sin(np\pi/q) = (-1)^{ap}\sin(bp\pi/q)$. Suppose, for example, $p$~even; +then, as $n$~increases from~$0$ to~$q - 1$, $\phi(n)$~takes the values +\[ +0,\quad +\sin(p\pi/q),\quad +\sin(2p\pi/q),\ \dots\quad +\sin\{(q - 1)p\pi/q\}. +\] +When $n$~increases from~$q$ to~$2q - 1$ these values are repeated; and so also +as $n$~goes from $2q$~to~$3q - 1$, $3q$~to~$4q - 1$, and so on. Thus the values of~$\phi(n)$ +form \emph{a perpetual cyclic repetition of a finite series of different values}. It is +evident that when this is the case $\phi(n)$~cannot tend to a limit, nor to~$+\infty$, +nor to~$-\infty$, as $n$~tends to infinity. + +The case in which $\theta$~is irrational is a little more difficult. It is discussed +in the next set of examples. +\end{Examples} + +\Paragraph{62. Oscillating Functions.} + \begin{Definition} +When $\phi(n)$ does +not tend to a limit, nor to~$+\infty$, nor to~$-\infty$, as $n$~tends to~$\infty$, we +say that $\phi(n)$ \Emph{oscillates} as $n$~tends to~$\infty$. +\end{Definition} + +A function $\phi(n)$ certainly oscillates if its values form, as +in the case considered in the last example above, a continual +repetition of a cycle of values. But of course it may oscillate +without possessing this peculiarity. Oscillation is defined in a +purely negative manner: a function oscillates when it does not do +certain other things. +\PageSep{122} + +The simplest example of an oscillatory function is given by +\[ +\phi(n) = (-1)^{n}, +\] +which is equal to~$+1$ when $n$~is even and to~$-1$ when $n$~is odd. +In this case the values recur cyclically. But consider +\[ +\phi(n) = (-1)^{n} + (1/n), +\] +the values of which are +\[ +-1 + 1,\quad + 1 + (1/2),\quad +-1 + (1/3),\quad + 1 + (1/4),\quad +-1 + (1/5),\ \dots. +\] +When $n$~is large every value is nearly equal to~$+1$ or~$-1$, and +obviously $\phi(n)$~does not tend to a limit or to~$+\infty$ or to~$-\infty$, and +therefore it oscillates: but the values do not recur. It is to be +observed that in this case every value of~$\phi(n)$ is numerically less +than or equal to~$3/2$. Similarly +\[ +\phi(n) = (-1)^{n} 100 + (1000/n) +\] +oscillates. When $n$~is large, every value is nearly equal to~$100$ +or to~$-100$. The numerically greatest value is~$900$ (for $n = 1$). +But now consider $\phi(n) = (-1)^{n}n$, the values of which are $-1$, $2$, +$-3$, $4$, $-5$,~\dots. This function oscillates, for it does not tend to a +limit, nor to~$+\infty$, nor to~$-\infty$. And in this case we cannot assign +any limit beyond which the numerical value of the terms does +not rise. The distinction between these two examples suggests a +further definition. + +\begin{Definition} +If $\phi(n)$ oscillates as $n$~tends to~$\infty$, then $\phi(n)$~will +be said to \Emph{oscillate finitely} or \Emph{infinitely} according as it is or is not +possible to assign a number~$K$ such that all the values of~$\phi(n)$ are +numerically less than~$K$, \ie\ $|\phi(n)| < K$ for all values of~$n$. +\end{Definition} + +These definitions, as well as those of \SecNo[§§]{58}~and~\SecNo{60}, are further +illustrated in the following examples. + +\begin{Examples}{XXIV.} +Consider the behaviour as $n$~tends to~$\infty$ of the +following functions: + +\Item{1.} $(-1)^{n}$, $5 + 3(-1)^{n}$, $(1\MC000\MC000/n) + (-1)^{n}$, $1\MC000\MC000(-1)^{n} + (1/n)$. + +\Item{2.} $(-1)^{n}n$, $1\MC000\MC000 + (-1)^{n}n$. + +\Item{3.} $1\MC000\MC000 - n$, $(-1)^{n}(1\MC000\MC000 - n)$. + +\Item{4.} $n\{1 + (-1)^{n}\}$. In this case the values of~$\phi(n)$ are +\[ +0,\quad 4,\quad 0,\quad 8,\quad 0,\quad 12,\quad 0,\quad 16,\ \dots. +\] +The odd terms are all zero and the even terms tend to~$+\infty$: $\phi(n)$~oscillates +infinitely. +\PageSep{123} + +\Item{5.} $n^{2} + (-1)^{n}2n$. The second term oscillates infinitely, but the first is +very much larger than the second when $n$~is large. In fact $\phi(n) \geq n^{2} - 2n$ and +$n^{2} - 2n = (n - 1)^{2} - 1$ is greater than any assigned value~$\Delta$ if $n > 1 + \sqrtp{\Delta + 1}$. +Thus $\phi(n) \to +\infty$. It should be observed that in this case $\phi(2k + 1)$~is +always less than~$\phi(2k)$, so that the function progresses to infinity by a continual +series of steps forwards and backwards. It does not however `oscillate' +according to our definition of the term. + +\Item{6.} $n^{2}\{1 + (-1)^{n}\}$, $(-1)^{n}n^{2} + n$, $n^{3} + (-1)^{n}n^{2}$. + +\Item{7.} $\sin n\theta\pi$. We have already seen (\Exs{xxiii}.~9) that $\phi(n)$~oscillates +finitely when $\theta$~is rational, unless $\theta$~is an integer, when $\phi(n)= 0$, $\phi(n) \to 0$. + +The case in which $\theta$~is irrational is a little more difficult. But it is not +difficult to see that $\phi(n)$~still oscillates finitely. We can without loss of +generality suppose $0 < \theta < 1$. In the first place $|\phi(n)| < 1$. Hence $\phi(n)$~must +oscillate finitely or tend to a limit. We shall consider whether the +second alternative is really possible. Let us suppose that +\[ +\lim \sin n\theta\pi = l. +\] +{\Loosen Then, however small $\DELTA$ may be, we can choose~$n_{0}$ so that $\sin n\theta\pi$ lies between +$l - \DELTA$ and $l + \DELTA$ for all values of~$n$ greater than or equal to~$n_{0}$. Hence +$\sin(n + 1)\theta\pi - \sin n\theta\pi$ is numerically less than~$2\DELTA$ for all such values of~$n$, +and so $|\sin \frac{1}{2}\theta\pi \cos(n + \frac{1}{2})\theta\pi| < \DELTA$.} + +Hence +\[ +\cos(n + \tfrac{1}{2})\theta\pi + = \cos n\theta\pi \cos\tfrac{1}{2}\theta\pi + - \sin n\theta\pi \sin\tfrac{1}{2}\theta\pi +\] +must be numerically less than~$\DELTA/|\sin\frac{1}{2}\theta\pi|$. Similarly +\[ +\cos(n - \tfrac{1}{2})\theta\pi + = \cos n\theta\pi \cos\tfrac{1}{2}\theta\pi + + \sin n\theta\pi \sin\tfrac{1}{2}\theta\pi +\] +must be numerically less than~$\DELTA/|\sin\frac{1}{2}\theta\pi|$; and so each of $\cos n\theta\pi \cos\frac{1}{2}\theta\pi$, +$\sin n\theta\pi \sin\frac{1}{2}\theta\pi$ must be numerically less than $\DELTA/|\sin\frac{1}{2}\theta\pi|$. That is to say, +$\cos n\theta\pi \cos\frac{1}{2}\theta\pi$ is very small if $n$~is large, and this can only be the case +if $\cos n\theta\pi$ is very small. Similarly $\sin n\theta\pi$ must be very small, so that $l$~must +be zero. But it is impossible that $\cos n\theta\pi$ and $\sin n\theta\pi$ can \emph{both} be +very small, as the sum of their squares is unity. Thus the hypothesis that +$\sin n\theta\pi$ tends to a limit~$l$ is impossible, and therefore $\sin n\theta\pi$ oscillates +as $n$~tends to~$\infty$. + +{\Loosen The reader should consider with particular care the argument +`$\cos n\theta\pi \cos\frac{1}{2}\theta\pi$ is very small, and this can only be the case if $\cos n\theta\pi$ +is very small'. Why, he may ask, should it not be the other factor $\cos\frac{1}{2}\theta\pi$ +which is `very small'? The answer is to be found, of course, in the meaning +of the phrase `very small' as used in this connection. When we say `$\phi(n)$~is +very small' for large values of~$n$, we mean that we can choose~$n_{0}$ so +that $\phi(n)$~is numerically smaller than \emph{any} assigned number, if \DPchg{$n$~is sufficiently +large}{$n \geq n_{0}$}. Such an assertion is palpably absurd when made of a \emph{fixed} number +such as~$\cos\frac{1}{2}\theta\pi$, which is not zero.} + +Prove similarly that $\cos n\theta\pi$ oscillates finitely, unless $\theta$~is an even integer. + +\Item{8.} $\sin n\theta\pi + (1/n)$, $\sin n\theta\pi + 1$, $\sin n\theta\pi + n$, $(-1)^{n} \sin n\theta\pi$. + +\Item{9.} $a\cos n\theta\pi + b\sin n\theta\pi$, $\sin^{2}n\theta\pi$, $a\cos^{2}n\theta\pi + b\sin^{2}n\theta\pi$. +\PageSep{124} + +\Item{10.} $a + bn + (-1)^{n} (c + dn) + e\cos n\theta\pi + f\sin n\theta\pi$. + +\Item{11.} $n\sin n\theta\pi$. If $\DPtypo{n}{\theta}$ is integral, then $\phi(n) = 0$, $\phi(n) \to 0$. If $\theta$~is rational +but not integral, or irrational, then $\phi(n)$~oscillates infinitely. + +\Item{12.} $n(a\cos^{2} n\theta\pi + b\sin^{2} n\theta\pi)$. In this case $\phi(n)$~tends to~$+\infty$ if $a$~and~$b$ +are both positive, but to~$-\infty$ if both are negative. Consider the special +cases in which $a = 0$, $b > 0$, or $a > 0$, $b = 0$, or $a = 0$, $b = 0$. If $a$~and~$b$ have +opposite signs $\phi(n)$~generally oscillates infinitely. Consider any exceptional +cases. + +\Item{13.} $\sin(n^{2}\theta\pi)$. If $\theta$~is integral, then $\phi(n) \to 0$. Otherwise $\phi(n)$~oscillates +finitely, as may be shown by arguments similar to though more complex +than those used in \Exs{xxiii}.~9 and \Exs[]{xxiv}.~7.\footnote + {See Bromwich's \textit{Infinite Series}, p.~485.} + +\Item{14.} $\sin(n!\, \theta\pi)$. If $\theta$~has a rational value~$p/q$, then $n!\, \theta$~is certainly +integral for all values of $n$ greater than or equal to~$q$. Hence $\phi(n) \to 0$. The +case in which $\theta$~is irrational cannot be dealt with without the aid of considerations +of a much more difficult character. + +\Item{15.} $\cos(n!\, \theta\pi)$, $a\cos^{2}(n!\, \theta\pi) + b\sin^{2}(n!\, \theta\pi)$, where $\theta$~is rational. + +\Item{16.} $an - [bn]$, $(-1)^{n}(an - [bn])$. + +\Item{17.} $[\sqrt{n}]$, $(-1)^{n}[\sqrt{n}]$, $\sqrt{n} - [\sqrt{n}]$. + +\Item{18.} \emph{The smallest prime factor of~$n$}. When $n$~is a prime, $\phi(n) = n$. When +$n$~is even, $\phi(n) = 2$. Thus $\phi(n)$ oscillates infinitely. + +\Item{19.} \emph{The largest prime factor of~$n$}. + +\Item{20.} \emph{The number of days in the year~$n$~\textsc{a.d.}} +\end{Examples} + +\begin{Examples}{XXV.} +\Item{1.} If $\phi(n) \to +\infty$ and $\psi(n) \geq \phi(n)$ for all values of~$n$, +then $\psi(n) \to +\infty$. + +\Item{2.} If $\phi(n) \to 0$, and $|\psi(n)| \leq |\phi(n)|$ for all values of~$n$, then $\psi(n) \to 0$. + +\Item{3.} If $\lim |\phi(n)| = 0$, then $\lim \phi(n) = 0$. + +\Item{4.} If $\phi(n)$ tends to a limit or oscillates finitely, and $|\psi(n)| \leq |\phi(n)|$ when +$n \geq n_{0}$, then $\psi(n)$~tends to a limit or oscillates finitely. + +\Item{5.} If $\phi(n)$ tends to~$+\infty$, or to~$-\infty$, or oscillates infinitely, and +\[ +|\psi(n)| \geq |\phi(n)| +\] +when $n \geq n_{0}$, then $\psi(n)$~tends to~$+\infty$ or to~$-\infty$ or oscillates infinitely. + +\Item{6.} `If $\phi(n)$~oscillates and, however great be~$n_{0}$, we can find values of~$n$ +greater than~$n_{0}$ for which $\psi(n) > \phi(n)$, and values of~$n$ greater than~$n_{0}$ for +which $\psi(n) < \phi(n)$, then $\psi(n)$ oscillates'. Is this true? If not give an +example to the contrary. + +\Item{7.} If $\phi(n) \to l$ as $n \to \infty$, then also $\phi(n + p) \to l$, $p$~being any fixed integer. +[This follows at once from the definition. Similarly we see that if $\phi(n)$~tends +to~$+\infty$ or~$-\infty$ or oscillates so also does~$\phi(n + p)$.] + +\Item{8.} The same conclusions hold (except in the case of oscillation) if $p$~varies +with~$n$ but is always numerically less than a fixed positive integer~$N$; or if $p$~varies +with~$n$ in any way, so long as it is always positive. +\PageSep{125} + +\Item{9.} Determine the least value of~$n_{0}$ for which it is true that +\[ +\Item{(\ia)}\ n^{2} + 2n > 999\MC999\quad (n \geq n_{0}),\qquad +\Item{(\ib)}\ n^{2} + 2n > 1\MC000\MC000\quad (n \geq n_{0}). +\] + +\Item{10.} Determine the least value of~$n_{0}$ for which it is true that +\[ +\Item{(\ia)}\ n + (-1)^{n} > 1000\quad (n \geq n_{0}),\qquad +\Item{(\ib)}\ n + (-1)^{n} > 1\MC000\MC000\quad (n \geq n_{0}). +\] + +\Item{11.} Determine the least value of~$n_{0}$ for which it is true that +\[ +\Item{(\ia)}\ n^{2} + 2n > \Delta\quad (n \geq n_{0}),\qquad +\Item{(\ib)}\ n + (-1)^{n} > \Delta\quad (n \geq n_{0}), +\] +$\Delta$~being any positive number. + +[(\ia)~$n_{0} = [\sqrtp{\Delta + 1}]$: (\ib)~$n_{0} = 1 + [\Delta]$ or $2 + [\Delta]$, according as $[\Delta]$~is odd or +even, \ie\ $n_{0} = 1 + [\Delta] + \frac{1}{2} \{1 + (-1)^{[\Delta]}\}$.] + +\Item{12.} Determine the least value of~$n_{0}$ such that +\[ +\Item{(\ia)}\ n/(n^{2} + 1) < .0001,\qquad +\Item{(\ib)}\ (1/n) + \{(-1)^{n}/n^{2}\} < .000\MS01, +\] +when $n \geq n_{0}$. [Let us take the latter case. In the first place +\[ +(1/n) + \{(-1)^{n}/n^{2}\} \leq (n + 1)/n^{2}, +\] +and it is easy to see that the least value of~$n_{0}$, such that $(n + 1)/n^{2} < .000\MS001$ +when $n \geq n_{0}$, is~$1\MC000\MC002$. But the inequality given is satisfied by $n = 1\MC000\MC001$, +and this is the value of~$n_{0}$ required.] +\end{Examples} + +\Paragraph{63. Some general theorems with regard to limits.} +\Topic{\Item{A.} The behaviour of the sum of two functions whose +behaviour is known.} + +\begin{Theorem}[I.] If $\phi(n)$~and~$\psi(n)$ tend to limits $a$,~$b$, then +$\phi(n) + \psi(n)$ tends to the limit $a + b$. +\end{Theorem} + +This is almost obvious.\footnote + {There is a certain ambiguity in this phrase which the reader will do well to + notice. When one says `such and such a theorem is almost obvious' one may + mean one or other of two things. One may mean `it is difficult to doubt the truth + of the theorem', `the theorem is such as common-sense instinctively accepts', as + it accepts, for example, the truth of the propositions `$2 + 2 = 4$' or `the base-angles + of an isosceles triangle are equal'. That a theorem is `obvious' in this sense does + not prove that it is true, since the most confident of the intuitive judgments of + common sense are often found to be mistaken; and even if the theorem is true, + the fact that it is also `obvious' is no reason for not proving it, if a proof can be + found. The object of mathematics is to prove that certain premises imply certain + conclusions; and the fact that the conclusions may be as `obvious' as the premises + never detracts from the necessity, and often not even from the interest of the proof. + + But sometimes (as for example here) we mean by `this is almost obvious' + something quite different from this. We mean `a moment's reflection should not + only convince the reader of the truth of what is stated, but should also suggest to + him the general lines of a rigorous proof'. And often, when a statement is + `obvious' in this sense, one may well omit the proof, not because the proof is in + any sense unnecessary, but because it is a waste of time and space to state in detail + what the reader can easily supply for himself.} +The argument which the reader will +\PageSep{126} +at once form in his mind is roughly this: `when $n$~is large, $\phi(n)$~is +nearly equal to~$a$ and $\psi(n)$ to~$b$, and therefore their sum is nearly +equal to $a + b$'. It is well to state the argument quite formally, +however. + +Let $\DELTA$ be any assigned positive number (\eg\ $.001$, $.000\MS000\MS1$,~\dots). +We require to show that a number~$n_{0}$ can be found such that +\[ +|\phi(n) + \psi(n) - a - b| < \DELTA, +\Tag{(1)} +\] +when $n \geq n_{0}$. Now by a proposition proved in \Ref{Chap.}{III} (more +generally indeed than we need here) the modulus of the sum of +two numbers is less than or equal to the sum of their moduli. +Thus +\[ +|\phi(n) + \psi(n) - a - b| \leq |\phi(n) - a| + |\psi(n) - b|. +\] +It follows that the desired condition will certainly be satisfied if +$n_{0}$~can be so chosen that +\[ +|\phi(n) - a| + |\psi(n) - b| < \DELTA, +\Tag{(2)} +\] +when $n \geq n_{0}$. But this is certainly the case. For since $\lim\phi(n) = a$ +we can, by the definition of a limit, find~$n_{1}$ so that $|\phi(n) - a| < \DELTA'$ +when $n \geq n_{1}$, and this however small $\DELTA'$ may be. Nothing prevents +our taking $\DELTA' = \frac{1}{2}\DELTA$, so that $|\phi(n) - a| < \frac{1}{2}\DELTA$ when $n \geq n_{1}$. Similarly +we can find~$n_{2}$ so that $|\psi(n) - b| < \frac{1}{2}\DELTA$ when $n \geq n_{2}$. Now take $n_{0}$ +to be \emph{the greater of the two numbers $n_{1}$,~$n_{2}$}. Then $|\phi(n) - a| < \frac{1}{2}\DELTA$ +and $|\psi(n) - b| < \frac{1}{2}\DELTA$ when $n \geq n_{0}$, and therefore \Eq{(2)}~is satisfied and +the theorem is proved. + +\begin{Remark} +The argument may be concisely stated thus: since $\lim\phi(n) = a$ and +$\lim\psi(n) = b$, we can choose $n_{1}$,~$n_{2}$ so that +\[ +|\phi(n) - a| < \tfrac{1}{2}\DELTA\quad (n \geq n_{1}),\qquad +|\psi(n) - b| < \tfrac{1}{2}\DELTA\quad (n \geq n_{2}); +\] +and then, if $n$~is not less than either $n_{1}$~or~$n_{2}$, +\[ +|\phi(n) + \psi(n) - a - b| + \leq |\phi(n) - a| + |\DPtypo{\phi}{\psi}(n) - b| < \DELTA; +\] +and therefore +\[ +\lim\{\phi(n) + \psi(n)\} = a + b. +\] +\end{Remark} + +\Paragraph{64. Results subsidiary to Theorem~I.} The reader should +have no difficulty in verifying the following subsidiary results. + +\begin{Result} +\Item{1.} If $\phi(n)$~tends to a limit, but $\psi(n)$~tends to~$+\infty$ or to~$-\infty$ +or oscillates finitely or infinitely, then $\phi(n) + \psi(n)$ behaves like~$\psi(n)$. +\end{Result} + +\begin{Result} +\Item{2.} {\Loosen If $\phi(n) \to +\infty$, and $\psi(n) \to +\infty$ or oscillates finitely, +then $\phi(n) + \psi(n) \to +\infty$.} +\end{Result} +\PageSep{127} + +In this statement we may obviously change $+\infty$ into~$-\infty$ +throughout. + +\begin{Result} +\Item{3.} If $\phi(n) \to \DPchg{\infty}{+\infty}$ and $\psi(n) \to -\infty$, then $\phi(n) + \psi(n)$ may +tend either to a limit or to~$+\infty$ or to~$-\infty$ or may oscillate either +finitely or infinitely. +\end{Result} + +\begin{Remark} +These five possibilities are illustrated in order by (i)~$\phi(n) = n$, $\psi(n) = -n$, +(ii)~$\phi(n) = n^{2}$, $\psi(n) = -n$, (iii)~$\phi(n) = n$, $\psi(n) = -n^{2}$, (iv)~$\phi(n) = n + (-1)^{n}$, +$\psi(n) = -n$, (v)~$\phi(n) = n^{2} + (-1)^{n}n$, $\psi(n) = -n^{2}$. The reader should construct +additional examples of each case. +\end{Remark} + +\begin{Result} +\Item{4.} If $\phi(n) \to +\infty$ and $\psi(n)$~oscillates infinitely, then +$\phi(n) + \psi(n)$ may tend to~$+\infty$ or oscillate infinitely, but cannot +tend to a limit, or to~$-\infty$, or oscillate finitely. +\end{Result} + +\begin{Remark} +For $\psi(n) = \{\phi(n) + \psi(n)\} - \phi(n)$; and, if $\phi(n) + \psi(n)$ behaved in any of the +three last ways, it would follow, from the previous results, that $\psi(n) \to -\infty$, +which is not the case. As examples of the two cases which are possible, +consider (i)~$\phi(n) = n^{2}$, $\psi(n) = (-1)^{n}n$, (ii)~$\phi(n) = n$, $\psi(n) = (-1)^{n}n^{2}$. Here +again the signs of~$+\infty$ and~$-\infty$ may be permuted throughout. +\end{Remark} + +\begin{Result} +\Item{5.} If $\phi(n)$ and $\psi(n)$ both oscillate finitely, then $\phi(n) + \psi(n)$ +must tend to a limit or oscillate finitely. +\end{Result} + +\begin{Remark} +As examples take +\[ +\Itemp{(i)} \phi(n) = (-1)^{n},\quad \psi(n) = (-1)^{n+1},\qquad +\Itemp{(ii)} \phi(n) = \psi(n) = (-1)^{n}. +\] +\end{Remark} + +\begin{Result} +\Item{6.} If $\phi(n)$ oscillates finitely, and $\psi(n)$~infinitely, then +$\phi(n) + \psi(n)$ oscillates infinitely. +\end{Result} + +\begin{Remark} +For $\phi(n)$ is in absolute value always less than a certain constant, say~$K$. +On the other hand $\psi(n)$, since it oscillates infinitely, must assume values +numerically greater than any assignable number (\eg\ $10K$, $100K$,~\dots). Hence +$\phi(n) + \psi(n)$ must assume values numerically greater than any assignable +number (\eg\ $9K$, $99K$,~\dots). Hence $\phi(n) + \psi(n)$ must either tend to~$+\infty$ or~$-\infty$ +or oscillate infinitely. But if it tended to~$+\infty$ then +\[ +\psi(n) = \{\phi(n) + \psi(n)\} - \phi(n) +\] +would also tend to~$+\infty$, in virtue of the preceding results. Thus $\phi(n) + \psi(n)$ +cannot tend to~$+\infty$, nor, for similar reasons, to~$-\infty$: hence it oscillates +infinitely. +\end{Remark} + +\begin{Result} +\Item{7.} If both $\phi(n)$ and $\psi(n)$ oscillate infinitely, then $\phi(n) + \psi(n)$ +may tend to a limit, or to~$+\infty$, or to~$-\infty$, or oscillate either finitely +or infinitely. +\end{Result} + +\begin{Remark} +Suppose, for instance, that $\phi(n) = (-1)^{n}n$, while $\psi(n)$~is in turn each of +the functions $(-1)^{n+1}n$, $\{1 + (-1)^{n+1}\}n$, $-\{1 + (-1)^{n}\}n$, $(-1)^{n+1}(n + 1)$, +$(-1)^{n}n$. We thus obtain examples of all five possibilities. +\end{Remark} +\PageSep{128} + +The results 1--7 cover all the cases which are really distinct. +Before passing on to consider the product of two functions, we +may point out that the result of Theorem~I may be immediately +extended to the sum of three or more functions which tend to +limits as $n\to\infty$. + +\Paragraph{65.} \Topic{\Item{B.} The behaviour of the product of two functions +whose behaviour is known.} We can now prove a similar +set of theorems concerning the product of two functions. The +principal result is the following. + +\begin{Theorem}[II.] +If $\lim\phi(n) = a$ and $\lim\psi(n) = b$, then +\[ +\lim\phi(n)\psi(n) = ab. +\] +\end{Theorem} + +Let +\[ +\phi(n) = a + \phi_{1}(n),\quad +\psi(n) = b + \psi_{1}(n), +\] +so that $\lim\phi_{1}(n) = 0$ and $\lim\psi_{1}(n) = 0$. Then +\[ +\phi(n)\psi(n) = ab + a\psi_{1}(n) + b\phi_{1}(n) + \phi_{1}(n)\psi_{1}(n). +\] +Hence the numerical value of the difference $\phi(n)\psi(n) - ab$ is +certainly not greater than the sum of the numerical values of +$a\psi_{1}(n)$, $b\phi_{1}(n)$, $\phi_{1}(n)\psi_{1}(n)$. From this it follows that +\[ +\lim\{\phi(n)\psi(n) - ab\} = 0, +\] +which proves the theorem. + +\begin{Remark} +The following is a strictly formal proof. We have +\[ +|\phi(n)\psi(n) - ab| + \leq |a\psi_{1}(n)| + |b\phi_{1}(n)| + |\phi_{1}(n)| |\psi_{1}(n)|. +\] +Assuming that neither $a$~nor~$b$ is zero, we may choose~$n_{0}$ so that +\[ +|\phi_{1}(n)| < \tfrac{1}{3}\DELTA/|b|,\quad +|\psi_{1}(n)| < \tfrac{1}{3}\DELTA/|a|, +\] +when $n \geq n_{0}$. Then +\[ +|\phi(n)\psi(n) - ab| + < \tfrac{1}{3}\DELTA + + \tfrac{1}{3}\DELTA + + \{\tfrac{1}{9}\DELTA^{2}/(|a||b|)\}, +\] +which is certainly less than~$\DELTA$ if $\DELTA < \frac{1}{3}|a||b|$. That is to say we can choose~$n_{0}$ +so that $|\phi(n)\psi(n) - ab| < \DELTA$ when $n \geq n_{0}$, and so the theorem follows. The +reader should supply a proof for the case in which at least one of $a$~and~$b$ is +zero. +\end{Remark} + +We need hardly point out that this theorem, like Theorem~I, +may be immediately extended to the product of any number of +functions of~$n$. There is also a series of subsidiary theorems +concerning products analogous to those stated in \SecNo[§]{64} for sums. +We must distinguish now \emph{six} different ways in which $\phi(n)$~may +behave as $n$~tends to~$\infty$. It may (1)~tend to a limit \emph{other than +\PageSep{129} +zero}, (2)~tend to zero, (3\ia)~tend to~$+\infty$, (3\ib)~tend to~$-\infty$, +(4)~oscillate finitely, (5)~oscillate infinitely. It is not necessary, as +a rule, to take account separately of (3\ia)~and~(3\ib), as the results +for one case may be deduced from those for the other by a change +of sign. + +\begin{Remark} +To state these subsidiary theorems at length would occupy more space +than we can afford. We select the two which follow as examples, leaving the +verification of them to the reader. He will find it an instructive exercise to +formulate some of the remaining theorems himself. + +\begin{Result} +\Itemp{(i)} If $\phi(n) \to +\infty$ and~$\psi(n)$~oscillates finitely, then $\phi(n)\psi(n)$ +must tend +to~$+\infty$ or to~$-\infty$ or oscillate infinitely. +\end{Result} + +Examples of these three possibilities may be obtained by taking $\phi(n)$ to +be~$n$ and $\psi(n)$ to be one of the three functions $2 + (-1)^{n}$, $-2 - (-1)^{n}$, $(-1)^{n}$. + +\begin{Result} +\Itemp{(ii)} If $\phi(n)$ and~$\psi(n)$ oscillate finitely, then $\phi(n)\psi(n)$ must tend to a +limit \(which may be zero\) or oscillate finitely. +\end{Result} + +{\Loosen For examples, take (\ia)~$\phi(n) = \psi(n) = (-1)^{n}$, (\ib)~$\phi(n) = 1 + (-1)^{n}$, +$\psi(n) = 1 - (-1)^{n}$, and (\ic)~$\phi(n) = \cos\frac{1}{3}n\pi$, $\psi(n) = \sin\tfrac{1}{3} n\pi$.} +\end{Remark} + +A particular case of Theorem~II which is important is that +in which $\psi(n)$~is constant. The theorem then asserts simply +that $\lim k\phi(n) = ka$ if $\lim\phi(n) = a$. To this we may join the +subsidiary theorem that if $\phi(n) \to +\infty$ then $k\phi(n) \to +\infty$ or +$k\phi(n) \to -\infty$, according as $k$~is positive or negative, unless $k = 0$, +when of course $k\phi(n) = 0$ for all values of~$n$ and $\lim k\phi(n) = 0$. +And if $\phi(n)$~oscillates finitely or infinitely, then so does $k\phi(n)$, +unless $k = 0$. + +\Paragraph{66.} \Topic{\Item{C.} The behaviour of the difference or quotient of +two functions whose behaviour is known.} There is, of +course, a similar set of theorems for the difference of two given +functions, which are obvious corollaries from what precedes. In +order to deal with the quotient +\[ +\frac{\phi(n)}{\psi(n)}, +\] +we begin with the following theorem. + +\begin{Theorem}[III.] +If $\lim\phi(n) = a$, and $a$~is not zero, then +\[ +\lim\frac{1}{\phi(n)} = \frac{1}{a}. +\] +\end{Theorem} + +Let +\[ +\phi(n) = a + \phi_{1}(n), +\] +\PageSep{130} +so that $\lim\phi_{1}(n) = 0$. Then +\[ +\left|\frac{1}{\phi(n)} - \frac{1}{a}\right| + = \frac{|\phi_{1}(n)|}{|a| |a + \phi_{1}(n)|}, +\] +and it is plain, since $\lim\phi_{1}(n) = 0$, that we can choose~$n_{0}$ so that +this is smaller than any assigned number~$\DELTA$ when $n \geq n_{0}$. + +From Theorems II~and~III we can at once deduce the principal +theorem for quotients, viz.\ + +\begin{Theorem}[IV.] +If $\lim\phi(n) = a$ and $\lim\psi(n) = b$, and $b$~is not +zero, then +\[ +\lim\frac{\phi(n)}{\psi(n)} = \frac{a}{b}. +\] +\end{Theorem} + +The reader will again find it instructive to formulate, prove, +and illustrate by examples some of the `subsidiary theorems' +corresponding to Theorems III~and~IV. + +\Paragraph{67.} +\begin{Theorem}[V.] +If $R\{\phi(n), \psi(n), \chi(n), \dots\}$ is any rational +function of $\phi(n)$, $\psi(n)$, $\chi(n)$,~\dots, \ie\ any function of the form +\[ +P\{\phi(n), \psi(n), \chi(n), \dots\}/Q\{\phi(n), \psi(n), \chi(n), \dots\}, +\] +where $P$~and~$Q$ denote polynomials in $\phi(n)$, $\psi(n)$, $\chi(n)$,~\dots: and if +\[ +\lim\phi(n) = a,\quad +\lim\psi(n) = b,\quad +\lim\chi(n) = c,\ \dots, +\] +and +\[ +Q(a, b, c, \dots) \neq 0; +\] +then +\[ +\lim R\{\phi(n), \psi(n), \chi(n), \dots\} = R(a, b, c, \dots). +\] +\end{Theorem} + +For $P$~is a sum of a finite number of terms of the type +\[ +A\{\phi(n)\}^{p} \{\psi(n)\}^{q} \dots, +\] +where $A$~is a constant and $p$,~$q$,~\dots\ positive integers. This term, +by Theorem~II (or rather by its obvious extension to the product +of any number of functions) tends to the limit $Aa^{p}b^{q}\dots$, and so $P$~tends +to the limit $P(a, b, c, \dots)$, by the similar extension of +Theorem~I\@. Similarly $Q$~tends to $Q(a, b, c, \dots)$; and the result +then follows from Theorem~IV. + +\Paragraph{68.} The preceding general theorem may be applied to the +following very important particular problem: \emph{what is the behaviour +of the most general rational function of~$n$, viz. +\[ +S(n) = \frac{a_{0}n^{p} + a_{1}n^{p-1} + \dots + a_{p}} + {b_{0}n^{q} + b_{1}n^{q-1} + \dots + b_{q}}, +\] +as $n$~tends to~$\infty$?}\footnote + {We naturally suppose that neither $a_{0}$~nor~$b_{0}$ is zero.} +\PageSep{131} + +In order to apply the theorem we transform $S(n)$ by writing +it in the form +\[ +n^{p-q}\left\{ + \biggl(a_{0} + \frac{a_{1}}{n} + \dots + \frac{a_{p}}{n^{p}}\biggr)\bigg/ + \biggl(b_{0} + \frac{b_{1}}{n} + \dots + \frac{b_{q}}{n^{q}}\biggr) +\right\}. +\] +The function in curly brackets is of the form $R\{\phi(n)\}$, where +$\phi(n) = 1/n$, and therefore tends, as $n$~tends to~$\infty$, to the limit +$R(0) = a_{0}/b_{0}$. Now $n^{p-q} \to 0$ if $p < q$; $n^{p-q} = 1$ and $n^{p-q} \to 1$ if +$p = q$; and $n^{p-q} \to +\infty$ if $p > q$. Hence, by Theorem~II, +\begin{gather*} +\lim S(n) = 0\quad (p < q), \\ +\lim S(n) = a_{0}/b_{0}\quad (p = q), \\ +S(n) \to +\infty\quad (p > q,\ \text{$a_{0}/b_{0}$ \emph{positive}}), \\ +S(n) \to -\infty\quad (p > q,\ \text{$a_{0}/b_{0}$ \emph{negative}}). +\end{gather*} + +\begin{Examples}{XXVI.} +\Item{1.} What is the behaviour of the functions +\[ +\left(\frac{n - 1}{n + 1}\right)^{2},\quad +(-1)^{n} \left(\frac{n - 1}{n + 1}\right)^{2},\quad +\frac{n^{2} + 1}{n},\quad +(-1)^{n} \frac{n^{2} + 1}{n}, +\] +as $n\to\infty$? + +\Item{2.} Which (if any) of the functions +\begin{gather*} +1/(\cos^{2}\tfrac{1}{2}n\pi + n\sin^{2}\tfrac{1}{2}n\pi),\quad +1/\{n(\cos^{2}\tfrac{1}{2}n\pi + n\sin^{2}\tfrac{1}{2}n\pi)\}, \\ + (n\cos^{2}\tfrac{1}{2}n\pi + \sin^{2}\tfrac{1}{2}n\pi)/ +\{n(\cos^{2}\tfrac{1}{2}n\pi + n\sin^{2}\tfrac{1}{2}n\pi)\} +\end{gather*} +tend to a limit as $n \to \infty$? + +\Item{3.} Denoting by~$S(n)$ the general rational function of~$n$ considered above, +show that in all cases +\[ +\lim\frac{S(n + 1)}{S(n)} = 1,\quad +\lim\frac{S\{n + (1/n)\}}{S(n)} = 1. +\] +\end{Examples} + +\Paragraph{69. Functions of~$n$ which increase steadily with~$n$.} A +special but particularly important class of functions of~$n$ is formed +by those whose variation as $n$~tends to~$\infty$ is always in the same +direction, that is to say those which always increase (or always +decrease) as $n$~increases. Since $-\phi(n)$ always increases if $\phi(n)$ +always decreases, it is not necessary to consider the two kinds of +functions separately; for theorems proved for one kind can at +once be extended to the other. + +\begin{Definition} The function $\phi(n)$ will be said to increase steadily +with~$n$ if $\phi(n + 1) \geq \phi(n)$ for all values of~$n$. +\end{Definition} +\PageSep{132} + +It is to be observed that we do not exclude the case in which +$\phi(n)$ has the \emph{same} value for several values of~$n$; all we exclude is +possible \emph{decrease}. Thus the function +\[ +\phi(n) = 2n + (-1)^{n}, +\] +whose values for $n = 0$, $1$, $2$, $3$, $4$,~\dots\ are +\[ +1,\ 1,\ 5,\ 5,\ 9,\ 9,\ \dots +\] +is said to increase steadily with~$n$. Our definition would indeed +include even functions which remain constant from some value of~$n$ +onwards; thus $\phi(n) = 1$ steadily increases according to our definition. +However, as these functions are extremely special ones, and as +there can be no doubt as to their behaviour as $n$~tends to~$\infty$, this +apparent incongruity in the definition is not a serious defect. + +There is one exceedingly important theorem concerning +functions of this class. + +\begin{Theorem} +If $\phi(n)$ steadily increases with~$n$, then either +\Inum{(i)}~$\phi(n)$ tends to a limit as $n$~tends to~$\infty$, or \Inum{(ii)}~$\phi(n)\to +\infty$. +\end{Theorem} + +That is to say, while there are in general \emph{five} alternatives as to +the behaviour of a function, there are \emph{two} only for this special +kind of function. + +This theorem is a simple corollary of Dedekind's Theorem +(\SecNo[§]{17}). We divide the real numbers~$\xi$ into two classes $L$~and~$R$, +putting $\xi$~in $L$~or~$R$ according as $\phi(n) \geq \xi$ for some value of~$n$ +(and so of course for all greater values), or $\phi(n) < \xi$ for all +values of~$n$. + +The class~$L$ certainly exists; the class~$R$ may or may not. +If it does not, then, given any number~$\Delta$, however large, $\phi(n) > \Delta$ +for all sufficiently large values of~$n$, and so +\[ +\phi(n) \to +\infty. +\] + +If on the other hand $R$~exists, the classes $L$~and~$R$ form a +section of the real numbers in the sense of \SecNo[§]{17}. Let $a$~be the +number corresponding to the section, and let $\DELTA$~be any positive +number. Then $\phi(n) < a + \DELTA$ for all values of~$n$, and so, since $\DELTA$~is +arbitrary, $\phi(n) \leq a$. On the other hand $\phi(n) > a - \DELTA$ for some +value of~$n$, and so for all sufficiently large values. Thus +\[ +a - \DELTA < \phi(n) \leq a +\] +\PageSep{133} +for all sufficiently large values of~$n$; \ie\ +\[ +\phi(n)\to a. +\] + +\begin{Remark} +It should be observed that in general $\phi(n) < a$ for all values of~$n$; for if +$\phi(n)$~is equal to~$a$ for any value of~$n$ it must be equal to~$a$ for all greater +values of~$n$. Thus $\phi(n)$~can never be equal to~$a$ except in the case in which +the values of~$\phi(n)$ are ultimately all the same. If this is so, $a$~is the largest +member of~$L$; otherwise $L$~has no largest member. +\end{Remark} + +\begin{Cor}[1.] +If $\phi(n)$ increases steadily with~$n$, then it will tend to a +limit or to~$+\infty$ according as it is or is not possible to find a number~$K$ +such that $\phi(n) < K$ for all values of~$n$. +\end{Cor} + +We shall find this corollary exceedingly useful later on. + +\begin{Cor}[2.] +If $\phi(n)$ increases steadily with~$n$, and $\phi(n) < K$ for +all values of~$n$, then $\phi(n)$~tends to a limit and this limit is less than +or equal to~$K$. +\end{Cor} + +{\Loosen It should be noticed that the limit may be equal to~$K$: if \eg\ +$\phi(n) = 3 - (1/n)$, then every value of~$\phi(n)$ is less than~$3$, but the +limit is equal to~$3$.} + +\begin{Cor}[3.] +If $\phi(n)$ increases steadily with~$n$, and tends to a limit, +then +\[ +\phi(n) \leq \lim\phi(n) +\] +for all values of~$n$. +\end{Cor} + +The reader should write out for himself the corresponding +theorems and corollaries for the case in which $\phi(n)$~\emph{decreases} as $n$~increases. + +\Paragraph{70.} The great importance of these theorems lies in the fact +that they give us (what we have so far been without) a means of +deciding, in a great many cases, whether a given function of~$n$ +does or does not tend to a limit as $n \to \infty$, \emph{without requiring us to +be able to guess or otherwise infer beforehand what the limit is}. If +we know what the limit, if there is one, must be, we can use the +test +\[ +|\phi(n) - l| < \DELTA\quad (n \geq n_{0}): +\] +as for example in the case of $\phi(n) = 1/n$, where it is obvious that +the limit can only be zero. But suppose we have to determine +whether +\[ +\phi(n) = \left(1 + \frac{1}{n}\right)^{n} +\] +\PageSep{134} +tends to a limit. In this case it is not obvious what the limit, if +there is one, will be: and it is evident that the test above, which +involves~$l$, cannot be used, at any rate directly, to decide whether +$l$~exists or not. + +\begin{Remark} +Of course the test can sometimes be used indirectly, to prove by means of +a \textit{reductio ad absurdum} that $l$~\emph{cannot} exist. If \eg\ $\phi(n) = (-1)^{n}$, it is clear +that $l$~would have to be equal to~$1$ and also equal to~$-1$, which is obviously +impossible. + +\Paragraph{71. Alternative proof of Weierstrass's Theorem of \SecNo[§]{19}.} The results +of \SecNo[§]{69} enable us to give an alternative proof of the important theorem +proved in \SecNo[§]{19}. + +If we divide~$PQ$ into two equal parts, one at least of them must contain +infinitely many points of~$S$. We select the one which does, or, if both do, we +select the left-hand half; and we denote the selected half by~$P_{1}Q_{1}$ (\Fig{28}). +If $P_{1}Q_{1}$ is the left-hand half, $P_{1}$~is the same point as~$P$. +%[Illustration: Fig. 28.] +\Figure[0.9\textwidth]{28}{p134} + +Similarly, if we divide $P_{1}Q_{1}$ into two halves, one at least of them must +contain infinitely many points of~$S$. We select the half $P_{2}Q_{2}$ which does so, +or, if both do so, we select the left-hand half. Proceeding in this way we can +define a sequence of intervals +\[ +PQ,\quad P_{1}Q_{1},\quad P_{2}Q_{2},\quad P_{3}Q_{3},\ \dots, +\] +each of which is a half of its predecessor, and each of which contains infinitely +many points of~$S$. + +The points $P$, $P_{1}$, $P_{2}$,~\dots\ progress steadily from left to right, and so $P_{n}$~tends +to a limiting position~$T$. Similarly $Q_{n}$~tends to a limiting position~$T'$. +But $TT'$~is plainly less than~$P_{n}Q_{n}$, whatever the value of~$n$; and $P_{n}Q_{n}$, being +equal to~$PQ/2^{n}$, tends to zero. Hence $T'$~coincides with~$T$, and $P_{n}$~and~$Q_{n}$ +both tend to~$T$. + +Then $T$~is a point of accumulation of~$S$. For suppose that $\xi$~is its +coordinate, and consider any interval of the type $\DPmod{(\xi - \DELTA, \xi + \DELTA)}{[\xi - \DELTA, \xi + \DELTA]}$. If $n$~is +sufficiently large, $P_{n}Q_{n}$ will lie entirely inside this interval.\footnote + {This will certainly be the case as soon as $PQ/2^{n} < \DELTA$.} +Hence +$\DPmod{(\xi - \DELTA, \xi + \DELTA)}{[\xi - \DELTA, \xi + \DELTA]}$ contains infinitely many points of~$S$. +\end{Remark} + +\Paragraph{72. The limit of~$x^{n}$ as $n$~tends to~$\infty$.} Let us apply the +results of \SecNo[§]{69} to the particularly important case in which +$\phi(n) = x^{n}$. If $x = 1$ then $\phi(n) = 1$, $\lim\phi(n) = 1$, and if $x = 0$ then +$\phi(n) = 0$, $\lim \phi(n) = 0$, so that these special cases need not detain us. +\PageSep{135} + +First, suppose $x$ positive. Then, since $\phi(n + 1) = x\phi(n)$, $\phi(n)$ +increases with~$n$ if $x > 1$, decreases as $n$~increases if $x < 1$. + +{\Loosen If $x > 1$, then $x^{n}$~must tend either to a limit (which must +obviously be greater than~$1$) or to~$+\infty$. Suppose it tends to a +limit~$l$. Then $\lim\phi(n + 1) = \lim\phi(n) = l$, by \Exs{xxv}.~7; but} +\[ +\lim\phi(n + 1) = \lim x\phi(n) = x\lim\phi(n) = xl, +\] +and therefore $l = xl$: and as $x$~and~$l$ are both greater than~$1$, this +is impossible. Hence +\[ +x^{n} \to +\infty\quad +(x > 1). +\] + +\begin{Remark} +\Par{Example.} The reader may give an alternative proof, showing by the +binomial theorem that $x^{n} > 1 + n\delta$ if $\delta$~is positive and $x = 1 + \delta$, and so that +\[ +x^{n} \to +\infty. +\] +\end{Remark} + +On the other hand $x^{n}$~is a decreasing function if $x < 1$, and +must therefore tend to a limit or to~$-\infty$. Since $x^{n}$~is positive +the second alternative may be ignored. Thus $\lim x^{n} = l$, say, and +as above $l = xl$, so that $l$~must be zero. Hence +\[ +\lim x^{n} = 0\quad +(0 < x < 1). +\] + +\begin{Remark} +\Par{Example.} Prove as in the preceding example that $(1/x)^{n}$ tends to~$+\infty$ if +$0 < x < 1$, and deduce that $x^{n}$~tends to~$0$. +\end{Remark} + +{\Loosen We have finally to consider the case in which $x$~is negative. +If $-1 < x < 0$ and $x = -y$, so that $0 < y < 1$, then it follows from +what precedes that $\lim y^{n} = 0$ and therefore $\lim x^{n} = 0$. If $x = -1$ +it is obvious that $x^{n}$~oscillates, taking the values $-1$,~$1$ alternatively. +Finally if $x < -1$, and $x = -y$, so that $y > 1$, then $y^{n}$~tends +to~$+\infty$, and therefore $x^{n}$~takes values, both positive and negative, +numerically greater than any assigned number. Hence $x^{n}$~oscillates +infinitely. To sum up:} +\begin{alignat*}{2} +&\phi(n) = x^{n} \to +\infty &&(x > 1),\\ +&\lim \phi(n) = 1 &&(x = 1),\\ +&\lim \phi(n) = 0 &&(-1 < x < 1),\\ +&\text{$\phi(n)$ \emph{oscillates finitely}} &&(x = -1),\\ +&\text{$\phi(n)$ \emph{oscillates infinitely}}\qquad &&(x < -1). +\end{alignat*} + +\begin{Examples}{XXVII.\protect\footnotemark} +\Item{1.} If $\phi(n)$~is positive and $\phi(n + 1) > K \phi(n)$, where +$K > 1$, for all values of~$n$, then $\phi(n) \to +\infty$.\footnotetext + {These examples are particularly important and several of them will be made + use of later in the text. They should therefore be studied very carefully.} +\PageSep{136} + +[For +\[ +\phi(n) > K\phi(n - 1) > K^{2}\phi(n - 2) \dots > K^{n-1}\phi(1), +\] +from which the conclusion follows at once, as $K^{n} \to\infty$.] + +\Item{2.} The same result is true if the conditions above stated are satisfied +only when $n \geq n_{0}$. + +\Item{3.} If $\phi(n)$~is positive and $\phi(n + 1) < K\phi(n)$, where $0 < K < 1$, then +$\lim\phi(n) = 0$. This result also is true if the conditions are satisfied only when +$n \geq n_{0}$. + +\Item{4.} If $|\phi(n + 1)| < K|\phi(n)|$ when $n \geq n_{0}$, and $0 < K < 1$, then $\lim\phi(n) = 0$. + +\Item{5.} If $\phi(n)$ is positive and $\lim\{\phi(n + 1)\}/\{\phi(n)\} = l > 1$, then $\phi(n) \to +\infty$. + +[For we can determine~$n_{0}$ so that $\{\phi(n + 1)\}/\{\phi(n)\} > K > 1$ when $n \geq n_{0}$: we +may, \eg, take $K$ \DPchg{half-way}{halfway} between $1$~and~$l$. Now apply Ex.~1.] + +\Item{6.} If $\lim\{\phi(n + 1)\}/\{\phi(n)\} = l$, where $l$~is numerically less than unity, +then $\lim\phi(n) = 0$. [This follows from Ex.~4 as Ex.~5 follows from Ex.~1.] + +\Item{7.} Determine the behaviour, as $n \to \infty$, of $\phi(n) = n^{r}x^{n}$, where $r$~is any +positive integer. + +[If $x = 0$ then $\phi(n) = 0$ for all values of~$n$, and $\phi(n) \to 0$. In all other cases +\[ +\frac{\phi(n + 1)}{\phi(n)} = \left(\frac{n + 1}{n}\right)^{r}x \to x. +\] +First suppose $x$~positive. Then $\phi(n) \to +\infty$ if $x > 1$ (Ex.~5) and $\phi(n) \to 0$ if +$x < 1$ (Ex.~6). If $x = 1$, then $\phi(n) = n^{r} \to +\infty$. Next suppose $x$~negative. +Then $|\phi(n)| = n^{r}|x|^{n}$ tends to~$+\infty$ if $|x| \geq 1$ and to~$0$ if $|x| < 1$. Hence +$\phi(n)$~oscillates infinitely if $x \leq -1$ and $\phi(n) \to 0$ if $-1 < x < 0$.] + +\Item{8.} Discuss $n^{-r}x^{n}$ in the same way. [The results are the same, except +that $\phi(n) \to 0$ when $x = 1$ or~$-1$.] + +\Item{9.} Draw up a table to show how $n^{k}x^{n}$ behaves as $n \to \infty$, for all real +values of~$x$, and all positive and negative integral values of~$k$. + +[The reader will observe that the value of~$k$ is immaterial except in the +special cases when $x = 1$ or~$-1$. Since $\lim\{(n + 1)/n\}^{k} = 1$, whether $k$~be +positive or negative, the limit of the ratio $\phi(n + 1)/\phi(n)$ depends only on~$x$, +and the behaviour of~$\phi(n)$ is in general dominated by the factor~$x^{n}$. The +factor~$n^{k}$ only asserts itself when $x$~is numerically equal to~$1$.] + +\Item{10.} Prove that if $x$~is positive then $\sqrt[n]{x} \to 1$ as $n \to \infty$. [Suppose, \eg, $x > 1$. +Then $x$,~$\sqrt{x}$, $\sqrt[3]{x}$,~\dots\ is a decreasing sequence, and $\sqrt[n]{x} > 1$ for all values of~$n$. +Thus $\sqrt[n]{x} \to l$, where $l \geq 1$. But if $l > 1$ we can find values of~$n$, as large as +we please, for which $\sqrt[n]{x} > l$ or $x > l^{n}$; and, since $l^{n} \to +\infty$ as $n \to \infty$, this +is impossible.] + +\Item{11.} $\sqrt[n]{n}\to 1$. [For $\sqrtp[n+1]{n + 1} < \sqrt[n]{n}$ if +$(n + 1)^{n} < n^{n+1}$ or $\{1 + (1/n)\}^{n} < n$, +which is certainly satisfied if $n \geq 3$ (see \SecNo[§]{73} for a proof). Thus $\sqrt[n]{n}$~decreases +as $n$~increases from $3$ onwards, and, as it is always greater than unity, it tends +to a limit which is greater than or equal to unity. But if $\sqrt[n]{n}\to l$, where $l > 1$, +then $n > l^{n}$, which is certainly untrue for sufficiently large values of~$n$, +since $l^{n}/n \to +\infty$ with~$n$ (Exs.~7,~8).] +\PageSep{137} + +\Item{12.} $\sqrtp[n]{n!} \to +\infty$. [However large~$\Delta$ may be, $n! > \Delta^{n}$ if $n$~is large enough. +For if $u_{n} = \Delta^{n}/n!$ then $u_{n+1}/u_{n} = \Delta/(n + 1)$, which tends to zero as $n \to \infty$, so +that $u_{n}$~does the same (Ex.~6).] + +\Item{13.} Show that if $-1 < x < 1$ then +\[ +u_{n} = \frac{m(m - 1) \dots (m - n + 1)}{n!} x^{n} = \binom{m}{n} x^{n} +\] +tends to zero as $n \to \infty$. + +[If $m$~is a positive integer, $u_{n} = 0$ for $n > m$. Otherwise +\[ +\frac{u_{n+1}}{u_{n}} = \frac{m - n}{n + 1}x \to -x, +\] +unless $x = 0$.] +\end{Examples} + +\Paragraph{73. The limit of $\left(1 + \dfrac{1}{n}\right)^{n}$.} A more difficult problem which +can be solved by the help of \SecNo[§]{69} arises when $\phi(n) = \{1 + 1/n\}^{n}$. + +It follows from the binomial theorem\footnote + {The binomial theorem for a positive integral exponent, which is what is used + here, is a theorem of elementary algebra. The other cases of the theorem belong + to the theory of infinite series, and will be considered later.} +that +{\setlength{\multlinegap}{0pt}% +\begin{multline*} +\begin{aligned} +\biggl(1 + \frac{1}{n}\biggr)^{n} + &= 1 + n · \frac{1}{n} + \frac{n(n - 1)}{1·2}\, \frac{1}{n^{2}} + \dots + + \frac{n(n - 1)\dots (n - n + 1)}{1·2\dots n}\, \frac{1}{n^{n}}\\ + &= 1 + 1 + \frac{1}{1·2} \biggl(1 - \frac{1}{n}\biggr) + + \frac{1}{1·2·3} \biggl(1 - \frac{1}{n}\biggr) \biggl(1- \frac{2}{n}\biggr) + \dots\\ +\end{aligned} \\ + + \frac{1}{1·2\dots n} + \biggl(1 - \frac{1}{n}\biggr) + \biggl(1 - \frac{2}{n}\biggr)\dots + \biggl(1 - \frac{n - 1}{n}\biggr). +\end{multline*}} + +The $(p + 1)$th~term in this expression, viz. +\[ +\frac{1}{1·2\dots p} + \left(1 - \frac{1}{n}\right) + \left(1 - \frac{2}{n}\right)\dots + \left(1 - \frac{p - 1}{n}\right), +\] +is positive and an increasing function of~$n$, and the number +of terms also increases with~$n$. Hence $\left(1 + \dfrac{1}{n}\right)^{n}$ increases with~$n$, +and so tends to a limit or to~$+\infty$, as $n \to \infty$. + +But +\begin{align*} +\left(1 + \frac{1}{n}\right)^{n} + &< 1 + 1 + \frac{1}{1·2} + \frac{1}{1·2·3} + \dots + \frac{1}{1·2·3 \dots n}\\ + &< 1 + 1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n-1}} < 3. +\end{align*} + +Thus $\left(1 + \dfrac{1}{n}\right)^{n}$ cannot tend to~$+\infty$, and so +\[ +\lim_{n \to\infty} \left(1 + \frac{1}{n}\right)^{n} = e, +\] +where $e$~is a number such that $2 < e \leq 3$. +\PageSep{138} + +\begin{Remark} +\Paragraph{74. Some algebraical lemmas.} It will be convenient to prove at +this stage a number of elementary inequalities which will be useful to us +later on. + +\Itemp{(i)} It is evident that if $\alpha > 1$ and $r$~is a positive integer then +\[ +r\alpha^{r} > \alpha^{r-1} + \alpha^{r-2} + \dots + 1. +\] +Multiplying both sides of this inequality by $\alpha - 1$, we obtain +\[ +r\alpha^{r}(\alpha - 1) > \alpha^{r} - 1; +\] +and adding $r(\alpha^{r} - 1)$ to each side, and dividing by $r(r + 1)$, we obtain +\[ +\frac{\alpha^{r+1} - 1}{r + 1} > \frac{\alpha^{r} - 1}{r}\quad +(\alpha > 1). +\Tag{(1)} +\] +Similarly we can prove that +\[ +\frac{1 - \beta^{r+1}}{r + 1} < \frac{1 - \beta^{r}}{r}\quad +(0 < \beta < 1). +\Tag{(2)} +\] + +It follows that if $r$~and~$s$ are positive integers, and $r > s$, then +\[ +\frac{\alpha^{r} - 1}{r} > \frac{a^{s} - 1}{s},\quad +\frac{1 - \beta^{r}}{r} < \frac{1 - \beta^{s}}{s}. +\Tag{(3)} +\] +Here $0 < \beta < 1 < \alpha$. In particular, when $s = 1$, we have +\[ +\alpha^{r} - 1 > r(\alpha - 1),\quad +1 - \beta^{r} < r(1 - \beta). +\Tag{(4)} +\] + +\Itemp{(ii)} The inequalities \Eq{(3)}~and~\Eq{(4)} have been proved on the supposition +that $r$~and~$s$ are positive integers. But it is easy to see that they hold under +the more general hypothesis that $r$~and~$s$ are any positive rational numbers. +Let us consider, for example, the first of the inequalities~\Eq{(3)}. Let $r = a/b$, +$s = c/d$, where $a$,~$b$, $c$,~$d$ are positive integers; so that $ad > bc$. If we put +$\alpha = \gamma^{bd}$, the inequality takes the form +\[ +(\gamma^{ad} - 1)/ad > (\gamma^{bc} - 1)/bc; +\] +and this we have proved already. The same argument applies to the remaining +inequalities; and it can evidently be proved in a similar manner that +\[ +\alpha^{s} - 1 < s(\alpha - 1),\quad +1 - \beta^{s} > s(1 - \beta), +\Tag{(5)} +\] +if $s$~is a positive rational number less than~$1$. + +\Itemp{(iii)} In what follows it is to be understood \emph{that all the letters denote +positive numbers, that $r$~and~$s$ are rational, and that $\alpha$~and~$r$ are greater +than $1$,~$\beta$ and $s$~less than~$1$}. Writing $1/\beta$ for~$\alpha$, and $1/\alpha$ for~$\beta$, in~\Eq{(4)}, we +obtain +\[ +\alpha^{r} - 1 < r\alpha^{r-1}(\alpha - 1),\quad +1 - \beta^{r} > r\beta^{r-1}(1 - \beta). +\Tag{(6)} +\] +Similarly, from~\Eq{(5)}, we deduce +\[ +\alpha^{s} - 1 > s\alpha^{s-1}(\alpha - 1),\quad +1 - \beta^{s} < s\beta^{s-1}(1 - \beta). +\Tag{(7)} +\] + +Combining \Eq{(4)}~and~\Eq{(6)}, we see that +\[ +r\alpha^{r-1}(\alpha - 1) > \alpha^{r} - 1 > r(\alpha - 1). +\Tag{(8)} +\] +\PageSep{139} +Writing $x/y$ for~$\alpha$, we obtain +\[ +rx^{r-1} (x - y) > x^{r} - y^{r} > ry^{r-1} (x - y) +\Tag{(9)} +\] +if $x > y > 0$. And the same argument, applied to \Eq{(5)}~and~\Eq{(7)}, leads to +\[ +sx^{s-1} (x - y) < x^{s} - y^{s} < sy^{s-1} (x - y). +\Tag{(10)} +\] +\end{Remark} + +\begin{Examples}{XXVIII.} +\Item{1.} Verify \Eq{(9)} for $r = 2$,~$3$, and \Eq{(10)} for $s = \frac{1}{2}$,~$\frac{1}{3}$. + +\Item{2.} Show that \Eq{(9)}~and~\Eq{(10)} are also true if $y > x > 0$. + +\Item{3.} Show that \Eq{(9)}~also holds for $r < 0$. [See Chrystal's \textit{Algebra}, vol.~ii, +pp.~43--45.] + +\Item{4.} If $\phi(n) \to l$, where $l > 0$, as $n \to \infty$, then $\phi^{k} \to l^{k}$, $k$~being any rational number. + +[We may suppose that $k > 0$, in virtue of Theorem~III of \SecNo[§]{66}; and that +$\frac{1}{2}l < \phi < 2l$, as is certainly the case from a certain value of $n$ onwards. If +$k > 1$, +\[ +k\phi^{k-1}(\phi - l) > \phi^{k} - l^{k} > kl^{k-1}(\phi - l) +\] +or +\[ +kl^{k-1}(l - \phi) > l^{k} - \phi^{k} > k\phi^{k-1}(l - \phi), +\] +according as $\phi > l$ or $\phi < l$. It follows that the ratio of $|\phi^{k} - l^{k}|$ and $|\phi - l|$ +lies between $k(\frac{1}{2}l)^{k-1}$ and $k(2l)^{k-1}$. The proof is similar when $0 < k < 1$. The +result is still true when $l = 0$, if $k > 0$.] + +\Item{5.} Extend the results of \Exs{xxvii}.\ 7,~8,~9 to the case in which $r$~or~$k$ +are any rational numbers. +\end{Examples} + +\begin{Remark} +\Paragraph{75. The limit of $n(\sqrt[n]{x} - 1)$.} If in the first inequality~\Eq{(3)} of \SecNo[§]{74} we +put $r = 1/(n - 1)$, $s = 1/n$, we see that +\[ +(n - 1)(\sqrt[n-1]{\alpha} - 1) > n(\sqrt[n]{\alpha} - 1) +\] +when $\alpha > 1$. Thus if $\phi(n) = n(\sqrt[n]{\alpha} - 1)$ then $\phi(n)$~decreases steadily as $n$~increases. +Also $\phi(n)$~is always positive. Hence $\phi(n)$~tends to a limit~$l$ as +$n \to \infty$, and $l \geq 0$. + +Again if, in the first inequality~\Eq{(7)} of \SecNo[§]{74}, we put $s = 1/n$, we obtain +\[ +n(\sqrt[n]{\alpha} - 1) + > \sqrt[n]{\alpha}\left(1 - \frac{1}{\alpha}\right) + > 1 - \frac{1}{\alpha}. +\] +Thus $l \geq 1 - (1/\alpha) > 0$. Hence, if $\alpha > 1$, we have +\[ +\lim_{n \to \infty} n(\sqrt[n]{\alpha} - 1) = f(\alpha), +\] +where $f(\alpha) > 0$. + +Next suppose $\beta < 1$, and let $\beta = 1/\alpha$; then $n(\sqrt[n]{\beta} - 1) = -n(\sqrt{\alpha} - 1)/\sqrt[n]{\alpha}$. Now +$n(\sqrt[n]{\alpha} - 1) \to f(\alpha)$, and (\Exs{xxvii}.~10) +\[ +\sqrt[n]{\alpha} \to 1. +\] +Hence, if $\beta = 1/\alpha < 1$, we have +\[ +n(\sqrt[n]{\beta} - 1) \to -f(\alpha). +\] +Finally, if $x = 1$, then $n(\sqrt[n]{x} - 1) = 0$ for all values of $n$. +\PageSep{140} + +Thus we arrive at the result: \emph{the limit +\[ +\lim n(\sqrt[n]{x} - 1) +\] +defines a function of~$x$ for all positive values of~$x$. This function~$f(x)$ +possesses the properties +\[ +f(1/x) = -f(x),\quad f(1) = 0, +\] +and is positive or negative according as $x > 1$ or $x < 1$.} Later on we +shall be able to identify this function with the \emph{Napierian logarithm} of~$x$. + +\Par{Example.} Prove that $f(xy) = f(x) + f(y)$. [Use the equations +\[ +f(xy) = \lim n(\DPtypo{\sqrt[n]{xy}}{\sqrtp[n]{xy}} - 1) + = \lim \{n(\sqrt[n]{x} - 1)\sqrt[n]{y} + n(\sqrt[n]{y} - 1)\}.] +\] +\end{Remark} + +\Paragraph{76. Infinite Series.} Suppose that $u(n)$~is any function of~$n$ +defined for all values of~$n$. If we add up the values of~$u(\nu)$ +for $\nu = 1$, $2$,~\dots~$n$, we obtain another function of~$n$, viz. +\[ +s(n) = u(1) + u(2) + \dots + u(n), +\] +also defined for all values of~$n$. It is generally most convenient +to alter our notation slightly and write this equation in the form +\[ +s_{n} = u_{1} + u_{2} + \dots + u_{n}, +\] +or, more shortly, +\[ +s_{n} = \sum_{\nu=1}^{n} u_{\nu}. +\] + +If now we suppose that $s_{n}$~tends to a limit~$s$ when $n$~tends +to~$\infty$, we have +\[ +\lim_{n\to\infty} \sum_{\nu=1}^{n} u_{\nu} = s. +\] +This equation is usually written in one of the forms +\[ +\sum_{\nu=1}^{\infty} u_{\nu} = s,\quad +u_{1} + u_{2} + u_{3} + \dots = s, +\] +the dots denoting the indefinite continuance of the series of~$u$'s. + +The meaning of the above equations, expressed roughly, is +that by adding more and more of the~$u$'s together we get nearer +and nearer to the limit~$s$. More precisely, if any small positive +number~$\DELTA$ is chosen, we can choose~$n_{0}(\DELTA)$ so that the sum of the first +$n_{0}(\DELTA)$~terms, or any of greater number of terms, lies between $s - \DELTA$ +and $s + \DELTA$; or in symbols +\[ +s - \DELTA < s_{n} < s + \DELTA, +\] +if $n \geq n_{0}(\DELTA)$. In these circumstances we shall call the series +\[ +u_{1} + u_{2} + \dots +\] +a \Emph{convergent infinite series}, and we shall call~$s$ the \emph{sum} of the +series, or the \emph{sum of all the terms} of the series. +\PageSep{141} + +Thus to say that the series $u_{1} + u_{2} + \dots$ \emph{converges and has the +sum~$s$}, or \emph{converges to the sum~$s$} or simply \emph{converges to~$s$}, is merely +another way of stating that the sum $s_{n} = u_{1} + u_{2} + \dots + u_{n}$ of the +first $n$~terms tends to the limit~$s$ as $n \to \infty$, and the consideration +of such infinite series introduces no new ideas beyond those with +which the early part of this chapter should already have made +the reader familiar. In fact the sum~$s_{n}$ is merely a function~$\phi(n)$, +such as we have been considering, expressed in a particular form. +Any function~$\phi(n)$ may be expressed in this form, by writing +\[ +\phi(n) = \phi(1) + \{\phi(2) - \phi(1)\} + \dots + \{\phi(n) - \phi(n - 1)\}; +\] +and it is sometimes convenient to say that $\phi(n)$~\emph{converges} (instead +of `tends') to the limit~$l$, say, as $n \to \infty$. + +If $s_{n} \to +\infty$ or $s_{n} \to -\infty$, we shall say that the series $u_{1} + u_{2} + \dots$ +is \Emph{divergent} or \emph{diverges to~$+\infty$}, or~$-\infty$, as the case may be. +These phrases too may be applied to any function~$\phi(n)$: thus if +$\phi(n) \to +\infty$ we may say that \emph{$\phi(n)$~diverges to~$+\infty$}. If $s_{n}$~does +not tend to a limit or to~$+\infty$ or to~$-\infty$, then it oscillates finitely or +infinitely: in this case we say that the series $u_{1} + u_{2} + \dots$ oscillates +finitely or infinitely.\footnote + {The reader should be warned that the words `divergent' and `oscillatory' + are used differently by different writers. The use of the words here agrees with + that of Bromwich's \textit{Infinite Series}. In Hobson's \textit{Theory of Functions of a Real + Variable} a series is said to oscillate only if it oscillates \emph{finitely}, series which + oscillate infinitely being classed as `divergent'. Many foreign writers use `divergent' + as meaning merely `not convergent'.} + +\Paragraph{77. General theorems concerning infinite series.} When +we are dealing with infinite series we shall constantly have +occasion to use the following general theorems. + +\Item{(1)} If $u_{1} + u_{2} + \dots$ is convergent, and has the sum~$s$, then +$a + u_{1} + u_{2} + \dots$ is convergent and has the sum $a + s$. Similarly +$a + b + c + \dots + k + u_{1} + u_{2} + \dots$ is convergent and has the sum +$a + b + c + \dots + k + s$. + +\Item{(2)} {\Loosen If $u_{1} + u_{2} + \dots$ is convergent and has the sum~$s$, then +$u_{m+1} + u_{m+2} + \dots$ is convergent and has the sum} +\[ +s - u_{1} - u_{2} - \dots - u_{m}. +\] + +\Item{(3)} If any series considered in (1)~or~(2) diverges or oscillates, +then so do the others. + +\Item{(4)} If $u_{1} + u_{2} + \dots$ is convergent and has the sum~$s$, then +$ku_{1} + ku_{2} + \dots$ is convergent and has the sum~$ks$. +\PageSep{142} + +\Item{(5)} If the first series considered in~(4) diverges or oscillates, +then so does the second, unless $k = 0$. + +\Item{(6)} If $u_{1} + u_{2} + \dots$ and $v_{1} + v_{2} + \dots$ are both convergent, then +the series $(u_{1} + v_{1}) + (u_{2} + v_{2}) + \dots$ is convergent and its sum is the +sum of the first two series. + +{\Loosen All these theorems are almost obvious and may be proved at +once from the definitions or by applying the results of \SecNo[§§]{63}--\SecNo{66} to +the sum $s_{n} = u_{1} + u_{2} + \dots + u_{n}$. Those which follow are of a somewhat +different character.} + +\begin{Result} +\Item{(7)} If $u_{1} + u_{2} + \dots$ is convergent, then $\lim u_{n} = 0$. +\end{Result} + +For $u_{n} = s_{n} - s_{n-1}$, and $s_{n}$~and~$s_{n-1}$ have the same limit~$s$. +Hence $\lim u_{n} = s - s = 0$. + +\begin{Remark} +The reader may be tempted to think that the converse of the theorem is +true and that if $\lim u_{n} = 0$ then the series~$\sum u_{n}$ must be convergent. That this +is not the case is easily seen from an example. Let the series be +\[ +1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + \dots +\] +so that $u_{n} = 1/n$. The sum of the first four terms is +\[ +1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + > 1 + \tfrac{1}{2} + \tfrac{2}{4} = 1 + \tfrac{1}{2} + \tfrac{1}{2}. +\] +The sum of the next four terms is $\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{4}{8} = \frac{1}{2}$; the sum of the next +eight terms is greater than $\frac{8}{16} = \frac{1}{2}$, and so on. The sum of the first +\[ +4 + 4 + 8 + 16 + \dots + 2^{n} = 2^{n+1} +\] +terms is greater than +\[ +2 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2} + \dots + \tfrac{1}{2} + = \tfrac{1}{2} (n + 3), +\] +and this increases beyond all limit with~$n$: hence the series diverges to~$+\infty$. +\end{Remark} + +\begin{Result} +\Item{(8)} If $u_{1} + u_{2} + u_{3} + \dots$ is convergent, then so is any series +formed by grouping the terms in brackets in any way to form new +single terms, and the sums of the two series are the same. +\end{Result} + +\begin{Remark} +The reader will be able to supply the proof of this theorem. Here again +the converse is not true. Thus $1 - 1 + 1 - 1 + \dots$ oscillates, while +\[ +(1 - 1) + (1 - 1) + \dots +\] +or $0 + 0 + 0 + \dots$ converges to~$0$. +\end{Remark} + +\begin{Result} +\Item{(9)} If every term~$u_{n}$ is positive \(or zero\), then the series~$\sum u_{n}$ +must either converge or diverge to~$+\infty$. If it converges, its sum +must be positive {\upshape(unless all the terms are zero, when of course its +sum is zero)}. +\end{Result} + +For $s_{n}$~is an increasing function of~$n$, according to the definition +of \SecNo[§]{69}, and we can apply the results of that section to~$s_{n}$. +\PageSep{143} + +\begin{Result} +\Item{(10)} If every term~$u_{n}$ is positive \(or zero\), then the necessary +and sufficient condition that the series~$\sum u_{n}$ should be convergent is +that it should be possible to find a number~$K$ such that the sum of +any number of terms is less than~$K$; and, if $K$ can be so found, then +the sum of the series is not greater than~$K$. +\end{Result} + +This also follows at once from \SecNo[§]{69}. It is perhaps hardly +necessary to point out that the theorem is not true if the condition +that every~$u_{n}$ is positive is not fulfilled. For example +\[ +1 - 1 + 1 - 1 + \dots +\] +obviously oscillates, $s_{n}$~being alternately equal to~$1$ and to~$0$. + +\begin{Result} +\Item{(11)} If $u_{1} + u_{2} + \dots$, $v_{1} + v_{2} + \dots$ are two series of positive \(or +zero\) terms, and the second series is convergent, and if $u_{n} \leq Kv_{n}$, +where $K$~is a constant, for all values of~$n$, then the first series is also +convergent, and its sum is less than or equal to\DPtypo{}{ $K$~times} that of the second. +\end{Result} + +For if $v_{1} + v_{2} + \dots = t$ then $v_{1} + v_{2} + \dots + v_{n} \leq t$ for all values of~$n$, +and so $u_{1} + u_{2} + \dots + u_{n} \leq Kt$; which proves the theorem. + +\begin{Result} +Conversely, if $\sum u_{n}$ is divergent, and $v_{n} \geq Ku_{n}$, then $\sum v_{n}$~is +divergent. +\end{Result} + +\Paragraph{78. The infinite geometrical series.} We shall now consider +the `geometrical' series, whose general term is $u_{n} = r^{n-1}$. In +this case +\[ +s_{n} = 1 + r + r^{2} + \dots + r^{n-1} = (1 - r^{n})/(1 - r), +\] +except in the special case in which $r = 1$, when +\[ +s_{n} = 1 + 1 + \dots + 1 = n. +\] +In the last case $s_{n} \to +\infty$. In the general case $s_{n}$~will tend to a +limit if and only if $r^{n}$ does so. Referring to the results of \SecNo[§]{72} +we see that +\begin{Result} +the series $1 + r + r^{2} + \dots$ is convergent and has the sum +$1/(1 - r)$ if and only if $-1 < r < 1$. +\end{Result} + +If $r \geq 1$, then $s_{n} \geq n$, and so $s_{n} \to +\infty$; \ie\ the series diverges to~$+\infty$. +If $r = -1$, then $s_{n} = 1$ or $s_{n} = 0$ according as $n$~is odd or +even: \ie\ $s_{n}$~oscillates finitely. If $r < -1$, then $s_{n}$~oscillates infinitely. +Thus, to sum up, +\begin{Result} +the series $1 + r + r^{2} + \dots$ diverges to~$+\infty$ if $r \geq 1$, +converges to $1/(1 - r)$ if $-1 < r < 1$, oscillates finitely if $r = -1$, +and oscillates infinitely if $r < -1$. +\end{Result} + +\begin{Examples}{XXIX.} +\Item{1.} \Topic{Recurring decimals.} The commonest example +of an infinite geometric series is given by an ordinary recurring decimal. +\PageSep{144} +Consider, for example, the decimal $.217\DPmod{\dot{1}\dot{3}}{\Repeat{13}}$. This stands, according to the +ordinary rules of arithmetic, for +\[ +\frac{2}{10} + \frac{1}{10^{2}} + \frac{7}{10^{3}} + + \frac{1}{10^{4}} + \frac{3}{10^{5}} + \frac{1}{10^{6}} + \frac{3}{10^{7}} + + \dots + = \frac{217}{1000} + + \frac{13}{10^{5}} \bigg/ \left(1 - \frac{1}{10^{2}}\right) + = \frac{2687}{12\MC375}. +\] +The reader should consider where and how any of the general theorems of +\SecNo[§]{77} have been used in this reduction. + +\Item{2.} Show that in general +\[ +.a_{1}a_{2}\dots a_{m} \DPmod{\dot{\alpha}_{1}\alpha_{2}\dots\dot{\alpha}_{n}} + {\Repeat{\alpha_{1}\alpha_{2}\dots \alpha_{n}}} + = \frac{a_{1}a_{2}\dots a_{m}\alpha_{1}\dots \alpha_{n} - a_{1}a_{2}\dots a_{n}} + {99\dots 900\dots 0}, +\] +the denominator containing~$n$ $9$'s and $m$~$0$'s. + +\Item{3.} Show that a pure recurring decimal is always equal to a proper +fraction whose denominator does not contain $2$~or~$5$ as a factor. + +\Item{4.} A decimal with $m$~non-recurring and $n$~recurring decimal figures is +equal to a proper fraction whose denominator is divisible by $2^{m}$~or~$5^{m}$ but by +no higher power of either. + +\Item{5.} The converses of Exs.~3,~4 are also true. Let $r = p/q$, and suppose first +that $q$~is prime to~$10$. If we divide all powers of~$10$ by~$q$ we can obtain at most +$q$~different remainders. It is therefore possible to find two numbers $n_{1}$~and~$n_{2}$, +where $\DPtypo{n_{2} > n_{1}}{n_{1} > n_{2}}$, such that $10^{n_{1}}$ and $10^{n_{2}}$ give the same remainder. Hence +$10^{n_{1}} - 10^{n_{2}} = 10^{n_{2}}(10^{n_{1}-n_{2}} - 1)$ is divisible by~$q$, and so $10^{n} - 1$, where $n = n_{1} - n_{2}$, +is divisible by~$q$. Hence $r$~may be expressed in the form~$P/(10^{n} - 1)$, or in the +form +\[ +\frac{P}{10^{n}} + \frac{P}{10^{2n}} + \dots, +\] +\ie\ as a pure recurring decimal with $n$~figures. If on the other hand $q = 2^{\alpha}5^{\beta}Q$, +where $Q$~is prime to~$10$, and $m$~is the greater of $\alpha$~and~$\beta$, then $10^{m}r$~has a denominator +prime to~$10$, and is therefore expressible as the sum of an integer +and a pure recurring decimal. But this is not true of~$10^{\mu}r$, for any value of~$\mu$ +less than~$m$; hence the decimal for~$r$ has exactly~$m$ non-recurring figures. + +\Item{6.} To the results of Exs.~2--5 we must add that of \Ex{i}.~3. Finally, if +we observe that +\[ +.\DPmod{\dot{9}}{\Repeat{9}} + = \frac{9}{10} + \frac{9}{10^{2}} + \frac{9}{10^{3}} + \dots + = 1, +\] +we see that every terminating decimal can also be expressed as a mixed +recurring decimal whose recurring part is composed entirely of~$9$'s. For +example, $.217 = .216\DPmod{\dot{9}}{\Repeat{9}}$. Thus every proper fraction can be expressed as a +recurring decimal, and conversely. + +\Item{7.} \Topic{Decimals in general. The expression of irrational numbers as +non-recurring decimals.} Any decimal, whether recurring or not, corresponds +to a definite number between $0$~and~$1$. For the decimal $.a_{1}a_{2}a_{3}a_{4}\dots$ stands +for the series +\[ +\frac{a_{1}}{10} + \frac{a_{2}}{10^{2}} + \frac{a_{3}}{10^{3}} + \dots. +\] +\PageSep{145} +Since all the digits~$a_{r}$ are positive, the sum~$s_{n}$ of the first $n$~terms of this +series increases with~$n$, and it is certainly not greater than~$.\DPmod{\dot{9}}{\Repeat{9}}$ or~$1$. Hence $s_{n}$~tends to a limit between $0$~and~$1$. + +Moreover no two decimals can correspond to the same number (except in +the special case noticed in Ex.~6). For suppose that $.a_{1}a_{2}a_{3} \dots$, $.b_{1}b_{2}b_{3} \dots$ are +two decimals which agree as far as the figures $a_{r-1}$,~$b_{r-1}$, while $a_{r} > b_{r}$. +Then $a_{r}\geq b_{r} + 1 > b_{r}.b_{r+1}b_{r+2} \dots$ (unless $b_{r+1}$, $b_{r+2}$,~\dots\ are all~$9$'s), and so +\[ +.a_{1}a_{2} \dots a_{r}a_{r+1} \dots > .b_{1}b_{2} \dots b_{r}b_{r+1} \dots. +\] +It follows that the expression of a rational fraction as a recurring decimal +(Exs.\ 2--6) is unique. It also follows that every decimal which does not +recur represents some \emph{irrational} number between $0$~and~$1$. Conversely, any +such number can be expressed as such a decimal. For it must lie in one of +the intervals +\[ +0,\ 1/10;\quad 1/10,\ 2/10;\ \dots;\quad 9/10,\ 1. +\] +If it lies between $r/10$ and $(r + 1)/10$, then the first figure is~$r$. By subdividing +this interval into $10$~parts we can determine the second figure; and so on. +But (Exs.~3,~4) the decimal cannot recur. Thus, for example, the decimal +$1.414\dots$, obtained by the ordinary process for the extraction of~$\sqrt{2}$, cannot +recur. + +\Item{8.} The decimals $.101\MS001\MS000\MS100\MS001\MS0\dots$ and $.202\MS002\MS000\MS200\MS002\MS0\dots$, in +which the number of zeros between two~$1$'s or $2$'s increases by one at each +stage, represent irrational numbers. + +\Item{9.} The decimal $.111\MS010\MS100\MS010\MS10\dots$, in which the $n$th~figure is~$1$ if $n$~is +prime, and zero otherwise, represents an irrational number. [Since the +number of primes is infinite the decimal does not terminate. Nor can it +recur: for if it did we could determine $m$~and~$p$ so that $m$,~$m + p$, $m + 2p$, +$m + 3p$,~\dots\ are all prime numbers; and this is absurd, since the series includes +$m + mp$.]\footnote + {All the results of \Exs{xxix} may be extended, with suitable modifications, to + decimals in any scale of notation. For a fuller discussion see Bromwich, \textit{Infinite + Series}, Appendix~I.} +\end{Examples} + +\begin{Examples}{XXX.} +\Item{1.} {\Loosen The series $r^{m} + r^{m+1} + \dots$ is convergent if $-1 < r < 1$, +and its sum is $1/(1 - r) - 1 - r - \dots - r^{m-1}$ (\SecNo[§]{77},~\Eq{(2)}).} + +\Item{2.} The series $r^{m} + r^{m+1} + \dots$ is convergent if $-1 < r < 1$, and its sum is +$r^{m}/(1 - r)$ (\SecNo[§]{77},~\Eq{(4)}). Verify that the results of Exs.\ 1~and~2 are in agreement. + +\Item{3.} Prove that the series $1 + 2r + 2r^{2} + \dots$ is convergent, and that its sum +is~$(1 + r)/(1 - r)$, ($\alpha$)~by writing it in the form $-1 + 2(1 + r + r^{2} + \dots)$, ($\beta$)~by +writing it in the form $1 + 2(r + r^{2} + \dots)$, ($\gamma$)~by adding the two series +$1 + r + r^{2} + \dots$, $r + r^{2} + \dots$. In each case mention which of the theorems of +\SecNo[§]{77} are used in your proof. +\PageSep{146} + +\Item{4.} Prove that the `arithmetic' series +\[ +a + (a + b) + (a + 2b) + \dots +\] +is always divergent, unless both $a$~and~$b$ are zero. Show that, if $b$ is not +zero, the series diverges to~$+\infty$ or to~$-\infty$ according to the sign of~$b$, while if +$b = 0$ it diverges to~$+\infty$ or~$-\infty$ according to the sign of~$a$. + +\Item{5.} What is the sum of the series +\[ +(1 - r) + (r - r^{2}) + (r^{2} - r^{3}) + \dots +\] +when the series is convergent? [The series converges only if $-1 < r \leq 1$. Its +sum is~$1$, except when $r = 1$, when its sum is~$0$.] + +\Item{6.} Sum the series +\[ +%[** TN: In-line equation in the original] +r^{2} + \frac{r^{2}}{1 + r^{2}} + \frac{r^{2}}{(1 + r^{2})^{2}} + \dots. +\] +[The series is always convergent. +Its sum is~$1 + r^{2}$, except when $r = 0$, when its sum is~$0$.] + +\Item{7.} If we assume that $1 + r + r^{2} + \dots$ is convergent then we can prove that its +sum is~$1/(1 - r)$ by means of \SecNo[§]{77}, \Eq{(1)}~and~\Eq{(4)}. For if $1 + r + r^{2} + \dots = s$ then +\[ +s = 1 + r(1 + r^{2} + \dots) = 1 + rs. +\] + +\Item{8.} Sum the series +\[ +r + \frac{r}{1 + r} + \frac{r}{(1 + r)^{2}} + \dots +\] +when it is convergent. [The series is convergent if $-1 < 1/(1 + r) < 1$, \ie\ if +$r < -2$ or if $r > 0$, and its sum is~$1 + r$. It is also convergent when $r = 0$, when +its sum is~$0$.] + +\Item{9.} Answer the same question for the series +\begin{align*} +& r - \frac{r}{1 + r} + \frac{r}{(1 + r)^{2}} - \dots, +&& r + \frac{r}{1 - r} + \frac{r}{(1 - r)^{2}} + \dots,\\ +& 1 - \frac{r}{1 + r} + \left(\frac{r}{1 + r}\right)^{2} - \dots, +&& 1 + \frac{r}{1 - r} + \left(\frac{r}{1 - r}\right)^{2} + \dots. +\end{align*} + +\Item{10.} Consider the convergence of the series +\begin{align*} +& (1 + r) + (r^{2} + r^{3}) + \dots, +&& (1 + r + r^{2}) + (r^{3} + r^{4} + r^{5}) + \dots,\\ +& 1 - 2r + r^{2} + r^{3} - 2r^{4} + r^{5} + \dots, +&& (1 - 2r + r^{2}) + (r^{3} - 2r^{4} + r^{5}) + \dots, +\end{align*} +and find their sums when they are convergent. + +\Item{11.} If $0 \leq a_{n} \leq 1$ then the series $a_{0} + a_{1}r + a_{2}r^{2} + \dots$ is convergent for +$0 \leq r < 1$, and its sum is not greater than~$1/(1 - r)$. + +\Item{12.} If in addition the series $a_{0} + a_{1} + a_{2} + \dots$ is convergent, then the series +$a_{0} + a_{1}r + a_{2}r^{2} + \dots$ is convergent for $0 \leq r \leq 1$, and its sum is not greater than +the lesser of $a_{0} + a_{1} + a_{2} + \dots$ and~$1/(1 - r)$. + +\Item{13.} The series +\[ +1 + \frac{1}{1} + \frac{1}{1·2} + \frac{1}{1·2·3} + \dots +\] +is convergent. [For $1/(1·2 \dots n) \leq 1/2^{n-1}$.] +\PageSep{147} + +\Item{14.} The series +\[ +1 + \frac{1}{1·2} + \frac{1}{1·2·3·4} + \dots,\quad +\frac{1}{1} + \frac{1}{1·2·3} + \frac{1}{1·2·3·4·5} + \dots +\] +are convergent. + +\Item{15.} The general harmonic series +\[ +\frac{1}{a} + \frac{1}{a + b} + \frac{1}{a + 2b} + \dots, +\] +where $a$~and~$b$ are positive, diverges to~$+\infty$. + +[For $u_{n} = 1/(a + nb) > 1/\{n(a + b)\}$. Now compare with $1 + \frac{1}{2} + \frac{1}{3} + \dots$.] + +\Item{16.} Show that the series +\[ +(u_{0} - u_{1}) + (u_{1} - u_{2}) + (u_{2} - u_{3}) + \dots +\] +is convergent if and only if $u_{n}$~tends to a limit as $n \to \infty$. + +\Item{17.} If $u_{1} + u_{2} + u_{3} + \dots$ is divergent then so is any series formed by +grouping the terms in brackets in any way to form new single terms. + +\Item{18.} Any series, formed by taking a selection of the terms of a convergent +series of positive terms, is itself convergent. +\end{Examples} + +\Paragraph{79. The representation of functions of a continuous +real variable by means of limits.} In the preceding sections +we have frequently been concerned with limits such as +\[ +\lim_{n \to \infty} \phi_{n}(x), +\] +and series such as +\[ +u_{1}(x) + u_{2}(x) + \dots + = \lim_{n \to \infty}\{u_{1}(x) + u_{2}(x) + \dots + u_{n}(x)\}, +\] +in which the function of~$n$ whose limit we are seeking involves, +besides~$n$, another variable~$x$. In such cases the limit is of course +a function of~$x$. Thus in \SecNo[§]{75} we encountered the function +\[ +f(x) = \lim_{n \to \infty} n(\sqrt[n]{x} - 1): +\] +and the sum of the geometrical series $1 + x + x^{2} + \dots$ is a function +of~$x$, viz.~the function which is equal to $1/(1 - x)$ if $-1 < x < 1$ and +is undefined for all other values of~$x$. + +Many of the apparently `arbitrary' or `unnatural' functions +considered in \Ref{Ch.}{II} are capable of a simple representation of this +kind, as will appear from the following examples. +\PageSep{148} + +\begin{Examples}{XXXI.} +\Item{1.} $\phi_{n}(x) = x$. Here $n$~does not appear at all in the +expression of~$\phi_{n}(x)$, and $\phi(x) = \lim\phi_{n}(x) = x$ for all values of~$x$. + +\Item{2.} $\phi_{n}(x) = x/n$. Here $\phi(x) = \lim\phi_{n}(x) = 0$ for all values of~$x$. + +\Item{3.} $\phi_{n}(x) = nx$. If $x > 0$, $\phi_{n}(x) \to +\infty$; if $x < 0$, $\phi_{n}(x) \to -\infty$: only when +$x = 0$ has $\phi_{n}(x)$ a limit (viz.~$0$) as $n \to \infty$. Thus $\phi(x) = 0$ when $x = 0$ and is +not defined for any other value of~$x$. + +\Item{4.} $\phi_{n}(x) = 1/nx$, $nx/(nx + 1)$. + +\Item{5.} $\phi_{n}(x) = x^{n}$. Here $\phi(x) = 0$, ($-1 < x < 1$); $\phi(x) = 1$, ($x = 1$); and $\phi(x)$ +is not defined for any other value of~$x$. + +\Item{6.} $\phi_{n}(x) = x^{n}(1 - x)$. Here $\phi(x)$~differs from the $\phi(x)$ of Ex.~5 in that +it has the value~$0$ when $x = 1$. + +\Item{7.} $\phi_{n}(x) = x^{n}/n$. Here $\phi(x)$ differs from the $\phi(x)$ of Ex.~6 in that it has +the value~$0$ when $x = -1$ as well as when $x = 1$. + +\Item{8.} $\phi_{n}(x) = x^{n}/(x^{n} + 1)$. [$\phi(x) = 0$, ($-1 < x < 1$); $\phi(x) = \frac{1}{2}$, ($x = 1$); $\phi(x) = 1$, +($x < -1$ or $x > 1$); and $\phi(x)$~is not defined when $x = -1$.] + +\Item{9.} $\phi_{n}(x) = x^{n}/(x^{n} - 1)$, $1/(x^{n} + 1)$, $1/(x^{n} - 1)$, $1/(x^{n} + x^{-n})$, $1/(x^{n} - x^{-n})$. + +\Item{10.} $\phi_{n}(x) = (x^{n} - 1)/(x^{n} + 1)$, $(nx^{n} - 1)/(nx^{n} + 1)$, $(x^{n} - n)/(x^{n} + n)$. [In the +first case $\phi(x) = 1$ when $|x| > 1$, $\phi(x) = -1$ when $|x| < 1$, $\phi(x) = 0$ when $x = 1$ +and $\phi(x)$~is not defined when $x = -1$. The second and third functions differ +from the first in that they are defined both when $x = 1$ and when $x = -1$: the +second has the value~$1$ and the third the value~$-1$ for both these values of~$x$.] + +\Item{11.} Construct an example in which $\phi(x) = 1$, ($|x| > 1$); $\phi(x) = -1$, +($|x| < 1$); and $\phi(x) = 0$, ($x = 1$ and $x = -1$). + +\Item{12.} $\phi_{n}(x) = x\{(x^{2n} - 1)/(x^{2n} + 1)\}^{2}$, $n/(x^{n} + x^{-n} + n)$. + +\Item{13.} $\phi_{n}(x) = \{x^{n}f(x) + g(x)\}/(x^{n} + 1)$. [Here $\phi(x) = f(x)$, ($|x| > 1$); +$\phi(x) = g(x)$, ($|x| < 1$); $\phi(x) = \frac{1}{2}\{f(x) + g(x)\}$, ($x = 1$); and $\phi(x)$~is undefined +when $x = -1$.] + +\Item{14.} $\phi_{n}(x) = (2/\pi) \arctan(nx)$. [$\phi(x) = 1$, ($x > 0$); $\phi(x) = 0$, ($x = 0$); +$\phi(x) = -1$, ($x < 0$). This function is important in the Theory of Numbers, +and is usually denoted by~$\sgn x$.] + +\Item{15.} $\phi_{n}(x) = \sin nx\pi$. [$\phi(x) = 0$ when $x$~is an integer; and $\phi(x)$~is +otherwise undefined (\Ex{xxiv}.~7).] + +\Item{16.} If $\phi_{n}(x) = \sin (n!\, x\pi)$ then $\phi(x) = 0$ for all rational values of~$x$ (\Ex{xxiv}.~14). +[The consideration of irrational values presents greater difficulties.] + +\Item{17.} $\phi_{n}(x) = (\cos^{2} x\pi)^{n}$. [$\phi(x) = 0$ except when $x$~is integral, when +$\phi(x) = 1$.] + +\Item{18.} If $N \geq 1752$ then the number of days in the year $N$~\textsc{a.d.}\ is +\[ +\lim \{365 + + (\cos^{2} \tfrac{1}{4} N\pi)^{n} - (\cos^{2} \tfrac{1}{100} N\pi)^{n} + + (\cos^{2} \tfrac{1}{400} N\pi)^{n}\}. +\] +\end{Examples} +\PageSep{149} + +\begin{Remark} +\Paragraph{80. The bounds of a bounded aggregate.} Let $S$~be any system or +aggregate of real numbers~$s$. If there is a number~$K$ such that $s \leq K$ for +every~$s$ of~$S$, we say that $S$~is \emph{bounded above}. If there is a number~$k$ such that +$s \geq k$ for every~$s$, we say that $S$~is \emph{bounded below}. If $S$~is both bounded above +and bounded below, we say simply that $S$~is \emph{bounded}. + +Suppose first that $S$~is bounded above (but not necessarily below). There +will be an infinity of numbers which possess the property possessed by~$K$; +any number greater than~$K$, for example, possesses it. We shall prove that +\emph{among these numbers there is a least},\footnote + {An infinite aggregate of numbers does not necessarily possess a least member. + The set consisting of the numbers + \[ + 1,\ \frac{1}{2},\ \frac{1}{3},\ \dots,\ \frac {1}{n},\ \dots, + \] + for example, has no least member.} +which we shall call~$M$. This number~$M$ +is not exceeded by any member of~$S$, but every number less than~$M$ is exceeded +by at least one member of~$S$. + +We divide the real numbers~$\xi$ into two classes $L$~and~$R$, putting $\xi$ into $L$~or~$R$ +according as it is or is not exceeded by members of~$S$. Then every~$\xi$ belongs +to one and one only of the classes $L$~and~$R$. Each class exists; for any +number less than any member of~$S$ belongs to~$L$, while $K$~belongs to~$R$. +Finally, any member of~$L$ is less than some member of~$S$, and therefore less +than any member of~$R$. Thus the three conditions of Dedekind's Theorem +(\SecNo[§]{17}) are satisfied, and there is a number~$M$ dividing the classes. + +The number~$M$ is the number whose existence we had to prove. In the +first place, $M$~cannot be exceeded by any member of~$S$. For if there were such +a member~$s$ of~$S$, we could write $s = M + \eta$, where $\eta$~is positive. The number +$M + \frac{1}{2}\eta$ would then belong to~$L$, because it is less than~$s$, and to~$R$, because it is +greater than~$M$; and this is impossible. On the other hand, any number less +than~$M$ belongs to~$L$, and is therefore exceeded by at least one member of~$S$. +Thus $M$~has all the properties required. + +This number~$M$ we call the \emph{upper bound} of~$S$, and we may enunciate the +following theorem. +\begin{Result} +Any aggregate~$S$ which is bounded above has an upper +bound~$M$. No member of~$S$ exceeds~$M$; but any number less than~$M$ is exceeded +by at least one member of~$S$. +\end{Result} + +In exactly the same way we can prove the corresponding theorem for an +aggregate bounded below (but not necessarily above). +\begin{Result} +Any aggregate~$S$ which +is bounded below has a lower bound~$m$. No member of~$S$ is less than~$m$; but +there is at least one member of~$S$ which is less than any number greater than~$m$. +\end{Result} + +It will be observed that, when $S$~is bounded above, $M \leq K$, and when $S$~is +bounded below, $m \geq k$. When $S$~is bounded, $k \leq m \leq M \leq K$. + +\Paragraph{81. The bounds of a bounded function.} Suppose that $\phi(n)$~is a function +of the positive integral variable~$n$. The aggregate of all the values~$\phi(n)$ +defines a set~$S$, to which we may apply all the arguments of \SecNo[§]{80}. If $S$~is +bounded above, or bounded below, or bounded, we say that $\phi(n)$~is bounded +\PageSep{150} +above, or bounded below, or bounded. If $\phi(n)$~is bounded above, that is to +say if there is a number~$K$ such that $\phi(n) \leq K$ for all values of~$n$, then there +is a number~$M$ such that + +\Itemp{(i)} \emph{$\phi(n) \leq M$ for all values of~$n$}; + +\Itemp{(ii)} \emph{if $\DELTA$ is any positive number then $\phi(n) > M - \DELTA$ for at least one value of~$n$.} +This number~$M$ we call the \Emph{upper bound} of~$\phi(n)$. Similarly, if $\phi(n)$~is +bounded below, that is to say if there is a number~$k$ such that $\phi(n) \leq k$ for all +values of~$n$, then there is a number~$m$ such that + +\Itemp{(i)} \emph{$\phi(n) \geq m$ for all values of $n$}; + +\Itemp{(ii)} \emph{if $\DELTA$ is any positive number then $\phi(n) < m + \DELTA$ for at least one value of~$n$.} +This number~$m$ we call the \Emph{lower bound} of~$\phi(n)$. + +If $K$~exists, $M \leq K$; if $k$~exists, $m \geq k$; and if both $k$~and~$K$ exist then +\[ +k \leq m \leq M \leq K. +\] + +\Paragraph{82. The limits of indetermination of a bounded function.} Suppose +that $\phi(n)$~is a bounded function, and $M$~and~$m$ its upper and lower bounds. +Let us take any real number~$\xi$, and consider now the relations of inequality +which may hold between~$\xi$ and the values assumed by~$\phi(n)$ for \emph{large} values +of~$n$. There are three mutually exclusive possibilities: + +\Item{(1)} $\xi \geq \phi(n)$ for all sufficiently large values of~$n$; + +\Item{(2)} $\xi \leq \phi(n)$ for all sufficiently large values of~$n$; + +\Item{(3)} $\xi < \phi(n)$ for an infinity of values of~$n$, and also $\xi > \phi(n)$ for an +infinity of values of~$n$. + +In case~(1) we shall say that $\xi$~is a \emph{superior} number, in case~(2) that it is +an \emph{inferior} number, and in case~(3) that it is an \emph{intermediate} number. It is +plain that no superior number can be less than~$m$, and no inferior number +greater than~$M$. + +Let us consider the aggregate of all superior numbers. It is bounded +below, since none of its members are less than~$m$, and has therefore a lower +bound, which we shall denote by~$\Lambda$. Similarly the aggregate of inferior +numbers has an upper bound, which we denote by~$\lambda$. + +We call $\Lambda$~and~$\lambda$ respectively the \emph{upper and lower limits of indetermination +of~$\phi(n)$ as $n$~tends to infinity}; and write +\[ +\Lambda = \limsup \phi(n),\quad +\lambda = \liminf \phi(n). +\] +These numbers have the following properties: + +\Item{(1)} $m \leq \lambda \leq \Lambda \leq M$; + +\Item{(2)} $\Lambda$~and~$\lambda$ are the upper and lower bounds of the aggregate of intermediate +numbers, if any such exist; + +\Item{(3)} if $\DELTA$ is any positive number, then $\phi(n) < \Lambda + \DELTA$ for all sufficiently large +values of~$n$, and $\phi(n) > \Lambda - \DELTA$ for an infinity of values of~$n$; + +\Item{(4)} {\Loosen similarly $\phi(n) > \lambda - \DELTA$ for all sufficiently large values of~$n$, and +$\phi(n) < \lambda + \DELTA$ for an infinity of values of~$n$;} +\PageSep{151} + +\Item{(5)} the necessary and sufficient condition that $\phi(n)$ should tend to a limit +is that $\Lambda = \lambda$, and in this case the limit is~$l$, the common value of $\lambda$~and~$\Lambda$. + +Of these properties, (1)~is an immediate consequence of the definitions; +and we can prove~(2) as follows. If $\Lambda = \lambda = l$, there can be at most one intermediate +number, viz.~$l$, and there is nothing to prove. Suppose then that +$\Lambda > \lambda$. Any intermediate number~$\xi$ is less than any superior and greater than +any inferior number, so that $\lambda \leq \xi \leq \Lambda$. But if $\lambda < \xi < \Lambda$ then $\xi$~must be +intermediate, since it is plainly neither superior nor inferior. Hence there are +intermediate numbers as near as we please to either $\lambda$~or~$\Lambda$. + +To prove~(3) we observe that $\Lambda + \DELTA$ is superior and $\Lambda - \DELTA$ intermediate or +inferior. The result is then an immediate consequence of the definitions; and +the proof of~(4) is substantially the same. + +Finally (5)~may be proved as follows. If $\Lambda = \lambda = l$, then +\[ +l - \DELTA < \phi(n) < l + \DELTA +\] +for every positive value of~$\DELTA$ and all sufficiently large values of~$n$, so that +$\phi(n)\to l$. Conversely, if $\phi(n) \to l$, then the inequalities above written hold +for all sufficiently large values of~$n$. Hence $l - \DELTA$ is inferior and $l + \DELTA$ superior, +so that +\[ +\lambda \geq l - \DELTA,\quad +\Lambda \leq l + \DELTA, +\] +and therefore $\Lambda - \lambda \leq 2\DELTA$. As $\Lambda - \lambda \geq 0$, this can only be true if $\Lambda = \lambda$. +\end{Remark} + +\begin{Examples}{XXXII.} +\Item{1.} Neither $\Lambda$~nor~$\lambda$ is affected by any alteration in +any finite number of values of~$\phi(n)$. + +\Item{2.} If $\phi(n) = a$ for all values of~$n$, then $m = \lambda = \Lambda = M = a$. + +\Item{3.} If $\phi(n) = 1/n$, then $m = \lambda = \Lambda = 0$ and $M = 1$. + +\Item{4.} If $\phi(n) = (-1)^{n}$, then $m = \lambda = -1$ and $\Lambda = M = 1$. + +\Item{5.} If $\phi(n) = (-1)^{n}/n$, then $m = -1$, $\lambda = \Lambda = 0$, $M = \frac{1}{2}$. + +\Item{6.} If $\phi(n) = (-1)^{n}\{1 + (1/n)\}$, then $m = -2$, $\lambda = -1$, $\Lambda = 1$, $M = \frac{3}{2}$. + +\Item{7.} Let $\phi(n) = \sin n\theta\pi$, where $\theta > 0$. If $\theta$~is an integer then $m = \lambda = \Lambda = M = 0$. +If $\theta$~is rational but not integral a variety of cases arise. Suppose, \eg, that +$\theta = p/q$, $p$~and~$q$ being positive, odd, and prime to one another, and $q > 1$. +Then $\phi(n)$~assumes the cyclical sequence of values +\[ +\sin(p\pi/q),\quad +\sin(2p\pi/q),\ \dots,\quad +\sin\{(2q - 1)p\pi/q\},\quad +\sin(2qp\pi/q),\ \dots. +\] +It is easily verified that the numerically greatest and least values of~$\phi(n)$ are +$\cos(\pi/2q)$ and $-\cos(\pi/2q)$, so that +\[ +m = \lambda = -\cos(\pi/2q),\quad +\Lambda = M = \cos(\pi/2q). +\] +The reader may discuss similarly the cases which arise when $p$~and~$q$ are +not both odd. + +The case in which $\theta$~is irrational is more difficult: it may be shown that +in this case $m = \lambda = -1$ and $\Lambda = M = 1$. It may also be shown that the values +of~$\phi(n)$ are scattered all over the interval $\DPmod{(-1, 1)}{[-1, 1]}$ in such a way that, if $\xi$~is +\PageSep{152} +\emph{any} number of the interval, then there is a sequence $n_{1}$, $n_{2}$,~\dots\ such that +$\phi(n_{k}) \to \xi$ as $k \to \infty$.\footnote + {A number of simple proofs of this result are given by Hardy and Littlewood, + ``Some Problems of Diophantine Approximation'', \textit{Acta Mathematica}, vol.~xxxvii.} + +The results are very similar when $\phi(n)$~is the fractional part of~$n\theta$. +\end{Examples} + +\begin{Remark} +\Paragraph{83. The general principle of convergence for a bounded function.} +The results of the preceding sections enable us to formulate a very important +necessary and sufficient condition that a bounded function~$\phi(n)$ should tend +to a limit, a condition usually referred to as \emph{the general principle of convergence} +to a limit. + +\begin{Theorem}[1.] +The necessary and sufficient condition that a bounded function~$\phi(n)$ +should tend to a limit is that, when any positive number~$\DELTA$ is given, it should +be possible to find a number~$n_{0}(\DELTA)$ such that +\[ +|\phi(n_{2}) - \phi(n_{1})| < \DELTA +\] +for all values of $n_{1}$~and~$n_{2}$ such that $n_{2} > n_{1} \geq n_{0}(\DELTA)$. +\end{Theorem} + +In the first place, the condition is \emph{necessary}. For if $\phi(n) \to l$ then we +can find~$n_{0}$ so that +\[ +l - \tfrac{1}{2}\DELTA < \phi(n) < l + \tfrac{1}{2}\DELTA +\] +when $n \geq n_{0}$, and so +\[ +|\phi(n_{2}) - \phi(n_{1})| < \DELTA +\Tag{(1)} +\] +when $n_{1} \geq n_{0}$ and $n_{2} \geq n_{0}$. + +In the second place, the condition is \emph{sufficient}. In order to prove this we +have only to show that it involves $\lambda = \Lambda$. But if $\lambda < \Lambda$ then there are, however +small $\DELTA$~may be, infinitely many values of~$n$ such that $\phi(n) < \lambda + \DELTA$ and +infinitely many such that $\phi(n) > \Lambda - \DELTA$; and therefore we can find values of +$n_{1}$~and~$n_{2}$, each greater than any assigned number~$n_{0}$, and such that +\[ +\phi(n_{2}) - \phi(n_{1}) > \Lambda - \lambda - 2\DELTA, +\] +which is greater than $\frac{1}{2}(\Lambda - \lambda)$ if $\DELTA$~is small enough. This plainly contradicts +the inequality~\Eq{(1)}. Hence $\lambda = \Lambda$, and so $\phi(n)$~tends to a limit. + +\Paragraph{84. Unbounded functions.} So far we have restricted ourselves to +bounded functions; but the `general principle of convergence' is the same +for unbounded as for bounded functions, and the words `\emph{a bounded function}' +may be omitted from the enunciation of Theorem~1. + +In the first place, if $\phi(n)$~tends to a limit~$l$ then it is certainly bounded; for +all but a finite number of its values are less than $l + \DELTA$ and greater than $l - \DELTA$. + +In the second place, if the condition of Theorem~1 is satisfied, we have +\[ +|\phi(n_{2}) - \phi(n_{1})| < \DELTA +\] +whenever $n_{1} \geq n_{0}$ and $n_{2} \geq n_{0}$. Let us choose some particular value~$n_{1}$ greater +than~$n_{0}$. Then +\[ +\phi(n_{1}) - \DELTA < \phi(n_{2}) < \phi(n_{1}) + \DELTA +\] +when $n_{2} \geq n_{0}$. Hence $\phi(n)$~is bounded; and so the second part of the proof of +the last section applies also. +\PageSep{153} + +The theoretical importance of the `general principle of convergence' can +hardly be overestimated. Like the theorems of \SecNo[§]{69}, it gives us a means of +deciding whether a function~$\phi(n)$ tends to a limit or not, without requiring +us to be able to tell beforehand what the limit, if it exists, must be; and +it has not the limitations inevitable in theorems of such a special character +as those of \SecNo[§]{69}. But in elementary work it is generally possible to dispense +with it, and to obtain all we want from these special theorems. And it will +be found that, in spite of the importance of the principle, practically no +applications are made of it in the chapters which follow.\footnote + {A few proofs given in \Ref{Ch.}{VIII} can be simplified by the use of the principle.} +We will only +remark that, if we suppose that +\[ +\phi(n) = s_{n} = u_{1} + u_{2} + \dots + u_{n}, +\] +we obtain at once a necessary and sufficient condition for the convergence of +an infinite series, viz: + +\begin{Theorem}[2.] +The necessary and sufficient condition for the convergence +of the series $u_{1} + u_{2} + \dots$ is that, given any positive number~$\DELTA$, it should be +possible to find~$n_{0}$ so that +\[ +|u_{n_{1}+1} + u_{n_{1}+2} + \dots + u_{n_{2}}| < \DELTA +\] +for all values of $n_{1}$~and~$n_{2}$ such that $n_{2} > n_{1} \geq n_{0}$. +\end{Theorem} +\end{Remark} + +\Paragraph{85. Limits of complex functions and series of complex +terms.} In this chapter we have, up to the present, concerned +ourselves only with real functions of~$n$ and series all of whose +terms are real. There is however no difficulty in extending our +ideas and definitions to the case in which the functions or the +terms of the series are complex. + +Suppose that $\phi(n)$~is complex and equal to +\[ +\rho(n) + i\sigma(n), +\] +where $\rho(n)$,~$\sigma(n)$ are real functions of~$n$. Then \emph{if $\rho(n)$~and~$\sigma(n)$ +converge respectively to limits $r$~and~$s$ as $n \to \infty$, we shall say that +$\phi(n)$~converges to the limit $l = r + is$, and write} +\[ +\lim\phi(n) = l. +\] +Similarly, when $u_{n}$~is complex and equal to $v_{n} + iw_{n}$, we shall say +that \emph{the series +\[ +u_{1} + u_{2} + u_{3} + \dots +\] +is convergent and has the sum $l = r + is$, if the series +\[ +v_{1} + v_{2} + v_{3} + \dots,\quad +w_{1} + w_{2} + w_{3} + \dots +\] +are convergent and have the sums $r$,~$s$ respectively}. +\PageSep{154} + +To say that $u_{1} + u_{2} + u_{3} + \dots$ is convergent and has the sum~$l$ +is of course the same as to say that the sum +\[ +s_{n} + = u_{1} + u_{2} + \dots + u_{n} + = (v_{1} + v_{2} + \dots + v_{n}) + + i(w_{1} + w_{2} + \dots + w_{n}) +\] +converges to the limit~$l$ as $n \to \infty$. + +In the case of real functions and series we also gave definitions +of \emph{divergence} and \emph{oscillation}, \emph{finite} or \emph{infinite}. But in the case +of complex functions and series, where we have to consider the +behaviour both of~$\rho(n)$ and of~$\sigma(n)$, there are so many possibilities +that this is hardly worth while. When it is necessary to make +further distinctions of this kind, we shall make them by stating +the way in which the real or imaginary parts behave when taken +separately. + +\Paragraph{86.} The reader will find no difficulty in proving such +theorems as the following, which are obvious extensions of +theorems already proved for real functions and series. + +\Item{(1)} If $\lim\phi(n) = l$ then $\lim\phi(n + p) = l$ for any fixed value +of~$p$. + +\Item{(2)} If $u_{1} + u_{2} + \dots$ is convergent and has the sum~$l$, then +$a + b + c + \dots + k + u_{1} + u_{2} + \dots$ is convergent and has the sum +$a + b + c + \dots + k + l$, and $u_{p+1} + u_{p+2} + \dots$ is convergent and has +the sum $l - u_{1} - u_{2} - \dots - u_{p}$. + +\Item{(3)} If $\lim\phi(n) = l$ and $\lim\psi(n) = m$, then +\[ +\lim\{\phi(n) + \psi(n)\} = l + m. +\] + +\Item{(4)} If $\lim\phi(n) = l$, then $\lim k\phi(n) = kl$. + +\Item{(5)} If $\lim\phi(n) = l$ and $\lim\psi(n) = m$, then $\lim \phi(n)\psi(n) = lm$. + +\Item{(6)} If $u_{1} + u_{2} + \dots$ converges to the sum~$l$, and $v_{1} + v_{2} + \dots$ to +the sum~$m$, then $(u_{1} + v_{1}) + (u_{2}+ v_{2}) + \dots$ converges to the sum~$l + m$. + +\Item{(7)} If $u_{1} + u_{2} + \dots$ converges to the sum~$l$ then $ku_{1} + ku_{2} + \dots$ +converges to the sum~$kl$. + +\Item{(8)} If $u_{1} + u_{2} + u_{3} + \dots$ is convergent then $\lim u_{n} = 0$. + +\Item{(9)} If $u_{1} + u_{2} + u_{3} + \dots$ is convergent, then so is any series +formed by grouping the terms in brackets, and the sums of the two +series are the same. +\PageSep{155} + +\begin{Remark} +As an example, let us prove theorem~(5). Let +\[ +\phi(n) = \rho(n) + i\sigma(n),\quad +\psi(n) = \rho'(n) + i\sigma'(n),\quad +l = r + is,\quad +m = r' + is'. +\] + +Then +\[ +\rho(n) \to r,\quad +\sigma(n) \to s,\quad +\rho'(n) \to r',\quad +\sigma'(n) \to s'. +\] + +But +\[ +\phi(n)\psi(n) = \rho\rho' - \sigma\sigma' + i(\rho\sigma' + \rho'\sigma), +\] +and +\[ +\rho\rho' - \sigma\sigma' \to rr' - ss',\quad +\rho\sigma' + \rho'\sigma \to rs' + r's; +\] +so that +\[ +\phi(n)\psi(n) \to rr' - ss' + i(rs' + r's), +\] +\ie +\[ +\phi(n)\psi(n) \to (r + is)(r' + is') = lm. +\] +\end{Remark} + +The following theorems are of a somewhat different character. + +\begin{Result} +\Item{(10)} In order that $\phi(n) = \rho(n) + i\sigma(n)$ should converge to +zero as $n \to \infty$, it is necessary and sufficient that +\[ +|\phi(n)| = \sqrtbr{\{\rho(n)\}^{2} + \{\sigma(n)\}^{2}} +\] +should converge to zero. +\end{Result} + +\begin{Remark} +If $\rho(n)$ and~$\sigma(n)$ both converge to zero then it is plain that $\sqrtp{\rho^{2} + \sigma^{2}}$ +does so. The converse follows from the fact that the numerical value of~$\rho$ or~$\sigma$ +cannot be greater than $\sqrtp{\rho^{2} + \sigma^{2}}$. +\end{Remark} + +\begin{Result} +\Item{(11)} More generally, in order that $\phi(n)$~should converge to a +limit~$l$, it is necessary and sufficient that +\[ +|\phi(n) - l| +\] +should converge to zero. +\end{Result} + +\begin{Remark} +For $\phi(n) - l$ converges to zero, and we can apply~(10). +\end{Remark} + +\begin{Result} +\Item{(12)} Theorems {\upshape1}~and~{\upshape2} of \SecNo[§§]{83}--\SecNo{84} are still true when +$\phi(n)$~and~$u_{n}$ are complex. +\end{Result} + +\begin{Remark} +We have to show that the necessary and sufficient condition that $\phi(n)$ +should tend to~$l$ is that +\[ +|\phi(n_{2}) - \phi(n_{1})| < \DELTA +\Tag{(1)} +\] +when $n_{2} > n_{1} \geq n_{0}$. + +If $\phi(n) \to l$ then $\rho(n) \to r$ and $\sigma(n) \to s$, and so we can find numbers $n_{0}'$ and +$n_{0}''$ depending on~$\DELTA$ and such that +\[ +|\rho(n_{2}) - \rho(n_{1})| < \tfrac{1}{2}\DELTA,\quad +|\sigma(n_{2}) - \sigma(n_{1})| < \tfrac{1}{2}\DELTA, +\] +{\Loosen the first inequality holding when $n_{2} > n_{1} \geq n_{0}'$, and the second when $n_{2} > n_{1} \geq n_{0}''$. +Hence} +\[ +|\phi(n_{2}) - \phi(n_{1})| + \leq |\rho(n_{2}) - \rho(n_{1})| + |\sigma(n_{2}) - \sigma(n_{1})| + < \DELTA +\] +when $n_{2} > n_{1} \geq n_{0}$, where $n_{0}$~is the greater of $n_{0}'$~and~$n_{0}''$. Thus the condition~\Eq{(1)} +is \emph{necessary}. To prove that it is \emph{sufficient} we have only to observe that +\[ +|\rho(n_{2}) - \rho(n_{1})| \leq |\phi(n_{2}) - \phi(n_{1})| < \DELTA +\] +when $n_{2} > n_{1} \geq n_{0}$. Thus $\rho(n)$~tends to a limit~$r$, and in the same way it may +be shown that $\sigma(n)$~tends to a limit~$s$. +\end{Remark} +\PageSep{156} + +\Paragraph{87. The limit of~$z^{n}$ as $n \to \infty$, $z$~being any complex +number.} Let us consider the important case in which $\phi(n) = z^{n}$. +This problem has already been discussed for real values of~$z$ in~\SecNo[§]{72}. + +If $z^{n} \to l$ then $z^{n+1} \to l$, by~\Eq{(1)} of \SecNo[§]{86}. But, by~\Eq{(4)} of \SecNo[§]{86}, +\[ +z^{n+1} = zz^{n} \to zl, +\] +and therefore $l = zl$, which is only possible if (\ia)~$l = 0$ or (\ib)~$z = 1$. +If $z = 1$ then $\lim z^{n} = 1$. Apart from this special case the limit, +if it exists, can only be zero. + +Now if $z = r(\cos\theta + i\sin\theta)$, where $r$~is positive, then +\[ +z^{n} = r^{n} (\cos n\theta + i\sin n\theta), +\] +so that $|z^{n}| = r^{n}$. Thus $|z^{n}|$~tends to zero if and only if $r < 1$; +and it follows from~\Eq{(10)} of \SecNo[§]{86} that +\[ +\lim z^{n} = 0 +\] +if and only if $r < 1$. In no other case does $z^{n}$~converge to a limit, +except when $z = 1$ and $z^n \to 1$. + +\Paragraph{88. The geometric series $1 + z + z^{2} + \dots$ when $z$~is +complex.} Since +\[ +s_{n} = 1 + z + z^{2} + \dots + z^{n-1} = (1 - z^{n})/(1 - z), +\] +{\Loosen unless $z = 1$, when the value of~$s_{n}$ is~$n$, it follows that \emph{the series +$1 + z + z^{2} + \dots$ is convergent if and only if $r = |z| < 1$. And its +sum when convergent is $1/(1 - z)$}.} + +Thus if $z = r(\cos\theta + i\sin\theta) = r\Cis\theta$, and $r < 1$, we have +\begin{align*} +1 + z + z^{2} + \dots &= 1/(1 - r\Cis\theta), +\intertext{or} +1 + r \Cis\theta + r^{2} \Cis 2\theta + \dots + &= 1/(1 - r\Cis\theta)\\ + &= (1 - r\cos\theta + ir\sin\theta)/(1 - 2r\cos\theta + r^{2}). +\end{align*} +Separating the real and imaginary parts, we obtain +\begin{align*} +1 + r\cos\theta + r^{2}\cos 2\theta + \dots + &= (1 - r\cos\theta)/(1 - 2r\cos\theta + r^{2}),\\ +r\sin\theta + r^{2}\sin 2\theta + \dots + &= r\sin\theta/(1 - 2r\cos\theta + r^{2}), +\end{align*} +provided $r < 1$. If we change~$\theta$ into~$\theta + \pi$, we see that these +results hold also for negative values of~$r$ numerically less than~$1$. +Thus they hold when $-1 < r < 1$. +\PageSep{157} + +\begin{Examples}{XXXIII.} +\Item{1.} Prove directly that $\phi(n) = r^{n} \cos n\theta$ converges +to~$0$ when $r < 1$ and to~$1$ when $r = 1$ and $\theta$~is a multiple of~$2\pi$. Prove further +that if $r = 1$ and $\theta$~is not a multiple of~$2\pi$, then $\phi(n)$~oscillates finitely; if +$r > 1$ and $\theta$~is a multiple of~$2\pi$, then $\phi(n) \to +\infty$; and if $r > 1$ and $\theta$~is not a +multiple of~$2\pi$, then $\phi(n)$~oscillates infinitely. + +\Item{2.} Establish a similar series of results for $\phi(n) = r^{n} \sin n\theta$. + +\Item{3.} Prove that +\begin{gather*} +z^{m} + z^{m+1} + \dots = z^{m}/(1 - z),\\ +z^{m} + 2z^{m+1} + 2z^{m+2} + \dots = z^{m}(1 + z)/(1 - z), +\end{gather*} +if and only if $|z| < 1$. Which of the theorems of \SecNo[§]{86} do you use? + +\Item{4.} Prove that if $-1 < r < 1$ then +\[ +1 + 2r\cos\theta + 2r^{2}\cos 2\theta + \dots + = (1 - r^{2})/(1 - 2r\cos\theta + r^{2}). +\] + +\Item{5.} The series +\[ +1 + \frac{z}{1 + z} + \left(\frac{z}{1 + z}\right)^{2} + \dots +\] +converges to the sum $1\bigg/\left(1 - \dfrac{z}{1 + z}\right) = 1 + z$ if $|z/(1 + z) | < 1$. Show that this +condition is equivalent to the condition that $z$~has a real part greater than~$-\frac{1}{2}$. +\end{Examples} + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER IV.} + +\begin{Examples}{} +\Item{1.} The function~$\phi(n)$ takes the values $1$, $0$, $0$, $0$, $1$, $0$, $0$, $0$, $1$,~\dots\ when +$n = 0$, $1$, $2$,~\dots. Express $\phi(n)$ in terms of~$n$ by a formula which does not +involve trigonometrical functions. [$\phi(n) = \frac{1}{4}\{1 + (-1)^{n} + i^{n} + (-i)^{n}\}$.] + +\Item{2.} If $\phi(n)$~steadily increases, and $\psi(n)$~steadily decreases, as $n$~tends to~$\infty$, +and if $\psi(n) > \phi(n)$ for all values of~$n$, then both $\phi(n)$~and~$\psi(n)$ tend to +limits, and $\lim\phi(n) \leq \lim\psi(n)$. [This is an \DPchg{intermediate}{immediate} corollary from~\SecNo[§]{69}.] + +\Item{3.} Prove that, if +\[ +\phi(n) = \left(1 + \frac{1}{n}\right)^{n},\quad +\psi(n) = \left(1 - \frac{1}{n}\right)^{-n}, +\] +then $\phi(n + 1) > \phi(n)$ and $\psi(n + 1) < \psi(n)$. [The first result has already been +proved in~\SecNo[§]{73}.] + +\Item{4.} Prove also that $\psi(n) > \phi(n)$ for all values of~$n$: and deduce (by means +of the preceding examples) that both $\phi(n)$~and~$\psi(n)$ tend to limits as $n$~tends +to~$\infty$.\footnote + {A proof that $\lim\{\psi(n) - \phi(n)\} = 0$, and that therefore each function tends to + the limit~$e$, will be found in Chrystal's \textit{Algebra}, vol.~ii, p.~78. We shall however + prove this in \Ref{Ch.}{IX} by a different method.} + +\Item{5.} The arithmetic mean of the products of all distinct pairs of positive +integers whose sum is~$n$ is denoted by~$S_{n}$. Show that $\lim(S_{n}/n^{2}) = 1/6$. +\MathTrip{1903.} +\PageSep{158} + +\Item{6.} Prove that if $x_{1} = \frac{1}{2}\{x + (A/x)\}$, $x_{2} = \frac{1}{2}\{x_{1} + (A/x_{1})\}$, and so on, $x$~and~$A$ +being positive, then $\lim x_{n} = \sqrt{A}$. + +[Prove first that $\dfrac{x_{n} - \sqrt{A}}{x_{n} + \sqrt{A}} = \biggl(\dfrac{x - \sqrt{A}}{x + \sqrt{A}}\biggr)^{2^{n}}$.] + +\Item{7.} If $\phi(n)$~is a positive integer for all values of~$n$, and tends to~$\infty$ with~$n$, +then $x^{\phi(n)}$~tends to~$0$ if $0 < x < 1$ and to~$+\infty$ if $x > 1$. Discuss the behaviour +of~$x^{\phi(n)}$, as $n \to \infty$, for other values of~$x$. + +\Item{8.\footnotemark} +If $a_{n}$~increases or decreases steadily as $n$~increases, then the same is +true of $(a_{1} + a_{2} + \dots + a_{n})/n$.\footnotetext + {Exs.\ 8--12 are taken from Bromwich's \textit{Infinite Series}.}% + + +\Item{9.} If $x_{n+1} = \sqrtp{k + x_{n}}$, and $k$~and~$x_{1}$ are positive, then the sequence $x_{1}$,~$x_{2}$, +$x_{3}$,~\dots\ is an increasing or decreasing sequence according as $x_{1}$~is less than or +greater than~$\alpha$, the positive root of the equation $x^{2} = x + k$; and in either case +$x_{n} \to \alpha$ as $n \to \infty$. + +\Item{10.} If $x_{n+1} = k/(1 + x_{n})$, and $k$~and~$x_{1}$ are positive, then the sequences +$x_{1}$,~$x_{3}$, $x_{5}$,~\dots\ and $x_{2}$,~$x_{4}$, $x_{6}$,~\dots\ are one an increasing and the other a decreasing +sequence, and each sequence tends to the limit~$\alpha$, the positive root of the +equation $x^{2} + x = k$. + +\Item{11.} The function~$f(x)$ is increasing and continuous (see \Ref{Ch.}{V}) for all +values of~$x$, and a sequence $x_{1}$,~$x_{2}$, $x_{3}$,~\dots\ is defined by the equation +$x_{n+1} = f(x_{n})$. Discuss on general graphical grounds the question as to +whether $x_{n}$~tends to a root of the equation $x = f(x)$. Consider in particular +the case in which this equation has only one root, distinguishing the cases in +which the curve $y = f(x)$ crosses the line $y = x$ from above to below and from +below to above. + +\Item{12.} If $x_{1}$,~$x_{2}$ are positive and $x_{n+1} = \frac{1}{2} (x_{n} + x_{n-1})$, then the sequences $x_{1}$,~$x_{3}$, +$x_{5}$,~\dots\ and $x_{2}$,~$x_{4}$, $x_{6}$,~\dots\ are one a decreasing and the other an increasing +sequence, and they have the common limit $\frac{1}{3}(x_{1} + 2x_{2})$. + +\Item{13.} Draw a graph of the function~$y$ defined by the equation +\[ +y = \lim_{n \to \infty} \frac{x^{2n} \sin\frac{1}{2}\pi x + x^{2}}{x^{2n} + 1}. +\] +\MathTrip{1901.} + +\Item{14.} The function +\[ +y = \lim_{n \to \infty} \frac{1}{1 + n\sin^{2} \pi x} +\] +is equal to~$0$ except when $x$~is an integer, and then equal to~$1$. The function +\[ +y = \lim_{n \to \infty} \frac{\psi(x) + n\phi(x) \sin^{2}\pi x}{1 + n\sin^{2}\pi x} +\] +is equal to~$\phi(x)$ unless $x$~is an integer, and then equal to~$\psi(x)$. + +\Item{15.} Show that the graph of the function +\[ +y = \lim_{n \to \infty} \frac{x^{n}\phi(x) + x^{-n}\psi(x)}{x^{n} + x^{-n}} +\] +\PageSep{159} +is composed of parts of the graphs of $\phi(x)$~and~$\psi(x)$, together with (as a rule) +two isolated points. Is $y$~defined when (\ia)~$x = 1$, (\ib)~$x = -1$, (\ic)~$x = 0$? + +\Item{16.} Prove that the function~$y$ which is equal to~$0$ when $x$~is rational, and +to~$1$ when $x$~is irrational, may be represented in the form +\[ +y = \lim_{m \to \infty} \sgn\{\sin^{2}(m!\, \pi x)\}, +\] +where +\[ +\sgn x = \lim_{n \to \infty} (2/\pi)\arctan(nx), +\] +{\Loosen as in \Ex{xxxi}.~14. [If $x$~is rational then $\sin^{2}(m!\, \pi x)$, and therefore +$\sgn\{\sin^{2}(m!\, \pi x)\}$, is equal to zero from a certain value of~$m$ onwards: if +$x$~is irrational then $\sin^{2}(m!\, \pi x)$ is always positive, and so $\sgn\{\sin^{2}(m!\, \pi x)\}$ +is always equal to~$1$.]} + +Prove that $y$~may also be represented in the form +\[ +1 - \lim_{m \to\infty} [\lim_{n \to\infty}\{\cos(m!\, \pi x)\}^{2n}]. +\] + +\Item{17.} Sum the series +\[ +\sum_{1}^{\infty} \frac{1}{\nu(\nu + 1)},\quad +\sum_{1}^{\infty} \frac{1}{\nu(\nu + 1)\dots(\nu + k)}. +\] + +[Since +\[ +%[** TN: Using \bigg to squeeze a bit of horizontal space] +\frac{1}{\nu(\nu + 1)\dots(\nu + k)} + = \frac{1}{k} \biggl\{\frac{1}{\nu(\nu + 1)\dots(\nu + k - 1)} + - \frac{1}{(\nu + 1)(\nu + 2)\dots(\nu + k)}\biggr\}, +\] +we have +\[ +\sum_{1}^{n} \frac{1}{\nu(\nu + 1)\dots(\nu + k)} + = \frac{1}{k}\left\{\frac{1}{1·2\dots k} + - \frac{1}{(n + 1)(n + 2)\dots (n + k)}\right\} +\] +and so +\[ +\sum_{1}^{\infty} \frac{1}{\nu(\nu + 1)\dots (\nu + k)} = \frac{1}{k(k!)}.] +\] + +\Item{18.} If $|z| < |\alpha|$, then +\begin{align*} +\frac{L}{z - \alpha} + = -&\frac{L}{\alpha}\left(1 + \frac{z}{\alpha} + \frac{z^{2}}{\alpha^{2}} + \dots\right); \\ +\intertext{and if $|z|>|\alpha|$, then} +\frac{L}{z - \alpha} + = &\frac{L}{z}\left(1 + \frac{\alpha}{z} + \frac{\alpha^{2}}{z^{2}} + \dots\right). +\end{align*} + +\Item{19.} \Topic{Expansion of $(Az + B)/(az^{2} + 2bz + c)$ in powers of~$z$.} Let $\alpha$,~$\beta$ +be the roots of $az^{2} + 2bz + c = 0$, so that $az^{2} + 2bz + c = a(z - \alpha)(z - \beta)$. We +shall suppose that $A$,~$B$, $a$,~$b$,~$c$ are all real, and $\alpha$~and~$\beta$ unequal. It is then +easy to verify that +\[ +\frac{Az + B}{az^{2} + 2bz + c} + = \frac{1}{a(\alpha - \beta)} + \left(\frac{A\alpha + B}{z - \alpha} + - \frac{A\beta + B}{z - \beta}\right). +\] +There are two cases, according as $b^{2} > ac$ or $b^{2} < ac$. + +\Item{(1)} If $b^{2} > ac$ then the roots $\alpha$,~$\beta$ are real and distinct. If $|z|$~is less than +either $|\alpha|$ or~$|\beta|$ we can expand $1/(z - \alpha)$ and $1/(z - \beta)$ in ascending powers of~$z$ +(Ex.~18). If $|z|$~is greater than either $|\alpha|$ or~$|\beta|$ we must expand in descending +powers of~$z$; while if $|z|$~lies between $|\alpha|$ and~$|\beta|$ one fraction must be expanded +in ascending and one in descending powers of~$z$. The reader should +write down the actual results. If $|z|$~is equal to~$|\alpha|$ or~$|\beta|$ then no such +expansion is possible. +\PageSep{160} + +\Item{(2)} If $b^{2} < ac$ then the roots are conjugate complex numbers (\Ref{Ch.}{III} +\SecNo[§]{43}), and we can write +\[ +\alpha = \rho\Cis\phi, \quad +\beta = \rho\Cis(-\phi), +\] +where $\rho^{2} = \alpha\beta = c/a$, $\rho\cos\phi = \frac{1}{2}(\alpha + \beta) = - b/a$, so that $\cos\phi = -\sqrtp{b^{2}/ac}$, +$\sin\phi = \sqrtb{1 - (b^{2}/ac)}$. + +If $|z| < \rho$ then each fraction may be expanded in ascending powers of~$z$. +The coefficient of~$z^{n}$ will be found to be +\[ +\frac{A\rho\sin n\phi + B\sin\{(n + 1)\phi\}}{a\rho^{n+1} \sin\phi}. +\] +If $|z| > \rho$ we obtain a similar expansion in descending powers, while if $|z| = \rho$ +no such expansion is possible. + +\Item{20.} Show that if $|z| < 1$ then +\[ +1 + 2z + 3z^{2} + \dots + (n + 1)z^{n} + \dots = 1/(1 - z)^{2}. +\] + +[The sum to $n$~terms is $\dfrac{1 - z^{n}}{(1 - z)^{2}} - \dfrac{nz^{n}}{1 - z}$.] + +\Item{21.} Expand $L/(z - \alpha)^{2}$ in powers of~$z$, ascending or descending according +as $|z| < |\alpha|$ or $|z| > |\alpha|$. + +\Item{22.} Show that if $b^{2} = ac$ and $|az| < |b|$ then +\[ +\frac{Az + B}{az^{2} + 2bz + c} = \sum_{0}^{\infty} p_{n}z^{n}, +\] +where $p_{n} = \{(-a)^{n}/b^{n+2}\} \{(n + 1)aB - nbA\}$; and find the corresponding expansion, +in descending powers of~$z$, which holds when $|az| > |b|$. + +\Item{23.} Verify the result of Ex.~19 in the case of the fraction $1/(1 + z^{2})$. [We +have $1/(1 + z^{2}) = \sum z^{n} \sin\{\frac{1}{2}(n + 1)\pi\} = 1 - z^{2} + z^{4} - \dots$.] + +\Item{24.} Prove that if $|z| < 1$ then +\[ +\frac{1}{1 + z + z^{2}} + = \frac{2}{\sqrt{3}} \sum_{0}^{\infty} z^{n} \sin\{\tfrac{2}{3}(n + 1)\pi\}. +\] + +\Item{25.} Expand $(1 + z)/(1 + z^{2})$, $(1 + z^{2})/(1 + z^{3})$ and $(1 + z + z^{2})/(1 + z^{4})$ in ascending +powers of~$z$. For what values of~$z$ do your results hold? + +\Item{26.} If $a/(a + bz + cz^{2}) = 1 + p_{1}z + p_{2}z^{2} + \dots$ then +\[ +1 + p_{1}^{2}z + p_{2}^{2}z^{2} + \dots + = \frac{a + cz}{a - cz}\, \frac{a^{2}}{a^{2} - (b^{2} - 2ac)z + c^{2}z^{2}}. +\] +\MathTrip{1900.} + +\Item{27.} If $\lim\limits_{n \to \infty} s_{n} = l$ then +\[ +\lim_{n \to \infty} \frac{s_{1} + s_{2} + \dots + s_{n}}{n} = l. +\] + +[Let $s_{n} = l + t_{n}$. Then we have to prove that $(t_{1} + t_{2} + \dots + t_{n})/n$ tends to +zero if $t_{n}$~does so. +\PageSep{161} + +We divide the numbers $t_{1}$, $t_{2}$,~\dots\Add{,} $t_{n}$ into two sets $t_{1}$, $t_{2}$,~\dots, $t_{p}$ and $t_{p+1}$, +$t_{p+2}$,~\dots, $t_{n}$. Here we suppose that $p$~is a function of~$n$ which tends to~$\infty$ +as $n \to \infty$, but \emph{more slowly than~$n$}, so that $p \to \infty$ and $p/n \to 0$: \eg\ we might +suppose $p$ to be the integral part of~$\sqrt{n}$. + +Let $\DPtypo{\epsilon}{\DELTA}$ be any positive number. However small $\DELTA$~may be, we can choose~$n_{0}$ +so that $t_{p+1}$, $t_{p+2}$,~\dots, $t_{n}$ are all numerically less than~$\frac{1}{2}\DELTA$ when $n \geq n_{0}$, and so +\[ +|(t_{p+1} + t_{p+2} + \dots + t_{n})/n| + < \tfrac{1}{2}\DELTA(n - p)/n < \tfrac{1}{2} \DELTA. +\] +But, if $A$~is the greatest of the moduli of all the numbers $t_{1}$, $t_{2}$,~\dots, we +have +\[ +|(t_{1} + t_{2} + \dots + t_{p})/n| < pA/n, +\] +and this also will be less than~$\frac{1}{2}\DELTA$ when $n \geq n_{0}$, if $n_{0}$~is large enough, since +$p/n \to 0$ as $n \to \infty$. Thus +\[ +|(t_{1} + t_{2} + \dots + t_{n})/n| + \leq |(t_{1} + t_{2} + \dots + t_{p})/n| + + |(t_{p+1} + \dots + t_{n})/n| < \DELTA +\] +when $n \geq n_{0}$; which proves the theorem. + +The reader, if he desires to become expert in dealing with questions about +limits, should study the argument above with great care. It is very often +necessary, in proving the limit of some given expression to be zero, to split it +into two parts which have to be proved to have the limit zero in slightly +different ways. When this is the case the proof is never very easy. + +The point of the proof is this: we have to prove that $(t_{1} + t_{2} + \dots + t_{n})/n$ is +small when $n$~is large, the~$t$'s being small when their suffixes are large. We +split up the terms in the bracket into two groups. The terms in the first +group are not all small, but their number is small compared with~$n$. The +number in the second group is \emph{not} small compared with~$n$, but the terms are +all small, and their number at any rate less than~$n$, so that their sum is small +compared with~$n$. Hence each of the parts into which $(t_{1} + t_{2} + \dots + t_{n})/n$ +has been divided is small when $n$~is large.] + +\Item{28.} If $\phi(n) - \phi(n - 1)\to l$ as $n \to \infty$, then $\phi(n)/n \to l$. + +[If $\phi(n) = s_{1} + s_{2} + \dots + s_{n}$ then $\phi(n) - \phi(n - 1) = s_{n}$, and the theorem reduces +to that proved in the last example.] + +\Item{29.} If $s_{n} = \frac{1}{2}\{1 - (-1)^{n}\}$, so that $s_{n}$~is equal to~$1$ or~$0$ according as $n$~is odd +or even, then $(s_{1} + s_{2} + \dots + s_{n})/n \to \frac{1}{2}$ as $n \to \infty$. + +[This example proves that the converse of Ex.~27 is not true: for $s_{n}$~oscillates +as $n \to \infty$.] + +\Item{30.} If $c_{n}$,~$s_{n}$ denote the sums of the first $n$~terms of the series +\[ +\tfrac{1}{2} + \cos\theta + \cos 2\theta + \dots,\quad +\sin\theta + \sin 2\theta + \dots, +\] +then +\[ +\lim (c_{1} + c_{2} + \dots + c_{n})/n = 0,\quad +\lim (s_{1} + s_{2} + \dots + s_{n})/n = \tfrac{1}{2} \cot\tfrac{1}{2} \theta. +\] +\end{Examples} +\PageSep{162} + + +\Chapter[LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE] +{V}{LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE. +CONTINUOUS AND DISCONTINUOUS FUNCTIONS} + +\Paragraph{89. Limits as $x$~tends to~$\infty$.} We shall now return to +functions of a continuous real variable. We shall confine ourselves +entirely to \emph{one-valued} functions,\footnote + {Thus $\sqrt{x}$ stands in this chapter for the one-valued function~$+\sqrt{x}$ and not (as + in \SecNo[§]{26}) for the two-valued function whose values are $+\sqrt{x}$~and~$-\sqrt{x}$.} +and we shall denote such +a function by~$\phi(x)$. We suppose $x$ to assume successively all +values corresponding to points on our fundamental straight line~$\Lambda$, +starting from some definite point on the line and progressing +always to the right. In these circumstances we say that \emph{$x$~tends +to infinity}, or \emph{to~$\infty$}, and write $x \to \infty$. The only difference +between the `tending of~$n$ to~$\infty$' discussed in the last chapter, and +this `tending of~$x$ to~$\infty$', is that $x$~assumes all values as it tends +to~$\infty$, \ie\ that the point~$P$ which corresponds to~$x$ coincides in +turn with every point of~$\Lambda$ to the right of its initial position, +whereas $n$~tended to~$\infty$ by a series of jumps. We can express this +distinction by saying that $x$~tends \emph{continuously} to~$\infty$. + +As we explained at the beginning of the last chapter, there is +a very close correspondence between functions of~$x$ and functions +of~$n$. Every function of~$n$ may be regarded as a selection from +the values of a function of~$x$. In the last chapter we discussed +the peculiarities which may characterise the behaviour of a +function~$\phi(n)$ as $n$~tends to~$\infty$. Now we are concerned with the +same problem for a function~$\phi(x)$; and the definitions and +theorems to which we are led are practically repetitions of those +of the last chapter. Thus corresponding to Def.~1 of \SecNo[§]{58} we +have: +\PageSep{163} + +\begin{Definition}[1.] +The function~$\phi(x)$ is said to tend to the limit~$l$ +as $x$~tends to~$\infty$ if, when any positive number~$\DELTA$, however small, is +assigned, a number~$x_{0}(\DELTA)$ can be chosen such that, for all values of~$x$ +equal to or greater than~$x_{0}(\DELTA)$, $\phi(x)$~differs from~$l$ by less than~$\DELTA$, +\ie\ if +\[ +|\phi(x) - l| < \DELTA +\] +when $x \geq x_{0}(\DELTA)$. +\end{Definition} + +When this is the case we may write +\[ +\lim_{x \to \infty} \phi(x) = l, +\] +or, when there is no risk of ambiguity, simply $\lim\phi(x) = l$, or +$\phi(x) \to l$. Similarly we have: + +\begin{Definition}[2.] +The function~$\phi(x)$ is said to tend to~$\infty$ with~$x$ +if, when any number~$\Delta$, however large, is assigned, we can choose +a number~$x_{0}(\Delta)$ such that +\[ +\phi(x) > \Delta +\] +when $x \geq x_{0}(\Delta)$. +\end{Definition} + +We then write +\[ +\phi(x) \to \infty. +\] +Similarly we define $\phi(x) \to -\infty$.\footnote + {We shall sometimes find it convenient to write~$+\infty$, $x \to +\infty$, $\phi(x) \to +\infty$ + instead of~$\infty$, $x \to \infty$, $\phi(x) \to \infty$.} +Finally we have: + +\begin{Definition}[3.] +If the conditions of neither of the two preceding +definitions are satisfied, then $\phi(x)$~is said to oscillate as $x$~tends +to~$\infty$. If $|\phi(x)|$~is less than some constant~$K$ when $x \geq x_{0}$,\footnote + {In the corresponding definition of \SecNo[§]{62}, we postulated that $|\phi(n)| < K$ for \emph{all} + values of~$n$, and not merely when $n \geq n_{0}$. But then the two hypotheses would have + been equivalent; for if $|\phi(n)| < K$ when $n \geq n_{0}$, then $|\phi(n)| < K'$ for all values + of~$n$, where $K'$~is the greatest of + \DPtypo{$\phi(1)$, $\phi(2)$,~\dots, $\phi(n_{0} - 1)$} + {$|\phi(1)|$, $|\phi(2)|$,~\dots, $|\phi(n_{0} - 1)|$} and~$K$. Here the + matter is not quite so simple, as there are infinitely many values of~$x$ less than~$x_{0}$.} +then +$\phi(x)$~is said to oscillate finitely, and otherwise infinitely. +\end{Definition} + +The reader will remember that in the last chapter we considered +very carefully various less formal ways of expressing the +facts represented by the formulae $\phi(n) \to l$, $\phi(n) \to \infty$. Similar +modes of expression may of course be used in the present case. +Thus we may say that $\phi(x)$~is small or nearly equal to~$l$ or large +when $x$~is large, using the words `small', `nearly', `large' in +a sense similar to that in which they were used in \Ref{Ch.}{IV}\@. +\PageSep{164} + +\begin{Examples}{XXXIV.} +\Item{1.} Consider the behaviour of the following functions +as $x \to \infty$: $1/x$, $1 + (1/x)$, $x^{2}$, $x^{k}$, $[x]$, $x - [x]$, $[x] + \sqrtb{x - [x]}$. + +The first four functions correspond exactly to functions of~$n$ fully discussed +in \Ref{Ch.}{IV}\@. The graphs of the last three were constructed in \Ref{Ch.}{II} +(\Exs{xvi}.\ 1,~2,~4), and the reader will see at once that $[x] \to \infty$, $x - [x]$ oscillates +finitely, and $[x] + \sqrtb{x - [x]} \to \infty$. + +One simple remark may be inserted here. The function $\phi(x) = x - [x]$ +oscillates between $0$~and~$1$, as is obvious from the form of its graph. It is +equal to zero whenever $x$~is an integer, so that the function~$\phi(n)$ derived +from it is always zero and so tends to the limit zero. The same is true if +\[ +\phi(x) = \sin x\pi,\quad +\phi(n) = \sin n\pi = 0. +\] +It is evident that $\phi(x) \to l$ or $\phi(x) \to \infty$ or $\phi(x) \to -\infty$ involves the corresponding +property for~$\phi(n)$, but that the converse is by no means always +true. + +\Item{2.} Consider in the same way the functions: +\[ +(\sin x\pi)/x,\quad +x\sin x\pi,\quad +(x\sin x\pi)^{2},\quad +\tan x\pi,\quad +a\cos^{2} x\pi + b\sin^{2} x\pi, +\] +illustrating your remarks by means of the graphs of the functions. + +\Item{3.} Give a geometrical explanation of Def.~1, analogous to the geometrical +explanation of \Ref{Ch.}{IV}, \SecNo[§]{59}. + +\Item{4.} If $\phi(x) \to l$, and $l$~is not zero, then $\phi(x)\cos x\pi$ and $\phi(x)\sin x\pi$ oscillate +finitely. If $\phi(x) \to \infty$ or $\phi(x) \to -\infty$, then they oscillate infinitely. The +graph of either function is a wavy curve oscillating between the curves +$y = \phi(x)$ and $y = -\phi(x)$. + +\Item{5.} Discuss the behaviour, as $x \to \infty$, of the function +\[ +y = f(x)\cos^{2} x\pi + F(x)\sin^{2} x\pi, +\] +where $f(x)$~and~$F(x)$ are some pair of simple functions (\eg\ $x$~and~$x^{2}$). [The +graph of~$y$ is a curve oscillating between the curves $y = f(x)$, $y = F(x)$.] +\end{Examples} + +\Paragraph{90. Limits as $x$~tends to~$-\infty$.} The reader will have no +difficulty in framing for himself definitions of the meaning of the +assertions `$x$~tends to~$-\infty$', or `$x \to -\infty$' and +\[ +\lim_{x \to -\infty} \phi(x) = l,\quad +\phi(x) \to \infty,\quad +\phi(x) \to -\infty. +\] +In fact, if $x = -y$ and $\phi(x) = \phi(-y) = \psi(y)$, then $y$~tends +to~$\infty$ as $x$~tends to~$-\infty$, and the question of the behaviour of~$\phi(x)$ +as $x$~tends to~$-\infty$ is the same as that of the behaviour of~$\psi(y)$ +as $y$~tends to~$\infty$. +\PageSep{165} + +\begin{Remark} +\Paragraph{91. Theorems corresponding to those of \Ref{Ch.}{IV}, \SecNo[§§]{63}--\SecNo{67}.} +The theorems concerning the sums, products, and quotients of functions +proved in \Ref{Ch.}{IV} are all true (with obvious verbal alterations which the +reader will have no difficulty in supplying) for functions of the continuous +variable~$x$. Not only the enunciations but the proofs remain substantially +the same. + +\Paragraph{92. Steadily increasing or decreasing functions.} The definition +which corresponds to that of \SecNo[§]{69} is as follows: \emph{the function~$\phi(x)$ will +be said to increase steadily with~$x$ if $\phi(x_{2}) \geq \phi(x_{1})$ whenever $x_{2} > x_{1}$}. In +many cases, of course, this condition is only satisfied from a definite value +of $x$ onwards, \ie\ when $x_{2} > x_{1} \geq x_{0}$. The theorem which follows in that section +requires no alteration but that of~$n$ into~$x$: and the proof is the same, except +for obvious verbal changes. + +{\Loosen If $\phi(x_{2}) > \phi(x_{1})$, the possibility of equality being excluded, whenever +$x_{2} > x_{1}$, then $\phi(x)$~will be said to be \emph{steadily increasing in the stricter sense}. +We shall find that the distinction is often important (cf.\ \SecNo[§§]{108}--\SecNo{109}).} + +The reader should consider whether or no\DPnote{** [sic], not "not"} the following functions +increase steadily with~$x$ (or at any rate increase steadily from a certain +value of $x$ onwards): $x^{2} - x$, $x + \sin x$, $x + 2\sin x$, $x^{2} + 2\sin x$, $[x]$, $[x] + \sin x$, +$[x] + \sqrtb{x - [x]}$. All these functions tend to~$\infty$ as $x \to \infty$. +\end{Remark} + +\Paragraph{93. Limits as $x$~tends to~$0$.} Let $\phi(x)$~be such a function +of~$x$ that $\lim\limits_{x \to \infty} \phi(x) = l$, and let $y = 1/x$. Then +\[ +\phi(x) = \phi(1/y) = \psi(y), +\] +say. As $x$~tends to~$\infty$, $y$~tends to the limit~$0$, and $\psi(y)$~tends to +the limit~$l$. + +Let us now dismiss~$x$ and consider $\psi(y)$ simply as a function +of~$y$. We are for the moment concerned only with those values +of~$y$ which correspond to large positive values of~$x$, that is to say +with small positive values of~$y$. And $\psi(y)$ has the property that +by making $y$ sufficiently small we can make $\psi(y)$ differ by as +little as we please from~$l$. To put the matter more precisely, +the statement expressed by $\lim\phi(x) = l$ means that, when any +positive number~$\DELTA$, however small, is assigned, we can choose~$x_{0}$ +so that $|\phi(x) - l| < \DELTA$ for all values of~$x$ greater than or equal +to~$x_{0}$. But this is the same thing as saying that we can choose +$y_{0} = 1/x_{0}$ so that $|\psi(y) - l| < \DELTA$ for all positive values of~$y$ less than +or equal to~$y_{0}$. + +We are thus led to the following definitions: +\PageSep{166} + +\begin{Defn} +\Item{A.} If, when any positive number~$\DELTA$, however small, is assigned, +we can choose~$y_{0}(\DELTA)$ so that +\[ +|\phi(y) - l| < \DELTA +\] +when $0 < y \leq y_{0}(\DELTA)$, then we say that $\phi(y)$~tends to the limit~$l$ as $y$~tends +to~$0$ by positive values, and we write +\[ +\lim_{y \to +0} \phi(y) = l. +\] +\end{Defn} + +\begin{Defn} +\Item{B.} If, when any number~$\Delta$, however large, is assigned, we can +choose $y_{0}(\Delta)$ so that +\[ +\phi(y) > \Delta +\] +when $0 < y \leq y_{0}(\Delta)$, then we say that $\phi(y)$~tends to~$\infty$ as $y$~tends +to~$0$ by positive values, and we write +\[ +\phi(y) \to \infty. +\] +\end{Defn} + +We define in a similar way the meaning of `$\phi(y)$~tends to +the limit~$l$ as $y$~tends to~$0$ by negative values', or `$\lim\phi(y) = l$ +when $y \to -0$'. We have in fact only to alter $0 < y \leq y_{0}(\DELTA)$ to +$-y_{0}(\DELTA) \leq y < 0$ in definition~A\@. There is of course a corresponding +analogue of definition~B, and similar definitions in which +\[ +\phi(y) \to -\infty +\] +as $y \to +0$ or $y \to -0$. + +If $\lim\limits_{y \to +0} \phi(y) = l$ \emph{and} $\lim\limits_{y \to -0} \phi(y) = l$, we write simply +\[ +\lim_{y \to 0} \phi(y) = l. +\] +This case is so important that it is worth while to give a formal +definition. + +\begin{Defn} +If, when any positive number~$\DELTA$, however small, is assigned, we +can choose $y_{0}(\DELTA)$ so that, for all values of~$y$ different from zero but +numerically less than or equal to~$y_{0}(\DELTA)$, $\phi(y)$~differs from~$l$ by less +than~$\DELTA$, then we say that $\phi(y)$~tends to the limit~$l$ as $y$~tends to~$0$, +and write +\[ +\lim_{y \to 0} \phi(y) = l. +\] +\end{Defn} + +So also, if $\phi(y) \to \infty$ as $y \to +0$ and also as $y \to -0$, we say +that $\phi(y) \to \infty$ as $y \to 0$. We define in a similar manner the +statement that $\phi(y) \to -\infty$ as $y \to 0$. +\PageSep{167} + +Finally, if $\phi(y)$~does not tend to a limit, or to~$\infty$, or to~$-\infty$, +as $y \to +0$, we say that $\phi(y)$~oscillates as $y \to +0$, finitely +or infinitely as the case may be; and we define oscillation as +$y \to -0$ in a similar manner. + +The preceding definitions have been stated in terms of a +variable denoted by~$y$: what letter is used is of course immaterial, +and we may suppose $x$ written instead of~$y$ throughout them. + +\Paragraph{94. Limits as $x$~tends to~$a$.} Suppose that $\phi(y) \to l$ as +$y \to 0$, and write +\[ +y = x - a,\quad +\phi(y) = \phi(x - a) = \psi(x). +\] +If $y \to 0$ then $x \to a$ and $\psi(x) \to l$, and we are naturally led to +write +\[ +\lim_{x \to a} \psi(x) = l, +\] +or simply $\lim\psi(x) = l$ or $\psi(x) \to l$, and to say that \emph{$\psi(x)$~tends to +the limit~$l$ as $x$~tends to~$a$}. The meaning of this equation may +be formally and directly defined as follows: +\begin{Defn}if, given~$\DELTA$, we can +always determine~$\EPSILON(\DELTA)$ so that +\[ +|\phi(x) - l| < \DELTA +\] +when $0 < |x - a| \leq \EPSILON(\DELTA)$, then +\[ +\lim_{x \to a} \phi(x) = l. +\] +\end{Defn} + +By restricting ourselves to values of~$x$ greater than~$a$, \ie\ by +replacing $0 < |x - a| \leq \EPSILON(\DELTA)$ by $a < x \leq a + \EPSILON(\DELTA)$, we define `$\phi(x)$~tends +to~$l$ when $x$~approaches~$a$ from the right', which we may +write as +\[ +\lim_{x \to a+0} \phi(x) = l. +\] +In the same way we can define the meaning of +\[ +\lim_{x \to a-0} \phi(x) = l. +\] +Thus $\lim\limits_{x \to a} \phi(x) = l$ is equivalent to the two assertions +\[ +\lim_{x \to a+0} \phi(x) = l,\quad +\lim_{x \to a-0} \phi(x) = l. +\] + +We can give similar definitions referring to the cases in which +$\phi(x) \to \infty$ or $\phi(x) \to -\infty$ as $x \to a$ through values greater or less +than~$a$; but it is probably unnecessary to dwell further on these +definitions, since they are exactly similar to those stated above in +\PageSep{168} +the special case when $a = 0$, and we can always discuss the +behaviour of~$\phi(x)$ as $x \to a$ by putting $x - a = y$ and supposing +that $y \to 0$. + +\begin{Remark} +\Paragraph{95. Steadily increasing or decreasing functions.} If there is a number~$\EPSILON$ +such that $\phi(x') \leq \phi(x'')$ whenever $a - \EPSILON < x' < x'' < a + \EPSILON$, then $\phi(x)$~will be +said to \emph{increase steadily in the neighbourhood of $x = a$}. + +Suppose first that $x < a$, and put $y = 1/(a - x)$. Then $y \to \infty$ as $x \to a-0$, +and $\phi(x) = \psi(y)$ is a steadily increasing function of~$y$, never greater than~$\phi(a)$. +It follows from \SecNo[§]{92} that $\phi(x)$~tends to a limit not greater than~$\phi(a)$. We +shall write +\[ +\lim_{x \to a+0} \phi(x) = \phi(a+0). +\] +We can define~$\phi(a-0)$ in a similar manner; and it is clear that +\[ +\phi(a-0) \leq \phi(a) \leq \phi(a+0). +\] +It is obvious that similar considerations may be applied to \emph{decreasing} +functions. + +{\Loosen If $\phi(x') < \phi(x'')$, the possibility of equality being excluded, whenever +$a - \EPSILON < x' < x'' < a + \EPSILON$, then $\phi(x)$~will be said to be \emph{steadily increasing in the +stricter sense}.} + +\Paragraph{96. Limits of indetermination and the principle of convergence.} +All of the argument of \SecNo[§§]{80}--\SecNo{84} may be applied to functions of a continuous +variable~$x$ which tends to a limit~$a$. In particular, if $\phi(x)$~is +\emph{bounded} in an interval including~$a$ (\ie\ if we can find $\EPSILON$,~$H$, and~$K$ so that +$H < \phi(x) < K$ when $a - \EPSILON \leq x \leq a + \EPSILON$).\footnote + {For some further discussion of the notion of \emph{a function bounded in an interval} + see \SecNo[§]{102}.} +then we can define $\lambda$~and~$\Lambda$, the lower and +upper limits of indetermination of~$\phi(x)$ as $x \to a$, and prove that the necessary +and sufficient condition that $\phi(x) \to l$ as $x \to a$ is that $\lambda = \Lambda = l$. We can also +establish the analogue of the principle of convergence, \ie\ prove that \emph{the +necessary and sufficient condition that $\phi(x)$ should tend to a limit as $x \to a$ is +that, when $\DELTA$~is given, we can choose~$\EPSILON(\DELTA)$ so that $|\phi(x_{2}) - \phi(x_{1})| < \DELTA$ when +$0 < |x_{2} - a| < |x_{1} - a| \leq \EPSILON(\DELTA)$}. +\end{Remark} + +\begin{Examples}{XXXV.} +\Item{1.} If +\[ +\phi(x) \to l,\quad +\psi(x) \to l', +\] +as $x \to a$, then $\phi(x) + \psi(x) \to l + l'$, $\phi(x)\psi(x) \to ll'$, and $\phi(x)/\psi(x) \to l/l'$, +unless in the last case $l' = 0$. + +[We saw in \SecNo[§]{91} that the theorems of \Ref{Ch.}{IV}, \SecNo[§§]{63}~\textit{et~seq.}\ hold also for +functions of~$x$ when $x \to \infty$ or $x \to -\infty$. By putting $x = 1/y$ we may extend +them to functions of~$y$, when $y \to 0$, and by putting $y = z - a$ to functions of~$z$, +when $z \to a$. +\PageSep{169} + +The reader should however try to prove them directly from the formal +definition given above. Thus, in order to obtain a strict direct proof of the +first result he need only take the proof of Theorem~I of \SecNo[§]{63} and write +throughout $x$~for~$n$, $a$~for~$\infty$ and $0 < |x - a| \leq \EPSILON$ for $n \geq n_{0}$.] + +\Item{2.} If $m$~is a positive integer then $x^{m} \to 0$ as $x \to 0$. + +\Item{3.} If $m$~is a negative integer then $x^{m} \to +\infty$ as $x \to +0$, while $x^{m} \to -\infty$ or +$x^{m} \to +\infty$ as $x \to -0$, according as $m$~is odd or even. If $m = 0$ then $x^{m} = 1$ +and $x^{m} \to 1$. + +\Item{4.} $\lim\limits_{x \to 0} (a + bx + cx^{2} + \dots + kx^{m}) = a$. + +\Item{5.} $\lim\limits_{x \to 0} \left\{(a + bx + \dots + kx^{m})/(\alpha + \beta x + \dots + \kappa x^{\mu})\right\} = a/\alpha$, unless $\alpha = 0$. If $\alpha = 0$ +and $a \neq 0$, $\beta \neq 0$, then the function tends to $+\infty$~or~$-\infty$, as $x \to +0$, according +as $a$~and~$\beta$ have like or unlike signs; the case is reversed if $x \to -0$. The +case in which both $a$~and~$\alpha$ vanish is considered in \Ex{xxxvi}.~5. Discuss the +cases which arise when $a \neq 0$ and more than one of the first coefficients in the +denominator vanish. + +\Item{6.} $\lim\limits_{x \to a} x^{m} = a^{m}$, if $m$~is any positive or negative integer, except when $a = 0$ +and $m$~is negative. [If $m > 0$, put $x = y + a$ and apply Ex.~4. When $m < 0$, +the result follows from Ex.~1 above. It follows at once that $\lim P(x) = P(a)$, +if $P(x)$~is any polynomial.] + +\Item{7.} $\lim\limits_{x \to a} R(x) = R(a)$, if $R$~denotes any rational function and $a$~is not one +of the roots of its denominator. + +\Item{8.} Show that $\lim\limits_{x \to a} x^{m} = a^{m}$ for all rational values of~$m$, except when $a = 0$ +and $m$~is negative. [This follows at once, when $a$~is positive, from the inequalities +\Eq{(9)}~or~\Eq{(10)} of \SecNo[§]{74}. For $|x^{m} - a^{m}| < H|x - a|$, where $H$~is the greater +of the absolute values of $mx^{m-1}$ and~$ma^{m-1}$ (cf.\ \Ex{xxviii}.~4). If $a$~is negative +we write $x = -y$ and $a = -b$. Then +\[ +\lim x^{m} = \lim (-1)^{m}y^{m} = (-1)^{m}b^{m} = a^{m}.] +\] +\end{Examples} + +\Paragraph{97.} The reader will probably fail to see at first that any proof +of such results as those of Exs.\ 4,~5, 6, 7,~8 above is necessary. +He may ask `why not simply put $x = 0$, or $x = a$? Of course +we then get $a$,~$a/\alpha$, $a^{m}$, $P(a)$,~$R(a)$'\Add{.} It is very important that he +should see exactly where he is wrong. We shall therefore consider +this point carefully before passing on to any further examples. + +The statement +\[ +\lim_{x \to 0} \phi(x) = l +\] +is a statement about the values of~$\phi(x)$ when $x$~has any value +\PageSep{170} +\emph{distinct from but differing by little from zero}.\footnote + {Thus in Def.~A of \SecNo[§]{93} we make a statement about values of~$y$ such that + $0 < y \leq y_{0}$, the first of these inequalities being inserted expressly in order to + exclude the value $y = 0$.} +It is \emph{not} a statement +about the \emph{value of~$\phi(x)$ when $x = 0$}. When we make the statement +we assert that, when $x$~is \emph{nearly} equal to zero, $\phi(x)$~is nearly +equal to~$l$. We assert nothing whatever about what happens +when $x$~is \emph{actually} equal to~$0$. So far as we know, $\phi(x)$~may +not be defined at all for $x = 0$; or it may have some value +other than~$l$. For example, consider the function defined for all +values of~$x$ by the equation $\phi(x) = 0$. It is obvious that +\[ +\lim\phi(x) = 0. +\Tag{(1)} +\] +{\Loosen Now consider the function~$\psi(x)$ which differs from~$\phi(x)$ only in +that $\psi(x) = 1$ when $x = 0$. Then} +\[ +\lim\psi(x) = 0, +\Tag{(2)} +\] +for, when $x$~is nearly equal to zero, $\psi(x)$~is not only nearly but +exactly equal to zero. But $\psi(0) = 1$. The graph of this function +consists of the axis of~$x$, with the point $x = 0$ left out, and one +isolated point, viz.\ the point $(0, 1)$. The equation~\Eq{(2)} expresses +the fact that if we move along the graph towards the axis of~$y$, +from either side, then the ordinate of the curve, being always equal +to zero, tends to the limit zero. This fact is in no way affected +by the position of the isolated point~$(0, 1)$. + +The reader may object to this example on the score of +artificiality: but it is easy to write down simple formulae representing +functions which behave precisely like this near $x = 0$. +One is +\[ +\psi(x) = [1 - x^{2}], +\] +where $[1 - x^{2}]$ denotes as usual the greatest integer not greater +than $1 - x^{2}$. For if $x = 0$ then $\psi(x) = [1] = 1$; while if $0 < x < 1$, +or $-1 < x < 0$, then $0 < 1 - x^{2} < 1$ and so $\psi(x) = [1 - x^{2}] = 0$. + +Or again, let us consider the function +\[ +y = x/x +\] +already discussed in \Ref{Ch.}{II}, \SecNo[§]{24},~\Eq{(2)}. This function is equal +to~$1$ for all values of~$x$ save $x = 0$. It is \emph{not} equal to~$1$ when +$x = 0$: it is in fact not defined at all for $x = 0$. For when we say +\PageSep{171} +that $\phi(x)$~is defined for $x = 0$ we mean (as we explained in \Ref{Ch.}{II}, +\textit{l.c.})\ that we can calculate its value for $x = 0$ by putting $x = 0$ +in the actual expression of~$\phi(x)$. In this case we cannot. When +we put $x = 0$ in~$\phi(x)$ we obtain~$0/0$, which is a meaningless +expression. The reader may object `divide numerator and denominator +by~$x$'. But he must admit that when $x = 0$ this is +impossible. Thus $y = x/x$ is a function which differs from $y = 1$ +solely in that it is not defined for $x = 0$. None the less +\[ +\lim(x/x) = 1, +\] +for $x/x$~is equal to~$1$ so long as $x$~differs from zero, however small +the difference may be. + +Similarly $\phi(x) = \{(x + 1)^{2} - 1\}/x = x + 2$ so long as $x$~is not +equal to zero, but is undefined when $x = 0$. None the less +$\lim\phi(x) = 2$. + +On the other hand there is of course nothing to prevent the +limit of~$\phi(x)$ as $x$~tends to zero from being equal to~$\phi(0)$, the value +of~$\phi(x)$ for $x = 0$. Thus if $\phi(x) = x$ then $\phi(0) = 0$ and $\lim\phi(x) = 0$. +This is in fact, from a practical point of view, \ie\ from the point +of view of what most frequently occurs in applications, the +ordinary case. + +\begin{Examples}{XXXVI.} +\Item{1.} $\lim\limits_{x \to a} (x^{2} - a^{2})/(x - a) = 2a$. + +\Item{2.} $\lim\limits_{x \to a} (x^{m} - a^{m})/(x - a) = ma^{m-1}$, if $m$~is any integer (zero included). + +\Item{3.} Show that the result of Ex.~2 remains true for all rational values +of~$m$, provided $a$~is positive. [This follows at once from the inequalities +\Eq{(9)}~and~\Eq{(10)} of~\SecNo[§]{74}.] + +\Item{4.} $\lim\limits_{x \to 1} (x^{7} - 2x^{5} + 1)/(x^{3} - 3x^{2} + 2) = 1$. [Observe that $x - 1$ is a factor of +both numerator and denominator.] + +\Item{5.} Discuss the behaviour of +\[ +\phi(x) = (a_{0}x^{m} + a_{1}x^{m+1} + \dots + a_{k}x^{m+k}) + /(b_{0}x^{n} + b_{1}x^{n+1} + \dots + b_{l}x^{n+l}) +\] +as $x$~tends to~$0$ by positive or negative values. + +[If $m > n$, $\lim\phi(x) = 0$. If $m = n$, $\lim\phi(x) = a_{0}/b_{0}$. If $m < n$ and $n - m$ is +even, $\phi(x) \to +\infty$ or $\phi(x) \to -\infty$ according as $a_{0}/b_{0} > 0$ or $a_{0}/b_{0} < 0$. If $m < n$ and +$n - m$ is odd, $\phi(x) \to +\infty$ as $x \to +0$ and $\phi(x) \to -\infty$ as $x \to -0$, or $\phi(x) \to -\infty$ +as $x \to +0$ and $\phi(x) \to +\infty$ as $x \to -0$, according as $a_{0}/b_{0} > 0$ or $a_{0}/b_{0} < 0$.] +\PageSep{172} + +\Item{6.} \Topic{Orders of smallness}. When $x$~is small $x^{2}$~is very much smaller, +$x^{3}$~much smaller still, and so on: in other words +\[ +\lim_{x\to 0} (x^{2}/x) = 0,\quad +\lim_{x\to 0} (x^{3}/x^{2}) = 0,\ \dots. +\] + +Another way of stating the matter is to say that, when $x$~tends to~$0$, +$x^{2}$, $x^{3}$,~\dots\ all also tend to~$0$, but $x^{2}$~tends to~$0$ more rapidly than~$x$, $x^{3}$~than~$x^{2}$, +and so on. It is convenient to have some scale by which to measure +the rapidity with which a function, whose limit, as $x$~tends to~$0$, is~$0$, +diminishes with~$x$, and it is natural to take the simple functions $x$,~$x^{2}$, $x^{3}$,~\dots\ +as the measures of our scale. + +We say, therefore, that \emph{$\phi(x)$~is of the first order of smallness} if $\phi(x)/x$ +tends to a limit other than~$0$ as $x$~tends to~$0$. Thus $2x + 3x^{2} + x^{7}$ is of the +first order of smallness, since $\lim(2x + 3x^{2} + x^{7})/x = 2$. + +Similarly we define the second, third, fourth,~\dots\ orders of smallness. It +must not be imagined that this scale of orders of smallness is in any way +complete. If it were complete, then every function~$\phi(x)$ which tends to zero +with~$x$ would be of either the first or second or some higher order of smallness. +This is obviously not the case. For example $\phi(x) = x^{7/5}$ tends to zero more +rapidly than~$x$ and less rapidly than~$x^{2}$. + +The reader may not unnaturally think that our scale might be made +complete by including in it \emph{fractional} orders of smallness. Thus we might +say that $x^{7/5}$~was of the $\frac{7}{5}$th~order of smallness. We shall however see later +on that such a scale of orders would still be altogether incomplete. And +as a matter of fact the \emph{integral} orders of smallness defined above are so +much more important in applications than any others that it is hardly +necessary to attempt to make our definitions more precise. + +\Topic{Orders of greatness.} Similar definitions are at once suggested to +meet the case in which $\phi(x)$~is large (positively or negatively) when $x$~is +small. We shall say that $\phi(x)$~is of the $k$th~order of greatness when $x$~is small +if $\phi(x)/x^{-k} = x^{k}\phi(x)$ tends to a limit different from~$0$ as $x$~tends to~$0$. + +These definitions have reference to the case in which $x \to 0$. There are of +course corresponding definitions relating to the cases in which $x \to \infty$ or $x \to a$. +Thus if $x^{k}\phi(x)$~tends to a limit other than zero, as $x \to \infty$, then we say that +$\phi(x)$~is of the $k$th~order of smallness when $x$~is large: while if $(x - a)^{k}\phi(x)$ +tends to a limit other than zero, as $x \to a$, then we say that $\phi(x)$~is of the $k$th~order +of greatness when $x$~is nearly equal to~$a$. + +\Item{7.\footnotemark} $\lim\sqrtp{1 + x} = \lim\sqrtp{1 - x} = 1$. [Put $1 + x = y$ or $1 - x = y$, and use +\Ex{xxxv}.~8.] + \footnotetext{In the examples which follow it is to be assumed that limits as $x \to 0$ are + required, unless (as in Exs.~19,~22) the contrary is explicitly stated.} + +\Item{8.} $\lim\{\sqrtp{1 + x} - \sqrtp{1 - x}\}/x = 1$. [Multiply numerator and denominator +by $\sqrtp{1 + x} + \sqrtp{1 - x}$.] +\PageSep{173} + +\Item{9.} Consider the behaviour of $\{\sqrtp{1 + x^{m}} - \sqrtp{1 - x^{m}}\}/x^{n}$ as $x \to 0$, $m$~and~$n$ +being positive integers. + +\Item{10.} $\lim\{\sqrtp{1 + x + x^{2}} - 1\}/x = \frac{1}{2}$. + +\Item{11.} $\lim\dfrac{\sqrtp{1 + x} - \sqrtp{1 + x^{2}}}{\sqrtp{1 - x^{2}} - \sqrtp{1 - x}} = 1$. + +\Item{12.} Draw a graph of the function +\[ +y = \biggl\{\frac{1}{x - 1} + + \frac{1}{x - \tfrac{1}{2}} + + \frac{1}{x - \tfrac{1}{3}} + + \frac{1}{x - \tfrac{1}{4}}\biggr\} \bigg/ + \biggl\{\frac{1}{x - 1} + + \frac{1}{x - \tfrac{1}{2}} + + \frac{1}{x - \tfrac{1}{3}} + + \frac{1}{x - \tfrac{1}{4}}\biggr\}. +\] + +Has it a limit as $x \to 0$? [Here $y = 1$ except for $x = 1$, $\frac{1}{2}$,~$\frac{1}{3}$,~$\frac{1}{4}$, when $y$~is +not defined, and $y \to 1$ as $x \to 0$.] + +\Item{13.} $\lim\dfrac{\sin x}{x} = 1$. + +[It may be deduced from the definitions of the trigonometrical ratios\footnote + {The proofs of the inequalities which are used here depend on certain properties + of the area of a sector of a circle which are usually taken as geometrically + intuitive; for example, that the area of the sector is greater than that of the + triangle inscribed in the sector. The justification of these assumptions must be + postponed to \Ref{Ch.}{VII}\@.} +that +if $x$~is positive and less than~$\frac{1}{2}\pi$ then +\[ +\sin x < x < \tan x +\] +or +\[ +\cos x < \frac{\sin x}{x} < 1 +\] +or +\[ +0 < 1 - \frac{\sin x}{x} < 1 - \cos x = 2\sin^{2} \tfrac{1}{2} x. +\] + +But $2\sin^{2} \frac{1}{2} x < 2(\frac{1}{2} x)^{2} \DPtypo{<}{=} \frac{1}{2} x^{2}$\Add{.} +Hence $\lim\limits_{x \to +0} \left(1 - \dfrac{\sin x}{x}\right) = 0$, and $\lim\limits_{x \to +0} \dfrac{\sin x}{x} = 1$. +As $\dfrac{\sin x}{x}$~is an even function, the result follows.] + +\Item{14.} $\lim \dfrac{1 - \cos x}{x^{2}} = \frac{1}{2}$. + +\Item{15.} $\lim \dfrac{\sin \alpha x}{x} = \alpha$. Is this true if $\alpha = 0$? + +\Item{16.} $\lim \dfrac{\arcsin x}{x} = 1$. [Put $x = \sin y$.] + +\Item{17.} $\lim \dfrac{\tan \alpha x}{x}= \alpha$,\quad $\lim\dfrac{\arctan \alpha x}{x} = \alpha$. + +\Item{18.} $\lim \dfrac{\cosec x - \cot x}{x} = \frac{1}{2}$. + +\Item{19.} $\lim\limits_{x \to 1} \dfrac{1 + \cos \pi x}{\tan^{2}\pi x} = \frac{1}{2}$. +\PageSep{174} + +\Item{20.} {\Loosen How do the functions $\sin(1/x)$, $(1/x)\sin(1/x)$, $x\sin(1/x)$ behave +as $x \to 0$? [The first oscillates finitely, the second infinitely, the third +tends to the limit~$0$. None is defined when $x = 0$. See \Exs{xv}.\ 6,~7,~8.]} + +\Item{21.} Does the function +\[ +y = \biggl(\sin \frac{1}{x}\biggr)\bigg/\biggr(\sin \frac{1}{x}\biggr) +\] +tend to a limit as $x$~tends to~$0$? [\emph{No}. The function is equal to~$1$ except when +$\sin(1/x) = 0$; \ie\ when $x = 1/\pi$, $1/2\pi$,~\dots, $-1/\pi$, $-1/2\pi$,~\dots. For these values the +formula for~$y$ assumes the meaningless form~$0/0$, and $y$~is therefore not defined +for an infinity of values of~$x$ near $x = 0$.] + +\Item{22.} Prove that if $m$ is any integer then $[x] \to m$ and $x - [x] \to 0$ as +$x \to m+0$, and $[x] \to m - 1$, $x - [x] \to 1$ as $x \to m-0$. +\end{Examples} + +\Paragraph{98. Continuous functions of a real variable.} The +reader has no doubt some idea as to what is meant by a \emph{continuous +curve}. Thus he would call the curve~$C$ in \Fig{29} continuous, +the curve~$C'$ generally continuous but discontinuous for $x = \xi'$ and +$x = \xi''$. +%[Illustration: Fig. 29.] +\Figure{29}{p174} + +Either of these curves may be regarded as the graph of a +function~$\phi(x)$. It is natural to call a function \emph{continuous} if its +graph is a continuous curve, and otherwise discontinuous. Let us +take this as a provisional definition and try to distinguish more +precisely some of the properties which are involved in it. + +In the first place it is evident that the property of the +function $y = \phi(x)$ of which $C$ is the graph may be analysed into +some property possessed by the curve at each of its points. +To be able to define continuity \emph{for all values of~$x$} we must first +define continuity \emph{for any particular value of~$x$}. Let us therefore +fix on some particular value of~$x$, say the value $x = \xi$ +\PageSep{175} +corresponding to the point~$P$ of the graph. What are the +characteristic properties of~$\phi(x)$ associated with this value of~$x$? + +In the first place \emph{$\phi(x)$~is defined for $x = \xi$}. This is obviously +essential. If $\phi(\xi)$~were not defined there would be a point +missing from the curve. + +Secondly \emph{$\phi(x)$~is defined for all values of~$x$ near $x = \xi$}; \ie\ we +can find an interval, including $x = \xi$ in its interior, for all points +of which $\phi(x)$~is defined. + +Thirdly \emph{if $x$~approaches the value~$\xi$ from either side then $\phi(x)$~approaches +the limit~$\phi(\xi)$}. + +The properties thus defined are far from exhausting those +which are possessed by the curve as pictured by the eye of +common sense. This picture of a curve is a generalisation from +particular curves such as straight lines and circles. But they are +the simplest and most fundamental properties: and the graph of +any function which has these properties would, so far as drawing +it is practically possible, satisfy our geometrical feeling of what a +continuous curve should be. We therefore select these properties +as embodying the mathematical notion of continuity. We are thus +led to the following + +\begin{Definition} +The function~$\phi(x)$ is said to be continuous for +$x = \xi$ if it tends to a limit as $x$~tends to~$\xi$ from either side, and +each of these limits is equal to~$\phi(\xi)$. +\end{Definition} + +We can now define \emph{continuity throughout an interval}. The +function~$\phi(x)$ is said to be continuous throughout a certain +interval of values of~$x$ if it is continuous for all values of~$x$ in that +interval. It is said to be \emph{continuous everywhere} if it is continuous +for every value of~$x$. Thus $[x]$~is continuous in the interval +$\DPmod{(\EPSILON, 1 - \EPSILON)}{[\EPSILON, 1 - \EPSILON]}$, where $\EPSILON$~is any positive number less than~$\frac{1}{2}$; and $1$ and~$x$ +are continuous everywhere.\PageLabel{175} + +If we recur\DPnote{** [sic]} to the definitions of a limit we see that our +definition is equivalent to `\emph{$\phi(x)$~is continuous for $x = \xi$ if, given~$\DELTA$, +we can choose~$\EPSILON(\DELTA)$ so that $|\phi(x) - \phi(\xi)| < \DELTA$ if $0 \leq |x - \xi| \leq \EPSILON(\DELTA)$}'. + +We have often to consider functions defined only in an interval +$\DPmod{(a, b)}{[a, b]}$. In this case it is convenient to make a slight and obvious +\PageSep{176} +change in our definition of continuity in so far as it concerns the +particular points $a$~and~$b$. We shall then say that $\phi(x)$~is continuous +for $x = a$ if $\phi(a + 0)$ exists and is equal to~$\phi(a)$, and for +$x = b$ if $\phi(b - 0)$ exists and is equal to~$\phi(b)$. + +\Paragraph{99.} The definition of continuity given in the last section may +be illustrated geometrically as follows. Draw the two horizontal +lines $y = \phi(\xi) - \DELTA$ and $y = \phi(\xi) + \DELTA$. Then $|\phi(x) - \phi(\xi)| < \DELTA$ expresses +the fact that the point on the curve corresponding to~$x$ lies +%[Illustration: Fig. 30.] +\ifthenelse{\boolean{Modernize}}{% +\Figure[\textwidth]{30}{p176}% +}{% +\Figure[\textwidth]{30}{p176_orig_notation}% +} +between these two lines. Similarly $|x - \xi| \leq \EPSILON$ expresses the fact +that $x$~lies in the interval $\DPmod{(\xi - \EPSILON, \xi + \EPSILON)}{[\xi - \EPSILON, \xi + \EPSILON]}$. Thus our definition asserts +that if we draw two such horizontal lines, no matter how close +together, we can always cut off a vertical strip of the plane by +two vertical lines in such a way that all that part of the curve +which is contained in the strip lies between the two horizontal +lines. This is evidently true of the curve~$C$ (\Fig{29}), whatever +value $\xi$ may have. + +We shall now discuss the continuity of some special types of +functions. Some of the results which follow were (as we pointed +out at the time) tacitly assumed in \Ref{Ch.}{II}\@. + +\begin{Examples}{XXXVII.} +\Item{1.} The sum or product of two functions continuous +at a point is continuous at that point. The quotient is also continuous +unless the denominator vanishes at the point. [This follows at once from +\Ex{xxxv}.~1.] + +\Item{2.} Any polynomial is continuous for all values of~$x$. Any rational +fraction is continuous except for values of~$x$ for which the denominator +vanishes. [This follows from \Exs{xxxv}.\ 6,~7.] +\PageSep{177} + +\Item{3.} $\sqrt{x}$~is continuous for all positive values of~$x$ (\Ex{xxxv}.~8). It is not +defined when $x < 0$, but is continuous for $x = 0$ in virtue of the remark made at +the end of \SecNo[§]{98}. The same is true of~$x^{m/n}$, where $m$~and~$n$ are any positive +integers of which $n$ is even. + +\Item{4.} The function~$x^{m/n}$, where $n$~is odd, is continuous for all values of~$x$. + +\Item{5.} $1/x$~is not continuous for $x = 0$. It has no value for $x = 0$, nor does it +tend to a limit as $x \to 0$. In fact $1/x \to +\infty$ or $1/x \to -\infty$ according as $x \to 0$ +by positive or negative values. + +\Item{6.} Discuss the continuity of~$x^{-m/n}$, where $m$~and~$n$ are positive integers, +for $x = 0$. + +\Item{7.} The standard rational function $R(x) = P(x)/Q(x)$ is discontinuous for +$x = a$, where $a$~is any root of $Q(x) = 0$. Thus $(x^{2} + 1)/(x^{2} - 3x + 2)$ is discontinuous +for $x = 1$. It will be noticed that in the case of rational functions a +discontinuity is always associated with (\ia)~a failure of the definition for a +particular value of~$x$ and (\ib)~a tending of the function to~$+\infty$ or~$-\infty$ as $x$~approaches +this value from either side. Such a particular kind of point of +discontinuity is usually described as an \Emph{infinity} of the function. An `infinity' +is the kind of discontinuity of most common occurrence in ordinary work. + +\Item{8.} Discuss the continuity of +\[ +\sqrtb{(x - a)(b - x)},\quad +\sqrtb[3]{(x - a)(b - x)},\quad +\sqrtb{(x - a)/(b - x)},\quad +\sqrtb[3]{(x - a)/(b - x)}\Add{.} +\] + +\Item{9.} $\sin x$ and $\cos x$ are continuous for all values of~$x$. + +[We have +\[ +\sin(x + h) - \sin x = 2\sin \tfrac{1}{2}h \cos(x + \tfrac{1}{2}h), +\] +which is numerically less than the numerical value of~$h$.] + +\Item{10.} For what values of~$x$ are $\tan x$, $\cot x$, $\sec x$, and $\cosec x$ continuous +or discontinuous? + +\Item{11.} If $f(y)$~is continuous for $y = \eta$, and $\phi(x)$~is a continuous function of~$x$ +which is equal to~$\eta$ when $x = \xi$, then $f\{\phi(x)\}$~is continuous for $x = \xi$. + +\Item{12.} If $\phi(x)$~is continuous for any particular value of~$x$, then any polynomial +in~$\phi(x)$, such as $a\{\phi(x)\}^{m} + \dots$, is so too. + +\Item{13.} Discuss the continuity of +\[ +1/(a\cos^{2} x + b\sin^{2} x),\quad +\sqrtp{2 + \cos x},\quad +\sqrtp{1 + \sin x},\quad +1/\sqrtp{1 + \sin x}. +\] + +\Item{14.} $\sin(1/x)$, $x\sin(1/x)$, and $x^{2}\sin(1/x)$ are continuous except for $x = 0$. + +\Item{15.} The function which is equal to $x\sin(1/x)$ except when $x = 0$, and to +zero when $x = 0$, is continuous for all values of~$x$. + +\Item{16.} $[x]$ and $x - [x]$ are discontinuous for all integral values of~$x$. + +\Item{17.} For what (if any) values of~$x$ are the following functions discontinuous: +$[x^{2}]$, $[\sqrt{x}\,]$, $\sqrtp{x - [x]}$, $[x] + \sqrtp{x - [x]}$, $[2x]$, $[x] + [-x]$? +\PageSep{178} + +\Item{18.} \Topic{Classification of discontinuities.} Some of the preceding examples +suggest a classification of different types of discontinuity. + +\SubItem{(1)} Suppose that $\phi(x)$~tends to a limit as $x \to a$ either by values less +than or by values greater than~$a$. Denote these limits, as in \SecNo[§]{95}, by $\phi(a - 0)$ +and $\phi(a + 0)$ respectively. Then, for continuity, it is necessary and sufficient +that $\phi(x)$~should be defined for $x = a$, and that $\phi(a - 0) = \phi(a) = \phi(a + 0)$. Discontinuity +may arise in a variety of ways. + +\Item{($\alpha$)} $\phi(a - 0)$ may be equal to $\phi(a + 0)$, but $\phi(a)$~may not be defined, or +may differ from $\phi(a - 0)$ and~$\phi(a + 0)$. Thus if $\phi(x) = x \sin(1/x)$ and $a = 0$, +$\phi(0 - 0) = \phi(0 + 0) = 0$, but $\phi(x)$~is not defined for $x = 0$. Or if $\phi(x) = [1 - x^{2}]$ +and $a = 0$, $\phi(0 - 0) = \phi(0 + 0) = 0$, but $\phi(0) = 1$. + +\Item{($\beta$)} {\Loosen$\phi(a - 0)$ and $\phi(a + 0)$ may be unequal. In this case $\phi(a)$~may be +equal to one or to neither, or be undefined. The first case is illustrated +by $\phi(x) = [x]$, for which $\phi(0 - 0) = -1$, $\phi(0 + 0) = \phi(0) = 0$; the second by +$\phi(x) = [x] - [-x]$, for which $\phi(0 - 0) = -1$, $\phi(0 + 0) = 1$, $\phi(0) = 0$; and the third +by $\phi(x) = [x] + x \sin(1/x)$, for which $\phi(0 - 0)= -1$, $\phi(0 + 0) = 0$, and $\phi(0)$~is +undefined.} + +In any of these cases we say that $\phi(x)$~has a \Emph{simple discontinuity} at +$x = a$. And to these cases we may add those in which $\phi(x)$~is defined only +on one side of $x = a$, and $\phi(a - 0)$ or~$\phi(a + 0)$, as the case may be, exists, but +$\phi(x)$~is either not defined when $x = a$ or has when $x = a$ a value different from +$\phi(a - 0)$ or~$\phi(a + 0)$. + +It is plain from \SecNo[§]{95} that \emph{a function which increases or decreases steadily +in the neighbourhood of $x = a$ can have at most a simple discontinuity for $x = a$}. + +\SubItem{(2)} It may be the case that only one (or neither) of $\phi(a - 0)$ and $\phi(a + 0)$ +exists, but that, supposing for example $\phi(a + 0)$ not to exist, $\phi(x) \to +\infty$ or +$\phi(x) \to -\infty$ as $x \to a+0$, so that $\phi(x)$~tends to a limit or to~$+\infty$ or to~$-\infty$ as +$x$~approaches~$a$ from either side. Such is the case, for instance, if $\phi(x) = 1/x$ or +$\phi(x) = 1/x^{2}$, and $a = 0$. In such cases we say (cf.\ Ex.~7) that $x = a$ is an \Emph{infinity} +of~$\phi(x)$. And again we may add to these cases those in which $\phi(x) \to +\infty$ +or $\phi(x) \to -\infty$ as $x \to a$ from one side, but $\phi(x)$~is not defined at all on the +other side of $x = a$. + +\SubItem{(3)} Any point of discontinuity which is not a point of simple discontinuity +nor an infinity is called a point of \Emph{oscillatory discontinuity}. Such +is the point $x = 0$ for the functions $\sin(1/x)$, $(1/x)\sin(1/x)$. + +\Item{19.} What is the nature of the discontinuities at $x = 0$ of the functions +$(\sin x)/x$, $[x] + [-x]$, $\cosec x$, $\sqrtp{1/x}$, $\sqrtp[3]{1/x}$, $\cosec(1/x)$, $\sin(1/x)/\sin(1/x)$? + +\Item{20.} The function which is equal to~$1$ when $x$~is rational and to~$0$ when +$x$~is irrational (\Ref{Ch.}{II}, \Ex{xvi}.~10) is discontinuous for all values of~$x$. So too +is any function which is defined only for rational or for irrational values of~$x$. +\PageSep{179} + +\Item{21.} {\Loosen The function which is equal to~$x$ when $x$~is irrational and to +$\sqrtb{(1 + p^{2})/(1 + q^{2})}$ when $x$~is a rational fraction~$p/q$ (\Ref{Ch.}{II}, \Ex{xvi}.~11) is +discontinuous for all negative and for positive rational values of~$x$, but +continuous for positive irrational values.} + +\Item{22.} For what points are the functions considered in \Ref{Ch.}{IV}, \Exs{xxxi} +discontinuous, and what is the nature of their discontinuities? [Consider, +\eg, the function $y = \lim x^{n}$ (Ex.~5). Here $y$~is only defined when $-1 < x \leq 1$: +it is equal to~$0$ when $-1 < x < 1$ and to~$1$ when $x = 1$. The points $x = 1$ and +$x = -1$ are points of simple discontinuity.] +\end{Examples} + +\Paragraph{100. The fundamental property of a continuous function.} +It may perhaps be thought that the analysis of the idea of a continuous +curve given in \SecNo[§]{98} is not the simplest or most natural +possible. Another method of analysing our idea of continuity is the +following. Let $A$~and~$B$ be two points on the graph of~$\phi(x)$ whose +coordinates are $x_{0}$,~$\phi(x_{0})$ and $x_{1}$,~$\phi(x_{1})$ respectively. Draw any +straight line~$\lambda$ which passes between $A$~and~$B$. Then common +sense certainly declares that if the graph of~$\phi(x)$ is continuous it +must cut~$\lambda$. + +If we consider this property as an intrinsic geometrical +property of continuous curves it is clear that there is no real +loss of generality in supposing $\lambda$ to be parallel to the axis of~$x$. +In this case the ordinates of $A$~and~$B$ cannot be equal: let us +suppose, for definiteness, that $\phi(x_{1}) > \phi(x_{0})$. And let $\lambda$ be the +line $y = \eta$, where $\phi(x_{0}) < \eta < \phi(x_{1})$. Then to say that the graph +of~$\phi(x)$ must cut~$\lambda$ is the same thing as to say that there is a +value of~$x$ between $x_{0}$~and~$x_{1}$ for which $\phi(x) = \eta$. + +We conclude then that a continuous function~$\phi(x)$ must +possess the following property: \emph{if +\[ +\phi(x_{0}) = y_{0},\quad +\phi(x_{1}) = y_{1}, +\] +and $y_{0} < \eta < y_{1}$, then there is a value of~$x$ between $x_{0}$~and~$x_{1}$ for which +$\phi(x) = \eta$}. In other words \emph{as $x$~varies from $x_{0}$ to~$x_{1}$, $y$~must assume +at least once every value between $y_{0}$~and~$y_{1}$}. + +We shall now prove that if $\phi(x)$~is a continuous function of~$x$ in +the sense defined in \SecNo[§]{98} then it does in fact possess this property. +There is a certain range of values of~$x$, to the right of~$x_{0}$, for which +$\phi(x) < \eta$. For $\phi(x_{0}) < \eta$, and so $\phi(x)$~is certainly less than~$\eta$ if +\PageSep{180} +$\phi(x) - \phi(x_{0})$ is numerically less than $\eta - \phi(x_{0})$. But since $\phi(x)$~is +continuous for $x = x_{0}$, this condition is certainly satisfied if $x$~is +near enough to~$x_{0}$. Similarly there is a certain range of values, +to the left of~$x_{1}$, for which $\phi(x) > \eta$. + +Let us divide the values of~$x$ between $x_{0}$~and~$x_{1}$ into two classes +$L$,~$R$ as follows: + +\Item{(1)} in the class~$L$ we put all values~$\xi$ of~$x$ such that $\phi(x) < \eta$ +when $x = \xi$ and for all values of~$x$ between $x_{0}$~and~$\xi$; + +\Item{(2)} in the class~$R$ we put all the other values of~$x$, \ie\ all +numbers~$\xi$ such that either $\phi(\xi) \geq \eta$ or there is a value of~$x$ between +$x_{0}$~and~$\xi$ for which $\phi(x) \geq \eta$. + +Then it is evident that these two classes satisfy all the +conditions imposed upon the classes $L$,~$R$ of \SecNo[§]{17}, and so constitute +a section of the real numbers. Let $\xi_{0}$ be the number corresponding +to the section. + +First suppose $\phi(\xi_{0}) > \eta$, so that $\xi_{0}$~belongs to the upper class: +and let $\phi(\xi_{0}) = \eta + k$, say. Then $\phi(\xi') < \eta$ and so +\[ +\phi(\xi_{0}) - \phi(\xi') > k, +\] +for all values of~$\xi'$ less than~$\xi_{0}$, which contradicts the condition of +continuity for $x = \xi_{0}$. + +Next suppose $\phi(\xi_{0}) = \eta - k < \eta$. Then, if $\xi'$~is any number +greater than~$\xi_{0}$, either $\phi(\xi') \geq \eta$ or we can find a number~$\xi''$ +between $\xi_{0}$~and~$\xi'$ such that $\phi(\xi'') \geq \eta$. In either case we can +find a number as near to~$\xi_{0}$ as we please and such that the corresponding +values of~$\phi(x)$ differ by more than~$k$. And this again +contradicts the hypothesis that $\phi(x)$~is continuous for $x = \xi_{0}$. + +Hence $\phi(\xi_{0}) = \eta$, and the theorem is established. It should +be observed that we have proved more than is asserted explicitly +in the theorem; we have proved in fact that $\xi_{0}$~is the \emph{least} value +of~$x$ for which $\phi(x) = \eta$. It is not obvious, or indeed generally +true, that there is a least among the values of~$x$ for which a +function assumes a given value, though this is true for continuous +functions. + +\begin{Remark} +It is easy to see that the converse of the theorem just proved is not +true. Thus such a function as the function~$\phi(x)$ whose graph is represented +\PageSep{181} +by \Fig{31} obviously assumes at least once every value between $\phi(x_{0})$ and~$\phi(x_{1})$: +yet $\phi(x)$~is discontinuous. Indeed it is not even true that $\phi(x)$~must +be continuous when it assumes each value \emph{once and once only}. Thus let $\phi(x)$ +be defined as follows from $x = 0$ to $x = 1$. If $x = 0$ let $\phi(x) = 0$; if $0 < x < 1$ +let $\phi(x) = 1 - x$; and if $x = 1$ let $\phi(x) = 1$. The graph of the function is +shown in \Fig{32}; it includes the points $O$,~$C$ but \emph{not} the points $A$,~$B$. It +is clear that, as $x$~varies from $0$ to~$1$, $\phi(x)$~assumes once and once only every +value between $\phi(0) = 0$ and $\phi(1) = 1$; but $\phi(x)$~is discontinuous for $x = 0$ and +$x = 1$. +%[Illustration: Fig. 31.] +%[Illustration: Fig. 32.] +\Figures{2.5in}{31}{p181a}{2in}{32}{p181b} + +As a matter of fact, however, the curves which usually occur in elementary +mathematics are composed of \emph{a finite number of pieces along which $y$~always +varies in the same direction}. It is easy to show that if $y = \phi(x)$ always varies +in the same direction, \ie\ steadily increases or decreases, as $x$~varies from $x_{0}$ to~$x_{1}$, + then the two notions of continuity are really equivalent, \ie\ that if +$\phi(x)$~takes every value between $\phi(x_{0})$ and~$\phi(x_{1})$ then it must be a continuous +function in the sense of \SecNo[§]{98}\Add{.} For let $\xi$ be any value of~$x$ between $x_{0}$ and~$x_{1}$. +As $x \to \xi$ through values less than~$\xi$, $\phi(x)$~tends to the limit~$\phi(\xi - 0)$ +(\SecNo[§]{95}). Similarly as $x \to \xi$ through values greater than~$\xi$, $\phi(x)$~tends to the +limit~$\phi(\xi + 0)$. The function will be continuous for $x = \xi$ if and only if +\[ +\phi(\xi - 0) = \phi(\xi) = \phi(\xi + 0)\Add{.} +\] +But if either of these equations is untrue, say the first, then it is evident that +$\phi(x)$~never assumes any value which lies between $\phi(\xi - 0)$ and~$\phi(\xi)$, which +is contrary to our assumption. Thus $\phi(x)$~must be continuous. The net +result of this and the last section is consequently to show that our common-sense +notion of what we mean by continuity is substantially accurate, and +capable of precise statement in mathematical terms. +\end{Remark} + +\Paragraph{101.} In this and the following paragraphs we shall state and +prove some general theorems concerning continuous functions. +\PageSep{182} + +\begin{Theorem}[1.] +Suppose that $\phi(x)$~is continuous for $x = \xi$, and +that $\phi(\xi)$~is positive. Then we can determine a positive number~$\EPSILON$ +such that $\phi(\xi)$~is positive throughout the interval $\DPmod{(\xi - \EPSILON, \xi + \EPSILON)}{[\xi - \EPSILON, \xi + \EPSILON]}$. +\end{Theorem} + +For, taking $\DELTA = \frac{1}{2}\phi(\xi)$ in the fundamental inequality of \PageRef{p.}{175}, +we can choose $\EPSILON$ so that +\[ +\DPtypo{\phi(x) - \phi(\xi)}{|\phi(x) - \phi(\xi)|} < \tfrac{1}{2}\phi(\xi) +\] +throughout $\DPmod{(\xi - \EPSILON, \xi + \EPSILON)}{[\xi - \EPSILON, \xi + \EPSILON]}$, and then +\[ +\phi(x) \geq \phi(\xi) - |\phi(x) - \phi(\xi)| > \tfrac{1}{2}\phi(\xi) > 0, +\] +so that $\phi(x)$~is positive. There is plainly a corresponding theorem +referring to negative values of~$\phi(x)$. + +\begin{Theorem}[2.] +If $\phi(x)$~is continuous for $x = \xi$, and $\phi(x)$~vanishes +for values of~$x$ as near to~$\xi$ as we please, or assumes, for values of~$x$ +as near to~$\xi$ as we please, both positive and negative values, then +$\phi(\xi) = 0$. +\end{Theorem} + +This is an obvious corollary of Theorem~1. If $\phi(\xi)$~is not zero, +it must be positive or negative; and if it were, for example, positive, +it would be positive for all values of~$x$ sufficiently near to~$\xi$, which +contradicts the hypotheses of the theorem. + +\Paragraph{102. The range of values of a continuous function.} Let +us consider a function~$\phi(x)$ about which we shall only assume at +present that it is defined for every value of~$x$ in an interval $\DPmod{(a, b)}{[a, b]}$. + +The values assumed by~$\phi(x)$ for values of~$x$ in~$\DPmod{(a, b)}{[a, b]}$ form an +aggregate~$S$ to which we can apply the arguments of \SecNo[§]{80}, as we +applied them in \SecNo[§]{81} to the aggregate of values of a function of~$n$. +If there is a number~$K$ such that $\phi(x) \leq K$, for all values of~$x$ in +question, we say that $\phi(x)$ is \emph{bounded above}. In this case $\phi(x)$ +possesses an \emph{upper bound}~$M$: no value of~$\phi(x)$ exceeds~$M$, but any +number less than~$M$ is exceeded by at least one value of~$\phi(x)$. +Similarly we define `\emph{bounded below}', `\emph{lower bound}', `\emph{bounded'}, as +applied to functions of a continuous variable~$x$. + +\begin{Theorem}[1.] +If $\phi(x)$ is continuous throughout~$\DPmod{(a, b)}{[a, b]}$, then it is +bounded in~$\DPmod{(a, b)}{[a, b]}$. +\end{Theorem} +\PageSep{183} + +We can certainly determine an interval~$\DPmod{(a, \xi)}{[a, \xi]}$, extending to +the right from~$a$, in which $\phi(x)$~is bounded. For since $\phi(x)$~is +continuous for $x = a$, we can, given any positive number~$\DELTA$ however +small, determine an interval~$\DPmod{(a, \xi)}{[a, \xi]}$ throughout which $\phi(x)$~lies +between $\phi(a) - \DELTA$ and $\phi(a) + \DELTA$; and obviously $\phi(x)$~is bounded in +this interval. + +Now divide the points~$\xi$ of the interval~$\DPmod{(a, b)}{[a, b]}$ into two classes +$L$,~$R$, putting~$\xi$ in~$L$ if $\phi(\xi)$~is bounded in~$\DPmod{(a, \xi)}{[a, \xi]}$, and in~$R$ if this +is not the case. It follows from what precedes that $L$~certainly +exists: what we propose to prove is that $R$~does not. Suppose +that $R$~does exist, and let $\beta$ be the number corresponding to the +section whose lower and upper classes are $L$~and~$R$. Since $\phi(x)$~is +continuous for $x = \beta$, we can, however small $\DELTA$ may be, determine +an interval $\DPmod{(\beta - \eta, \beta + \eta)}{[\beta - \eta, \beta + \eta]}$\footnote + {If $\beta = b$ we must replace this interval by $\DPmod{(\beta - \eta, \beta)}{[\beta - \eta, \beta]}$, and $\beta + \eta$ by~$\beta$, throughout + the argument which follows.} +throughout which +\[ +\phi(\beta) - \DELTA < \phi(x) < \phi(\beta) + \DELTA. +\] +Thus $\phi(x)$~is bounded in $\DPmod{(\beta - \eta, \beta + \eta)}{[\beta - \eta, \beta + \eta]}$. Now $\beta - \eta$ belongs to~$L$. +Therefore $\phi(x)$~is bounded in~$\DPmod{(a, \beta - \eta)}{[a, \beta - \eta]}$: and therefore it is +bounded in the whole interval $\DPmod{(a, \beta + \eta)}{[a, \beta + \eta]}$. But $\beta + \eta$ belongs to~$R$ +and so $\phi(x)$~is \emph{not} bounded in~$\DPmod{(a, \beta + \eta)}{[a, \beta + \eta]}$. This contradiction +shows that $R$~does not exist. And so $\phi(x)$~is bounded in the +whole interval $\DPmod{(a, b)}{[a, b]}$\Add{.} + +\begin{Theorem}[2.] +If $\phi(x)$~is continuous throughout~$\DPmod{(a, b)}{[a, b]}$, and $M$~and~$m$ +are its upper and lower bounds, then $\phi(x)$~assumes the values +$M$~and~$m$ at least once each in the interval. +\end{Theorem} + +For, given any positive number~$\DELTA$, we can find a value of~$x$ for +which $M - \phi(x) < \DELTA$ or $1/\{M - \phi(x)\} > 1/\DELTA$. Hence $1/\{M - \phi(x)\}$ +is not bounded, and therefore, by Theorem~1, is not continuous. +But $M - \phi(x)$ is a continuous function, and so $1/\{M - \phi(x)\}$ is +continuous at any point at which its denominator does not vanish +(\Ex{xxxvii}.~1). There must therefore be one point at which +the denominator vanishes: at this point $\phi(x) = M$. Similarly it +may be shown that there is a point at which $\phi(x) = m$. + +The proof just given is somewhat subtle and indirect, and it +may be well, in view of the great importance of the theorem, +to indicate alternative lines of proof. It will however be convenient +to postpone these for a moment.\footnote + {See \SecNo[§]{104}.} +\PageSep{184} + +\begin{Examples}{XXXVIII.} +\Item{1.} {\Loosen If $\phi(x) = 1/x$ except when $x = 0$, and $\phi(x) = 0$ +when $x = 0$, then $\phi(x)$~has neither an upper nor a lower bound in any +interval which includes $x = 0$ in its interior, as \eg\ the interval~$\DPmod{(-1, +1)}{[-1, +1]}$.} + +\Item{2.} If $\phi(x) = 1/x^{2}$ except when $x = 0$, and $\phi(x) = 0$ when $x = 0$, then $\phi(x)$~has +the lower bound~$0$, but no upper bound, in the interval~$\DPmod{(-1, +1)}{[-1, +1]}$. + +\Item{3.} Let $\phi(x) = \sin(1/x)$ except when $x = 0$, and $\phi(x) = 0$ when $x = 0$. Then +$\phi(x)$~is discontinuous for $x = 0$. In any interval~$\DPmod{(-\DELTA, +\DELTA)}{[-\DELTA, +\DELTA]}$ the lower bound is~$-1$ +and the upper bound~$+1$, and each of these values is assumed by~$\phi(x)$ an +infinity of times. + +\Item{4.} Let $\phi(x) = x - [x]$. This function is discontinuous for all integral +values of~$x$. In the interval~$\DPmod{(0, 1)}{[0, 1]}$ its lower bound is~$0$ and its upper bound~$1$. +It is equal to~$0$ when $x = 0$ or $x = 1$, but it is never equal to~$1$. Thus $\phi(x)$~never +assumes a value equal to its upper bound. + +\Item{5.} Let $\phi(x) = 0$ when $x$~is irrational, and $\phi(x) = q$ when $x$~is a rational +fraction~$p/q$. Then $\phi(x)$~has the lower bound~$0$, but no upper bound, in any +interval~$\DPmod{(a, b)}{[a, b]}$. But if $\phi(x) = (-1)^{p}q$ when $x = p/q$, then $\phi(x)$~has neither an +upper nor a lower bound in any interval. +\end{Examples} + +\Paragraph{103. The oscillation of a function in an interval.} Let +$\phi(x)$ be any function bounded throughout~$\DPmod{(a, b)}{[a, b]}$, and $M$~and~$m$ +its upper and lower bounds. We shall now use the notation +$M(a, b)$, $m(a, b)$ for $M$,~$m$, in order to exhibit explicitly the dependence +of $M$~and~$m$ on $a$~and~$b$, and we shall write +\[ +O(a, b) = M(a, b) - m(a, b). +\] + +This number~$O(a, b)$, the difference between the upper and +lower bounds of~$\phi(x)$ in~$\DPmod{(a, b)}{[a, b]}$, we shall call the \Emph{oscillation} \emph{of~$\phi(x)$ +in~$\DPmod{(a, b)}{[a, b]}$}. The simplest of the properties of the functions $M(a, b)$, +$m(a, b)$, $O(a, b)$ are as follows. + +\begin{Result} +\Item{(1)} If $a \leq c \leq b$ then $M(a, b)$~is equal to the greater of $M(a, c)$ +and~$M(c, b)$, and $m(a, b)$ to the lesser of $m(a, c)$ and~$m(c, b)$. +\end{Result} + +\begin{Result} +\Item{(2)} $M(a, b)$~is an increasing, $m(a, b)$~a decreasing, and $O(a, b)$ +an increasing function of~$b$. +\end{Result} + +\begin{Result} +\Item{(3)} $O(a, b) \leq O(a, c) + O(c, b)$. +\end{Result} + +The first two theorems are almost immediate consequences of +our definitions. Let $\mu$~be the greater of $M(a, c)$ and~$M(c, b)$, and +let $\DELTA$ be any positive number. Then $\phi(x) \leq \mu$ throughout $\DPmod{(a, c)}{[a, c]}$ +and~$\DPmod{(c, b)}{[c, b]}$, and therefore throughout~$\DPmod{(a, b)}{[a, b]}$; and $\phi(x) > \mu - \DELTA$ +somewhere in~$\DPmod{(a, c)}{[a, c]}$ or in~$\DPmod{(c, b)}{[c, b]}$, and therefore somewhere in~$\DPmod{(a, b)}{[a, b]}$. +\PageSep{185} +Hence $M(a, b) = \mu$. The proposition concerning~$m$ may be proved +similarly. Thus (1)~is proved, and (2)~is an obvious corollary. + +Suppose now that $M_{1}$~is the greater and $M_{2}$~the less of $M(a, c)$ +and~$M(c, b)$, and that $m_{1}$~is the less and $m_{2}$~the greater of $m(a, c)$ +and~$m(c, b)$. Then, since $c$~belongs to both intervals, $\phi(c)$~is not +greater than~$M_{2}$ nor less than~$m_{2}$. Hence $M_{2} \geq m_{2}$, whether these +numbers correspond to the same one of the intervals $\DPmod{(a, c)}{[a, c]}$ and +$\DPmod{(c, b)}{[c, b]}$ or not, and +\[ +O(a, b) = M_{1} - m_{1} \leq M_{1} + M_{2} - m_{1} - m_{2}. +\] +But +\[ +O(a, c) + O(c, b) = M_{1} + M_{2} - m_{1} - m_{2}; +\] +and (3)~follows. + +\begin{Remark} +\Paragraph{104. Alternative proofs of Theorem~2 of \SecNo[§]{102}.} The most straightforward +proof of Theorem~2 of \SecNo[§]{102} is as follows. Let $\xi$~be any number of +the interval~$\DPmod{(a, b)}{[a, b]}$. The function $M(a, \xi)$ increases steadily with~$\xi$ and never +exceeds~$M$. We can therefore construct a section of the numbers~$\xi$ by +putting~$\xi$ in~$L$ or in~$R$ according as $M(a, \xi) < M$ or $M(a, \xi) = M$. Let $\beta$~be +the number corresponding to the section. If $a < \beta < b$, we have +\[ +M(a, \beta - \eta) < M,\quad +M(a, \beta + \eta) = M +\] +for all positive values of~$\eta$, and so +\[ +M(\beta - \eta, \beta + \eta) = M, +\] +by~\Eq{(1)} of \SecNo[§]{103}. Hence $\phi(x)$~assumes, for values of~$x$ as near as we please to~$\beta$, +values as near as we please to~$M$, and so, since $\phi(x)$~is continuous, $\phi(\beta)$ +must be equal to~$M$. + +If $\beta = a$ then $M(a, a + \eta) = M$. And if $\beta = b$ then $M(a, b - \eta) < M$, and +so $M(b - \eta, b) = M$. In either case the argument may be completed as +before. + +The theorem may also be proved by the method of repeated bisection +used in \SecNo[§]{71}. If $M$~is the upper bound of~$\phi(x)$ in an interval~$PQ$, and $PQ$~is +divided into two equal parts, then it is possible to find a half~$P_{1}Q_{1}$ in which +the upper bound of~$\phi(x)$ is also~$M$. Proceeding as in \SecNo[§]{71}, we construct a +sequence of intervals $PQ$, $P_{1}Q_{1}$, $P_{2}Q_{2}$,~\dots\ in each of which the upper bound +of~$\phi(x)$ is~$M$. These intervals, as in \SecNo[§]{71}, converge to a point~$T$, and it is +easily proved that the value of~$\phi(x)$ at this point is~$M$. +\end{Remark} + +\Paragraph{105. Sets of intervals on a line. The Heine-Borel +Theorem.} We shall now proceed to prove some theorems concerning +the oscillation of a function which are of a somewhat +abstract character but of very great importance, particularly, as +we shall see later, in the theory of integration. These theorems +depend upon a general theorem concerning intervals on a line. +\PageSep{186} + +Suppose that we are given a \emph{set of intervals} in a straight +line, that is to say an aggregate each of whose members is an +interval~$\DPmod{(\alpha, \beta)}{[\alpha, \beta]}$. We make no restriction as to the nature of +these intervals; they may be finite or infinite in number; they +may or may not overlap;\footnote + {The word \emph{overlap} is used in its obvious sense: two intervals overlap if they + have points in common which are not end points of either. Thus $\DPmod{(0, \frac{2}{3})}{[0, \frac{2}{3}]}$ and~$\DPmod{(\frac{1}{3}, 1)}{[\frac{1}{3}, 1]}$ + overlap. A pair of intervals such as $\DPmod{(0, \frac{1}{2})}{[0, \frac{1}{2}]}$ and~$\DPmod{(\frac{1}{2}, 1)}{[\frac{1}{2}, 1]}$ may be said to \emph{abut}.} +and any number of them may be +included in others. + +\begin{Remark} +It is worth while in passing to give a few examples of sets of intervals to +which we shall have occasion to return later.\PageLabel{186} + +\Itemp{(i)} If the interval~$\DPmod{(0, 1)}{[0, 1]}$ is divided into $n$~equal parts then the $n$~intervals +thus formed define a finite set of non-overlapping intervals which just cover +up the line. + +\Itemp{(ii)} We take every point~$\xi$ of the interval~$\DPmod{(0, 1)}{[0, 1]}$, and associate with~$\xi$ the +interval $\DPmod{(\xi - \EPSILON, \xi + \EPSILON)}{[\xi - \EPSILON, \xi + \EPSILON]}$, where $\EPSILON$~is a positive number less than~$1$, except that +with~$0$ we associate $\DPmod{(0, \EPSILON)}{[0, \EPSILON]}$ and with~$1$ we associate $\DPmod{(1 - \EPSILON, 1)}{[1 - \EPSILON, 1]}$, and in general we +reject any part of any interval which projects outside the interval~$\DPmod{(0, 1)}{[0, 1]}$. We +thus define an infinite set of intervals, and it is obvious that many of them +overlap with one another. + +\Itemp{(iii)} We take the rational points~$p/q$ of the interval~$\DPmod{(0, 1)}{[0, 1]}$, and associate +with~$p/q$ the interval +\[ +\DPmod{\left(\frac{p}{q} - \frac{\EPSILON}{q^{3}}, + \frac{p}{q} + \frac{\EPSILON}{q^{3}}\right)} + {\left[\frac{p}{q} - \frac{\EPSILON}{q^{3}}, + \frac{p}{q} + \frac{\EPSILON}{q^{3}}\right]}, +\] +where $\EPSILON$~is positive and less than~$1$. We regard~$0$ as~$0/1$ and~$1$ as~$1/1$: in +these two cases we reject the part of the interval which lies outside~$\DPmod{(0, 1)}{[0, 1]}$. We +obtain thus an infinite set of intervals, which plainly overlap with one another, +since there are an infinity of rational points, other than~$p/q$, in the interval +associated with~$p/q$. +\end{Remark} + +\begin{ParTheorem}{The Heine-Borel Theorem.} +Suppose that we are given an +interval~$\DPmod{(a, b)}{[a, b]}$, and a set of intervals~$I$ each of whose members is +included in~$\DPmod{(a, b)}{[a, b]}$. Suppose further that $I$~possesses the following +properties: + +\Itemp{(i)} every point of~$\DPmod{(a, b)}{[a, b]}$, other than $a$~and~$b$, lies inside\footnote + {That is to say `in and not at an end of'.} +at +least one interval of~$I$; + +\Itemp{(ii)} $a$~is the left-hand end point, and $b$~the right-hand end +point, of at least one interval of~$I$. + +Then it is possible to choose \Emph{a finite number} of intervals from +the set~$I$ which form a set of intervals possessing the properties \Inum{(i)}~and~\Inum{(ii)}. +\end{ParTheorem} +\PageSep{187} + +We know that $a$~is the left-hand end point of at least one +interval of~$I$, say~$\DPmod{(a, a_{1})}{[a, a_{1}]}$. We know also that $a_{1}$~lies inside at least +one interval of~$I$, say~$\DPmod{(a_{1}', a_{2})}{[a_{1}', a_{2}]}$. Similarly $a_{2}$~lies inside an interval +$\DPmod{(a_{2}', a_{3})}{[a_{2}', a_{3}]}$ of~$I$. It is plain that this argument may be repeated indefinitely, +unless after a finite number of steps $a_{n}$~coincides with~$b$. + +If $a_{n}$~does coincide with~$b$ after a finite number of steps then +there is nothing further to prove, for we have obtained a finite set +of intervals, selected from the intervals of~$I$, and possessing the +properties required. If $a_{n}$~never coincides with~$b$, then the points +$a_{1}$,~$a_{2}$, $a_{3}$,~\dots\ must (since each lies to the right of its predecessor) +tend to a limiting position, but this limiting position may, so far +as we can tell, lie anywhere in~$\DPmod{(a, b)}{[a, b]}$. + +Let us suppose now that the process just indicated, starting +from~$a$, is performed in all possible ways, so that we obtain all +possible sequences of the type $a_{1}$,~$a_{2}$, $a_{3}$,~\dots. Then we can prove +that \emph{there must be at least one such sequence which arrives at~$b$ +after a finite number of steps}. +%[Illustration: Fig. 33.] +\Figure[\textwidth]{33}{p187} + +There are two possibilities with regard to any point~$\xi$ between +$a$~and~$b$. Either (i)~$\xi$~lies to the left of \emph{some} point~$a_{n}$ of \emph{some} +sequence or (ii)~it does not. We divide the points~$\xi$ into two +classes $L$~and~$R$ according as to whether (i)~or~(ii) is true. The +class~$L$ certainly exists, since all points of the interval $\DPmod{(a, a_{1})}{[a, a_{1}]}$ +belong to~$L$. We shall now prove that $R$~does not exist, so that +every point~$\xi$ belongs to~$L$. + +If $R$~exists then $L$~lies entirely to the left of~$R$, and the classes +$L$,~$R$ form a section of the real numbers between $a$~and~$b$, to +which corresponds a number~$\xi_{0}$. The point~$\xi_{0}$ lies inside an interval +of~$I$, say~$\DPmod{(\xi', \xi'')}{[\xi', \xi'']}$, and $\xi'$~belongs to~$L$, and so lies to the left of +some term~$a_{n}$ of some sequence. But then we can take $\DPmod{(\xi', \xi'')}{[\xi', \xi'']}$ +as the interval $\DPmod{(a_{n}', a_{n+1})}{[a_{n}', a_{n+1}]}$ associated with~$a_{n}$ in our construction +of the sequence $a_{1}$,~$a_{2}$, $a_{3}$,~\dots; and all points to the left of~$\xi''$ +lie to the left of~$a_{n+1}$. There are therefore points of~$L$ to the +right of~$\xi_{0}$, and this contradicts the definition of~$R$. It is +therefore impossible that $R$~should exist. +\PageSep{188} + +Thus every point~$\xi$ belongs to~$L$. Now $b$~is the right-hand +end point of an interval of~$I$, say~$\DPmod{(b_{1}, b)}{[b_{1}, b]}$, and $b_{1}$~belongs to~$L$. +Hence there is a member~$a_{n}$ of a sequence $a_{1}$,~$a_{2}$, $a_{3}$,~\dots\ such that +$a_{n} > b_{1}$. But then we may take the interval $\DPmod{(a_{n}', a_{n+1})}{[a_{n}', a_{n+1}]}$ corresponding +to~$a_{n}$ to be~$\DPmod{(b_{1}, b)}{[b_{1}, b]}$, and so we obtain a sequence in which +the term after the~$n$th coincides with~$b$, and therefore a finite set +of intervals having the properties required. Thus the theorem is +proved. + +\begin{Remark} +It is instructive to consider the examples of \PageRef{p.}{186} in the light of this +theorem. + +\Itemp{(i)} Here the conditions of the theorem are not satisfied\Add{:} the points +$1/n$,~$2/n$, $3/n$,~\dots\ do not lie inside any interval of~$I$\Add{.} + +\Itemp{(ii)} Here the conditions of the theorem are satisfied. The set of +intervals +\[ +\DPmod{(0, 2\EPSILON)}{[0, 2\EPSILON]}, \quad +\DPmod{(\EPSILON, 3\EPSILON)}{[\EPSILON, 3\EPSILON]}, \quad +\DPmod{(2\EPSILON, 4\EPSILON)}{[2\EPSILON, 4\EPSILON]}, \ \dots, \quad +\DPmod{(1 - 2\EPSILON, 1)}{[1 - 2\EPSILON, 1]}, +\] +associated with the points $\EPSILON$,~$2\EPSILON$, $3\EPSILON$,~\dots, $1 - \EPSILON$, possesses the properties required. + +\Itemp{(iii)} In this case we can prove, by using the theorem, that there are, +if $\EPSILON$~is small enough, points of~$\DPmod{(0, 1)}{[0, 1]}$ which do not lie in any interval of~$I$. + +If every point of~$\DPmod{(0, 1)}{[0, 1]}$ lay inside an interval of~$I$ (with the obvious +reservation as to the end points), then we could find a finite number of intervals +of~$I$ possessing the same property and having therefore a total length greater +than~$1$. Now there are two intervals, of total length~$2\EPSILON$, for which $q = 1$, and +$q - 1$~intervals, of total length $2\EPSILON(q - 1)/q^{3}$, associated with any other value +of~$q$. The sum of any finite number of intervals of~$I$ can therefore not be +greater than $2\EPSILON$~times that of the series +\[ +1 + \frac{1}{2^{3}} + \frac{2}{3^{3}} + \frac{3}{4^{3}} + \dots, +\] +which will be shown to be convergent in \Ref{Ch.}{VIII}\@. Hence it follows that, if +$\EPSILON$~is small enough, the supposition that every point of~$\DPmod{(0, 1)}{[0, 1]}$ lies inside an +interval of~$I$ leads to a contradiction. + +The reader may be tempted to think that this proof is needlessly +elaborate, and that the existence of points of the interval, not in any interval +of~$I$, follows at once from the fact that the sum of all these intervals is less +than~$1$. But the theorem to which he would be appealing is (when the set of +intervals is infinite) far from obvious, and can only be proved rigorously by +some such use of the Heine-Borel Theorem as is made in the text. +\end{Remark} + +\Paragraph{106.} We shall now apply the Heine-Borel Theorem to the +proof of two important theorems concerning the oscillation of a +continuous function. +\PageSep{189} + +\begin{Theorem}[I.] +If $\phi(x)$~is continuous throughout the interval +$\DPmod{(a, b)}{[a, b]}$, then we can divide $\DPmod{(a, b)}{[a, b]}$ into a finite number of sub-intervals +$\DPmod{(a, x_{1})}{[a, x_{1}]}$, $\DPmod{(x_{1}, x_{2})}{[x_{1}, x_{2}]}$,~\dots\Add{,} $\DPmod{(x_{n}, b)}{[x_{n}, b]}$, in each of which the oscillation of~$\phi(x)$ is +less than an assigned positive number~$\DELTA$. +\end{Theorem} + +Let $\xi$ be any number between $a$~and~$b$. Since $\phi(x)$~is continuous +for $x = \xi$, we can determine an interval $\DPmod{(\xi - \EPSILON, \xi + \EPSILON)}{[\xi - \EPSILON, \xi + \EPSILON]}$ such +that the oscillation of~$\phi(x)$ in this interval is less than~$\DELTA$. It is +indeed obvious that there are an infinity of such intervals corresponding +to every~$\xi$ and every~$\DELTA$, for if the condition is satisfied for +any particular value of~$\EPSILON$, then it is satisfied \textit{a~fortiori} for any smaller +value. What values of~$\EPSILON$ are admissible will naturally depend upon~$\xi$; +we have at present no reason for supposing that a value of~$\EPSILON$ +admissible for one value of~$\xi$ will be admissible for another. We +shall call the intervals thus associated with~$\xi$ \emph{the $\DELTA$-intervals of~$\xi$}. + +If $\xi = a$ then we can determine an interval $\DPmod{(a, a + \EPSILON)}{[a, a + \EPSILON]}$, and so an +infinity of such intervals, having the same property. These we +call the $\DELTA$-intervals of~$a$, and we can define in a similar manner the +$\DELTA$-intervals of~$b$. + +Consider now the set~$I$ of intervals formed by taking all the +$\DELTA$-intervals of all points of~$\DPmod{(a, b)}{[a, b]}$. It is plain that this set satisfies +the conditions of the Heine-Borel Theorem; every point interior +to the interval is interior to at least one interval of~$I$, and $a$~and~$b$ +are end points of at least one such interval. We can therefore +determine a set~$I'$ which is formed by a finite number of intervals +of~$I$, and which possesses the same property as $I$~itself. + +The intervals which compose the set~$I'$ will in general overlap +as in \Fig{34}. But their end +%[Illustration: Fig. 34.] +\Figure[2.5in]{34}{p189} +points obviously divide up +$\DPmod{(a, b)}{[a, b]}$ into a finite set of intervals~$I''$ +each of which is +included in an interval of~$I'$, and in each of which the oscillation +of~$\phi(x)$ is less than~$\DELTA$. Thus Theorem~I is proved. + +\begin{Theorem}[II.] +Given any positive number~$\DELTA$, we can find a +number~$\eta$ such that, if the interval $\DPmod{(a, b)}{[a, b]}$ is divided in any manner +into sub-intervals of length less than~$\eta$, then the oscillation of~$\phi(x)$ +in each of them will be less than~$\DELTA$. +\end{Theorem} +\PageSep{190} + +Take $\DELTA_{1} < \frac{1}{2}\DELTA$, and construct, as in Theorem~I, a finite set of sub-intervals~$j$ +in each of which the oscillation of~$\phi(x)$ is less than~$\DELTA_{1}$. +Let $\eta$ be the length of the least of these sub-intervals~$j$. If +now we divide $\DPmod{(a, b)}{[a, b]}$ into parts each of length less than~$\eta$, then any +such part must lie entirely within at most two successive sub-intervals~$j$. +Hence, in virtue of~(3) of \SecNo[§]{103}, the oscillation of~$\phi(x)$, +in one of the parts of length less than~$\eta$, cannot exceed twice the +greatest oscillation of~$\phi(x)$ in a sub-interval~$j$, and is therefore +less than~$2\DELTA_{1}$, and therefore than~$\DELTA$. + +This theorem is of fundamental importance in the theory of +definite integrals (\Ref{Ch.}{VII}). It is impossible, without the use of +this or some similar theorem, to prove that a function continuous +throughout an interval necessarily possesses an integral over that +interval. + +\Paragraph{107. Continuous functions of several variables.} The +notions of continuity and discontinuity may be extended to +functions of several independent variables (\Ref{Ch.}{II}, \SecNo[§§]{31}~\textit{et~seq.}). +Their application to such functions however, raises questions +much more complicated and difficult than those which we have +considered in this chapter. It would be impossible for us to +discuss these questions in any detail here; but we shall, in the +sequel, require to know what is meant by a continuous function of +two variables, and we accordingly give the following definition. +It is a straightforward generalisation of the last form of the definition +of~\SecNo[§]{98}. + +\begin{Defn} +The function $\phi(x, y)$ of the two variables $x$~and~$y$ is said to be +\Emph{continuous} for $x = \xi$, $y = \eta$ if, given any positive number~$\DELTA$, however +small, we can choose~$\EPSILON(\DELTA)$ so that +\[ +|\phi(x, y) - \phi(\xi, \eta) | < \DELTA +\] +when $0 \leq |x - \xi| \leq \EPSILON(\DELTA)$ and $0 \leq |y - \eta| \leq \EPSILON(\DELTA)$; that is to say if we +can draw a square, whose sides are parallel to the axes of coordinates +and of length~$2\EPSILON(\DELTA)$, whose centre is the point~$(\xi, \eta)$, and which is such +that the value of~$\phi(x, y)$ at any point inside it or on its boundary +differs from~$\phi(\xi, \eta)$ by less than~$\DELTA$.\footnote + {The reader should draw a figure to illustrate the definition.} +\end{Defn} + +This definition of course presupposes that $\phi(x, y)$~is defined at +all points of the square in question, and in particular at the point +\PageSep{191} +$(\xi, \eta)$. Another method of stating the definition is this: \emph{$\phi(x, y)$~is +continuous for $x = \xi$, $y = \eta$ if $\phi(x, y) \to \phi(\xi, \eta)$ when $x \to \xi$, $y \to \eta$ +in any manner}. This statement is apparently simpler; but it +contains phrases the precise meaning of which has not yet been +explained and can only be explained by the help of inequalities +like those which occur in our original statement. + +It is easy to prove that the sums, the products, and in general +the quotients of continuous functions of two variables are themselves +continuous. A polynomial in two variables is continuous for +all values of the variables; and the ordinary functions of~$x$ and~$y$ +which occur in every-day analysis are \emph{generally} continuous, \ie\ +are continuous except for pairs of values of $x$~and~$y$ connected by +special relations. + +\begin{Remark} +The reader should observe carefully that to assert the continuity of~$\phi(x, y)$ +with respect to the two variables $x$~and~$y$ is to assert much more +than its continuity with respect to each variable considered separately. It is +plain that if $\phi(x, y)$~is continuous with respect to $x$~and~$y$ then it is certainly +continuous with respect to~$x$ (or~$y$) when any fixed value is assigned to~$y$ +(or~$x$). But the converse is by no means true. Suppose, for example, that +\[ +\phi(x, y) = \frac{2xy}{x^{2} + y^{2}} +\] +when neither $x$~nor~$y$ is zero, and $\phi(x, y) = 0$ when either $x$ or~$y$ is zero. Then +if $y$~has any fixed value, zero or not, $\phi(x, y)$~is a continuous function of~$x$, +and in particular continuous for $x = 0$; for its value when $x = 0$ is zero, and it +tends to the limit zero as $x \to 0$. In the same way it may be shown that +$\phi(x, y)$~is a continuous function of~$y$. But $\phi(x, y)$~is \emph{not} a continuous function +of $x$~\emph{and}~$y$ for $x = 0$, $y = 0$. Its value when $x = 0$, $y = 0$ is zero; but if $x$~and~$y$ +tend to zero along the straight line~$y = ax$, then +\[ +\phi(x, y) = \frac{2a}{1 + a^{2}},\quad +\lim\phi(x, y) = \frac{2a}{1 + a^{2}}, +\] +which may have any value between $-1$~and~$1$. + +\Paragraph{108. Implicit functions.} We have already, in \Ref{Ch.}{II}, met with +the idea of an \emph{implicit function}. Thus, if $x$~and~$y$ are connected by the +relation +\[ +y^{5} - xy - y - x = 0, +\Tag{(1)} +\] +then $y$~is an `implicit function' of~$x$. + +But it is far from obvious that such an equation as this does really define +a function~$y$ of~$x$, or several such functions. In \Ref{Ch.}{II} we were content to +take this for granted. We are now in a position to consider whether the +assumption we made then was justified. +\PageSep{192} + +We shall find the following terminology useful. Suppose that it is possible +to surround a point~$(a, b)$, as in \SecNo[§]{107}, with a square throughout which +a certain condition is satisfied. We shall call such a square a \emph{neighbourhood} +of~$(a, b)$, and say that the condition in question is satisfied \emph{in the neighbourhood +of~$(a, b)$}, or \emph{near~$(a, b)$}, meaning by this simply that it is possible to find +\emph{some} square throughout which the condition is satisfied. It is obvious that +similar language may be used when we are dealing with a single variable, the +square being replaced by an interval on a line. + +\begin{Theorem} +If \Itemp{(i)} $f(x, y)$~is a continuous function of $x$~and~$y$ in the +neighbourhood of~$(a, b)$, + +\Itemp{(ii)} $f(a, b) = 0$, + +\Itemp{(iii)} $f(x, y)$ is, for all values of~$x$ in the neighbourhood of~$a$, a steadily +increasing function of~$y$, in the stricter sense of~\SecNo[§]{95}, + +%[** TN: Paragraph break in the original] +then \Inum{(1)}~there is a unique function $y = \phi(x)$ which, when substituted in the +equation $f(x, y) = 0$, satisfies it identically for all values of~$x$ in the neighbourhood +of~$a$, + +\Inum{(2)}~$\phi(x)$ is continuous for all values of~$x$ in the neighbourhood of~$a$. +\end{Theorem} + +In the figure the square represents a `neighbourhood' of~$(a, b)$ throughout +which the conditions (i)~and~(iii) are +satisfied, and $P$~the point~$(a, b)$. If we +%[Illustration: Fig. 35.] +\Figure[2in]{35}{p192} +take $Q$~and~$R$ as in the figure, it follows from~(iii) +that $f(x, y)$ is positive at~$Q$ and negative +at~$R$. This being so, and $f(x, y)$ being continuous +at~$Q$ and at~$R$, we can draw lines $QQ'$ +and~$RR'$ parallel to~$OX$, so that $R'Q'$~is parallel +to~$OY$ and $f(x, y)$ is positive at all points of~$QQ'$ +and negative at all points of~$RR'$. In particular +$f(x, y)$ is positive at~$Q'$ and negative at~$R'$, +and therefore, in virtue of (iii)~and \SecNo[§]{100}, +vanishes once and only once at a point~$P'$ on~$R'Q'$. +The same construction gives us a unique point at which $f(x, y) = 0$ +on each ordinate\DPnote{** TN: i.e., vertical segment} between $RQ$~and~$R'Q'$. It is obvious, moreover, that the +same construction can be carried out to the left of~$RQ$. The aggregate of +points such as~$P'$ gives us the graph of the required function $y = \phi(x)$. + +It remains to prove that $\phi(x)$~is continuous. This is most simply effected +by using the idea of the `limits of indetermination' of~$\phi(x)$ as $x \to a$ (\SecNo[§]{96}). +Suppose that $x \to a$, and let $\lambda$~and~$\Lambda$ be the limits of indetermination of~$\phi(x)$ +as $x \to a$. It is evident that the points $(a, \lambda)$ and~$(a, \Lambda)$ lie on~$QR$. Moreover, +we can find a sequence of values of~$x$ such that $\phi(x) \to \lambda$ when $x \to a$ through +the values of the sequence; and since $f\{x, \phi(x)\} = 0$, and $f(x, y)$~is a continuous +function of $x$~and~$y$, we have +\[ +f(a, \lambda) = 0. +\] +Hence $\lambda = b$; and similarly $\Lambda = b$. Thus $\phi(x)$~tends to the limit~$b$ as $x \to a$, +and so $\phi(x)$~is continuous for $x = a$. It is evident that we can show in +\PageSep{193} +exactly the same way that $\phi(x)$~is continuous for any value of~$x$ in the +neighbourhood of~$a$. + +It is clear that the truth of the theorem would not be affected if we were +to change `increasing' to `decreasing' in condition~(iii). + +As an example, let us consider the equation~\Eq{(1)}, taking $a = 0$, $b = 0$. It is +evident that the conditions (i)~and~(ii) are satisfied. Moreover +\[ +f(x, y) - f(x, y') + = (y - y') (y^{4} + y^{3}y' + y^{2}y'^{2} + yy'^{3} + y'^{4} - x - 1) +\] +has, when $x$,~$y$, and~$y'$ are sufficiently small, the sign opposite to that of~$y - y'$. +Hence condition~(iii) (with `decreasing' for `increasing') is satisfied. +It follows that there is one and only one continuous function~$y$ which +satisfies the equation~\Eq{(1)} identically and vanishes with~$x$. + +The same conclusion would follow if the equation were +\[ +y^{2} - xy - y - x = 0. +\] +The function in question is in this case +\[ +y = \tfrac{1}{2}\{1 + x - \sqrtp{1 + 6 x + x^{2}}\}, +\] +where the square root is positive. The second root, in which the sign of the +square root is changed, does not satisfy the condition of vanishing with~$x$. + +There is one point in the proof which the reader should be careful to observe. +We supposed that the hypotheses of the theorem were satisfied `in +the neighbourhood of~$(a, b)$', that is to say throughout a certain square +$\xi - \EPSILON \leq x \leq \xi + \EPSILON$, $\eta - \EPSILON \leq y \leq \eta + \EPSILON$. The conclusion holds `in the neighbourhood +of $x = a$', that is to say throughout a certain interval $\xi - \EPSILON_{1} \leq x \leq \xi + \EPSILON_{1}$. There +is nothing to show that the~$\EPSILON_{1}$ of the conclusion is the~$\EPSILON$ of the hypotheses, and +indeed this is generally untrue. + +\Paragraph{109. Inverse Functions.} Suppose in particular that $f(x, y)$~is of the +form $F(y) - x$. We then obtain the following theorem. + +\begin{Result} +If $F(y)$ is a function of~$y$, continuous and steadily increasing \(or decreasing\), +in the stricter sense of \SecNo[§]{95}, in the neighbourhood of $y = b$, and $F(b) = a$, then +there is a unique continuous function $y = \phi(x)$ which is equal to~$b$ when $x = a$ +and satisfies the equation $F(y) = x$ identically in the neighbourhood of $x = a$. +\end{Result} + +The function thus defined is called the \emph{inverse function of~$F(y)$}. + +Suppose for example that $y^{3} = x$, $a = 0$, $b = 0$. Then all the conditions of +the theorem are satisfied. The inverse function is $x = \sqrt[3]{y}$. + +If we had supposed that $y^{2} = x$ then the conditions of the theorem would +not have been satisfied, for $y^{2}$~is not a steadily increasing function of~$y$ in any +interval which includes $y = 0$: it decreases when $y$~is negative and increases +when $y$~is positive. And in this case the conclusion of the theorem does not +hold, for $y^{2} = x$ defines \emph{two} functions of~$x$, viz.\ $y = \sqrt{x}$ and $y = -\sqrt{x}$, both of +which vanish when $x = 0$, and each of which is defined only for positive values +of~$x$, so that the equation has sometimes two solutions and sometimes none. +The reader should consider the more general equations +\[ +y^{2n} = x, \quad y^{2n+1} = x, +\] +\PageSep{194} +in the same way. Another interesting example is given by the equation +\[ +y^{5} - y - x = 0, +\] +already considered in \Ex{xiv}.~7. + +Similarly the equation +\[ +\sin y = x +\] +has just one solution which vanishes with~$x$, viz.\ the value of~$\arcsin x$ which +vanishes with~$x$. There are of course an infinity of solutions, given by the +other values of~$\arcsin x$ (cf.\ \Ex{xv}.~10), which do not satisfy this condition. + +So far we have considered only what happens in the neighbourhood of a +particular value of~$x$. Let us suppose now that $F(y)$~is positive and steadily +increasing (or decreasing) throughout an interval~$\DPmod{(a, b)}{[a, b]}$. Given any point~$\xi$ +of~$\DPmod{(a, b)}{[a, b]}$, we can determine an interval~$i$ including~$\xi$, and a unique and continuous +inverse function~$\phi_{i} (x)$ defined throughout~$i$. + +From the set~$I$ of intervals~$i$ we can, in virtue of the Heine-Borel Theorem, +pick out a finite sub-set covering up the whole interval~$\DPmod{(a, b)}{[a, b]}$; and it is plain +that the finite set of functions~$\phi_{i} (x)$, corresponding to the sub-set of intervals~$i$ +thus selected, define together a unique inverse function~$\phi(x)$ continuous +throughout~$\DPmod{(a, b)}{[a, b]}$. + +We thus obtain the theorem: \begin{Result}if $x = F(y)$, where $F(y)$~is continuous and +increases steadily and strictly from~$A$ to~$B$ as $x$~increases from~$a$ to~$b$, then there +is a unique inverse function $y = \phi(x)$ which is continuous and increases steadily +and strictly from~$a$ to~$b$ as $x$~increases from~$A$ to~$B$. +\end{Result} + +It is worth while to show how this theorem can be obtained directly without +the help of the more difficult theorem of \SecNo[§]{108}. Suppose that $A < \xi < B$, +and consider the class of values of~$y$ such that (i)~$a <y < b$ and (ii)~$F(y) \leq \xi$. +This class has an upper bound~$\eta$, and plainly $F(\eta) \leq \xi$. If $F(\eta)$~were less +than~$\xi$, we could find a value of~$y$ such that $y > \eta$ and $F(y) < \xi$, and $\eta$~would +not be the upper bound of the class considered. Hence $F(\eta) = \xi$. The +equation $F(y) = \xi$ has therefore a unique solution $y = \eta = \phi(\xi)$, say; and +plainly $\eta$~increases steadily and continuously with~$\xi$, which proves the theorem. +\end{Remark} + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER V.} + +\begin{Examples}{} +\Item{1.} Show that, if neither $a$~nor~$b$ is zero, then +\[ +ax^{n} + bx^{n-1} + \dots + k = ax^{n} (1 + \epsilon_{x}), +\] +where $\epsilon_{x}$~is of the first order of smallness when $x$~is large. + +\Item{2.} If $P(x) = ax^{n} + bx^{n-1} + \dots + k$, and $a$~is not zero, then as $x$~increases +$P(x)$~has ultimately the sign of~$a$; and so has $P(x + \lambda) - P(x)$, where $\lambda$~is +any constant. + +\Item{3.} Show that in general +\[ +(ax^{n} + bx^{n-1} + \dots + k)/(Ax^{n} + Bx^{n-1} + \dots + K) + = \alpha + (\beta/x) (1 + \epsilon_{x}), +\] +where $\alpha = a/A$, $\beta = (bA - aB)/A^{2}$, and $\epsilon_{x}$~is of the first order of smallness when +$x$~is large. Indicate any exceptional cases. +\PageSep{195} + +\Item{4.} Express +\[ +(ax^{2} + bx + c)/(Ax^{2} + Bx + C) +\] +in the form +\[ +\alpha + (\beta/x) + (\gamma/x^{2})(1 + \epsilon_{x}), +\] +where $\epsilon_{x}$~is of the first order of smallness when $x$~is large. + +\Item{5.} Show that +\[ +\lim_{x\to\infty}\sqrt{x}\{\sqrtp{x + a} - \sqrt{x}\} = \tfrac{1}{2} a. +\] + +[Use the formula $\sqrtp{x + a} - \sqrt{x} = a/\{\sqrtp{x + a} + \sqrt{x}\}$.] + +\Item{6.} Show that $\sqrtp{x + a} = \sqrt{x} + \frac{1}{2}(a/\sqrt{x}) (1 + \epsilon_{x})$, where $\epsilon_{x}$~is of the first order +of smallness when $x$~is large. + +\Item{7.} Find values of $\alpha$~and~$\beta$ such that $\sqrtp{a x^{2} + 2bx + c} - \alpha x - \beta$ has the limit +zero as $x \to \infty$; and prove that $\lim x\{\sqrtp{ax^{2} + 2bx + c} - \alpha x - \beta\} = (ac - b^{2})/2a$. + +\Item{8.} Evaluate +\[ +\lim_{x \to\infty} x\left\{\sqrtbr{x^{2} + \sqrtp{x^{4} + 1}} - x\sqrt{2}\right\}. +\] + +\Item{9.} Prove that $(\sec x - \tan x) \to 0$ as $x \to \frac{1}{2}\pi$. + +\Item{10.} Prove that $\phi(x) = 1 - \cos(1 - \cos x)$ is of the fourth order of smallness +when $x$~is small; and find the limit of $\phi(x)/x^{4}$ as $x \to 0$. + +\Item{11.} Prove that $\phi(x) = x\sin(\sin x) - \sin^{2}x$ is of the sixth order of smallness +when $x$~is small; and find the limit of $\phi(x)/x^{6}$ as $x \to 0$. + +\Item{12.} From a point~$P$ on a radius~$OA$ of a circle, produced beyond the circle, +a tangent~$PT$ is drawn to the circle, touching it in~$T$, and $TN$~is drawn perpendicular +to~$OA$. Show that $NA/AP \to 1$ as $P$~moves up to~$A$. + +\Item{13.} Tangents are drawn to a circular arc at its middle point and its +extremities; $\Delta$~is the area of the triangle formed by the chord of the arc and +the two tangents at the extremities, and $\Delta'$~the area of that formed by the +three tangents. Show that $\Delta/\Delta' \to 4$ as the length of the arc tends to zero. + +\Item{14.} For what values of~$a$ does $\{a + \sin(1/x)\}/x$ tend to (1)~$\infty$, (2)~$-\infty$, +as $x \to 0$? [To~$\infty$ if~$a > 1$, to~$-\infty$ if~$a < -1$: the function oscillates if +$-1 \leq a \leq 1$.] + +\Item{15.} If $\phi(x) = 1/q$ when $x = p/q$, and $\phi(x) = 0$ when $x$~is irrational, then +$\phi(x)$~is continuous for all irrational and discontinuous for all rational values +of~$x$. + +\Item{16.} Show that the function whose graph is drawn in \Fig{32} may be represented +by either of the formulae +\[ +1 - x + [x] - [1 - x],\quad +1 - x - \lim_{n\to\infty} (\cos^{2n+1}\pi x). +\] + +\Item{17.} Show that the function~$\phi(x)$ which is equal to~$0$ when $x = 0$, to~$\frac{1}{2} - x$ +when $0 < x < \frac{1}{2}$, to~$\frac{1}{2}$ when $x = \frac{1}{2}$, to~$\frac{3}{2} - x$ +when $\frac{1}{2}< x < 1$, and to~$1$ when +$x = 1$, assumes every value between $0$~and~$1$ once and once only as $x$~increases +from $0$~to~$1$, but is discontinuous for $x = 0$, $x = \frac{1}{2}$, and $x = 1$. Show also that +the function may be represented by the formula +\[ +\tfrac{1}{2} - x - \tfrac{1}{2}[2x] - \tfrac{1}{2}[1 - 2x]. +\] +\PageSep{196} + +\Item{18.} Let $\phi(x) = x$ when $x$~is rational and $\phi(x) = 1 - x$ when $x$~is irrational. +Show that $\phi(x)$~assumes every value between $0$ and~$1$ once and once only as $x$~increases +from $0$ to~$1$, but is discontinuous for every value of~$x$ except $x = \frac{1}{2}$. + +\Item{19.} As $x$~increases from~$-\frac{1}{2}\pi$ to~$\frac{1}{2}\pi$, $y = \sin x$ is continuous and steadily +increases, in the stricter sense, from~$-1$ to~$1$. Deduce the existence of a +function $x = \arcsin y$ which is a continuous and steadily increasing function +of~$y$ from $y = -1$ to~$y = 1$. + +\Item{20.} Show that the numerically least value of~$\arctan y$ is continuous for +all values of~$y$ and increases steadily from $-\frac{1}{2}\pi$ to~$\frac{1}{2}\pi$ as $y$~varies through all +real values. + +\Item{21.} Discuss, on the lines of \SecNo[§§]{108}--\SecNo{109}, the solution of the equations +\[ +y^{2} - y - x = 0,\quad +y^{4} - y^{2} - x^{2} = 0,\quad +y^{4} - y^{2} + x^{2} = 0 +\] +in the neighbourhood of $x = 0$, $y = 0$. + +\Item{22.} If $ax^{2} + 2bxy + cy^{2} + 2dx + 2ey = 0$ and $\Delta = 2bde - ae^{2} - cd^{2}$, then one +value of~$y$ is given by $y = \alpha x + \beta x^{2} + (\gamma + \epsilon_{x}) x^{3}$, where +\[ +\alpha = -d/e,\quad +\beta = \Delta/2e^{3},\quad +\gamma = (cd - be) \Delta/2e^{5}, +\] +and $\DPtypo{e_{x}}{\epsilon_{x}}$~is of the first order of smallness when $x$~is small. + +[If $y - \alpha x = \eta $ then +\[ +-2e\eta + = ax^{2} + 2bx(\eta + \alpha x) + c(\eta + \alpha x)^{2} + = Ax^{2} + 2Bx \eta + C\eta^{2}, +\] +say. It is evident that $\eta$~is of the second order of smallness, $x\eta$~of the third, +and $\eta^{2}$~of the fourth; and $-2e\eta = Ax^{2} - (AB/e) x^{3}$, the error being of the fourth +order.] + +\Item{23.} If $x = ay + by^{2} + cy^{3}$ then one value of~$y$ is given by +\[ +y = \alpha x + \beta x^{2} + (\gamma + \epsilon_{x}) x^{3}, +\] +where $\alpha = 1/a$, $\beta = -b/a^{3}$, $\gamma = (2b^{2} - ac)/a^{5}$, and $\epsilon_{x}$~is of the first order of smallness +when $x$~is small. + +\Item{24.} If $x = ay + by^{n}$, where $n$~is an integer greater than unity, then one +value of~$y$ is given by $y = \alpha x + \beta x^{n} + (\gamma + \epsilon_{x}) x^{2n-1}$, where $\alpha = 1/a$, $\beta = -b/a^{n+1}$, +$\gamma = nb^{2}/a^{2n+1}$, and $\epsilon_{x}$~is of the $(n - 1)$th~order of smallness when $x$~is small. + +\Item{25.} Show that the least positive root of the equation $xy = \sin x$ is a continuous +function of~$y$ throughout the interval $\DPmod{(0, 1)}{[0, 1]}$, and decreases steadily +from $\pi$ to~$0$ as $y$~increases from $0$ to~$1$. [The function is the inverse of +$(\sin x)/x$: apply~\SecNo[§]{109}.] + +\Item{26.} The least positive root of $xy = \tan x$ is a continuous function of~$y$ +throughout the interval $\DPmod{(1, \infty)}{[1, \infty)}$, and increases steadily from $0$ to~$\frac{1}{2}\pi$ as $y$~increases +from $1$ towards~$\infty$. +\end{Examples} +\PageSep{197} + + +\Chapter{VI}{DERIVATIVES AND INTEGRALS} + +\Paragraph{110. Derivatives or Differential Coefficients.} Let us return +to the consideration of the properties which we naturally associate +with the notion of a curve. The first and most obvious property +is, as we saw in the last chapter, that which gives a curve its +appearance of connectedness, and which we embodied in our definition +of a continuous function. + +The ordinary curves which occur in elementary geometry, such +as straight lines, circles and conic sections, have of course many +other properties of a general character. The simplest and most +noteworthy of these is perhaps that they have a definite \emph{direction} +at every point, or what is the same thing, that at every point of +the curve we can draw a \emph{tangent} to it. The reader will probably +remember that in elementary geometry the tangent to a curve at~$P$ +is defined to be `the limiting position of the chord~$PQ$, when $Q$~moves +up towards coincidence with~$P$'. Let us consider what is +implied in the assumption of the existence of such a limiting +position. + +In the figure (\Fig{36}) $P$~is a fixed point on the curve, and $Q$ +a variable point; $PM$,~$QN$ are parallel to~$OY$ and $PR$ to~$OX$. +We denote the coordinates of~$P$ by $x$,~$y$ and those of~$Q$ by +$x + h$,~$y + k$: $h$~will be positive or negative according as $N$~lies to +the right or left of~$M$. + +We have assumed that there is a tangent to the curve at~$P$, +or that there is a definite `limiting position' of the chord~$PQ$. +Suppose that $PT$, the tangent at~$P$, makes an angle~$\psi$ with~$OX$. +Then to say that $PT$~is the limiting position of~$PQ$ is equivalent +to saying that the limit of the angle $QPR$ is~$\psi$, when $Q$~approaches~$P$ +\PageSep{198} +along the curve from either side. We have now to distinguish +two cases, a general case and an exceptional one. +%[Illustration: Fig. 36.] +\Figure[3in]{36}{p198} + +The general case is that in which $\psi$~is not equal to~$\frac{1}{2}\pi$, so that +$PT$~is not parallel to~$OY$. In this case $RPQ$ tends to the limit~$\psi$, +and +\[ +RQ/PR = \tan RPQ +\] +tends to the limit $\tan \psi$. Now +\[ +RQ/PR = (NQ - MP)/MN = \{\phi(x + h) - \phi(x)\}/h; +\] +and so +\[ +\lim_{h \to 0} \frac{\phi(x + h) - \phi(x)}{h} = \tan\psi. +\Tag{(1)} +\] + +The reader should be careful to note that in all these equations +all lengths are regarded as affected with the proper sign, +so that (\eg)\ $RQ$~is negative in the figure when $Q$~lies to the left +of~$P$; and that the convergence to the limit is unaffected by the +sign of~$h$. + +Thus the assumption that the curve which is the graph of~$\phi(x)$ +has a tangent at~$P$, which is not perpendicular to the axis of~$x$, +implies that $\phi(x)$~has, for the particular value of~$x$ corresponding +to~$P$, the property that \emph{$\{\phi(x + h) - \phi(x)\}/h$ tends to a limit when +$h$~tends to zero}. + +\begin{Remark} +This of course implies that both of +\[ +\{\phi(x + h) - \phi(x)\}/h,\quad +\{\phi(x - h) - \phi(x)\}/(-h) +\] +tend to limits when $h \to 0$ by positive values only, and that the two limits +are equal. If these limits exist but are not equal, then the curve $y = \phi(x)$ +has an angle at the particular point considered, as in \Fig{37}. +\end{Remark} + +Now let us suppose that the curve has (like the circle or +ellipse) a tangent at every point of its length, or at any rate every +\PageSep{199} +portion of its length which corresponds to a certain range of +variation of~$x$. Further let us suppose this tangent never perpendicular +to the axis of~$x$: in the case of a circle this would of +course restrict us to considering an arc less than a semicircle. +Then an equation such as~\Eq{(1)} holds for all values of~$x$ which fall +inside this range. To each such value of~$x$ corresponds a value of +$\tan\psi$: $\tan\psi$~is a function of~$x$, which is defined for all values of~$x$ +in the range of values under consideration, and which may be +calculated or \emph{derived} from the original function~$\phi(x)$. We shall +call this function the \Emph{derivative} or \emph{derived function} of~$\phi(x)$, and +we shall denote it by +\[ +\phi'(x). +\] + +Another name for the derived function of~$\phi(x)$ is the \Emph{differential +coefficient} of~$\phi(x)$; and the operation of calculating +$\phi'(x)$ from~$\phi(x)$ is generally known as \Emph{differentiation}. This +terminology is firmly established for historical reasons: see~\SecNo[§]{115}. + +Before we proceed to consider the special case mentioned +above, in which $\psi = \frac{1}{2}\pi$, we shall illustrate our definition by some +general remarks and particular illustrations. + +\Paragraph{111. Some general remarks.} \Item{(1)} The existence of a derived +function~$\phi'(x)$ for all values of~$x$ in the interval $a \leq x \leq b$ implies +that $\phi(x)$~is continuous at every point of this interval. For it is +evident that $\{\phi(x + h) - \phi(x)\}/h$ cannot tend to a limit unless +$\lim\phi(x + h) = \phi(x)$, and it is this which is the property denoted +by continuity. + +\Item{(2)} It is natural to ask whether the converse is true, \ie\ +whether every continuous curve has a +definite tangent at every point, and +%[Illustration: Fig. 37.] +\Figure[2in]{37}{p199} +every function a differential coefficient +for every value of~$x$ for which it is +continuous.\footnote + {We leave out of account the exceptional case (which we have still to examine) + in which the curve is supposed to have a tangent perpendicular to~$OX$: apart from + this possibility the two forms of the question stated above are equivalent.} +The answer is obviously +\emph{No}: it is sufficient to consider the +curve formed by two straight lines +meeting to form an angle (\Fig{37}). +\PageSep{200} +The reader will see at once that in this case $\{\phi(x + h) - \phi(x)\}/h$ +has the limit $\tan\beta$ when $h \to 0$ by positive values and the limit +$\tan\alpha$ when $h \to 0$ by negative values. + +\begin{Remark} +This is of course a case in which a curve might reasonably be said to have +\emph{two} directions at a point. But the following example, although a little more +difficult, shows conclusively that there are cases in which a continuous curve +cannot be said to have either one direction or several directions at one of its +points. Draw the graph (\Fig{14}, \PageRef{p.}{53}) of the function $x\sin(1/x)$. The +function is not defined for $x = 0$, and so is discontinuous for $x = 0$. On +the other hand the function defined by the equations +\[ +\phi(x) = x\sin(1/x)\quad (x \neq 0),\qquad +\phi(x) = 0\quad (x = 0) +\] +is continuous for $x = 0$ (\Exs{xxxvii}.~14,~15), and the graph of this +function is a continuous curve. + +But $\phi(x)$~has no derivative for $x = 0$. For $\phi'(0)$~would be, by definition, +$\lim\{\phi(h) - \phi(0)\}/h$ or $\lim\sin(1/h)$; and no such limit exists. + +It has even been shown that a function of~$x$ may be continuous and yet +have no derivative for \emph{any} value of~$x$, but the proof of this is much more +difficult. The reader who is interested in the question may be referred to +Bromwich's \textit{Infinite Series}, pp.~490--1, or Hobson's \textit{Theory of Functions +of a Real Variable}, pp.~620--5. +\end{Remark} + +\Item{(3)} The notion of a derivative or differential coefficient was +suggested to us by geometrical considerations. But there is +nothing geometrical in the notion itself. The derivative $\phi'(x)$ of +a function $\phi(x)$ may be defined, without any reference to any kind +of geometrical representation of~$\phi(x)$, by the equation +\[ +\phi'(x) = \lim_{h \to 0} \frac{\phi(x + h) - \phi(x)}{h}; +\] +and $\phi(x)$~has or has not a derivative, for any particular value of~$x$, +according as this limit does or does not exist. The geometry of +curves is merely one of many departments of mathematics in which +the idea of a derivative finds an application. + +\begin{Remark} +Another important application is in dynamics. Suppose that a particle is +moving in a straight line in such a way that at time~$t$ its distance from a fixed +point on the line is $s = \phi(t)$. Then the `velocity of the particle at time~$t$' is +by definition the limit of +\[ +\frac{\phi(t + h) - \phi(t)}{h} +\] +as $h \to 0$. The notion of `velocity' is in fact merely a special case of +that of the derivative of a function. +\end{Remark} +\PageSep{201} + +\begin{Examples}{XXXIX.} +\Item{1.} If $\phi(x)$~is a constant then $\phi'(x) = 0$. Interpret +this result geometrically. + +\Item{2.} If $\phi(x) = ax + b$ then $\phi'(x) = a$. Prove this (i)~from the formal definition +and (ii)~by geometrical considerations. + +\Item{3.} If $\phi(x) = x^{m}$, where $m$~is a positive integer, then $\phi'(x) = mx^{m-1}$. + +[For +\begin{align*} +\phi'(x) &= \lim \frac{(x + h)^{m} - x^{m}}{h}\\ + &= \lim \left\{mx^{m-1} + \frac{m(m - 1)}{1·2} x^{m-2} h + \dots + h^{m-1}\right\}. +\end{align*} + +The reader should observe that this method cannot be applied to~$x^{p/q}$, +where $p/q$~is a rational fraction, as we have no means of expressing $(x + h)^{p/q}$ +as a finite series of powers of~$h$. We shall show later on (\SecNo[§]{118}) that the result +of this example holds for all rational values of~$m$. Meanwhile the reader +will find it instructive to determine $\phi'(x)$ when $m$~has some special fractional +value (\eg~$\frac{1}{2}$), by means of some special device.] + +\Item{4.} {\Loosen If $\phi(x) = \sin x$, then $\phi'(x) = \cos x$; and if $\phi(x) = \cos x$, then +$\phi'(x) = -\sin x$.} + +[For example, if $\phi(x) = \sin x$, we have +\[ +\{\phi(x + h) - \phi(x)\}/h + = \{2\sin \tfrac{1}{2}h \cos(x + \tfrac{1}{2}h)\}/h, +\] +the limit of which, when $h \to 0$, is $\cos x$, since $\lim\cos(x + \frac{1}{2}h) = \cos x$ (the cosine +being a continuous function) and $\lim\{(\sin \frac{1}{2}h)/\frac{1}{2}h\} = 1$ (\Ex{xxxvi}.~13).] + +\Item{5.} \Topic{Equations of the tangent and normal to a curve $y = \phi(x)$.} The +tangent to the curve at the point $(x_{0}, y_{0})$ is the line through $(x_{0}, y_{0})$ which +makes with~$OX$ an angle~$\psi$, where $\tan\psi = \phi'(x_{0})$. Its equation is therefore +\[ +y - y_{0} = (x - x_{0}) \phi'(x_{0}); +\] +and the equation of the normal (the perpendicular to the tangent at the +point of contact) is +\[ +(y - y_{0}) \phi'(x_{0}) + x - x_{0} = 0. +\] +We have assumed that the tangent is not parallel to the axis of~$y$. In +this special case it is obvious that the tangent and normal are $x = x_{0}$ and +$y = y_{0}$ respectively. + +\Item{6.} Write down the equations of the tangent and normal at any point of +the parabola $x^{2} = 4ay$. Show that if $x_{0} = 2a/m$, $y_{0} = a/m^{2}$, then the tangent +at $(x_{0}, y_{0})$ is $x = my + (a/m)$. +\end{Examples} + +\Paragraph{112.} We have seen that if $\phi(x)$~is not continuous for a value +of~$x$ then it cannot possibly have a derivative for that value of~$x$. +Thus such functions as $1/x$ or $\sin(1/x)$, which are not defined for +$x = 0$, and so necessarily discontinuous for $x = 0$, cannot have +derivatives for $x = 0$. Or again the function~$[x]$, which is discontinuous +for every integral value of~$x$, has no derivative for any +such value of~$x$. +\PageSep{202} + +\begin{Remark} +\Par{Example.} Since $[x]$~is constant between every two integral values of~$x$, +its derivative, whenever it exists, has the value zero. Thus the derivative +of~$[x]$, which we may represent by~$[x]'$, is a function equal to zero for +all values of~$x$ save integral values and undefined for integral values. It +is interesting to note that the function $1 - \dfrac{\sin\pi x}{\sin\pi x}$ has exactly the same +properties. +\end{Remark} + +We saw also in \Ex{xxxvii}.~7 that the types of discontinuity +which occur most commonly, when we are dealing with the very +simplest and most obvious kinds of functions, such as polynomials +or rational or trigonometrical functions, are associated with a +relation of the type +\[ +\phi(x) \to +\infty +\] +or $\phi(x) \to -\infty$. In all these cases, as in such cases as those considered +above, there is no derivative for certain special values of $x$. +%[Illustration: Fig. 38.] +\Figure{38}{p202} +In fact, as was pointed out in \SecNo[§]{111},~\Eq{(1)}, \emph{all discontinuities of~$\phi(x)$ are +also discontinuities of~$\phi'(x)$}. But the converse is not true, as we +may easily see if we return to the geometrical point of view of \SecNo[§]{110} +and consider the special case, hitherto left aside, in which the graph +of~$\phi(x)$ has a tangent parallel to~$OY$. This case may be subdivided +into a number of cases, of which the most typical are shown in +\Fig{38}. In cases (\ic)~and~(\id) the function is two valued on one side +of~$P$ and not defined on the other. In such cases we may consider +the two sets of values of~$\phi(x)$, which occur on one side of~$P$ or the +other, as defining distinct functions $\phi_{1}(x)$ and~$\phi_{2}(x)$, the upper +part of the curve corresponding to~$\phi_{1}(x)$. +\PageSep{203} + +The reader will easily convince himself that in~(\ia) +\[ +\{\phi(x + h) - \phi(x)\}/h \to +\infty, +\] +as $h \to 0$, and in~(\ib) +\[ +\{\phi(x + h) - \phi(x)\}/h \to -\infty; +\] +while in~(\ic) +\[ +\{\phi_{1}(x + h) - \phi_{1}(x)\}/h \to +\infty,\quad +\{\phi_{2}(x + h) - \phi_{2}(x)\}/h \to -\infty, +\] +and in~(\id) +\[ +\{\phi_{1}(x + h) - \phi_{1}(x)\}/h \to -\infty,\quad +\{\phi_{2}(x + h) - \phi_{2}(x)\}/h \to +\infty, +\] +though of course in~(\ic) only positive and in~(\id) only negative +values of~$h$ can be considered, a fact which by itself would preclude +the existence of a derivative. + +We can obtain examples of these four cases by considering the +functions defined by the equations +\[ +\Item{(\ia)}\ y^{3} = x,\quad +\Item{(\ib)}\ y^{3} = -x,\quad +\Item{(\ic)}\ y^{2} = x,\quad +\Item{(\id)}\ y^{2} = -x, +\] +the special value of~$x$ under consideration being $x = 0$. + +\Paragraph{113. Some general rules for differentiation.} Throughout +the theorems which follow we assume that the functions +$f(x)$~and~$F(x)$ have derivatives $f'(x)$~and~$F'(x)$ for the values of~$x$ +considered. + +\begin{Result} +\Item{(1)} If $\phi(x) = f(x) + F(x)$, then $\phi(x)$ has a derivative +\[ +\phi'(x) = f'(x) + F'(x). +\] +\end{Result} + +\begin{Result} +\Item{(2)} If $\phi(x) = kf(x)$, where $k$~is a constant, then $\phi(x)$~has a +derivative +\[ +\phi'(x) = kf'(x). +\] +\end{Result} + +We leave it as an exercise to the reader to deduce these results +from the general theorems stated in \Ex{xxxv}.~1. + +\begin{Result} +\Item{(3)} If $\phi(x) = f(x)F(x)$, then $\phi(x)$~has a derivative +\[ +\phi'(x) = f(x)F'(x) + f'(x)F(x). +\] +\end{Result} + +For +\begin{align*} +%[** TN: Re-aligned] +\phi'(x) + &= \lim\frac{f(x + h)F(x + h) - f(x)F(x)}{h}\\ + &= \lim\left\{f(x + h)\frac{F(x + h) - F(x)}{h} + + F(x)\frac{f(x + h) - f(x)}{h}\right\}\\ + &=f(x)F'(x) + F(x)f'(x). +\end{align*} +\PageSep{204} + +\begin{Result} +\Item{(4)} If $\phi(x) = \dfrac{1}{f(x)}$, then $\phi(x)$~has a derivative +\[ +\phi'(x) = -\frac{f'(x)}{\{f(x)\}^{2}}. +\] +\end{Result} + +In this theorem we of course suppose that $f(x)$~is not equal to +zero for the particular value of~$x$ under consideration. Then +\[ +\phi'(x) + = \lim \frac{1}{h} \left\{\frac{f(x) - f(x + h)}{f(x + h)f(x)}\right\} + = -\frac{f'(x)}{\{f(x)\}^{2}}. +\] + +\begin{Result} +\Item{(5)} If $\phi(x) = \dfrac{f(x)}{F(x)}$, then $\phi(x)$~has a derivative +\[ +\phi'(x) = \frac{f'(x)F(x) - f(x)F'(x)}{\{F(x)\}^{2}}. +\] +\end{Result} + +This follows at once from (3)~and~(4). + +\begin{Result} +\Item{(6)} If $\phi(x) = F\{f(x)\}$, then $\phi(x)$~has a derivative +\[ +\phi'(x) = F'\{f(x)\} f'(x). +\] +\end{Result} + +For let +\[ +f(x) = y,\quad +f(x + h) = y + k. +\] +Then $k \to 0$ as $h \to 0$, and $k/h \to f'(x)$. And +\begin{align*} +%[** TN: Not strictly correct: k can be zero infinitely often as h -> 0] +\phi'(x) + & = \lim \frac{F\{f(x + h)\} - F\{f(x)\}}{h}\\ + & = \lim \left\{\frac{F(y + k) - F(y)}{k}\right\} + × \lim \left(\frac{k}{h}\right)\\ + & = F'(y)f'(x). +\end{align*} + +This theorem includes (2)~and~(4) as special cases, as we see on +taking $F(x) = kx$ or $F(x) = 1/x$. Another interesting special case +is that in which $f(x) = ax + b$: the theorem then shows that the +derivative of~$F(ax + b)$ is~$aF'(ax + b)$. + +Our last theorem requires a few words of preliminary explanation. +Suppose that $x = \psi(y)$, where $\psi(y)$~is continuous and +steadily increasing (or decreasing), in the stricter sense of \SecNo[§]{95}, in +a certain interval of values of~$y$. Then we may write $y = \phi(x)$, +where $\phi$~is the `inverse' function (\SecNo[§]{109}) of~$\psi$. + +\begin{Result} +\Item{(7)} If $y = \phi(x)$, where $\phi$~is the inverse function of~$\psi$, so that +$x = \psi(y)$, and $\psi(y)$~has a derivative~$\psi'(y)$ which is not equal to +zero, then $\phi(x)$~has a derivative +\[ +\phi'(x) = \frac{1}{\psi'(y)}. +\] +\end{Result} +\PageSep{205} + +For if $\phi(x + h) = y + k$, then $k \to 0$ as $h \to 0$, and +\[ +\phi'(x) + = \lim_{h \to 0} \frac{\phi(x + h) - \phi(x)}{(x + h) - x} + = \lim_{k \to 0} \frac{(y + k) - y}{\psi(y + k) - \psi(y)} + = \frac{1}{\psi'(y)}. +\] +The last function may now be expressed in terms of~$x$ by means +of the relation $y = \phi(x)$, so that $\phi'(x)$~is the reciprocal of~$\psi'\{\phi(x)\}$. +This theorem enables us to differentiate any function if we know +the derivative of the inverse function. + +\Paragraph{114. Derivatives of complex functions.} So far we have +supposed that $y = \phi(x)$ is a purely \emph{real} function of~$x$. If $y$~is a +complex function $\phi(x) + i\psi(x)$, then we define the derivative of~$y$ +as being $\phi'(x) + i\psi'(x)$. The reader will have no difficulty in +seeing that Theorems~(1)--(5) above retain their validity when +$\phi(x)$~is complex. Theorems (6)~and~(7) have also analogues for +complex functions, but these depend upon the general notion of +a `function of a complex variable', a notion which we have encountered +at present only in a few particular cases. + +\Paragraph{115. The notation of the differential calculus.} We have +already explained that what we call a \emph{derivative} is often called a +\emph{differential coefficient}. Not only a different name but a different +notation is often used; the derivative of the function $y = \phi(x)$ +is often denoted by one or other of the expressions +\[ +D_{x}y,\quad +\frac{dy}{dx}. +\] +Of these the last is the most usual and convenient: the reader +must however be careful to remember that $dy/dx$ does not mean +`a certain number~$dy$ divided by another number~$dx$': it means +`the result of a certain operation~$D_{x}$ or~$d/dx$ applied to $y = \phi(x)$', +the operation being that of forming the quotient $\{\phi(x + h) - \phi(x)\}/h$ +and making $h \to 0$. + +\begin{Remark} +Of course a notation at first sight so peculiar would not have been +adopted without some reason, and the reason was as follows. The denominator~$h$ +of the fraction $\{\phi(x + h) - \phi(x)\}/h$ is the difference of the values $x+h$,~$x$ +of the independent variable~$x$; similarly the numerator is the difference of +the corresponding values $\phi(x + h)$,~$\phi(x)$ of the dependent variable~$y$. These +differences may be called the \emph{increments} of $x$~and~$y$ respectively, and denoted +by $\delta x$~and~$\delta y$. Then the fraction is~$\delta y/\delta x$, and it is for many purposes +convenient to denote the limit of the fraction, which is the same thing as~$\phi'(x)$, +\PageSep{206} +by~$dy/dx$. But this notation must for the present be regarded as +purely symbolical. The $dy$~and~$dx$ which occur in it cannot be separated, +and standing by themselves they would mean nothing: in particular $dy$~and~$dx$ +do not mean $\lim\delta y$ and~$\lim\delta x$, these limits being simply equal to zero. +The reader will have to become familiar with this notation, but so long as it +puzzles him he will be wise to avoid it by writing the differential coefficient in +the form~$D_{x}y$, or using the notation $\phi(x)$,~$\phi'(x)$, as we have done in the +preceding sections of this chapter. + +In \Ref{Ch.}{VII}, however, we shall show how it is possible to define the symbols +$dx$~and~$dy$ in such a way that they have an independent meaning and that +the derivative~$dy/dx$ is actually their quotient. +\end{Remark} + +The theorems of \SecNo[§]{113} may of course at once be translated into +this notation. They may be stated as follows: + +%[** TN: The conclusions below are aligned on their equals signs in the orig.] +\begin{Result} +\Item{(1)} if $y = y_{1} + y_{2}$, then +\[ +\frac{dy}{dx} = \frac{dy_{1}}{dx} + \frac{dy_{2}}{dx}; +\] + +\Item{(2)} if $y = ky_{1}$, then +\[ +\frac{dy}{dx} = k\frac{dy_{1}}{dx}; +\] + +\Item{(3)} if $y = y_{1}y_{2}$, then +\[ +\frac{dy}{dx} = y_{1}\frac{dy_{2}}{dx} + y_{2}\frac{dy_{1}}{dx}; +\] + +\Item{(4)} if $y = \dfrac{1}{y_{1}}$, then +\[ +\frac{dy}{dx} = -\frac{1}{y_{1}^{2}}\, \frac{dy_{1}}{dx}; +\] + +\Item{(5)} if $y = \dfrac{y_{1}}{y_{2}}$, then +\[ +\frac{dy}{dx} += \biggl(y_{2}\frac{dy_{1}}{dx} - y_{1}\frac{dy_{2}}{dx}\biggr) \bigg/ y_{2}^{2}; +\] + +\Item{(6)} if $y$~is a function of~$x$, and $z$~a function of~$y$, then +\[ +\frac{dz}{dx} = \frac{dz}{dy}\, \frac{dy}{dx}; +\] +\CenterLine{\Item{(7)}}{$\dfrac{dy}{dx} = 1 \bigg/ \biggl(\dfrac{dx}{dy}\biggr)$.} +\end{Result} + +\begin{Examples}{XL.} +\Item{1.} If $y = y_{1}y_{2}y_{3}$ then +\[ +\frac{dy}{dx} + = y_{2}y_{3}\, \frac{dy_{1}}{dx} + + y_{3}y_{1}\, \frac{dy_{2}}{dx} + + y_{1}y_{2}\, \frac{dy_{3}}{dx}, +\] +and if $y = y_{1}y_{2} \dots y_{n}$ then +\[ +\frac{dy}{dx} + = \sum_{r=1}^{n} y_{1}y_{2} \dots y_{r-1}y_{r+1} \dots y_{n}\, \frac{dy_{r}}{dx}. +\] +In particular, if $y = z^{n}$, then $dy/dx = nz^{n-1}(dz/dx)$; and if $y = x^{n}$, then +$dy/dx = nx^{n-1}$, as was proved otherwise in \Ex{xxxix}.~3. +\PageSep{207} + +\Item{2.} If $y = y_{1}y_{2}\dots y_{n}$ then +\[ +\frac{1}{y}\, \frac{dy}{dx} + = \frac{1}{y_{1}}\, \frac{dy_{1}}{dx} + + \frac{1}{y_{2}}\, \frac{dy_{2}}{dx} + \dots + + \frac{1}{y_{n}}\, \frac{dy_{n}}{dx}. +\] +In particular, if $y = z^{n}$, then $\dfrac{1}{y}\, \dfrac{dy}{dx} = \dfrac{n}{z}\, \dfrac{dz}{dx}$. +\end{Examples} + +\Paragraph{116. Standard forms.} We shall now investigate more +systematically the forms of the derivatives of a few of the +the simplest types of functions. + +\Topic{\Item{A.} Polynomials.} If $\phi(x) = a_{0}x^{n} + a_{1}x^{n-1} + \dots + a_{n}$, then +\[ +\phi'(x) = na_{0}x^{n-1} + (n - 1)a_{1}x^{n-2} + \dots + a_{n-1}. +\] +It is sometimes more convenient to use for the standard form of a +polynomial of degree~$n$ in~$x$ what is known as the \emph{binomial form}, +viz. +\[ +a_{0}x^{n} + + \binom{n}{1} a_{1}x^{n-1} + + \binom{n}{2} a_{2}x^{n-2} + \dots + a_{n}. +\] +In this case +\[ +\phi'(x) = n \left\{ + a_{0}x^{n-1} + + \binom{n - 1}{1} a_{1}x^{n-2} + + \binom{n - 1}{2} a_{2}x^{n-3} + \dots + + a_{n-1} +\right\}. +\] + +The binomial form of~$\phi(x)$ is often written symbolically as +\[ +(a_{0}, a_{1}, \dots, a_{n} \btw x, 1)^{n}; +\] +and then +\[ +\phi'(x) = n(a_{0}, a_{1}, \dots, a_{n-1} \btw x, 1)^{n-1}. +\] + +We shall see later that $\phi(x)$~can always be expressed as the +product of $n$~factors in the form +\[ +\phi(x) = a_{0}(x - \alpha_{1})(x - \alpha_{2}) \dots (x - \alpha_{n}), +\] +where the~$\alpha$'s are real or complex numbers. Then +\[ +\phi'(x) = a_{0}\tsum (x - \alpha_{2})(x - \alpha_{3}) \dots (x - \alpha_{n}), +\] +the notation implying that we form all possible products of $n - 1$ +factors, and add them all together. This form of the result holds +even if several of the numbers~$\alpha$ are equal; but of course then +some of the terms on the right-hand side are repeated. The +reader will easily verify that if +\[ +\phi(x) = a_{0}(x - \alpha_{1})^{m_{1}} + (x - \alpha_{2})^{m_{2}}\dots + (x - \alpha_{\nu})^{m_{\nu}}, +\] +then +\[ +%[** TN: Implicit indexing; notation matches the original.] +\phi'(x) = a_{0} \tsum m_{1}(x - \alpha_{1})^{m_{1}-1} + (x - \alpha_{2})^{m_{2}}\dots + (x - \alpha_{\nu})^{m_{\nu}}. +\] +\PageSep{208} + +\begin{Examples}{XLI.} +\Item{1.} Show that if $\phi(x)$~is a polynomial then $\phi'(x)$~is +the coefficient of~$h$ in the expansion of~$\phi(x + h)$ in powers of~$h$. + +\Item{2.} If $\phi(x)$~is divisible by~$(x - \alpha)^{2}$, then $\phi'(x)$~is divisible by~$x - \alpha$: and +generally, if $\phi(x)$~is divisible by~$(x - \alpha)^{m}$, then $\phi'(x)$~is divisible by~$(x - \alpha)^{m-1}$. + +\Item{3.} Conversely, if $\phi(x)$ and~$\phi'(x)$ are \emph{both} divisible by~$x - \alpha$, then $\phi(x)$~is +divisible by~$(x - \alpha)^{2}$; and if $\phi(x)$~is divisible by~$x - \alpha$ and $\phi'(x)$ by~$(x - \alpha)^{m-1}$, +then $\phi(x)$~is divisible by~$(x - \alpha)^{m}$. + +\Item{4.} Show how to determine as completely as possible the multiple roots +of $P(x) = 0$, where $P(x)$~is a polynomial, with their degrees of multiplicity, +by means of the elementary algebraical operations. + +[If $H_{1}$~is the highest common factor of $P$~and~$P'$, $H_{2}$~the highest common +factor of $H_{1}$ and~$P''$, $H_{3}$ that of $H_{2}$ and~$P'''$, and so on, then the roots of +$H_{1}H_{3}/H_{2}^{2} = 0$ are the \emph{double} roots of $P = 0$, the roots of $H_{2}H_{4}/H_{3}^{2} = 0$ the \emph{treble} +roots, and so on. But it may not be possible to complete the solution of +$H_{1}H_{3}/H_{2}^{2} = 0$, $H_{2}H_{4}/H_{3}^{2} = 0$,~\dots. Thus if $P(x) = (x - 1)^{3}(x^{5} - x - 7)^{2}$ then +$H_{1}H_{3}/H_{2}^{2} = x^{5} - x - 7$ and $H_{2}H_{4}/H_{3}^{2} = x - 1$; and we cannot solve the first +equation.] + +\Item{5.} Find all the roots, with their degrees of multiplicity, of +\[ +x^{4} + 3x^{3} - 3x^{2} - 11x - 6 = 0,\quad +x^{6} + 2x^{5} - 8x^{4} - 14x^{3} + 11x^{2} + 28x + 12 = 0. +\] + +\Item{6.} If $ax^{2} + 2bx + c$ has a double root, \ie\ is of the form $a(x - \alpha)^{2}$, then +$2(ax + b)$~must be divisible by~$x - \alpha$, so that $\alpha = -b/a$. This value of~$x$ must +satisfy $ax^{2} + 2bx + c = 0$. Verify that the condition thus arrived at is +$ac - b^{2} = 0$. + +\Item{7.} The equation $1/(x - a) + 1/(x - b) + 1/(x - c) = 0$ can have a pair of +equal roots only if $a = b = c$. \MathTrip{1905.} + +\Item{8.} Show that +\[ +ax^{3} + 3bx^{2} + 3cx + d = 0 +\] +has a double root if $G^{2} + 4H^{3} = 0$, where $H = ac - b^{2}$, $G = a^{2}d - 3abc + 2b^{3}$. + +[Put $ax + b = y$, when the equation reduces to $y^{3} + 3Hy + G = 0$. This +must have a root in common with $y^{2} + H = 0$.] + +\Item{9.} The reader may verify that if $\alpha$,~$\beta$, $\gamma$,~$\delta$ are the roots of +\[ +ax^{4} + 4bx^{3} + 6cx^{2} + 4dx + e = 0, +\] +then the equation whose roots are +\[ +\tfrac{1}{12}a \{ + (\alpha - \beta)(\gamma - \delta) - (\gamma - \alpha)(\beta - \delta) +\}, +\] +and two similar expressions formed by permuting $\alpha$,~$\beta$,~$\gamma$ cyclically, is +\[ +4\theta^{3} - g_{2}\theta - g_{3} = 0, +\] +where +\[ +g_{2} = ae - 4bd + 3c^{2},\quad +g_{3} = ace + 2bcd - ad^{2} - eb^{2} - c^{3}. +\] +It is clear that if two of $\alpha$,~$\beta$, $\gamma$,~$\delta$ are equal then two of the roots of this cubic +will be equal. Using the result of Ex.~8 we deduce that $g_{2}^{3} - 27g_{3}^{2} = 0$. +\PageSep{209} + +\begin{Result} +\Item{10.} \Topic{Rolle's Theorem for polynomials.} If $\phi(x)$~is any polynomial, +then between any pair of roots of $\phi(x) = 0$ lies a root of $\phi'(x) = 0$. +\end{Result} + +A general proof of this theorem, applying not only to polynomials but to +other classes of functions, will be given later. The following is an algebraical +proof valid for polynomials only. We suppose that $\alpha$,~$\beta$ are two successive +roots, repeated respectively $m$~and~$n$ times, so that +\[ +\phi(x) = (x - \alpha)^{m} (x - \beta)^{n} \theta(x), +\] +where $\theta(x)$~is a polynomial which has the same sign, say the positive sign, for +$\alpha \leq x \leq \beta$. Then +{\footnotesize\begin{align*} +\phi'(x) + &= (x - \alpha)^{m} (x - \beta)^{n} \theta'(x) + + \{m(x - \alpha)^{m-1} (x - \beta)^{n} + + n(x - \alpha)^{m} (x - \beta)^{n-1}\} \theta(x)\\ + &= (x - \alpha)^{m-1} (x - \beta)^{n-1} + [(x - \alpha) (x - \beta) \theta'(x) + + \{m(x - \beta) + n(x - \alpha)\} \theta(x)]\\ + &= (x - \alpha)^{m-1} (x - \beta)^{n-1} F(x), +\end{align*}}% +say. Now $F(\alpha) = m(\alpha - \beta) \theta(\alpha)$ and $F(\beta) = n(\beta - \alpha) \theta(\beta)$, which have opposite +signs. Hence $F(x)$, and so~$\phi'(x)$, vanishes for some value of~$x$ between +$\alpha$~and~$\beta$\Add{.} +\end{Examples} + +\Paragraph{117.} \Topic{\Item{B.} Rational Functions.} If +\[ +R(x) = \frac{P(x)}{Q(x)}, +\] +where $P$~and~$Q$ are polynomials, it follows at once from \SecNo[§]{113},~(5) that +\[ +R'(x) = \frac{P'(x)Q(x) - P(x)Q'(x)}{\{Q(x)\}^{2}}, +\] +and this formula enables us to write down the derivative of any +rational function. The form in which we obtain it, however, may or +may not be the simplest possible. It will be the simplest possible if +$Q(x)$ and~$Q'(x)$ have no common factor, \ie\ if $Q(x)$~has no repeated +factor. But if $Q(x)$~has a repeated factor then the expression +which we obtain for~$R'(x)$ will be capable of further reduction. + +It is very often convenient, in differentiating a rational +function, to employ the method of partial fractions. We shall +suppose that~$Q(x)$, as in \SecNo[§]{116}, is expressed in the form +\[ +a_{0}(x - \alpha_{1})^{m_{1}} + (x - \alpha_{2})^{m_{2}}\dots + (x - \alpha_{\nu})^{m_{\nu}}. +\] +Then it is proved in treatises on Algebra\footnote + {See, \eg, Chrystal's \textit{Algebra}, vol.~i, pp.~151~\textit{et~seq.}} +that $R(x)$~can be +expressed in the form +\begin{align*} +\Pi(x) &+ \frac{A_{1, 1}}{x - \alpha_{1}} + + \frac{A_{1, 2}}{(x - \alpha_{1})^{2}} + \dots + + \frac{A_{1, m_{1}}}{(x - \alpha_{1})^{m_{1}}}\\ + &+ \frac{A_{2, 1}}{x - \alpha_{2}} + + \frac{A_{2, 2}}{(x - \alpha_{2})^{2}} + \dots + + \frac{A_{2, m_{2}}}{(x - \alpha_{2})^{m_{2}}} + \dots, +\end{align*} +\PageSep{210} +where $\Pi(x)$~is a polynomial; \ie\ as the sum of a polynomial and +the sum of a number of terms of the type +\[ +\frac{A}{(x - \alpha)^{p}}, +\] +where $\alpha$~is a root of $Q(x) = 0$. We know already how to find the +derivative of the polynomial: and it follows at once from Theorem~(4) +of \SecNo[§]{113}, or, if $\alpha$~is complex, from its extension indicated in \SecNo[§]{114}, +that the derivative of the rational function last written is +\[ +-\frac{pA(x -\alpha)^{p-1}}{(x - \alpha)^{2p}} + = -\frac{pA}{(x - \alpha)^{p+1}}. +\] + +We are now able to write down the derivative of the general +rational function~$R(x)$, in the form +\[ +\Pi'(x) - \frac{A_{1, 1}}{(x - \alpha_{1})^{2}} + - \frac{2A_{1, 2}}{(x - \alpha_{1})^{3}} - \dots + - \frac{A_{2, 1}}{(x - \alpha_{2})^{2}} + - \frac{2A_{2, 2}}{(x - \alpha_{2})^{3}} - \dots. +\] +Incidentally we have proved that \emph{the derivative of~$x^{m}$ is~$mx^{m-1}$, +for all integral values of~$m$ positive or negative}. + +The method explained in this section is particularly useful +when we have to differentiate a rational function several times +(see \Exs{xlv}). + +\begin{Examples}{XLII.} +\Item{1.} Prove that +\[ +\frac{d}{dx}\left(\frac{x}{1 + x^{2}}\right) + = \frac{1 - x^{2}}{(1 + x^{2})^{2}},\quad +\frac{d}{dx}\left(\frac{1 - x^{2}}{1 + x^{2}}\right) + = -\frac{4x}{(1 + x^{2})^{2}}. +\] + +\Item{2.} Prove that +\[ +\frac{d}{dx}\left(\frac{ax^{2} + 2bx + c}{Ax^{2} + 2Bx + C}\right) + = \frac{(ax + b) (Bx + C) - (bx + c) (Ax + B)}{(Ax^{2} + 2Bx + C)^{2}}. +\] + +\Item{3.} If $Q$~has a factor $(x - \alpha)^{m}$ then the denominator of~$R'$ (when $R'$~is +reduced to its lowest terms) is divisible by~$(x - \alpha)^{m+1}$ but by no higher power +of~$x - \alpha$. + +\Item{4.} In no case can the denominator of~$R'$ have a \emph{simple} factor~$x - \alpha$. +Hence no rational function (such as~$1/x$) whose denominator contains any +simple factor can be the derivative of another rational function. +\end{Examples} + +\Paragraph{118.} \Topic{\Item{C.} Algebraical Functions.} The results of the preceding +sections, together with Theorem~(6) of \SecNo[§]{113}, enable us to +obtain the derivative of any explicit algebraical function whatsoever. + +The most important such function is~$x^{m}$, where $m$~is a rational +number. We have seen already (\SecNo[§]{117}) that the derivative of this +\PageSep{211} +function is~$mx^{m-1}$ when $m$~is an integer positive or negative; and +we shall now prove that this result is true for all rational values +of~$m$. Suppose that $y = x^{m} = x^{p/q}$, where $p$~and~$q$ are integers and +$q$~positive; and let $z = x^{1/q}$, so that $x = z^{q}$ and $y = z^{p}$. Then +\[ +\frac{dy}{dx} + = \biggl(\frac{dy}{dz}\biggr) \bigg/ \biggl(\frac{dx}{dz}\biggr) + = \frac{p}{q} z^{p-q} = mx^{m-1}. +\] + +This result may also be deduced as a corollary from \Ex{xxxvi}.~3. +For, if $\phi(x) = x^{m}$, we have +\begin{align*} +\phi'(x) + &= \lim_{h \to 0} \frac{(x + h)^{m} - x^{m}}{h}\\ + &= \lim_{\xi \to x} \frac{\xi^{m} - x^{m}}{\xi - x} + = mx^{m-1}. +\end{align*} +It is clear that the more general formula +\[ +\frac{d}{dx} (ax + b)^{m} = ma(ax + b)^{m-1} +\] +holds also for all rational values of~$m$. + +The differentiation of \emph{implicit} algebraical functions involves +certain theoretical difficulties to which we shall return in \Ref{Ch.}{VII}\@. +But there is no practical difficulty in the actual calculation of the +derivative of such a function: the method to be adopted will be +illustrated sufficiently by an example. Suppose that $y$~is given by +the equation +\[ +x^{3} + y^{3} - 3axy = 0. +\] +Differentiating with respect to~$x$ we find +\[ +x^{2} + y^{2} \frac{dy}{dx} - a\left(y + x \frac{dy}{dx}\right) = 0 +\] +and so +\[ +\frac{dy}{dx} = -\frac{x^{2} - ay}{y^{2} - ax}. +\] + +\begin{Examples}{XLIII.} +\Item{1.} Find the derivatives of +\[ +\bigsqrtp{\frac{1 + x}{1 - x}},\quad +\bigsqrtp{\frac{ax + b}{cx + d}},\quad +\bigsqrtp{\frac{ax^{2} + 2bx + c}{Ax^{2} + 2Bx + C}},\quad +(ax + b)^{m} (cx + d)^{n}. +\] + +\Item{2.} Prove that +\[ +\frac{d}{dx}\left\{\frac{x}{\sqrtp{a^{2} + x^{2}}}\right\} + = \frac{a^{2}}{(a^{2} + x^{2})^{(3/2)}},\quad +\frac{d}{dx}\left\{\frac{x}{\sqrtp{a^{2} - x^{2}}}\right\} + = \frac{a^{2}}{(a^{2} - x^{2})^{3/2}}. +\] + +\Item{3.} Find the differential coefficient of $y$ when +\[ +\Itemp{(i)} ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0,\quad +\Itemp{(ii)} x^{5} + y^{5} - 5ax^{2}y^{2} = 0. +\] +\end{Examples} +\PageSep{212} + +\Paragraph{119.} \Topic{\Item{D.} Transcendental Functions.} We have already +proved (\Ex{xxxix}.~4) that +\[ +D_{x} \sin x = \cos x, \quad +D_{x} \cos x = -\sin x. +\] + +By means of Theorems (4)~and~(5) of \SecNo[§]{113}, the reader will +easily verify that +\begin{alignat*}{2} +D_{x} \tan x &= \sec^{2} x, & D_{x} \cot x &= -\cosec^{2} x,\\ +D_{x} \sec x &= \tan x \sec x, \quad & D_{x} \cosec x &= -\cot x\cosec x. +\end{alignat*} +And by means of Theorem~(7) we can determine the derivatives +of the ordinary inverse trigonometrical functions. The reader +should verify the following formulae: +\begin{alignat*}{2} +D_{x} \arcsin x &= ±1/\sqrtp{1 - x^{2}}, & +D_{x} \arccos x &= \mp 1/\sqrtp{1 - x^{2}},\\ +% +D_{x} \arctan x &= 1/(1 + x^{2}), & +D_{x} \arccot x &= -1/(1 + x^{2}),\\ +D_{x} \arcsec x &= ± 1/\{x\sqrtp{x^{2} - 1}\}, \quad & +D_{x} \arccosec x &= \mp 1/\{x\sqrtp{x^{2} - 1}\}. +\end{alignat*} +In the case of the inverse sine and cosecant the ambiguous sign +is the same as that of~$\cos(\arcsin x)$, in the case of the inverse +cosine and secant the same as that of~$\sin(\arccos x)$. + +The more general formulae +\[ +D_{x} \arcsin(x/a) = ±1/\sqrtp{a^{2} - x^{2}},\quad +D_{x} \arctan(x/a) = a/(x^{2} + a^{2}), +\] +which are also easily derived from Theorem~(7) of \SecNo[§]{113}, are also +of considerable importance. In the first of them the ambiguous +sign is the same as that of~$a\cos\{\arcsin(x/a)\}$, since +\[ +a\sqrtb{1 - (x^{2}/a^{2})} = ±\sqrtp{a^{2} - x^{2}} +\] +according as $a$~is positive or negative. + +Finally, by means of Theorem~(6) of \SecNo[§]{113}, we are enabled to +differentiate composite functions involving symbols both of algebraical +and trigonometrical functionality, and so to write down +the derivative of any such function as occurs in the following +examples. + +\begin{Examples}{XLIV.\protect\footnotemark} +\Item{1.} Find the derivatives of\footnotetext + {In these examples $m$~is a rational number and $a$, $b$,~\dots, $\alpha$, $\beta$~\dots\ have such + values that the functions which involve them are real.} + +\begin{gather*} +\cos^{m} x, \quad \sin^{m} x, \quad +\cos x^{m}, \quad \sin x^{m}, \quad +\cos (\sin x), \quad \sin (\cos x),\\ +\sqrtp{a^{2}\cos^{2} x + b^{2}\sin^{2} x}, \quad +\frac{\cos x\sin x}{\sqrtp{a^{2}\cos^{2} x + b^{2}\sin^{2} x}},\\ +x\arcsin x + \sqrtp{1 - x^{2}}, \quad +(1 + x)\arctan\sqrt{x} - \sqrt{x}. +\end{gather*} +\PageSep{213} + +\Item{2.} Verify by differentiation that $\arcsin x + \arccos x$ is constant for all +values of~$x$ between $0$~and~$1$, and $\arctan x + \arccot x$ for all positive values +of~$x$. + +\Item{3.} Find the derivatives of +\[ +\arcsin\sqrtp{1 - x^{2}},\quad +\arcsin\{2x\sqrtp{1 - x^{2}}\},\quad +\arctan \left(\frac{a + x}{1 - ax}\right). +\] +How do you explain the simplicity of the results? + +\Item{4.} Differentiate +\[ + \frac{1}{\sqrtp{ac - b^{2}}} \arctan \frac{ax + b}{\sqrtp{ac - b^{2}}},\quad +-\frac{1}{\sqrtp{-a}} \arcsin\frac{ax + b}{\sqrtp{b^{2} - ac}}. +\] + +\Item{5.} Show that each of the functions +\[ +2\arcsin \bigsqrtp{\frac{x - \beta}{\alpha - \beta}},\quad +2\arctan \bigsqrtp{\frac{x - \beta}{\alpha - x}},\quad +\arcsin \frac{2\sqrtb{(\alpha - x)(x - \beta)}}{\alpha - \beta} +\] +has the derivative +\[ +\frac{1}{\sqrtb{(\alpha - x)(x - \beta)}}. +\] + +\Item{6.} Prove that +\[ +\frac{d}{d\theta}\left\{ + \arccos \bigsqrtp{\frac{\cos 3\theta}{\cos^{3}\theta}} +\right\} + = \bigsqrtp{\frac{3}{\cos\theta \cos 3\theta}}. +\] +\MathTrip{1904.} + +\Item{7.} Show that +\[ +\frac{1}{\sqrtp{C(Ac - aC)}}\, \frac{d}{dx} \left[ + \arccos \bigsqrtb{\frac{C(ax^{2} + c)}{c(Ax^{2} + C)}} +\right] + = \frac{1}{(Ax^{2} + C) \sqrtp{ax^{2} + c}}. +\] + +\Item{8.} Each of the functions +\[ +\frac{1}{\sqrtp{a^{2} - b^{2}}} + \arccos \left(\frac{a\cos x + b}{a + b\cos x}\right),\quad +\frac{2}{\sqrtp{a^{2} - b^{2}}} + \arctan \left\{\bigsqrtp{\frac{a - b}{a + b }} \tan \tfrac{1}{2}x\right\} +\] +has the derivative~$1/(a + b\cos x)$. + +\Item{9.} If $X = a + b\cos x + c\sin x$, and +\[ +y = \frac{1}{\sqrtp{a^{2} - b^{2} -c^{2}}} + \arccos \frac{aX - a^{2} + b^{2} + c^{2}}{X \sqrtp{b^{2} + c^{2}}}, +\] +then $dy/dx = 1/X$. + +\Item{10.} Prove that the derivative of $F[f\{\phi(x)\}]$ is $F'[f\{\phi(x)\}]\, f'\{\phi(x)\}\phi'(x)$, +and extend the result to still more complicated cases. + +\Item{11.} If $u$~and~$v$ are functions of~$x$, then +\[ +D_{x} \arctan(u/v) = (vD_{x}u - uD_{x}v)/(u^{2} + v^{2}). +\] + +\Item{12.} The derivative of $y = (\tan x + \sec x)^{m}$ is $my\sec x$. + +\Item{13.} The derivative of $y = \cos x + i\sin x$ is~$iy$. + +\Item{14.} Differentiate $x\cos x$, $(\sin x)/x$. Show that the values of~$x$ for which +the tangents to the curves $y = x\cos x$, $y = (\sin x)/x$ are parallel to the axis of~$x$ +are roots of $\cot x = x$, $\tan x = x$ respectively. +\PageSep{214} + +\Item{15.} It is easy to see (cf.\ \Ex{xvii}.~5) that the equation $\sin x = ax$, where $a$~is +positive, has no real roots except $x = 0$ if $a \geq 1$, and if $a < 1$ a finite number of +roots which increases as $a$~diminishes. Prove that the values of~$a$ for which +the number of roots changes are the values of~$\cos\xi$, where $\xi$~is a positive root +of the equation $\tan\xi = \xi$. [The values required are the values of~$a$ for which +$y = ax$ touches $y = \sin x$.] + +\Item{16.} If $\phi(x) = x^{2}\sin(1/x)$ when $x \neq 0$, and $\phi(0) = 0$, then +\[ +\phi'(x) = 2x\sin(1/x) - \cos(1/x) +\] +when $x\neq 0$, and $\phi'(0) = 0$. And $\phi'(x)$~is discontinuous for $x = 0$ (cf.\ \SecNo[§]{111},~(2)). + +\Item{17.} Find the equations of the tangent and normal at the point $(x_{0}, y_{0})$ +of the circle $x^{2} + y^{2} = a^{2}$. + +[Here $y = \sqrtp{a^{2} - x^{2}}$, $dy/dx = -x/\sqrtp{a^{2} - x^{2}}$, and the tangent is +\[ +y - y_{0} = (x - x_{0}) \left\{-x_{0}/\sqrtp{a^{2} - x_{0}^{2}}\right\}, +\] +which may be reduced to the form $xx_{0} + yy_{0} = a^{2}$. The normal is $xy_{0} - yx_{0} = 0$, +which of course passes through the origin.] + +\Item{18.} Find the equations of the tangent and normal at any point of the +ellipse $(x/a)^{2} + (y/b)^{2} = 1$ and the hyperbola $(x/a)^{2} - (y/b)^{2} = 1$. + +\Item{19.} The equations of the tangent and normal to the curve $x = \phi(t)$, +$y = \psi(t)$, at the point whose parameter is~$t$, are +\[ +\frac{x - \phi(t)}{\phi'(t)} = \frac{y - \psi(t)}{\psi'(t)},\quad +\{x - \phi(t)\} \phi'(t) + \{y - \psi(t)\} \psi'(t) = 0. +\] +\end{Examples} + +\Paragraph{120. Repeated differentiation.} We may form a new function~$\phi''(x)$ +from~$\phi'(x)$ just as we formed~$\phi'(x)$ from~$\phi(x)$. This +function is called the \emph{second derivative} or \emph{second differential +coefficient} of~$\phi(x)$. The second derivative of $y = \phi(x)$ may also +be written in any of the forms +\[ +D_{x}^{2}y,\quad +\left(\frac{d}{dx}\right)^{2}y,\quad +\frac{d^{2}y}{dx^{2}}. +\] + +In exactly the same way we may define the \emph{$n$th~derivative or +$n$th~differential coefficient of $y = \phi(x)$}, which may be written in any +of the forms +\[ +\phi^{(n)}(x),\quad +D_{x}^{n}y,\quad +\left(\frac{d}{dx}\right)^{n}y,\quad +\frac{d^{n}y}{dx^{n}}. +\] +But it is only in a few cases that it is easy to write down a +general formula for the $n$th~differential coefficient of a given +function. Some of these cases will be found in the examples +which follow. +\PageSep{215} + +\begin{Examples}{XLV.} +\Item{1.} If $\phi(x) = x^{m}$ then +\[ +\phi^{(n)}(x) = m(m - 1) \dots (m - n + 1)x^{m-n}. +\] +This result enables us to write down the $n$th~derivative of any polynomial. + +\Item{2.} If $\phi(x) = (ax + b)^{m}$ then +\[ +\phi^{(n)}(x) = m(m - 1) \dots (m - n + 1)a^{n}(ax + b)^{m-n}. +\] +In these two examples $m$~may have any rational value. If $m$~is a positive +integer, and $n > m$, then $\phi^{(n)}(x) = 0$. + +\Item{3.} The formula +\[ +\left(\frac{d}{dx}\right)^{n} \frac{A}{(x - \alpha)^{p}} + = (-1)^{n} \frac{p(p + 1) \dots (p + n - 1)A}{(x - \alpha)^{p+n}} +\] +enables us to write down the $n$th~derivative of any rational function expressed +in the standard form as a sum of partial fractions. + +\Item{4.} Prove that the $n$th~derivative of $1/(1 - x^{2})$ is +\[ +\tfrac{1}{2}(n!) \{(1 - x)^{-n-1} + (-1)^{n}(1 + x)^{-n-1}\}. +\] + +\Item{5.} \Topic{Leibniz' Theorem.} If $y$~is a product~$uv$, and we can form the +first $n$~derivatives of $u$ and~$v$, then we can form the $n$th~derivative of~$y$ by +means of \emph{Leibniz' Theorem}, which gives the rule +\[ +(uv)_{n} = u_{n}v + + \binom{n}{1}u_{n-1}v_{1} + + \binom{n}{2}u_{n-2}v_{2} + \dots + + \binom{n}{r}u_{n-r}v_{r} + \dots + uv_{n}, +\] +where suffixes indicate differentiations, so that $u_{n}$, for example, denotes the +$n$th~derivative of~$u$. To prove the theorem we observe that +\begin{align*} +(uv)_{1} &= u_{1}v + uv_{1},\\ +(uv)_{2} &= u_{2}v + 2u_{1}v_{1} + uv_{2}, +\end{align*} +and so on. It is obvious that by repeating this process we arrive at a +formula of the type +\[ +(uv)_{n} = u_{n}v + + a_{n, 1} u_{n-1} v_{1} + + a_{n, 2} u_{n-2} v_{2} + \dots + + a_{n, r} u_{n-r} v_{r} + \dots + uv_{n}. +\] + +Let us assume that $a_{n, r} = \dbinom{n}{r}$ for $r = 1$, $2$,~\dots\Add{,} $n - 1$, and show that if this +is so then $a_{n+1, r} = \dbinom{n + 1}{r}$ for $r = 1$, $2$,~\dots~$n$. It will then follow by the +principle of mathematical induction that $a_{n, r} = \dbinom{n}{r}$ for all values of $n$ and~$r$ +in question. + +When we form $(uv)_{n+1}$ by differentiating $(uv)_{n}$ it is clear that the coefficient +of~$u_{n+1-r}v_{r}$ is +\[ +a_{n, r} + a_{n, r-1} = \binom{n}{r} + \binom{n}{r - 1} = \binom{n + 1}{r}. +\] +This establishes the theorem. +\PageSep{216} + +\Item{6.} The $n$th~derivative of~$x^{m}f(x)$ is +\begin{multline*} +\frac{m!}{(m - n)!} x^{m-n} f(x) + n \frac{m!}{(m - n + 1)!} x^{m-n+1} f'(x)\\ + + \frac{n(n - 1)}{1·2}\, \frac{m!}{(m - n + 2)!} x^{m-n+2} f''(x) + \dots, +\end{multline*} +the series being continued for $n + 1$~terms or until it terminates. + +\Item{7.} Prove that $D_{x}^{n}\cos x = \cos(x + \frac{1}{2}n\pi)$, $D_{x}^{n}\sin x = \sin(x + \frac{1}{2}n\pi)$\Add{.} + +\Item{8.} If $y = A\cos mx + B\sin mx$ then $D_{x}^{2} y + m^{2} y = 0$. And if +\[ +y = A\cos mx + B\sin mx + P_{n}(x), +\] +where $P_{n}(x)$~is a polynomial of degree~$n$, then $D_{x}^{n+3} y + m^{2} D_{x}^{n+1} y = 0$. + +\Item{9.} If $x^{2} D_{x}^{2}y + x D_{x} y + y = 0$ then +\[ +x^{2} D_{x}^{n+2} y + (2n + 1)x D_{x}^{n+1} y + (n^{2} + 1) D_{x}^{n} y = 0. +\] + +[Differentiate $n$~times by \DPchg{Leibnitz'}{Leibniz'} Theorem.] + +\Item{10.} If $U_{n}$~denotes the $n$th~derivative of $(Lx + M)/(x^{2} - 2Bx + C)$, then +\[ +\frac{x^{2} - 2Bx + C}{(n + 1)(n + 2)} U_{n+2} + + \frac{2(x - B)}{n + 1} U_{n+1} + U_{n} = 0. +\] +\MathTrip{1900.} + +[First obtain the equation when $n = 0$; then differentiate $n$~times by +\DPchg{Leibnitz'}{Leibniz'} Theorem.] + +\Item{11.} \Topic{The $n$th~derivatives of $a/(a^{2} + x^{2})$ and $x/(a^{2} + x^{2})$.} Since +\[ +\frac{a}{a^{2} + x^{2}} + = \frac{1}{2i} \left(\frac{1}{x - ai} - \frac{1}{x + ai}\right), \quad +\frac{x}{a^{2} + x^{2}} + = \frac{1}{2} \left(\frac{1}{x - ai} + \frac{1}{x + ai}\right), +\] +we have +\[ +D_{x}^{n} \left(\frac{a}{a^{2} + x^{2}}\right) + = \frac{(-1)^{n} n!}{2i} \left\{ + \frac{1}{(x - ai)^{n+1}} - \frac{1}{(x + ai)^{n+1}} +\right\}, +\] +{\Loosen and a similar formula for $D_{x}^{n}\{x/(a^{2} + x^{2})\}$. If $\rho = \sqrtp{x^{2} + a^{2}}$, and $\theta$~is the +numerically smallest angle whose cosine and sine are $x/\rho$~and~$a/\rho$, then +$x + ai = \rho\Cis\theta$ and $x - ai = \rho\Cis(-\theta )$, and so} +\begin{align*} +D_{x}^{n} \{a/(a^{2} + x^{2})\} + &= \{(-1)^{n} n!/2i\} \rho^{-n-1} + [\Cis \{(n + 1)\theta\} - \Cis \{-(n + 1)\theta\}]\\ + &= (-1)^{n} n!\, (x^{2} + a^{2})^{-(n+1)/2} \sin \{(n + 1) \arctan(a/x)\}. +\end{align*} +Similarly +\[ +D_{x}^{n} \{x/(a^{2} + x^{2})\} + = (-1)^{n} n!\, (x^{2} + a^{2})^{-(n+1)/2} \cos \{(n + 1) \arctan (a/x)\}. +\] + +\Item{12.} Prove that +\begin{align*} +D_{x}^{n} \{(\cos x)/x\} + &= \{P_{n} \cos(x + \tfrac{1}{2}n\pi) + + Q_{n} \sin(x + \tfrac{1}{2}n\pi)\}/x^{n+1},\\ +D_{x}^{n} \{(\sin x)/x\} + &= \{P_{n} \sin(x + \tfrac{1}{2}n\pi) + - Q_{n} \cos(x + \tfrac{1}{2}n\pi)\}/x^{n+1}, +\end{align*} +where $P_{n}$ and~$Q_{n}$ are polynomials in~$x$ of degree $n$~and~$n-1$ respectively. + +\Item{13.} Establish the formulae +\begin{gather*} +%[** TN: Set on one line in the orignal] +\frac{dx}{dy} = 1 \bigg/\biggl(\frac{dy}{dx}\biggr),\quad +\frac{d^{2} x}{dy^{2}} + = -\frac{d^{2} y}{dx^{2}} \bigg/ \biggl(\frac{dy}{dx}\biggr)^{3},\\ +\frac{d^{3} x}{dy^{3}} + = -\biggl\{\frac{d^{3} y}{dx^{3}}\, \frac{dy}{dx} + - 3\biggl(\frac{d^{2} y}{dx^{2}}\biggr)\biggr\} \bigg/ + \biggl(\frac{dy}{dx}\biggr)^{5}. +\end{gather*} +\PageSep{217} + +\Item{14.} If $yz = 1$ and $y_{r} = (1/r!) D_{x}^{r}y$, $z_{s} = (1/s!) D_{x}^{s}z$, then +\[ +\frac{1}{z^{3}} +\begin{vmatrix} +z & z_{1}& z_{2}\\ +z_{1}& z_{2}& z_{3}\\ +z_{2}& z_{3}& z_{4} +\end{vmatrix} += \frac{1}{y^{2}} +\begin{vmatrix} +y_{2}& y_{3}\\ +y_{3}& y_{4} +\end{vmatrix}. +\] +\MathTrip{1905.} + +\Item{15.} If +\[ +W(y, z, u) = +\begin{vmatrix} +y & z & u\\ +y' & z' & u'\\ +y''& z''& u'' +\end{vmatrix}, +\] +dashes denoting differentiations with +respect to~$x$, then +\[ +W(y, z, u) = y^{3}\, W\left(1, \frac{z}{y}, \frac{u}{y}\right). +\] + +\Item{16.} If +\[ +ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, +\] +then +\[ +dy/dx = -(ax + hy + g)/(hx + by + f) +\] +and +\[ +d^{2}y/dx^{2} = (abc + 2fgh - af^{2} - bg^{2} - ch^{2})/(hx + by + f)^{3}. +\] +\end{Examples} + +\Paragraph{121. Some general theorems concerning derived functions.} +In all that follows we suppose that $\phi(x)$~is a function of~$x$ +which has a derivative~$\phi'(x)$ for all values of~$x$ in question. This +assumption of course involves the continuity of~$\phi(x)$. + +\begin{ParTheorem}{The meaning of the sign of~$\phi'(x)$. \normalfont\textsc{Theorem~A\@.}} +If +$\phi'(x_{0}) > 0$ then $\phi(x) < \phi(x_{0})$ for all values of~$x$ less than~$x_{0}$ but +sufficiently near to~$x_{0}$, and $\phi(x) > \phi(x_{0})$ for all values of~$x$ greater +than~$x_{0}$ but sufficiently near to~$x_{0}$. +\end{ParTheorem} + +For $\{\phi(x_{0} + h) - \phi(x_{0})\}/h$ converges to a positive limit~$\phi'(x_{0})$ as +$h \to 0$. This can only be the case if $\phi(x_{0} + h) - \phi(x_{0})$ and~$h$ have +the same sign for sufficiently small values of~$h$, and this is precisely +what the theorem states. Of course from a geometrical point of +view the result is intuitive, the inequality $\phi'(x) > 0$ expressing +the fact that the tangent to the curve $y = \phi(x)$ makes a positive +acute angle with the axis of~$x$. The reader should formulate for +himself the corresponding theorem for the case in which $\phi'(x) < 0$. + +An immediate deduction from Theorem~A is the following +important theorem, generally known as Rolle's Theorem. In view +of the great importance of this theorem it may be well to repeat +that its truth depends on the assumption of the existence of the +derivative~$\phi'(x)$ for all values of~$x$ in question. + +\begin{Theorem}[B\@.] +If $\phi(a) = 0$ and $\phi(b) = 0$, then there must be at +least one value of~$x$ which lies between $a$ and~$b$ and for which +$\phi'(x) = 0$. +\end{Theorem} + +There are two possibilities: the first is that $\phi(x)$~is equal to +\PageSep{218} +zero throughout the whole interval~$\DPmod{(a, b)}{[a, b]}$. In this case $\phi'(x)$~is +also equal to zero throughout the interval. If on the other hand +$\phi(x)$~is not always equal to zero, then there must be values of~$x$ +for which $\phi(x)$~is positive or negative. Let us suppose, for +example, that $\phi(x)$~is sometimes positive. Then, by Theorem~2 of +\SecNo[§]{102}, there is a value~$\xi$ of~$x$, not equal to $a$~or~$b$, and such that $\phi(\xi)$~is +at least as great as the value of~$\phi(x)$ at any other point in +the interval. And $\phi'(\xi)$~must be equal to zero. For if it were +positive then $\phi(x)$ would, by Theorem~A, be greater than~$\phi(\xi)$ for +values of~$x$ greater than~$\xi$ but sufficiently near to~$\xi$, so that there +would certainly be values of~$\phi(x)$ greater than~$\phi(\xi)$. Similarly we +can show that $\phi'(\xi)$ cannot be negative. + +\begin{Cor}[1.] +If $\phi(a) = \phi(b) = k$, then there must be a value of~$x$ +between $a$~and~$b$ such that $\phi'(x) = 0$. +\end{Cor} + +We have only to put $\phi(x) - k = \psi(x)$ and apply Theorem~B +to~$\psi(x)$. + +\begin{Cor}[2.] +If $\phi'(x) > 0$ for all values of~$x$ in a certain interval, +then $\phi(x)$~is an increasing function of~$x$, in the stricter sense of \SecNo[§]{95}, +throughout that interval. +\end{Cor} + +Let $x_{1}$ and~$x_{2}$ be two values of~$x$ in the interval in question, +and $x_{1} < x_{2}$. We have to show that $\phi(x_{1}) < \phi(x_{2})$. In the first +place $\phi(x_{1})$~cannot be equal to~$\phi(x_{2})$; for, if this were so, there +would, by Theorem~B, be a value of~$x$ between $x_{1}$ and~$x_{2}$ for which +$\phi'(x) = 0$. Nor can $\phi(x_{1})$~be greater than~$\phi(x_{2})$. For, since $\phi'(x_{1})$~is +positive, $\phi(x)$~is, by Theorem~A, greater than~$\phi(x_{1})$ when $x$~is +greater than~$x_{1}$ and sufficiently near to~$x_{1}$. It follows that there is +a value~$x_{3}$ of~$x$ between $x_{1}$ and$~x_{2}$ such that $\phi(x_{3}) = \phi(x_{1})$; and so, +by Theorem~B, that there is a value of~$x$ between $x_{1}$ and~$x_{3}$ for +which $\phi'(x) = 0$. + +\begin{Cor}[3.] +The conclusion of Cor.~\Inum{2} still holds if the interval~$\DPmod{(a, b)}{[a, b]}$ +considered includes a finite number of exceptional values of~$x$ +for which $\phi'(x)$~does not exist, or is not positive, provided $\phi(x)$~is +continuous even for these exceptional values of~$x$. +\end{Cor} + +It is plainly sufficient to consider the case in which there is +one exceptional value of~$x$ only, and that corresponding to an end +of the interval, say to~$a$. If $a < x_{1} < x_{2} < b$, we can choose~$a + \EPSILON$ +so that $a + \EPSILON < x_{1}$, and $\phi'(x) > 0$ throughout $\DPmod{(a + \EPSILON, b)}{[a + \EPSILON, b]}$, so that +$\phi(x_{1}) < \phi(x_{2})$, by Cor.~2. All that remains is to prove that +\PageSep{219} +$\phi(a) < \phi(x_{1})$. Now $\phi(x_{1})$~decreases steadily, and in the stricter +sense, as $x_{1}$~decreases towards~$a$, and so +\[ +\phi(a) = \phi(a + 0) = \lim_{x_{1}\to a+0} \phi(x_{1}) < \phi(x_{1}). +\] + +\begin{Cor}[4.] +If $\phi'(x) > 0$ throughout the interval~$\DPmod{(a, b)}{[a, b]}$, and $\phi(a) \geq 0$, +then $\phi(x)$~is positive throughout the interval~$\DPmod{(a, b)}{[a, b]}$. +\end{Cor} + +\begin{Remark} +The reader should compare the second of these corollaries very carefully +with Theorem~A\@. If, as in Theorem~A, we assume only that $\phi'(x)$~is positive +\emph{at a single point $x = x_{0}$}, then we can prove that $\phi(x_{1}) < \phi(x_{2})$ when $x_{1}$~and~$x_{2}$ +are sufficiently near to~$x_{0}$ and $x_{1} < x_{0} < x_{2}$. For $\phi(x_{1}) < \phi(x_{0})$ and $\phi(x_{2}) > \phi(x_{0})$, +by Theorem~A\@. But this does not prove that there is any interval including~$x_{0}$ +throughout which $\phi(x)$~is a steadily increasing function, for the assumption +that $x_{1}$~and~$x_{2}$ lie on opposite sides of~$x_{0}$~is essential to our conclusion. We +shall return to this point, and illustrate it by an actual example, in a moment~(\SecNo[§]{124}). +\end{Remark} + +\Paragraph{122. Maxima and Minima.} We shall say that the value~$\phi(\xi)$ +assumed by~$\phi(x)$ when $x = \xi$ is a \emph{maximum} if $\phi(\xi)$~is greater than +any other value assumed by~$\phi(x)$ in the immediate neighbourhood +of $x = \xi$, \ie\ if we can find an interval $\DPmod{(\xi - \EPSILON, \xi + \EPSILON)}{[\xi - \EPSILON, \xi + \EPSILON]}$ of values of~$x$ +such that $\phi(\xi) > \phi(x)$ when $\xi - \EPSILON < x < \xi$ and when $\xi < x < \xi + \EPSILON$; +and we define a \emph{minimum} in a similar manner. Thus in the figure +the points~$A$ correspond to maxima, the points~$B$ to minima of +%[Illustration: Fig. 39.] +\Figure[\textwidth]{39}{p219} +the function whose graph is there shown. It is to be observed that +the fact that $A_{3}$~corresponds to a maximum and $B_{1}$~to a minimum +is in no way inconsistent with the fact that the value of the +function is greater at~$B_{1}$ than at~$A_{3}$. + +\begin{Theorem}[C\@.] +A \Emph{necessary} condition for a maximum or +minimum value of~$\phi(x)$ at $x = \xi$ is that $\phi'(\xi) = 0$.\footnote + {A function which is continuous but has no derivative may have maxima and + minima. We are of course assuming the existence of the derivative.} +\end{Theorem} +\PageSep{220} + +This follows at once from Theorem~A\@. That the condition is not +\emph{sufficient} is evident from a glance at the point~$C$ in the figure. +Thus if $y = x^{3}$ then $\phi'(x) = 3x^{2}$, which vanishes when $x = 0$. But +$x = 0$ does not give either a maximum or a minimum of~$x^{3}$, as is +obvious from the form of the graph of~$x^{3}$ (\Fig{10}, \PageRef{p.}{45}). + +But \emph{there will certainly be a maximum at $x = \xi$ if $\phi'(\xi) = 0$, +$\phi'(x) > 0$ for all values of~$x$ less than but near to~$\xi$, and $\phi'(x) < 0$ +for all values of~$x$ greater than but near to~$\xi$}: and if the signs +of these two inequalities are reversed there will certainly be a +minimum. For then we can (by Cor.~3 of \SecNo[§]{121}) determine an +interval $\DPmod{(\xi - \EPSILON, \xi)}{[\xi - \EPSILON, \xi]}$ throughout which $\phi(x)$~increases with~$x$, and an +interval~$\DPmod{(\xi, \xi + \EPSILON)}{[\xi, \xi + \EPSILON]}$ throughout which it decreases as $x$~increases: +and obviously this ensures that $\phi(\xi)$~shall be a maximum. + +This result may also be stated thus. If the sign of~$\phi'(x)$ +changes at $x = \xi$ from positive to negative, then $x = \xi$ gives +a maximum of~$\phi(x)$: and if the sign of~$\phi'(x)$ changes in the +opposite sense, then $x = \xi$ gives a minimum. + +\Paragraph{123.} There is another way of stating the conditions for a +maximum or minimum which is often useful. Let us assume +that $\phi(x)$~has a second derivative~$\phi''(x)$: this of course does not +follow from the existence of~$\phi'(x)$, any more than the existence of~$\phi'(x)$ +follows from that of~$\phi(x)$. But in such cases as we are +likely to meet with at present the condition is generally satisfied. + +\begin{Theorem}[D\@.] +If $\phi'(\xi) = 0$ and $\phi''(\xi) \neq 0$, then $\phi(x)$~has a +maximum or minimum at $x = \xi$, a maximum if $\phi''(\xi) < 0$, a +minimum if $\phi''(\xi) > 0$. +\end{Theorem} + +Suppose, \eg, that $\phi''(\xi) < 0$. Then, by Theorem~A, $\phi'(x)$~is +negative when $x$~is less than~$\xi$ but sufficiently near to~$\xi$, and +positive when $x$~is greater than~$\xi$ but sufficiently near to~$\xi$. Thus +$x = \xi$ gives a maximum. + +\begin{Remark} +\Paragraph{124.} In what has preceded (apart from the last paragraph) we have +assumed simply that $\phi(x)$~has a derivative for all values of~$x$ in the interval +under consideration. If this condition is not fulfilled the theorems cease to +be true. Thus Theorem~B fails in the case of the function +\[ +y = 1 - \sqrtp{x^{2}}, +\] +\PageSep{221} +where the square root is to be taken positive. The graph of this function is +shown in \Fig{40}. Here $\phi(-1) = \phi(1) = 0$: but $\phi'(x)$, as is evident from the +figure, is equal to~$1$ if $x$~is negative and to~$-1$ if $x$~is positive, and never +%[Illustration: Fig. 40.] +\Figure[2.75in]{40}{p221} +vanishes. There is no derivative for $x = 0$, and no tangent to the graph +at~$P$. And in this case $x = 0$ obviously gives a maximum of~$\phi(x)$, but +$\phi'(0)$, as it does not exist, cannot be +equal to zero, so that the test for a +maximum fails. + +The bare existence of the derivative~$\phi'(x)$, +however, is all that we have assumed. +And there is one assumption +in particular that we have not made, +and that is that \emph{$\phi'(x)$~itself is a continuous +function}. This raises a rather +subtle but still a very interesting point. +\emph{Can} a function~$\phi(x)$ have a derivative +for all values of~$x$ which is not itself continuous? In other words can a +curve have a tangent at every point, and yet the direction of the tangent +not vary continuously? The reader, if he considers what the question means +and tries to answer it in the light of common sense, will probably incline +to the answer \emph{No}. It is, however, not difficult to see that this answer is +wrong. + +Consider the function~$\phi(x)$ defined, when $x \neq 0$, by the equation +\[ +\phi(x) = x^{2}\sin(1/x); +\] +and suppose that $\phi(0) = 0$. Then $\phi(x)$~is continuous for all values of~$x$. +If $x \neq 0$ then +\[ +\phi'(x) = 2x \sin(1/x) - \cos(1/x); +\] +while +\[ +\phi'(0) = \lim_{h \to 0} \frac{h^{2}\sin(1/h)}{h} = 0. +\] +Thus $\phi'(x)$~exists for all values of~$x$. But $\phi'(x)$~is discontinuous for $x = 0$; +for $2x\sin(1/x)$~tends to~$0$ as $x \to 0$, and $\cos(1/x)$~oscillates between the limits +of indetermination $-1$~and~$1$, so that $\phi'(x)$~oscillates between the same +limits. + +What is practically the same example enables us also to illustrate the +point referred to at the end of \SecNo[§]{121}. Let +\[ +\phi(x) = x^{2}\sin(1/x) + ax, +\] +where $0 < a < 1$, when $x \neq 0$, and $\phi(0) = 0$. Then $\phi'(0) = a > 0$. Thus the +conditions of Theorem~A of \SecNo[§]{121} are satisfied. But if $x \neq 0$ then +\[ +\phi'(x) = 2x\sin(1/x) - \cos(1/x) + a, +\] +which oscillates between the limits of indetermination $a - 1$ and~$a + 1$ as $x \to 0$. +As $a - 1 < 0$, we can find values of~$x$, as near to~$0$ as we like, for which +$\phi'(x) < 0$; and it is therefore impossible to find any interval, including $x = 0$, +throughout which $\phi(x)$~is a steadily increasing function of~$x$. +\PageSep{222} + +It is, however, impossible that $\phi'(x)$~should have what was called in +\Ref{Ch.}{V} (\Ex{xxxvii}.~18) a `simple' discontinuity; \eg\ that $\phi'(x) \to a$ when +$x \to +0$, $\phi'(x) \to b$ when $x \to -0$, and $\phi'(0) = c$, unless $a = b = c$, in which case +$\phi'(x)$~is continuous for $x = 0$. For a proof see \SecNo[§]{125}, \Ex{xlvii}.~3. +\end{Remark} + +\begin{Examples}{XLVI.} +\Item{1.} Verify Theorem~B when $\phi(x) = (x - a)^{m} (x - b)^{n}$ or +$\phi(x) = (x - a)^{m} (x - b)^{n} (x - c)^{p}$, where $m$,~$n$,~$p$ are positive integers and $a < b < c$. + +[The first function vanishes for $x = a$ and $x = b$. And +\[ +\phi'(x) = (x - a)^{m-1} (x - b)^{n-1} \{(m + n)x - mb - na\} +\] +vanishes for $x = (mb + na)/(m + n)$, which lies between $a$~and~$b$. In the +second case we have to verify that the quadratic equation +\[ +(m + n + p)x^{2} - \{m(b + c) + n(c + a) + p(a + b)\}x + mbc + nca + pab = 0 +\] +has roots between $a$~and~$b$ and between $b$~and~$c$.] + +\Item{2.} Show that the polynomials +\[ +2x^{3} + 3x^{2} - 12x + 7,\quad +3x^{4} + 8x^{3} - 6x^{2} - 24x + 19 +\] +are positive when $x > 1$. + +\Item{3.} Show that $x - \sin x$ is an increasing function throughout any interval +of values of~$x$, and that $\tan x - x$ increases as $x$~increases from $-\frac{1}{2}\pi$ to~$\frac{1}{2}\pi$. +For what values of~$a$ is $ax - \sin x$ a steadily increasing or decreasing function +of~$x$? + +\Item{4.} Show that $\tan x - x$ also increases from $x = \frac{1}{2}\pi$ to $x = \frac{3}{2}\pi$, from $x = \frac{3}{2}\pi$ +to $x = \frac{5}{2}\pi$, and so on, and deduce that there is one and only one root of the +equation $\tan x = x$ in each of these intervals (cf.\ \Ex{xvii}.~4). + +\Item{5.} {\Loosen Deduce from Ex.~3 that $\sin x - x < 0$ if $x > 0$, from this that +$\cos x - 1 + \frac{1}{2}x^{2} > 0$, and from this that $\sin x - x + \frac{1}{6} x^{3} > 0$. And, generally, +prove that if} +\begin{align*} +C_{2m} & = \cos x - 1 + \frac{x^{2}}{2!} - \dots - (-1)^{m} \frac{x^{2m}}{\DPchg{2m!}{(2m)!}},\\ +S_{2m+1}& = \sin x - x + \frac{x^{3}}{3!} - \dots - (-1)^{m} \frac{x^{2m+1}}{(2m+1)!}, +\end{align*} +and $x> 0$, then $C_{2m}$~and~$S_{2m+1}$ are positive or negative according as $m$~is odd +or even. + +\Item{6.} If $f(x)$~and~$f''(x)$ are continuous and have the same sign at every +point of an interval~$\DPmod{(a, b)}{[a, b]}$, then this interval can include at most one root of +either of the equations $f(x) = 0$, $f'(x) = 0$. + +\Item{7.} The functions $u$,~$v$ and their derivatives $u'$,~$v'$ are continuous +throughout a certain interval of values of~$x$, and $uv' - u'v$ never vanishes +at any point of the interval. Show that between any two roots of $u = 0$ +lies one of $v = 0$, and conversely. Verify the theorem when $u = \cos x$, $v = \sin x$. + +[If $v$~does not vanish between two roots of $u = 0$, say $\alpha$~and~$\beta$, then the +function~$u/v$ is continuous throughout the interval~$\DPmod{(\alpha, \beta)}{[\alpha, \beta]}$ and vanishes at its +extremities. Hence $(u/v)' = (u'v - uv')/v^{2}$ must vanish between $\alpha$~and~$\beta$, which +contradicts our hypothesis.] +\PageSep{223} + +\Item{8.} Determine the maxima and minima (if any) of $(x - 1)^{2} (x + 2)$, $x^{3} - 3x$, +$2x^{3} - 3x^{2} - 36x + 10$, $4x^{3} - 18x^{2} + 27x - 7$, $3x^{4} - 4x^{3} + 1$, $x^{5} - 15x^{3} + 3$. In each +case sketch the form of the graph of the function. + +[Consider the last function, for example. Here $\phi'(x) = 5x^{2} (x^{2} - 9)$, which +vanishes for $x = -3$, $x = 0$, and $x = 3$. It is easy to see that $x = -3$ gives a +maximum and $x = 3$ a minimum, while $x = 0$ gives neither, as $\phi'(x)$~is negative +on both sides of $x = 0$.] + +\Item{9.} Discuss the maxima and minima of the function $(x - a)^{m} (x - b)^{n}$, where +$m$~and~$n$ are any positive integers, considering the different cases which occur +according as $m$~and~$n$ are odd or even. Sketch the graph of the function. + +\Item{10.} Discuss similarly the function $(x - a) (x - b)^{2} (x - c)^{3}$, distinguishing +the different forms of the graph which correspond to different hypotheses as +to the relative magnitudes of $a$,~$b$,~$c$. + +\Item{11.} Show that $(ax + b)/(cx + d)$ has no maxima or minima, whatever +values $a$,~$b$, $c$,~$d$ may have. Draw a graph of the function. + +\Item{12.} Discuss the maxima and minima of the function +\[ +y = (ax^{2} + 2bx + c)/(Ax^{2} + 2Bx + \DPtypo{c}{C}), +\] +when the denominator has complex roots. + +[We may suppose $a$~and~$A$ positive. The derivative vanishes if +\[ +(ax + b)(Bx + C) - (Ax + B)(bx + c) = 0. +\Tag{(1)} +\] +This equation must have real roots. For if not the derivative would always +have the same sign, and this is impossible, since $y$~is continuous for all values +of~$x$, and $y \to a/A$ as $x \to +\infty$ or $x \to -\infty$. It is easy to verify that the curve +cuts the line $y = a/A$ in one and only one point, and that it lies above this +line for large positive values of~$x$, and below it for large negative values, or +\textit{vice versa}, according as $b/a > B/A$ or $b/a < B/A$. Thus the algebraically +greater root of~\Eq{(1)} gives a maximum if $b/a > B/A$, a minimum in the contrary +case.] + +\Item{13.} The maximum and minimum values themselves are the values of~$\lambda$ +for which $ax^{2} + 2bx + c - \lambda(Ax^{2} + 2Bx + C)$ is a perfect square. [This is the +condition that $y = \lambda$ should touch the curve.] + +\Item{14.} In general the maxima and maxima of $R(x) = P(x)/Q(x)$ are among +the values of~$\lambda$ obtained by expressing the condition that $P(x) - \lambda Q(x) = 0$ +should have a pair of equal roots. + +\Item{15.} If $Ax^{2} + 2Bx + C = 0$ has real roots then it is convenient to proceed as +follows. We have +\[ +y - (a/A) = (\lambda x + \mu)/\{A(Ax^{2} + 2Bx + C)\}, +\] +where $\lambda = bA - aB$, $\mu = cA - aC$. Writing further $\xi$ for $\lambda x + \mu$ and $\eta$ for +$(A/\lambda^{2})(Ay - a)$, we obtain an equation of the form +\[ +\eta = \xi/\{(\xi - p)(\xi - q)\}. +\] +\PageSep{224} + +This transformation from $(x, y)$ to $(\xi, \eta)$ amounts only to a shifting of the +origin, keeping the axes parallel to themselves, a change of scale along each +axis, and (if $\lambda < 0$) a reversal in direction of the axis of abscissae; and so a +minimum of~$y$, considered as a function of~$x$, corresponds to a minimum of~$\eta$ +considered as a function of~$\xi$, and \textit{vice versa}, and similarly for a maximum. + +The derivative of~$\eta$ with respect to~$\xi$ vanishes if +\[ +(\xi - p)(\xi - q) - \xi(\xi - p) - \xi(\xi - q) = 0, +\] +or if $\xi^{2} = pq$. Thus there are two roots of the derivative if $p$~and~$q$ have the +same sign, none if they have opposite signs. In the latter case the form of +% [** TN: Figure labels italicized in the original] +the graph of~$\eta$ is as shown in \Fig{41a}. +%[Illustration: Fig. 41a.] +%[Illustration: Fig. 41b.] +%[Illustration: Fig. 41c.] +\begin{figure}[hbt!] +\centering + \begin{minipage}{0.3\textwidth} + \centering + \Graphic{1.5in}{p224a} + \caption{Fig.~41a.} + \label{fig:41a} + \end{minipage}\hfill + \begin{minipage}{0.3\textwidth} + \centering + \Graphic{1.5in}{p224b} + \caption{Fig.~41b.} + \label{fig:41b} + \end{minipage}\hfill + \begin{minipage}{0.3\textwidth} + \centering + \Graphic{1.5in}{p224c} + \caption{Fig.~41c.} + \label{fig:41c} + \end{minipage} +\end{figure} + +When $p$ and $q$ are positive the general form of the graph is as shown in +Fig~41b, and it is easy to see that $\xi = \sqrtp{pq}$ gives a maximum and $\xi = -\sqrtp{pq}$ +a minimum.\footnote + {The maximum is $-1/(\sqrt{p} - \sqrt{q})^{2}$, the minimum $-1/(\sqrt{p} + \sqrt{q})^{2}$, of which the + latter is the greater.} + +In the particular case in which $p = q$ the +function is +\[ +\eta = \xi/(\xi - p)^{2}, +\] +and its graph is of the form shown in \Fig{41c}. + +The preceding discussion fails if $\lambda = 0$, \ie\ +if $a/A = b/B$. But in this case we have +\begin{align*} +y - (a/A) &= \mu/\{A(Ax^{2} + 2Bx + C)\}\\ + &= \mu/\{A^{2}(x - x_{1})(x - x_{2})\}, +\end{align*} +say, and $dy/dx = 0$ gives the single value $x = \frac{1}{2}(x_{1} + x_{2})$. On drawing a graph +it becomes clear that this value gives a maximum or minimum according as +$\mu$~is positive or negative. The graph shown in \Fig{42} corresponds to the +former case. +%[Illustration: Fig. 42.] +\Figure[1.5in]{42}{p224d} + +{\Loosen[A full discussion of the general function $y = (ax^{2} + 2bx + c)/(Ax^{2} + 2Bx + C)$, +by purely algebraical methods, will be found in Chrystal's \textit{Algebra}, vol~i, +pp.~464--7.]} + +\Item{16.} Show that $(x - \alpha)(x - \beta)/(x - \gamma)$ assumes all real values as $x$~varies, if +$\gamma$~lies between $\alpha$ and~$\beta$, and otherwise assumes all values except those included +in an interval of length $4\sqrtp{|\alpha - \gamma||\beta - \gamma|}$. +\PageSep{225} + +\Item{17.} Show that +\[ +y = \frac{x^{2} + 2x + c}{x^{2} + 4x + 3c} +\] +can assume any real value if $0 < c < 1$, and draw a graph of the function in +this case. \MathTrip{1910.} + +\Item{18.} Determine the function of the form $(ax^{2} + 2bx + c)/(Ax^{2} + 2Bx + C)$ +which has turning values (\ie\ maxima or minima) $2$~and~$3$ when $x = 1$ and +$x = -1$ respectively, and has the value~$2.5$ when $x = 0$. \MathTrip{1908.} + +\Item{19.} The maximum and minimum of $(x + a) (x + b)/(x - a) (x - b)$, where $a$~and~$b$ are positive, are +\[ +-\left(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}}\right)^{2},\quad +-\left(\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}}\right)^{2}. +\] + +\Item{20.} The maximum value of $(x - 1)^{2}/(x + 1)^{3}$ is~$\frac{2}{27}$. + +\Item{21.} Discuss the maxima and minima of +\begin{gather*} +x(x - 1)/(x^{2} + 3x + 3),\quad x^{4}/(x - 1)(x - 3)^{3},\\ +(x - 1)^{2}(3x^{2} - 2x - 37)/(x + 5)^{2}(3x^{2} - 14x - 1). +\end{gather*} +\longpage +\MathTrip{1898.} + +[If the last function be denoted by~$P(x)/Q(x)$, it will be found that +\[ +P'Q - PQ' = 72(x - 7)(x - 3)(x - 1)(x + 1)(x + 2)(x + 5).] +\] + +\Item{22.} Find the maxima and minima of $a\cos x + b\sin x$. Verify the result +by expressing the function in the form~$A\cos(x - a)$. + +\Item{23.} Find the maxima and minima of +\[ +a^{2}\cos^{2} x + b^{2}\sin^{2} x,\quad +A\cos^{2}x + 2H\cos x\sin x + B\sin^{2} x. +\] + +\Item{24.} Show that $\sin(x + a)/\sin(x + b)$ has no maxima or minima. Draw +a graph of the function. + +\Item{25.} Show that the function +\[ +\frac{\sin^{2}x}{\sin(x + a)\sin(x + b)}\quad +(0 < a < b < \pi) +\] +has an infinity of minima equal to~$0$ and of maxima equal to +\[ +-4\sin a\sin b/\sin^{2}(a - b). +\] +\MathTrip{1909.} + +\Item{26.} The least value of $a^{2}\sec^{2}x + b^{2}\cosec^{2}x$ is $(a + b)^{2}$. + +\Item{27.} Show that $\tan 3x \cot 2x$ cannot lie between $\frac{1}{9}$~and~$\frac{3}{2}$. + +\Item{28.} Show that, if the sum of the lengths of the hypothenuse\DPnote{** [sic], variant spelling} and another +side of a right-angled triangle is given, then the area of the triangle is a +maximum when the angle between those sides is~$60°$. \MathTrip{1909.} + +\Item{29.} A line is drawn through a fixed point~$(a, b)$ to meet the axes $OX$,~$OY$ +in $P$~and~$Q$. Show that the minimum values of $PQ$, $OP + OQ$, and $OP·OQ$ +are respectively $(a^{2/3} + b^{2/3})^{3/2}$, $(\sqrt{a} + \sqrt{b})^{2}$, and~$4ab$. +\PageSep{226} + +\Item{30.} A tangent to an ellipse meets the axes in $P$~and~$Q$. Show that the +least value of~$PQ$ is equal to the sum of the semiaxes of the ellipse. + +\Item{31.} Find the lengths and directions of the axes of the conic +\[ +ax^{2} + 2hxy + by^{2} = 1. +\] + +[The length~$r$ of the \DPchg{semidiameter}{semi-diameter} which makes an angle~$\theta$ with the axis +of~$x$ is given by +\[ +1/r^{2} = a\cos^{2} \theta + 2h\cos\theta \sin\theta + b\sin^{2} \theta. +\] +The condition for a maximum or minimum value of~$r$ is $\tan 2\theta = 2h/(a - b)$. +Eliminating~$\theta$ between these two equations we find +\[ +\{a - (1/r^{2})\} \{b - (1/r^{2})\} = h^{2}.] +\] + +\Item{32.} The greatest value of~$x^{m}y^{n}$, where $x$~and~$y$ are positive and +$x + y = k$, is +\[ +m^{m} n^{n} k^{m+n}/(m + n)^{m+n}. +\] + +\Item{33.} {\Loosen The greatest value of $ax + by$, where $x$~and~$y$ are positive and +$x^{2} + xy + y^{2} = 3\kappa^{2}$, is} +\[ +2\kappa \sqrtp{a^{2} - ab + b^{2}}. +\] + +[If $ax + by$ is a maximum then $a + b(dy/dx) = 0$. The relation between $x$~and~$y$ +gives $(2x + y) + (x + 2y)(dy/dx) = 0$. Equate the two values of~$dy/dx$.] + +\Item{34.} If $\theta$ and~$\phi$ are acute angles connected by the relation $a \sec\theta + b \sec\phi = c$, +where $a$,~$b$,~$c$ are positive, then $a\cos\theta + b\cos\phi$ is a minimum when $\theta = \phi$. +\end{Examples} + +\Paragraph{125. The Mean Value Theorem.} We can proceed now to +the proof of another general theorem of extreme importance, a +theorem commonly known as `\emph{The Mean Value Theorem}\Add{'} or `\emph{The +Theorem of the Mean}'. + +\begin{Theorem} +If $\phi(x)$ has a derivative for all values of~$x$ in the +interval~$\DPmod{(a, b)}{[a, b]}$, then there is a +value~$\xi$ of~$x$ between $a$~and~$b$, +such that +\[ +\phi(b) - \phi(a) = (b - a)\phi'(\xi). +\] +\end{Theorem} + +Before we give a strict proof +of this theorem, which is perhaps +the most important theorem in +the Differential Calculus, it will +be well to point out its obvious +geometrical meaning. This is +simply (see \Fig{43}) that if the +curve~$APB$ has a tangent at all points of its length then there +%[Illustration: Fig. 43.] +\Figure[2in]{43}{p226} +\PageSep{227} +must be a point, such as~$P$, where the tangent is parallel to~$AB$. +For $\phi'(\xi)$~is the tangent of the angle which the tangent at~$P$ +makes with~$OX$, and $\{\phi(b) - \phi(a)\}/(b - a)$ the tangent of the angle +which $AB$ makes with~$OX$. + +It is easy to give a strict analytical proof. Consider the +function +\[ +\phi(b) - \phi(x) - \frac{b - x}{b - a}\{\phi(b) - \phi(a)\}, +\] +which vanishes when $x = a$ and $x = b$. It follows from Theorem~B +of \SecNo[§]{121} that there is a value~$\xi$ for which its derivative vanishes. +But this derivative is +\[ +\frac{\phi(b) - \phi(a)}{b - a} - \phi'(x); +\] +which proves the theorem. It should be observed that it has not +been assumed in this proof that $\phi'(x)$~is continuous. + +It is often convenient to express the Mean Value Theorem in +the form +\[ +\phi(b) = \phi(a) + (b - a) \phi'\{a + \theta(b - a)\}, +\] +where $\theta$~is a number lying between $0$ and~$1$. Of course $a + \theta(b - a)$ +is merely another way of writing `some number~$\xi$ between $a$~and~$b$'. +If we put $b = a + h$ we obtain +\[ +\phi(a + h) = \phi(a) + h\phi'(a + \theta h), +\] +which is the form in which the theorem is most often quoted. + +\begin{Examples}{XLVII.} +\Item{1.} Show that +\[ +\phi(b) - \phi(x) - \frac{b - x}{b - a}\{\phi(b) - \phi(a)\} +\] +is the difference between the ordinates of a point on the curve and the +corresponding point on the chord. + +\Item{2.} Verify the theorem when $\phi(x) = x^{2}$ and when $\phi(x) = x^{3}$. + +[In the latter case we have to prove that $(b^{3} - a^{3})/(b - a) = 3\xi^{2}$, where +$a < \xi < b$; \ie\ that if $\frac{1}{3}(b^{2} + ab + a^{2}) = \xi^{2}$ then $\xi$~lies between $a$ and~$b$.] + +\Item{3.} Establish the theorem stated at the end of \SecNo[§]{124} by means of the Mean +Value Theorem. + +{\Loosen[Since $\phi'(0) = c$, we can find a small positive value of~$x$ such that +$\{\phi(x) - \phi(0)\}/x$ is nearly equal to~$c$; and therefore, by the theorem, a small +positive value of~$\xi$ such that $\phi'(\xi)$~is nearly equal to~$c$, which is inconsistent +with $\lim\limits_{x \to +0} \phi'(x) = a$, unless $a = c$. Similarly $b = c$.]} +\PageSep{228} + +\Item{4.} Use the Mean Value Theorem to prove Theorem~(6) of \SecNo[§]{113}, assuming +that the derivatives which occur are continuous. + +[The derivative of~$F\{f(x)\}$ is by definition +\[ +\lim \frac{F\{f(x + h)\} - F\{f(x)\}}{h}. +\] +But, by the Mean Value Theorem, $f(x + h) = f(x) + hf'(\xi)$, where $\xi$~is a number +lying between $x$ and~$x + h$. And +\[ +F\{f(x) + hf'(\xi)\} = F\{f(x)\} + hf'(\xi)\, F'(\xi_{1}), +\] +where $\xi_{1}$~is a number lying between $f(x)$ and~$f(x) + hf'(\xi)$. Hence the derivative +of~$F\{f(x)\}$ is +\[ +\lim f'(\xi)\, F'(\xi_{1}) = f'(x)\, F'\{f(x)\}, +\] +since $\xi \to x$ and $\xi_{1} \to f(x)$ as $h \to 0$.] +\end{Examples} + +\Paragraph{126.} The Mean Value Theorem furnishes us with a proof of a +result which is of great importance in what follows: \begin{Result}if $\phi'(x) = 0$, +throughout a certain interval of values of~$x$, then $\phi(x)$~is constant +throughout that interval. +\end{Result} + +For, if $a$~and~$b$ are any two values of~$x$ in the interval, then +\[ +\phi(b) - \phi(a) = (b - a) \phi'\{a + \theta(b - a)\} = 0. +\] +An immediate corollary is that if $\phi'(x) = \psi'(x)$, throughout a +certain interval, then the functions $\phi(x)$ and~$\psi(x)$ differ throughout +that interval by a constant. + +\Paragraph{127. Integration.} We have in this chapter seen how we can +find the derivative of a given function~$\phi(x)$ in a variety of cases, +including all those of the commonest occurrence. It is natural to +consider the converse question, that of \emph{determining a function +whose derivative is a given function}. + +Suppose that $\psi(x)$~is the given function. Then we wish to +determine a function such that $\phi'(x) = \psi(x)$. A little reflection +shows us that this question may really be analysed into three +parts. + +\Item{(1)} In the first place we want to know whether such a +function as $\phi(x)$ \emph{actually exists}. This question must be carefully +distinguished from the question as to whether (supposing that +there is such a function) we can find any simple formula to +express it. + +\Item{(2)} We want to know whether it is possible that more than +one such function should exist, \ie\ we want to know whether our +\PageSep{229} +problem is one which admits of a \emph{unique} solution or not; and +if not, we want to know whether there is any simple relation +between the different solutions which will enable us to express all +of them in terms of any particular one. + +\Item{(3)} If there is a solution, we want to know \emph{how to find an +actual expression for it}. + +It will throw light on the nature of these three distinct questions +if we compare them with the three corresponding questions +which arise with regard to the differentiation of functions. + +\Item{(1)} A function~$\phi(x)$ may have a derivative for all values of~$x$, +like~$x^{m}$, where $m$~is a positive integer, or~$\sin x$. It may generally, +but not always have one, like $\sqrt[3]{x}$ or~$\tan x$ or~$\sec x$. Or again +it may never have one: for example, the function considered in +\Ex{xxxvii}.~20, which is nowhere continuous, has obviously no +derivative for any value of~$x$. Of course during this chapter we +have confined ourselves to functions which are continuous except for +some special values of~$x$. The example of the function~$\sqrt[3]{x}$, however, +shows that a continuous function may not have a derivative +for some special value of~$x$, in this case $x = 0$. Whether there +are continuous functions which \emph{never} have derivatives, or continuous +curves which never have tangents, is a further question +which is at present beyond us. Common-sense says \emph{No}: but, as +we have already stated in \SecNo[§]{111}, this is one of the cases in which +higher mathematics has proved common-sense to be mistaken. + +But at any rate it is clear enough that the question `has $\phi(x)$ +a derivative~$\phi'(x)$?'\ is one which has to be answered differently +in different circumstances. And we may expect that the converse +question `is there a function~$\phi(x)$ of which $\psi(x)$~is the derivative?'\ +will have different answers too. We have already seen +that there are cases in which the answer is \emph{No}: thus if $\psi(x)$~is +the function which is equal to $a$,~$b$, or~$c$ according as $x$~is less than, +equal to, or greater than~$0$, then the answer is \emph{No} (\Ex{xlvii}.~3), +unless $a = b = c$. + +This is a case in which the given function is discontinuous. +In what follows, however, we shall always suppose $\psi(x)$~continuous. +And then the answer is~\emph{Yes}: \emph{if $\psi(x)$~is continuous then there is +always a function~$\phi(x)$ such that $\phi'(x) = \psi(x)$}. The proof of this +will be given in \Ref{Ch.}{VII}\@. +\PageSep{230} + +\Item{(2)} The second question presents no difficulties. In the case +of differentiation we have a direct definition of the derivative +which makes it clear from the beginning that there cannot +possibly be more than one. In the case of the converse problem +the answer is almost equally simple. It is that if $\phi(x)$~is one +solution of the problem then $\phi(x) + C$ is another, for any value of +the constant~$C$, and that all possible solutions are comprised in +the form $\phi(x) + C$. This follows at once from \SecNo[§]{126}. + +\Item{(3)} The practical problem of actually finding~$\phi'(x)$ is a fairly +simple one in the case of any function defined by some finite combination +of the ordinary functional symbols. The converse problem +is much more difficult. The nature of the difficulties will appear +more clearly later on. + +\begin{Definitions} +If $\psi(x)$ is the derivative of~$\phi(x)$, then we call +$\phi(x)$ an \Emph{integral} or \Emph{integral function} of~$\psi(x)$. The operation +of forming~$\psi(x)$ from~$\phi(x)$ we call \Emph{integration}. +\end{Definitions} + +We shall use the notation +\[ +\phi(x) = \int \psi(x)\, dx. +\] +It is hardly necessary to point out that $\int\dots dx$ like $d/dx$ must, at +present at any rate, be regarded purely as a symbol of operation: +the~$\int$ and the~$dx$ no more mean anything when taken by themselves +than do the~$d$ and~$dx$ of the other operative symbol~$d/dx$. + +\Paragraph{128. The practical problem of integration.} The results +of the earlier part of this chapter enable us to write down at once +the integrals of some of the commonest functions. Thus +\[ +\int x^{m}\, dx = \frac{x^{m+1}}{m + 1},\quad +\int \cos x\, dx = \sin x,\quad +\int \sin x\, dx = -\cos x. +\Tag{(1)} +\] + +These formulae must be understood as meaning that the +function on the right-hand side is \emph{one} integral of that under +the sign of integration. The \emph{most general} integral is of course +obtained by adding to the former a constant~$C$, known as the +\Emph{arbitrary constant} of integration. +\PageSep{231} + +There is however one case of exception to the first formula, that +in which $m = -1$. In this case the formula becomes meaningless, +as is only to be expected, since we have seen already (\Ex{xlii}.~4) +that $1/x$ cannot be the derivative of any polynomial or rational +fraction. + +That there really is a function~$F(x)$ such that $D_{x}F(x) = 1/x$ +will be proved in the next chapter. For the present we shall be +content to assume its existence. This function~$F(x)$ is certainly +not a polynomial or rational function; and it can be proved that +it is not an algebraical function. It can indeed be proved that +$F(x)$~is an essentially new function, independent of any of the +classes of functions which we have considered yet, that is to say +incapable of expression by means of any finite combination of the +functional symbols corresponding to them. The proof of this is +unfortunately too detailed and tedious to be inserted in this book; +but some further discussion of the subject will be found in \Ref{Ch.}{IX}, +where the properties of~$F(x)$ are investigated systematically. + +Suppose first that $x$~is positive. Then we shall write +\[ +\int \frac{dx}{x} = \log x, +\Tag{(2)} +\] +and we shall call the function on the right-hand side of this +equation \Emph{the logarithmic function}: it is defined so far only for +positive values of~$x$. + +Next suppose $x$~negative. Then $-x$~is positive, and so $\log(-x)$ +is defined by what precedes. Also +\[ +\frac{d}{dx} \log(-x) = \frac{-1}{-x} = \frac{1}{x}, +\] +so that, when $x$~is negative, +\[ +\int \frac{dx}{x} = \log(-x). +\Tag{(3)} +\] + +The formulae \Eq{(2)}~and~\Eq{(3)} may be united in the formulae +\[ +\int \frac{dx}{x} = \log(±x) = \log|x|, +\Tag{(4)} +\] +where the ambiguous sign is to be chosen so that $±x$~is positive: +these formulae hold for all real values of~$x$ other than $x = 0$. +\PageSep{232} + +\begin{Remark} +The most fundamental of the properties of~$\log x$ which will be proved in +\Ref{Ch.}{IX} are expressed by the equations +\[ +\log 1 = 0,\quad +\log (1/x) = -\log x,\quad +\log xy = \log x + \log y, +\] +of which the second is an obvious deduction from the first and third. It is +not really necessary, for the purposes of this chapter, to assume the truth of +any of these formulae; but they sometimes enable us to write our formulae +in a more compact form than would otherwise be possible. + +It follows from the last of the formulae that $\log x^{2}$~is equal to~$2\log x$ if $x > 0$ +and to~$2\log(-x)$ if $x < 0$, and in either case to~$2\log |x|$. Either of the +formulae~\Eq{(4)} is therefore equivalent to the formula +\[ +\int \frac{dx}{x} = \tfrac{1}{2}\log x^{2}. +\Tag{(5)} +\] +\end{Remark} + +The five formulae \Eq{(1)}--\Eq{(3)} are the five most fundamental +\emph{standard forms} of the Integral Calculus. To them should be +added two more, viz. +\[ +\int \frac{dx}{1 + x^{2}} = \arctan x,\quad +\int \frac{x}{\sqrtp{1 - x^{2}}} = ±\arcsin x.\footnotemark +\Tag{(6)} +\] +\footnotetext{See \SecNo[§]{119} for the rule for determining the ambiguous sign.}% + +\Paragraph{129. Polynomials.} All the general theorems of \SecNo[§]{113} may of +course also be stated as theorems in integration. Thus we have, +to begin with, the formulae +\begin{gather*} +\int \{f(x) + F(x)\}\, dx = \int f(x) dx + \int F(x)\, dx, +\Tag{(1)}\\ +\int kf(x)\, dx = k\int f(x)\, dx. +\Tag{(2)} +\end{gather*} + +Here it is assumed, of course, that the arbitrary constants are +adjusted properly. Thus the formula~\Eq{(1)} asserts that the sum of +\emph{any} integral of~$f(x)$ and \emph{any} integral of~$F(x)$~is \emph{an} integral of +$f(x) + F(x)$. + +These theorems enable us to write down at once the integral +of any function of the form $\sum A_{\nu} f_{\nu}(x)$, the sum of a finite number +of constant multiples of functions whose integrals are known. In +particular we can write down the integral of any \emph{polynomial}: +thus +\[ +\int (a_{0}x^{n} + a_{1}x^{n-1} + \dots + a_{n})\, dx + = \frac{a_{0}x^{n+1}}{n + 1} + \frac{a_{1}x^{n}}{n} + \dots + a_{n}x. +\] +\PageSep{233} + +\Paragraph{130. Rational Functions.} After integrating polynomials +it is natural to turn our attention next to \emph{rational functions}. +Let us suppose $R(x)$ to be any rational function expressed in the +standard form of \SecNo[§]{117}, viz.\ as the sum of a polynomial~$\Pi(x)$ and +a number of terms of the form~$A/(x - \alpha)^{p}$. + +We can at once write down the integrals of the polynomial +and of all the other terms except those for which $p = 1$, since +\[ +\int \frac{A}{(x - \alpha)^{p}}\, dx + = -\frac{A}{p - 1}\, \frac{1}{(x - \alpha)^{p-1}}, +\] +whether $\alpha$~be real or complex (\SecNo[§]{117}). + +The terms for which $p = 1$ present rather more difficulty. +It follows immediately from Theorem~(6) of \SecNo[§]{113} that +\[ +\int F'\{f(x)\}\, f'(x)\, dx = F\{f(x)\}. +\Tag{(3)} +\] +In particular, if we take $f(x) = ax + b$, where $a$~and~$b$ are real, +and write $\phi(x)$ for~$F(x)$ and $\psi(x)$ for~$F'(x)$, so that $\phi(x)$~is an +integral of~$\psi(x)$, we obtain +\[ +\int \psi(ax + b)\, dx = \frac{1}{a}\phi(ax + b). +\Tag{(4)} +\] + +Thus, for example, +\[ +\int \frac{dx}{ax + b} = \frac{1}{a} \log|ax + b|, +\] +and in particular, if $\alpha$~is real, +\[ +\int \frac{dx}{x - \alpha} = \log|x - \alpha|. +\] +We can therefore write down the integrals of all the terms +in~$R(x)$ for which $p = 1$ and $\alpha$~is real. There remain the terms for +which $p = 1$ and $\alpha$~is complex. + +In order to deal with these we shall introduce a restrictive +hypothesis, viz.\ that all the coefficients in~$R(x)$ are real. Then if +$\alpha = \gamma + \delta i$ is a root of $Q(x) = 0$, of multiplicity~$m$, so is its conjugate +$\bar{\alpha} = \gamma - \delta i$; and if a partial fraction $A_{p}/(x - \alpha)^{p}$ occurs in +the expression of~$R(x)$, so does $\bar{A}_{p}/(x - \bar{\alpha})^{p}$, where $\bar{A}_{p}$~is conjugate +to~$A_{p}$. This follows from the nature of the algebraical processes +by means of which the partial fractions can be found, and which +are explained at length in treatises on Algebra.\footnote + {See, for example, Chrystal's \textit{Algebra}, vol.~i, pp.~151--9.} +\PageSep{234} + +Thus, if a term $(\lambda + \mu i)/(x - \gamma - \delta i)$ occurs in the expression +of~$R(x)$ in partial fractions, so will a term $(\lambda - \mu i)/(x - \gamma + \delta i)$; +and the sum of these two terms is +\[ +\frac{2\{\lambda(x - \gamma) - \mu\delta\}}{(x - \gamma)^{2} + \delta^{2}}. +\] +This fraction is in reality the most general fraction of the form +\[ +\frac{Ax + B}{ax^{2} + 2bx + c}, +\] +where $b^{2} < ac$. The reader will easily verify the equivalence of +the two forms, the formulae which express $\lambda$,~$\mu$, $\gamma$,~$\delta$ in terms of +$A$,~$B$, $a$,~$b$,~$c$ being +\[ +\lambda = A/2a,\quad +\mu = -D/(2a\sqrt{\Delta}),\quad +\gamma = -b/a,\quad +\delta = \sqrt{\Delta}/a, +\] +where $\Delta = ac - b^{2}$, and $D = aB - bA$.\PageLabel{234} + +If in~\Eq{(3)} we suppose $F\{f(x)\}$~to be~$\log |f(x)|$, we obtain +\[ +\int \frac{f'(x)}{f(x)}\, dx = \log |f(x)|; +\Tag{(5)} +\] +and if we further suppose that $f(x) = (x - \lambda)^{2} + \mu^{2}$, we obtain +\[ +\int \frac{2(x - \lambda)}{(x - \lambda)^{2} + \mu^{2}}\, dx + = \log\{(x - \lambda)^{2} + \mu^{2}\}. +\] +And, in virtue of the equations~\Eq{(6)} of \SecNo[§]{128} and \Eq{(4)}~above, we +have +\[ +\int \frac{-2\delta\mu}{(x - \lambda)^{2} + \mu^{2}}\, dx + = -2\delta \arctan \left(\frac{x - \lambda}{\mu}\right). +\] + +These two formulae enable us to integrate the sum of the two +terms which we have been considering in the expression of~$R(x)$; +and we are thus enabled to write down the integral of any real +rational function, if all the factors of its denominator can be determined. +The integral of any such function is composed of \begin{Result}the sum +of a polynomial, a number of rational functions of the type +\[ +-\frac{A}{p - 1}\, \frac{1}{(x - \alpha)^{p-1}}, +\] +a number of logarithmic functions, and a number of inverse tangents. +\end{Result} + +It only remains to add that if $\alpha$~is complex then the rational +function just written always occurs in conjunction with another in +which $A$ and~$\alpha$ are replaced by the complex numbers conjugate to +them, and that the sum of the two functions is a real rational function. +\PageSep{235} + +\begin{Examples}{XLVIII.} +\Item{1.} Prove that +\[ +\int \frac{Ax + B}{ax^{2} + 2bx + c}\, dx + = \frac{A}{2a} \log |X| + \frac{D}{2a \sqrtp{-\Delta}} + \log \left|\frac{ax + b - \sqrtp{-\Delta}}{ax + b + \sqrtp{-\Delta}}\right| +\] +(where $X = ax^{2} + bx + c$) if $\Delta < 0$, and +\[ +\int \frac{Ax + B}{ax^{2} + 2bx + c}\, dx + = \frac{A}{2a} \log |X| + \frac{D}{2a \sqrt{\Delta}} + \arctan \left(\frac{ax + b}{\sqrt{\Delta}}\right) +\] +if $\Delta > 0$, $\Delta$ and~$D$ having the same meanings as on \PageRef{p.}{234}. + +\Item{2.} In the particular case in which $ac = b^{2}$ the integral is +\[ +-\frac{D}{a(ax + b)} + \frac{A}{a} \log |ax + b|. +\] + +\Item{3.} Show that if the roots of $Q(x) = 0$ are all real and distinct, and $P(x)$~is +of lower degree than~$Q(x)$, then +\[ +\int R(x)\, dx = \tsum \frac{P(\alpha)}{Q'(\alpha)} \log |x - \alpha|, +\] +the summation applying to all the roots~$\alpha$ of $Q(x) = 0$. + +[The form of the fraction corresponding to~$\alpha$ may be deduced from the +facts that +\[ +\frac{Q(x)}{x - \alpha} \to Q'(\alpha),\quad +(x - \alpha) R(x) \to \frac{P(\alpha)}{Q'(\alpha)}, +\] +as $x \to \alpha$.] + +\Item{4.} If all the roots of~$Q(x)$ are real and $\alpha$~is a double root, the other roots +being simple roots, and $P(x)$~is of lower degree than~$Q(x)$, then the integral +is $A/(x - \alpha) + A'\log |x - \alpha| + \sum B\log |x - \beta|$, where +\[ +A = -\frac{2P(\alpha)}{Q''(\alpha)},\quad +A' = \frac{2\{3P'(\alpha) Q''(\alpha) - P(a) Q'''(\alpha)\}} + {3\{Q''(\alpha)\}^{2}},\quad +B = \frac{P(\beta)}{Q'(\beta)}, +\] +and the summation applies to all roots~$\beta$ of $Q(x) = 0$ other than~$\alpha$. + +\Item{5.} Calculate +\[ +\int \frac{dx}{\{(x - 1) (x^{2} + 1)\}^{2}}. +\] + +[The expression in partial fractions is +\[ +\frac{1}{4(x - 1)^{2}} + - \frac{1}{2(x - 1)} + - \frac{i}{8(x - i)^{2}} + + \frac{2 - i}{8(x - i)} + + \frac{i}{8(x + i)^{2}} + + \frac{2 + i}{8(x + i)}, +\] +and the integral is +\[ +-\frac{1}{4(x - 1)} - \frac{1}{4(x^{2} + 1)} + - \tfrac{1}{2} \log |x - 1| + + \tfrac{1}{4} \log (x^{2} + 1) + + \tfrac{1}{4} \arctan x.] +\] + +\Item{6.} Integrate +\begin{gather*} +\frac{x}{(x - a)(x - b)(x - c)},\quad +\frac{x}{(x - a)^{2}(x - b)},\quad +\frac{x}{(x - a)^{2} (x - b)^{2}},\quad +\frac{x}{(x - a)^{3}},\\ +% +\frac{x}{(x^{2} + a^{2}) (x^{2} + b^{2})},\quad +\frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b)^{2}},\quad +\frac{x^{2} - a^{2}}{x^{2}(x^{2} + a^{2})},\quad +\frac{x^{2} - a^{2}}{x(x^{2} + a^{2})^{2}}. +\end{gather*} +\PageSep{236} + +\Item{7.} Prove the formulae: +\begin{alignat*}{3} +\int \frac{dx}{1 + x^{4}} + &= \frac{1}{4\sqrt{2}} \biggl\{% + &&\log \biggl(\frac{1 + x\sqrt{2} + x^{2}}{1 - x\sqrt{2} + x^{2}}\biggr) + &&+ 2\arctan \biggl(\frac{x\sqrt{2}}{1 - x^{2}}\biggr)\biggr\},\\ +% +\int \frac{x^{2}\, dx}{1 + x^{4}} + &= \frac{1}{4\sqrt{2}} \biggl\{% + &-&\log \biggl(\frac{1 + x\sqrt{2} + x^{2}}{1 - x\sqrt{2} + x^{2}}\biggr) + &&+ 2\arctan \biggl(\frac{x\sqrt{2}}{1 - x^{2}}\biggr)\biggr\},\\ +% +\int \frac{dx}{1 + x^{2} + x^{4}} + &= \frac{1}{4\sqrt{3}}\biggl\{% + &\sqrt{3}&\log \biggl(\frac{1 + x + x^{2}}{1 - x + x^{2}}\biggr) + &&+ 2\arctan \biggl(\frac{x\sqrt{3}}{1 - x^{2}}\biggr)\biggr\}. +\end{alignat*} +\end{Examples} + +\begin{Remark} +\Paragraph{131. Note on the practical integration of rational functions.} +The analysis of \SecNo[§]{130} gives us a general method by which we can find the +integral of any real rational function~$R(x)$, \emph{provided we can solve the equation +$Q(x) = 0$}. In simple cases (as in Ex.~5 above) the application of the method +is fairly simple. In more complicated cases the labour involved is sometimes +prohibitive, and other devices have to be used. It is not part of the +purpose of this book to go into practical problems of integration in detail. +The reader who desires fuller information may be referred to Goursat's \textit{Cours +d'Analyse}, second~ed., vol.~i, pp.~246~\textit{et~seq.}, Bertrand's \textit{Calcul Intégral}, and +Dr~Bromwich's tract \textit{Elementary Integrals} (Bowes and Bowes,~1911). + +If the equation $Q(x) = 0$ cannot be solved algebraically, then the method of +partial fractions naturally fails and recourse must be had to other methods.\footnote + {See the author's tract ``The integration of functions of a single variable'' + (\textit{Cambridge Tracts in Mathematics}, No.~2,\PageLabel{236} second edition, 1915). This does not + often happen in practice.} +\end{Remark} + +\Paragraph{132. Algebraical Functions.} We naturally pass on next to +the question of the integration of \emph{algebraical} functions. We have +to consider the problem of integrating~$y$, where $y$~is an algebraical +function of~$x$. It is however convenient to consider an apparently +more general integral, viz. +\[ +\int R(x, y)\, dx, +\] +where $R(x, y)$~is any rational function of $x$~and~$y$. The greater +generality of this form is only apparent, since (\Ex{xiv}.~6) the +function~$R(x, y)$ is itself an algebraical function of~$x$. The choice +of this form is in fact dictated simply by motives of convenience: +such a function as +\[ +\frac{px + q + \sqrtp{ax^{2} + 2bx + c}} + {px + q - \sqrtp{ax^{2} + 2bx + c}} +\] +is far more conveniently regarded as a rational function of $x$ and +the simple algebraical function $\sqrtp{ax^{2} + 2bx + c}$, than directly as +itself an algebraical function of~$x$. +\PageSep{237} + +\Paragraph{133. Integration by substitution and rationalisation.} +It follows from equation~\Eq{(3)} of \SecNo[§]{130} that if $\ds\int \psi(x)\, dx = \phi(x)$ then +\[ +\int \psi\{f(t)\}\, f'(t)\, dt = \phi\{f(t)\}. +\Tag{(1)} +\] + +This equation supplies us with a method for determining the +integral of~$\psi(x)$ in a large number of cases in which the form of +the integral is not directly obvious. It may be stated as a rule as +follows: \emph{put $x = f(t)$, where $f(t)$~is any function of a new variable~$t$ +which it may be convenient to choose; multiply by~$f'(t)$, and +determine \(if possible\) the integral of $\psi\{f(t)\}\, f'(t)$; express the +result in terms of~$x$}. It will often be found that the function of~$t$ +to which we are led by the application of this rule is one whose +integral can easily be calculated. This is always so, for example, +if it is a rational function, and it is very often possible to choose +the relation between $x$ and~$t$ so that this shall be the case. Thus +the integral of~$R(\sqrt{x})$, where $R$~denotes a rational function, is +reduced by the substitution $x = t^{2}$ to the integral of~$2tR(t^{2})$, +\ie\ to the integral of a rational function of~$t$. This method of +integration is called \Emph{integration by rationalisation}, and is of +extremely wide application. + +Its application to the problem immediately under consideration +is obvious. \begin{Result}If we can find a variable~$t$ such that $x$~and~$y$ are both +rational functions of~$t$, say $x = R_{1}(t)$, $y = R_{2}(t)$, then +\[ +\int R(x, y)\, dx = \int R\{R_{1}(t), R_{2}(t)\}\, R_{1}'(t)\, dt, +\] +and the latter integral, being that of a rational function of~$t$, can be +calculated by the methods of~\SecNo[§]{130}. +\end{Result} + +It would carry us beyond our present range to enter upon any +general discussion as to when it is and when it is not possible to +find an auxiliary variable~$t$ connected with $x$~and~$y$ in the manner +indicated above. We shall consider only a few simple and interesting +special cases. + +\Paragraph{134. Integrals connected with conics.} Let us suppose +that $x$~and~$y$ are connected by an equation of the form +\[ +ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0; +\] +in other words that the graph of~$y$, considered as a function of~$x$ +\PageSep{238} +is a conic. Suppose that $(\xi, \eta)$ is any point on the conic, and +let $x - \xi = X$, $y - \eta = Y$. If the relation between $x$~and~$y$ is +expressed in terms of $X$~and~$Y$, it assumes the form +\[ +aX^{2} + 2hXY + bY^{2} + 2GX + 2FY = 0, +\] +where $F = h\xi + b\eta + f$, $G = a\xi + h\eta + g$. In this equation put +$Y = tX$. It will then be found that $X$~and~$Y$ can both be +expressed as rational functions of~$t$, and therefore $x$~and~$y$ can +be so expressed, the actual formulae being +\[ +x - \xi = -\frac{2 (G + Ft)}{a + 2ht + bt^{2}},\quad +y - \eta = -\frac{2t(G + Ft)}{a + 2ht + bt^{2}}. +\] +Hence the process of rationalisation described in the last section +can be carried out. + +The reader should verify that +\[ +hx + by + f = -\tfrac{1}{2}(a + 2ht + bt^{2}) \frac{dx}{dt}, +\] +so that +\[ +\int \frac{dx}{hx + by + f}= -2\int \frac{dt}{a + 2ht + bt^{2}}. +\] + +When $h^{2} > ab$ it is in some ways advantageous to proceed as +follows. The conic is a hyperbola whose asymptotes are parallel +to the lines +\[ +ax^{2} + 2hxy + by^{2} = 0, +\] +or +\[ +b(y - \mu x) (y - \mu' x) = 0, +\] +say\Add{.} If we put $y - \mu x = t$, we obtain +\[ +y - \mu x = t,\quad +y - \mu' x = -\frac{2gx + 2fy + c}{bt}, +\] +and it is clear that $x$~and~$y$ can be calculated from these equations +as rational functions of~$t$. We shall illustrate this process by an +application to an important special case. + +\begin{Remark} +\Paragraph{135. The integral $\ds\int \frac{dx}{\sqrtp{ax^{2} + 2bx + c}}$.} {\Loosen Suppose in particular that +$y^{2} = ax^{2} + 2bx + c$, where $a > 0$. It will be found that, if we put $y + x\sqrt{a} = t$, +we obtain} +\[ +2\frac{dx}{dt} = \frac{(t^{2} + c)\sqrt{a} + 2bt}{(t\sqrt{a} + b)^{2}},\quad +2y = \frac{(t^{2} + c)\sqrt{a} + 2bt}{t\sqrt{a} + b}, +\] +and so +\[ +\int \frac{dx}{y} + = \int \frac{dt}{t\sqrt{a} + b} + = \frac{1}{\sqrt{a}} \log \left|x\sqrt{a} + y + \frac{b}{\sqrt{a}}\right|. +\Tag{(1)} +\] +\PageSep{239} +If in particular $a = 1$, $b = 0$, $c = a^{2}$, or $a = 1$, $b = 0$, $c = -a^{2}$, we obtain +\[ +\int \frac{dx}{\sqrtp{x^{2} + a^{2}}} = \log \{x + \sqrtp{x^{2} + a^{2}}\},\quad +\int \frac{dx}{\sqrtp{x^{2} - a^{2}}} = \log |x + \sqrtp{x^{2} - a^{2}}|, +\Tag{(2)} +\] +equations whose truth may be verified immediately by differentiation. With +these formulae should be associated the third formula +\[ +\int \frac{dx}{\sqrtp{a^{2} - x^{2}}} = \arcsin(x/a), +\Tag{(3)} +\] +which corresponds to a case of the general integral of this section in which +$a < 0$. In~\Eq{(3)} it is supposed that $a > 0$; if $a < 0$ then the integral is $\arcsin(x/|a|)$ +(cf.\ \SecNo[§]{119}). In practice we should evaluate the general integral by reducing it +(as in the next section) to one or other of these standard forms. + +The formula~\Eq{(3)} appears very different from the formulae~\Eq{(2)}: the reader +will hardly be in a position to appreciate the connection between them until +he has read \Ref{Ch.}{X}\@. +\end{Remark} + +\Paragraph{136. The integral $\ds\int \frac{\lambda x + \mu}{\sqrtp{ax^{2} + 2bx + c}}\, dx$.} This integral can +be integrated in all cases by means of the results of the preceding +sections. It is most convenient to proceed as follows. Since +\begin{gather*} +\lambda x + \mu = (\lambda/a) (ax + b) + \mu - (\lambda b/a),\\ +\int \frac{ax + b}{\sqrtp{ax^{2} + 2bx + c}}\, dx = \sqrtp{ax^{2} + 2bx + c}, +\end{gather*} +we have +\[ +\int \frac{(\lambda x + \mu)\, dx}{\sqrtp{ax^{2} + 2bx + c}} + = \frac{\lambda}{a} \sqrtp{ax^{2} + 2bx + c} + + \left(\mu - \frac{\lambda b}{a}\right) + \int \frac{dx}{\sqrtp{ax^{2} + 2bx + c}}. +\] + +In the last integral $a$~may be positive or negative. If $a$~is +positive we put $x\sqrt{a} + (b/\sqrt{a}) = t$, when we obtain +\[ +\frac{1}{\sqrt{a}} \int \frac{dt}{\sqrtp{t^{2} + \kappa}}, +\] +{\Loosen where $\kappa = (ac - b^{2})/a$. If $a$~is negative we write~$A$ for~$-a$ and +put $x\sqrt{A} - (b/\sqrt{A}) = t$, when we obtain} +\[ +\frac{1}{\sqrtp{-a}}\int \frac{dt}{\sqrtp{-\kappa - t^{2}}}. +\] + +It thus appears that in any case the calculation of the integral +may be made to depend on that of the integral considered in +\SecNo[§]{135}, and that this integral may be reduced to one or other +of the three forms +\[ +\int \frac{dt}{\sqrtp{t^{2} + a^{2}}},\quad +\int \frac{dt}{\sqrtp{t^{2} - a^{2}}},\quad +\int \frac{dt}{\sqrtp{a^{2} - t^{2}}}. +\] +\PageSep{240} + +\begin{Remark} +\Paragraph{137. The integral $\ds\int (\lambda x + \mu) \sqrtp{ax^{2} + 2bx + c}\, dx$.} In exactly the same +way we find +{\setlength{\multlinegap}{0pt}% +\begin{multline*}%[** TN: Re-broken] +\int(\lambda x + \mu) \sqrtp{ax^{2} + 2bx + c}\, dx \\ + = \left(\frac{\lambda}{3a}\right) (ax^{2} + 2bx + c)^{3/2} + + \left(\Add{\mu} - \frac{\lambda b}{a}\right) \int \sqrtp{ax^{2} + 2bx + c}\, dx; +\end{multline*}}% +and the last integral may be reduced to one or other of the three forms +\[ +\int \sqrtp{t^{2} + a^{2}}\, dt,\quad +\int \sqrtp{t^{2} - a^{2}}\, dt,\quad +\int \sqrtp{a^{2} - t^{2}}\, dt. +\] +In order to obtain these integrals it is convenient to introduce at this point +another general theorem in integration. +\end{Remark} + +\Paragraph{138. Integration by parts.} The theorem of \emph{integration by +parts} is merely another way of stating the rule for the differentiation +of a product proved in \SecNo[§]{113}. It follows at once from +Theorem~(3) of \SecNo[§]{113} that +\[ +\int f'(x)F(x)\, dx = f(x)F(x) - \int f(x)F'(x)\, dx. +\] +It may happen that the function which we wish to integrate is +expressible in the form~$f'(x)F(x)$, and that $f(x)F'(x)$ can be +integrated. Suppose, for example, that $\phi(x) = x\psi(x)$, where $\psi(x)$~is +the second derivative of a known function~$\chi(x)$. Then +\[ +\int\phi(x)\, dx + = \int x\chi''(x)\, dx + = x\chi'(x) - \int \chi'(x)\, dx + = x\chi'(x) - \chi(x). +\] + +\begin{Remark} +We can illustrate the working of this method of integration by applying +it to the integrals of the last section. Taking +\[ +f(x) = ax + b,\quad +F(x) = \sqrtp{ax^{2} + 2bx + c} = y, +\] +we obtain +\begin{align*} +%[** TN: Set on one line in the original] +a\int y\, dx + &= (ax + b)y - \int \frac{(ax + b)^{2}}{y}\, dx \\ + &= (ax + b)y - a\int y\, dx + (ac - b^{2}) \int \frac{dx}{y}, +\end{align*} +so that +\[ +\int y\, dx = \frac{(ax + b)y}{2a} + \frac{ac - b^{2}}{2a} \int \frac{dx}{y}; +\] +and we have seen already (\SecNo[§]{135}) how to determine the last integral. +\end{Remark} + +\begin{Examples}{XLIX.} +\Item{1\Add{.}} Prove that if $a > 0$ then +\begin{align*} +\int \sqrtp{x^{2} + a^{2}}\, dx + &= \tfrac{1}{2}x \sqrtp{x^{2} + a^{2}} + + \tfrac{1}{2}a^{2} \log \{x + \sqrtp{x^{2} + a^{2}}\},\\ +\int \sqrtp{x^{2} - a^{2}}\, dx + &= \tfrac{1}{2}x \sqrtp{x^{2} - a^{2}} + - \tfrac{1}{2}a^{2} \log |x + \sqrtp{x^{2} - a^{2}}|,\\ +\int \sqrtp{a^{2} - x^{2}}\, dx + &= \tfrac{1}{2}x \sqrtp{a^{2} - x^{2}} + + \tfrac{1}{2}a^{2} \arcsin(x/a). +\end{align*} +\PageSep{241} + +\Item{2.} Calculate the integrals $\ds\int \frac{dx}{\sqrtp{a^{2} - x^{2}}}$, $\ds\int \sqrtp{a^{2} - x^{2}}\, dx$ by means of the +substitution $x = a\sin\theta$, and verify that the results agree with those obtained +in \SecNo[§]{135} and Ex.~1. + +\Item{3.} Calculate $\ds\int x(x + a)^{m}\, dx$, where $m$~is any rational number, in three +ways, viz.\ (i)~by integration by parts, (ii)~by the substitution $(x + a)^{m} = t$, and +(iii)~by writing $(x + a) - a$ for~$x$; and verify that the results agree. + +\Item{4.} Prove, by means of the substitutions $ax + b = 1/t$ and $x = 1/u$, that (in +the notation of \SecNo[§§]{130}~and~\SecNo{138}) +\[ +\int \frac{dx}{y^{3}} = \frac{ax + b}{\Delta y},\quad +\int \frac{x\, dx}{y^{3}} = -\frac{bx + c}{\Delta y}. +\] + +\Item{5.} Calculate $\ds\int \frac{dx}{\sqrtb{(x - a) (b - x)}}$, where $b > a$, in three ways, viz.\ (i)~by +the methods of the preceding sections, (ii)~by the substitution $(b - x)/(x - a) = t^{2}$, +and (iii)~by the substitution $x = a\cos^{2}\theta + b\sin^{2}\theta$; and verify that the results +agree. + +\Item{6.} Integrate $\sqrtb{(x - a) (b - x)}$ and $\sqrtb{(b - x)/(x - a)}$. + +\Item{7.} Show, by means of the substitution $2x + a + b = \frac{1}{2}(a - b) \{t^{2} + (1/t)^{2}\}$, +or by multiplying numerator and denominator by $\sqrtp{x + a} -\sqrtp{x + b}$, that if $a > b$ then +\[ +\int \frac{dx}{\sqrtp{x + a} + \sqrtp{x + b}} + = \tfrac{1}{2}\sqrtp{a - b} \left(t + \frac{1}{3t^{3}}\right). +\] + +\Item{8.} Find a substitution which will reduce $\ds\int \frac{dx}{(x + a)^{3/2} + (x - a)^{3/2}}$ to the +integral of a rational function. \MathTrip{1899.} + +\Item{9.} {\Loosen Show that $\ds\int R\{x, \sqrtp[n]{ax + b}\}\, dx$ is reduced, by the substitution +$ax + b = y^{n}$, to the integral of a rational function.} + +\Item{10.} Prove that +\[ +\int f''(x) F(x)\, dx = f'(x) F(x) - f(x) F'(x) + \int f(x) F''(x)\, dx +\] +and generally +{\setlength{\multlinegap}{\parindent}% +\begin{multline*} +%[** TN: Set on one line in the original] +\int f^{(n)}(x) F(x)\, dx \\ + = f^{(n-1)}(x) F(x) - f^{(n-2)}(x) F'(x) + \dots + + (-1)^{n} \int f(x) F^{(n)}(x)\, dx. +\end{multline*}} + +\Item{11.} The integral $\ds\int (1 + x)^{p} x^{q}\, dx$, where $p$~and~$q$ are rational, can be found +in three cases, viz.\ (i)~if $p$~is an integer, (ii)~if $q$~is an integer, and (iii)~if $p + q$~is an integer. [In case~(i) put $x = u^{s}$, where $s$~is the denominator of~$q$; +in case~(ii) put $1 + x = t^{s}$, where $s$~is the denominator of~$p$; and in case~(iii) put +$1 + x = xt^{s}$, where $s$~is the denominator of~$p$.] +\PageSep{242} + +\Item{12.} The integral $\ds\int x^{m}(ax^{n} + b)^{q}\, dx$ can be reduced to the preceding +integral by the substitution $ax^{n} = bt$. [In practice it is often most convenient +to calculate a particular integral of this kind by a `formula of +reduction' (cf.\ \MiscEx{VI}~39).] + +\Item{13.} The integral $\ds\int R\{x, \sqrtp{ax + b}, \sqrtp{cx + d}\}\, dx$ can be reduced to that of +a rational function by the substitution +\[ +4x = -(b/a) \{t + (1/t)\}^{2} - (d/c)\{t - (1/t)\}^{2}. +\] + +\Item{14.} Reduce $\ds\int R(x, y)\, dx$, where $y^{2}(x - y) = x^{2}$, to the integral of a rational +function. [Putting $y = tx$ we obtain $x = 1/\{t^{2}(1 - t)\}$, $y = 1/\{t(1 - t)\}$.] + +\Item{15.} {\Loosen Reduce the integral in the same way when (\ia)~$y(x - y)^{2} = x$, +(\ib)~$(x^{2} + y^{2})^{2} = a^{2}(x^{2} - y^{2})$. [In case~(\ia) put $x - y = t$: in case~(b) put +$x^{2} + y^{2} = t(x - y)$, when we obtain} +\[ +%[** TN: Set in-line in the original] +x = a^{2}t(t^{2} + a^{2})/(t^{4} + a^{4}),\quad +y = a^{2}t(t^{2} - a^{2})/(t^{4} + a^{4}).] +\] + +\Item{16.} If $y(x - y)^{2} = x$ then +\[ +\int \frac{dx}{x - 3y} = \tfrac{1}{2} \log\{(x - y)^{2} - 1\}. +\] + +\Item{17.} If $(x^{2} + y^{2})^{2} = 2c^{2}(x^{2} - y^{2})$ then +\[ +\int \frac{dx}{y(x^{2} + y^{2} + c^{2})} + = - \frac{1}{c^{2}}\log\left(\frac{x^{2} + y^{2}}{x - y}\right). +\] +\end{Examples} + +\begin{Remark} +\Paragraph{139. The general integral $\ds\int R(x, y)\, dx$, where $y^{2} = ax^{2} + 2bx + c$.} +The most general integral, of the type considered in \SecNo[§]{134}, and associated with +the special conic $y^{2} = ax^{2} + 2bx + c$, is +\[ +\int R(x, \sqrt{X})\, dx, +\Tag{(1)} +\] +where $X = y^{2} = ax^{2} + 2bx + c$. We suppose that $R$~is a \emph{real} function. + +The subject of integration is of the form~$P/Q$, where $P$~and~$Q$ are polynomials +in $x$~and~$\sqrt{X}$. It may therefore be reduced to the form +\[ +\frac{A + B\sqrt{X}}{C + D\sqrt{X}} + = \frac{(A + B\sqrt{X})(C - D\sqrt{X})}{C^{2} - D^{2}X} + = E + F\sqrt{X}, +\] +where $A$, $B$,~\dots\ are rational functions of~$x$. The only new problem which +arises is that of the integration of a function of the form~$F\sqrt{X}$, or, what is +the same thing,~$G/\sqrt{X}$, where $G$~is a rational function of~$x$. And the integral +\[ +\int \frac{G}{\sqrt{X}}\, dx +\Tag{(2)} +\] +can always be evaluated by splitting up $G$ into partial fractions. When we +do this, integrals of three different types may arise. + +\Itemp{(i)} In the first place there may be integrals of the type +\[ +\int \frac{x^{m}}{\sqrt{X}}\, dx, +\Tag{(3)} +\] +\PageSep{243} +where $m$~is a positive integer. The cases in which $m = 0$ or $m = 1$ have been +disposed of in \SecNo[§]{136}. In order to calculate the integrals corresponding to +larger values of~$m$ we observe that +\[ +\frac{d}{dx}(x^{m-1}\sqrt{X}) + = (m - 1)x^{m-2} \sqrt{X} + \frac{(ax + b) x^{m-1}}{\sqrt{X}} + = \frac{\alpha x^{m} + \beta x^{m-1} + \gamma x^{m-2}}{\sqrt{X}}, +\] +where $\alpha$,~$\beta$,~$\gamma$ are constants whose values may be easily calculated. It is clear +that, when we integrate this equation, we obtain a relation between three +successive integrals of the type~\Eq{(3)}. As we know the values of the integral +for $m = 0$ and $m = 1$, we can calculate in turn its values for all other values of~$m$. + +\Itemp{(ii)} In the second place there may be integrals of the type +\[ +\int \frac{dx}{(x - p)^{m}\sqrt{X}}, +\Tag{(4)} +\] +where $p$~is real. If we make the substitution $x - p = 1/t$ then this integral is +reduced to an integral in~$t$ of the type~\Eq{(3)}. + +\Itemp{(iii)} Finally, there may be integrals corresponding to complex roots of the +denominator of~$G$. We shall confine ourselves to the simplest case, that in +which all such roots are simple roots. In this case (cf.~\SecNo[§]{130}) a pair of conjugate +complex roots of~$G$ gives rise to an integral of the type +\[ +\int \frac{Lx + M}{(Ax^{2} + 2Bx + C) \sqrt{ax^{2} + 2bx + c}}\, dx. +\Tag{(5)} +\] + +In order to evaluate this integral we put +\[ +x = \frac{\mu t + \nu}{t + 1}, +\] +where $\mu$~and~$\nu$ are so chosen that +\[ +a\mu\nu + b(\mu + \nu) + c = 0,\quad +A\mu\nu + B(\mu + \nu) + C = 0; +\] +so that $\mu$~and~$\nu$ are the roots of the equation +\[ +(aB - bA)\xi^{2} - (cA - aC)\xi + (bC - cB) = 0. +\] +This equation has certainly real roots, for it is the same equation as +equation~\Eq{(1)} of \Ex{xlvi}.~12; and it is therefore certainly possible to find +real values of $\mu$~and~$\nu$ fulfilling our requirements. + +It will be found, on carrying out the substitution, that the integral~\Eq{(5)} +assumes the form +\[ +H\int \frac{t\, dt}{(\alpha t^{2} + \beta)\sqrtp{\gamma t^{2} + \delta}} + + K\int \frac{dt}{(\alpha t^{2} + \beta)\sqrtp{\gamma t^{2} + \delta}}. +\Tag{(6)} +\] +The second of these integrals is rationalised by the substitution +\[ +\frac{t}{\sqrtp{\gamma t^{2} + \delta}} = u, +\] +which gives +\[ +\int \frac{dt}{(\alpha t^{2} + \beta) \sqrtp{\gamma t^{2} + \delta}} + = \int \frac{du}{\beta + (\alpha\delta - \beta\gamma) u^{2}}. +\] +\PageSep{244} +Finally, if we put $t = 1/u$ in the first of the integrals~\Eq{(6)}, it is transformed into +an integral of the second type, and may therefore be calculated in the manner +just explained, viz.\ by putting $u/\sqrtp{\gamma + \delta u^{2}} = u$, \ie\ $1/\sqrtp{\gamma t^{2} + \delta} = v$.\footnote + {The method of integration explained here fails if $a/A = b/B$; but then the + integral may be reduced by the substitution $ax + b = t$. For further information + concerning the integration of algebraical functions see Stolz, \textit{Grundzüge der + Differential-und-integralrechnung}, vol.~i, pp.~331~\textit{et~seq.}; Bromwich, \textit{Elementary + Integrals} (Bowes and Bowes, 1911). An alternative method of reduction has been + given by Sir~G. Greenhill: see his \textit{A Chapter in the Integral Calculus}, pp.~12 \textit{et~seq.}, + and the author's tract quoted on \PageRef{p.}{236}.} +\end{Remark} + +\begin{Examples}{L.} +\Item{1.} Evaluate +\[ +\int \frac{dx}{x \sqrtp{x^{2} + 2x + 3}},\quad +\int \frac{dx}{(x - 1) \sqrtp{x^{2} + 1}},\quad +\int \frac{dx}{(x + 1) \sqrtp{1 + 2x - x^{2}}}. +\] + +\Item{2.} Prove that +\[ +\int \frac{dx}{(x - p) \sqrtb{(x - p) (x - q)}} + = \frac{2}{q - p} \bigsqrtp{\frac{x - q}{x - p}}. +\] + +\Item{3.} If $ag^{2} + ch^{2} = -\nu < 0$ then +\[ +\int \frac{dx}{(hx + g) \sqrtp{ax^{2} + c}} + = -\frac{1}{\sqrt{\nu}} \arctan\left[ + \frac{\sqrtb{\nu(ax^{2} + c)}}{ch - agx} + \right]. +\] + +\Item{4.} Show that $\ds\int \frac{dx}{(x - x_{0})y}$, where $y^{2} = ax^{2} + 2bx + c$, may be expressed in one +or other of the forms +\[ +-\frac{1}{y_{0}} \log\left| + \frac{axx_{0} + b(x + x_{0}) + c + yy_{0}}{x - x_{0}} +\right|,\quad +\frac{1}{z_{0}} \arctan \left\{ + \frac{axx_{0} + b(x + x_{0}) + c}{yz_{0}} +\right\}, +\] +according as $ax_{0}^{2} + 2bx_{0} + c$ is positive and equal to~$y_{0}^{2}$ or negative and equal +to~$-z_{0}^{2}$. + +\Item{5.} Show by means of the substitution $y = \sqrtp{ax^{2} + 2bx + c}/(x - p)$ that +\[ +\int \frac{dx}{(x - p) \sqrtp{ax^{2} + 2bx + c}} + = \int \frac{dy}{\sqrtp{\lambda y^{2} - \mu}}, +\] +where $\lambda = ap^{2} + 2bp + c$, $\mu = ac - b^{2}$. [This method of reduction is elegant but +less straightforward than that explained in \SecNo[§]{139}.] + +\Item{6.} Show that the integral +\[ +\int \frac{dx}{x \sqrtp{3x^{2} + 2x + 1}} +\] +is rationalised by the substitution $x = (1 + y^{2})/(3 - y^{2})$. \MathTrip{1911.} + +\Item{7.} Calculate +\[ +\int \frac{(x + 1)\, dx}{(x^{2} + 4) \sqrtp{x^{2} + 9}}. +\] +\PageSep{245} + +\Item{8.} Calculate +\[ +\int \frac{dx}{(5x^{2} + 12x + 8) \sqrtp{5x^{2} + 2x - 7}}. +\] + +{\Loosen[Apply the method of \SecNo[§]{139}. The equation satisfied by $\mu$~and~$\nu$ is +$\xi^{2} + 3\xi + 2 = 0$, so that $\mu = -2$, $\nu = -1$, and the appropriate substitution is +$x = -(2t + 1)/(t + 1)$. This reduces the integral to} +\[ +-\int \frac{dt}{(4t^{2} + 1) \sqrtp{9t^{2} - 4}} + - \int \frac{t\, dt}{(4t^{2} + 1) \sqrtp{9t^{2} - 4}}. +\] +The first of these integrals may be rationalised by putting $t/\sqrtp{9t^{2} - 4} = u$ and +the second by putting $1/\sqrtp{9t^{2} - 4} = v$.] + +\Item{9.} Calculate +\[ +\int \frac{(x + 1)\, dx}{(2x^{2} - 2x + 1) \sqrtp{3x^{2} - 2x + 1}},\quad +\int \frac{(x - 1)\, dx}{(2x^{2} - 6x + 5) \sqrtp{7x^{2} - 22x + 19}}. +\] +\MathTrip{1911.} + +\Item{10.} Show that the integral $\ds\int R(x, y)\, dx$, where $y^{2} = ax^{2} + 2bx + c$, is rationalised +by the substitution $t = (x - p)/(y + q)$, where $(p, q)$~is any point on the +conic $y^{2} = ax^{2} + 2bx + c$. [The integral is of course also rationalised by the +substitution $t = (x - p)/(y - q)$: cf.~\SecNo[§]{134}.] +\end{Examples} + +\Paragraph{140. Transcendental Functions.} Owing to the immense +variety of the different classes of transcendental functions, the +theory of their integration is a good deal less systematic than +that of the integration of rational or algebraical functions. We +shall consider in order a few classes of transcendental functions +whose integrals can always be found. + +\Paragraph{141. Polynomials in cosines and sines of multiples of~$x$.} +We can always integrate any function which is the sum of a +finite number of terms such as +\[ +A\cos^{m} ax \sin^{m'} ax \cos^{n} bx \sin^{n'} bx\dots, +\] +where $m$,~$m'$, $n$,~$n'$,~\dots\ are positive integers and $a$,~$b$,~\dots\ any real +numbers whatever. For such a term can be expressed as the +sum of a finite number of terms of the types +\[ +\alpha\cos\{(pa + qb + \dots)x\},\quad +\beta \sin\{(pa + qb + \dots)x\} +\] +and the integrals of these terms can be written down at once. +\PageSep{246} + +\begin{Examples}{LI.} +\Item{1.} Integrate $\sin^{3} x \cos^{2} 2x$. In this case we use the +formulae +\[ +\sin^{3} x = \tfrac{1}{4}(3\sin x - \sin 3x),\quad +\cos^{2} 2x = \tfrac{1}{2}(1 + \cos 4x). +\] +Multiplying these two expressions and replacing $\sin x\cos 4x$, for example, +by $\frac{1}{2}(\sin 5x - \sin 3x)$, we obtain +\begin{multline*} +\tfrac{1}{16}\int (7\sin x - 5\sin 3x + 3\sin 5x - \sin 7x)\, dx\\ + = - \tfrac{7}{16}\cos x + \tfrac{5}{48}\cos 3x + - \tfrac{3}{80}\cos 5x + \tfrac{1}{112}\cos 7x. +\end{multline*} + +The integral may of course be obtained in different forms by different +methods. For example +\[ +\int \sin^{3}x \cos^{2}2x\, dx + = \int(4\cos^{4}x - 4\cos^{2}x + 1) (1 - \cos^{2} x)\sin x\, dx, +\] +which reduces, on making the substitution $\cos x = t$, to +\[ +\int(4t^{6} - 8t^{4} + 5t^{2} - 1)\, dt + = \tfrac{4}{7}\cos^{7} x - \tfrac{8}{5}\cos^{5}x + + \tfrac{5}{3}\cos^{3}x - \cos x. +\] +It may be verified that this expression and that obtained above differ only by +a constant. + +\Item{2.} Integrate by any method $\cos ax \cos bx$, $\sin ax \sin bx$, $\cos ax \sin bx$, +$\cos^{2}x$, $\sin^{3}x$, $\cos^{4}x$, $\cos x \cos 2x \cos 3x$, $\cos^{3}2x \sin^{2}3x$, $\cos^{5}x \sin^{7}x$. [In cases of +this kind it is sometimes convenient to use a formula of reduction (\MiscEx{VI}~39).] +\end{Examples} + +\Paragraph{142. The integrals $\ds\int x^{n}\cos x\, dx$, $\ds\int x^{n}\sin x\, dx$ and associated +integrals.} The method of integration by parts enables us to +generalise the preceding results. For +\begin{alignat*}{3} +\int x^{n}\cos x\, dx &= & &x^{n}\sin x &&- n\int x^{n-1}\sin x\, dx,\\ +\int x^{n}\sin x\, dx &= &-&x^{n}\cos x &&+ n\int x^{n-1}\cos x\, dx, +\end{alignat*} +and clearly the integrals can be calculated completely by a +repetition of this process whenever $n$~is a positive integer. It +follows that we can always calculate $\ds\int x^{n}\cos ax\, dx$ and $\ds\int x^{n}\sin ax\, dx$ +if $n$~is a positive integer; and so, by a process similar to that of +the preceding paragraph, we can calculate +\[ +\int P(x, \cos ax, \sin ax, \cos bx, \sin bx, \dots)\, dx, +\] +where $P$~is any polynomial. +\PageSep{247} + +\begin{Examples}{LII.} +\Item{1.} Integrate $x\sin x$, $x^{2}\cos x$, $x^{2}\cos^{2}x$, $x^{2}\sin^{2}x \sin^{2} 2x$, +$x\sin^{2}x \cos^{4}x$, $x^{3}\sin^{3}\frac{1}{3}x$. + +\Item{2.} Find polynomials $P$~and~$Q$ such that +\[ +\int\{(3x - 1)\cos x + (1 - 2x)\sin x\}\, dx = P\cos x + Q\sin x. +\] + +\Item{3.} Prove that $\ds\int x^{n}\cos x\, dx = P_{n}\cos x + Q_{n}\sin x$, where +\[ +P_{n} = nx^{n-1} - n(n - 1)(n - 2) x^{n-3} + \dots,\quad +Q_{n} = x^{n} - n(n - 1) x^{n-2} + \dots. +\] +\end{Examples} + +\Paragraph{143. Rational Functions of $\cos x$ and $\sin x$.} The integral +of any rational function of $\cos x$~and $\sin x$ may be calculated by +the substitution $\tan \frac{1}{2}x = t$. For +\[ +\cos x = \frac{1 - t^{2}}{1 + t^{2}},\quad +\sin x = \frac{2t}{1 + t^{2}},\quad +\frac{dx}{dt} = \frac{2}{1 + t^{2}}, +\] +so that the substitution reduces the integral to that of a rational +function of~$t$. + +\begin{Examples}{LIII.} +\Item{1.} Prove that +\[ +\int \sec x\, dx = \log |\sec x + \tan x|,\quad +\int \cosec x\, dx = \log |\tan \tfrac{1}{2}x|. +\] + +[Another form of the first integral is $\log |\tan(\frac{1}{4}\pi + \frac{1}{2}x)|$; a third form is +$\frac{1}{2}\log |(1 + \sin x)/(1 - \sin x)|$.] + +\Item{2.} $\ds\int \tan x\, dx = -\log |\cos x|$, $\ds\int \cot x\, dx = \log |\sin x|$, $\ds\int\sec^{2} x\, dx = \tan x$, +$\ds\int \cosec^{2} x\, dx = -\cot x$, $\ds\int \tan x\sec x\, dx = \sec x$, $\ds\int \cot x \cosec x\, dx = -\cosec x$. + +[These integrals are included in the general form, but there is no need to +use a substitution, as the results follow at once from \SecNo[§]{119} and equation~\Eq{(5)} +of~\SecNo[§]{130}.] + +\Item{3.} Show that the integral of $1/(a + b\cos x)$, where $a + b$~is positive, may +be expressed in one or other of the forms +\[ +\frac{2}{\sqrtp{a^{2} - b^{2}}} + \arctan \left\{t\bigsqrtp{\frac{a - b}{a + b}}\right\},\quad +\frac{1}{\sqrtp{b^{2} - a^{2}}} + \log \left|\frac{\sqrtp{b + a} + t\sqrtp{b - a}} + {\sqrtp{b + a} - t\sqrtp{b - a}}\right|, +\] +where $t = \tan\frac{1}{2}x$, according as $a^{2} > b^{2}$ or $a^{2} < b^{2}$. If $a^{2} = b^{2}$ then the integral +reduces to a constant multiple of that of $\sec^{2}\frac{1}{2}x$ or $\cosec^{2}\frac{1}{2}x$, and its value +may at once be written down. Deduce the forms of the integral when $a + b$ +is negative. + +\Item{4.} Show that if $y$~is defined in terms of~$x$ by means of the equation +\[ +(a + b\cos x)(a - b\cos y) = a^{2} - b^{2}, +\] +where $a$~is positive and $a^{2} > b^{2}$, then as $x$~varies from $0$~to~$\pi$ one value of~$y$ +also varies from $0$~to~$\pi$. Show also that +\[ +\sin x = \frac{\sqrtp{a^{2} - b^{2}} \sin y}{a - b\cos y},\quad +\frac{\sin x}{a + b\cos x}\, \frac{dx}{dy} = \frac{\sin y}{a - b\cos y}; +\] +\PageSep{248} +and deduce that if $0 < x < \pi$ then +\[ +\int \frac{dx}{a + b\cos x} + = \frac{1}{\sqrtp{a^{2} - b^{2}}} + \arccos \left(\frac{a\cos x + b}{a + b\cos x}\right). +\] + +Show that this result agrees with that of Ex.~3. + +\Item{5.} Show how to integrate $1/(a + b\cos x + c\sin x)$. [Express $b\cos x + c\sin x$ +in the form $\sqrtp{b^{2} + c^{2}} \cos(x - \alpha)$.] + +\Item{6.} Integrate $(a + b\cos x + c\sin x)/(\alpha + \beta\cos x + \gamma\sin x)$\Add{.} + +[Determine $\lambda$,~$\mu$,~$\nu$ so that +\[ +a + b\cos x + c\sin x + = \lambda + \mu(\alpha + \beta\cos x + \gamma\sin x) + + \nu(-\beta\sin x + \gamma\cos x). +\] +Then the integral is +\[ +\mu x + \nu \log |\alpha + \beta\cos x + \gamma\sin x| + + \lambda \int \frac{dx}{\alpha + \beta\cos x + \gamma\sin x}.] +\] + +\Item{7.} Integrate $1/(a\cos^{2} x + 2b\cos x\sin x + c\sin^{2} x)$. [The subject of integration +may be expressed in the form $1/(A + B\cos 2x + C\sin 2x)$, where +$A = \frac{1}{2}(a + c)$, $B = \frac{1}{2}(a - c)$, $C = b$: but the integral may be calculated more +simply by putting $\tan x = t$, when we obtain +\[ +\int \frac{\sec^{2} x\, dx}{a + 2b\tan x + c\tan^{2} x} + = \int \frac{dt}{a + 2bt + ct^{2}}.] +\] +\end{Examples} + +\Paragraph{144. Integrals involving $\arcsin x$, $\arctan x$, and $\log x$.} The +integrals of the inverse sine and tangent and of the logarithm can +easily be calculated by integration by parts. Thus +\begin{align*} +\int \arcsin x\, dx + &= x\arcsin x - \int \frac{x\, dx}{\sqrtp{1 - x^{2}}} + = x\arcsin x + \sqrtp{1 - x^{2}},\\ +% +\int \arctan x\, dx + &= x\arctan x - \int \frac{x\, dx}{1 + x^{2}} + = x\arctan x - \tfrac{1}{2} \log(1 + x^{2}),\\ +% +\int \log x\, dx + &= x\log x - \int dx = x(\log x - 1). +\end{align*} + +It is easy to see that if we can find the integral of $y = f(x)$ +then we can always find that of $x = \phi(y)$, where $\phi$~is the function +inverse to~$f$. For on making the substitution $y = f(x)$ we obtain +\[ +\int \phi(y)\, dy = \int xf'(x)\, dx = xf(x) - \int f(x)\, dx. +\] +The reader should evaluate the integrals of $\arcsin y$ and $\arctan y$ +in this way. + +Integrals of the form +\[ +\int P(x, \arcsin x)\, dx,\quad +\int P(x, \log x)\, dx, +\] +\PageSep{249} +where $P$~is a polynomial, can always be calculated. Take the +first form, for example. We have to calculate a number of integrals +of the type $\ds\int x^{m} (\arcsin x)^{n}\, dx$. Making the substitution $x = \sin y$, +we obtain $\ds\int y^{n}\sin^{m}y \cos y\, dy$, which can be found by the method of +\SecNo[§]{142}. In the case of the second form we have to calculate a number +of integrals of the type $\ds\int x^{m} (\log x)^{n}\, dx$. Integrating by parts we +obtain +\[ +\int x^{m}(\log x)^{n}\, dx + = \frac{x^{m+1} (\log x)^{n}}{m + 1} + - \frac{n}{m + 1} \int x^{m}(\log x)^{n-1}\, dx, +\] +and it is evident that by repeating this process often enough we +shall always arrive finally at the complete value of the integral. + +\Paragraph{145. Areas of plane curves.} One of the most important +applications of the processes of integration which have been +explained in the preceding sections is to the calculation of \emph{areas} +of plane curves. Suppose that $P_{0}PP'$ (\Fig{44}) is the graph of +a continuous curve $y = \phi(x)$ which lies wholly above the axis of~$x$, +$P$~being the point $(x, y)$ and $P'$~the point $(x + h, y + k)$, and $h$~being +either positive or negative (positive in the figure). +%[Illustration: Fig. 44a.] +%[Illustration: Fig. 44.] +\Figures{2.25in}{44}{p249a}{1.5in}{44a}{p249b} + +The reader is of course familiar with the idea of an `area', and +in particular with that of an area such as~$ONPP_{0}$. This idea we +shall at present take for granted. It is indeed one which needs +and has received the most careful mathematical analysis: later on +we shall return to it and explain precisely what is meant by +\PageSep{250} +ascribing an `area' to such a region of space as~$ONPP_{0}$. For the +present we shall simply assume that any such region has associated +with it a definite positive number $(ONPP_{0})$ which we call its +area, and that these areas possess the obvious properties indicated +by common sense, \eg\ that +\[ +(PRP') + (NN'RP) = (NN'P'P),\quad +(N_{1}NPP_{1}) < (ONPP_{0}), +\] +and so on. + +Taking all this for granted it is obvious that the area $ONPP_{0}$ +is a function of~$x$; we denote it by~$\Phi(x)$. Also $\Phi(x)$~is a +\emph{continuous} function. For +\begin{align*} +\Phi(x + h) - \Phi(x) + &= (NN'P'P)\\ + &= (NN'RP) + (PRP') = h\phi(x) + (PRP'). +\end{align*} + +As the figure is drawn, the area~$PRP'$ is less than~$hk$. This is +not however necessarily true in general, because it is not necessarily +the case (see for example \Fig{44a}) that the arc~$PP'$ +should rise or fall steadily from $P$ to~$P'$. But the area~$PRP'$ +is always less than~$|h|\lambda(h)$, where $\lambda(h)$~is the greatest distance of +any point of the arc~$PP'$ from~$PR$. Moreover, since $\phi(x)$~is a +continuous function, $\lambda(h) \to 0$ as $h \to 0$. Thus we have +\[ +\Phi(x + h) - \Phi(x) = h\{\phi(x) + \mu(h)\}, +\] +where $|\mu(h)| < \lambda(h)$ and $\lambda(h) \to 0$ as $h \to 0$. From this it follows +at once that $\Phi(x)$~is continuous. Moreover +\[ +\Phi'(x) + = \lim_{h \to 0} \frac{\Phi(x + h) - \Phi(x)}{h} + = \lim_{h \to 0} \{\phi(x) + \mu(h)\} + = \phi(x). +\] +Thus \emph{the ordinate of the curve is the derivative of the area, and the +area is the integral of the ordinate}. + +We are thus able to formulate a rule for determining the +area~$ONPP_{0}$. \emph{Calculate $\Phi(x)$, the integral of~$\phi(x)$. This involves +an arbitrary constant, which we suppose so chosen that $\Phi(0) = 0$. +Then the area required is~$\Phi(x)$.} + +\begin{Remark} +If it were the area~$N_{1}NPP_{1}$ which was wanted, we should of course determine +the constant so that $\Phi(x_{1}) = 0$, where $x_{1}$~is the abscissa of~$P_{1}$. If the +curve lay below the axis of~$x$, $\Phi(x)$~would be negative, and the area would be +the absolute value of~$\Phi(x)$. +\end{Remark} +\PageSep{251} + +\Paragraph{146. Lengths of plane curves.} The notion of the length +of a curve, other than a straight line, is in reality a more difficult +one even than that of an area. In fact the assumption that $P_{0}P$ +(\Fig{44}) has a definite length, which we may denote by~$S(x)$, +does not suffice for our purposes, as did the corresponding assumption +about areas. We cannot even prove that $S(x)$~is continuous, +\ie\ that $\lim\{S(P') - S(P)\} = 0$. This looks obvious +enough in the larger figure, but less so in such a case as is shown +in the smaller figure. Indeed it is not possible to proceed further, +with any degree of rigour, without a careful analysis of precisely +what is meant by the length of a curve. + +It is however easy to see what the \emph{formula} must be. Let +us suppose that the curve has a tangent whose direction varies +continuously, so that $\phi'(x)$~is continuous. Then the assumption +that the curve has a length leads to the equation +\[ +\{S(x + h) - S(x)\}/h = \{PP'\}/h = (PP'/h) × (\{PP'\}/PP'), +\] +where $\{PP'\}$~is the arc whose chord is~$PP'$. Now +\[ +PP' + \sqrtp{PR^{2} + RP'^{2}} = h\bigsqrtp{1 + \frac{k^{2}}{h^{2}}}, +\] +and +\[ +k = \phi(x + h) - \phi(x) = h\phi'(\xi), +\] +where $\xi$~lies between $x$ and~$x + h$. Hence +\[ +\lim (PP'/h) = \lim \sqrtb{1 + [\phi'(\xi)]^{2}} = \sqrtb{1 + [\phi'(x)]^{2}}. +\] +If also we assume that +\[ +\lim \{PP'\}/PP' = 1, +\] +we obtain the result +\[ +S'(x) = \lim \{S(x + h) - S(x)\}/h = \sqrtb{1 + [\phi'(x)]^{2}} +\] +and so +\[ +S(x) = \int \sqrtb{1 + [\phi'(x)]^{2}}\, dx. +\] + +\begin{Examples}{LIV.} +\Item{1.} Calculate the area of the segment cut off from the +parabola $y = x^{2}/4a$ by the ordinate $x = \xi$, and the length of the arc which +bounds it. + +\Item{2.} Answer the same questions for the curve $ay^{2} = x^{3}$, showing that the +length of the arc is +\[ +\frac{8a}{27} \left\{\left(1 + \frac{9\xi}{4a}\right)^{3/2} - 1\right\}. +\] + +\Item{3.} Calculate the areas and lengths of the circles $x^{2} + y^{2} = a^{2}$, $x^{2} + y^{2} = 2ax$ +by means of the formulae of \SecNo[§§]{145}--\SecNo{146}. +\PageSep{252} + +\Item{4.} Show that the area of the ellipse $(x^{2}/a^{2}) + (y^{2}/b^{2}) = 1$ is~$\pi ab$. + +\Item{5.} Find the area bounded by the curve $y = \sin x$ and the segment of the +axis of~$x$ from $x = 0$ to $x = 2\pi$. [Here $\Phi(x) = -\cos x$, and the difference +between the values of $-\cos x$ for $x = 0$ and $x = 2\pi$ is zero. The explanation of +this is of course that between $x = \pi$ and $x = 2\pi$ the curve lies below the axis +of~$x$, and so the corresponding part of the area is counted negative in applying +the method. The area from $x = 0$ to $x = \pi$ is $-\cos \pi + \cos 0 = 2$; and the +whole area required, when every part is counted positive, is twice this, +\ie\ is~$4$.] + +\Item{6.} Suppose that the coordinates of any point on a curve are expressed +as functions of a parameter~$t$ by equations of the type $x = \phi(t)$, $y = \psi(t)$, +$\phi$~and~$\psi$ being functions of~$t$ with continuous derivatives. Prove that +if $x$~steadily increases as $t$~varies from $t_{0}$ to~$t_{1}$, then the area of the region +bounded by the corresponding portion of the curve, the axis of~$x$, and the two +ordinates corresponding to $t_{0}$ and~$t_{1}$, is, apart from sign, $A(t_{1}) - A(t_{0})$, where +\[ +A(t) = \int \psi(t)\phi'(t)\, dt = \int y \frac{dx}{dt}\, dt. +\] + +\Item{7.} Suppose that $C$~is a closed curve formed of a single loop and not +met by any parallel to either axis in more than two points. And suppose +that the coordinates of any point~$P$ on the curve can be expressed as in Ex.~6 +in terms of~$t$, and that, as $t$~varies from $t_{0}$ to~$t_{1}$, $P$~moves in the same +direction round the curve and returns after a single circuit to its original +position. Show that the area of the loop is equal to the difference of the +initial and final values of any one of the integrals +\[ +-\int y \frac{dx}{dt}\, dt,\quad + \int x \frac{dy}{dt}\, dt,\quad +\tfrac{1}{2} \int \left(x \frac{dy}{dt} - y \frac{dx}{dt}\right) dt, +\] +this difference being of course taken positively. + +\Item{8.} Apply the result of Ex.~7 to determine the areas of the curves +given by +\[ +\Itemp{(i)} +\frac{x}{a} = \frac{1 - t^{2}}{1 + t^{2}},\quad +\frac{y}{a} = \frac{2t}{1 + t^{2}},\qquad +\Itemp{(ii)} +x = a\cos^{3} t,\quad +y = b\sin^{3} t. +\] + +\Item{9.} Find the area of the loop of the curve $x^{3} + y^{3} = 3axy$. [Putting +$y = tx$ we obtain $x = 3at/(1 + t^{3})$, $y = 3at^{2}/(1 + t^{3})$. As $t$~varies from~$0$ towards~$\infty$ +the loop is described once. Also +\[ +\tfrac{1}{2} \int \left(y \frac{dx}{dt} - x \frac{dy}{dt}\right)\, dt + = -\tfrac{1}{2} \int x^{2} \frac{d}{dt}\left(\frac{y}{x}\right)\, dt + = -\tfrac{1}{2} \int \frac{9a^{2}t^{2}}{(1 + t^{3})^{2}}\, dt + = \frac{3a^{2}}{2(1 + t^{3})}, +\] +which tends to~$0$ as $t \to \infty$. Thus the area of the loop is~$\frac{3}{2}a^{2}$.] + +\Item{10.} Find the area of the loop of the curve $x^{5} + y^{5} = 5ax^{2}y^{2}$. + +\Item{11.} Prove that the area of a loop of the curve $x = a\sin 2t$, $y = a\sin t$ is~$\frac{4}{3}a^{2}$. \MathTrip{1908.} +\PageSep{253} + +\Item{12.} The arc of the ellipse given by $x = a\cos t$, $y = b\sin t$, between the +points $t = t_{1}$ and $t = t_{2}$, is $F(t_{2}) - F(t_{1})$, where +\[ +F(t) = a\int \sqrtp{1 - e^{2}\sin^{2} t}\, dt, +\] +$e$~being the eccentricity. [This integral cannot however be evaluated in +terms of such functions as are at present at our disposal.] + +\Item{13.} \Topic{Polar coordinates.} Show that the area bounded by the curve +$r = f(\theta)$, where $f(\theta)$~is a one-valued function of~$\theta$, and the radii $\theta = \theta_{1}$, $\theta = \theta_{2}$, is +$F(\theta_{2}) - F(\theta_{1})$, where $\ds F(\theta) = \tfrac{1}{2} \int r^{2}\, d\theta$. And the length of the corresponding +arc of the curve is $\Phi(\theta_{2}) - \Phi(\theta_{1})$, where +\[ +\Phi(\theta) + = \bigint \bigsqrtb{r^{2} + \biggl(\frac{dr}{d\theta}\biggr)^{2}}\, d\theta. +\] + +Hence determine (i)~the area and perimeter of the circle $r = 2a\sin\theta$; +(ii)~the area between the parabola $r = \frac{1}{2}l\sec^{2} \frac{1}{2}\theta$ and its latus rectum, and the +length of the corresponding arc of the parabola; (iii)~the area of the limaçon +$r = a + b\cos\theta$, distinguishing the cases in which $a > b$, $a = b$, and $a < b$; +and (iv)~the areas of the ellipses $1/r^{2} = a\cos^{2} \theta + 2h\cos\theta\sin\theta + b\sin^{2} \theta$ and +$l/r = 1 + e\cos\theta$. [In the last case we are led to the integral $\ds \int \frac{d\theta}{(1 + e\cos\theta)^{2}}$, +which may be calculated (cf.\ \Ex{liii}.~4) by the help of the substitution +\[ +(1 + e\cos\theta) (1 - e\cos\phi) = 1 - e^{2}.] +\] + +\Item{14.} Trace the curve $2\theta = (a/r) + (r/a)$, and show that the area bounded +by the radius vector $\theta = \beta$, and the two branches which touch at the point +$r = a$, $\theta = 1$, is $\frac{2}{3} a^{2}(\beta^{2} - 1)^{3/2}$. \MathTrip{1900.} + +\Item{15.} A curve is given by an equation $p = f(r)$, $r$~being the radius vector +and $p$~the perpendicular from the origin on to the tangent. Show that the +calculation of the area of the region bounded by an arc of the curve and two +radii vectores depends upon that of the integral $\frac{1}{2} \ds \int \frac{pr\, dr}{\sqrtp{r^{2} - p^{2}}}$. +\end{Examples} + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER VI.} + +\begin{Examples}{} +\Item{1.} A function~$f(x)$ is defined as being equal to $1 + x$ when $x \leq 0$, to~$x$ when +$0 < x < 1$, to $2 - x$ when $1 \leq x \leq 2$, and to $3x - x^{2}$ when $x > 2$. Discuss the +continuity of~$f(x)$ and the existence and continuity of~$f'(x)$ for $x = 0$, $x = 1$, +and $x = 2$. \MathTrip{1908.} + +\Item{2.} Denoting $a$, $ax + b$, $ax^{2} + 2bx + c$,~\dots\ by $u_{0}$,~$u_{1}$, $u_{2}$,~\dots, show that +$u_{0}^{2} u_{3} - 3u_{0} u_{1} u_{2} + 2u_{1}^{3}$ and $u_{0} u_{4} - 4u_{1} u_{3} + 3u_{2}^{2}$ are independent of~$x$. +\PageSep{254} + +\Item{3.} If $a_{0}$, $a_{1}$,~\dots, $a_{2n}$ are constants and $U_{r} = (a_{0}, a_{1}, \dots, a_{r} \btw x, 1)^{r}$, then +\[ +U_{0}U_{2n} - 2nU_{1}U_{2n-1} + + \frac{2n(2n - 1)}{1·2} U_{2}U_{2n-2} - \dots + U_{2n}U_{0} +\] +is independent of~$x$. \MathTrip{1896.} + +[Differentiate and use the relation $U_{r}' = rU_{r-1}$.] + +\Item{4.} The first three derivatives of the function $\arcsin(\mu\sin x) - x$, where +$\mu > 1$, are positive when $0 \leq x \leq \frac{1}{2} \pi$. + +\Item{5.} The constituents of a determinant are functions of~$x$. Show that its +differential coefficient is the sum of the determinants formed by differentiating +the constituents of one row only, leaving the rest unaltered. + +\Item{6.} If $f_{1}$, $f_{2}$, $f_{3}$, $f_{4}$ are polynomials of degree not greater than~$4$, then +\[ +\begin{vmatrix} +f_{1}& f_{2}& f_{3}& f_{4}\\ +f_{1}'& f_{2}'& f_{3}'& f_{4}'\\ +f_{1}''& f_{2}''& f_{3}''& f_{4}''\\ +f_{1}'''& f_{2}'''& f_{3}'''& f_{4}''' +\end{vmatrix} +\] +is also a polynomial of degree not greater than~$4$. [Differentiate five times, +using the result of Ex.~5, and rejecting vanishing determinants.] + +\Item{7.} If $y^{3} + 3yx + 2x^{3} = 0$ then $x^{2}(1 + x^{3})y'' - \frac{3}{2}xy' + y = 0$. \MathTrip{1903.} + +\Item{8.} Verify that the differential equation $y = \phi\{\psi(y_{1})\} + \phi\{x - \psi(y_{1})\}$, +where $y_{1}$~is the derivative of~$y$, and $\psi$~is the function inverse to~$\phi'$, is +satisfied by $y = \phi(c) + \phi(x - c)$ or by $y = 2\phi(\frac{1}{2}x)$. + +\Item{9.} Verify that the differential equation $y = \{x/\psi(y_{1})\} \phi\{\psi(y_{1})\}$, where the +notation is the same as that of Ex.~8, is satisfied by $y = c\phi(x/c)$ or by $y = \beta x$, +where $\beta = \phi(\alpha)/\alpha$ and $\alpha$~is any root of the equation +$\phi(\alpha) - \alpha\phi'(\alpha) = 0$. + +\Item{10.} If $ax + by + c = 0$ then $y_{2} = 0$ (suffixes denoting differentiations with +respect to~$x$). We may express this by saying that \emph{the general differential +equation of all straight lines is $y_{2} = 0$}. Find the general differential equations +of (i)~all circles with their centres on the axis of~$x$, (ii)~all parabolas with +their axes along the axis of~$x$, (iii)~all parabolas with their axes parallel to +the axis of~$y$, (iv)~all circles, (v)~all parabolas, (vi)~all conics. + +[The equations are (i)~$1 + y_{1}^{2} + yy_{2} = 0$, (ii)~$y_{1}^{2} + yy_{2} = 0$, (iii)~$y_{3} = 0$, +(iv)~$(1 + y_{1}^{2}) y_{3} = 3y_{1} y_{2}^{2}$, (v)~$5y_{3}^{2} = 3y_{2} y_{4}$, +(vi)~$9y_{2}^{2} y_{5} - 45y_{2} y_{3} y_{4} + 40y_{3}^{3} = 0$. +In each case we have only to write down the general equation of the curves +in question, and differentiate until we have enough equations to eliminate all +the arbitrary constants.] + +\Item{11.} Show that the general differential equations of all parabolas and of +all conics are respectively +\[ +D_{x}^{2} (y_{2}^{-2/3}) = 0,\quad +D_{x}^{3} (y_{2}^{-2/3}) = 0. +\] +\PageSep{255} + +[The equation of a conic may be put in the form +\[ +y = ax + b ± \sqrtp{px^{2} + 2qx + r}. +\] +From this we deduce +\[ +y_{2} = ±(pr - q^{2})/(px^{2} + 2qx + r)^{3/2}. +\] +If the conic is a parabola then $p = 0$.] + +\Item{12.} Denoting +$\dfrac{dy}{dx}$, +$\dfrac{1}{2!}\, \dfrac{d^{2}y}{dx^{2}}$, +$\dfrac{1}{3!}\, \dfrac{d^{3}y}{dx^{3}}$, +$\dfrac{1}{4!}\, \dfrac{d^{4}y}{dx^{4}}$,~\dots\ +by $t$, $a$, $b$, $c$,~\dots\ and +$\dfrac{dx}{dy}$, +$\dfrac{1}{2!}\, \dfrac{d^{2}x}{dy^{2}}$, +$\dfrac{1}{3!}\, \dfrac{d^{3}x}{dy^{3}}$, +$\dfrac{1}{4!}\, \dfrac{d^{4}x}{dy^{4}}$,~\dots\ +by $\tau$, $\alpha$, $\beta$, $\gamma$,~\dots, show that +\[ +4ac - 5b^{2} = (4\alpha\gamma - 5\beta^{2})/\tau^{8},\quad +bt - a^{2} = - (\beta\tau - \alpha^{2})/\tau^{6}. +\] +Establish similar formulae for the functions $a^{2}d - 3abc - 2b^{3}$, $(1 + t^{2})b - 2a^{2}t$, +$2ct - 5ab$. + +\Item{13.} Prove that, if $y_{k}$~is the $k$th~derivative of $y = \sin(n\arcsin x)$, then +\[ +(1 - x^{2})y_{k+2} - (2k + 1)xy_{k+1} + (n^{2} - k^{2})y_{k} = 0. +\] + +[Prove first when $k = 0$, and differentiate $k$~times by Leibniz' Theorem.] + +\Item{14.} Prove the formula +\[ +vD_{x}^{n}u = D_{x}^{n}(uv) - nD_{x}^{n-1}(uD_{x}v) + + \frac{n(n - 1)}{1·2} D_{x}^{n-2}(uD_{x}^{2}v) - \dots +\] +where $n$~is any positive integer. [Use the method of induction.] + +\Item{15.} A curve is given by +\[ +x = a(2\cos t + \cos 2t),\quad +y = a(2\sin t - \sin 2t). +\] + +Prove (i)~that the equations of the tangent and normal, at the point~$P$ +whose parameter is~$t$, are +\[ +x\sin \tfrac{1}{2} t + y\cos \tfrac{1}{2} t = a\sin \tfrac{3}{2} t,\quad +x\cos \tfrac{1}{2} t - y\sin \tfrac{1}{2} t = 3a\cos \tfrac{3}{2} t; +\] +(ii)~that the tangent at~$P$ meets the curve in the points $Q$,~$R$ whose parameters +are $-\frac{1}{2} t$ and $\pi - \frac{1}{2} t$; (iii)~that $QR = 4a$; (iv)~that the tangents at $Q$ +and~$R$ are at right angles and intersect on the circle $x^{2} + y^{2} = a^{2}$; (v)~that the +normals at $P$,~$Q$, and~$R$ are concurrent and intersect on the circle $x^{2} + y^{2} = 9a^{2}$; +(vi)~that the equation of the curve is +\[ +(x^{2} + y^{2} + 12ax + 9a^{2})^{2} = 4a(2x + 3a)^{3}. +\] + +Sketch the form of the curve. + +\Item{16.} Show that the equations which define the curve of Ex.~15 may +be replaced by $\xi/a = 2u + (1/u^{2})$, $\eta/a = (2/u) + u^{2}$, where $\xi = x + yi$, $\eta = x - yi$, +$u = \Cis t$. Show that the tangent and normal, at the point defined by~$u$, are +\[ +u^{2}\xi - u\eta = a(u^{3} - 1),\quad +u^{2}\xi + u\eta = 3a(u^{3} + 1), +\] +and deduce the properties (ii)--(v) of Ex.~15. + +\Item{17.} Show that the condition that $x^{4} + 4px^{3} - 4qx - 1 = 0$ should have +equal roots may be expressed in the form $(p + q)^{2/3} - (p - q)^{2/3} = 1$. +\MathTrip{1898.} +\PageSep{256} + +\Item{18.} The roots of a cubic $f(x) = 0$ are $\alpha$,~$\beta$,~$\gamma$ in ascending order of magnitude. +Show that if $\DPmod{(\alpha, \beta)}{[\alpha, \beta]}$ and~$\DPmod{(\beta, \gamma)}{[\beta, \gamma]}$ are each divided into six equal sub-intervals, +then a root of $f'(x) = 0$ will fall in the fourth interval from~$\beta$ on each side. +What will be the nature of the cubic in the two cases when a root of $f'(x) = 0$ +falls at a point of division? \MathTrip{1907.} + +\Item{19.} {\Loosen Investigate the maxima and minima of~$f(x)$, and the real roots of +$f(x) = 0$, $f(x)$~being either of the functions} +\[ +x - \sin x - \tan\alpha (1 - \cos x),\quad +x - \sin x - (\alpha - \sin\alpha) - \tan \tfrac{1}{2}\alpha (\cos\alpha - \cos x), +\] +and $\alpha$~an angle between $0$~and~$\pi$. Show that in the first case the condition for +a double root is that $\tan\alpha - \alpha$ should be a multiple of~$\pi$. + +\Item{20.} {\Loosen Show that by choice of the ratio~$\lambda : \mu$ we can make the roots of +$\lambda(ax^{2} + bx + c) + \mu(a'x^{2} + b'x + c') = 0$ real and having a difference of any magnitude, +unless the roots of the two quadratics are all real and interlace; and +that in the excepted case the roots are always real, but there is a lower limit +for the magnitude of their difference. \MathTrip{1895.}} + +[Consider the form of the graph of the function $(ax^{2} + bx + c)/(a'x^{2} + b'x + c')$: +cf.\ \Exs{xlvi}.\ 12~\textit{et~seq.}] + +\Item{21.} Prove that +\[ +\pi < \frac{\sin \pi x}{x(1 - x)} \leq 4 +\] +when $0 < x < 1$, and draw the graph of the function. + +\Item{22.} Draw the graph of the function +\[ +\pi \cot\pi x - \frac{1}{x} - \frac{1}{x - 1}. +\] + +\Item{23.} Sketch the general form of the graph of~$y$, given that +\[ +\frac{dy}{dx} = \frac{(6x^{2} + x - 1) (x - 1)^{2} (x + 1)^{3}}{x^{2}}. +\] +\MathTrip{1908.} + +\Item{24.} A sheet of paper is folded over so that one corner just reaches the +opposite side. Show how the paper must be folded to make the length of the +crease a maximum. + +\Item{25.} The greatest acute angle at which the ellipse $(x^{2}/a^{2}) + (y^{2}/b^{2}) = 1$ can +be cut by a concentric circle is $\arctan\{(a^{2} - b^{2})/2ab\}$. \MathTrip{1900.} + +\Item{26.} In a triangle the area~$\Delta$ and the semi-perimeter~$s$ are fixed. Show that +any maximum or minimum of one of the sides is a root of the equation +$s(x - s) x^{2} + 4\Delta^{2} = 0$. Discuss the reality of the roots of this equation, and +whether they correspond to maxima or minima. + +[The equations $a + b + c = 2s$, $s(s - a)(s - b)(s - c) = \Delta^{2}$ determine $a$~and~$b$ +as functions of~$c$. Differentiate with respect to~$c$, and suppose that $da/dc = 0$. +It will be found that $b = c$, $s - b = s - c = \frac{1}{2} a$, from which we deduce that +$s(a - s)a^{2} + 4\Delta^{2} = 0$. +\PageSep{257} + +This equation has three real roots if $s^{4} > 27\Delta^{2}$, and one in the contrary +case. In an equilateral triangle (the triangle of minimum perimeter for a +given area) $s^{4} = 27\Delta^{2}$; thus it is impossible that $s^{4} < 27\Delta^{2}$. Hence the +equation in~$a$ has three real roots, and, since their sum is positive and their +product negative, two roots are positive and the third negative. Of the two +positive roots one corresponds to a maximum and one to a minimum.] + +\Item{27.} The area of the greatest equilateral triangle which can be drawn +with its sides passing through three given points $A$,~$B$,~$C$ is +\[ +2\Delta + \frac{a^{2} + b^{2} + c^{2}}{2\sqrt{3}}, +\] +$a$,~$b$,~$c$ being the sides and $\Delta$~the area of~$ABC$. \MathTrip{1899.} + +\Item{28.} If $\Delta$,~$\Delta'$ are the areas of the two maximum isosceles triangles which +can be described with their vertices at the origin and their base angles on the +cardioid $r = a(1 + \cos\theta)$, then $256\Delta\Delta' = 25a^{4}\sqrt{5}$. \MathTrip{1907.} + +\Item{29.} Find the limiting values which $(x^{2} - 4y + 8)/(y^{2} - 6x + 3)$ approaches +as the point~$(x, y)$ on the curve $x^{2}y - 4x^{2} - 4xy + y^{2} + 16x - 2y - 7 = 0$ approaches +the position~$(2, 3)$. \MathTrip{1903.} + +{\Loosen[If we take $(2, 3)$ as a new origin, the equation of the curve becomes +$\xi^{2} \eta - \xi^{2} + \eta^{2} = 0$, and the function given becomes $(\xi^{2} + 4\xi - 4\eta)/(\eta^{2} + 6\eta - 6\xi)$. If +we put $\eta = t\xi$, we obtain $\xi = (1 - t^{2})/t$, $\eta = 1 - t^{2}$. The curve has a loop branching +at the origin, which corresponds to the two values $t = -1$ and $t= 1$. Expressing +the given function in terms of~$t$, and making $t$~tend to $-1$~or~$1$, we obtain the +limiting values $-\frac{3}{2}$,~$-\frac{2}{3}$.]} + +\Item{30.} If +%[** TN: Displayed in the original] +$f(x) = \dfrac{1}{\sin x - \sin a} - \dfrac{1}{(x - a)\cos a}$, +then +\[ +\frac{d}{da}\{\lim_{x \to a} f(x)\} - \lim_{x \to a}f'(x) + = \tfrac{3}{4} \sec^{3} a - \tfrac{5}{12} \sec a. +\] +\longpage\MathTrip{1896.} + +\Item{31.} Show that if $\phi(x) = 1/(1 + x^{2})$ then $\phi^{n} (x) = Q_{n}(x)/(1 + x^{2})^{n+1}$, where +$Q_{n}(x)$~is a polynomial of degree~$n$. Show also that + +\Itemp{(i)} $Q_{n+1} = (1 + x^{2}) Q_{n}' - 2(n + 1) x Q_{n}$, + +\Itemp{(ii)} $Q_{n+2} + 2(n + 2) x Q_{n+1} + (n + 2)(n + 1)(1 + x^{2})Q_{n} = 0$, + +\Itemp{(iii)} $(1 + x^{2}) Q_{n}'' - 2nx Q_{n}' + n(n + 1)Q_{n} = 0$, + +\Itemp{(iv)} $Q_{n} = (-1)^{n} n!\left\{(n + 1)x^{n} - \dfrac{(n + 1)n(n - 1)}{3!} x^{n-2} + \dots\right\}$, + +\Itemp{(v)} all the roots of $Q_{n} = 0$ are real and separated by those of $Q_{n-1} = 0$. + +\Item{32.} If $f(x)$, $\phi(x)$, $\psi(x)$ have derivatives when $a \leq x \leq b$, then there is +a value of~$\xi$ lying between $a$~and~$b$ and such that +\[ +\begin{vmatrix} +f(a) & \phi(a) & \psi(a)\\ +f(b) & \phi(b) & \psi(b)\\ +f'(\xi)& \phi'(\xi)& \psi'(\xi) +\end{vmatrix} +=0. +\] +\PageSep{258} + +[Consider the function formed by replacing the constituents of the third +row by $f(x)$,~$\phi(x)$,~$\psi(x)$. This theorem reduces to the Mean Value Theorem +(\SecNo[§]{125}) when $\phi(x) = x$ and $\psi(x) = 1$.] + +\Item{33.} Deduce from Ex.~32 the formula +\[ +\frac{f(b) - f(a)}{\phi(b) - \phi(a)} = \frac{f'(\xi)}{\phi'(\xi)}\Add{.} +\] + +\Item{34.} If $\phi'(x) \to a$ as $x \to \infty$, then $\phi(x)/x \to a$. If $\phi'(x) \to \infty$ then +$\phi(x) \to \infty$. [Use the formula $\phi(x) - \phi(x_{0}) = (x - x_{0})\phi'(\xi)$, +where $x_{0} < \xi < x$.] + +\Item{35.} If $\phi(x) \to a$ as $x \to \infty$, then $\phi'(x)$~cannot tend to any limit other than +zero. + +\Item{36.} If $\phi(x) + \phi'(x) \to a$ as $x \to \infty$, then $\phi(x) \to a$ and $\phi'(x) \to 0$. + +[Let $\phi(x) = a + \psi(x)$, so that $\psi(x) + \psi'(x) \to 0$. If $\psi'(x)$~is of constant +sign, say positive, for all sufficiently large values of~$x$, then $\psi(x)$~steadily +increases and must tend to a limit~$l$ or to~$\infty$. If $\psi(x) \to \infty$ then $\psi'(x) \to -\infty$, +which contradicts our hypothesis. If $\psi(x) \to l$ then $\psi'(x) \to -l$, and this +is impossible (Ex.~35) unless $l = 0$. Similarly we may dispose of the case in +which $\psi'(x)$~is ultimately negative. If $\psi(x)$~changes sign for values of~$x$ which +surpass all limit, then these are the maxima and minima of~$\psi(x)$. If $x$~has +a large value corresponding to a maximum or minimum of~$\psi(x)$, then +$\psi(x) + \psi'(x)$ is small and $\psi'(x) = 0$, so that $\psi(x)$~is small. \textit{A~fortiori} are the +other values of~$\psi(x)$ small when $x$~is large. + +For generalisations of this theorem, and alternative lines of proof, see a +paper by the author entitled ``Generalisations of a limit theorem of Mr~Mercer,'' +in volume~43 of the \textit{Quarterly Journal of Mathematics}. The simple proof +sketched above was suggested by Prof.~E.~W. Hobson.] + +\Item{37.} Show how to reduce +$\ds\int R\left\{x, \bigsqrtp{\frac{ax + b}{mx + n}}, \bigsqrtp{\frac{cx + d}{mx + n}}\right\} dx$ to +the integral of a rational function. [Put $mx + n = 1/t$ and use \Ex{xlix}.~13.] + +\Item{38.} Calculate the integrals: +\begin{gather*} +\int \frac{dx}{(1 + x^{2})^{3}},\quad +\int \bigsqrtp{\frac{x - 1}{x + 1}}\, \frac{dx}{x},\quad +\int \frac{x\, dx}{\sqrtp{1 + x} - \sqrtp[3]{1 + x}},\displaybreak[1]\\ +% +\int \bigsqrtb{a^{2} + \bigsqrtp{b^{2} + \frac{c}{x}}}\, dx,\quad +\int \cosec^{3}x\, dx,\quad +\int \frac{5\cos x + 6}{2\cos x + \sin x + 3}\, dx,\displaybreak[1]\\ +% +\int \frac{dx}{(2 - \sin^{2}x) (2 + \sin x - \sin^{2} x)},\quad +\int \frac{\cos x\sin x \, dx}{\cos^{4}x + \sin^{4}x},\quad +\int \cosec x \sqrtp{\sec 2x}\, dx,\displaybreak[1]\\ +% +%[** TN: Slightly wide, but visually harmless] +\int \frac{dx}{\sqrtb{(1 + \sin x) (2 + \sin x)}},\quad +\int \frac{x + \sin x}{1 + \cos x}\, dx,\quad +\int \arcsec x\, dx,\quad +\int (\arcsin x)^{2}\, dx,\displaybreak[1]\\ +% +\int x\arcsin x\, dx,\quad +\int \frac{x\arcsin x}{\sqrtp{1 - x^{2}}}\, dx,\quad +\int \frac{\arcsin x}{x^{3}}\, dx,\quad +\int \frac{\arcsin x}{(1 + x)^{2}}\, dx,\displaybreak[1]\\ +% +%[** TN: Slightly wide, but visually harmless] +\int \frac{\arctan x}{x^{2}}\, dx,\quad +\int \frac{\arctan x}{(1 + x^{2})^{3/2}}\, dx,\quad +\int \frac{\log(\alpha^{2} + \beta^{2}x^{2})}{x^{2}}\, dx,\quad +\int \frac{\log(\alpha + \beta x)}{(a + bx)^{2}}\, dx. +\end{gather*} +\PageSep{259} + +\Item{39.} \Topic{Formulae of reduction.} \Itemp{(i)} Show that +\begin{multline*} +%[** TN: Re-broken] +2(n - 1)(q - \tfrac{1}{4}p^{2}) \int \frac{dx}{(x^{2} + px + q)^{n}} \\ + = \frac{x + \frac{1}{2}p}{(x^{2} + px + q)^{n-1}} + + (2n - 3) \int \frac{dx}{(x^{2} + px + q)^{n-1}}. +\end{multline*} + +[Put $x + \frac{1}{2}p = t$, $q - \frac{1}{4}p^{2} = \lambda$: then we obtain +\begin{align*} +\int \frac{dt}{(t^{2} + \lambda)^{n}} + &= \frac{1}{\lambda} \int \frac{dt}{(t^{2} + \lambda)^{n-1}} + - \frac{1}{\lambda} \int \frac{t^{2}\, dt}{(t^{2} + \lambda)^{n}} \\ +% + &= \frac{1}{\lambda} \int \frac{dt}{(t^{2} + \lambda)^{n-1}} + + \frac{1}{2\lambda(n-1)} + \int t \frac{d}{dt} \left\{\frac{1}{(t^{2} + \lambda)^{n-1}}\right\} dt, +\end{align*} +and the result follows on integrating by parts. + +A formula such as this is called a \emph{formula of reduction}. It is most useful +when $n$~is a positive integer. We can then express $\ds\int \frac{dx}{(x^{2} + px + q)^{n}}$ in terms +of $\ds\int \frac{dx}{(x^{2} + px + q)^{n-1}}$, and so evaluate the integral for every value of~$n$ in +turn.] + +\Itemp{(ii)} Show that if $I_{p, q} = \ds\int x^{p}(1 + x)^{q}\, dx$ then +\[ +(p + 1) I_{p, q} = x^{p+1}(1 + x)^{q} - qI_{p+1, q-1}, +\] +and obtain a similar formula connecting $I_{p, q}$ with~$I_{p-1, q+1}$. Show also, by +means of the substitution $x = -y/(1 + y)$, that +\[ +I_{p, q} = (-1)^{p+1} \int y^{p} (1 + y)^{-p-q-2}\, dy. +\] + +\Itemp{(iii)} Show that if $X = a + bx$ then +\begin{align*} +\int xX^{-1/3}\, dx &= -3(3a - 2bx) X^{2/3}/10b^{2}, \\ +\int x^{2}X^{-1/3}\, dx &= 3(9a^{2} - 6abx + 5b^{2}x^{2}) X^{2/3}/40b^{3}\DPchg{.}{,}\\ +% +\int xX^{-1/4}\, dx &= -4(4a - 3bx) X^{3/4}/21b^{2},\\ +\int x^{2}X^{-1/4}\, dx &= 4(32a^{2} - 24abx + 21b^{2}x^{2}) X^{3/4}/231b^{3}. +\end{align*} + +\Itemp{(iv)} If $I_{m, n} = \ds\int \frac{x^{m}\, dx}{(1 + x^{2})^{n}}$ then +\[ +2(n - 1)I_{m, n} = -x^{m-1} (1 + x^{2})^{-(n-1)} + (m - 1)I_{m-2, n-1}. +\] + +\Itemp{(v)} If $I_{n} = \ds\int x^{n} \cos\beta x\, dx$ and $J_{n} = \ds\int x^{n} \sin\beta x\, dx$ then +\[ +\beta I_{n} = x^{n} \sin\beta x - nJ_{n-1},\quad +\beta J_{n} = -x^{n} \cos\beta x + nI_{n-1}. +\] +\PageSep{260} + +\Itemp{(vi)} If $I_{n} = \ds\int \cos^{n} x\, dx$ and $J_{n} = \ds\int \sin^{n} x\, dx$ then +\[ +nI_{n} = \sin x\cos^{n-1} x + (n - 1) I_{n-2},\quad +nJ_{n} = -\cos x\sin^{n-1} x + (n - 1) J_{n-2}. +\] + +\Itemp{(vii)} If $I_{n} = \ds\int \tan^{n}x\, dx$ then $(n - 1)(I_{n} + I_{n-2}) = \tan^{n-1}x$. + +\Itemp{(viii)} If $I_{m, n} = \ds\int \cos^{m}x \sin^{n}x\, dx$ then +\begin{alignat*}{2} +(m+n)I_{m, n} + &= -&&\cos^{m+1}x \sin^{n-1}x + (n - 1) I_{m, n-2}\\ + &= &&\cos^{m-1}x \sin^{n+1}x + (m - 1) I_{m-2, n}. +\end{alignat*} + +[We have +\begin{align*} +(m+1)I_{m, n} + &= -\int \sin^{n-1}x \frac{d}{dx} (\cos^{m+1}x)\, dx\\ + &= -\cos^{m+1}x \sin^{n-1}x + (n - 1)\int \cos^{m+2}x \sin^{n-2}x\, dx\\ + &= -\cos^{m+1}x \sin^{n-1}x + (n - 1)(I_{m, n-2} - I_{m, n}), +\end{align*} +which leads to the first reduction formula.] + +\Itemp{(ix)} Connect $I_{m, n} = \ds\int \sin^{m}x \sin nx\, dx$ with~$I_{m-2, n}$. \MathTrip{1897.} + +\Itemp{(x)} If $I_{m, n} = \ds\int x^{m} \cosec^{n}x\, dx$ then +\begin{multline*} +(n - 1)(n - 2)I_{m, n} = (n - 2)^{2}I_{m, n-2} + m(m - 1)I_{m-2, n-2}\\ + -x^{m-1} \cosec^{n-1}x \{m\sin x + (n - 2) x\cos x\}. +\end{multline*} +\MathTrip{1896.} + +\Itemp{(xi)} If $I_{n} = \ds\int (a + b\cos x)^{-n}\, dx$ then +\[ +(n - 1)(a^{2} - b^{2}) I_{n} + = -b\sin x (a + b\cos x)^{-(n-1)} + (2n - 3)aI_{n-1} - (n - 2)I_{n-2}. +\] + +\Itemp{(xii)} If $I_{n} = \ds\int (a\cos^{2} x + 2h\cos x\sin x + b\sin^{2}x)^{-n}\, dx$ then +\[ +4n(n + 1)(ab - h^{2})I_{n+2} - 2n(2n + 1)(a + b)I_{n+1} + 4n^{2}I_{n} + = -\frac{d^{2} I_{n}}{dx^{2}}. +\] +\MathTrip{1898.} + +\Itemp{(xiii)} If $I_{m, n} = \ds\int x^{m}(\log x)^{n}\, dx$ then +\[ +%[** TN: In-line in the original] +(m + 1)I_{m, n} = x^{m+1}(\log x)^{n} - nI_{m, n-1}. +\] + +\Item{40.} If $n$~is a positive integer then the value of $\ds\int x^{m}(\log x)^{n}\, dx$ is +\[ +x^{m+1} \left\{\frac{(\log x)^{n}}{m + 1} + - \frac{n(\log x)^{n-1}}{(m + 1)^{2}} + + \frac{n(n - 1)(\log x)^{n-2}}{(m + 1)^{3}} - \dots + + \frac{(-1)^{n}n!}{(m + 1)^{n+1}}\right\}. +\] + +\Item{41.} Show that the most general function~$\phi(x)$, such that $\phi'' + a^{2}\phi = 0$ for +all values of~$x$, may be expressed in either of the forms $A\cos ax + B\sin ax$, +$\rho\cos(ax + \epsilon)$, where $A$,~$B$, $\rho$,~$\epsilon$ are constants. [Multiplying by~$2\phi'$ and +\PageSep{261} +integrating we obtain $\phi'^{2} + a^{2}\phi^{2} = a^{2}b^{2}$, where $b$~is a constant, from which we +deduce that $ax = \ds\int \frac{d\phi}{\sqrtp{b^{2} - \phi^{2}}}$.] + +\Item{42.} Determine the most general functions $y$~and~$z$ such that $y' + \omega z = 0$, +and $z' - \omega y = 0$, where $\omega$~is a constant and dashes denote differentiation with +respect to~$x$. + +\Item{43.} The area of the curve given by +\[ +x = \cos\phi + \frac{\sin\alpha \sin\phi}{1 - \cos^{2}\alpha \sin^{2}\phi},\quad +y = \sin\phi - \frac{\sin\alpha \cos\phi}{1 - \cos^{2}\alpha \sin^{2}\phi}, +\] +where $\alpha$~is a positive acute angle, is $\frac{1}{2}\pi(1 + \sin\alpha)^{2}/\sin\alpha$. +\MathTrip{1904.} + +\Item{44.} The projection of a chord of a circle of radius~$a$ on a diameter is of +constant length~$2a\cos\beta$; show that the locus of the middle point of the chord +consists of two loops, and that the area of either is $a^{2}(\beta - \cos\beta\sin\beta)$. +\MathTrip{1903.} + +\Item{45.} Show that the length of a quadrant of the curve $(x/a)^{2/3} + (y/b)^{2/3} = 1$ is +$(a^{2} + ab + b^{2})/(a + b)$. \MathTrip{1911.} + +\Item{46.} A point $A$~is inside a circle of radius~$a$, at a distance~$b$ from the +centre. Show that the locus of the foot of the perpendicular drawn from +$A$ to a tangent to the circle encloses an area $\pi(a^{2} + \frac{1}{2}b^{2})$. +\MathTrip{1909.} + +\Item{47.} Prove that if $(a, b, c, f, g, h \btw x, y, 1)^{2} = 0$ is the equation of a conic, then +\[ +\int \frac{dx}{(lx + my + n)(hx + by + f)} = \alpha\log \frac{PT}{PT'} + \beta, +\] +where $PT$,~$PT'$ are the perpendiculars from a point~$P$ of the conic on the +tangents at the ends of the chord $lx + my + n = 0$, and $\alpha$,~$\beta$ are constants. +\MathTrip{1902.} + +\Item{48.} Show that +\[ +\int \frac{ax^{2} + 2bx + c}{(Ax^{2} + 2Bx + C)^{2}}\, dx +\] +will be a rational function of~$x$ if and only if one or other of $AC - B^{2}$ and +$aC + cA - 2bB$ is zero.\footnote + {See the author's tract quoted on \PageRef{p.}{236}.} + +\Item{49.} Show that the necessary and sufficient condition that +\[ +\int \frac{f(x)}{\{F(x)\}^{2}}\, dx, +\] +where $f$~and~$F$ are polynomials of which the latter has no repeated factor, +should be a rational function of~$x$, is that $f'F' - fF''$ should be divisible by~$F$. +\MathTrip{1910.} + +\Item{50.} Show that +\[ +\int \frac{\alpha\cos x + \beta\sin x + \gamma}{(1 - e\cos x)^{2}}\, dx +\] +is a rational function of $\cos x$ and~$\sin x$ if and only if $\alpha e + \gamma = 0$; and determine +the integral when this condition is satisfied. +\MathTrip{1910.} +\end{Examples} +\PageSep{262} + + +\Chapter[ADDITIONAL THEOREMS IN THE CALCULUS] +{VII}{ADDITIONAL THEOREMS IN THE DIFFERENTIAL AND \\ +INTEGRAL CALCULUS} + +\Paragraph{147. Higher Mean Value Theorems.} In the preceding +chapter (\SecNo[§]{125}) we proved that if $f(x)$~has a derivative~$f'(x)$ +throughout the interval $\DPmod{(a, b)}{[a, b]}$ then +\[ +f(b) - f(a) = (b - a) f'(\xi), +\] +where $a < \xi < b$; or that, if $f(x)$~has a derivative throughout +$\DPmod{(a, a + h)}{[a, a + h]}$, then +\[ +f(a + h) - f(a) = hf'(a + \theta_{1} h), +\Tag{(1)} +\] +where $0 < \theta_{1} < 1$. This we proved by considering the function +\[ +f(b) - f(x) - \frac{b - x}{b - a} \{f(b) - f(a)\} +\] +which vanishes when $x = a$ and when $x = b$. + +Let us now suppose that $f(x)$~has also a second derivative~$f''(x)$ +throughout $\DPmod{(a, b)}{[a, b]}$, an assumption which of course involves +the continuity of the first derivative~$f'(x)$, and consider the +function +\[ +f(b) - f(x) - (b - x) f'(x) + - \left(\frac{b - x}{b - a}\right)^{2} \{f(b) - f(a) - (b - a)f'(a)\}. +\] +This function also vanishes when $x = a$ and when $x = b$; and its +derivative is +\[ +\frac{2(b - x)}{(b - a)^{2}} + \{f(b) - f(a) - (b - a) f'(a) - \tfrac{1}{2}(b - a)^{2}f''(x)\}, +\] +and this must vanish (\SecNo[§]{121}) for some value of~$x$ between $a$ and~$b$ +(exclusive of $a$~and~$b$). Hence there is a value~$\xi$ of~$x$, between +\PageSep{263} +$a$ and~$b$, and therefore capable of representation in the form +$a + \theta_{2}(b - a)$, where $0 < \theta_{2} < 1$, for which +\[ +f(b) = f(a) + (b - a)f'(a) + \tfrac{1}{2}(b - a)^{2}f''(\xi). +\] + +If we put $b = a + h$ we obtain the equation +\[ +f(a + h) = f(a) + hf'(a) + \tfrac{1}{2}h^{2} f''(a + \theta_{2}h), +\Tag{(2)} +\] +which is the standard form of what may be called the \emph{Mean Value +Theorem of the second order}. + +The analogy suggested by \Eq{(1)}~and~\Eq{(2)} at once leads us to +formulate the following theorem: + +\begin{ParTheorem}{Taylor's or the General Mean Value Theorem.} +If +$f(x)$~is a function of~$x$ which has derivatives of the first $n$ orders +throughout the interval $\DPmod{(a, b)}{[a, b]}$, then +\begin{multline*} +f(b) = f(a) + (b - a)f'(a) + \frac{(b - a)^{2}}{2!} f''(a) + \dots\\ + + \frac{(b - a)^{n-1}}{(n - 1)!} f^{(n-1)}(a) + + \frac{(b - a)^{n}}{n!}f^{(n)}(\xi), +\end{multline*} +where $a < \xi < b$; and if $b = a + h$ then +\begin{multline*} +f(a + h) = f(a) + hf'(a) + \tfrac{1}{2} h^{2}f''(a) + \dots\\ + + \frac{h^{n-1}}{(n - 1)!} f^{(n-1)}(a) + + \frac{h^{n}}{n!} f^{(n)}(a + \theta_{n}h), +\end{multline*} +where $0 < \theta_{n} < 1$. +\end{ParTheorem} + +The proof proceeds on precisely the same lines as were adopted +before in the special cases in which $n = 1$ and $n = 2$. We consider +the function +\[ +F_{n}(x) - \left(\frac{b - x}{b - a}\right)^{n} F_{n}(a), +\] +where +\begin{multline*} +F_{n}(x) = f(b) - f(x) - (b - x)f'(x) - \frac{(b - x)^{2}}{2!} f''(x) - \dots\\ + - \frac{(b - x)^{n-1}}{(n - 1)!} f^{(n-1)}(x). +\end{multline*} +This function vanishes for $x = a$ and $x = b$; its derivative is +\[ +\frac{n(b - x)^{n-1}}{(b - a)^{n}} + \left\{F_{n}(a) - \frac{(b - a)^{n}}{n!} f^{(n)}(x)\right\}; +\] +and there must be some value of~$x$ between $a$ and~$b$ for which +the derivative vanishes. This leads at once to the desired result. +\PageSep{264} + +In view of the great importance of this theorem we shall give +at the end of this chapter another proof, not essentially distinct +from that given above, but different in form and depending on +the method of integration by parts. + +\begin{Examples}{LV.} +\Item{1.} Suppose that $f(x)$~is a polynomial of degree~$r$. +Then $f^{(n)}(x)$~is identically zero when $n > r$, and the theorem leads to the +algebraical identity +\[ +f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!} f''(a) + \dots + + \frac{h^{r}}{r!} f^{(r)}(a). +\] + +\Item{2.} By applying the theorem to $f(x) = 1/x$, and supposing $x$ and~$x + h$ +positive, obtain the result +\[ +\frac{1}{x + h} = \frac{1}{x} - \frac{h}{x^{2}} + \frac{h^{2}}{x^{3}} - \dots + + \frac{(-1)^{n-1} h^{n-1}}{x^{n}} + + \frac{(-1)^{n} h^{n}}{(x + \theta_{n} h)^{n+1}}. +\] + +[Since +\[ +\frac{1}{x + h} = \frac{1}{x} - \frac{h}{x^{2}} + \frac{h^{2}}{x^{3}} - \dots + + \frac{(-1)^{n-1} h^{n-1}}{x^{n}} + + \frac{(-1)^{n} h^{n}}{x^{n}(x + h)},\quad%[** TN: Quick spacing hack] +\] +we can verify the result by showing that $x^{n}(x + h)$ can be put in the form +$(x + \theta_{n}h)^{n+1}$, or that $x^{n+1} < x^{n}(x + h) < (x + h)^{n+1}$, as is evidently the case.] + +\Item{3.} Obtain the formula +\begin{multline*} +\sin(x + h) + = \sin x + h\cos x - \frac{h^{2}}{2!}\sin x - h^{3}\frac{3!}\cos x + \dots\\ + + (-1)^{n-1}\frac{h^{2n-1}}{(2n - 1)!}\cos x + + (-1)^{n} h^{2n}\frac{2n!}\sin(x + \theta_{2n} h), +\end{multline*} +the corresponding formula for $\cos(x + h)$, and similar formulae involving +powers of~$h$ extending up to~$h^{2n+1}$. + +\Item{4.} Show that if $m$~is a positive integer, and $n$~a positive integer not +greater than~$m$, then +\[ +(x + h)^{m} = x^{m} + \binom{m}{1}x^{m-1} h + \dots + + \binom{m}{n - 1}x^{m-n+1} h^{n-1} + + \binom{m}{n}(x + \theta_{n} h)^{m-n} h^{n}. +\] +Show also that, if the interval $\DPmod{(x, x + h)}{[x, x + h]}$ does not include $x = 0$, the formula +holds for all real values of~$m$ and all positive integral values of~$n$; and that, +even if $x < 0 < x + h$ or $x + h < 0 < x$, the formula still holds if $m - n$~is +positive. + +\Item{5.} The formula $f(x + h) = f(x) + hf'(x + \theta_{1}h)$ is not true if $f(x) = 1/x$ and +$x < 0 < x + h$. [For $f(x + h) - f(x) > 0$ and $hf'(x + \theta_{1} h) = -h/(x + \theta_{1} h)^{2} < 0$; it +is evident that the conditions for the truth of the Mean Value Theorem are +not satisfied.] + +\Item{6.} If $x = -a$, $h = 2a$, $f(x) = x^{1/3}$, then the equation +\[ +f(x + h) = f(x) + hf'(x + \theta_{1} h) +\] +is satisfied by $\theta_{1} = \frac{1}{2} ± \frac{1}{18}\sqrt{3}$. [This example shows that the result of the +theorem may hold even if the conditions under which it was proved are +not satisfied.] +\PageSep{265} + +\Item{7.} \Topic{Newton's method of approximation to the roots of equations.} Let +$\xi$~be an approximation to a root of an algebraical equation $f(x) = 0$, the actual +root being~$\xi + h$. Then +\[ +0 = f(\xi + h) = f(\xi) + hf'(\xi) + \tfrac{1}{2} h^{2}f''(\xi + \theta_{2}h), +\] +so that +\[ +h = -\frac{f(\xi)}{f'(\xi)} + - \tfrac{1}{2} h^{2} \frac{f''(\xi + \theta_{2}h)}{f'(\xi)}. +\] + +It follows that in general a better approximation than $x = \xi$ is +\[ +x = \xi - \frac{f(\xi)}{f'(\xi)}. +\] +If the root is a simple root, so that $f'(\xi + h) \neq 0$, we can, when $h$~is small +enough, find a positive constant~$K$ such that $|f'(x)| > K$ for all the values of~$x$ +which we are considering, and then, if $h$~is regarded as of the first order of +smallness, $f(\xi)$~is of the first order of smallness, and the error in taking +$\xi - \{f(\xi)/f'(\xi)\}$ as the root is of the second order. + +\Item{8.} Apply this process to the equation $x^{2} = 2$, taking $\xi = 3/2$ as the first +approximation. [We find $h = -1/12$, $\xi + h = 17/12 = 1.417\dots$, which is quite a +good approximation, in spite of the roughness of the first. If now we repeat +the process, taking $\xi = 17/12$, we obtain $\xi + h = 577/408 = 1.414\MS215\dots$, which +is correct to $5$~places of decimals.\Add{]} + +\Item{9.} By considering in this way the equation $x^{2} - 1 - y = 0$, where $y$~is +small, show that $\sqrtp{1 + y} = 1 + \frac{1}{2} y - \{\frac{1}{4}y^{2}/(2 + y)\}$ approximately, the error being +of the fourth order. + +\Item{10.} Show that the error in taking the root to be $\xi - (f/f') - \frac{1}{2}(f^{2}f''/f'^{3})$, +where $\xi$~is the argument of every function, is in general of the third order. + +\Item{11.} The equation $\sin x = \alpha x$, where $\alpha$~is small, has a root nearly equal to~$\pi$. +Show that $(1 - \alpha)\pi$~is a better approximation, and $(1 - \alpha + \alpha^{2})\pi$ a better +still. [The method of Exs.~7--10 does not depend on $f(x) = 0$ being an +algebraical equation, so long as $f'$~and~$f''$ are continuous.] + +\Item{12.} Show that the limit when $h \to 0$ of the number~$\theta_{n}$ which occurs in +the general Mean Value Theorem is~$1/(n + 1)$, provided that $f^{(n+1)}(x)$~is +continuous. + +[For $f(x + h)$~is equal to each of +\[ +f(x) + \dots + \frac{h^{n}}{n!} f^{(n)}(x + \theta_{n}h),\quad +f(x) + \dots + \frac{h^{n}}{n!} f^{(n)}(x) + + \frac{h^{n+1}}{(n + 1)!} f^{(n+1)}(x + \theta_{n+1}h), +\] +where $\theta_{n+1}$ as well as~$\theta_{n}$ lies between $0$~and~$1$. Hence +\[ +f^{(n)}(x + \theta_{n}h) + = f^{(n)}(x) + \frac{hf^{(n+1)}(x + \theta_{n+1}h)}{n + 1}\Add{.} +\] +But if we apply the original Mean Value Theorem to the function~$f^{(n)}(x)$, +taking $\theta_{n}h$ in place of~$h$, we find +\[ +f^{(n)}(x + \theta_{n}h) + = f^{(n)}(x) + \theta_{n}hf^{(n+1)}(x + \theta\theta_{n}h), +\] +\PageSep{266} +where $\theta$ also lies between $0$~and~$1$. Hence +\[ +\theta_{n} f^{(n+1)}(x + \theta\theta_{n} h) + = \frac{f^{(n+1)}(x + \theta_{n+1} h)}{n + 1}, +\] +from which the result follows, since $f^{(n+1)}(x + \theta\theta_{n} h)$ and $f^{(n+1)}(x + \theta_{n+1} h)$ tend +to the same limit~$f^{(n+1)}(x)$ as $h \to 0$.] + +\Item{13.} Prove that $\{f(x + 2h) - 2f(x + h) + f(x)\}/h^{2} \to f''(x)$ as $h \to 0$, provided +that $f''(x)$~is continuous. [Use equation~\Eq{(2)} of~\SecNo[§]{147}.] + +%[** TN: [sic] "the" f^{(n)}(x)] +\Item{14.} Show that, if the $f^{(n)}(x)$ is continuous for $x = 0$, then +\[ +f(x) = a_{0} + a_{1}x + a_{2}x^{2} + \dots + (a_{n} + \epsilon_{x}) x^{n}, +\] +where $a_{r} = f^{(r)}(0)/r!$ and $\epsilon_{x} \to 0$ as $x \to 0$.\footnote + {It is in fact sufficient to suppose that \emph{$f^{(n)}(0)$~exists}. See R.~H. Fowler, ``The + elementary differential geometry of plane curves'' (\textit{Cambridge Tracts in Mathematics}, + No.~20, p.~104).\PageLabel{266}} + +\Item{15.} Show that if +\[ +a_{0} + a_{1}x + a_{2}x^{2} + \dots + (a_{n} + \epsilon_{x}) x^{n} = +b_{0} + b_{1}x + b_{2}x^{2} + \dots + (b_{n} + \eta_{x}) x^{n}, +\] +where $\epsilon_{x}$ and~$\eta_{x}$ tend to zero as $x \to 0$, then $a_{0} = b_{0}$, $a_{1} = b_{1}$,~\dots, $a_{n} = b_{n}$. [Making +$x \to 0$ we see that $a_{0} = b_{0}$. Now divide by~$x$ and afterwards make $x \to 0$. +We thus obtain $a_{1} = b_{1}$; and this process may be repeated as often as is +necessary. It follows that if $f(x) = a_{0} + a_{1}x + a_{2}x^{2} + \dots + (a_{n} + \epsilon_{x}) x^{n}$, and the +first~$n$ derivatives of~$f(x)$ are continuous, then $a_{r} = f^{(r)}(0)/r!$.] +\end{Examples} + +\Paragraph{148. Taylor's Series.} Suppose that $f(x)$~is a function all +of whose differential coefficients are continuous in an interval +$\DPmod{(a - \eta, a + \eta)}{[a - \eta, a + \eta]}$ surrounding the point $x = a$. Then, if $h$~is numerically +less than~$\eta$, we have +\[ +f(a + h) = f(a) + hf'(a) + \dots + + \frac{h^{n-1}}{(n - 1)!} f^{(n-1)}(a) + + \frac{h^{n}}{n!} f^{(n)}(a + \theta_{n} h), +\] +where $0 < \theta_{n} < 1$, for all values of~$n$. Or, if +\[ +S_{n} = \sum_{0}^{n-1} \frac{h^{\nu}}{\nu!} f^{(\nu)}(a),\quad +R_{n} = \frac{h^{n}}{n!} f^{(n)}(a + \theta_{n} h), +\] +we have +\[ +f(a + h) - S_{n} = R_{n}. +\] + +Now let us suppose, in addition, that we can prove that +$R_{n} \to 0$ as $n \to \infty$. Then +\[ +f(a + h) = \lim_{n\to\infty} S_{n} + = f(a) + hf'(a) + \frac{h^{2}}{2!} f''(a) + \dots. +\] + +This expansion of~$f(a + h)$ is known as \Emph{Taylor's Series}. +When $a = 0$ the formula reduces to +\[ +f(h) = f(0) + hf'(0) + \frac{h^{2}}{2!} f''(0) + \dots, +\] +\PageSep{267} +which is known as \Emph{Maclaurin's Series}. The function~$R_{n}$ is known +as \Emph{Lagrange's form of the remainder}. + +\begin{Remark} +The reader should be careful to guard himself against supposing that the +continuity of all the derivatives of~$f(x)$ is a sufficient condition for the validity +of Taylor's series. A direct discussion of the behaviour of~$R_{n}$ is always +essential. +\end{Remark} + +\begin{Examples}{LVI.} +\Item{1.} Let $f(x) = \sin x$. Then all the derivatives of~$f(x)$ +are continuous for all values of~$x$. Also $|f^{n}(x)| \leq 1$ for all values of $x$~and~$n$. +Hence in this case $|R_{n}| \leq h^{n}/n!$, which tends to zero as $n \to \infty$ (\Ex{xxvii}.~12) +whatever value $h$ may have. It follows that +\[ +\sin(x + h) = \sin x + h\cos x - \frac{h^{2}}{2!}\sin x + - \frac{h^{3}}{3!}\cos x + \frac{h^{4}}{4!}\sin x + \dots, +\] +for all values of $x$~and~$h$. In particular +\[ +\sin h = h - \frac{h^{3}}{3!} + \frac{h^{5}}{5!} - \dots, +\] +for all values of~$h$. Similarly we can prove that +\[ +\cos(x + h) = \cos x - h\sin x - \frac{h^{2}}{2!}\cos x + + \frac{h^{3}}{3!} \sin x + \dots,\quad +\cos h = 1 - \frac{h^{2}}{2!} + \frac{h^{4}}{4!} - \dots. +\] + +\Item{2.} \Topic{The Binomial Series.} Let $f(x) = (1 + x)^{m}$, where $m$~is any rational +number, positive or negative. Then $f^{(n)}(x) = m(m - 1) \dots (m - n + 1) (1 + x)^{m-n}$ +and Maclaurin's Series takes the form +\[ +(1 + x)^{m} = 1 + \binom{m}{1}x + \binom{m}{2}x^{2} + \dots. +\] + +When $m$~is a positive integer the series terminates, and we obtain the +ordinary formula for the Binomial Theorem with a positive integral exponent. +In the general case +\[ +R_{n} = \frac{x^{n}}{n!} f^{(n)}(\theta_{n}x) + = \binom{m}{n}x^{n}(1 + \theta_{n}x)^{m-n}, +\] +and in order to show that Maclaurin's Series really represents $(1 + x)^{m}$ for +any range of values of~$x$ when $m$~is not a positive integer, we must show that +$R_{n} \to 0$ for every value of~$x$ in that range. This is so in fact if $-1 < x < 1$, +and may be proved, when $0\leq x < 1$, by means of the expression given above +for~$R_{n}$, since $(1 + \theta_{n}x)^{m-n} < 1$ if $n > m$, and $\dbinom{m}{n} x^{n} \to 0$ as $n \to \infty$ (\Ex{xxvii}.~13). +But a difficulty arises if $-1 < x < 0$, since $1 + \theta_{n}x < 1$ and $(1 + \theta_{n}x)^{m-n} > 1$ +if $n > m$; knowing only that $0 < \theta_{n} < 1$, we cannot be assured that $1 + \theta_{n}x$~is not +quite small and $(1 + \theta _{n}x)^{m-n}$ quite large. + +In fact, in order to prove the Binomial Theorem by means of Taylor's +Theorem, we need some different form for~$R_{n}$, such as will be given later~(\SecNo[§]{162}). +\end{Examples} +\PageSep{268} + +\Paragraph{149. Applications of Taylor's Theorem.} \Topic{\Item{A.} Maxima +and minima.} Taylor's Theorem may be applied to give greater +theoretical completeness to the tests of \Ref{Ch.}{VI}, \SecNo[§§]{122}--\SecNo{123}, +though the results are not of much practical importance. It +will be remembered that, assuming that $\phi(x)$~has derivatives of +the first two orders, we stated the following as being sufficient +conditions for a maximum or minimum of~$\phi(x)$ at $x = \xi$: \emph{for a +maximum}, $\phi'(\xi) = 0$, $\phi''(\xi) < 0$; \emph{for a minimum}, $\phi'(\xi) = 0$, $\phi''(\xi) > 0$. +It is evident that these tests fail if $\phi''(\xi)$ as well as $\phi'(\xi)$ is zero. + +Let us suppose that the first~$n$ derivatives +\[ +\phi'(x),\quad \phi''(x),\ \dots,\quad \phi^{(n)}(x) +\] +are continuous, and that all save the last vanish when $x = \xi$. Then, +for sufficiently small values of~$h$, +\[ +\phi(\xi + h) - \phi(\xi) = \frac{h^{n}}{n!} \phi^{(n)} (\xi + \theta_{n} h). +\] +In order that there should be a maximum or a minimum this +expression must be of constant sign for all sufficiently small +values of~$h$, positive or negative. This evidently requires that $n$~should +be even. And if $n$~is even there will be a maximum or a +minimum according as $\phi^{(n)}(\xi)$~is negative or positive. + +Thus we obtain the test: \begin{Result}if there is to be a maximum or +minimum the first derivative which does not vanish must be an even +derivative, and there will be a maximum if it is negative, a minimum +if it is positive. +\end{Result} + +\begin{Examples}{LVII.} +\Item{1.} Verify the result when $\phi(x) = (x - a)^{m}$, $m$~being a +positive integer, and $\xi = a$. + +\Item{2.} Test the function $(x - a)^{m} (x - b)^{n}$, where $m$~and~$n$ are positive integers, +for maxima and minima at the points $x = a$, $x = b$. Draw graphs of the +different possible forms of the curve $y = (x - a)^{m} (x - b)^{n}$. + +\Item{3.} Test the functions $\sin x - x$, $\sin x - x + \dfrac{x^{3}}{6}$, +$\sin x - x + \dfrac{x^{3}}{6} - \dfrac{x^{5}}{120}$,~\dots, +$\cos x - 1$, $\cos x - 1 + \dfrac{x^{2}}{2}$, $\cos x - 1 + \dfrac{x^{2}}{2} - \dfrac{x^{4}}{24}$,~\dots\ +for maxima or minima at $x = 0$. +\end{Examples} + +\Paragraph{150.} \Topic{\Item{B.} The calculation of certain limits.} Suppose +that $f(x)$ and~$\phi(x)$ are two functions of~$x$ whose derivatives $f'(x)$ +and~$\phi'(x)$ are continuous for $x = \xi$ and that $f(\xi)$ and~$\phi(\xi)$ are +both equal to zero. Then the function +\[ +\psi(x) = f(x)/\phi(x) +\] +\PageSep{269} +is not defined when $x = \xi$. But of course it may well tend to a +limit as $x \to \xi$. + +Now +\[ +f(x) = f(x) - f(\xi) = (x - \xi)f'(x_{1}), +\] +where $x_{1}$~lies between $\xi$ and~$x$; and similarly $\phi(x) = (x - \xi)\phi'(x_{2})$, +where $x_{2}$~also lies between $\xi$ and~$x$. Thus +\[ +\psi(x) = f'(x_{1})/\phi'(x_{2}). +\] +We must now distinguish four cases. + +\Item{(1)} If neither $f'(\xi)$ nor $\phi'(\xi)$ is zero, then +\[ +f(x)/\phi(x) \to f'(\xi)/\phi'(\xi). +\] + +\Item{(2)} If $f'(\xi) = 0$, $\phi'(\xi) \neq 0$, then +\[ +f(x)/\phi(x) \to 0. +\] + +\Item{(3)} If $f'(\xi) \neq 0$, $\phi'(\xi)= 0$, then $f(x)/\phi(x)$ becomes numerically +very large as $x \to \xi$: but whether $f(x)/\phi(x)$ tends to $\infty$~or~$-\infty$, +or is sometimes large and positive and sometimes large and +negative, we cannot say, without further information as to the way +in which $\phi'(x) \to 0$ as $x \to \xi$. + +\Item{(4)} If $f'(\xi) = 0$, $\phi'(\xi) = 0$, then we can as yet say nothing about +the behaviour of~$f(x)/\phi(x)$ as $x \to 0$. + +But in either of the last two cases it may happen that $f(x)$ +and $\phi(x)$ have continuous second derivatives. And then +\begin{align*} +f(x) &= f(x) - f(\xi) - (x - \xi)f'(\xi) + = \tfrac{1}{2}(x - \xi)^{2} f''(x_{1}),\\ +\phi(x) &= \phi(x) - \phi(\xi) - (x - \xi)\phi'(\xi) + = \tfrac{1}{2}(x - \xi)^{2} \phi''(x_{2}), +\end{align*} +where again $x_{1}$ and~$x_{2}$ lie between $\xi$ and~$x$; so that +\[ +\psi(x)= f''(x_{1})/\phi''(x_{2}). +\] +We can now distinguish a variety of cases similar to those +considered above. In particular, if neither second derivative +vanishes for $x = \xi$, we have +\[ +f(x)/\phi(x) \to f''(\xi)/\phi''(\xi). +\] + +It is obvious that this argument can be repeated indefinitely, +and we obtain the following theorem: \begin{Result}suppose that $f(x)$ and $\phi(x)$ +and their derivatives, so far as may be wanted, are continuous for +$x = \xi$. Suppose further that $f^{(p)}(x)$ and~$\phi^{(q)}(x)$ are the first +derivatives of $f(x)$ and $\phi(x)$ which do not vanish when $x = \xi$. Then + +\Item{(1)} if $p = q$, $f(x)/\phi(x) \to f^{(p)}(\xi)/\phi^{(p)}(\xi)$; + +\Item{(2)} if $p > q$, $f(x)/\phi(x) \to 0$; +\PageSep{270} + +\Item{(3)} {\Loosen if $p < q$, and $q - p$~is even, either $f(x)/\phi(x) \to +\infty$ or +$f(x)/\phi(x) \to -\infty$, the sign being the same as that of~$f^{(p)}(\xi)/\phi^{(q)}(\xi)$;} + +\Item{(4)} {\Loosen if $p < q$ and $q - p$~is odd, either $f(x)/\phi(x) \to +\infty$ or +$f(x)/\phi(x) \to -\infty$, as $x \to \xi+0$, the sign being the same as that of +$f^{(p)}(\xi)/\phi^{(q)}(\xi)$, while if $x \to \xi - 0$ the sign must be reversed.} +\end{Result} + +This theorem is in fact an immediate corollary from the +equations +\[ +f(x) = \frac{(x - \xi)^{p}}{p!}f^{(p)}(x_{1}),\quad +\phi(x) = \frac{(x - \xi)^{q}}{q!}\phi^{(q)}(x_{2}). +\] + +\begin{Examples}{LVIII.} +\Item{1.} Find the limit of +\[ +\{x - (n + 1)x^{n+1} + nx^{n+2}\}/(1 - x)^{2}, +\] +as $x \to 1$. [Here the functions and their first derivatives vanish for $x = 1$, +and $f''(1) = n(n + 1)$, $\phi''(1) = 2$.] + +\Item{2.} Find the limits as $x \to 0$ of +\[ +(\tan x - x)/(x - \sin x),\quad +(\tan nx - n\tan x)/(n\sin x - \sin nx). +\] + +\Item{3.} Find the limit of $x\{\sqrtp{x^{2} + a^{2}} - x\}$ as $x \to \infty$. [Put $x = 1/y$.] + +\Item{4.} Prove that +\[ +\lim_{x \to n} (x - n)\cosec x\pi = \frac{(-1)^{n}}{\pi},\quad +\lim_{x \to n} \frac{1}{x - n} \left\{ + \cosec x\pi - \frac{(-1)^{n}}{(x - n)\pi} +\right\} = \frac{(-1)^{n}\pi}{6}, +\] +$n$~being any integer; and evaluate the corresponding limits involving $\cot x\pi$. + +\Item{5.} Find the limits as $x \to 0$ of +\[ +\frac{1}{x^{3}}\left(\cosec x - \frac{1}{x} - \frac{x}{6}\right),\quad +\frac{1}{x^{3}}\left(\cot x - \frac{1}{x} + \frac{x}{3}\right). +\] + +\Item{6.} $(\sin x\arcsin x - x^{2})/x^{6} \to \frac{1}{18}$, $(\tan x\arctan x - x^{2})/x^{6} \to \frac{2}{9}$, as $x \to 0$. +\end{Examples} + +\Paragraph{151.} \Topic{\Item{C.} The contact of plane curves.} Two curves are +said to \emph{intersect} (or \emph{cut}) at a point if the point lies on each of them. +They are said to \emph{touch} at the point if they have the same tangent +at the point. + +Let us suppose now that $f(x)$,~$\phi(x)$ are two functions which +possess derivatives of all orders continuous for $x = \xi$, and let us +consider the curves $y = f(x)$, $y = \phi(x)$. In general $f(\xi)$ and~$\phi(\xi)$ +will not be equal. In this case the abscissa $x = \xi$ does not correspond +to a point of intersection of the curves. If however +\PageSep{271} +$f(\xi) = \phi(\xi)$, the curves intersect in the point $x = \xi$, $y = f(\xi) = \phi(\xi)$. +Let us suppose this to be the case. Then +in order that the curves should not only +cut but touch at this point it is obviously +necessary and sufficient that the first +derivatives $f'(x)$,~$\phi'(x)$ should also have +the same value when $x = \xi$. + +The contact of the curves in this +case may be regarded from a different +point of view. In the figure the two +%[Illustration: Fig. 45.] +\Figure[2.25in]{45}{p271} +curves are drawn touching at~$P$, and $QR$~is +equal to $\phi(\xi + h) - f(\xi + h)$, or, since $\phi(\xi) = f(\xi)$, $\phi'(\xi) = f'(\xi)$, to +\[ +\tfrac{1}{2} h^{2}\{\phi''(\xi + \theta h) - f''(\xi + \theta h)\}, +\] +where $\theta$~lies between $0$ and~$1$. Hence +\[ +\lim \frac{QR}{h^{2}} = \tfrac{1}{2}\{\phi''(\xi) - f''(\xi)\}, +\] +when $h \to 0$. In other words, when the curves touch at the point +whose abscissa is~$\xi$, \emph{the difference of their ordinates at the point +whose abscissa is $\xi + h$ is at least of the second order of smallness +when $h$~is small}. + +\begin{Remark} +The reader will easily verify that $\lim (QR/h) = \phi'(\xi) - f'(\xi)$ when the curves +cut and do not touch, so that $QR$~is then of the first order of smallness only. +\end{Remark} + +It is evident that the degree of smallness of~$QR$ may be taken +as a kind of measure of the \emph{closeness of the contact} of the curves. +It is at once suggested that if the first $n - 1$ derivatives of $f$ +and~$\phi$ have equal values when $x = \xi$, then $QR$~will be of +$n$th~order of smallness; and the reader will have no difficulty +in proving that this is so and that +\[ +\lim \frac{QR}{h^{n}} = \frac{1}{n!}\{\phi^{(n)}(\xi) - f^{(n)}(\xi)\}. +\] +We are therefore led to frame the following definition: + +\begin{Defn} +\Emph{Contact of the $n$th~order.} If $f(\xi) = \phi(\xi)$, $f'(\xi) = \phi'(\xi)$,~\dots, +$f^{(n)}(\xi) = \phi^{(n)}(\xi)$, but $f^{(n+1)}(\xi) \neq \phi^{(n+1)}(\xi)$, then the curves +$y = f(x)$, $y = \phi(x)$ will be said to have contact of the $n$th~order +at the point whose abscissa is~$\xi$. +\end{Defn} + +The preceding discussion makes the notion of contact of +the $n$th~order dependent on the choice of axes, and fails entirely +\PageSep{272} +when the tangent to the curves is parallel to the axis of~$y$. We can +deal with this case by taking $y$~as the independent and $x$~as the +dependent variable. It is better, however, to consider $x$~and~$y$ as +functions of a parameter~$t$. An excellent account of the theory will +be found in Mr~Fowler's tract referred to on \PageRef{p.}{266}, or in de~la~Vallée +Poussin's \textit{Cours d'Analyse}, vol.~ii, pp.~396~\textit{et~seq.} + +\begin{Examples}{LIX.} +\Item{1.} Let $\phi(x) = ax + b$, so that $y = \phi(x)$ is a straight line. +The conditions for contact at the point for which $x = \xi$ are $f(\xi) = a\xi + b$, +$f'(\xi) = a$. If we determine $a$~and~$b$ so as to satisfy these equations we find +$a = f'(\xi)$, $b = f(\xi) - \xi f'(\xi)$, and the equation of the tangent to $y = f(x)$ at the +point $x = \xi$ is +\[ +y = xf'(\xi) + \{f(\xi) - \xi f'(\xi)\}, +\] +or $y - f(\xi) = (x - \xi)f'(\xi)$. Cf.\ \Ex{xxxix}.~5. + +\Item{2.} The fact that the line is to have simple contact with the curve +completely determines the line. In order that the tangent should have +\emph{contact of the second order} with the curve we must have $f''(\xi) = \phi''(\xi)$, \ie\ +$f''(\xi) = 0$. A point at which the tangent to a curve has contact of the +second order is called a \Emph{point of inflexion}. +%[** TN: Differs from the modern definition] + +\Item{3.} Find the points of inflexion on the graphs of the functions $3x^{4} - 6x^{3} + 1$, +$2x/(1 + x^{2})$, $\sin x$, $a\cos^{2}x + b\sin^{2}x$, $\tan x$, $\arctan x$. + +\Item{4.} Show that the conic $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$ cannot have a +point of inflexion. [Here $ax + hy + g + (hx + by + f)y_{1} = 0$ and +\[ +a + 2hy_{1} + by_{1}^{2} + (hx + by + f)y_{2} = 0, +\] +suffixes denoting differentiations. Thus at a point of inflexion +\[ +a + 2hy_{1} + by_{1}^{2} = 0, +\] +or +\[ +a(hx + by + f)^{2} - 2h(ax + hy + g)(hx + by + f) + b(ax + hy + g)^{2} = 0, +\] +or +\[ +(ab - h^{2})\{ax^{2} + 2hxy + by^{2} + 2gx + 2fy\} + af^{2} - 2fgh + bg^{2} = 0. +\] +But this is inconsistent with the equation of the conic unless +\[ +af^{2} - 2fgh + bg^{2} = c(ab - h^{2}) +\] +or $abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0$; and this is the condition that the conic +should degenerate into two straight lines.] + +\Item{5.} The curve $y = (ax^{2} + 2bx + c)/(\alpha x^{2} + 2\beta x + \gamma)$ has one or three points of +inflexion according as the roots of $\alpha x^{2} + 2\beta x + \gamma = 0$ are real or complex. + +[The equation of the curve can, by a change of origin (cf.\ \Ex{xlvi}.~15), be +reduced to the form +\[ +\eta = \xi/(A\xi^{2} + 2B\xi + C) = \xi/\{A(\xi - p)(\xi - q)\}, +\] +where $p$,~$q$ are real or conjugate. The condition for a point of inflexion will +be found to be $\xi^{3} - 3pq\xi + pq(p + q) = 0$, which has one or three real roots +according as $\DPtypo{\{pq(p - q)\}}{\{pq(p - q)\}^{2}}$ is positive or negative, \ie\ according as $p$~and~$q$ are +real or conjugate.] +\PageSep{273} + +\Item{6.} Discuss in particular the curves $y = (1 - x)/(1 + x^{2})$, $y = (1 - x^{2})/(1 + x^{2})$, +$y = (1 + x^{2})/(1 - x^{2})$. + +\Item{7.} Show that when the curve of Ex.~5 has three points of inflexion, they +lie on a straight line. [The equation $\xi^{3} - 3pq\xi + pq(p + q) = 0$ can be put in +the form $(\xi - p)(\xi - q)(\xi + p + q) + (p - q)^{2}\xi = 0$, so that the points of inflexion +lie on the line $\xi + A(p - q)^{2}\eta + p + q = 0$ or $A\xi - 4(AC - B^{2})\eta = 2B$.] + +\Item{8.} Show that the curves $y = x\sin x$, $y = (\sin x)/x$ have each infinitely +many points of inflexion. + +\Item{9.} \Topic{Contact of a circle with a curve. Curvature.\footnote + {A much fuller discussion of the theory of curvature will be found in Mr~Fowler's +%[** TN: Reference on page 272 of orig. points to page 266.] + tract referred to on \PageRef{p.}{\DPchg{272}{266}}.}} +The general +equation of a circle, viz. +\[ +(x - a)^{2} + (y - b)^{2} = r^{2}, +\Tag{(1)} +\] +contains three arbitrary constants. Let us attempt to determine them so +that the circle has contact of as high an order as possible with the curve +$y = f(x)$ at the point $(\xi, \eta)$, where $\eta = f(\xi)$. We write $\eta_{1}$,~$\eta_{2}$ for $f'(\xi)$,~$f''(\xi)$. +Differentiating the equation of the circle twice we obtain +\begin{align} +(x - a) + (y - b)y_{1} &= 0, +\Tag{(2)}\\ +1 + y_{1}^{2} + (y - b)y_{2} &= 0. +\Tag{(3)} +\end{align} + +If the circle touches the curve then the equations \Eq{(1)}~and~\Eq{(2)} are satisfied +when $x = \xi$, $y = \eta$, $y_{1} = \eta_{1}$. This gives $(\xi - a)/\eta_{1} = -(\eta - b) = r/\sqrtp{1 + \eta_{1}^{2}}$. If +the contact is of the second order then the equation~\Eq{(3)} must also be satisfied +when $y_{2} = \eta_{2}$. Thus $b = \eta + \{(1 + \eta_{1}^{2})/\eta_{2}\}$; and hence we find +\[ +a = \xi - \frac{\eta_{1}(1 + \eta_{1}^{2})}{\eta_{2}},\quad +b = \eta + \frac{1 + \eta_{1}^{2}}{\eta_{2}},\quad +r = \frac{(1 + \eta_{1}^{2})^{3/2}}{\eta_{2}}. +\] + +The circle which has contact of the second order with the curve at the point +$(\xi, \eta)$ is called the \Emph{circle of curvature}, and its radius the \Emph{radius of curvature}. +The \Emph{measure of curvature} (or simply the \emph{curvature}) is the reciprocal of the +radius: thus the measure of curvature is $f''(\xi)/\{1 + [f'(\xi)]^{2}\}^{3/2}$, or +\[ +\frac{d^{2}\eta}{d\xi^{2}} \bigg/ + \biggl\{1 + \biggl(\frac{d\eta}{d\xi}\biggr)^{2}\biggr\}^{3/2}. +\] + +\Item{10.} Verify that the curvature of a circle is constant and equal to the +reciprocal of the radius; and show that the circle is the only curve whose +curvature is constant. + +\Item{11.} {\Loosen Find the centre and radius of curvature at any point of the conics +$y^{2} = 4ax$, $(x/a)^{2} + (y/b)^{2} = 1$.} + +\Item{12.} In an ellipse the radius of curvature at~$P$ is~$CD^{3}/ab$, where $CD$~is +the semi-diameter conjugate to~$CP$. +\PageSep{274} + +\Item{13.} Show that in general a conic can be drawn to have contact of the +fourth order with the curve $y = f(x)$ at a given point~$P$. + +[Take the general equation of a conic, viz. +\[ +ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0, +\] +and differentiate four times with respect to~$x$. Using suffixes to denote +differentiation we obtain +\begin{align*} +ax + hy + g + (hx + by + f) y_{1} &= 0,\\ +a + 2hy_{1} + by_{1}^{2} + (hx + by + f) y_{2} &= 0,\\ +3(h + by_{1}) y_{2} + (hx + by + f) y_{3} &= 0,\\ +4(h + by_{1}) y_{3} +3by_{2}^{2} + (hx + by + f) y_{4} &= 0. +\end{align*} +If the conic has contact of the fourth order, then these five equations must +be satisfied by writing $\xi$, $\eta$, $\eta_{1}$, $\eta_{2}$, $\eta_{3}$, $\eta_{4}$, for $x$, $y$, $y_{1}$, $y_{2}$, $y_{3}$, $y_{4}$. We have thus +just enough equations to determine the ratios $a : b : c : f : g : h$.] + +\Item{14.} An infinity of conics can be drawn having contact of the third order +with the curve at~$P$. Show that their centres all lie on a straight line. + +[Take the tangent and normal as axes. Then the equation of the conic is +of the form $2y = ax^{2} + 2hxy + by^{2}$, and when $x$~is small one value of~$y$ may be +expressed (\Ref{Ch.}{V}, \MiscEx{V}~22) in the form +\[ +y = \tfrac{1}{2}ax^{2} + \left(\tfrac{1}{2}ah + \epsilon_{x}\right) x^{3}, +\] +where $\epsilon_{x} \to 0$ with~$x$. But this expression must be the same as +\[ +y = \tfrac{1}{2}f''(0) x^{2} + \{\tfrac{1}{6}f'''(0) + \epsilon'_{x}\} x^{3}, +\] +where $\epsilon'_{x} \to 0$ with~$x$, and so $a = f''(0)$, $h = f'''(0)/3f''(0)$, in virtue of the result +of \Ex{lv}.~15. But the centre lies on the line $ax + hy = 0$.] + +\Item{15.} Determine a parabola which has contact of the third order with the +ellipse $(x/a)^{2} + (y/b)^{2} = 1$ at the extremity of the major axis. + +\Item{16.} The locus of the centres of conics which have contact of the third +order with the ellipse $(x/a)^{2} + (y/b)^{2} = 1$ at the point $(a\cos\alpha, b\sin\alpha)$ is the +diameter $x/(a\cos\alpha) = y/(b\sin\alpha)$. [For the ellipse itself is one such conic.] +\end{Examples} + +\Paragraph{152. Differentiation of functions of several variables.} +So far we have been concerned exclusively with functions of a +single variable~$x$, but there is nothing to prevent us applying the +notion of differentiation to functions of several variables $x$, $y$,~\dots. + +Suppose then that $f(x, y)$~is a function of two\footnote + {The new points which arise when we consider functions of several variables + are illustrated sufficiently when there are two variables only. The generalisations + of our theorems for three or more variables are in general of an obvious character.} +real variables +$x$~and~$y$, and that the limits +\[ +\lim_{h\to 0}\frac{f(x + h, y) - f(x, y)}{h},\quad +\lim_{k\to 0}\frac{f(x, y + k) - f(x, y)}{k} +\] +\PageSep{275} +exist for all values of $x$~and~$y$ in question, that is to say that +$f(x, y)$ possesses a derivative~$df/dx$ or~$D_{x}f(x, y)$ with respect to~$x$ +and a derivative~$df/dy$ or~$D_{y}f(x, y)$ with respect to~$y$. It is usual +to call these derivatives the \emph{partial differential coefficients} of~$f$, and +to denote them by +\[ +\frac{\dd f}{\dd x},\quad +\frac{\dd f}{\dd y} +\] +or +\[ +f_{x}'(x, y),\quad +f_{y}'(x, y) +\] +or simply $f_{x}'$,~$f_{y}'$ or $f_{x}$,~$f_{y}$. The reader must not suppose, however, +that these new notations imply any essential novelty of idea: +`partial differentiation' with respect to~$x$ is exactly the same +process as ordinary differentiation, the only novelty lying in the +presence in~$f$ of a second variable~$y$ independent of~$x$. + +In what precedes we have supposed $x$~and~$y$ to be two real +variables entirely independent of one another. If $x$~and~$y$ were +connected by a relation the state of affairs would be very different. +In this case our definition of~$f_{x}'$ would fail entirely, as we could +not change~$x$ into~$x + h$ without at the same time changing~$y$. +But then $f(x, y)$ would not really be a function of two variables +at all. A function of two variables, as we defined it in \Ref{Ch.}{II}, +is essentially a function of two \emph{independent} variables. If $y$~depends +on~$x$, $y$~is a function of~$x$, say $y = \phi(x)$; and then +\[ +f(x, y) = f\{x, \phi(x)\} +\] +is really a function of the single variable~$x$. Of course we may also +represent it as a function of the single variable~$y$. Or, as is often +most convenient, we may regard $x$~and~$y$ as functions of a third +variable~$t$, and then $f(x, y)$, which is of the form $f\{\phi(t), \psi(t)\}$, +is a function of the single variable~$t$. + +\begin{Examples}{LX.} +\Item{1.} {\Loosen Prove that if $x = r\cos\theta$, $y = r\sin\theta$, so that $r = \sqrtp{x^{2} + y^{2}}$, +$\theta = \arctan(y/x)$, then} +\begin{align*} + \frac{\dd r}{\dd x} &= \frac{x}{\sqrtp{x^{2} + y^{2}}}, +&\frac{\dd r}{\dd y} &= \frac{y}{\sqrtp{x^{2} + y^{2}}}, +&\frac{\dd \theta}{\dd x} &= -\frac{y}{x^{2} + y^{2}}, +&\frac{\dd \theta}{\dd y} &= \frac{x}{x^{2} + y^{2}},\\ +% + \frac{\dd x}{\dd r} &= \cos\theta, +&\frac{\dd y}{\dd r} &= \sin\theta, +&\frac{\dd x}{\dd \theta} &= -r\sin\theta, +&\frac{\dd y}{\dd \theta} &= r\cos\theta. +\end{align*} + +\Item{2.} Account for the fact that +$\dfrac{\dd r}{\dd x}\neq 1\bigg/\biggl(\dfrac{\dd x}{\dd r}\biggr)$ and +$\dfrac{\dd \theta}{\dd x}\neq 1\bigg/\biggl(\dfrac{\dd x}{\dd \theta}\biggr)$. [When +we were considering a function~$y$ of one variable~$x$ it followed from the +definitions that $dy/dx$ and~$dx/dy$ were reciprocals. This is no longer the +\PageSep{276} +case when we are dealing with functions of two variables. Let $P$ (\Fig{46}) +be the point $(x, y)$ or $(r, \theta)$. To find $\dd r/\dd x$ we must increase~$x$, say by an +increment $MM_{1} = \delta x$, while keeping $y$~constant. This brings~$P$ to~$P_{1}$. If +along~$OP_{1}$ we take $OP' = OP$, the increment of~$r$ is $P'P_{1} = \delta r$, say; and +$\dd r/\dd x = \lim(\delta r/\delta x)$. If on the other hand we want to calculate $\dd x/\dd r$, $x$~and~$y$ +%[Illustration: Fig. 46.] +\Figure[2.25in]{46}{p276} +being now regarded as functions of $r$~and~$\theta$, +we must increase~$r$ by~$\Delta r$, say, +keeping $\theta$~constant. This brings~$P$ to~$P_{2}$, +where $PP_{2} = \Delta r$: the corresponding +increment of~$x$ is $MM_{1} = \Delta x$, say; and +\[ +\dd x/\dd r = \lim(\Delta x/\Delta r). +\] +Now $\Delta x = \delta x$:\footnote + {Of course the fact that $\Delta x = \delta x$ is due merely to the particular value of~$\Delta r$ + that we have chosen (viz.~$PP_{2}$). Any other choice would give us values of $\Delta x$,~$\Delta r$ + proportional to those used here.} +but $\Delta r \neq \delta r$. Indeed it is +easy to see from the figure that +\[ +\lim (\delta r/\delta x) = \lim (P'P_{1}/PP_{1}) = \cos\theta, +\] +but +\[ +\lim (\Delta r/\Delta x) = \lim (PP_{2}/PP_{1}) = \sec\theta, +\] +so that +\[ +\lim (\delta r/\Delta r) = \cos^{2}\theta. +\] + +The fact is of course that \emph{$\dd x/\dd r$ and +$\dd r/\dd x$ are not formed upon the same hypothesis as to the variation of~$P$.}] + +\Item{3.} Prove that if $z = f(ax + by)$ then $b(\dd z/\dd x) = a(\dd z/\dd y)$. + +\Item{4.} Find $\dd X/\dd x$, $\dd X/\dd y$,~\dots\ when $X + Y = x$, $Y = xy$. Express $x$,~$y$ as +functions of $X$,~$Y$ and find $\dd x/\dd X$, $\dd x/\dd Y$,~\dots. + +\Item{5.} Find $\dd X/\dd x$,~\dots\ when $X + Y + Z = x$, $Y + Z = xy$, $Z = xyz$; express +$x$,~$y$,~$z$ in terms of $X$,~$Y$,~$Z$ and find $\dd x/\dd X$,~\dots. + +[There is of course no difficulty in extending the ideas of the last section +to functions of any number of variables. But the reader must be careful to +impress on his mind that the notion of the partial derivative of a function of +several variables is only determinate when \emph{all} the independent variables are +specified. Thus if $u = x + y + z$, $x$,~$y$, and~$z$ being the independent variables, +then $\dd u/\dd x = 1$. But if we regard $u$ as a function of the variables $x$, $x + y = \eta$, +and $x + y + z = \zeta$, so that $u = \zeta$, then $\dd u/\dd x = 0$.] +\end{Examples} + +\Paragraph{153. Differentiation of a function of two functions.} +There is a theorem concerning the differentiation of a function +of \emph{one} variable, known generally as the \Emph{Theorem of the Total +Differential Coefficient}, which is of very great importance and +depends on the notions explained in the preceding section regarding +functions of \emph{two} variables. This theorem gives us a rule +for differentiating +\[ +f\{\phi(t), \psi(t)\}, +\] +with respect to~$t$. +\PageSep{277} + +Let us suppose, in the first instance, that $f(x, y)$ is a function +of the two variables $x$~and~$y$, and that $f_{x}'$,~$f_{y}'$ are continuous +functions of both variables (\SecNo[§]{107}) for all of their values which +come in question. And now let us suppose that the variation of +$x$~and~$y$ is restricted in that $(x, y)$ lies on a curve +\[ +x = \phi(t),\quad +y = \psi(t), +\] +where $\phi$ and~$\psi$ are functions of~$t$ with continuous differential +coefficients $\phi'(t)$,~$\psi' (t)$. Then $f(x, y)$ reduces to a function of the +single variable~$t$, say~$F(t)$. The problem is to determine~$F'(t)$. + +Suppose that, when $t$~changes to~$t + \tau$, $x$~and~$y$ change to +$x + \xi$ and $y + \eta$. Then by definition +\begin{align*} +%[** TN: Third line not aligned in the original] +\frac{dF(t)}{dt} + &= \lim_{\tau\to 0} + \frac{1}{\tau}[f\{\phi(t + \tau), \psi(t + \tau)\} - f\{\phi(t), \psi(t)\}]\\ + &= \lim \frac{1}{\tau}\{f(x + \xi, y + \eta) - f(x, y)\} \\ + &= \lim \left[ + \frac{f(x + \xi, y + \eta) - f(x, y + \eta)}{\xi}\, \frac{\xi}{\tau} + + \frac{f(x, y + \eta) - f(x, y)}{\eta}\, \frac{\eta}{\tau} +\right]. +\end{align*} + +But, by the Mean Value Theorem, +\begin{align*} +\{f(x + \xi, y + \eta) - f (x, y + \eta)\}/\xi + &= f_{x}'(x + \theta\xi, y + \eta),\\ +\{f(x, y + \eta) - f(x, y)\}/\eta + &= f_{y}'(x, y + \theta'\eta), +\end{align*} +where $\theta$~and~$\theta'$ each lie between $0$ and~$1$. As $\tau \to 0$, $\xi \to 0$ and +$\eta \to 0$, and $\xi/\tau \to \phi'(t)$, $\eta/\tau \to \psi'(t)$: also +\[ +f_{x}'(x + \theta\xi, y + \eta) \to f_{x}'(x, y),\quad +f_{y}'(x, y + \theta'\eta) \to f_{y}'(x, y). +\] +Hence +\[ +F'(t) = D_{t}f \{\phi(t), \psi(t)\} + = f_{x}'(x, y)\phi'(t) + f_{y}'(x, y)\psi'(t), +\] +where we are to put $x = \phi(t)$, $y = \psi(t)$ after carrying out the +differentiations with respect to $x$~and~$y$. This result may also be +expressed in the form +\[ +\frac{df}{dt} + = \frac{\dd f}{\dd x}\, \frac{dx}{dt} + + \frac{\dd f}{\dd y}\, \frac{dy}{dt}\Add{.} +\] + +\begin{Examples}{LXI.} +\Item{1.} Suppose $\phi(t) = (1 - t^{2})/(1 + t^{2})$, $\psi(t) = 2t/(1 + t^{2})$, so +that the locus of~$(x, y)$ is the circle $x^{2} + y^{2} = 1$. Then +\begin{align*} +\phi'(t) &= -4t/(1 + t^{2})^{2},\quad \psi'(t) = 2(1 - t^{2})/(1 + t^{2})^{2},\\ +F'(t) &= \{-4t/(1 + t^{2})^{2}\}f_{x}' + \{2(1 - t^{2})/(1 + t^{2})^{2}\}f_{y}', +\end{align*} +where $x$~and~$y$ are to be put equal to $(1 - t^{2})/(1 + t^{2})$ and $2t/(1 + t^{2})$ after +carrying out the differentiations. +\PageSep{278} + +{\Loosen We can easily verify this formula in particular cases. Suppose, \eg, +that $f(x, y) = x^{2} + y^{2}$. Then $f_{x}' = 2x$, $f_{y}' = 2y$, and it is easily verified that +$F'(t) = 2x\phi'(t) + 2y\psi'(t) = 0$, which is obviously correct, since $F(t) = 1$.} + +\Item{2.} Verify the theorem in the same way when (\ia)~$x = t^{m}$, $y = 1 - t^{m}$, +$f(x, y) = x + y$; (\ib)~$x = a\cos t$, $y = a\sin t$, $f(x, y) = x^{2} + y^{2}$. + +\Item{3.} One of the most important cases is that in which $t$ is $x$~itself. We +then obtain +\[ +D_{x}f\{x, \psi(x)\} = D_{x}f(x, y) + D_{y}f(x, y)\psi'(x). +\] +where $y$~is to be replaced by~$\psi(x)$ after differentiation. + +It was this case which led to the introduction of the notation $\dd f/\dd x$, $\dd f/\dd y$. +For it would seem natural to use the notation~$df/dx$ for \emph{either} of the functions +$D_{x}f\{x, \psi(x)\}$ and $D_{x}f(x, y)$, in one of which $y$~is put equal to~$\psi(x)$ before +and in the other after differentiation. Suppose for example that $y = 1 - x$ +and $f(x, y) = x + y$. Then $D_{x}f(x, 1 - x) = D_{x}1 = 0$, but $D_{x}f(x, y) = 1$. + +The distinction between the two functions is adequately shown by +denoting the first by~$df/dx$ and the second by~$\dd f/\dd x$, in which case the +theorem takes the form +\[ +\frac{df}{dx} = \frac{\dd f}{\dd x} + \frac{\dd f}{\dd y}\, \frac{dy}{dx}; +\] +though this notation is also open to objection, in that it is a little misleading +to denote the functions $f\{x, \psi(x)\}$ and $f(x, y)$, whose forms as functions of~$x$ +are quite different from one another, by the same letter~$f$ in $df/dx$ and~$\dd f/\dd x$. + +\Item{4.} If the result of eliminating~$t$ between $x = \phi(t)$, $y = \psi(t)$ is $f(x, y) = 0$, +then +\[ +\frac{\dd f}{\dd x}\, \frac{dx}{dt} + \frac{\dd f}{\dd y}\, \frac{dy}{dt} = 0. +\] + +\Item{5.} If $x$~and~$y$ are functions of~$t$, and $r$~and~$\theta$ are the polar coordinates of +$(x, y)$, then $r' = (xx' + yy')/r$, $\theta' = (xy' - yx')/r^{2}$, dashes denoting differentiations +with respect to~$t$. +\end{Examples} + +\Paragraph{154. The Mean Value Theorem for functions of two +variables.} Many of the results of the last chapter depended +upon the Mean Value Theorem, expressed by the equation +\[ +\phi(x + h) - \phi(x) = hf'(x + \theta h), +\] +or as it may be written, if $y = \phi(x)$, +\[ +\delta y = f'(x + \theta\, \delta x)\, \delta x. +\] + +Now suppose that $z = f(x, y)$ is a function of the two independent +variables $x$~and~$y$, and that $x$~and~$y$ receive increments +$h$,~$k$ or $\delta x$,~$\delta y$ respectively: and let us attempt to express the +corresponding increment of~$z$, viz. +\[ +\delta z = f(x + h, y + k) - f(x, y), +\] +in terms of $h$,~$k$ and the derivatives of~$z$ with respect to $x$~and~$y$. +\PageSep{279} + +Let $f(x + ht, y + kt) = F(t)$. Then +\[ +f(x + h, y + k) - f(x, y) = F(1) - F(0) = F'(\theta), +\] +where $0 < \theta < 1$. But, by \SecNo[§]{153}, +\begin{align*} +F' (t) + &= D_{t} f(x + ht, y + kt)\\ + &= hf_{x}'(x + ht, y + kt) + kf_{y}'(x + ht, y + kt). +\end{align*} +Hence finally +\[ +\delta z = f(x + h, y + k) - f(x, y) + = hf_{x}'(x + \theta h, y + \theta k) + kf_{y}'(x + \theta h, y + \theta k), +\] +which is the formula desired. Since $f_{x}'$,~$f_{y}'$ are supposed to be +continuous functions of $x$~and~$y$, we have +\begin{align*} +f_{x}'(x + \theta h, y + \theta k) &= f_{x}'(x, y) + \epsilon_{h, k},\\ +f_{y}'(x + \theta h, y + \theta k) &= f_{y}'(x, y) + \eta_{h, k}, +\end{align*} +where $\epsilon_{h, k}$ and~$\eta_{h, k}$ tend to zero as $h$~and~$k$ tend to zero. Hence +the theorem may be written in the form +\[ +\delta z = (f_{x}' + \epsilon)\, \delta x + (f_{y}' + \eta)\, \delta y, +\Tag{(1)} +\] +where $\epsilon$~and~$\eta$ are small when $\delta x$~and~$\delta y$ are small. + +The result embodied in~\Eq{(1)} may be expressed by saying that the +equation +\[ +\delta z = f_{x}'\, \delta x + f_{y}'\, \delta y +\] +is \emph{approximately} true; \ie\ that the difference between the two +sides of the equation is small in comparison with the larger of $\delta x$ +and~$\delta y$.\footnote + {Or with $|\delta x| + |\delta y|$ or $\sqrtp{\delta x^{2} + \delta y^{2}}$.} +We must say `\emph{the larger of $\delta x$~and~$\delta y$}' because one of +them might be small in comparison with the other; we might +indeed have $\delta x = 0$ or $\delta y = 0$. + +\begin{Remark} +It should be observed that if any equation of the form $\delta z = \lambda\, \delta x + \mu\, \delta y$ +is `approximately true' in this sense, we must have $\lambda = f_{x}'$, $\mu = f_{y}'$. For we +have +\[ +\delta z - f_{x}'\, \delta x - f_{y}'\, \delta y + = \epsilon\, \delta x + \eta\, \delta y,\quad +\delta z - \lambda\, \delta x - \mu\, \delta y + = \epsilon'\, \delta x + \eta'\, \delta y +\] +where $\epsilon$,~$\eta$, $\epsilon'$,~$\eta'$ all tend to zero as $\delta x$~and~$\delta y$ tend to zero; and so +\[ +(\lambda - f_{x}')\, \delta x + (\mu - f_{y}')\, \delta y + = \rho\, \delta x + \rho'\, \delta y +\] +where $\rho$~and~$\rho'$ tend to zero. Hence, if $\zeta$~is any assigned positive number, we +can choose~$\sigma$ so that +\[ +|(\lambda - f_{x}')\, \delta x + (\mu - f_{y}')\, \delta y| + < \zeta(|\delta x| + |\delta y|) +\] +for all values of $\delta x$ and~$\delta y$ numerically less than~$\sigma$. Taking $\delta y = 0$ we obtain +$|(\lambda - f_{x}')\, \delta x| < \zeta|\delta x|$, or $|\lambda - f_{x}'| < \zeta$, and, as $\zeta$~may be as small as we please, +this can only be the case if $\lambda = f_{x}'$. Similarly $\mu = f_{y}'$. +\end{Remark} +\PageSep{280} + +\Paragraph{155. Differentials.} In the applications of the Calculus, +especially in geometry, it is usually most convenient to work with +equations expressed not, like equation~\Eq{(1)} of \SecNo[§]{154}, in terms of the +increments $\delta x$,~$\delta y$,~$\delta z$ of the functions $x$,~$y$,~$z$, but in terms of what +are called their \emph{differentials} $dx$,~$dy$,~$dz$. + +Let us return for a moment to a function $y = f(x)$ of a single +variable~$x$. If $f'(x)$~is continuous then +\[ +\delta y = \{f'(x) + \epsilon\}\, \delta x, +\Tag{(1)} +\] +where $\epsilon \to 0$ as $\delta x \to 0$: in other words the equation +\[ +\delta y = f'(x)\, \delta x +\Tag{(2)} +\] +is `approximately' true. We have up to the present attributed +no meaning of any kind to the symbol~$dy$ standing by itself. We +now agree to \emph{define}~$dy$ by the equation +\[ +dy = f'(x)\, \delta x. +\Tag{(3)} +\] + +If we choose for~$y$ the particular function~$x$, we obtain +\[ +dx = \delta x, +\Tag{(4)} +\] +so that +\[ +dy = f'(x)\, dx. +\Tag{(5)} +\] +If we divide both sides of~\Eq{(5)} by~$dx$ we obtain +\[ +\frac{dy}{dx} = f'(x), +\Tag{(6)} +\] +where $dy/dx$ denotes not, as heretofore, the differential coefficient +of~$y$, but the quotient of the differentials $dy$,~$dx$. The symbol +$dy/dx$ thus acquires a double meaning; but there is no inconvenience +in this, since \Eq{(6)}~is true whichever meaning we choose. + +\begin{Remark} +The equation~\Eq{(5)} has two apparent advantages over~\Eq{(2)}. It is exact and +not merely approximate, and its truth does not depend on any assumption as +to the continuity of~$f'(x)$. On the other hand it is precisely the fact that we +can, under certain conditions, pass from the exact equation~\Eq{(5)} to the approximate +equation~\Eq{(2)}, which gives the former its importance. The advantages of +the `differential' notation are in reality of a purely technical character. These +technical advantages are however so great, especially when we come to deal +with functions of several variables, that the use of the notation is almost +inevitable. + +When $f'(x)$~is continuous, we have +\[ +\lim \frac{dy}{\delta y} = 1 +\] +when $\delta x \to 0$. This is sometimes expressed by saying that $dy$~is the \emph{principal +part} of~$\delta y$ when $\delta x$~is small, just as we might say that $ax$~is the `principal +part' of $ax + bx^{2}$ when $x$~is small. +\end{Remark} +\PageSep{281} + +We pass now to the corresponding definitions connected with +a function~$z$ of two independent variables $x$~and~$y$. We define the +differential~$dz$ by the equation +\[ +dz = f_{x}'\, \delta x + f_{y}'\, \delta y. +\Tag{(7)} +\] +Putting $z = x$ and $z = y$ in turn, we obtain +\begin{align*} +dx &= \delta x,\quad +dy = \delta y, +\Tag{(8)} +\intertext{so that} +dz &= f_{x}'\, dx + f_{y}'\, dy, +\Tag{(9)} +\end{align*} +which is the exact equation corresponding to the approximate +equation~\Eq{(1)} of \SecNo[§]{154}. Here again it is to be observed that the +former is of importance only for reasons of practical convenience +in working and because the latter can in certain circumstances be +deduced from it. + +\begin{Remark} +One property of the equation~\Eq{(9)} deserves special remark. We saw in +\SecNo[§]{153} that if $z = f(x, y)$, $x$~and~$y$ being not independent but functions of a +single variable~$t$, so that $z$~is also a function of $t$~alone, then +\[ +\frac{dz}{dt} + = \frac{\dd f}{\dd x}\, \frac{dx}{dt} + + \frac{\dd f}{\dd y}\, \frac{dy}{dt}. +\] +Multiplying this equation by~$dt$ and observing that +\[ +dx = \frac{dx}{dt}\, dt,\quad +dy = \frac{dy}{dt}\, dt,\quad +dz = \frac{dz}{dt}\, dt, +\] +we obtain +\[ +dz = f_{x}'\, dx + f_{y}'\, dy, +\] +which is the same in form as~\Eq{(9)}. Thus \emph{the formula which expresses~$dz$ in terms +of $dx$~and~$dy$ is the same whether the variables $x$~and~$y$ are independent or not}. +This remark is of great importance in applications. + +It should also be observed that if $z$~is a function of the two independent +variables $x$~and~$y$, and +\[ +dz = \lambda\, dx + \mu\, dy, +\] +then $\lambda = f_{x}'$, $\mu = f_{y}'$. This follows at once from the last paragraph of~\SecNo[§]{154}. + +It is obvious that the theorems and definitions of the last three sections +are capable of immediate extension to functions of any number of variables. +\end{Remark} + +\begin{Examples}{LXII.} +\Item{1.} The area of an ellipse is given by $A = \pi ab$, where +$a$,~$b$ are the semiaxes. Prove that +\[ +\frac{dA}{A} = \frac{da}{a} + \frac{db}{b}, +\] +and state the corresponding approximate equation connecting the increments +of the axes and the area. +\PageSep{282} + +\Item{2.} Express $\Delta$, the area of a triangle~$ABC$, as a function of (i)~$a$, $B$,~$C$, +(ii)~$A$, $b$,~$c$, and (iii)~$a$, $b$,~$c$, and establish the formulae +\begin{gather*} +\frac{d\Delta}{\Delta} + = 2\frac{da}{a} + \frac{c\, dB}{a\sin B} + \frac{b\, dC}{a\sin C},\quad +\frac{d\Delta}{\Delta} + = \cot A\, dA + \frac{db}{b} + \frac{dc}{c},\\ +d\Delta = R(\cos A\, da + \cos B\, db + \cos C\, dc), +\end{gather*} +%[** TN: Sole instance of circumcircle, not hyphenated in the original] +where $R$~is the radius of the circumcircle. + +\Item{3.} The sides of a triangle vary in such a way that the area remains +constant, so that $a$~may be regarded as a function of $b$~and~$c$. Prove that +\[ +\frac{\dd a}{\dd b} = -\frac{\cos B}{\cos A},\quad +\frac{\dd a}{\dd c} = -\frac{\cos C}{\cos A}. +\] + +[This follows from the equations +\[ +da = \frac{\dd a}{\dd b}\, db + \frac{\dd a}{\dd c}\, dc,\quad +\cos A\, da + \cos B\, db + \cos C\, dc = 0.\Add{]} +\] + +\Item{4.} If $a$,~$b$,~$c$ vary so that $R$~remains constant, then +\[ +\frac{da}{\cos A} + \frac{db}{\cos B} + \frac{dc}{\cos C} = 0, +\] +and so +\[ +\frac{\dd a}{\dd b} = -\frac{\cos A}{\cos B},\quad +\frac{\dd a}{\dd c} = -\frac{\cos A}{\cos C}. +\] + +[Use the formulae $a = 2R\sin A$,~\dots, and the facts that $R$ and $A + B + C$ are +constant.] + +\Item{5.} If $z$~is a function of $u$~and~$v$, which are functions of $x$~and~$y$, then +\[ +\frac{\dd z}{\dd x} = \frac{\dd z}{\dd u}\, \frac{\dd u}{\dd x} + + \frac{\dd z}{\dd v}\, \frac{\dd v}{\dd x},\quad +\frac{\dd z}{\dd y} = \frac{\dd z}{\dd u}\, \frac{\dd u}{\dd y} + + \frac{\dd z}{\dd v}\, \frac{\dd v}{\dd y}. +\] + +[We have +\[ +dz = \frac{\dd z}{\dd u}\, du + \frac{\dd z}{\dd v}\, dv,\quad +du = \frac{\dd u}{\dd x}\, dx + \frac{\dd u}{\dd y}\, dy,\quad +dv = \frac{\dd v}{\dd x}\, dx + \frac{\dd v}{\dd y}\, dy. +\] +Substitute for $du$~and~$dv$ in the first equation and compare the result with +the equation +\[ +dz = \frac{\dd z}{\dd x}\, dx + \frac{\dd z}{\dd y}\, dy.] +\] + +\Item{6.} Let $z$~be a function of $x$~and~$y$, and let $X$,~$Y$,~$Z$ be defined by the +equations +\[ +x = a_{1} X + b_{1} Y + c_{1} Z,\quad +y = a_{2} X + b_{2} Y + c_{2} Z,\quad +z = a_{3} X + b_{3} Y + c_{3} Z. +\] +Then $Z$~may be expressed as a function of $X$~and~$Y$. Express $\dd Z/\dd X$, +$\dd Z/\dd Y$ in terms of $\dd z/\dd x$, $\dd z/\dd y$. [Let these differential coefficients be denoted +by $P$,~$Q$ and $p$,~$q$. Then $dz - p\, dx - q\, dy = 0$, or +\[ +(c_{1} p + c_{2} q - c_{3})\, dZ + +(a_{1} p + a_{2} q - a_{3})\, dX + +(b_{1} p + b_{2} q - b_{3})\, dY = 0. +\] +\PageSep{283} +Comparing this equation with $dZ - P\, dX - Q\, dY = 0$ we see that +\[ +P = -\frac{a_{1}p + a_{2}q - a_{3}}{c_{1}p + c_{2}q - c_{3}},\quad +Q = -\frac{b_{1}p + b_{2}q - b_{3}}{c_{1}p + c_{2}q - c_{3}}.] +\] + +\Item{7.} If +\[ +(a_{1} x + b_{1} y + c_{1} z)p + (a_{2} x + b_{2} y + c_{2} z)q + = a_{3} x + b_{3} y + c_{3} z, +\] +then +\[ +(a_{1} X + b_{1} Y + c_{1} Z) P + (a_{2} X + b_{2} Y + c_{2} Z) Q + = a_{3} X + b_{3} Y + c_{3} Z. +\] +\MathTrip{1899.} + +\Item{8.} \Topic{Differentiation of implicit functions.} Suppose that $f(x, y)$ and its +derivative $f_{y}'(x, y)$ are continuous in the neighbourhood of the point $(a, b)$, +and that +\[ +f(a, b) = 0,\quad +f_{b}'(a, b) \neq 0. +\] +Then we can find a neighbourhood of~$(a, b)$ throughout which $f_{y}'(x, y)$ has +always the same sign. Let us suppose, for example, that $f_{y}'(x, y)$~is positive +near $(a, b)$. Then $f(x, y)$~is, for any value of~$x$ sufficiently near to~$a$, and for +values of~$y$ sufficiently near to~$b$, an increasing function of~$y$ in the stricter +sense of \SecNo[§]{95}. It follows, by the theorem of \SecNo[§]{108}, that there is a unique +continuous function~$y$ which is equal to~$b$ when $x = a$ and which satisfies the +equation $f(x, y) = 0$ for all values of~$x$ sufficiently near to~$a$. + +Let us now suppose that $f(x, y)$ possesses a derivative $f_{x}'(x, y)$ which is +also continuous near $(a, b)$. If $f(x, y) = 0$, $x = a + h$, $y = b + k$, we have +\[ +0 = f(x, y) - f(a, b) = (f_{a}' + \epsilon) h + (f_{b}' + \eta) k, +\] +where $\DPtypo{}{\epsilon}$ and~$\eta$ tend to zero with $h$~and~$k$. Thus +\[ +\frac{k}{h} = -\frac{f_{a}' + \epsilon}{f_{b}' + \eta} \to -\frac{f_{a}'}{f_{b}'}, +\] +or +\[ +\frac{dy}{dx} = -\frac{f_{a}'}{f_{b}'}. +\] + +\Item{9.} The equation of the tangent to the curve $f(x, y) = 0$, at the point +$x_{0}$,~$y_{0}$, is +\[ +(x - x_{0}) f_{x_{0}}'(x_{0}, y_{0}) + (y - y_{0}) f_{y_{0}}'(x_{0}, y_{0}) = 0. +\] +\end{Examples} + +\Paragraph{156. Definite Integrals and Areas.} It will be remembered +that, in \Ref{Ch.}{VI}, \SecNo[§]{145}, we assumed that, if $f(x)$~is a continuous +function of~$x$, and $PQ$~is the +%[Illustration: Fig. 47.] +\Figure[2.5in]{47}{p283} +graph of $y = f(x)$, then the +region~$PpqQ$ shown in \Fig{47} +has associated with it a definite +number which we call its \emph{area}. +It is clear that, if we denote $Op$~and~$Oq$ by $a$~and~$x$, and +allow $x$ to vary, this area is a +function of~$x$, which we denote +by~$F(x)$. +\PageSep{284} + +Making this assumption, we proved in \SecNo[§]{145} that $F'(x) = f(x)$, +and we showed how this result might be used in the calculation +of the areas of particular curves. But we have still to justify +the fundamental assumption that there is such a number as the +area~$F(x)$. + +We know indeed what is meant by the area of a \emph{rectangle}, +and that it is measured by the product of its sides. Also the +properties of triangles, parallelograms, and polygons proved by +Euclid enable us to attach a definite meaning to the areas of +such figures. But nothing which we know so far provides us with +a direct definition of the area of a figure bounded by curved lines. +We shall now show how to give a definition of~$F(x)$ which will +enable us to \emph{prove} its existence.\footnote + {The argument which follows is modelled on that given in Goursat's \textit{Cours + d'Analyse} (second edition), vol.~i, pp.~171~\textit{et~seq.}; but Goursat's treatment is much + more general.} + +Let us suppose $f(x)$ continuous throughout the interval~$\DPmod{(a, b)}{[a, b]}$, +and let us divide up the interval into a number of sub-intervals +by means of the points of division $x_{0}$,~$x_{1}$, $x_{2}$,~\dots, $x_{n}$, where +\[ +a = x_{0} < x_{1} < \dots < x_{n-1} < x_{n} = b. +\] +Further, let us denote by~$\delta_{\nu}$ the interval $\DPmod{(x_{\nu}, x_{\nu+1})}{[x_{\nu}, x_{\nu+1}]}$, and by~$m_{\nu}$ the +lower bound (\SecNo[§]{102}) of~$f(x)$ in~$\delta_{\nu}$, and let us write +\[ +%[** TN: Hardy now means the *length* of \delta_{\nu}] +s = m_{0}\delta_{0} + m_{1}\delta_{1} + \dots + m_{n}\delta_{n} + = \tsum m_{\nu}\delta_{\nu}, +\] +say. + +{\Loosen It is evident that, if $M$~is the upper bound of~$f(x)$ in~$\DPmod{(a, b)}{[a, b]}$, then +$s \leq M(b - a)$. The aggregate of values of~$s$ is therefore, in the +language of \SecNo[§]{80}, bounded above, and possesses an upper bound +which we will denote by~$j$. No value of~$s$ exceeds~$j$, but there are +values of~$s$ which exceed any number less than~$j$.} + +In the same way, if $M_{\nu}$~is the upper bound of~$f(x)$ in~$\delta_{\nu}$, we can +define the sum +\[ +S = \tsum M_{\nu}\delta_{\nu}. +\] + +{\Loosen It is evident that, if $m$~is the lower bound of~$f(x)$ in~$\DPmod{(a, b)}{[a, b]}$, then +$S \geq m(b - a)$. The aggregate of values of~$S$ is therefore bounded +below, and possesses a lower bound which we will denote by~$J$. +No value of~$S$ is less than~$J$, but there are values of~$S$ less than any +number greater than~$J$.} +\PageSep{285} + +\begin{Remark} +It will help to make clear the significance of the sums $s$ and~$S$ if +we observe that, in the simple case +in which $f(x)$~increases steadily +from $x = a$ to $x = b$, $m_{\nu}$~is $f(x_{\nu})$ +and $M_{\nu}$~is $f(x_{\nu+1})$. In this case $s$~is +the total area of the rectangles +shaded in \Fig{48}, and $S$~is the +%[Illustration: Fig. 48.] +\Figure[2.25in]{48}{p285} +area bounded by a thick line. In +general $s$ and~$S$ will still be areas, +composed of rectangles, respectively +included in and including the curvilinear +region whose area we are +trying to define. +\end{Remark} + +We shall now show that \emph{no +sum such as~$s$ can exceed any +sum such as~$S$.} Let $s$,~$S$ be the sums corresponding to one mode of +subdivision, and $s'$,~$S'$ those corresponding to another. We have +to show that $s \leq S'$ and $s' \leq S$. + +We can form a third mode of subdivision by taking as dividing +points all points which are such for either $s$,~$S$ or $s'$,~$S'$. Let $\mathbf{s}$,~$\mathbf{S}$ +be the sums corresponding to this third mode of subdivision. +Then it is easy to see that +\[ +\mathbf{s} \geq s,\quad \mathbf{s} \geq s',\quad +\mathbf{S} \leq S,\quad \mathbf{S} \leq S'. +\Tag{(1)} +\] +For example, $\mathbf{s}$~differs from~$s$ in that at least one interval~$\delta_{\nu}$ which +occurs in~$s$ is divided into a number of smaller intervals +\[ +\delta_{\nu, 1},\ \delta_{\nu, 2},\ \dots,\ \delta_{\nu, p}, +\] +so that a term $m_{\nu}\delta_{\nu}$ of~$s$ is replaced in~$\mathbf{s}$ by a sum +\[ +m_{\nu, 1}\delta_{\nu, 1} + m_{\nu, 2}\delta_{\nu, 2} + \dots + m_{\nu, p}\delta_{\nu, p}, +\] +where $m_{\nu, 1}$, $m_{\nu, 2}$,~\dots\ are the lower bounds of~$f(x)$ in $\delta_{\nu, 1}$, $\delta_{\nu, 2}$,~\dots. +But evidently $m_{\nu, 1} \geq m_{\nu}$, $m_{\nu, 2} \geq m_{\nu}$,~\dots, so that the sum just written +is not less than~$m_{\nu}\delta_{\nu}$. Hence $\mathbf{s} \geq s;$ and the other inequalities~\Eq{(1)} +can be established in the same way. But, since $\mathbf{s} \leq \mathbf{S}$, it follows +that +\[ +s \leq \mathbf{s} \leq \mathbf{S} \leq S', +\] +which is what we wanted to prove. + +It also follows that $j \leq J$. For we can find an~$s$ as near to~$j$ +as we please and an~$S$ as near to~$J$ as we please,\footnote + {The $s$ and the~$S$ do not in general correspond to the same mode of subdivision.} +and so $j > J$ +would involve the existence of an~$s$ and an~$S$ for which $s > S$. +\PageSep{286} + +So far we have made no use of the fact that $f(x)$~is continuous. +We shall now show that $j = J$, and that the sums $s$,~$S$ tend to the +limit~$J$ when the points of division~$x_{\nu}$ are multiplied indefinitely +in such a way that all the intervals~$\delta_{\nu}$ tend to zero. More precisely, +we shall show that, \begin{Result}given any positive number~$\epsilon$, it is possible +to find~$\delta$ so that +\[ +0 \leq J - s < \epsilon,\quad +0 \leq S - J < \epsilon +\] +whenever $\delta_{\nu} < \delta$ for all values of~$\nu$. +\end{Result} + +There is, by Theorem~II of \SecNo[§]{106}, a number~$\delta$ such that +\[ +M_{\nu} - m_{\nu} < \epsilon/(b - a), +\] +whenever every~$\delta_{\nu}$ is less than~$\delta$. Hence +\[ +S - s = \tsum (M_{\nu} - m_{\nu})\, \delta_{\nu} < \epsilon. +\] +But +\[ +S - s = (S - J) + (J - j) + (j - s); +\] +and all the three terms on the right-hand side are positive, and +therefore all less than~$\epsilon$. As $J - j$~is a constant, it must be zero. +Hence $j = J$ and $0 \leq j - s < \epsilon$, $0 \leq S - J < \epsilon$, as was to be proved. + +We define the area of~$PpqQ$ as being \emph{the common limit of $s$~and~$S$, +that is to say~$J$}. It is easy to give a more general form to this +definition. Consider the sum +\[ +\sigma = \tsum f_{\nu}\delta_{\nu} +\] +where $f_{\nu}$~denotes the value of~$f(x)$ at any point in~$\delta_{\nu}$. Then $\sigma$ +plainly lies between $s$~and~$S$, and so tends to the limit~$J$ when the +intervals~$\delta_{\nu}$ tend to zero. We may therefore define the area as +the limit of~$\sigma$. + +\Paragraph{157. The definite integral.} Let us now suppose that $f(x)$~is +a continuous function, so that the region bounded by the curve +$y = f(x)$, the ordinates $x = a$ and $x = b$, and the axis of~$x$, has a +definite area. We proved in \Ref{Ch.}{VI}, \SecNo[§]{145}, that if $F(x)$~is an +`integral function' of~$f(x)$, \ie\ if +\[ +F'(x) = f(x),\quad +F(x) = \int f(x)\, dx, +\] +then the area in question is $F(b) - F(a)$. + +As it is not always practicable actually to determine the form +of~$F(x)$, it is convenient to have a formula which represents the +area~$PpqQ$ and contains no explicit reference to~$F(x)$. We shall +write +\[ +(PpqQ) = \int_{a}^{b} f(x)\, dx. +\] +\PageSep{287} + +The expression on the right-hand side of this equation may +then be regarded as being defined in either of two ways. We +may regard it as simply an abbreviation for $F(b) - F(a)$, where +$F(x)$~is some integral function of~$f(x)$, whether an actual formula +expressing it is known or not; or we may regard it as the value of +the area~$PpqQ$, as directly defined in~\SecNo[§]{156}. + +The number +\[ +\int_{a}^{b} f(x)\, dx +\] +is called a \Emph{definite integral}; $a$~and~$b$ are called its \Emph{lower and +upper limits}; $f(x)$~is called the \Emph{subject of integration} or +\Emph{integrand}; and the interval~$\DPmod{(a, b)}{[a, b]}$ the \Emph{range of integration}. +The definite integral depends on $a$~and~$b$ and the form of the +function~$f(x)$ only, and is not a function of~$x$. On the other hand +the integral function +\[ +F(x) = \int f(x)\, dx +\] +is sometimes called the \Emph{indefinite integral} of~$f(x)$. + +\begin{Remark} +The distinction between the definite and the indefinite integral is merely +one of point of view. The definite integral $\ds\int_{a}^{b} f(x)\, dx = F(b) - F(a)$ is a +function of~$b$, and may be regarded as a particular integral function of~$f(b)$. +On the other hand the indefinite integral~$F(x)$ can always be expressed by +means of a definite integral, since +\[ +F(x) = F(a) + \int_{a}^{x} f(t)\, dt. +\] + +But when we are considering `indefinite integrals' or `integral functions' +we are usually thinking of \emph{a relation between two functions}, in virtue of which +one is the derivative of the other. And when we are considering a `definite +integral' we are not as a rule concerned with any possible variation of the +limits. Usually the limits are constants such as $0$ and~$1$; and +\[ +\int_{0}^{1} f(x)\, dx = F(1) - F(0) +\] +is not a function at all, but a mere number. + +It should be observed that the integral $\ds\int_{a}^{x} f(t)\, dt$, having a differential +coefficient~$f(x)$, is \textit{a~fortiori} a continuous function of~$x$. + +Since $1/x$~is continuous for all positive values of~$x$, the investigations of +the preceding paragraphs supply us with a proof of the actual existence of the +function~$\log x$, which we agreed to assume provisionally in~\SecNo[§]{128}. +\end{Remark} +\PageSep{288} + +\Paragraph{158. Area of a sector of a circle. The circular functions.} +The theory of the trigonometrical functions $\cos x$, $\sin x$, etc., as +usually presented in text-books of elementary trigonometry, rests +on an unproved assumption. An \emph{angle} is the configuration formed +by two straight lines $OA$,~$OP$; there is no particular difficulty in +translating this `geometrical' definition into purely analytical +terms. The assumption comes at the next stage, when it is assumed +that \emph{angles are capable of numerical measurement}, that is to say +%[Illustration: Fig. 49.] +\Figure[2in]{49}{p288} +that there is a real number~$x$ associated +with the configuration, just as there is +a real number associated with the region~$PpqQ$ +of \Fig{47}. This point once admitted, +$\cos x$ and $\sin x$ may be defined +in the ordinary way, and there is no +further difficulty of principle in the +elaboration of the theory. The whole +difficulty lies in the question, \emph{what is the~$x$ +which occurs in $\cos x$ and $\sin x$}? To answer this question, we +must define the measure of an angle, and we are now in a position +to do so. The most natural definition would be this: suppose that +$AP$~is an arc of a circle whose centre is~$O$ and whose radius is +unity, so that $OA = OP = 1$. Then $x$, the measure of the angle, is +\emph{the length of the arc~$AP$}. This is, in substance, the definition +adopted in the text-books, in the accounts which they give of the +theory of `circular measure'. It has however, for our present purpose, +a fatal defect; for we have not proved that the arc of a curve, +even of a circle, possesses a length. The notion of the length of a +curve is capable of precise mathematical analysis just as much as +that of an area; but the analysis, although of the same general +character as that of the preceding sections, is decidedly more +difficult, and it is impossible that we should give any general +treatment of the subject here. + +We must therefore found our definition on the notion not of +length but of \emph{area}. We define the measure of the angle~$AOP$ as +\emph{twice the area of the sector~$AOP$ of the unit circle}. + +Suppose, in particular, that $OA$~is $y = 0$ and that $OP$~is $y = mx$, +where $m > 0$. The area is a function of~$m$, which we may denote +by~$\phi(m)$. If we write~$\mu$ for $(1 + m^{2})^{-\frac{1}{2}}$, $P$~is the point $(\mu, m\mu)$, and +\PageSep{289} +we have +\[ +\phi(m) = \tfrac{1}{2} m\mu^{2} + \int_{\mu}^{1} \sqrtp{1 - x^{2}}\, dx. +\] +Differentiating with respect to~$m$, we find +\[ +\phi'(m) = \frac{1}{2(1 + m^{2})},\quad +\phi(m) = \tfrac{1}{2} \int_{0}^{m} \frac{dt}{1 + t^{2}}. +\] +Thus the analytical equivalent of our definition would be to define +$\arctan m$ by the equation +\[ +\arctan m = \int_{0}^{m} \frac{dt}{1 + t^{2}}; +\] +and the whole theory of the circular functions could be worked out +from this starting point, just as the theory of the logarithm is +worked out from a similar definition in \Ref{Ch.}{IX}\@. See \Ref{Appendix}{III}\@. + +\begin{Examples}{LXIII.} \Topic{Calculation of the definite from the indefinite +integral.} +\Item{1.} Show that +\[ +\int_{a}^{b} x^{n}\, dx = \frac{b^{n+1} - a^{n+1}}{n + 1}, +\] +and in particular that +\[ +\int_{0}^{1} x^{n}\, dx = \frac{1}{n + 1}. +\] + +\Item{2.} $\ds\int_{a}^{b} \cos mx\, dx = \frac{\sin mb - \sin ma}{m}$, +$\ds\int_{a}^{b} \sin mx\, dx = \frac{\cos ma - \cos mb}{m}$. + +\Item{3.} $\ds\int_{a}^{b}\frac{dx}{1 + x^{2}} = \arctan b - \arctan a$, +$\ds\int_{0}^{1}\frac{dx}{1 + x^{2}} = \tfrac{1}{4}\pi$. + +[There is an apparent difficulty here owing to the fact that $\arctan x$~is a +many valued function. The difficulty may be avoided by observing that, in +the equation +\[ +\int_{0}^{x} \frac{dt}{1 + t^{2}} = \arctan x, +\] +$\arctan x$ must denote an angle lying between $-\frac{1}{2}\pi$ and~$\frac{1}{2}\pi$. For the integral +vanishes when $x = 0$ and increases steadily and continuously as $x$~increases. +Thus the same is true of~$\arctan x$, which therefore tends to~$\tfrac{1}{2}\pi$ as $x \to \infty$. +In the same way we can show that $\arctan x \to -\frac{1}{2}\pi$ as $x \to -\infty$. Similarly, +in the equation +\[ +\int_{0}^{x} \frac{dt}{\sqrtp{1 - t^{2}}} = \arcsin x, +\] +where $-1 < x < 1$, $\arcsin x$ denotes an angle lying between $-\frac{1}{2}\pi$ and $\frac{1}{2}\pi$. +Thus, if $a$~and~$b$ are both numerically less than unity, we have +\[ +\int_{a}^{b} \frac{dx}{\sqrtp{1 - x^{2}}} = \arcsin b - \arcsin a.] +\] + +\Item{4.} $\ds\int_{0}^{1} \frac{dx}{1 - x + x^{2}} = \frac{2\pi}{3\sqrt3}$, +$\ds\int_{0}^{1} \frac{dx}{1 + x + x^{2}} = \frac{\pi}{3\sqrt3}$\Add{.} +\PageSep{290} + +\Item{5.} $\ds\int_{0}^{1} \frac{dx}{1 + 2x\cos\alpha + x^{2}} = \frac{\alpha}{2\sin\alpha}$ if $-\pi < \alpha < \pi$, except when $\alpha = 0$, when the +value of the integral is~$\frac{1}{2}$, which is the limit of~$\frac{1}{2}\alpha\cosec\alpha$ as $\alpha \to 0$. + +\Item{6.} $\ds\int_{0}^{\DPtypo{}{1}} \sqrtp{1 - x^{2}}\, dx = \tfrac{1}{4}\pi$, +$\ds\int_{0}^{a} \sqrtp{a^{2} - x^{2}}\, dx = \tfrac{1}{4}\pi a^{2}$\quad $(a > 0)$. + +\Item{7.} $\ds\int_{0}^{\pi} \frac{dx}{a + b\cos x} = \frac{\pi}{\sqrt{a^{2} - b^{2}}}$, if $a > |b|$. [For the form of the indefinite +integral see \Exs{liii}.\ 3,~4. If $|a| < |b|$ then the subject of integration has an +infinity between $0$ and~$\pi$. What is the value of the integral when $a$~is +negative and $-a > |b|$?] + +\Item{8.} $\ds\int_{0}^{\frac{1}{2}\pi} \frac{dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x} = \frac{\pi}{2ab}$, if $a$~and~$b$ are positive. What is the +value of the integral when $a$~and~$b$ have opposite signs, or when both are +negative? + +\Item{9.} \Topic{Fourier's integrals.} Prove that if $m$~and~$n$ are positive integers then +\[ +\int_{0}^{2\pi} \cos mx \sin nx\, dx +\] +is always equal to zero, and +\[ +\int_{0}^{2\pi} \cos mx \cos nx\, dx,\quad +\int_{0}^{2\pi} \sin mx \sin nx\, dx +\] +are equal to zero unless $m = n$, when each is equal to~$\pi$. + +\Item{10.} Prove that $\ds\int_{0}^{\pi} \cos mx \cos nx\, dx$ and $\ds\int_{0}^{\pi} \sin mx \sin nx\, dx$ are each equal +to zero except when $m = n$, when each is equal to~$\frac{1}{2}\pi$; and that +\[ +\int_{0}^{\pi} \cos mx \sin nx\, dx = \frac{2n}{n^{2} - m^{2}},\quad +\int_{0}^{\pi} \cos mx \sin nx\, dx = 0, +\] +according as $n - m$~is odd or even. +\end{Examples} + +\Paragraph{159. Calculation of the definite integral from its definition +as the limit of a sum.} In a few cases we can evaluate a +definite integral by direct calculation, starting from the definitions +of \SecNo[§§]{156}~and~\SecNo{157}. As a rule it is much simpler to use the +indefinite integral, but the reader will find it instructive to work +through a few examples. + +\begin{Examples}{LXIV.} +\Item{1.} Evaluate $\ds\int_{a}^{b} x\, dx$ by dividing $\DPmod{(a, b)}{[a, b]}$ into $n$~equal +parts by the points of division $a = x_{0}$, $x_{1}$, $x_{2}$,~\dots, $x_{n} = b$, and calculating the +limit as $n \to \infty$ of +\[ +(x_{1} - x_{0})f(x_{0}) + (x_{2} - x_{1})f(x_{1}) + \dots + (x_{n} - x_{n-1})f(x_{n-1}). +\] +\PageSep{291} + +[This sum is +\begin{gather*} +\frac{b - a}{n}\left[ + a + \left(a + \frac{b - a}{n}\right) + \left(a + 2\frac{b - a}{n}\right) + + \dots + \left\{a + (n - 1)\frac{b - a}{n}\right\} + \right]\\ + = \frac{b - a}{n}\left[ + na + \frac{b - a}{n} \{1 + 2 + \dots + (n - 1)\} + \right] + = (b - a)\left\{a + (b - a)\frac{n(n - 1)}{2n^{2}}\right\}, +\end{gather*} +which tends to the limit $\frac{1}{2} (b^{2} - a^{2})$ as $n \to \infty$. Verify the result by graphical +reasoning.] + +\Item{2.} Calculate $\ds\int_{a}^{b} x^{2}\, dx$ in the same way. + +\Item{3.} Calculate $\ds\int_{a}^{b} x\, dx$, where $0 < a < b$, by dividing $\DPmod{(a, b)}{[a, b]}$ into $n$~parts by +the points of division $a$, $ar$, $ar^{2}$,~\dots\Add{,} $ar^{n-1}$, $ar^{n}$, where $r^{n} = b/a$. Apply the same +method to the more general integral $\ds\int_{a}^{b} x^{m}\, dx$. + +\Item{4.} Calculate $\ds\int_{a}^{b}\cos mx\, dx$ and $\ds\int_{a}^{b}\sin mx\, dx$ by the method of Ex.~1. + +\Item{5.} Prove that $n\sum\limits_{r=0}^{n-1} \dfrac{1}{n^{2} + r^{2}} \to \tfrac{1}{4}\pi$ as $n \to \infty$. + +[This follows from the fact that +\[ +\frac{n}{n^{2}} + \frac{n}{n^{2} + 1^{2}} + \dots + \frac{n}{n^{2} + (n - 1)^{2}} + = \sum_{r=0}^{n-1} \frac{(1/n)}{1 + (r/n)^{2}}, +\] +which tends to the limit $\ds\int_{0}^{1} \frac{dx}{1 + x^{2}}$ as $n \to \infty$, in virtue of the direct definition +of the integral.] + +\Item{6.} Prove that $\dfrac{1}{n^{2}} \sum\limits_{r=0}^{n-1} \sqrtp{n^{2} - r^{2}} \to \tfrac{1}{4}\pi$. +[The limit is $\ds\int_{0}^{1} \sqrtp{1 - x^{2}}\, dx$.] +\end{Examples} + +\Paragraph{160. General properties of the definite integral.} The +definite integral possesses the important properties expressed +by the following equations.\footnote + {All functions mentioned in these equations are of course continuous, as the + definite integral has been defined for continuous functions only.} +\CenterLine{\Item{(1)}}{$\ds\int_{a}^{b} f(x)\, dx = -\int_{b}^{a} f(x)\, dx$.} + +This follows at once from the definition of the integral by means of the +integral function~$F(x)$, since $F(b) - F(a) = -\{F(a) - F(b)\}$. It should be +observed that in the direct definition it was presupposed that the upper +limit is greater than the lower; thus this method of definition does +not apply to the integral $\ds\int_{b}^{a} f(x)\, dx$ when $a < b$. If we adopt this definition +as fundamental we must extend it to such cases by regarding the equation~\Eq{(1)} +as a definition of its right-hand side. +\PageSep{292} +\CenterLine{\Item{(2)}}{$\ds\int_{a}^{a} f(x)\, dx = 0$.} +\CenterLine{\Item{(3)}} + {$\ds\int_{a}^{b}f(x)\, dx + \int_{b}^{c}f(x)\, dx = \int_{a}^{c}f(x)\, dx$.} +\CenterLine{\Item{(4)}} + {$\ds\int_{a}^{b}kf(x)\, dx = k \int_{a}^{b}f(x)\, dx$.} +\CenterLine{\Item{(5)}}{$\ds\int_{a}^{b}\{f(x) + \phi(x)\}\, dx += \int_{a}^{b}f(x)\, dx + \int_{a}^{b}\phi(x)\, dx$.} + +\begin{Remark} +The reader will find it an instructive exercise to write out formal proofs +of these properties, in each case giving a proof starting from ($\alpha$)~the definition +by means of the integral function and ($\beta$)~the direct definition. +\end{Remark} + +The following theorems are also important. + +\begin{Result} +\Item{(6)} If $f(x) \geq 0$ when $a \leq x \leq b$, then $\ds\int_{a}^{b}f(x)\, dx \geq 0$. +\end{Result} + +\begin{Remark} +We have only to observe that the sum~$s$ of \SecNo[§]{156} cannot be negative. It +will be shown later (\MiscEx{VII}~41) that the value of the integral cannot be +zero unless $f(x)$~is always equal to zero: this may also be deduced from the +second corollary of~\SecNo[§]{121}. +\end{Remark} + +\begin{Result} +\Item{(7)} If $H \leq f(x) \leq K$ when $a \leq x \leq b$, then +\[ +H(b - a) \leq \int_{a}^{b}f(x)\, dx \leq K(b - a). +\] +\end{Result} + +\begin{Remark} +This follows at once if we apply~(6) to $f(x) - H$ and $K - f(x)$. +\end{Remark} + +\begin{Result} +\CenterLine{\Item{(8)}}{$\ds\int_{a}^{b}f(x)\, dx = (b-a)f(\xi)$,} +where $\xi$ lies between $a$ and~$b$. +\end{Result} + +\begin{Remark} +This follows from~(7). For we can take $H$ to be the least and $K$~the +greatest value of~$f(x)$ in~$\DPmod{(a, b)}{[a, b]}$. Then the integral is equal to~$\eta(b - a)$, where +$\eta$~lies between $H$ and~$K$. But, since $f(x)$~is continuous, there must be a +value of~$\xi$ for which $f(\xi) = \eta$~(\SecNo[§]{100}). + +If $F(x)$~is the integral function, we can write the result of~(8) in the form +\[ +F(b) - F(a) = (b - a)F'(\xi), +\] +so that (8)~appears now to be only another way of stating the Mean Value +Theorem of \SecNo[§]{125}. We may call~(8) the \Emph{First Mean Value Theorem for +Integrals}. +\end{Remark} +\PageSep{293} + +\begin{Result} +\Item{(9)} \Topic{The Generalised Mean Value Theorem for integrals.} +If $\phi(x)$~is positive, and $H$ and~$K$ are defined as in~\Eq{(7)}, then +\[ +H\int_{a}^{b} \phi(x)\, dx + \leq \int_{a}^{b} f(x)\phi(x)\, dx + \leq K\int_{a}^{b} \phi(x)\, dx; +\] +and +\[ +\int_{a}^{b} f(x)\phi(x)\, dx = f(\xi) \int_{a}^{b} \phi(x)\, dx, +\] +where $\xi$~is defined as in~\Eq{(8)}. +\end{Result} + +\begin{Remark} +This follows at once by applying Theorem~\Eq{(6)} to the integrals +\[ +\int_{a}^{b} \{f(x) - H\}\phi(x)\, dx,\quad +\int_{a}^{b} \{K - f(x)\}\phi(x)\, dx. +\] +The reader should formulate for himself the corresponding result which +holds when $\phi(x)$~is always negative. +\end{Remark} + +\begin{Result} +\Itemp{(10)} \Topic{The Fundamental Theorem of the Integral Calculus.} +The function +\[ +F(x) = \int_{a}^{x} f(t)\, dt +\] +has a derivative equal to $f(x)$. +\end{Result} + +This has been proved already in \SecNo[§]{145}, but it is convenient to +restate the result here as a formal theorem. It follows as a +corollary, as was pointed out in \SecNo[§]{157}, that \emph{$F(x)$~is a continuous +function of~$x$}. + +\begin{Examples}{LXV.} +\Item{1.} Show, by means of the direct definition of the +definite integral, and equations \Eq{(1)}--\Eq{(5)} above, that +\CenterLine{\Itemp{(i)}}{$\ds\int_{-a}^{a} \phi(x^{2})\, dx = 2\int_{0}^{a} \phi(x^{2})\, dx$,\quad +$\ds\int_{-a}^{a} x\phi(x^{2})\, dx = 0$;} +% +\CenterLine{\Itemp{(ii)}}{$\ds\int_{0}^{\frac{1}{2}\pi} \phi(\cos x)\, dx += \int_{0}^{\frac{1}{2} \pi} \phi(\sin x)\, dx += \tfrac{1}{2} \int_{0}^{\pi} \phi(\sin x)\, dx$;} +% +\CenterLine{\Itemp{(iii)}}{$\ds\int_{0}^{m\pi} \phi(\cos^{2} x)\, dx = m\int_{0}^{\pi} \phi(\cos^{2} x)\, dx$,} +$m$~being an integer. [The truth of these equations will appear geometrically +intuitive, if the graphs of the functions under the sign of integration are +sketched.] + +\Item{2.} Prove that $\ds\int_{0}^{\pi} \frac{\sin nx}{\sin x}\, dx$ is equal to~$\pi$ or to~$0$ according as $n$~is odd or +or even. [Use the formula $(\sin nx)/(\sin x) = 2\cos \{(n - 1)x\} + 2\cos \{(n - 3)x\} + \dots$, +the last term being $1$~or $2\cos x$.] + +\Item{3.} Prove that $\ds\int_{0}^{\pi} \sin nx \cot x\, dx$ is equal to~$0$ or to~$\pi$ according as $n$~is odd +or even. +\PageSep{294} + +\Item{4.} If $\phi(x) = a_{0} + a_{1}\cos x + b_{1}\sin x + a_{2}\cos 2x + \dots + a_{n}\cos nx + b_{n}\sin nx$, +and $k$~is a positive integer not greater than~$n$, then +\[ +\int_{0}^{2\pi} \phi(x)\, dx = 2\pi a_{0},\quad +\int_{0}^{2\pi} \cos kx \phi(x)\, dx = \pi a_{k},\quad +\int_{0}^{2\pi} \sin kx \phi(x)\, dx = \pi b_{k}. +\] +If $k > n$ then the value of each of the last two integrals is zero. [Use +\Ex{lxiii}.~9.] + +\Item{5.} If $\phi(x) = a_{0} + a_{1} \cos x + a_{2}\cos 2x + \dots + a_{n}\cos nx$, and $k$~is a positive +integer not greater than~$n$, then +\[ +\int_{0}^{\pi} \phi(x)\, dx = \pi a_{0},\quad +\int_{0}^{\pi} \cos kx \phi(x)\, dx = \tfrac{1}{2}\pi a_{k}. +\] +If $k > n$ then the value of the last integral is zero. [Use \Ex{lxiii}.~10.] + +\Item{6.} Prove that if $a$ and~$b$ are positive then +\[ +%[** TN: In-line in the original] +\int_{0}^{2\pi} \frac{dx}{a^{2}\cos^{2} x + b^{2}\sin^{2} x} = \frac{2\pi}{ab}. +\] + +%[** TN: No paragraph break in the original] +[Use \Ex{lxiii}.~8 and Ex.~1 above.] + +\Item{7.} If $f(x) \leq \phi(x)$ when $a \leq x \leq b$, then +$\ds\int_{a}^{b} f\, dx \leq \int_{a}^{b}\phi\, dx$. + +\Item{8.} Prove that +\begin{alignat*}{2} +%[** TN: Set on one line in the original] +0 &< \int_{0}^{\frac{1}{2}\pi} \sin^{n+1}x\, dx + &&< \int_{0}^{\frac{1}{2}\pi} \sin^{n}x\, dx,\\ +0 &< \int_{0}^{\frac{1}{4}\pi} \tan^{n+1}x\, dx + &&< \int_{0}^{\frac{1}{4}\pi} \tan^{n}x\, dx. +\end{alignat*} + +\Item{9.\footnotemark} If $n > 1$ then +\[ +%[** TN: In-line in the original] +.5 < \int_{0}^{\frac{1}{2}} \frac{dx}{\sqrtp{1 - x^{2n}}} < .524. +\] +\footnotetext{Exs.~9--13 are taken from Prof.\ Gibson's \textit{Elementary Treatise on the Calculus}.}% + +[The first inequality follows +from the fact that $\sqrtp{1 - x^{2n}} < 1$, the second from the fact that +$\sqrtp{1 - x^{2n}} > \sqrtp{1 - x^{2}}$.] %[** TN: Displayed in the original] + +\Item{10.} Prove that +\[ +%[** TN: In-line in the original] +\tfrac{1}{2} < \int_{0}^{1} \frac{dx}{\sqrtp{4 - x^{2} + x^{3}}} + < \tfrac{1}{6}\pi. +\] + +\Item{11.} Prove that $(3x + 8)/16 < 1/\sqrtp{4 - 3x + x^{3}} < 1/\sqrtp{4 - 3x}$ if $0 < x < 1$, +and hence that +\[ +%[** TN: In-line in the original] +\tfrac{19}{32} < \int_{0}^{1} \frac{dx}{\sqrtp{4 - 3x + x^{3}}} < \tfrac{2}{3}. +\] + +\Item{12.} Prove that +\[ +%[** TN: In-line in the original] +.573 < \int_{1}^{2} \frac{dx}{\sqrtp{4 - 3x + x^{3}}} < .595. +\] + +[Put $x = 1 + u$: then replace +$2 + 3u^{2} + u^{3}$ by $2 + 4u^{2}$ and by $2 + 3u^{2}$.] + +\Item{13.} If $\alpha$~and~$\phi$ are positive acute angles then +\[ +\phi < \int_{0}^{\phi} \frac{dx}{\sqrtp{1 - \sin^{2}\alpha \sin^{2} x}} + < \frac{\phi}{\sqrtp{1 - \sin^{2}\alpha \sin^{2}\phi}}. +\] +If $\alpha = \phi = \frac{1}{6}\pi$, then the integral lies between $.523$ and~$.541$. + +\Item{14.} Prove that +\[ +%[** TN: In-line in the original] +\left|\int_{a}^{b} f(x)\, dx\right| \leq \int_{a}^{b}|f(x)|\, dx. +\] + +[If $\sigma$~is the sum considered at the end of \SecNo[§]{156}, and $\sigma'$~the corresponding +sum formed from the function~$|f(x)|$, then $|\sigma| \leq \sigma'$.] + +%[** TN: Left vertical bar around integral missing in original] +\Item{15.} If $|f(x)| \leq M$, then +\[ +%[** TN: In-line in the original] +\left|\int_{a}^{b} f(x)\phi(x)\, dx\right| \leq M\int_{a}^{b}|\phi(x)|\, dx. +\] +\end{Examples} +\PageSep{295} + +\Paragraph{161. Integration by parts and by substitution.} It +follows from \SecNo[§]{138} that +\[ +\int_{a}^{b} f(x)\phi'(x)\, dx + = f(b)\phi(b) - f(a)\phi(a) - \int_{a}^{b} f'(x)\phi(x)\, dx. +\] +This formula is known as the formula for \Emph{integration of a +definite integral by parts}. + +Again, we know (\SecNo[§]{133}) that if $F(t)$~is the integral function of~$f(t)$, +then +\[ +\int f\{\phi(x)\}\phi'(x)\, dx = F\{\phi(x)\}. +\] +Hence, if $\phi(a) = c$, $\phi(b) = d$, we have +\[ +\int_{c}^{d} f(t)\, dt + = F(d) - F(c) + = F\{\phi(b)\} - F\{\phi(a)\} + = \int_{a}^{b} f\{\phi(x)\}\phi'(x)\, dx; +\] +which is the formula for the transformation of a definite integral +by \Emph{substitution}. + +The formulae for integration by parts and for transformation +often enable us to evaluate a definite integral without the labour +of actually finding the integral function of the subject of integration, +and sometimes even when the integral function cannot be +found. Some instances of this will be found in the following +examples. That the value of a definite integral may sometimes +be found without a knowledge of the integral function is only to +be expected, for the fact that we cannot determine the general +form of a function~$F(x)$ in no way precludes the possibility that +we may be able to determine the difference $F(b) - F(a)$ between +two of its particular values. But as a rule this can only be +effected by the use of more advanced methods than are at +present at our disposal. + +\begin{Examples}{LXVI.} +\Item{1.} Prove that +\[ +\int_{a}^{b} x f''(x)\, dx = \{bf'(b) - f(b)\} - \{af'(a) - f(a)\}. +\] + +\Item{2.} More generally, +\[ +\int_{a}^{b} x^{m} f^{(m+1)}(x)\, dx = F(b) - F(a), +\] +where +\begin{multline*} +%[** TN: Set on one line in the original] +F(x) = x^{m} f^{(m)}(x) + - mx^{m-1} f^{(m-1)}\DPtypo{x}{(x)} \\ + + m(m - 1)x^{m-2} f^{(m-2)}\DPtypo{x}{(x)} - \dots + + (-1)^{m} m!\, f(x). +\end{multline*} + +\Item{3.} Prove that +\[ +\int_{0}^{1} \arcsin x\, dx = \tfrac{1}{2}\pi - 1,\quad +\int_{0}^{1}x\arctan x\, dx = \tfrac{1}{4}\pi - \tfrac{1}{2}. +\] +\PageSep{296} + +\Item{4.} Prove that if $a$~and~$b$ are positive then +\[ +\int_{0}^{\frac{1}{2}\pi} + \frac{x\cos x\sin x\, dx}{(a^{2}\cos^{2}x + b^{2}\sin^{2}x)^{2}} + = \frac{\pi}{4ab^{2}(a + b)}. +\] + +[Integrate by parts and use \Ex{lxiii}.~8.] + +\Item{5.} If +\[ +f_{1}(x) = \int_{0}^{x}f(t)\, dt,\quad +f_{2}(x) = \int_{0}^{x}f_{1}(t)\, dt,\ \dots,\quad +f_{k}(x) = \int_{0}^{x} f_{k-1}(t)\, dt, +\] +then +\[ +f_{k}(x) = \frac{1}{(k - 1)!} \int_{0}^{x} f(t)(x - t)^{k-1}\, dt. +\] + +[Integrate repeatedly by parts.] + +\Item{6.} Prove by integration by parts that if +\[ +%[** TN: In-line in the original] +u_{m, n} = \int_{0}^{1} x^{m} (1 - x)^{n}\, dx, +\] +where $m$~and~$n$ are positive integers, then $(m + n + 1) u_{m, n} = nu_{m, n-1}$, and deduce that +\[ +u_{m, n} = \frac{m!\, n!}{(m + n + 1)!}. +\] + +\Item{7.} Prove that if +%[** TN: In-line in the original] +\[ +u_{n} = \int_{0}^{\frac{1}{4}\pi} \tan^{n}x\, dx +\] +then $u_{n} + u_{n-2} = 1/(n - 1)$. Hence +evaluate the integral for all positive integral values of~$n$. + +[Put $\tan^{n}x = \tan^{n-2}x(\sec^{2}x - 1)$ and integrate by parts.] + +\Item{8.} Deduce from the last example that $u_{n}$~lies between $1/\{2(n - 1)\}$ and +$1/\{2(n + 1)\}$. + +\Item{9.} Prove that if +\[ +%[** TN: In-line in the original] +u_{n} = \int_{0}^{\frac{1}{2}\pi} \sin^{n} x\, dx +\] +then $u_{n} = \{(n - 1)/n\} u_{n-2}$. [Write +$\sin^{n-1}x\sin x$ for $\sin^{n}x$ and integrate by parts.] + +\Item{10.} Deduce that $u_{n}$~is equal to +\[ +\frac{2·4·6 \dots (n - 1)}{3·5·7 \dots n},\quad +\tfrac{1}{2}\pi \frac{1·3·5 \dots (n - 1)}{2·4·6 \dots n}, +\] +according as $n$~is odd or even. + +\Item{11.} \Topic{The Second Mean Value Theorem.} If $f(x)$~is a function of~$x$ +which has a differential coefficient of constant sign for all values of~$x$ from +$x = a$ to $x = b$, then there is a number~$\xi$ between $a$~and~$b$ such that +\[ +\int_{a}^{b} f(x)\phi(x)\, dx + = f(a) \int_{a}^{\xi} \phi(x)\, dx + + f(b) \int_{\xi}^{b} \phi(x)\, dx. +\] + +[Let $\ds\int_{a}^{x}\phi(t)\, dt = \Phi(x)$. Then +\begin{align*} +\int_{a}^{b} f(x)\phi(x)\, dx + = \int_{a}^{b} f(x)\Phi'(x)\, dx + &= f(b)\Phi(b) - \int_{a}^{b} f'(x)\Phi(x)\, dx\\ + &= f(b)\Phi(b) - \Phi(\xi) \int_{a}^{b}f'(x)\, dx, +\end{align*} +by the generalised Mean Value Theorem of \SecNo[§]{160}: \ie +\[ +\int_{a}^{b} f(x)\phi(x)\, dx = f(b)\Phi(b) + \{f(a) - f(b)\}\Phi(\xi), +\] +which is equivalent to the result given.] +\PageSep{297} + +\Item{12.} \Topic{Bonnet's form of the Second Mean Value Theorem.} If $f'(x)$~is +of constant sign, and $f(b)$ and $f(a) - f(b)$ have the same sign, then +\[ +\int_{a}^{b} f(x)\phi(x)\, dx = f(a) \int_{a}^{X} \phi(x)\, dx, +\] +where $X$~lies between $a$ and~$b$. [For $f(b)\Phi(b) + \{f(a) - f(b)\}\Phi(\xi) = \mu f(a)$, +where $\mu$~lies between $\Phi(\xi)$ and~$\Phi(b)$, and so is the value of~$\Phi(x)$ for a value +of~$x$ such as~$X$. The important case is that in which $0 \leq f(b) \leq f(x) \leq f(a)$.] + +Prove similarly that if $f(a)$ and $f(b) - f(a)$ have the same sign, then +\[ +\int_{a}^{b} f(x)\phi(x)\, dx = f(b) \int_{X}^{b} \phi(x)\, dx, +\] +where $X$~lies between $a$ and~$b$. [Use the function $\Psi(\xi) = \ds\int_{\xi}^{b} \phi(x)\, dx$. It +will be found that the integral can be expressed in the form +\[ +f(a)\DPtypo{\psi(a)}{\Psi(a)} + \{f(b) - f(a)\}\Psi(\xi). +\] +The important case is that in which $0 \leq f(a) \leq f(x) \leq f(b)$.] + +\Item{13.} Prove that +\[ +%[** TN: In-line in the original] +\left|\int_{X}^{X'} \frac{\sin x}{x}\, dx\right| < \frac{2}{X} +\] +if $X' > X > 0$. [Apply the first +formula of Ex.~12, and note that the integral of $\sin x$ over any interval whatever +is numerically less than~$2$.] + +\Item{14.} Establish the results of \Ex{lxv}.~1 by means of the rule for substitution. +[In (i)~divide the range of integration into the two parts $\DPmod{(-a, 0)}{[-a, 0]}$, +$\DPmod{(0, a)}{[0, a]}$, and put $x = -y$ in the first. In (ii)~use the substitution $x = \frac{1}{2}\pi - y$ to +obtain the first equation: to obtain the second divide the range $\DPmod{(0, \pi)}{[0, \pi]}$ into +two equal parts and use the substitution $x = \frac{1}{2}\pi + y$. In (iii)~divide the range +into $m$~equal parts and use the substitutions $x = \pi + y$, $x = 2\pi + y$,~\dots.] + +%[** TN: Integrals in next five examples are set in-line in the original] +\Item{15.} Prove that +\[ +\int_{a}^{b} F(x)\, dx = \int_{a}^{b} F(a + b - x)\, dx. +\] + +\Item{16.} Prove that +\[ +\int_{0}^{\frac{1}{2}\pi} \cos^{m} x\sin^{m} x\, dx + = 2^{-m} \int_{0}^{\frac{1}{2}\pi} \cos^{m} x\, dx. +\] + +\Item{17.} Prove that +\[ +\int_{0}^{\pi} x\phi(\sin x)\, dx + = \tfrac{1}{2}\pi \int_{0}^{\pi} \phi(\sin x)\, dx. +\] + +[Put $x = \pi - y$.] + +\Item{18.} Prove that +\[ +\int_{0}^{\pi} \frac{x\sin x}{1 + \cos^{2} x}\, dx = \tfrac{1}{4}\pi^{2}. +\] + +\Item{19.} Show by means of the transformation $x = a\cos^{2}\theta + b\sin^{2}\theta$ that +\[ +\int_{a}^{b} \sqrtb{(x - a)(b - x)}\, dx = \tfrac{1}{8}\pi (b - a)^{2}. +\] + +\Item{20.} Show by means of the substitution $(a + b\cos x) (a - b\cos y) = a^{2} - b^{2}$ +that +\[ +\int_{0}^{\pi} (a + b\cos x)^{-n}\, dx + = (a^{2} - b^{2})^{-(n - \frac{1}{2})} \int_{0}^{\pi} (a - b\cos y)^{n-1}\, dy, +\] +when $n$~is a positive integer and $a > |b|$, and evaluate the integral when +$n = 1$, $2$,~$3$. +\PageSep{298} + +\Item{21.} If $m$~and~$n$ are positive integers then +\[ +\int_{a}^{b} (x - a)^{m} (b - x)^{n}\, dx + = (b - a)^{m+n+1} \frac{m!\, n!}{(m + n + 1)!}. +\] + +[Put $x = a + (b - a)y$, and use Ex.~6.] +\end{Examples} + +\Paragraph{162. Proof of Taylor's Theorem by Integration by +Parts.} We shall now give the alternative form of the proof of +Taylor's Theorem to which we alluded in \SecNo[§]{147}. + +Let $f(x)$~be a function whose first $n$~derivatives are continuous, +and let +\[ +F_{n}(x) = f(b) - f(x) - (b - x)f'(x) - \dots + - \frac{(b - x)^{n-1}}{(n - 1)!} f^{(n-1)}(x). +\] + +Then +\[ +F_{n}'(x) = -\frac{(b - x)^{n-1}}{(n - 1)!} f^{(n)}(x), +\] +and so +\[ +F_{n}(a) = F_{n}(b) - \int_{a}^{b}F_{n}'(x)\, dx + = \frac{1}{(n - 1)!} \int_{a}^{b} (b - x)^{n-1} f^{(n)}(x)\, dx. +\] +If now we write $a + h$ for~$b$, and transform the integral by putting +$x = a + th$, we obtain +\[ +f(a + h) = f(a) + hf'(a) + \dots + \frac{h^{n-1}}{(n - 1)!} f^{(n-1)}(a) + R_{n}, +\Tag{(1)} +\] +where +\[ +R_{n} = \frac{h^{n}}{(n - 1)!} \int_{0}^{1} (1 - t)^{n-1} f^{(n)}(a + th)\, dt. +\Tag{(2)} +\] + +Now, if $p$~is any positive integer not greater than~$n$, we have, +by Theorem~(9) of \SecNo[§]{160}, +\begin{align*} +\int_{0}^{1} (1 - t)^{n-1} f^{(n)}(a + th)\, dt + &= \int_{0}^{1}(1 - t)^{n-p} (1 - t)^{p-1} f^{(n)}(a + th)\, dt \\ + &= (1 - \theta)^{n-p} f^{(n)}(a + \theta h) \int_{0}^{1} (1 - t)^{p-1}\, dt, +\end{align*} +where $0 < \theta < 1$. Hence +\[ +R_{n} = \frac{(1 - \theta)^{n-p} f^{(n)}(a + \theta h)h^{n}}{p(n - 1)!}. +\Tag{(3)} +\] + +If we take $p = n$ we obtain Lagrange's form of~$R_{n}$ (\SecNo[§]{148}). If +on the other hand we take $p = 1$ we obtain \Emph{Cauchy's form}, viz. +\[ +R_{n} = \frac{(1 - \theta)^{n-1} f^{(n)}(a + \theta h) h^{n}}{(n - 1)!}.\footnotemark +\Tag{(4)} +\] +\footnotetext{The method used in \SecNo[§]{147} can also be modified so as to obtain these + alternative forms of the remainder.} +\PageSep{299} + +\begin{Remark} +\Paragraph{163. Application of Cauchy's form to the Binomial Series.} If +$f(x) = (1 + x)^{m}$, where $m$~is not a positive integer, then Cauchy's form of the +remainder is +\[ +R_{n} = \frac{m(m - 1)\dots (m - n + 1)}{1·2\dots (n - 1)}\, + \frac{(1 - \theta )^{n-1} x^{n}}{(1 + \theta x)^{n-m}}. +\] + +Now $(1 - \theta)/(1 + \theta x)$ is less than unity, so long as $-1 < x < 1$, whether +$x$~is positive or negative; and $(1 + \theta x)^{m-1}$ is less than a constant~$K$ for +all values of~$n$, being in fact less than $(1 + |x|)^{m-1}$ if $m > 1$ and than +$(1 - |x|)^{m-1}$ if $m < 1$\Add{.} Hence +\[ +|R_{n}| < K |m| \left|\binom{m - 1}{n - 1}\right| |x^{n}| = \rho_{n}, +\] +say\Add{.} But $\rho_{n} \to 0$ as $n \to \infty$, by \Ex{xxvii}.~13, and so $R_{n} \to 0$. The truth of the +Binomial Theorem is thus established for all rational values of~$m$ and all +values of~$x$ between $-1$ and~$1$. It will be remembered that the difficulty in +using Lagrange's form, in \Ex{lvi}.~2, arose in connection with negative +values of~$x$. +\end{Remark} + +\Paragraph{164. Integrals of complex functions of a real variable.} +So far we have always supposed that the subject of integration in +a definite integral is real. We define the integral of a complex +function $f(x) = \DPtypo{\psi}{\phi}(x) + i\psi(x)$ of the real variable~$x$, between the +limits $a$~and~$b$, by the equations +\[ +\int_{a}^{b} f(x)\, dx + = \int_{a}^{b} \{\phi(x) + i\psi(x)\}\, dx + = \int_{a}^{b} \phi(x)\, dx + i \int_{a}^{b} \psi(x)\, dx; +\] +and it is evident that the properties of such integrals may be +deduced from those of the real integrals already considered. + +There is one of these properties that we shall make use of +later on. It is expressed by the inequality +\[ +\left|\int_{a}^{b} f(x)\, dx\right| \leq \int_{a}^{b} |f(x)|\, dx.\footnotemark +\Tag{(1)} +\] +\footnotetext{The corresponding inequality for a real integral was proved in \Ex{lxv}.~14.}% +This inequality may be deduced without difficulty from the +definitions of \SecNo[§§]{156}~and~\SecNo{157}. If $\delta_{\nu}$~has the same meaning as in +\SecNo[§]{156}, $\phi_{\nu}$~and~$\psi_{\nu}$ are the values of $\phi$~and~$\psi$ at a point of~$\delta_{\nu}$, and +$f_{\nu} = \phi_{\nu} + i\psi_{\nu}$, then we have +\begin{align*} +\int_{a}^{b} f\, dx + = \int_{a}^{b} \phi\, dx + i \int_{a}^{b} \psi\, dx + &= \lim \tsum \phi_{\nu}\, \delta_{\nu} + i \lim \tsum \psi_{\nu}\, \delta_{\nu} \\ + &= \lim \tsum (\phi_{\nu} + i\psi_{\nu})\, \delta_{\nu} + = \lim \tsum f_{\nu}\, \delta_{\nu}, +\end{align*} +and so +\[ +\int_{a}^{b} f\, dx + = |\lim \tsum f_{\nu}\, \delta_{\nu}| + = \lim |\tsum f_{\nu}\, \delta_{\nu}|; +\] +\PageSep{300} +while +\[ +\int_{a}^{b} |f|\, dx = \lim \tsum |f_{\nu}|\, \delta_{\nu}. +\] +The result now follows at once from the inequality +\[ +|\tsum f_{\nu}\, \delta_{\nu}| \leq \tsum |f_{\nu}|\, \delta_{\nu}. +\] + +It is evident that the formulae \Eq{(1)}~and~\Eq{(2)} of \SecNo[§]{162} remain +true when $f$~is a complex function $\phi + i\psi$. + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER VII.} + +%[** TN: Several displayed integrals are in-line in the original] +\begin{Examples}{} +\Item{1.} Verify the terms given of the following Taylor's Series: +\begin{alignat*}{2} +&\Item{(1)} & +\tan x &= x + \tfrac{1}{3} x^{3} + \tfrac{2}{15} x^{5} + \dots, \\ +&\Item{(2)} & +\sec x &= 1 + \tfrac{1}{2} x^{2} + \tfrac{5}{24} x^{4} + \dots, \\ +&\Item{(3)}\quad & +x\cosec x &= 1 + \tfrac{1}{6} x^{2} + \tfrac{7}{360} x^{4} + \dots, \\ +&\Item{(4)} & +x\cot x &= 1 - \tfrac{1}{3} x^{2} - \tfrac{1}{45} x^{4} - \dots. +\end{alignat*} + +\Item{2.} Show that if $f(x)$ and its first $n + 2$ derivatives are continuous, and +$f^{(n+1)}(0) \neq 0$, and $\theta_{n}$~is the value of~$\theta$ which occurs in Lagrange's form of the +remainder after $n$~terms of Taylor's Series, then +\[ +\theta_{n} = \frac{1}{n + 1} + + \frac{n}{2(n + 1)^{2}(n + 2)} + \left\{\frac{f^{(n+2)}(0)}{f^{(n+1)}(0)} + \epsilon_{x}\right\}x, +\] +where $\epsilon_{x} \to 0$ as $x \to 0$. [Follow the method of \Ex{lv}.~12.] + +\Item{3.} Verify the last result when $f(x) = 1/(1+ x)$. [Here $(1 + \theta_{n}x)^{n+1} = 1 + x$.] + +\Item{4.} Show that if $f(x)$~has derivatives of the first three orders then +\[ +f(b) = f(a) + \tfrac{1}{2}(b - a) \{f'(a) + f'(b)\} + - \tfrac{1}{12}(b - a)^{3} f'''(\alpha), +\] +where $a < \alpha < b$. [Apply to the function +\begin{multline*} +f(x) - f(a) - \tfrac{1}{2}(x - a) \{f'(a) + f'(x)\}\\ + - \left(\frac{x - a}{b - a}\right)^{3} + [f(b) - f(a) - \tfrac{1}{2}(b - a) \{f'(a) + f'(b)\}] +\end{multline*} +arguments similar to those of \SecNo[§]{147}.] + +\Item{5.} Show that under the same conditions +\[ +f(b) = f(a) + (b - a) f'\{\tfrac{1}{2}(a + b)\} + + \tfrac{1}{24}(b - a)^{3}f'''(\alpha). +\] + +\Item{6.} Show that if $f(x)$ has derivatives of the first five orders then +\[ +f(b) = f(a) + \tfrac{1}{6}(b - a) [f'(a) + f'(b) + 4f'\{\tfrac{1}{2}(a + b)\}] + - \tfrac{1}{2880}(b - a)^{5} f^{(5)}(\DPtypo{a}{\alpha}). +\] + +\Item{7.} Show that under the same conditions +\[ +f(b) = f(a) + \tfrac{1}{2}(b - a) \{f'(a) + f'(b)\} + - \tfrac{1}{12}(b - a)^{2} \{f''(b) - f''(a)\} + + \tfrac{1}{720}(b - a)^{5} f^{(5)}(\alpha). +\] +\PageSep{301} + +\Item{8.} Establish the formulae +\CenterLine{\Itemp{(i)}}{$\ds +\begin{vmatrix} + f(a) & f(b)\\ + g(a) & g(b) +\end{vmatrix} += (b - a) +\begin{vmatrix} + f(a) & f'(\beta)\\ + g(a) & g'(\beta) +\end{vmatrix}$,} +where $\beta$~lies between $a$ and~$b$, and +\CenterLine{\Itemp{(ii)}}{$\ds +\begin{vmatrix} + f(a) & f(b) & f(c)\\ + g(a) & g(b) & g(c)\\ + h(a) & h(b) & h(c) +\end{vmatrix} += \tfrac{1}{2} (b - c)(c - a)(a - b) +\begin{vmatrix} + f(a) & f'(\beta) & f''(\gamma)\\ + g(a) & g'(\beta) & g''(\gamma)\\ + h(a) & h'(\beta) & h''(\gamma) +\end{vmatrix}$,} +where $\beta$ and $\gamma$ lie between the least and greatest of $a$,~$b$,~$c$. [To prove~(ii) +consider the function +\[ +\phi(x) = +\begin{vmatrix} + f(a) & f(b) & f(x)\\ + g(a) & g(b) & g(x)\\ + h(a) & h(b) & h(x) +\end{vmatrix} +- \frac{(x - a)(x - b)}{(c - a)(c - b)} +\begin{vmatrix} + f(a) & f(b) & f(c)\\ + g(a) & g(b) & g(c)\\ + h(a) & h(b) & h(c) +\end{vmatrix}\Add{,} +\] +which vanishes when $x = a$, $x = b$, and $x = c$. Its first derivative, by Theorem~B +of \SecNo[§]{121}, must vanish for two distinct values of~$x$ lying between the least and +greatest of $a$,~$b$,~$c$; and its second derivative must therefore vanish for a value~$\gamma$ +of~$x$ satisfying the same condition. We thus obtain the formula +\[ +\begin{vmatrix} + f(a) & f(b) & f(c)\\ + g(a) & g(b) & g(c)\\ + h(a) & h(b) & h(c) +\end{vmatrix} += \tfrac{1}{2}(c - a)(c - b) +\begin{vmatrix} + f(a) & f(b) & f''(\gamma)\\ + g(a) & g(b) & g''(\gamma)\\ + h(a) & h(b) & h''(\gamma) +\end{vmatrix}. +\] +The reader will now complete the proof without difficulty.] + +\Item{9.} If $F(x)$~is a function which has continuous derivatives of the first $n$~orders, +of which the first~$n - 1$ vanish when $x = 0$, and $A \leq F^{(n)}(x) \leq B$ when +$0 \leq x \leq h$, then $A(x^{n}/n!) \leq F(x) \leq B(x^{n}/n!)$ when $0 \leq x \leq h$. + +Apply this result to +\[ +f(x) - f(0) - xf'(0) - \dots - \frac{x^{n-1}}{(n - 1)!} f^{(n-1)}(0), +\] +and deduce Taylor's Theorem. + +\Item{10.} If $\Delta_{h}\phi(x) = \phi(x) - \phi(x + h)$, $\Delta_{h}^{2}\phi(x) = \Delta_{h}\{\Delta_{h}\phi(x)\}$, and so on, and +$\phi(x)$~has derivatives of the first $n$~orders, then +\[ +\Delta_{h}^{n}\phi(x) + = \sum_{r=0}^{n}(-1)^{r} \binom{n}{r} \phi(x + rh) + = (-h)^{n} \phi^{(n)}(\xi), +\] +where $\xi$~lies between $x$ and~$x + nh$. Deduce that if $\phi^{(n)}(x)$~is continuous then +$\{\Delta_{h}^{n}\phi(x)\}/h^{n} \to (-1)^{n}\phi^{(n)}(x)$ as $h \to 0$. [This result has been stated already +when $n = 2$, in \Ex{lv}.~13.] + +\Item{11.} Deduce from Ex.~10 that $x^{n-m}\, \Delta_{h}^{n} x^{m} \to m(m - 1) \dots (m - n + 1)h^{n}$ as +$x \to \infty$, $m$~being any rational number and $n$~any positive integer. In +particular prove that +\[ +x\sqrt{x} \{\sqrt{x} - 2\sqrtp{x + 1} + \sqrtp{x + 2}\} \to -\tfrac{1}{4}. +\] +\PageSep{302} + +\Item{12.} Suppose that $y = \phi(x)$ is a function of~$x$ with continuous derivatives +of at least the first four orders, and that $\phi(0) = 0$, $\phi'(0) = 1$, so that +\[ +y = \phi(x) = x + a_{2}x^{2} + a_{3}x^{3} + (a_{4} + \epsilon_{x})x^{4}, +\] +where $\epsilon_{x} \to 0$ as $x \to 0$. Establish the formula +\[ +x = \psi(y) + = y - a_{2}y^{2} + (2a_{2}^{2} - a_{3})y^{3} + - (5a_{2}^{3} - 5a_{2}a_{3} + a_{4} + \epsilon_{y})y^{4}, +\] +where $\epsilon_{y} \to 0$ as $y \to 0$, for that value of~$x$ which vanishes with~$y$; and prove +that +\[ +\frac{\phi(x)\psi(x) - x^{2}}{x^{4}} \to a_{2}^{2} +\] +as $x \to 0$. + +\Item{13.} The coordinates $(\xi, \eta)$ of the centre of curvature of the curve $x = f(t)$, +$y = F(t)$, at the point $(x, y)$, are given by +\[ +-(\xi - x)/y' = (\eta - y)/x' = (x'^{2} + y'^{2})/(x'y'' - x''y'); +\] +and the radius of curvature of the curve is +\[ +(x'^{2} + y'^{2})^{3/2}/(x'y'' - x''y'), +\] +dashes denoting differentiations with respect to~$t$. + +\Item{14.} The coordinates $(\xi, \eta)$ of the centre of curvature of the curve +$27ay^{2} = 4x^{3}$, at the point $(x, y)$, are given by +\[ +3a(\xi + x) + 2x^{2} = 0, \quad +\eta = 4y + (9ay)/x.\quad +\] +\MathTrip{1899.} + +\Item{15.} Prove that the circle of curvature at a point $(x, y)$ will have contact +of the third order with the curve if $(1 + y_{1}^{2})y_{3} = 3y_{1}y_{2}^{2}$ at that point. Prove +also that the circle is the only curve which possesses this property at every +point; and that the only points on a conic which possess the property +are the extremities of the axes. [Cf.\ \Ref{Ch.}{VI}, \MiscEx{VI}\ 10~(iv).] + +\Item{16.} The conic of closest contact with the curve $y = ax^{2} + bx^{3} + cx^{4} + \dots + kx^{n}$, +at the origin, is $a^{3}y = a^{4}x^{2} + a^{2}bxy + (ac - b^{2})y^{2}$. Deduce that the conic of +closest contact at the point $(\xi, \eta)$ of the curve $y = f(x)$ is +\[ +18\eta_{2}^{3}T + = 9\eta_{2}^{4}(x - \xi)^{2} + + 6\eta_{2}^{2}\eta_{3}(x - \xi)T + + (3\eta_{2}\eta_{4} - 4\eta_{3}^{2})T^{2}, +\] +where $T = (y - \eta) - \eta_{1}(x - \xi)$. +\MathTrip{1907.} + +\Item{17.} \Topic{Homogeneous functions.\footnote +{In this and the following examples the reader is to assume the continuity of + all the derivatives which occur.}} +If $u = x^{n} f(y/x, z/x, \dots)$ then $u$~is unaltered, +save for a factor~$\lambda^{n}$, when $x$,~$y$, $z$,~\dots\ are all increased in the ratio $\lambda : 1$. +In these circumstances $u$~is called a \emph{homogeneous function of degree~$n$} in the +variables $x$,~$y$, $z$,~\dots. Prove that if $u$~is homogeneous and of degree~$n$ then +\[ +x\frac{\dd u}{\dd x} + y\frac{\dd u}{\dd y} + z\frac{\dd u}{\dd z} + \dots = nu. +\] +This result is known as \Emph{Euler's Theorem} on homogeneous functions. + +\Item{18.} If $u$~is homogeneous and of degree~$n$ then $\dd u/\dd x$, $\dd u/\dd y$,~\dots\ are +homogeneous and of degree $n - 1$. +\PageSep{303} + +\Item{19.} Let $f(x, y) = 0$ be an equation in $x$~and~$y$ (\eg\ $x^{n} + y^{n} - x = 0$), and let +$F(x, y, z) = 0$ be the form it assumes when made homogeneous by the introduction +of a third variable~$z$ in place of unity (\eg\ $x^{n} + y^{n} - xz^{n-1} = 0$). Show +that the equation of the tangent at the point $(\xi, \eta)$ of the curve $f(x, y) = 0$ is +\[ +xF_{\xi} + yF_{\eta} + zF_{\zeta} = 0, +\] +where $F_{\xi}$,~$F_{\eta}$,~$F_{\zeta}$ denote the values of $F_{x}$,~$F_{y}$,~$F_{z}$ when $x = \xi$, $y = \eta$, $z = \zeta = 1$. + +\Item{20.} \Topic{Dependent and independent functions. Jacobians or functional +determinants.} Suppose that $u$~and~$v$ are functions of $x$~and~$y$ connected by +an identical relation +\[ +\phi(u, v) = 0. +\Tag{(1)} +\] + +Differentiating~\Eq{(1)} with respect to $x$~and~$y$, we obtain +\[ +\frac{\dd \phi}{\dd u}\, \frac{\dd u}{\dd x} + +\frac{\dd \phi}{\dd v}\, \frac{\dd v}{\dd x} = 0,\quad +\frac{\dd \phi}{\dd u}\, \frac{\dd u}{\dd y} + +\frac{\dd \phi}{\dd v}\, \frac{\dd v}{\dd y} = 0, +\Tag{(2)} +\] +and, eliminating the derivatives of~$\phi$, +\[ +J = +\begin{vmatrix} + u_{x} & u_{y}\\ + v_{x} & v_{y} +\end{vmatrix} += u_{x}v_{y} - u_{y}v_{x} = 0, +\Tag{(3)} +\] +where $u_{x}$,~$u_{y}$, $v_{x}$,~$v_{y}$ are the derivatives of $u$~and~$v$ with respect to $x$~and~$y$. +This condition is therefore \emph{necessary} for the existence of a relation such +as~\Eq{(1)}. It can be proved that the condition is also \emph{sufficient}; for this we must +refer to Goursat's \textit{Cours d' Analyse}, vol.~i, pp.~125~\textit{et~seq.} + +Two functions $u$~and~$v$ are said to be \emph{dependent} or \emph{independent} according +as they are or are not connected by such a relation as~\Eq{(1)}. It is usual to call~$J$ +the \emph{Jacobian} or \emph{functional determinant} of $u$~and~$v$ with respect to $x$~and~$y$, +and to write +\[ +J = \frac{\dd(u, v)}{\dd(x, y)}. +\] + +Similar results hold for functions of any number of variables. Thus three +functions $u$,~$v$,~$w$ of three variables $x$,~$y$,~$z$ are or are not connected by a +relation $\phi(u, v, w) = 0$ according as +\[ +J = +\begin{vmatrix} + u_{x} & u_{y} & u_{z}\\ + v_{x} & v_{y} & v_{z}\\ + w_{x} & w_{y} & w_{z} +\end{vmatrix} += \frac{\dd(u, v, w)}{\dd(x, y, z)} +\] +does or does not vanish for all values of $x$,~$y$,~$z$. + +\Item{21.} Show that $ax^{2} + 2hxy + by_{2}$ and $Ax^{2} + 2Hxy + By^{2}$ are independent +unless $a/A = h/H = b/B$. + +\Item{22.} Show that $ax^{2} + by^{2} + cz^{2} + 2fyz + 2gzx + 2hxy$ can be expressed as a +product of two linear functions of $x$,~$y$, and~$z$ if and only if +\[ +abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0. +\] + +[Write down the condition that $px + qy + rz$ and $p'x + q'y + r'z$ should be +connected with the given function by a functional relation.] +\PageSep{304} + +\Item{23.} If $u$~and~$v$ are functions of $\xi$~and~$\eta$, which are themselves functions +of $x$~and~$y$, then +\[ +\frac{\dd(u, v)}{\dd(x, y)} + = \frac{\dd(u, v)}{\dd(\xi, \eta)}\, \frac{\dd(\xi, \eta)}{\dd(x, y)}. +\] +Extend the result to any number of variables. + +\Item{24.} Let $f(x)$~be a function of~$x$ whose derivative is~$1/x$ and which vanishes +when $x = 1$. Show that if $u = f(x) + f(y)$, $v = xy$, then $u_{x}v_{y} - u_{y}v_{x} = 0$, and hence +that $u$~and~$v$ are connected by a functional relation. By putting $y = 1$, show +that this relation must be $f(x) + f(y) = f(xy)$. Prove in a similar manner that +if the derivative of~$f(x)$ is $1/(1 + x^{2})$, and $f(0) = 0$, then $f(x)$~must satisfy the +equation +\[ +f(x) + f(y) = f\left(\frac{x + y}{1 - xy}\right). +\] + +\Item{25.} Prove that if $\ds f(x) = \int_{0}^{x} \frac{dt}{\sqrtp{1 - t^{4}}}$ then +\[ +f(x) + f(y) = f\left\{ + \frac{x\sqrtp{1 - y^{4}} + y\sqrtp{1 - x^{4}}}{1 + x^{2}y^{2}} +\right\}. +\] + +\Item{26.} Show that if a functional relation exists between +\[ +u = f(x) + f(y) + f(z),\quad +v = f(y)f(z) + f(z)f(x) + f(x)f(y),\quad +w = f(x)f(y)f(z), +\] +then $f$~must be a constant. [The condition for a functional relation will be +found to be +\[ +f'(x)f'(y)f'(z) \{f(y) - f(z)\} \{f(z) - f(x)\} \{f(x) - f(y)\} = 0.] +\] + +\Item{27.} If $f(y, z)$, $f(z, x)$, and $f(x, y)$ are connected by a functional relation +then $f(x, x)$~is independent of~$x$. \MathTrip{1909.} + +\Item{28.} If $u = 0$, $v = 0$, $w = 0$ are the equations of three circles, rendered +homogeneous as in Ex.~19, then the equation +\[ +\frac{\dd(u, v, w)}{\dd(x, y, z)} = 0 +\] +represents the circle which cuts them all orthogonally. \MathTrip{1900.} + +\Item{29.} If $A$,~$B$,~$C$ are three functions of~$x$ such that +\[ +\begin{vmatrix} + A & A' & A''\\ + B & B' & B''\\ + C & C' & C'' +\end{vmatrix} +\] +vanishes identically, then we can find constants $\lambda$,~$\mu$,~$\nu$ such that $\lambda A + \mu B + \nu C$ +vanishes identically; and conversely. [The converse is almost obvious. To +prove the direct theorem let $\alpha = BC' - B'C$,~\dots. Then $\alpha' = BC'' - B''C$,~\dots, +and it follows from the vanishing of the determinant that $\beta\gamma' - \beta'\gamma = 0$,~\dots; +and so that the ratios $\alpha : \beta : \gamma$ are constant. But $\alpha A + \beta B + \gamma C = 0$.] + +\Item{30.} Suppose that three variables $x$,~$y$,~$z$ are connected by a relation in +virtue of which (i)~$z$~is a function of $x$~and~$y$, with derivatives $z_{x}$\Add{,}~$z_{y}$, and (ii)~$x$ +is a function of $y$~and~$z$, with derivatives $x_{y}$,~$x_{z}$. Prove that +\[ +x_{y} = - z_{y}/z_{x},\quad +x_{z} = 1/z_{x}. +\] +\PageSep{305} + +[We have +\[ +dz = z_{x}\, dx + z_{y}\, dy,\quad +dx = x_{y}\, dy + x_{z}\, dz\Add{.} +\] +The result of substituting for~$dx$ in the first equation is +\[ +dz = (z_{x} x_{y} + z_{y})\, dy + z_{x}x_{z}\, dz, +\] +which can be true only if $z_{x} x_{y} + z_{y} = 0$, $z_{x} x_{z} = 1$.] + +\Item{31.} Four variables $x$, $y$, $z$, $u$ are connected by two relations in virtue of +which any two can be expressed as functions of the others. Show that +\[ +y_{z}^{u}z_{x}^{u}x_{y}^{u} = -y_{z}^{x}z_{x}^{y}x_{y}^{z} = 1,\quad +x_{z}^{u}z_{x}^{y} + y_{z}^{u}z_{y}^{x} = 1, +\] +where $y_{z}^{u}$~denotes the derivative of~$y$, when expressed as a function of $z$~and~$u$, +with respect to~$z$. \MathTrip{1897.} + +\Item{32.} Find $A$, $B$, $C$, $\lambda$ so that the first four derivatives of +\[ +\int_{a}^{a+x} f(t)\, dt - x[Af(a) + Bf(a + \lambda x) + Cf(a + x)] +\] +vanish when $x = 0$; and $A$, $B$, $C$, $D$, $\lambda$,~$\mu$ so that the first six derivatives of +\[ +\int_{a}^{a+x} f(t)\, dt + - x[Af(a) + Bf(a + \lambda x) + Cf(a + \mu x) + Df(a + x)] +\] +vanish when $x = 0$. + +\Item{33.} If $a > 0$, $ac - b^{2} > 0$, and $x_{1} > x_{0}$, then +\[ +\int_{x_{0}}^{x_{1}} \frac{dx}{ax^{2} + 2bx + c} + = \frac{1}{\sqrtp{ac - b^{2}}} \arctan\left\{ + \frac{(x_{1} - x_{0}) \sqrtp{ac - b^{2}}} + {ax_{1}x_{0} + b(x_{1} + x_{0}) + c} + \right\}, +\] +the inverse tangent lying between $0$ and~$\pi$.\footnote + {In connection with Exs.~33--35, 38, and~40 see a paper by Dr~Bromwich + in vol.~xxxv of the \textit{Messenger of Mathematics}.} + +\Item{34.} Evaluate the integral $\ds\int_{-1}^{1} \frac{\sin\alpha\, dx}{1 - 2x\cos\alpha + x^{2}}$. For what values of~$\alpha$ is +the integral a discontinuous function of~$\alpha$? \MathTrip{1904.} + +{\Loosen[The value of the integral is~$\frac{1}{2}\pi$ if $2n\pi < \alpha < (2n + 1)\pi$, and $-\frac{1}{2}\pi$ if +$(2n - 1)\pi < \alpha < 2n\pi$, $n$~being any integer; and $0$~if $\alpha$~is a multiple of~$\pi$.]} + +\Item{35.} If $ax^{2} + 2bx + c > 0$ when $x_{0} \leq x \leq x_{1}$, $f(x) = \sqrtp{ax^{2} + 2bx + c}$, and +\[ +y = f(x),\quad +y_{0} = f(x_{0}),\quad +y_{1} = f(x_{1}),\quad +X = (x_{1} - x_{0})/(y_{1} + y_{0}), +\] +then +\[ +\int_{x_{0}}^{x_{1}} \frac{dx}{y} + = \frac{1}{\sqrt{a}} \log \frac{1 + X\sqrt{a}}{1 - X\sqrt{a}},\quad +\frac{-2}{\sqrtp{-a}} \arctan\{X\sqrtp{-a}\}, +\] +according as $a$~is positive or negative. In the latter case the inverse +tangent lies between $0$ and~$\frac{1}{2}\pi$. [It will be found that the substitution $t = \dfrac{x - x_{0}}{y + y_{0}}$ reduces the integral to the form $\ds 2\int_{0}^{X} \frac{dt}{1 - at^{2}}$.] + +\Item{36.} Prove that +\[ +\int_{0}^{a} \frac{dx}{x + \sqrtp{a^{2} - x^{2}}} = \tfrac{1}{4}\pi. +\] +\MathTrip{1913.} + +\Item{37.} If $a > 1$ then +\[ +\int_{-1}^{1} \frac{\sqrtp{1 - x^{2}}}{a - x}\, dx = \pi\{a - \sqrtp{a^{2} - 1}\}. +\] +\PageSep{306} + +\Item{38.} If $p > 1$, $0 < q < 1$, then +\[ +\int_{0}^{1} \frac{dx}{\sqrtbr{\{1 + (p^{2} - 1)x\}\{1 - (1 - q^{2}) x\}}} + = \frac{2\omega}{(p + q)\sin\omega}, +\] +where $\omega$~is the positive acute angle whose cosine is $(1 + pq)/(p + q)$. + +\Item{39.} If $a > b > 0$, then +\[ +%[** TN: In-line in the original] +\int_{0}^{2\pi} \frac{\sin^{2}\theta\, d\theta}{a - b\cos\theta} + = \frac{2\pi}{b^{2}} \{a - \sqrtp{a^{2} - b^{2}}\}. +\] +\MathTrip{1904.} + +\Item{40.} Prove that if $a > \sqrtp{b^{2} + c^{2}}$ then +\[ +\int_{0}^{\pi} \frac{d\theta}{a + b\cos\theta + c\sin\theta} + = \frac{2}{\sqrtp{a^{2} - b^{2} - c^{2}}} + \arctan \left\{\frac{\sqrtp{a^{2} - b^{2} - c^{2}}}{c}\right\}, +\] +the inverse tangent lying between $0$ and~$\pi$. + +\Item{41.} If $f(x)$~is continuous and never negative, and $\ds\int_{a}^{b} f(x)\, dx = 0$, then +$f(x) = 0$ for all values of~$x$ between $a$ and~$b$. [If $f(x)$~were equal to a positive +number~$k$ when $x = \xi$, say, then we could, in virtue of the continuity of~$f(x)$, +find an interval $\DPmod{(\xi - \delta, \xi + \delta)}{[\xi - \delta, \xi + \delta]}$ throughout which $f(x) > \frac{1}{2}k$; and then the value +of the integral would be greater than~$\delta k$.] + +\Item{42.} \Topic{Schwarz's inequality for integrals.} Prove that +\[ +\left(\int_{a}^{b} \phi\psi\, dx\right)^{2} + \leq \int_{a}^{b} \phi^{2}\, dx \int_{a}^{b} \psi^{2}\, dx. +\] + +[Use the definitions of \SecNo[§§]{156}~and~\SecNo{157}, and the inequality +\[ +\left(\tsum\phi_{\nu}\psi_{\nu}\, \delta_{\nu}\right)^{2} + \leq \tsum\phi_{\nu}^{2}\, \delta_{\nu} \tsum\psi_{\nu}^{2}\, \delta_{\nu} +\] +(\Ref{Ch.}{I}, \MiscEx{I}~10).] + +\Item{43.} If +\[ +%[** TN: In-line in the original] +P_{n}(x) = \frac{1}{(\beta - \alpha)^{n} n!} + \left(\frac{d}{dx}\right)^{n} \{(x - \alpha)(\beta - x)\}^{n}, +\] +then $P_{n}(x)$~is a polynomial of degree~$n$, which possesses the property that +\[ +\int_{\alpha}^{\beta} P_{n}(x)\theta(x)\, dx = 0 +\] +if $\theta(x)$~is any polynomial of degree less than~$n$. [Integrate by parts $m + 1$ +times, where $m$~is the degree of~$\theta(x)$, and observe that $\theta^{(m+1)}(x) = 0$.] + +\Item{44.} Prove that +\[ +%[** TN: In-line in the original] +\int_{\alpha}^{\beta} P_{m}(x) P_{n}(x)\, dx = 0 +\] +if $m \neq n$, but that if $m = n$ then the +value of the integral is $(\beta - \alpha)/(2n + 1)$. + +\Item{45.} If $Q_{n}(x)$~is a polynomial of degree $n$, which possesses the property that +\[ +%[** TN: In-line in the original] +\int_{\alpha}^{\beta} Q_{n}(x)\theta(x)\, dx = 0 +\] +if $\theta(x)$~is any polynomial of degree less than~$n$, then +$Q_{n}(x)$~is a constant multiple of~$P_{n}(x)$. + +[We can choose~$\kappa$ so that $Q_{n} - \kappa P_{n}$ is of degree~$n - 1$: then +\[ +\int_{\alpha}^{\beta} Q_{n}(Q_{n} - \kappa P_{n})\, dx = 0,\quad +\int_{\alpha}^{\beta} P_{n}(Q_{n} - \kappa P_{n})\, dx = 0, +\] +\PageSep{307} +and so +\[ +\int_{\alpha}^{\beta} (Q_{n} - \kappa P_{n})^{2}\, dx = 0. +\] +Now apply Ex.~41.] + +\Item{46.} \Topic{Approximate Values of definite integrals.} Show that the error +in taking $\tfrac{1}{2}(b - a) \{\phi(a) + \phi(b)\}$ as the value of the integral $\ds\int_{a}^{b} \phi(x)\, dx$ is less +than $\tfrac{1}{12}M(b - a)^{3}$, where $M$~is the maximum of~$|\phi''(x)|$ in the interval $\DPmod{(a, b)}{[a, b]}$; +and that the error in taking $(b - a)\phi\{\tfrac{1}{2}(a + b)\}$ is less than $\tfrac{1}{24}M(b - a)^{3}$. [Write +$f'(x)= \phi(x)$ in Exs.\ 4~and~5.] Show that the error in taking +\[ +\tfrac{1}{6}(b - a)[\phi(a) + \phi(b) + 4\phi\{\tfrac{1}{2}(a + b)\}] +\] +as the value is less than $\tfrac{1}{2880}M(b - a)^{5}$, where $M$~is the maximum of~$\phi^{(4)}(x)$. +[Use Ex.~6. This rule, which gives a very good approximation, is known as +\Emph{Simpson's Rule}. It amounts to taking one-third of the first approximation +given above and two-thirds of the second.] + +Show that the approximation assigned by Simpson's Rule is the area +bounded by the lines $x = a$, $x = b$, $y = 0$, and a parabola with its axis parallel +to~$OY$ and passing through the three points on the curve $y = \phi(x)$ whose +abscissae are $a$,~$\tfrac{1}{2}(a + b)$,~$b$. + +It should be observed that if $\phi(x)$~is any cubic polynomial then $\phi^{(4)}(x) = 0$, +and Simpson's Rule is exact. That is to say, given three points whose +abscissae are $a$,~$\tfrac{1}{2}(a + b)$,~$b$, we can draw through them an infinity of curves +of the type $y = \alpha + \beta x + \gamma x^{2} + \delta x^{3}$; and all such curves give the same area. For +one curve $\delta = 0$, and this curve is a parabola. + +\Item{47.} If $\phi(x)$~is a polynomial of the fifth degree, then +\[ +\int_{0}^{1} \phi(x)\, dx + = \tfrac{1}{18}\{5\phi(\alpha) + 8\phi(\tfrac{1}{2}) + 5\phi(\beta)\}, +\] +$\alpha$~and~$\beta$ being the roots of the equation $x^{2} - x + \frac{1}{10} = 0$. \MathTrip{1909.} + +\Item{48.} {\Loosen Apply Simpson's Rule to the calculation of~$\pi$ from the formula +$\ds\tfrac{1}{4}\pi = \int_{0}^{1} \frac{dx}{1 + x^{2}}$. [The result is~$.7833\dots$. If we divide the integral into two, +from $0$ to~$\tfrac{1}{2}$ and $\tfrac{1}{2}$ to~$1$, and apply Simpson's Rule to the two integrals +separately, we obtain $.785\MS391\MS6\dots$. The correct value is~$.785\MS398\MS1\dots$.]} + +\Item{49.} Show that +\[ +8.9 < \int_{3}^{5} \sqrtp{4 + x^{2}}\, dx < 9. +\] +\MathTrip{1903.} + +\Item{50.} Calculate the integrals +\[ +\int_{0}^{1} \frac{dx}{1 + x},\quad +\int_{0}^{1} \frac{dx}{\sqrtp{1 + x^{4}}},\quad +\int_{0}^{\pi} \sqrtp{\sin x}\, dx,\quad +\int_{0}^{\pi} \frac{\sin x}{x}\, dx, +\] +to two places of decimals. [In the last integral the subject of integration is +not defined when $x = 0$: but if we assign to it, when $x = 0$, the value~$1$, it +becomes continuous throughout the range of integration.] +\end{Examples} +\PageSep{308} + + +\Chapter[THE CONVERGENCE OF INFINITE SERIES, ETC.] +{VIII}{THE CONVERGENCE OF INFINITE SERIES AND \\ +INFINITE INTEGRALS} + +\Paragraph{165.} \First{In} \Ref{Ch.}{IV} we explained what was meant by saying +that an infinite series is \emph{convergent}, \emph{divergent}, or \emph{oscillatory}, and +illustrated our definitions by a few simple examples, mainly +derived from the geometrical series +\[ +1 + x + x^{2} + \dots +\] +and other series closely connected with it. In this chapter we +shall pursue the subject in a more systematic manner, and prove +a number of theorems which enable us to determine when the +simplest series which commonly occur in analysis are convergent. + +We shall often use the notation +\[ +u_{m} + u_{m+1} + \dots + u_{n} = \sum_{m}^{n} \phi(\nu), +\] +and write $\sum\limits_{0}^{\infty} u_{n}$, or simply $\sum u_{n}$, for the infinite series $u_{0} + u_{1} + u_{2} + \dots$.\footnote + {It is of course a matter of indifference whether we denote our series by + $u_{1} + u_{2} + \dots$ (as in \Ref{Ch.}{IV}) or by $u_{0} + u_{1} + \dots$ (as here). Later in this chapter we + shall be concerned with series of the type $a_{0} + a_{1}x + a_{2}x^{2} + \dots$: for these the latter + notation is clearly more convenient. We shall therefore adopt this as our standard + notation. But we shall not adhere to it systematically, and we shall suppose that $u_{1}$~is + the first term whenever this course is more convenient. It is more convenient, + for example, when dealing with the series $1 + \frac{1}{2} + \frac{1}{3} + \dots$, to suppose that $u_{n} = 1/n$ + and that the series begins with~$u_{1}$, than to suppose that $u_{n} = 1/(n + 1)$ and that the + series begins with~$u_{0}$. This remark applies, \eg, to \Ex{lxviii}.~4.} + +\Paragraph{166. Series of Positive Terms.} The theory of the convergence +of series is comparatively simple when all the terms of +the series considered are positive.\footnote + {Here and in what follows `positive' is to be regarded as including zero.} +We shall consider such series +\PageSep{309} +first, not only because they are the easiest to deal with, but also +because the discussion of the convergence of a series containing +negative or complex terms can often be made to depend upon +a similar discussion of a series of positive terms only. + +When we are discussing the convergence or divergence of a +series we may disregard any finite number of terms. Thus, when +a series contains a finite number only of negative or complex terms, +we may omit them and apply the theorems which follow to the +remainder. + +\Paragraph{167\Add{.}} It will be well to recall the following fundamental +theorems established in~\SecNo[§]{77}. + +\begin{Result} +\Item{A.} A series of positive terms must be convergent or diverge +to~$\infty$, and cannot oscillate. +\end{Result} + +\begin{Result} +\Item{B.} The necessary and sufficient condition that $\sum u_{n}$ should be +convergent is that there should be a number~$K$ such that +\[ +u_{0} + u_{1} + \dots + u_{n} < K +\] +for all values of~$n$. +\end{Result} + +\begin{Result} +\Topic{\Item{C.} The comparison theorem.} If $\sum u_{n}$~is convergent, and +$v_{n} \leq u_{n}$ for all values of~$n$, then $\sum v_{n}$~is convergent, and $\sum v_{n} \leq \sum u_{n}$. +More generally, if $v_{n} \leq Ku_{n}$, where $K$~is a constant, then $\sum v_{n}$ +is convergent and $\sum v_{n} \leq K \sum u_{n}$. And if $\sum u_{n}$~is divergent, and +$v_{n} \geq Ku_{n}$, then $\sum v_{n}$~is divergent.\footnote + {The last part of this theorem was not actually stated in \SecNo[§]{77}, but the reader + will have no difficulty in supplying the proof.} +\end{Result} + +Moreover, in inferring the convergence or divergence of~$\sum v_{n}$ +by means of one of these tests, it is sufficient to know that the +test is satisfied for \emph{sufficiently large} values of~$n$, \ie\ for all values +of~$n$ greater than a definite value~$n_{0}$. But of course the conclusion +that $\sum v_{n} \leq K \sum u_{n}$ does not necessarily hold in this case. + +A particularly useful case of this theorem is + +\begin{Result} +\Item{D.} If $\sum u_{n}$~is convergent \(divergent\) and $u_{n}/v_{n}$~tends to a limit +other than zero as $n \to \infty$, then $\sum v_{n}$~is convergent \(divergent\). +\end{Result} + +\Paragraph{168. First applications of these tests.} The one important +fact which we know at present, as regards the convergence of any +\PageSep{310} +special class of series, is that $\sum r^{n}$~is convergent if $r < 1$ and +divergent if $r \geq 1$.\footnote + {We shall use $r$ in this chapter to denote a number which is always positive + or zero.} +It is therefore natural to try to apply +Theorem~C, taking $u_{n} = r^{n}$. We at once find + +\begin{Result} +\Item{1.} The series~$\sum v_{n}$ is convergent if $v_{n} \leq Kr^{n}$, where $r < 1$, for all +sufficiently large values of~$n$. +\end{Result} + +When $K = 1$, this condition may be written in the form $v_{n}^{1/n} \leq r$. +Hence we obtain what is known as \Emph{Cauchy's test} for the convergence +of a series of positive terms; viz. + +\begin{Result} +\Item{2.} The series~$\sum v_{n}$ is convergent if $v_{n}^{1/n} \leq r$, where $r < 1$, for +all sufficiently large values of~$n$. +\end{Result} + +There is a corresponding test for divergence, viz. + +\begin{Result} +\Item{2\ia.} The series~$\sum v_{n}$ is divergent if $v_{n}^{1/n} \geq 1$ for an infinity of +values of~$n$. +\end{Result} + +This hardly requires proof, for $v_{n}^{1/n} \geq 1$ involves $v_{n} \geq 1$. The +two theorems 2~and~2\ia\ are of very wide application, but for +some purposes it is more convenient to use a different test of +convergence, viz. + +\begin{Result} +\Item{3.} The series~$\sum v_{n}$ is convergent if $v_{n+1}/v_{n} \leq r$, $r < 1$, for +all sufficiently large values of~$n$. +\end{Result} + +To prove this we observe that if $v_{n+1}/v_{n} \leq r$ when $n \geq n_{0}$ then +\[ +v_{n} = \frac{v_{n}}{v_{n-1}}\, + \frac{v_{n-1}}{v_{n-2}} \dots + \frac{v_{n_{0}+1}}{v_{n_{0}}}\, v_{n_{0}} + \leq \frac{v_{n_{0}}}{r^{n_{0}}} r^{n}; +\] +and the result follows by comparison with the convergent series~$\sum r^{n}$. +This test is known as \Emph{d'Alembert's test}. We shall see later that +it is less general, theoretically, than Cauchy's, in that Cauchy's +test can be applied whenever d'Alembert's can, and sometimes +when the latter cannot. Moreover the test for divergence which +corresponds to d'Alembert's test for convergence is much less +general than the test given by Theorem~2\ia. It is true, as the +reader will easily prove for himself, that if $v_{n+1}/v_{n} \geq r \geq 1$ for all +values of~$n$, or all sufficiently large values, then $\sum v_{n}$~is divergent. +But it is not true (see \Ex{lxvii}.~9) that this is so if only +$v_{n+1}/v_{n} \geq r \geq 1$ for an \emph{infinity} of values of~$n$, whereas in Theorem~2\ia\ +\PageSep{311} +our test had only to be satisfied for such an infinity of values. +None the less d'Alembert's test is very useful in practice, because +when $v_{n}$~is a complicated function $v_{n+1}/v_{n}$~is often much less +complicated and so easier to work with. + +In the simplest cases which occur in analysis it often happens +that $v_{n+1}/v_{n}$ or $v_{n}^{1/n}$ tends to a limit as $n \to \infty$.\footnote + {It will be proved in \Ref{Ch.}{IX} (\Ex{lxxxvii}.~36) that if $v_{n+1}/v_{n} \to l$ then $v_{n}^{1/n} \to l$. + That the converse is not true may be seen by supposing that $v_{n} = 1$ when $n$~is odd + and $v_{n} = 2$ when $n$~is even.} +When this limit +is less than~$1$, it is evident that the conditions of Theorems 2~or~3 +above are satisfied. Thus + +\begin{Result} +\Item{4.} If $v_{n}^{1/n}$ or $v_{n+1}/v_{n}$ tends to a limit less than unity as $n \to \infty$, +then the series~$\sum v_{n}$ is convergent. +\end{Result} + +It is almost obvious that if either function tend\DPnote{** [sic]} to a limit +greater than unity, then $\sum v_{n}$~is divergent. We leave the formal +proof of this as an exercise to the reader. But when $v_{n}^{1/n}$ or +$v_{n+1}/v_{n}$ tends to~$1$ these tests generally fail completely, and they +fail also when $v_{n}^{1/n}$ or $v_{n+1}/v_{n}$ oscillates in such a way that, while +always less than~$1$, it assumes for an infinity of values of~$n$ values +approaching indefinitely near to~$1$. And the tests which involve +$v_{n+1}/v_{n}$ fail even when that ratio oscillates so as to be sometimes +less than and sometimes greater than~$1$. When $v_{n}^{1/n}$~behaves in +this way Theorem~2\ia\ is sufficient to prove the divergence of the +series. But it is clear that there is a wide margin of cases in +which some more subtle tests will be needed. + +\begin{Examples}{LXVII.} +\Item{1.} Apply Cauchy's and d'Alembert's tests (as\PageLabel{311} +specialised in 4~above) to the series $\sum n^{k} r^{n}$, where $k$~is a positive rational +number. + +[Here $v_{n+1}/v_{n} = \{(n + 1)/n\}^{k} r \to r$, so that d'Alembert's test shows at once +that the series is convergent if $r < 1$ and divergent if $r > 1$. The test fails if +$r = 1$: but the series is then obviously divergent. Since $\lim n^{1/n} = 1$ (\Ex{xxvii}.~11), +Cauchy's test leads at once to the same conclusions.] + +\Item{2.} Consider the series $\sum(An^{k} + Bn^{k-1} + \dots + K) r^{n}$. [We may suppose $A$ +positive. If the coefficient of~$r^{n}$ is denoted by~$P(n)$, then $P(n)/n^{k} \to A$ and, +by D~of \SecNo[§]{167}, the series behaves like $\sum n^{k} r^{n}$.] + +\Item{3.} Consider +\[ +\sum \frac{An^{k} + Bn^{k-1} + \dots + K} + {\alpha n^{l} + \beta n^{l-1} + \dots + \kappa} r^{n}\quad +(A > 0,\ \alpha > 0). +\] + +[The series behaves like $\sum n^{k-l} r^{n}$. The case in which $r = 1$, $k < l$ requires +further consideration.] +\PageSep{312} + +\Item{4.} We have seen (\Ref{Ch.}{IV}, \MiscEx{IV}~17) that the series +\[ +\sum \frac{1}{n(n + 1)},\quad +\sum \frac{1}{n(n + 1)\dots (n + p)} +\] +are convergent. Show that Cauchy's and d'Alembert's tests both fail when +applied to them. [For $\lim u_{n}^{1/n} = \lim (u_{n+1}/u_{n}) = 1$.] + +\Item{5.} Show that the series~$\sum n^{-p}$, where $p$~is an integer not less than~$2$, is +convergent. [Since $\lim \{n(n + 1)\dots (n + p - 1)\}/n^{p} = 1$, this follows from the +convergence of the series considered in Ex.~4. It has already been shown +in \SecNo[§]{77},~\Eq{(7)} that the series is divergent if $p = 1$, and it is obviously divergent if +$p \leq 0$.] + +\Item{6.} Show that the series +\[ +\sum \frac{An^{k} + Bn^{k-1} + \dots + K} + {\alpha n^{l} + \beta n^{l-1} + \dots + \kappa} +\] +is convergent if $l > k + 1$ and divergent if $l \leq k + 1$. + +\Item{7.} If $m_{n}$~is a positive integer, and $m_{n+1} > m_{n}$, then the series $\sum r^{m_{n}}$ is convergent +if $r < 1$ and divergent if $r \geq 1$. For example the series $1 + r + r^{4} + r^{9} + \dots$ +is convergent if $r < 1$ and divergent if $r \geq 1$. + +\Item{8.} Sum the series $1 + 2r + 2r^{4} + \dots$ to $24$~places of decimals when $r = .1$ +and to $2$~places when $r = .9$. [If $r = .1$, then the first $5$~terms give the +sum $1.200\MS200\MS002\MS000\MS000\MS2$, and the error is +\[ +2r^{25} + 2r^{36} + \dots + < 2r^{25} + 2r^{36} + 2r^{47} + \dots + = 2r^{25}/(1 - r^{11}) + < 3/10^{25}. +\] +If $r = .9$, then the first $8$~terms give the sum $5.458\dots$, and the error is less +than $2r^{64}/(1 - r^{17}) < .003$.] + +\Item{9\Add{.}} If $0 < a < b < 1$, then the series $a + b + a^{2} + b^{2} + a^{3} + \dots$ is convergent. +Show that Cauchy's test may be applied to this series, but that d'Alembert's +test fails. [For +\[ +v_{2n+1}/v_{2n} = (b/a)^{n+1} \to \infty,\quad +v_{2n+2}/v_{2n+1} = b(a/b)^{n+2} \to 0.] +\] + +\Item{10.} The series $1 + r + \dfrac{r^{2}}{2!} + \dfrac{r^{3}}{3!} + \dots$ and $1 + r + \dfrac{r^{2}}{2^{2}} + \dfrac{r^{3}}{3^{3}} + \dots$ are convergent +for all positive values of~$r$. + +\Item{11.} If $\sum u_{n}$~is convergent then so are $\sum u_{n}^{2}$ and $\sum u_{n}/(1 + u_{n})$. + +\Item{12.} If $\sum u_{n}^{2}$~is convergent then so is $\sum u_{n}/n$. [For $2u_{n}/n \leq u_{n}^{2} + (1/n^{2})$ and +$\sum (1/n^{2})$~is convergent.] + +\Item{13.} Show that +\[ +%[** TN: In-line in the original] +1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots + = \frac{3}{4}\left(1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots \right) +\] +and +\[ +1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + + \frac{1}{5^{2}} + \frac{1}{6^{2}} + + \frac{1}{7^{2}} + \frac{1}{9^{2}} + \dots + = \frac{15}{16} \left(1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots\right). +\] +\PageSep{313} + +[To prove the first result we note that +\begin{align*} +1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + &= \left(1 + \frac{1}{2^{2}}\right) + + \left(\frac{1}{3^{2}} + \frac{1}{4^{2}}\right) + \dots\\ + &= 1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots + + \frac{1}{2^{2}} \left(1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots\right), +\end{align*} +by theorems \Eq{(8)}~and~\Eq{(6)} of~\SecNo[§]{77}.] + +\Item{14.} Prove by a \textit{reductio ad absurdum} that $\sum (1/n)$~is divergent. [If the +series were convergent we should have, by the argument used in Ex.~13, +\[ +1 + \tfrac{1}{2} + \tfrac{1}{3} + \dots + = (1 + \tfrac{1}{3} + \tfrac{1}{5} + \dots) + + \tfrac{1}{2} (1 + \tfrac{1}{2} + \tfrac{1}{3}+ \dots), +\] +or +\[ +\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{6} + \dots + = 1 + \tfrac{1}{3} + \tfrac{1}{5} + \dots +\] +which is obviously absurd, since every term of the first series is less than the +corresponding term of the second.]\PageLabel{313} +\end{Examples} + +\Paragraph{169.} Before proceeding further in the investigation of tests +of convergence and divergence, we shall prove an important general +theorem concerning series of positive terms. + +\begin{ParTheorem}{Dirichlet's Theorem.\protect\footnotemark} +The sum of a series of positive\footnotetext + {This theorem seems to have first been stated explicitly by Dirichlet in 1837. + It was no doubt known to earlier writers, and in particular to Cauchy.} +terms is the same in whatever order the terms are taken. +\end{ParTheorem} + +This theorem asserts that if we have a convergent series of +positive terms, $u_{0} + u_{1} + u_{2} + \dots$ say, and form any other series +\[ +v_{0} + v_{1} + v_{2} + \dots +\] +out of the same terms, by taking them in any new order, then the +second series is convergent and has the same sum as the first. +Of course no terms must be omitted: every~$u$ must come somewhere +among the~$v'$s, and \textit{vice versa}. + +The proof is extremely simple. Let $s$~be the sum of the series +of~$u'$s. Then the sum of any number of terms, selected from the~$u'$s, +is not greater than~$s$. But every~$v$ is a~$u$, and therefore the +sum of any number of terms selected from the~$v'$s is not greater +than~$s$. Hence $\sum v_{n}$~is convergent, and its sum~$t$ is not greater +than~$s$. But we can show in exactly the same way that $s \leq t$. +Thus $s = t$. + +\Paragraph{170. Multiplication of Series of Positive Terms.} An +immediate corollary from Dirichlet's Theorem is the following +theorem: \begin{Result}if $u_{0} + u_{1} + u_{2} + \dots$ and $v_{0} + v_{1} + v_{2} + \dots$ are two convergent +\PageSep{314} +series of positive terms, and $s$~and~$t$ are their respective sums, +then the series +\[ +u_{0} v_{0} + (u_{1} v_{0} + u_{0} v_{1}) + + (u_{2} v_{0} + u_{1} v_{1} + u_{0} v_{2}) + \dots +\] +is convergent and has the sum~$st$. +\end{Result} + +Arrange all the possible products of pairs~$u_{m}v_{n}$ in the form of +a doubly infinite array +\[ +\begin{array}{c|c|c|c|cc} +u_{0}v_{0}& u_{1}v_{0}& u_{2}v_{0}& u_{3}v_{0}& \dots\Strut \\ +\cline{1-1} +\TEntry{u_{0}v_{1}}& u_{1}v_{1}& u_{2}v_{1}& u_{3}v_{1}& \dots\Strut \\ +\cline{1-2} +\TEntry{u_{0}v_{2}}& \TEntry{u_{1}v_{2}}& u_{2}v_{2}& u_{3}v_{2}& \dots\Strut \\ +\cline{1-3} +\TEntry{u_{0}v_{3}}& \TEntry{u_{1}v_{3}}& \TEntry{u_{2}v_{3}}& u_{3}v_{3}& \dots\Strut \\ +\cline{1-4} +\TEntry{\dots}& \TEntry{\dots}& \TEntry{\dots}& \TEntry{\dots}& \dots\rlap{\;.}\Strut +\end{array} +\] +We can rearrange these terms in the form of a simply infinite +series in a variety of ways. Among these are the following. + +\Item{(1)} We begin with the single term~$u_{0}v_{0}$ for which $m + n = 0$; +then we take the two terms $u_{1}v_{0}$,~$u_{0}v_{1}$ for which $m + n = 1$; then +the three terms $u_{2}v_{0}$,~$u_{1}v_{1}$,~$u_{0}v_{2}$ for which $m + n = 2$; and so on. +We thus obtain the series +\[ +u_{0}v_{0} + (u_{1}v_{0} + u_{0}v_{1}) + + (u_{2}v_{0} + u_{1}v_{1} + u_{0}v_{2}) + \dots +\] +of the theorem. + +\Item{(2)} We begin with the single term~$u_{0}v_{0}$ for which both +suffixes are zero; then we take the terms $u_{1}v_{0}$,~$u_{1}v_{1}$,~$u_{0}v_{1}$ which +involve a suffix~$1$ but no higher suffix; then the terms $u_{2}v_{0}$, $u_{2}v_{1}$, +$u_{2}v_{2}$, $u_{1}v_{2}$,~$u_{0}v_{2}$ which involve a suffix~$2$ but no higher suffix; and +so on. The sums of these groups of terms are respectively equal to +\begin{multline*} +u_{0}v_{0},\quad +(u_{0} + u_{1})(v_{0} + v_{1}) - u_{0}v_{0},\\ +(u_{0} + u_{1} + u_{2})(v_{0} + v_{1} + v_{2}) - (u_{0} + u_{1})(v_{0} + v_{1}),\ +\dots +\end{multline*} +and the sum of the first $n + 1$ groups is +\[ +(u_{0} + u_{1} + \dots + u_{n})(v_{0} + v_{1} + \dots + v_{n}), +\] +and tends to~$st$ as $n \to \infty$. When the sum of the series is formed +in this manner the sum of the first one, two, three,~\dots\ groups +comprises all the terms in the first, second, third,~\dots\ rectangles +indicated in the diagram above. + +The sum of the series formed in the second manner is~$st$. +But the first series is (when the brackets are removed) a rearrangement +of the second; and therefore, by Dirichlet's Theorem, it converges +to the sum~$st$. Thus the theorem is proved. +\PageSep{315} + +\begin{Examples}{LXVIII.} +\Item{1\Add{.}} Verify that if $r < 1$ then +\[ +1 + r^{2} + r + r^{4} + r^{6} + r^{3} + \dots + = 1 + r + r^{3} + r^{2} + r^{5} + r^{7} + \dots + = 1/(1 - r). +\] + +\Item{2.\footnote + {In Exs.~2--4 the series considered are of course series of positive terms.}} +If either of the series $u_{0} + u_{1} + \dots$, $v_{0} + v_{1} + \dots$ is divergent, then so is +the series $u_{0}v_{0} + (u_{1}v_{0} + u_{0}v_{1}) + (u_{2}v_{0} + u_{1}v_{1} + u_{0}v_{2}) + \dots$, except in the trivial +case in which every term of one series is zero. + +\Item{3.} If the series $u_{0} + u_{1} + \dots$, $v_{0} + v_{1} + \dots$, $w_{0} + w_{1} + \dots$ converge to sums +$r$,~$s$,~$t$, then the series $\sum \lambda_{k}$, where $\lambda_{k} = \sum u_{m}v_{n}w_{p}$, the summation being extended +to all sets of values of $m$,~$n$,~$p$ such that $m + n + p = k$, converges to the +sum~$rst$. + +\Item{4.} If $\sum u_{n}$ and~$\sum v_{n}$ converge to sums $s$ and~$t$, then the series~$\sum w_{n}$, where +$w_{n} = \sum u_{l} v_{m}$, the summation extending to all pairs $l$,~$m$ for which $lm = n$, +converges to the sum~$st$. +\end{Examples} + +\Paragraph{171. Further tests for convergence and divergence.} +The examples on pp.~\PgNo[ref]{311}--\PgNo[ref]{313} suffice to show that there are +simple and interesting types of series of positive terms which +cannot be dealt with by the general tests of \SecNo[§]{168}. In fact, if +we consider the simplest type of series, in which $u_{n+1}/u_{n}$~tends +to a limit as $n \to \infty$, \emph{the tests of \SecNo[§]{168} generally fail when this limit +is~$1$}. Thus in \Ex{lxvii}.~5 these tests failed, and we had to fall +back upon a special device, which was in essence that of using +the series of \Ex{lxvii}.~4 as our comparison series, instead of +the geometric series. + +\begin{Remark} +The fact is that the geometric series, by comparison with which the tests +of \SecNo[§]{168} were obtained, is not only convergent but \emph{very rapidly} convergent, +far more rapidly than is necessary in order to ensure convergence. The tests +derived from comparison with it are therefore naturally very crude, and much +more delicate tests are often wanted. + +We proved in \Ex{xxvii}.~7 that $n^{k}r^{n} \to 0$ as $n \to \infty$, provided $r < 1$, whatever +value $k$ may have; and in \Ex{lxvii}.~1 we proved more than this, +viz.\ that the series $\sum n^{k}r^{n}$ is convergent. It follows that the sequence +$r$,~$r^{2}$, $r^{3}$,~\dots, $r^{n}$,~\dots, where $r < 1$, diminishes more rapidly than the sequence +$1^{-k}$,~$2^{-k}$, $3^{-k}$,~\dots, $n^{-k}$,~\dots. This seems at first paradoxical if $r$~is not much less +than unity, and $k$~is large. Thus of the two sequences +\[ +\tfrac{2}{3},\quad \tfrac{4}{9},\quad \tfrac{8}{27},\ \dots;\qquad +1,\quad \tfrac{1}{4096},\quad \tfrac{1}{531\MC441},\ \dots +\] +whose general terms are $(\frac{2}{3})^{n}$ and~$n^{-12}$, the second seems at first sight to +decrease far more rapidly. But this is far from being the case; if only we +go far enough into the sequences we shall find the terms of the first sequence +very much the smaller. For example, +\[ +(2/3)^{4} = 16/81 < 1/5,\quad +(2/3)^{12} < (1/5)^{3} < (1/10)^{2},\quad +(2/3)^{1000} < (1/10)^{166}, +\] +while +\[ +1000^{-12} = 10^{-36}; +\] +\PageSep{316} +so that the $1000$th~term of the first sequence is less than the $10^{130}$th~part of +the corresponding term of the second sequence. Thus the series $\sum (2/3)^{n}$ is +far more rapidly convergent than the series $\sum n^{-12}$, and even this series is +very much more rapidly convergent than~$\sum n^{-2}$.\footnote + {Five terms suffice to give the sum of~$\sum n^{-12}$ correctly to $7$~places of decimals, + whereas some $10,000,000$ are needed to give an equally good approximation to $\sum n^{-2}$. + A large number of numerical results of this character will be found in Appendix~III + (compiled by Mr~J. Jackson) to the author's tract `Orders of Infinity' (\textit{Cambridge + Math.\ Tracts}, No.~12).} +\end{Remark} + +\Paragraph{172.} We shall proceed to establish two further tests for the +convergence or divergence of series of positive terms, \Emph{Maclaurin's +(or Cauchy's) Integral Test} and \Emph{Cauchy's Condensation +Test}, which, though very far from being completely general, are +sufficiently general for our needs in this chapter. + +In applying either of these tests we make a further assumption +as to the nature of the function~$u_{n}$, about which we have so far +assumed only that it is positive. We assume that \emph{$u_{n}$~decreases +steadily with $n$}: \ie\ that $u_{n+1} \leq u_{n}$ for all values of~$n$\DPtypo{.}{,} or at any rate +all sufficiently large values. + +\begin{Remark} +This condition is satisfied in all the most important cases. From one +point of view it may be regarded as no restriction at all, so long as we are +dealing with series of positive terms: for in virtue of Dirichlet's theorem +above we may rearrange the terms without affecting the question of convergence +or divergence; and there is nothing to prevent us rearranging the +terms \emph{in descending order of magnitude}, and applying our tests to the series of +decreasing terms thus obtained. +\end{Remark} + +But before we proceed to the statement of these two tests, +we shall state and prove a simple and important theorem, which +we shall call \Emph{Abel's Theorem}.\footnote + {This theorem was discovered by Abel but forgotten, and rediscovered by + Pringsheim.} +This is a \emph{one-sided} theorem in +that it gives a sufficient test for divergence only and not for +convergence, but it is essentially of a more elementary character +than the two theorems mentioned above. + +\begin{Remark} +\Paragraph{173. Abel's (or Pringsheim's) Theorem.} \begin{Result}If $\sum u_{n}$~is a convergent series of +positive and decreasing terms, then $\lim nu_{n} = 0$. +\end{Result} + +%[** TN: Keeping notation \delta] +Suppose that $nu_{n}$ does not tend to zero. Then it is possible to find a +positive number~$\delta$ such that $nu_{n} \geq \delta$ for an infinity of values of~$n$. Let $n_{1}$ be +the first such value of~$n$; $n_{2}$~the next such value of~$n$ which is more than +\PageSep{317} +twice as large as~$n_{1}$; $n_{3}$~the next such value of~$n$ which is more than twice +as large as~$n_{2}$; and so on. Then we have a sequence of numbers $n_{1}$,~$n_{2}$, $n_{3}$,~\dots\ +such that $n_{2} > 2n_{1}$, $n_{3} > 2n_{2}$,~\dots\ and so $n_{2} - n_{1} > \frac{1}{2}n_{2}$, +$n_{3} - \DPtypo{n_{1}}{n_{2}} > \frac{1}{2}n_{3}$,~\dots; +and also $n_{1}u_{n_{1}} \geq \delta$, $n_{2}u_{n_{2}} \geq \delta$,~\dots. But, since $u_{n}$~decreases as $n$~increases, +we have +\begin{gather*} +u_{0} + u_{1} + \dots + u_{n_{1} - 1} \geq n_{1}u_{n_{1}} \geq \delta,\\ +u_{n_{1}} + \dots + u_{n_{2} - 1} + \geq (n_{2} - n_{1})u_{n_{2}} > \tfrac{1}{2} n_{2}u_{n_{2}} + \geq \tfrac{1}{2} \delta,\\ +u_{n_{2}} + \dots + u_{n_{3} - 1} + \geq (n_{3} - n_{2})u_{n_{3}} > \tfrac{1}{2} n_{3}u_{n_{3}} + \geq \tfrac{1}{2} \delta, +\end{gather*} +and so on. Thus we can bracket the terms of the series~$\sum u_{n}$ so as to obtain +a new series whose terms are severally greater than those of the divergent +series +\[ +\delta + \tfrac{1}{2} \delta + \tfrac{1}{2} \delta + \dots; +\] +and therefore $\sum u_{n}$~is divergent. +\end{Remark} + +\begin{Examples}{LXIX.} +\Item{1.} Use Abel's theorem to show that $\sum (1/n)$ and +$\sum \{1/(an + b)\}$ are divergent. [Here $nu_{n} \to 1$ or $nu_{n} \to 1/a$.] + +\Item{2.} Show that Abel's theorem is not true if we omit the condition that $u_{n}$~decreases +as $n$~increases. [The series +\[ +1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + + \frac{1}{4} + + \frac{1}{5^{2}} + \frac{1}{6^{2}} + \frac{1}{7^{2}} + \frac{1}{8^{2}} + + \frac{1}{9} + + \frac{1}{10^{2}} + \dots, +\] +in which $u_{n} = 1/n$ or $1/n^{2}$, according as $n$~is or is not a perfect square, is +convergent, since it may be rearranged in the form +\[ +\frac{1}{2^{2}} + \frac{1}{3^{2}} + + \frac{1}{5^{2}} + \frac{1}{6^{2}} + \frac{1}{7^{2}} + \frac{1}{8^{2}} + + \frac{1}{10^{2}} + + \dots + \left(1 + \frac{1}{4} + \frac{1}{9} + \dots\right), +\] +and each of these series is convergent. But, since $nu_{n} = 1$ whenever $\DPtypo{u}{n}$~is a +perfect square, it is clearly not true that $nu_{n} \to 0$.] + +\Item{3.} \emph{The converse of Abel's theorem is not true}, \ie\ it is not true that, if $u_{n}$~decreases +with~$n$ and $\lim nu_{n} = 0$, then $\sum u_{n}$~is convergent. + +[Take the series $\sum(1/n)$ and multiply the first term by~$1$, the second by~$\frac{1}{2}$, +the next two by~$\frac{1}{3}$, the next four by~$\frac{1}{4}$, the next eight by~$\frac{1}{5}$, and so on. On +grouping in brackets the terms of the new series thus formed we obtain +\[ +1 + \tfrac{1}{2} · \tfrac{1}{2} + + \tfrac{1}{3} \left(\tfrac{1}{3} + \tfrac{1}{4}\right) + + \tfrac{1}{4} \left(\tfrac{1}{5} + \tfrac{1}{6} + \tfrac{1}{7} + \tfrac{1}{8}\right) + \dots; +\] +and this series is divergent, since its terms are greater than those of +\[ +1 + \tfrac{1}{2} · \tfrac{1}{2} + + \tfrac{1}{3} · \tfrac{1}{2} + + \tfrac{1}{4} · \tfrac{1}{2} + \dots, +\] +which is divergent. But it is easy to see that the terms of the series +\[ +1 + \tfrac{1}{2} · \tfrac{1}{2} + + \tfrac{1}{3} · \tfrac{1}{3} + + \tfrac{1}{3} · \tfrac{1}{4} + + \tfrac{1}{4} · \tfrac{1}{5} + + \tfrac{1}{4} · \tfrac{1}{6} + \dots +\] +satisfy the condition that $nu_{n} \to 0$. In fact $nu_{n} = 1/\nu$ if $2^{\nu-2} < n \leq 2^{\nu-1}$, and +$\nu \to \infty$ as $n \to \infty$.] +\end{Examples} +\PageSep{318} + +\Paragraph{174. Maclaurin's (or Cauchy's) Integral Test.\protect\footnotemark} +If $u_{n}$~decreases + \footnotetext{The test was discovered by Maclaurin and rediscovered by Cauchy, to whom + it is usually attributed.}% +steadily as $n$~increases, we can write $u_{n} = \phi(n)$ and +suppose that $\phi(n)$~is the value assumed, when $x = n$, by a continuous +and steadily decreasing function~$\phi(x)$ of the continuous +variable~$x$. Then, If $\nu$~is any positive integer, we have +\[ +\phi(\nu - 1) \geq \phi(x) \geq \phi(\nu) +\] +when $\nu - 1 \leq x \leq \nu$. Let +\[ +v_{\nu} + = \phi(\nu - 1) - \int_{\nu-1}^{\nu} \phi(x)\, dx + = \int_{\nu-1}^{\nu} \{\phi(\nu - 1) - \phi(x)\}\, dx, +\] +so that +\[ +0 \leq v_{\nu} \leq \phi(\nu - 1) - \phi(\nu). +\] +Then $\sum v_{\nu}$~is a series of positive terms, and +\[ +v_{2} + v_{3} + \dots + v_{n} \leq \phi(1) - \phi(n) \leq \phi(1). +\] +Hence $\sum v_{\nu}$~is convergent, and so $v_{2} + v_{3} + \dots + v_{n}$ or +\[ +\sum_{1}^{n-1} \phi(\nu) - \int_{1}^{n} \phi(x)\, dx +\] +tends to a positive limit as $n \to \infty$. + +Let us write +\[ +\Phi(\xi) = \int_{1}^{\xi} \phi(x)\, dx, +\] +so that $\Phi(\xi)$~is a continuous and steadily increasing function of~$\xi$. +Then +\[ +u_{1} + u_{2} + \dots + u_{n-1} - \Phi(n) +\] +tends to a positive limit, not greater than~$\phi(1)$, as $n \to \infty$. Hence +$\sum u_{\nu}$~is convergent or divergent according as $\Phi(n)$~tends to a limit +or to infinity as $n \to \infty$, and therefore, since $\Phi(n)$~increases steadily, +according as $\Phi(\xi)$~tends to a limit or to infinity as $\xi \to \infty$. Hence +\begin{Result}if $\phi(x)$~is a function of~$x$ which is positive and continuous for all +values of~$x$ greater than unity, and decreases steadily as $x$~increases, +then the series +\[ +\phi(1) + \phi(2) + \dots +\] +does or does not converge according as +\[ +\Phi(\xi) = \int_{1}^{\xi} \phi(x)\, dx +\] +does or does not tend to a limit~$l$ as $\xi \to \infty$; and, in the first case, +the sum of the series is not greater than $\phi(1) + l$. +\end{Result} +\PageSep{319} + +\begin{Remark} +The sum must in fact be less than~$\phi(1) + l$. For it follows from \Eq{(6)}~of +\SecNo[§]{160}, and \Ref{Ch.}{VII}, \MiscEx{VII}~41, that $v_{\nu} < \phi(\nu - 1) - \phi(\nu)$, unless $\phi(x) = \phi(\nu)$ +throughout the interval $\DPmod{(\nu - 1, \nu)}{[\nu - 1, \nu]}$; and this cannot be true for all values of~$\nu$. +\end{Remark} + +\begin{Examples}{LXX.} +\Item{1.} Prove that +\[ +\sum_{1}^{\infty} \frac{1}{n^{2} + 1} < \tfrac{1}{2} + \tfrac{1}{4}\pi\Add{.} +\] + +\Item{2.} Prove that +\[ +-\tfrac{1}{2} \pi < \sum_{1}^{\infty} \frac{a}{a^{2} + n^{2}} < \tfrac{1}{2} \pi. +\] +\MathTrip{1909.} + +\Item{3.} Prove that if $m > 0$ then +\[ +\frac{1}{m^{2}} + \frac{1}{(m + 1)^{2}} + \frac{1}{(m + 2)^{2}} + \dots + < \frac{m + 1}{m}\Add{.} +\] +\end{Examples} + +\Paragraph{175. The series $\sum n^{-s}$.} By far the most important application +of the Integral Test is to the series +\[ +1^{-s} + 2^{-s} + 3^{-s} + \dots + n^{-s} + \dots, +\] +where $s$~is any rational number. We have seen already (\SecNo[§]{77} and +\Exs{lxvii}.~14, \Exs[]{lxix}.~1) that the series is divergent when $s = 1$. + +If $s \leq 0$ then it is obvious that the series is divergent. If +$s > 0$ then $u_{n}$ decreases as $n$~increases, and we can apply the test. +Here +\[ +\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x^{s}} = \frac{\xi^{1-s} - 1}{1 - s}, +\] +unless $s = 1$. If $s > 1$ then $\xi^{1-s} \to 0$ as $\xi \to \infty$, and +\[ +\Phi(\xi) \to \frac{1}{(s - 1)} = l, +\] +say. And if $s < 1$ then $\xi^{1-s} \to \infty$ as $\xi \to \infty$, and so $\Phi(\xi) \to \infty$. +Thus \begin{Result}the series $\sum n^{-s}$ is convergent if $s > 1$, divergent if $s \leq 1$, and in +the first case its sum is less than $s/(s - 1)$. +\end{Result} + +\begin{Remark} +So far as divergence for $s < 1$ is concerned, this result might have +been derived at once from comparison with~$\sum (1/n)$, which we already know +to be divergent. + +It is however interesting to see how the Integral Test may be applied to +the series~$\sum (1/n)$, when the preceding analysis fails. In this case +\[ +\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x}, +\] +and it is easy to see that $\Phi(\xi) \to \infty$ as $\xi \to \infty$. For if $\xi > 2^{n}$ then +\[ +\Phi(\xi) > \int_{1}^{2^{n}} \frac{dx}{x} + = \int_{1}^{2} \frac{dx}{x} + + \int_{2}^{4} \frac{dx}{x} + \dots + + \DPtypo{\int_{2^{n}}^{2^{n-1}}}{\int_{2^{n-1}}^{2^{n}}} \frac{dx}{x}. +\] +\PageSep{320} +But by putting $x = 2^{r}u$ we obtain +\[ +\int_{2^{r}}^{2^{r+1}} \frac{dx}{x} = \int_{1}^{2} \frac{du}{u}, +\] +and so $\ds\Phi(\xi) > n\int_{1}^{2} \frac{du}{u}$, which shows that $\Phi(\xi) \to \infty$ as $\xi \to \infty$. +\end{Remark} + +\begin{Examples}{LXXI.} +\Item{1.} Prove by an argument similar to that used above, +and without integration, that $\ds\Phi(\xi) = \int_{1}^{\xi} \frac{dx}{x^{s}}$, where $s < 1$, tends to infinity with~$\xi$. + +\Item{2.} The series $\sum n^{-2}$, $\sum n^{-3/2}$, $\sum n^{-11/10}$ are convergent, and their sums are +not greater than $2$,~$3$,~$11$ respectively. The series $\sum n^{-1/2}$, $\sum n^{-10/11}$ are +divergent. + +\Item{3.} The series $\sum n^{s}/(n^{t} + a)$, where $a > 0$, is convergent or divergent according +as $t > 1 + s$ or $t \leq 1 + s$. [Compare with~$\sum n^{s-t}$.] + +\Item{4.} Discuss the convergence or divergence of the series +\[ +\tsum(a_{1}n^{s_{1}} + a_{2}n^{s_{2}} + \dots + a_{k}n^{s_{k}})/ + (b_{1}n^{t_{1}} + b_{2}n^{t_{2}} + \dots + b_{l}n^{t_{l}}), +\] +where all the letters denote positive numbers and the $s$'s and~$t$'s are rational +and arranged in descending order of magnitude. + +\Item{5.} Prove that +\begin{gather*} +2\sqrt{n} - 2 + < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} + < 2\sqrt{n} - 1, \\ +\tfrac{1}{2} \pi + < \frac{1}{2\sqrt{1}} + \frac{1}{3\sqrt{2}} + \frac{1}{4\sqrt{3}} + \dots + < \tfrac{1}{2}(\pi + 1). +\end{gather*} +\MathTrip{1911.} + +\Item{6.} If $\phi(n) \to l > 1$ then the series $\sum n^{-\phi(n)}$ is convergent. If $\phi(n) \to l < 1$ +then it is divergent. +\end{Examples} + +\Paragraph{176. Cauchy's Condensation Test.} The second of the +two tests mentioned in \SecNo[§]{172} is as follows: \begin{Result}if $u_{n} = \phi(n)$ is a +decreasing function of~$n$, then the series $\sum \phi(n)$~is convergent or +divergent according as $\sum 2^{n}\phi(2^{n})$~is convergent or divergent. +\end{Result} + +We can prove this by an argument which we have used +already (\SecNo[§]{77}) in the special case of the series $\sum(1/n)$. In the +first place +\begin{gather*} +\phi(3) + \phi(4) \geq 2\phi(4), \\ +\phi(5) + \phi(6) + \dots + \phi(8) \geq 4\phi(8), \\ +%[** TN: Hard-coded width] +\DotRow{2.5in} \\ +\phi(2^{n} + 1) + \phi(2^{n} + 2) + \dots + \phi(2^{n+1}) \geq 2^{n}\phi(2^{n+1}). +\end{gather*} +If $\sum 2^{n}\phi(2^{n})$ diverges then so do $\sum 2^{n+1}\phi(2^{n+1})$ and $\sum 2^{n}\phi(2^{n+1})$, +and then the inequalities just obtained show that $\sum\phi(n)$~diverges. +\PageSep{321} + +On the other hand +\[ +\phi(2) + \phi(3) \leq 2\phi(2),\quad +\phi(4) + \phi(5) + \dots + \phi(7) \leq 4\phi(4), +\] +and so on. And from this set of inequalities it follows that +if $\sum 2^{n}\phi(2^{n})$ converges then so does $\sum \phi(n)$. Thus the theorem is +established. + +For our present purposes the field of application of this test is +practically the same as that of the Integral Test. It enables us +to discuss the series $\sum n^{-s}$ with equal ease. For $\sum n^{-s}$ will converge +or diverge according as $\sum 2^{n}2^{-ns}$ converges or diverges, \ie\ according +as $s > 1$ or $s \leq 1$. + +\begin{Examples}{LXXII.} +\Item{1.} Show that if $a$~is any positive integer greater +than~$1$ then $\sum \phi(n)$~is convergent or divergent according as $\sum a^{n}\phi(a^{n})$ is +convergent or divergent. [Use the same arguments as above, taking groups +of $a$,~$a^{2}$, $a^{3}$,~\dots\ terms.] + +\Item{2.} If $\sum 2^{n}\phi(2^{n})$ converges then it is obvious that $\lim 2^{n}\phi(2^{n}) = 0$. Hence +deduce Abel's Theorem of~\SecNo[§]{173}. +\end{Examples} + +\Paragraph{177. Infinite Integrals.} The Integral Test of \SecNo[§]{174} shows +that, if $\phi(x)$~is a positive and decreasing function of~$x$, then the +series $\sum \phi(n)$ is convergent or divergent according as the integral +function~$\Phi(x)$ does or does not tend to a limit as $x \to \infty$. Let +us suppose that it does tend to a limit, and that +\[ +\lim_{x \to \infty} \int_{1}^{x} \phi(t)\, dt = l. +\] +Then we shall say that \emph{the integral +\[ +\int_{1}^{\infty} \phi(t)\, dt +\] +is \Emph{convergent}, and has the value~$l$}; and we shall call the +integral an \Emph{infinite integral}. + +So far we have supposed $\phi(t)$ positive and decreasing. But it +is natural to extend our definition to other cases. Nor is there +any special point in supposing the lower limit to be unity. We +are accordingly led to formulate the following definition: + +\begin{Defn} +If $\phi(t)$~is a function of~$t$ continuous when $t \geq a$, and +\[ +\lim_{x \to \infty} \int_{a}^{x} \phi(t)\, dt = l, +\] +\PageSep{322} +then we shall say that the infinite integral +\[ +\int_{a}^{\infty}\phi(t)\, dt +\Tag{(1)} +\] +is convergent and has the value~$l$. +\end{Defn} + +The ordinary integral between limits $a$~and~$A$, as defined in +\Ref{Ch.}{VII}, we shall sometimes call in contrast a \Emph{finite} integral. + +On the other hand, when +\[ +\int_{a}^{x}\phi(t)\, dt \to \infty, +\] +we shall say that the integral \emph{diverges} to~$\infty$, and we can give a +similar definition of divergence to~$-\infty$. Finally, when none of +these alternatives occur, we shall say that the integral \emph{oscillates}, +\emph{finitely} or \emph{infinitely}, as $x \to \infty$. + +These definitions suggest the following remarks. + +\begin{Remark} +\Itemp{(i)} If we write +\[ +\int_{a}^{x}\phi(t)\, dt = \Phi(x), +\] +then the integral converges, diverges, or oscillates according as $\Phi(x)$~tends to +a limit, tends to~$\infty$ (or to~$-\infty$), or oscillates, as $x \to \infty$. If $\Phi(x)$ tends to a +limit, which we may denote by~$\Phi(\infty)$, then the value of the integral is~$\Phi(\infty)$. +More generally, if $\Phi(x)$~is any integral function of~$\phi(x)$, then the value of the +integral is $\Phi(\infty) - \Phi(a)$. + +\Itemp{(ii)} In the special case in which $\phi(t)$~is always positive it is clear +that $\Phi(x)$~is an increasing function of~$x$. Hence the only alternatives are +convergence and divergence to~$\infty$. + +\Itemp{(iii)} The integral~\Eq{(1)} of course depends on~$a$, but is quite independent of~$t$, +and is in no way altered by the substitution of any other letter for~$t$ (cf.~\SecNo[§]{157}). + +\Itemp{(iv)} Of course the reader will not be puzzled by the use of the term +\emph{infinite integral} to denote something which has a definite value such as +$2$ or~$\frac{1}{2}\pi$. The distinction between an infinite integral and a finite integral +is similar to that between an infinite series and a finite series: no one supposes +that an infinite series is necessarily divergent. + +\Itemp{(v)} The integral $\ds\int_{a}^{x} \phi(t)\, dt$ was defined in \SecNo[§§]{156}~and~\SecNo{157} as a \emph{simple} +limit, \ie\ the limit of a certain finite sum. The infinite integral is therefore +\emph{the limit of a limit}, or what is known as a \emph{repeated} limit. The notion of the +infinite integral is in fact essentially more complex than that of the finite +integral, of which it is a development. +\PageSep{323} + +\Itemp{(vi)} The Integral Test of \SecNo[§]{174} may now be stated in the form: \begin{Result}if $\phi(x)$~is +positive and steadily decreases as $x$~increases, then the infinite series $\sum\phi(n)$ and the +infinite integral $\ds\int_{1}^{\infty} \phi(x)\, dx$ converge or diverge together. +\end{Result} + +\Itemp{(vii)} The reader will find no difficulty in formulating and proving theorems +for infinite integrals analogous to those stated in \Eq{(1)}--\Eq{(6)} of \SecNo[§]{77}. Thus the +result analogous to~\Eq{(2)} is that \begin{Result}if $\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent, and $b > a$, then +$\ds\int_{b}^{\infty} \phi(x)\, dx$ is convergent and +\[ +\int_{a}^{\infty} \phi(x)\, dx + = \int_{a}^{b} \phi(x)\, dx + \int_{b}^{\infty}\phi(x)\, dx. +\] +\end{Result} +\end{Remark} + +\Paragraph{178. The case in which $\phi(x)$~is positive.} It is natural +to consider what are the general theorems, concerning the convergence +or divergence of the infinite integral~\Eq{(1)} of \SecNo[§]{177}, +analogous to theorems A--D of~\SecNo[§]{167}. That A~is true of integrals +as well as of series we have already seen in \SecNo[§]{177},~\Eq{(ii)}. Corresponding +to~B we have the theorem that \begin{Result}the necessary and sufficient +condition for the convergence of the integral~\Eq{(1)} is that it should be +possible to find a constant~$K$ such that +\[ +\int_{a}^{x} \phi(t)\, dt < K +\] +for all values of~$x$ greater than~$a$. +\end{Result} + +Similarly, corresponding to~C, we have the theorem: \begin{Result}if +$\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent, and $\psi(x) \leq K\phi(x)$ for all values of~$x$ +greater than~$a$, then $\ds\int_{a}^{\infty} \psi(x)\, dx$ is convergent and +%[** TN: Code hack; place envt. end here to avoid paragraph break below.] +\end{Result} +\[ +\int_{a}^{\infty} \psi(x)\, dx \leq K\int_{a}^{\infty} \phi(x)\, dx. +\] +We leave it to the reader to formulate the corresponding test for +divergence. + +We may observe that \DPchg{D'Alembert's}{d'Alembert's} test (\SecNo[§]{168}), depending +as it does on the notion of successive terms, has no analogue for +integrals; and that the analogue of Cauchy's test is not of much +importance, and in any case could only be formulated when we +have investigated in greater detail the theory of the function +\PageSep{324} +$\phi(x) = r^{x}$, as we shall do in \Ref{Ch.}{IX}\@. The most important special +tests are obtained by comparison with the integral +\[ +\int_{a}^{\infty} \frac{dx}{x^{s}}\quad (a > 0), +\] +whose convergence or divergence we have investigated in \SecNo[§]{175}, +and are as follows: \begin{Result}if $\phi(x) < Kx^{-s}$, where $s > 1$, when $x \geq a$, then +$\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent; and if $\phi(x) > Kx^{-s}$, where $s \leq 1$, when +$x \geq a$, then the integral is divergent; and in particular, if +$\lim x^{s}\phi(x) = l$, where $l > 0$, then the integral is convergent or +divergent according as $s > 1$ or $s \leq 1$. +\end{Result} + +\begin{Remark} +There is one fundamental property of a convergent infinite series in +regard to which the analogy between infinite series and infinite integrals +breaks down. If $\sum \phi(n)$~is convergent then $\phi(n) \to 0$; but it is \emph{not} always +true, even when $\phi(x)$~is always positive, that if $\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent +then $\phi(x) \to 0$. + +Consider for example the function~$\phi(x)$ whose graph is indicated by the +thick line in the figure. Here the height of the peaks corresponding to the +points $x = 1$, $2$,~$3$,~\dots\ is in each case unity, and the breadth of the peak corresponding +%[Illustration: Fig. 50.] +\Figure[3in]{50}{p324} +to $x = n$ is~$2/(n + 1)^{2}$. The area of the peak is~$1/(n + 1)^{2}$, and it is +evident that, for any value of~$\xi$, +\[ +\int_{0}^{\xi} \phi(x)\, dx < \sum_{0}^{\infty} \frac{1}{(n + 1)^{2}}, +\] +so that $\ds\int_{0}^{\infty} \phi(x)\, dx$ is convergent; but it is not true that $\phi(x) \to 0$\Add{.} +\end{Remark} + +\begin{Examples}{LXXIII.} +\Item{1.} The integral +\[ +\int_{a}^{\infty} \frac{\alpha x^{r} + \beta x^{r-1} + \dots + \lambda} + {Ax^{s} + Bx^{s-1} + \dots + L}\, dx, +\] +where $\alpha$ and~$A$ are positive and $a$~is greater than the greatest root of the +denominator, is convergent if $s > r + 1$ and otherwise divergent. +\PageSep{325} + +\Item{2.} Which of the integrals +%[** TN: All are displayed on one line in the original] +$\ds\int_{a}^{\infty} \frac{dx}{\sqrt{x}}$, +$\ds\int_{a}^{\infty} \frac{dx}{x^{4/3}}$, +\[ +\int_{a}^{\infty} \frac{dx}{c^{2} + x^{2}},\quad +\int_{a}^{\infty} \frac{x\, dx}{c^{2} + x^{2}},\quad +\int_{a}^{\infty} \frac{x^{2}\, dx}{c^{2} + x^{2}},\quad +\int_{a}^{\infty} \frac{x^{2}\, dx}{\alpha + 2\beta x^{2} + \gamma x^{4}} +\] +are convergent? In the first two integrals it is supposed that $a > 0$, and +in the last that $a$~is greater than the greatest root (if any) of the denominator. + +\Item{3.} The integrals +\[ +\int_{a}^{\xi} \cos x\, dx,\quad +\int_{a}^{\xi} \sin x\, dx,\quad +\int_{a}^{\xi} \cos(\alpha x + \beta)\, dx +\] +oscillate finitely as $\xi \to \infty$. + +\Item{4.} The integrals +\[ +\int_{a}^{\xi} x\cos x\, dx,\quad +\int_{a}^{\xi} x^{2}\sin x\, dx\quad +\int_{a}^{\xi} x^{n} \cos(\alpha x + \beta)\, dx, +\] +where $n$~is any positive integer, oscillate infinitely as $\xi \to \infty$. + +\Item{5.} \Topic{Integrals to~$-\infty$.} If $\ds\int_{\xi}^{a} \phi(x)\, dx$ tends to a limit~$l$ as $\xi \to -\infty$, then we +say that $\ds\int_{-\infty}^{a} \phi(x)\, dx$ is convergent and equal to~$l$. Such integrals possess +properties in every respect analogous to those of the integrals discussed in the +preceding sections: the reader will find no difficulty in formulating them. + +\Item{6.} \Topic{Integrals from~$-\infty$ to~$+\infty$.} If the integrals +\[ +\int_{-\infty}^{a} \phi(x)\, dx,\quad +\int_{a}^{\infty} \phi(x)\, dx +\] +are both convergent, and have the values $k$,~$l$ respectively, then we say that +\[ +\int_{-\infty}^{\infty} \phi(x)\, dx +\] +is convergent and has the value $k + l$. + +\Item{7.} Prove that +\[ +\int_{-\infty}^{0} \frac{dx}{1 + x^{2}} + = \int_{0}^{\infty} \frac{dx}{1 + x^{2}} + = \tfrac{1}{2} \int_{-\infty}^{\infty} \frac{dx}{1 + x^{2}} + = \tfrac{1}{2}\pi. +\] + +\Item{8.} Prove generally that +\[ +\int_{-\infty}^{\infty} \phi(x^{2})\, dx = 2\int_{0}^{\infty} \phi(x^{2})\, dx, +\] +provided that the integral $\ds\int_{0}^{\infty} \phi(x^{2})\, dx$ is convergent. + +\Item{9.} Prove that if $\ds\int_{0}^{\infty} x\phi(x^{2})\, dx$ is convergent then $\ds\int_{-\infty}^{\infty} x\phi(x^{2})\, dx = 0$. +\PageSep{326} + +\Item{10.} \Topic{Analogue of Abel's Theorem of \SecNo[§]{173}.} \emph{If $\phi(x)$~is positive and +steadily decreases, and $\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent, then $x\phi(x) \to 0$.} Prove this +(\ia)~by means of Abel's Theorem and the Integral Test and (\ib)~directly, by +arguments analogous to those of~\SecNo[§]{173}. + +\Item{11.} If $a = x_{0} < x_{1} < x_{2} < \dots$ and $x_{n} \to \infty$, and $\ds u_{n}= \int_{x_{n}}^{x_{n+1}} \phi(x)\, dx$, then the +convergence of $\ds\int_{a}^{\infty} \phi(x)\, dx$ involves that of $\sum u_{n}$. If $\phi(x)$~is always positive +the converse statement is also true. [That the converse is not true in +general is shown by the example in which $\phi(x) = \cos x$, $x_{n} = n\pi$.] +\end{Examples} + +\Paragraph{179. Application to infinite integrals of the rules for +substitution and integration by parts.} The rules for the +transformation of a definite integral which were discussed in +\SecNo[§]{161} may be extended so as to apply to infinite integrals. + +\Item{(1)} \Topic{Transformation by substitution.} Suppose that +\[ +\int_{a}^{\infty} \phi(x)\, dx +\Tag{(1)} +\] +is convergent. Further suppose that, for any value of~$\xi$ greater +than~$a$, we have, as in~\SecNo[§]{161}, +\[ +\int_{a}^{\xi} \phi(x)\, dx = \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt, +\Tag{(2)} +\] +where $a = f(b)$, $\xi = f(\tau)$. Finally suppose that the functional +relation $x = f(t)$ is such that $x \to \infty$ as $t \to \infty$. Then, making $\tau$ +and so~$\xi$ tend to~$\infty$ in~\Eq{(2)}, we see that the integral +\[ +\int_{b}^{\infty} \phi\{f(t)\}f'(t)\, dt +\Tag{(3)} +\] +is convergent and equal to the integral~\Eq{(1)}. + +On the other hand it may happen that $\xi \to \infty$ as $\tau \to -\infty$ +or as $\tau \to c$. In the first case we obtain +\begin{alignat*}{2} +%[** TN: Unaligned in the original] +\int_{a}^{\infty} \phi(x)\, dx + &= &&\lim_{\tau\to-\infty} \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt\\ + &= -&&\lim_{\tau\to-\infty} \int_{\tau}^{b} \phi\{f(t)\}f'(t)\, dt + = -\int_{-\infty}^{b} \phi\{f(t)\}f'(t)\, dt. +\end{alignat*} +In the second case we obtain +\[ +\int_{a}^{\infty} \phi(x)\, dx + = \lim_{\tau\to c} \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt. +\Tag{(4)} +\] +We shall return to this equation in~\SecNo[§]{181}. +\PageSep{327} + +There are of course corresponding results for the integrals +\[ +\int_{-\infty}^{a} \phi(x)\, dx,\quad +\int_{-\infty}^{\infty} \phi(x)\, dx, +\] +which it is not worth while to set out in detail: the reader will +be able to formulate them for himself. + +\begin{Examples}{LXXIV.} +\Item{1.} Show, by means of the substitution $x = t^{\alpha}$, +that if $s > 1$ and $\alpha >0$ then +\[ +\int_{1}^{\infty} x^{-s}\, dx = \alpha\int_{1}^{\infty} t^{\alpha(1-s) - 1}\, dt; +\] +and verify the result by calculating the value of each integral directly. + +\Item{2.} If $\ds\int_{a}^{\infty} \phi(x)\, dx$ is convergent then it is equal to one or other of +\[ + \alpha\int_{(a-\beta)/\alpha}^{\infty} \phi(\alpha t + \beta)\, dt,\quad +-\alpha\int_{-\infty}^{(a-\beta)/\alpha} \phi(\alpha t + \beta)\, dt, +\] +according as $\alpha$~is positive or negative. + +\Item{3.} If $\phi(x)$~is a positive and steadily decreasing function of~$x$, and $\alpha$~and~$\beta$ +are any positive numbers, then the convergence of the series $\sum \phi(n)$ implies +and is implied by that of the series $\sum \phi(\alpha n + \beta)$. + +[It follows at once, on making the substitution $x = \alpha t + \beta$, that the +integrals +\[ +\int_{a}^{\infty} \phi(x)\, dx,\quad +\int_{(a-\beta)/\alpha}^{\infty} \phi(\alpha t + \beta)\, dt +\] +converge or diverge together. Now use the Integral Test.] + +\Item{4.} Show that +\[ +%[** TN: In-line in the original] +\int_{1}^{\infty} \frac{dx}{(1 + x)\sqrt{x}} = \tfrac{1}{2} \pi. +\] + +%[** TN: Added paragraph break] +[Put $x = t^{2}$.] + +\Item{5.} Show that +\[ +\int_{0}^{\infty} \frac{\sqrt{x}}{(1 + x)^{2}}\, dx = \tfrac{1}{2}\pi. +\] + +[Put $x = t^{2}$ and integrate by parts.] + +\Item{6.} If $\phi(x) \to h$ as $x \to \infty$, and $\phi(x) \to k$ as $x \to -\infty$, then +\[ +\int_{-\infty}^{\infty} \{\phi(x - a) - \phi(x - b)\}\, dx = -(a - b)(h - k). +\] + +[For +\begin{alignat*}{2} +%[** TN: Re-broken] +\int_{-\xi'}^{\xi} \{\phi(x - a) - \phi(x - b)\}\, dx + &= \int_{-\xi'}^{\xi} \phi(x - a)\, dx &&- \int_{-\xi'}^{\xi} \phi(x - b)\, dx\\ + &= \int_{-\xi'-a}^{\xi-a} \phi(t)\, dt &&- \int_{-\xi'-b}^{\xi-b} \phi(t)\, dt\\ + &= \int_{-\xi'-a}^{-\xi'-b} \phi(t)\, dt &&- \int_{\xi-a}^{\xi-b} \phi(t)\, dt. +\end{alignat*} +\PageSep{328} +The first of these two integrals may be expressed in the form +\[ +(a - b) k + \int_{-\xi'-a}^{-\xi'-b} \rho\, dt, +\] +where $\rho \to 0$ as $\xi' \to \infty$, and the modulus of the last integral is less than or +equal to $|a - b| \kappa$, where $\kappa$~is the greatest value of $\rho$ throughout the interval +$\DPmod{(-\xi' - a, -\xi' - b)}{[-\xi' - a, -\xi' - b]}$. Hence +\[ +\int_{-\xi'-a}^{-\xi'-b} \phi(t)\, dt \to (a - b) k. +\] +The second integral may be discussed similarly.] +\end{Examples} + +\Item{(2)} \Topic{Integration by parts.} The formula for integration by +parts (\SecNo[§]{161}) is +\[ +\int_{a}^{\xi} f(x)\phi'(x)\, dx + = f(\xi)\phi(\xi) - f(a)\phi(a) - \int_{a}^{\xi} f'(x)\phi(x)\, dx. +\] + +Suppose now that $\xi \to \infty$. Then if any two of the three terms +in the above equation which involve~$\xi$ tend to limits, so does the +third, and we obtain the result +\[ +\int_{a}^{\infty} f(x)\phi'(x)\, dx + = \lim_{\xi\to\infty} f(\xi)\phi(\xi) - f(a)\phi(a) + - \int_{a}^{\infty} f'(x)\phi(x)\, dx. +\] +There are of course similar results for integrals to~$-\infty$, or from +$-\infty$ to~$\infty$. + +\begin{Examples}{LXXV.} +\Item{1.} Show that +\[ +%[** TN: In-line in the original] +\int_{0}^{\infty} \frac{x}{(1 + x)^{3}}\, dx + = \tfrac{1}{2} \int_{0}^{\infty} \frac{dx}{(1 + x)^{2}} + = \tfrac{1}{2}. +\] + +\Item{2.} $\ds\int_{0}^{\infty} \frac{x^{2}}{(1 + x)^{4}}\, dx = \tfrac{2}{3} \int_{0}^{\infty} \frac{x}{(1 + x)^{3}}\, dx = \tfrac{1}{3}$. + +\Item{3.} If $m$ and~$n$ are positive integers, and +\[ +%[** TN: Two equations not displayed in the original] +I_{m, n} = \int_{0}^{\infty} \frac{x^{m}\, dx}{(1 + x)^{m+n}}, +\] +then +\[ +I_{m, n} = \{m/(m + n - 1)\} I_{m-1, n}. +\] +Hence prove that $I_{m, n} = m!\, (n - 2)!/(m + n - 1)!$. + +\Item{4.} Show similarly that if +\[ +%[** TN: Not displayed in the original] +I_{m, n} = \int_{0}^{\infty} \frac{x^{2m+1}\, dx}{(1 + x^{2})^{m+n}} +\] +then +\[ +I_{m, n} = \{m/(m + n - 1)\} I_{m-1, n},\quad +2I_{m, n} = m!\, (n - 2)!/(m + n - 1)!. +\] +Verify the result by applying the substitution $x = t^{2}$ to the result of Ex.~3. +\end{Examples} + +\Paragraph{180. Other types of infinite integrals.} It was assumed, +in the definition of the ordinary or finite integral given in +\Ref{Ch.}{VII}, that (1)~the range of integration is finite and (2)~the +subject of integration is continuous. + +It is possible, however, to extend the notion of the `definite +integral' so as to apply to many cases in which these conditions +\PageSep{329} +are not satisfied. The `infinite' integrals which we have discussed +in the preceding sections, for example, differ from those of \Ref{Ch.}{VII} +in that the range of integration is infinite. We shall now suppose +that it is the second of the conditions (1),~(2) that is not satisfied. +It is natural to try to frame definitions applicable to some such +cases at any rate. There is only one such case which we shall +consider here. We shall suppose that $\phi(x)$~is continuous throughout +the range of integration $\DPmod{(a, A)}{[a, A]}$ except for a finite number of values +of~$x$, say $x = \xi_{1}$, $\xi_{2}$,~\dots, and that $\phi(x) \to \infty$ or $\phi(x) \to -\infty$ as $x$~tends +to any of these exceptional values from either side. + +It is evident that we need only consider the case in which $\DPmod{(a, A)}{[a, A]}$ +contains \emph{one} such point~$\xi$. When there is more than one such +point we can divide up~$\DPmod{(a, A)}{[a, A]}$ into a finite number of sub-intervals +each of which contains only one; and, if the value of the integral +over each of these sub-intervals has been defined, we can then +define the integral over the whole interval as being the sum of +the integrals over each sub-interval. Further, we can suppose +that the one point~$\xi$ in~$\DPmod{(a, A)}{[a, A]}$ comes at one or other of the +limits $a$,~$A$. For, if it comes between $a$ and~$A$, we can then +define $\ds\int_{a}^{A} \phi(x)\, dx$ as +\[ +\int_{a}^{\xi} \phi(x)\, dx + \int_{\xi}^{A} \phi(x)\, dx, +\] +assuming each of these integrals to have been satisfactorily defined. +We shall suppose, then, that $\xi = a$; it is evident that the +definitions to which we are led will apply, with trifling changes, to +the case in which $\xi = A$. + +Let us then suppose $\phi(x)$ to be continuous throughout $\DPmod{(a, A)}{[a, A]}$ +except for $x = a$, while $\phi(x) \to \infty$ as $x \to a$ through values greater +than~$a$. A typical example of such a function is given by +\[ +\phi(x) = (x - a)^{-s}, +\] +where $s > 0$; or, in particular, if $a = 0$, by $\phi(x) = x^{-s}$. Let us +therefore consider how we can define +\[ +\int_{0}^{A} \frac{dx}{x^{s}}, +\Tag{(1)} +\] +when $s > 0$. +\PageSep{330} + +The integral $\ds\int_{1/A}^{\infty} y^{s-2}\, dy$ is convergent if $s < 1$ (\SecNo[§]{175}) and means +$\lim\limits_{\eta\to\infty} \ds\int_{1/A}^{\eta} y^{s-2}\, dy$. But if we make the substitution $y = 1/x$, we +obtain +\[ +\int_{1/A}^{\eta} y^{s-2}\, dy = \int_{1/\eta}^{A} x^{-s}\, dx. +\] +Thus $\lim\limits_{\eta\to\infty} \ds\int_{1/\eta}^{A} x^{-s}\, dx$, or, what is the same thing, +\[ +% [** TN: Keeping notation \epsilon] +\lim_{\epsilon\to +0} \int_{\epsilon}^{A} x^{-s}\, dx, +\] +exists provided that $s < 1$; and it is natural to define the value of +the integral~\Eq{(1)} as being equal to this limit. Similar considerations +lead us to define $\ds\int_{a}^{A} (x - a)^{-s}\, dx$ by the equation +\[ +\int_{a}^{A} (x - a)^{-s}\, dx + = \lim_{\epsilon\to +0} \int_{a+\epsilon}^{A} (x - a)^{-s}\, dx. +\] + +We are thus led to the following general definition: \begin{Defn}if the integral +\[ +\int_{a+\epsilon}^{A} \phi(x)\, dx +\] +tends to a limit~$l$ as $\epsilon \to +0$, we shall say that the integral +\[ +\int_{a}^{A} \phi(x)\, dx +\] +is convergent and has the value~$l$. +\end{Defn} + +Similarly, when $\phi(x) \to \infty$ as $x$~tends to the upper limit~$A$, we +define $\ds\int_{a}^{A} \phi(x)\, dx$ as being +\[ +\lim_{\epsilon \to +0} \int_{a}^{A-\epsilon} \phi(x)\, dx: +\] +and then, as we explained above, we can extend our definitions to +cover the case in which the interval $\DPmod{(a, A)}{[a, A]}$ contains any finite +number of infinities of~$\phi(x)$. + +An integral in which the subject of integration tends to~$\infty$ +or to~$-\infty$ as $x$~tends to some value or values included in the range +of integration will be called an \emph{infinite integral of the second kind}: +the \emph{first kind} of infinite integrals being the class discussed in +\SecNo[§§]{177}~\textit{et~seq.} Nearly all the remarks (i)--(vii) made at the end of +\SecNo[§]{177} apply to infinite integrals of the second kind as well as to +those of the first. +\PageSep{331} + +\begin{Remark} +\Paragraph{181.} We may now write the equation~\Eq{(4)} of \SecNo[§]{179} in the form +\[ +\int_{a}^{\infty} \phi(x)\, dx = \int_{b}^{c} \phi\{f(t)\}f'(t)\, dt. +\Tag{(1)} +\] +The integral on the right-hand side is defined as the limit, as $\tau \to c$, of the +corresponding integral over the range $\DPmod{(b, \tau)}{[b, \tau]}$, \ie\ as an infinite integral of the +second kind. And when $\phi\{f(t)\}f'(t)$ has an infinity at $t = c$ the integral is +essentially an infinite integral. Suppose for example, that $\phi(x) = (1 + x)^{-m}$, +where $1 < m <2$, and $a = 0$, and that $f(t) = t/(1 - t)$. Then $b = 0$, $c = 1$, and \Eq{(1)}~becomes +\[ +\int_{0}^{\infty} \frac{dx}{(1 + x)^{m}} = \int_{0}^{1} (1 - t)^{m-2}\, dt; +\Tag{(2)} +\] +and the integral on the right-hand side is an infinite integral of the second +kind. + +On the other hand it may happen that $\phi\{f(t)\}f'(t)$ is continuous for $t = c$. +In this case +\[ +\int_{b}^{c} \phi\{f(t)\}f'(t)\, dt +\] +is a finite integral, and +\[ +\lim_{\tau \to c} \int_{b}^{\tau} \phi\{f(t)\}f'(t)\, dt + = \int_{b}^{c} \phi\{f(t)\}f'(t)\, dt, +\] +in virtue of the corollary to Theorem~\Eq{(10)} of \SecNo[§]{160}. In this case the +substitution $x = f(t)$ transforms an infinite into a finite integral. This case +arises if $m \geq 2$ in the example considered a moment ago. +\end{Remark} + +\begin{Examples}{LXXVI.} +\Item{1.} If $\phi(x)$~is continuous except for $x = a$, while +$\phi(x) \to \infty$ as $x \to a$, then the necessary and sufficient condition that $\ds\int_{a}^{A} \phi(x)\, dx$ +should be convergent is that we can find a constant~$K$ such that +\[ +\int_{a+\epsilon}^{A} \phi(x)\, dx < K +\] +for all values of~$\epsilon$, however small (cf.~\SecNo[§]{178}). + +It is clear that we can choose a number~$A'$ between $a$ and~$A$, such that +$\phi(x)$~is positive throughout $\DPmod{(a, A')}{[a, A']}$. If $\phi(x)$~is positive throughout the +whole interval $\DPmod{(a, A)}{[a, A]}$ then we can of course identify $A'$ and~$A$. Now +\[ +\int_{a-\epsilon}^{A} \phi(x)\, dx + = \int_{a-\epsilon}^{A'} \phi(x)\, dx + \int_{A'}^{A} \phi(x)\, dx. +\] +The first integral on the right-hand side of the above equation increases +as $\epsilon$~decreases, and therefore tends to a limit or to~$\infty$; and the truth of the +result stated becomes evident. + +If the condition is not satisfied then $\ds\int_{a-\epsilon}^{A} \phi(x)\, dx \to \infty$. We shall then say +that the integral $\ds\int_{a}^{A} \phi(x)\, dx$ \Emph{diverges} to~$\infty$. It is clear that, if $\phi(x) \to \infty$ +as $x \to a + 0$, then convergence and divergence to~$\infty$ are the only alternatives +for the integral. We may discuss similarly the case in which $\phi(x) \to -\infty$. +\PageSep{332} + +%[** TN: Several displayed integrals are in-line in the original] +\Item{2.} Prove that +\[ +\int_{a}^{A} (x - a)^{-s}\, dx = \frac{(A - a)^{1-s}}{1 - s} +\] +if $s < 1$, while the integral is divergent if $s \geq 1$. + +\Item{3.} If $\phi(x) \to \infty$ as $x \to a + 0$ and $\phi(x) < K(x - a)^{-s}$, where $s < 1$, then +$\ds\int_{a}^{A} \phi(x)\, dx$ is convergent; and if $\phi(x) > K(x - a)^{-s}$, where $s \geq 1$, then the +integral is divergent. [This is merely a particular case of a general comparison +theorem analogous to that stated in~\SecNo[§]{178}.] + +\Item{4.} Are the integrals +\begin{gather*} +\int_{a}^{A} \frac{dx}{\sqrtb{(x - a)(A - x)}},\quad +\int_{a}^{A} \frac{dx}{(A - x)\sqrtp[3]{x - a}},\quad +\int_{a}^{A} \frac{dx}{(A - x)\sqrtp[3]{A - x}},\\ +\int_{a}^{A} \frac{dx}{\sqrtp{x^{2} - a^{2}}},\quad +\int_{a}^{A} \frac{dx}{\sqrtp[3]{A^{3} - x^{3}}},\quad +\int_{a}^{A} \frac{dx}{x^{2} - a^{2}},\quad +\int_{a}^{A} \frac{dx}{A^{3} - x^{3}} +\end{gather*} +convergent or divergent? + +\Item{5.} The integrals +\[ +%[** TN: Integrals in next ten questions in-line in the original] +\int_{-1}^{1}\frac{dx}{\sqrt[3]{x}},\quad +\int_{a-1}^{a+1} \frac{dx}{\sqrtp[3]{x - a}} +\] +are convergent, and the value of +each is zero. + +\Item{6.} The integral +\[ +\int_{0}^{\pi} \frac{dx}{\sqrtp{\sin x}} +\] +is convergent. [The subject of integration +tends to~$\infty$ as $x$~tends to either limit.] + +\Item{7.} The integral +\[ +\int_{0}^{\pi} \frac{dx}{(\sin x)^{s}} +\] +is convergent if and only if $s < 1$. + +\Item{8.} The integral +\[ +\int_{0}^{\frac{1}{2}\pi} \frac{x^{s}}{(\sin x)^{t}}\, dx +\] +is convergent if $t < s + 1$. + +\Item{9.} Show that +\[ +\int_{0}^{h} \frac{\sin x}{x^{p}}\, dx, +\] +where $h > 0$, is convergent if $p < 2$. Show also +that, if $0 < p < 2$, the integrals +\[ +\int_{0}^{\pi} \frac{\sin x}{x^{p}} dx,\quad +\int_{\pi}^{2\pi} \frac{\sin x}{x^{p}}\, dx,\quad +\int_{2\pi}^{3\pi} \frac{\sin x}{x^{p}}\, dx,\ \dots +\] +alternate in sign and steadily decrease in absolute value. [Transform the +integral whose limits are $k\pi$ and~$(k + 1)\pi$ by the substitution $x = k\pi + y$.] + +\Item{10.} Show that +\[ +\int_{0}^{h} \frac{\sin x}{x^{p}}\, dx, +\] +where $0 < p < 2$, attains its greatest value +when $h = \pi$. \MathTrip{1911.} + +\Item{11.} The integral +\[ +\int_{0}^{\frac{1}{2} \pi}(\cos x)^{l}(\sin x)^{m}\, dx +\] +is convergent if and only if $l > -1$, +$m > -1$. + +\Item{12.} Such an integral as +\[ +\int_{0}^{\infty} \frac{x^{s-1}\, dx}{1 + x}, +\] +where $s < 1$, does not fall directly +under any of our previous definitions. For the range of integration is infinite +\PageSep{333} +and the subject of integration tends to~$\infty$ as $x \to +0$. It is natural to +define this integral as being equal to the sum +\[ +\int_{0}^{1} \frac{x^{s-1}\, dx}{1 + x} + + \int_{1}^{\infty} \frac{x^{s-1}\, dx}{1 + x}, +\] +provided that these two integrals are both convergent. + +{\Loosen The first integral is a convergent infinite integral of the second kind +if $0 < s < 1$. The second is a convergent infinite integral of the first kind if +$s < 1$. It should be noted that when $s > 1$ the first integral is an ordinary +finite integral; but then the second is divergent. Thus the integral from~$0$ to~$\infty$ +is convergent if and only if $0 < s < 1$.} + +\Item{13.} Prove that +\[ +\int_{0}^{\infty} \frac{x^{s-1}}{1 + x^{t}}\, dx +\] +is convergent if and only if $0 < s < t$. + +\Item{14.} The integral +\[ +\int_{0}^{\infty} \frac{x^{s-1} - x^{t-1}}{1 - x}\, dx +\] +is convergent if and only if $0 < s < 1$, +$0 < t < 1$. [It should be noticed that the subject of integration is undefined +when $x = 1$; but $(x^{s-1} - x^{t-1})/(1 - x) \to t - s$ as $x \to 1$ from either side; so that +the subject of integration becomes a continuous function of~$x$ if we assign to it +the value $t - s$ when $x = 1$. + +It often happens that the subject of integration has a discontinuity which +is due simply to a failure in its definition at a particular point in the range +of integration, and can be removed by attaching a particular value to it at +that point. In this case it is usual to suppose the definition of the subject +of integration completed in this way. Thus the integrals +\[ +\int_{0}^{\frac{1}{2} \pi} \frac{\sin mx}{x}\, dx,\quad +\int_{0}^{\frac{1}{2} \pi} \frac{\sin mx}{\sin x}\, dx +\] +are ordinary finite integrals, if the subjects of integration are regarded as +having the value~$m$ when $x = 0$.] + +\Item{15.} \Topic{Substitution and integration by parts.} The formulae for transformation +by substitution and integration by parts may of course be extended +to infinite integrals of the second as well as of the first kind. The reader +should formulate the general theorems for himself, on the lines of~\SecNo[§]{179}. + +\Item{16.} Prove by integration by parts that if $s > 0$, $t > 1$, then +\[ +\int_{0}^{1} x^{s-1}(1 - x)^{t-1}\, dx + = \frac{t - 1}{s} \int_{0}^{1} x^{s} (1 - x)^{t-2}\, dx. +\] + +\Item{17.} If $s > 0$ then +\[ +\int_{0}^{1} \frac{x^{s-1}\, dx}{1 + x} + = \int_{1}^{\infty} \frac{t^{-s}\, dt}{1 + t}. +\] + +%[** TN: Added paragraph break] +[Put $x = 1/t$.] + +\Item{18.} If $0 < s < 1$ then +\[ +\int_{0}^{1} \frac{x^{s-1} + x^{-s}}{1 + x}\, dx + = \int_{0}^{\infty} \frac{t^{-s}\, dt}{1 + t} + = \int_{0}^{\infty} \frac{t^{s-1}\, dt}{1 + t}. +\] + +\Item{19.} If $a + b > 0$ then +\[ +\int_{b}^{\infty} \frac{dx}{(x + a)\sqrtp{x - b}} = \frac{\pi}{\sqrtp{a + b}}. +\] +\MathTrip{1909.} +\PageSep{334} + +\Item{20.} Show, by means of the substitution $x = t/(1 - t)$, that if $l$~and~$m$ are +both positive then +\[ +\int_{0}^{\infty} \frac{x^{l-1}}{(1 + x)^{l+m}}\, dx + = \int_{0}^{1} t^{l-1} (1 - t)^{m-1}\, dt. +\] + +\Item{21.} Show, by means of the substitution $x = pt/(p + 1 - t)$, that if $l$,~$m$, and~$p$ +are all positive then +\[ +\int_{0}^{1} x^{l-1} (1 - x)^{m-1}\, \frac{dx}{(x + p)^{l + m}} + = \frac{1}{(1 + p)^{l} p^{m}} \int_{0}^{1} t^{l-1} (1 - t)^{m-1}\, dt. +\] + +\Item{22.} Prove that +\[ +%[** TN: In-line in the original] +\int_{a}^{b} \frac{dx}{\sqrtb{(x - a)(b - x)}} = \pi\quad\text{and}\quad +\int_{a}^{b} \frac{x\, dx}{\sqrtb{(x - a)(b - x)}} = \tfrac{1}{2} \pi (a + b), +\] +(i)~by means of the substitution $x = a + (b - a)t^{2}$, (ii)~by means of the substitution +$(b - x)/(x - a) = t$, and (iii)~by means of the substitution $x = a\cos^{2} t + b\sin^{2} t$. + +\Item{23.} If $s > -1$ then +\[ +\int_{0}^{\frac{1}{2} \pi} (\sin\theta)^{s}\, d\theta + = \int_{0}^{1} \frac{x^{s}\, dx}{\sqrtp{1 - x^{2}}} + = \tfrac{1}{2} \int_{0}^{1} \frac{x^{\frac{1}{2}(s-1)}\, dx}{\sqrtp{1 - x}} + = \tfrac{1}{2} \int_{0}^{1} (1 - x)^{\frac{1}{2}(s-1)} \frac{dx}{\sqrt{x}}. +\] + +\Item{24.} Establish the formulae +\begin{align*} +&\int_{0}^{1} \frac{f(x)\, dx}{\sqrtp{1 - x^{2}}} + = \int_{0}^{\frac{1}{2}\pi} f(\sin\theta)\, d\theta,\\ +% +&\int_{a}^{b} \frac{f(x)\, dx}{\sqrtb{(x - a)(b - x)}} + = 2\int_{0}^{\frac{1}{2}\pi} f(a\cos^{2}\theta + b\sin^{2}\theta)\, d\theta,\\ +% +&\int_{-a}^{a} f\left\{\bigsqrtp{\frac{a - x}{a + x}}\right\} dx + = 4a\int_{0}^{\frac{1}{2}\pi} f(\tan\theta) \cos\theta \sin\theta\, d\theta\Add{.} +\end{align*} + +\Item{25.} Prove that +\[ +\int_{0}^{1} \frac{dx}{(1 + x)(2 + x) \sqrtb{x(1 - x)}} + = \pi\left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{6}}\right)\Add{.} +\] + +%[** Added paragraph break] +[Put $x = \sin^{2}\theta$ and use \Ex{lxiii}.~8.] \MathTrip{1912.} %[** TN: Dot added after "Math"] +\end{Examples} + +\begin{Remark} +\Paragraph{182.} Some care has occasionally to be exercised in applying the rule +for transformation by substitution. The following example affords a good +illustration of this. + +Let +\[ +J = \int_{1}^{7} (x^{2} - 6x + 13)\, dx. +\] +We find by direct integration that $J = 48$. Now let us apply the substitution +\[ +y = x^{2} - 6x + 13, +\] +which gives $x = 3 ± \sqrtp{y - 4}$. Since $y = 8$ when $x = 1$ and $y = 20$ when $x = 7$, we +appear to be led to the result +\[ +J = \int_{8}^{20} y\frac{dx}{dy}\, dy + = ±\tfrac{1}{2}\int_{8}^{20} \frac{y\, dy}{\sqrtp{y - 4}}. +\] +The indefinite integral is +\[ +\tfrac{1}{3}(y - 4)^{3/2} + 4(y - 4)^{1/2}, +\] +and so we obtain the value~$±\frac{80}{3}$, which is certainly wrong whichever sign we +choose. +\PageSep{335} + +The explanation is to be found in a closer consideration of the relation +between $x$~and~$y$. The function $x^{2} - 6x + 13$ has a minimum for $x = 3$, when +$y = 4$. As $x$~increases from $1$ to~$3$, $y$~decreases from $8$ to~$4$, and $dx/dy$~is +negative, so that +\[ +\frac{dx}{dy} = -\frac{1}{2\sqrtp{y - 4}}. +\] +As $x$~increases from $3$ to~$7$, $y$~increases from $4$ to~$20$, and the other sign must +be chosen. Thus +\[ +J = \int_{1}^{7} y\, dx + = \int_{8}^{4} \left\{-\frac{y}{2\sqrtp{y - 4}}\right\} dy + + \int_{4}^{20} \frac{y}{2\sqrtp{y - 4}}\, dy, +\] +a formula which will be found to lead to the correct result. + +{\Loosen Similarly, if we transform the integral $\ds\int_{0}^{\pi} dx = \pi$ by the substitution +$x = \arcsin y$, we must observe that $dx/dy = 1/\sqrtp{1 - y^{2}}$ or $dx/dy = -1/\sqrtp{1 - y^{2}}$ +according as $0 \leq x < \frac{1}{2}\pi$ or $\frac{1}{2}\pi < x \leq \pi$.} + +\Par{Example.} Verify the results of transforming the integrals +\[ +\int_{0}^{1} (4x^{2} - x + \tfrac{1}{16})\, dx,\quad +\int_{0}^{\pi} \cos^{2}x\, dx +\] +by the substitutions $4x^{2} - x + \frac{1}{16} = y$, $x = \arcsin y$ respectively. +\end{Remark} + +\Paragraph{183. Series of positive and negative terms.} Our definitions +of the sum of an infinite series, and the value of an infinite +integral, whether of the first or the second kind, apply to series +of terms or integrals of functions whose values may be either +positive or negative. But the special tests for convergence or +divergence which we have established in this chapter, and the +examples by which we have illustrated them, have had reference +almost entirely to the case in which all these values are positive. +Of course the case in which they are all negative is not essentially +different, as it can be reduced to the former by changing $u_{n}$ into +$-u_{n}$ or $\phi(x)$ into~$-\phi(x)$. + +In the case of a series it has always been explicitly or tacitly +assumed that any conditions imposed upon~$u_{n}$ may be violated for +a finite number of terms: all that is necessary is that such a +condition (\eg\ that all the terms are positive) should be satisfied +\emph{from some definite term onwards}. Similarly in the case of an +infinite integral the conditions have been supposed to be satisfied +\emph{for all values of~$x$ greater than some definite value}, or for all values +of~$x$ within some definite interval $\DPmod{(a, a + \delta)}{[a, a + \delta]}$ which includes the +\PageSep{336} +value~$a$ near which the subject of integration tends to infinity. +Thus our tests apply to such a series as +\[ +\sum \frac{n^{2} - 10}{n^{4}}, +\] +since $n^{2} - 10 > 0$ when $n \geq 4$, and to such integrals as +\[ +\int_{1}^{\infty} \frac{3x - 7}{(x + 1)^{3}}\, dx,\quad +\int_{0}^{1} \frac{1 - 2x}{\sqrt{x}}\, dx, +\] +since $3x - 7 > 0$ when $x > \frac{7}{3}$, and $1 - 2x > 0$ when $0 < x < \frac{1}{2}$. + +But when the changes of sign of~$u_{n}$ \emph{persist throughout the series}, +\ie~when the number of both positive and negative terms is infinite, +as in the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$; or when $\phi(x)$~continually +changes sign as $x \to \infty$, as in the integral +\[ +\int_{1}^{\infty} \frac{\sin x}{x^{s}}\, dx, +\] +or as $x \to a$, where $a$~is a point of discontinuity of~$\phi(x)$, as in +the integral +\[ +\int_{a}^{A} \sin\left(\frac{1}{x - a}\right) \frac{dx}{x - a}; +\] +then the problem of discussing convergence or divergence becomes +more difficult. For now we have to consider the possibility of +oscillation as well as of convergence or divergence. + +We shall not, in this volume, have to consider the more +general problem for integrals. But we shall, in the ensuing +chapters, have to consider certain simple examples of series containing +an infinite number of both positive and negative terms. + +\Paragraph{184. Absolutely Convergent Series.} Let us then consider +a series $\sum u_{n}$ in which any term may be either positive or +negative. Let +\[ +|u_{n}| = \alpha_{n}, +\] +so that $\alpha_{n} = u_{n}$ if $u_{n}$~is positive and $\alpha_{n} = -u_{n}$ if $u_{n}$~is negative. +Further, let $v_{n} = u_{n}$ or $v_{n} = 0$, according as $u_{n}$~is positive or negative, +and $w_{n} = -u_{n}$ or $w_{n} = 0$, according as $u_{n}$~is negative or positive; +or, what is the same thing, let $v_{n}$ or~$w_{n}$ be equal to~$\alpha_{n}$ according +as $u_{n}$~is positive or negative, the other being in either case equal +to zero. Then it is evident that $v_{n}$ and~$w_{n}$ are always positive, and +that +\[ +u_{n} = v_{n} - w_{n},\quad +\alpha_{n} = v_{n} + w_{n}. +\] +\PageSep{337} + +\begin{Remark} +If, for example, our series is $1 - (1/2)^{2} + (1/3)^{2} - \dots$, then $u_{n} = (-1)^{n-1}/n^{2}$ +and $\alpha_{n} = 1/n^{2}$, while $v_{n} = 1/n^{2}$ or $v_{n} = 0$ according as $n$~is odd or even and +$w_{n} = 1/n^{2}$ or $w_{n} = 0$ according as $n$~is even or odd. +\end{Remark} + +We can now distinguish two cases. + +\Item{A.} Suppose that the series $\sum \alpha_{n}$~is convergent. This is the +case, for instance, in the example above, where $\sum \alpha_{n}$ is +\[ +1 + (1/2)^{2} + (1/3)^{2} + \dots. +\] +Then both $\sum v_{n}$ and $\sum w_{n}$ are convergent: for (\Ex{xxx}.~18) any +series selected from the terms of a convergent series of positive +terms is convergent. And hence, by theorem~\Eq{(6)} of \SecNo[§]{77}, $\sum u_{n}$ or +$\sum(v_{n} - w_{n})$ is convergent and equal to $\sum v_{n} - \sum w_{n}$. + +We are thus led to formulate the following definition. + +\begin{Definition} +When $\sum \alpha_{n}$ or $\sum |u_{n}|$ is convergent, the series $\sum u_{n}$ +is said to be \Emph{absolutely convergent}. +\end{Definition} + +And what we have proved above amounts to this: \begin{Result}if $\sum u_{n}$~is +absolutely convergent then it is convergent; so are the series formed +by its positive and negative terms taken separately; and the sum of +the series is equal to the sum of the positive terms plus the sum +of the negative terms. +\end{Result} + +\begin{Remark} +The reader should carefully guard himself against supposing that the +statement `an absolutely convergent series is convergent' is a mere tautology. +When we say that $\sum u_{n}$~is `absolutely convergent' we do \emph{not} assert directly +that $\sum u_{n}$~is convergent: we assert the convergence of \emph{another} series $\sum |u_{n}|$, +and it is by no means evident \textit{a~priori} that this precludes oscillation on +the part of~$\sum u_{n}$. +\end{Remark} + +\begin{Examples}{LXXVII.} +\Item{1.} Employ the `general principle of convergence' +(\SecNo[§]{84}) to prove the theorem that an absolutely convergent series is convergent. +[Since $\sum |u_{n}|$ is convergent, we can, when any positive number~$\DELTA$ is +assigned, choose~$n_{0}$ so that +\[ +|u_{n_{1}+1}| + |u_{n_{1}+2}| + \dots + |u_{n_{2}}| < \DELTA +\] +when $n_{2} > n_{1} \geq n_{0}$. \textit{A~fortiori} +\[ +|u_{n_{1}+1} + u_{n_{1}+2} + \dots + u_{n_{2}}| < \DELTA, +\] +and therefore $\sum u_{n}$~is convergent.] + +\Item{2.} If $\sum a_{n}$ is a convergent series of positive terms, and $|b_{n}|\leq Ka_{n}$, then +$\sum b_{n}$ is absolutely convergent. + +\Item{3.} If $\sum a_{n}$ is a convergent series of positive terms, then the series $\sum a_{n}x^{n}$ is +absolutely convergent when $-1 \leq x \leq 1$. +\PageSep{338} + +\Item{4.} If $\sum a_{n}$ is a convergent series of positive terms, then the series $\sum a_{n} \cos n\theta$, +$\sum a_{n}\sin n\theta$ are absolutely convergent for all values of~$\theta$. [Examples are +afforded by the series $\sum r^{n}\cos n\theta$, $\sum r^{n}\sin n\theta$ of~\SecNo[§]{88}.] + +\Item{5.} Any series selected from the terms of an absolutely convergent series +is absolutely convergent. [For the series of the moduli of its terms is a +selection from the series of the moduli of the terms of the original series.] + +\Item{6.} Prove that if $\sum |u_{n}|$~is convergent then +\[ +|\tsum u_{n}| \leq \tsum |u_{n}|, +\] +and that the only case to which the sign of equality can apply is that in +which every term has the same sign. +\end{Examples} + +\Paragraph{185. Extension of Dirichlet's Theorem to absolutely +convergent series.} Dirichlet's Theorem (\SecNo[§]{169}) shows that the +terms of a series of positive terms may be rearranged in any way +without affecting its sum. It is now easy to see that any absolutely +convergent series has the same property. For let $\sum u_{n}$ be +so rearranged as to become $\sum u'_{n}$, and let $\alpha'_{n}$,~$v'_{n}$,~$w'_{n}$ be formed +from~$u'_{n}$ as $\alpha_{n}$,~$v_{n}$,~$w_{n}$ were formed from~$u_{n}$. Then $\sum \alpha'_{n}$ is convergent, +as it is a rearrangement of~$\sum \alpha_{n}$, and so are $\sum v'_{n}$, $\sum w'_{n}$, +which are rearrangements of $\sum v_{n}$, $\sum w_{n}$. Also, by Dirichlet's +Theorem, $\sum v'_{n} = \sum v_{n}$ and $\sum w'_{n} = \sum w_{n}$ and so +\[ +\tsum u'_{n} + = \tsum v'_{n} - \tsum w'_{n} + = \tsum v_{n} - \tsum w_{n} + = \tsum u_{n}. +\] + +\Paragraph{186. Conditionally convergent series.} \Item{B.} We have +now to consider the second case indicated above, viz.\ that in +which the series of moduli $\sum \alpha_{n}$ diverges to~$\infty$. + +\begin{Definition} +If $\sum u_{n}$ is convergent, but $\sum |u_{n}|$ divergent, the +original series is said to be \Emph{conditionally convergent}. +\end{Definition} + +In the first place we note that, if $\sum u_{n}$ is conditionally convergent, +then the series $\sum v_{n}$, $\sum w_{n}$ of \SecNo[§]{184} must both diverge to~$\infty$. +For they obviously cannot both converge, as this would involve +the convergence of $\sum(v_{n} + w_{n})$ or~$\sum \alpha_{n}$. And if one of them, say +$\sum w_{n}$, is convergent, and $\sum v_{n}$ divergent, then +\[ +\sum_{0}^{N} u_{n} = \sum_{0}^{N} v_{n} - \sum_{0}^{N} w_{n}, +\Tag{(1)} +\] +and therefore tends to~$\infty$ with~$N$, which is contrary to the +hypothesis that $\sum u_{n}$ is convergent. + +Hence $\sum v_{n}$, $\sum w_{n}$ are both divergent. It is clear from equation~\Eq{(1)} +above that the sum of a conditionally convergent series +\PageSep{339} +is the limit of the difference of two functions each of which tends +to~$\infty$ with~$n$. It is obvious too that $\sum u_{n}$ no longer possesses the +property of convergent series of positive terms (\Ex{xxx}.~18), and +all absolutely convergent series (\Ex{lxxvii}.~5), that any selection +from the terms itself forms a convergent series. And it seems more +than likely that the property prescribed by Dirichlet's Theorem +will not be possessed by conditionally convergent series; at any +rate the proof of \SecNo[§]{185} fails completely, as it depended essentially +on the convergence of $\sum v_{n}$ and $\sum w_{n}$ separately. We shall see in a +moment that this conjecture is well founded, and that the theorem +is not true for series such as we are now considering. + +\Paragraph{187. Tests of convergence for conditionally convergent +series.} It is not to be expected that we should be able to find +tests for conditional convergence as simple and general as those +of \SecNo[§§]{167}~\textit{et~seq.} It is naturally a much more difficult matter to +formulate tests of convergence for series whose convergence, as is +shown by equation~\Eq{(1)} above, depends essentially on the cancelling +of the positive by the negative terms. In the first instance \emph{there +are no comparison tests for convergence of conditionally convergent +series}. + +For suppose we wish to infer the convergence of $\sum v_{n}$ from +that of $\sum u_{n}$. We have to compare +\[ +v_{0} + v_{1} + \dots + v_{n},\quad +u_{0} + u_{1} + \dots + u_{n}. +\] +If every~$u$ and every~$v$ were positive, and every~$v$ less than the +corresponding~$u$, we could at once infer that +\[ +v_{0} + v_{1} + \dots + v_{n} < u_{0} + \dots + u_{n}, +\] +and so that $\sum v_{n}$ is convergent. If the~$u$'s only were positive and +every~$v$ \emph{numerically} less than the corresponding~$u$, we could infer +that +\[ +|v_{0}| + |v_{1}| + \dots + |v_{n}| < u_{0} + \dots + u_{n}, +\] +and so that $\sum v_{n}$ is absolutely convergent. But in the general case, +when the $u$'s and~$v$'s are both unrestricted as to sign, all that we +can infer is that +\[ +|v_{0}| + |v_{1}| + \dots + |v_{n}| < |u_{0}| + \dots + |u_{n}|. +\] +This would enable us to infer the absolute convergence of $\sum v_{n}$ +from the absolute convergence of~$\sum u_{n}$; but if $\sum u_{n}$ is only conditionally +convergent we can draw no inference at all. +\PageSep{340} + +\begin{Remark} +\Par{Example.} We shall see shortly that the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ is convergent. +But the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots$ is divergent, although each of its +terms is numerically less than the corresponding term of the former series. +\end{Remark} + +It is therefore only natural that such tests as we can obtain +should be of a much more special character than those given in +the early part of this chapter. + +\Paragraph{188. Alternating Series.} The simplest and most common +conditionally convergent series are what is known as \emph{alternating +series}, series whose terms are alternately positive and negative. +The convergence of the most important series of this type is +established by the following theorem. + +\begin{Result} +If $\phi(n)$~is a positive function of~$n$ which tends \Emph{steadily} to +zero as $n \to \infty$, then the series +\[ +\phi(0) - \phi(1) + \phi(2) - \dots +\] +is convergent, and its sum lies between $\phi(0)$ and $\phi(0) - \phi(1)$. +\end{Result} + +Let us write $\phi_{0}$, $\phi_{1}$,~\dots\ for $\phi(0)$, $\phi(1)$,~\dots; and let +\[ +s_{n} = \phi_{0} - \phi_{1} + \phi_{2} - \dots + (-1)^{n}\phi_{n}. +\] +Then +\[ +s_{2n+1} - s_{2n-1} = \phi_{2n} - \phi_{2n+1}\geq 0,\quad +s_{2n} - s_{2n-2} = -(\phi_{2n-1} - \phi_{2n}) \leq 0. +\] +{\Loosen Hence $s_{0}$, $s_{2}$, $s_{4}$,~\dots, $s_{2n}$,~\dots\ is a decreasing sequence, and therefore +tends to a limit or to~$-\infty$, and $s_{1}$, $s_{3}$, $s_{5}$,~\dots, $s_{2n+1}$,~\dots\ is an increasing +sequence, and therefore tends to a limit or to~$\infty$. But +$\lim (s_{2n+1} - s_{2n}) = \lim (-1)^{2n+1} \phi_{2n+1} = 0$, from which it follows that +both sequences must tend to limits, and that the two limits must +be the same. That is to say, the sequence $s_{0}$, $s_{1}$,~\dots, $s_{n}$,~\dots\ tends to +a limit. Since $s_{0} = \phi_{0}$, $s_{1} = \phi_{0} - \phi_{1}$, it is clear that this limit lies +between $\phi_{0}$ and~$\phi_{0} - \phi_{1}$.} + +\begin{Examples}{LXXVIII.} +\Item{1.} The series +\begin{gather*} +1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots,\quad +1 - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} + \dots,\\ +\sum \frac{(-1)^{n}}{(n + a)},\quad +\sum \frac{(-1)^{n}}{\sqrtp{n + a}},\quad +\sum \frac{(-1)^{n}}{(\sqrt{n} + \sqrt{a})},\quad +\sum \frac{(-1)^{n}}{(\sqrt{n} + \sqrt{a})^{2}}, +\end{gather*} +where $a > 0$, are conditionally convergent. + +\Item{2.} The series $\sum(-1)^{n}(n + a)^{-s}$, where $a > 0$, is absolutely convergent if +$s > 1$, conditionally convergent if $0 < s \leq 1$, and oscillatory if $s \leq 0$. +\PageSep{341} + +\Item{3.} The sum of the series of \SecNo[§]{188} lies between $s_{n}$ and~$s_{n+1}$ for all values +of~$n$; and the error committed by taking the sum of the first $n$ terms instead +of the sum of the whole series is numerically not greater than the modulus of +the $(n + 1)$th~term. + +\Item{4.} Consider the series +\[ +\sum \frac{(-1)^{n}}{\sqrt{n} + (-1)^{n}}, +\] +which we suppose to begin with the term for which $n = 2$, to avoid any +difficulty as to the definitions of the first few terms. This series may be +written in the form +\[ +\sum \left[\left\{ + \frac{(-1)^{n}}{\sqrt{n} + (-1)^{n}} + - \frac{(-1)^{n}}{\sqrt{n}}\right\} + + \frac{(-1)^{n}}{\sqrt{n}}\right] +\] +or +\[ +\sum \left\{\frac{(-1)^{n}}{\sqrt{n}} - \frac{1}{n + (-1)^{n}\sqrt{n}}\right\} + = \sum (\psi_{n} - \chi_{n}), +\] +say. The series $\sum \psi_{n}$ is convergent; but $\sum \chi_{n}$~is divergent, as all its terms are +positive, and $\lim n\chi_{n} = 1$. Hence the original series is divergent, although it +is of the form $\phi_{2} - \phi_{3} + \phi_{4} - \dots$, where $\phi_{n} \to 0$. This example shows that the +condition that $\phi_{n}$~should tend \emph{steadily} to zero is essential to the truth of the +theorem. The reader will easily verify that $\sqrtp{2n + 1} - 1 < \sqrtp{2n} + 1$, so that +this condition is not satisfied. + +\Item{5.} If the conditions of \SecNo[§]{188} are satisfied except that $\phi_{n}$~tends steadily +to a positive limit~$l$, then the series $\sum (-1)^{n}\phi_{n}$ oscillates finitely. + +\Item{6.} \Topic{Alteration of the sum of a conditionally convergent series by +rearrangement of the terms.} Let $s$~be the sum of the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$, +and $s_{2n}$~the sum of its first $2n$ terms, so that $\lim s_{2n} = s$. + +Now consider the series +\[ +1 + \tfrac{1}{3} - \tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{7} - \tfrac{1}{4} + \dots +\Tag{(1)} +\] +in which two positive terms are followed by one negative term, and let $t_{3n}$ +denote the sum of the first $3n$~terms. Then +\begin{align*} +t_{3n} + &= 1 + \frac{1}{3} + \dots + \frac{1}{4n-1} + - \frac{1}{2} - \frac{1}{4} - \dots - \frac{1}{2n}\\ + &= s_{2n} + \frac{1}{2n + 1} + \frac{1}{2n + 3} + \dots + \frac{1}{4n - 1}. +\end{align*} + +Now +\[ +\lim \left[\frac{1}{2n + 1} - \frac{1}{2n + 2} + \frac{1}{2n + 3} - \dots + + \frac{1}{4n - 1} - \frac{1}{4n}\right] = 0, +\] +{\Loosen since the sum of the terms inside the bracket is clearly less than +$n/(2n + 1)(2n + 2)$; and} +\[ +\lim \left(\frac{1}{2n + 2} + \frac{1}{2n + 4} + \dots + \frac{1}{4n}\right) + = \tfrac{1}{2} \lim \frac{1}{n} \sum_{r=1}^{n} \frac{1}{1 + (r/n)} + = \tfrac{1}{2} \int_{1}^{2} \frac{dx}{x}, +\] +by \SecNo[§§]{156} and~\SecNo{158}. Hence +\[ +\lim t_{3n} = s + \tfrac{1}{2} \int_{1}^{2} \frac{dx}{x}, +\] +\PageSep{342} +and it follows that the sum of the series~\Eq{(1)} is not~$s$, but the right-hand side of +the last equation. Later on we shall give the actual values of the sums of the +two series: see \SecNo[§]{213} and \Ref{Ch.}{IX}, \MiscEx{IX}~19. + +It can indeed be proved that a conditionally convergent series can always +be so rearranged as to converge to any sum whatever, or to diverge to~$\infty$ or +to~$-\infty$. For a proof we may refer to Bromwich's \textit{Infinite Series}, p.~68. + +\Item{7.} The series +\[ +1 + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{2}} + + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{7}} - \frac{1}{\sqrt{4}} + \dots +\] +diverges to~$\infty$. [Here +\[ +t_{3n} = s_{2n} + \frac{1}{\sqrtp{2n + 1}} + \frac{1}{\sqrtp{2n + 3}} + \dots + + \frac{1}{\sqrtp{4n - 1}} + > s_{2n} + \frac{n}{\sqrtp{4n - 1}}, +\] +where $s_{2n} = 1 - \dfrac{1}{\sqrt{2}} + \dots - \dfrac{1}{\DPtypo{\sqrt{2n}}{\sqrtp{2n}}}$, which tends to a limit as $n \to \infty$.] +\end{Examples} + +\begin{Remark} +\Paragraph{189. Abel's and Dirichlet's Tests of Convergence.} A more general +test, which includes the test of \SecNo[§]{188} as a particular test case, is the following. + +\begin{ParTheorem}{Dirichlet's Test.} +If $\phi_{n}$~satisfies the same conditions as in \SecNo[§]{188}, and $\sum a_{n}$ +is any series which converges or oscillates finitely, then the series +\[ +a_{0}\phi_{0} + a_{1}\phi_{1} + a_{2}\phi_{2} + \dots +\] +is convergent. +\end{ParTheorem} + +The reader will easily verify the identity +\[ +a_{0}\phi_{0} + a_{1}\phi_{1} + \dots + a_{n}\phi_{n} + = s_{0}(\phi_{0} - \phi_{1}) + + s_{1}(\phi_{1} - \phi_{2}) + \dots + + s_{n-1}(\phi_{n-1} - \phi_{n}) + s_{n}\phi_{n}, +\] +where $s_{n} = a_{0} + a_{1} + \dots + a_{n}$. Now the series $(\phi_{0} - \phi_{1}) + (\phi_{1} - \phi_{2}) + \dots$ is convergent, +since the sum to $n$~terms is $\phi_{0} - \phi_{n}$ and $\lim \phi_{n} = 0$; and all its +terms are positive. Also since $\sum a_{n}$, if not actually convergent, at any rate +oscillates finitely, we can determine a constant~$K$ so that $|s_{\nu}| < K$ for all +values of~$\nu$. Hence the series +\[ +\tsum s_{\nu}(\phi_{\nu} - \phi_{\nu+1}) +\] +is absolutely convergent, and so +\[ +s_{0}(\phi_{0} - \phi_{1}) + + s_{1}(\phi_{1} - \phi_{2}) + \dots + + s_{n-1}(\phi_{n-1} - \phi_{n}) +\] +tends to a limit as $n \to \infty$. Also $\phi_{n}$, and therefore $s_{n}\phi_{n}$, tends to zero\Add{.} +And therefore +\[ +a_{0}\phi_{0} + a_{1}\phi_{1} + \dots + a_{n}\phi_{n} +\] +tends to a limit, \ie\ the series $\sum a_{\nu}\phi_{\nu}$ is convergent. + +\Topic{Abel's Test.} There is another test, due to Abel, which, though of less +frequent application than Dirichlet's, is sometimes useful. + +Suppose that $\phi_{n}$, as in Dirichlet's Test, is a positive and decreasing +function of~$n$, but that its limit as $n \to \infty$ is not necessarily zero. Thus we +postulate less about~$\phi_{n}$, but to make up for this we postulate more about +$\sum a_{n}$, viz.\ that it is \emph{convergent}. Then we have the theorem: \begin{Result}if $\phi_{n}$~is a positive +and decreasing function of~$n$, and $\sum a_{n}$~is convergent, then $\sum a_{n}\phi_{n}$~is convergent. +\end{Result} + +For $\phi_{n}$~has a limit as $n \to \infty$, say~$l$: and $\lim (\phi_{n} - l) = 0$. Hence, by +Dirichlet's Test, $\sum a_{n}(\phi_{n} - l)$ is convergent; and as $\sum{a_{n}}$~is convergent it +follows that $\sum a_{n}\phi_{n}$ is convergent. +\PageSep{343} + +This theorem may be stated as follows: \begin{Result}a convergent series remains convergent +if we multiply its terms by any sequence of positive and decreasing +factors. +\end{Result} +\end{Remark} + +\begin{Examples}{LXXIX.} +\Item{1.} Dirichlet's and Abel's Tests may also be established +by means of the general principle of convergence (\SecNo[§]{84}). Let us suppose, +for example, that the conditions of Abel's Test are satisfied. We have +identically +{\setlength{\multlinegap}{0pt} +\begin{multline*} +a_{m}\phi_{m} + a_{m+1}\phi_{m+1} + \dots + a_{n}\phi_{n} + = s_{m, m}(\phi_{m} - \phi_{m+1}) + s_{m, m+1}(\phi_{m+1} - \phi_{m+2})\\ + + \dots + s_{m, n-1}(\phi_{n-1} - \phi_{n}) + s_{m, n}\phi_{n}\dots, +\Tag{(1)} +\end{multline*}}% +where +\[ +s_{m, \nu} = a_{m} + a_{m+1} + \dots + a_{\nu}. +\] + +The left-hand side of~\Eq{(1)} therefore lies between $h\phi_{m}$ and~$H\phi_{m}$, where $h$~and~$H$ +are the algebraically least and greatest of $s_{m, m}$, $s_{m, m+1}$,~\dots, $s_{m, n}$. But, +given any positive number~$\DELTA$, we can choose~$m_{0}$ so that $|s_{m, \nu}| < \DELTA$ when $m \geq m_{0}$, +and so +\[ +|a_{m}\phi_{m} + a_{m+1}\phi_{m+1} + \dots + a_{n}\phi_{n}| + < \DELTA \phi_{m} \leq \DELTA \phi_{1} +\] +when $n > m \geq m_{0}$. Thus the series $\sum a_{n}\phi_{n}$ is convergent. + +\Item{2.} The series $\sum \cos n\theta$ and $\sum \sin n\theta$ oscillate finitely when $\theta$~is not a +multiple of~$\pi$. For, if we denote the sums of the first $n$ terms of the two +series by $s_{n}$ and~$t_{n}$, and write $z = \Cis\theta$, so that $|z| = 1$ and $z \neq 1$, we have +\[ +|s_{n} + it_{n}| + = \left|\frac{1 - z^{n}}{1 - z}\right| + \leq \frac{1 + |z^{n}|}{|1 - z|} + \leq \frac{2}{|1 - z|}; +\] +and so $|s_{n}|$ and~$|t_{n}|$ are also not greater than~$2/|1 - z|$. That the series are +not actually convergent follows from the fact that their $n$th~terms do not tend +to zero (\Exs{xxiv}.~7,~8). + +The sine series converges to zero if $\theta$~is a multiple of~$\pi$. The cosine +series oscillates finitely if $\theta$~is an odd multiple of~$\pi$ and diverges if $\theta$~is an +even multiple of~$\pi$. + +It follows that \emph{if $\theta_{n}$~is a positive function of~$n$ which tends steadily to +zero as $n \to \infty$, then the series +\[ +\tsum \phi_{n} \cos n\theta,\quad +\tsum \phi_{n} \sin n\theta +\] +are convergent}, except perhaps the first series when $\theta$~is a multiple of~$2\pi$. In +this case the first series reduces to~$\sum \phi_{n}$, which may or may not be convergent: +the second series vanishes identically. If $\sum \phi_{n}$~is convergent then both +series are absolutely convergent (\Ex{lxxvii}.~4) for all values of~$\theta$, and the +whole interest of the result lies in its application to the case in which +$\sum \phi_{n}$~is divergent. And in this case the series above written are conditionally +and \emph{not} absolutely convergent, as will be proved in \Ex{lxxix}.~6. +If we put $\theta = \pi$ in the cosine series we are led back to the result of \SecNo[§]{188}, +since $\cos n\pi = (-1)^{n}$. + +\Item{3.} The series $\sum n^{-s} \cos n\theta$, $\sum n^{-s} \sin n\theta$ are convergent if $s > 0$, unless (in +the case of the first series) $\theta$~is a multiple of~$2\pi$ and $0 < s \leq 1$. +\PageSep{344} + +\Item{4.} The series of Ex.~3 are in general absolutely convergent if $s > 1$, +conditionally convergent if $0 < s \leq 1$, and oscillatory if $s \leq 0$ (finitely if $s = 0$ +and infinitely if $s < 0$). Mention any exceptional cases. + +\Item{5.} If $\sum a_{n}n^{-s}$~is convergent or oscillates finitely, then $\sum a_{n}n^{-t}$~is convergent +when $t > s$. + +\Item{6.} If $\phi_{n}$~is a positive function of~$n$ which tends steadily to~$0$ as $n \to \infty$, +and $\sum \phi_{n}$~is divergent, then the series $\sum \phi_{n} \cos n\theta$, $\sum \phi_{n} \sin n\theta$ are \emph{not} absolutely +convergent, except the sine-series when $\theta$~is a multiple of~$\pi$. [For suppose, +\eg, that $\sum \phi_{n} |\cos n\theta|$ is convergent. Since $\cos^{2} n\theta \leq |\cos n\theta|$, it follows that +$\sum \phi_{n} \cos^{2} n\theta$ or +\[ +\tfrac{1}{2} \tsum \phi_{n} (1 + \cos 2n\theta) +\] +is convergent. But this is impossible, since $\sum \phi_{n}$~is divergent and $\sum \phi_{n} \cos 2n\theta$, +by Dirichlet's Test, convergent, unless $\theta$~is a multiple of~$\pi$. And in this +case it is obvious that $\sum \phi_{n} |\cos n\theta|$ is divergent. The reader should write +out the corresponding argument for the sine-series, noting where it fails +when $\theta$~is a multiple of~$\pi$.] +\end{Examples} + +\Paragraph{190. Series of complex terms.} So far we have confined +ourselves to series all of whose terms are real. We shall now +consider the series +\[ +\tsum u_{n} = \tsum (v_{n} + iw_{n}), +\] +where $v_{n}$ and~$w_{n}$ are real. The consideration of such series does +not, of course, introduce anything really novel. The series is +convergent if, and only if, the series +\[ +\tsum v_{n},\quad +\tsum w_{n} +\] +are separately convergent. There is however one class of such +series so important as to require special treatment. Accordingly +we give the following definition, which is an obvious extension of +that of~\SecNo[§]{184}. + +\begin{Definition} +The series $\sum u_{n}$, where $u_{n} = v_{n} + iw_{n}$, is said to be +\Emph{absolutely convergent} if the series $\sum v_{n}$ and $\sum w_{n}$ are absolutely +convergent. +\end{Definition} + +\begin{Theorem} +The necessary and sufficient condition for the absolute +convergence of~$\sum u_{n}$ is the convergence of $\sum |u_{n}|$ or $\sum \sqrtp{v_{n}^{2} + w_{n}^{2}}$. +\end{Theorem} + +For if $\sum u_{n}$~is absolutely convergent, then both of the series +$\sum |v_{n}|$, $\sum |w_{n}|$ are convergent, and so $\sum \{|v_{n}| + |w_{n}|\}$ is convergent: but +\[ +|u_{n}| = \sqrtp{v_{n}^{2} + w_{n}^{2}} \leq |v_{n}| + |w_{n}|, +\] +\PageSep{345} +and therefore $\sum |u_{n}|$~is convergent. On the other hand +\[ +|v_{n}| \leq \sqrtp{v_{n}^{2} + w_{n}^{2}},\quad +|w_{n}| \leq \sqrtp{v_{n}^{2} + w_{n}^{2}}, +\] +so that $\sum |v_{n}|$ and $\sum |w_{n}|$ are convergent whenever $\sum |u_{n}|$~is convergent. + +It is obvious that \emph{an absolutely convergent series is convergent}, +since its real and imaginary parts converge separately. And +Dirichlet's Theorem (\SecNo[§§]{169},~\SecNo{185}) may be extended at once to +absolutely convergent complex series by applying it to the +separate series $\sum v_{n}$ and~$\sum w_{n}$. + +\begin{Remark} +The convergence of an absolutely convergent series may also be deduced +directly from the general principle of convergence (cf.\ \Ex{lxxvii}.~1). We leave +this as an exercise to the reader. +\end{Remark} + +\Paragraph{191. Power Series.} One of the most important parts of +the theory of the ordinary functions which occur in elementary +analysis (such as the sine and cosine, and the logarithm and +exponential, which will be discussed in the next chapter) is that +which is concerned with their expansion in series of the form +$\sum a_{n}x^{n}$. Such a series is called a \Emph{power series} in~$x$. We have +already come across some cases of expansion in series of this kind +in connection with Taylor's and Maclaurin's series (\SecNo[§]{148}). There, +however, we were concerned only with a real variable~$x$. We shall +now consider a few general properties of power series in~$z$, where +$z$~is a complex variable. + +\begin{Result} +\Item{A.} A power series $\sum a_{n}z^{n}$ may be convergent for all values of~$z$, +for a certain region of values, or for no values except $z = 0$. +\end{Result} + +It is sufficient to give an example of each possibility. + +\begin{Remark} +\Item{1.} \emph{The series $\sum \dfrac{z^{n}}{n!}$ is convergent for all values of~$\DPtypo{x}{z}$.} For if $u_{n} = \dfrac{z^{n}}{n!}$ then +\[ +|u_{n+1}|/|u_{n}| = |z|/(n + 1) \to 0 +\] +as $n \to \infty$, whatever value $z$ may have. Hence, by d'Alembert's Test, $\sum |u_{n}|$~is +convergent for all values of~$z$, and the original series is absolutely convergent +for all values of~$z$. We shall see later on that a power series, when +convergent, is \emph{generally} absolutely convergent. + +\Item{2.} \emph{The series $\sum n!\, z^{n}$ is not convergent for any value of~$z$ except $z = 0$.} +For if $u_{n} = n!\, z^{n}$ then $|u_{n+1}|/|u_{n}| = (n + 1)|z|$, which tends to~$\infty$ with~$n$, unless +$z = 0$. Hence (cf.\ \Exs{xxvii}.\ 1,~2,~5) the modulus of the $n$th~term tends to~$\infty$ +with~$n$; and so the series cannot converge, except when $z = 0$. It is obvious +that any power series converges when $z = 0$. +\PageSep{346} + +\Item{3.} \emph{The series $\sum z^{n}$ is always convergent when $|z| < 1$, and never convergent +when $|z| \geq 1$.} This was proved in \SecNo[§]{88}. Thus we have an actual example of +each of the three possibilities. +\end{Remark} + +\Paragraph{192.} \begin{Result}\Item{B.} If a power series $\sum a_{n}z^{n}$ is convergent for a particular +value of~$z$, say $z_{1} = r_{1}(\cos\theta_{1} + i\sin\theta_{1})$, then it is absolutely +convergent for all values of~$z$ such that $|z| < r_{1}$. +\end{Result} + +For $\lim a_{n}z_{1}^{n} = 0$, since $\sum a_{n}z_{1}^{n}$~is convergent, and therefore we +can certainly find a constant~$K$ such that $|a_{n}z_{1}^{n}| < K$ for all +values of~$n$. But, if $|z| = r < r_{1}$, we have +\[ +|a_{n}z^{n}| + = |a_{n}z_{1}^{n}| \left(\frac{r}{r_{1}}\right)^{n} + < K \left(\frac{r}{r_{1}}\right)^{n}, +\] +and the result follows at once by comparison with the convergent +geometrical series $\sum (r/r_{1})^{n}$. + +In other words, if the series converges at~$P$ \emph{then it converges +absolutely at all points nearer to the origin than~$P$}. + +\begin{Remark} +\Par{Example.} Show that the result is true even if the series oscillates +finitely when $z = z_{1}$. [If $s_{n} = a_{0} + a_{1}z_{1} + \dots + a_{n}z_{1}^{n}$ then we can find~$K$ so that +$|s_{n}| < K$ for all values of~$n$. But $|a_{n}z_{1}^{n}| = |s_{n} - s_{n-1}| \leq |s_{n-1}| + |s_{n}| < 2K$, +and the argument can be completed as before.] +\end{Remark} + +\Paragraph{193. The region of convergence of a power series. +The circle of convergence.} Let $z = r$ be any point on the +positive real axis. If the power series converges when $z = r$ then +it converges absolutely at all points inside the circle $|z| = r$. In +particular it converges for all real values of~$z$ less than~$r$. + +Now let us divide the points~$r$ of the positive real axis into +two classes, the class at which the series converges and the class +at which it does not. The first class must contain at least the +one point $z = 0$. The second class, on the other hand, need not +exist, as the series may converge for all values of~$z$. Suppose +however that it does exist, and that the first class of points +does include points besides $z = 0$. Then it is clear that every +point of the first class lies to the left of every point of the second +class. Hence there is a point, say the point $z = R$, which divides +the two classes, and may itself belong to either one or the other. +\emph{Then the series is absolutely convergent at all points inside the +circle $|z| = R$.} +\PageSep{347} + +For let $P$~be any such point. We can draw a circle, whose +centre is~$O$ and whose radius is +%[Illustration: Fig. 51.] +\Figure[2.5in]{51}{p347} +less than~$R$, so as to include~$P$ +inside it. Let this circle cut~$OA$ +in~$Q$. Then the series is convergent +at~$Q$, and therefore, by +Theorem~B, absolutely convergent +at~$P$. + +On the other hand the series +cannot converge at any point~$P'$ +\emph{outside} the circle. For if it converged at~$P'$ it would converge +absolutely at all points nearer to~$O$ than~$P$; and this is absurd, +as it does not converge at any point between $A$ and~$Q'$ (\Fig{51}). + +So far we have excepted the cases in which the power series +(1)~does not converge at any point on the positive real axis +except $z = 0$ or (2)~converges at all points on the positive real +axis. It is clear that in case~(1) the power series converges +nowhere except when $z = 0$, and that in case~(2) it is absolutely +convergent everywhere. Thus we obtain the following result: \begin{Result}a +power series either + +\Item{(1)} converges for $z = 0$ and for no other value of~$z$; or + +\Item{(2)} converges absolutely for all values of~$z$; or + +\Item{(3)} \Hang[3.5em] converges absolutely for all values of~$z$ within a certain +circle of radius~$R$, and does not converge for any value +of~$z$ outside this circle. +\end{Result} + +In case~(3) the circle is called the \Emph{circle of convergence} +and its radius the \Emph{radius of convergence} of the power series. + +It should be observed that this general result gives absolutely +no information about the behaviour of the series \emph{on} the circle of +convergence. The examples which follow show that as a matter +of fact there are very diverse possibilities as to this. + +\begin{Examples}{LXXX.} +\Item{1.} The series $1 + az + a^{2}z^{2} + \dots$, where $a > 0$, has a +radius of convergence equal to~$1/a$. It does not converge anywhere on its +circle of convergence, diverging when $z = 1/a$ and oscillating finitely at all other +points on the circle. + +\Item{2.} The series $\dfrac{z}{1^{2}} + \dfrac{z^{2}}{2^{2}} + \dfrac{z^{3}}{3^{2}} + \dots$ has its radius of convergence equal to~$1$; +it converges absolutely at all points on its circle of convergence. +\PageSep{348} + +\Item{3.} More generally, if $|a_{n+1}|/|a_{n}| \to \lambda$, or $|a_{n}|^{1/n} \to \lambda$, as $n \to \infty$, then the +series $a_{0} + a_{1}z + a_{2}z^{2} + \dots$ has $1/\lambda$ as its radius of convergence. In the first case +\[ +\lim |a_{n+1}z^{n+1}|/|a_{n}z^{n}| = \lambda |z|, +\] +which is less or greater than unity according as $|z|$~is less or greater than~$1/\lambda$, +so that we can use \DPchg{D'Alembert's}{d'Alembert's} Test (\SecNo[§]{168},~3). In the second case we +can use Cauchy's Test (\SecNo[§]{168},~2) similarly. + +\Item{4.} \Topic{The logarithmic series.} The series +\[ +z - \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} - \dots +\] +is called (for reasons which will appear later) the `logarithmic' series. It +follows from Ex.~3 that its radius of convergence is unity. + +When $z$~is on the circle of convergence we may write $z = \cos\theta + i\sin\theta$, +and the series assumes the form +\[ + \cos\theta - \tfrac{1}{2} \cos 2\theta + \tfrac{1}{3} \cos 3\theta - \dots ++ i(\sin\theta - \tfrac{1}{2} \sin 2\theta + \tfrac{1}{3} \sin 3\theta - \dots). +\] + +The real and imaginary parts are both convergent, though not absolutely +convergent, unless $\theta$~is an odd multiple of~$\pi$ (\Exs{lxxix}.~3,~4). If $\theta$~is an odd +multiple of~$\pi$ then $z = -1$, and the series assumes the form $-1 - \frac{1}{2} - \frac{1}{3} - \dots$, +and so diverges to~$-\infty$. Thus the logarithmic series converges at all points +of its circle of convergence except the point $z = -1$. + +\Item{5.} \Topic{The binomial series.} Consider the series +\[ +1 + mz + \frac{m(m - 1)}{2!} z^{2} + \frac{m(m - 1)(m - 2)}{3!} z^{3} + \dots\Add{.} +\] +If $m$~is a positive integer then the series terminates. In general +\[ +\frac{|a_{n+1}|}{|a_{n}|} = \frac{|m - n|}{n + 1} \to 1, +\] +so that the radius of convergence is unity. We shall not discuss here the +question of its convergence on the circle, which is a little more difficult.\footnote + {See Bromwich, \textit{Infinite Series}, pp.~225 \textit{et~seq.}; Hobson, \textit{Plane Trigonometry} + (3rd~edition), pp.~268~\textit{et~seq.}} +\end{Examples} + +\begin{Remark} +\Paragraph{194. Uniqueness of a power series.} If $\sum a_{n} z^{n}$ is a power series which +is convergent for some values of~$z$ at any rate besides $z = 0$, and $f(z)$~is its +sum, then it is easy to see that $f(z)$~can be expressed in the form +\[ +a_{0} + a_{1}z + a_{2}z^{2} + \dots + (a_{n} + \epsilon_{z})z^{n}, +\] +where $\epsilon_{z} \to 0$ as $|z| \to 0$. For if $\mu$~is any number less than the radius of convergence +of the series, and $|z| < \mu$, then $|a_{n}| \mu^{n} < K$, where $K$~is a constant +(cf.~\SecNo[§]{192}), and so +\begin{align*} +\left|f(z) - \sum_{0}^{n} a_{\nu}z^{\nu}\right| + &\leq |a_{n+1}| |z^{n+1}| + |a_{n+2}| |z^{n+2}| + \dots\\ + &< K \left(\frac{|z|}{\mu}\right)^{n+1} + \left(1 + \frac{|z|}{\mu} + \frac{|z|^{2}}{\mu^{2}} + \dots\right) + = \frac{K |z|^{n+1}}{\mu^{n} (\mu - |z|)}, +\end{align*} +\PageSep{349} +where $K$~is a number independent of~$z$. It follows from \Ex{lv}.~15 that +if $\sum a_{n}z^{n} = \sum b_{n}z^{n}$ for all values of~$z$ whose modulus is less than some +number~$\mu$, then $a_{n} = b_{n}$ for all values of~$n$. This result is capable of considerable +generalisations into which we cannot enter now. It shows that \emph{the same +function~$f(z)$ cannot be represented by two different power series}. +\end{Remark} + +\Paragraph{195. Multiplication of Series.} We saw in \SecNo[§]{170} that if +$\sum u_{n}$ and $\sum v_{n}$ are two convergent series of positive terms, then +$\sum u_{n} × \sum v_{n} = \sum w_{n}$, where +\[ +w_{n} = u_{0}v_{n} + u_{1}v_{n-1} + \dots + u_{n}v_{0}. +\] +We can now extend this result to all cases in which $\sum u_{n}$ and $\sum v_{n}$ +are \emph{absolutely} convergent; for our proof was merely a simple +application of Dirichlet's Theorem, which we have already extended +to all absolutely convergent series. + +\begin{Examples}{LXXXI.} +\Item{1.} If $|z|$~is less than the radius of convergence +of either of the series $\sum a_{n}z^{n}$, $\sum b_{n}z^{n}$, then the product of the two series is +$\sum c_{n}z^{n}$, where $c_{n} = a_{0}b_{n} + a_{1}b_{n-1} + \dots + a_{n}b_{0}$. + +\Item{2.} {\Loosen If the radius of convergence of $\sum a_{n}z^{n}$ is~$R$, and $f(z)$~is the sum of +the series when $|z| < R$, and $|z|$~is less than either $R$ or unity, then +$f(z)/(1 - z) = \sum s_{n}z^{n}$, where $s_{n} = a_{0} + a_{1} + \dots + a_{n}$.} + +\Item{3.} Prove, by squaring the series for $1/(1 - z)$, that $1/(1 - z)^{2} = 1 + 2z + 3z^{2} + \dots$ +if $|z| < 1$. + +\Item{4.} Prove similarly that $1/(1 - z)^{3} = 1 + 3z + 6z^{2} + \dots$, the general term +being $\frac{1}{2}(n + 1)(n + 2)z^{n}$. + +\Item{5.} \Topic{The Binomial Theorem for a negative integral exponent.} If +$|z| < 1$, and $m$~is a positive integer, then +\[ +\frac{1}{(1 - z)^{m}} + = 1 + mz + \frac{m(m + 1)}{1·2} z^{2} + \dots + + \frac{m(m + 1) \dots (m + n - 1)}{1·2 \dots n} z^{n} + \dots. +\] + +[Assume the truth of the theorem for all indices up to~$m$. Then, by Ex.~2, +$1/(1 - z)^{m+1} = \sum s_{n}z^{n}$, where +\begin{align*} +%[** TN: Set on a single line in the original] +s_{n} + &= 1 + m + \frac{m(m + 1)}{1·2} + \dots + + \frac{m(m + 1) \dots (m + n - 1)}{1·2 \dots n} \\ + &= \frac{(m + 1)(m + 2) \dots (m + n)}{1·2 \dots n}, +\end{align*} +as is easily proved by induction.] + +\Item{6.} Prove by multiplication of series that if +\[ +f(m, z) = 1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots, +\] +and $|z| < 1$, then $f(m, z)f(m', z) = f(m + m', z)$. [This equation forms the basis of +Euler's proof of the Binomial Theorem. The coefficient of~$z^{n}$ in the product +series is +\[ +\binom{m'}{n} + + \binom{m}{1} \binom{m'}{n - 1} + + \binom{m}{2} \binom{m'}{n - 2} + \dots + + \binom{m}{n - 1} \binom{m'}{1} + + \binom{m}{n}. +\] +\PageSep{350} + +This is a polynomial in $m$~and~$m'$: but when $m$~and~$m'$ are positive +integers this polynomial must reduce to $\dbinom{m + m'}{k}$ in virtue of the Binomial +Theorem for a positive integral exponent, and if two such polynomials are +equal for all positive integral values of $m$~and~$m'$ then they must be equal +identically.] + +\Item{7.} If $f(z) = 1 + z + \dfrac{z^{2}}{2!} + \dots$ then $f(z)f(z') = f(z + z')$. [For the series for~$f(z)$ +is absolutely convergent for all values of~$z$: and it is easy to see that if +$u_{n} = \dfrac{z^{n}}{n!}$, $v_{n} = \dfrac{z'^{n}}{n!}$, then $w_{n} = \dfrac{(z + z')^{n}}{n!}$.] + +\Item{8.} If +\[ +C(z) = 1 - \frac{z^{2}}{2!} + \frac{z^{4}}{4!} - \dots,\quad +S(z) = z - \frac{z^{3}}{3!} + \frac{z^{5}}{5!} - \dots, +\] +then +\[ +C(z + z') = C(z)C(z') - S(z)S(z'),\quad +S(z + z') = S(z)C(z') + C(z)S(z'), +\] +and +\[ +\{C(z)\}^{2} + \{S(z)\}^{2} = 1. +\] + +\Item{9.} \Topic{Failure of the Multiplication Theorem.} That the theorem is not +always true when $\sum u_{n}$ and $\sum v_{n}$ are not \emph{absolutely} convergent may be seen by +considering the case in which +\[ +u_{n} = v_{n} = \frac{(-1)^{n}}{\sqrtp{n + 1}}. +\] +Then +\[ +w_{n} = (-1)^{n} \sum_{r=0}^{n} \frac{1}{\sqrtb{(r + 1)(n + 1 - r)}}. +\] +But $\sqrtb{(r + 1)(n + 1 - r)} \leq \frac{1}{2}(n + 2)$, and so $|w_{n}| > (2n + 2)/(n + 2)$, which tends +to~$2$; so that $\sum w_{n}$~is certainly not convergent. +\end{Examples} + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER VIII.} + +\begin{Examples}{} +\Item{1.} Discuss the convergence of the series $\sum n^{k}\{\sqrtp{n + 1} - 2\sqrt{n} + \sqrtp{n - 1}\}$, +where $k$~is real. \MathTrip{1890.} + +\Item{2.} Show that +\[ +\tsum n^{r} \Delta^{k}(n^{s}), +\] +where +\[ +\Delta u_{n} = u_{n} - u_{n+1},\quad +\Delta^{2} u_{n} = \Delta(\Delta u_{n}), +\] +and so on, is convergent if and only if $k > r + s + 1$, except when $s$~is a positive +integer less than~$k$, when every term of the series is zero. + +[The result of \Ref{Ch.}{VII}, \MiscEx{VII}~11, shows that $\Delta^{k}(n^{s})$~is in general of +order~$n^{s-k}$.] + +\Item{3.} Show that +\[ +\sum_{1}^{\infty} \frac{n^{2} + 9n + 5}{(n + 1)(2n + 3)(2n + 5)(n + 4)} + = \frac{5}{36}. +\] +\MathTrip{1912.} + +[Resolve the general term into partial fractions.] +\PageSep{351} + +\Item{4.} Show that, if $R(n)$~is any rational function of~$n$, we can determine +a polynomial~$P(n)$ and a constant~$A$ such that $\sum \{R(n) - P(n) - (A/n)\}$ is +convergent. Consider in particular the cases in which $R(n)$~is one of +the functions $1/(an + b)$, $(an^{2} + 2bn + c)/(\alpha n^{2} + 2\beta n + \gamma)$. + +\Item{5.} Show that the series +\[ +1 - \frac{1}{1 + z} + + \frac{1}{2} - \frac{1}{2 + z} + + \frac{1}{3} - \frac{1}{3 + z} + \dots +\] +is convergent provided only that $z$~is not a negative integer. + +\Item{6.} Investigate the convergence or divergence of the series +\begin{gather*} +%[** TN: Set on one line in the original] +\sum \sin\frac{a}{n},\quad +\sum \frac{1}{n} \sin\frac{a}{n},\quad +\sum (-1)^{n} \sin\frac{a}{n},\\ +\sum \left(1 - \cos\frac{a}{n}\right),\quad +\sum (-1)^{n} n\left(1 - \cos\frac{a}{n}\right), +\end{gather*} +where $a$~is real. + +\Item{7.} Discuss the convergence of the series +\[ +\sum_{1}^{\infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right) + \frac{\sin(n\theta + \alpha)}{n}, +\] +where $\theta$~and~$\alpha$ are real. \MathTrip{1989.} + +\Item{8.} Prove that the series +\[ +1 - \tfrac{1}{2} - \tfrac{1}{3} + \tfrac{1}{4} + \tfrac{1}{5} + \tfrac{1}{6} + - \tfrac{1}{7} - \tfrac{1}{8} - \tfrac{1}{9} - \tfrac{1}{10} + \dots, +\] +in which successive terms of the same sign form groups of $1$, $2$, $3$, $4$,~\dots\ terms, +is convergent; but that the corresponding series in which the groups contain +$1$, $2$, $4$, $8$,~\dots\ terms oscillates finitely. \MathTrip{1908.} + +\Item{9.} If $u_{1}$, $u_{2}$, $u_{3}$,~\dots\ is a decreasing sequence of positive numbers whose +limit is zero, then the series +\[ +u_{1} - \tfrac{1}{2}(u_{1} + u_{2}) + \tfrac{1}{3}(u_{1} + u_{2} + u_{3}) - \dots, +\quad +u_{1} - \tfrac{1}{3}(u_{1} + u_{3}) + \tfrac{1}{5}(u_{1} + u_{3} + u_{5}) - \dots +\] +are convergent. [For if $(u_{1} + u_{2} + \dots + u_{n})/n = v_{n}$ then $v_{1}$, $v_{2}$, $v_{3}$,~\dots\ is also a +decreasing sequence whose limit is zero (\Ref{Ch.}{IV}, \MiscExs{IV}~8,~27). This +shows that the first series is convergent; the second we leave to the reader. +In particular the series +\[ +1 - \tfrac{1}{2}\left(1 + \tfrac{1}{2}\right) + + \tfrac{1}{3}\left(1 + \tfrac{1}{2} + \tfrac{1}{3}\right) - \dots,\quad +1 - \tfrac{1}{3}\left(1 + \tfrac{1}{\DPtypo{5}{3}}\right) + + \tfrac{1}{5}\left(1 + \tfrac{1}{3} + \tfrac{1}{5}\right) - \dots +\] +are convergent.] + +\Item{10.} If $u_{0} + u_{1} + u_{2} + \dots$ is a divergent series of positive and decreasing +terms, then +\[ +(u_{0} + u_{2} + \dots + u_{2n})/(u_{1} + u_{3} + \dots + u_{2n+1}) \to 1. +\] + +\Item{11.} Prove that if $\alpha > 0$ then $\lim\limits_{p\to\infty} \sum\limits_{n=0}^{\infty} (p + n)^{-1-\alpha} = 0$. + +\Item{12.} Prove that $\lim\limits_{\alpha\to 0+} \alpha \sum\limits_{1}^{\infty} n^{-1-\alpha} = 1$. [It follows from \SecNo[§]{174} that +\[ +0 < 1^{-1-\alpha} + 2^{-1-\alpha} + \dots + (n - 1)^{-1-\alpha} + - \int_{1}^{n} x^{-1-\alpha}\, dx \leq 1, +\] +and it is easy to deduce that $\sum n^{-1-\alpha}$ lies between $1/\alpha$ and~$(1/\alpha) + 1$.] +\PageSep{352} + +\Item{13.} Find the sum of the series $\sum\limits_{1}^{\infty} u_{n}$, where +\[ +u_{n} = \frac{x^{n} - x^{-n-1}}{(x^{n} + x^{-n})(x^{n+1} + x^{-n-1}) } + = \frac{1}{x - 1} + \left(\frac{1}{x^{n} + x^{-n}} - \frac{1}{x^{n+1} + x^{-n-1}}\right), +\] +for all real values of~$x$ for which the series is convergent. \MathTrip{1901.} + +[If $|x|$~is not equal to unity then the series has the sum $x/\{(x - 1)(x^{2} + 1)\}$. +If $x = 1$ then $u_{n} = 0$ and the sum is~$0$. If $x = -1$ then $u_{n} = \frac{1}{2}(-1)^{n+1}$ and +the series oscillates finitely.] + +\Item{14.} Find the sums of the series +\[ +\frac{z}{1 + z} + \frac{2z^{2}}{1 + z^{2}} + \frac{4z^{4}}{1 + z^{4}} + \dots,\quad +\frac{z}{1 - z^{2}} + \frac{z^{2}}{1 - z^{4}} + \frac{z^{4}}{1 - z^{8}} + \dots +\] +(in which all the indices are powers of~$2$), whenever they are convergent. + +[The first series converges only if $|z| < 1$, its sum then being~$z/(1 - z)$; the +second series converges to~$z/(1 - z)$ if $|z| < 1$ and to~$1/(1 - z)$ if $|z| > 1$.] + +\Item{15.} If $|a_{n}| \leq 1$ for all values of~$n$ then the equation +\[ +0 = 1 + a_{1}z + a_{2}z^{2} + \dots +\] +cannot have a root whose modulus is less than~$\frac{1}{2}$, and the only case in which +it can have a root whose modulus is equal to~$\frac{1}{2}$ is that in which $a_{n} = -\Cis(n\theta)$, +when $z = \frac{1}{2} \Cis(-\theta)$ is a root. + +\Item{16.} \Topic{Recurring Series.} A power series $\sum a_{n}z^{n}$ is said to be a \emph{recurring +series} if its coefficients satisfy a relation of the type +\[ +a_{n} + p_{1}a_{n-1} + p_{2}a_{n-2} + \dots + p_{k}a_{n-k} = 0, +\Tag{(1)} +\] +where $n \geq k$ and $p_{1}$, $p_{2}$,~\dots, $p_{k}$ are independent of~$n$. Any recurring series is +the expansion of a rational function of~$z$. To prove this we observe in the +first place that the series is certainly convergent for values of~$z$ whose modulus +is sufficiently small. For let $G$ be the greater of the two numbers +\[ +1,\quad +|p_{1}| + |p_{2}| + \dots + |p_{k}|. +\] +Then it follows from the equation~\Eq{(1)} that $|a_{n}| \leq G\alpha_{n}$, where $\alpha_{n}$~is the +modulus of the numerically greatest of the preceding coefficients; and from +this that $|a_{n}| < KG^{n}$, where $K$~is independent of~$n$. Thus the recurring series +is certainly convergent for values of~$z$ whose modulus is less than~$1/G$. + +But if we multiply the series $f(z) = \sum a_{n}z^{n}$ by $p_{1}z$, $p_{2}z^{2}$,~\dots\Add{,} $p_{k}z^{k}$, and add +the results, we obtain a new series in which all the coefficients after +the~$(k - 1)$th vanish in virtue of the relation~\Eq{(1)}, so that +\[ +(1 + p_{1}z + p_{2}z^{2} + \dots + p_{k}z^{k})f(z) + = P_{0} + P_{1}z + \dots + P_{k-1}z^{k-1}, +\] +where $P_{0}$, $P_{1}$,~\dots, $P_{k-1}$ are constants. The polynomial $1 + p_{1}z + p_{2}z^{2} + \dots + p_{k}z^{k}$ +is called the \emph{scale of relation} of the series. + +Conversely, it follows from the known results as to the expression of any +rational function as the sum of a polynomial and certain partial fractions of +the type~$A/(z - a)^{p}$, and from the Binomial Theorem for a negative integral +\PageSep{353} +exponent, that any rational function whose denominator is not divisible by~$z$ +can be expanded in a power series convergent for values of~$z$ whose modulus is +sufficiently small, in fact if $|z| < \rho$, where $\rho$~is the least of the moduli of the roots +of the denominator (cf.\ \Ref{Ch.}{IV}, \MiscExs{IV}\ 18~\textit{et~seq.}). And it is easy to see, +by reversing the argument above, that the series is a recurring series. Thus +\begin{Result}the necessary and sufficient condition that a power series should be a recurring +series is that it should be the expansion of such a rational function of~$z$. +\end{Result} + +\Item{17.} \Topic{Solution of Difference-Equations.} A relation of the type of~\Eq{(1)} +in Ex.~16 is called a \emph{linear difference-equation in~$a_{n}$ with constant coefficients}. +Such equations may be solved by a method which will be sufficiently explained +by an example. Suppose that the equation is +\[ +a_{n} - a_{n-1} - 8a_{n-2} + 12a_{n-3} = 0. +\] +Consider the recurring power series $\sum a_{n}z^{n}$. We find, as in Ex.~16, that its +sum is +\[ +\frac{a_{0} + (a_{1} - a_{0}) z + (a_{2} - a_{1} - 8a_{0}) z^{2}} + {1 - z - 8z^{2} + 12z^{3}} + = \frac{A_{1}}{1 - 2z} + \frac{A_{2}}{(1 - 2z)^{2}} + \frac{B}{1 + 3z}, +\] +where $A_{1}$,~$A_{2}$, and~$B$ are numbers easily expressible in terms of $a_{0}$,~$a_{1}$, and~$a_{2}$. +Expanding each fraction separately we see that the coefficient of~$z^{n}$ is +\[ +a_{n} = 2^{n}\{A_{1} + (n + 1) A_{2}\} + (-3)^{n} B. +\] +The values of $A_{1}$,~$A_{2}$,~$B$ depend upon the first three coefficients $a_{0}$,~$a_{1}$,~$a_{2}$, +which may of course be chosen arbitrarily. + +\Item{18.} The solution of the difference-equation $u_{n} - 2\cos\theta u_{n-1} + u_{n-2} = 0$ is +$u_{n} = A\cos n\theta + B\sin n\theta$, where $A$~and~$B$ are arbitrary constants. + +\Item{19.} If $u_{n}$~is a polynomial in~$n$ of degree~$k$, then $\sum u_{n} z^{n}$~is a recurring +series whose scale of relation is $(1 - z)^{k+1}$. \MathTrip{1904.} + +\Item{20.} Expand $9/\{(z - 1)(z + 2)^{2}\}$ in ascending powers of~$z$. +\MathTrip{1913.} + +\Item{21.} Prove that if $f(n)$~is the coefficient of~$z^{n}$ in the expansion of~$z/(1 + z + z^{2})$ +in powers of~$z$, then +\[ +\Item{(1)}\ f(n) + f(n - 1) + f(n - 2) = 0,\quad +\Item{(2)}\ f(n) = (\omega_{3}^{n} - \omega_{3}^{2n})/(\omega_{3} - \omega_{3}^{2}), +\] +where $\omega_{3}$~is a complex cube root of unity. Deduce that $f(n)$~is equal to $0$ +or $1$ or~$-1$ according as $n$~is of the form $3k$ or $3k + 1$ or~$3k + 2$, and verify +this by means of the identity $z/(1 + z + z^{2}) = z(1 - z)/(1 - z^{3})$. + +\Item{22.} A player tossing a coin is to score one point for every head he turns +up and two for every tail, and is to play on until his score reaches or passes +a total~$n$. Show that his chance of making exactly the total~$n$ is $\frac{1}{3}\{2 + (-\frac{1}{2})^{n}\}$. \MathTrip{1898.} + +[If $p_{n}$~is the probability then $p_{n} = \frac{1}{2} (p_{n-1} + p_{n-2})$. \DPtypo{also}{Also} $p_{0} = 1$, $p_{1} = \frac{1}{2}$.] +\PageSep{354} + +\Item{23.} Prove that +\[ +\frac{1}{a + 1} + \frac{1}{a + 2} + \dots + \frac{1}{a + n} + = \binom{n}{1}\frac{1}{a + 1} + - \binom{n}{2}\frac{1!}{(a + 1)(a + 2)} + \dots +\] +if $n$~is a positive integer and $a$~is not one of the numbers $-1$, $-2$,~\dots,~$-n$. + +[This follows from splitting up each term on the right-hand side into partial +fractions. When $a > -1$, the result may be deduced very simply from the +equation +\[ +\int_{0}^{1} x^{a}\frac{1 - x^{n}}{1 - x}\, dx + = \int_{0}^{1} (1 - x)^{a}\{1 - (1 - x)^{n}\}\frac{dx}{x} +\] +by expanding $(1 - x^{n})/(1 - x)$ and $1 - (1 - x)^{n}$ in powers of~$x$ and integrating +each term separately. The result, being merely an algebraical identity, must +be true for all values of~$a$ save $-1$, $-2$,~\dots,~$-n$.] + +\Item{24.} Prove by multiplication of series that +\[ +\sum_{0}^{\infty} \frac{z^{n}}{n!} \sum_{1}^{\infty} \frac{(-1)^{n-1}z^{n}}{n·n!} + = \sum_{1}^{\infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right) \frac{z^{n}}{n!}. +\] + +[The coefficient of~$z^{n}$ will be found to be +\[ +\frac{1}{n!}\left\{ + \binom{n}{1} - \frac{1}{2}\binom{n}{2} + \frac{1}{3}\binom{n}{3} - \dots +\right\}. +\] +Now use Ex.~23, taking $a = 0$.] + +\Item{25.} If $A_{n} \to A$ and $B_{n} \to B$ as $n \to \infty$, then +\[ +(A_{1}B_{n} + A_{2}B_{n-1} + \dots + A_{n}B_{1})/n \to AB. +\] + +[Let $A_{n} = A + \epsilon_{n}$. Then the expression given is equal to +\[ +A \frac{B_{1} + B_{2} + \dots + B_{n}}{n} + + \frac{\epsilon_{1}B_{n} + \epsilon_{2}B_{n-1} + \dots + \epsilon_{n}B_{1}}{n}. +\] + +The first term tends to~$AB$ (\Ref{Ch.}{IV}, \MiscEx{IV}~27). The modulus of +the second is less than $\beta\{|\epsilon_{1}| + |\epsilon_{2}| + \dots + |\epsilon_{n}|\}/n$, where $\beta$~is any number +greater than the greatest value of~$|B_{\nu}|$: and this expression tends to zero.] + +\Item{26.} Prove that if $c_{n} = a_{1}b_{n} + a_{2}b_{n-1} + \dots + a_{n}b_{1}$ and +\[ +A_{n} = a_{1} + a_{2} + \dots + a_{n},\quad +B_{n} = b_{1} + b_{2} + \dots + b_{n},\quad +C_{n} = c_{1} + c_{2} + \dots + c_{n}, +\] +then +\[ +C_{n} = a_{1}B_{n} + a_{2}B_{n-1} + \dots + a_{n}B_{1} + = b_{1}A_{n} + b_{2}A_{n-1} + \dots + b_{n}A_{1} +\] +and +\[ +C_{1} + C_{2} + \dots + C_{n} = A_{1}B_{n} + A_{2}B_{n-1} + \dots + A_{n}B_{1}. +\] + +Hence prove that if the series $\sum a_{n}$, $\sum b_{n}$ are convergent and have the sums +$A$,~$B$, so that $A_{n} \to A$, $B_{n} \to B$, then +\[ +(C_{1} + C_{2} + \dots + C_{n})/n \to AB. +\] +Deduce that \emph{if $\sum c_{n}$ is convergent then its sum is~$AB$}. This result is known as +\Emph{Abel's Theorem on the multiplication of Series}. We have already seen +that we can multiply the series $\sum a_{n}$, $\sum b_{n}$ in this way if both series are +\emph{absolutely} convergent: Abel's Theorem shows that we can do so even if +one or both are not absolutely convergent, \emph{provided only that the product series +is convergent}. +\PageSep{355} + +\Item{27.} Prove that +\begin{align*} +\tfrac{1}{2} \left(1 - \tfrac{1}{2} + \tfrac{1}{3} - \dots\right)^{2} + &= \tfrac{1}{2} - \tfrac{1}{3} \left(1 + \tfrac{1}{2}\right) + + \tfrac{1}{4} \left(1 + \tfrac{1}{2} + \tfrac{1}{3}\right) - \dots,\\ +% +\tfrac{1}{2} \left(1 - \tfrac{1}{3} + \tfrac{1}{5} - \dots\right)^{2} + &= \tfrac{1}{2} - \tfrac{1}{4} \left(1 + \tfrac{1}{3}\right) + + \tfrac{1}{6} \left(1 + \tfrac{1}{3} + \tfrac{1}{5}\right) - \dots. +\end{align*} + +[Use Ex.~9 to establish the convergence of the series.] + +\Item{28.} For what values of $m$~and~$n$ is the integral $\ds\int_{0}^{\pi} \sin^{m} x (1 - \cos x)^{n}\, dx$ +convergent? [If $m + 1$ and~$m + 2n + 1$ are positive.] + +\Item{29.} Prove that if $a > 1$ then +\[ +\int_{-1}^{1} \frac{dx}{(a - x) \sqrtp{1 - x^{2}}} + = \frac{\pi}{\sqrtp{a^{2} - 1}}. +\] + +\Item{30.} Establish the formulae +\begin{alignat*}{2} +\int_{0}^{\infty} F\{\sqrtp{x^{2} + 1} + x\}\, dx + &= \tfrac{1}{2}\int_{1}^{\infty} &&\left(1 + \frac{1}{y^{2}}\right) F(y) \, dy,\\ +% +\int_{0}^{\infty} F\{\sqrtp{x^{2} + 1} - x\}\, dx + &= \tfrac{1}{2}\int_{0}^{1} &&\left(1 + \frac{1}{y^{2}}\right) F(y)\, dy. +\end{alignat*} +In particular, prove that if $n > 1$ then +\[ +\int_{0}^{\infty} \frac{dx}{\{\sqrtp{x^{2} + 1} + x\}^{n}} + = \int_{0}^{\infty} \{\sqrtp{x^{2} + 1} - x\}^{n}\, dx + = \frac{n}{n^{2} - 1}. +\] + +[In this and the succeeding examples it is of course supposed that the +arbitrary functions which occur are such that the integrals considered have a +meaning in accordance with the definitions of \SecNo[§§]{177}~\textit{et~seq.}] + +\Item{31.} Show that if $2y = ax - (b/x)$, where $a$~and~$b$ are positive, then $y$~increases +steadily from $-\infty$ to~$\infty$ as $x$~increases from $0$ to~$\infty$. Hence show that +\begin{align*} +\int_{0}^{\infty} f\left\{\tfrac{1}{2}\left(ax + \frac{b}{x}\right)\right\} dx + &= \frac{1}{a} \int_{-\infty}^{\infty} f\{\sqrtp{y^{2} + ab}\} + \left\{1 + \frac{y}{\sqrtp{y^{2} + ab}}\right\} dy\\ + &= \frac{2}{a} \int_{0}^{\infty} f\{\sqrtp{y^{2} + ab}\}\, dy. +\end{align*} + +\Item{32.} Show that if $2y = ax + (b/x)$, where $a$~and~$b$ are positive, then two +values of~$x$ correspond to any value of~$y$ greater than~$\sqrtp{ab}$. Denoting the +greater of these by~$x_{1}$ and the less by~$x_{2}$, show that, as $y$~increases +from~$\sqrtp{ab}$ +towards~$\infty$, $x_{1}$~increases from~$\sqrtp{b/a}$ towards~$\infty$, and $x_{2}$~decreases +from~$\sqrtp{b/a}$ to~$0$. Hence show that +\begin{align*} +\int_{\sqrtp{b/a}}^{\infty} f(y)\, dx_{1} + &= \frac{1}{a} \int_{\sqrtp{ab}}^{\infty} f(y) + \left\{\frac{y}{\sqrtp{y^{2} - ab}} + 1\right\} dy,\\ +% +\int_{0}^{\sqrtp{b/a}} f(y)\, dx_{2} + &= \frac{1}{a} \int_{\sqrtp{ab}}^{\infty} f(y) + \left\{\frac{y}{\sqrtp{y^{2} - ab}} - 1\right\} dy, +\end{align*} +and that +\[ +\int_{0}^{\infty} f\left\{\tfrac{1}{2}\left(ax + \frac{b}{x}\right)\right\} dx + = \frac{2}{a} \int_{\sqrtp{ab}}^{\infty} \frac{yf(y)}{\sqrtp{y^{2} - ab}}\, dy + = \frac{2}{a} \int_{0}^{\infty} f\{\sqrtp{z^{2} + ab}\}\, dz. +\] +\PageSep{356} + +\Item{33.} Prove the formula +\[ +\int_{0}^{\pi} f(\sec\tfrac{1}{2}x + \tan\tfrac{1}{2}x)\frac{dx}{\sqrtp{\sin x}} + = \int_{0}^{\pi} f(\cosec x)\frac{dx}{\sqrtp{\sin x}}. +\] + +\Item{34.} If $a$~and~$b$ are positive, then +\[ +\int_{0}^{\infty} \frac{dx}{(x^{2} + a^{2})(x^{2} + b^{2})} + = \frac{\pi}{2ab(a + b)},\quad +\int_{0}^{\infty} \frac{x^{2}\, dx}{(x^{2} + a^{2})(x^{2} + b^{2})} + = \frac{\pi}{2(a + b)}. +\] +Deduce that if $\alpha$,~$\beta$, and~$\gamma$ are positive, and $\beta^{2} \geq \alpha\gamma$, then +\[ +\int_{0}^{\infty} \frac{dx}{\alpha x^{4} + 2\beta x^{2} + \gamma} + = \frac{\pi}{2\sqrtp{2\gamma A}}, \quad +\int_{0}^{\infty} \frac{x^{2}\, dx}{\alpha x^{4} + 2\beta x^{2} + \gamma} + = \frac{\pi}{2\sqrtp{2\alpha A}}, +\] +where $A = \beta + \sqrtp{\alpha\gamma}$. Also deduce the last result from Ex.~31, by putting +$f(y) = 1/(c^{2} + y^{2})$. The last two results remain true when $\beta^{2} < \alpha\gamma$, but their +proof is then not quite so simple. + +\Item{35.} Prove that if $b$~is positive then +\[ +\int_{0}^{\infty} \frac{x^{2}\, dx}{(x^{2} - a^{2})^{2} + b^{2}x^{2}} + = \frac{\pi}{2b},\quad +\int_{0}^{\infty} \frac{x^{4}\, dx}{\{(x^{2} - a^{2})^{2} + b^{2}x^{2}\}^{2}} + = \frac{\pi}{4b^{3}}. +\] + +\Item{36.} Extend Schwarz's inequality (\Ref{Ch.}{VII}, \MiscEx{VII}~42) to infinite +integrals of the first and second kinds. + +\Item{37.} Prove that if $\phi(x)$~is the function considered at the end of~\SecNo[§]{178} +then +\[ +\int_{0}^{\infty} \phi(x)\, dx = \sum_{0}^{\infty} \frac{1}{(n + 1)^{2}}. +\] + +\Item{38.} Prove that +\begin{alignat*}{2} +\int_{1}^{\infty} dx \left(\int_{1}^{\infty} \frac{x - y}{(x + y)^{3}}\, dy\right) + &= -1, \quad & +\int_{1}^{\infty} dy \left(\int_{1}^{\infty} \frac{x - y}{(x + y)^{3}}\, dx\right) + &= 1;\\ +% +\int_{1}^{\infty} dx \left(\int_{1}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}\, dy\right) + &= -\tfrac{1}{4}\pi, \quad & +\int_{1}^{\infty} dy \left(\int_{1}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}\, dx\right) + &= \tfrac{1}{4}\pi. +\end{alignat*} + +Establish similar results in which the limits of integration are $0$~and~$1$. +\MathTrip{1913.} +\end{Examples} +\PageSep{357} + + +\Chapter[THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS] +{IX}{THE LOGARITHMIC AND EXPONENTIAL FUNCTIONS \\ +OF A REAL VARIABLE} + +\Paragraph{196.} \First{The} number of essentially different types of functions +with which we have been concerned in the foregoing chapters +is not very large. Among those which have occurred the most +important for ordinary purposes are polynomials, rational functions, +algebraical functions, explicit or implicit, and trigonometrical +functions, direct or inverse. + +We are however far from having exhausted the list of functions +which are important in mathematics. The gradual expansion of +the range of mathematical knowledge has been accompanied by +the introduction into analysis of one new class of function after +another. These new functions have generally been introduced +because it appeared that some problem which was occupying the +attention of mathematicians was incapable of solution by means of +the functions already known. The process may fairly be compared +with that by which the irrational and complex numbers were first +introduced, when it was found that certain algebraical equations +could not be solved by means of the numbers already recognised. +One of the most fruitful sources of new functions has been the +problem of \emph{integration}. Attempts have been made to integrate +some function~$f(x)$ in terms of functions already known. These +attempts have failed; and after a certain number of failures it +has begun to appear probable that the problem is insoluble. +Sometimes it has been \emph{proved} that this is so; but as a rule such +a strict proof has not been forthcoming until later on. Generally +it has happened that mathematicians have taken the impossibility +for granted as soon as they have become reasonably convinced +of it, and have introduced a new function~$F(x)$ \emph{defined} by its +\PageSep{358} +possessing the required property, viz.\ that $F'(x) = f(x)$. Starting +from this definition, they have investigated the properties of~$F(x)$; +and it has then appeared that $F(x)$~has properties which +no finite combination of the functions previously known could +possibly have; and thus the correctness of the assumption that +the original problem could not possibly be solved has been +established. One such case occurred in the preceding pages, +when in \Ref{Ch.}{VI} we defined the function~$\log x$ by means of the +equation +\[ +\log x = \int \frac{dx}{x}. +\] + +\begin{Remark} +Let us consider what grounds we have for supposing $\log x$ to be a really +new function. We have seen already (\Ex{xlii}.~4) that it cannot be a rational +function, since the derivative of a rational function is a rational function +whose denominator contains only repeated factors. The question whether it +can be an algebraical or trigonometrical function is more difficult. But it is +very easy to become convinced by a few experiments that differentiation will +never get rid of algebraical irrationalities. For example, the result of +differentiating $\sqrtp{1 + x}$ any number of times is always the product of $\sqrtp{1 + x}$ +by a rational function, and so generally. The reader should test the +correctness of the statement by experimenting with a number of examples. +Similarly, if we differentiate a function which involves $\sin x$ or $\cos x$, one +or other of these functions persists in the result. + +We have, therefore, not indeed a strict proof that $\log x$~is a new function---that +we do not profess to give\footnotemark---but a reasonable presumption that it is. +We shall therefore treat it as such, and we shall find on examination that its +properties are quite unlike those of any function which we have as yet +encountered. +\end{Remark} +\footnotetext{For such a proof see the author's tract quoted on \PageRef{p.}{236}.} + +\Paragraph{197. Definition of $\log x$.} We define $\log x$, the logarithm of~$x$, +by the equation +\[ +\log x = \int_{1}^{x} \frac{dt}{t}. +\] +We must suppose that $x$~is positive, since (\Ex{lxxvi}.~2) the +integral has no meaning if the range of integration includes +the point $x = 0$. We might have chosen a lower limit other +than~$1$; but $1$~proves to be the most convenient. With this +definition $\log 1 = 0$. + +We shall now consider how $\log x$ behaves as $x$~varies from $0$ +towards~$\infty$. It follows at once from the definition that $\log x$~is a +\PageSep{359} +continuous function of~$x$ which increases steadily with~$x$ and has +a derivative +\[ +D_{x} \log x = 1/x; +\] +and it follows from \SecNo[§]{175} that $\log x$ tends to~$\infty$ as $x \to \infty$. + +If $x$~is positive but less than~$1$, then $\log x$~is negative. For +\[ +\log x = \int_{1}^{x} \frac{dt}{t} = -\int_{x}^{1} \frac{dt}{t} < 0. +\] +Moreover, if we make the substitution $t = 1/u$ in the integral, we +obtain +\[ +\log x = \int_{1}^{x} \frac{dt}{t} + = -\int_{1}^{1/x} \frac{du}{u} + = -\log(1/x). +\] +Thus $\log x$ tends steadily to~$-\infty$ as $x$~decreases from $1$ to~$0$. + +The general form of the graph of the logarithmic function is +shown in \Fig{52}. Since the derivative of~$\log x$ is~$1/x$, the slope of +%[Illustration: Fig. 52.] +\Figure[2.75in]{52}{p359} +the curve is very gentle when $x$~is very large, and very steep +when $x$~is very small. + +\begin{Examples}{LXXXII.} +\Item{1.} Prove from the definition that if $u > 0$ then +\[ +u/(1 + u) < \log(1 + u) < u. +\] + +[For $\ds\log(1 + u) = \int_{0}^{u} \frac{dt}{1 + t}$, and the subject of integration lies between $1$ and +$1/(1 + u)$.] + +\Item{2.} Prove that $\log(1 + u)$ lies between $u - \dfrac{u^{2}}{2}$ and $u - \dfrac{u^{2}}{2(1 + u)}$ when $u$~is +positive. [Use the fact that $\ds\log(1 + u) = u - \int_{0}^{u} \frac{t\, dt}{1 + t}$.] + +\Item{3.} If $0 < u < 1$ then $u < -\log(1 - u) < u/(1 - u)$. + +\Item{4.} Prove that +\[ +\lim_{x\to 1} \frac{\log x}{x - 1} = \lim_{t\to 0} \frac{\log (1 + t)}{t} = 1. +\] + +%[** TN: Indented for consistency; no indent in the original] +[Use Ex.~1.] +\end{Examples} +\PageSep{360} + +\Paragraph{198. The functional equation satisfied by $\log x$.} \emph{The +function $\log x$ satisfies the functional equation} +\[ +f(xy) = f(x) + f(y). +\Tag{(1)} +\] +For, making the substitution $t = yu$, we see that +\begin{align*} +\log xy + &= \int_{1}^{xy} \frac{dt}{t} + = \int_{1/y}^{x} \frac{du}{u} + = \int_{1}^{x} \frac{du}{u} - \int_{1}^{1/y} \frac{du}{u}\\ + &= \log x - \log(1/y) = \log x + \log y, +\end{align*} +which proves the theorem. + +\begin{Examples}{LXXXIII.} +\Item{1.} It can be shown that there is no solution of +the equation~\Eq{(1)} which possesses a differential coefficient and is fundamentally +distinct from $\log x$. For when we differentiate the functional equation, first +with respect to~$x$ and then with respect to~$y$, we obtain the two equations +\[ +yf'(xy) = f'(x),\quad +xf'(xy) = f'(y); +\] +and so, eliminating $f'(xy)$, $xf'(x) = yf'(y)$. But if this is true for every pair +of values of $x$~and~$y$, then we must have $xf'(x) = C$, or $f'(x) = C/x$, where $C$~is +a constant. Hence +\[ +f(x) = \int \frac{C}{x}\, dx + C' = C\log x + C', +\] +and it is easy to see that $C' = 0$. Thus there is no solution fundamentally +distinct from~$\log x$, except the trivial solution $f(x) = 0$, obtained by taking +$C = 0$. + +\Item{2.} Show in the same way that there is no solution of the equation +\[ +f(x) + f(y) = f\left(\frac{x + y}{1 - xy}\right) +\] +which possesses a differential coefficient and is fundamentally distinct from +$\arctan x$. +\end{Examples} + +\Paragraph{199. The manner in which $\log x$ tends to infinity with~$x$.} +It will be remembered that in \Ex{xxxvi}.~6 we defined certain +different ways in which a function of~$x$ may tend to infinity with~$x$, +distinguishing between functions which, when $x$~is large, are of +the first, second, third,~\dots\ orders of greatness. A function~$f(x)$ +was said to be of the $k$th~order of greatness when $f(x)/x^{k}$ tends to +a limit different from zero as $x$~tends to infinity. + +It is easy to define a whole series of functions which tend to +infinity with~$x$, but whose order of greatness is smaller than the first. +Thus $\sqrt{x}$, $\sqrt[3]{x}$, $\sqrt[4]{x}$,~\dots\ are such functions. We may say generally +that $x^{\alpha}$, where $\alpha$~is any positive rational number, is of the $\alpha$th~order +of greatness when $x$~is large. We may suppose $\alpha$ as small +\PageSep{361} +as we please, \DPtypo{e.g.}{\eg}\ less than~$.000\MS000\MS1$. And it might be thought +that by giving $\alpha$ all possible values we should exhaust the +possible `orders of infinity' of~$f(x)$. At any rate it might be +supposed that if $f(x)$~tends to infinity with~$x$, however slowly, we +could always find a value of~$\alpha$ so small that $x^{\alpha}$~would tend to +infinity more slowly still; and, conversely, that if $f(x)$~tends to +infinity with~$x$, however rapidly, we could always find a value +of~$\alpha$ so great that $x^{\alpha}$~would tend to infinity more rapidly still. + +Perhaps the most interesting feature of the function $\log x$ is its +behaviour as $x$~tends to infinity. It shows that the presupposition +stated above, which seems so natural, is unfounded. \emph{The logarithm +of~$x$ tends to infinity with~$x$, but more slowly than \Emph{any} positive power +of~$x$, integral or fractional.} In other words $\log x \to \infty$ but +\[ +\frac{\log x}{x^{\alpha}} \to 0 +\] +for \emph{all} positive values of~$\alpha$. This fact is sometimes expressed +loosely by saying that the `order of infinity of~$\log x$ is infinitely +small'; but the reader will hardly require at this stage to be warned +against such modes of expression. + +\Paragraph{200. Proof that $(\log x)/x^{\alpha} \to 0$ as $x \to \infty$.} Let $\beta$~be any +positive number. Then $1/t < 1/t^{1-\beta}$ when $t > 1$, and so +\[ +\log x = \int_{1}^{x} \frac{dt}{t} < \int_{1}^{x} \frac{dt}{t^{1-\beta}}, +\] +or +\[ +\log x < (x^{\beta} - 1)/\beta < x^{\beta}/\beta, +\] +when $x > 1$. Now if $\alpha$~is any positive number we can choose a +smaller positive value of~$\beta$. And then +\[ +0 < (\log x)/x^{\alpha} < x^{\beta-\alpha}/\beta \quad (x > 1). +\] +But, since $\alpha > \beta$, $x^{\beta-\alpha}/\beta \to 0$ as $x \to \infty$, and therefore +\[ +(\log x)/x^{\alpha} \to 0. +\] + +\Paragraph{201. The behaviour of $\log x$ as $x \to +0$.} Since +\[ +(\log x)/x^{\alpha} = -y^{\alpha} \log y +\] +if $x = 1/y$, it follows from the theorem proved above that +\[ +\lim_{y\to +0} y^{\alpha} \log y = -\lim_{x\to +\infty} (\log x)/x^{\alpha} = 0. +\] +Thus $\log x$ tends to~$-\infty$ and $\log(1/x) = -\log x$ to~$\infty$ as $x$~tends +to zero by positive values, but $\log(1/x)$ tends to~$\infty$ more slowly +than any positive power of~$1/x$, integral or fractional. +\PageSep{362} + +\begin{Remark} +\Paragraph{202. Scales of infinity. The logarithmic scale.} Let us consider once +more the series of functions +\[ +x,\quad +\sqrt{x},\quad +\sqrt[3]{x},\ \dots,\quad +\sqrt[n]{x},\ \dots, +\] +which possesses the property that, if $f(x)$ and~$\phi(x)$ are any two of the +functions contained in it, then $f(x)$ and~$\phi(x)$ both tend to~$\infty$ as $x \to \infty$, while +$f(x)/\phi(x)$ tends to $0$ or to~$\infty$ according as $f(x)$~occurs to the right or the +left of~$\phi(x)$ in the series. We can now continue this series by the insertion +of new terms to the right of all those already written down. We can begin +with $\log x$, which tends to infinity more slowly than any of the old terms. +Then $\sqrtp{\log x}$ tends to~$\infty$ more slowly than~$\log x$, $\sqrtp[3]{\log x}$ than~$\sqrtp{\log x}$, and +so on. Thus we obtain a series +\[ +x,\quad \sqrt{x},\quad \sqrt[3]{x},\ \dots,\quad \sqrt[n]{x},\ \dots\quad +\log x,\quad \sqrtp{\log x},\quad +\sqrtp[3]{\log x},\ \dots\quad +\sqrtp[n]{\log x},\ \dots +\] +formed of two simply infinite series arranged one after the other. But this +is not all. Consider the function $\log\log x$, the logarithm of~$\log x$. Since +$(\log x)/x^{\alpha} \to 0$, for all positive values of~$\alpha$, it follows on putting $x = \log y$ that +\[ +(\log\log y)/(\log y)^{\alpha} = (\log x)/x^{\alpha} \to 0. +\] +Thus $\log\log y$ tends to~$\infty$ with~$y$, but more slowly than any power of~$\log y$. +Hence we may continue our series in the form +\begin{gather*} +%[** TN: Set on one line in the original] +x,\quad \sqrt{x},\quad \sqrt[3]{x},\ \dots\qquad +\log x,\quad \sqrtp{\log x},\quad \sqrtp[3]{\log x},\ \dots\\ +\log\log x,\quad \sqrtp{\log\log x},\ \dots\quad \sqrtp[n]{\log\log x},\ \dots; +\end{gather*} +and it will by now be obvious that by introducing the functions $\log\log\log x$, +$\log\log\log\log x$,~\dots\ we can prolong the series to any extent we like. By +putting $x = 1/y$ we obtain a similar scale of infinity for functions of~$y$ which +tend to~$\infty$ as $y$~tends to~$0$ by positive values.\footnote + {For fuller information as to `scales of infinity' see the author's tract `Orders + of Infinity', \textit{Camb.\ Math.\ Tracts}, No.~12.\PageLabel{362}} +\end{Remark} + +\begin{Examples}{LXXXIV.} +\Item{1.} Between any two terms $f(x)$,~$F(x)$ of the series +we can insert a new term~$\phi(x)$ such that $\phi(x)$~tends to~$\infty$ more slowly than~$f(x)$ +and more rapidly than~$F(x)$. [Thus between $\sqrt{x}$ and~$\sqrt[3]{x}$ we could insert~$x^{5/12}$: +between $\sqrtp{\log x}$ and~$\sqrtp[3]{\log x}$ we could insert $(\log x)^{5/12}$. And, generally, +$\phi(x) = \sqrtb{f(x) F(x)}$ satisfies the conditions stated.] + +\Item{2.} Find a function which tends to~$\infty$ more slowly than~$\sqrt{x}$, but more +rapidly than~$x^{\alpha}$, where $\alpha$~is any rational number less than~$1/2$. [$\sqrt{x}/(\log x)$~is +such a function; or $\sqrt{x}/(\log x)^{\beta}$, where $\beta$~is any positive rational number.] + +\Item{3.} Find a function which tends to~$\infty$ more slowly than~$\sqrt{x}$, but more +rapidly than~$\sqrt{x}/(\log x)^{\alpha}$, where $\alpha$~is any rational number. [The function +$\sqrt{x}/(\log\log x)$ is such a function. It will be gathered from these examples that +\emph{incompleteness} is an inherent characteristic of the logarithmic scale of infinity.] + +\Item{4.} How does the function +\[ +f(x) = \{x^{\alpha} (\log x)^{\alpha'} (\log\log x)^{\alpha''}\}/ + \{x^{\beta} (\log x)^{\beta'} (\log\log x)^{\beta''}\} +\] +behave as $x$~tends to~$\infty$? [If $\alpha \neq \beta$ then the behaviour of +\[ +f(x) = x^{\alpha-\beta} (\log x)^{\alpha'-\beta'} (\log\log x)^{\alpha''-\beta''} +\] +\PageSep{363} +is dominated by that of~$x^{\alpha-\beta}$. If $\alpha = \beta$ then the power of~$x$ disappears and +the behaviour of~$f(x)$ is dominated by that of $(\log x)^{\alpha'-\beta'}$, unless $\alpha' = \beta'$, when +it is dominated by that of $(\log\log x)^{\alpha''-\beta''}$. Thus $f(x) \to \infty$ if $\alpha > \beta$, or +$\alpha = \beta$, $\alpha' > \beta'$, or $\alpha = \beta$, $\alpha' = \beta'$, $\alpha'' > \beta''$, and $f(x) \to 0$ if $\alpha < \beta$, or $\alpha = \beta$, $\alpha' < \beta'$, or +$\alpha = \beta$, $\alpha' = \beta'$, $\alpha'' < \beta''$.] + +\Item{5.} Arrange the functions $x/\sqrtp{\log x}$, $x\sqrtp{\log x}/\log\log x$, $x\log\log x/\sqrtp{\log x}$, +$(x\log\log\log x)/\sqrtp{\log\log x}$ according to the rapidity with which they tend +to infinity as $x \to \infty$. + +\Item{6.} Arrange +\[ +\log\log x/(x\log x),\quad +(\log x)/x,\quad +x\log\log x/\sqrtp{x^{2} + 1},\quad +\{\sqrtp{x + 1}\}/x(\log x)^{2} +\] +according to the rapidity with which they tend to zero as $x \to \infty$. + +\Item{7.} Arrange +\[ +x\log\log(1/x),\quad +\sqrt{x}/\{\log(1/x)\},\quad +\sqrtb{x\sin x\log(1/x)},\quad +(1 - \cos x)\log(1/x) +\] +according to the rapidity with which they tend to zero as $x \to +0$. + +\Item{8.} Show that +\[ +D_{x}\log\log x = 1/(x\log x),\quad +D_{x}\log\log\log x = 1/(x\log x\log\log x), +\] +and so on. + +\Item{9.} Show that +\[ +D_{x}(\log x)^{\alpha} = \alpha/\{x(\log x)^{1-\alpha}\},\quad +D_{x}(\log\log x)^{\alpha} = \alpha/\{x\log x(\log\log x)^{1-\alpha}\}, +\] +and so on. +\end{Examples} + +\Paragraph{203. The number $e$\Add{.}} We shall now introduce a number, +usually denoted by~$e$, which is of immense importance in higher +mathematics. It is, like~$\pi$, one of the fundamental constants +of analysis. + +We define~$e$ as \emph{the number whose logarithm is~$1$}. In other +words $e$~is defined by the equation +\[ +1 = \int_{1}^{e} \frac{dt}{t}. +\] +Since $\log x$~is an increasing function of~$x$, in the stricter sense of +\SecNo[§]{95}, it can only pass once through the value~$1$. Hence our +definition does in fact define one definite number. + +Now $\log xy = \log x + \log y$ and so +\[ +\log x^{2} = 2\log x,\quad +\log x^{3} = 3\log x,\ \dots,\quad +\log x^{n} = n\log x, +\] +where $n$~is any positive integer. Hence +\[ +\log e^{n} = n\log e = n. +\] +\PageSep{364} +Again, if $p$~and~$q$ are any positive integers, and $e^{p/q}$~denotes the +positive $q$th~root of~$e^{p}$, we have +\[ +p = \log e^{p} = \log(e^{p/q})^{q} = q\log e^{p/q}, +\] +so that $\log e^{p/q} = p/q$. Thus, if $y$~has any positive rational value, +and $e^{y}$~denotes the positive $y$th~power of~$e$, we have +\[ +\log e^{y} = y, +\Tag{(1)} +\] +and $\log e^{-y} = -\log e^{y} = -y$. Hence the equation~\Eq{(1)} is true for +all rational values of~$y$, positive or negative. In other words the +equations +\[ +y = \log x,\quad +x = e^{y} +\Tag{(2)} +\] +are consequences of one another so long as $y$~is rational and $e^{y}$~has +its positive value. At present we have not given any definition +of a power such as~$e^{y}$ in which the index is irrational, and the +function~$e^{y}$ is defined for rational values of~$y$ only. + +\Par{Example.} Prove that $2 < e < 3$. [In the first place it is evident that +\[ +\int_{1}^{2} \frac{dt}{t} < 1, +\] +and so $2 < e$. Also +\[ +\int_{1}^{3} \frac{dt}{t} = \int_{1}^{2} \frac{dt}{t} + \int_{2}^{3} \frac{dt}{t} + = \int_{0}^{1} \frac{du}{2 - u} + \int_{0}^{1} \frac{du}{2 + u} + = 4\int_{0}^{1} \frac{du}{4 - u^{2}} > 1, +\] +so that $e < 3$.] + +\Paragraph{204. The exponential function.} We now define the \emph{exponential +function}~$e^{y}$ for all real values of~$y$ as the inverse of +the logarithmic function. In other words we write +\[ +x = e^{y} +\] +if $y = \log x$. + +We saw that, as $x$~varies from $0$ towards~$\infty$, $y$~increases +steadily, in the stricter sense, from $-\infty$ towards~$\infty$. Thus to +one value of~$x$ corresponds one value of~$y$, and conversely. Also $y$~is +a continuous function of~$x$, and it follows from \SecNo[§]{109} that $x$~is +likewise a continuous function of~$y$. + +\begin{Remark} +It is easy to give a direct proof of the continuity of the exponential function. +For if $x = e^{y}$ and $x + \xi = e^{y+\eta}$ then +\[ +\eta = \int_{x}^{x+\xi} \frac{dt}{t}. +\] +Thus $|\eta|$~is greater than $\xi/(x + \xi)$ if $\xi > 0$, and than $|\xi|/x$ if $\xi < 0$; and if $\eta$~is +very small $\xi$~must also be very small. +\end{Remark} +\PageSep{365} + +Thus $e^{y}$~is a positive and continuous function of~$y$ which +increases steadily from $0$ towards~$\infty$ as $y$~increases from $-\infty$ +towards~$\infty$. Moreover $e^{y}$~is the positive $y$th~power of the number~$e$, +in accordance with the elementary definitions, whenever $y$~is +a rational number. In particular $e^{y} = 1$ when $y = 0$. The general +form of the graph of~$e^{y}$ is as shown in \Fig{53}. +%[Illustration: Fig. 53.] +\Figure[3in]{53}{p365} + +\Paragraph{205. The principal properties of the exponential +function.} \Item{(1)}~If $x = e^{y}$, so that $y = \log x$, then $dy/dx = 1/x$ +and +\[ +\frac{dx}{dy} = x = e^{y}. +\] +Thus \emph{the derivative of the exponential function is equal to the +function itself}. More generally, if $x = e^{ay}$ then $dx/dy = ae^{ay}$. + +\Item{(2)} \emph{The exponential function satisfies the functional equation} +\[ +f(y + z) = f(y)f(z). +\] + +This follows, when $y$ and~$z$ are rational, from the ordinary rules +of indices. If $y$ or~$z$, or both, are irrational then we can choose two +sequences $y_{1}$, $y_{2}$,~\dots, $y_{n}$,~\dots\ and $z_{1}$, $z_{2}$,~\dots, $z_{n}$,~\dots\ of rational numbers +such that $\lim y_{n} = y$, $\lim z_{n} = z$. Then, since the exponential +function is continuous, we have +\[ +e^{y} × e^{z} = \lim e^{y_{n}} × \lim e^{z_{n}} = \lim e^{y_{n}+z_{n}} = e^{y+z}. +\] +In particular $e^{y} × e^{-y} = e^{0} = 1$, or $e^{-y} = 1/e^{y}$. + +We may also deduce the functional equation satisfied by~$e^{y}$ +from that satisfied by~$\log x$. For if $y_{1} = \log x_{1}$, $y_{2} = \log x_{2}$, so that +$x_{1} = e^{y_{1}}$, $x_{2} = e^{y_{2}}$, then $y_{1} + y_{2} = \log x_{1} + \log x_{2} = \log x_{1}x_{2}$ and +\[ +e^{y_{1}+y_{2}} = e^{\log x_{1}x_{2}} = x_{1}x_{2} = e^{y_{1}} × e^{y_{2}}. +\] +\PageSep{366} + +\begin{Examples}{LXXXV.} +\Item{1.} If $dx/dy = ax$ then $x = Ke^{ay}$, where $K$~is a +constant. + +\Item{2.} There is no solution of the equation $f(y + z) = f(y)f(z)$ fundamentally +distinct from the exponential function. [We assume that $f(y)$~has a differential +coefficient. Differentiating the equation with respect to $y$~and~$z$ in turn, we +obtain +\[ +f'(y + z) = f'(y)f(z),\quad +f'(y + z) = f(y)f'(z) +\] +and so $f'(y)/f(y) = f'(z)/f(z)$, and therefore each is constant. Thus if $x = f(y)$ +then $dx/dy = ax$, where $a$~is a constant, so that $x = Ke^{ay}$ (Ex.~1).] + +\Item{3.} Prove that $(e^{ay} - 1)/y \to a$ as $y \to 0$. [Applying the Mean Value +Theorem, we obtain $e^{ay} - 1 = aye^{a\eta}$, where $0 < |\eta| < |y|$.] +\end{Examples} + +\Paragraph{206.} \begin{Result}\Item{(3)} The function~$e^{y}$ tends to infinity with~$y$ more rapidly +than any power of~$y$, or +\[ +\lim y^{\alpha}/e^{y} = \lim e^{-y}y^{\alpha} = 0 +\] +as $y \to \infty$, for all values of~$\alpha$ however great. +\end{Result} + +We saw that $(\log x)/x^{\beta} \to 0$ as $x \to \infty$, for any positive value +of~$\beta$ however small. Writing $\alpha$ for~$1/\beta$, we see that $(\log x)^{\alpha}/x \to 0$ +for any value of~$\alpha$ however large. The result follows on putting +$x = e^{y}$. It is clear also that $e^{\gamma y}$~tends to~$\infty$ if $\gamma > 0$, and to~$0$ +if $\gamma < 0$, and in each case more rapidly than any power of~$y$. + +\begin{Remark} +From this result it follows that we can construct a `scale of infinity' +similar to that constructed in \SecNo[§]{202}, but extending in the opposite direction; +\ie\ a scale of functions which tend to~$\infty$ more and more rapidly as $x \to \infty$.\footnote + {The exponential function was introduced by inverting the equation $y = \log x$ + into $x = e^{y}$; and we have accordingly, up to the present, used $y$~as the independent + and $x$~as the dependent variable in discussing its properties. We shall now revert + to the more natural plan of taking~$x$ as the independent variable, except when it is + necessary to consider a pair of equations of the type $y = \log x$, $x = e^{y}$ simultaneously, + or when there is some other special reason to the contrary.} +The scale is +\[ +x,\quad x^{2},\quad x^{3},\ \dots\quad +e^{x},\quad e^{2x},\ \dots\quad +e^{x^{2}},\ \dots,\quad e^{x^{3}},\ \dots,\quad e^{e^{x}},\ \dots, +\] +where of course $e^{x^{2}}$,~\dots, $e^{e^{x}}$,~\dots\ denote $e^{(x^{2})}$,~\dots, $e^{(e^{x})}$,~\dots. + +The reader should try to apply the remarks about the logarithmic scale, +made in \SecNo[§]{202} and \Exs{lxxxiv}, to this `exponential scale' also. The two scales +may of course (if the order of one is reversed) be combined into one scale +\[ +\dots\ \log\log x,\ \dots\quad \log x,\ \dots\quad +x,\ \dots\quad e^{x},\ \dots\quad e^{e^{x}},\ \dots. +\] +\end{Remark} + +\Paragraph{207. The general power~$a^{x}$.} The function~$a^{x}$ has been +defined only for rational values of~$x$, except in the particular case +\PageSep{367} +when $a = e$. We shall now consider the case in which $a$~is any +positive number. Suppose that $x$~is a positive rational number~$p/q$. +Then the positive value~$y$ of the power~$a^{p/q}$ is given by +$y^{q} = a^{p}$; from which it follows that +\[ +q\log y = p\log a,\quad +\log y = (p/q)\log a = x\log a, +\] +and so +\[ +y = e^{x\log a}. +\] +{\Loosen We take this as our \emph{definition} of~$a^{x}$ when $x$~is irrational. Thus +$10^{\sqrt{2}} = e^{\sqrt{2}\log 10}$. It is to be observed that $a^{x}$, when $x$~is irrational, +is defined only for positive values of~$a$, and is itself essentially +positive; and that $\log a^{x} = x\log a$. The most important properties +of the function~$a^{x}$ are as follows.} + +\Item{(1)} Whatever value $a$ may have, $a^{x} × a^{y} = a^{x+y}$ and $(a^{x})^{y} = a^{xy}$. +In other words the laws of indices hold for irrational no less than +for rational indices. For, in the first place, +\[ +a^{x} × a^{y} = e^{x\log a} × e^{y\log a} = e^{(x+y)\log a} = a^{x+y}; +\] +and in the second +\[ +(a^{x})^{y} = e^{y\log a^{x}} = e^{xy\log a} = a^{xy}. +\] + +\Item{(2)} If $a > 1$ then $a^{x} = e^{x\log a} = e^{\alpha x}$, where $\alpha$~is positive. The +graph of~$a^{x}$ is in this case similar to that of~$e^{x}$, and $a^{x} \to \infty$ +as $x \to \infty$, more rapidly than any power of~$x$. + +If $a < 1$ then $a^{x} = e^{x\log a} = e^{-\beta x}$, where $\beta$~is positive. The graph +of~$a^{x}$ is then similar in shape to that of~$e^{x}$, but reversed as regards +right and left, and $a^{x} \to 0$ as $x \to \infty$, more rapidly than any +power of~$1/x$. + +\Item{(3)} $a^{x}$~is a continuous function of~$x$, and +\[ +D_{x} a^{x} = D_{x} e^{x\log a} = e^{x\log a} \log a = a^{x} \log a. +\] + +\Item{(4)} $a^{x}$~is also a continuous function of~$a$, and +\[ +D_{a} a^{x} = D_{a} e^{x\log a} = e^{x\log a} (x/a) = xa^{x-1}. +\] + +\Item{(5)} $(a^{x} - 1)/x \to \log a$ as $x \to 0$. This of course is a mere +corollary from the fact that $D_{x}a^{x} = a^{x}\log a$, but the particular +form of the result is often useful; it is of course equivalent to the +result (\Ex{lxxxv}.~3) that $(e^{\alpha x} - 1)/x \to \alpha$ as $x \to 0$. + +\begin{Remark} +In the course of the preceding chapters a great many results involving +the function~$a^{x}$ have been stated with the limitation that $x$~is rational. The +definition and theorems given in this section enable us to remove this +restriction. +\end{Remark} +\PageSep{368} + +\Paragraph{208. The representation of~$e^{x}$ as a limit.} In \Ref{Ch.}{IV}, +\SecNo[§]{73}, we proved that $\{1 + (1/n)\}^{n}$ tends, as $n \to \infty$, to a limit +which we denoted provisionally by~$e$. We shall now identify this +limit with the number~$e$ of the preceding sections. We can +however establish a more general result, viz.\ that expressed by +the equations +\[ +\lim_{n\to\infty} \left(1 + \frac{x}{n}\right)^{n} + = \lim_{n\to\infty} \left(1 - \frac{x}{n}\right)^{-n} + = e^{x}. +\Tag{(1)} +\] +As the result is of very great importance, we shall indicate alternative +lines of proof. + +\Item{(1)} Since +\[ +\frac{d}{dt} \log(1 + xt) = \frac{x}{1 + xt}, +\] +it follows that +\[ +\lim_{h\to 0} \frac{\log(1 + xh)}{h} = x. +\] +If we put $h = 1/\xi$, we see that +\[ +\lim \xi \log\left(1 + \frac{x}{\xi}\right) = x +\] +as $\xi \to \infty$ or $\xi \to -\infty$. Since the exponential function is continuous +it follows that +\[ +\left(1 + \frac{x}{\xi}\right)^{\xi} = e^{\xi\log\{1+(x/\xi)\}} \to e^{x} +\] +as $\xi \to \infty$ or $\xi \to -\infty$: \ie\ that +\[ +\lim_{\xi\to\infty} \left(1 + \frac{x}{\xi}\right)^{\xi} + = \lim_{\xi\to -\infty} \left(1 + \frac{x}{\xi}\right)^{\xi} + = e^{x}. +\Tag{(2)} +\] + +If we suppose that $\xi \to \infty$ or $\xi \to -\infty$ through integral values +only, we obtain the result expressed by the equations~\Eq{(1)}. + +\begin{Remark} +\Item{(2)} If $n$~is any positive integer, however large, and $x > 1$, we have +\[ +\int_{1}^{x} \frac{dt}{t^{1+(1/n)}} + < \int_{1}^{x} \frac{dt}{t} + < \int_{1}^{x} \frac{dt}{t^{1-(1/n)}}, +\] +or +\[ +n(1 - x^{-1/n}) < \log x < n(x^{1/n} - 1). +\Tag{(3)} +\] +Writing $y$ for~$\log x$, so that $y$~is positive and $x = e^{y}$, we obtain, after some +simple transformations, +\[ +\left(1 + \frac{y}{n}\right)^{n} < x < \left(1 - \frac{y}{n}\right)^{-n}. +\Tag{(4)} +\] +Now let +\[ +1 + \frac{y}{n} = \eta_{1},\quad +1 - \frac{y}{n} = \frac{1}{\eta_{2}}. +\] +\PageSep{369} +Then $0 < \eta_{1} < \eta_{2}$, at any rate for sufficiently large values of~$n$; and, by~\Eq{(9)} +of \SecNo[§]{74}, +\[ +\eta_{2}^{n} - \eta_{1}^{n} + < n\eta_{2}^{n-1} (\eta_{2} - \eta_{1}) + = y^{2}\eta_{2}^{n}/n, +\] +which evidently tends to $0$ as $n \to \infty$. The result now follows from the +inequalities~\Eq{(4)}. The more general result~\Eq{(2)} may be proved in the same way, +if we replace~$1/n$ by a continuous variable~$h$. + +\Paragraph{209. The representation of $\log x$ as a limit.} We can also prove +(cf.\ \SecNo[§]{75}) that +\[ +\lim n(1 - x^{-1/n}) = \lim n(x^{1/n} - 1) = \log x. +\] + +For +\[ +n(x^{1/n} - 1) - n(1 - x^{-1/n}) = n(x^{1/n} - 1)(1 - x^{-1/n}), +\] +which tends to zero as $n \to \infty$, since $n(x^{1/n} - 1)$ tends to a limit (\SecNo[§]{75}) and +$x^{-1/n}$ to~$1$ (\Ex{xxvii}.~10). The result now follows from the inequalities~\Eq{(3)} of +\SecNo[§]{208}. +\end{Remark} + +\begin{Examples}{LXXXVI.} +\Item{1.} Prove, by taking $y = 1$ and $n = 6$ in the inequalities~\Eq{(4)} +of \SecNo[§]{208}, that $2.5 < e < 2.9$. + +\Item{2.} Prove that if $t > 1$ then $(t^{1/n} - t^{-1/n})/(t - t^{-1}) < 1/n$, and so that if +$x > 1$ then +\[ +\int_{1}^{x} \frac{dt}{t^{1-(1/n)}} - \int_{1}^{x} \frac{dt}{t^{1+(1/n)}} + < \frac{1}{n} \int_{1}^{x} \left(t - \frac{1}{t}\right) \frac{dt}{t} + = \frac{1}{n} \left(x + \frac{1}{x} - 2\right). +\] +Hence deduce the results of \SecNo[§]{209}. + +\Item{3.} If $\xi_{n}$~is a function of~$n$ such that $n\xi_{n} \to l$ as $n \to \infty$, then $(1 + \xi_{n})^{n} \to e^{l}$. +[Writing $n\log(1 + \xi_{n})$ in the form +\[ +l \left(\frac{n\xi_{n}}{l}\right) \frac{\log(1 + \xi_{n})}{\xi_{n}}, +\] +and using \Ex{lxxxii}.~4, we see that $n\log(1 + \xi_{n})\to l$.] + +\Item{4.} If $n\xi_{n} \to \infty$, then $(1 + \xi_{n})^{n} \to \infty$; and if $1 + \xi_{n} > 0$ and $n\xi_{n} \to -\infty$, then +\[ +(1 + \xi_{n})^{n} \to 0. +\] + +\Item{5.} Deduce from~\Eq{(1)} of \SecNo[§]{208} the theorem that $e^{y}$~tends to infinity more +rapidly than any power of~$y$. +\end{Examples} + +\Paragraph{210. Common logarithms.} The reader is probably familiar +with the idea of a logarithm and its use in numerical calculation. +He will remember that in elementary algebra $\log_{a} x$, the logarithm +of~$x$ to the base~$a$, is defined by the equations +\[ +x = a^{y},\quad +y = \log_{a} x. +\] +This definition is of course applicable only when $y$~is rational, +though this point is often passed over in silence. +\PageSep{370} + +Our logarithms are therefore logarithms to the base~$e$. For +numerical work logarithms to the base~$10$ are used. If +\[ +y = \log x = \log_{e} x,\quad +z = \log_{10} x, +\] +then $x = e^{y}$ and also $x = 10^{z} = e^{z\log 10}$, so that +\[ +\log_{10} x = (\log_{e} x)/(\log_{e} 10). +\] +Thus it is easy to pass from one system to the other when once +$\log_{e} 10$ has been calculated. + +It is no part of our purpose in this book to go into details +concerning the practical uses of logarithms. If the reader is +not familiar with them he should consult some text-book on +Elementary Algebra or Trigonometry.\footnote + {See for example Chrystal's \textit{Algebra}, vol.~i, ch.~\textsc{xxi}. The value of~$\log_{e} 10$ is + $2.302\dots$ and that of its reciprocal~$.434\dots$.} + +\begin{Examples}{LXXXVII.} +\Item{1.} Show that +\[ +D_{x} e^{ax}\cos bx = re^{ax} \cos(bx + \theta),\quad +D_{x} e^{ax}\sin bx = re^{ax} \sin(bx + \theta) +\] +where $r = \sqrtp{a^{2} + b^{2}}$, $\cos\theta = a/r$, $\sin\theta = b/r$. Hence determine the $n$th~derivatives +{\Loosen of the functions $e^{ax}\cos bx$, $e^{ax}\sin bx$, and show in particular that +$D_{x}^{n} e^{ax} = a^{n} e^{ax}$.} + +\Item{2.} Trace the curve $y = e^{-ax}\sin bx$, where $a$~and~$b$ are positive. Show +that $y$~has an infinity of maxima whose values form a geometrical progression +and which lie on the curve +\[ +y = \frac{b}{\sqrtp{a^{2} + b^{2}}}\, e^{-ax}. +\] +\MathTrip{1912.} + +\Item{3.} \Topic{Integrals containing the exponential function.} Prove that +\begin{align*} +%[** TN: Set on one line in the original] +\int e^{ax}\cos bx\, dx &= \frac{a\cos bx + b\sin bx}{a^{2} + b^{2}}\, e^{ax}, \\ +\int e^{ax}\sin bx\, dx &= \frac{a\sin bx - b\cos bx}{a^{2} + b^{2}}\, e^{ax}. +\end{align*} + +[Denoting the two integrals by $I$,~$J$, and integrating by parts, we obtain +\[ +aI = e^{ax}\cos bx + bJ,\quad +aJ = e^{ax}\sin bx - bI. +\] +Solve these equations for $I$~and~$J$.] + +\Item{4.} Prove that the successive areas bounded by the curve of Ex.~2 and the +positive half of the axis of~$x$ form a geometrical progression, and that their +sum is +\[ +\frac{b}{a^{2} + b^{2}}\, \frac{1 + e^{-a\pi/b}}{1 - e^{\DPtypo{}{-}a\pi/b}}. +\] + +\Item{5.} Prove that if $a > 0$ then +\[ +\int_{0}^{\infty} e^{-ax}\cos bx\, dx = \frac{a}{a^{2} + b^{2}},\quad +\int_{0}^{\infty} e^{-ax}\sin bx\, dx = \frac{b}{a^{2} + b^{2}}. +\] +\PageSep{371} + +\Item{6.} If $I_{n} = \ds\int e^{ax}x^{n}\, dx$ then $aI_{n} = e^{ax}x^{n} - nI_{n-1}$. [Integrate by parts. It +follows that $I_{n}$~can be calculated for all positive integral values of~$n$.] + +\Item{7.} Prove that, if $n$~is a positive integer, then +\[ +\int_{0}^{\xi} e^{-x}x^{n}\, dx + = n!\, e^{-\xi} \left( + e^{\xi} - 1 - \xi - \frac{\xi^{2}}{2!} - \dots - \frac{\xi^{n}}{n!} + \right) +\] +and +\[ +\int_{0}^{\infty} e^{-x}x^{n}\, dx = n!. +\] + +\Item{8.} {\Loosen Show how to find the integral of any rational function of~$e^{x}$. [Put +$x = \log u$, when $e^{x} = u$, $dx/du = 1/u$, and the integral is transformed into that +of a rational function of~$u$.]} + +\Item{9.} Integrate +\[ +\frac{e^{2x}}{(c^{2}e^{x} + a^{2}e^{-x})(c^{2}e^{x} + b^{2}e^{-x})}, +\] +distinguishing the cases in which $a$~is and is not equal to~$b$. + +\Item{10.} Prove that we can integrate any function of the form $P(x, e^{ax}, e^{bx}, \dots)$, +where $P$~denotes a polynomial. [This follows from the fact that $P$~can be +expressed as the sum of a number of terms of the type $Ax^{m}e^{kx}$, where $m$~is a +positive integer.] + +\Item{11.} Show how to integrate any function of the form +\[ +P(x,\ e^{ax},\ e^{bx},\ \dots,\ \cos lx,\ \cos mx,\ \dots,\ \sin lx,\ \sin mx,\ \dots). +\] + +\Item{12.} Prove that $\ds\int_{a}^{\infty} e^{-\lambda x} R(x)\, dx$, where $\lambda > 0$ and $a$~is greater than the +greatest root of the denominator of~$R(x)$, is convergent. [This follows from +the fact that $e^{\lambda x}$~tends to infinity more rapidly than any power of~$x$.] + +\Item{13.} Prove that $\ds\int_{-\infty}^{\infty} e^{-\lambda x^{2} + \mu x}\, dx$, where $\lambda > 0$, is convergent for all values of~$\mu$, +and that the same is true of $\ds\int_{-\infty}^{\infty} e^{-\lambda x^{2} + \mu x} x^{n}\, dx$, where $n$~is any positive +integer. + +\Item{14.} Draw the graphs of $e^{x^{2}}$, $e^{-x^{2}}$, $xe^{x}$, $xe^{-x}$, $xe^{x^{2}}$, $xe^{-x^{2}}$, and $x\log x$, determining +any maxima and minima of the functions and any points of inflexion +on their graphs. + +\Item{15.} Show that the equation $e^{ax} = bx$, where $a$~and~$b$ are positive, has two +real roots, one, or none, according as $b > ae$, $b = ae$, or $b < ae$. [The tangent +to the curve $y = e^{ax}$ at the point $(\xi, e^{a\xi})$ is +\[ +y - e^{a\xi} = ae^{a\xi}(x - \xi), +\] +which passes through the origin if $a\xi = 1$, so that the line $y = aex$ touches the +curve at the point $(1/a, e)$. The result now becomes obvious when we draw +the line $y = bx$. The reader should discuss the cases in which $a$~or~$b$ or both +are negative.] +\PageSep{372} + +\Item{16.} Show that the equation $e^{x} = 1 + x$ has no real root except $x = 0$, and +that $e^{x} = 1 + x + \frac{1}{2}x^{2}$ has three real roots. + +\Item{17.} Draw the graphs of the functions +\begin{gather*} +\log(x + \sqrtp{x^{2} + 1}),\quad +\log\left(\frac{1 + x}{1 - x}\right),\quad +e^{-ax}\cos^{2}bx,\\ +e^{-(1/x)^{2}},\quad +e^{-(1/x)^{2}}\sqrtp{1/x},\quad +e^{-\cot x},\quad +e^{-\cot^{2} x}. +\end{gather*} + +\Item{18.} Determine roughly the positions of the real roots of the equations +\[ +\log(x + \sqrtp{x^{2} + 1}) = \frac{x}{100},\quad +e^{x} - \frac{2 + x}{2 - x} = \frac{1}{10\MC000},\quad +e^{x}\sin x = 7,\quad +e^{x^{2}}\sin x = 10\MC000. +\] + +\Item{19.} \Topic{The hyperbolic functions.} The hyperbolic functions $\cosh x$,\footnote + {`Hyperbolic cosine': for an explanation of this phrase see Hobson's \textit{Trigonometry}, + ch.~\textsc{xvi}.} +$\sinh x$,~\dots\ are defined by the equations +\begin{gather*} +\cosh x = \tfrac{1}{2}(e^{x} + e^{-x}),\quad +\sinh x = \tfrac{1}{2}(e^{x} - e^{-x}), \displaybreak[1]\\ +% +\tanh x = (\sinh x)/(\cosh x),\quad +\coth x = (\cosh x)/(\sinh x), \displaybreak[1]\\ +% +\sech x = 1/(\cosh x),\quad +\cosech x = 1/(\sinh x). +\end{gather*} +Draw the graphs of these functions. + +\Item{20.} Establish the formulae +\begin{gather*} +\cosh(-x) = \cosh x,\quad +\sinh(-x) = -\sinh x,\quad +\tanh(-x) = -\tanh x, \displaybreak[1]\\ +% +\cosh^{2} x - \sinh^{2} x = 1,\quad +\sech^{2} x + \tanh^{2} x = 1,\quad +\coth^{2} x - \cosech^{2} x = 1, \displaybreak[1]\\ +% +\cosh 2x = \cosh^{2} x + \sinh^{2} x,\quad +\sinh 2x = 2\sinh x\cosh x, \displaybreak[1]\\ +% +\begin{alignedat}{2} +\cosh(x + y) &= \cosh x\cosh y &&+ \sinh x\sinh y,\\ +\sinh(x + y) &= \sinh x\cosh y &&+ \cosh x\sinh y. +\end{alignedat} +\end{gather*} + +\Item{21.} Verify that these formulae may be deduced from the corresponding +formulae in $\cos x$ and $\sin x$, by writing $\cosh x$ for $\cos x$ and $i\sinh x$ for~$\sin x$. + +[It follows that the same is true of all the formulae involving $\cos nx$ and +$\sin nx$ which are deduced from the corresponding elementary properties of +$\cos x$ and~$\sin x$. The reason of this analogy will appear in \Ref{Ch.}{X}\@.] + +\Item{22.} Express $\cosh x$ and $\sinh x$ in terms (\ia)~of $\cosh 2x$ (\ib)~of $\sinh 2x$. +Discuss any ambiguities of sign that may occur. \MathTrip{1908.} + +\Item{23.} Prove that +\begin{gather*} +D_{x}\cosh x = \sinh x,\quad +D_{x}\sinh x = \cosh x,\\ +% +D_{x}\tanh x = \sech^{2}x,\quad +D_{x}\coth x = -\cosech^{2}x,\\ +% +D_{x}\sech x = -\sech x\tanh x,\quad +D_{x}\cosech x = -\cosech x\coth x,\\ +% +D_{x}\log \cosh x = \tanh x,\quad +D_{x}\log|\sinh x| = \coth x,\\ +% +D_{x}\arctan e^{x} = \tfrac{1}{2}\sech x,\quad +D_{x}\log |\tanh \tfrac{1}{2} x| = \cosech x. +\end{gather*} + +[All these formulae may of course be transformed into formulae in integration.] +\PageSep{373} + +\Item{24.} Prove that $\cosh x > 1$ and $-1 < \tanh x < 1$. + +\Item{25.} Prove that if $y = \cosh x$ then $x = \log\{y ± \sqrtp{y^{2} - 1}\}$, if $y = \sinh x$ then +$x = \log\{y + \sqrtp{y^{2} + 1}\}$, and if $y = \tanh x$ then $x = \frac{1}{2}\log\{(1 + y)/(1 - y)\}$. Account +for the ambiguity of sign in the first case. + +\Item{26.} We shall denote the functions inverse to $\cosh x$, $\sinh x$, $\tanh x$ by +$\argcosh x$, $\argsinh x$, $\argtanh x$. Show that $\argcosh x$ is defined only when +$x \geq 1$, and is in general two-valued, while $\argsinh x$ is defined for all real +values of~$x$, and $\argtanh x$ when $-1 < x < 1$, and both of the two latter +functions are one-valued. Sketch the graphs of the functions. + +\Item{27.} Show that if $-\frac{1}{2}\pi < x < \frac{1}{2}\pi$ and $y$~is positive, and $\cos x\cosh y = 1$, then +\[ +y = \log(\sec x + \tan x),\quad +D_{x} y = \sec x,\quad +D_{y} x = \sech y. +\] + +\Item{28.} Prove that if $a > 0$ then $\ds\int \frac{dx}{\sqrtp{x^{2} + a^{2}}} = \argsinh(x/a)$, and $\ds\int \frac{dx}{\sqrtp{x^{2} - a^{2}}}$ is +equal to $\argcosh(x/a)$ or to $-\argcosh(-x/a)$, according as $x > 0$ or $x < 0$. + +\Item{29.} Prove that if $a > 0$ then $\ds\int \frac{dx}{x^{2} - a^{2}}$ is equal to $-(1/a)\argtanh(x/a)$ or +to $-(1/a)\argcoth(x/a)$, according as $|x|$~is less than or greater than~$a$. [The +results of Exs.\ 28~and~29 furnish us with an alternative method of writing +a good many of the formulae of \Ref{Ch.}{VI}\@.] + +\Item{30.} Prove that +\begin{alignat*}{3} +\int \frac{dx}{\sqrtb{(x - a)(x - b)}} + &= &&2\log\{\sqrtp{x - a} + \sqrtp{x - b}\} &&(a < b < x),\\ +\int \frac{dx}{\sqrtb{(a - x)(b - x)}} + &= -&&2\log\{\sqrtp{a - x} + \sqrtp{b - x}\}\quad &&(x < a < b),\\ +\int \frac{dx}{\sqrtb{(x - a)(b - x)}} + &= &&2\arctan\bigsqrtp{\frac{x - a}{b - x}} &&(a < x < b). +\end{alignat*} + +\Item{31.} Prove that +\[ +\int_{0}^{1} x \log(1 + \tfrac{1}{2}x)\, dx + = \tfrac{3}{4} - \tfrac{3}{2}\log\tfrac{3}{2} + < \tfrac{1}{2} \int_{0}^{1} x^{2}\, dx + = \tfrac{1}{6}. +\] +\MathTrip{1913.} + +\Item{32.} Solve the equation $a\cosh x + b\sinh x = c$, where $c > 0$, showing that it +has no real roots if $b^{2} + c^{2} - a^{2} < 0$, while if $b^{2} + c^{2} - a^{2} > 0$ it has two, one, or +no real roots according as $a + b$ and $a - b$ are both positive, of opposite signs, +or both negative. Discuss the case in which $b^{2} + c^{2} - a^{2} = 0$. + +\Item{33.} Solve the simultaneous equations $\cosh x\cosh y = a$, $\sinh x\sinh y = b$. + +\Item{34.} $x^{1/x} \to 1$ as $x \to \infty$. [For $x^{1/x} = e^{(\log x)/x}$, and $(\log x)/x \to 0$. Cf.\ +\Ex{xxvii}.~11.] Show also that the function~$x^{1/x}$ has a maximum when +$x = e$, and draw the graph of the function for positive values of~$x$. + +\Item{35.} $x^{x} \to 1$ as $x \to +0$. +\PageSep{374} + +\Item{36.} If $\{f(n + 1)\}/\{f(n)\} \to l$, where $l > 0$, as $n \to \infty$, then $\sqrtb[n]{f(n)} \to l$. +[For $\log f(n + 1) - \log f(n) \to \log l$, and so $(1/n)\log f(n) \to \log l$ (\Ref{Ch.}{IV}, \MiscEx{IV}~27).] + +\Item{37.} $\sqrt[n]{n!}/n \to 1/e$ as $n \to \infty$. + +[If $f(n) = n^{-n} n!$ then $\{f(n + 1)\}/\{f(n)\} = \{1 + (1/n)\}^{-n} \to 1/e$. Now use +Ex.~36.] + +\Item{38.} $\sqrt[n]{(2n)!/(n!)^{2}} \to 4$ as $n \to \infty$. + +\Item{39.} Discuss the approximate solution of the equation $e^{x} = x^{1\MC000\MC000}$. + +[It is easy to see by general graphical considerations that the equation +has two positive roots, one a little greater than~$1$ and one very large,\footnote + {The phrase `very large' is of course not used here in the technical sense + explained in \Ref{Ch.}{IV}\@. It means `a good deal larger than the roots of such equations + as usually occur in elementary mathematics'. The phrase `a little greater than' + must be interpreted similarly.} +and one +negative root a little greater than~$-1$. To determine roughly the size of the +large positive root we may proceed as follows. If $e^{x} = x^{1\MC000\MC000}$ then +\[ +x = 10^{6} \log x,\quad +\log x = 13.82 + \log\log x,\quad +\log\log x = 2.63 + \log \left(1 + \frac{\log\log x}{13.82}\right), +\] +roughly, since $13.82$ and $2.63$ are approximate values of $\log 10^{6}$ and $\log\log 10^{6}$ +respectively. It is easy to see from these equations that the ratios $\log x : 13.82$ +and $\log\log x : 2.63$ do not differ greatly from unity, and that +\[ +x = 10^{6}(13.82 + \log\log x) = 10^{6}(13.82 + 2.63) = 16\MC450\MC000 +\] +gives a tolerable approximation to the root, the error involved being roughly +measured by $10^{6}(\log\log x - 2.63)$ or $(10^{6} \log\log x)/13.82$ or $(10^{6} × 2.63)/13.82$, +which is less than~$200,000$. The approximations are of course very rough, +but suffice to give us a good idea of the scale of magnitude of the root.] + +\Item{40.} Discuss similarly the equations +\[ +%[** TN: In-line in the original] +e^{x} = 1\MC000\MC000 x^{1\MC000\MC000},\quad +e^{x^{2}} = x^{1\MC000\MC000\MC000}. +\] +\end{Examples} + +\Paragraph{211. Logarithmic tests of convergence for series and +integrals.} We showed in \Ref{Ch.}{VIII} (\SecNo[§§]{175}~\textit{et~seq.})\ that +\[ +\sum_{1}^{\infty} \frac{1}{n^{s}},\quad +\int_{a}^{\infty} \frac{dx}{x^{s}}\qquad (a > 0) +\] +are convergent if $s > 1$ and divergent if $s \leq 1$. Thus $\sum (1/n)$~is +divergent, but $\sum n^{-1-\alpha}$~is convergent for all positive values of~$\alpha$. + +We saw however in \SecNo[§]{200} that with the aid of logarithms we +can construct functions which tend to zero, as $n \to \infty$, more +rapidly than~$1/n$, yet less rapidly than~$n^{-1-\alpha}$, however small $\alpha$ may +be, provided of course that it is positive. For example $1/(n\log n)$ +is such a function, and the question as to whether the series +\[ +\sum \frac{1}{n\log n} +\] +\PageSep{375} +is convergent or divergent cannot be settled by comparison with +any series of the type $\sum n^{-s}$. + +The same is true of such series as +\[ +\sum \frac{1}{n(\log n)^{2}},\quad +\sum \frac{\log\log n}{n\sqrtp{\log n}}. +\] +It is a question of some interest to find tests which shall enable +us to decide whether series such as these are convergent or +divergent; and such tests are easily deduced from the Integral +Test of~\SecNo[§]{174}. + +For since +\[ +D_{x}(\log x)^{1-s} = \frac{1 - s}{x(\log x)^{s}},\quad +D_{x}\log\log x = \frac{1}{x\log x}, +\] +we have +\[ +\int_{a}^{\xi} \frac{dx}{x(\log x)^{s}} + = \frac{(\log\xi)^{1-s} - (\log a)^{1-s}}{1 - s},\quad +\int_{\DPtypo{}{a}}^{\xi} \frac{dx}{x\log x} = \log\log \xi - \log\log a, +\] +if $a > 1$. The first integral tends to the limit $-(\log a)^{1-s}/(1 - s)$ +as $\xi \to \infty$, if $s > 1$, and to~$\infty$ if $s < 1$. The second integral tends +to~$\infty$. Hence \begin{Result}the series and integral\PageLabel{375} +\[ +\sum_{n_{0}}^{\infty} \frac{1}{n(\log n)^{s}},\quad +\int_{a}^{\infty} \frac{dx}{x(\log x)^{s}}, +\] +where $n_{0}$ and $a$ are greater than unity, are convergent if $s > 1$, +divergent if $s \leq 1$. +\end{Result} + +It follows, of course, that $\sum \phi(n)$~is convergent if $\phi(n)$~is +positive and less than $K/\{n(\log n)^{s}\}$, where $s > 1$, for all values of $n$ +greater than some definite value, and divergent if $\phi(n)$~is positive +and greater than $K/(n\log n)$ for all values of $n$ greater than some +definite value. And there is a corresponding theorem for integrals +which we may leave to the reader. + +\begin{Examples}{LXXXVIII.} +\Item{1.} The series +\[ +\sum \frac{1}{n(\log n)^{2}},\quad +\sum \frac{(\log n)^{100}}{n^{101/100}},\quad +\sum \frac{n^{2} - 1}{n^{2} + 1}\, \frac{1}{n(\log n)^{7/6}} +\] +are convergent. [The convergence of the first series is a direct consequence +of the theorem of the preceding section. That of the second follows from +the fact that $(\log n)^{100}$~is less than~$n^{\beta}$ for sufficiently large values of~$n$, however +small $\beta$ may be, provided that it is positive. And so, taking $\beta = 1/200$, +$(\log n)^{100} n^{-101/100}$ is less than~$n^{-201/200}$ for sufficiently large values of~$n$. The +convergence of the third series follows from the comparison test at the end of +the last section.] +\PageSep{376} + +\Item{2.} The series +\[ +\sum \frac{1}{n(\log n)^{6/7}},\quad +\sum \frac{1}{n^{100/101}(\log n)^{100}},\quad +\sum \frac{n\log n}{(n\log n)^{2} + 1} +\] +are divergent. + +\Item{3.} The series +\[ +\sum \frac{(\log n)^{p}}{n^{1+s}},\quad +\sum \frac{(\log n)^{p} (\log\log n)^{q}}{n^{1+s}},\quad +\sum \frac{(\log\log n)^{p}}{n(\log n)^{1+s}}, +\] +where $s > 0$, are convergent for all values of $p$~and~$q$; similarly the series +\[ +\sum \frac{1}{n^{1-s}(\log n)^{p}},\quad +\sum \frac{1}{n^{1-s}(\log n)^{p}(\log\log n)^{q}},\quad +\sum \frac{1}{n(\log n)^{1-s}(\log\log n)^{p}} +\] +are divergent. + +\Item{4.} The question of the convergence or divergence of such series as +\[ +\sum \frac{1}{n\log n\log\log n},\quad +\sum \frac{\log\log\log n}{n\log n\sqrtp{\log\log n}} +\] +cannot be settled by the theorem of \PageRef{p.}{375}, since in each case the function +under the sign of summation tends to zero more rapidly than $1/(n\log n)$ yet +less rapidly than $n^{-1}(\log n)^{-1-\alpha}$, where $\alpha$~is any positive number however +small. For such series we need a still more delicate test. The reader should +be able, starting from the equations +\begin{align*} +D_{x}(\log_{k}x)^{1-s} + &= \frac{1 - s}{x \log x \log_{2}x \dots \log_{k-1} x (\log_{k}x)^{s}},\\ +D_{x}\log_{k+1}x + &= \frac{1}{x \log x \log_{2}x \dots \log_{k-1}x \log_{k}x}, +\end{align*} +where $\log_{2}x = \log\log x$, $\log_{3} x = \log\log\log x$,~\dots, to prove the following +theorem: \emph{the series and integral +\[ +\sum_{n_{0}}^{\infty} \frac{1}{n \log n \log_{2}n \dots \log_{k-1}n (\log_{k}n)^{s}},\quad +\int_{a}^{\infty} \frac{dx}{x \log x \log_{2}x \dots \log_{k-1}x (\log_{k}x)^{s}} +\] +are convergent if $s > 1$ and divergent if $s \leq 1$}, {\Loosen$n_{0}$~and~$a$ being any numbers +sufficiently great to ensure that $\log_{k}n$ and $\log_{k}x$ are positive when $n \geq n_{0}$ +or $x \geq a$. These values of $n_{0}$ and~$a$ increase very rapidly as $k$~increases: +thus $\log x > 0$ requires $x > 1$, $\log_{2}x > 0$ requires $x > e$, $\DPtypo{\log\log x}{\log_{3}x} > 0$ requires +$x > e^{e}$, and so on; and it is easy to see that $e^{e} > 10$, $e^{e^{e}} > e^{10} > 20,000$, +$e^{e^{e^{e}}} > e^{20,000} > 10^{8000}$.} + +The reader should observe the extreme rapidity with which the higher +exponential functions, such as $e^{e^{x}}$ and~$e^{e^{e^{x}}}$, increase with~$x$. The same +remark of course applies to such functions as $a^{a^{x}}$ and~$a^{a^{a^{x}}}$, where $a$~has +any value greater than unity\Add{.} It has been computed that $9^{9^{9}}$~has $369,693,100$ +figures, while $10^{10^{10}}$ has of course $10,000,000,000$. Conversely, the rate of +increase of the higher logarithmic functions is extremely slow. Thus to make +$\log\log\log\log x > 1$ we have to suppose $x$~a number with over $8000$~figures.\footnote + {See the footnote to \PageRef{p.}{362}.} +\PageSep{377} + +\Item{5.} Prove that the integral $\ds\int_{0}^{a} \frac{1}{x} \left\{\log \left(\frac{1}{x}\right)\right\}^{s} dx$, where $0 < a < 1$, is convergent +if $s < -1$, divergent if $s \geq -1$. [Consider the behaviour of +\[ +\int_{\epsilon}^{a} \frac{1}{x} \left\{\log \left(\frac{1}{x}\right)\right\}^{s} dx +\] +as $\epsilon \to +0$. This result also may be refined upon by the introduction of +higher logarithmic factors.] + +\Item{6.} Prove that $\ds\int_{0}^{1} \frac{1}{x} \left\{\log \left(\frac{1}{x}\right)\right\}^{s} dx$ has no meaning for any value of~$s$. +[The last example shows that $s < -1$ is a necessary condition for convergence +at the lower limit: but $\{\log(1/x)\}^{s}$ tends to~$\infty$ like $(1 - x)^{s}$, as $x \to 1 - 0$, if $s$~is +negative, and so the integral diverges at the upper limit when $s < -1$.] + +\Item{7.} {\Loosen The necessary and sufficient conditions for the convergence of +$\ds\int_{0}^{1} x^{a-1} \left\{\log \left(\frac{1}{x}\right)\right\}^{s} dx$ are $a > 0$, $s > -1$.} +\end{Examples} + +\begin{Examples}{LXXXIX.} +\Item{1.} \Topic{Euler's limit.} Show that +\[ +\phi(n) = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n - 1} - \log n +\] +tends to a limit~$\gamma$ as $n \to \infty$, and that $0 < \gamma \leq 1$. [This follows at once from +\SecNo[§]{174}. The value of~$\gamma$ is in fact~$.577\dots$, and $\gamma$~is usually called \Emph{Euler's +constant}.] + +\Item{2.} If $a$ and~$b$ are positive then +\[ +\frac{1}{a} + \frac{1}{a + b} + \frac{1}{a + 2b} + \dots + + \frac{1}{a + (n - 1) b} - \frac{1}{b}\log \DPtypo{(a + nb}{(a + nb)} +\] +tends to a limit as $n \to \infty$. + +\Item{3.} If $0 < s < 1$ then +\[ +\phi(n) = 1 + 2^{-s} + 3^{-s} + \dots + (n - 1)^{-s} - \frac{n^{1-s}}{1 - s} +\] +tends to a limit as $n \to \infty$. + +\Item{4.} Show that the series +\[ +\frac{1}{1} + + \frac{1}{2(1 + \frac{1}{2})} + + \frac{1}{3(1 + \frac{1}{2} + \frac{1}{3})} + \dots +\] +is divergent. [Compare the general term of the series with $1/(n\log n)$.] +Show also that the series derived from $\sum n^{-s}$, in the same way that the above +series is derived from~$\sum (1/n)$, is convergent if $s > 1$ and otherwise divergent. + +\Item{5.} Prove generally that if $\sum u_{n}$~is a series of positive terms, and +\[ +s_{n} = u_{1} + u_{2} + \dots + u_{n}, +\] +then $\sum (u_{n}/s_{n-1})$~is convergent or divergent according as $\sum u_{n}$~is convergent or +\PageSep{378} +divergent. [If $\sum u_{n}$~is convergent then $s_{n-1}$~tends to a positive limit~$l$, and so +$\sum (u_{n}/s_{n-1})$~is convergent. If $\sum u_{n}$~is divergent then $s_{n-1} \to \infty$, and +\[ +u_{n}/s_{n-1} > \log\{1 + (u_{n}/s_{n-1})\} = \log (s_{n}/s_{n-1}) +\] +(\Ex{lxxxii}.~1); and it is evident that +\[ +\log(s_{2}/s_{1}) + \log(s_{3}/s_{2}) + \dots + \log(s_{n}/s_{n-1}) + = \log(s_{n}/s_{1}) +\] +tends to~$\infty$ as $n \to \infty$.] + +\Item{6.} Prove that the same result holds for the series $\sum (u_{n}/s_{n})$. [The proof +is the same in the case of convergence. If $\sum u_{n}$~is divergent, and $u_{n} < s_{n-1}$ +from a certain value of~$n$ onwards, then $s_{n} < 2s_{n-1}$, and the divergence of +$\sum (u_{n}/s_{n})$ follows from that of $\sum (u_{n}/s_{n-1})$. If on the other hand $u_{n} \geq s_{n-1}$ for +an infinity of values of~$n$, as might happen with a rapidly divergent series, +then $u_{n}/s_{n} \geq \frac{1}{2}$ for all these values of~$n$.] + +\Item{7.} Sum the series $1 - \frac{1}{2} + \frac{1}{3} - \dots$. [We have +\[ +1 + \frac{1}{2} + \dots + \frac{1}{2n} = \log(2n + 1) + \gamma + \epsilon_{n}, +\quad +2\left(\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2n}\right) + = \log(n + 1) + \gamma + \epsilon_{n}', +\] +by Ex.~1, $\gamma$~denoting Euler's constant, and $\epsilon_{n}$,~$\epsilon_{n}'$ being numbers which tend +to zero as $n \to \infty$. Subtracting and making $n \to \infty$ we see that the sum of the +given series is~$\log 2$. See also~\SecNo[§]{213}.] + +\Item{8.} Prove that the series +\[ +\sum_{0}^{\infty} (-1)^{n}\left(1 + \frac{1}{2} + \dots + \frac{1}{n + 1} - \log n - C\right) +\] +oscillates finitely except when $C = \gamma$, when it converges. +\end{Examples} + +\Paragraph{212. Series connected with the exponential and logarithmic +functions. Expansion of~$e^{x}$ by Taylor's Theorem.} +Since all the derivatives of the exponential function are equal +to the function itself, we have +\[ +e^{x} = 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n-1}}{(n - 1)!} + + \frac{x^{n}}{n!} e^{\theta x} +\] +where $0 < \theta < 1$. But $x^{n}/n! \to 0$ as $n \to \infty$, whatever be the value of~$x$ +(\Ex{xxvii}.~12); and $e^{\theta x} < e^{x}$. Hence, making $n$~tend to~$\infty$, we have +\[ +e^{x} = 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!} + \dots. +\Tag{(1)} +\] + +The series on the right-hand side of this equation is known as +the \Emph{exponential series}. In particular we have +\[ +e = 1 + 1 + \frac{1}{2!} + \dots + \frac{1}{n!} + \dots; +\Tag{(2)} +\] +and so +\[ +\left(1 + 1 + \frac{1}{2!} + \dots + \frac{1}{n!} + \dots\right)^{x} + = 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!} + \dots, +\Tag{(3)} +\] +\PageSep{379} +a result known as the \Emph{exponential theorem}. Also +\[ +a^{x} = e^{x\log a} = 1 + (x\log a) + \frac{(x\log a)^{2}}{2!} + \dots +\Tag{(4)} +\] +for all positive values of~$a$. + +\begin{Remark} +The reader will observe that the exponential series has the property of +reproducing itself when every term is differentiated, and that no other series +of powers of~$x$ would possess this property: for some further remarks in this +connection see \Ref{Appendix}{II}\@. + +The power series for~$e^{x}$ is so important that it is worth while to investigate +it by an alternative method which does not depend upon Taylor's Theorem. +Let +\[ +E_{n}(x) = 1 + x + \frac{x^{2}}{2!} + \dots + \frac{x^{n}}{n!}, +\] +and suppose that $x > 0$. Then +\[ +\left(1 + \frac{x}{n}\right)^{n} + = 1 + n\left(\frac{x}{n}\right) + + \frac{n(n - 1)}{1·2} \left(\frac{x}{n}\right)^{2} + \dots + + \frac{n(n - 1)\dots 1}{1·2\dots n} \left(\frac{x}{n}\right)^{n}\Add{,} +\] +which is less than~$E_{n}(x)$. And, provided $n > x$, we have also, by the binomial +theorem for a negative integral exponent, +\[ +\left(1 - \frac{x}{n}\right)^{-n} + = 1 + n\left(\frac{x}{n}\right) + + \frac{n(n + 1)}{1·2} \left(\frac{x}{n}\right)^{2} + \dots + > E_{n}(x). +\] +Thus +\[ +\left(1 + \frac{x}{n}\right)^{n} < E_{n}(x) < \left(1 - \frac{x}{n}\right)^{-n}. +\] +{\Loosen But (\SecNo[§]{208}) the first and last functions tend to the limit~$e^{x}$ as $n \to \infty$, and +therefore $E_{n}(x)$~must do the same. From this the equation~\Eq{(1)} follows when +$x$~is positive; its truth when $x$~is negative follows from the fact that the +exponential series, as was shown in \Ex{lxxxi}.~7, satisfies the functional +equation $f(x)f(y) = f(x + y)$, so that $f(x)f(-x) = f(0) = 1$.} +\end{Remark} + +\begin{Examples}{XC.} +\Item{1.} Show that +\[ +\cosh x = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots,\quad +\sinh x = x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots. +\] + +\Item{2.} If $x$~is positive then the greatest term in the exponential series is the +$([x] + 1)$-th, unless $x$~is an integer, when the preceding term is equal to it. + +\Item{3.} Show that $n! > (n/e)^{n}$. [For $n^{n}/n!$~is one term in the series for~$e^{n}$.] + +\Item{4.} Prove that $e^{n} = (n^{n}/n!)(2 + S_{1} + S_{2})$, where +\[ +S_{1} = \frac{1}{1 + \nu} + \frac{1}{(1 + \nu)(1 + 2\nu)} + \dots,\quad +S_{2} = (1 - \nu) + (1 - \nu)(1 - 2\nu) + \dots, +\] +and $\nu = 1/n$; and deduce that $n!$~lies between $2(n/e)^{n}$ and~$2(n + 1)(n/e)^{n}$. + +\Item{5.} Employ the exponential series to prove that $e^{x}$~tends to infinity more +rapidly than any power of~$x$. [Use the inequality $e^{x} > x^{n}/n!$.] +\PageSep{380} + +\Item{6.} Show that $e$~is not a rational number. [If $e = p/q$, where $p$ and~$q$ are +integers, we must have +\[ +\frac{p}{q} = \DPtypo{}{1 + {}} 1 + \frac{1}{2!}+\frac{1}{3!} + \dots + \frac{1}{q!} + \dots +\] +or, multiplying up by~$q!$, +\[ +q! \left(\frac{p}{q} - 1 - 1 - \frac{1}{2!} - \dots - \frac{1}{q!}\right) + = \frac{1}{q + 1} + \frac{1}{(q + 1)(q + 2)} + \dots +\] +and this is absurd, since the left-hand side is integral, and the right-hand +side less than $\{1/(q + 1)\} + \{1/(q + 1)\}^{2} + \dots = 1/q$.] + +\Item{7.} Sum the series $\sum\limits_{0}^{\infty} P_{r}(n)\dfrac{x^{n}}{n!}$, where $P_{r}(n)$~is a polynomial of degree~$r$ +in~$n$. [We can express $P_{r}(n)$ in the form +\[ +A_{0} + A_{1}n + A_{2}n(n - 1) + \dots + A_{r}n(n - 1) \dots (n - r + 1), +\] +and +\begin{align*} +\sum_{0}^{\infty} P_{r}(n) \frac{x^{n}}{n!} + &= A_{0}\sum_{0}^{\infty}\frac{x^{n}}{n!} + + A_{1}\sum_{1}^{\infty}\frac{x^{n}}{(n - 1)!} + \dots + + A_{r}\sum_{r}^{\infty}\frac{x^{n}}{(n - r)!}\\ + &= (A_{0} + A_{1}x + A_{2}x^{2} + \dots + A_{r}x^{r})e^{x}.] +\end{align*} + +\Item{8.} Show that +\[ +\sum_{1}^{\infty} \frac{n^{3}}{n!} x^{n} = (x + 3x^{2} + x^{3})e^{x},\quad +\sum_{1}^{\infty} \frac{n^{4}}{n!} x^{n} = (x + 7x^{2} + 6x^{3} + x^{4})e^{x}; +\] +and that if $S_{n} = 1^{3} + 2^{3} + \dots + n^{3}$ then +\[ +\sum_{1}^{\infty} S_{n}\frac{x^{n}}{n!} + = \tfrac{1}{4}(4x + 14x^{2} + 8x^{3} + x^{4})e^{x}. +\] +In particular the last series is equal to zero when $x = -2$. \MathTrip{1904.} + +\Item{9.} Prove that $\sum (n/n!) = e$, $\sum (n^{2}/n!) = 2e$, $\sum (n^{3}/n!) = 5e$, and that $\sum (n^{k}/n!)$, +where $k$~is any positive integer, is a positive integral multiple of~$e$. + +\Item{10.} Prove that $\sum\limits_{1}^{\infty} \dfrac{(n - 1)x^{n}}{(n + 2)n!} = \left\{(x^{2} - 3x + 3)e^{x} + \frac{1}{2}x^{2} - 3\right\}/x^{2}$. + +[Multiply numerator and denominator by~$n + 1$, and proceed as in Ex.~7.] + +\Item{11.} Determine $a$,~$b$,~$c$ so that $\{(x + a)e^{x} + (bx + c)\}/x^{3}$ tends to a limit +as $x \to 0$, evaluate the limit, and draw the graph of the function $e^{x} + \dfrac{bx + c}{x + a}$. + +\Item{12.} Draw the graphs of $1 + x$, $1 + x + \frac{1}{2}x^{2}$, $1 + x + \frac{1}{2}x^{2} + \frac{1}{6}x^{3}$, and compare +them with that of~$e^{x}$. + +\Item{13.} Prove that $e^{-x} - 1 + x - \dfrac{x^{n}}{2!} + \dots - (-1)^{n}\dfrac{x^{n}}{n!}$ is positive or negative +according as $n$~is odd or even. Deduce the exponential theorem. +\PageSep{381} + +\Item{14.} If +\[ +X_{0} = e^{x},\quad +X_{1} = e^{x} - 1,\quad +X_{2} = e^{x} - 1 - x,\quad +X_{3} = e^{x} - 1 - x - (x^{2}/2!),\ \dots, +\] +then $dX_{\nu}/dx = X_{\nu-1}$. Hence prove that if $t > 0$ then +\[ +X_{1}(t) = \int_{0}^{t} X_{0}\, dx < te^{t},\quad +X_{2}(t) = \int_{0}^{t} X_{1}\, dx < \int_{0}^{t} xe^{x}\, dx + < e^{t} \int_{0}^{t} x\, dx = \frac{t^{2}}{2!} e^{t}, +\] +and generally $X_{\nu}(t) < \dfrac{t^{\nu}}{\nu!} e^{t}$. Deduce the exponential theorem. + +\Item{15.} Show that the expansion in powers of~$p$ of the positive root of +$x^{2+p} = a^{2}$ begins with the terms +\[ +a\{1 - \tfrac{1}{2} p\log a + \tfrac{1}{8} p^{2}\log a (2 + \log a)\}. +\] +\MathTrip{1909.} +\end{Examples} + +\Paragraph{213. The logarithmic series.} Another very important +expansion in powers of~$x$ is that for~$\log(1 + x)$. Since +\[ +\log(1 + x) = \int_{0}^{x} \frac{dt}{1 + t}, +\] +and $1/(1 + t) = 1 - t + t^{2} - \dots$ if $t$~is numerically less than unity, it is +natural to expect\footnote + {See \Ref{Appendix}{II} for some further remarks on this subject.} +that $\log(1 + x)$ will be equal, when $-1 < x < 1$, +to the series obtained by integrating each term of the series +$1 - t + t^{2} - \dots$ from $t = 0$ to $t = x$, \ie\ to the series $x - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} - \dots$. +And this is in fact the case. For +\[ +1/(1 + t) + = 1 - t + t^{2} - \dots + (-1)^{m-1} t^{m-1} + \frac{(-1)^{m} t^{m}}{1 + t}, +\] +and so, if $x > -1$, +\[ +\log(1 + x) = \int_{0}^{x} \frac{dt}{1 + t} + = x - \frac{x^{2}}{2} + \dots + (-1)^{m-1} \frac{x^{m}}{m} + (-1)^{m} R_{m}, +\] +where +\[ +R_{m} = \int_{0}^{x} \frac{t^{m}\, dt}{1 + t}. +\] + +We require to show that the limit of~$R_{m}$, when $m$~tends to~$\infty$, +is zero. This is almost obvious when $0 < x \leq 1$; for then $R_{m}$~is +positive and less than +\[ +\int_{0}^{x} t^{m}\, dt = \frac{x^{m+1}}{m + 1}, +\] +and therefore less than $1/(m + 1)$. If on the other hand $-1 < x < 0$, +we put $t = -u$ and $x = -\xi$, so that +\[ +R_{m} = (-1)^{m} \int_{0}^{\xi} \frac{u^{m}\, du}{1 - u}, +\] +\PageSep{382} +which shows that $R_{m}$~has the sign of~$(-1)^{m}$. Also, since the +greatest value of~$1/(1 - u)$ in the range of integration is~$1/(1 - \xi)$, +we have +\[ +0 < |R_{m}| < \frac{1}{1 - \xi} \int_{0}^{\xi} u^{m}\, du + = \frac{\xi^{m}}{(m + 1)(1 - \xi)} + < \frac{1}{(m + 1)(1 - \xi)}: +\] +and so $R_{m} \to 0$. + +Hence +\[ +\log(1 + x) = x - \tfrac{1}{2} x^{2} + \tfrac{1}{3} x^{3} - \dots, +\] +provided that $-1 < x \leq 1$. If $x$~lies outside these limits the series +is not convergent. If $x = 1$ we obtain +\[ +\log 2 = 1 - \tfrac{1}{2} + \tfrac{1}{3} - \dots, +\] +a result already proved otherwise (\Ex{lxxxix}.~7). + +\Paragraph{214. The series for the inverse tangent.} It is easy to +prove in a similar manner that +\begin{align*} +\arctan x = \int_{0}^{x} \frac{dt}{1 + t^{2}} + &= \int_{0}^{x}(1 - t^{2} + t^{4} - \dots)\, dt\\ + &= x - \tfrac{1}{3} x^{3} + \tfrac{1}{5} x^{5} - \dots, +\end{align*} +provided that $-1 \leq x \leq 1$. The only difference is that the proof is +a little simpler; for, since $\arctan x$~is an odd function of~$x$, we need +only consider positive values of~$x$. And the series is convergent +when $x = -1$ as well as when $x = 1$. We leave the discussion to the +reader. The value of~$\arctan x$ which is represented by the series +is of course that which lies between $-\frac{1}{4}\pi$ and~$\frac{1}{4}\pi$ when $-1 \leq x \leq 1$, +and which we saw in \Ref{Ch.}{VII} (\Ex{lxiii}.~3) to be the value +represented by the integral. If $x = 1$, we obtain the formula +\[ +\tfrac{1}{4}\pi = 1 - \tfrac{1}{3} + \tfrac{1}{5} - \dots. +\] + +\begin{Examples}{XCI.} +\Item{1.} $\log \left(\dfrac{1}{1 - x}\right) = x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \dots$ if $-1 \leq x < 1$. + +\Item{2.} $\argtanh x = \frac{1}{2} \log\left(\dfrac{1 + x}{1 - x}\right) = x + \frac{1}{3} x^{3} + \frac{1}{5} x^{5} + \dots$ if $-1 < x < 1$. + +\Item{3.} Prove that if $x$~is positive then +\[ +\log(1 + x) = \frac{x}{1 + x} + + \tfrac{1}{2} \left(\frac{x}{1 + x}\right)^{2} + + \tfrac{1}{3} \left(\frac{x}{1 + x}\right)^{3} + \dots. +\] +\MathTrip{1911.} + +\Item{4.} Obtain the series for $\log(1 + x)$ and $\arctan x$ by means of Taylor's +theorem. + +[A difficulty presents itself in the discussion of the remainder in the +\PageSep{383} +first series when $x$~is negative, if Lagrange's form $R_{n} = (-1)^{n-1} x^{n}/\{n(1 + \theta x)^{n}\}$ +is used; Cauchy's form, viz. +\[ +R_{n} = (-1)^{n-1} (1 - \theta)^{n-1} x^{n}/(1 + \theta x)^{n}, +\] +should be used (cf.\ the corresponding discussion for the Binomial Series, +\Ex{lvi}.~2 and~\SecNo[§]{163}). + +In the case of the second series we have +\begin{align*} +D_{x}^{n} \arctan x + &= D_{x}^{n-1} \{1/(1 + x^{2})\}\\ + &= (-1)^{n-1} (n - 1)! (x^{2} + 1)^{-n/2} \sin \{n\arctan(1/x)\} +\end{align*} +(\Ex{xlv}.~11), and there is no difficulty about the remainder, which is obviously +not greater in absolute value than~$1/n$.\footnotemark] + \footnotetext{The formula for $D_{x}^{n} \arctan x$ fails when $x = 0$, as $\arctan(1/x)$ is then + undefined. It is easy to see (cf.\ \Ex{xlv}.~11) that $\arctan(1/x)$~must then be + interpreted as meaning~$\frac{1}{2}\pi$.} + +\Item{5.} If $y > 0$ then +\[ +\log y = 2 \left\{\frac{y - 1}{y + 1} + + \frac{1}{3} \left(\frac{y - 1}{y + 1}\right)^{3} + + \frac{1}{5} \left(\frac{y - 1}{y + 1}\right)^{5} + \dots\right\}. +\] + +[Use the identity $y = \biggl(1 + \dfrac{y - 1}{y + 1}\biggr) \bigg/ \biggl(1 - \dfrac{y - 1}{y + 1}\biggr)$. This series may be used to +calculate~$\log 2$, a purpose for which the series $1 - \frac{1}{2} + \frac{1}{3} - \dots$, owing to the +slowness of its convergence, is practically useless. Put $y = 2$ and find $\log 2$ +to $3$~places of decimals.] + +\Item{6.} Find $\log 10$ to $3$~places of decimals from the formula +\[ +\log 10 = 3\log 2 + \log(1 + \tfrac{1}{4}). +\] + +\Item{7.} Prove that +\[ +\log \left(\frac{x + 1}{x}\right) + = 2\left\{\frac{1}{2x + 1} + \frac{1}{3(2x + 1)^{3}} + \frac{1}{5(2x + 1)^{5}} + \dots\right\} +\] +if $x > 0$, and that +\[ +\log \frac{(x - 1)^{2}(x + 2)}{(x + 1)^{2}(x - 2)} + = 2\left\{\frac{2}{x^{3} - 3x} + + \frac{1}{3}\left(\frac{2}{x^{3} - 3x}\right)^{3} + + \frac{1}{5}\left(\frac{2}{x^{3} - 3x}\right)^{5} + \dots\right\} +\] +if $x > 2$. Given that $\log 2 = .693\MS147\MS1\dots$ and $\log 3 = 1.098\MS612\MS3\dots$, show, by +putting $x = 10$ in the second formula, that $\log 11 = 2.397\MS895\dots$. +\MathTrip{1912.} + +\Item{8.} Show that if $\log 2$, $\log 5$, and $\log 11$ are known, then the formula +\[ +\log 13 = 3\log 11 + \log 5 - 9\log 2 +\] +gives $\log 13$ with an error practically equal to~$.000\MS15$. \MathTrip{1910.} + +\Item{9.} Show that +\[ +\tfrac{1}{2} \log 2 = 7a + 5b + 3c,\quad +\tfrac{1}{2} \log 3 = 11a + 8b + 5c,\quad +\tfrac{1}{2} \log 5 = 16a + 12b + 7c, +\] +where $a = \argtanh(1/31)$, $b = \argtanh(1/49)$, $c = \argtanh(1/161)$. + +[These formulae enable us to find $\log 2$, $\log 3$, and $\log 5$ rapidly and with +any degree of accuracy.] +\PageSep{384} + +\Item{10.} Show that +\[ +\tfrac{1}{4}\pi = \arctan(1/2) + \arctan(1/3) = 4\arctan(1/5) - \arctan(1/239), +\] +and calculate~$\pi$ to $6$~places of decimals. + +\Item{11.} Show that the expansion of $(1 + x)^{1+x}$ in powers of~$x$ begins with the +terms $1 + x + x^{2} + 1/2 x^{3}$. \MathTrip{1910.} + +\Item{12.} Show that +\[ +\log_{10} e - \sqrtb{x(x + 1)} \log_{10}\left(\frac{1 + x}{x}\right) + = \frac{\log_{10} e}{24x^{2}}, +\] +approximately, for large values of~$x$. Apply the formula, when $x = 10$, to +obtain an approximate value of~$\log_{10} e$, and estimate the accuracy of the result. +\MathTrip{1910.} + +\Item{13.} Show that +\[ +\frac{1}{1 - x} \log\left(\frac{1}{1 - x}\right) + = x + \left(1 + \tfrac{1}{2}\right)x^{2} + + \left(1 + \tfrac{1}{2} + \tfrac{1}{3}\right)x^{3} + \dots, +\] +if $-1 < x < 1$. [Use \Ex{lxxxi}.~2.] + +\Item{14.} {\Loosen Using the logarithmic series and the facts that $\log_{10} 2.3758 = .375\MS809\MS9\dots$ +and $\log_{10} e = .4343\dots$, show that an approximate solution of the equation +$x = 100 \log_{10}x$ is~$237.581\MS21$.} \MathTrip{1910.} + +\Item{15.} Expand $\log\cos x$ and $\log(\sin x/x)$ in powers of~$x$ as far as~$x^{4}$, and +verify that, to this order, +\[ +\log\sin x + = \log x - \tfrac{1}{45} \log\cos x + \tfrac{64}{45}\log\cos \tfrac{1}{2}x. +\] +\MathTrip{1908.} + +\Item{16.} Show that +\[ +%[** TN: In-line in the original] +\int_{0}^{x} \frac{dt}{1 + t^{4}} = x - \tfrac{1}{5}x^{5} + \tfrac{1}{9}x^{9} - \dots +\] +if $-1 \leq x \leq 1$. Deduce that +\[ +1 - \tfrac{1}{5} + \tfrac{1}{9} - \dots + = \{\pi + 2\log(\sqrt{2} + 1)\}/4\sqrt{2}. +\] +\MathTrip{1896.} + +[Proceed as in \SecNo[§]{214} and use the result of \Ex{xlviii}.~7.] + +\Item{17.} Prove similarly that +\[ +\tfrac{1}{3} - \tfrac{1}{7} + \tfrac{1}{11} - \dots + = \int_{0}^{1} \frac{t^{2}\, dt}{1 + t^{4}} + = \{\pi - 2\log(\sqrt{2} + 1)\}/4\sqrt{2}. +\] + +\Item{18.} Prove generally that if $a$ and~$b$ are positive integers then +\[ +\frac{1}{a} - \frac{1}{a + b} + \frac{1}{a + 2b} - \dots + = \int_{0}^{1} \frac{t^{a-1}\, dt}{1 + t^{b}}, +\] +and so that the sum of the series can be found. Calculate in this way the +sums of $1 - \frac{1}{4} + \frac{1}{7} - \dots$ and $\frac{1}{2} - \frac{1}{5} + \frac{1}{8} - \dots$. +\end{Examples} + +\Paragraph{215. The Binomial Series.} We have already (\SecNo[§]{163}) +investigated the Binomial Theorem +\[ +(1 + x)^{m} = 1 + \binom{m}{1}x + \binom{m}{2}x^{2} + \dots, +\] +\PageSep{385} +assuming that $-1 < x < 1$ and that $m$~is rational. When $m$~is +irrational we have +\begin{gather*} +(1 + x)^{m} = e^{m\log(1+ x)},\\ +D_{x}(1 + x)^{m} = \{m/(1 + x)\} e^{m\log(1 + x)} = m(1 + x)^{m-1}, +\end{gather*} +so that the rule for the differentiation of~$(1 + x)^{m}$ remains the +same, and the proof of the theorem given in \SecNo[§]{163} retains its +validity. We shall not discuss the question of the convergence +of the series when $x = 1$ or $x = -1$.\footnote + {See Bromwich, \textit{Infinite Series}, pp.~150~\textit{et~seq.}; Hobson, \textit{Plane Trigonometry} + (3rd~edition), p.~271.} + +\begin{Examples}{XCII.} +\Item{1.} Prove that if $-1 < x < 1$ then +\[ +\frac{1}{\sqrtp{1 + x^{2}}} = 1 - \frac{1}{2}x^{2} + \frac{1·3}{2·4}x^{4} - \dots,\quad +\frac{1}{\sqrtp{1 - x^{2}}} = 1 + \frac{1}{2}x^{2} + \frac{1·3}{2·4}x^{4} + \dots. +\] + +\Item{2.} \Topic{Approximation to quadratic and other surds.} {\Loosen Let $\sqrt{M}$ be a +quadratic surd whose numerical value is required. Let $N^{2}$ be the square +nearest to~$M$; and let $M = N^{2} + x$ or $M = N^{2} - x$, $x$~being positive. Since $x$~cannot +be greater than~$N$, $x/N^{2}$~is comparatively small and the surd +$\sqrt{M} = N\sqrtb{1 ± (x/N^{2})}$ can be expressed in a series} +\[ += N\left\{ + 1 ± \frac{1}{2}\left(\frac{x}{N^{2}}\right) + - \frac{1·1}{2·4}\left(\frac{x}{N^{2}}\right)^{2} ± \dots +\right\}, +\] +which is at any rate fairly rapidly convergent, and may be very rapidly so. +Thus +\[ +\sqrt{67} = \sqrtp{64 + 3} + = 8\left\{ + 1 + \frac{1}{2}\left(\frac{3}{64}\right) + - \frac{1·1}{2·4}\left(\frac{3}{64}\right)^{2} + \dots +\right\}. +\] + +Let us consider the error committed in taking~$8\frac{3}{16}$ (the value given by +the first two terms) as an approximate value. After the second term the +terms alternate in sign and decrease. Hence the error is one of excess, and +is less than~$3^{2}/64^{2}$, which is less than~$.003$. + +\Item{3\Add{.}} If $x$~is small compared with~$N^{2}$ then +\[ +\sqrtp{N^{2} + x} = N + \frac{x}{4N} + \frac{Nx}{2(2N^{2} + x)}, +\] +the error being of the order~$x^{4}/N^{7}$. Apply the process to~$\sqrt{907}$. + +[Expanding by the binomial theorem, we have +\[ +\sqrtp{N^{2} + x} + = N + \frac{x}{2N} - \frac{x^{2}}{8N^{3}} + \frac{x^{3}}{16N^{5}}, +\] +the error being less than the numerical value of the next term, viz.\ +$5x^{4}/128N^{7}$. Also +\[ +\frac{Nx}{2(2N^{2} + x)} + = \frac{x}{4N} \left(1 + \frac{x}{2N^{2}}\right)^{-1} + = \frac{x}{4N} - \frac{x^{2}}{8N^{3}} + \frac{x^{3}}{16N^{5}}, +\] +the error being less than~$x^{4}/32N^{7}$. The result follows. The same method +may be applied to surds other than quadratic surds, \eg\ to~$\sqrt[3]{1031}$.] +\PageSep{386} + +\Item{4.} If $M$~differs from~$N^{3}$ by less than $1$~per~cent.\ of either then $\sqrt[3]{M}$~differs +from $\frac{2}{3}N + \frac{1}{3}(M/N^{2})$ by less than $N/90\MC000$. \MathTrip{1882.} + +\Item{5.} If $M = N^{4} + x$, and $x$~is small compared with~$N$, then a good approximation +for~$\sqrt[4]{M}$ is +\[ +\frac{51}{56} N + \frac{5}{56}\, \frac{M}{N^{3}} + \frac{27Nx}{14(7M + 5N^{4})}. +\] +Show that when $N = 10$, $x = 1$, this approximation is accurate to $16$~places +of decimals. \MathTrip{1886.} + +\Item{6.} Show how to sum the series +\[ +\sum_{0}^{\infty} P_{r}(n) \binom{m}{n} x^{n}, +\] +where $P_{r}(n)$~is a polynomial of degree~$r$ in~$n$. + +[Express $P_{r}(n)$ in the form $A_{0} + A_{1}n + A_{2}n(n - 1) + \dots$ as in \Ex{xc}.~7.] + +\Item{7.} Sum the series $\sum\limits_{0}^{\infty} n \dbinom{m}{n} x^{n}$, $\sum\limits_{0}^{\infty} n^{2} \dbinom{m}{n} x^{n}$ and prove that +\[ +\sum_{0}^{\infty} n^{3} \binom{m}{n} x^{n} + = \{m^{3}x^{3} + m(3m - 1)x^{2} + mx\}(1 + x)^{m-3}. +\] +\end{Examples} + +\begin{Remark} +\Paragraph{216. An alternative method of development of the theory of the +exponential and logarithmic functions.} We shall now give an outline of +a method of investigation of the properties of $e^{x}$ and $\log x$ entirely different +in logical order from that followed in the preceding pages. This method +starts from the exponential series $1 + x + \dfrac{x^{2}}{2!} + \dots$. We know that this series +is convergent for all values of~$x$, and we may therefore define the function +$\exp x$ by the equation +\[ +\exp x = 1 + x + \frac{x^{2}}{2!} + \dots. +\Tag{(1)} +\] + +We then prove, as in \Ex{lxxxi}.~7, that +\[ +\exp x × \exp y = \exp(x + y). +\Tag{(2)} +\] + +Again +\[ +\frac{\exp h - 1}{h} + = 1 + \frac{h}{2!} + \frac{h^{2}}{3!} + \dots + = 1 + \rho(h), +\] +where $\rho(h)$~is numerically less than +\[ +|\tfrac{1}{2}h| + |\tfrac{1}{2}h|^{2} + |\tfrac{1}{2}h|^{3} + \dots + = |\tfrac{1}{2}h|/(1 - |\tfrac{1}{2}h|), +\] +so that $\rho(h) \to 0$ as $h \to 0$. And so +\[ +\frac{\exp(x + h) - \exp x}{h} + = \exp x \left(\frac{\exp h - 1}{h}\right) \to \exp x +\] +as $h \to 0$, or +\[ +D_{x} \exp x = \exp x. +\Tag{(3)} +\] +Incidentally we have proved that $\exp x$ is a continuous function. + +We have now a choice of procedure. Writing $y = \exp x$ and observing +that $\exp 0 = 1$, we have +\[ +\frac{dy}{dx} = y,\quad +x = \int_{1}^{y} \frac{dt}{t}, +\] +\PageSep{387} +and, if we define the logarithmic function as the function inverse to the +exponential function, we are brought back to the point of view adopted earlier +in this chapter. + +But we may proceed differently. From~\Eq{(2)} it follows that if $n$~is a positive +integer then +\[ +(\exp x)^{n} = \exp nx,\quad +(\exp 1)^{n} = \exp n. +\] +If $x$~is a positive rational fraction~$m/n$, then +\[ +\{\exp(m/n)\}^{n} = \exp m = (\exp 1)^{m}, +\] +and so $\exp(m/n)$~is equal to the positive value of~$(\exp 1)^{m/n}$. This result may +be extended to negative rational values of~$x$ by means of the equation +\[ +\exp x \exp(-x) = 1; +\] +and so we have +\[ +\exp x = (\exp 1)^{x} = e^{x}, +\] +say, where +\[ +e = \exp 1 = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots, +\] +for all rational values of~$x$. Finally we define $e^{x}$, when $x$~is irrational, as +being equal to~$\exp x$. The logarithm is then defined as the function inverse +to $\exp x$ or~$e^{x}$. + +\Par{Example.} Develop the theory of the binomial series +\[ +1 + \binom{m}{1} x + \binom{m}{2} x^{2} + \dots = f(m, x), +\] +where $-1 < x < 1$, in a similar manner, starting from the equation +\[ +f(m, x) f(m', x) = f(m + m'\Add{,} x) +\] +(\Ex{lxxxi}.~6). +\end{Remark} + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER IX\protect\footnotemark} + +\footnotetext{A considerable number of these examples are taken from Bromwich's \textit{Infinite Series}.} + +\begin{Examples}{} +\Item{1.} Given that $\log_{10} e = .4343$ and that $2^{10}$ and $3^{21}$ are nearly equal to powers +of~$10$, calculate $\log_{10}2$ and $\log_{10}3$ to four places of decimals. \MathTrip{1905.} + +\Item{2.} Determine which of $(\frac{1}{2}e)^{\sqrt{3}}$ and $(\sqrt{2})^{\frac{1}{2}\pi}$ is the greater. [Take logarithms +and observe that $\sqrt{3}/(\sqrt{3} + \frac{1}{4}\pi) < \frac{2}{5} \sqrt{3} < .6929 < \log 2$.] + +\Item{3.} {\Loosen Show that $\log_{10}n$ cannot be a rational number if $n$~is any positive +integer not a power of~$10$. [If $n$~is not divisible by~$10$, and $\log_{10}n = p/q$, we +have $10^{p} = n^{q}$, which is impossible, since $10^{p}$~ends with~$0$ and $n^{q}$~does not. +If $n = 10^{a}N$, where $N$~is not divisible by~$10$, then $\log_{10}N$ and therefore} +\[ +\log_{10}n = a + \log_{10}N +\] +cannot be rational.] +\PageSep{388} + +\Item{4.} For what values of~$x$ are the functions $\log x$, $\log\log x$, $\log\log\log x$,~\dots\ +(\ia)~equal to~$0$ (\ib)~equal to~$1$ (\ic)~not defined? Consider also the same question +for the functions $lx$, $llx$, $lllx$,~\dots, where $lx = \log |x|$. + +\Item{5.} Show that +\[ +\log x - \binom{n}{1} \log(x + 1) + \binom{n}{2} \log(x + 2) - \dots + + (-1)^{n} \log(x + n) +\] +is negative and increases steadily towards $0$ as $x$~increases from $0$ towards~$\infty$. + +[The derivative of the function is +\[ +\sum_{0}^{n} (-1)^{r} \binom{n}{r} \frac{1}{x + r} + = \frac{n!}{x(x + 1) \dots (x + n)}, +\] +as is easily seen by splitting up the right-hand side into partial fractions. +This expression is positive, and the function itself tends to zero as $x \to \infty$, +since +\[ +\log(x + r) = \log x + \epsilon_{x}, +\] +where $\epsilon_{x} \to 0$, and $1 - \dbinom{n}{1} + \dbinom{n}{2} - \dots = 0$.] + +\Item{6.} Prove that +\[ +\left(\frac{d}{dx}\right)^{n} \frac{\log x}{x} + = \frac{(-1)^{n} n!}{x^{n+1}} \left(\log x - 1 - \frac{1}{2} - \dots - \frac{1}{n}\right). +\] +\MathTrip{1909.} + +\Item{7.} If $x > -1$ then $x^{2} > (1 + x) \{\log(1 + x)\}^{2}$. \MathTrip{1906.} + +[Put $1 + x = e^{\xi}$, and use the fact that $\sinh \xi > \xi$ when $\xi > 0$.] + +\Item{8.} Show that $\{\log(1 + x)\}/x$ and $x/\{(1 + x)\log(1 + x)\}$ both decrease steadily +as $x$~increases from $0$ towards~$\infty$. + +\Item{9.} Show that, as $x$~increases from $-1$ towards~$\infty$, the function +$(1 + x)^{-1/x}$ assumes once and only once every value between $0$ and~$1$. \MathTrip{1910.} + +\Item{10.} Show that $\dfrac{1}{\log(1 + x)} - \dfrac{1}{x} \to \dfrac{1}{2}$ as $x \to 0$. + +\Item{11.} Show that $\dfrac{1}{\log(1 + x)} - \dfrac{1}{x}$ decreases steadily from $1$ to~$0$ as $x$~increases +from $-1$ towards~$\infty$. [The function is undefined when $x = 0$, but if we +attribute to it the value~$\frac{1}{2}$ when $x = 0$ it becomes continuous for $x = 0$. Use +Ex.~7 to show that the derivative is negative.] + +\Item{12.} Show that the function $(\log \xi - \log x)/(\xi - x)$, where $\xi$~is positive, +decreases steadily as $x$~increases from $0$ to~$\xi$, and find its limit as $x \to \xi$. + +\Item{13.} Show that $e^{x} > Mx^{N}$, where $M$~and~$N$ are large positive numbers, \DPtypo{f}{if} +$x$~is greater than the greater of $2\log M$ and~$16N^{2}$. + +[It is easy to prove that $\log x < 2\sqrt{x}$; and so the inequality given is +certainly satisfied if +\[ +x > \log M + 2N\sqrt{x}, +\] +and therefore certainly satisfied if $\frac{1}{2}x > \log M$, $\frac{1}{2}x > 2N\sqrt{x}$.] +\PageSep{389} + +\Item{14.} If $f(x)$ and $\phi(x)$ tend to infinity as $x \to \infty$, and $f'(x)/\phi'(x) \to \infty$, +then $f(x)/\phi(x) \to \infty$. [Use the result of \Ref{Ch.}{VI}, \MiscEx{VI}~33.] By taking +$f(x) = x^{\alpha}$, $\phi(x) = \log x$, prove that $(\log x)/x^{\alpha} \to 0$ for all positive values of~$\alpha$. + +\Item{15.} If $p$ and~$q$ are positive integers then +\[ +\frac{1}{pn + 1} + \frac{1}{pn + 2} + \dots + \frac{1}{qn} + \to \log\left(\frac{q}{p}\right) +\] +as $n \to \infty$. [Cf.\ \Ex{lxxviii}.~6.] + +\Item{16.} Prove that if $x$~is positive then $n\log\{\frac{1}{2}(1 + x^{1/n})\} \to -\frac{1}{2}\log x$ as +$n \to \infty$. [We have +\[ +n\log\{\tfrac{1}{2}(1 + x^{1/n})\} + = n\log\{1 - \tfrac{1}{2}(1 - x^{1/n})\} + = \tfrac{1}{2}n(1 - x^{1/n}) \frac{\log(1 - u)}{u} +\] +where $u = \frac{1}{2}(1 - x^{1/n})$. Now use \SecNo[§]{209} and \Ex{lxxxii}.~4.] + +\Item{17.} Prove that if $a$ and~$b$ are positive then +\[ +\{\tfrac{1}{2}(a^{1/n} + b^{1/n})\}^{n} \to \sqrtp{ab}. +\] + +%[** TN: No paragraph break in the original] +[Take logarithms and use Ex.~16.] + +\Item{18.} Show that +\[ +1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{2n - 1} + = \tfrac{1}{2}\log n + \log 2 + \tfrac{1}{2} \gamma + \epsilon_{n}, +\] +where $\gamma$~is Euler's constant (\Ex{lxxxix}.~1) and $\epsilon_{n} \to 0$ as $n \to \infty$. + +\Item{19.} Show that +\[ +1 + \tfrac{1}{3} - \tfrac{1}{2} + \tfrac{1}{5} + + \tfrac{1}{7} - \tfrac{1}{4} + \tfrac{1}{9} + \dots + = \tfrac{3}{2} \log 2, +\] +the series being formed from the series $1 - \frac{1}{2} + \frac{1}{3} - \dots$ by taking alternately two +positive terms and then one negative. [The sum of the first $3n$ terms is +\begin{multline*} +1 + \frac{1}{3} + \frac{1}{5} + \dots + \frac{1}{4n - 1} + - \frac{1}{2} \left(1 + \frac{1}{2} + \dots + \frac{1}{n}\right)\\ + = \tfrac{1}{2}\log 2n + \log 2 + \tfrac{1}{2}\gamma + \epsilon_{n} + - \tfrac{1}{2}(\log n + \gamma + \epsilon_{n}'), + \end{multline*} +where $\epsilon_{n}$ and~$\epsilon'_{n}$ tend to~$0$ as $n \to \infty$. (Cf.\ \Ex{lxxviii}.~6).] + +\Item{20.} Show that $1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \dots = \frac{1}{2}\log 2$. + +\Item{21.} Prove that +\[ +\sum_{1}^{n} \frac{1}{\nu(36\nu^{2} - 1)} + = -3 + 3\Sigma_{3n+1} - \Sigma_{n} - S_{n} +\] +where $S_{n} = 1 + \dfrac{1}{2} + \dots + \dfrac{1}{n}$, $\Sigma_{n} = 1 + \dfrac{1}{3} + \dots + \dfrac{1}{2n - 1}$. Hence prove that the sum +of the series when continued to infinity is +\[ +-3 + \tfrac{3}{2}\log 3 + 2\log 2. +\] +\MathTrip{1905.} + +\Item{22.} Show that +\[ +\sum_{1}^{\infty} \frac{1}{n(4n^{2} - 1)} = 2\log 2 - 1, \quad +\sum_{1}^{\infty} \frac{1}{n(9n^{2} - 1)} = \tfrac{3}{2}(\log 3 - 1). +\] +\PageSep{390} + +\Item{23.} Prove that the sums of the four series +\[ +\sum_{1}^{\infty} \frac{1}{4n^{2} - 1},\quad +\sum_{1}^{\infty} \frac{(-1)^{n-1}}{4n^{2} - 1},\quad +\sum_{1}^{\infty} \frac{1}{(2n + 1)^{2} - 1},\quad +\sum_{1}^{\infty} \frac{(-1)^{n-1}}{(2n + 1)^{2} - 1} +\] +are $\frac{1}{2}$, $\frac{1}{4}\pi - \frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{2}\log 2 - \frac{1}{4}$ respectively. + +\Item{24.} Prove that $n!\, (a/n)^{n}$ tends to~$0$ or to~$\infty$ according as $a < e$ or $a > e$. + +[If $u_{n} = n!\, (a/n)^{n}$ then $u_{n+1}/u_{n} = a\{1 + (1/n)\}^{-n} \to a/e$. It can be shown +that the function tends to~$\infty$ when $a = e$: for a proof, which is rather beyond +the scope of the theorems of this chapter, see Bromwich's \textit{Infinite Series}, +pp.~461~\textit{et~seq.}] + +\Item{25.} Find the limit as $x \to \infty$ of +\[ +\left(\frac{a_{0} + a_{1} x + \dots + a_{r} x^{r}} + {b_{0} + b_{1} x + \dots + b_{r} x^{r}}\right)^{\lambda_{0}+\lambda_{1}x}, +\] +distinguishing the different cases which may arise. \MathTrip{1886.} + +\Item{26.} Prove that +\[ +\sum \log \left(1 + \frac{x}{n}\right)\quad (x > 0) +\] +diverges to~$\infty$. [Compare with $\sum (x/n)$.] Deduce that if $x$~is positive then +\[ +(1 + x)(2 + x) \dots (n + x)/n! \to \infty +\] +as $n \to \infty$. [The logarithm of the function is $\sum\limits_{1}^{n} \log \left(1 + \dfrac{x}{\nu}\right)$.] + +\Item{27.} Prove that if $x > -1$ then +\begin{multline*} +\frac{1}{(x + 1)^{2}} + = \frac{1}{(x + 1) (x + 2)} + + \frac{1!}{(x + 1) (x + 2) (x + 3)}\\ + + \frac{2!}{(x + 1) (x + 2) (x + 3) (x + 4)} + \dots. +\end{multline*} +\MathTrip{1908.} + +[The difference between $1/(x + 1)^{2}$ and the sum of the first $n$ terms of the +series is +\[ +\frac{1}{(x + 1)^{2}}\, \frac{n!}{(x + 2) (x + 3) \dots (x + n + 1)}.] +\] + +\Item{28.} No equation of the type +\[ +Ae^{\alpha x} + Be^{\beta x} + \dots = 0, +\] +where $A$, $B$,~\dots\ are polynomials and $\alpha$, $\beta$,~\dots\ different real numbers, can hold +for all values of~$x$. [If $\alpha$~is the algebraically greatest of $\alpha$, $\beta$,~\dots, then the term~$Ae^{\alpha x}$ +outweighs all the rest as $x \to \infty$.] + +\Item{29.} Show that the sequence +\[ +a_{1} = e,\quad +a_{2} = e^{e^{2}},\quad +a_{3} = e^{e^{e^{3}}},\ \dots +\] +tends to infinity more rapidly than any member of the exponential scale. + +[Let $e_{1}(x) = e^{x}$, $e_{2}(x) = e^{e_{1}(x)}$, and so on. Then, if $e_{k}(x)$~is any member of the +exponential scale, $a_{n} > e_{k}(n)$ when $n > k$.] +\PageSep{391} + +\Item{30.} Prove that +\[ +\frac{d}{dx} \{\phi(x)\}^{\psi(x)} + = \frac{d}{dx} \{\phi(x)\}^{\alpha} + \frac{d}{dx} \{\beta^{\psi(x)}\} +\] +where $\alpha$~is to be put equal to~$\psi(x)$ and $\beta$ to~$\phi(x)$ after differentiation. +Establish a similar rule for the differentiation of $\phi(x)^{[\{\psi(x)\}^{\chi(x)}]}$. + +\Item{31.} Prove that if $D_{x}^{n} e^{-x^{2}} = e^{-x^{2}} \phi_{n}(x)$ then (i)~$\phi_{n}(x)$ is a polynomial of +degree~$n$, (ii)~$\phi_{n+1} = -2x\phi_{n} + \phi_{n}'$, and (iii)~all the roots of $\phi_{n} = 0$ are real and +distinct, and separated by those of $\phi_{n-1} = 0$. [To prove~(iii) assume the truth +% [** TN: Typo in original; fixed while swapping roles of n and \kappa] +of the result for $\DPtypo{n}{\kappa} = 1$, $2$,~\dots\Add{,} $\DPtypo{\kappa}{n}$, and consider the signs of~$\DPtypo{\phi_{\kappa+1}}{\phi_{n+1}}$ for the $n$~values +of~$x$ for which $\DPtypo{\phi_{\kappa}}{\phi_{n}} = 0$ and for large (positive or negative) values of~$x$.] + +\Item{32.} The general solution of $f(xy) = f(x)f(y)$, where $f$~is a differentiable +function, is~$x^{a}$, where $a$~is a constant: and that of +\[ +f(x + y) + f(x - y) = 2f(x)f(y) +\] +is $\cosh ax$ or $\cos ax$, according as $f''(0)$~is positive or negative. [In proving +the second result assume that $f$~has derivatives of the first three orders. +Then +\[ +2f(x) + y^{2}\{f''(x) + \epsilon_{y}\} + = 2f(x)[f(0) + yf'(0) + \tfrac{1}{2} y^{2}\{f''(0) + \epsilon_{y}'\}], +\] +where $\epsilon_{y}$ and~$\epsilon_{y}'$ tend to zero with~$y$. It follows that $f(0) = 1$, $f'(0) = 0$, +$f''(x) = f''(0)f(x)$, so that $a = \sqrtb{f''(0)}$ or $a = \sqrtb{-f''(0)}$.] + +\Item{33.} How do the functions $x^{\sin(1/x)}$, $x^{\sin^{2}(1/x)}$, $x^{\cosec(1/x)}$ behave as $x \to +0$? + +\Item{34.} Trace the curves $y = \tan x e^{\tan x}$, $y = \sin x \log \tan \frac{1}{2}x$. + +\Item{35.} The equation $e^{x} = ax + b$ has one real root if $a < 0$ or $a = 0$, $b > 0$. If +$a > 0$ then it has two real roots or none, according as $a\log a > b - a$ or +$a\log a < b - a$. + +\Item{36.} Show by graphical considerations that the equation $e^{x} = ax^{2} + 2bx + c$ +has one, two, or three real roots if $a > 0$, none, one, or two if $a < 0$; and show +how to distinguish between the different cases. + +\Item{37.} Trace the curve $y = \dfrac{1}{x} \log\left(\dfrac{e^{x} - 1}{x}\right)$, showing that the point $(0, \frac{1}{2})$ is +a centre of symmetry, and that as $x$~increases through all real values, $y$~steadily +increases from $0$ to~$1$. Deduce that the equation +\[ +\frac{1}{x} \log\left(\frac{e^{x} - 1}{x}\right) = \alpha +\] +has no real root unless $0 < \alpha < 1$, and then one, whose sign is the same as +that of $\alpha - \frac{1}{2}$. [In the first place +\[ +y - \tfrac{1}{2} + = \frac{1}{x} \left\{\log\left(\frac{e^{x} - 1}{x}\right) - \log e^{\frac{1}{2} x}\right\} + = \frac{1}{x} \log\left(\frac{\sinh \frac{1}{2}x}{\frac{1}{2}x}\right) +\] +is clearly an odd function of~$x$. Also +\[ +\frac{dy}{dx} + = \frac{1}{x^{2}} \left\{\tfrac{1}{2} x\coth \tfrac{1}{2}x - 1 + - \log\left(\frac{\sinh \frac{1}{2}x}{\frac{1}{2}x}\right)\right\}. +\] +\PageSep{392} +The function inside the large bracket tends to zero as $x \to 0$; and its +derivative is +\[ +\frac{1}{x} \left\{1 - \left(\frac{\frac{1}{2}x}{\sinh \frac{1}{2}x}\right)^2\right\}, +\] +which has the sign of~$x$. Hence $dy/dx > 0$ for all values of~$x$.] + +\Item{38.} Trace the curve $y = e^{1/x} \sqrtp{x^{2} + 2x}$, and show that the equation +\[ +e^{1/x} \sqrtp{x^{2} + 2x} = \alpha +\] +has no real roots if $\alpha$~is negative, one negative root if +\[ +%[** TN: In-line in the original] +0 < \alpha < a = e^{1/\sqrt{2}} \sqrtp{2 + 2\sqrt{2}}, +\] +and two positive roots and one negative if $\alpha > a$. + +\Item{39.} Show that the equation $f_{n}(x) = 1 + x + \dfrac{x^{2}}{2!} + \dots + \dfrac{x^{n}}{n!} = 0$ has one real +root if $n$~is odd and none if $n$~is even. + +[Assume this proved for $n = 1$, $2$,~\dots~$2k$. Then $f_{2k+1}(x) = 0$ has at least +one real root, since its degree is odd, and it cannot have more since, if it +had, $f'_{2k+1}(x)$ or~$f_{2k}(x)$ would have to vanish once at least. Hence $f_{2k+1}(x) = 0$ +has just one root, and so $f_{2k+2}(x) = 0$ cannot have more than two. If it has +two, say $\alpha$~and~$\beta$, then $f'_{2k+2}(x)$ or~$f_{2k+1}(x)$ must vanish once at least between +$\alpha$~and~$\beta$, say at~$\gamma$. And +\[ +f_{2k+2}(\gamma) = f_{2k+1}(\gamma) + \frac{\gamma^{2k+2}}{(2k + 2)!} > 0. +\] +But $f_{2k+2}(x)$~is also positive when $x$~is large (positively or negatively), and +a glance at a figure will show that these results are contradictory. Hence +$f_{2k+2}(x) = 0$ has no real roots.] + +\Item{40.} Prove that if $a$~and~$b$ are positive and nearly equal then +\[ +\log \frac{a}{b} = \frac{1}{2}(a - b) \left(\frac{1}{a} + \frac{1}{b}\right), +\] +approximately, the error being about $\frac{1}{6}\{(a - b)/a\}^{3}$. [Use the logarithmic +series. This formula is interesting historically as having been employed by +Napier for the numerical calculation of logarithms.] + +\Item{41.} Prove by multiplication of series that if $-1 < x < 1$ then +\begin{align*} +\tfrac{1}{2}\{\log(1 + x)\}^{2} + &= \tfrac{1}{2} x^{2} + - \tfrac{1}{3}(1 + \tfrac{1}{2})x^{3} + + \tfrac{1}{4}(1 + \tfrac{1}{2} + \tfrac{1}{3})x^{4} - \dots,\\ +\tfrac{1}{2}(\arctan x)^{2} + &= \tfrac{1}{2} x^{2} + - \tfrac{1}{4}(1 + \tfrac{1}{3})x^{4} + + \tfrac{1}{6}(1 + \tfrac{1}{3} + \tfrac{1}{5})x^{6} - \dots. +\end{align*} + +\Item{42.} Prove that +\[ +(1 + \alpha x)^{1/x} + = e^{\alpha}\{1 - \tfrac{1}{2} a^{2}x + + \tfrac{1}{24}(8 + 3a)a^{3}x^{2}(1 + \epsilon_{x})\}, +\] +where $\epsilon_{x} \to 0$ with~$x$. + +\Item{43.} The first $n + 2$ terms in the expansion of $\log\left(1 + x + \dfrac{x^{2}}{2!} + \dots + \dfrac{x^{n}}{n!}\right)$ in +powers of~$x$ are +\[ +x - \frac{x^{n+1}}{n!} + \left\{\frac{1}{n + 1} + - \frac{x}{1!\, (n + 2)} + + \frac{x^{2}}{2!\, (n + 3)} - \dots + + (-1)^{n} \frac{x^{n}}{n!\, (2n + 1)} + \right\}. +\] +\MathTrip{1899.} +\PageSep{393} + +\Item{44.} Show that the expansion of +\[ +\exp \left(-x - \frac{x^{2}}{2} - \dots - \frac{x^{n}}{n}\right) +\] +in powers of~$x$ begins with the terms +\[ +1 - x + \frac{x^{n+1}}{n + 1} + - \sum_{s=1}^{n} \frac{x^{n+s+1}}{(n + s)(n + s + 1)}. +\] +\MathTrip{1909.} + +\Item{45.} Show that if $-1 < x < 1$ then +\begin{align*} +\frac{1}{3}x + \frac{1·4}{3·6}2^{2}x^{2} + \frac{1·4·7}{3·6·9}3^{2}x^{3} + \dots + &= \frac{x(x + 3)}{9(1 - x)^{7/3}},\\ +\frac{1}{3}x + \frac{1·4}{3·6}2^{3}x^{2} + \frac{1·4·7}{3·6·9}3^{3}x^{3} + \dots + &= \frac{x(x^{2} + 18x + 9)}{27(1 - x)^{10/3}}. +\end{align*} + +[Use the method of \Ex{xcii}.~6. The results are more easily obtained by +differentiation; but the problem of the differentiation of an infinite series is +beyond our range.] + +\Item{46.} Prove that +\begin{align*} +\int_{0}^{\infty} \frac{dx}{(x + a)(x + b)} + &= \frac{1}{a - b} \log\left(\frac{a}{b}\right), \\ +\int_{0}^{\infty} \frac{dx}{(x + a)(x + b)^{2}} + &= \frac{1}{(a - b)^{2}b}\left\{a - b - b\log\left(\frac{a}{b}\right)\right\},\\ +\int_{0}^{\infty} \frac{x\, dx}{(x + a)(x + b)^{2}} + &= \frac{1}{(a - b)^{2}} \left\{a\log\left(\frac{a}{b}\right) - a + b\right\},\\ +\int_{0}^{\infty} \frac{dx}{(x + a)(x^{2} + b^{2})} + &= \frac{1}{(a^{2} + b^{2})b} \left\{\tfrac{1}{2}\pi a - b\log\left(\frac{a}{b}\right)\right\},\\ +\int_{0}^{\infty} \frac{x\, dx}{(x + a)(x^{2} + b^{2})} + &= \frac{1}{a^{2} + b^{2}} \left\{\tfrac{1}{2}\pi b + a\log\left(\frac{a}{b}\right)\right\}, +\end{align*} +provided that $a$~and~$b$ are positive. Deduce, and verify independently, that +each of the functions +\[ +a - 1 - \log a,\quad +a\log a - a + 1,\quad +\tfrac{1}{2}\pi a - \log a,\quad +\tfrac{1}{2}\pi + a\log a +\] +is positive for all positive values of~$a$. + +\Item{47.} Prove that if $\alpha$,~$\beta$,~$\gamma$ are all positive, and $\beta^{2} > \alpha\gamma$, then +\[ +\int_{0}^{\infty} \frac{dx}{\alpha x^{2} + 2\beta x + \gamma} + = \frac{1}{\sqrtp{\beta^{2} - \alpha\gamma}} + \log \left\{\frac{\beta + \sqrtp{\beta^{2} - \alpha\gamma}} + {\sqrtp{\alpha\gamma}} + \right\}; +\] +while if $\alpha$~is positive and $\alpha\gamma > \beta^{2}$ the value of the integral is +\[ +\frac{1}{\sqrtp{\alpha\gamma - \beta^{2}}} + \arctan \left\{\frac{\sqrtp{\alpha\gamma - \beta^{2}}}{\beta}\right\}, +\] +that value of the inverse tangent being chosen which lies between $0$ and~$\pi$. +Are there any other really different cases in which the integral is convergent? + +\Item{48.} Prove that if $a > -1$ then +\[ +\int_{1}^{\infty} \frac{dx}{(x + a)\sqrtp{x^{2} - 1}} + = \int_{0}^{\infty} \frac{dt}{\cosh t + a} + = 2\int_{1}^{\infty}\frac{du}{u^{2} + 2au + 1}; +\] +\PageSep{394} +and deduce that the value of the integral is +\[ +\frac{2}{\sqrtp{1 - a^{2}}} \arctan \bigsqrtp{\frac{1 - a}{1 + a}} +\] +if $-1 < a < 1$, and +\[ +\frac{1}{\sqrtp{a^{2} - 1}} + \log\frac{\sqrtp{a + 1} + \sqrtp{a - 1}} + {\sqrtp{a + 1} - \sqrtp{a - 1}} + = \frac{2}{\sqrtp{a^{2} - 1}} \argtanh \bigsqrtp{\frac{a - 1}{a + 1}} +\] +if $a > 1$. Discuss the case in which $a = 1$. + +\Item{49.} Transform the integral $\ds\int_{0}^{\infty} \frac{dx}{(x + a) \sqrtp{x^{2} + 1}}$, where $a > 0$, in the same +ways, showing that its value is +\[ +\frac{1}{\sqrtp{a^{2} + 1}} + \log\frac{a + 1 + \sqrtp{a^{2} + 1}}{a + 1 - \sqrtp{a^{2} + 1}} + = \frac{2}{\sqrtp{a^{2} + 1}} \argtanh \frac{\sqrtp{a^{2} + 1}}{a + 1}\Add{.} +\] + +\Item{50.} Prove that +\[ +\int_{0}^{1} \arctan x\, dx = \tfrac{1}{4}\pi - \tfrac{1}{2}\log 2. +\] + +\Item{51.} If $0 < \alpha < 1$, $0 < \beta < 1$, then +\[ +\int_{-1}^{1} \frac{dx}{\sqrtb{(1 - 2\alpha x + \alpha^{2})(1 - 2\beta x + \beta^{2})}} + = \frac{1}{\sqrtp{\alpha\beta}} + \log \frac{1 + \sqrtp{\alpha\beta}}{1 - \sqrtp{\alpha\beta}}. +\] + +\Item{52.} Prove that if $a > b > 0$ then +\[ +\int_{-\infty}^{\infty} \frac{d\theta}{a\cosh \theta + b\sinh \theta} + = \frac{\pi}{\sqrtp{a^{2} - b^{2}}}\Add{.} +\] + +\Item{53.} Prove that +\[ +\int_{0}^{1} \frac{\log x}{1 + x^{2}}\, dx + = -\int_{1}^{\infty} \frac{\log x}{1 + x^{2}}\, dx,\quad +\int_{0}^{\infty} \frac{\log x}{1 + x^{2}}\, dx = 0\Add{,} +\] +and deduce that if $a > 0$ then +\[ +\int_{0}^{\infty} \frac{\log x}{a^{2} + x^{2}}\, dx = \frac{\pi}{2a}\log a. +\] + +[Use the substitutions $x = 1/t$ and $x = au$.] + +\Item{54.} Prove that +\[ +%[** TN: In-line in the original] +\int_{0}^{\infty} \log \left(1 + \frac{a^{2}}{x^{2}}\right) dx = \pi a +\] +if $a > 0$. [Integrate by parts.] +\end{Examples} +\PageSep{395} + + +\Chapter{X}{THE GENERAL THEORY OF THE LOGARITHMIC, EXPONENTIAL, +AND CIRCULAR FUNCTIONS} + +\Paragraph{217. Functions of a complex variable.} In \Ref{Ch.}{III} we +defined the complex variable +\[ +z = x + iy,\footnotemark +\] +\footnotetext{In this chapter we shall generally find it convenient to write $x + iy$ rather + than $x + yi$.}% +and we considered a few simple properties of some classes of +expressions involving~$z$, such as the polynomial~$P(z)$. It is +natural to describe such expressions as \emph{functions} of~$z$, and in +fact we did describe the quotient $P(z)/Q(z)$, where $P(z)$ and~$Q(z)$ +are polynomials, as a `rational function'. We have however given +no general definition of what is meant by a function of~$z$. + +It might seem natural to define a function of~$z$ in the same +way as that in which we defined a function of the real variable~$x$, +\ie\ to say that $Z$~is a function of~$z$ if any relation subsists +between $z$ and~$Z$ in virtue of which a value or values of~$Z$ corresponds +to some or all values of~$z$. But it will be found, on closer +examination, that this definition is not one from which any profit +can be derived. For if $z$~is given, so are $x$~and~$y$, and conversely: +to assign a value of~$z$ is precisely the same thing as to assign a +pair of values of $x$~and~$y$. Thus a `function of~$z$', according to +the definition suggested, is precisely the same thing as \emph{a complex +function +\[ +f(x, y) + ig(x, y), +\] +of the two real variables $x$~and~$y$}. For example +\[ +x - iy,\quad +xy,\quad +|z| = \sqrtp{x^{2} + y^{2}},\quad +\am z = \arctan(y/x) +\] +are `functions of~$z$'. The definition, although perfectly legitimate, +\PageSep{396} +is futile because it does not really define a new idea at all. It is +therefore more convenient to use the expression `function of the +complex variable~$z$' in a more restricted sense, or in other words +to pick out, from the general class of complex functions of the +two real variables $x$~and~$y$, a special class to which the expression +shall be restricted. But if we were to attempt to explain how +this selection is made, and what are the characteristic properties +of the special class of functions selected, we should be led far +beyond the limits of this book. We shall therefore not attempt +to give any general definitions, but shall confine ourselves entirely +to special functions defined directly. + +\Paragraph{218.} We have already defined \emph{polynomials} in~$z$ (\SecNo[§]{39}), +\emph{rational functions} of~$z$ (\SecNo[§]{46}), and \emph{roots} of~$z$ (\SecNo[§]{47}). There is +no difficulty in extending to the complex variable the definitions +of \emph{algebraical functions}, explicit and implicit, which we gave +(\SecNo[§§]{26}--\SecNo{27}) in the case of the real variable~$x$. In all these cases +we shall call the complex number~$z$, the argument (\SecNo[§]{44}) of the +point~$z$, the \emph{argument} of the function~$f(z)$ under consideration. +The question which will occupy us in this chapter is that of defining +and determining the principal properties of the logarithmic, exponential, +and trigonometrical or circular functions of~$z$. These +functions are of course so far defined for real values of~$z$ only, the +logarithm indeed for positive values only. + +We shall begin with the logarithmic function. It is natural +to attempt to define it by means of some extension of the definition +\[ +\log x = \int_{1}^{x} \frac{dt}{t}\quad (x > 0); +\] +and in order to do this we shall find it necessary to consider +briefly some extensions of the notion of an integral. + +\Paragraph{219. Real and complex curvilinear integrals.} Let $AB$ +be an arc~$C$ of a curve defined by the equations +\[ +x = \phi(t),\quad +y = \psi(t), +\] +where $\phi$ and~$\psi$ are functions of~$t$ with continuous differential +coefficients $\phi'$ and~$\psi'$; and suppose that, as $t$~varies from $t_{0}$ to~$t_{1}$, +the point~$(x, y)$ moves along the curve, in the same direction, from +$A$ to~$B$. +\PageSep{397} + +Then we define the \emph{curvilinear integral} +\[ +\int_{C} \{g(x, y)\, dx + h(x, y)\, dy\}, +\Tag{(1)} +\] +{\Loosen where $g$ and~$h$ are continuous functions of $x$~and~$y$, as being equivalent +to the ordinary integral obtained by effecting the formal +substitutions $x = \phi(t)$, $y = \psi(t)$, \ie\ to} +\[ +\int_{t_{0}}^{t_{1}} \{g(\phi, \psi) \phi' + h(\phi, \psi) \psi'\}\, dt. +\] +We call $C$ the \emph{path of integration}. + +Let us suppose now that +\[ +z = x + iy = \phi(t) + i\psi(t), +\] +so that $z$~describes the curve~$C$ in Argand's diagram as $t$~varies. +Further let us suppose that +\[ +f(z) = u + iv +\] +is a polynomial in~$z$ or rational function of~$z$. + +Then we define +\[ +\int_{C} f(z)\, dz +\Tag{(2)} +\] +as meaning +\[ +\int_{C} (u + iv) (dx + i\, dy), +\] +which is itself defined as meaning +\[ +\int_{C} (u\, dx - v\, dy) + i\int_{C} (v\, dx + u\, dy). +\] + +\Paragraph{220. The definition of $\Log \zeta$.} Now let $\zeta = \xi + i\eta$ be any +complex number. We define~$\Log \zeta$, the general logarithm of~$\zeta$, +by the equation +\[ +\Log \zeta = \int_{C} \frac{dz}{z}, +\] +where $C$~is a curve which starts from~$1$ and ends at~$\zeta$ and does +not pass through the origin. Thus (\Fig{54}) the paths (\ia),~(\ib),~(\ic) +are paths such as are contemplated in the definition. The value +of~$\Log z$ is thus defined when the particular path of integration +has been chosen. But at present it is not clear how far the value +of~$\Log z$ resulting from the definition depends upon what path is +chosen. Suppose for example that $\zeta$~is real and positive, say +\PageSep{398} +equal to~$\xi$. Then one possible path of integration is the straight +line from $1$ to~$\xi$, a path which we may suppose to be defined by +%[Illustration: Fig. 54.] +\Figure[3in]{54}{p398} +the equations $x = t$, $y = 0$. In this case, and with this particular +choice of the path of integration, we have +\[ +\Log \xi = \int_{1}^{\xi} \frac{dt}{t}, +\] +so that $\Log \xi$~is equal to~$\log \xi$, the logarithm of~$\xi$ according to the +definition given in the last chapter. Thus one value at any rate +of~$\Log \xi$, when $\xi$~is real and positive, is~$\log \xi$. But in this case, as +in the general case, the path of integration can be chosen in an +infinite variety of different ways. There is nothing to show that +\emph{every} value of~$\Log \xi$ is equal to~$\log \xi$; and in point of fact we +shall see that this is not the case. This is why we have adopted +the notation $\Log \zeta$,~$\Log \xi$ instead of $\log \zeta$,~$\log \xi$. $\Log \xi$~is (possibly +at any rate) a many valued function, and $\log \xi$~is only one of its +values. And in the general case, so far as we can see at present, +three alternatives are equally possible, viz.\ that + +\Item{(1)} \Hang[4em] we may always get the same value of~$\Log \zeta$, by whatever +path we go from $1$ to~$\zeta$; + +\Item{(2)} \Hang[4em] we may get a different value corresponding to every +different path; + +\Item{(3)} \Hang[4em] we may get a number of different values each of which +corresponds to a whole class of paths: + +\noindent and the truth or falsehood of any one of these alternatives is in +no way implied by our definition. +\PageSep{399} + +\Paragraph{221. The values of $\Log \zeta$.} Let us suppose that the polar +coordinates of the point $z = \zeta$ are $\rho$~and~$\phi$, so that +\[ +\zeta = \rho(\cos\phi + i\sin\phi). +\] +We suppose for the present that $-\pi < \phi < \pi$, while $\rho$~may have +any positive value. Thus $\zeta$~may have any value other than zero +or a real negative value. + +The coordinates $(x, y)$ of any point on the path~$C$ are functions +of~$t$, and so also are its polar coordinates~$(r, \theta)$. Also +\begin{align*} +\Log \zeta + &= \int_{C} \frac{dz}{z} + = \int_{C} \frac{dx + i\, dy}{x + iy} \\ + &= \int_{t_{0}}^{t_{1}} \frac{1}{x + iy} \left(\frac{dx}{dt} + i\frac{dy}{dt}\right) dt, +\end{align*} +in virtue of the definitions of \SecNo[§]{219}. But $x = r\cos\theta$, $y = r\sin\theta$, and +\begin{align*} +\frac{dx}{dt} + i\frac{dy}{dt} + &= \left(\cos\theta\, \frac{dr}{dt} - r\sin\theta\, \frac{d\theta}{dt}\right) + + i\left(\sin\theta\, \frac{dr}{dt} + r\cos\theta\, \frac{d\theta}{dt}\right) +\\ + &= (\cos\theta + i\sin\theta) \left(\frac{dr}{dt} + ir\frac{d\theta}{dt}\right); +\end{align*} +so that +\[ +\Log \zeta + = \int_{t_{0}}^{t_{1}} \frac{1}{r}\, \frac{dr}{dt}\, dt + + i\int_{t_{0}}^{t_{1}} \frac{d\theta}{dt}\, dt + = [\log r] + i[\theta], +\] +where $[\log r]$~denotes the difference between the values of~$\log r$ at +the points corresponding to $t = t_{1}$ and $t = t_{0}$, and $[\theta]$~has a similar +meaning. + +It is clear that +\[ +[\log r] = \log \rho - \log 1 = \log \rho; +\] +but the value of~$[\theta]$ requires a little more consideration. Let us +suppose first that the path of integration is the straight line from +$1$ to~$\zeta$. The initial value of~$\theta$ is the amplitude of~$1$, or rather +%[Illustration: Fig. 55.] +\Figure[1.75in]{55}{p399} +one of the amplitudes of~$1$, viz.\ +$2k\pi$, where $k$~is any integer. Let +us suppose that initially $\theta = 2k\pi$. +It is evident from the figure that +$\theta$~increases from $2k\pi$ to~$2k\pi + \phi$ +as $t$~moves along the line. Thus +\[ +[\theta] = (2k\pi + \phi) - 2k\pi = \phi, +\] +and, when the path of integration +is a straight line, $\Log \zeta = \log \rho + i\phi$. +\PageSep{400} + +We shall call this particular value of~$\Log \zeta$ the \Emph{principal +value}. When $\zeta$~is real and positive, $\zeta = \rho$ and $\phi = 0$, so that the +principal value of~$\Log \zeta$ is the ordinary logarithm~$\log \zeta$. Hence it +will be convenient in general to denote the principal value of~$\Log \zeta$ +by~$\log \zeta$. Thus +\[ +\log \zeta = \log \rho + i\phi, +\] +and the principal value is characterised by the fact that its +imaginary part lies between $-\pi$ and~$\pi$. + +Next let us consider any path (such as those shown in \Fig{56}) +such that the area or areas included +%[Illustration: Fig. 56.] +\Figure[2.75in]{56}{p400a} +between the path and the straight +line from~$1$ to~$\zeta$ does not include +the origin. It is easy to see that +$[\theta]$~is still equal to~$\phi$. Along the +curve shown in the figure by a +continuous line, for example, $\theta$, +initially equal to~$2k\pi$, first decreases +to the value +\[ +2k\pi - XOP +\] +and then increases again, being equal to~$2k\pi$ at~$Q$, and finally +to~$2k\pi + \phi$. The dotted curve shows a similar but slightly more +complicated case in which the straight line and the curve bound +two areas, neither of which includes the origin. Thus \begin{Result}if the path +of integration is such that the closed curve formed by it and the +line from~$1$ to~$\zeta$ does not include the origin, then +\[ +\Log \zeta = \log \zeta = \log \rho + i\phi. +\] +\end{Result} + +On the other hand it is easy +to construct paths of integration +such that $[\theta]$~is not equal to~$\phi$. +Consider, for example, the curve +indicated by a continuous line in +\Fig{57}. If $\theta$~is initially equal +to~$2k\pi$, it will have increased +%[Illustration: Fig. 57.] +\Figure[2.75in]{57}{p400b} +by~$2\pi$ when we get to~$P$ and +by~$4\pi$ when we get to~$Q$; and its +final value will be~$2k\pi + 4\pi + \phi$, +so that $[\theta] = 4\pi + \phi$ and +\[ +\Log \zeta = \log \rho + i(4\pi + \phi). +\] +\PageSep{401} + +In this case the path of integration winds twice round the +origin in the positive sense. If we had taken a path winding +$k$~times round the origin we should have found, in a precisely +similar way, that $[\theta] = 2k\pi+ \phi$ and +\[ +\Log \zeta = \log \rho + i(2k\pi + \phi). +\] +Here $k$~is positive. By making the path wind round the origin +in the opposite direction (as shown in the dotted path in \Fig{57}), +we obtain a similar series of values in which $k$~is negative. +Since $|\zeta | = \rho$, and the different angles~$2k\pi + \phi$ are the different +values of~$\am \zeta$, we conclude that every value of~$\log |\zeta| + i\am \zeta$ is +a value of~$\Log \zeta$; and it is clear from the preceding discussion +that every value of~$\Log \zeta$ must be of this form. + +We may summarise our conclusions as follows: \begin{Result}the general +value of $\Log \zeta$ is +\[ +\log |\zeta| + i\am \zeta = \log \rho + i(2k\pi + \phi), +\] +where $k$~is any positive or negative integer. The value of~$k$ is +determined by the path of integration chosen. If this path is a +straight line then $k = 0$ and\PageLabel{401} +\[ +\Log \zeta = \log \zeta = \log \rho + i\phi. +\] +\end{Result} + +In what precedes we have used~$\zeta$ to denote the argument of +the function~$\Log \zeta$, and $(\xi, \eta)$ or $(\rho, \phi)$ to denote the coordinates of~$\zeta$; +and $z$, $(x, y)$, $(r, \theta)$ to denote an arbitrary point on the path of +integration and its coordinates. There is however no reason now +why we should not revert to the natural notation in which $z$~is used +as the argument of the function~$\Log z$, and we shall do this in +the following examples. + +\begin{Examples}{XCIII.} +\Item{1.} We supposed above that $-\pi < \theta < \pi$, and so +excluded the case in which $z$~is \emph{real and negative}. In this case the straight +line from~$1$ to~$z$ passes through~$0$, and is therefore not admissible as a path of +integration. Both $\pi$ and~$-\pi$ are values of~$\am z$, and $\theta$~is equal to one or +other of them: also $r = -z$. The values of~$\Log z$ are still the values +of~$\log |z| + i\am z$, viz.\ +\[ +\log (-z) + (2k + 1)\pi i, +\] +where $k$~is an integer. The values~$\log (-z) + \pi i$ and~$\log (-z) - \pi i$ correspond +to paths from~$1$ to~$z$ lying respectively entirely above and entirely below the +real axis. Either of them may be taken as the principal value of~$\Log z$, as +convenience dictates. We shall choose the value~$\log (-z) + \pi$ i corresponding +to the first path. +\PageSep{402} + +\Item{2.} The real and imaginary parts of any value of~$\Log z$ are both continuous +functions of $x$~and~$y$, except for $x = 0$, $y = 0$. + +\Item{3.} \Topic{The functional equation satisfied by~$\Log z$.} The function~$\Log z$ +satisfies the equation +\[ +\Log z_{1} z_{2} = \Log z_{1} + \Log z_{2}, +\Tag{(1)} +\] +in the sense that \emph{every} value of either side of this equation is \emph{one} of the values +of the other side. This follows at once by putting +\[ +z_{1} = r_{1}(\cos\theta_{1} + i\sin\theta_{1}),\quad +z_{2} = r_{2}(\cos\theta_{2} + i\sin\theta_{2}), +\] +and applying the formula of \PageRef{p.}{401}. It is however not true that +\[ +\log z_{1}z_{2} = \log z_{1} + \log z_{2} +\Tag{(2)} +\] +in all circumstances. If, \eg, +\[ +z_{1} = z_{2} = \tfrac{1}{2}(-1 + i\sqrt{3}) + = \cos \tfrac{2}{3}\pi + i \sin \tfrac{2}{3}\pi, +\] +then $\log z_{1} = \log z_{2} = \frac{2}{3}\pi i$, and $\log z_{1} + \log z_{2} = \frac{4}{3}\pi i$, which is one of the values of +$\Log z_{1}z_{2}$, but not the principal value. In fact $\log z_{1}z_{2} = -\frac{2}{3}\pi i$. + +An equation such as~\Eq{(1)}, in which every value of either side is a value +of the other, we shall call a \emph{complete} equation, or an equation which is +\emph{completely true}. + +\Item{4.} The equation $\Log z^{m} = m\Log z$, where $m$~is an integer, is not completely +true: every value of the right-hand side is a value of the left-hand side, but +the converse is not true. + +\Item{5.} The equation $\Log (1/z) = -\Log z$ is completely true. It is also true +that $\log (1/z) = -\log z$, except when $z$~is real and negative. + +\Item{6.} The equation +\[ +\log \left(\frac{z - a}{z - b}\right) = \log (z - a) - \log (z - b) +\] +is true if $z$~lies outside the region bounded by the line joining the points $z = a$, +$z = b$, and lines through these points parallel to~$OX$ and extending to infinity +in the negative direction. + +\Item{7.} The equation +\[ +\log \left(\frac{a - z}{b - z}\right) + = \log \left(1 - \frac{a}{z}\right) - \log \left(1 - \frac{b}{z}\right) +\] +is true if $z$~lies outside the triangle formed by the three points $O$,~$a$,~$b$. + +\Item{8.} Draw the graph of the function $\Imag(\Log x)$ of the real variable~$x$. [The +graph consists of the positive halves of the lines $y = 2k\pi$ and the negative +halves of the lines $y = (2k + 1)\pi$.] + +\Item{9.} The function~$f(x)$ of the real variable~$x$, defined by +\[ +\pi f(x) = p\pi + (q - p)\Imag(\log x), +\] +is equal to~$p$ when $x$~is positive and to~$q$ when $x$~is negative. +\PageSep{403} + +\Item{10.} The function~$f(x)$ defined by +\[ +\pi f(x) = p\pi + (q - p)\Imag\{\log(x - 1)\} + (r - q)\Imag(\log x) +\] +is equal to~$p$ when $x > 1$, to~$q$ when $0 < x < 1$, and to~$r$ when $x < 0$. + +\Item{11.} For what values of~$z$ is (i)~$\log z$ (ii)~any value of~$\Log z$ (\ia)~real or +(\ib)~purely imaginary? + +\Item{12.} If $z = x + iy$ then $\Log\Log z = \log R + i(\Theta + 2k'\pi)$, where +\[ +R^{2} = (\log r)^{2} + (\theta + 2k\pi)^{2} +\] +and $\Theta$~is the least positive angle determined by the equations +\[ +\cos\Theta : \sin\Theta : 1 :: +\log r : \theta + 2k\pi: \sqrtb{(\log r)^{2} + (\theta + 2k\pi)^{2}}. +\] +Plot roughly the doubly infinite set of values of $\Log\Log(1 + i\sqrt{3})$, indicating +which of them are values of $\log\Log(1 + i \sqrt{3})$ and which of $\Log\log(1 + i\sqrt{3})$. +\end{Examples} + +\Paragraph{222. The exponential function.} In \Ref{Ch.}{IX} we defined +a function~$e^{y}$ of the real variable~$y$ as the inverse of the function +$y = \log x$. It is naturally suggested that we should define a function +of the complex variable~$z$ which is the inverse of the function~$\Log z$. + +\begin{Definition} +If any value of~$\Log z$ is equal to~$\zeta$, we call $z$ the +exponential of~$\zeta$ and write +\[ +z = \exp \zeta. +\] +\end{Definition} + +Thus $z = \exp \zeta$ if $\zeta = \Log z$. It is certain that to any given +value of~$z$ correspond infinitely many different values of~$\zeta$. It +would not be unnatural to suppose that, conversely, to any given +value of~$\zeta$ correspond infinitely many values of~$z$, or in other words +that $\exp \zeta$~is an infinitely many-valued function of~$\zeta$. This is +however not the case, as is proved by the following theorem. + +\begin{Theorem} +The exponential function $\exp \zeta$ is a one-valued +function of~$\zeta$. +\end{Theorem} + +For suppose that +\[ +z_{1} = r_{1}(\cos\theta_{1} + i\sin\theta_{1}),\quad +z_{2} = r_{2}(\cos\theta_{2} + i\sin\theta_{2}) +\] +are both values of~$\exp \zeta$. Then +\[ +\zeta = \Log z_{1} = \Log z_{2}, +\] +and so +\[ +\log r_{1} + i(\theta_{1} + 2m\pi) = \log r_{2} + i(\theta_{2} + 2n\pi), +\] +where $m$~and~$n$ are integers. This involves +\[ +\log r_{1} = \log r_{2},\quad +\theta_{1} + 2m\pi = \theta_{2} + 2n\pi. +\] +Thus $r_1 = r_2$, and $\theta_{1}$~and~$\theta_{2}$ differ by a multiple of~$2\pi$. Hence +$z_{1} = z_{2}$. +\PageSep{404} + +\begin{Corollary} +If $\zeta$~is real then $\exp \zeta = e^{\zeta}$, the real exponential +function of~$\zeta$ defined in \Ref{Ch.}{IX}\@. +\end{Corollary} + +For if $z = e^{\zeta}$ then $\log z = \zeta$, \ie\ one of the values of~$\Log z$ is~$\zeta$. +Hence $z = \exp \zeta$. + +\Paragraph{223. The value of $\exp \zeta$.} Let $\zeta = \xi + i\eta$ and +\[ +z = \exp \zeta = r(\cos\theta + i\sin\theta). +\] +Then +\[ +\xi + i\eta = \Log z = \log r + i(\theta + 2m\pi), +\] +where $m$~is an integer. Hence $\xi = \log r$, $\eta = \theta + 2m\pi$, or +\[ +r = e^{\xi},\quad +\theta = \eta - 2m\pi; +\] +and accordingly +\[ +\exp (\xi + i\eta) = e^{\xi} (\cos\eta + i\sin\eta). +\] + +If $\eta = 0$ then $\exp \xi = e^{\xi}$, as we have already inferred in \SecNo[§]{222}. +It is clear that both the real and the imaginary parts of $\exp (\xi + i\eta)$ +are continuous functions of $\xi$~and~$\eta$ for all values of $\xi$~and~$\eta$. + +{\Loosen\Paragraph{224. The functional equation satisfied by $\exp \zeta$.} Let +$\zeta_{1} = \xi_{1} + i\eta_{1}$, $\zeta_{2} = \xi_{2} + i\eta_{2}$. Then} +\begin{align*} +\exp \zeta_{1} × \exp \zeta_{2} + &= e^{\xi_{1}} (\cos\eta_{1} + i\sin\eta_{1}) + × e^{\xi_{2}} (\cos\eta_{2} + i\sin\eta_{2}) \\ + &= e^{\xi_{1}+\xi_{2}} \{\cos(\eta_{1} + \eta_{2}) + i\sin(\eta_{1} + \eta_{2})\} \\ + &= \exp(\zeta_{1} + \zeta_{2}). +\end{align*} +{\Loosen The exponential function therefore satisfies the functional relation +$f(\zeta_{1} + \zeta_{2}) = f(\zeta_{1}) f(\zeta_{2})$, an equation which we have proved already +(\SecNo[§]{205}) to be true for real values of $\zeta_{1}$~and~$\zeta_{2}$.} + +\Paragraph{225. The general power~$a^{\zeta}$.} It might seem natural, as +$\exp \zeta = e^{\zeta}$ when $\zeta$~is real, to adopt the same notation when $\zeta$~is +complex and to drop the notation $\exp \zeta$ altogether. We shall not +follow this course because we shall have to give a more general +definition of the meaning of the symbol~$e^{\zeta}$: we shall find then +that $e^{\zeta}$~represents a function with infinitely many values of which +$\exp \zeta$~is only one. + +We have already defined the meaning of the symbol~$a^{\zeta}$ in a +considerable variety of cases. It is defined in elementary Algebra +in the case in which $a$~is real and positive and $\zeta$~rational, or $a$~real +and negative and $\zeta$~a rational fraction whose denominator is odd. +According to the definitions there given $a^{\zeta}$~has at most two values. +\PageSep{405} +In \Ref{Ch.}{III} we extended our definitions to cover the case in which +$a$~is any real or complex number and $\zeta$~any rational number~$p/q$; +and in \Ref{Ch.}{IX} we gave a new definition, expressed by the equation +\[ +a^{\zeta} = e^{\zeta\log a}, +\] +which applies whenever $\zeta$~is real and $a$~real and positive. + +Thus we have, in one way or another, attached a meaning to +such expressions as +\[ +3^{1/2},\quad +(-1)^{1/3},\quad +(\sqrt{3} + \tfrac{1}{2}i)^{-1/2},\quad +(3.5)^{1+\sqrt{2}}; +\] +but we have as yet given no definitions which enable us to attach +any meaning to such expressions as +\[ +(1 + i)^{\sqrt{2}},\quad +2^{i},\quad +(3 + 2i)^{2+3i}. +\] +We shall now give a general definition of~$a^{\zeta}$ which applies to all +values of $a$ and~$\zeta$, real or complex, with the one limitation that +$a$~must not be equal to zero. + +\begin{Definition} +The function~$a^{\zeta}$ is defined by the equation +\[ +a^{\zeta} = \exp (\zeta\Log a) +\] +where $\Log a$~is any value of the logarithm of~$a$. +\end{Definition} + +We must first satisfy ourselves that this definition is consistent +with the previous definitions and includes them all as particular +cases. + +\Item{(1)} If $a$~is positive and $\zeta$~real, then one value of~$\zeta\Log a$, viz.\ +$\zeta\log a$, is real: and $\exp (\zeta\log a) = e^{\zeta\log a}$, which agrees with the +definition adopted in \Ref{Ch.}{IX}\@. The definition of \Ref{Ch.}{IX} is, as +we saw then, consistent with the definition given in elementary +Algebra; and so our new definition is so too. + +\Item{(2)} If $a = e^{\tau} (\cos\psi + i\sin\psi)$, then +\begin{gather*} +\Log a = \tau + i(\psi + 2m\pi), \\ +\exp \{(p/q)\Log a\} = e^{p\tau/q} \Cis \{(p/q)(\psi + 2m\pi)\}, +\end{gather*} +where $m$~may have any integral value. It is easy to see that if $m$~assumes +all possible integral values then this expression assumes $q$ +and only~$q$ different values, which are precisely the values of~$a^{p/q}$ +found in \SecNo[§]{48}. Hence our new definition is also consistent with +that of \Ref{Ch.}{III}\@. +\PageSep{406} + +\Paragraph{226. The general value of~$a^{\zeta}$.} Let +\[ +\zeta = \xi + i\eta,\quad +a = \sigma(\cos\psi + i\sin\psi) +\] +where $-\pi < \psi \leq \pi$, so that, in the notation of \SecNo[§]{225}, $\sigma = e^{\tau}$ or +$\tau = \log \sigma$. + +Then +\[ +\zeta \Log a = (\xi + i\eta)\{\log \sigma + i(\psi + 2m\pi)\} = L + iM, +\] +where +\[ +L = \xi \log \sigma - \eta(\psi + 2m\pi),\quad +M = \eta\log \sigma + \xi (\psi + 2m\pi); +\] +and +\[ +a^{\zeta} = \exp(\zeta\Log a) = e^{L}(\cos M + i\sin M). +\] +Thus the general value of~$a^{\zeta}$ is +\[ +e^{\xi\log \sigma - \eta(\psi+2m\pi)} + [\cos\{\eta\log \sigma + \xi(\psi + 2m\pi)\} + + i\sin\{\eta\log \sigma + \xi(\psi + 2m\pi)\}]. +\] + +In general $a^{\zeta}$~is an infinitely many-valued function. For +\[ +|a^{\zeta}| = e^{\xi\log \sigma - \eta(\psi+2m\pi)} +\] +has a different value for every value of~$m$, unless $\eta = 0$. If on the +other hand $\eta = 0$, then the moduli of all the different values of~$a^{\zeta}$ +are the same. But any two values differ unless their amplitudes +are the same or differ by a multiple of~$2\pi$. This requires that +$\xi(\psi + 2m\pi)$ and $\xi(\psi + 2n\pi)$, where $m$~and~$n$ are different integers, +shall differ, if at all, by a multiple of~$2\pi$. But if +\[ +\xi(\psi + 2m\pi) - \xi(\psi + 2n\pi) = 2k\pi, +\] +then $\xi = k/(m - n)$ is rational. We conclude that \emph{$a^{\zeta}$~is infinitely +many-valued unless $\zeta$~is real and rational}. On the other hand we +have already seen that, when $\zeta$~is real and rational, $a^{\zeta}$~has but a +finite number of values. + +\begin{Remark} +The \emph{principal value} of $a^{\zeta} = \exp (\zeta\Log a)$ is obtained by giving $\Log a$ its +principal value, \ie\ by supposing $m = 0$ in the general formula. Thus the +principal value of~$a^{\zeta}$ is +\[ +e^{\xi\log \sigma - \eta\psi} + \{\cos(\eta\log \sigma + \xi\psi) + i\sin(\eta\log \sigma + \xi\psi)\}. +\] + +Two particular cases are of especial interest. If $a$~is real and positive +and $\zeta$~real, then $\sigma = a$, $\psi = 0$, $\xi = \zeta$, $\eta = 0$, and the principal value of~$a^{\zeta}$ is~$e^{\zeta\log a}$, +which is the value defined in the last chapter. If $|a| = 1$ and $\zeta$~is +real, then $\sigma = 1$, $\xi = \zeta$, $\eta = 0$, and the principal value of $(\cos\psi + i\sin\psi)^{\zeta}$ is +$\cos\zeta\psi + i\sin\zeta\psi$. This is a further generalisation of De~Moivre's Theorem +(\SecNo[§§]{45},~\SecNo{49}). +\end{Remark} +\PageSep{407} + +\begin{Examples}{XCIV.} +\Item{1.} Find all the values of~$i^{i}$. [By definition +\[ +i^{i} = \exp (i\Log i). +\] +But +\[ +i = \cos \tfrac{1}{2}\pi + i\sin \tfrac{1}{2}\pi,\quad +\Log i = (2k + \tfrac{1}{2})\pi i, +\] +where $k$~is any integer. Hence +\[ +i^{i} = \exp\{-(2k + \tfrac{1}{2})\pi\} = e^{-(2k + \frac{1}{2})\pi}. +\] +All the values of~$i^{i}$ are therefore real and positive.] + +\Item{2.} Find all the values of $(1 + i)^{i}$, $i^{1+i}$, $(1 + i)^{1+i}$. + +\Item{3.} The values of~$a^{\zeta}$, when plotted in the Argand diagram, are the vertices +of an equiangular polygon inscribed in an equiangular spiral whose angle is +independent of~$a$. \MathTrip{1899.} + +[If $a^{\zeta} = r(\cos\theta + i\sin\theta)$ we have +\[ +r = e^{\xi\log \sigma - \eta(\psi + 2m\pi)},\quad +\theta = \eta\log \sigma + \xi(\psi + 2m\pi); +\] +and all the points lie on the spiral $r = \sigma^{(\xi^{2} + \eta^{2})/\xi} e^{-\eta \theta/\xi}$.] + +\Item{4.} \Topic{The function~$e^{\zeta}$.} If we write~$e$ for~$a$ in the general formula, so that +$\log \sigma = 1$, $\psi = 0$, we obtain +\[ +e^{\zeta} = e^{\xi-2m\pi\eta} \{\cos(\eta + 2m\pi\xi) + i\sin(\eta + 2m\pi\xi)\}. +\] +The principal value of~$e^{\zeta}$ is $e^{\xi}(\cos\eta + i\sin\eta)$, which is equal to~$\exp \zeta$ (\SecNo[§]{223}). +In particular, if $\zeta$~is real, so that $\eta = 0$, we obtain +\[ +e^{\zeta} (\cos 2m\pi\zeta + i\sin 2m\pi\zeta) +\] +as the general and $e^{\zeta}$~as the principal value, $e^{\zeta}$~denoting here the positive +value of the exponential defined in \Ref{Ch.}{IX}\@. + +\Item{5.} Show that $\Log e^{\zeta} = (1 + 2m\pi i)\zeta + 2n\pi i$, where $m$~and~$n$ are any integers, +and that in general $\Log a^{\zeta}$~has a double infinity of values. + +\Item{6.} The equation $1/a^{\zeta} = a^{-\zeta}$ is completely true (\Ex{xciii}.~3): it is also true +of the principal values. + +\Item{7.} The equation $a^{\zeta} × b^{\zeta} = (ab)^{\zeta}$ is completely true but not always true of +the principal values. + +\Item{8.} The equation $a^{\zeta} × a^{\zeta'} = a^{\zeta+\zeta'}$ is not completely true, but is true of the +principal values. [Every value of the right-hand side is a value of the left-hand +side, but the general value of $a^{\zeta} × a^{\zeta'}$, viz. +\[ +\exp \{\zeta(\log a + 2m\pi i) + \zeta'(\log a + 2n\pi i)\}, +\] +is not as a rule a value of~$a^{\zeta+\zeta'}$ unless $m = n$.] + +\Item{9.} What are the corresponding results as regards the equations +\[ +\Log a^{\zeta} = \zeta\Log a,\quad +(a^{\zeta})^{\zeta'} = (a^{\zeta'})^{\zeta} = a^{\zeta\zeta'}? +\] + +\Item{10.} For what values of~$\zeta$ is (\ia)~any value (\ib)~the principal value of~$e^{\zeta}$ +(i)~real (ii)~purely imaginary (iii)~of unit modulus? +\PageSep{408} + +\Item{11.} The necessary and sufficient conditions that all the values of~$a^{\zeta}$ should +be real are that $2\xi$~and~$\{\eta\log |a| + \xi\am a\}/\pi$, where $\am a$~denotes any value of +the amplitude, should both be integral. What are the corresponding conditions +that all the values should be of unit modulus? + +\Item{12.} The general value of~$|x^{i} + x^{-i}|$, where $x > 0$, is +\[ +e^{-(m-n)\pi} \sqrtbr{2\{\cosh 2(m + n)\pi + \cos(2\log x)\}}. +\] + +\Item{13.} Explain the fallacy in the following argument: since $e^{2m\pi i} = e^{2n\pi i} = 1$, +where $m$~and~$n$ are any integers, therefore, raising each side to the power~$i$ +we obtain $e^{-2m\pi} = e^{-2n\pi}$. + +\Item{14.} In what circumstances are any of the values of~$x^{x}$, where $x$~is real, +themselves real? [If $x > 0$ then +\[ +x^{x} = \exp (x\Log x) = \exp (x\log x) \Cis 2m\pi x, +\] +the first factor being real. The principal value, for which $m = 0$, is always +real. + +If $x$~is a rational fraction~$p/(2q + 1)$, or is irrational, then there is no other +real value. But if $x$~is of the form~$p/2q$, then there is one other real value, +viz.\ $-\exp (x\log x)$, given by $m = q$. + +If $x = -\xi < 0$ then +\[ +x^{x} = \exp \{-\xi\Log (-\xi)\} + = \exp (-\xi\log \xi) \Cis\{-(2m + 1)\pi\xi\}. +\] +The only case in which any value is real is that in which $\xi = p/(2q + 1)$, when +$m = q$ gives the real value +\[ +\exp (-\xi\log \xi) \Cis (-p\pi) = (-1)^{p} \xi^{-\xi}. +\] +The cases of reality are illustrated by the examples +\[ +(\tfrac{1}{3})^{1/3} = \sqrt[3]{\tfrac{1}{3}},\quad +(\tfrac{1}{2})^{\frac{1}{2}} = ±\sqrt{\tfrac{1}{2}},\quad +(-\tfrac{2}{3})^{-\frac{2}{3}} = \sqrt[3]{\tfrac{9}{4}},\quad +(-\tfrac{1}{3})^{-\frac{1}{3}} = -\sqrt[3]{3}.] +\] + +\Item{15.} \Topic{Logarithms to any base.} We may define $\zeta = \Log_{a} z$ in two different +ways. We may say (i)~that $\zeta = \Log_{a} z$ if the \emph{principal} value of~$a^{\zeta}$ is equal to~$z$; +or we may say (ii)~that $\zeta = \Log_{a} z$ if \emph{any} value of~$a^{\zeta}$ is equal to~$z$. + +Thus if $a = e$ then $\zeta = \Log_{e} z$, according to the first definition, if the +principal value of~$e^{\zeta}$ is equal to~$z$, or if $\exp \zeta = z$; and so $\Log_{e} z$~is identical +with~$\Log z$. But, according to the second definition, $\zeta = \Log_{e} z$ if +\[ +e^{\zeta} = \exp (\zeta\Log e) = z,\quad +\zeta\Log e = \Log z, +\] +or $\zeta = (\Log z)/(\Log e)$, any values of the logarithms being taken. Thus +\[ +\zeta = \Log_{e} z = \frac{\log |z| + (\am z + 2m\pi)i}{1 + 2n\pi i}, +\] +so that $\zeta$~is a doubly infinitely many-valued function of~$z$. And generally, +according to this definition, $\Log_{a} z = (\Log z)/(\Log a)$. + +\Item{16.} $\Log_{e} 1 = 2m\pi i/(1 + 2n\pi i)$, $\Log_{e}(-1) = (2m + 1)\pi i/(1 + 2n\pi i)$, where $m$~and~$n$ +are any integers. +\end{Examples} +\PageSep{409} + +\Paragraph{227. The exponential values of the sine and cosine.} +From the formula +\[ +\exp (\xi + i\eta) = \exp \xi(\cos\eta + i\sin\eta), +\] +we can deduce a number of extremely important subsidiary +formulae. Taking $\xi = 0$, we obtain $\exp (i\eta) = \cos\eta + i\sin\eta$; and, +changing the sign of~$\eta$, $\exp (-i\eta) = \cos\eta - i\sin\eta$. Hence +\begin{alignat*}{3} +\cos\eta &= &&\tfrac{1}{2} &&\{\exp (i\eta) + \exp (-i\eta)\},\\ +\sin\eta &= -&&\tfrac{1}{2}i&&\{\exp (i\eta) - \exp (-i\eta)\}. +\end{alignat*} +We can of course deduce expressions for any of the trigonometrical +ratios of~$\eta$ in terms of~$\exp (i\eta)$. + +\Paragraph{228. Definition of $\sin\zeta$ and~$\cos\zeta$ for all values of~$\zeta$.} +We saw in the last section that, when $\zeta$~is real, +\begin{alignat*}{3} +\cos\zeta &= &&\tfrac{1}{2} &&\{\exp (i\zeta) + \exp (-i\zeta)\}, +\Tag{(1a)}\\ +\sin\zeta &= -&&\tfrac{1}{2}i&&\{\exp (i\zeta) - \exp (-i\zeta)\}. +\Tag{(1b)} +\end{alignat*} + +The left-hand sides of these equations are defined, by the ordinary +geometrical definitions adopted in elementary Trigonometry, only +for real values of~$\zeta$. The right-hand sides have, on the other +hand, been defined for all values of~$\zeta$, real or complex. We are +therefore naturally led to adopt the formulae~\Eq{(1)} as the \emph{definitions} +of $\cos \zeta$ and~$\sin \zeta$ for all values of~$\zeta$. These definitions agree, in +virtue of the results of \SecNo[§]{227}, with the elementary definitions for +real values of~$\zeta$. + +Having defined $\cos \zeta$ and~$\sin \zeta$, we define the other trigonometrical +ratios by the equations +\[ +\tan \zeta = \frac{\sin \zeta}{\cos \zeta},\quad +\cot \zeta = \frac{\cos \zeta}{\sin \zeta},\quad +\sec \zeta = \frac{1}{\cos \zeta},\quad +\cosec \zeta = \frac{1}{\sin \zeta}. +\Tag{(2)} +\] +It is evident that $\cos \zeta$ and~$\sec \zeta$ are even functions of~$\zeta$, and +$\sin \zeta$, $\tan \zeta$, $\cot \zeta$, and~$\cosec \zeta$ odd functions. Also, if $\exp (i\zeta) = t$, +we have +\begin{gather*} +\cos \zeta = \tfrac{1}{2} \{t + (1/t)\},\quad +\sin \zeta = -\tfrac{1}{2}i \{t - (1/t)\},\\ +\cos^{2} \zeta + \sin^{2} \zeta + = \tfrac{1}{4}[\{t + (1/t)\}^{2} - \{t - (1/t)\}^{2}] = 1. +\Tag{(3)} +\end{gather*} + +We can moreover express the trigonometrical functions of +$\zeta + \zeta'$ in terms of those of $\zeta$~and~$\zeta'$ by precisely the same formulae +\PageSep{410} +as those which hold in elementary trigonometry. For if $\exp (i\zeta) = t$, +$\exp (i\zeta') = t'$, we have +\begin{align*} +%[** TN: Set on two lines in the original, not aligned] +\cos (\zeta + \zeta') + &= \tfrac{1}{2} \left(tt' + \frac{1}{tt'}\right) \\ + &= \tfrac{1}{4} \left\{ + \left(t + \frac{1}{t}\right) \left(t' + \frac{1}{t'}\right) + + \left(t - \frac{1}{t}\right) \left(t' - \frac{1}{t'}\right)\right\}\\ + &= \cos\zeta \cos\zeta' - \sin\zeta \sin\zeta'; +\Tag{(4)} +\end{align*} +and similarly we can prove that +\[ +\sin (\zeta + \zeta') = \sin\zeta \cos\zeta' + \cos\zeta \sin\zeta'. +\Tag{(5)} +\] +In particular +\[ +\cos(\zeta + \tfrac{1}{2}\pi) = -\sin\zeta,\quad +\sin(\zeta + \tfrac{1}{2}\pi) = \cos\zeta. +\Tag{(6)} +\] + +All the ordinary formulae of elementary Trigonometry are +algebraical corollaries of the equations~\Eq{(2)}--\Eq{(6)}; and so all such +relations hold also for the generalised trigonometrical functions +defined in this section. + +\begin{Remark} +\Paragraph{229. The generalised hyperbolic functions.} In \Ex{lxxxvii}.~19, we +defined $\cosh \zeta$ and~$\sinh \zeta$, for real values of~$\zeta$, by the equations +\[ +\cosh\zeta = \tfrac{1}{2} \{\exp \zeta + \exp (-\zeta)\},\quad +\sinh\zeta = \tfrac{1}{2} \{\exp \zeta - \exp (-\zeta)\}. +\Tag{(1)} +\] + +We can now extend this definition to complex values of the variable; +\ie\ we can agree that the equations~\Eq{(1)} are to define $\cosh \zeta$ and~$\sinh \zeta$ for +all values of~$\zeta$ real or complex. The reader will easily verify the following +relations: +\[ +\cos i\zeta = \cosh \zeta,\quad +\sin i\zeta = i\sinh \zeta,\quad +\cosh i\zeta = \cos \zeta,\quad +\sinh i\zeta = i\sin \zeta. +\] + +We have seen that any elementary trigonometrical formula, such as +the formula $\cos 2\zeta = \cos^{2} \zeta - \sin^{2} \zeta$, remains true when $\zeta$~is allowed to assume +complex values. It remains true therefore if we write $\cos i\zeta$ for~$\cos \zeta$, $\sin i\zeta$ +for~$\sin \zeta$ and $\cos 2i\zeta$ for~$\cos 2\zeta$; or, in other words, if we write $\cosh \zeta$ for~$\cos \zeta$, +$i\sinh \zeta$ for~$\sin \zeta$, and $\cosh 2\zeta$ for~$\cos 2\zeta$. Hence +\[ +\cosh 2\zeta = \cosh^{2} \zeta + \sinh^{2} \zeta. +\] +The same process of transformation may be applied to any trigonometrical +identity. It is of course this fact which explains the correspondence noted +in \Ex{lxxxvii}.~21 between the formulae for the hyperbolic and those for the +ordinary trigonometrical functions. + +\Paragraph{230. Formulae for $\cos(\xi + i\eta)$, $\sin(\xi + i\eta)$,~etc.} It follows from the +addition formulae that +\begin{alignat*}{4} +\cos (\xi + i\eta) + &= \cos\xi \cos i\eta &&- \sin\xi \sin i\eta + &&= \cos\xi \cosh \eta &&- i\sin\xi \sinh \eta,\\ +\sin (\xi + i\eta) + &= \sin\xi \cos i\eta &&+ \cos\xi \sin i\eta + &&= \sin\xi \cosh \eta &&+ i\cos\xi \sinh \eta. +\end{alignat*} +These formulae are true for all values of $\xi$ and~$\eta$. The interesting case +is that in which $\xi$~and~$\eta$ are real. They then give expressions for the real and +imaginary parts of the cosine and sine of a complex number. +\end{Remark} +\PageSep{411} + +\begin{Examples}{XCV.} +\Item{1.} Determine the values of~$\zeta$ for which $\cos\zeta$ and~$\sin\zeta$ +are (i)~real (ii)~purely imaginary. [For example $\cos\zeta$~is real when $\eta = 0$ or +when $\xi$~is any multiple of~$\pi$.] + +\Item{2.} +\begin{alignat*}{2} +|\cos (\xi + i\eta)| + &= \sqrtp{\cos^{2} \xi + \sinh^{2} \eta} + &&= \sqrtb{\tfrac{1}{2} (\cosh 2\eta + \cos 2\xi)}, \\ +|\sin (\xi + i\eta)| + &= \sqrtp{\sin^{2} \xi + \sinh^{2} \eta} + &&= \sqrtb{\tfrac{1}{2} (\cosh 2\eta - \cos 2\xi)}. +\end{alignat*} + +[Use (\eg)\ the equation $|\cos(\xi + i\eta)| = \sqrtb{\cos(\xi + i\eta) \cos(\xi - i\eta)}$.] + +\Item{3.} +$\tan (\xi + i \eta) + = \dfrac{\sin 2\xi + i\sinh 2\eta}{\cosh 2\eta + \cos 2\xi}$,\quad +$\cot (\xi + i \eta) + = \dfrac{\sin 2\xi - i\sinh 2\eta}{\cosh 2\eta - \cos 2\xi}$. + +[For example +\[ +\tan (\xi + i\eta) + = \frac{\sin (\xi + i\eta) \cos (\xi - i\eta)} + {\cos (\xi + i\eta) \cos (\xi - i\eta)} + = \frac{\sin 2\xi + \sin 2i\eta}{\cos 2\xi + \cos 2i\eta}, +\] +which leads at once to the result given.] + +\Item{4.} +\begin{align*} +\sec (\xi + i \eta) + &= \frac{\cos\xi \cosh\eta + i\sin\xi \sinh\eta} + {\frac{1}{2} (\cosh 2\eta + \cos 2\xi)}, \\ +\cosec (\xi + i \eta) + &= \frac{\sin\xi \cosh\eta - i\cos\xi \sinh\eta} + {\frac{1}{2} (\cosh 2\eta - \cos 2\xi)}. +\end{align*} + +\Item{5.} If $|\cos (\xi + i\eta)| = 1$ then $\sin^{2} \xi = \sinh^{2} \eta$, and if $|\sin (\xi + i\eta)| = 1$ then +$\cos^{2} \xi = \sinh^{2} \eta$. + +\Item{6.} If $|\cos (\xi + i\eta)| = 1$, then +\[ +\sin \{\am \cos (\xi + i\eta)\} = ±\sin^{2} \xi = ±\sinh^{2} \eta. +\] + +\Item{7.} Prove that $\Log \cos (\xi + i\eta) = A + iB$, where +\[ +A = \tfrac{1}{2} \log \{\tfrac{1}{2} (\cosh 2\eta + \cos 2\xi)\} +\] +and $B$~is any angle such that +\[ +\frac{\cos B}{\cos\xi \cosh\eta} + = -\frac{\sin B}{\sin\xi \sinh\eta} + = \frac{1}{\sqrtb{\frac{1}{2} (\cosh 2\eta + \cos 2\xi)}}. +\] +Find a similar formula for $\Log \sin (\xi + i\eta)$. + +\Item{8.} \Topic{Solution of the equation $\cos\zeta = a$, where $a$~is real.} Putting +$\zeta = \xi + i\eta$, and equating real and imaginary parts, we obtain +\[ +\cos\xi \cosh\eta = a,\quad +\sin\xi \sinh\eta = 0. +\] +Hence either $\eta = 0$ or $\xi$~is a multiple of~$\pi$. If (i)~$\eta = 0$ then $\cos\xi = a$, which is +impossible unless $-1 \leq a \leq 1$. This hypothesis leads to the solution +\[ +\zeta = 2k\pi ± \arccos a, +\] +where $\arccos a$ lies between $0$ and~$\frac{1}{2}\pi$. If (ii)~$\xi = m\pi$ then $\cosh\eta = (-1)^{m}a$, so +that either $a \geq 1$ and $m$~is even, or $a \leq -1$ and $m$~is odd. If $a = ± 1$ then $\eta = 0$, +and we are led back to our first case. If $|a| > 1$ then $\cosh\eta = |a|$, and we +are led to the solutions +\begin{alignat*}{4} +\zeta &=& 2k &\pi ± i\log \{ &&a + \sqrt{a^{2} - 1}\}\quad &&(a > 1), \\ +\zeta &=&(2k + 1) &\pi ± i\log \{-&&a + \sqrt{a^{2} - 1}\}\quad &&(a < -1). +\end{alignat*} +For example, the general solution of $\cos\zeta = -\frac{5}{3}$ is $\zeta = (2k + 1)\pi ± i\log 3$. +\PageSep{412} + +\Item{9.} Solve $\sin\zeta = \alpha$, where $\alpha$~is real. + +\Item{10.} \Topic{Solution of $\cos\zeta = \alpha + i\beta$, where $\beta \neq 0$.} We may suppose $\beta > 0$, +since the results when $\beta < 0$ may be deduced by merely changing the sign of~$i$. +In this case +\[ +\cos\xi \cosh\eta = \alpha,\quad +\sin\xi \sinh\eta = -\beta, +\Tag{(1)} +\] +and +\[ +(\alpha/\cosh\eta)^{2} + (\beta/\sinh\eta)^{2} = 1. +\] + +If we put $\cosh^{2} \eta = x$ we find that +\[ +x^{2} - (1 + \alpha^{2} + \beta^{2})x + \alpha^{2} = 0 +\] +or $x = (A_{1} ± A_{2})^{2}$, where +\[ +A_{1} = \tfrac{1}{2}\sqrtb{(\alpha + 1)^{2} + \beta^{2}},\quad +A_{2} = \tfrac{1}{2}\sqrtb{(\alpha - 1)^{2} + \beta^{2}}. +\] +Suppose $\alpha > 0$. Then $A_{1} > A_{2} > 0$ and $\cosh\eta = A_{1} ± A_{2}$. Also +\[ +\cos\xi = \alpha/(\cosh\eta) = A_{1} \mp A_{2}, +\] +and since $\cosh\eta > \cos\xi$ we must take +\[ +\cosh\eta = A_{1} + A_{2},\quad +\cos\xi = A_{1} - A_{2}. +\] +The general solutions of these equations are +\[ +\xi = 2k\pi ± \arccos M,\quad +\eta = ±\log \{L + \sqrtp{L^{2} - 1}\}, +\Tag{(2)} +\] +where $L = A_{1} + A_{2}$, $M = A_{1} - A_{2}$, and $\arccos M$ lies between $0$ and~$\frac{1}{2}\pi$. + +The values of $\eta$ and~$\xi$ thus found above include, however, the solutions of +the equations +\[ +\cos\xi \cosh\eta = \alpha,\quad +\sin\xi \sinh\eta = \beta, +\Tag{(3)} +\] +as well as those of the equations~\Eq{(1)}, since we have only used the second of +the latter equations after squaring it. To distinguish the two sets of +solutions we observe that the sign of~$\sin\xi$ is the same as the ambiguous sign +in the first of the equations~\Eq{(2)}, and the sign of~$\sinh\eta$ is the same as the +ambiguous sign in the second. Since $\beta > 0$, these two signs must be different. +Hence the general solution required is +\[ +\zeta = 2k\pi ± [\arccos M - i\log \{L + \sqrtp{L^{2} - 1}\}]. +\] + +\Item{11.} Work out the cases in which $\alpha < 0$ and $\alpha = 0$ in the same way. + +\Item{12.} If $\beta = 0$ then $L = \frac{1}{2}|\alpha + 1| + \frac{1}{2}|\alpha - 1|$ and $M = \frac{1}{2}|\alpha + 1| - \frac{1}{2}|\alpha - 1|$. +Verify that the results thus obtained agree with those of Ex.~8. + +\Item{13.} {\Loosen Show that if $\alpha$~and~$\beta$ are positive then the general solution of +$\sin\zeta = \alpha + i\beta$ is} +\[ +\zeta = k\pi +(-1)^{k} [\arcsin M + i\log \{L + \sqrtp{L^{2} - 1}\}], +\] +where $\arcsin M$ lies between $0$ and~$\frac{1}{2}\pi$. Obtain the solution in the other +possible cases. + +\Item{14.} Solve $\tan\zeta = \alpha$, where $\alpha$~is real. [All the roots are real.] +\PageSep{413} + +\Item{15.} Show that the general solution of $\tan \zeta = \alpha + i\beta$, where $\beta \neq 0$, is +\[ +\zeta = k\pi + \tfrac{1}{2}\theta + \tfrac{1}{4} i\log\left\{ + \frac{\alpha^{2} + (1 + \beta)^{2}} + {\alpha^{2} + (1 - \beta)^{2}} + \right\}, +\] +where $\theta$~is the numerically least angle such that +\[ +\cos \theta : \sin \theta : 1 :: +1 - \alpha^{2} - \beta^{2} : 2\alpha : + \sqrtb{(1 - \alpha^{2} - \beta^{2})^{2} + 4\alpha^{2}}. +\] + +\Item{16.} If $z = \xi\exp(\frac{1}{4}\pi i)$, where $\xi$~is real, and $c$~is also real, then the modulus +of $\cos 2\pi z - \cos 2\pi c$ is +\[ +\begin{aligned}[b] + \surd[\tfrac{1}{2}\{1 + \cos 4\pi c + \cos(2\pi\xi\sqrt{2}) &+ \cosh(2\pi\xi\sqrt{2}) \\ + &- 4\cos 2\pi c \cos(\pi\xi\sqrt{2}) \cosh(\pi\xi\sqrt{2})\}] +\end{aligned}. +\] + +\Item{17.} Prove that +\begin{gather*} +|\exp \exp(\xi + i\eta)| = \exp(\exp\xi \cos\eta), \\ +\begin{aligned} +\Real \{\cos\cos(\xi + i\eta)\} + &= \cos(\cos\xi \cosh\eta) \cosh(\sin\xi \sinh\eta),\\ +\Imag \{\sin\sin(\xi + i\eta)\} + &= \cos(\sin\xi \cosh\eta) \sinh(\cos\xi \sinh\eta). +\end{aligned} +\end{gather*} + +\Item{18.} Prove that $|\exp\zeta|$~tends to~$\infty$ if $\zeta$~moves away towards infinity along +any straight line through the origin making an angle less than~$\frac{1}{2}\pi$ with~$OX$, +and to~$0$ if $\zeta$~moves away along a similar line making an angle greater than~$\frac{1}{2}\pi$ +with~$OX$. + +\Item{19.} Prove that $|\cos\zeta|$ and $|\sin\zeta|$ tend to~$\infty$ if $\zeta$~moves away towards +infinity along any straight line through the origin other than either half of +the real axis. + +\Item{20.} Prove that $\tan\zeta$ tends to~$-i$ or to~$i$ if $\zeta$~moves away to infinity +along the straight line of Ex.~19, to $-i$~if the line lies above the real axis and +to~$i$ if it lies below. +\end{Examples} + +\begin{Remark} +\Paragraph{231. The connection between the logarithmic and the inverse +trigonometrical functions.} We found in \Ref{Ch.}{VI} that the integral of a +rational or algebraical function $\phi(x, \alpha, \beta, \dots)$, where $\alpha$,~$\beta$,~\dots\ are constants, +often assumes different forms according to the values of $\alpha$,~$\beta$,~\dots; sometimes +it can be expressed by means of logarithms, and sometimes by means of +inverse trigonometrical functions. Thus, for example, +\[ +\int \frac{dx}{x^{2} + \alpha} + = \frac{1}{\sqrt{\alpha}} \arctan \frac{x}{\sqrt{\alpha}} +\Tag{(1)} +\] +if $\alpha > 0$, but +\[ +\int \frac{dx}{x^{2} + \alpha} + = \frac{1}{2\sqrtp{-\alpha}} + \log \left|\frac{x - \sqrtp{-\alpha}}{x + \sqrtp{-\alpha}}\right| +\Tag{(2)} +\] +if $\alpha < 0$. These facts suggest the existence of some functional connection +between the logarithmic and the inverse circular functions. That there +is such a connection may also be inferred from the facts that we have expressed +the circular functions of~$\zeta$ in terms of~$\exp i\zeta$, and that the logarithm +is the inverse of the exponential function. + +Let us consider more particularly the equation +\[ +\int \frac{dx}{x^{2} - \alpha^{2}} + = \frac{1}{2\alpha} \log \left(\frac{x - \alpha}{x + \alpha}\right), +\] +\PageSep{414} +which holds when $\alpha$~is real and $(x - \alpha)/(x + \alpha)$ is positive. If we could write +$i\alpha$ instead of~$\alpha$ in this equation, we should be led to the formula +\[ +\arctan \left(\frac{x}{\alpha}\right) + = \frac{1}{2i} \log\left(\frac{x - i\alpha}{x + i\alpha}\right) + C, +\Tag{(3)} +\] +where $C$~is a constant, and the question is suggested whether, now that we +have defined the logarithm of a complex number, this equation will not be +found to be actually true. + +Now (\SecNo[§]{221}) +\[ +\Log(x ± i\alpha) = \tfrac{1}{2} \log(x^{2} + \alpha^{2}) ± i(\phi + 2k\pi), +\] +{\Loosen where $k$~is an integer and $\phi$~is the numerically least angle such that +$\cos\phi = x/\sqrtp{x^{2} + \alpha^{2}}$ and $\sin\phi = \alpha/\sqrtp{x^{2} + \alpha^{2}}$. Thus} +\[ +\frac{1}{2i} \Log\left(\frac{x - i\alpha}{x + i\alpha}\right) = -\phi - l\pi, +\] +where $l$~is an integer, and this does in fact differ by a constant from any +value of~$\arctan(x/\alpha)$. + +The standard formula connecting the logarithmic and inverse circular +functions is +\[ +\arctan x = \frac{1}{2i} \Log\left(\frac{1 + ix}{1 - ix}\right), +\Tag{(4)} +\] +where $x$~is real. It is most easily verified by putting $x = \tan y$, when the right-hand +side reduces to +\[ +\frac{1}{2i} \Log\left(\frac{\cos y + i\sin y}{\cos y - i\sin y}\right) + = \frac{1}{2i} \Log(\exp 2iy) = y + k\pi, +\] +where $k$~is any integer, so that the equation~\Eq{(4)} is `completely' true (\Ex{xciii}.~3). +The reader should also verify the formulae +\[ +\arccos x = -i \Log\{x ± i\sqrtp{1 - x^{2}}\},\quad +\arcsin x = -i \Log\{ix ± \sqrtp{1 - x^{2}}\}, +\Tag{(5)} +\] +where $-1 \leq x \leq 1$: each of these formulae also is `completely' true. + +\Par{Example.} Solving the equation +\[ +\cos u = x = \tfrac{1}{2}\{y + (1/y)\}, +\] +where $y = \exp(iu)$, with respect to~$y$, we obtain $y = x ± i\sqrtp{1 - x^{2}}$. Thus: +\[ +u = -i \Log y = -i \Log\{x ± i\sqrtp{1 - x^{2}}\}, +\] +which is equivalent to the first of the equations~\Eq{(5)}. Obtain the remaining +equations \Eq{(4)}~and~\Eq{(5)} by similar reasoning. +\end{Remark} + +\Paragraph{232. The power series for $\exp z$.\protect\footnotemark} +We saw in \SecNo[§]{212}\footnotetext + {It will be convenient now to use~$z$ instead of~$\zeta$ as the argument of the + exponential function.} +that when $z$~is real +\[ +\exp z = 1 + z +\frac{z^{2}}{2!} + \dots. +\Tag{(1)} +\] +Moreover we saw in \SecNo[§]{191} that the series on the right-hand side +\PageSep{415} +remains convergent (indeed absolutely convergent) when $z$~is complex. +It is naturally suggested that the equation~\Eq{(1)} also remains +true, and we shall now prove that this is the case. + +Let the sum of the series~\Eq{(1)} be denoted by~$F(z)$. The series +being absolutely convergent, it follows by direct multiplication (as +in \Ex{lxxxi}.~7) that $F(z)$~satisfies the functional equation +\[ +F(z) F(h) = F(z + h). +\Tag{(2)} +\] +Now let $z = iy$, where $y$~is real, and $F(z) = f(y)$. Then +\[ +f(y) f(k) = f(y + k); +\] +and so +\[ +\frac{f(y + k) - f(y)}{k} = f(y) \left\{\frac{f(k) - 1}{k}\right\}. +\] + +But +\[ +\frac{f(k) - 1}{k} + = i\left\{1 + \frac{ik}{2!} + \frac{(ik)^{2}}{3!} + \dots\right\}; +\] +and so, if $|k| < 1$, +\[ +\left|\frac{f(k) - 1}{k} - i\right| + < \left(\frac{1}{2!} + \frac{1}{3!} + \dots\right)|k| + < (e - 2)|k|. +\] +Hence $\{f(k) - 1\}/k\to i$ as $k \to 0$, and so +\[ +f'(y) = \lim_{k \to 0} \frac{f(y + k) - f(y)}{k} = if(y). +\Tag{(3)} +\] + +Now +\[ +% [** TN: 2! visually consistent, both mathematically correct] +f(y) = F(iy) = 1 + (iy) + \frac{(iy)^{2}}{\DPchg{2}{2!}} + \dots + = \phi(y) + i\psi(y), +\] +where $\phi(y)$~is an even and $\psi(y)$~an odd function of~$y$, and so +\begin{align*} +|f(y)| &= \sqrtbr{\{\phi(y)\}^{2} + \{\psi(y)\}^{2}}\\ + &= \sqrtbr{\{\phi(y) + i\psi(y)\}\{\phi(y) - i\psi(y)\}}\\ + &= \sqrtb{F(iy) F(-iy)} = \sqrtb{F(0)} = 1; +\end{align*} +and therefore +\[ +f(y) = \cos Y + i \sin Y, +\] +where $Y$~is a function of~$y$ such that $-\pi < Y \leq \pi$. Since $f(y)$~has +a differential coefficient, its real and imaginary parts $\cos Y$ and~$\sin Y$ +have differential coefficients, and are \textit{a~fortiori} continuous functions +of~$y$. Hence $Y$~is a continuous function of~$y$. Suppose that $Y$ +changes to~$Y + K$ when $y$~changes to~$y + k$. Then $K$~tends to +zero with~$k$, and +\[ +\frac{K}{k} + = \biggl\{\frac{\cos(Y + K) - \cos Y}{k}\biggr\} \bigg/ + \biggl\{\frac{\cos(Y + K) - \cos Y}{K}\biggr\}. +\] +Of the two quotients on the right-hand side the first tends to a +\PageSep{416} +limit when $k \to 0$, since $\cos Y$~has a differential coefficient with +respect to~$y$, and the second tends to the limit~$-\sin Y$. Hence $K/k$~tends +to a limit, so that $Y$~has a differential coefficient with respect +to~$y$. + +Further +\[ +f'(y) = (-\sin Y + i\cos Y) \frac{dY}{dy}. +\] +But we have seen already that +\[ +f'(y) = if(y) = -\sin Y + i\cos Y. +\] +Hence +\[ +\frac{dY}{dy} = 1,\quad +Y = y + C, +\] +where $C$~is a constant, and +\[ +f(y) = \cos(y + C) + i\sin(y + C). +\] + +{\Loosen But $f(0) = 1$ when $y = 0$, so that $C$~is a multiple of~$2\pi$, and +$f(y) = \cos y + i\sin y$. Thus $F(iy) = \cos y + i\sin y$ for all real +values of~$y$. And, if $x$~also is real, we have} +\[ +F(x + iy) = F(x) F(iy) = \exp x(\cos y + i\sin y) = \exp(x + iy), +\] +or +\[ +\exp z = 1 + z + \frac{z^{2}}{2!} + \dots, +\] +for all values of~$z$. + +\Paragraph{233. The power series for $\cos z$ and~$\sin z$.} From the +result of the last section and the equations~\Eq{(1)} of \SecNo[§]{228} it follows +at once that +\[ +\cos z = 1 - \frac{z^{2}}{2!} + \frac{z^{4}}{4!} - \dots,\quad +\sin z = z - \frac{z^{3}}{3!} + \frac{z^{5}}{5!} - \dots +\] +for all values of~$z$. These results were proved for real values of~$z$ +in \Ex{lvi}.~1. + +\begin{Examples}{XCVI.} +\Item{1.} Calculate $\cos i$ and $\sin i$ to two places of decimals +by means of the power series for $\cos z$ and~$\sin z$. + +\Item{2.} Prove that $|\cos z| \leq \cosh|z|$ and $|\sin z| \leq \sinh|z|$. + +\Item{3.} Prove that if $|z| < 1$ then $|\cos z| < 2$ and $|\sin z| < \frac{6}{5}|z|$. + +\Item{4.} Since $\sin 2z = 2\sin z \cos z$ we have +\[ +(2z) - \frac{(2z)^{3}}{3!} + \frac{(2z)^{5}}{5!} - \dots + = 2\left(z - \frac{z^{3}}{3!} + \dots\right) + \left(1 - \frac{z^{2}}{2!} + \dots\right). +\] +Prove by multiplying the two series on the right-hand side (\SecNo[§]{195}) and +equating coefficients (\SecNo[§]{194}) that +\[ +\binom{2n + 1}{1} + \binom{2n + 1}{3} + \dots + \binom{2n + 1}{2n + 1} = 2^{2n}. +\] +Verify the result by means of the binomial theorem. Derive similar identities +from the equations +\[ +\cos^{2}z + \sin^{2}z = 1,\quad +\cos2z = 2\cos^{2}z - 1 = 1 - 2\sin^{2}z. +\] +\PageSep{417} + +\Item{5.} Show that +\[ +\exp\{(1 + i)z\} + = \sum_{0}^{\infty} 2^{\frac{1}{2}n} \exp(\tfrac{1}{4}n\pi i) \frac{z^{n}}{n!}. +\] + +\Item{6.} Expand $\cos z \cosh z$ in powers of~$z$. [We have +\begin{align*} +\cos z \cosh z + i\sin z \sinh z + &= \cos\{(1 - i)z\} + = \tfrac{1}{2} [\exp\{(1 + i)z\} + \exp\{-(1 + i)z\}]\\ + &= \tfrac{1}{2} \sum_{0}^{\infty} 2^{\frac{1}{2}n} + \{1 + (-1)^{n}\} \exp(\tfrac{1}{4}n\pi i) \frac{z^{n}}{n!}, +\end{align*} +and similarly +\[ +\cos z \cosh z - i\sin z \sinh z = \cos (1 + i)z + = \tfrac{1}{2} \sum_{0}^{\infty} 2^{\frac{1}{2}n} + \{1 + (-1)^{n}\} \exp(-\tfrac{1}{4}n\pi i) \frac{z^{n}}{n!}. +\] +Hence +\[ +\cos z \cosh z + = \tfrac{1}{2} \sum_{0}^{\infty} 2^{\frac{1}{2}n}\{1 + (-1)^{n}\} \cos \tfrac{1}{4}n\pi \frac{z^{n}}{n!} + = 1 - \frac{2^{2}z^{4}}{4!} + \frac{2^{4}z^{8}}{8!} - \dots.] +\] + +\Item{7.} Expand $\sin z \sinh z$, $\cos z \sinh z$, and $\sin z \cosh z$ in powers of~$z$. + +\Item{8.} Expand $\sin^{2} z$ and $\sin^{3} z$ in powers of~$z$. [Use the formulae +\[ +\sin^{2} z = \tfrac{1}{2} (1 - \cos 2z),\quad +\sin^{3} z = \tfrac{1}{4} (3\sin z - \sin 3z),\ \dots. +\] +It is clear that the same method may be used to expand $\cos^{n} z$ and~$\sin^{n} z$, +where $n$~is any integer.] + +\Item{9.} Sum the series +\[ +C = 1 + \frac{\cos z}{1!} + \frac{\cos 2z}{2!} + \frac{\cos 3z}{3!} +\dots,\quad +S = \frac{\sin z}{1!} + \frac{\sin 2z}{2!} + \frac{\sin 3z}{3!} + \dots. +\] + +[Here +\begin{align*} +C + iS &= 1 + \dfrac{\exp(iz)}{1!} + \dfrac{\exp(2iz)}{2!} + \dots + = \exp\{\exp(iz)\} \\ + &= \exp(\cos z) \{\cos(\sin z) + i\sin(\sin z)\}, +\end{align*} +and similarly +\[ +C - iS = \exp\{\exp(-iz)\} = \exp(\cos z)\{\cos(\sin z) - i\sin(\sin z)\}. +\] +Hence +\[ +C = \exp(\cos z)\cos(\sin z),\quad +S = \exp(\cos z)\sin(\sin z).] +\] + +\Item{10.} Sum +\[ +1 + \frac{a\cos z}{1!} + \frac{a^{2}\cos 2z}{2!} + \dots,\quad +\frac{a\sin z}{1!} + \frac{a^{2}\sin 2z}{2!} + \dots. +\] + +\Item{11.} Sum +\[ +1 - \frac{\cos 2z}{2!} + \frac{\cos 4z}{4!} - \dots,\quad +\frac{\cos z}{1!} - \frac{\cos 3z}{3!} + \dots +\] +and the corresponding series involving sines. + +\Item{12.} Show that +\[ +1 + \frac{\cos 4z}{4!} + \frac{\cos 8z}{8!} + \dots + = \tfrac{1}{2}\{\cos(\cos z) \cosh(\sin z) + \cos(\sin z) \cosh(\cos z)\}. +\] + +\Item{13.} Show that the expansions of $\cos(x + h)$ and $\sin(x + h)$ in powers of~$h$ +(\Ex{lvi}.~1) are valid for all values of $x$~and~$h$, real or complex. +\end{Examples} + +\Paragraph{234. The logarithmic series.} We found in \SecNo[§]{213} that +\[ +\log(1 + z) = z - \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} - \dots +\Tag{(1)} +\] +when $z$~is real and numerically less than unity. The series on the +right-hand side is convergent, indeed absolutely convergent, when +\PageSep{418} +$z$~has any complex value whose modulus is less than unity. It is +naturally suggested that the equation~\Eq{(1)} remains true for such +complex values of~$z$. That this is true may be proved by a +modification of the argument of \SecNo[§]{213}. We shall in fact prove +rather more than this, viz.\ that \Eq{(1)}~is true for all values of~$z$ such +that $|z| \leq 1$, with the exception of the value~$-1$. + +It will be remembered that $\log(1 + z)$~is the principal value of +$\Log(1 + z)$, and that +\[ +\log(1 + z) = \int_{C} \frac{du}{u}, +\] +where $C$~is the straight line joining the points $1$ and~$1 + z$ in the +plane of the complex variable~$u$. We may suppose that $z$~is not +real, as the formula~\Eq{(1)} has been proved already for real values +of~$z$. + +If we put +\[ +z = r(\cos\theta + i\sin\theta) = \zeta r, +\] +so that $|r| \leq 1$, and +\[ +u = 1 + \zeta t, +\] +then $u$~will describe~$C$ as $t$~increases from $0$ to~$r$. And +\begin{align*} +\int_{C} \frac{du}{u} + &= \int_{0}^{r} \frac{\zeta\, dt}{1 + \zeta t} \\ + &= \int_{0}^{r} \left\{\zeta - \zeta^{2} t + \zeta^{3} t^{2} - \dots + + (-1)^{m-1} \zeta^{m} t^{m-1} + + \frac{(-1)^{m} \zeta^{m+1} t^{m}}{1 + \zeta t}\right\} dt \\ + &= \zeta r - \frac{(\zeta r)^{2}}{2} + \frac{(\zeta r)^{3}}{3} - \dots + + (-1)^{m-1} \frac{(\zeta r)^{m}}{m} + R_{m} \\ + &= z - \frac{z^{2}}{2} + \frac{z^{3}}{3} - \dots + + (-1)^{m-1} \frac{z^{m}}{m} + R_{m}, +\Tag{(2)} +\end{align*} +where +\[ +R_{m} = (-1)^{m} \zeta^{m+1} \int_{0}^{r} \frac{t^{m}\, dt}{1 + \zeta t}. +\Tag{(3)} +\] + +It follows from \Eq{(1)}~of \SecNo[§]{164} that +\[ +|R_{m}| \leq \int_{0}^{r} \frac{t^{m}\, dt}{|1 + \zeta t|}. +\Tag{(4)} +\] +Now $|1 + \zeta t|$ or $|u|$~is never less than~$\varpi$, the perpendicular from~$O$ +on to the line~$C$.\footnote + {Since $z$~is not real, $C$~cannot pass through~$O$ when produced. The reader is + recommended to draw a figure to illustrate the argument.} +Hence +\[ +|R_{m}| \leq \frac{1}{\varpi} \int_{0}^{r} t^{m}\, dt + = \frac{r^{m+1}}{(m + 1) \varpi} + \leq \frac{1}{(m + 1) \varpi}, +\] +\PageSep{419} +and so $R_{m} \to 0$ as $m \to \infty$. It follows from~\Eq{(2)} that +\[ +\log(1 + z) = z - \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} - \dots. +\Tag{(5)} +\] + +We have of course shown in the course of our proof that the +series is convergent: this however has been proved already +(\Ex{lxxx}.~4). The series is in fact absolutely convergent when +$\DPtypo{z|}{|z|} < 1$ and conditionally convergent when $|z| = 1$. + +Changing $z$ into~$-z$ we obtain +\[ +\log \left(\frac{1}{1 - z}\right) + = -\log(1 - z) + = z + \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} + \dots. +\Tag{(6)} +\] + +\Paragraph{235.} Now +\begin{align*} +\log(1 + z) + &= \log\{(1 + r \cos\theta) + ir\sin\theta\} \\ + &= \tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) + + i\arctan \left(\frac{r\sin\theta}{1 + r\cos\theta}\right). +\end{align*} +That value of the inverse tangent must be taken which lies +between $-\frac{1}{2}\pi$ and~$\frac{1}{2}\pi$. For, since $1 + z$~is the vector represented +by the line from $-1$ to~$z$, the principal value of~$\am(1 + z)$ always +lies between these limits when $z$~lies within the circle $|z| = 1$.\footnote + {See the preceding footnote.} + +Since $z^{m} = r^{m}(\cos m\theta + i\sin m\theta)$, we obtain, on equating the +real and imaginary parts in equation~\Eq{(5)} of~\SecNo[§]{234}, +\begin{align*} +\tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) + &= r\cos\theta - \tfrac{1}{2}r^{2} \cos 2\theta + + \tfrac{1}{3}r^{3} \cos 3\theta - \dots, \\ +\arctan \left(\frac{r\sin\theta}{1 + r\cos\theta}\right) + &= r\sin\theta - \tfrac{1}{2}r^{2} \sin 2\theta + + \tfrac{1}{3}r^{3} \sin 3\theta - \dots. +\end{align*} +These equations hold when $0 \leq r \leq 1$, and for all values of~$\theta$, except +that, when $r = 1$, $\theta$~must not be equal to an odd multiple of~$\pi$. +It is easy to see that they also hold when $-1 \leq r \leq 0$, except that, +when $r = -1$, $\theta$~must not be equal to an even multiple of~$\pi$. + +A particularly interesting case is that in which $r = 1$. In +this case we have +\begin{align*} +\log(1 + z) = \log(1 + \Cis\theta) + &= \tfrac{1}{2} \log(2 + 2\cos\theta) + + i\arctan\left(\frac{\sin\theta}{1 + \cos\theta}\right) \\ + &= \tfrac{1}{2} \log(4\cos^{2} \tfrac{1}{2}\theta) + \tfrac{1}{2}i\theta, +\end{align*} +if $-\pi < \theta < \pi$, and so +\begin{alignat*}{4} +\cos\theta &- \tfrac{1}{2} \cos 2\theta &&+ \tfrac{1}{3} \cos 3\theta &&- \dots + &&= \tfrac{1}{2} \log(4\cos^{2} \tfrac{1}{2}\theta), \\ +\sin\theta &- \tfrac{1}{2} \sin 2\theta &&+ \tfrac{1}{3} \sin 3\theta &&- \dots + &&= \tfrac{1}{2} \theta. +\end{alignat*} +\PageSep{420} +The sums of the series, for other values of~$\theta$, are easily found from +the consideration that they are periodic functions of~$\theta$ with the +period~$2\pi$. Thus the sum of the cosine series is $\frac{1}{2} \log(4\cos^{2} \frac{1}{2}\theta)$ for +all values of~$\theta$ save odd multiples of~$\pi$ (for which values the series +is divergent), while the sum of the sine series is $\frac{1}{2} (\theta - 2k\pi)$ if +$(2k - 1)\pi < \theta < (2k + 1)\pi$, and zero if $\theta$~is an odd multiple of~$\pi$. +The graph of the function represented by the sine series is shown +in \Fig{58}. The function is discontinuous for $\theta = (2k + 1)\pi$. +%[Illustration: Fig. 58.] +\Figure{58}{p420} + +\begin{Remark} +If we write $iz$ and~$-iz$ for~$z$ in~\Eq{(5)}, and subtract, we obtain +\[ +\frac{1}{2i} \log\left(\frac{1 + iz}{1 - iz}\right) + = z - \tfrac{1}{3}z^{3} + \tfrac{1}{5}z^{5} - \dots. +\] +If $z$~is real and numerically less than unity, we are led, by the results of +\SecNo[§]{231}, to the formula +\[ +\arctan z = z - \tfrac{1}{3}z^{3} + \tfrac{1}{5}z^{5} - \dots, +\] +already proved in a different manner in~\SecNo[§]{214}. +\end{Remark} + +\begin{Examples}{XCVII.} +\Item{1.} Prove that, in any triangle in which $a > b$, +\[ +\log c = \log a - \frac{b}{a} \cos C - \frac{b^{2}}{2a^{2}} \cos 2C - \dots. +\] + +[Use the formula $\log c = \frac{1}{2} \log(a^{2} + b^{2} - 2ab\cos C )$.] + +\Item{2.} Prove that if $-1 < r < 1$ and $-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi$ then +\[ +r\sin 2\theta + - \tfrac{1}{2}r^{2} \sin 4\theta + + \tfrac{1}{3}r^{3} \sin 6\theta - \dots + = \theta - \arctan \left\{\left(\frac{1 - r}{1 + r}\right) \tan\theta\right\}, +\] +the inverse tangent lying between $-\frac{1}{2}\pi$ and~$\frac{1}{2}\pi$. Determine the sum of the +series for all other values of~$\theta$. + +\Item{3.} Prove, by considering the expansions of $\log(1 + iz)$ and $\log(1 - iz)$ in +powers of~$z$, that if $-1 < r < 1$ then +\begin{gather*} +\begin{alignedat}{4} +r\sin\theta &+ \tfrac{1}{2}r^{2} \cos 2\theta + &&- \tfrac{1}{3}r^{3} \sin 3\theta + &&- \tfrac{1}{4}r^{4} \cos 4\theta + \dots + &&= \tfrac{1}{2} \log(1 + 2r \sin\theta + r^{2}),\\ +r\cos\theta &+ \tfrac{1}{2}r^{2} \sin 2\theta + &&- \tfrac{1}{3}r^{3} \cos 3\theta + &&- \tfrac{1}{4}r^{4} \sin 4\theta + \dots + &&= \arctan \left(\frac{r\cos\theta}{1 - r\sin\theta}\right), +\end{alignedat} \displaybreak[1] \\ +\begin{alignedat}{2} +r\sin\theta &- \tfrac{1}{3}r^{3} \sin 3\theta + \dots + &&= \tfrac{1}{4} \log\left(\frac{1 + 2r \sin\theta + r^{2}} + {1 - 2r \sin\theta + r^{2}}\right),\\ +r\cos\theta &- \tfrac{1}{3}r^{3} \cos 3\theta + \dots + &&= \tfrac{1}{2} \arctan \left(\frac{2r\cos\theta}{1 - r^{2}}\right), +\end{alignedat} +\end{gather*} +the inverse tangents lying between $-\frac{1}{2}\pi$ and~$\frac{1}{2}\pi$. +\PageSep{421} + +\Item{4.} Prove that +\begin{alignat*}{3} +\cos\theta \cos\theta &- \tfrac{1}{2} \cos 2\theta \cos^{2}\theta + &&+ \tfrac{1}{3} \cos 3\theta \cos^{3} \theta - \dots + &&= \tfrac{1}{2} \log(1 + 3\cos^{2} \theta),\\ +\sin\theta \sin\theta &- \tfrac{1}{2} \sin 2\theta \sin^{2}\theta + &&+ \tfrac{1}{3} \sin 3\theta \sin^{3} \theta - \dots + &&= \arccot (1 + \cot\theta + \cot^{2}\theta), +\end{alignat*} +the inverse cotangent lying between $-\frac{1}{2}\pi$ and~$\frac{1}{2}\pi$; and find similar expressions +for the sums of the series +\[ +\cos\theta \sin\theta - \tfrac{1}{2} \cos 2\theta \sin^{2}\theta + \dots,\quad +\sin\theta \cos\theta - \tfrac{1}{2} \sin 2\theta \cos^{2}\theta + \dots. +\] +\end{Examples} + +\Paragraph{236. Some applications of the logarithmic series. The +exponential limit.} Let $z$~be any complex number, and $h$~a real +number small enough to ensure that $|hz| < 1$. Then +\[ +\log(1 + hz) = hz - \tfrac{1}{2}(hz)^{2} + \tfrac{1}{3}(hz)^{3} - \dots, +\] +and so +\[ +\frac{\log(1 + hz)}{h} = z + \phi(h, z), +\] +where +\begin{gather*} +\phi(h, z) = -\tfrac{1}{2}hz^{2} + + \tfrac{1}{3}h^{2}z^{3} + - \tfrac{1}{4}h^{3}z^{4} + \dots,\\ +|\phi(h, z)| < |hz^{2}| (1 + |hz| + |h^{2}z^{2}| + \dots) + = \frac{|hz^{2}|}{1 - |hz|}, +\end{gather*} +so that $\phi(h, z) \to 0$ as $h \to 0$. It follows that +\[ +\lim_{h\to 0} \frac{\log(1 + hz)}{h} = z. +\Tag{(1)} +\] + +If in particular we suppose $h = 1/n$, where $n$~is a positive integer, +we obtain +\[ +\lim_{n\to \infty} n\log \left(1 + \frac{z}{n}\right) = z, +\] +and so +\[ +\lim_{n\to \infty} \left(1 + \frac{z}{n}\right)^{n} + = \lim_{n\to \infty} \exp\left\{n\log\left(1 + \frac{z}{n}\right)\right\} + = \exp z. +\Tag{(2)} +\] +This is a generalisation of the result proved in \SecNo[§]{208} for real +values of~$z$. + +From~\Eq{(1)} we can deduce some other results which we shall +require in the next section. If $t$ and~$h$ are real, and $h$~is sufficiently +small, we have +\[ +\frac{\log(1 + tz + hz) - \log(1 + tz)}{h} + = \frac{1}{h}\log\left(1 + \frac{hz}{1 + tz}\right) +\] +which tends to the limit $z/(1 + tz)$ as $h \to 0$. Hence +\[ +\frac{d}{dt} \{\log(1 + tz)\} = \frac{z}{1 + tz}. +\Tag{(3)} +\] +\PageSep{422} + +We shall also require a formula for the differentiation of +$(1 + tz)^{m}$, where $m$~is any number real or complex, with respect +to~$t$. We observe first that, if $\phi(t) = \psi(t) + i\chi(t)$ is a complex +function of~$t$, whose real and imaginary parts $\phi(t)$ and~$\chi(t)$ +possess derivatives, then +\begin{align*} +\frac{d}{dt}(\exp\phi) + &= \frac{d}{dt}\{(\cos\chi + i\sin\chi) \exp\psi\}\\ + &= \{(\cos\chi + i\sin\chi) \psi' + (-\sin\chi + i\cos\chi)\chi'\} \exp\psi\\ + &= (\psi' + i\chi')(\cos\chi + i\sin\chi) \exp\psi\\ + &= (\psi' + i\chi') \exp(\psi + i\chi) = \phi' \exp\phi, +\end{align*} +so that the rule for differentiating~$\exp\phi$ is the same as when $\phi$~is +real. This being so we have +\begin{align*} +\frac{d}{dt}(1 + tz)^{m} + &= \frac{d}{dt} \exp\{m\log(1 + tz)\}\\ + &= \frac{mz}{1 + tz} \exp\{m\log(1 + tz)\}\\ + &= mz(1 + tz)^{m-1}. +\Tag{(4)} +\end{align*} +Here both $(1 + tz)^{m}$ and~$(1 + tz)^{m-1}$ have their principal values\Add{.} + +\Paragraph{237. The general form of the Binomial Theorem.} We +have proved already (\SecNo[§]{215}) that the sum of the series +\[ +1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots +\] +is $(1 + z)^{m} = \exp\{m\log(1 + z)\}$, for all real values of~$m$ and all real +values of~$z$ between $-1$ and~$1$. If $a_{n}$~is the coefficient of~$z^{n}$ then +\[ +\left|\frac{a_{n+1}}{a_{n}}\right| = \left|\frac{m - n}{n + 1}\right| \to 1, +\] +whether $m$~is real or complex. Hence (\Ex{lxxx}.~3) the series +is always convergent if the modulus of~$z$ is less than unity, and we +shall now prove that its sum is still $\exp\{m\log(1 + z)\}$, \ie\ the +principal value of~$(1 + z)^{m}$. + +It follows from \SecNo[§]{236} that if $t$~is real then +\[ +\frac{d}{dt}(1 + tz)^{m} = mz(1 + tz)^{m-1}, +\] +\PageSep{423} +$z$ and~$m$ having any real or complex values and each side having +its principal value. Hence, if $\phi(t) = (1 + tz)^{m}$, we have +\[ +\phi^{(n)}(t) = m(m - 1) \dots (m - n + 1)z^{n} (1 + tz)^{m-n}. +\] +This formula still holds if $t = 0$, so that +\[ +\frac{\phi^{n}(0)}{n!} = \binom{m}{n} z^{n}. +\] + +Now, in virtue of the remark made at the end of \SecNo[§]{164}, we have +\[ +\phi(1) = \phi(0) + \phi'(0) + \frac{\phi''(0)}{2!} + \dots + + \frac{\phi^{(n-1)}(0)}{(n - 1)!} + R_{n}, +\] +where +\[ +R_{n} = \frac{1}{(n - 1)!}\int_{0}^{1} (1 - t)^{n-1} \phi^{(n)}(t)\, dt. +\] +But if $z = r(\cos\theta + i\sin\theta)$ then +\[ +|1 + tz| = \sqrtp{1 + 2tr\cos\theta + t^{2}r^{2}} \geq 1 - tr, +\] +and therefore +\begin{align*} +|R_{n}| + &< \frac{|m(m - 1) \dots (m - n + 1)|}{(n - 1)!}\, + r^{n} \int_{0}^{1} \frac{(1 - t)^{n-1}}{(1 - tr)^{n-m}}\, dt\\ + &< \frac{|m(m - 1) \dots (m - n + 1)|}{(n - 1)!}\, + \frac{(1 - \theta)^{n-1} r^{n}}{(1 - \theta r)^{n-m}}, +\end{align*} +where $0 < \theta < 1$; so that (cf.\ \SecNo[§]{163}) +\[ +|R_{n}| < K\frac{|m(m - 1) \dots (m - n + 1)|}{(n - 1)!}\, r^{n} = \rho_{n}, +\] +say. But +\[ +\frac{\rho_{n+1}}{\rho_{n}} = \frac{|m - n|}{n}r \to r, +\] +and so (\Ex{xxvii}.~6) $\rho_{n} \to 0$, and therefore $R_{n} \to 0$, as $n \to \infty$. +Hence we arrive at the following theorem. + +\begin{Theorem} +The sum of the binomial series\PageLabel{423} +\[ +1 + \binom{m}{1} z + \binom{m}{2} z^{2} + \dots +\] +is $\exp\{m\log(1 + z)\}$, where the logarithm has its principal value, +for all values of~$m$, real or complex, and all values of~$z$ such that +$\DPtypo{z|}{|z|} < 1$. +\end{Theorem} + +A more complete discussion of the binomial series, taking +account of the more difficult case in which $|z| = 1$, will be found +on pp.~225~\textit{et~seq.}\ of Bromwich's \textit{Infinite Series}. +\PageSep{424} + +\begin{Examples}{XCVIII.} +\Item{1.} Suppose $m$~real. Then since +\[ +\log(1 + z) = \tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) + + i\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right), +\] +we obtain +\begin{align*} +\sum_{0}^{\infty} \binom{m}{n} z^{n} + &= \exp\{\tfrac{1}{2}m \log(1 + 2r\cos\theta + r^{2})\} + \Cis \left\{m\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right)\right\} \\ + &= (1 + 2r\cos\theta + r^{2})^{\frac{1}{2}m} + \Cis \left\{m\arctan\left(\frac{r\sin\theta}{1 + r\cos\theta}\right)\right\}, +\end{align*} +all the inverse tangents lying between $-\frac{1}{2}\pi$ and~$\frac{1}{2}\pi$. In particular, if we +suppose $\theta = \frac{1}{2}\pi$, $z = ir$, and equate the real and imaginary parts, we obtain +\begin{align*} +1 - \binom{m}{2} r^{2} + \binom{m}{4} r^{4} - \dots + &= (1 + r^{2})^{\frac{1}{2}m} \cos(m\arctan r), \\ +\binom{m}{1} r - \binom{m}{3} r^{3} + \binom{m}{5} r^{5} - \dots + &= (1 + r^{2})^{\frac{1}{2}m} \sin(m\arctan r). +\end{align*} + +\Item{2.} Verify the formulae of Ex.~1 when $m = 1$, $2$, $3$. [Of course when $m$~is +a positive integer the series is finite.] + +\Item{3.} Prove that if $0 \leq r < 1$ then +\begin{align*} +1 - \frac{1·3}{2·4} r^{2} + \frac{1·3·5·7}{2·4·6·8} r^{4} - \dots + &= \bigsqrtb{\frac{\sqrtp{1 + r^{2}} + 1}{2(1 + r^{2})}}, \\ +\frac{1}{2} r - \frac{1·3·5}{2·4·6} r^{3} + \frac{1·3·5·7·9}{2·4·6·8·10} r^{5} - \dots + &= \bigsqrtb{\frac{\sqrtp{1 + r^{2}} - 1}{2(1 + r^{2})}}. +\end{align*} + +[Take $m = -\frac{1}{2}$ in the last two formulae of Ex.~1.] + +\Item{4.} Prove that if $-\frac{1}{4}\pi < \theta < \frac{1}{4}\pi$ then +\begin{align*} +\cos m\theta &= \cos^{m} \theta \left\{1 - \binom{m}{2} \tan^{2} \theta + + \binom{m}{4} \tan^{4} \theta - \dots\right\}, \\ +\sin m\theta &= \cos^{m} \theta \left\{\binom{m}{1} \tan\theta + - \binom{m}{3} \tan^{3} \theta + \dots\right\}, +\end{align*} +for all real values of~$m$. [These results follow at once from the equations +\[ +\cos m\theta + i\sin m\theta + = (\cos\theta + i\sin\theta )^{m} + = \cos^{m} \theta(1 + i\tan\theta)^{m}.] +\] + +\Item{5.} We proved (\Ex{lxxxi}.~6), by direct multiplication of series, that +$f(m, z) = \sum\dbinom{m}{n} z^{n}$, where $|z| < 1$, satisfies the functional equation +\[ +f(m, z) f(m', z) = f(m + m', z). +\] +Deduce, by an argument similar to that of \SecNo[§]{216}, and without assuming the +general result of \PageRef{p.}{423}, that if $m$~is real and rational then +\[ +f(m, z) = \exp\{m\log(1 + z)\}. +\] + +\Item{6.} If $z$~and~$\mu$ are real, and $-1 < z < 1$, then +\[ +\sum \binom{i\mu}{n} z^{n} = \cos\{\mu\log(1 + z)\} + i\sin\{\mu\log(1 + z)\}. +\] +\end{Examples} +\PageSep{425} + + +\Section{MISCELLANEOUS EXAMPLES ON CHAPTER X.} + +\begin{Examples}{} +\Item{1.} Show that the real part of $i^{\log(1+i)}$ is +\[ +e^{(4k+1)\pi^{2}/8 } \cos \{\tfrac{1}{4}(4k + 1)\pi\log 2\}, +\] +where $k$~is any integer. + +\Item{2.} If $a\cos\theta + b\sin\theta + c = 0$, where $a$,~$b$,~$c$ are real and $c^{2} > a^{2} + b^{2}$, then +\[ +\theta = m\pi + \alpha + ± i\log \frac{|c| + \sqrtp{c^{2} - a^{2} - b^{2}}}{\sqrtp{a^{2} + b^{2}}}, +\] +where $m$~is any odd or any even integer, according as $c$~is positive or negative, +and $\alpha$~is an angle whose cosine and sine are $a/\sqrtp{a^{2} + b^{2}}$ and $b/\sqrtp{a^{2} + b^{2}}$. + +\Item{3.} Prove that if $\theta$~is real and $\sin\theta \sin\phi = 1$ then +\[ +\phi = (k + \tfrac{1}{2})\pi ± i\log \cot \tfrac{1}{2}(k\pi + \theta), +\] +where $k$~is any even or any odd integer, according as $\sin\theta$~is positive or +negative. + +\Item{4.} Show that if $x$~is real then +\begin{gather*} +\frac{d}{dx} \exp\{(a + ib)x\} = (a + ib) \exp\{(a + ib) x\}, \\ +\int \exp \{(a + ib)x\}\, dx = \frac{\exp{(a + ib)x}}{a + ib}. +\end{gather*} +Deduce the results of \Ex{lxxxvii}.~3. + +\Item{5.} Show that if $a > 0$ then $\ds\int_{0}^{\infty} \exp\{-(a + ib)x\}\, dx = \frac{1}{a + ib}$, and deduce the +results of \Ex{lxxxvii}.~5. + +\Item{6.} Show that if $(x/a)^{2} + (y/b)^{2} = 1$ is the equation of an ellipse, and $f(x, y)$ +denotes the terms of highest degree in the equation of any other algebraic +curve, then the sum of the eccentric angles of the points of intersection of the +ellipse and the curve differs by a multiple of~$2\pi$ from +\[ +-i\{\log f(a, ib) - \log f(a, -ib)\}. +\] + +[The eccentric angles are given by $f(a\cos\alpha, b\sin\alpha) + \dots = 0$ or by +\[ +f\left\{\tfrac{1}{2} a \left(u + \frac{1}{u}\right),\ + -\tfrac{1}{2} ib \left(u - \frac{1}{u}\right) \right\} + \dots = 0, +\] +where $u = \exp i\alpha$; and $\sum\alpha$~is equal to one of the values of~$-i\Log P$, where $P$~is +the product of the roots of this equation.] + +\Item{7.} Determine the number and approximate positions of the roots of the +equation $\tan z = az$, where $a$~is real. + +[We know already (\Ex{xvii}.~4) that the equation has infinitely many real +roots. Now let $z = x + iy$, and equate real and imaginary parts. We obtain +\[ +\sin 2x/(\cos 2x + \cosh 2y) = ax,\quad +\sinh 2y/(\cos 2x + \cosh 2y) = ay, +\] +so that, unless $x$ or~$y$ is zero, we have +\[ +(\sin 2x)/2x = (\sinh 2y)/2y. +\] +\PageSep{426} +This is impossible, the left-hand side being numerically less, and the right-hand +side numerically greater than unity. Thus $x = 0$ or $y = 0$. If $y = 0$ we +come back to the real roots of the equation. If $x = 0$ then $\tanh y = ay$. It is +easy to see that this equation has no real root other than zero if $a \leq 0$ or +$a \geq 1$, and two such roots if $0 < a < 1$. Thus there are two purely imaginary +roots if $0 < a < 1$; otherwise all the roots are real.] + +\Item{8.} The equation $\tan z = az + b$, where $a$ and~$b$ are real and $b$~is not equal +to zero, has no complex roots if $a \leq 0$. If $a > 0$ then the real parts of all the +complex roots are numerically greater than~$|b/2a|$. + +\Item{9.} The equation $\tan z = a/z$, where $a$~is real, has no complex roots, but +has two purely imaginary roots if $a < 0$. + +\Item{10.} The equation $\tan z = a\tanh cz$, where $a$ and~$c$ are real, has an infinity +of real and of purely imaginary roots, but no complex roots. + +\Item{11.} Show that if $x$~is real then +\[ +e^{ax} \cos bx = \sum_{0}^{\infty} \frac{x^{n}}{n!} \left\{ + a^{n} - \binom{n}{2} a^{n-2} b^{2} + \binom{n}{4} a^{n-4} b^{4} - \dots +\right\}, +\] +where there are $\frac{1}{2}(n + 1)$ or~$\frac{1}{2}(n + 2)$ terms inside the large brackets. Find +a similar series for~$e^{ax} \sin bx$. + +\Item{12.} If $n\phi(z, n) \to z$ as $n \to \infty$, then $\{1 + \phi(z, n)\}^{n} \to \exp z$. + +\Item{13.} If $\phi(t)$~is a complex function of the real variable~$t$, then +\[ +\frac{d}{dt} \log \phi(t) = \frac{\phi'(t)}{\phi(t)}. +\] + +%[** TN: Paragraph break added] +[Use the formulae +\[ +\phi = \psi + i\chi,\quad +\log \phi = \tfrac{1}{2}\log(\psi^{2} + \chi^{2}) + i\arctan(\chi/\psi).] +\] + +\Item{14.} \Topic{Transformations.} In \Ref{Ch.}{III} (\Exs{xxi}.\ 21~\textit{et~seq.}, and \MiscExs{III}\ +22~\textit{et seq.})\ we considered some simple examples of the geometrical relations +between figures in the planes of two variables $z$,~$Z$ connected by a relation +$z = f(Z)$. We shall now consider some cases in which the relation involves +logarithmic, exponential, or circular functions. + +Suppose firstly that +\[ +z = \exp(\pi Z/a),\quad +Z = (a/\pi) \Log z +\] +where $a$~is positive. To one value of~$Z$ corresponds one of~$z$, but to one of~$z$ +infinitely many of~$Z$. If $x$,~$y$, $r$,~$\theta$ are the coordinates of~$z$ and $X$,~$Y$, $R$,~$\Theta$ +those of~$Z$, we have the relations +\begin{alignat*}{2} +x &= e^{\pi X/a} \cos(\pi Y/a),\qquad & y &= e^{\pi X/a} \sin(\pi Y/a),\\ +X &= (a/\pi) \log r, & Y &= (a\theta/\pi) + 2ka, +\end{alignat*} +where $k$~is any integer. If we suppose that $-\pi < \theta \leq \pi$, and that $\Log z$~has its +principal value~$\log z$, then $k = 0$, and $Z$~is confined to a strip of its plane parallel +to the axis~$OX$ and extending to a distance~$a$ from it on each side, one point +\PageSep{427} +of this strip corresponding to one of the whole $z$-plane, and conversely. By +taking a value of~$\Log z$ other than the principal value we obtain a similar +relation between the $z$-plane and another strip of breadth~$2a$ in the $Z$-plane. + +To the lines in the $Z$-plane for which $X$~and~$Y$ are constant correspond the +circles and radii vectores in the $z$-plane for which $r$~and~$\theta$ are constant. To +one of the latter lines corresponds the whole of a parallel to~$OX$, but to a +circle for which $r$~is constant corresponds only a part, of length~$2a$, of a +parallel to~$OY$. To make $Z$~describe the whole of the latter line we must +make $z$ move continually round and round the circle. + +\Item{15.} Show that to a straight line in the $Z$-plane corresponds an equiangular +spiral in the $z$-plane. + +\Item{16.} Discuss similarly the transformation $z = c\cosh(\pi Z/a)$, showing in +particular that the whole $z$-plane corresponds to any one of an infinite +number of strips in the $Z$-plane, each parallel to the axis $OX$ and of +breadth~$2a$. Show also that to the line $X = X_{0}$ corresponds the ellipse +\[ +\left\{\frac{x}{c\cosh(\pi X_{0}/a)}\right\}^{2} + +\left\{\frac{y}{c\sinh(\pi X_{0}/a)}\right\}^{2} = 1, +\] +and that for different values of~$X_{0}$ these ellipses form a confocal system; and +that the lines $Y = Y_{0}$ correspond to the associated system of confocal hyperbolas. +Trace the variation of~$z$ as $Z$~describes the whole of a line $X = X_{0}$ or +$Y = Y_{0}$. How does $Z$~vary as $z$~describes the degenerate ellipse and hyperbola +formed by the segment between the foci of the confocal system and the +remaining segments of the axis of~$x$? + +\Item{17.} Verify that the results of Ex.~16 are in agreement with those of Ex.~14 +and those of \Ref{Ch.}{III}, \MiscEx{III}~25. [The transformation $z = c\cosh(\pi Z/a)$ +may be regarded as compounded from the transformations +\[ +z = cz_{1},\quad +z_{1} = \tfrac{1}{2}\{z_{2} + (1/z_{2})\},\quad +z_{2} = \exp(\pi Z/a).] +\] + +\Item{18.} Discuss similarly the transformation $z = c\tanh(\pi Z/a)$, showing that +to the lines $X = X_{0}$ correspond the coaxal circles +\[ +\{x - c\coth(2\pi X_{0}/a)\}^{2} + y^{2} = c^{2}\cosech^{2}(2\pi X_{0}/a), +\] +and to the lines $Y = Y_{0}$ the orthogonal system of coaxal circles. + +\Item{19.} \Topic{The Stereographic and Mercator's Projections.} The points of a +unit sphere whose centre is the origin are projected from the south pole (whose +coordinates are $0$,~$0$,~$-1$) on to the tangent plane at the north pole. The +coordinates of a point on the sphere are $\xi$,~$\eta$,~$\zeta$, and Cartesian axes $OX$,~$OY$ +are taken on the tangent plane, parallel to the axes of $\xi$ and~$\eta$. Show that +the coordinates of the projection of the point are +\[ +x = 2\xi/(1 + \zeta),\quad +y = 2\eta/(1 + \zeta), +\] +and that $x + iy = 2\tan \frac{1}{2}\theta \Cis\phi$, where $\phi$~is the longitude (measured from the +plane $\eta = 0$) and $\theta$~the north polar distance of the point on the sphere. +\PageSep{428} + +This projection gives a map of the sphere on the tangent plane, generally +known as the \emph{Stereographic Projection}. If now we introduce a new complex +variable +\[ +Z = X + iY = -i\log \tfrac{1}{2}z = -i\log \tfrac{1}{2}(x + iy) +\] +so that $X = \phi$, $Y = \log \cot \frac{1}{2}\theta$, we obtain another map in the plane of~$Z$, +usually called \emph{Mercator's Projection}. In this map parallels of latitude and +longitude are represented by straight lines parallel to the axes of $X$ and $Y$ +respectively. + +\Item{20.} Discuss the transformation given by the equation +\[ +z = \Log \left(\frac{Z - a}{Z - b}\right), +\] +showing that the straight lines for which $x$~and~$y$ are constant correspond to +two orthogonal systems of coaxal circles in the $Z$-plane. + +\Item{21.} Discuss the transformation +\[ +z = \Log \left\{\frac{\sqrtp{Z - a} + \sqrtp{Z - b}}{\sqrtp{b - a}}\right\}, +\] +showing that the straight lines for which $x$~and~$y$ are constant correspond to +sets of confocal ellipses and hyperbolas whose foci are the points $Z = a$ and +$Z = b$. + +[We have +\begin{alignat*}{2} +\sqrtp{Z - a} + \sqrtp{Z - b} &= \sqrtp{b - a}\, \exp(& &x + iy), \\ +\sqrtp{Z - a} - \sqrtp{Z - b} &= \sqrtp{b - a}\, \exp(&-&x - iy); +\end{alignat*} +and it will be found that +\[ +|Z - a| + |Z - b| = |b - a|\cosh 2x,\quad +|Z - a| - |Z - b| = |b - a|\cos 2y.] +\] + +\Item{22.} \Topic{The transformation $z = Z^{i}$.} If $z = Z^{i}$, where the imaginary power +has its principal value, we have +\[ +\exp(\log r + i\theta) = z = \exp(i\log Z) = \exp(i\log R - \Theta), +\] +so that $\log r = -\Theta$, $\theta = \log R + 2k\pi$, where $k$~is an integer. As all values of~$k$ +give the same point~$z$, we shall suppose that $k = 0$, so that +\[ +\log r = -\Theta,\quad +\theta = \log R. +\Tag{(1)} +\] + +The whole plane of~$Z$ is covered when $R$~varies through all positive +values and $\Theta$~from $-\pi$ to~$\pi$: then $r$~has the range $\exp(-\pi)$ to~$\exp\pi$ and $\theta$~ranges +through all real values. Thus the $Z$-plane corresponds to the ring +bounded by the circles $r = \exp(-\pi)$, $r = \exp\pi$; but this ring is covered +infinitely often. If however $\theta$~is allowed to vary only between $-\pi$ and~$\pi$, +so that the ring is covered only once, then $R$~can vary only from $\exp(-\pi)$ to~$\exp \pi$, +so that the variation of~$Z$ is restricted to a ring similar in all respects +to that within which $z$~varies. Each ring, moreover, must be regarded as +having a barrier along the negative real axis which~$z$ (or~$Z$) must not cross, as +its amplitude must not transgress the limits $-\pi$ and~$\pi$. +\PageSep{429} + +We thus obtain a correspondence between two rings, given by the pair of +equations +\[ +z = Z^{i},\quad +Z = z^{-i}, +\] +where each power has its principal value. To circles whose centre is the +origin in one plane correspond straight lines through the origin in the other. + +\Item{23.} Trace the variation of~$z$ when $Z$, starting at the point~$\exp \pi$, moves +round the larger circle in the positive direction to the point~$-\exp \pi$, along +the barrier, round the smaller circle in the negative direction, back along the +barrier, and round the remainder of the larger circle to its original position. + +\Item{24.} Suppose each plane to be divided up into an infinite series of rings +by circles of radii +\[ +\dots,\quad e^{-(2n+1)\pi},\ \dots,\quad +e^{-\pi},\quad e^{\pi},\quad e^{3\pi},\ \dots,\quad +e^{(2n+1)\pi},\ \dots. +\] +Show how to make any ring in one plane correspond to any ring in the +other, by taking suitable values of the powers in the equations $z = Z^{i}$, $Z = z^{-i}$. + +\Item{25.} If $z = Z^{i}$, any value of the power being taken, and $Z$~moves along an +equiangular spiral whose pole is the origin in its plane, then $z$~moves along an +equiangular spiral whose pole is the origin in its plane. + +\Item{26.} How does $Z = z^{ai}$, where $a$~is real, behave as $z$~approaches the origin +along the real axis\DPtypo{.}{?} [$Z$~moves round and round a circle whose centre is the +origin (the unit circle if $z^{ai}$~has its principal value), and the real and imaginary +parts of~$Z$ both oscillate finitely.] + +\Item{27.} Discuss the same question for $Z = z^{a+bi}$, where $a$~and~$b$ are any real +numbers. + +\Item{28.} Show that the region of convergence of a series of the type $\sum\limits_{-\infty}^{\infty} a_{n}z^{nai}$, +where $a$~is real, is an angle, \ie\ a region bounded by inequalities of the type +$\theta_{0} < \am z < \theta_{1}$ [The angle may reduce to a line, or cover the whole plane.] + +\Item{29.} \Topic{Level Curves.} If $f(z)$~is a function of the complex variable~$z$, we +call the curves for which $|f(z)|$~is constant the \emph{level curves} of~$f(z)$. Sketch +the forms of the level curves of +\begin{alignat*}{2} +z - a \quad& \text{(\emph{concentric circles})}, \qquad& +(z - a)(z - b) \quad& \text{(\emph{Cartesian ovals})}, \\ +(z - a)/(z - b) \quad& \text{(\emph{coaxal circles})}, \qquad& +\exp z \quad& \text{(\emph{straight lines})}. +\end{alignat*} + +\Item{30.} Sketch the forms of the level curves of $(z - a)(z - b)(z - c)$, +$(1 + z\sqrt{3} + z^{2})/z$. [Some of the level curves of the latter function are drawn in +\Fig{59}, the curves marked \textsc{i}--\textsc{vii} corresponding to the values +\[ +.10,\quad 2 - \sqrt{3} = .27,\quad +.40,\quad 1.00,\quad 2.00,\quad +2 + \sqrt{3} = 3.73,\quad 4.53 +\] +of~$|f(z)|$. The reader will probably find but little difficulty in arriving at a +general idea of the forms of the level curves of any given rational function; +but to enter into details would carry us into the general theory of functions +of a complex variable.] +\PageSep{430} +%[Illustration: Fig. 59.] +%[Illustration: Fig. 60.] +%[Illustration: Fig. 61.] +\ifthenelse{\boolean{ForPrinting}}{% +\begin{figure}[p!] +\centering +\Graphic{0.9\textwidth}{p430a} +\caption{Fig.~59.} +\label{fig:59} +\vfill +\begin{minipage}{0.45\textwidth} +\centering +\Graphic{2in}{p430b} +\caption{Fig.~60.} +\label{fig:60} +\end{minipage} +\begin{minipage}{0.45\textwidth} +\centering +\Graphic{2in}{p430c} +\caption{Fig.~61.} +\label{fig:61} +\end{minipage} +\end{figure} +}{% Else not ForPrinting +\Figure[\textwidth]{59}{p430a} +\Figures{2in}{60}{p430b}{2in}{61}{p430c} +} +\PageSep{431} + +\Item{31.} Sketch the forms of the level curves of (i)~$z\exp z$, (ii)~$\sin z$. [See +\Fig{60}, which represents the level curves of~$\sin z$. The curves marked \textsc{i}--\textsc{viii} +correspond to $k = .35$, $.50$, $.71$, $1.00$, $1.41$, $2.00$, $2.83$,~$4.00$.] + +\Item{32.} Sketch the forms of the level curves of~$\exp z - c$, where $c$~is a real +constant. [\Fig{61} shows the level curves of $|\exp z - 1|$, the curves \textsc{i}--\textsc{vii} +corresponding to the values of~$k$ given by $\log k = -1.00$, $-.20$, $-.05$, $0.00$, +$.05$, $.20$,~$1.00$.] + +\Item{33.} The level curves of~$\sin z - c$, where $c$~is a positive constant, are +sketched in Figs.~62,~63. [The nature of the curves differs according as +to whether $c < 1$ or~$c > 1$. In \Fig{62} we have taken $c = .5$, and the curves +\textsc{i}--\textsc{viii} correspond to $k = .29$, $.37$, $.50$, $.87$, $1.50$, $2.60$, $4.50$,~$7.79$. In \Fig{63} +we have taken $c = 2$, and the curves \textsc{i}--\textsc{vii} correspond to $k = .58$, $1.00$, $1.73$, +$3.00$, $5.20$, $9.00$,~$15.59$. If $c = 1$ then the curves are the same as those of +\Fig{60}, except that the origin and scale are different.] +%[Illustration: Fig. 62.] +%[Illustration: Fig. 63.] +\Figures{2.25in}{62}{p431a}{2.25in}{63}{p431b} + +\Item{34.} Prove that if $0 < \theta < \pi$ then +\begin{alignat*}{3} +\cos\theta &+ \tfrac{1}{3} \cos 3\theta &&+ \tfrac{1}{5} \cos 5\theta &&+ \dots + = \tfrac{1}{4} \log \cot^{2}\tfrac{1}{2}\theta,\\ +\sin\theta &+ \tfrac{1}{3} \sin 3\theta &&+ \tfrac{1}{5} \sin 5\theta &&+ \dots + = \tfrac{1}{4}\pi, +\end{alignat*} +and determine the sums of the series for all other values of~$\theta$ for which they +are convergent. [Use the equation +\[ +z + \tfrac{1}{3}z^{3} + \tfrac{1}{5}z^{5} + \dots + = \tfrac{1}{2} \log \left(\frac{1 + z}{1 - z}\right) +\] +where $z = \cos\theta + i\sin\theta$. When $\theta$~is increased by~$\pi$ the sum of each series +simply changes its sign. It follows that the first formula holds for all values +of~$\theta$ save multiples of~$\pi$ (for which the series diverges), while the sum of the +second series is~$\frac{1}{4}\pi$ if $2k\pi < \theta < (2k + 1)\pi$, $-\frac{1}{4}\pi$ if $(2k + 1)\pi < \theta < (2k + 2)\pi$, +and $0$ if $\theta$~is a multiple of~$\pi$.] +\PageSep{432} + +\Item{35.} Prove that if $0 < \theta < \frac{1}{2}\pi$ then +\begin{alignat*}{3} +\cos\theta &- \tfrac{1}{3} \cos 3\theta &&+ \tfrac{1}{5} \cos 5\theta &&- \dots + = \tfrac{1}{4}\pi,\\ +\sin\theta &- \tfrac{1}{3} \sin 3\theta &&+ \tfrac{1}{5} \sin 5\theta &&- \dots + = \tfrac{1}{4} \log (\sec\theta + \tan\theta)^{2}; +\end{alignat*} +and determine the sums of the series for all other values of~$\theta$ for which they +are convergent. + +\Item{36.} Prove that +\[ +\cos\theta \cos\alpha + + \tfrac{1}{2} \cos 2\theta \cos 2\alpha + + \tfrac{1}{3} \cos 3\theta \cos 3\alpha + \dots + = -\tfrac{1}{4} \log \{4(\cos\theta - \cos\alpha)^{2}\}, +\] +unless $\theta - \alpha$ or $\theta + \alpha$ is a multiple of~$2\pi$. + +\Item{37.} Prove that if neither $a$ nor~$b$ is real then +\[ +\int_{0}^{\infty} \frac{dx}{(x - a)(x - b)} + = -\frac{\log(-a) - \log(-b)}{a - b}, +\] +each logarithm having its principal value. Verify the result when $a = ci$, +$b = -ci$, where $c$~is positive. Discuss also the cases in which $a$ or~$b$ or both +are real and negative. + +\Item{38.} Prove that if $\alpha$ and~$\beta$ are real, and $\beta > 0$, then +\[ +\int_{0}^{\infty} \frac{d}{x^{2} - (\alpha + i\beta)^{2}} + = \frac{\pi i}{2(\alpha + i\beta)}. +\] +What is the value of the integral when $\beta < 0$? + +\Item{39.} Prove that, if the roots of $Ax^{2} + 2Bx + C = 0$ have their imaginary +parts of opposite signs, then +\[ +\int_{-\infty}^{\infty} \frac{dx}{Ax^{2} + 2Bx + C} + = \frac{\pi i}{\sqrtp{B^{2} - AC}}, +\] +the sign of $\sqrtp{B^{2} - AC}$ being so chosen that the real part of $\{\sqrtp{B^{2} - AC}\}/Ai$ +is positive. +\end{Examples} +\PageSep{433} + + +\BackMatter +\Appendix{I}{(To Chapters III, IV, V)}{The Proof that every Equation has a Root} + +\First{Let} +\[ +Z = P(z) = \alpha_{0} z^{n} + \alpha_{1} z^{n-1} + \dots + \alpha_{n} +\] +be a polynomial in~$z$, with real or complex coefficients. We can represent +the values of $z$ and~$Z$ by points in two planes, which we may call the $z$-plane +and the $Z$-plane respectively. It is evident that if $z$~describes a closed path~$\gamma$ +in the $z$-plane, then $Z$~describes a corresponding closed path~$\Gamma$ in the $Z$-plane. +We shall assume for the present that the path~$\Gamma$ does not pass through the +origin. + +To any value of~$Z$ correspond an infinity of values of~$\am Z$, differing by +multiples of~$2\pi$, and each of these values varies continuously as $Z$~describes~$\Gamma$.\footnote + {It is here that we assume that $\Gamma$~does not pass through the origin.} +We can select a particular value of~$\am Z$ corresponding to each point +%[Illustration: Fig. A.] +%[Illustration: Fig. B.] +\Figures{2.25in}{A}{p433a}{2.25in}{B}{p433b} +of~$\Gamma$, by first selecting a particular value corresponding to the initial value +of~$Z$, and then following the continuous variation of this value as $Z$~moves +along~$\Gamma$. We shall, in the argument which follows, use the phrase `the +amplitude of~$Z$' and the formula~$\am Z$ to denote the particular value of the +amplitude of~$Z$ thus selected. Thus $\am Z$~denotes a one-valued and continuous +function of $X$~and~$Y$, the real and imaginary parts of~$Z$. +\PageSep{434} + +When $Z$, after describing~$\Gamma$, returns to its original position, its amplitude +may be the same as before, as will certainly be the case if $\Gamma$~does not enclose +the origin, like path~(\ia) in \Fig{B}, or it may differ from its original value by +any multiple of~$2\pi$. Thus if its path is like~(\ib) in \Fig{B}, winding once round +the origin in the positive direction, then its amplitude will have increased +by~$2\pi$. These remarks apply, not merely to~$\Gamma$, but to any closed contour in +the $Z$-plane which does not pass through the origin. Associated with any +such contour there is a number which we may call `the increment of~$\am Z$ +when $Z$~describes the contour', a number independent of the initial choice of +a particular value of the amplitude of~$Z$. + +We shall now prove that \begin{Result}if the amplitude of~$Z$ is not the same when $Z$~returns +to its original position, then the path of~$z$ must contain inside or on +it at least one point at which $Z = 0$. +\end{Result} + +We can divide~$\gamma$ into a number of smaller contours by drawing parallels +to the axes at a distance~$\delta_{1}$ from one another, as in \Fig{C}\@.\footnote + {There is no difficulty in giving a definite rule for the construction of these + parallels: the most obvious course is to draw all the lines $x = k\delta_{1}$, $y = k\delta_{1}$, where + $k$~is an integer positive or negative.} +If there is, +on the boundary of any one of these contours, a point at which $Z = 0$, +what we wish to prove is already established. We may therefore suppose +%[Illustration: Fig. C.] +%[Illustration: Fig. D.] +\Figures{2.5in}{C}{p434a}{2in}{D}{p434b} +that this is not the case. Then the increment of~$\am Z$, when $z$~describes~$\gamma$, +is equal to the sum of all the increments of~$\am Z$ obtained by supposing +$z$~to describe each of these smaller contours separately in the same sense as~$\gamma$. +For if $z$~describes each of the smaller contours in turn, in the same sense, +it will ultimately (see \Fig{D}) have described the boundary of~$\gamma$ once, and +each part of each of the dividing parallels twice and in opposite directions. +Thus $PQ$~will have been described twice, once from $P$ to~$Q$ and once from $Q$ +to~$P$. As $z$~moves from $P$ to~$Q$, $\am Z$~varies continuously, since $Z$~does not +pass through the origin; and if the increment of~$\am Z$ is in this case~$\theta$, then +its increment when $z$~moves from $Q$ to~$P$ is~$-\theta$; so that, when we add +up the increments of~$\am Z$ due to the description of the various parts of the +smaller contours, all cancel one another, save the increments due to the +description of parts of $\gamma$~itself. +\PageSep{435} + +Hence, if $\am Z$~is changed when $z$~describes~$\gamma$, there must be \emph{at least one} +of the smaller contours, say~$\gamma_{1}$, such that $\am Z$~is changed when $z$~describes~$\gamma_{1}$. +This contour may be a square whose sides are parts of the auxiliary +parallels, or may be composed of parts of these parallels and parts of the +boundary of~$\gamma$. In any case every point of the contour lies in or on the +boundary of a square~$\Delta_{1}$ whose sides are parts of the auxiliary parallels and +of length~$\delta_{1}$. + +We can now further subdivide~$\gamma_{1}$ by the help of parallels to the axes at a +smaller distance~$\delta_{2}$ from one another, and we can find a contour~$\gamma_{2}$, entirely +included in a square~$\Delta_{2}$, of side~$\delta_{2}$ and itself included in~$\Delta_{1}$ such that $\am Z$~is +changed when $z$~describes the contour. + +Now let us take an infinite sequence of decreasing numbers $\delta_{1}$, $\delta_{2}$,~\dots, +$\delta_{m}$,~\dots, whose limit is zero.\footnote + {We may, \eg, take $\delta_{m} = \delta_{1}/2^{m-1}$.} +By repeating the argument used above, we can +determine a series of squares $\Delta_{1}$, $\Delta_{2}$,~\dots, $\Delta_{m}$,~\dots\ and a series of contours $\gamma_{1}$, +$\gamma_{2}$,~\dots, $\gamma_{m}$,~\dots\ such that (i)~$\Delta_{m+1}$~lies entirely inside~$\Delta_{m}$, (ii)~$\gamma_{m}$~lies entirely +inside~$\Delta_{m}$, (iii)~$\am Z$~is changed when $z$~describes~$\gamma_{m}$. + +If $(x_{m}, y_{m})$ and~$(x_{m} + \delta_{m}, y_{m} + \delta_{m})$ are the lower left-hand and upper right-hand +corners of~$\Delta_{m}$, it is clear that $x_{1}$, $x_{2}$,~\dots, $x_{m}$,~\dots\ is an increasing and +$x_{1} + \delta_{1}$, $x_{2} + \delta_{2}$,~\dots, $x_{m} + \delta_{m}$,~\dots\ a decreasing sequence, and that they have a +common limit~$x_{0}$. Similarly $y_{m}$~and~$y_{m} + \delta_{m}$ have a common limit~$y_{0}$, and +$(x_{0}, y_{0})$~is the one and only point situated inside every square~$\Delta_{m}$. However +small $\delta$ may be, we can draw a square which includes~$(x_{0}, y_{0})$, and whose +sides are parallel to the axes and of length~$\delta$, and inside this square a closed +contour such that $\am Z$~is changed when $z$~describes the contour. + +It can now be shown that +\[ +P(x_{0} + iy_{0}) = 0. +\] +For suppose that $P(x_{0} + iy_{0}) = a$, where $|a| = \rho > 0$. Since $P(x + iy)$~is a continuous +function of $x$~and~$y$, we can draw a square whose centre is~$(x_{0}, y_{0})$ +and whose sides are parallel to the axes, and which is such that +\[ +|P(x + iy) - P(x_{0} + iy_{0})| < \tfrac{1}{2}\rho +\] +at all points~$x + iy$ inside the square or on its boundary. At all such points +\[ +P(x + iy) = a + \phi, +\] +where $|\phi| < \frac{1}{2}\rho$. Now let us take any closed contour lying entirely inside +this square. As $z$~describes this contour, $Z = a + \phi$ also describes a closed +contour. But the latter contour evidently lies inside the circle whose centre +is~$a$ and whose radius is~$\frac{1}{2}\rho$, and this circle does not include the origin. +Hence the amplitude of~$Z$ is unchanged. + +But this contradicts what was proved above, viz.\ that inside each square~$\Delta_{m}$ +we can find a closed contour the description of which by~$z$ changes~$\am Z$\Add{.} +Hence $P(x_{0} + iy_{0}) = 0$. +\PageSep{436} + +All that remains is to show that we can always find \emph{some} contour such that +$\am Z$~is changed when $z$~describes~$\gamma$. Now +\[ +Z = a_{0} z^{n} \left(1 + \frac{a_{1}}{a_{0}z} + \frac{a_{2}}{a_{0} z^{2}} + \dots + + \frac{a_{n}}{a_{0} z^{n}}\right). +\] +We can choose $R$ so that +\[ +\frac{|a_{1}|}{|a_{0}| R} + +\frac{|a_{2}|}{|a_{0}| R^{2}} + \dots + +\frac{|a_{n}|}{|a_{0}| R^{n}} < \delta, +\] +where $\delta$~is any positive number, however small; and then, if $\gamma$~is the circle +whose centre is the origin and whose radius is~$R$, we have +\[ +Z = a_{0} z^{n} (1 + \rho), +\] +where $|\rho| < \delta$, at all points on~$\gamma$. We can then show, by an argument +similar to that used above, that $\am(1 + \rho)$~is unchanged as $z$~describes +$\gamma$~in the positive sense, while $\am z^{n}$ on the other hand is increased by~$2n\pi$. +Hence $\am Z$~is increased by~$2n\pi$, and the proof that $Z = 0$ has a root is +completed. + +We have assumed throughout the argument that neither~$\Gamma$, nor any of the +smaller contours into which it is resolved, passes through the origin. This +assumption is obviously legitimate, for to suppose the contrary, at any stage +of the argument, is to admit the truth of the theorem. + +We leave it as an exercise to the reader to infer, from the discussion +which precedes and that of \SecNo[§]{43}, that \begin{Result}when $z$~describes any contour~$\gamma$ in the +positive sense the increment of~$\am Z$ is~$2k\pi$, where $k$~is the number of roots +of $Z = 0$ inside~$\gamma$, multiple roots being counted multiply. +\end{Result} + +There is another proof, proceeding on different lines, which is often given. +It depends, however, on an extension to functions of two or more variables of +the results of \SecNo[§§]{102}~\textit{et~seq.} + +We define, precisely on the lines of \SecNo[§]{102}, the \emph{upper and lower bounds} of a +function~$f(x, y)$, for all pairs of values of $x$ and~$y$ corresponding to any point +of any region in the plane of~$(x, y)$ bounded by a closed curve. And we +can prove, much as in \SecNo[§]{102}, that a continuous function~$f(x, y)$ attains its +upper and lower bounds in any such region. + +Now +\[ +|Z| = |P(x + iy)| +\] +is a positive and continuous function of $x$ and~$y$. If $m$~is its lower bound for +points on and inside~$\gamma$, then there must be a point~$z_{0}$ for which $|Z| = m$, and +this must be the \emph{least} value assumed by~$|Z|$. If $m = 0$, then $P(z_{0}) = 0$, and +we have proved what we want. We may therefore suppose that $m > 0$. + +The point~$z_{0}$ must lie either inside or on the boundary of~$\gamma$: but if $\gamma$~is +a circle whose centre is the origin, and whose radius~$R$ is large enough, then +the last hypothesis is untenable, since $|P(z)| \to \infty$ as $|z| \to \infty$. We may +therefore suppose that $z_{0}$~lies inside~$\gamma$. +\PageSep{437} + +If we put $z = z_{0} + \zeta$, and rearrange $P(z)$ according to powers of~$\zeta$, we obtain +\[ +P(z) = P(z_{0}) + A_{1}\zeta + A_{2}\zeta^{2} + \dots + A_{n}\zeta^{n}, +\] +say. Let $A_{k}$ be the first of the coefficients which does not vanish, and let +$|A_{k}| = \mu$, $|\zeta| = \rho$. We can choose~$\rho$ so small that +\[ +|A_{k+1}|\rho + |A_{k+2}|\rho^{2} + \dots + |A_{n}|\rho^{n-k} < \tfrac{1}{2}\mu. +\] +Then +\[ +|P(z) - P(z_{0}) - A_{k}\zeta^{k}| < \tfrac{1}{2}\mu\rho^{k}, +\] +and +\[ +|P(z)| < |P(z_{0} + A_{k}\zeta^{k}| + \tfrac{1}{2}\mu\rho^{k}. +\] + +Now suppose that $z$~moves round the circle whose centre is~$z_{0}$ and radius~$\rho$. +Then +\[ +P(z_{0}) + A_{k}\zeta^{k} +\] +moves $k$~times round the circle whose centre is~$P(z_{0})$ and radius $|A_{k}\zeta^{k}| = \mu\rho^{k}$, +and passes $k$~times through the point in which this circle is intersected by +the line joining~$P(z_{0})$ to the origin. Hence there are $k$~points on the circle +described by~$z$ at which $|P(z_{0}) + A_{k}\zeta^{k}| = |P(z_{0})| - \mu\rho^{k}$ and so +\[ +|P(z)| < |P(z_{0})| - \mu\rho^{k} + \tfrac{1}{2}\mu\rho^{k} + = m - \tfrac{1}{2}\mu\rho^{k} + < m; +\] +and this contradicts the hypothesis that $m$~is the lower bound of~$|P(z)|$. + +It follows that $m$~must be zero and that $P(z_{0}) = 0$. + + +\Section{EXAMPLES ON APPENDIX I} + +\begin{Examples}{} +\Item{1.} Show that the number of roots of $f(z) = 0$ which lie within a closed +contour which does not pass through any root is equal to the increment of +\[ +\{\log f(z)\}/2\pi i +\] +when $z$~describes the contour. + +\Item{2.} Show that if $R$~is any number such that +\[ +\frac{|a_{1}|}{R} + \frac{|a_{2}|}{R^{2}} + \dots + \frac{|a_{n}|}{R^{n}} < 1, +\] +then all the roots of $z^{n} + a_{1}z^{n-1} + \dots + a_{n} = 0$ are in absolute value less than~$R$. +In particular show that all the roots of $z^{5} - 13z -7 = 0$ are in absolute +value less than~$2\frac{1}{67}$. + +\Item{3.} Determine the numbers of the roots of the equation $z^{2p} + az + b = 0$ +where $a$~and~$b$ are real and $p$~odd, which have their real parts positive and +negative. Show that if $a > 0$, $b > 0$ then the numbers are $p - 1$ and $p + 1$; if +$a < 0$, $b > 0$ they are $p + 1$ and $p - 1$; and if $b < 0$ they are $p$~and~$p$. Discuss +the particular cases in which $a = 0$ or $b = 0$. Verify the results when $p = 1$. + +[Trace the variation of $\am(z^{2p} + az + b)$ as $z$~describes the contour formed +by a large semicircle whose centre is the origin and whose radius is~$R$, and +the part of the imaginary axis intercepted by the semicircle.] + +\Item{4.} Consider similarly the equations +\[ +z^{4q} + az + b = 0,\quad +z^{4q-1} + az + b = 0,\quad +z^{4q+1} + az + b = 0. +\] +\PageSep{438} + +\Item{5.} Show that if $\alpha$~and~$\beta$ are real then the numbers of the roots of the +equation $z^{2n} + \alpha^{2} z^{2n-1} + \beta^{2} = 0$ which have their real parts positive and +negative are $n - 1$ and $n + 1$, or $n$~and~$n$, according as $n$~is odd or even. +\MathTrip{1891.} + +\Item{6.} Show that when $z$~moves along the straight line joining the points +$z = z_{1}$, $z = z_{2}$, from a point near~$z_{1}$ to a point near~$z_{2}$, the increment of +\[ +\am \left(\frac{1}{z - z_{1}} + \frac{1}{z - z_{2}}\right) +\] +is nearly equal to~$\pi$. + +\Item{7.} A contour enclosing the three points $z = z_{1}$, $z = z_{2}$, $z = z_{3}$ is defined by +parts of the sides of the triangle formed by $z_{1}$,~$z_{2}$,~$z_{3}$, and the parts exterior +to the triangle of three small circles with their centres at those points. +Show that when $z$~describes the contour the increment of +\[ +\am \left(\frac{1}{z - z_{1}} + \frac{1}{z - z_{2}} + \frac{1}{z - z_{3}}\right) +\] +is equal to~$-2\pi$. + +\Item{8.} Prove that a closed oval path which surrounds all the roots of a cubic +equation $f(z) = 0$ also surrounds those of the derived equation $f'(z) = 0$. [Use +the equation +\[ +f'(z) = f(z) \left( + \frac{1}{z - z_{1}} + \frac{1}{z - z_{2}} + \frac{1}{z - z_{3}} +\right), +\] +where $z_{1}$,~$z_{2}$,~$z_{3}$ are the roots of $f(z) = 0$, and the result of Ex.~7.] + +\Item{9.} Show that the roots of $f'(z) = 0$ are the foci of the ellipse which touches +the sides of the triangle $(z_{1}, z_{2}, z_{3})$ at their middle points. [For a proof see +Cesàro's \textit{Elementares Lehrbuch der algebraischen Analysis}, p.~352.] + +\Item{10.} Extend the result of Ex.~8 to equations of any degree. + +\Item{11.} If $f(z)$ and~$\phi(z)$ are two polynomials in~$z$, and $\gamma$~is a contour which +does not pass through any root of~$f(z)$, and $|\phi(z)| < |f(z)|$ at all points on~$\gamma$, +then the numbers of the roots of the equations +\[ +f(z) = 0,\quad +f(z) + \phi(z) = 0 +\] +which lie inside~$\gamma$ are the same. + +\Item{12.} Show that the equations +\[ +e^{z} = az,\quad +e^{z} = az^{2},\quad +e^{z} = az^{3}, +\] +where $a > e$, have respectively (i)~one positive root (ii)~one positive and one +negative root and (iii)~one positive and two complex roots within the circle +$|z| = 1$. \MathTrip{1910.} +\end{Examples} + +\PageSep{439} + + +\Appendix{II}{(To Chapters IX, X)}{A Note on Double Limit Problems} + +\First{In} the course of Chapters IX~and~X we came on several occasions into +contact with problems of a kind which invariably puzzle beginners and +are indeed, when treated in their most general forms, problems of great +difficulty and of the utmost interest and importance in higher mathematics. + +Let us consider some special instances. In \SecNo[§]{213} we proved that +\[ +\log(1 + x) = x - \tfrac{1}{2}x^{2} + \tfrac{1}{3}x^{3} - \dots, +\] +where $-1 < x \leq 1$, by integrating the equation +\[ +1/(1 + t) = 1 - t + t^{2} - \dots +\] +between the limits $0$ and~$x$. What we proved amounted to this, that +\[ +\int_{0}^{x} \frac{dt}{1 + t} + = \int_{0}^{x} dt - \int_{0}^{x} t\, dt + \int_{0}^{x} t^{2}\, dt - \dots; +\] +{\Loosen or in other words that \emph{the integral of the sum of the infinite series $1 - t + t^{2} - \dots$, +taken between the limits $0$ and~$x$, is equal to the sum of the integrals of its +terms taken between the same limits}. Another way of expressing this fact is to +say that the operations of summation from $0$ to~$\infty$, and of integration from +$0$ to~$x$, are \emph{commutative} when applied to the function $(-1)^{n}t^{n}$, \ie\ that it does +not matter in what order they are performed on the function.} + +Again, in \SecNo[§]{216}, we proved that the differential coefficient of the exponential +function +\[ +\exp x = 1 + x + \frac{x^{2}}{2!} + \dots +\] +is itself equal to $\exp x$, or that +\[ +D_{x} \left(1 + x + \frac{x^{2}}{2!} + \dots\right) + = D_{x}1 + D_{x}x + D_{x} \frac{x^{2}}{2!} + \dots; +\] +\PageSep{440} +that is to say that \emph{the differential coefficient of the sum of the series is equal +to the sum of the differential coefficients of its terms}, or that the operations of +summation from $0$ to~$\infty$ and of differentiation with respect to~$x$ are commutative +when applied to~$x^{n}/n!$. + +Finally we proved incidentally in the same section that the function +$\exp x$ is a continuous function of~$x$, or in other words that +\[ +\lim_{x\to\xi} \left(1 + x + \frac{x^{2}}{2!} + \dots\right) + = 1 + \xi + \frac{\xi^{2}}{2!} + \dots + = \lim_{x\to\xi} 1 + \lim_{x\to\xi} x + \lim_{x\to\xi} \frac{x^{2}}{2!} + \dots; +\] +\ie\ that the limit of the sum of the series is equal to the sum of the limits of +the terms, or that the sum of the series is continuous for $x = \xi$, or that the +operations of summation from $0$ to~$\infty$ and of making $x$~tend to~$\xi$ are commutative +when applied to~$x^{n}/n!$. + +In each of these cases we gave a special proof of the correctness of the +result. We have not proved, and in this volume shall not prove, any general +theorem from which the truth of any one of them could be inferred immediately. +In \Ex{xxxvii}.~1 we saw that the sum of a finite number of continuous +terms is itself continuous, and in \SecNo[§]{113} that the differential coefficient +of the sum of a finite number of terms is equal to the sum of their differential +coefficients; and in \SecNo[§]{160} we stated the corresponding theorem for integrals. +Thus we have proved that in certain circumstances the operations symbolised +by +\[ +\lim_{x\to\xi} \dots,\quad +D_{x} \dots,\quad +\int \dots\, dx +\] +are commutative with respect to the operation of summation of a \emph{finite} number +of terms. And it is natural to suppose that, in certain circumstances which +it should be possible to define precisely, they should be commutative also with +respect to the operation of summation of an \emph{infinite} number. It is natural to +suppose so: but that is all that we have a right to say at present. + +A few further instances of commutative and non-commutative operations +may help to elucidate these points. + +\Item{(1)} Multiplication by~$2$ and multiplication by~$3$ are always commutative, +for +\[ +2 × 3 × x = 3 × 2 × x +\] +for all values of~$x$. + +\Item{(2)} The operation of taking the real part of~$z$ is never commutative with +that of multiplication by~$i$, except when $z = 0$; for +\[ +i × \Re(x + iy) = ix,\quad +\Re\{i × (x + iy)\} = -y. +\] + +\Item{(3)} The operations of proceeding to the limit zero with each of two +variables $x$~and~$y$ may or may not be commutative when applied to a +function~$f(x, y)$. Thus +\[ +\lim_{x\to 0} \{\lim_{y\to 0} (x + y)\} = \lim_{x\to 0} x = 0,\quad +\lim_{y\to 0} \{\lim_{x\to 0} (x + y)\} = \lim_{y\to 0} y = 0; +\] +\PageSep{441} +but on the other hand +\begin{alignat*}{2} +\lim_{x\to 0} \left(\lim_{y\to 0} \frac{x - y}{x + y}\right) + &= \lim_{x\to 0} \frac{x}{x} &&= \lim_{x\to 0} 1 = 1,\\ +\lim_{y\to 0} \left(\lim_{x\to 0} \frac{x - y}{x + y}\right) + &= \lim_{y\to 0}\frac{-y}{y} &&= \lim_{y\to 0} (-1) = -1. +\end{alignat*} + +\Item{(4)} The operations $\sum\limits_{1}^{\infty} \dots$, $\lim\limits_{x\to 1} \dots$ may or may not be commutative. Thus +if $x \to 1$ through values less than~$1$ then +\begin{alignat*}{2} +\lim_{x\to 1} \left\{\sum_{1}^{\infty} \frac{(-1)^{n}}{n}x^{n}\right\} + &= \lim_{x\to 1}\log(1 + x) &&= \log 2,\\ +\sum_{1}^{\infty} \left\{\lim_{x\to 1} \frac{(-1)^{n}}{n}x^{n}\right\} + &= \quad \sum_{1}^{\infty} \frac{(-1)^{n}}{n} &&= \log 2; +\end{alignat*} +but on the other hand +\begin{align*} +\lim_{x\to 1} \left\{\sum_{1}^{\infty} (x^{n} - x^{n+1})\right\} + &= \lim_{x\to 1} \{(1 - x) + (x - x^{2}) + \dots\} + = \lim_{x\to 1} 1 = 1,\\ +\sum_{1}^{\infty} \left\{\lim_{x\to 1} (x^{n} - x^{n+1})\right\} + &= \sum_{1}^{\infty} (1 - 1) = 0 + 0 + 0 + \dots = 0. +\end{align*} + +The preceding examples suggest that there are three possibilities with +respect to the commutation of two given operations, viz.:\ (1)~the operations +may \emph{always} be commutative; (2)~they may \emph{never} be commutative, \emph{except in +very special circumstances}; (3)~they may be commutative \emph{in most of the ordinary +cases which occur practically}. + +The really important case (as is suggested by the instances which we +gave from \Ref{Ch.}{IX}) is that in which each operation is one which involves +a passage to the limit, such as a differentiation or the summation of an +infinite series: such operations are called \emph{limit operations}. The general +question as to the circumstances in which two given limit operations are +commutative is one of the most important in all mathematics. But to +attempt to deal with questions of this character by means of general theorems +would carry us far beyond the scope of this volume. + +We may however remark that the answer to the general question is on +the lines suggested by the examples above. If $L$~and~$L'$ are two limit +operations then the numbers $LL'z$ and~$L'Lz$ are not \emph{generally} equal, in the +strict theoretical sense of the word `general'. We can always, by the exercise +of a little ingenuity, find~$z$ so that $LL'z$ and~$L'Lz$ shall differ from one another. +But they \emph{are} equal generally, if we use the word in a more practical sense, +viz.\ as meaning `in a great majority of such cases as are likely to occur +naturally' or in \emph{ordinary} cases. +\PageSep{442} + +Of course, in an exact science like pure mathematics, we cannot be satisfied +with an answer of this kind; and in the higher branches of mathematics the +detailed investigation of these questions is an absolute necessity. But for +the present the reader may be content if he realises the point of the remarks +which we have just made. In practice, a result obtained by assuming that +two limit-operations are commutative is \emph{probably} true: it at any rate affords +a valuable \emph{suggestion} as to the answer to the problem under consideration. +But an answer thus obtained must, in default of a further study of the general +question or a special investigation of the particular problem, such as we gave +in the instances which occurred in \Ref{Ch.}{IX}, be regarded as suggested only and +not proved. + +Detailed investigations of a large number of important double limit +problems will be found in Bromwich's \textit{Infinite Series}. +\PageSep{443} + + +\Appendix{III}{(To \SecNo[§]{158} and \Ref{Chapter}{IX})}{The circular functions} + +\First{The} reader will find it an instructive exercise to work out the theory of +the circular functions, starting from the definition +\CenterDef[\footnotemark]{\Item{(1)}}{$y = y(x) = \arctan x = \ds\int_{0}^{x} \frac{dt}{1 + t^{2}}$.} +\footnotetext{These letters at the end of a line indicate that the formulae which it contains + are definitions.} + +The equation~\Eq{(1)} defines a unique value of~$y$ corresponding to every real +value of~$x$. As $y$~is continuous and strictly increasing, there is an inverse +function $x = x(y)$, also continuous and steadily increasing. We write +\CenterDef{\Item{(2)}}{$x = x(y) = \tan y$.} + +If we define~$\pi$ by the equation +\CenterDef{\Item{(3)}}{$\frac{1}{2}\pi = \ds\int_{0}^{\infty} \frac{dt}{1 + t^{2}}$,} +then this function is defined for $-\frac{1}{2}\pi < y < \frac{1}{2}\pi$. + +We write further +\CenterDef{\Item{(4)}}{$\cos y = \dfrac{1}{\sqrt{1 + x^{2}}},\quad \sin y = \dfrac{x}{\sqrt{1 + x^{2}}}$,} +where the square root is positive; and we define $\cos y$ and~$\sin y$, when $y$~is $-\frac{1}{2}\pi$ +or~$\frac{1}{2}\pi$, so that the functions shall remain continuous for those values of~$y$. +Finally we define $\cos y$ and~$\sin y$, outside the interval $\DPmod{(-\frac{1}{2}\pi, \frac{1}{2}\pi)}{[-\frac{1}{2}\pi, \frac{1}{2}\pi]}$, by +%[** TN: Set on one line in the original] +\CenterDef{\Item{(5)}}{$\begin{alignedat}{2}\tan(y + \pi) &= &&\tan y,\\ \cos(y + \pi) &= -&&\cos y, \\ \sin(y + \pi) &= -&&\sin y.\end{alignedat}$} + +We have thus defined $\cos y$ and~$\sin y$ for all values of~$y$, and $\tan y$~for all +values of~$y$ other than odd multiples of~$\frac{1}{2}\pi$. The cosine and sine are continuous +for all values of~$y$, the tangent except at the points where its definition fails. + +The further development of the theory depends merely on the addition +formulae. Write +\[ +x = \frac{x_{1} + x_{2}}{1 - x_{1}x_{2}}, +\] +and transform the equation~\Eq{(1)} by the substitution +\[ +t = \frac{x_{1} + u}{1 - x_{1}u},\quad +u = \frac{t - x_{1}}{1 + x_{1}t}. +\] + +We find +\begin{align*} +\arctan \frac{x_{1} + x_{2}}{1 - x_{1}x_{2}} + &= \int_{-x_{1}}^{x_{2}} \frac{du}{1 + u^{2}} + = \int_{0}^{x_{1}} \frac{du}{1 + u^{2}} + \int_{0}^{x_{2}} \frac{du}{1 + u^{2}} \\ + &= \arctan x_{1} + \arctan x_{2}. +\end{align*} +\PageSep{444} + +From this we deduce +\CenterLine{\Item{(6)}}{$\tan (y_{1} + y_{2}) = \dfrac{\tan y_{1} + \tan y_{2}}{1 - \tan y_{1}\tan y_{2}}$,} +an equation proved in the first instance only when $y_{1}$,~$y_{2}$, and~$y_{1} + y_{2}$ lie in +$\DPmod{(-\frac{1}{2}\pi, \frac{1}{2}\pi)}{[-\frac{1}{2}\pi, \frac{1}{2}\pi]}$, but immediately extensible to all values of $y_{1}$~and~$y_{2}$ by means of +the equations~\Eq{(5)}. + +From~\Eq{(4)} and~\Eq{(6)} we deduce +\[ +\cos(y_{1} + y_{2}) = ±(\cos y_{1}\cos y_{2} - \sin y_{1}\sin y_{2}). +\] +{\Loosen To determine the sign put $y_{2} = 0$. The equation reduces to $\cos y_{1} = ±\cos y_{1}$, +which shows that the positive sign must be chosen for at least one value of~$y_{2}$, +viz.\ $y_{2} = 0$. It follows from considerations of continuity that the positive sign +must be chosen in all cases. The corresponding formula for $\sin(y_{1} + y_{2})$ may +be deduced in a similar manner.} + +The formulae for differentiation of the circular functions may now be deduced +in the ordinary way, and the power series derived from Taylor's +Theorem. + +An alternative theory of the circular functions is based on the theory of +infinite series. An account of this theory, in which, for example, $\cos x$~is +defined by the equation +\[ +\cos x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \dots +\] +will be found in Whittaker and Watson's \textit{Modern Analysis} (Appendix~A). +\PageSep{445} + + +\Appendix{IV}{}{The infinite in analysis and geometry} + +\First{Some}, though not all, systems of analytical geometry contain `infinite' +elements, the line at infinity, the circular points at infinity, and so on. The +object of this brief note is to point out that these concepts are in no way +dependent upon the analytical doctrine of limits. + +In what may be called `common Cartesian geometry', a \emph{point} is \emph{a pair of +real numbers $(x, y)$}. A \emph{line} is the class of points which satisfy a linear relation +$ax + by + c=0$, in which $a$~and~$b$ are not both zero. There are no infinite elements, +and two lines may have no point in common. + +In a system of real homogeneous geometry a point is \emph{a class of triads of +real numbers $(x, y, z)$}, not all zero, triads being classed together when their +constituents are proportional. A line is a class of points which satisfy a linear +relation $ax + by + cz = 0$, where $a$,~$b$,~$c$ are not all zero. In some systems one +point or line is on exactly the same footing as another. In others certain +`special' points and lines are regarded as peculiarly distinguished, and it is on +the relations of other elements to these special elements that emphasis is laid. +Thus, in what may be called `real homogeneous Cartesian geometry', those +points are special for which $z = 0$, and there is one special line, viz.\ the line +$z = 0$. This special line is called `the line at infinity'. + +This is not a treatise on geometry, and there is no occasion to develop the +matter in detail. The point of importance is this. The infinite of analysis +is a `limiting' and not an `actual' infinite. The symbol~`$\infty$' has, throughout +this book, been regarded as an `incomplete symbol', a symbol to which no +independent meaning has been attached, though one has been attached to +certain phrases containing it. But \emph{the infinite of geometry is an actual and +not a limiting infinite}. The `line at infinity' is a line in precisely the same +sense in which other lines are lines. + +{\Loosen It is possible to set up a correlation between `homogeneous' and `common' +Cartesian geometry in which all elements of the first system, \emph{the special +elements excepted}, have correlates in the second. The line $ax + by + cz = 0$, for +example, corresponds to the line $ax + by + c = 0$. Every point of the first line +has a correlate on the second, except one, viz.\ the point for which $z = 0$. +When $(x, y, z)$ varies on the first line, in such a manner as to tend in the limit +to the special point for which $z = 0$, the corresponding point on the second line +varies so that its distance from the origin tends to infinity. This correlation +is historically important, for it is from it that the vocabulary of the subject +has been derived, and it is often useful for purposes of illustration. It is however +no more than an illustration, and no rational account of the geometrical +infinite can be based upon it. The confusion about these matters so prevalent +among students arises from the fact that, in the commonly used text books of +analytical geometry, the illustration is taken for the reality.} +\PageSep{446} +\clearpage +\thispagestyle{empty} +\null\vfill +\begin{center} +\footnotesize +CAMBRIDGE: PRINTED BY \\ +J.\ B.\ PEACE, M.A., \\ +AT THE UNIVERSITY PRESS +\end{center} +\vfill +%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%% +\FlushRunningHeads +\vfill +\begin{center} +\TranscribersNote[Modification Note]{% +\ChangeNote\bigskip + +\ifthenelse{\boolean{Modernize}}{\ModernizationNote}{} +} +\end{center} + +\PGLicense +\begin{PGtext} +End of the Project Gutenberg EBook of A Course of Pure Mathematics, by +G. H. 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