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| author | Roger Frank <rfrank@pglaf.org> | 2025-10-15 02:48:13 -0700 |
|---|---|---|
| committer | Roger Frank <rfrank@pglaf.org> | 2025-10-15 02:48:13 -0700 |
| commit | 5c5e4d98db9e19fea18282dcf193c6b29d26cc05 (patch) | |
| tree | cd45c72fdb9d24912cb7691a12b3cae31666b7d1 /29785-t | |
Diffstat (limited to '29785-t')
55 files changed, 31983 insertions, 0 deletions
diff --git a/29785-t/29785-t.tex b/29785-t/29785-t.tex new file mode 100644 index 0000000..daf7978 --- /dev/null +++ b/29785-t/29785-t.tex @@ -0,0 +1,14424 @@ +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% The Project Gutenberg EBook of First Course in the Theory of Equations, by +% Leonard Eugene Dickson % +% % +% 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.org % +% % +% % +% Title: First Course in the Theory of Equations % +% % +% Author: Leonard Eugene Dickson % +% % +% Release Date: August 25, 2009 [EBook #29785] % +% Most recently updated: June 11, 2021 % +% % +% Language: English % +% % +% Character set encoding: UTF-8 % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF EQUATIONS *** % +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{29785} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% %% +%% inputenc: Standard DP encoding. Required. %% +%% fontenc: Font encoding, for bold smallcaps. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. Required. %% +%% amssymb: AMS mathematics symbols. Required. %% +%% %% +%% calc: Infix arithmetic for length calculations. Required. %% +%% ifthen: Logical conditionals. Required. %% +%% %% +%% alltt: Fixed-width font environment for PG stanzas. Required. %% +%% %% +%% textcase: \MakeTextUppercase, for running heads. Required. %% +%% %% +%% array: Enhanced tabular environment. Required. %% +%% arydshln: Dashed lines in tabular/array environments. Required. %% +%% %% +%% indentfirst: Indent sectional units' first paragraph. Required. %% +%% %% +%% multicol: Multi-column environment. Required. %% +%% makeidx: Indexing capabilities. Required. %% +%% %% +%% wrapfig: Wrap text around figures. Required. %% +%% fancyhdr: Enhanced running headers and footers. Required. %% +%% %% +%% tikz: Replacement for picture environment. Required. %% +%% %% +%% geometry: Enhanced page layout package. Required. %% +%% hyperref: Hypertext embellishments for pdf output. Required. %% +%% %% +%% %% +%% Producer's Comments: %% +%% %% +%% The book comprises multiple sub-page units of varying font %% +%% size, figures wrapped by text, and tall, table-like %% +%% calculations. Vertical spacing is loose, so only a few %% +%% locations need visual checking. %% +%% %% +%% %% +%% Things to Check: %% +%% %% +%% * Table of contents chapter headers not separated from %% +%% first line of section headers. %% +%% * Wrapped images on page 64 are both positioned on the page. %% +%% * Wrapped images on page 73 are all positioned on the page. %% +%% * Tall tables on pages 98 ff. are not too badly spaced. %% +%% %% +%% %% +%% Notes on spelling: %% +%% %% +%% All changes to the text are [** PP: Noted] in this file. %% +%% The old spellings constructible and parallelopiped have been %% +%% retained, as have the variants coordinates and coördinates, %% +%% 2-rowed and two-rowed, etc. %% +%% %% +%% Spellcheck: OK %% +%% Smoothreading pool: Yes %% +%% lacheck: OK, about 100 false positives %% +%% %% +%% Lprep/gutcheck: OK %% +%% PDF pages: 207 %% +%% PDF page size: US Letter (8.5" x 11") %% +%% PDF bookmarks: created %% +%% PDF document info: filled in %% +%% ToC page numbers: OK %% +%% Images: 26 picture environments (written with tikz macros) %% +%% If compiling with tikz 2.00 of greater, may need to replace %% +%% the option [cap=round] with [line cap=round] in each illo. %% +%% %% +%% Summary of log file: %% +%% * Three underfull hboxes -- two from fixed-width boxes on title %% +%% page, one from illo on PDF page 64. %% +%% * One underfull vbox %% +%% %% +%% %% +%% Compile History: %% +%% %% +%% May, 2009: adhere (Andrew D. Hwang) %% +%% texlive2007, GNU/Linux %% +%% %% +%% Command block: %% +%% %% +%% pdflatex x3 %% +%% makeindex %% +%% pdflatex %% +%% %% +%% %% +%% August 2009: pglatex. %% +%% Compile this project with: %% +%% pdflatex 29785-t.tex ..... THREE times %% +%% makeindex 29785-t.idx %% +%% pdflatex 29785-t.tex %% +%% %% +%% pdfTeXk, Version 3.141592-1.40.3 (Web2C 7.5.6) %% +%% %% +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +% %%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +\listfiles +\documentclass[12pt,leqno,letterpaper]{book}[2005/09/16] + +\usepackage[utf8]{inputenc}[2006/05/05] +\usepackage[T1]{fontenc}[2005/09/27] + +\usepackage{amsmath}[2000/07/18] +\usepackage{amssymb}[2002/01/22] + +\usepackage{calc}[2005/08/06] + +\usepackage{ifthen}[2001/05/26] + +\usepackage{alltt}[1997/06/16] + +\usepackage{textcase}[2004/10/07] % for running heads $n$ + +\usepackage{array}[2005/08/23] +\usepackage{arydshln}[2004/08/31] + +\usepackage{indentfirst}[1995/11/23] + +\usepackage{multicol}[2006/05/18] +\usepackage{makeidx}[2000/03/29] + +\usepackage{wrapfig}[2003/01/31] +\usepackage{fancyhdr} + +\usepackage{tikz}[2006/10/17] + +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% %% Interlude: Set up PRINTING (default) or SCREEN VIEWING %% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +% ForPrinting=true (default) false +% Letterpaper Cropped pages +% Asymmetric margins Symmetric margins +% Black hyperlinks Blue hyperlinks +\newboolean{ForPrinting} +\newboolean{ExplainSymbols} + +% --> USER-CHANGEABLE Conditional Compilation Switches <-- + +% COMMENT the next line for a SCREEN-OPTIMIZED VERSION of the text % +\setboolean{ForPrinting}{true} + +% UNCOMMENT to add verbal descriptions to index ``Symbol'' entries +%\setboolean{ExplainSymbols}{true} + + +% Initialize values to ForPrinting=false +\newcommand{\Margins}{hmarginratio=1:1} % Symmetric margins +\newcommand{\HLinkColor}{blue} % Hyperlink color +\newcommand{\PDFPageLayout}{SinglePage} +\newcommand{\TransNote}{Transcriber's Note} +\newcommand{\TransNoteCommon} +{% + \bigskip + + Table of contents entries and running heads have been normalized. + \bigskip + + Archaic spellings (constructible, parallelopiped) and variants + (coordinates/coördinates, two-rowed/$2$-rowed, etc.)\ have been + retained from the original. + \bigskip + + Minor typographical corrections, and minor changes to the + presentational style, have been made without comment. Figures may + have been relocated slightly with respect to the surrounding text. +} + +\newcommand{\TransNoteText} +{% + This PDF file is formatted for screen viewing, but may be easily + formatted for printing. Please consult the preamble of the \LaTeX\ + source file for instructions. + + \TransNoteCommon +} + +% Re-set if ForPrinting=true +\ifthenelse{\boolean{ForPrinting}}{% + \renewcommand{\Margins}{hmarginratio=2:3} % Asymmetric margins + \renewcommand{\HLinkColor}{black} % Hyperlink color + \renewcommand{\PDFPageLayout}{TwoPageRight} + \renewcommand{\TransNote}{Transcriber's Note} + \renewcommand{\TransNoteText}{% + This PDF file is formatted for printing, but may be easily formatted + for screen viewing. Please see the preamble of the \LaTeX\ source + file for instructions. + + \TransNoteCommon + } +}{} +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% %% End of PRINTING/SCREEN VIEWING code; back to packages %% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\usepackage[body={5.5in,8.25in},\Margins]{geometry}[2002/07/08] + +\providecommand{\ebook}{00000}% Overridden during white-washing +\usepackage[pdftex, + hyperfootnotes=false, + pdftitle={The Project Gutenberg eBook \#\ebook: + First Course in the Theory of Equations}, + pdfauthor={Leonard Eugene Dickson}, + pdfkeywords={Peter Vachuska, Dave Morgan, Andrew D. Hwang, + Project Gutenberg Online Distributed Proofreading Team}, + 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] + +%%%% Re-crop screen-formatted version, omit blank verso pages %%%% +\ifthenelse{\boolean{ForPrinting}} + {} + {\hypersetup{pdfpagescrop = 90 75 522 775} + % If ForPrinting=false, don't skip to recto + \renewcommand{\cleardoublepage}{\clearpage} +} + +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% %%%%%%%%%%%%%%%%%% END OF PACKAGE SETUP %%%%%%%%%%%%%%%%%% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + + + +% Miscellaneous style parameters +% No header rule +\renewcommand{\headrulewidth}{0pt} + +% Miscellaneous fonts +\newcommand{\Titling}{\bfseries\scshape} +\newcommand{\Running}{\itshape\footnotesize}% Running heads + +% Loosen up page spacing +\setlength{\parsep}{1ex plus 0.5ex minus 1ex} +\setlength{\partopsep}{0.5ex plus 1.5ex minus 0.25ex} +\setlength{\itemsep}{1ex plus 1ex minus 1ex} + + +% The book contains sets of Exercises, some of which have selected +% answers in the back. The heading on a section of Answers is the +% page number or range on which the Exercise set is found. The +% Exercises environment accepts an optional argument (#1, empty by +% default, or the string MISCELLANEOUS for sets of Exercises at the +% ends of chapters) and one mandatory argument (#2, a tag Pagexx +% specifying the book folio number on which the set begins). +% +% If #2 is non-empty, the Exercises environment sets labels keyed to +% Pagexx at its start and finish. +% +% Each Answers environment in the back accepts a Pagexx argument. To +% set the section heading, the macro \AH compares the \pageref{}s for +% the labels of the corresponding Exercises. If they're equal, it +% prints the section heading Page xx; otherwise it prints Pages xx--yy. +% +% An Exercises environment contains one or more list-like Problems +% environments, which controls the typesetting of the actual material, +% particularly the number of columns (via an optional argument). + +% variables to store state data for Exercises and Answers environments +\newcommand{\AnsPg}{} +\newboolean{AnsArg} + +\newboolean{InMulticols}% true iff we're in a multicolumn envt + + +% List environment initializer for Answers section +\newcommand{\ListInitAns}{% + \setlength{\leftmargin}{\parindent}%0pt + \setlength{\labelwidth}{\parindent}% + \setlength{\labelsep}{0.5em}% + \setlength{\itemindent}{0pt}%\parindent}% + \renewcommand{\makelabel}{\boldlabel\hss}% +} + + +% Reset number of columns *within an Answers environment* +\newcommand{\ResetCols}[1] +{% + \ifthenelse{\boolean{InMulticols}} + {\end{multicols}} + {} + \ifthenelse{\equal{#1}{1}} + {\setboolean{InMulticols}{false}} + {\setboolean{InMulticols}{true} + \begin{multicols}{#1}[\raggedcolumns]} +} + + +% Answer Heading +% #1 has the form Pagexx. Print appropriate heading according to +% whether corresponding Exercise set starts and ends on the same +% page or not. +\newcommand{\AH}[1] +{% +\ifthenelse{\equal{\pageref{exer:#1:start}}{\pageref{exer:#1:end}}} + {\subsection*{\centering\normalfont\normalsize\bfseries% + Page \pageref{exer:#1:start}}} + {\subsection*{\centering\normalfont\normalsize\bfseries% + Pages \pageref{exer:#1:start}--\pageref{exer:#1:end}}} +% drop anchor so corresponding Exercises can link back to us + \label{ans:#1} +} + + +% \begin{Exercises}[cols]{Pagexx} +% Set section title and reduce font size +\newenvironment{Exercises}[2][]{% + \vspace{0pt plus 36pt}% Encourage a page break; N.B. hard-coded vspace +% If we're passed a Pagexx argument (#2 non-empty), set opening label... + \ifthenelse{\equal{#2}{}} + {\setboolean{AnsArg}{false}}% store argument for later use... + {\setboolean{AnsArg}{true}% + \phantomsection\label{exer:#2:start}% + \renewcommand{\AnsPg}{exer:#2:end}}% + +% write section heading; optional arg provides own space for simplicity + \ifthenelse{\boolean{AnsArg}}% We have an Answers section + {\section*{\centering\normalfont\normalsize\bfseries% + \hyperref[ans:#2]{#1EXERCISES}}\small}% Link section heading to answers + {\section*{\centering\normalfont\normalsize\bfseries #1EXERCISES}\small} +}{% End of Exercises + \normalfont\normalsize% +% and set closing label if necessary. + \ifthenelse{\boolean{AnsArg}}{\phantomsection\label{\AnsPg}}{}% +} + + +% #1 = number of columns, 1 by default +\newenvironment{Problems}[1][]{% + \begin{list}{}{\ListInit}% + \ifthenelse{\equal{#1}{}}{\setboolean{InMulticols}{false}}% + {\setboolean{InMulticols}{true}\begin{multicols}{#1}}% +}{% + \ifthenelse{\boolean{InMulticols}}{\end{multicols}}{}% + \setboolean{InMulticols}{false} + \end{list}% +} + + +% Manage multicolumn state, then call \AH +\newenvironment{Answers}[2][] +{% + \begin{list}{}{\ListInitAns}% + \ifthenelse{\equal{#1}{1}} + {\AH{#2}\setboolean{InMulticols}{false}}% Update state; else + {% + \setboolean{InMulticols}{true} + \ifthenelse{\equal{#1}{}}% Two columns by default + {\begin{multicols}{2}[\raggedcolumns\AH{#2}]}% Else + {\begin{multicols}{#1}[\raggedcolumns\AH{#2}]}% + } +}{% + \ifthenelse{\boolean{InMulticols}}{\end{multicols}}{}% + \end{list}% + \setboolean{InMulticols}{false} +} + + +% Document divisions: \Chapter (+ \Preface and \Appendix), \Section +% \Chapter{I}{Title} +\newcommand{\Chapter}[3][] +{% + % Chapter I has a half-title; unlike in original, chapters start recto + \ifthenelse{\not\equal{#2}{I}}% stuff done for everyone but Chapter I + {% + \clearpage + \fancyhead{} % clear stale headings + \cleardoublepage + \ChapSkip + % Encourage page break before ToC heading; N.B. hard-coded vspace + \addtocontents{toc}{\protect\vspace{0pt plus 24pt}} + }{}% else do nothing + + % Set up running heads + \pagestyle{fancy} + \fancyfoot{} + \ifthenelse{\boolean{ForPrinting}} + {\fancyhead[LE,RO]{\thepage}% + \fancyhead[RE]{[\small\scshape Ch.~#2}}% + % For screen-formatted version + {\fancyhead[R]{\thepage}% + \fancyhead[LE]{[\small\scshape Ch.~#2]}}% + + \thispagestyle{empty} + + \section*{\centering\large CHAPTER~#2\TitleSkip% + \normalfont\normalsize\scshape #3} + + % cross-referencing + \label{chap:#2} + + \ifthenelse{\equal{#1}{}}% Optional argument omitted? + {% Use full title + \pdfbookmark[0]{#3}{#3}% + \fancyhead[CE]{\Running\MakeTextUppercase{#3}}% Odd heads set by Paragraph + % ToC entry + \addtocontents{toc}{\protect\subsubsection*{\scshape\protect\centering% + CHAPTER #2\protect\\ \protect\hyperref[chap:#2]{\protect\small #3}}}% + }% + {% Else use short title from optional argument + \pdfbookmark[0]{#1}{#1}% + \fancyhead[CE]{\Running\MakeTextUppercase{#1}}% + \addtocontents{toc}{\protect\subsubsection*{\scshape\protect\centering% + CHAPTER #2\protect\\ \protect\hyperref[chap:#2]{\protect\small #1}}}% + } + + \setcounter{footnote}{0} +} + +\newcommand{\Preface} +{% + \cleardoublepage + \ChapSkip + \section*{\centering\large PREFACE} + \begin{center}\TB\end{center} + + \pagestyle{fancy} + \fancyfoot{} + \fancyhead{} + \setlength{\headheight}{15pt} + \ifthenelse{\boolean{ForPrinting}} + {\fancyhead[LE,RO]{\thepage}}% Print-formatted + {\fancyhead[R]{\thepage}}% Screen-formatted + + \fancyhead[C]{\Running PREFACE} + \thispagestyle{empty} +} + +\newcommand{\Appendix} +{% + \clearpage + \fancyhead{} % clear stale headings + \cleardoublepage + \ChapSkip + + \section*{\centering\large APPENDIX} + \begin{center}\TB\end{center} + \subsection*{\centering\normalfont\normalsize\scshape% + THE FUNDAMENTAL THEOREM OF ALGEBRA} + + \pdfbookmark[0]{Appendix}{Appendix}% + \label{appendix}% + + \ifthenelse{\boolean{ForPrinting}} + {\fancyhead[LE,RO]{\thepage}}% Print-formatted + {\fancyhead[R]{\thepage}}% Screen-formatted + + \fancyhead[CE]{\Running APPENDIX} + \fancyhead[CO]{\Running FUNDAMENTAL THEOREM OF ALGEBRA} + \thispagestyle{empty} + % ToC entry + \addtocontents{toc}{\protect\subsubsection*{\scshape\protect\centering + APPENDIX\protect\\ + \protect\hyperref[appendix]{\protect\small Fundamental Theorem of Algebra}}} + + \setcounter{footnote}{0} +} + +% \Section[ToC entry]{number}{title} +% Run-in heading, ToC entry defaults to title. +% If the optional argument is present, the title must provide its own +% trailing punctuation, and the optional argument must not contain a +% trailing period. If the optional argument is absent, the \Section +% command adds a trailing period to the title. +\newcommand{\Section}[3][] +{% + \phantomsection% + \ifthenelse{\equal{#1}{}}%optional argument absent + {% + \paragraph{\indent#2~#3.}% add trailing . + \addtocontents{toc}{\protect\small% + \hyperref[section:#2]{#3},~\pageref*{section:#2}.}% + \fancyhead[CO]{\Running\MakeTextUppercase{#3}}% + }{% else optional argument present + \paragraph{\indent#2~#3}% #3 provides punctuation as needed + \addtocontents{toc}{\protect\small% + \hyperref[section:#2]{#1},~\pageref*{section:#2}.}% + \fancyhead[CO]{\Running\MakeTextUppercase{#1}}% + } + + % anchor and running head + \label{section:#2}% + \ifthenelse{\boolean{ForPrinting}} + {\fancyhead[LO]{\small §#2]}} + {\fancyhead[LO]{\small [§#2]}} +} + +% Other non-repeated document units (Contents, Index, Answers) +\newcommand{\PrintContents} +{% + \clearpage + \fancyhead{} + \fancyhead[LE,RO]{\thepage} + \fancyhead[C]{\Running CONTENTS} + \tableofcontents + \clearpage + \fancyhead{} +} + +\newcommand{\PrintIndex} +{% + \cleardoublepage + \phantomsection + \addcontentsline{toc}{chapter} + {\texorpdfstring{\protect\small\protect\scshape Index}{Index}} + + \fancyhead{} + \fancyhead[LE,RO]{\thepage} + \fancyhead[C]{\Running INDEX} + \printindex +} + +\newcommand{\PrintAnswers} +{% + \cleardoublepage + \section*{\centering\large ANSWERS} + + \fancyhead{} + \fancyhead[LE,RO]{\thepage} + \fancyhead[C]{\Running ANSWERS} + \thispagestyle{empty} + % ToC entry + \addcontentsline{toc}{chapter} + {\texorpdfstring{\protect\small\protect\scshape Answers}{Answers}} +} + + +% Logical units +\newenvironment{Proof} + {\normalfont\normalsize{\itshape Proof}.\quad} + {\normalfont} + +% Optional numerical argument +\newenvironment{Example}[1][] + {\par\BeforeSkip\normalfont\small% + \ifthenelse{\equal{#1}{}} + {\scshape Example.}% + {\scshape Example~#1}% + \quad\upshape} + {\normalfont\normalsize} + +% Optional argument of arbitrary text and format +\newenvironment{Theorem}[1][] + {\par\BeforeSkip\normalfont\normalsize% + \ifthenelse{\equal{#1}{}} + {\scshape Theorem.}% + {\scshape #1}% + \quad\itshape} + {\normalfont} + +% Optional numerical argument +\newenvironment{Lemma}[1][] + {\par\BeforeSkip\normalfont\normalsize% + \ifthenelse{\equal{#1}{}} + {\scshape Lemma.}% + {\scshape Lemma~#1}% + \quad\itshape} + {\normalfont} + + +% Unnumbered environments: Criterion, Corollary, Solution, Note, Remark +\newenvironment{Criterion} + {\par\BeforeSkip\normalfont\normalsize\scshape Criterion.\quad\itshape} + {\normalfont} + +\newenvironment{Corollary} + {\par\BeforeSkip\normalfont\normalsize\scshape Corollary.\quad\itshape} + {\normalfont} + +\newenvironment{Solution} + {\par\BeforeSkip\normalfont\small\itshape Solution.\quad\upshape} + {\normalfont\normalsize} + +\newenvironment{Note} + {\normalfont\small\scshape Note.\quad\upshape} + {\normalfont\normalsize} + +% Unmarked segments in smaller font +\newenvironment{Remark} + {\normalfont\small\upshape} + {\normalfont\normalsize} + +% For italicized statements with run-in Pargraph headings +% Make parentheses upright? 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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.org + + +Title: First Course in the Theory of Equations + +Author: Leonard Eugene Dickson + +Release Date: August 25, 2009 [EBook #29785] +Most recently updated: June 11, 2021 + +Language: English + +Character set encoding: UTF-8 + +*** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF EQUATIONS *** +\end{PGtext} +\end{minipage} +\end{center} +\vfill + +\clearpage + +% Credits and transcriber's note +\phantomsection +\pdfbookmark[0]{Transcriber's Note}{Transcriber's Note} + +\begin{center} +\begin{minipage}{\textwidth} +\begin{PGtext} +Produced by Peter Vachuska, Andrew D. Hwang, Dave Morgan, +and the Online Distributed Proofreading Team at +http://www.pgdp.net +\end{PGtext} +\end{minipage} +\end{center} +\vfill + +\begin{minipage}{0.85\textwidth} +\small +\subsection*{\centering\normalfont\Titling\TransNote} + +\raggedright +\TransNoteText +\end{minipage} + +%% -----File: 001.png---Folio xx------- + +\frontmatter +\pagenumbering{roman} +\pagestyle{empty} + +\settowidth{\TmpLen}{{\Large\Titling JOHN WILEY \& SONS, Inc.}} +\begin{center} +\Titling +{\Huge FIRST COURSE}\\[0.25in] +{\Large IN THE}\\[0.25in] +{\Huge THEORY OF EQUATIONS}\\[0.75in] +BY\TitleSkip +{\large LEONARD EUGENE DICKSON, Ph.D.} + +{\tiny CORRESPONDANT DE L'INSTITUT DE FRANCE}\\[-4pt] +{\tiny PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF CHICAGO} + +\vfill +NEW YORK\TitleSkip +% [** PP: Next line is underfull, but deliberately spaced out] +\makebox[\TmpLen][s]{\large JOHN WILEY \& SONS, Inc.}\\[0pt] +{\small London: CHAPMAN \& HALL, Limited} +\normalfont +\end{center} + +%% -----File: 002.png---Folio xx------- + +\clearpage + +\settowidth{\TmpLen}{\itshape This book or any part thereof must not} +\begin{center} +\null\vfill +\textsc{Copyright, 1922,}\TitleSkip +\textsc{by}\TitleSkip +{\Titling LEONARD EUGENE DICKSON}\TitleSkip +\rule{0.5in}{0.5pt}\TitleSkip +\textit{All Rights Reserved} + +\makebox[\TmpLen][s]{\itshape This book or any part thereof must not}\\ +% [** PP: Next line is underfull, but deliberately spaced out] +\makebox[\TmpLen][s]{\itshape be reproduced in any form without} \\ +\makebox[\TmpLen][s]{\itshape the written permission of the publisher.} +\vfill + +Printed in U. S. A.\\[0.25in] +{\normalfont\tiny\sffamily +PRESS OF \\[0.5ex] +BRAUNWORTH\quad \&\quad CO.,\quad INC. \\[0.5ex] % [** PP: Added .] +BOOK\quad MANUFACTURERS \\[0.5ex] +BROOKLYN, NEW YORK \\ +} +\end{center} + +\iffalse +11/30 % [** PP: Omitted printer's mark] +\fi + +%% -----File: 003.png---Folio xx------- + + +\Preface + +The theory of equations is not only a necessity in the subsequent +mathematical courses and their applications, but furnishes an illuminating +sequel to geometry, algebra and analytic geometry. Moreover, +it develops anew and in greater detail various fundamental ideas of +calculus for the simple, but important, case of polynomials. The +theory of equations therefore affords a useful supplement to differential +calculus whether taken subsequently or simultaneously. + +It was to meet the numerous needs of the student in regard to his +earlier and future mathematical courses that the present book was +planned with great care and after wide consultation. It differs essentially +from the author's \textit{Elementary Theory of Equations}, both in regard to +omissions and additions, and since it is addressed to younger students +and may be used parallel with a course in differential calculus. Simpler +and more detailed proofs are now employed. The exercises are simpler, +more numerous, of greater variety, and involve more practical applications. + +This book throws important light on various elementary topics. +For example, an alert student of geometry who has learned how to bisect +any angle is apt to ask if every angle can be trisected with ruler and +compasses and if not, why not. After learning how to construct regular +polygons of $3$, $4$, $5$, $6$, $8$ and~$10$ sides, he will be inquisitive about the +missing ones of~$7$ and~$9$ sides. The teacher will be in a comfortable position +if he knows the facts and what is involved in the simplest discussion to +date of these questions, as given in \ChapRef{III}. Other chapters throw +needed light on various topics of algebra. In particular, the theory +of graphs is presented in \ChapRef{V} in a more scientific and practical +manner than was possible in algebra and analytic geometry. + +There is developed a method of computing a real root of an equation +with minimum labor and with certainty as to the accuracy of all the +decimals obtained. We first find by Horner's method successive transformed +%% -----File: 004.png---Folio xx------- +equations whose number is half of the desired number of significant +figures of the root. The final equation is reduced to a linear equation +by applying to the constant term the correction computed from the +omitted terms of the second and higher degrees, and the work is completed +by abridged division. The method combines speed with control of +accuracy. + +Newton's method, which is presented from both the graphical and +the numerical standpoints, has the advantage of being applicable also to +equations which are not algebraic; it is applied in detail to various such +equations. + +In order to locate or isolate the real roots of an equation we may +employ a graph, provided it be constructed scientifically, or the theorems +of Descartes, Sturm, and Budan, which are usually neither stated, nor +proved, correctly. + +The long chapter on determinants is independent of the earlier chapters. +The theory of a general system of linear equations is here presented +also from the standpoint of matrices. + +For valuable suggestions made after reading the preliminary manuscript +of this book, the author is greatly indebted to Professor Bussey +of the University of Minnesota, Professor Roever of Washington University, +Professor Kempner of the University of Illinois, and Professor +Young of the University of Chicago. The revised manuscript was much +improved after it was read critically by Professor Curtiss of Northwestern +University. The author's thanks are due also to Professor Dresden of +the University of Wisconsin for various useful suggestions on the +proof-sheets. + +{\small\textsc{Chicago, 1921.}} + +%% -----File: 005.png---Folio xx------- + +\PrintContents + +\iffalse%%%% COMMENTED HARD-CODED TABLE OF CONTENTS %%%% +\begin{center} +\section*{CONTENTS} + +{\tiny Numbers refer to pages.} + +\rule{3em}{0.75pt} + +\subsection*{CHAPTER I} + +\textsc{Complex Numbers} +\end{center} + +Square roots, 1. Addition, multiplication and division of complex numbers, +2. Cube roots of unity, 3. Geometrical representation, 3. Product +and quotient, 4. De Moivre's theorem, 5. Cube roots, 5. $n$th roots, 7. +Roots of unity, 8. Primitive roots of unity, 9. + +\begin{center} +\subsection*{CHAPTER II} + +\textsc{Elementary Theorems on the Roots of an Equation} +\end{center} + +Quadratic equation, 11. Remainder theorem, 12. Synthetic division, 13. +Factored form of a polynomial, 15. Multiple roots, 16. Identical polynomials, +16. Relations between the roots and the coefficients, 17. Imaginary +roots occur in pairs, 19. Upper limit to the real roots, 21. Integral roots, +24. Rational roots, 27. + +\begin{center} +\subsection*{CHAPTER III} + +\textsc{Constructions with Ruler and Compasses} +\end{center} + +Graphical solution of a quadratic equation, 29. Analytic criterion for +constructibility, 30. Cubic equations with a constructible root, 32. Trisection +of an angle, 34. Duplication of a cube, 35. Regular polygons of +7, 9, 17, and $n$ sides, 35-44. Reciprocal equations, 37. + +\begin{center} +\subsection*{CHAPTER IV} + +\textsc{Cubic and Quartic Equations} +\end{center} + +Algebraic solution of a cubic, 45. Discriminant, 47. Number of real roots +of a cubic, 48. Trigonometric solution of a cubic, 49. Ferrari's and Descartes' +solutions of a quartic, 50. Resolvent cubic, 51. Discriminant of a quartic, 51. + +\begin{center} +\subsection*{CHAPTER V} + + +\textsc{The Graph of an Equation} +\end{center} + +Use of graphs, 55. Caution in plotting, 55. Bend points, 56. Derivatives, +58. Horizontal tangents, 60. Multiple roots, 60. Ordinary and inflexion +%% -----File: 006.png---Folio xx------- +* tangents, 62. Real roots of a cubic equation, 65. Continuity, 66. Condition +for a root between $a$ and $b$, 67. Sign of a polynomial at infinity, 68. +Rolle's theorem, 69. + +\begin{center} +\subsection*{CHAPTER VI} + +\textsc{Isolation of the Real Roots} +\end{center} + +Descartes' rule of signs, 71. Sturm's method, 75. Sturm's functions for +the general quartic equation, 80. Budan's theorem, 83. + +\begin{center} +\subsection*{CHAPTER VII} + +\textsc{Solution of Numerical Equations} +\end{center} + +Horner's method, 86. Newton's method, algebraic and graphical discussion, +systematic computation, also for functions not polynomials, 90. Imaginary +roots, 98. + +\begin{center} +\subsection*{CHAPTER VIII} + +\textsc{Determinants; Systems of Linear Equations} +\end{center} + +Solution of 2 or 3 linear equations by determinants, 101. Even and odd +arrangements, 103. Definition of a determinant of order $n$, 105. Interchange +of rows and columns, 106. Interchange of two columns or two rows, 107. +Minors, 109. Expansion, 109. Removal of factors, 111. Sum of determinants, +112. Addition of columns or rows, 113. Rank, 116. System +of $n$ linear equations in n unknowns, 114, 116. Homogeneous equations, 119. +System of $m$ linear equations in $n$ unknowns, matrix and augmented matrix, +120. Complementary minors, 122. Laplace's development, 122. Product of +determinants, 124. + + +\begin{center} +\subsection*{CHAPTER IX} + +\textsc{Symmetric Functions} +\end{center} + +Sigma functions, 128. Elementary symmetric functions, 128. Fundamental +theorem, 129. Rational functions symmetric in all but one of the +roots, 132. Sums of like powers of the roots, Newton's identities, 134. +Waring's formula, 136. Computation of symmetric functions, 141. + +\begin{center} +\subsection*{CHAPTER X} + +\textsc{Elimination, Resultants and Discriminants} +\end{center} + +Methods of Sylvester, Euler, and Bézout, 143. Discriminants, 152. + +\begin{center} +\subsection*{APPENDIX} + +\textsc{The Fundamental Theorem of Algebra} +\end{center} + +\textsc{Answers} \dotfill 159 + +\textsc{Index} \dotfill 167 +\fi%%%% END OF COMMENTED HARD-CODED TABLE OF CONTENTS %%%% + +%% -----File: 007.png---Folio 1------- + +\mainmatter +\pagenumbering{arabic} + +\ChapSkip +\begin{center} +\textbf{\LARGE First Course in}\\[0.25in] +\textbf{\LARGE The Theory of Equations} +\medskip + +\TB +\end{center} + +\Chapter{I}{Complex Numbers} + + +\Section{1.}{Square Roots} If $p$ is a positive real number, the symbol~$\sqrt{p}$ is +used to denote the positive square root of~$p$. It is most easily computed +by logarithms. +\index{Square roots}% + +We shall express the square roots of negative numbers in terms of the +symbol~$i$ such that the relation $i^2 = -1$ holds. Consequently we denote +the roots of $x^2 = -1$ by~$i$ and~$-i$. The roots of $x^2 = -4$ are written in the +form $± 2i$ in preference to $±\sqrt{-4}$. In general, if $p$ is positive, the roots +of $x^2=-p$ are written in the form $±\sqrt{p} i$ in preference to $±\sqrt{-p}$. + +\begin{Remark} +The square of either root is thus $(\sqrt{p})^2 i^2 = -p$. Had we used the less desirable +notation $±\sqrt{-p}$ for the roots of $x^2 = -p$, we might be tempted to find the square of +either root by multiplying together the values under the radical sign and conclude +erroneously that +\[ +\sqrt{-p}\,\sqrt{-p} = \sqrt{p^2} = +p. +\] +To prevent such errors we use $\sqrt{p}\,i$ and not~$\sqrt{-p}$. +\end{Remark} + + +% [** PP: ToC entry reads Addition, multiplication and division of complex numbers] +\Section{2.}{Complex Numbers} If $a$ and~$b$ are any two real numbers and +\index{Complex number}% +$i^2 = -1$, $a+bi$ is called a \emph{complex number}\footnote + {Complex numbers are essentially couples of real numbers. For a treatment from + this standpoint and a treatment based upon vectors, see the author's \textit{Elementary Theory + of Equations}, p.~21, p.~18.} +and $a-bi$ its \emph{conjugate}. Either +\index{Conjugate}% +is said to be \emph{zero} if $a = b = 0$. Two complex numbers $a+bi$ and $c+di$ +are said to be \emph{equal} if and only if $a = c$ and $b = d$. In particular, $a+bi=0$ +%% -----File: 008.png---Folio 2------- +if and only if $a = b = 0$. If $b \ne 0$, $a+bi$ is said to be \emph{imaginary}. In particular, +$bi$ is called a \emph{pure imaginary}. +\index{Pure imaginary}% +\index{Imaginary}% + +Addition of complex numbers is defined by +\[ +(a+bi) + (c+di) = (a+c) + (b+d)i. +\] +The inverse operation to addition is called subtraction, and consists in +finding a complex number~$z$ such that +\[ +(c+di) + z = a+bi. +\] +In notation and value, $z$ is +\[ +(a+bi) - (c+di) = (a-c) + (b-d)i. +\] + +Multiplication is defined by +\[ +(a+bi)(c+di) = ac-bd+(ad+bc)i, +\] +and hence is performed as in formal algebra with a subsequent reduction +by means of $i^2 = -1$. For example, +\[ +(a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2. +\] + +Division is defined as the operation which is inverse to multiplication, +and consists in finding a complex number~$q$ such that $(a+bi)q = e+fi$. +Multiplying each member by $a-bi$, we find that~$q$ is, in notation and +value, +\[ +\frac{e+fi}{a+bi} = \frac{(e+fi)(a-bi)}{a^2+b^2} + = \frac{ae+bf}{a^2+b^2} + \frac{af-be}{a^2+b^2} i. +\] +Since $a^2+b^2 = 0$ implies $a = b = 0$ when $a$ and~$b$ are real, we conclude that +division except by zero is possible and unique. + + +\begin{Exercises}{Page2} +Express as complex numbers +\begin{Problems}[2] + +\item[1.] $\sqrt{-9}$. + +\item[2.] $\sqrt{4}$. + +\ResetCols{2} + +\item[3.] $(\sqrt{25} + \sqrt{-25})\sqrt{-16}$. + +\item[4.] $-\frac{2}{3}$. + +\ResetCols{4} + +\item[5.] $8 + 2\sqrt{3}\vphantom{\dfrac{1}{1}}$. + +\item[6.] $\dfrac{3 + \sqrt{-5}}{2 + \sqrt{-1}}$. + +\item[7.] $\dfrac{3 + 5i}{2 - 3i}$. + +\item[8.] $\dfrac{a + bi}{a - bi}$. + +\ResetCols{1} + +\item[9.] Prove that the sum of two conjugate complex numbers is real and that their +difference is a pure imaginary. + +\item[10.] Prove that the conjugate of the sum of two complex numbers is equal to the +sum of their conjugates. Does the result hold true if each word sum is replaced by the +word difference? + +%% -----File: 009.png---Folio 3------- + +\item[11.] Prove that the conjugate of the product (or quotient) of two complex numbers +is equal to the product (or quotient) of their conjugates. + +\item[12.] Prove that, if the product of two complex numbers is zero, at least one of them +is zero. + +\item[13.] Find two pairs of real numbers $x$, $y$ for which +\[ +(x+yi)^2 = -7+24i. +\] +\end{Problems} + +As in Ex.~13, express as complex numbers the square roots of +\begin{Problems}[3] + +\item[14.] {$-11+60i$.} + +\item[15.] {$5-12i$.} + +\item[16.] {$4cd+(2c^2-2d^2)i$.} + +\end{Problems} + +\end{Exercises} + + +\Section{3.}{Cube Roots of Unity} Any complex number~$x$ whose cube is equal +to unity is called a \emph{cube root of unity}. Since +\index{Cube root!of unity}% +\[ +x^3-1 = (x-1) (x^2+x+1), +\] +the roots of $x^3=1$ are~$1$ and the two numbers~$x$ for which +\[ +x^2+x+1=0,\qquad +(x + \tfrac{1}{2})^2 = -\tfrac{3}{4}, \qquad + x + \tfrac{1}{2} = ±\tfrac{1}{2} \sqrt{3}i. +\] +Hence there are three cube roots of unity, viz., +\[ +1, \qquad +\omega = -\tfrac{1}{2} + \tfrac{1}{2}\sqrt{3}i,\qquad +\omega' = -\tfrac{1}{2} - \tfrac{1}{2}\sqrt{3}i. +\] + +In view of the origin of~$\omega$, we have the important relations +\[ +\omega^2 + \omega+1 = 0, \quad \omega^3 = 1. +\] +Since $\omega \omega' = 1$ and $\omega^3 = 1$, it follows that $\omega' = \omega^2$, $\omega = \omega'^2$. + + +\Section[Geometrical Representation] +{4.}{Geometrical Representation of Complex Numbers.} Using rectangular +axes of coördinates, $OX$ and~$OY$, we represent the complex number +$a+bi$ by the point~$A$ having the coördinates $a$, $b$ (Fig.~1). +\index{Complex number!geometrical representation}% + +%[Illustration: \textsc{Fig}. 1] +\begin{wrapfigure}{r}{2.25in} +\hfill\Input{009a} +\end{wrapfigure} +The positive number $r = \sqrt{a^2+b^2}$ giving +the length of~$OA$ is called the \emph{modulus} (or +\emph{absolute value}) of~$a+bi$. The angle $\theta = XOA$, +measured counter-clockwise from~$OX$ to~$OA$, +is called the \emph{amplitude} (or \emph{argument}) of~$a+bi$. +\index{Absolute value}% +\index{Amplitude}% +\index{Argument}% +\index{Modulus}% +Thus $\cos \theta = a/r$, $\sin \theta = b/r$, whence +\[ +a+bi = r(\cos\theta + i\sin\theta). +\Tag{1} +\] +The second member is called the \emph{trigonometric form} of~$a+bi$. +\index{Complex number!trigonometric form}% + +For the amplitude we may select, instead of~$\theta$, any of the angles $\theta±360°$, +$\theta±720°$, etc. + +%% -----File: 010.png---Folio 4------- + +Two complex numbers are equal if and only if their moduli are equal +and an amplitude of the one is equal to an amplitude of the other. + +%[Illustration: \textsc{Fig}. 2] +\begin{Remark} +\begin{wrapfigure}{l}{2.125in} +\Input{010a} +\end{wrapfigure} +For example, the cube roots of unity are~$1$ and +\index{Cube root!of unity}% +\begin{align*} +\omega &= -\tfrac{1}{2} + \tfrac{1}{2}\sqrt{3} i \\ + &= \cos 120° + i \sin 120°, \\ +\omega^2 &= -\tfrac{1}{2} - \tfrac{1}{2}\sqrt{3} i \\ + &= \cos 240° + i \sin 240°, +\end{align*} +and are represented by the points marked $1$, $\omega$, $\omega^2$ +at the vertices of an equilateral triangle inscribed +in a circle of radius unity and center at the origin~$O$ +(Fig.~2). The indicated amplitudes of~$\omega$ and~$\omega^2$ +are~$120°$ and~$240°$ respectively, while the modulus +of each is~$1$. + +The modulus of~$-3$ is $3$ and its amplitude is~$180°$ or~$180°$ plus or minus the product +of~$360°$ by any positive whole number. +\end{Remark} + + +% [** PP: ToC entry reads ``Product and quotient'', splitting] +\Section[Product] +{5.}{Product of Complex Numbers.} By actual multiplication, +\begin{align*} +&\quad +\bigl[r (\cos\theta + i \sin\theta)\bigr] +\bigl[r'(\cos\alpha + i \sin\alpha)\bigr] \\ +&= rr' +\bigl[ (\cos\theta \cos\alpha - \sin\theta \sin\alpha) + + i(\sin\theta \cos\alpha + \cos\theta \sin\alpha)\bigr] \\ +&= rr' +\bigl[ \cos(\theta + \alpha) + i \sin(\theta + \alpha)], \quad +\text{by trigonometry.} +\end{align*} +Hence \textit{the modulus of the product of two complex numbers is equal to the +product of their moduli, while the amplitude of the product is equal to the +sum of their amplitudes.} + +\begin{Remark} +For example, the square of $\omega = \cos 120°+ i \sin 120°$ has the modulus~$1$ and the amplitude +$120°+120°$ and hence is $\omega^2 = \cos 240°+i \sin 240°$. Again, the product of~$\omega$ and~$\omega^2$ +has the modulus~$1$ and the amplitude $120°+ 240°$ and hence is $\cos 360°+ i \sin 360°$, +which reduces to~$1$. This agrees with the known fact that $\omega^3 = 1$. +\end{Remark} + +Taking $r = r' = 1$ in the above relation, we obtain the useful formula +\[ +(\cos \theta + i \sin \theta) +(\cos \alpha + i \sin \alpha) + = \cos (\theta + \alpha) + i \sin (\theta + \alpha). +\Tag{2} +\] + +% [** PP: Split ToC entry, part 2] +\Section[Quotient] +{6.}{Quotient of Complex Numbers.} Taking $\alpha = \beta - \theta$ in~\Eq{2} and dividing +the members of the resulting equation by $\cos\theta + i \sin\theta$, we get +\[ +\frac{\cos \beta + i \sin \beta} + {\cos \theta + i \sin \theta} + = \cos(\beta - \theta) + i \sin(\beta - \theta). +\] +%% -----File: 011.png---Folio 5------- +Hence \textit{the amplitude of the quotient of $R(\cos \beta+i \sin \beta)$ by $r(\cos \theta+i \sin \theta)$ +is equal to the difference $\beta - \theta$ of their amplitudes, while the modulus of the +quotient is equal to the quotient~$R/r$ of their moduli.} + +The case $\beta = 0$ gives the useful formula +\[ +\frac{1}{\cos\theta + i \sin\theta} = \cos\theta - i \sin\theta. +\] + + +\Section{7.}{De Moivre's Theorem} +\index{De Moivre's!theorem}% +\begin{Thm} +If $n$ is any positive whole number, +\[ +(\cos\theta + i \sin\theta)^n = \cos n\theta + i \sin n\theta. +\Tag{3} +\] +\end{Thm} + +This relation is evidently true when $n = 1$, and when $n = 2$ it follows +from formula~\Eq{2} with $\alpha = \theta$. To proceed by mathematical induction, +suppose that our relation has been established for the values $1, 2, \dotsc, m$ +of~$n$. We can then prove that it holds also for the next value $m+1$ of~$n$. +For, by hypothesis, we have +\[ +(\cos\theta + i \sin\theta)^m = \cos m\theta + i \sin m\theta. +\] +Multiply each member by $\cos\theta + i \sin\theta$, and for the product on the right +substitute its value from~\Eq{2} with $\alpha = m \theta$. Thus +\begin{align*} +(\cos\theta + i \sin\theta)^{m+1} + &= (\cos\theta + i \sin\theta)(\cos m\theta + i \sin m\theta), \\ + &= \cos(\theta + m \theta) + i \sin(\theta + m \theta), +\end{align*} +which proves~\Eq{3} when $n = m+1$. Hence the induction is complete. + +\begin{Remark} +Examples are furnished by the results at the end of~§5: +\begin{align*} +(\cos 120° + i \sin 120°)^2 = \cos 240° + i \sin 240°, \\ +(\cos 120° + i \sin 120°)^3 = \cos 360° + i \sin 360°. +\end{align*} +\end{Remark} + + +\Section{8.}{Cube Roots} To find the cube roots of a complex number, we first +express the number in its trigonometric form. For example, +\index{Cube root}% +\[ +4\sqrt{2} + 4\sqrt{2} i = 8(\cos 45° + i \sin 45°). +\] +If it has a cube root which is a complex number, the latter is expressible +in the trigonometric form +\[ +r(\cos\theta + i \sin\theta). +\Tag{4} +\] +The cube of the latter, which is found by means of~\Eq{3}, must be equal +to the proposed number, so that +\[ +r^3(\cos 3\theta + i \sin 3\theta) = 8(\cos 45° + i \sin 45°). +\] +%% -----File: 012.png---Folio 6------- +The moduli $r^3$ and~$8$ must be equal, so that the positive real number~$r$ +is equal to~$2$. Furthermore, $3 \theta$ and~$45°$ have equal cosines and equal +sines, and hence differ by an integral multiple of~$360°$. Hence $3 \theta = 45°+ +k·360°$, or $\theta = 15°+k·120°$, where $k$ is an integer.\footnote + {Here, as elsewhere when the contrary is not specified, zero and negative as well + as positive whole numbers are included under the term ``integer.''} +Substituting this +value of~$\theta$ and the value~$2$ of~$r$ in~\Eq{4}, we get the desired cube roots. The +values $0$, $1$, $2$ of~$k$ give the distinct results +\begin{alignat*}{2} % [** PP: Re-breaking, aligning] +R_1 &= 2(\cos 15° &+{}& i \sin 15°), \\ +R_2 &= 2(\cos 135° &+{}& i \sin 135°),\\ +R_3 &= 2(\cos 255° &+{}& i \sin 255°). +\end{alignat*} + +Each new integral value of~$k$ leads to a result which is equal to~$R_1$, +$R_2$ or~$R_3$. In fact, from $k = 3$ we obtain~$R_1$, from $k = 4$ we obtain~$R_2$, from +$k = 5$ we obtain~$R_3$, from $k = 6$ we obtain~$R_1$ again, and so on periodically. + + +\begin{Exercises}{Page6} + +\begin{Problems} +\item[1.] Verify that $R_2 = \omega R_1$, $R_3 = \omega^2 R_1$. Verify that $R_1$ is a cube root of $8 (\cos 45°+ +i \sin 45°)$ by cubing~$R_1$ and applying De Moivre's theorem. Why are the new expressions +for~$R_2$ and~$R_3$ evidently also cube roots? + +\item[2.] Find the three cube roots of~$-27$; those of~$-i$; those of~$\omega$. + +\item[3.] Find the two square roots of~$i$; those of~$-i$; those of~$\omega$. + +\item[4.] Prove that the numbers $\cos\theta + i \sin\theta$ and no others are represented by points +on the circle of radius unity whose center is the origin. + +\item[5.] If $a+bi$ and $c+di$ are represented by the points~$A$ and~$C$ in Fig.~3, prove that +their sum is represented by the fourth vertex~$S$ of the parallelogram two of whose sides +are~$OA$ and~$OC$. Hence show that the modulus of the sum of two complex numbers +is equal to or less than the sum of their moduli, and is equal to or greater than the difference +of their moduli. +\index{Complex number!geometrical representation}% +%[Illustration: \textsc{Fig}. 3] +%[Illustration: \textsc{Fig}. 4] +\begin{figure*}[hbt] +\begin{center} +\Input{012a}\hfil +\Input{012b} +\end{center} +\end{figure*} + +%% -----File: 013.png---Folio 7------- + +\item[6.] Let $r$ and~$r'$ be the moduli and $\theta$ and~$\alpha$ the amplitudes of two complex numbers +represented by the points $A$ and~$C$ in Fig.~4. Let~$U$ be the point on the $x$-axis one +unit to the right of the origin~$O$. Construct triangle $OCP$ similar to triangle $OUA$ and +similarly placed, so that corresponding sides are $OC$ and~$OU, CP$ and~$UA$, $OP$ and~$OA$, +while the vertices $O$, $C$, $P$ are in the same order (clockwise or counter-clockwise) as +the corresponding vertices $O$, $U$, $A$. Prove that~$P$ represents the product~(§5) of the +complex numbers represented by $A$ and~$C$. + +\item[7.] If $a+bi$ and $e+fi$ are represented by the points $A$ and~$S$ in Fig.~3, prove that +the complex number obtained by subtracting $a+bi$ from $e+fi$ is represented by the point~$C$. +Hence show that the absolute value of the difference of two complex numbers is +equal to or less than the sum of their absolute values, and is equal to or greater than +the difference of their absolute values. + +\item[8.] By modifying Ex.~6, show how to construct geometrically the quotient of two +complex numbers. +\index{Complex number!geometrical representation}% +\end{Problems} +\end{Exercises} + + +% [** PP: ToC entry matches unit title, running head is as given] +\Section[Roots of Complex Numbers] +{9.}{$n$th~Roots.} As illustrated in~§8, it is evident that the $n$th roots +of any complex number $\rho(\cos A + i \sin A)$ are the products of the $n$th +roots of $\cos A + i \sin A$ by the positive real $n$th root of the positive real +number~$\rho$ (which may be found by logarithms). +\index{Rootz@{Roots, $n$th}}% [** PP: Rootz places entry after Roots] + +Let an $n$th root of $\cos A + i \sin A$ be of the form +\[ +r(\cos \theta+i \sin \theta). +\tag{4}%[ ** PP: [sic], equation repeated] +\] +Then, by De Moivre's theorem, +\[ +r^n(\cos n\theta + i \sin n\theta) = \cos A + i \sin A. +\] +The moduli $r^n$ and~$1$ must be equal, so that the positive real number~$r$ +is equal to~$1$. Since $n\theta$ and~$A$ have equal sines and equal cosines, they +differ by an integral multiple of~$360°$. Hence $n\theta = A + k·360°$, where~$k$ +is an integer. Substituting the resulting value of~$\theta$ and the value~$1$ of~$r$ +in~\Eq{4}, we get +\[ + \cos\left(\frac{A + k·360°}{n}\right) + + i \sin\left(\frac{A + k·360°}{n}\right). +\Tag{5} +\] + +For each integral value of~$k$, \Eq{5}~is an answer since its $n$th power reduces +to $\cos A + i \sin A$ by DeMoivre's theorem. Next, the value~$n$ of~$k$ gives +the same answer as the value~$0$ of~$k$; the value $n+1$ of~$k$ gives the same +answer as the value~$1$ of~$k$; and in general the value $n+m$ of~$k$ gives the +same answer as the value~$m$ of~$k$. Hence we may restrict attention to +the values $0, 1, \dotsc, n-1$ of~$k$. Finally, the answers~\Eq{5} given by these +%% -----File: 014.png---Folio 8------- +values $0, 1,\ldots, n-1$ of $k$ are all distinct, since they are represented by +points whose distance from the origin is the modulus~$1$ and whose amplitudes +are +\[ +\frac{A}{n},\qquad +\frac{A}{n} + \frac{360°}{n},\qquad +\frac{A}{n} + \frac{2·360°}{n},\dotsc, +\frac{A}{n} + \frac{(n-1)360°}{n}, +\] +so that these $n$ points are equally spaced points on a circle of radius unity. +Special cases are noted at the end of~§10. Hence +\begin{Thm}% +any complex number +different from zero has exactly $n$ distinct complex $n$th roots. +\end{Thm} + + +\Section{10.}{Roots of Unity} The trigonometric form of~$1$ is $\cos 0° + i \sin 0°$. +Hence by~§9 with $A=0$, the $n$ distinct $n$th roots of unity are +\index{Roots of unity}% +\[ +\cos\frac{2k \pi}{n} + i \sin\frac{2k \pi}{n}\quad +(k=0, 1, \dotsc, n-1), +\Tag{6} +\] +where now the angles are measured in radians (an angle of $180$~degrees +being equal to $\pi$~radians, where $\pi= 3.1416$, approximately). For $k = 0$, +\Eq{6} reduces to~$1$, which is an evident $n$th root of unity. For $k = 1$, \Eq{6}~is +\[ +R = \cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n}. +\Tag{7} +\] + +By De Moivre's theorem, the general number~\Eq{6} is equal to the +$k$th power of~$R$. Hence the $n$ distinct $n$th roots of unity are +\[ +R,\ R^2,\ R^3,\dotsc,\ R^{n-1},\ R^n = 1. +\Tag{8} +\] + +As a special case of the final remark in~§9, the $n$~complex numbers~\Eq{6}, +and therefore the numbers~\Eq{8}, are represented geometrically by the +vertices of a regular polygon of $n$~sides inscribed in the circle of radius +unity and center at the origin with one vertex on the positive $x$-axis. +\index{Regular polygon}% + +%[Illustraton: \textsc{Fig}. 5] +% [** PP: wrapfigure doesn't play nicely with Remark environment] +\noindent\raisebox{-12pt}{\Input{014a}}\hfill +\begin{minipage}[b]{\linewidth-1.75in} +\begin{Remark} +\hspace*{1.5em}% We're in a minipage +For $n =3$, the numbers~\Eq{8} are $\omega$, $\omega^2$, $1$, which are represented +in Fig.~2 by the vertices of an equilateral triangle. + +\hspace*{1.5em}% +For $n = 4$, $R = \cos\pi/2 + i \sin\pi/2 = i$. The four fourth roots +of unity~\Eq{8} are $i$, $i^2=-1$, $i^3=-i$, $i^4 = 1$, which are represented +by the vertices of a square inscribed in a circle of +radius unity and center at the origin~$O$ (Fig.~5). +\end{Remark} +\end{minipage} + +%% -----File: 015.png---Folio 9------- + +\begin{Exercises}{Page9} + +\begin{Problems} +\item[1.] Simplify the trigonometric forms~\Eq{6} of the four fourth roots of unity. Check +the result by factoring $x^4-1$. + +\item[2.] For $n=6$, show that $R = -\omega^2$. The sixth roots of unity are the three cube roots +of unity and their negatives. Check by factoring $x^6-1$. + +\item[3.] From the point representing $a+bi$, how do you obtain that representing $-(a+bi)$? +Hence derive from Fig.~2 and Ex.~2 the points representing the six sixth roots of unity. +Obtain this result another way. + +\item[4.] Find the five fifth roots of~$-1$. + +\item[5.] Obtain the trigonometric forms of the nine ninth roots of unity. Which of +them are cube roots of unity? + +\item[6.] Which powers of a ninth root~\Eq{7} of unity are cube roots of unity? + +\end{Problems} +\end{Exercises} + + +\Section[Primitive Roots of Unity] +{11.}{Primitive $n$th Roots of Unity.} An $n$th root of unity is called +\emph{primitive} if $n$ is the smallest positive integral exponent of a power of it +that is equal to unity. Thus $\rho$ is a primitive $n$th root of unity if and only +if $\rho^n=1$ and $\rho^l \neq 1$ for all positive integers $l<n$. +\index{Primitive root of unity}% + +Since only the last one of the numbers~\Eq{8} is equal to unity, the number~$R$, +defined by~\Eq{7}, is a primitive $n$th root of unity. We have shown that +the powers~\Eq{8} of~$R$ give all of the $n$th roots of unity. Which of these +powers of~$R$ are primitive $n$th roots of unity? + +\begin{Remark} +For $n=4$, the powers~\Eq{8} of $R=i$ were seen to be +\[ +i^1 = i,\ +i^2 = -1,\ +i^3 = -i,\ +i^4 = 1. +\] +The first and third are primitive fourth roots of unity, and their exponents $1$ and~$3$ +are relatively prime to~$4$, i.e., each has no divisor $>1$ in common with~$4$. But the +second and fourth are not primitive fourth roots of unity (since the square of $-1$ and the +first power of~$1$ are equal to unity), and their exponents $2$ and~$4$ have the divisor~$2$ in +common with $n=4$. These facts illustrate and prove the next theorem for the case +$n=4$. +\end{Remark} + +\begin{Theorem} +The primitive $n$th roots of unity are those of the numbers~\Eq{8} +whose exponents are relatively prime to $n$. +\end{Theorem} +\index{Relatively prime}% + +\begin{Proof} +If $k$ and~$n$ have a common divisor $d$ $(d>1)$, $R^k$ is not a primitive +$n$th root of unity, since +\[ +(R^k)^{\frac{n}{d}} = (R^n)^{\frac{k}{d}} = 1, +\] +and the exponent $n/d$ is a positive integer less than~$n$. + +%% -----File: 016.png---Folio 10------- + +But if $k$ and~$n$ are relatively prime, i.e., have no common divisor $>1$, +$R^k$ is a primitive $n$th root of unity. To prove this, we must show that +$(R^k)^{l} \ne 1$ if $l$~is a positive integer~$<n$. By De Moivre's theorem, +\index{Relatively prime}% +\[ +R^{kl} = \cos\frac{2kl\pi}{n} + + i \sin\frac{2kl\pi}{n}. +\] +If this were equal to unity, $2kl\pi/n$ would be a multiple of~$2\pi$, and hence +$kl$ a multiple of~$n$. Since $k$ is relatively prime to~$n$, the second factor~$l$ +would be a multiple of~$n$, whereas $0 < l < n$. +\end{Proof} + + +\begin{Exercises}{Page10} + +\begin{Problems} + +\item[1.] Show that the primitive cube roots of unity are $\omega$ and~$\omega^{2}$. + +\item[2.] For $R$ given by~\Eq{7}, prove that the primitive $n$th +roots of unity are (i)~for $n=6$, +$R$, $R^5$; (ii)~for $n=8$, $R$, $R^3$, $R^5$, $R^7$; (iii)~for $n=12$, $R$, $R^5$, $R^7$, $R^{11}$. + +\item[3.] When $n$ is a prime, prove that any $n$th root of unity, other than~$1$, is primitive. + +\item[4.] Let $R$ be a primitive $n$th root \Eq{7} of unity, where $n$ is a product of two different +primes $p$ and~$q$. Show that $R, \dotsc, R^n$ are primitive with the exception of $R^p$, $R^{2p}, \dotsc, +R^{qp}$, whose $q$th powers are unity, and $R^q$, $R^{2q}, \dotsc, R^{pq}$, whose $p$th powers are unity. +These two sets of exceptions have only $R^{pq}$ in common. Hence there are exactly +$pq - p - q + 1$ primitive $n$th roots of unity. + +\item[5.] Find the number of primitive $n$th roots of unity if $n$ is a square of a prime~$p$. + +\item[6.] Extend Ex.~4 to the case in which $n$ is a product of three distinct primes. + +\item[7.] If $R$ is a primitive $15$th root \Eq{7} of unity, verify that $R^3$, $R^6$, $R^9$, $R^{12}$ are the primitive +fifth roots of unity, and $R^5$ and~$R^{10}$ are the primitive cube roots of unity. Show +that their eight products by pairs give all the primitive $15$th roots of unity. + +\item[8.] If $\rho$ is any primitive $n$th root of unity, prove that $\rho$, $\rho^2, \dots, \rho^n$ are distinct and +give all the $n$th roots of unity. Of these show that $\rho^k$ is a primitive $n$th root of unity +if and only if $k$ is relatively prime to~$n$. + +\item[9.] Show that the six primitive $18$th roots of unity are the negatives of the primitive +ninth roots of unity. +\end{Problems} +\end{Exercises} + +%% -----File: 017.png---Folio 11------- + + +\Chapter[Theorems on Roots of Equations] +{II}{Elementary Theorems on the Roots of an Equation} + +\Section{12.}{Quadratic Equation} If $a$, $b$, $c$ are given numbers, $a \ne 0$, +\index{Quadratic equation}% +\[ +ax^2 + bx + c = 0 \quad (a \ne 0) +\Tag{1} +\] +is called a \emph{quadratic equation} or equation of the second degree. The +reader is familiar with the following method of solution by ``completing +the square.'' Multiply the terms of the equation by~$4a$, and transpose +the constant term; then +\[ +4a^{2}x^2 + 4abx = -4ac. +\] +Adding $b^2$ to complete the square, we get +\[ +(2ax + b)^2 = \Delta,\qquad \Delta = b^2 - 4ac, +\] +\[ +x_{1} = \frac{-b + \sqrt{\Delta}}{2a}\qquad +x_{2} = \frac{-b - \sqrt{\Delta}}{2a} +\Tag{2} +\] + +By addition and multiplication, we find that +\[ +x_{1} + x_{2} = \frac{-b}{a},\qquad +x_{1} x_{2} = \frac{ c}{a}. +\Tag{3} +\] +Hence for all values of the variable~$x$, +\[ +a(x - x_1)(x - x_2) + \equiv ax^2 - a(x_1 + x_2)x + ax_1 x_2 + \equiv ax^2 + bx + c, +\Tag{4} +\] +the sign $\equiv$ being used instead of~$=$ since these functions of~$x$ are \emph{identically +equal}, i.e., the coefficients of like powers of~$x$ are the same. We speak +of $a(x - x_1)(x - x_2)$ as the \emph{factored form} of the quadratic function $ax^2 + bx + c$, +and of $x - x_1$ and $x - x_2$ as its \emph{linear factors}. +\index{Factored form}% +\index{Identity}% +\index{Linear factors}% +\index{Symbol!a@{$\equiv$\IndAdd{identically equal to}}}% [** PP: Manually alphabetized] + +In \Eq{4} we assign to~$x$ the values $x_1$ and~$x_2$ in turn, and see that +\[ +0 = ax_1^2 + bx_1 + c,\qquad +0 = ax_2^2 + bx_2 + c. +\] + +Hence the values \Eq{2} are actually the roots of equation~\Eq{1}. + +We call $\Delta = b^2 - 4ac$ the \emph{discriminant} of the function $ax^2 + bx + c$ or +\index{Discriminant!of quadratic}% +of the corresponding equation~\Eq{1}. If $\Delta = 0$, the roots~\Eq{2} are evidently +equal, so that, by~\Eq{4}, $ax^2 + bx + c$ is the square of $\sqrt{a}(x - x_1)$, and conversely. +%% -----File: 018.png---Folio 12------- +We thus obtain the useful result that $ax^2 + bx + c$ \emph{is a perfect +square (of a linear function of $x$) if and only if $b^2 = 4ac$ (\emph{i.e.}, if its discriminant +is zero)}. +\index{Quadratic function a square}% + +Consider a \emph{real} quadratic equation, i.e., one whose coefficients $a$, $b$, $c$ +are all real numbers. Then if $\Delta$ is positive, the two roots~\Eq{2} are real. +But if $\Delta$ is negative, the roots are conjugate imaginaries~(§2). + +When the coefficients of a quadratic equation \Eq{1} are any complex +numbers, $\Delta$ has two complex square roots~(§9), so that the roots \Eq{2} of +\Eq{1} are complex numbers, which need not be conjugate. + +\begin{Remark} +For example, the discriminant of $x^2 - 2x + c$ is $\Delta = 4(1 - c)$. If $c = 1$, then $\Delta = 0$ and +$x^2 - 2x + 1 \equiv (x - 1)^2$ is a perfect square, and the roots $1$, $1$ of $x^2 - 2x + 1 = 0$ are equal. +If $c = 0$, $\Delta = 4$ is positive and the roots $0$ and~$2$ of $x^2 - 2x \equiv x(x - 2) = 0$ are real. If $c = 2$, +$\Delta = -4$ is negative and the roots $1 ± \sqrt{-1}$ of $x^2 - 2x + 2 = 0$ are conjugate complex +numbers. The roots of $x^2 - x + 1 + i = 0$ are $i$ and~$1 - i$, and are not conjugate. +\index{Discriminant!of quadratic}% +\end{Remark} + + +% [** PP: No ToC entry in original] +\Section[Polynomial] +{13.}{Integral Rational Function, Polynomial.} If $n$ is a positive integer +and $c_0$, $c_1, \dotsc, c_n$ are constants (real or imaginary), +\index{Integral!rational function}% +\index{Polynomial}% +\[ +f(x) \equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_{n-1} x + c_n +\] +is called a \emph{polynomial} in~$x$ of \emph{degree}~$n$, or also an \emph{integral rational function} +of~$x$ of degree~$n$. It is given the abbreviated notation~$f(x)$, just as the +logarithm of $x + 2$ is written $\log(x + 2)$. +\index{Symbol!b@{$f(x)$\IndAdd{polynomial}}}% [** PP: Manually alphabetized] + +If $c_0 \ne 0$, $f(x) = 0$ is an equation of degree~$n$. If $n = 3$, it is often called +a \emph{cubic equation}; and, if $n = 4$, a \emph{quartic equation}. For brevity, we often +speak of an equation all of whose coefficients are real as a \emph{real equation}. +\index{Real equation}% + + +\Section[Remainder Theorem] +{14.}{The Remainder Theorem.} +\index{Remainder theorem}% +\begin{Thm} +If a polynomial $f(x)$ be divided by +$x - c$ until a remainder independent of~$x$ is obtained, this remainder is equal +to~$f(c)$, which is the value of~$f(x)$ when $x = c$. +\end{Thm} + +Denote the remainder by~$r$ and the quotient by~$q(x)$. Since the +dividend is~$f(x)$ and the divisor is~$x - c$, we have +\[ +f(x) \equiv (x-c)q(x) + r, +\] +identically in~$x$. Taking $x = c$, we obtain $f(c) = r$. + +If $r = 0$, the division is exact. Hence we have proved also the following +useful theorem. + +\begin{Theorem}[The Factor Theorem.] +If $f(c)$ is zero, the polynomial $f(x)$ has the +factor $x - c$. In other words, if $c$ is a root of $f(x) = 0$, $x - c$ is a factor of~$f(x)$. +\end{Theorem} +\index{Factor theorem}% + +%% -----File: 019.png---Folio 13------- + +For example, $2$~is a root of $x^3 - 8 = 0$, so that $x - 2$ is a factor of $x^3 - 8$. Another +illustration is furnished by formula (4). % [** PP: Added period] + + +\begin{Exercises}{Page13} + +Without actual division find the remainder when +\begin{Problems} +\item[1.] $x^4 - 3x^2 - x - 6$ is divided by $x + 3$. + +\item[2.] $x^3 - 3x^2 + 6x - 5$ is divided by $x - 3$. +\end{Problems} + +Without actual division show that +\begin{Problems} +\item[3.] $18x^{10} + 19x^5 + 1$ is divisible by $x + 1$. + +\item[4.] $2x^4 - x^3 - 6x^2 + 4x - 8$ is divisible by $x - 2$ and $x + 2$. + +\item[5.] $x^4 - 3x^3 + 3x^2 - 3x + 2$ is divisible by $x - 1$ and $x - 2$. + +\item[6.] $r^3 - 1$, $r^4 - 1$, $r^5 - 1$ are divisible by $r - 1$. + +\item[7.] By performing the indicated multiplication, verify that +\[ +r^n - 1 \equiv (r - 1)(r^{n-1} + r^{n-2} + \dotsb + r + 1). +\] + +\item[8.] In the last identity replace $r$ by~$x/y$, multiply by~$y^n$, and derive +\[ +x^n - y^n \equiv (x-y)(x^{n-1} + x^{n-2}y + \dotsb + xy^{n-2} + y^{n-1}). +\] + +\item[9.] In the identity of Exercise~8 replace $y$ by $-y$, and derive +\begin{align*} +x^n + y^n + &\equiv (x+y)(x^{n-1} - x^{n-2} y + \dotsb - xy^{n-2} + y^{n-1}), + \quad \text{$n$~odd}; \\ +x^n - y^n + &\equiv (x+y)(x^{n-1} - x^{n-2} y + \dotsb + xy^{n-2} - y^{n-1}), + \quad\text{$n$~even}. +\end{align*} +\end{Problems} + +Verify by the Factor Theorem that $x + y$ is a factor. +\begin{Problems} +\item[10.] If $a$, $ar$, $ar^2, \dotsc, ar^{n-1}$ are $n$ numbers in \emph{geometrical progression} (the ratio of any +term to the preceding being a constant $r \ne 1$), prove by Exercise~7 that their sum is +equal to +\index{Geometrical!progression}% +\[ +\frac{a(r^n - 1)}{r - 1}. +\] + +\item[11.] At the end of each of $n$ years a man deposits in a savings bank $a$~dollars. With +annual compound interest at~4\%, show that his account at the end of $n$~years will be +\index{Compound interest}% +\[ +\frac{a}{.04} \bigl\{(1.04)^n - 1\bigr\} +\] +dollars. Hint: The final deposit draws no interest; the prior deposit will amount to +$a(1.04)$ dollars; the deposit preceding that will amount to $a(1.04)^2$ dollars, etc. Hence +apply Exercise~10 for $r = 1.04$. +\end{Problems} +\end{Exercises} + + +\Section{15.}{Synthetic Division} The labor of computing the value of a polynomial +in~$x$ for an assigned value of~$x$ may be shortened by a simple device. +To find the value of +\index{Synthetic division}% +\[ +x^4 + 3x^3 - 2x - 5 +\] +%% -----File: 020.png---Folio 14------- +for $x = 2$, note that $x^4 = x·x^3 = 2x^3$, so that the sum of the first two terms +of the polynomial is~$5x^3$. To $5x^3 = 5·2^2x$ we add the next term~$-2x$ and +obtain~$18x$ or~$36$. Combining~$36$ with the final term~$-5$, we obtain the +desired value~$31$. + +This computation may be arranged systematically as follows. After +supplying zero coefficients of missing powers of~$x$, we write the coefficients +in a line, ignoring the powers of~$x$. +\[ +\begin{array}{rRRRRPc} +1 & 3 & 0 & -2 & -5 && \Lcol{2} \\ +\cline{7-7} + & 2 & 10 & 20 & 36 && \\ +\cline{1-5} +1 & 5 & 10 & 18 & 31 && \Strut +\end{array} +\] +First we bring down the first coefficient~$1$. Then we multiply it by the +given value~$2$ and enter the product~$2$ directly under the second coefficient~$3$, +add and write the sum~$5$ below. Similarly, we enter the product of +$5$ by~$2$ under the third coefficient~$0$, add and write the sum~$10$ below; etc. +The final number~$31$ in the third line is the value of the polynomial when +$x = 2$. The remaining numbers in this third line are the coefficients, in +their proper order, of the quotient +\[ +x^3 + 5x^2 + 10x + 18, +\] +which would be obtained by the ordinary long division of the given polynomial +by $x - 2$. + +We shall now prove that this process, called \emph{synthetic division}, enables +us to find the quotient and remainder when any polynomial $f(x)$ is divided +by $x - c$. Write +\index{Quotient by synthetic division}% +\begin{align*} +f(x) &\equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_n, \\ +\intertext{and let the constant remainder be $r$ and the quotient be} +q(x) &\equiv b_0 x^{n-1} + b_1 x^{n-2} + \dotsb + b_{n-1}. +\end{align*} + +By comparing the coefficients of $f(x)$ with those in +\begin{multline*} +(x - c)q(x) + r + \equiv b_0 x^n + + (b_1 - cb_0) x^{n-1} \\ + + (b_2 - cb_1) x^{n-2} + \dotsb + + (b_{n-1} - cb_{n-2}) x + + r - cb_{n-1}, +\end{multline*} +we obtain relations which become, after transposition of terms, +\[ +b_0 = a_0,\ +b_1 = a_1 + cb_0,\ +b_2 = a_2 + cb_1, \dotsc,\ +b_{n-1} = a_{n-1} + cb_{n-2},\ +r = a_n + cb_{n-1}. +\] +%% -----File: 021.png---Folio 15------- +The steps in the work of computing the~$b$'s may be tabulated as follows: +\[ +\begin{array}{rRRRWWPc} +a_0 & a_1 & a_2 & \cdots & a_{n-1} & a_n && \Lcol{c} \\ +\cline{8-8} + & cb_0 & cb_1 & \cdots & cb_{n-2} & cb_{n-1} && \\ +\cline{1-6} +b_0 & b_1 & b_2 & \cdots & b_{n-1}, & r && \Strut +\end{array} +\] +In the second space below $a_0$ we write $b_0$ (which is equal to $a_0$). We +multiply $b_0$ by $c$ and enter the product directly under $a_1$, add and write +the sum $b_1$ below it. Next we multiply $b_1$ by $c$ and enter the product +directly under $a_2$, add and write the sum $b_2$ below it; etc. + + +\begin{Exercises}{Page15} + +Work each of the following exercises by synthetic division. +\begin{Problems} +\item[1.] Divide $x^3 + 3x^2 - 2x - 5$ by $x-2$. + +\item[2.] Divide $2x^5 - x^3 + 2x - 1$ by $x+2$. + +\item[3.] Divide $x^3 + 6x^2 + 10x - 1$ by $x - 0.09$. + +\item[4.] Find the quotient of $x^3 - 5x^2 - 2x + 24$ by $x-4$, and then divide the quotient by +$x-3$. What are the roots of $x^3 - 5x^2 - 2x + 24 = 0$? + +\item[5.] Given that $x^4 - 2x^3 - 7x^2 + 8x + 12 = 0$ has the roots $-1$ and~$2$, find the quadratic +equation whose roots are the remaining two roots of the given equation, and find these +roots. + +\item[6.] If $x^4 - 2x^3 - 12x^2 + 10x + 3 = 0$ has the roots $1$ and~$-3$, find the remaining two roots. + +\item[7.] Find the quotient of $2x^4 - x^3 - 6x^2 + 4x - 8$ by $x^2 - 4$. + +\item[8.] Find the quotient of $x^4 - 3x^3 + 3x^2 - 3x + 2$ by $x^2 - 3x + 2$. + +\item[9.] Solve Exercises 1, 2, 3, 6, 7 of~§14 by synthetic division. +\end{Problems} +\end{Exercises} + + +\Section{16.}{Factored Form of a Polynomial} Consider a polynomial %[** PP: Typo polynominal] +\index{Factored form}% +\[ +f(x) \equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_n \quad (c_0 \ne 0), +\] +whose leading coefficient $c_0$ is not zero. If $f(x) = 0$ has the root~$\alpha_1$, which +may be any complex number, the Factor Theorem shows that $f(x)$ has the +factor $x - \alpha_1$, so that +\[ +f(x) \equiv (x - \alpha_1)Q(x),\quad +Q(x) \equiv c_0 x^{n-1} + c_1' x^{n-2} + \dotsb + c_{\alpha-1}'. +\] +If $Q(x) = 0$ has the root~$\alpha_2$, then +\[ +Q(x) \equiv (x - \alpha_2)Q_1(x),\quad +f(x) \equiv (x - \alpha_1)(x - \alpha_2)Q_1(x). +\] +If $Q_1(x) = 0$ has the root~$\alpha_3$, etc., we finally get +\[ +f(x) \equiv c_0(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n). +\Tag{5} +\] + +We shall deduce several important conclusions from the preceding +discussion. First, suppose that the equation $f(x)=0$ of degree~$n$ is known +%% -----File: 022.png---Folio 16------- +to have $n$~distinct roots $\alpha_1, \dotsc, \alpha_n$. In $f(x) \equiv (x-\alpha_1)Q(x)$ take $x=\alpha_2$; +then $0=(\alpha_2-\alpha_1)Q(\alpha_2)$, whence $Q(\alpha_2)=0$ and $Q(x)=0$ has the root $\alpha_2$. +Similarly, $Q_1(x)=0$ has the root $\alpha_3$, etc. Thus all of the assumptions +(each introduced by an ``if'') made in the above discussion have been +justified and we have the conclusion~\Eq{5}. Hence \emph{if an equation $f(x)=0$ +of degree~$n$ has $n$~distinct roots $\alpha_1, \dotsc, \alpha_n$, $f(x)$ can be expressed in the +factored form~\Eq{5}}. +\index{Number!of roots}% + +It follows readily that the equation can not have a root~$\alpha$ different +from $\alpha_1, \dotsc, \alpha_n$. For, if it did, the left member of~\Eq{5} is zero when +$x=\alpha$ and hence one of the factors of the right member must then be zero, +say $\alpha-\alpha_j = 0$, whence the root~$\alpha$ is equal to~$\alpha_j$. We have now proved +the following important result. + +\begin{Theorem} +An equation of degree $n$ cannot have more than $n$ distinct roots. +\end{Theorem} + + +\Section[Multiple Roots] +{17.}{Multiple Roots.\protect\footnotemark}\addtocounter{footnote}{1}% + \footnotetext{Multiple roots are treated by calculus in~§58.}% +\addtocounter{footnote}{-1}% +\index{Multiple roots}% +Equalities may occur among the $\alpha$'s in~\Eq{5}. +Suppose that exactly $m_1$ of the $\alpha$'s (including~$\alpha_1$) are equal to~$\alpha_1$; that +$\alpha_2 \ne \alpha_1$, while exactly~$m_2$ of the~$\alpha$'s are equal to~$\alpha_2$; etc. Then \Eq{5} becomes +\[ +f(x) \equiv + c_0(x-\alpha_1)^{m_1} + (x-\alpha_2)^{m_2} \dotsm + (x-\alpha_k)^{m_k}, \quad m_1 + m_2 + \dotsb + m_k = n, +\Tag{6} +\] +where $\alpha_1, \dotsc, \alpha_k$ are distinct. We then call~$\alpha_1$ a \emph{root of multiplicity}~$m_1$ +of $f(x)= 0$, $\alpha_2$ a root of multiplicity~$m_2$, etc. In other words, $\alpha_1$ is a root +of multiplicity~$m_1$ of $f(x)=0$ if $f(x)$ is exactly divisible by $(x-\alpha_1)^{m_1}$, but is +not divisible by $(x-\alpha_1)^{m_1+1}$. We call $\alpha_1$ also an $m_1$-\emph{fold root}. In the +particular cases $m_1=1$, $2$, and~$3$, we also speak of $\alpha_1$ as a \emph{simple root}, \emph{double +root}, and \emph{triple root}, respectively. For example, $4$~is a simple root, $3$~a +\index{Double root|see{Discriminant}}% +\index{Multiplicity of root}% +\index{Simple root}% +\index{Triple root}% +double root, $-2$~a triple root, and $6$ a root of multiplicity~$4$ (or a $4$-fold +root) of the equation +\[ +7(x-4)(x-3)^2(x+2)^3(x-6)^4 = 0 +\] +of degree~$10$ which has no further root. This example illustrates the +next theorem, which follows from~\Eq{6} exactly as the theorem in~§16 +followed from~\Eq{5}. + +\begin{Theorem} +An equation of degree~$n$ cannot have more than $n$~roots, +a root of multiplicity~$m$ being counted as $m$~roots. +\end{Theorem} + + +\Section{18.}{Identical Polynomials} +\index{Identical polynomials}% +\begin{Thm} +If two polynomials in~$x$, +\[ +a_0 x^n + a_1 x^{n-1} + \dotsb + a_n,\qquad +b_0 x^n + b_1 x^{n-1} + \dotsb + b_n, +\] +%% -----File: 023.png---Folio 17------- +each of degree~$n$, are equal in value for more than $n$~distinct values of~$x$, they +are term by term identical, i.e., $a_0 = b_0$, $a_1 = b_1, \dotsc, a_n = b_n$. +\end{Thm} + +For, taking their difference and writing $c_0 = a_0 - b_0, \dotsc, c_n = a_n - b_n$, +we have +\index{Number!of roots}% +\[ +c_0 x^n + c_1 x^{n-1} + \dotsb + c_n = 0 +\] +for more than $n$~distinct values of~$x$. If $c_0 \ne 0$, we would have a contradiction +with the theorem in~§16. Hence $c_0 = 0$. If $c_1 \ne 0$, we would have +a contradiction with the same theorem with $n$~replaced by~$n-1$. Hence +$c_1 = 0$, etc. Thus $a_0 = b_0$, $a_1 = b_1$, etc. + + +\begin{Exercises}{Page17} + +\begin{Problems} +\item[1.] Find a cubic equation having the roots $0$, $1$, $2$. + +\item[2.] Find a quartic equation having the roots $±1$, $±2$. + +\item[3.] Find a quartic equation having the two double roots $3$ and~$-3$. + +\item[4.] Find a quartic equation having the root~$2$ and the triple root~$1$. + +\item[5.] What is the condition that $ax^2+bx+c=0$ shall have a double root? + +\item[6.] If $a_0 x^n + \dotsb + a_n = 0$ has more than $n$~distinct roots, each coefficient is zero. + +\item[7.] Why is there a single answer to each of Exercises 1--4, if the coefficient of the +highest power of the unknown be taken equal to unity? State and answer the corresponding +general question. +\end{Problems} +\end{Exercises} + + +\Section[Fundamental Theorem of Algebra] +{19.}{The Fundamental Theorem of Algebra.} +\index{Fundamental theorem of algebra}% +\begin{Thm} +Every algebraic equation +with complex coefficients has a complex \(real or imaginary\) root. +\end{Thm} + +This theorem, which is proved in the Appendix, implies that +\begin{Thm}% +every +equation of degree~$n$ has exactly $n$~roots if a root of multiplicity~$m$ be counted +as $m$~roots. +\end{Thm}% +In other words, +\begin{Thm}% +\index{Integral!rational function}% +\index{Linear factors}% +every integral rational function of degree~$n$ +is a product of $n$~linear factors. +\end{Thm}% +For, in~§16, equations $f(x)=0$, $Q(x)=0$, +$Q_1(x)=0, \dotsc$ each has a root, so that \Eq{5} and~\Eq{6} hold. + + +% [** PP: ToC entry matches unit title; using running head] +\Section[Relations between Roots and Coefficients] +{20.}{Relations between the Roots and the Coefficients.} In~§12 we +found the sum and the product of the two roots of any quadratic equation +and then deduced the factored form of the equation. We now apply +the reverse process to any equation +\index{Relations between roots and coefficients}% +\begin{align*} +f(x) &\equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_n = 0\qquad (c_0 \ne 0), +\Tag{7} \\ +\intertext{whose factored form is} +f(x) &\equiv c_0(x-\alpha_1)(x-\alpha_2) \dotsm (x-\alpha_n). +\Tag{8} +\end{align*} +Our next step is to find the expanded form of this product. The following +special products may be found by actual multiplication: +%% -----File: 024.png---Folio 18------- +\begin{align*}% [** PP: Aligning, breaking second line] +(x - \alpha_1)(x - \alpha_2) + &\equiv x^2 - (\alpha_1 + \alpha_2)x + \alpha_1\alpha_2, \\ +%[** PP: Typo (x_1 - \alpha_1)] +(x - \alpha_1)(x - \alpha_2)(x - \alpha_3) + &\equiv x^3 - (\alpha_1 + \alpha_2 + \alpha_3)x^2 \\ + &\qquad + (\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3)x + - \alpha_1\alpha_2\alpha_3. +\end{align*} +These identities are the cases $n = 2$ and $n = 3$ of the following general +formula: +\begin{multline*} +%[** PP: Typo (x_1 - \alpha_1)] +(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n) + \equiv x^n + - (\alpha_1 + \dotsb + \alpha_n)x^{n-1} \\ + + (\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3 + \dotsb + + \alpha_{n-1}\alpha_n)x^{n-2} \\ + - (\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \dotsb + + \alpha_{n-2}\alpha_{n-1}\alpha_n)x^{n-3} \\ + + \dotsb + (-1)^n \alpha_1\alpha_2 \dotsm \alpha_n, +\Tag{9} +\end{multline*} +the quantities in parentheses being described in the theorem below. If +we multiply each member of~\Eq{9} by $x - \alpha_{n+1}$, it is not much trouble to verify +that the resulting identity can be derived from~\Eq{9} by changing $n$ into +$n+1$, so that \Eq{9} is proved true by mathematical induction. Hence the +quotient of \Eq{7} by $c_0$ is term by term identical with~\Eq{9}, so that +\[ +\begin{aligned} +\alpha_1 + \alpha_2 + \dotsb + \alpha_n &= -c_1 / c_0, \\ +% +\alpha_1\alpha_2 + \alpha_1\alpha_3 + + \alpha_2\alpha_3 + \dotsb + + \alpha_{n-1}\alpha_n &= c_2 / c_0, \\ +% +\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \dotsb + + \alpha_{n-2}\alpha_{n-1}\alpha_n &= -c_3 / c_0, \\ +&\vdots\\ %<tb> +\alpha_1\alpha_2 \dotsm \alpha_{n-1}\alpha_n &= (-1)^n c_n / c_0. +\end{aligned} +\Tag{10} +\] +These results may be expressed in the following words: + +\begin{Theorem} +If $\alpha_1, \dotsc, \alpha_n$ are the roots of equation~\Eq{7}, the sum of the +roots is equal to~$-c_1 / c_0$, the sum of the products of the roots taken two at +a time is equal to~$c_2 / c_0$, the sum of the products of the roots taken three at a +time is equal to~$-c_3 / c_0$, etc.; finally, the product of all the roots is equal to~$(-1)^n c_n / c_0$. +\index{Product of roots}% +\index{Sum of!products of roots}% +\index{Sum of!roots}% +\end{Theorem} + +Since we may divide the terms of our equation~\Eq{7} by~$c_0$, the essential +part of our theorem is contained in the following simpler statement: + +\begin{Corollary} +In an equation in~$x$ of degree~$n$, in which the coefficient +of~$x^n$ is unity, the sum of the $n$~roots is equal to the negative of the coefficient +of~$x^{n-1}$, the sum of the products of the roots two at a time is equal to the coefficient +of~$x^{n-2}$, etc.; finally the product of all the roots is equal to the constant +term or its negative, according as $n$ is even or odd. +\end{Corollary} + +\begin{Remark} +For example, in a cubic equation having the roots $2$, $2$, $5$, and having unity as the +coefficient of~$x^3$, the coefficient of~$x$ is $2·2 + 2·5 + 2·5 = 24$. +\end{Remark} + +%% -----File: 025.png---Folio 19------- + + +\begin{Exercises}{Page19} + +\begin{Problems} +\item[1.] Find a cubic equation having the roots $1$, $2$, $3$. + +\item[2.] Find a quartic equation having the double roots $2$ and~$-2$. + +\item[3.] Solve $x^4 - 6x^3 + 13x^2 - 12x + 4 = 0$, which has two double roots. + +\item[4.] Prove that one root of $x^3 + px^2 + qx + r = 0$ is the negative of another root if and +only if $r = pq$. + +\item[5.] Solve $4x^3 - 16x^2 - 9x + 36 = 0$, given that one root is the negative of another. + +\item[6.] Solve $x^3 - 9x^2 + 23x - 15 = 0$, given that one root is the triple of another. + +\item[7.] Solve $x^4 - 6x^3 + 12x^2 - 10x + 3 = 0$, which has a triple root. + +\item[8.] Solve $x^3 - 14x^2 - 84x + 216 = 0$, whose roots are in geometrical progression, i.e., +with a common ratio $r$ [say $m/r$, $m$, $mr$]. +\index{Geometrical!progression}% + +\item[9.] Solve $x^3 - 3x^2 - 13x + 15 = 0$, whose roots are in arithmetical progression, i.e., +with a common difference $d$ [say $m-d$, $m$, $m+d$]. +\index{Arithmetical progression}% + +\item[10.] Solve $x^4 - 2x^3 - 21x^2 + 22x + 40 = 0$, whose roots are in arithmetical progression. +[Denote them by $c-3b$, $c-b$, $c+b$, $c+3b$, with the common difference $2b$]. % [** PP: Added period] + +\item[11.] Find a quadratic equation whose roots are the squares of the roots of +$x^2-px+q = 0$. + +\item[12.] Find a quadratic equation whose roots are the cubes of the roots of $x^2 - px + q = 0$. +Hint: $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$. + +\item[13.] If $\alpha$ and~$\beta$ are the roots of $x^2 - px + q = 0$, find an equation whose roots are (i)~$\alpha^2 / \beta$; +and~$\beta^2 / \alpha$; (ii)~$\alpha^3\beta$ and~$\alpha\beta^3$; (iii)~$\alpha+1 / \beta$ and~$\beta + 1 / \alpha$. + +\item[14.] Find a necessary and sufficient condition that the roots, taken in some order, +of $x^3 + px^2 + qx + r = 0$ shall be in geometrical progression. + +\item[15.] Solve $x^3 - 28x + 48 = 0$, given that two roots differ by~$2$. +\end{Problems} +\end{Exercises} + + +\Section{21.}{Imaginary Roots occur in Pairs} The two roots of a real quadratic +equation whose discriminant is negative are conjugate imaginaries~(§12). +This fact illustrates the following useful result. +\index{Imaginary!roots}% +\index{Surd roots in pairs}% [** PP: Original entry points to page 20] + +\begin{Theorem} +If an algebraic equation with real coefficients has the root $a+bi$, +where $a$ and~$b$ are real and $b \ne 0$, it has also the root~$a-bi$. +\end{Theorem} + +Let the equation be $f(x)= 0$ and divide $f(x)$ by +\[ +(x-a)^2 + b^2 \equiv (x - a-bi)(x - a+bi) +\Tag{11} +\] +until we reach a remainder $rx + s$ whose degree in~$x$ is less than the degree +of the divisor. Since the coefficients of the dividend and divisor are all +real, those of the quotient~$Q(x)$ and remainder are real. We have +\[ +f(x) \equiv Q(x)\bigl\{(x-a)^2 + b^2\bigr\} + rx + s, +\] +identically in~$x$. This identity is true in particular when $x = a+bi$, so +that +\[ +0 = r(a+bi)+s = ra+s+rbi. +\] +%% -----File: 026.png---Folio 20------- +Since all of the letters, other than~$i$, denote real numbers, we have~(§2) +$ra + s = 0$, $rb = 0$. But $b \ne 0$. Hence $r = 0$, and then $s = 0$. Hence $f(x)$ +is exactly divisible by the function~\Eq{11}, so that $f(x) = 0$ has the root $a - bi$. + +The theorem may be applied to the real quotient~$Q(x)$. We obtain +the + +\begin{Corollary} +If a real algebraic equation has an imaginary root of +multiplicity~$m$, the conjugate imaginary of this root is a root of multiplicity~$m$. +\index{Multiplicity of root}% +\end{Corollary} + +Counting a root of multiplicity~$m$ as $m$~roots, we see that a real equation +cannot have an odd number of imaginary roots. Hence by~§19, \emph{a real +equation of odd degree has at least one real root}. +\index{Real equation}% + +Of the $n$~linear factors of a real integral rational function of degree~$n$ +(§19), those having imaginary coefficients may be paired as in~\Eq{11}. +Hence +\begin{Thm}% +\index{Integral!rational function}% +every integral rational function with real coefficients can be expressed +as a product of real linear and real quadratic factors. +\end{Thm} + + +\begin{Exercises}{Page20} + +\begin{Problems} +\item[1.] Solve $x^3 - 3x^2 - 6x - 20 = 0$, one root being $-1 + \sqrt{-3}$. + +\item[2.] Solve $x^4 - 4x^3 + 5x^2 - 2x - 2 = 0$, one root being $1-i$. + +\item[3.] Find a cubic equation with real coefficients two of whose roots are $1$ and $3+2i$. + +\item[4.] If a real cubic equation $x^3 - 6x^2 + \dotsb = 0$ has the root $1 +\sqrt{-5}$, what are the +remaining roots? Find the complete equation. + +\item[5.] If an equation with \emph{rational} coefficients has a root $a + \sqrt{b}$, where $a$ and $b$ are +rational, but $\sqrt{b}$ is irrational, prove that it has the root $a - \sqrt{b}$. [Use the method of~§21.] + +\item[6.] Solve $x^4 - 4x^3 + 4x - 1 = 0$, one root being $2 + \sqrt{3}$. + +\item[7.] Solve $x^3 - (4 + \sqrt{3})x^2 + (5 + 4\sqrt{3})x - 5\sqrt{3} = 0$, having the root $\sqrt{3}$. + +\item[8.] Solve the equation in Ex.~7, given that it has the root $2+i$. + +\item[9.] Find a cubic equation with rational coefficients having the roots $\frac{1}{2}, \frac{1}{2} + \sqrt{2}$. + +\item[10.] Given that $x^4 - 2x^3 - 5x^2 - 6x + 2 = 0$ has the root $2 - \sqrt{3}$, find another root and +by means of the sum and the product of the four roots deduce, without division, the +quadratic equation satisfied by the remaining two roots. + +\item[11.] Granted that a certain cubic equation has the root~$2$ and no real root different +from~$2$, does it have two imaginary roots? + +\item[12.] Granted that a certain quartic equation has the roots $2 ± 3i$, and no imaginary +roots different from them, does it have two real roots? + +\item[13.] By means of the proof of Ex.~5, may we conclude as at the end of~§21 that +every integral rational function with rational coefficients can be expressed as a product +of linear and quadratic factors with rational coefficients? +\end{Problems} +\end{Exercises} + +%% -----File: 027.png---Folio 21------- + + +\Section{22.}{Upper Limit to the Real Roots} Any number which exceeds +\index{Upper limit to roots|(}% +all real roots of a real equation is called an \emph{upper limit to the real roots}. +We shall prove two theorems which enable us to find readily upper limits +to the real roots. For some equations Theorem~I gives a better (smaller) +upper limit than Theorem~II; for other equations, the reverse is true. +Evidently any positive number is an upper limit to the real roots of an +equation having no negative coefficients. + + +\begin{Theorem}[Theorem~I.] +If, in a real equation +\[ +f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_n = 0 \qquad (a_0 > 0), +\] +the first negative coefficient is preceded by $k$ coefficients which are positive or +zero, and if $G$~denotes the greatest of the numerical values of the negative +coefficients, then each real root is less than $1 + \sqrt[k]{G / a_0}$. +\end{Theorem} + +\begin{Remark} +For example, in $x^5 + 4x^4 - 7x^2 - 40x + 1 = 0$, $G = 40$ and $k = 3$ since we must supply +the coefficient zero to the missing power~$x^3$. Thus the theorem asserts that each root +is less than $1 + \sqrt[3]{40}$ and therefore less than~$4.42$. Hence $4.42$ is an upper limit to the +roots. +\end{Remark} + +\begin{Proof} +For positive values of~$x$, $f(x)$ will be reduced in value or remain +unchanged if we omit the terms $a_1 x^{n-1}, \dotsc, a_{k-1}x^{n-k+1}$ (which are +positive or zero), and if we change each later coefficient $a_k, \dotsc, a_n$ to~$-G$. +Hence +\[ +f(x) \geqq a_0 x^n - G(x^{n-k} + x^{n-k-1} + \dotsb + x + 1). +\] +But, by Ex.~7 of~§14, +\[ +x^{n-k} + \dotsb + x + 1 \equiv \frac{x^{n-k+1} - 1}{x - 1}, +\] +if $x \ne 1$. Furthermore, +\[ +a_0 x^n - G\left(\frac{x^{n-k+1} - 1}{x - 1}\right) + \equiv \frac{x^{n-k+1}\{a_0 x^{k-1}(x-1) - G\} + G}{x - 1}. +\] +Hence, if $x>1$, +\begin{align*} +f(x) &> \frac{x^{n-k+1}\bigl\{a_0 x^{k-1}(x-1) - G\bigr\}}{x - 1}, \\ +f(x) &> \frac{x^{n-k+1}\bigl\{a_0 (x-1)^k - G\bigr\}}{x - 1}. +\end{align*} +Thus, for $x>1$, $f(x)>0$ and $x$ is not a root if $a_0 (x-1)^k - G \geqq 0$, which +is true if $x \geqq 1 + \sqrt[k]{G / a_0}$. +\end{Proof} + +%% -----File: 028.png---Folio 22------- + + +% [** PP: No ToC entry] +\Section{23.}{Another Upper Limit to the Roots}\hfill\break +% [** PP: Format Theorem heading manually to avoid \BeforeSkip] +{\indent\normalfont\normalsize\scshape Theorem~II.}\quad +\begin{Thm}% +If, in a real algebraic equation in which the coefficient +of the highest power of the unknown is positive, the numerical value of each +negative coefficient be divided by the sum of all the positive coefficients which +precede it, the greatest quotient so obtained increased by unity is an upper +limit to the roots. +\end{Thm} + +\begin{Remark} +For the example in~§22, the quotients are $7/(1+4)$ and~$40/5$, so that Theorem~II +asserts that $1+8$ or~$9$ is an upper limit to the roots. Theorem~I gave the better upper +limit~$4.42$. But for $x^3 + 8x^2 - 9x + c^2 = 0$, Theorem~I gives the upper limit~$4$, while +Theorem~II gives the better upper limit~$2$. + +We first give the proof for the case of the equation +\[ +f(x) \equiv p_4 x^4 - p_3 x^3 + p_2 x^2 - p_1 x + p_0 = 0 +\] +in which each $p_i$ is positive. In view of the identities +\[ +x^4 \equiv (x-1) (x^3 + x^2 + x + 1) + 1,\qquad +x^2 \equiv (x-1) (x+1) + 1, +\] +$f(x)$ is equal to the sum of the terms +\begin{alignat*}{3} +p_4 (x-1) x^3 &+{}& p_4 (x-1)x^2 + p_4 (x-1)x &+{}& p_4 (x-1) &+ p_4, \\ + {} - p_3 x^3 & & + p_2 (x-1)x &+{}& p_2 (x-1) &+ p_2, \\ + & & {} - p_1 x & & &+ p_0. +\end{alignat*} +If $x>1$, negative terms occur only in the first and third columns, while the sum of the +terms in each of these two columns will be $\geqq 0$ if +\[ +p_4 (x-1) - p_3 \geqq 0,\quad (p_4 + p_2 )(x-1) - p_1 \geqq 0. +\] +Hence $f(x) > 0$ and $x$ is not a root if +\[ +x \geqq 1 + \frac{p_3}{p_4},\quad +x \geqq 1 + \frac{p_1}{p_4 + p_2}. +\] +This proves the theorem for the present equation. + +Next, let $f(x)$ be modified by changing its constant term to~$-p_0$. We modify the +above proof by employing the sum $(p_4 + p_2)x - p_0$ of all the terms in the corresponding +last two columns. This sum will be $>0$ if $x > p_0 / (p_4 + p_2)$, which is true if +\[ +x \geqq 1 + \frac{p_0}{p_4 + p_2}. +\] +\end{Remark} + +To extend this method of proof to the general case +\[ +f(x) \equiv a_n x^n + \dotsb + a_0\qquad (a_n > 0), +\] +we have only to employ suitable general notations. Let the negative +coefficients be $a_{k_1}, \dotsc, a_{k_t} $, where $k_1 > k_2 > \dots > k_t$. For each positive +%% -----File: 029.png---Folio 23------- +integer~$m$ which is $\leqq n$ and distinct from $k_1, \dotsc, k_t$, we replace~$x^m$ +by the equal value +\[ +d(x^{m-1} + x^{m-2} + \dotsb + x + 1) + 1 +\] +where $d \equiv x-1$. Let $F(x)$ denote the polynomial in~$x$, with coefficients +involving~$d$, which is obtained from~$f(x)$ by these replacements. Let +$x>1$, so that $d$ is positive. Thus the terms $a_{k_i} x^{k_i}$ are the only negative +quantities occurring in~$F(x)$. If $k_i > 0$, the terms of~$F(x)$ which involve +explicitly the power $x^{k_i}$ are $a_{k_i} x^{k_i}$ and the $a_m d x^{k_i}$ for the various positive +coefficients~$a_m$ which precede~$a_{k_i}$. The sum of these terms will be $\geqq 0$ +if $a_{k_i} + d \sum a_m \geqq 0$, i.e., if +\[ +x \geqq 1 + \frac{-a_{k_i}}{\sum a_m}. +\] + +There is an additional case if $k_t = 0$, i.e., if $a_0$ is negative. Then the +terms of~$F(x)$ not involving~$x$ explicitly are $a_0$ and the $a_m (d+1)$ for the +various positive coefficients~$a_m$. Their sum, $a_0 + x \sum a_m$, will be $> 0$ if +\[ +x > \frac{-a_0}{\sum a_m}, +\] +which is true if +\[ +x \geqq 1 + \frac{-a_0}{\sum a_m}. +\] +\index{Upper limit to roots|)}% + + +\begin{Exercises}{Page23} + +Apply the methods of both §22 and~§23 to find an upper limit to the roots~of +\begin{Problems} +\item[1.] $4x^5 - 8x^4 + 22x^3 + 98x^2 - 73x + 5 = 0$. + +\item[2.] $x^4 - 5x^3 + 7x^2 - 8x + 1 = 0$. + +\item[3.] $x^7 + 3x^6 - 4x^5 + 5x^4 - 6x^3 - 7x^2 - 8 = 0$. + +\item[4.] $x^7 + 2x^5 + 4x^4 - 8x^2 - 32 = 0$. + +\item[5.] A lower limit to the negative roots of $f(x) = 0$ may be found by applying our +theorems to $f(-x) = 0$, i.e., to the equation derived from $f(x) = 0$ by replacing~$x$ by~$-x$. +Find a lower limit to the negative roots in Exs.\ 2, 3,~4. +\index{Lower limit to roots}% + +\item[6.] Prove that every real root of a real equation $f(x) = 0$ is less than $1 + g / a_0$ if $a_0 > 0$, +where $g$ denotes the greatest of the numerical values of $a_1, \dotsc, a_n$. Hint: if $x>0$, +\[ +a_0 x^n + a_1 x^{n-1} + \dotsb \geqq a_0 x^n - g(x^{n-1} + \dotsb + x + 1). +\] +Proceed as in~§22 with $k = 1$. + +\item[7.] Prove that $1 + g \div |a_0|$ is an upper limit for the moduli of all complex roots of any +equation $f(x)=0$ with complex coefficients, where $g$ is the greatest of the values $|a_1|, +\dotsc, |a_n|$, and $|a|$ denotes the modulus of~$a$. Hint: use Ex.~5 of~§8. +\index{Symbol!c@{$\lvert a\rvert$\IndAdd{modulus}}}% [** PP: Manually alphabetized] +\end{Problems} +\end{Exercises} + +%% -----File: 030.png---Folio 24------- + + +\Section{24.}{Integral Roots} +\index{Integral!roots|(}% +\begin{Thm} +For an equation all of whose coefficients are integers, +any integral root is an exact divisor of the constant term. +\end{Thm} + +For, if~$x$ is an integer such that +\[ +a_0 x^n + \dotsb + a_{n-1}x + a_n = 0, +\Tag{12} +\] +where the $a$'s are all integers, then, by transposing terms, we obtain +\[ +x(-a_0x^{n-1} - \dotsb - a_{n-1}) = a_n. +\] +Thus $x$ is an exact divisor of~$a_n$ since the quotient is the integer given by +the quantity in parenthesis. + +\begin{Example}[1.] +Find all the integral roots of +\[ +x^3 + x^2 - 3x + 9 = 0. +\] +\end{Example} + +\begin{Solution} +The exact divisors of the constant term~$9$ are $±1$, $±3$,~$±9$. By trial, +no one of $±1$, $3$ is a root. Next, we find that~$-3$ is a root by synthetic division~(§15): +\[ +\begin{array}{rRRRPc} +1 & 1 & -3 & 9 && \Lcol{-3} \\ +\cline{6-6} + & -3 & 6 & -9 && \\ +\cline{1-4} +1 & -2 & 3 & 0 && \Strut +\end{array} +\] + +Hence the quotient is $x^2 - 2x + 3$, which is zero for $x = 1 ±\sqrt{-2}$. Thus~$-3$ is the +only integral root. +\end{Solution} + +When the constant term has numerous exact divisors, some device +may simplify the application of the theorem. + +\begin{Example}[2.\protect\footnotemark] Find all the integral roots of +\[ +y^3 + 12y^2 - 32y - 256 = 0. +\] + \footnotetext{This problem is needed for the solution~(§48) of a certain quartic equation.} +\end{Example} + +\begin{Solution} +Since all the terms except~$y^3$ are divisible by~$2$, an integral root~$y$ must +be divisible by~$2$. Since all the terms except~$y^3$ are now divisible by~$2^4$, we have $y = 4z$, +where $z$ is an integer. Removing the factor~$2^6$ from the equation in~$z$, we obtain +\[ +z^3 + 3z^2 - 2z - 4 = 0. +\] +An integral root must divide the constant term~$4$. Hence, if there are any integral +roots, they occur among the numbers $±1$, $±2$, $±4$. By trial, $-1$ is found to be a +root: +\[ +\begin{array}{rRRRPc} +1 & 3 & -2 & -4 && \Lcol{-1} \\ +\cline{6-6} + & -1 & -2 & 4 && \\ +\cline{1-4} +1 & 2 & -4 & 0 && \Strut +\end{array} +\] +%% -----File: 031.png---Folio 25------- +Hence the quotient is $z^2 + 2z - 4$, which is zero for $z = -1 ± \sqrt{5}$. Thus $y = 4z = -4$ +is the only integral root of the proposed equation. +\end{Solution} + + +\begin{Exercises}{Page25} + +Find all the integral roots of +\begin{Problems}[2] +\item[1.] $x^3 + 8x^2 + 13x + 6 = 0$. + +\item[2.] $x^3 - 5x^2 - 2x + 24 = 0$. + +\ResetCols{2} + +\item[3.] $x^3 - 10x^2 + 27x - 18 = 0$. + +\item[4.] $x^4 + 4x^3 + 8x + 32 = 0$. + +\ResetCols{2} + +\item[5.] The equation in Ex.~4 of~§23. + +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{25.}{Newton's Method for Integral Roots} In~§24 we proved that +an integral root~$x$ of equation~\Eq{12} having integral coefficients must be +an exact divisor of~$a_n$. Similarly, if we transpose all but the last two +terms of~\Eq{12}, we see that $a_{n-1}x + a_n$ must be divisible by~$x^2$, and hence +$a_{n-1} + a_n/x$ divisible by~$x$. By transposing all but the last three terms +of~\Eq{12}, we see that their sum must be divisible by~$x^3$, and hence $a_{n-2} + +(a_{n-1} + a_n/x)/x$ divisible by~$x$. We thus obtain a series of conditions +of divisibility which an integral root must satisfy. The final sum +$a_0 + a_{1}/x + \dotsb$ must not merely be divisible by~$x$, but be actually zero, +since it is the quotient of the function~\Eq{12} by~$x^n$. + +In practice, we must test in turn the various divisors~$x$ of~$a_n$. If a +chosen~$x$ is not a root, that fact will be disclosed by one of the conditions +mentioned. Newton's method is quicker than synthetic division since +it usually detects early and throws out wrong guesses as to a root, whereas +in synthetic division the decision comes only at the final step. + +\begin{Remark} +For example, the divisor~$-3$ of the constant term of +\[ +f(x) \equiv x^4 - 9x^3 + 24x^2 - 23x + 15 = 0 +\Tag{13} +\] +is not a root since $-23 + 15/(-3) = -28$ is not divisible by~$-3$. To show that none +of the tests fails for~$3$, so that $3$ is a root, we may arrange the work systematically as +follows: +\[ +\begin{array}{rRRRRPl} + 1 & -9 & 24 & -23 & 15 && \Lcol{\quad3} \\ +-1 & 6 & -6 & 5 & && \Lcol{\text{(divisor)}} \\ +\cline{1-4}\cline{7-7} + 0 & -3 & 18 & -18 & && \Strut +\end{array} +\Tag{14} +\] +First we divide the final coefficient~$15$ by~$3$, place the quotient~$5$ directly under the coefficient~$-23$, +and add. Next, we divide this sum~$-18$ by~$3$, place the quotient~$-6$ +directly under the coefficient~$24$, and add. After two more such steps we obtain the +sum zero, so that $3$ is a root. + +It is instructive to obtain the preceding process by suitably modifying synthetic +division. First, we replace~$x$ by~$1/y$ in~\Eq{13}, multiply each term by~$y^4$, and obtain +\[ +15y^4 - 23y^3 + 24y^2 - 9y + 1 = 0. +\] +%% -----File: 032.png---Folio 26------- +We may test this for the root $y = \frac{1}{3}$, which corresponds to the root $x = 3$ of~\Eq{13}, by +ordinary synthetic division: +\[ +\begin{array}{rRRRRPl} +15 & -23 & 24 & -9 & 1 && \Lcol{\quad\frac{1}{3}} \\ + & 5 & -6 & 6 & -1 && \Lcol{\text{(multiplier)}} \\ +\cline{1-5}\cline{7-7} +15 & -18 & 18 & -3 & 0 && \Strut +\end{array} +\] +The coefficients in the last two lines (after omitting~$15$) are the same as those of the last +two lines in~\Eq{14} read in reverse order. This should be the case since we have here +multiplied the same numbers by~$\frac{1}{3}$ that we divided by~$3$ in~\Eq{14}. The numbers in +the present third line are the coefficients of the quotient~(§15). Since we equate the +quotient to zero for the applications, we may replace these coefficients by the numbers +in the second line which are the products of the former numbers by~$\frac{1}{3}$. The numbers +in the second line of~\Eq{14} are the negatives of the coefficients of the quotient of~$f(x)$ +by $x-3$. +\end{Remark} + +\begin{Example} +Find all the integral roots of equation~\Eq{13}. +\end{Example} + +\begin{Solution} +For a negative value of~$x$, each term is positive. Hence all the real +roots are positive. By~§23, $10$ is an upper limit to the roots. By~§24, any integral +root is an exact divisor of the constant term~$15$. Hence the integral roots, if any, occur +among the numbers $1$, $3$,~$5$. Since $f(1) = 8$, $1$~is not a root. By~\Eq{14}, $3$~is a root. Proceeding +similarly with the quotient by~$x-3$, whose coefficients are the negatives of the +numbers in the second line of~\Eq{14}, we find that $5$ is a root. +\end{Solution} + + +\begin{Exercises}{}% [** PP: No answers] + +\begin{Problems} +\item[1.] Solve Exs.~1--4 of~§24 by Newton's method. + +\item[2.] Prove that, in extending the process~\Eq{14} to the general equation~\Eq{12}, we may +employ the final equations in~§15 with $r = 0$ and write +\[ +\begin{array}{rRRRWWNPl} + a_0 & a_1 & a_2 & \ldots & a_{n-2} & a_{n-1} & a_n && \Lcol{\quad c} \\ +-b_0 & -b_1 & -b_2 & \ldots & -b_{n-2} & -b_{n-1} & && \Lcol{\text{(divisor)}} \\ +\cline{1-6}\cline{9-9} + 0 & -c{b_0} & -cb_1 & \ldots & -cb_{n-3} & -cb_{n-2} & && \Strut +\end{array} +\] +Here the quotient, $-b_{n-1}$, of~$a_n$ by~$c$ is placed directly under $a_{n-1}$, and added to it to +yield the sum $-cb_{n-2}$, etc. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{26.}{Another Method for Integral Roots} An integral divisor~$d$ of the constant term is not a root if $d-m$ is not a divisor of~$f(m)$, where $m$ is +any chosen integer. For, if $d$ is a root of $f(x)=0$, then +\[ +f(x) \equiv (x-d)Q(x), +\] +where $Q(x)$ is a polynomial having integral coefficients~(§15). Hence +$f(m) = (m-d) q$, where $q$ is the integer~$Q(m)$. + +%% -----File: 033.png---Folio 27------- + +\begin{Remark} +In the example of~§25, take $d = 15$, $m=1$. Since $f(1)=8$ is not divisible by $15-1=14$, $15$ is not an integral root. + +Consider the more difficult example +\[ +f(x) \equiv x^3 - 20x^2 + 164x - 400 = 0, +\] +whose constant term has many divisors. There is evidently no negative root, while +$21$ is an upper limit to the roots. The positive divisors less than~$21$ of $400 = 2^4 5^2$ are +$d = 1$, $2$, $4$, $8$, $16$, $5$, $10$,~$20$. First, take $m=1$ and note that $f(1) = -255 = -3·5·17$. The +corresponding values of $d-1$ are $0$, $1$, $3$, $7$, $15$, $4$, $9$,~$19$; of these, $7$, $4$, $9$,~$19$ are not divisors +of~$f(1)$, so that $d = 8$, $5$, $10$ and~$20$ are not roots. Next, take $m=2$ and note that $f(2)= -144$ +is not divisible by $16-2 = 14$. Hence $16$ is not a root. Incidentally, $d=1$ and +$d=2$ were excluded since $f(d)\neq 0$. There remains only $d=4$, which is a root. +\end{Remark} + +In case there are numerous divisors within the limits to the roots, it +is usually a waste of time to list all these divisors. For, if a divisor is +found to be a root, it is preferable to employ henceforth the quotient, +as was done in the example in~§25. + + +\begin{Exercises}{Page27} + +Find all the integral roots of +\begin{Problems} +\item[1.] $x^4 - 2x^3 - 21x^2 + 22x + 40 = 0$. + +\item[2.] $y^3 - 9y^2 - 24y + 216 = 0$. + +\item[3.] $x^4 - 23x^3 + 187x^2 - 653x + 936 = 0$. + +\item[4.] $x^5 + 47x^4 + 423x^3 + 140x^2 + 1213x - 420 = 0$. + +\item[5.] $x^5 - 34x^3 + 29x^2 + 212x - 300 = 0$. +\end{Problems} +\end{Exercises} +\index{Integral!roots|)}% + + +\Section{27.}{Rational Roots} +\index{Rational roots}% +\begin{Thm} +If an equation with integral coefficients +\[ +c_0 x^n + c_1 x^{n-1} + \dotsb + c_{n-1} x + c_n = 0 +\Tag{15} +\] +has the rational root~$a/b$, where $a$ and~$b$ are integers without a common divisor +$>1$, then $a$ is an exact divisor of~$c_n$, and~$b$ is an exact divisor of~$c_0$. +\end{Thm} + +Insert the value $a/b$ of~$x$ and multiply all terms of the equation by~$b^n$. +We obtain +\[ +c_0 a^n + c_1 a^{n-1} b + \dotsb + c_{n-1} ab^{n-1} + c_n b^n = 0. +\] +Since $a$ divides all the terms preceding the last term, it divides that term. +But $a$ has no divisor in common with~$b^n$; hence $a$ divides~$c_n$. Similarly, +$b$~divides all the terms after the first term and hence divides~$c_0$. + +\begin{Example} +Find all the rational roots of +\[ +2x^3 - 7x^2 + 10x - 6 = 0. +\] +\end{Example} + +%% -----File: 034.png---Folio 28------- + +\begin{Solution} +By the theorem, the denominator of any rational root~$x$ is a divisor of~$2$. +Hence $y = 2x$ is an integer. Multiplying the terms of our equation by~$4$, we obtain +\index{Transformed equation}% +\[ +y^3 - 7y^2 + 20y - 24 = 0. +\] +There is evidently no negative root. By either of the tests in §§22,~23, an upper limit +to the positive roots of our equation in~$x$ is $1+7/2$, so that $y<9$. Hence the only +possible values of an integral root~$y$ are $1$, $2$, $3$, $4$, $6$,~$8$. Since~$1$ and~$2$ are not roots, we try~$3$: +\[ +\begin{array}{rRRRPc} + 1 & -7 & 20 & -24 && \Lcol{3} \\ +\cline{6-6} +-1 & 4 & -8 & && \\ +\cline{1-3} + 0 & -3 & 12 & && \Strut +\end{array} +\] +Hence $3$ is a root and the remaining roots satisfy the equation $y^2 - 4y + 8 = 0$ and are +$2±2i$. Thus the only rational root of the proposed equation is $x=3/2$. +\end{Solution} + +If $c_0=1$, then $b=±1$ and~$a/b$ is an integer. Hence we have the +\begin{Corollary} +Any rational root of an equation with integral coefficients, +that of the highest power of the unknown being unity, is an integer. +\end{Corollary} + +Given any equation with integral coefficients +\[ +a_0 y^n + a_1 y^{n-1} + \dotsb + a_n = 0, +\] +we multiply each term by~${a_0}^{n-1}$, write $a_0 y = x$, and obtain an equation~\Eq{15} +with integral coefficients, in which the coefficient~$c_0$ of~$x^n$ is now unity. +By the Corollary, each rational root~$x$ is an integer. Hence we need only +find all the integral roots~$x$ and divide them by~$a_0$ to obtain all the rational +roots~$y$ of the proposed equation. + +Frequently it is sufficient (and of course simpler) to set $ky = x$, where +$k$ is a suitably chosen integer less than~$a_0$. + + +\begin{Exercises}{Page28} + +Find all of the rational roots of +\begin{Problems}[2] + +\item[1.] $y^4 -\frac{40}{3}y^3 + \frac{130}{3}y^2 - 40y + 9 = 0$. + +\item[2.] $6y^3 - 11y^2 + 6y - 1 = 0$. + +\ResetCols{1} + +\item[3.] $108y^3 - 270y^2 - 42y + 1 = 0$. [Use $k = 6$.] + +\item[4.] $32y^3 - 6y - 1 = 0$. [Use the least~$k$.] + +\ResetCols{2} + +\item[5.] $96y^3 - 16y^2 - 6y + 1 = 0$. + +\item[6.] $24y^3 - 2y^2 - 5y + 1 = 0$. + +\ResetCols{2} + +\item[7.] $y^3 - \frac{1}{2}y^2 - 2y + 1 = 0$. + +\item[8.] $y^3 - \frac{2}{3}y^2 + 3y - 2 = 0$. + +\ResetCols{1} + +\item[9.] Solve Exs.~2--6 by replacing $y$ by~$1/x$. +\end{Problems} + +Find the equations whose roots are the products of 6 by the roots of +\begin{Problems}[2] + +\item[10.] $y^2 - 2y - \frac{1}{3} = 0$. + +\item[11.] $y^3 - \frac{1}{2}y^2 - \frac{1}{3}y + \frac{1}{4} = 0$. +\end{Problems} +\end{Exercises} + +%% -----File: 035.png---Folio 29------- + + +\Chapter{III}{Constructions with Ruler and Compasses} +\index{Geometrical!construction|(}% [** PP: Using index subitem] + +% [** PP: No ToC entry] +\Section{28.}{Impossible Constructions} We shall prove that it is not possible, +by the methods of Euclidean geometry, to trisect all angles, or to construct +a regular polygon of~$7$ or $9$~sides. The proof, which is beyond the +scope of elementary geometry, is based on principles of the theory of +equations. Moreover, the discussion will show that a regular polygon +of $17$~sides can be constructed with ruler and compasses, a fact not suspected +during the twenty centuries from Euclid to Gauss. + +\Section{29.}{Graphical Solution of a Quadratic Equation} If $a$ +and~$b$ are constructible, and +\index{Quadratic equation!graphical solution}% +\begin{flalign*}% [** PP: Hack to center over the narrowed text block] +\makebox[\linewidth-3.25in][c]{$x^2 - ax + b = 0$}&& & && +\Tag{1} +\end{flalign*} +%Illustration: \textsc{Fig}.~6 +\begin{wrapfigure}{r}{2.75in} +\hfil\Input{035a} +\end{wrapfigure} +has real coefficients and real roots, the roots +can be constructed with ruler and compasses +as follows. Draw a circle having as a diameter +the line~$BQ$ joining the points $B = (0, 1)$ +and $Q = (a, b)$ in Fig.~6. Then \emph{the +abscissas $ON$ and~$OM$ of the points of intersection +of this circle with the $x$-axis are the +roots of~\Eq{1}}. + +For, the center of the circle is $\bigl(a/2, (b+1)/2\bigr)$; the square of~$BQ$ is +$a^2 + (b-1)^2$; hence the equation of the circle is +\[ +\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b+1}{2}\right)^2 + = \frac{a^2 + (b-1)^2}{4}. +\] + +This is found to reduce to~\Eq{1} when $y = 0$, which proves the theorem. + +When the circle is tangent to the $x$-axis, so that $M$ and~$N$ coincide, +the two roots are equal. When the circle does not cut the $x$-axis, or +when $Q$ coincides with~$B$, the roots are imaginary. + +Another construction follows from~§30. + +%% -----File: 036.png---Folio 30------- + +\begin{Exercises}{Page30} + +Solve graphically: +\begin{Problems}[3] + +\item[1.] $x^2 - 5x + 4 = 0$. + +\item[2.] $x^2 + 5x + 4 = 0$. + +\item[3.] $x^2 + 5x - 4 = 0$. + +\ResetCols{3} + +\item[4.] $x^2 - 5x - 4 = 0$. + +\item[5.] $x^2 - 4x + 4 = 0$. + +\item[6.] $x^2 - 3x + 4 = 0$. + +\end{Problems} +\end{Exercises} + + +\Section{30.}{Analytic Criterion for Constructibility} The first step in our +consideration of a problem proposed for construction consists in formulating +the problem analytically. In some instances elementary algebra +suffices for this formulation. For example, in the ancient problem of +the duplication of a cube, we take as a unit of length a side of the given +cube, and seek the length~$x$ of a side of another cube whose volume is +double that of the given cube; hence +\[ +x^3 = 2. +\Tag{2} +\] + +But usually it is convenient to employ analytic geometry as in~§29; +a point is determined by its coordinates $x$ and~$y$ with reference to fixed +rectangular axes; a straight line is determined by an equation of the +first degree, a circle by one of the second degree, in the coordinates of the +general point on it. Hence we are concerned with certain numbers, +some being the coordinates of points, others being the coefficients of equations, +and still others expressing lengths, areas or volumes. These numbers +may be said to define analytically the various geometric elements +involved. + +\begin{Criterion} +A proposed construction is possible by ruler and compasses +if and only if the numbers which define analytically the desired geometric +elements can be derived from those defining the given elements by a +finite number of rational operations and extractions of real square +roots. +\end{Criterion} +\index{Square roots}% + +\begin{Remark} +In §29 we were given the numbers $a$ and~$b$, and constructed lines of lengths +\[ +\tfrac{1}{2}(a ± \sqrt{a^2 -4b)}. +\] +\end{Remark} + +\begin{Proof} +First, we grant the condition stated in the criterion and prove +that the construction is possible with ruler and compasses. For, a rational +function of given quantities is obtained from them by additions, subtractions, +multiplications, and divisions. The construction of the sum +or difference of two segments is obvious. The construction, by means +of parallel lines, of a segment whose length~$p$ is equal to the product $a·b$ +of the lengths of two given segments is shown in Fig.~7; that for the quotient +%% -----File: 037.png---Folio 31------- +$q = a/b$ in Fig.~8. Finally, a segment of length $s = \sqrt{n}$ may be constructed, +as in Fig.~9, by drawing a semicircle on a diameter composed +of two segments of lengths $1$ and~$n$, and then drawing a perpendicular +to the diameter at the point which separates the two segments. Or we +may construct a root of $x^2 - n = 0$ by~§29. +%[Illustration: Fig. 7] +%[Illustration: Fig. 8] +%[Illustration: Fig. 9] +\begin{figure*}[hbt] +\begin{center} +\Input{037a}\hfil +\Input{037b}\hfil\hfil +\Input{037c} +\end{center} +\end{figure*} + +Second, suppose that the proposed construction is possible with ruler +and compasses. The straight lines and circles drawn in making the construction +are located by means of points either initially given or obtained +as the intersections of two straight lines, a straight line and a circle, or +two circles. Since the axes of coordinates are at our choice, we may +assume that the $y$-axis is not parallel to any of the straight lines employed +in the construction. Then the equation of any one of our lines is +\[ +y = mx + b. +\Tag{3} +\] + +Let $y = m'x + b'$ be the equation of another of our lines which intersects~\Eq{3}. +The coordinates of their point of intersection are +\[ +x = \frac{b' - b}{m - m'},\qquad +y = \frac{mb' - m'b}{m - m'}, +\] +which are rational functions of the coefficients of the equations of the +two lines. + +Suppose that a line~\Eq{3} intersects the circle +\[ +(x - c)^2 + (y - d)^2 = r^2, +\] +with the center $(c, d)$ and radius~$r$. To find the coordinates of the points +of intersection, we eliminate~$y$ between the equations and obtain a quadratic +equation for~$x$. Thus $x$ (and hence also $mx + b$ or~$y$) involves no +%% -----File: 038.png---Folio 32------- +irrationality other than a real square root, besides real irrationalities +present in $m$, $b$, $c$, $d$,~$r$. +\index{Square roots}% + +Finally, the intersections of two circles are given by the intersections +of one of them with their common chord, so that this case reduces to the +preceding. +\end{Proof} + +\begin{Remark} +For example, a side of a regular pentagon inscribed in a circle of radius unity is +(Ex.~2 of~§37) +\[ +s = \tfrac{1}{2}\sqrt{10 - 2\sqrt{5}}, +\Tag{4} +\] +which is a number of the type mentioned in the criterion. Hence a regular pentagon +can be constructed by ruler and compasses (see the example above +quoted). +\end{Remark} + + +\Section{31.}{Cubic Equations with a Constructible Root} We saw that the +\index{Cubic equation}% +problem of the duplication of a cube led to a cubic equation~\Eq{2}. We +shall later show that each of the problems, to trisect an angle, and to construct +regular polygons of $7$ and~$9$ sides with ruler and compasses, leads +to a cubic equation. We shall be in a position to treat all of these problems +as soon as we have proved the following general result. + +\begin{Theorem} +It is not possible to construct with ruler and compasses a +line whose length is a root or the negative of a root of a cubic equation with +rational coefficients having no rational root. +\end{Theorem} + +Suppose that $x_1$ is a root of +\[ +x^3 + \alpha x^2 + \beta x + \gamma = 0 \qquad +\text{($\alpha$, $\beta$, $\gamma$ rational)} +\Tag{5} +\] +such that a line of length $x_1$ or~$-x_1$ can be constructed with ruler and compasses; +we shall prove that one of the roots of~\Eq{5} is rational. We have +only to discuss the case in which $x_1$ is irrational. + +By the criterion in~§30, since the given numbers in this problem are +$\alpha$, $\beta$, $\gamma$, all rational, $x_1$ can be obtained by a finite number of rational +operations and extractions of real square roots, performed upon rational +numbers or numbers derived from them by such operations. Thus $x_1$ +involves one or more real square roots, but no further irrationalities. + +As in the case of~\Eq{4}, there may be superimposed radicals. Such a +two-story radical which is not expressible as a rational function, with +rational coefficients, of a finite number of square roots of positive rational +numbers is said to be a radical of \emph{order}~2. In general, an $n$-story radical +is said to be of order~$n$ if it is not expressible as a rational function, with +\index{Order of radical}% +rational coefficients, of radicals each with fewer than $n$~superimposed +radicals, the innermost ones affecting positive rational numbers. + +%% -----File: 039.png---Folio 33------- + +We agree to simplify $x_1$ by making all possible replacements of certain +types that are sufficiently illustrated by the following numerical examples. + +If $x_1$ involves $\sqrt{3}$, $\sqrt{5}$, and~$\sqrt{15}$, we agree to replace $\sqrt{15}$ by~$\sqrt{3}·\sqrt{5}$. +If $x_1 = s - 7t$, where $s$ is given by~\Eq{4} and +\[ +t = \tfrac{1}{2} \sqrt{10 + 2\sqrt{5}}, +\] +so that $st = \sqrt{5}$, we agree to write $x_1$ in the form $s - 7\sqrt{5}/s$, which involves +a single radical of order~$2$ and no new radical of lower order. Finally, +we agree to replace $\sqrt{4 - 2\sqrt{3}}$ by its simpler form $\sqrt{3}-1$. + +After all possible simplifications of these types have been made, the +resulting expressions have the following properties (to be cited as our +agreements): no one of the radicals of highest order~$n$ in~$x_1$ is equal to +a rational function, with rational coefficients, of the remaining radicals +of order~$n$ and the radicals of lower orders, while no one of the radicals +of order $n-1$ is equal to a rational function of the remaining radicals of +order $n-1$ and the radicals of lower orders, etc. + +Let $\sqrt{k}$ be a radical of highest order~$n$ in~$x_1$. Then +\[ +x_1 = \frac{a + b \sqrt{k}}{c + d \sqrt{k}}, +\] +where $a$, $b$, $c$, $d$ do not involve $\sqrt{k}$, but may involve other radicals. If +$d = 0$, then $c \neq 0$ and we write $e$ for $a/c$, $f$ for $b/c$, and get +\begin{flalign*} +&& x_1 &= e + f\sqrt{k}, && \Rightmark{(f \neq 0)} +\Tag{6} +\end{flalign*} +where neither $e$ nor~$f$ involves~$\sqrt{k}$. If $d \neq 0$, we derive~\Eq{6} by multiplying +the numerator and denominator of the fraction for~$x_1$ by $c - d\sqrt{k}$, which +is not zero since $\sqrt{k} = c/d$ would contradict our above agreements. + +By hypothesis, \Eq{6} is a root of equation~\Eq{5}. After expanding the +powers and replacing the square of $\sqrt{k}$ by~$k$, we see that +\[ +(e + f \sqrt{k})^3 + + \alpha(e + f \sqrt{k})^2 + + \beta(e + f \sqrt{k}) + + \gamma = A + B\sqrt{k}, +\Tag{7} +\] +where $A$ and~$B$ are certain polynomials in $e$, $f$,~$k$ and the rational numbers +$\alpha$, $\beta$,~$\gamma$. Thus $A + B \sqrt{k} = 0$. If $B \neq 0$, $\sqrt{k} = -A/B$ is a rational function, +with rational coefficients, of the radicals, other than~$\sqrt{k}$, in~$x_1$, contrary +to our agreements. Hence $B = 0$ and therefore $A = 0$. + +When $e - f \sqrt{k}$ is substituted for~$x$ in the cubic function~\Eq{5}, the result +%% -----File: 040.png---Folio 34------- +is the left member of~\Eq{7} with $\sqrt{k}$ replaced by~$-\sqrt{k}$, and hence the result +is $A - B\sqrt{k}$. But $A = B = 0$. This shows that +\[ +x_2 = e - f \sqrt{k} +\Tag{8} +\] +is a new root of our cubic equation. Since the sum of the three roots +is equal to~$-\alpha$ by~§20, the third root is +\[ +x_3 = -\alpha - x_1 - x_2 = -\alpha - 2e. +\Tag{9} +\] + +Now $\alpha$ is rational. If also $e$ is rational, $x_3$ is a rational root and we have +reached our goal. We next make the assumption that $e$ is irrational +and show that it leads to a contradiction. Since $e$ is a component part +of the constructible root~\Eq{6}, its only irrationalities are square roots. +Let $\sqrt{s}$ be one of the radicals of highest order in~$e$. By the argument +which led to~\Eq{6}, we may write $e = e' + f'\sqrt{s}$, whence, by~\Eq{9}, +\begin{flalign*} +&& x_3 &= g + h \sqrt{s},&& \Rightmark{(h \neq 0)} +\Tag{9'} +\end{flalign*} +where neither $g$ nor~$h$ involves~$\sqrt{s}$. Then by the argument which led +to~\Eq{8}, $g - h \sqrt{s}$ is a root, different from~$x_3$, of our cubic equation, and hence +is equal to $x_1$ or~$x_2$ since there are only three roots~(§16). Thus +\[ +g - h \sqrt{s} = e ± f \sqrt{k}. +\] + +By definition, $\sqrt{s}$ is one of the radicals occurring in $e$. Also, by~\Eq{9'}, +every radical occurring in $g$ or~$h$ occurs in $x_3$ and hence in $e = \frac{1}{2}(-\alpha - x_3)$, +by~(9), $\alpha$ being rational. Hence $\sqrt{k}$ is expressible rationally in terms +of the remaining radicals occurring in $e$ and~$f$, and hence in~$x_1$, whose value +is given by~\Eq{6}. But this contradicts one of our agreements. + + +\Section{32.}{Trisection of an Angle} For a given angle~$A$, we can construct +\index{Trisection of angle}% +with ruler and compasses a line of length $\cos A$ or $-\cos A$, namely the +adjacent leg of a right triangle, with hypotenuse unity, formed by dropping +a perpendicular from a point in one side of $A$ to the other, produced if +necessary. If it were possible to trisect angle~$A$, i.e., construct the angle +$A/3$ with ruler and compasses, we could as before construct a line whose +length is $±\cos(A/3)$. Hence if we show that this last cannot be done +when the only given geometric elements are the angle~$A$ and a line of +unit length, we shall have proved that the angle~$A$ cannot be trisected. +We shall give the proof for $A = 120°$. + +We employ the trigonometric identity +\[ +\cos A = 4 \cos^3 \frac{A}{3} - 3 \cos \frac{A}{3}. +\] +%% -----File: 041.png---Folio 35------- +Multiply each term by~$2$ and write $x$ for $2\cos(A/3)$. Thus +\[ +x^3 - 3x = 2\cos A. +\Tag{10} +\] + +For $A = 120°$, $\cos A = -\frac{1}{2}$ and~\Eq{10} becomes +\[ +x^3 - 3x + 1 = 0. +\Tag{11} +\] + +Any rational root is an integer~(§27) which is an exact divisor of the +constant term~(§24). By trial, neither $+1$ nor~$-1$ is a root. Hence +\Eq{11} has no rational root. Hence~(§31) +\begin{Thm}% +it is not possible to trisect all +angles with ruler and compasses. +\end{Thm} + +\begin{Remark} +Certain angles, like $90°$, $180°$, can be trisected. When $A=180°$, the equation +\Eq{10} becomes $x^3 - 3x = -2$ and has the rational root $x = 1$. It is the rationality of a +root which accounts for the possibility of trisecting this special angle~$180°$. +\end{Remark} + + +\Section[Duplication of a Cube] +{33.}{Regular Polygon of $9$~Sides, Duplication of a Cube.} Since angle +\index{Duplication of cube}% +$120°$ cannot be trisected with ruler and compasses~(§32), angle~$40°$ cannot +be so constructed in terms of angle~$120°$ and the line of unit length +as the given geometric elements. Since the former of these elements +and its cosine are constructible when the latter is given, we may take +the line of unit length as the only given element. In a regular polygon +of $9$~sides, the angle subtended at the center by one side is $\frac{1}{9}·360° = 40°$. +Hence \emph{a regular polygon of $9$~sides cannot be constructed with ruler and compasses}. +\index{Regular polygon!09@{$9$~sides}}% +Here, as in similar subsequent statements where the given +elements are not specified, the only such element is the line of unit length. + +A rational root of $x^3 = 2$ is an integer~(§27) which is an exact divisor +of~$2$. The cubes of $±1$ and~$±2$ are distinct from~$2$. Hence there is no +rational root. Hence (§§30,~31) +\begin{Thm}% +it is not possible to duplicate a cube with +ruler and compasses. +\end{Thm} + + +% [** PP: ToC entry reads Regular Polygons of 7, 9, 17, and n Sides] +\Section{34.}{Regular Polygon of $7$~Sides} If we could construct with ruler +and compasses an angle~$B$ containing $360/7$ degrees, we could so construct +a line of length $x = 2 \cos B$. Since $7B = 360°$, $\cos 3B = \cos 4B$. +But +\index{Regular polygon!07@{$7$~sides}|(}% +\begin{align*} +2 \cos 3B &= 2(4 \cos^3 B - 3 \cos B) = x^3 - 3x, \\ +2 \cos 4B &= 2(2 \cos^2 2B - 1) = 4(2 \cos^2 B - 1)^2 - 2 = (x^2 - 2)^2 - 2. +\end{align*} +Hence +\[ +0 = x^4 - 4x^2 + 2 - (x^3 - 3x) = (x - 2)(x^3 + x^2 - 2x - 1). +\] +But $x = 2$ would give $\cos B = 1$, whereas $B$ is acute. Hence +\[ +x^3 + x^2 - 2x - 1 = 0. +\Tag{12} +\] + +%% -----File: 042.png---Folio 36------- + +Since this has no rational root, \emph{it is impossible to construct a regular +polygon of $7$~sides with ruler and compasses}. + + +% [** PP: No ToC entry] +\Section{35.}{Regular Polygon of $7$~Sides and Roots of Unity} If +\index{Roots of unity}% +\[ +R = \cos\frac{2\pi}{7} + i \sin\frac{2\pi}{7}, +\] +we saw in~§10 that $R$, $R^2$, $R^3$, $R^4$, $R^5$, $R^6$, $R^7 = 1$ give all the roots of $y^7 = 1$ +and are complex numbers represented by the vertices of a regular polygon +of $7$~sides inscribed in a circle of radius unity and center at the origin of +coordinates. By~§6, +\[ +\frac{1}{R} = \cos\frac{2\pi}{7} -i \sin\frac{2\pi}{7},\qquad +R + \frac{1}{R} = 2\cos\frac{2\pi}{7}. +\] + +We saw in~§34 that $2 \cos(2\pi/7)$ is one of the roots of the cubic +equation~\Eq{12}. This equation can be derived in a new manner by utilizing +the preceding remarks on $7$th roots of unity. Our purpose is not primarily +to derive~\Eq{12} again, but to illustrate some principles necessary in the +general theory of the construction of regular polygons. + +Removing from $y^7 - 1$ the factor $y-1$, we get +\[ +y^6 + y^5 + y^4 + y^3 + y^2 + y + 1 = 0, +\Tag{13} +\] +whose roots are $R$, $R^2, \dotsc, R^6$. Since we know that $R+1/R$ is one of +the roots of the cubic equation~\Eq{12}, it is a natural step to make the substitution +\[ +y+ \frac{1}{y} = x +\Tag{14} +\] +in~\Eq{13}. After dividing its terms by~$y^3$, we have +\[ +\left(y^3 + \frac{1}{y^3}\right) + +\left(y^2 + \frac{1}{y^2}\right) + +\left(y + \frac{1}{y }\right) + 1=0. +\Tag{13'} +\] +By squaring and cubing the members of~\Eq{14}, we see that +\[ +y^2 + \frac{1}{y^2} = x^2 - 2,\qquad +y^3 + \frac{1}{y^3} = x^3 - 3x. +\Tag{15} +\] +Substituting these values in~\Eq{13'}, we obtain +\[ +x^3 + x^2 - 2x - 1 = 0. +\tag{12}% [** PP: [sic], equation repeated] +\] +That is, the substitution~\Eq{14} converts equation~\Eq{13} into~\Eq{12}. +\index{Regular polygon!07@{$7$~sides}|)}% + +%% -----File: 043.png---Folio 37------- + +If in~\Eq{14} we assign to~$y$ the six values $R, \dotsc, R^6$, we obtain only +three distinct values of~$x$: +\[ +x_1 = R + \frac{1}{R} = R + R^6, \quad +x_2 = R^2 + \frac{1}{R^2} = R^2 + R^5, \quad +x_3 = R^3 + \frac{1}{R^3} = R^3 + R^4. +\Tag{16} +\] + +In order to illustrate a general method of the theory of regular polygons, +we start with the preceding sums of the six roots in pairs and find +the cubic equation having these sums as its roots. For this purpose we +need to calculate +\[ +x_1 + x_2 + x_3, \qquad +x_1 x_2 + x_1 x_3 + x_2 x_3, \qquad +x_1 x_2 x_3. +\] +First, by~\Eq{16}, +\[ +x_1 + x_2 + x_3 = R + R^2 + \dotsb + R^6 = -1, +\] +since $R, \dotsc, R^6$ are the roots of~\Eq{13}. Similarly, +\begin{gather*} +x_1 x_2 + x_1 x_3 + x_2 x_3 = 2(R + R^2 + \dotsb +R^6) = -2, \\ +x_1 x_2 x_3 = 2 + R + R^2 + \dotsb + R^6 = 1. +\end{gather*} + +Consequently~(§20), the cubic having $x_1$, $x_2$, $x_3$ as roots is~\Eq{12}. + + +\Section{36.}{Reciprocal Equations} Any algebraic equation such that the +reciprocal of each root is itself a root of the same multiplicity is called a +\emph{reciprocal equation}. +\index{Reciprocal equation}% + +\begin{Remark} +The equation $y^7 - 1 = 0$ is a reciprocal equation, since if $r$ is any root, $1/r$ is evidently +also a root. Since~\Eq{13} has the same roots as this equation, with the exception of unity +which is its own reciprocal, \Eq{13} is also a reciprocal equation. +\end{Remark} + +If $r$ is any root $\neq 0$ of any equation +\[ +f(y) \equiv y^n + \dotsb + c = 0, +\] +$1/r$ is a root of $f(1/y)=0$ and hence of +\[ +y^n f \left(\frac{1}{y}\right) \equiv 1 + \dotsb + cy^n = 0. +\] +If the former is a reciprocal equation, it has also the root~$1/r$, so that every +root of the former is a root of the latter equation. Hence, by~§18, the +left member of the latter is identical % [** PP: Typo indentical] +with $cf(y)$. Equating the constant +terms, we have $c^2 = 1$, $c= ±1$. Hence +\[ +y^n f\left(\frac{1}{y}\right) \equiv ±f(y). +\Tag{17} +\] +%% -----File: 044.png---Folio 38------- +Thus if $p_i y^{n-i}$ is a term of~$f(y)$, also $±p_i y^i$ is a term. Hence +\[ +f(y) \equiv y^n ± 1 + p_1(y^{n-1} ± y) + p_2 (y^{n-2} ± y^2) + \dotsb. +\Tag{18'} +\] + +If $n$ is \emph{odd}, $n = 2t+1$, the final term is $p_t(y^{t+1} ± y^t)$, and $y ± 1$ is a factor +of $f(y)$. In view of~\Eq{17}, the quotient +\[ +Q(y) \equiv \frac{f(y)}{y±1} +\] +has the property that +\[ +y^{n-1} Q \left(\frac{1}{y}\right) \equiv Q(y). +\] +Comparing this with~\Eq{17}, which implied~\Eq{18'}, we see that $Q(y)=0$ is +a reciprocal equation of the type +\[ +y^{2t} + 1 + + c_1 (y^{2t-1} + y) + + c_2 (y^{2t-2} + y^2) + \dotsb + + c_{t-1} (y^{t+1} + y^{t-1}) + + c_t y^t = 0. +\Tag{18} +\] + +If $n$ is \emph{even}, $n = 2t$, and if the upper sign holds in~\Eq{17}, then~\Eq{18'} is +of the form~\Eq{18}. Next, let the lower sign hold in~\Eq{17}. Since a term +$p_t y^t$ would imply a term~$-p_t y^t$, we have $p_t = 0$. The final term in~\Eq{18'} +is therefore $p_{t-1} (y^{t+1} - y^{t-1})$. Hence $f(y)$ has the factor $y^2-1$. The +quotient $q(y) \equiv f(y)/(y^2-1)$ has the property that +\[ +y^{n-2} q \left(\frac{1}{y}\right) \equiv q(y). +\] +Comparing this with~\Eq{17} as before, we see that $q(y)=0$ is of the form~\Eq{18} +where now $2t = n -2$. Hence, at least after removing one or both +of the factors $y±1$, \emph{any reciprocal equation may be given the form~\Eq{18}}. + +The method by which~\Eq{13} was reduced to a cubic equation may be +used to reduce any equation~\Eq{18} to an equation in~$x$ of half the degree. +First, we divide the terms of~\Eq{18} by~$y^t$ and obtain +\index{Cubic equation}% [** PP: Original entry points to Folio 40] +\[ +\left(y^t + \frac{1}{y^t}\right) + + c_1 \left(y^{t-1} + \frac{1}{y^{t-1}}\right) + \dotsb + + c_{t-1} \left(y + \frac{1}{y}\right) + c_t = 0. +\] +Next, we perform the substitution~\Eq{14} by either of the following methods: +We may make use of the relation +\[ +y^k + \frac{1}{y^k} + = x \left(y^{k-1} + \frac{1}{y^{k-1}}\right) + - \left(y^{k-2} + \frac{1}{y^{k-2}}\right) +\] +to compute the values of $y^k + 1/y^k$ in terms of~$x$, starting with the special +%% -----File: 045.png---Folio 39------- +cases \Eq{14} and~\Eq{15}. For example, +\[ +\begin{split} +y^4 + \frac{1}{y^4} + &= x \left(y^3 + \frac{1}{y^3}\right) + - \left(y^2 + \frac{1}{y^2}\right) \\ + &= x (x^3-3x) - (x^2-2) = x^4 - 4x^2 + 2. % [** PP: Added .] +\end{split} +\] +Or we may employ the explicit formula~\Eq{19} of~§107 for the sum $y^k + 1/y^k$ +of the $k$th~powers of the roots~$y$ and~$1/y$ of $y^2 - xy + 1 = 0$. + + +% [** PP: No separate ToC entry] +\Section[Regular Polygon of $9$~Sides] +{37.}{Regular Polygon of $9$~Sides and Roots of Unity.} If +\index{Regular polygon!09@{$9$~sides}}% +\index{Roots of unity}% +\[ +R = \cos\frac{2\pi}{9} + i \sin\frac{ 2\pi}{9}, +\] +the powers $R$, $R^2$, $R^4$, $R^5$, $R^7$, $R^8$, are the primitive ninth roots of unity~(§11). +They are therefore the roots of +\[ +\frac{y^9 - 1}{y^3 - 1} = y^6 + y^3 + 1 = 0. +\Tag{19} +\] +Dividing the terms of this reciprocal equation by~$y^3$ and applying the second +relation~\Eq{15}, we obtain our former cubic equation~\Eq{11}. + + +\begin{Exercises}{Page40} + +\begin{Problems} +\item[1.] Show by~\Eq{16} that the roots of~\Eq{12} are $2\cos 2\pi/7$, $2\cos 4\pi/7$, $2\cos 6\pi/7$. + +\begin{minipage}[b]{\linewidth-2.25in} +\item[2.] The imaginary fifth roots of unity satisfy +$y^4 + y^3 + y^2 + y + 1 = 0$, which by the substitution~\Eq{14} becomes +$x^2 + x - 1 = 0$. It has the root +\[ +R + \frac{1}{R} = 2 \cos\frac{2\pi}{5} = \frac{1}{2}(\sqrt{5}-1). +\] +In a circle of radius unity and center~$O$ draw two perpendicular +diameters $AOA'$, $BOB'$. With the middle +point~$M$ of~$OA'$ as center and radius~$MB$ draw a circle +cutting~$OA$ at~$C$ (Fig.~10). Show that $OC$ and~$BC$ +are the sides~$s_{10}$ and~$s_5$ of the inscribed regular decagon +and pentagon respectively. Hints: +\end{minipage}% +% Illustration: \textsc{Fig}. 10 +\hfill\Input{045a} +\index{Regular!decagon}% +\index{Regular!pentagon}% +\begin{align*} +MB &= \tfrac{1}{2}\sqrt{5},\qquad +OC = \tfrac{1}{2}(\sqrt{5} - 1),\qquad +BC = \sqrt{1+OC^2} = \tfrac{1}{2}\sqrt{10 - 2\sqrt{5}}, \\ +s_{10} &= 2 \sin 18° = 2 \cos\frac{2\pi}{5} = OC, \\ +{s_5}^2 &= (2 \sin 36°)^2 = 2\left(1 - \cos\frac{2\pi}{5}\right) + = \frac{1}{4}(10 - 2\sqrt{5}), \qquad s_5 = BC. +\end{align*} + +%% -----File: 046.png---Folio 40------- + +\item[3.] If $R$ is a root of~\Eq{19} verify as at the end of~§35 that $R+R^8$, $R^2+R^7$, and $R^4+R^5$ +are the roots of~\Eq{11}. + +\item[4.] Hence show that the roots of~\Eq{11} are $2\cos 2\pi/9$, $2\cos 4\pi/9$, $2\cos 8\pi/9$. + +\item[5.] Reduce $y^{11} = 1$ to an equation of degree~$5$ in~$x$. + +\item[6.] Solve $y^5 - 7y^4 + y^3 - y^2 + 7y - 1 = 0$ by radicals. [One root is~$1$.] + +\item[7.] After finding so easily in \ChapRef{I} the trigonometric forms of the complex roots +of unity, why do we now go to so much additional trouble to find them algebraically? + +\item[8.] Prove that every real root of $x^4 + ax^2 + b = 0$ can be constructed with ruler and +compasses, given lines of lengths $a$ and~$b$. + +\item[9.] Show that the real roots of $x^3 - px - q = 0$ are the abscissas of the intersections +of the parabola $y = x^2$ and the circle through the origin with the center +$(\frac{1}{2}q, \frac{1}{2} + \frac{1}{2}p)$. +\end{Problems} + +Prove that it is impossible, with ruler and compasses: +\begin{Problems} +\item[10.] To construct a straight line representing the distance from the circular base +of a hemisphere to the parallel plane which bisects the hemisphere. + +\item[11.] To construct lines representing the lengths of the edges of an existing rectangular +parallelopiped having a diagonal of length~$5$, surface area~$24$, and volume~$1$, $2$, $3$, or~$5$. + +\item[12.] To trisect an angle whose cosine is $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, $\frac{1}{8}$ or~$p/q$, where $p$ and~$q$ ($q>1$) are +integers without a common factor, and $q$ is not divisible by a cube. +\index{Trisection of angle}% +\end{Problems} + +Prove algebraically that it is possible, with ruler and compasses: +\begin{Problems} +\item[13.] To trisect an angle whose cosine is $(4a^3 - 3ab^2)/b^3$, where the integer~$a$ is numerically +less than the integer~$b$; for example, $\cos^{-1} 11/16$ if $a = -1$, $b = 4$. + +\item[14.] To construct the legs of a right triangle, given its area and hypotenuse. + +\item[15.] To construct the third side of a triangle, given two sides and its area. + +\item[16.] To locate the point~$P$ on the side $BC=1$ of a given square $ABCD$ such that +the straight line $AP$ cuts $DC$ produced at a point~$Q$ for which the +length of $PQ$ is a given +number~$g$. Show that $y=BP$ is a root of a reciprocal quartic equation, and solve it +when $g = 10$. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{38.}{The Periods of Roots of Unity} Before taking up the regular +\index{Roots of unity!periods of}% +polygon of $17$~sides, we first explain another method of finding the pairs +of imaginary seventh roots of unity $R$ and~$R^6$, $R^2$ and~$R^5$, $R^3$ and~$R^4$, +employed in~\Eq{16}. To this end we seek a positive integer~$g$ such that +the six roots can be arranged in the order +\[ +R,\quad R^g,\quad R^{g^2},\quad R^{g^3},\quad R^{g^4},\quad R^{g^5}, +\Tag{20} +\] +where each term is the $g$th power of its predecessor. Trying $g = 2$, we find +that the fourth term would then be $R^8 = R$. Hence $g\neq 2$. Trying $g = 3$, +we obtain +\[ +R,\quad R^3,\quad R^2,\quad R^6,\quad R^4,\quad R^5, +\Tag{21} +\] +where each term is the cube of its predecessor. + +%% -----File: 047.png---Folio 41------- + +To define three \emph{periods}, each of two terms, +\[ +R + R^6,\qquad +R^2 + R^5,\qquad +R^3 + R^4, +\Tag{16'} +\] +we select the first term~$R$ of~\Eq{21} and the third term $R^6$ after it and add +them, then the second term $R^3$ and the third term $R^4$ after it, and finally +$R^2$ and the third term $R^5$ after it. + +We may also define two periods, each of three terms, +\[ +z_1 = R + R^2 + R^4,\qquad +z_2 = R^3 + R^6 + R^5, +\] +by taking alternate terms in~\Eq{21}. + +\begin{Remark} +Since $z_1 + z_2 = -1$, $z_1 z_2 = 3 + R + \dotsb + R^6 = 2$, $z_1$ and $z_2$ are the roots of $z^2 + z + 2 = 0$. +Then $R$, $R^2$, $R^4$ are the roots of $w^3 - z_1w^2 + z_2w - 1 = 0$. +\end{Remark} + + +% [** PP: No separate ToC entry] +\Section{39.}{Regular Polygon of $17$~Sides} Let $R$ be a root $\neq 1$ of $x^{17} = 1$. +Then +\index{Regular polygon!17@{$17$~sides}|(}% +\[ +\frac{R^{17} - 1}{R - 1} = R^{16} + R^{15} + \dotsb + R + 1 = 0. +\] +As in §38, we may take $g=3$ and arrange the roots $R, \dotsc, R^{16}$ so that +each is the cube of its predecessor: +\[ +R,\ R^3,\ R^9,\ R^{10},\ +R^{13},\ R^5,\ R^{15},\ R^{11},\ +R^{16},\ R^{14},\ R^8,\ R^7,\ +R^4,\ R^{12},\ R^2,\ R^6. +\] + +Taking alternate terms, we get the two periods, each of eight terms, +\begin{align*} +y_1 &= R + R^9 + R^{13} + R^{15} + R^{16} + R^8 + R^4 + R^2, \\ +y_2 &= R^3 + R^{10} + R^5 + R^{11} + R^{14} + R^7 + R^{12} + R^6. +\end{align*} +Hence $y_1 + y_2 = -1$. We find that $y_1 y_2 = 4(R + \dotsb + R^{16}) = -4$. Thus +\[ +y_1,\ y_2 \quad\text{satisfy}\quad y^2 + y - 4 = 0. +\Tag{22} +\] + +Taking alternate terms in~$y_1$, we obtain the two periods +\[ +z_1 = R + R^{13} + R^{16} + R^4, \qquad +z_2 = R^9 + R^{15} + R^8 + R^2. +\] +Taking alternate terms in~$y_2$, we get the two periods +\[ +w_1 = R^3 + R^5 + R^{14} + R^{12}, \qquad +w_2 = R^{10} + R^{11} + R^7 + R^6. +\] +Thus $z_1 + z_2 = y_1$, $w_1 + w_2 = y_2$. We find that $z_1 z_2 = w_1 w_2 = -1$. Hence +\begin{align*} +z_1,\ z_2 &\quad\text{satisfy}\quad z^2 - y_1 z - 1 = 0, +\Tag{23} \\ +w_1,\ w_2 &\quad\text{satisfy}\quad w^2 - y_2 w - 1 = 0. +\Tag{24} +\end{align*} + +%% -----File: 048.png---Folio 42------- + +Taking alternate terms in~$z_1$, we obtain the periods +\[ +v_1 = R + R^{16}, \qquad v_2 = R^{13} + R^4. +\] +Now, $v_1 + v_2 = z_1$, $v_1v_2 = w_1$. Hence +\begin{align*} +v_1,\ v_2 &\quad\text{satisfy}\quad v^2 - z_1v + w_1 = 0, +\Tag{25} \\ +R,\ R^{16} &\quad\text{satisfy}\quad \rho^2 - v_1\rho + 1 = 0. +\Tag{26} +\end{align*} + +Hence we can find $R$ by solving a series of quadratic equations. Which +of the sixteen values of~$R$ we shall thus obtain depends upon which root +of~\Eq{22} is called $y_1$ and which~$y_2$, and similarly in \Eq{23}--\Eq{26}. We shall now +show what choice is to be made in each such case in order that we shall +finally get the value of the particular root +\[ +R = \cos\frac{2\pi}{17} + i \sin\frac{2\pi}{17}. +\] +Then +\begin{alignat*}{4} +\frac{1}{R} + &= \cos \frac{2\pi}{17} - i \sin \frac{2\pi}{17}, &\qquad +v_1 &= R &+{}& \frac{1}{R} &&= 2 \cos \frac{2\pi}{17}, \\ +% +R^4 &= \cos \frac{8\pi}{17} + i \sin \frac{8\pi}{17}, & +v_2 &= R^4 &+{}& \frac{1}{R^4} &&= 2 \cos \frac{8\pi}{17}. +\end{alignat*} +Hence $v_1 > v_2 > 0$, and therefore $z_1 = v_1 + v_2 > 0$. Similarly, +\begin{align*} +w_1 &= R^3 + \frac{1}{R^3} + R^5 + \frac{1}{R^5} + = 2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17} + = 2 \cos \frac{6\pi}{17} - 2 \cos \frac{7\pi}{17} > 0, \\ +% +y_2 &= 2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17} + + 2 \cos \frac{12\pi}{17} + 2 \cos \frac{14\pi}{17} < 0, +\end{align*} +since only the first cosine in $y_2$ is positive and it is numerically less than +the third. But $y_1 y_2 = -4$. Hence $y_1>0$. Thus \Eq{22}--\Eq{24} give +\begin{align*} +y_1 &= \tfrac{1}{2}( \sqrt{17}-1), & +y_2 &= \tfrac{1}{2}(-\sqrt{17}-1), \\ +% +z_1 &= \tfrac{1}{2}y_1 + \sqrt{1 + \tfrac{1}{4}y_1^2}, & +w_1 &= \tfrac{1}{2}y_2 + \sqrt{1 + \tfrac{1}{4}y_2^2}. +\end{align*} + +We may readily construct segments of these lengths. Evidently +$\sqrt{17}$ is the length of the hypotenuse of a right triangle whose legs are of +lengths $1$ and~$4$, while for the radical in~$z_1$ we employ legs of lengths $1$ +and~$\frac{1}{2}y_1$. We thus obtain segments representing the coefficients of the +%% -----File: 049.png---Folio 43------- +quadratic equation~\Eq{25}. Its roots may be constructed as in~§29. The +larger root is +\[ +v_1 = 2 \cos\frac{2\pi}{17}. +\] +Hence we can construct angle $2\pi/17$ with ruler and compasses, and therefore +a regular polygon of $17$~sides. + + +% [** PP: No ToC entry] +\Section{40.}{Construction of a Regular Polygon of $17$~Sides} In a circle of +radius unity, construct two perpendicular +diameters $AB$, $CD$, +and draw tangents at $A$, $D$, +which intersect at~$S$ (Fig.~11). +Find the point~$E$ in~$AS$ for which +$AE = \frac{1}{4} AS$, by means of two bisections. +Then +\[ +AE = \tfrac{1}{4}, \qquad +OE = \tfrac{1}{4} \sqrt{17}. +\] +%[Illustration: \textsc{Fig.} 11] +\begin{figure*}[b] +\begin{center} +\Input{049a} +\end{center} +\end{figure*} + +\noindent Let the circle with center~$E$ +and radius~$OE$ cut~$AS$ at~$F$ and~$F'$. +Then +\begin{align*} +AF &= EF - EA = OE - \tfrac{1}{4} = \tfrac{1}{2} y_1, \\ +AF' &= EF' + EA = OE + \tfrac{1}{4} = -\tfrac{1}{2} y_2, \\ +OF &= \sqrt{OA^2 + AF^2} = \sqrt{1 + \tfrac{1}{4} y_1^2}, \qquad +OF' = \sqrt{1 + \tfrac{1}{4} y_2^2}. +\end{align*} +Let the circle with center~$F$ and radius~$FO$ cut~$AS$ at~$H$, outside of~$F'F$; +that with center~$F'$ and radius~$F'O$ cut~$AS$ at~$H'$ between $F'$ and~$F$. Then +\begin{align*} +AH &= AF + FH = AF + OF + = \tfrac{1}{2} y_1 + \sqrt{1 + \tfrac{1}{4} y_1^2} = z_1, \\ +AH' &= F'H' - F'A = OF'- AF'= w_1. +\end{align*} + +It remains to construct the roots of equation~\Eq{25}. This will be done +as in~§29. Draw $HTQ$ parallel to $AO$ and intersecting $OC$ produced at~$T$. +Make $TQ = AH'$. Draw a circle having as diameter the line $BQ$ +joining $B = (0,1)$ with $Q = (z_1, w_1)$. The abscissas $ON$ and $OM$ of the intersections +of this circle with the $x$-axis $OT$ are the roots of~\Eq{25}. Hence +the larger root~$v_1$ is $OM = 2 \cos(2\pi/17)$. + +%% -----File: 050.png---Folio 44------- + +Let the perpendicular bisector $LP$ of $OM$ cut the initial circle of unit +radius at~$P$. Then +\[ +\cos LOP = OL = \cos\frac{2\pi}{17},\qquad LOP=\frac{2\pi}{17}. +\] + +Hence the chord~$CP$ is a side of the inscribed regular polygon of $17$~sides, +constructed with ruler and compasses. +\index{Regular polygon!17@{$17$~sides}|)}% + + +% [** PP: No separate ToC entry] +\Section{41.}{Regular Polygon of $n$ Sides} If $n$ be a prime such that $n-1$ is +\index{Regular polygon!n@{$n$~sides}}% +\index{Roots of unity}% +a power $2^h$ of~$2$ (as is the case when $n = 3$, $5$, $17$), the $n-1$ imaginary $n$th +roots of unity can be separated into 2~sets each of $2^{h-1}$ roots, each of these +sets subdivided into 2~sets each of $2^{h-2}$ roots, etc., until we reach the pairs +$R$, $1/R$ and $R^2$, $1/R^2$, etc., and in fact\footnote + {See the author's article ``Constructions with ruler and compasses; regular polygons,'' + in \textit{Monographs on Topics of Modern Mathematics}, Longmans, Green and Co., +1911, p.~374.} +in such a manner that we have a +series of quadratic equations, the coefficients of any one of which depend +only upon the roots of quadratic equations preceding it in the series. +Note that this was the case for $n = 17$ and for $n = 5$. It is in this manner +that it can be proved that the roots of $x^n = 1$ can be found in terms of +square roots, so that a regular polygon of $n$~sides can be inscribed by ruler +and compasses, provided $n$ be a prime of the form $2^h + 1$. + +If $n$ be a product of distinct primes of this form, or $2^k$ times such a +product (for example, $n = 15$, $30$ or~$6$), or if $n = 2^m$ ($m > 1$), it follows readily +(see Ex.~1 below) that we can inscribe with ruler and compasses a regular +polygon of $n$~sides. But this is impossible for all other values of~$n$. + + +\begin{Exercises}{Page44} + +\begin{Problems} +\item[1.] If $a$ and~$b$ are relatively prime numbers, so that their greatest common divisor +is unity, we can find integers $c$ and~$d$ such that $ac + bd = 1$. Show that, if regular polygons +of $a$ and~$b$ sides can be constructed and hence angles $2\pi/a$ and $2\pi/b$, a regular +polygon of $a·b$ sides can be derived. + +\item[2.] If $p = 2^h + 1$ is a prime, $h$ is a power of~$2$. For $h = 2^0$, $2^1$, $2^2$, $2^3$, the values of~$p$ +are $3$, $5$, $17$, $257$ and are primes. [Show that $h$ cannot have an odd factor other than +unity.] + +\item[3.] For $13$th roots of unity find the least~$g$~(§38), write out the three periods each +of four terms, and find the cubic equation having them as roots. +\index{Cubic equation}% + +\item[4.] For the primitive ninth roots of unity find the least~$g$ and write out the three +periods each of two terms. +\end{Problems} + +Solve the following reciprocal equations: +\index{Reciprocal equation}% +\begin{Problems}[2] +\item[5.] $y^4 + 4y^3 - 3y^2 + 4y + 1 = 0$. + +\item[6.] $y^5 - 4y^4 + y^3 + y^2 - 4y + 1 = 0$. + +\ResetCols{2} + +\item[7.] $2y^6 - 5y^5 + 4y^4 - 4y^2 + 5y - 2 = 0$. + +\item[\qquad8.] $y^5 + 1 = 31(y + 1)^5$. +\end{Problems} +\end{Exercises} +\index{Geometrical!construction|)}% [** PP: Using index subitem] + +%% -----File: 051.png---Folio 45------- + + +\Chapter[Cubic and Quartic Equations] +{IV}{Solution of Cubic and Quartic Equations; Their Discriminants} +\index{Cubic equation}% + +% [** PP: No ToC entry] +\Section{42.}{Reduced Cubic Equation} If, in the general cubic equation +\index{Cubic equation!reduced}% +\[ +x^3 + bx^2 + cx + d = 0, +\Tag{1} +\] +we set $x = y-b/3$, we obtain the \emph{reduced cubic equation} +\[ +y^3 + py + q = 0, +\Tag{2} +\] +lacking the square of the unknown~$y$, where +\[ +p = c - \frac{b^2}{3}, \qquad +q = d - \frac{bc}{3} + \frac{2b^3}{27}. +\Tag{3} +\] + +After finding the roots $y_1$, $y_2$, $y_3$ of~\Eq{2}, we shall know the roots of~\Eq{1}: +\[ +x_1 = y_1 - \frac{b}{3}, \qquad +x_2 = y_2 - \frac{b}{3}, \qquad +x_3 = y_3 - \frac{b}{3}. +\Tag{4} +\] + + +\Section[Algebraic Solution of a Cubic] +{43.}{Algebraic Solution of the Reduced Cubic Equation.} We shall +employ the method which is essentially the same as that given by Vieta +in~1591. We make the substitution +\[ +y = z - \frac{p}{3z} +\Tag{5} +\] +in~\Eq{2} and obtain +\[ +z^3 - \frac{p^3}{27z^3} + q = 0, +\] +since the terms in $z$ cancel, and likewise the terms in~$1/z$. Thus +\[ +z^6 + qz^3 - \frac{p^3}{27} = 0. +\Tag{6} +\] +Solving this as a quadratic equation for~$z^3$, we obtain +\[ +z^3 = -\frac{q}{2} ±\sqrt{R},\qquad +R = \left(\frac{p}{3}\right)^3 + \left(\frac{q}{2}\right)^2. +\Tag{7} +\] + +%% -----File: 052.png---Folio 46------- + +By §8, any number has three cube roots, two of which are the products +of the remaining one by the imaginary cube roots of unity: +\[ +\omega = -\tfrac{1}{2} + \tfrac{1}{2} \sqrt{3}i,\qquad +\omega^2 = -\tfrac{1}{2} - \tfrac{1}{2} \sqrt{3}i. +\Tag{8} +\] +We can choose particular cube roots +\[ +A = \sqrt[3]{-\frac{q}{2} + \sqrt{R}},\qquad +B = \sqrt[3]{-\frac{q}{2} - \sqrt{R}}, +\Tag{9} +\] +such that $AB = -p/3$, since the product of the numbers under the cube +root radicals is equal to~$(-p/3)^3$. Hence the six values of~$z$ are +\[ +A,\quad \omega A,\quad \omega^2 A,\quad +B,\quad \omega B,\quad \omega^2 B. +\] +These can be paired so that the product of the two in each pair is~$-p/3$: %[** PP: Punctuation indistinct] +\[ +AB = -\frac{p}{3},\qquad +\omega A·\omega^2 B = -\frac{p}{3},\qquad +\omega^2 A·\omega B = -\frac{p}{3}. +\] +Hence with any root~$z$ is paired a root equal to~$-p/(3z)$. By~\Eq{5}, the sum +of the two is a value of~$y$. Hence the \emph{three} values of~$y$ are +\[ +y_1 = A + B,\qquad +y_2 = \omega A + \omega^2 B,\qquad +y_3 = \omega^2 A + \omega B. +\Tag{10} +\] + +It is easy to verify that these numbers are actually roots of~\Eq{2}. For +example, since $\omega^3 = 1$, the cube of $y_2$ is +\[ +A^3 + B^3 + 3\omega A^2 B + 3\omega^2 AB^2 + = -q - p(\omega A + \omega^2 B) = -q - py_2, +\] +by~\Eq{9} and $AB = -p/3$. + +The numbers~\Eq{10} are known as \emph{Cardan's formulas} for the roots of a +\index{Cardan's formulas}% +reduced cubic equation~\Eq{2}. The expression $A + B$ for a root was first +published by Cardan in his \textit{Ars Magna} of~1545, although he had obtained +it from Tartaglia under promise of secrecy. + +\begin{Example} +Solve $y^3 - 15y - 126 = 0$. +\end{Example} + +\begin{Solution} +The substitution~\Eq{5} is here $y = z + 5/z$. We get +\[ +z^6 - 126z^3 + 125 = 0,\qquad +z^3 = 1 \text{ or } 125. +\] +The pairs of values of~$z$ whose product is $5$ are $1$ and~$5$, $\omega$ and $5\omega^2$, $\omega^2$ and $5\omega$. Their +sums $6$, $\omega + 5\omega^2$, and $\omega^2 + 5\omega$ give the three roots. +\end{Solution} + +\begin{Exercises}{Page46} + +Solve the equations: + +\begin{Problems}[2] + +\item[1.] $y^3 - 18y + 35 = 0$. + +\item[2.] $x^3 + 6x^2 + 3x + 18 = 0$. + +\ResetCols{2} + +\item[3.] $y^3 - 2y + 4 = 0$. + +\item[4.] $28x^3 + 9x^2 - 1 = 0$. +\end{Problems} +\end{Exercises} + +%% -----File: 053.png---Folio 47------- + +\Section{44.}{Discriminant} The product of the squares of the differences of +the roots of any equation in which the coefficient of the highest power of +the unknown is unity shall be called the \emph{discriminant} of the equation. +For the reduced cubic~\Eq{2}, the discriminant is +\index{Discriminant!of cubic}% +\[ +(y_1 - y_2)^2 (y_1 - y_3)^2 (y_2 - y_3)^2 = -4p^3 - 27q^2, +\Tag{11} +\] +a result which should be memorized in view of its important applications. +It is proved by means of~\Eq{10} and $\omega^3 = 1$, $\omega^2 + \omega + 1 = 0$, as follows: +\begin{gather*} +y_1 - y_2 = (1-\omega)(A-\omega^2 B), \qquad +y_1 - y_3 = (1-\omega^2)(A-\omega B), \\ +y_2 - y_3 = (\omega -\omega^2) (A-B), \\ +(1-\omega)(1-\omega^2) = 3, \quad +\omega - \omega^2 = \sqrt{3}i. +\end{gather*} +Since $1$, $\omega$, $\omega^2$ are the cube roots of unity, +\[ +(x-1)(x-\omega)(x-\omega^2) \equiv x^3 - 1, +\] +identically in~$x$. Taking $x = A/B$, we see that +\[ +(A-B)(A-\omega B)(A-\omega^2 B) = A^3 - B^3 = 2 \sqrt{R}, +\] +by~\Eq{9}. Hence +\[ +(y_1-y_2)(y_1-y_3)(y_2-y_3) = 6\sqrt{3}\sqrt{R}i. +\] +Squaring, we get~\Eq{11}, since $-108R = -4p^3 - 27q^2$ by~\Eq{7}. For later +use, we note that the discriminant of the reduced cubic is equal to $-108 R$. + +\emph{The discriminant $\Delta$ of the general cubic~\Eq{1} is equal to the discriminant +of the corresponding reduced cubic~\Eq{2}.} For, by~\Eq{4}, +\[ +x_1 - x_2 = y_1 - y_2, \qquad +x_1 - x_3 = y_1 - y_3, \qquad +x_2 - x_3 = y_2 - y_3. +\] + +Inserting in~\Eq{11} the values of $p$ and~$q$ given by~\Eq{3}, we get +\[ +\Delta = 18bcd - 4b^3 d + b^2 c^2 - 4c^3 - 27d^2. +\Tag{12} +\] + +\begin{Remark} +It is sometimes convenient to employ a cubic equation +\[ +ax^3 + bx^2 + cx +d = 0 \quad (a \neq 0), +\Tag{13} +\] +in which the coefficient of $x^3$ has not been made unity by division. The product~$P$ +of the squares of the differences of its roots is evidently derived from~\Eq{12} by replacing +$b$, $c$, $d$ by $b/a$, $c/a$, $d/a$. Hence +\[ +a^4 P = 18 abcd - 4b^3 d + b^2 c^2 - 4ac^3 - 27a^2 d^2. +\Tag{14} +\] +This expression (and not $P$ itself) is called the discriminant of~\Eq{13}. +\end{Remark} + +%% -----File: 054.png---Folio 48------- + + +\Section[Number of Real Roots of a Cubic] +{45.}{Number of Real Roots of a Cubic Equation.} +\index{Cubic equation!number of real roots}% [** PP: Added ``of''] +\index{Number!of roots}% +\begin{Thm} +A cubic equation +with real coefficients has three distinct real roots if its discriminant~$\Delta$ is positive, +a single real root and two conjugate imaginary roots if $\Delta$ is negative, and at +least two equal real roots if $\Delta$ is zero. +\end{Thm} + +If the roots $x_1$, $x_2$, $x_3$ are all real and distinct, the square of the difference +of any two is positive and hence $\Delta$ is positive. + +If $x_1$ and $x_2$ are conjugate imaginaries and hence $x_3$ is real~(§21), +$(x_1 -x_2)^2$ is negative. Since $x_1 - x_3$ and $x_2 - x_3$ are conjugate imaginaries, +their product is positive. Hence $\Delta$ is negative. + +If $x_1 = x_2$, $\Delta$ is zero. If $x_2$ were imaginary, its conjugate would be +equal to~$x_3$ by~§21, and $x_2$, $x_3$ would be the roots of a real quadratic +equation. The remaining factor $x - x_1$ of the cubic would have real +coefficients, whereas $x_1 = x_2$ is imaginary. Hence the equal roots must +be real. + +Our theorem now follows from these three results by formal logic. +For example, if $\Delta$ is positive, the roots are all real and distinct, since +otherwise either two would be imaginary and $\Delta$ would be negative, or two +would be equal and $\Delta$ would be zero. + + +\begin{Exercises}{Page48} + +Compute the discriminant and find the number of real roots of +\begin{Problems}[2] +\item[1.] $y^3 - 2y - 4 = 0$. + +\item[2.] $y^3 - 15y + 4 = 0$. + +\ResetCols{2} + +\item[3.] $y^3 - 27y + 54 = 0$. + +\item[4.] $x^3 + 4x^2 - 11x + 6 = 0$. + +\ResetCols{1} + +\item[5.] Show by means of~§21 that a double root of a real cubic is real. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{46.}{Irreducible Case} When the roots of a real cubic equation are +all real and distinct, the discriminant $\Delta$ is positive and $R = -\Delta/108$ is +negative, so that Cardan's formulas present the values of the roots in a +\index{Cardan's formulas}% +\index{Cube root}% +\index{Irreducible case}% +form involving cube roots of imaginaries. This is called the irreducible +case since it may be shown that a cube root of a general complex number +cannot be expressed in the form $a + bi$, where $a$ and $b$ involve only real +radicals.\footnote + {Author's \textit{Elementary Theory of Equations,} pp.~35, 36.} % [** PP: Added . after pp] +While we cannot always find these cube roots algebraically, +we have learned how to find them trigonometrically~(§8). + +\begin{Example} +Solve the cubic equation~\Eq{2} when $p = -12,\, q = -8\sqrt{2}$. +\end{Example} + +\begin{Solution} +By~\Eq{7}, $R = -32$ Hence formulas~\Eq{9} become +\[ +A = \sqrt[3]{4\sqrt{2} + 4\sqrt{2}i},\qquad +B = \sqrt[3]{4\sqrt{2} - 4\sqrt{2}i}. +\] +%% -----File: 055.png---Folio 49------- +The values of~$A$ were found in~§8. The values of~$B$ are evidently the conjugate imaginaries +of the values of~$A$. Hence the roots are +\[ +4\cos{15°},\quad +4\cos{135°},\quad +4\cos{255°}. +\] +\end{Solution} + + +\begin{Exercises}{Page49} +\begin{Problems}[2] + +\item[1.] Solve $y^3 -15y+4=0$. + +\item[2.] Solve $y^3 -2y-1=0$. + +\ResetCols{2} + +\item[3.] Solve $y^3 -7y+7=0$. + +\item[4.] Solve $x^3+ 3x^2 -2x-5=0$. + +\ResetCols{2} + +\item[5.] Solve $x^3 +x^2 -2x-1=0$. + +\item[6.] Solve $x^3 +4x^2 -7=0$. + +\end{Problems} +\end{Exercises} + + +\Section[Trigonometric Solution of a Cubic] +{47.}{Trigonometric Solution of a Cubic Equation with $\Delta>0$.} When +\index{Cubic equation!trigonometric solution}% +the roots of a real cubic equation are all real, i.e., if $R$ is negative, they +can be computed simultaneously by means of a table of cosines with much +less labor than required by Cardan's formulas. To this end we write +the trigonometric identity +\[ +\cos 3A = 4\cos^3 A - 3\cos A +\] +in the form +\[ +z^3 - \tfrac{3}{4}z - \tfrac{1}{4}\cos 3A = 0\qquad (z = \cos A). +\] +In the given cubic $y^3 + py + q = 0$ take $y=nz$; then +\[ +z^3 + \frac{p}{n^2}z + \frac{q}{n^3} = 0, +\] +which will be identical with the former equation in~$z$ if +\[ +n = \sqrt{-\tfrac{4}{3}p},\quad +\cos{3A} = -\tfrac{1}{2}q ÷ \sqrt{-p^{3}/27}. +\] +Since $R = p^3/27 + q^2/4$ is negative, $p$ must be negative, so that $n$ is real +and the value of $\cos{3A}$ is real and numerically less than unity. Hence +we can find $3A$ from a table of cosines. The three values of $z$ are then +\[ +\cos A,\qquad +\cos(A + 120°),\qquad +\cos(A+240°). +\] +Multiplying these by~$n$, we obtain the three roots~$y$ correct to a number +of decimal places which depends on the tables used. + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] For $y^3 - 2y - 1 =0$, show that $n^2 =8/3$, $\cos{3A} = \sqrt{27/32}$, $3A=23° 17' 0''$, +$\cos A = 0.99084$, $\cos (A+120°) = -0.61237$, $\cos (A+240°) = -0.37847$, and that the +roots~$y$ are $1.61804$, $-1$, $-0.61804$. + +\item[2.] Solve Exs.\ 1, 3, 4, 5, 6 of~§46 by trigonometry. +\end{Problems} +\end{Exercises} + +%% -----File: 056.png---Folio 50------- + + +% [** PP: ToC reads ``Ferrari's and Descartes' Solution of a Quartic''] +\Section{48.}{Ferrari's Solution of the Quartic Equation} The general quartic +equation +\index{Quartic equation|(}% +\[ +x^4 +bx^3 +cx^2 +dx+e=0, +\Tag{15} +\] +or equation of degree four, becomes after transposition of terms +\[ +x^4 + bx^3 = -cx^2 - dx - e. +\] +The left member contains two of the terms of the square of $x^2 +\tfrac{1}{2}bx$. +Hence by completing the square, we get +\[ +(x^2 + \tfrac{1}{2}bx)^2 = (\tfrac{1}{4}b^2 -c)x^2 - dx - e. +\] + +Adding $(x^2+ \frac{1}{2}bx)y+ \frac{1}{4}y^2$ to each member, we obtain +\[ +(x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y)^2 + = (\tfrac{1}{4}b^2 - c + y)x^2 + + (\tfrac{1}{2}by - d)x + + \tfrac{1}{4}y^2 - e. +\Tag{16} +\] +The second member is a perfect square of a linear function of $x$ if and +only if its discriminant is zero~(§12): +\[ +(\tfrac{1}{2}by - d)^2 - 4(\tfrac{1}{4}b^2 - c + y)(\tfrac{1}{4}y^2 - e) = 0, +\] +which may be written in the form +\[ +y^3 - cy^2 + (bd - 4e)y - b^{2}e + 4ce - d^2 = 0. +\Tag{17} +\] + +Choose any root~$y$ of this \emph{resolvent cubic equation~\Eq{17}}. Then the +right member of~\Eq{16} is the square of a linear function, say $mx+n$. Thus +\index{Resolvent cubic}% +\[ +x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y = mx+n \quad\text{or}\quad +x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y = -mx-n. +\Tag{18} +\] +The roots of these quadratic equations are the four roots of~\Eq{16} and +hence of the equivalent equation~\Eq{15}. This method of solution is due +to Ferrari (1522--1565). + +\begin{Example} +Solve $x^4 +2x^3 -12x^2 -10x+3 = 0$. +\end{Example} + +\begin{Solution} +Here $b=2$, $c=-12$, $d=-10$, $e = 3$. Hence~\Eq{17} becomes +\[ +y^3 + 12y^2 - 32y - 256 = 0, +\] +which by Ex.~2 of~§24 has the root $y=-4$. Our quartic may be written in the form +\[ +(x^2 + x)^2 = 13x^2 + 10x-3. +\] +Adding $(x^2 +x)(-4)+4$ to each member, we get +\begin{gather*} +(x^2 + x - 2)^2 = 9x^2 + 6x + 1 = (3x + 1)^2, \\ +x^2 + x - 2 = ±(3x + 1),\qquad +x^2 - 2x - 3 = 0 \text{ or } x^2 + 4x - 1 = 0, +\end{gather*} +whose roots are $3$, $-1$, $-2± \sqrt{5}$. As a check, note that the sum of the roots is~$-2$. +\end{Solution} + +%% -----File: 057.png---Folio 51------- + + +\begin{Exercises}{Page51} + +\begin{Problems} +\item[1.] Solve $x^4 - 8x^3 + 9x^2 + 8x - 10 = 0$. Note that~\Eq{17} is $(y - 9) (y^2 - 24) = 0$. + +\item[2.] Solve $x^4 - 2x^3 - 7x^2 + 8x + 12 = 0$. Since the right member of~\Eq{16} is +$(8 + y) (x^2 - x) + \frac{1}{4} y^2 - 12$, use $y = -8$. + +\item[3.] Solve $x^4 - 3x^2 + 6x - 2 = 0$. + +\item[4.] Solve $x^4 - 2x^2 - 8x - 3 = 0$. + +\item[5.] Solve $x^4 - 10x^2 - 20x - 16 = 0$. + +\end{Problems} +\end{Exercises} + + +\Section[Resolvent Cubic] +{49.}{Roots of the Resolvent Cubic Equation.} Let $y_1$ be the root~$y$ +which was employed in~§48. Let $x_1$ and~$x_2$ be the roots of the first +quadratic equation~\Eq{18}, and $x_3$ and~$x_4$ the roots of the second. Then +\index{Resolvent cubic}% +\[ +x_1 x_2 = \tfrac{1}{2} y_1 - n,\qquad +x_3 x_4 = \tfrac{1}{2} y_1 + n,\qquad +x_1 x_2 + x_3 x_4 = y_1. +\] +If, instead of $y_1$, another root $y_2$ or~$y_3$ of the resolvent cubic~\Eq{17} had been +employed in~§48, quadratic equations different from~\Eq{18} would have +been obtained, such, however, that their four roots are $x_1$, $x_2$, $x_3$, $x_4$, paired +in a new manner. The root which is paired with $x_1$ is $x_2$ or $x_3$ or~$x_4$. It +is now plausible that the values of the three $y$'s are +\[ +y_1 = x_1 x_2 + x_3 x_4,\qquad +y_2 = x_1 x_3 + x_2 x_4,\qquad +y_3 = x_1 x_4 + x_2 x_3. +\Tag{19} +\] + +To give a more formal proof that the $y$'s given by~\Eq{19} are the roots +of~\Eq{17}, we employ~(§20) +\begin{gather*} +x_1 + x_2 + x_3 + x_4 = -b,\qquad +x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 = -d, \\ +x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 = c,\qquad + x_1 x_2 x_3 x_4 = e. +\end{gather*} + +From these four relations we conclude that +\begin{gather*} +y_1 + y_2 + y_3 = c,\\ +\begin{split} +y_1 y_2 + y_1 y_3 + y_2 y_3 + &= (x_1 + x_2 + x_3 + x_4) (x_1 x_2 x_3 + \dotsb + x_2 x_3 x_4) + - 4x_1 x_2 x_3 x_4 \\ +&= bd - 4e, +\end{split} \\ +\begin{split} +y_1 y_2 y_3 + &= (x_1 x_2 x_3 + \dotsb )^2 + + x_1 x_2 x_3 x_4 \bigl\{(x_1 + \dotsb )^2 - 4( x_1 x_2 + \dotsb) \bigr\} \\ + &= d^2 + e ( b^2 - 4c ). +\end{split} +\end{gather*} +Hence (§20) $y_1$, $y_2$, $y_3$ are the roots of the cubic equation~\Eq{17}. + + +% [** PP: No ToC entry] +\Section{50.}{Discriminant} The discriminant $\Delta$ of the quartic equation~\Eq{15} +\index{Discriminant!of quartic}% +is defined to be the product of the squares of the differences of its roots: +\[ +\Delta = ( x_1 - x_2 )^2 ( x_1 - x_3 )^2 ( x_1 - x_4 )^2 + ( x_2 - x_3 )^2 ( x_2 - x_4 )^2 ( x_3 - x_4 )^2. +\] + +%% -----File: 058.png---Folio 52------- + +The fact that $\Delta$ is equal to the discriminant of the resolvent cubic +equation~\Eq{17} follows at once from~\Eq{19}, by which +\begin{align*} +y_1 - y_2 &= (x_1-x_4)(x_2-x_3),\qquad + y_1-y_3 = (x_1-x_3)(x_2-x_4), \\ +% +y_2 - y_3 &= (x_1-x_2)(x_3-x_4),\qquad + (y_1-y_2)^2 (y_1-y_3)^2 (y_2- y_3)^2 = \Delta. +\end{align*} +Hence (§44) $\Delta$ is equal to the discriminant $-4p^3 - 27q^2$ of the reduced +cubic $Y^3 + pY + q = 0$, obtained from~\Eq{17} by setting $y = Y + c/3$. Thus +\[ +p = bd - 4e - \tfrac{1}{3} c^2,\qquad +q = -b^2 e + \tfrac{1}{3} bcd + \tfrac{8}{3} ce - d^2 - \tfrac{2}{27} c^3. +\Tag{20} +\] + +\begin{Theorem} +The discriminant of any quartic equation~\Eq{15} is equal to +the discriminant of its resolvent cubic equation and therefore is equal to the +discriminant $-4p^3 - 27q^2$ of the corresponding reduced cubic $Y^3 + pY + q = 0$, +whose coefficients have the values~\Eq{20}. +\end{Theorem} + + +\begin{Exercises}{} +\index{Number!of roots}% + +\begin{Problems} +\item[1.] Find the discriminant of $x^4 - 3x^3 + x^2 + 3x - 2 = 0$ and show that the equation +has a multiple root. + +\item[2.] Show by its discriminant that $x^4 - 8x^3 + 22x^2 - 24x + 9 = 0$ has a multiple root. + +\item[3.] If a real quartic equation has two pairs of conjugate imaginary roots, show that +its discriminant~$\Delta$ is positive. Hence prove that, if $\Delta<0$, there are exactly two real +roots. + +\item[4.] Hence show that $x^4 - 3x^3 + 3x^2 - 3x + 2 = 0$ has two real and two imaginary roots. +\end{Problems} +\end{Exercises} + + +% [** PP: No separate ToC entry] +\Section{51.}{Descartes' Solution of the Quartic Equation} Replacing $x$ by +$z - b/4$ in the general quartic~\Eq{15}, we obtain the \emph{reduced} quartic equation +\[ +z^4 + qz^2 + rz + s = 0, +\Tag{21} +\] +lacking the term with~$z^3$. We shall prove that we can express the left +member of~\Eq{21} as the product of two quadratic factors +\[ +(z^2 + 2kz + l)(z^2 - 2kz + m) + = z^4 + (l + m - 4k^2)z^2 + 2k(m - l)z + lm. +\] +The conditions are +\[ +l + m - 4k^2 = q,\qquad +2k(m-l)=r,\qquad +lm = s. +\] +If $k\neq0$, the first two give +\[ +2l = q + 4k^2 - \frac{r}{2k},\qquad +2m = q + 4k^2 + \frac{r}{2k}. +\] +Inserting these values in $2l·2m = 4s$, we obtain +\[ +64k^6 + 32qk^4 + 4(q^2 - 4s)k^2 - r^2 = 0. +\Tag{22} +\] +%% -----File: 059.png---Folio 53------- +The latter may be solved as a cubic equation for~$k^2$. Any root $k^2 \neq 0$ +gives a pair of quadratic factors of~\Eq{21}: +\[ +z^2 ± 2kz + \tfrac{1}{2}q + 2k^2 \mp \frac{r}{4k}. +\Tag{23} +\] +The four roots of these two quadratic functions are the four roots of~\Eq{21}. +This method of Descartes (1596--1650) therefore succeeds unless every +root of~\Eq{22} is zero, whence $q = s = r = 0$, so that \Eq{12} is the trivial equation +$z^4 = 0$. + +\begin{Remark} +For example, consider $z^4 - 3z^2 + 6z - 2 = 0$. Then \Eq{22} becomes +\[ +64k^6 - 3·32k^4 + 4·17k^2 - 36 = 0. +\] +The value $k^2 = 1$ gives the factors $z^2 + 2z - 1$, $z^2 - 2z + 2$. Equating these to zero, we +find the four roots $-1 ± \sqrt{2}$, $1± \sqrt{-1}$. +\end{Remark} + + +% [** PP: No ToC entry] +\Section{52.}{Symmetrical Form of Descartes' Solution} To obtain this symmetrical +form, we use all three roots $k_1^2$, $k_2^2$, $k_3^2$ of~\Eq{22}. Then +\[ +k_1^2 + k_2^2 + k_3^2 = -\tfrac{1}{2}q,\qquad +k_1^2 k_2^2 k_3^2 = \frac{r^2}{64}. +\] +It is at our choice as to which square root of~$k_1^2$ is denoted by~$+k_1$ and +which by~$-k_1$, and likewise as to $±k_2$, $±k_3$. For our purposes any +choice of these signs is suitable provided the choice give +\[ +k_1 k_2 k_3 = -\frac{r}{8}. +\Tag{24} +\] + +Let $k_1 \neq 0$. The quadratic function~\Eq{23} is zero for $k = k_1$ if +\[ +(z ± k_1)^2 = -\frac{q}{2} - k_1^2 ± \frac{r}{4k_1} + = k_2^2 + k_3^2 \mp \frac{8k_1k_2k_3}{4k_1} + = (k_2 \mp k_3)^2. +\] +Hence the four roots of the quartic equation~\Eq{21} are +\[ + k_1 + k_2 + k_3,\qquad + k_1 - k_2 - k_3,\qquad +-k_1 + k_2 - k_3,\qquad +-k_1 - k_2 + k_3. +\Tag{25} +\] + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Solve Exs.~4, 5 of~§48 by the method of Descartes. + +\item[2.] By writing $y_1$, $y_2$, $y_3$ for the roots $k_1^2$, $k_2^2$, $k_3^2$ of +\[ +64y^3 + 32qy^2 + 4(q^2 - 4s)y - r^2 = 0, +\Tag{26} +\] +show that the four roots of~\Eq{21} are the values of +\[ +z = \sqrt{y_1} + \sqrt{y_2} + \sqrt{y_3} +\Tag{27} +\] +%% -----File: 060.png---Folio 54------- +for all combinations of the square roots for which +\[ +\sqrt{y_1}·\sqrt{y_2}·\sqrt{y_3} = -\frac{r}{8}. +\Tag{28} +\] + +\item[3.] Euler (1707--1783) solved~\Eq{21} by assuming that it has a root of the form~\Eq{27}. +Square~\Eq{27}, transpose the terms free of radicals, square again, replace the last factor +of $8\sqrt{y_1 y_2 y_3}\,(\sqrt{y_1} + \sqrt{y_2} + \sqrt{y_3})$ by~$z$, and identify the resulting quartic in~$z$ with~\Eq{21}. +Show that $y_1$, $y_2$, $y_3$ are the roots of~\Eq{26} and that relation~\Eq{28} holds. + +\item[4.] Find the six differences of the roots~\Eq{25} and verify that the discriminant $\Delta$ of~\Eq{21} +is equal to the quotient of the discriminant of~\Eq{26} by~$4^6$. + +\item[5.] In the theory of the inflexion points of a plane cubic curve there occurs the +equation +\[ +z^4 - Sz^2 - \tfrac{4}{3}Tz - \tfrac{1}{12}S^2 = 0. +\] +Show that~\Eq{26} now becomes +\[ +\left(y - \frac{S}{6}\right)^3 = C,\qquad +C \equiv \left(\frac{T}{6}\right)^2 - \left(\frac{S}{6}\right)^3, +\] +and that the roots of the quartic equation are +\[ +±\sqrt{\tfrac{1}{6}S + \sqrt[3]{C}} +±\sqrt{\tfrac{1}{6}S + \omega\sqrt[3]{C}} +±\sqrt{\tfrac{1}{6}S + \omega^{2}\sqrt[3]C}, +\] +where $\omega$ is an imaginary cube root of unity and the signs are to be chosen so that the product of the three summands is equal to~$+\tfrac{1}{6}T$. +\end{Problems} +\end{Exercises} + + +\begin{Exercises}[MISCELLANEOUS~]{Page54} + +\begin{Problems} +\item[1.] Find the coordinates of the single real point of intersection of the parabola +$y = x^2$ and the hyperbola $xy - 4x + y + 6 = 0$. + +\item[2.] Show that the abscissas of the points of intersection of $y=x^2$ and +$ax^2 - xy + y^2 - x - (a+5)y - 6 = 0$ +are the roots of $x^4 - x^3 - 5x^2 - x - 6 = 0$. Compute the discriminant +of the latter and show that only two of the four points of intersection are real. + +\item[3.] Find the coordinates of the two real points in Ex.~2. + +\item[4.] A right prism of height~$h$ has a square base whose side is~$b$ and whose diagonal +is therefore $b\sqrt{2}$. If $v$ denotes the volume and $d$ a diagonal of the prism, $v = hb^2$ and +$d^2 = h^2 + (b\sqrt{2})^2$. Multiply the last equation by~$h$ and replace $hb^2$ by~$v$. Hence +$h^3 - d^2h + 2v = 0$. +Its discriminant is zero if $d = 3\sqrt{3}$, $v = 27$; find~$h$. + +\item[5.] Find the admissible values of~$h$ in Ex.~4 when $d = 12$, $v = 332.5$. + +\item[6.] Find a necessary and sufficient condition that quartic equation~\Eq{15} shall have +one root the negative of another root. + +Hint: $(x_1 + x_2)(x_3 + x_4) = q - y_1$. Hence substitute $q$ for~$y$ in~\Eq{17}. +\index{Quartic equation|)}% + +\item[7.] In the study of parabolic orbits occurs the equation +\[% [** PP: Displayed for better line breaking.] +\tan\tfrac{1}{2}v + \tfrac{1}{3}\tan^3 \tfrac{1}{2}v = t. +\] +Prove that there is a single real root and that it has the same sign as~$t$. + +\item[8.] In the problem of three astronomical bodies occurs the equation $x^3 + ax + 2 = 0$. +Prove that it has three real roots if and only if $a\leqq{-3}$. +\end{Problems} +\end{Exercises} + +%% -----File: 061.png---Folio 55------- + + +% [** PP: Not matching running head] +\Chapter{V}{The Graph of an Equation} + +\index{Graphs|(}% +\Section[Use of Graphs] +{53.}{Use of Graphs in the Theory of Equations.} To find geometrically +the real roots of a real equation $f(x)=0$, we construct a graph of $y=f(x)$ +and measure the distances from the origin~$O$ to the intersections of the +graph and the $x$-axis, whose equation is $y=0$. + +%%{Illustration} \textsc{Fig. 12} +\begin{wrapfigure}[16]{r}{2.25in} +\quad\Input{061a} +\end{wrapfigure} +For example to find geometrically the real +roots of +\[ +x^2 - 6x - 3 = 0, +\Tag{1} +\] +we equate the left member to~$y$ and make a +graph of +\[ +y = x^2 - 6x - 3. +\Tag{1'} +\] +We obtain the parabola in Fig.~12. Of the +points shown, $P$ has the \emph{abscissa} +\index{Abscissa}% +$x = OQ = 4$ +and the \emph{ordinate} +\index{Ordinate}% +$y = -QP = -11$. From the +points of intersection of $y = 0$ (the $x$-axis $OX$) +with the parabola, we obtain the approximate +values $6.46$ and $-0.46$ of the roots of~\Eq{1}. + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Find graphically the real roots of $x^2 - 6x + 7 = 0$. +\index{Quadratic equation!graphical solution}% +\end{Problems} + +Hint: For each $x$, $y = x^2 - 6x + 7$ exceeds the $y$ in~\Eq{1'} by~$10$, so that the new graph +is obtained by shifting the parabola in Fig.~12 upward $10$~units, leaving the axes~$OX$ +and~$OY$ unchanged. What amounts to the same thing, but is simpler to do, we leave +the parabola and~$OY$ unchanged, and move the axis~$OX$ downward $10$~units. + +\begin{Problems} +\item[2.] Discuss graphically the reality of the roots of $x^2 - 6x + 12 = 0$. + +\item[3.] Find graphically the roots of $x^2 - 6x + 9 = 0$. +\end{Problems} +\end{Exercises} + +%%Illustration \textsc{Fig}.~13 +\begin{wrapfigure}[21]{l}{1.25in} +\Input{062a} +\end{wrapfigure} +\Section{54.}{Caution in Plotting} If the example set were +\index{Plotting}% +\[ +y = 8x^4 - 14x^3 - 9x^2 + 11x - 2, +\Tag{2} +\] +one might use successive integral values of~$x$, obtain the points $(-2, 180)$, +%% -----File: 062.png---Folio 56------- +$(-1, 0)$, $(0, -2)$, $(1, -6)$, $(2, 0)$, $(3, 220)$, all but the +first and last of which are shown (by crosses) in Fig.~13, +and be tempted to conclude that the graph is a +\Shape{U}-shaped curve approximately like that in Fig.~12 +and that there are just two real roots, $-1$ and~$2$, of +\[ +8x^4 - 14x^3 - 9x^2 + 11x - 2 = 0. +\Tag{2'} +\] +But both of these conclusions would be false. In fact, +the graph is a \Shape{W}-shaped curve (Fig.~13) and the +additional real roots are $\frac{1}{4}$ and~$\frac{1}{2}$. + +This example shows that it is often necessary to +employ also values of~$x$ which are not integers. The +purpose of the example was, however, not to point +out this obvious fact, but rather to emphasize the +chance of serious error in sketching a curve through a +number of points, however numerous. The true curve +between two points below the $x$-axis may not cross the +$x$-axis, or may have a peak and actually cross the $x$-axis +twice, or may be an \Shape{M}-shaped curve crossing it four +times, etc. + +%%Illustration \textsc{Fig}.~14 +\begin{wrapfigure}{r}{1.5in} +\Input{062b} +\end{wrapfigure} +For example, the graph (Fig.~14) of +\[ +y = x^3 + 4x^2 - 11 +\Tag{3} +\] +crosses the $x$-axis only once; but this fact cannot +be established by a graph located by a number +of points, however numerous, whose abscissas +are chosen at random. + +We shall find that correct conclusions regarding +the number of real roots may be deduced +from a graph whose bend points~(§55) have been +located. + + +\Section{55.}{Bend Points} A point (like $M$ or~$M'$ in +Fig.~14) is called a \emph{bend point} of the graph of +\index{Bend point}% +$y=f(x)$ if the tangent to the graph at that point +is horizontal and if all of the adjacent points of +the graph lie below the tangent or all above the +tangent. The first, but not the second, condition +%% -----File: 063.png---Folio 57------- +is satisfied by the point~$O$ of the graph of $y = x^3$ given in Fig.~15 (see~§57). +In the language of the calculus, $f(x)$ has a (relative) maximum or +minimum value at the abscissa of a bend point on the graph of $y=f(x)$. +\index{Maximum}% +\index{Minimum}% +%[Illustration: Fig. 15] +%[Illustration: Fig. 16] +\begin{figure*}[hbt] +\begin{center} +\Input{063a}\hfil +\Input{063b} +\end{center} +\end{figure*} + +Let $P = (x, y)$ and $Q = (x+h, Y)$ be two points on the graph, sketched +in Fig.~16, of $y=f(x)$. By the \emph{slope} of a straight line is meant the tangent +of the angle between the line and the $x$-axis, measured counter-clockwise +from the latter. In Fig.~16, the slope of the straight line~$PQ$ is +\index{Slope}% +\[ +\frac{Y - y}{h} = \frac{f(x+h) - f(x)}{h}. +\Tag{4} +\] + +For equation~\Eq{3}, $f(x) = x^3 + 4x^2 - 11$. Hence +\begin{align*} +f(x+h) &= (x+h)^3 + 4(x+h)^2 - 11 \\ + &= x^3 + 4x^2 - 11 + (3x^2 +8x)h + (3x+4)h^2 + h^3. +\end{align*} +The slope~\Eq{4} of the secant~$PQ$ is therefore here +\index{Derivative|(}% +\[ +3x^2 + 8x + (3x+4)h + h^2. +\] +Now let the point~$Q$ move along the graph toward~$P$. Then $h$ approaches +the value zero and the secant~$PQ$ approaches the tangent at~$P$. The +slope of the tangent at~$P$ is therefore the corresponding limit $3x^2 + 8x$ +of the preceding expression. We call $3x^2 + 8x$ the \emph{derivative} of $x^3 + 4x^2 - 11$. + +%% -----File: 064.png---Folio 58------- + +In particular, if $P$ is a bend point, the slope of the (horizontal) tangent +at~$P$ is zero, whence $3x^2 + 8x = 0$, $x = 0$ or $x = -\tfrac{8}{3}$. Equation~\Eq{3} gives +the corresponding values of~$y$. The resulting points +\[ +M = (0, -11),\qquad M' = (-\tfrac{8}{3}, -\tfrac{41}{27}) +\] +are easily shown to be bend points. Indeed, for $x>0$ and for $x$ between +$-4$ and~$0$, $x^2(x+4)$ is positive, and hence $f(x) > -11$ for such values of~$x$, +so that the function~\Eq{3} has a relative minimum at $x = 0$. Similarly, +there is a relative maximum at $x = -\tfrac{8}{3}$. We may also employ the general +method of~§59 to show that $M$ and~$M'$ are bend points. Since these bend +points are both below the $x$-axis we are now certain that the graph +crosses the $x$-axis only once. + +The use of the bend points insures greater accuracy to the graph than +the use of dozens of points whose abscissas are taken at random. + +\Section{56.}{Derivatives} We shall now find the slope of the tangent to the +graph of $y=f(x)$, where $f(x)$ is any polynomial +\[ +\Tag{5} +f(x) = a_0 x^n + a_1 x^{n-1} + \dotsb + a_{n-1} x + a_n. +\] +We need the expansion of $f(x+h)$ in powers of~$x$. By the binomial +theorem, +\begin{align*} +a_0 (x+h)^n + &= a_0 x^n + na_0 x^{n-1} h + \frac{n(n-1)}{2} a_0 x^{n-2} h^2 + \dotsb, \\ +a_1 (x+h)^{n-1} + &= a_1 x^{n-1} + (n-1)a_1 x^{n-2} h + + \frac{(n-1)(n-2)}{2} a_1 x^{n-3} h^2 + \dotsb, \\ +\multispan{2}{\dotfill} \\ +a_{n-2} (x+h)^2 + &= a_{n-2} x^2 +2a_{n-2} xh + a_{n-2} h^2, \\ +a_{n-1} (x+h) + &= a_{n-1} x + a_{n-1} h, \\ +a_n &= a_n. +\end{align*} +The sum of the left members is evidently $f(x+h)$. On the right, the +sum of the first terms (i.e., those free of~$h$) is~$f(x)$. The sum of the coefficients +of~$h$ is denoted by~$f'(x)$, the sum of the coefficients of $\tfrac{1}{2} h^2$ is denoted +by $f''(x), \dotsc$, the sum of the coefficients of +\[ +\frac{h^k}{1·2\dotsm k} +\] +%% -----File: 065.png---Folio 59------- +is denoted by~$f^{(k)}(x)$. Thus +\begin{align*} +f'(x) &= na_0 x^{n-1} + (n-1)a_1 x^{n-2} + \dotsb + 2a_{n-2} x + a_{n-1}, +\Tag{6} \\ +f''(x) &= n(n-1)a_0 x^{n-2} + (n-1)(n-2)a_1 x^{n-3} + \dotsb + 2a_{n-2}, +\Tag{7} +\intertext{etc. Hence we have} +f(x+h) &= f(x) + f'(x)h + f''(x) \frac{h^2}{1·2} + f'''(x) \frac{h^3}{1·2·3} +\Tag{8} \\ + & \qquad + \dotsb + f^{(r)}(x) \frac{h^r}{r!} + + \dotsb + f^{(n)}(x) \frac{h^n}{n!}, +\end{align*} +where $r!$ is the symbol, read $r$~\emph{factorial}, for the product $1·2·3\dotsm(r-1)r$. +Here $r$ is a positive integer, but we include the case $r = 0$ by the definition, +$0! = 1$. +\index{Symbol!d@{$r$"!\IndAdd{factorial}}}% [** PP: Manually alphabetized] + +This formula~\Eq{8} is known as \emph{Taylor's theorem} for the present case of +\index{Taylor's theorem}% +a polynomial~$f(x)$ of degree~$n$. We call $f'(x)$ the (\emph{first}) \emph{derivative of~$f(x)$}, +and $f''(x)$ the \emph{second derivative} of~$f(x)$, etc. +Concerning the fact that +$f''(x)$ is equal to the first derivative of $f'(x)$ and that, in general, the $k$th +derivative $f^{(k)}(x)$ of $f(x)$ is equal to the first derivative of $f^{(k-1)}(x)$, see +Exs.~6--9 of the next set. +\index{Symbol!e@{$f^{(k)}(x)$\IndAdd{$k$th derivative}}}% [** PP: Manually alphabetized] + +In view of~\Eq{8}, the limit of~\Eq{4} as $h$~approaches zero is $f'(x)$. Hence +\begin{Thm}% +$f'(x)$ is the slope of the tangent to the graph of $y=f(x)$ at the point~$(x, y)$. +\end{Thm} +\index{Slope}% + +In \Eq{5} and~\Eq{6}, let every $a$ be zero except~$a_0$. Thus the derivative of +$a_0 x^n$ is $na_0 x^{n-1}$, and hence is obtained by multiplying the given term by +its exponent~$n$ and then diminishing its exponent by unity. For example, +the derivative of~$2x^3$ is~$6x^2$. + +Moreover, the derivative of $f(x)$ is equal to the sum of the derivatives +of its separate terms. Thus the derivative of $x^3 + 4x^2 - 11$ is $3x^2 + 8x$, +as found also in~§55. + + +\begin{Exercises}{Page59} + +\begin{Problems} +\item[1.] Show that the slope of the tangent to $y = 8x^3 - 22x^2 + 13x - 2$ at $(x, y)$ is +$24x^2 - 44x + 13$, and that the bend points are $(0.37, 0.203)$, $(1.46, -5.03)$, approximately. +Draw the graph. + +\item[2.] Prove that the bend points of $y = x^3 - 2x - 5$ are $(.82, -6.09)$, $(-.82$, $-3.91)$, % [** PP: Allow line break between coordinates] +approximately. Draw the graph and locate the real roots. + +\item[3.] Find the bend points of $y = x^3 + 6x^2 + 8x + 8$. Locate the real roots. + +\item[4.] Locate the real roots of $f(x) = x^4 + x^3 - x - 2 = 0$. + +Hints: The abscissas of the bend points are the roots of $f'(x) = 4x^3 + 3x^2 - 1 = 0$. +The bend points of $y = f'(x)$ are $(0, -1)$ and $(-\frac{1}{2}, -\frac{3}{4})$, so that $f'(x)= 0$ has a single real +root (it is just less than $\frac{1}{2}$). The single bend point of $y=f(x)$ is $(\frac{1}{2}, -\frac{37}{16})$, approximately. + +%% -----File: 066.png---Folio 60------- + +\item[5.] Locate the real roots of $x^6 - 7x^4 - 3x^2 + 7 = 0$. + +\item[6.] Prove that $f''(x)$, given by~\Eq{7}, is equal to the first derivative of~$f'(x)$. + +\item[7.] If $f(x) = f_1(x) + f_2(x)$, prove that the $k$th derivative of~$f$ is equal to the sum of +the $k$th derivatives of $f_1$ and~$f_2$. Use~\Eq{8}. + +\item[8.] Prove that $f^{(k)}(x)$ is equal to the first derivative of $f^{(k-1)}(x)$. Hint: prove this +for $f = ax^m$; then prove that it is true for $f=f_1 + f_2$ if true for $f_1$ and~$f_2$. + +\item[9.] Find the third derivative of $x^6 + 5x^4$ by forming successive first derivatives; +also that of $2x^5 - 7x^3 + x$. + +\item[10.] Prove that if $g$ and~$k$ are polynomials in~$x$, the derivative of $gk$ is $g'k + gk'$. Hint: +multiply the members of $g(x+h) = g(x) + g'(x)h + \dotsb$ and $k(x+h) = k(x) + k'(x)h + \dotsb$ +and use~\Eq{8} for $f = gk$. + +\end{Problems} +\end{Exercises} + + +\Section{57.}{Horizontal Tangents} If $(x, y)$ is a bend point of the graph of +\index{Tangents}% +$y=f(x)$, then, by definition, the slope of the tangent at $(x, y)$ is zero. +Hence~(§56), the abscissa~$x$ is a root of $f'(x)=0$. In Exs.~1--5 of the +preceding set, it was true that, conversely, any real root of $f'(x)=0$ is +the abscissa of a bend point. However, this is not always the case. +We shall now consider in detail an example illustrating this fact. The +example is the one merely mentioned in~§55 to indicate the need of the +second requirement made in our definition of a bend point. + +The graph (Fig.~15) of $y = x^3$ has no bend point since $x^3$ increases when +$x$~increases. Nevertheless, the derivative $3x^2$ of~$x^3$ is zero for the real +value $x = 0$. The tangent to the curve at $(0, 0)$ is the horizontal line +$y=0$. It may be thought of as the limiting position of a secant through~$O$ +which meets the curve in two further points, seen to be equidistant +from~$O$. When one, and hence also the other, of the latter points approaches~$O$, +the secant approaches the position of tangency. In this sense the +tangent at~$O$ is said to meet the curve in three coincident points, their +abscissas being the three coinciding roots of $x^3 = 0$. In the language of~§17, +$x^3 = 0$ has the triple root $x = 0$. The subject of bend points, to which +we recur in~§59, has thus led us to a digression on the important subject +of multiple roots. +\index{Derivative|)}% + + +\Section{58.}{Multiple Roots} In~\Eq{8} replace $x$ by~$\alpha$, and $h$ by $x-\alpha$. Then +\index{Multiple roots}% +\begin{align*} +f(x) &= f(\alpha) + f'(\alpha) (x-\alpha) + + f''(\alpha) \frac{(x-\alpha)^2}{1·2} + + f'''(\alpha) \frac{(x-\alpha)^3}{1·2·3} + \dotsb +\Tag{9} \\ +&\phantom{{} = f(\alpha)} + {} + f^{(m-1)}(\alpha) \frac{(x-\alpha)^{m-1}}{(m-1)!} + + f^{(m)}(\alpha) \frac{(x-\alpha)^m}{m!} + \dotsb. +\end{align*} +By definition~(§17) $\alpha$ is a root of $f(x)=0$ of multiplicity~$m$ if $f(x)$ is exactly +%% -----File: 067.png---Folio 61------- +divisible by $(x-\alpha)^m$, but not by $(x-\alpha)^{m+1}$. Hence \emph{$\alpha$~is a root of multiplicity~$m$ +of $f(x) = 0$ if and only if} +\index{Multiplicity of root}% +\[ +f(\alpha) = 0,\quad +f'(\alpha) = 0,\quad +f''(\alpha) = 0, \dotsc,\quad +f^{(m-1)}(\alpha) = 0,\quad +f^{(m)}(\alpha) \ne 0. +\Tag{10} +\] + +\begin{Remark} +For example, $x^4 + 2x^3 =0$ has the triple root $x = 0$ since $0$ is a root, and since the +first and second derivatives $4x^3 +6x^2$ and $12x^2 +12x$ are zero for $x = 0$, while the third +derivative $24x + 12$ is not zero for $x = 0$. +\end{Remark} + +If in~\Eq{9} we replace $f$ by~$f'$ and hence $f^{(k)}$ by~$f^{(k+1)}$, or if we differentiate +every term with respect to~$x$, we see by either method that +\begin{multline*} +f'(x) = f'(\alpha) + f''(\alpha)(x-\alpha) + \dotsb + + f^{(m-1)} (\alpha)\frac{(x-\alpha)^{m-2}}{(m-2)!} \\ + + f^{(m)}(\alpha)\frac{(x-\alpha)^{m-1}}{(m-1)!} + \dotsb. +\Tag{11} +\end{multline*} + +Let $f(x)$ and $f'(x)$ have the common factor $(x-\alpha)^{m-1}$, but not the common +factor $(x-\alpha)^m$, where $m>1$. Since~\Eq{11} has the factor $(x-\alpha)^{m-1}$, we +have $f'(\alpha) = 0, \dotsc, f^{(m-1)}(\alpha) = 0$. Since also $f(x)$ has the factor $x-\alpha$, +evidently $f(\alpha)=0$. Then, by~\Eq{9}, $f(x)$ has the factor $(x-\alpha)^m$, which, +by hypothesis, is not also a factor of $f'(x)$. Hence, in~\Eq{11}, $f^{(m)}(\alpha)\ne 0$. +Thus, by~\Eq{10}, $\alpha$ is a root of $f(x)=0$ of multiplicity~$m$. + +Conversely, let $\alpha$ be a root of $f(x)=0$ of multiplicity~$m$. Then relations~\Eq{10} +hold, and hence, by~\Eq{11}, $f'(x)$ is divisible by $(x-\alpha)^{m-1}$, but +not by $(x-\alpha)^m$. Thus $f(x)$ and $f'(x)$ have the common factor $(x-\alpha)^{m-1}$, +but not the common factor $(x-\alpha)^m$. + +We have now proved the following useful result. + +\begin{Theorem} +If $f(x)$ and $f'(x)$ have a greatest common divisor $g(x)$ +\index{Greatest common divisor}% +involving~$x$, a root of $g(x)=0$ of multiplicity $m-1$ is a root of $f(x)=0$ of +multiplicity~$m$, and conversely any root of $f(x) =0$ of multiplicity~$m$ is a root +of $g(x)=0$ of multiplicity $m-1$. +\end{Theorem} + +In view of this theorem, the problem of finding all the multiple roots +of $f(x)=0$ and the multiplicity of each multiple root is reduced to the +problem of finding the roots of $g(x)=0$ and the multiplicity of each. + +\begin{Remark} +For example, let $f(x) = x^3 - 2x^2 - 4x + 8$. Then +\[ +f'(x) = 3x^2 - 4x - 4,\qquad +9f(x) = f'(x)(3x-2) - 32(x-2). +\] +Since $x - 2$ is a factor of $f'(x)$, it may be taken to be the greatest common divisor of~$f(x)$ +and~$f'(x)$, the choice of the constant factor~$c$ in $c(x-2)$ being here immaterial. Hence~$2$ +is a double root of $f(x)=0$, while the remaining root~$-2$ is a simple root. +\end{Remark} + +%% -----File: 068.png---Folio 62------- + +\begin{Exercises}{Page62} + +\begin{Problems} +\item[1.] Prove that $x^3 - 7x^2 + 15x - 9 = 0$ has a double root. + +\item[2.] Show that $x^4 - 8x^2 + 16 = 0$ has two double roots. + +\item[3.] Prove that $x^4 - 6x^2 - 8x - 3 = 0$ has a triple root. + +\item[4.] Test $x^4 - 8x^3 + 22x^2 - 24x + 9 = 0$ for multiple roots. + +\item[5.] Test $x^3 - 6x^2 + 11x - 6 = 0$ for multiple roots. + +\item[6.] Test $x^4 - 9x^3 + 9x^2 + 81x - 162 = 0$ for multiple roots. +\end{Problems} +\end{Exercises} + + +\Section{59.}{Ordinary and Inflexion Tangents} The equation of the straight +\index{Inflexion|(}% +\index{Tangents}% +line through the point $(\alpha, \beta)$ with the slope~$s$ is $y-\beta = s(x -\alpha)$. The slope +of the tangent to the graph of $y=f(x)$ at the point $(\alpha, \beta)$ on it is $s=f'(\alpha)$ +by~§56. Also, $\beta=f(\alpha)$. Hence the equation of the tangent is +\[ +y = f(\alpha) + f'(\alpha)(x-\alpha). +\Tag{12} +\] + +By subtracting the members of this equation from the corresponding +members of equation~\Eq{9}, we see that the abscissas~$x$ of the points of intersection +of the graph of $y=f(x)$ with its tangent satisfy the equation +\begin{multline*} +f''(\alpha)\frac{(x-\alpha)^2}{2!} + + f'''(\alpha)\frac{(x-\alpha)^3}{3!} + \dotsb + + f^{(m-1)}(\alpha)\frac{(x-\alpha)^{m-1}}{(m-1)!} \\ + + f^{(m)}(\alpha)\frac{(x-\alpha)^m}{m!} + \dotsb = 0. +\end{multline*} +Here the term containing $f^{(m-1)}(\alpha)$ must evidently be suppressed if $m = 2$, +since the term containing $f^{(m)}(\alpha)$ then coincides with the first term. + +If $\alpha$ is a root of multiplicity~$m$ of this equation, i.e., if the left member +is divisible by $(x -\alpha)^m$, but not by $(x-\alpha)^{m+1}$, the point $(\alpha, \beta)$ is counted +as $m$~coincident points of intersection of the curve with its tangent (just +as in the case of $y=x^3$ and its tangent $y = 0$ in~§57). This will be the case +if and only if +\[ +f''(\alpha)=0,\qquad +f'''(\alpha) =0, \dotsc,\qquad +f^{(m-1)}(\alpha)=0,\qquad +f^{(m)}(\alpha)\ne 0, +\Tag{13} +\] +in which $m>1$ and, as explained above, only the final relation $f''(\alpha)\ne 0$ +is retained if $m= 2$. If $m=3$, the conditions are $f''(\alpha)= 0$, $f^{(3)}(\alpha) \ne 0$. + +\begin{Remark} +For example, if $f(x) = x^4$ and $\alpha = 0$, then $f''(0) =f'''(0) =0$, $f^{(4)} (0) = 24\ne 0$, so that +$m=4$. The graph of $y = x^4$ is a \Shape{U}-shaped curve, whose intersection with the tangent +(the $x$-axis) at $(0, 0)$ is counted as four coincident points of intersection. +\end{Remark} + +Given $f(x)$ and $\alpha$, we can find, as in the preceding example, the value +of~$m$ for which relations~\Eq{13} hold. We then apply the +%% -----File: 069.png---Folio 63------- +\begin{Theorem} +If $m$ is even \($m>0$\), the points of the curve in the vicinity +of the point of tangency $(\alpha, \beta)$ are all on the same side of the tangent, which +is then called an \emph{\textbf{ordinary tangent}}. But if $m$ is odd \($m>1$\), the curve crosses +the tangent at the point of tangency $(\alpha, \beta)$, and this point is called an \emph{\textbf{inflexion +point}}, while the tangent is called an \emph{\textbf{inflexion tangent}}. +\end{Theorem} + +\begin{Remark} +For example, in Fig.~15, $OX$ is an inflexion tangent, while the tangent at any point +except~$O$ is an ordinary tangent. In Figs.~18, 19, 20, the tangents at the points marked +by crosses are ordinary tangents, but the tangent at the point midway between them +and on the $y$-axis is an inflexion tangent. +\end{Remark} + +To simplify the proof, we first take as new axes lines parallel to the +old axes and intersecting at $(\alpha, \beta)$. In other words, we set $x-\alpha = X$, +$y-\beta=Y$, where $X$, $Y$ are the coordinates of $(x, y)$ referred to the new +axes. Since $\beta = f(\alpha)$, the tangent~\Eq{12} becomes $Y = f'(\alpha)X$, while, by~\Eq{9}, +$y = f(x) = \beta + f'(\alpha)(x-\alpha) + \dotsb$ becomes +\[ +Y = f'(\alpha)X + f''(\alpha)\frac{X^2}{2} + \dotsb + = f'(\alpha)X + f^{(m)}(\alpha)\frac{X^m}{m!} + \dotsb, +\] +after omitting terms which are zero by~\Eq{13}. + +%Illustration: Fig. 17 +\begin{wrapfigure}[15]{r}{2.375in} +\hfil\Input{069a} +\end{wrapfigure} +To simplify further the algebraic work, +we pass to oblique axes,\footnote + {Since the earlier $x$, $y$ do not occur in~\Eq{14} and the new equation of the tangent, + we shall designate the final coordinates by $x$, $y$ without confusion.} +the new $y$-axis +coinciding with the $Y$-axis, while the new +$x$-axis is the tangent, the angle between +which and the $X$-axis is designated by $\theta$. +Then +\[ +\tan\theta=f'(\alpha). +\] +By Fig.~17, +\[ +X = x\cos\theta,\qquad +Y-y = f'(\alpha)X. +\] + +Hence when expressed in terms of the +new coordinates $x$, $y$, the tangent is $y = 0$, while the equation~\Eq{14} of the +curve becomes +\[ +y = cx^m + dx^{m+1} + \dotsb,\qquad +c = \frac{f^{(m)}(\alpha)\cos^m \theta}{m!} \ne 0. +\] + +For $x$ sufficiently small numerically, whether positive or negative, +the sum of the terms after $cx^m$ is insignificant in comparison with $cx^{m}$, +%% -----File: 070.png---Folio 64------- +so that $y$ has the same sign as $cx^m$~(§64). Hence, if $m$ is even, the points +of the curve in the vicinity of the origin and on both sides of it are all +on the same side of the $x$-axis, i.e., the tangent. But, if $m$ is odd, the points +with small positive abscissas~$x$ lie on one side of the $x$-axis and those with +numerically small negative abscissas lie on the opposite side. + +Our transformations of coordinates changed the equations of the +curve and of its tangent, but did not change the curve itself and its tangent. +Hence our theorem is proved. + +By our theorem, $\alpha$ is the abscissa of an inflexion point of the graph +of $y=f(x)$ if and only if conditions~\Eq{13} hold with $m$~odd ($m>1$). These +conditions include neither $f(\alpha) = 0$ nor $f'(\alpha)=0$, in contrast with~\Eq{10}. +In the theory of equations we are primarily interested in the abscissas +$\alpha$ of only those points of inflexion whose inflexion tangents are horizontal, +and are interested in them, because we must exclude such roots $\alpha$ of +$f'(x)=0$ when seeking the abscissas of bend points, which are the important +points for our purposes. A point on the graph at which the tangent is +both horizontal and an ordinary tangent is a bend point by the definition +in~§55. Hence if we apply our theorem to the special case $f'(\alpha)=0$, +we obtain the following + +\begin{Criterion} +Any root $\alpha$ of $f'(x) = 0$ is the abscissa of a bend point +of the graph of $y = f(x)$ or of a point with a horizontal inflexion tangent according +as the value of~$m$ for which relations~\Eq{13} hold is even or odd. +\index{Bend point}% +\end{Criterion} + +\begin{Remark} +For example, if $f(x) = x^4$, then $\alpha = 0$ and $m = 4$, so that $(0, 0)$ is a bend point of the +\Shape{U}-shaped graph of $y=x^4$. If $f(x)=x^3$, then $\alpha = 0$ and $m = 3$, so that $(0, 0)$ is a point +with a horizontal inflexion tangent $(OX$ in Fig.~15) of the graph of $y = x^3$. +\end{Remark} + + +\begin{Exercises}{Page64} + +\begin{Problems} + +\item[1.] If $f(x) = 3x^5 + 5x^3 + 4$, the only real root of $f'(x)=0$ is $x = 0$. Show that $(0, 4)$ +is an inflexion point, and thus that there is no bend point and hence that $f(x)=0$ has a +single real root. + +\item[2.] Prove that $x^3 - 3x^2 + 3x + c = 0$ has an inflexion point, but no bend point. + +\item[3.] Show that $x^5 - 10x^3 - 20x^2 - 15x + c = 0$ has two bend points and no horizontal +inflexion tangents. + +\item[4.] Prove that $3x^5 - 40x^3 + 240x + c = 0$ has no bend point, but has two horizontal +inflexion tangents. + +\item[5.] Prove that any function $x^3 - 3\alpha x^2 + \dotsb$ of the third degree can be written in +\index{Cubic equation!reduced}% [** PP: Not using page range] +the form $f(x) = (x-\alpha)^3 + ax + b$. The straight line having the equation $y = ax+b$ meets +the graph of $y=f(x)$ in three coincident points with the abscissa $\alpha$ and hence is an +inflexion tangent. If we take new axes of coordinates parallel to the old and intersecting +at the new origin $(\alpha, 0)$, i.e., if we make the transformation $x = X+\alpha$, $y = Y$, +%% -----File: 071.png---Folio 65------- +of coordinates, we see that the equation $f(x)=0$ becomes a reduced cubic equation +$X^3 + pX + q = 0$~(§42). + +\item[6.] Find the inflexion tangent to $y = x^3 + 6x^2 - 3x + 1$ and transform +$x^3 + 6x^2 - 3x + 1 = 0$ into a reduced cubic equation. +\end{Problems} +\end{Exercises} +\index{Inflexion|)}% [** PP: Original page range is 62--64] + + +\Section[Real Roots of a Cubic Equation] +{60.}{Real Roots of a Real Cubic Equation.} It suffices to consider +\begin{flalign*} +&& f(x) &= x^3 - 3lx + q && \Rightmark{(l \neq 0),} +\end{flalign*} +in view of Ex.~5 above. Then $f' = 3 (x^2 - l)$, $f'' = 6x$. If $l<0$, there is +no bend point and the cubic equation $f(x)=0$ has a single real root. +If $l>0$, there are two bend points +\[ +( \sqrt{l}, q - 2l\sqrt{l}),\qquad +(-\sqrt{l}, q + 2l\sqrt{l}), +\] +which are shown by crosses in Figs.~18--20 for the graph of $y=f(x)$ in the +\index{Cubic equation!graph of}% +\index{Cubic equation!number of real roots}% [** PP: Added ``of''] +three possible cases specified by the inequalities shown below the figures. +For a large positive~$x$, the term~$x^3$ in~$f(x)$ predominates, so that the graph +contains a point high up in the first quadrant, +thence extends downward to the +right-hand bend point, then ascends to +the left-hand bend point, and finally descends. +As a check, the graph contains +a point far down in the third quadrant, +since for $x$ negative, but sufficiently large +numerically, the term $x^3$ predominates and the sign of~$y$ is negative. +% [** Illustrations] +% \caption{$q \geq 2l\sqrt{l}$} FIG. 18 +% \caption{$q \leq -2l\sqrt{l}$} FIG. 19 +\begin{figure*}[htb] +\begin{center} +\Input{071a}\hfill +\Input{071b} +\end{center} +\end{figure*} + +% [** Illustration] +% \caption{$-2l\sqrt{l} < q < 2l\sqrt{l}$} Fig. 20 +\begin{wrapfigure}[9]{r}{2.6in} +\hfill\raisebox{\baselineskip}{\Input{071c}} +\end{wrapfigure} +If the equality sign holds in Fig.~18 or Fig.~19, a necessary and sufficient +condition for which is $q^2 = 4l^3$, one of the bend points is on the $x$-axis, and +the cubic equation has a double root. The inequalities in Fig.~20 hold +if and only if $q^2 < 4l^3$, which implies that $l>0$. Hence \emph{$x^3 - 3lx + q = 0$ +has three distinct real roots if and only if $q^2 < 4l^3$, a single real root if and +only if $q^2 > 4l^3$, a double root \(necessarily real\) if and only if $q^2 = 4l^3$ and $l\neq 0$, +and a triple root if $q^2 = 4l^3 = 0$}. +\index{Discriminant!of cubic}% + +%% -----File: 072.png---Folio 66------- + + +\begin{Exercises}{Page66} + +Find the bend points, sketch the graph, and find the number of real roots of +\begin{Problems}[2] +\item[1.] $x^3 + 2x - 4 = 0$. + +\item[2.] $x^3 - 7x + 7 = 0$. + +\item[3.] $x^3 - 2x - 1 = 0$. + +\item[4.] $x^3 + 6x^2 - 3x + 1 = 0$. +\end{Problems} +\begin{Problems} +\item[5.] Prove that the inflexion point of $y = x^3 - 3lx + q$ is $(0, q)$. + +\item[6.] Show that the theorem in the text is equivalent to that in~§45. + +\item[7.] Prove that, if $m$ and~$n$ are positive odd integers and $m>n$, $x^m + px^n + q = 0$ has +no bend point and hence has a single real root if $p>0$; but, if $p<0$, it has just two +bend points which are on the same side or opposite sides of the $x$-axis according as +\[ +\left(\frac{np}{m}\right)^m + \left(\frac{nq}{m-n}\right)^{m-n} +\] +is positive or negative, so that the number of real roots is $1$ or~$3$ in the respective cases. + +\item[8.] Draw the graph of $y = x^4 - x^2$. By finding its intersections with the line $y = mx + b$, solve $x^4 - x^2 - mx - b= 0$. + +\item[9.] Prove that, if $p$ and~$q$ are positive, $x^{2m} - px^{2n} + q = 0$ has four distinct real roots, +two pairs of equal roots, or no real root, according as +\[ +\left(\frac{np}{m}\right)^m - \left(\frac{nq}{m-n}\right)^{m-n} > 0, +\quad\text{${} = 0$,\quad or\quad ${} < 0$}. +\] + +\item[10.] Prove that no straight line crosses the graph of $y = f(x)$ in more than $n$~points if +the degree~$n$ of the real polynomial $f(x)$ exceeds unity. [Apply~§16.] This fact serves as a check on the accuracy of a graph. +\end{Problems} +\end{Exercises} + + +\Section[Continuity] +{61.}{Definition of Continuity of a Polynomial.} Hitherto we have +located certain points of the graph of $y=f(x)$, where $f(x)$ is a polynomial +in~$x$ with real coefficients, and taken the liberty to join them by a continuous +curve. +\index{Continuity}% + +A polynomial $f(x)$ with real coefficients shall be called \emph{continuous at} +$x = a$, where $a$ is a real constant, if the difference +\[ +D = f(a+h) - f(a) +\] +is numerically less than any assigned positive number~$p$ for all real values +of~$h$ sufficiently small numerically. + + +% [** PP: No ToC entry] +\Section[Continuity of Polynomials] +{62.}{Any Polynomial $f(x)$ with real Coefficients is continuous at $x = a$, +where $a$ is any real Constant.} Taylor's formula~\Eq{8} gives +\index{Polynomial}% +\[ +D = f'(a) h + \frac{f''(a)}{1·2} h^2 + \dotsb + + \frac{f^{(n)}(a)}{1·2\dotsm n} h^n. +\] +This polynomial is a special case of +\[ +F = a_1 h + a_2 h^2 + \dotsb + a_n h^n. +\] +%% -----File: 073.png---Folio 67------- +We shall prove that, \emph{if $a_1, \dotsc, a_n$ are all real, $F$ is numerically less than +any assigned positive number~$p$ for all real values of~$h$ sufficiently small +numerically}. Denote by~$g$ the greatest numerical value of $a_1, \dotsc, a_n$. +If $h$ is numerically less than~$k$, where $k<1$, we see that $F$ is numerically less than +\[ +g(k + k^2 + \dotsb + k^n) < g\frac{k}{1-k} < p,\qquad +\text{if } k < \frac{p}{p + g}. +\] +Hence a real polynomial $f(x)$ is continuous at every real value of~$x$. But +the function $\tan x$ is not continuous at $x=90°$~(§63). + + +\Section[Condition for a Root Between $a$ and~$b$] +{63.}{Root between $a$ and~$b$ if $f(a)$ and~$f(b)$ have opposite Signs.} +\begin{Thm} +If +the coefficients of a polynomial $f(x)$ are real and if $a$ and~$b$ are real numbers +such that $f(a)$ and~$f(b)$ have opposite signs, the equation $f(x) = 0$ has at least +one real root between $a$ and~$b$; in fact, an odd number of such roots, if an +$m$-fold root is counted as $m$~roots. +\end{Thm} + +%[Illustration: \textsc{Fig}. 21] [** PP: Move to top of paragraph] +\begin{wrapfigure}[15]{r}{1.3in} +\hfill\Input{073a} +\end{wrapfigure} +The only argument\footnote + {An arithmetical proof based upon a refined theory of irrational numbers is given + in Weber's \textit{Lehrbuch der Algebra}, ed.~2, vol.~1, p.~123.} +given here (other than that in Ex.~5 below) is +one based upon geometrical intuition. We are stating that, if the points +\[ +\bigl(a, f(a)\bigr),\qquad +\bigl(b, f(b)\bigr) +\] +lie on opposite sides of the $x$-axis, the graph of $y=f(x)$ crosses the $x$-axis +once, or an odd number of times, between the vertical lines through +these two points. Indeed, the part of the graph between +these verticals is a continuous curve having one +and only one point on each intermediate vertical line, +since the function has a single value for each value +of~$x$. + +This would not follow for the graph of $y^2 = x$, which +is a parabola with the $x$-axis as its axis. It may not +cross the $x$-axis between the two initial vertical lines, +but cross at a point to the left of each. + +A like theorem does not hold for $f(x) = \tan x$, when +$x$ is measured in radians and $0 < a < \pi/2 < b < \pi$, since +$\tan x$ is not continuous at $x=\pi/2$. When $t$ increases +from $a$ to $\pi/2$, $\tan x$ increases without limit. When +$x$ decreases from $b$ to~$\pi/2$, $\tan x$ decreases without +limit. There is no root between $a$ and~$b$ of $\tan x = 0$. + +%% -----File: 074.png---Folio 68------- + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Prove that $8x^3 - 4x^2 - 18x + 9 = 0$ has a root between $0$ and~$1$, one between $1$ and~$2$, +and one between $-2$ and~$-1$. + +\item[2.] Prove that $16x^4 - 24x^2 + 16x - 3 = 0$ has a triple root between $0$ and~$1$, and a +simple root between $-2$ and~$-1$. + +\item[3.] Prove that if $a < b < c \dotsb < l$, and $\alpha$, $\beta, \dotsc, \lambda$ are positive, these quantities +being all real, +\[ +\frac{\alpha}{x-a} + +\frac{\beta}{x-b} + +\frac{\gamma}{x-c} + \dotsb + +\frac{\lambda}{x-l} + t = 0 +\] +has a real root between $a$ and~$b$, one between $b$ and~$c, \dotsc$ one between $k$ and~$l$, and +if $t$ is negative one greater than~$l$, but if $t$ is positive one less than~$a$. + +\item[4.] Verify that the equation in Ex.~3 has no imaginary root by substituting $r+si$ +and $r-si$ in turn for~$x$, and subtracting the results. + +\item[5.] Admitting that an equation $f(x) \equiv x^n + \dotsb = 0$ with real coefficients has $n$~roots, +show algebraically that there is a real root between $a$ and~$b$ if $f(a)$ and~$f(b)$ have opposite +signs. Note that a pair of conjugate imaginary roots $c ± di$ are the roots of +\[ +(x-c)^2 + d^2 = 0 +\] +and that this quadratic function is positive if $x$ is real. Hence if $x_1, \dotsc, x_r$ are the +real roots and +\[ +\phi(x) \equiv (x-x_1) \dotsm (x-x_r), +\] +then $\phi(a)$ and~$\phi(b)$ have opposite signs. Thus $a-x_i$ and $b-x_i$ have opposite signs for +at least one real root~$x_i$. (Lagrange.) +\end{Problems} +\end{Exercises} + + +\Section[Sign of a Polynomial at Infinity] +{64.}{Sign of a Polynomial.} Given a polynomial +\index{Polynomial!sign of}% +\index{Sign of polynomial}% +\begin{flalign*} +&& f(x) &= a_0 x^n + a_1 x^{n-1} + \dotsb + a_n && \Rightmark{(a_0 \neq 0)} +\end{flalign*} +with real coefficients, we can find a positive number~$P$ such that $f(x)$ has +the same sign as $a_0 x^n$ when $x>P$. In fact, +\[ +f(x) = x^n (a_0 + \phi),\qquad +\phi = \frac{a_1}{x} + \frac{a_2}{x^2} + \dotsb + \frac{a_n}{x^n}. +\] +By the result in~§62, the numerical value of $\phi$ is less than that of~$a_0$ +when $1/x$ is positive and less than a sufficiently small positive number, +say $1/P$, and hence when $x>P$. Then $a_0 + \phi$ has the same sign as~$a_0$, +and hence $f(x)$ the same sign as~$a_0 x^n$. + +The last result holds also when $x$ is a negative number sufficiently large +numerically. For, if we set $x=-X$, the former case shows that $f(-X)$ +has the same sign as $(-1)^n a_0 X^n$ when $X$ is a sufficiently large positive +number. + +%% -----File: 075.png---Folio 69------- + +We shall therefore say briefly that, for $x = +\infty$, $f(x)$ has the same +sign as~$a_0$; while, for $x = -\infty$, $f(x)$ has the same sign as $a_0$ if $n$ is even, +but the sign opposite to $a_0$ if $n$ is odd. + + +\begin{Exercises}{} +\index{Number!of roots}% + +\begin{Problems} +\item[1.] Prove that $x^3 + ax^2 + bx - 4 = 0$ has a positive real root [use $x=0$ and $x=+\infty$]. + +\item[2.] Prove that $x^3 + ax^2 + bx + 4 = 0$ has a negative real root [use $x=0$ and $x=-\infty$]. + +\item[3.] Prove that if $a_0 > 0$ and $n$ is odd, $a_0x^n + \dotsb + a_n = 0$ has a real root of sign opposite +to the sign of $a_n$ [use $x = -\infty$, $0$, $+\infty$]. + +\item[4.] Prove that $x^4 + ax^3 + bx^2 + cx - 4 = 0$ has a positive and a negative root. + +\item[5.] Show that any equation of even degree $n$ in which the coefficient of $x^n$ and the +constant term are of opposite signs has a positive and a negative root. +\end{Problems} +\end{Exercises} + + +\Section{65.}{Rolle's Theorem} +\index{Rolle's theorem}% +\index{Root@{Root between $a$ and~$b$}}% +\begin{Thm} +Between two consecutive real roots $a$ and~$b$ of $f(x)=0$, +there is an odd number of real roots of $f'(x) = 0$, a root of multiplicity~$m$ +being counted as $m$~roots. +\end{Thm} + +Let +\begin{flalign*} +&& f(x) &\equiv (x-a)^r(x-b)^s Q(x), && \Rightmark{a<b,} +\end{flalign*} +where $Q(x)$ is a polynomial divisible by neither $x-a$ nor $x-b$. Then +by the rule for the derivative of a product (§56, Ex.~10), +\index{Derivative}% +\[ +\frac{(x-a)(x-b)f'(x)}{f(x)} + \equiv r(x-b) + s(x-a) + (x-a)(x-b) \frac{Q'(x)}{Q(x)}. +\] +The second member has the value $r(a-b) < 0$ for $x = a$ and the value +$s(b-a) > 0$ for $x=b$, and hence vanishes an odd number of times between +$a$ and $b$~(§63). But, in the left member, $(x-a)(x-b)$ and~$f(x)$ remain +of constant sign between $a$ and~$b$, since $f(x) = 0$ has no root between $a$ +and~$b$. Hence $f'(x)$ vanishes an odd number of times. + +\begin{Corollary} +Between two consecutive %[** PP: Typo consective] +real roots $\alpha$ and~$\beta$ of $f'(x) = 0$ +there occurs at most one real root of $f(x) = 0$. +\end{Corollary} + +For, if there were two such real roots $a$ and~$b$ of $f(x) = 0$, the theorem +shows that $f'(x) = 0$ would have a real root between $a$ and~$b$ and hence between +$\alpha$ and~$\beta$, contrary to hypothesis. + +Applying also~§63 we obtain the +\begin{Criterion} +If $\alpha$ and~$\beta$ are consecutive real roots of $f'(x) = 0$, then $f(x) = 0$ +has a single real root between $\alpha$ and~$\beta$ if $f(\alpha)$ and~$f(\beta)$ have opposite signs, +but no root if they have like signs. At most one real root of $f(x) = 0$ is greater +than the greatest real root of $f'(x) = 0$, and at most one real root of $f(x) = 0$ is +less than the least real root of $f'(x) = 0$. +\end{Criterion} + +%% -----File: 076.png---Folio 70------- + +If $f(\alpha) = 0$ for our root $\alpha$ of $f'(x) = 0, \alpha$ is a multiple root of $f(x) = 0$ and it would be removed before the criterion is applied. + +\begin{Example} +For $f(x) = 3x^5 - 25x^3 + 60x - 20$, +\[ +\tfrac{1}{15}f'(x) = x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4). +\] +Hence the roots of $f'(x)=0$ are $± 1,\, ± 2$. Now +{\footnotesize +\[ +f(-\infty) = -\infty,\ +f(-2) = -36,\ +f(-1) = -58,\ +f(1) = 18,\ +f(2) = -4,\ +f(+\infty) = +\infty. +\]}% +Hence there is a single real root in each of the intervals +\[ +(-1, 1),\quad (1, 2),\quad (2, +\infty), +\] +and two imaginary roots. The three real roots are positive. +\end{Example} + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Prove that $x^5 - 5x + 2 = 0$ has 1~negative, 2~positive and 2~imaginary roots. + +\item[2.] Prove that $x^6 + x - 1 = 0$ has 1~negative, 1~positive and 4~imaginary roots. + +\item[3.] Show that $x^5 - 3x^3 + 2x^2 - 5 = 0$ has two imaginary roots, and a real root in each +of the intervals $(-2, -1.5)$, $(-1.5, -1)$, $(1, 2)$. + +\item[4.] Prove that $4x^5 - 3x^4 - 2x^2 + 4x - 10 = 0$ has a single real root. + +\item[5.] Show that, if $f^{(k)}(x) = 0$ has imaginary roots, $f(x) = 0$ has imaginary roots. + +\item[6.] Derive Rolle's theorem from the fact that there is an odd number of bend points +between $a$ and~$b$, the abscissa of each being a root of $f'(x) = 0$ of odd multiplicity, while +the abscissa of an inflexion point with a horizontal tangent is a root of $f'(x) = 0$ of even +multiplicity. +\end{Problems} +\end{Exercises} +\index{Graphs|)}% + +%% -----File: 077.png---Folio 71------- + + +% [** PP: ToC entry reads Isolation of the Real Roots] +\Chapter[Isolation of Real Roots] +{VI}{Isolation of the Real Roots of a Real Equation} + +% [** PP: No ToC entry] +\Section{66.}{Purpose and Methods of Isolating the Real Roots} In the next +chapter we shall explain processes of computing the real roots of a given +real equation to any assigned number of decimal places. Each such +method requires some preliminary information concerning the root to +be computed. For example, it would be sufficient to know that the root +is between $4$ and~$5$, provided there be no other root between the same +limits. But in the contrary case, narrower limits are necessary, such +as $4$ and~$4.3$, with the further fact that only one root is between these new +limits. Then that root is said to be \emph{isolated}. +\index{Isolation of roots}% [** PP: Original has no page range] + +\begin{Remark} +If an equation has a single positive root and a single negative root, the real roots +are isolated, since there is a single root between $-\infty$ and~$0$, and a single one between +$0$ and~$+\infty$. However, for the practical purpose of their computation, we shall need +narrower limits, sufficient to fix the first significant figure of each root, for example +$-40$ and~$-30$, or $20$ and~$30$. +\end{Remark} + +We may isolate the real roots of $f(x)=0$ by means of the graph of +$y=f(x)$. But to obtain a reliable graph, we saw in \ChapRef{V} that we +must employ the bend points, whose abscissas occur among the roots of % [** PP: Added `of'] +$f'(x)=0$. Since the latter equation is of degree $n-1$ when $f(x)=0$ is of +degree~$n$, this method is usually impracticable when $n$ exceeds~$3$. The +method based on Rolle's theorem~(§65) is open to the same objection. + +The most effective method is that due to Sturm~(§68). We shall, +however, begin with Descartes' rule of signs since it is so easily applied. +Unfortunately it rarely tells us the exact number of real roots. + + +\Section{67.}{Descartes' Rule of Signs} Two consecutive terms of a real polynomial +\index{Descartes' rule of signs}% +or equation are said to present a \emph{variation of sign} if their coefficients +have unlike signs. By the variations of sign of a real polynomial or equation +we mean all the variations presented by consecutive terms. +\index{Variation of sign}% + +\begin{Remark} +Thus, in $x^5 - 2x^3 - 4x^2 + 3 = 0$, the first two terms present a variation of sign, and +likewise the last two terms. The number of variations of sign of the equation is two. +\end{Remark} + +%% -----File: 078.png---Folio 72------- + +\begin{Theorem}[Descartes' Rule] +\index{Number!of roots|(}% +The number of positive real roots of an equation +with real coefficients is either equal to the number of its variations of sign +or is less than that number by a positive even integer. A root of multiplicity~$m$ +is here counted as $m$~roots. +\end{Theorem} + +\begin{Remark} +For example, $x^6 - 3x^2 + x + 1 = 0$ has either two or no positive roots, the exact number +not being found. But $3x^3 - x - 1 = 0$ has exactly one positive root, which is a simple +root. +\end{Remark} + +Descartes' rule will be derived in §73 as a corollary to Budan's theorem. +The following elementary proof\footnote + {The proofs given in college algebras are mere verifications of special cases.} +was communicated to the author by +Professor D.~R. Curtiss. + +Consider any real polynomial +\[ +f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_l x^{n-l} +\qquad (a_0 \ne 0,\ a_l \ne 0). +\] + +Let $r$ be a positive real number. By actual multiplication, +\[ +F(x) \equiv (x-r)f(x) + \equiv A_0 x^{n+1} + A_1 x^n + \dotsb + A_{l+1}x^{n-l}, +\] +where +\[ +A_0 = a_0,\quad +A_1 = a_1 - ra_0,\quad +A_2 = a_2 - ra_1, \dotsc, +A_l = a_l - ra_{l-1},\quad +A_{l+1} = -r a_l. +\] +In $f(x)$ let $a_{k_1}$ be the first non-vanishing coefficient of different sign from~$a_0$, +let $a_{k_2}$ be the first non-vanishing coefficient following $a_{k_1}$ and of the +same sign as~$a_0$, etc., the last such term, $a_{k_v}$, being either $a_l$ or of the same +sign as~$a_l$. Evidently $v$ is the number of variations of sign of~$f(x)$. + +\begin{Remark} +For example, if $f(x) \equiv 2x^6 + 3x^5 - 4x^4 - 6x^3 + 7x$, we have $v=2$, $a_{k_1} = a_2 = -4$, $a_{k_2} = a_5 = 7$. +Note that $a_4 = 0$ since $x^2$ is absent. +\end{Remark} + +The numbers $A_0, A_{k_1}, \dotsc, A_{k_v}, A_{l+1}$ are all different from zero and +have the same signs as $a_0, a_{k_1}, \dotsc, a_{k_v}, -a_l$, respectively. This is +obviously true for $A_0 = a_0$ and $A_{l+1}= -ra_l$. Next, $A_{k_i}$ is the sum of the +non-vanishing number $a_{k_i}$ and the number $-ra_{k_i - 1}$, which is either zero +or else of the same sign as $a_{k_i}$ since $a_{k_i - 1}$ is either zero or of opposite sign +to~$a_{k_i}$. Hence the sum~$A_{k_i}$ is not zero and has the same sign as~$a_{k_i}$. + +By hypothesis, each of the numbers $a_0, a_{k_1}, \dotsc, a_{k_v}$ after the first +is of opposite sign to its predecessor, while $-a_l$ is of opposite sign to~$a_{k_v}$. +Hence each term after the first in the sequence $A_0, A_{k_1}, \dotsc, A_{k_v}, A_{l+1}$ +is of opposite sign to its predecessor. Thus these terms present $v+1$ +variations of sign. We conclude that $F(x)$ has at least one more variation +of sign than $f(x)$. But we may go further and prove the following + +%% -----File: 079.png---Folio 73------- + +\begin{Lemma} +The number of variations of sign of~$F(x)$ is equal to that of +$f(x)$ increased by some positive odd integer. +\end{Lemma} + +For, the sequence $A_0, A_1, \dotsc, A_{k_1}$ has an odd number of variations +of sign since its first and last terms are of opposite sign; and similarly +for the $v$~sequences +\[ +\begin{array}{c@{\quad}c@{\,}c@{\,}c} +A_{k_1}, & A_{k_1 + 1}, & \dotsc, & A_{k_2}; \\ +\Dots{4} \\ +A_{k_v}, & A_{k_v + 1}, & \dotsc, & A_{l+1}. +\end{array} +\] +The total number of variations of sign of the entire sequence $A_0, A_1,\dotsc, +A_{l+1}$ is evidently the sum of the numbers of variations of sign for the +$v+1$ partial sequences indicated above, and is thus the sum of $v+1$ positive +odd integers. Since each such odd integer may be expressed as $1$ +plus~$0$ or a positive even integer, the sum mentioned is equal to $v+1$ plus~$0$ +or a positive even integer, i.e., to $v$ plus a positive odd integer. + +To prove Descartes' rule of signs, consider first the case in which $f(x)= 0$ +has no positive real roots, i.e., no real root between $0$ and~$+\infty$. Then +$f(0)$ and $f(\infty$) are of the same sign~(§63), and hence the first and last +coefficients of $f(x)$ are of the same sign.\footnote + {In case $f(x)$ has a factor $x^{n-l}$, we use the polynomial $f(x)/x^{n-l}$ instead of $f(x)$ in + this argument.} +Thus $f(x)$ has either no variations +of sign or an even number of them, as Descartes' rule requires. + +Next, let $f(x)= 0$ have the positive real roots $r_1,\dotsc, r_k$ and no others. +A root of multiplicity~$m$ occurs here $m$~times, so that the $r$'s need not be +distinct. Then +\[ +f(x) \equiv (x - r_1)\dotsm (x - r_k)\phi(x), +\] +where $\phi(x)$ is a polynomial with real coefficients such that $\phi(x)=0$ has +no positive real roots. We saw in the preceding paragraph that $\phi(x)$ +has either no variations of sign or an even number of them. By the +Lemma, the product $(x - r_k)\phi(x)$ has as the number of its variations of +sign the number for $\phi(x)$ increased by a positive odd integer. Similarly +when we introduce each new factor $x - r_i$. Hence the number of variations +of sign of the final product $f(x)$ is equal to that of $\phi(x)$ increased +by $k$~positive odd integers, i.e., by $k$ plus $0$ or a positive even integer. +Since $\phi(x)$ has either no variations of sign or an even number of them, +the number of variations of sign of $f(x)$ is $k$ plus $0$ or a positive even integer, +a result equivalent to our statement of Descartes' rule. + +%% -----File: 080.png---Folio 74------- + +If $-p$ is a negative root of $f(x)= 0$, then $p$ is a positive root of $f(-x)= 0$. +Hence we obtain the + +\begin{Corollary} +\index{Number!of negative roots}% +The number of \emph{negative} roots of $f(x)=0$ is either equal +to the number of variations of sign of $f(-x)$ or is less than that number +by a positive even integer. +\end{Corollary} + +\begin{Remark} +For example, $x^4 + 3x^3 + x - 1 = 0$ has a single negative root, which is a simple root, +since $x^4 - 3x^3 - x - 1 = 0$ has a single positive root. +\end{Remark} + +As indicated in Exs.~10, 11 below, Descartes' rule may be used to isolate +the roots. + + +\begin{Exercises}{Page74} + +Prove by Descartes' rule the statements in Exs.~1--8, 12,~15. +\begin{Problems} + +\item[1.] An equation all of whose coefficients are of like sign has no positive root. Why +is this self-evident? + +\item[2.] There is no negative root of an equation, like $x^5 - 2x^4 - 3x^2 + 7x - 5 = 0$, in which +the coefficients of the odd powers of~$x$ are of like sign, and the coefficients of the even +powers (including the constant term) are of the opposite sign. Verify by taking $x= -p$, +where $p$ is positive. + +\item[3.] $x^3 + a^2 x + b^2 = 0$ has two imaginary roots if $b\ne 0$. + +\item[4.] For $n$~even, $x^n - 1 = 0$ has only two real roots. + +\item[5.] For $n$~odd, $x^n - 1 = 0$ has only one real root. + +\item[6.] For $n$~even, $x^n + 1 = 0$ has no real root; for $n$~odd, only one. + +\item[7.] $x^4 + 12x^2 + 5x - 9 = 0$ has just two imaginary roots. + +\item[8.] $x^4 + a^2 x^2 + b^2 x - c^2 = 0$ ($c\ne 0$) has just two imaginary roots. + +\item[9.] Descartes' rule enables us to find the exact number of positive roots only when +all the coefficients are of like sign or when +\[ +f(x) = x^n + p_1 x^{n-1} + \dotsb + p_{n-s} x^s + - p_{n-s+1} x^{s-1} - \dotsb - p_n = 0, +\] +each $p_i$ being $\geqq 0$. Without using that rule, show that the latter equation has one +and only one positive root~$r$. Hints: There is a positive root~$r$ by~§63 ($a=0$, $b=\infty$). +Denote by~$P(x)$ the quotient of the sum of the positive terms by~$x^s$, and by $-N(x)$ +that of the negative terms. Then $N(x)$ is a sum of powers of~$1/x$ with positive coefficients. +\begin{align*} +\text{If}\quad x>r,\qquad P(x)>P(r),\qquad N(x)<N(r),\qquad f(x)>0; \\ +\text{If}\quad x<r,\qquad P(x)<P(r),\qquad N(x)>N(r),\qquad f(x)<0. +\end{align*} + +\item[10.] Prove that we obtain an upper limit to the number of real roots of $f(x)=0$ +between $a$ and~$b$, if we set +\[ +x = \frac{a+by}{1+y}\qquad +\left(\therefore y=\frac{x-a}{b-x}\right), +\] +multiply by $(1+y)^n$, and apply Descartes' rule to the resulting equation in~$y$. + +%% -----File: 081.png---Folio 75------- + +\item[11.] Show by the method of Ex.~10 that there is a single root between $2$ and~$4$ of +$x^3 + x^2 - 17x + 15 = 0$. Here we have $27y^3 + 3y^2 - 23y - 7 = 0$. + +\item[12.] In the astronomical problem of three bodies occurs the equation +\[ +r^5 + (3 - \mu)r^4 + (3 - 2\mu )r^3 - \mu r^2 - 2\mu r - \mu = 0, +\] +where $0 < \mu < 1$. Why is there a single positive real root? + +\item[13.] Prove that $x^5 + x^3 - x^2 + 2x - 3 = 0$ has four imaginary roots by applying Descartes' +rule to the equation in~$y$ whose roots are the squares of the roots of the former. +Transpose the odd powers, square each new member, and replace $x^2$ by~$y$. + +\item[14.] As in Ex.~13 prove that $x^3 + x^2 + 8x + 6 = 0$ has imaginary roots. + +\item[15.] If a real equation $f(x)=0$ of degree~$n$ has $n$~real roots, the number of positive +roots is exactly equal to the number~$V$ of variations of sign. Hint: consider also +$f(-x)$. + +\item[16.] Show that $x^3 - x^2 + 2x + 1 = 0$ has no positive root. Hint: multiply by $x + 1$. +\end{Problems} +\end{Exercises} + + +\Section{68.}{Sturm's Method} Let $f(x) = 0$ be an equation with real coefficients, +\index{Sturm's functions|(}% +and $f'(x)$ the first derivative of~$f(x)$. The first step of the usual process +of finding the greatest common divisor of $f(x)$ and $f'(x)$, if it exists, consists +\index{Greatest common divisor}% +in dividing~$f$ by~$f'$ until we obtain a remainder $r(x)$, whose degree +is less than that of~$f'$. Then, if $q_1$ is the quotient, we have $f = q_1 f' + r$. +Instead of dividing~$f'$ by~$r$, as in the greatest common divisor process, and +proceeding further in that manner, we write $f_2 = -r$, divide~$f'$ by~$f_2$, and +denote by~$f_3$ the remainder with its sign changed. Thus +\[ +f = q_1 f' - f_2,\qquad +f' = q_2 f_2 - f_3,\qquad +f_2 = q_3 f_3 - f_4,\dotsc. +\] + +The latter equations, in which each remainder is exhibited as the negative +of a polynomial~$f_i$, yield a modified process, just as effective as the +usual process, of finding the greatest common divisor~$G$ of $f(x)$ and~$f'(x)$ if +it exists. + +Suppose that $-f_4$ is the first constant remainder. If $f_4 = 0$, then $f_3 = G$, +since $f_3$ divides~$f_2$ and hence also $f'$ and~$f$ (as shown by using our above +equations in reverse order); while, conversely, any common divisor of +$f$ and~$f'$ divides $f_2$ and hence also~$f_3$. + +But if $f_4$ is a constant $\ne 0$, $f$ and~$f'$ have no common divisor involving~$x$. +This case arises if and only if $f(x) = 0$ has no multiple root~(§58), +and is the only case considered in~§§69--71. + +Before stating Sturm's theorem in general, we shall state it for a +numerical case and illustrate its use. + +%% -----File: 082.png---Folio 76------- + + +\begin{Example} $f(x) =x^3 +4x^2 -7$. Then $f'=3x^2 +8x$, +\begin{alignat*}{3} +f &= (\tfrac{1}{3}x + \tfrac{4}{9})f'- f_2, &&\qquad +f_2 &\equiv{}& \tfrac{32}{9}x + 7, \\ +f' &= (\tfrac{27}{32}x + \tfrac{603}{1024})f_2 - f_3, &&\qquad +f_3 &={}& \tfrac{4221}{1024}. +\end{alignat*} + +For\footnote + {Before going further, check that the preceding relations hold when $x= 1$ by inserting +the computed values of $f$, $f'$, $f_2$ for $x =1$. Experience shows that most students make some +error in finding $f_2, f_3, \dotsc$, so that checking is essential.} +$x = 1$, the signs of $f$, $f'$, $f_2$, $f_3$, are ${}-{}+{}+{}+{}$, showing a single variation of +consecutive signs. For $x = 2$, the signs are ${}+{}+{}+{}+{}$, showing no variation of sign. +Sturm's theorem states that there is a \emph{single} real root between $1$ and~$2$. For $x= -\infty$, +the signs are ${}-{}+{}-{}+{}$, showing 3~variations of sign. The theorem states that there +are $3-1=2$ real roots between $-\infty$ and~$1$. Similarly, +\[ +\begin{array}{c@{\quad}|c|c} + x & \text{Signs} & \text{Variations} \\ +\hline +-1 & {}-{}-{}+{}+{} & 1 \\ +-2 & {}+{}-{}-{}+{} & 2 \\ +-3 & {}+{}+{}-{}+{} & 2 \\ +-4 & {}-{}+{}-{}+{} & 3 +\end{array} +\] +Hence there is a single real root between $-2$ and~$-1$, and a single one between $-4$ +and~$-3$. Each real root has now been \emph{isolated} since we have found two numbers +such that a single real root lies between these two numbers or is equal to one of them. + +Some of the preceding computation was unnecessary. After isolating a root between +$-2$ and~$-1$, we know that the remaining root is isolated between $-\infty$ and~$-2$. But +before we can compute it by Horner's method, we need closer limits for it. For that +purpose it is unnecessary to find the signs of all four functions, but merely the sign +of~$f$~(§63). +\end{Example} + + +\Section{69.}{Sturm's Theorem} +\begin{Thm} +Let $f(x) =0$ be an equation with real coefficients +and without multiple roots. Modify the usual process of seeking the greatest +common divisor of $f(x)$ and its first derivative\footnote + {The notation $f_1$ instead of the usual~$f'$, and similarly $f_0$ instead of~$f$, is used to regularize + the notation of all the~$f$'s, and enables us to write any one of the equations~\Eq{1} + in the single notation~\Eq{3}.} +$f_1(x)$ by exhibiting each +remainder as the negative of a polynomial~$f_i$: +\[ +f = q_1 f_1 - f_2,\ +f_1 = q_2 f_2 - f_3,\ +f_2 = q_3 f_3 - f_4, \dotsc,\ +f_{n-2} = q_{n-1}f_{n-1} - f_n, +\Tag{1} +\] +where\footnote + {If the division process did not yield ultimately a constant remainder $\ne 0$, $f$ and~$f_1$ + would have a common factor involving~$x$, and hence $f(x) =0$ a multiple root.} +$f_n$ is a constant $\ne 0$. If $a$ and~$b$ are real numbers, $a<b$, neither +%% -----File: 083.png---Folio 77------- +a root of $f(x) = 0$, the number of real roots of $f(x) = 0$ between $a$ and~$b$ is equal +to the excess of the number of variations of sign of +\[ +f(x),\quad f_1(x),\quad f_2(x), \dotsc, f_{n-1}(x),\quad f_n +\Tag{2} +\] +for $x = a$ over the number of variations of sign for $x = b$. Terms which vanish +are to be dropped out before counting the variations of sign. +\end{Thm} + +For brevity, let $V_x$ denote the number of variations of sign of the +numbers~\Eq{2} when $x$ is a particular real number not a root of $f(x)= 0$. + +First, if $x_1$ and~$x_2$ are real numbers such that no one of the continuous +functions~\Eq{2} vanishes for a value of~$x$ between $x_1$ and~$x_2$ or for $x = x_1$ or +$x = x_2$, the values of any one of these functions for $x = x_1$ and $x = x_2$ are +both positive or both negative~(§63), and therefore $V_{x_1} = V_{x_2}$. + +Second, let $\rho$ be a root of $f_i(x) = 0$, where $1 \leqq i < n$. Then +\[ +f_{i-1}(x) = q_i f_i(x) - f_{i+1}(x) +\Tag{3} +\] +and the equations~\Eq{1} following this one show that $f_{i-1}(x)$ and~$f_i(x)$ have +no common divisor involving~$x$ (since it would divide the constant~$f_n$). +By hypothesis, $f_i(x)$ has the factor $x-\rho$. Hence $f_{i-1}(x)$ does not have +this factor $x-\rho$. Thus, by~\Eq{3}, +\[ +f_{i-1}(\rho) = -f_{i+1}(\rho) \ne 0. +\] +Hence, if $p$ is a sufficiently small positive number, the values of +\[ +f_{i-1}(x),\quad f_i(x),\quad f_{i+1}(x) +\] +for $x = \rho - p$ show just one variation of sign, since the first and third +values are of opposite sign, and for $x = \rho + p$ show just one variation of +sign, and therefore show no change in the number of variations of sign +for the two values of~$x$. + +It follows from the first and second cases that $V_\alpha = V_\beta$ if $\alpha$ and~$\beta$ are +real numbers for neither of which any one of the functions~\Eq{2} vanishes +and such that no root of $f(x) = 0$ lies between $\alpha$ and~$\beta$. + +Third, let $r$ be a root of $f(x) = 0$. By Taylor's theorem~\Eq{8} of~§56, +\begin{align*} +f(r - p) &= -pf'(r) + \tfrac{1}{2} p^2 f''(r) - \dotsb, \\ +f(r + p) &= \Neg pf'(r) + \tfrac{1}{2} p^2 f''(r) + \dotsb. +\end{align*} +If $p$ is a sufficiently small positive number, each of these polynomials in~$p$ +has the same sign as its first term. For, after removing the factor~$p$, +%% -----File: 084.png---Folio 78------- +we obtain a quotient of the form $a_0 + s$, where $s = a_1 p + a_2 p^2 + \dotsb$ is +numerically less than~$a_0$ for all values of~$p$ sufficiently small~(§62). Hence +if $f'(r)$ is positive, $f(r-p)$ is negative and $f(r+p)$ is positive, so that the +terms $f(x)$, $f_1(x)\equiv f'(x)$ have the signs ${}-{}+{}$ for $x = r-p$ and the signs +${}+{}+{}$ for $x = r+p$. If $f'(r)$ is negative, these signs are ${}+{}-{}$ and~${}-{}-{}$ +respectively. In each case, $f(x)$, $f_1(x)$ show one more variation of sign +for $x = r-p$ than for $x = r+p$. Evidently $p$ may be chosen so small that +no one of the functions $f_1(x), \dotsc, f_n$ vanishes for either $x=r-p$ or $x=r+p$, +and such that $f_1(x)$ does not vanish for a value of~$x$ between $r-p$ and~$r+p$, +so that $f(x) = 0$ has the single real root~$r$ between these limits~(§65). +Hence by the first and second cases, $f_1, \dotsc, f_n$ show the same number +of variations of sign for $x = r-p$ as for $x = r+p$. Thus, for the entire +series of functions~\Eq{2}, we have +\[ +V_{r-p} - V_{r+p} = 1. +\Tag{4} +\] + +The real roots of $f(x)= 0$ within the main interval from $a$ to~$b$ (i.e., the % [** PP: Not italicizing i.e.] +aggregate of numbers between $a$ and~$b$) separate it into intervals. By +the earlier result, $V_x$ has the same value for all numbers in the same +interval. By the present result~\Eq{4}, the value~$V_x$ in any interval exceeds +the value for the next interval by unity. Hence $V_a$ exceeds~$V_b$ by the +number of real roots between $a$ and~$b$. +\index{Interval}% + +\begin{Corollary} +If $a<b$, then $V_a\geqq V_{b}$. +\end{Corollary} + +A violation of this Corollary usually indicates an error in the computation +of Sturm's functions~\Eq{2}. + + +\begin{Exercises}{Page78} + +Isolate by Sturm's theorem the real roots of +\begin{Problems}[2] +\item[1.] $x^3 +2x +20 = 0$. + +\item[2.] $x^3 +x-3 = 0$. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{70.}{Simplifications of Sturm's Functions} In order to avoid fractions, +we may first multiply $f(x)$ by a \emph{positive} constant before dividing it by +by $f_1(x)$, and similarly multiply~$f_1$ by a positive constant before dividing it +by~$f_2$, etc. Moreover, we may remove from any~$f_i$ any factor~$k_i$ which is +either a positive constant or a polynomial in~$x$ positive for\footnote + {Usually we would require that $k_i$ be positive for all values of~$x$, since we usually +wish to employ the limits $-\infty$ and~$+\infty$.} +$a\leqq x \leqq b$, +and use the remaining factor~$F_i$ as the next divisor. + +To prove that Sturm's theorem remains true when these modified +%% -----File: 085.png---Folio 79------- +functions $f$, $F_1, \dotsc, F_m$ are employed in place of functions~\Eq{2}, consider +the equations replacing~\Eq{1}: +\begin{align*} +f_1 = k_1 F_1,\qquad +c_2 f &= q_1 F_1 - k_2 F_2,\qquad c_3 F_1 = q_2 F_2 - k_3 F_3, \\ +c_4 F_2 &= q_3 F_3 - k_4 F_4, \dotsc, +c_m F_{m-2} = q_{m-1} F_{m-1} - k_m F_m, +\end{align*} +in which $c_2, c_3, \dotsc$ are positive constants and $F_m$ is a constant $\ne 0$. A +common divisor (involving~$x$) of~$F_{i-1}$ and~$F_i$ would divide $F_{i-2},\dotsc, +F_2, F_1$, $f$,~$f_1$, whereas $f(x)=0$ has no multiple roots. Hence if $\rho$ is a root +of $F_i(x)=0$, then $F_{i-1}(\rho)\ne 0$ and +\[ +c_{i+1} F_{i-1}(\rho) = -k_{i+1}(\rho) F_{i+1}(\rho),\qquad +c_{i+1}>0,\qquad k_{i+1}(\rho)>0. +\] +Thus $F_{i-1}$ and~$F_{i+1}$ have opposite signs for $x = \rho$. We proceed as in~§69. + + +\begin{Example}[1.] +If $f(x) = x^3 + 6x - 10$, $f_1 = 3(x^2 + 2)$ is always positive. Hence we may +employ $f$ and~$F_1 = 1$. For $x =-\infty$, there is one variation of sign; for $x =+\infty$, no +variation. Hence there is a single real root; it lies between $1$ and~$2$. +\end{Example} + +\begin{Example}[2.] +\index{Cubic equation!number of real roots}% [** PP: Added ``of''] +If $f(x) = 2x^4 - 13x^2 - 10x - 19$, we may take +\[ +f_1 = 4x^3 - 13x - 5. +\] +Then +\[ +2f = xf_1 - f_2,\qquad +f_2 = 13x^2 + 15x + 38 = 13(x + \tfrac{15}{26})^2 + \tfrac{1751}{52}. +\] +Since $f_2$ is always positive, we need go no further (we may take $F_2 = 1$). For $x =-\infty$, +the signs are ${}+{}-{}+{}$; for $x = +\infty$, ${}+{}+{}+{}$. Hence there are two real roots. The +signs for $x = 0$ are ${}-{}-{}+{}$. Hence one real root is positive and the other negative. +\end{Example} + + +\begin{Exercises}{Page79} + +Isolate by Sturm's theorem the real roots of +\begin{Problems}[2] + +\item[1.] $x^3 + 3x^2 - 2x - 5 = 0$. + +\item[2.] $x^4 + 12x^2 + 5x - 9 = 0$. + +\ResetCols{2} + +\item[3.] $x^3 - 7x - 7 = 0$. + +\item[4.] $3x^4 - 6x^2 + 8x - 3 = 0$. + +\ResetCols{1} + +\item[5.] $x^6 + 6x^5 - 30x^2 - 12x - 9 = 0$ [stop with~$f_2$]. + +\item[6.] $x^4 - 8x^3 + 25x^2 - 36x + 8 = 0$. + +\item[7.] For $f = x^3 + px + q$ ($p\ne 0$), show that $f_1 = 3x^2 + p$, $f_2 = -2px - 3q$, +\[ +4p^2 f_1 = (-6px + 9q)f_2 - f_3,\quad +f_3 = -4p^3 - 27q^2, +\] +so that $f_3$ is the discriminant~$\Delta$~(§44). Let $[p]$ denote the sign of~$p$. Then the signs +of $f$, $f_1$, $f_2$, $f_3$ are +\begin{align*} +&{}-{}+{}+ [p]\ [\Delta]\quad \text{for $x = -\infty$}, \\ +&{}+{}+{}- [p]\ [\Delta]\quad \text{for $x = +\infty$}. +\end{align*} +For $\Delta$ negative there is a single real root. For $\Delta$ positive and therefore $p$~negative, +there are three distinct real roots. For $\Delta = 0$, $f_2$~is a divisor of~$f_1$ and~$f$, so that +$x = -3q/(2p)$ is a double root. + +%% -----File: 086.png---Folio 80------- + +\item[8.] Prove that if one of Sturm's functions has $p$~imaginary roots, the initial equation +has at least $p$~imaginary roots. + +\item[9.] State Sturm's theorem so as to include the possibility of~$a$, or~$b$, or both $a$ and~$b$ being roots of $f(x)=0$. +\end{Problems} +\end{Exercises} + + +% [** PP: ToC reads Sturm's Functions for a General Quartic Equation] +\Section{71.}{Sturm's Functions for a Quartic Equation} For the reduced quartic +equation $f(z) =0$, +\index{Quartic equation|(}% +\[ +\left\{ +\begin{aligned} +f &= z^4 + qz^2 + rz + s, \\ +f_1 &= 4z^3 + 2qz + r, \\ +f_2 &= -2qz^2 - 3rz - 4s. +\end{aligned} +\right. +\Tag{5} +\] +Let $q\ne 0$ and divide $q^2 f_1$ by~$f_2$. The negative of the remainder is +\[ +f_3 = Lz - 12rs - rq^2,\qquad +L = 8qs - 2q^3 - 9r^2. +\Tag{6} +\] +Let $L\ne 0$. Then $f_4$ is a constant which is zero if and only if $f=0$ has +multiple roots, i.e., if its discriminant~$\Delta$ is zero. We therefore desire~$f_4$ % [** PP: Not italicizing i.e.] +expressed as a multiple of~$\Delta$. By~§50, +\[ +\Delta = -4P^3 - 27Q^2,\qquad +P = -4s - \frac{q^2}{3},\qquad +Q = \tfrac{8}{3}qs - r^2 - \tfrac{2}{27}q^3. +\Tag{7} +\] +We may employ $P$ and~$Q$ to eliminate +\[ +4s = -P - \frac{q^2}{3},\qquad +r^2 = -Q - \tfrac{2}{3}qP - \tfrac{8}{27}q^3. +\Tag{8} +\] +We divide $L^2 f_2$ by +\[ +f_3 = Lz + 3rP,\qquad +L = 9Q + 4qP. +\Tag{9} +\] +The negative of the remainder\footnote + {Found directly by the Remainder Theorem~(§14) by inserting the root $z = -3rP/L$ + of $f_3=0$ into $L^2 f_2$.} +is +\[ +18r^2 qP^2 - 9r^2 LP + 4sL^2 = q^2 \Delta. +\Tag{10} +\] +The left member is easily reduced to~$q^2\Delta$. Inserting the values~\Eq{8} and +replacing $L^2$ by $L(9Q + 4qP)$, we get +\[ +-18qQP^2 - 12q^2 P^3 - \tfrac{16}{3}q^4 P^2 + 2qP^2 L + + \tfrac{4}{3}q^3 PL - 3q^2 QL. +\] +Replacing $L$ by its value~\Eq{9}, we get~$q^2\Delta$. Hence we may take +\[ +f_4 = \Delta. +\Tag{11} +\] +Hence if $qL\Delta\ne 0$, we may take \Eq{5}, \Eq{9}, \Eq{11} as Sturm's functions. + +%% -----File: 087.png---Folio 81------- + +Denote the sign of~$q$ by~$[q]$. The signs of Sturm's functions are +\[ +\begin{array}{rrrrr@{\quad}l} ++ & - & -[q] & -[L] & [\Delta] & \text{for $x = -\infty$}, \\ ++ & + & -[q] & [L] & [\Delta] & \text{for $x = +\infty$}. +\end{array} +\] + +First, let $\Delta > 0$. If $q$ is negative and $L$ is positive, the signs are +${}+{}-{}+{}-{}+{}$ and~${}+{}+{}+{}+{}+{}$, so that there are four real roots. In each +of the remaining three cases for $q$ and~$L$, there are two variations of sign +in either of the two series and hence there is no real root. + +Next, let $\Delta < 0$. In each of the three cases in which $q$ and~$L$ are not +both positive, there are three variations of sign in the first series and one +variation in the second, and hence just two real roots. If $q$ and~$L$ are +both positive, the number of variations is~$1$ in the first series and~$3$ in the +second, so that this case is excluded by the Corollary to Sturm's theorem. +To give a direct proof, note that, by the value of~$L$ in~\Eq{6}, $L>0$, $q>0$ +imply $4s > q^2$, i.e., $s>0$, and hence, by~\Eq{7}, $P$~is negative, so that each term +of~\Eq{10} is $\geqq 0$, whence $\Delta > 0$. % [** PP: Added space] + +Hence, if $qL\Delta \ne 0$, there are four distinct real roots if and only if $\Delta$ +and~$L$ are positive, and $q$~negative; two distinct real and two imaginary +roots if and only if $\Delta$ is negative. + +Combining this result with that in Ex.~4 below, we obtain the + +\begin{Theorem} +If the discriminant~$\Delta$ of $z^4 + qz^2 + rz + s = 0$ is negative, there +are two distinct real roots and two imaginary roots; if $\Delta > 0$, $q<0$, $L>0$, +four distinct real roots; if $\Delta > 0$ and either $q\geqq 0$ or $L\leqq 0$, no real roots. +Here $L = 8qs - 2q^3 - 9r^2$. +\end{Theorem} +\index{Discriminant!of quartic}% +\index{Quartic equation|)}% + +Our discussion furnished also the series of Sturm functions, which +may be used in isolating the roots. + + +\begin{Exercises}{} + +\begin{Problems} + +\item[1.] If $q\Delta\ne 0$, $L = 0$, then $f_3 = 3rP$ is not zero (there being no multiple root) and its +sign is immaterial in determining the number of real roots. Prove that there are just +two real roots if $q<0$, and none if $q>0$. By~\Eq{10}, $q$ has the same sign as~$\Delta$. + +\item[2.] If $r\Delta\ne 0$, $q = 0$, obtain~$-f_3$ by substituting $z = -4s/(3r)$ in~$f_1$. Show that we may +take $f_3 = r\Delta$ and that there are just two real roots if $\Delta < 0$, and no real +roots if $\Delta >0$. + +\item[3.] If $\Delta \ne 0$, $q = r = 0$, prove that there are just two real roots if $\Delta<0$, and no real +roots if $\Delta > 0$. Since $\Delta = 256s^3$, check by solving $z^4 + s = 0$. + +\item[4.] If $\Delta \ne 0$, $qL = 0$, there are just two real roots if $\Delta < 0$, and no real roots if $\Delta > 0$. +[Combine the results in Exs.~1--3.] + +\item[5.] Apply the theorem to Exs.~2, 4, 6 of~§70. + +\item[6.] Isolate the real roots of Exs.~3, 4, 5 of~§48. +\end{Problems} +\end{Exercises} + +%% -----File: 088.png---Folio 82------- + + +% [** PP: No ToC entry] +\Section[Sturm's Theorem for Multiple Roots] +{72.}{Sturm's Theorem for the Case of Multiple Roots.} We might +\index{Multiple roots}% +remove the multiple roots by dividing $f(x)$ by\footnote + {The degree of~$f(x)$ is not~$n$, + nor was it necessarily~$n$ in~§69.} +$f_n(x)$, the greatest common +divisor of $f(x)$ and $f_1 = f'(x)$; but this would involve considerable +work, besides wasting the valuable information in hand. As before, we +suppose $f(a)$ and $f(b)$ different from zero. We have equations~\Eq{1} in +which $f_n$ is now not a constant. + +\begin{Thm}% +The difference $V_a - V_b$ is the number of real roots between $a$ and~$b$, each +multiple root being counted only once. +\end{Thm} + +If $\rho$ is a root of $f_i(x) = 0$, but not a multiple root of $f(x)= 0$, then $f_{i-1}(\rho)$ +is not zero. For, if it were zero, $x-\rho$ would by~\Eq{1} be a common factor +of $f$ and~$f_1$. We may now proceed as in the second case in~§69. + +The third case requires a modified proof only when $r$ is a multiple root. +Let $r$ be a root of multiplicity~$m$, $m\geqq 2$. Then $f(r)$, $f'(r), \dotsc, f^{(m-1)}(r)$ +are zero and, by Taylor's theorem, +\begin{align*} +f(r+p) &= \frac{p^m}{1·2\dotsm m} f^{(m)}(r) + \dotsb, \\ +f'(r+p) &= \frac{p^{m-1}}{1·2\dotsm (m-1)} f^{(m)}(r) + \dotsb. +\end{align*} +These have like signs if $p$ is a positive number so small that the signs of +the polynomials are those of their first terms. Similarly, $f(r-p)$ and +$f'(r-p)$ have opposite signs. Hence $f$ and~$f_1$ show one more variation +of sign for $x = r-p$ than for $x = r+p$. Now $(x-r)^{m-1}$ is a factor of~$f$ +and~$f_1$ and hence, by~\Eq{1}, of $f_2, \dotsc, f_n$. Let their quotients by this +factor be $\phi, \phi_1, \dotsc, \phi_n$. Then equations~\Eq{1} hold after the $f'$s are replaced +by the~$\phi$'s. Taking $p$ so small that $\phi_1(x) = 0$ has no root between $r-p$ +and $r+p$, we see by the first and second cases in~§69 that $\phi_1, \dotsc, \phi_n$ +show the same number of variations of sign for $x = r-p$ as for $x = r+p$. +The same is true for $f_1, \dotsc, f_n$ since the products of $\phi_1, \dotsc, \phi_n$ by +$(x-r)^{m-1}$ have for a given~$x$ the same signs as $\phi_1, \dotsc, \phi_n $ or the same +signs as $-\phi_1, \dotsc, -\phi_n$. But the latter series evidently shows the +same number of variations of sign as $\phi_1, \dotsc, \phi_n$. Hence~\Eq{4} is proved +and consequently the present theorem. +\index{Sturm's functions|)}% + +%% -----File: 089.png---Folio 83------- + + +\begin{Exercises}{Page83} + +\begin{Problems} + +\item[1.] For $f = x^4 - 8x^2 + 16$, prove that $F_1 = x^3 - 4x$, $F_2 = x^2 - 4$, $F_1 = xF_2$. Hence $n = 2$. +Verify that $V_{-\infty} = 2$, $V_{\infty} = 0$, and that there are just two real roots, each a double +root. +\end{Problems} + +Discuss similarly the following equations. +\begin{Problems}[3] +\item[2.] $x^4 - 5x^3 + 9x^2 - 7x + 2 = 0$. + +\item[3.] $x^4 + 2x^3 - 3x^2 - 4x + 4 = 0$. + +\item[4.] $x^4 - x^2 - 2x + 2 = 0$. +\end{Problems} +\end{Exercises} + + +\Section{73.}{Budan's Theorem} +\index{Budan's theorem}% +\index{Derivative}% +\begin{Thm}% +Let $a$ and~$b$ be real numbers, $a<b$, neither\footnote + {In case $a$ or~$b$ + is a root of $f(x)=0$, the theorem holds if we count the number of + roots $>a$ and $\leqq b$. This inclusive theorem has been proved, by means of Rolle's + theorem, by A.~Hurwitz, \textit{Mathematische Annalen}, Vol.~71, 1912, p.~584, who extended + Budan's theorem from the case of a polynomial to a function~$f(x)$ which is real and + regular for $a \leqq x < b$.} +a root of $f(x) = 0$, an equation of degree~$n$ with real coefficients. Let $V_a$ +denote the number of variations of sign of +\[ +f(x),\qquad f'(x),\qquad f''(x),\qquad \dotsc,\qquad f^{(n)}(x) +\Tag{12} +\] +for $x = a$, after vanishing terms have been deleted. Then $V_a - V_b$ is either +the number of real roots of $f(x) = 0$ between $a$ and~$b$ or exceeds the number +of those roots by a positive even integer. A root of multiplicity~$m$ is here +counted as $m$~roots. +\end{Thm} + +\begin{Remark} +For example, if $f(x) = x^3 - 7x - 7$, then $f' = 3x^2 - 7$, $f'' = 6x$, $f''' = 6$. Their values +for $x = 3$, $4$, $-2$, $-1$ are tabulated below. +\[ +\begin{array}{r|rrrr|c} + x & f & f' & f'' & f''' & \text{Variations} \\ +\hline + 3 & -1 & 20 & 18 & 6 & 1 \\ + 4 & 29 & 41 & 24 & 6 & 0 \\ +-2 & -1 & 5 & -12 & 6 & 3 \\ +-1 & -1 & -4 & -6 & 6 & 1 +\end{array} +\] +Hence the theorem shows that there is a single real root between $3$ and~$4$, and two +or no real roots between $-2$ and~$-1$. The theorem does not tell us the exact number +of roots between the latter limits. To decide this ambiguity, note that $f(-3/2) = +1/8$, +so that there is a single real root between $-2$ and~$-1.5$, and a single one between +$-1.5$ and~$-1$. +\end{Remark} + +The proof is quite simple if no term of the series~\Eq{12} vanishes for +$x = a$ or for $x = b$ and if no two consecutive terms vanish for the same +value of~$x$ between $a$ and~$b$. Indeed, if no one of the terms vanishes for +$x_1 \leqq x\leqq x_2$, then $V_{x_1} = V_{x_2}$, since any term has the same sign for $x = x_1$ +as for $x = x_2$. Next, let $r$ be a root of $f^{(i)}(x)=0$, $a<r<b$. By hypothesis, +%% -----File: 090.png---Folio 84------- +the first derivative $f^{(i+1)}(x)$ of $f^{(i)}(x)$ is not zero for $x=r$. As in the third +step (now actually the case $i = 0$) in~§69, $f^{(i)}(x)$ and $f^{(i+1)}(x)$ show one +more variation of sign for $x = r-p$ than for $x = r+p$, where $p$ is a sufficiently +small positive number. If $i>0$, $f^{(i)}$ is preceded by a term $f^{(i-1)}$ in~\Eq{12}. +By hypothesis, $f^{(i-1)}(x) \ne 0$ for $x=r$ and hence has the same sign for +$x = r - p$ and $x = r + p$ when $p$ is sufficiently small. For these values of~$x$, +$f^{(i)}(x)$ has opposite signs. Hence $f^{(i-1)}$ and $f^{(i)}$ show one more or one less +variation of sign for $x = r-p$ than for $x = r+p$, so that $f^{(i-1)}$, $f^{(i)}$, $f^{(i+1)}$ show +two more variations or the same number of variations of sign. + +Next, let no term of the series~\Eq{12} vanish for $x = a$ or for $x = b$, but +let several successive % [** PP: Typo succssive] +terms +\[ +f^{(i)}(x),\qquad f^{(i+1)}(x), \dotsc, f^{(i+j-1)}(x) +\Tag{13} +\] +all vanish for a value~$r$ of~$x$ between $a$ and~$b$, while $f^{(i+j)}(r)$ is not zero, +but is say positive.\footnote + {If negative, all signs in the table below are to be changed; but the conclusion holds.} +Let $I_1$ be the interval between $r-p$ and~$r$, and $I_2$ +the interval between $r$ and~$r+p$. Let the positive number~$p$ be so small +that no one of the functions~\Eq{13} or $f^{(i+j)}(x)$ is zero in these intervals, so +that the last function remains positive. Hence $f^{(i+j-1)}(x)$ increases with~$x$ +(since its derivative is positive) and is therefore negative in~$I_1$ +and positive in~$I_2$. Thus $f^{(i+j-2)}(x)$ decreases in~$I_1$ and increases in~$I_2$ +and hence is positive in each interval. In this manner we may verify the +signs in the following table: +\[ +\begin{array}{c|cccccccc} +\multicolumn{2}{r}{f^{(i)}} & f^{(i+1)} & f^{(i+2)} & \ldots & + f^{(i+j-3)} & f^{(i+j-2)} & f^{(i+j-1)} & f^{(i+j)} \\ +% +I_1 & (-)^{j} & (-)^{j-1} & (-)^{j-2} & \ldots & + - & + & - & + \\ +I_2 & + & + & + & \ldots & + & + & + & + +\end{array} +\] +Hence these functions show $j$~variations of sign in~$I_1$ and none in~$I_2$. + +If $i>0$, the first term of~\Eq{13} is preceded by a function $f^{(i-1)}(x)$ which +is not zero for $x = r$, and hence not zero in $I_1$ or $I_2$ if $p$ is sufficiently small. +If $j$ is even, the signs of $f^{(i-1)}$ and $f^{(i)}$ are ${}+{}+{}$ or ${}-{}+{}$ in both $I_1$ and~$I_2$, +showing no loss in the number of variations of sign. If $j$ is odd, their +signs are +\[ +\begin{array}{c|ccc} +I_1 & {}+{}-{} & \raisebox{-12pt}{\smash{\text{or}}} & {}-{}-{} \\ +I_2 & {}+{}+{} & & {}-{}+{} +\end{array} +\] +so that there is a loss or gain of a single variation of sign. Hence +\[ +f^{(i-1)}, \qquad +f^{(i)}, \qquad +f^{(i+1)} \quad \dotsc,\qquad +f^{(i+j)} +\] +%% -----File: 091.png---Folio 85------- +show a loss of $j$~variations of sign if $j$ is even, and a loss of $j±1$ if $j$ is odd, +and hence always a loss of an even number $\geqq 0$ of variations of sign. + +If $i=0$, $f^{(i)}\equiv f$ has $r$ as a $j$-fold root and the functions in the table show +$j$~more variations of sign for $x = r-p$ than for $x = r+p$. + +Thus, when no one of the functions~\Eq{12} vanishes for $x=a$ or for $x=b$, +the theorem follows as at the end of~§69 (with unity replaced by the +multiplicity of a root). + +Finally, let one of the functions~\Eq{12}, other than $f(x)$ itself, vanish for +$x= a$ or for $x = b$. If $\delta$ is a sufficiently small positive number, all of the +$N$~roots of $f(x)=0$ between $a$ and~$b$ lie between $a+\delta$ and $b-\delta$, and for +the latter values no one of the functions~\Eq{12} is zero. By the above +proof, +\begin{gather*} +V_{a+\delta} - V_{b-\delta} = N + 2t, \\ +V_a - V_{a + \delta} = 2j,\qquad +V_{b - \delta} - V_b = 2s, +\end{gather*} +where $t$, $j$, $s$ are integers $\geqq 0$. Hence $V_a - V_b = N+2(t+j+s)$. + +Descartes' rule of signs~(§67) is a corollary to Budan's theorem. Consider +any equation with real coefficients +\index{Descartes' rule of signs}% +\[ +f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_{n-1}x + a_n = 0, +\] +having $a_n \ne 0$. For $x = 0$ the functions~\Eq{12} have the same signs as +\[ +a_n,\qquad a_{n-1},\qquad \dotsc, \quad a_1,\qquad a_0. +\] +Hence $V_0$ is equal to the number~$V$ of variations of sign of~$f(x)$. + +For $x= +\infty$, the functions all have the same sign, which is that of~$a_0$. +Thus $V_0 - V_{\infty} = V$ is either the number of positive roots or exceeds that +number by a positive even integer. Finally, Descartes' rule holds if +$a_n = 0$, as shown by removing the factors~$x$. +\index{Number!of roots|)}% + + +\begin{Exercises}{Page85} + +Isolate by Budan's theorem the real roots of +\begin{Problems}[2] + +\item[1.] $x^3 -x^2 -2x+1=0$. + +\item[2.] $x^3 +3x^2 -2x-5=0$. + +\ResetCols{1} + +\item[3.] Prove that if $f{(a)}\ne 0$, $V_a$ equals the number of real roots $>a$ or exceeds that number by an even integer. + +\item[4.] Prove that there is no root greater than a number making each of the functions~\Eq{12} +positive, if the leading coefficient of $f(x)$ is positive. (Newton.) + +\item[5.] Hence verify that $x^4 -4x^3 - 3x + 23 = 0$ has no root~$>4$. + +\item[6.] Show that $x^4 - 4x^3 + x^2 + 6x + 2 = 0$ has no root~$>3$. +\end{Problems} +\end{Exercises} + +%% -----File: 092.png---Folio 86------- + + +\Chapter{VII}{Solution of Numerical Equations} +\index{Solution of numerical equations|(}% + + +\Section[Horner's Method] +{74.}{Horner's Method.\protect\footnotemark}\addtocounter{footnote}{1}% + \footnotetext{W.~G. Horner, London Philosophical Transactions, 1819. Earlier (1804) by + P.~Ruffini. See Bulletin American Math.\ Society, May, 1911.} +\addtocounter{footnote}{-1}% +\index{Horner's method}% +\index{Synthetic division|(}% +After we have isolated a real root of a real +equation by one of the methods in \ChapRef{VI}, we can compute the root +to any desired number of decimal places either by Horner's method, +which is available only for polynomial equations, or by Newton's method~(§75), +which is applicable also to logarithmic, trigonometric, and other +equations. + +To find the root between $2$ and~$3$ of +\[ +x^3 - 2x - 5 = 0, +\Tag{1} +\] +set $x = 2+p$. Direct substitution gives the \emph{transformed equation} for~$p$: +\index{Transformed equation}% +\[ +p^3 + 6p^2 + 10p - 1 = 0. +\Tag{2} +\] +The method just used is laborious especially for equations of high degree. +We next explain a simpler method. Since $p = x -2$, +\[ +x^3 - 2x - 5 \equiv (x-2)^3 + 6(x-2)^2 + 10(x-2) - 1, +\] +identically in~$x$. Hence $-1$ is the remainder obtained when the given +polynomial $x^3 - 2x - 5$ is divided by $x-2$. By inspection, the quotient~$Q$ +is equal to +\[ +(x-2)^2 + 6(x-2) + 10. +\] +Hence $10$ is the remainder obtained when $Q$ is divided by $x-2$. The +new quotient is equal to $(x-2) + 6$, and another division gives the +remainder~$6$. Hence to find the coefficients $6$, $10$, $-1$ of the terms following~$p^3$ +in the transformed equation~\Eq{2}, we have only to divide the given +polynomial $x^3 - 2x - 5$ by $x-2$, the quotient~$Q$ by $x-2$, etc., and take +the remainders in reverse order. However, when this work is performed +by synthetic division~(§15) as tabulated below, no reversal of order is +%% -----File: 093.png---Folio 87------- +necessary, since the coefficients then appear on the page in their desired +order. +\[ +\begin{array}{rRRRPc} +1 & 0 & -2 & -5 && \Lcol{2} \\ +\cline{6-6} + & 2 & 4 & 4 && \\ +\cline{1-4} +1 & 2 & \Rcol{2} & \mathbf{-1} && \\ + & 2 & \Rcol{8} & && \\ +\cline{1-3} +1 & \Rcol{4} & \mathbf{10} & && \\ + & \Rcol{2} & & && \\ +\cline{1-2} +1 & \mathbf{6} & & && +\end{array} +\] + +Thus $1$, $6$, $10$, $-1$ are the coefficients of the desired equation~\Eq{2}. + +To obtain an approximation to the decimal~$p$, we ignore for the moment +the terms involving $p^3$ and~$p^2$; then by $10p - 1 = 0$, $p=0.1$. But this +value is too large since the terms ignored are all positive. For $p=0.09$, +the polynomial in~\Eq{2} is found to be negative, while for $p=0.1$ it was just +seen to be positive. Hence $p = 0.09+h$, where $h$ is of the denomination +thousandths. The coefficients $1$, $6.27, \dotsc$ of the transformed equation +for~$h$ appear in heavy type just under the first zigzag line in the following +scheme: +\[ +\begin{array}{r<{\qquad}l Pr@{}l Pr@{}l r@{}l} +1 & 6 && 10 & && -1 & && \Lcol{0.09} \\ +\cline{10-10} + & 0.09 && 0 & .5481 && 0 & .949329 && \\ +\cline{1-8} +1 & 6.09 && 10 & .5481 &\Lcol{}& \mathbf{-0} & \mathbf{.050671} && \\ + & 0.09 && 0 & .5562 &\Lcol{}& & && \\ +\cline{1-5} +1 & 6.18 &\Lcol{}& \mathbf{11} & \mathbf{.1043} & & &\Rcol{}& & +\smash[b]{\raisebox{-1.5ex}{$\dfrac{0.05}{11.1}$}} \\ + & 0.09 &\Lcol{}& & & & &\Rcol{}& & \\ +\cline{1-2} +1 & \textbf{6.27} && & & & &\Rcol{}& $=\,$&$0.004$\\ +\cline{9-10} + & 0.004 && 0 & .025096 && 0 & .044517584 & & \\ +\cline{1-8} +1 & 6.274 &&11 & .129396 &\Lcol{}& \mathbf{-0} & \mathbf{.006153416} && \\ + & 0.004 && 0 & .025112 &\Lcol{}& & && \\ +\cline{1-5} +1 & 6.278 &\Lcol{}& \mathbf{11} & \mathbf{.154508} && & && \\ + & 0.004 &\Lcol{}& & && & && \\ +\cline{1-2} +1 & \textbf{6.282} && & && & && +\end{array} +\] +Hence $x=2.094+t$, where $t$ is a root of +\[ +t^3 + 6.282t^2 + 11.154508t - 0.006153416 = 0. +\] +By the last two terms, $t$ is between $0.0005$ and~$0.0006$. Then the value +%% -----File: 094.png---Folio 88------- +of $C\equiv t^3 +6.282t^2$ is found to lie between $0.00000157$ and~$0.00000227$. +Hence we may ignore~$C$ provided the constant term be reduced by an +amount between these limits. Whichever of the two limits we use, we +obtain the same dividend below correct to 6~decimal places. +\[ +\begin{array}{r>{\quad}r<{\quad}l@{}l} +\multicolumn{1}{r|}{1\xatop{1}.\xatop{1}\xatop{5}4508} +& 0.006151 & \multicolumn{1}{|l@{}}{0.000551} & =t \\ +\cline{1-1} \cline{3-3} +& 5577 & \\[-8pt] +& \settowidth{\TmpLen}{5577}\rule{\TmpLen}{0.5pt} & \\ +& 574 & \\ +& 558 & \\[-8pt] +& \settowidth{\TmpLen}{558}\rule{\TmpLen}{0.5pt} & \\ +& 16 & \\ +& 11 & \\[-8pt] +& \settowidth{\TmpLen}{11}\rule{\TmpLen}{0.5pt} & \\ +& 5 & +\end{array} +\] +Since the quotient is $0.0005+$, only two decimal places of the divisor are +used, except to see by inspection how much is to be carried when making +the first multiplication. Hence we mark a cross above the figure~$5$ in +the hundredths place of the divisor and use only~$11.15$. Before making +the multiplication by the second significant figure~$5$ of the quotient~$t$, +we mark a cross over the figure~$1$ in the tenths place of the divisor and +hence use only~$11.1$. Thus $x = 2.0945514+$, with doubt only as to whether +the last figure should be $4$ or~$5$. + +If we require a greater number of decimal places, it is not necessary +to go back and construct a new transformed equation from the equation +in~$t$. We have only to revise our preceding dividend on the basis of our +present better value of~$t$. We now know that $t$ is between $0.000551$ and +$0.000552$. To compute the new value of the correction~$C$, in which we +may evidently ignore~$t^3$, we use logarithms. +\[ +% [** PP: Alignment: log integer mantissa = integer mantissa, x 2] +\begin{array}{rr@{}l@{}>{{}}c<{{}}@{}r@{}l@{}>{\qquad}rr@{}l@{}>{{}}c<{{}}@{}r@{}l@{}} +\log & 5&.51 & = & &.74115 & \log & 5&.52 & = && .74194 \\ +\therefore +\log & 5&.51^2 & = & 1&.48230 &\therefore + \log & 5&.52^2 & = &1&.48388 \\ +\log & 6&.282 & = & &.79810 & \log & 6&.282 & = & &.79810 \\ +\cline{5-6} \cline{11-12} +\log & 190&.72 & = & 2&.28040 & \log &191&.42 & = &2&.28198 \Strut +\end{array} +\] +Hence $C$ is between $0.000001907$ and $0.000001915$. Whichever of the two +limits we use, we obtain the same new dividend below correct to 8~decimal +places. +%% -----File: 095.png---Folio 89------- +\[ +\begin{array}{r>{\quad}r<{\quad}l@{}l} +\multicolumn{1}{r|}{1\xatop{1}.\xatop{1}\xatop{5}\xatop{4}\xatop{5}08} +& 0.00615150 & \multicolumn{1}{|l@{}}{0.00055148} \\ +\cline{1-1} \cline{3-3} +& 557725 & \\[-8pt] +& \settowidth{\TmpLen}{557725}\rule{\TmpLen}{0.5pt} & \\ +& 57425 & \\ +& 55773 & \\[-8pt] +& \settowidth{\TmpLen}{55773}\rule{\TmpLen}{0.5pt} & \\ +& 1652 & \\ +& 1115 & \\[-8pt] +& \settowidth{\TmpLen}{1115}\rule{\TmpLen}{0.5pt} & \\ +& 537 & \\ +& 446 & \\[-8pt] +& \settowidth{\TmpLen}{446}\rule{\TmpLen}{0.5pt} & \\ +& 91 & \\ +& 89 & \\[-8pt] +& \settowidth{\TmpLen}{89}\rule{\TmpLen}{0.5pt} & \\ +& 2 & +\end{array} +\] +Hence, finally, $x=2.094551482$, with doubt only as to the last figure. + + +\begin{Exercises}{Page89} + +(The number of transformations made by synthetic division should be about half +the number of significant figures desired for a root.) + +By one of the methods in \ChapRef{VI}, isolate each real root of the following equations, +and compute each real root to 5~decimal places. +\begin{Problems}[2] + +\item[1.] $x^3 +2x+20=0$. + +\item[2.] $x^3 +3x^2 -2x-5=0$. + +\ResetCols{2} + +\item[3.] $x^3 +x^2 -2x-1=0$. + +\item[4.] $x^4 +4x^3 -17.5x^2 -18x+58.5=0$. + +\ResetCols{2} + +\item[5.] $x^4 -11,727x+40,385=0$. + +\item[6.] $x^3 =10$. +\end{Problems} + +Find to 7~decimal places all the real roots of +\begin{Problems}[2] +\item[7.] $x^3 +4x^2 -7=0$. + +\item[8.] $x^3 -7x-7=0$. +\end{Problems} + +Find to 8~decimal places +\begin{Problems} +\item[9.] The root between $2$ and~$3$ of $x^3 -x-9=0$ (make only 3~transformations). + +\item[10.] The real cube root of~$7.976$. + +\item[11.] The abscissa of the real point of intersection of the conics $y=x^2$, $xy+x+3y-6=0$. + +\item[12.] Find to 3~decimal places the abscissas of the points of intersection of $x^2+y^2=9$, +$y=x^2-x$. + +\item[13.] A sphere two feet in diameter is formed of a kind of wood a cubic foot of which +weighs two-thirds as much as a cubic foot of water (i.e., the specific gravity of the wood +is~$2/3$). Find to four significant figures the depth~$h$ to which the floating sphere +will sink in water. +\index{Specific gravity}% + +Hints: The volume of a sphere of radius~$r$ is~$\tfrac{4}{3}\pi r^3$. Hence our sphere whose radius +%% -----File: 096.png---Folio 90------- +is $1$~foot weighs as much as $\tfrac{4}{3}\pi·\tfrac{2}{3}$ cubic feet of water. The volume of the submerged +portion of the sphere is $\pi h^2 (r-\tfrac{1}{3}h)$ cubic feet. Since this is also the volume of the displaced +water, its value for $r=1$ must equal $\tfrac{4}{3}\pi·\tfrac{2}{3}$. Hence +$h^3 - 3h^2 + \tfrac{8}{3} = 0$. + +\item[14.] If the specific gravity of cork is~$1/4$, find to four significant figures how far a +cork sphere two feet in diameter will sink in water. + +\item[15.] Compute $\cos 20°$ to four decimal places by use of +\[ +\cos 3A = 4\cos^3 A - 3\cos A,\qquad +\cos 60° = \tfrac{1}{2}. +\] + +\item[16.] Three intersecting edges of a rectangular parallelopiped are of lengths $6$,~$8$, +and $10$~feet. If the volume is increased by $300$~cubic feet by equal elongations of the +edges, find the elongation to three decimal places. + +\item[17.] Given that the volume of a right circular cylinder is $\alpha\pi$ and the total area of +its surface is~$2\beta\pi$, prove that the radius~$r$ of its base is a root of $r^3 - \beta r + \alpha = 0$. If $\alpha = 56$, +$\beta = 28$, find to four decimal places the two positive roots~$r$. The corresponding altitude +is~$\alpha/r^2$. + +\item[18.] What rate of interest is implied in an offer to sell a house for \$2700 cash, or +in annual installments each of \$1000 payable 1,~2, and 3~years from date? +\index{Compound interest|(}% + +Hint: The amount of \$2700 with interest for 3~years should be equal to the sum +of the first payment with interest for 2~years, the amount of the second payment with +interest for 1~year, and the third payment. Hence if $r$ is the rate of interest and we +write~$x$ for~$1+r$, we have +\[ +2700x^3 = 1000x^2 + 1000x + 1000. +\] + +\item[19.] Find the rate of interest implied in an offer to sell a house for \$3500 cash, or in +annual installments each of \$1000 payable 1,~2, 3, and 4~years from date. + +\item[20.] Find the rate of interest implied in an offer to sell a house for \$3500 cash, or +\$4000 payable in annual installments each of \$1000, the first payable now. +\index{Compound interest|)}% +\end{Problems} +\end{Exercises} + + +% [** PP: Splitting combined ToC entries] +\Section{75.}{Newton's Method} Prior to 1676, Newton\footnote + {Isaac Newton, \textit{Opuscula},~I, 1794, p.~10, p.~37.} +\index{Newton's!method of solution|(}% +had already found +the root between $2$ and~$3$ of equation~\Eq{1}. He replaced~$x$ by~$2+p$ and +obtained~\Eq{2}. Since $p$ is a decimal, he neglected the terms in $p^3$ and~$p^2$, +and hence obtained $p=0.1$, approximately. Replacing $p$ by $0.1 + q$ in~\Eq{2}, +he obtained +\[ +q^3 + 6.3q^2 + 11.23q + 0.061 = 0. +\] +Dividing $-0.061$ by~$11.23$, he obtained $-0.0054$ as the approximate +value of~$q$. Neglecting $q^3$ and replacing $q$ by $-0.0054 + r$, he obtained +\[ +6.3r^2 + 11.16196r + 0.000541708 = 0. +\] +Dropping $6.3r^2$, he found $r$ and hence +\[ +x = 2 + 0.1 - 0.0054 - 0.00004853 = 2.09455147, +\] +%% -----File: 097.png---Folio 91------- +of which all figures but the last are correct~(§74). But the method will +not often lead so quickly to so accurate a value of the root. + +Newton used the close approximation $0.1$ to~$p$, in spite of the fact +that this value exceeds the root~$p$ and hence led to a negative correction +at the next step. This is in contrast with Horner's method in which each +correction is positive, so that each approximation must be chosen less +than the root, as $0.09$ for~$p$. + +Newton's method may be presented in the following general form, +which is applicable to any equation $f(x) = 0$, whether $f(x)$ is a polynomial +or not. Given an approximate value~$a$ of a real root, we can usually +find a closer approximation $a+h$ to the root by neglecting the powers +$h^2$, $h^3, \dotsc$ of the small number~$h$ in Taylor's formula~(§56) +\[ +f(a+h) = f(a) + f'(a)h + f''(a) \frac{h^2}{2} + \dotsb +\] +and hence by taking +\[ +f(a) + f'(a)h = 0,\qquad h = \frac{-f(a)}{f'(a)}. +\] +We then repeat the process with $a_1 = a+h$ in place of the former~$a$. + +Thus in Newton's example, $f(x) = x^3 - 2x - 5$, we have, for $a=2$, +\begin{align*} +h &= \frac{-f(2)}{f'(2)} + = \frac{1}{10},\qquad + a_1 = a+h = 2.1, \\ +% +h_1 &= \frac{-f(2.1)}{f'(2.1)} + = \frac{-0.061}{11.23} + = -0.0054\dotsc. +\end{align*} + + +% [** PP: Split ToC entry, part 2] +\Section[Algebraic and Graphical Discussion] +{76.}{Graphical Discussion of Newton's Method.} Using rectangular +coördinates, consider the graph of $y = f(x)$ and the point~$P$ on it with the +abscissa $OQ = a$ (Fig.~22). Let the tangent at~$P$ meet the $x$-axis at~$T$ +%[** Illustrations FIG. 22 FIG. 23] +\begin{figure*}[hbt] +\begin{center} +\Input{097a}\hfill +\Input{097b} +\end{center} +\end{figure*} +%% -----File: 098.png---Folio 92------- +and let the graph meet the $x$-axis at~$S$. Take $h = QT$, the subtangent. +Then +\begin{align*} +QP=f(a),\qquad f'(a) + &= \tan XTP = \frac{-f(a)}{h}, \\ +h &= \frac{-f(a)}{f'(a)}. +\end{align*} +In the graph in Fig.~22, $OT=a+h$ is a better approximation to the +root $OS$ than $OQ=a$. The next step (indicated by dotted lines) gives a +still better approximation~$OT_1$. + +If, however, we had begun with the abscissa~$a$ of a point~$P_1$ in Fig.~22 +near a bend point, the subtangent would be very large and the method +would probably fail to give a better approximation. Failure is certain +if we use a point~$P_2$ such that a single bend point lies between it and~$S$. + +We are concerned with the approximation to a root previously isolated +as the only real root between two given numbers $\alpha$ and~$\beta$. These should +be chosen so nearly equal that $f'(x)=0$ has no real root between $\alpha$ and~$\beta$, +and hence $f(x)=y$ has no bend point between $\alpha$ and~$\beta$. Further, if $f''(x)=0$ +has a root between our limits, our graph will have an inflexion point with +an abscissa between $\alpha$ and~$\beta$, and the method will likely fail (Fig.~23). + +Let, therefore, neither $f'(x)$ nor $f''(x)$ vanish between $\alpha$ and~$\beta$. Since +$f''$ preserves its sign in the interval from $\alpha$ to~$\beta$, while $f$ changes in sign, +$f''$ and~$f$ will have the same sign for one end point. According as the +abscissa of this point is $\alpha$ or~$\beta$, we take $a=\alpha$ or $a=\beta$ for the first step of +Newton's process. In fact, the tangent at one of the end points meets +the $x$-axis at a point~$T$ with an abscissa within the interval from $\alpha$ to~$\beta$. +If $f'(x)$ is positive in the interval, so that the tangent makes an acute +angle with the $x$-axis, we have Fig.~24 or Fig.~25; if $f'$ is negative, Fig.~26 or Fig.~22. +%[** Illustration \textsc{Fig.~24} \textsc{Fig.~25} \textsc{Fig.~26}] +\begin{figure*}[hbt] +\begin{center} +\Input{098a}\hfill +\Input{098b}\hfill +\Input{098c} +\end{center} +\end{figure*} + +%% -----File: 099.png---Folio 93------- + +\begin{Remark} +In Newton's example, the graph between the points with the abscissas $\alpha = 2$ and +$\beta = 3$ is of the type in Fig.~24, but more nearly like a vertical straight line. In view +of this feature of the graph, we may safely take $a=\alpha$, as did Newton, although our +general procedure would be to take $a = \beta$. The next step, however, accords with our +present process; we have $\alpha=2$, $\beta=2.1$ in Fig.~24 and hence we now take $a=\beta$, getting +\[ +\frac{0.061}{11.23} = 0.0054 +\] +as the subtangent, and hence $2.1 - 0.0054$ as the approximate root. +\end{Remark} + +If we have secured (as in Fig.~24 or Fig.~26) a better upper limit to the +root than~$\beta$, we may take the abscissa~$c$ of the intersection of the chord +$AB$ with the $x$-axis as a better lower limit than~$\alpha$. By similar triangles, +\[ +-f(\alpha) : c - \alpha = f(\beta) : \beta - c, +\] +whence +\[ +c = \frac{\alpha f(\beta) - \beta f(\alpha)}{f(\beta) - f(\alpha)}. +\Tag{3} +\] +This method of finding the value of~$c$ intermediate to $\alpha$ and~$\beta$ is called the +method of interpolation (\emph{regula falsi}). +\index{Interpolation}% +\index{Regula falsi}% + +\begin{Remark} +In Newton's example, $\alpha= 2$, $\beta=2.1$, +\[ +f(\alpha) = -1,\qquad +f(\beta) = 0.061,\qquad +c = 2.0942. +\] +\end{Remark} + +The advantage of having $c$ at each step is that we know a close limit +of the error made in the approximation to the root. + +We may combine the various possible cases discussed into one: + +\begin{Thm}% +If $f(x)=0$ has a single real root between $\alpha$ and~$\beta$, and $f'(x) = 0$, $f''(x) = 0$ +have no real root between $\alpha$ and~$\beta$, and if we designate by~$\beta$ that one of the +numbers $\alpha$ and~$\beta$ for which $f(\beta)$ and $f''(\beta)$ have the same sign, then the root +lies in the narrower interval from~$c$ to $\beta - f(\beta)/f'(\beta)$, where~$c$ is given by~\Eq{3}. +\end{Thm} + +It is possible to prove\footnote + {Weber's \textit{Algebra}, 2d~ed.,~I, pp.~380--382; \textit{Kleines Lehrbuch der Algebra}, 1912, p.~163.} +this theorem algebraically and to show that by +repeated applications of it we can obtain two limits $\alpha'$, $\beta'$ between which +the root lies, such that $\alpha' - \beta'$ is numerically less than any assigned positive +number. Hence the root can be found in this manner to any desired accuracy. + + +\begin{Example} +$f(x) = x^3 - 2x^2 - 2$,\quad $\alpha = 2\tfrac{1}{4}$,\quad $\beta = 2\tfrac{1}{2}$. Then +\[ +f(\alpha) = -\tfrac{47}{64},\qquad f(\beta) = \tfrac{9}{8}. +\] +%% -----File: 100.png---Folio 94------- +Neither of the roots $0$, $4/3$ of $f'(x)=0$ lies between $\alpha$ and~$\beta$, so that $f(x)=0$ has a single +real root between these limits~(§65). Nor is the root~$\tfrac{2}{3}$ of $f''(x)=0$ within these limits. +The conditions of the theorem are therefore satisfied. For $\alpha<x<\beta$, the graph is of +the type in Fig.~24. We find that approximately +\begin{gather*} +c = \tfrac{559}{238} = 2.3487,\qquad +\beta_1 = \beta - \frac{f(\beta)}{f'(\beta)} = 2.3714, \\ +\beta_1 - \frac{f(\beta_{1})}{f'(\beta_{1})} = 2.3597. +\end{gather*} +For $x = 2.3593$, $f(x) = -0.00003$. We therefore have the root to four decimal places. +For $a = 2.3593$, +\[ +f'(a) = 7.2620,\qquad +a - \frac{f(a)}{f'(a)} = 2.3593041, +\] +which is the value of the root correct to 7~decimal places. We at once verify that the +result is greater than the root in view of our work and Fig.~24, while if we change the +final digit from $1$ to~$0$, $f(x)$ is negative. +\end{Example} + + +\begin{Exercises}{Page94} + +\begin{Problems} +\item[1.] For $f(x) = x^4 + x^3 - 3x^2 - x - 4$, show by Descartes' rule of signs that $f'(x)=0$ +and $f''(x)=0$ each have a single positive root and that neither has a root between $1$ +and~$2$. Which of the values $1$ and~$2$ should be taken as~$\beta$? + +\item[2.] When seeking a root between $2$ and~$3$ of $x^3 - x - 9 = 0$, which value should be +taken as~$\beta$? +\end{Problems} +\end{Exercises} + + +% [** PP: Split ToC entry, part 3] +\Section[Systematic Computation] +{77.}{Systematic Computation of Roots by Newton's Method.} By way +of illustration we shall compute to 7~decimal places a positive root of +\[ +f(x) = x^4 + x^3 - 3x^2 - x - 4 = 0. +\] + +Since $f(1) = -6$, $f(2) = 6$, there is a real root between $1$ and~$2$. Since +\[ +f'(x) = 4x^3 + 3x^2 - 6x - 1,\qquad +f'(1) = 0,\qquad +f'(2) = 31, +\] +the graph of $y=f(x)$ is approximately horizontal near $(1, -6)$ and approximately +vertical near $(2, 6)$. Hence the root is much nearer to $2$ than to~$1$. +Thus in applying Newton's method we employ $a = 2$ as the first approximation +to the root. The correction~$h$ is then +\[ +h = \frac{-f(2)}{f'(2)} = \frac{-6}{31} = -0.2 \ldots . +\] +The work of performing the substitutions $x = 2+d$, $d = -0.2 + e, \dotsc$, +to find the transformed equations satisfied by $d$, $e, \dotsc$, is done by synthetic +%% -----File: 101.png---Folio 95------- +division, exactly as in Horner's method, except that some of the +multipliers are now negative: see \hyperref[table:1]{Table~1}, +page~\pageref*{table:1}. +% The structure below is so large it's almost certain to cause a +% horribly-underfull page. Instead, float it, setting up bidirectional +% anchors, and slightly change the wording of the previous sentence. +\phantomsection\label{tableanchor:1}% + +\begin{table*}[hp] +% ** PP: Dear future maintainer of this table code: +% The P column specifier has fixed width 1/8in and serves as +% padding. The idea is to set columns of numbers in stanzas rlP, +% with no inter-column space, and with the decimal points in the +% left-aligned column. \Lcol and \Rcol are wrappers for center-aligned +% columns with a vertical bar at the corresponding side. The floating +% sub-tables at the right side are rather terribly hacked; to get the +% ``underlines'' of correct width, entries are wrapped in left-aligned +% makebox commands. In short, this structure is a nest of hacks. +% +{\small% Coax width into text block +\begin{gather*}% Commit further abuse of semantics.... +\begin{array}{r<{\qquad} r@{}l@{}P<{\qquad} *{3}{>{\quad}r@{}l@{}P} r@{}lP} + 1 & 1 &&& -3 &&& -1 &&& -4 && &\Lcol{2} & \\ +\cline{14-14} + & 2 &&& 6 &&& 6 &&& 10 && &&\Strut \\ +\cline{1-11} + 1 & 3 &&& 3 &&& 5 &\Rcol{}&& \mathbf{6} && &&\Strut \\ + & 2 &&& 10 &&& 26&\Rcol{}&& && && \\ +\cline{1-9} + 1 & 5 &&& 13 &\Rcol{}&& \mathbf{31} &&& && &&\Strut \\ + & 2 &&& 14 &\Rcol{}&& &&& && && \\ +\cline{1-6} + 1 & 7 &\Rcol{}&& \mathbf{27} &&& &&& && &&\Strut \\ + & 2 &\Rcol{}&& &&& &&& && && \\ +\cline{1-3} +\mathbf{1} & \mathbf{9}\Strut + &&& &&& &&& && &\Lcol{\makebox[1em][l]{$-0.2$}} \\ +\cline{14-15} + & -0&.2&& -1&.76&& + -5&.048&& -5&.1904& &&\Strut \\ +\cline{2-12} + & 8&.8&& 25&.24&& + 25&.952&\Lcol{}& + \mathbf{0}&\mathbf{.8096}& &&\Strut \\ +% + & -0&.2 && -1&.72 && + -4&.704 &\Lcol{}& && && \\ +\cline{2-9} +% + & 8&.6 && 23&.52 &\Rcol{}& + \mathbf{21}&\mathbf{.248} && && && \Strut \\ +% + & -0&.2 && -1&.68 &\Rcol{}& + & && && && \\ +\cline{2-7} + & 8&.4 &\Rcol{}& + \mathbf{21}&\mathbf{.84}& &&& && && + \Lcol{\makebox[0pt][l]{\smash[b]{\raisebox{-1ex}{$\dfrac{-0.8096}{21.248}$}}}} \\ +% + &-0&.2 &\Rcol{}& + & & &&& && && \Lcol{} \\ +\cline{2-4} + \mathbf{1} & \mathbf{8}&\mathbf{.2}&& + & & &&& && && \Lcol{\makebox[0pt][l]{$= -0.04$}}&&\ \\ +\cline{14-16} + & -0&.04 && -0&.3264 & + & -0&.860544 & &-0&.81549824 && & \\ +\cline{2-12} + & 8&.16 && 21&.5136 & + & 20&.387456 &\Rcol{}& + \mathbf{-0}&\mathbf{.00589824}\Strut && & \\ +% + & -0&.04 && -0&.3248 & + & -0&.847552 &\Rcol{}& + & && & \\ +\cline{2-10} + & 8&.12 && 21&.1888 + &\Rcol{}& \mathbf{19}&\mathbf{.539904} &&&&&&\Strut \\ +% + & -0&.04 && -0&.3232 + &\Rcol{}& & &&&&&& \\ +\cline{2-7} + & 8&.08 &\Rcol{}& \mathbf{20}&\mathbf{.8656} + & & & &&&&&&\Strut \\ + & -0&.04 &\Rcol{}&& &&& &&& && \\ +\cline{2-4} + \mathbf{1} & \mathbf{8}&\mathbf{.04} && + & & & &&&\Strut&& +\smash{\makebox[0pt][r]{$g=\dfrac{0.005898}{19.54} = .000302$}} && +\end{array} \\[24pt] +\text{\hyperref[tableanchor:1]{\scshape Table~1.}}% Caption...! +\end{gather*}}% End of \small +\phantomsection\label{table:1}% +\end{table*} + +The root is $2 - 0.2 - 0.04 + 0.000302 = 1.760302$, in which the last +figure is in slight doubt. Indeed, it can be proved that \emph{if the final fraction~$g$, +when expressed as a decimal, has $k$~zeros between the decimal point and the +first significant figure, the division may be safely carried to $2k$~decimal places}. +In our example $k=3$, so that we retained 6~decimal places in~$g$. +\index{Synthetic division|)}% + +To proceed independently of this rule, we note that $g$ is obviously +between $0.00030$ and $0.00031$. Then the value of $g^4 + 8.04g^3 + 20.8656g^2$ +%% -----File: 102.png---Folio 96------- +is found to lie between $0.000001878$ and $0.000002006$. Whichever of these +limits we use as a correction by which to reduce the constant term, we +obtain the same dividend below correct to 6~decimal places. +\[ +\begin{array}{r>{\quad}r<{\quad}l@{}l} +\multicolumn{1}{r|}{\xatop{1}\xatop{9}.\xatop{5}\xatop{3}9904} +& 0.005896 & \multicolumn{1}{|l@{}}{0.0003017} \\ +\cline{1-1} \cline{3-3} + & 005862 & \\[-8pt] +& \settowidth{\TmpLen}{005862}\rule{\TmpLen}{0.5pt} & \\ + & 34 & \\ + & 20 & \\[-8pt] +& \settowidth{\TmpLen}{20}\rule{\TmpLen}{0.5pt} & \\ + & 14 & \\ + & 14 & \\[-8pt] +& \settowidth{\TmpLen}{14}\rule{\TmpLen}{0.5pt} & \\ +\end{array} +\] +Hence the root is $1.7603017$ to 7~decimal places. + + +\begin{Exercises}{Page96} + +\begin{Problems} +\item[1.] Find to 8~decimal places the root between $2$ and~$3$ of $x^3 - x - 9 = 0$. + +\item[2.] Find to 7~decimal places the root between $2$ and~$3$ of $x^3 - 2x^2 - 2 = 0$. + +\item[3.] Find the real cube root of $7.976$ to 6~decimal places. + +\item[4.] Explain by Taylor's expansion of $f(2+d)$ why the values of +\[ +f(2),\qquad f'(2),\qquad +\tfrac{1}{2}f''(2),\qquad +\frac{1}{2·3} f'''(2),\qquad +\frac{1}{2·3·4} f''''(2) +\] +are in reverse order the coefficients of the transformed equation +\[ +d^4 + 9d^3 + 27d^2 + 31d + 6 = 0, +\] +obtained in the Example in the text, and printed in heavy type. + +\item[5.] The method commonly used to find the positive square root of~$n$ by a computing +machine consists in dividing~$n$ by an assumed approximate value~$a$ of the square root +and taking half the sum of~$a$ and the quotient as a better approximation. Show that +the latter agrees with the value of $a+h$ given by applying Newton's method to +$f(x) = x^2-n$. +\index{Square roots}% +\end{Problems} +\end{Exercises} + + +% [** PP: Split ToC entry, part 4] +\Section[For Functions not Polynomials] +{78.}{Newton's Method for Functions not Polynomials.}\hfill\break +% [** PP: Format Example manually to avoid \BeforeSkip] +{\indent\normalfont\small\scshape Example~1.\quad\upshape +Find the angle~$x$ at the center of a circle subtended by a chord which +cuts off a segment whose area is one-eighth of that of the circle.} + +\begin{Solution} +If $x$ is measured in radians and if $r$ is the radius, the area of the segment +is equal to the left member of +\[ +\tfrac{1}{2} r^2(x - \sin x) = \tfrac{1}{8} \pi r^2, +\] +whence +\[ +x - \sin x = \tfrac{1}{4} \pi. +\] +%% -----File: 103.png---Folio 97------- +By means of a graph of $y = \sin x$ and the straight line represented by $y = x-\tfrac{1}{4}\pi$, we +see that the abscissa of their point of intersection is approximately $1.78$~radians or~$102°$. +Thus $a = 102°$ is a first approximation to the root of +\[ +f(x) \equiv x - \sin x - \tfrac{1}{4} \pi = 0. +\] +By Newton's method a better approximation is $a+h$, where\footnote + {The derivative of $\sin x$ is~$\cos x$. We need the limit of +\index{Derivative}% + \[ + \frac{\sin(x+2k) - \sin x}{2k} + \equiv \frac{2\cos \tfrac{1}{2}(2x + 2k) \sin \tfrac{1}{2}(2k)}{2k} + \equiv \frac{\cos(x+k) \sin k}{k} + \] + as $2k$ approaches zero. Since the ratio of $\sin k$ to~$k$ approaches~$1$, the limit is $\cos x$.} +\[ +h = \frac{-f(a)}{f'(a)} + = \frac{-a + \sin a + \tfrac{1}{4} \pi}{1 - \cos a}. +\] +\begin{gather*} +\begin{array}{@{}r@{}c@{}l@{}l@{}} + \sin 102° = & & 0.9781 & \\ +\tfrac{1}{4}(3.1416) = & & 0.7854 & \\ +\cline{3-3} + & & 1.7635 & \\ +& \\ + 102° &{}={} & 1.7802 & \text{ radians} \\ +\cline{2-3} + &{}-{} & 0.0167 & +\end{array} +\qquad +\begin{aligned} + \cos 102° &= -0.2079 \\ +1 - \cos 102° &= \Neg 1.2079 \\ +& \\ +h &= \frac{-0.0167}{1.2079} = -0.0138 \\ +a_1 &= a+h = 1.7664 +\end{aligned} \\ % Top-level end-of-line +h_1 = \frac{-f(a_1)}{f'(a_1)} + = \frac{-1.7664 + 0.9809 + 0.7854}{1.1944} + = -0.0001. +\end{gather*} +Hence $x = a_1 + h_1 = 1.7663$~radians, or $101° 12'$. % [** PP: Added period] +\end{Solution} + + +\begin{Example}[2.\protect\footnotemark] +\index{Interpolation}% + \footnotetext{This Ex.~2, which should be contrasted with Ex.~3, is solved by interpolation + since that method is simpler than Newton's method in this special case.} +Solve $x - \log x = 7$, the logarithm being to base~$10$. +\end{Example} + +\begin{Solution} +Evidently $x$ exceeds~$7$ by a positive decimal which is the value of~$\log x$. +Hence in a table of common logarithms, we seek a number~$x$ between $7$ and~$8$ whose +logarithm coincides approximately with the decimal part of~$x$. We read off the values +in the second column. +\[ +\begin{array}{c|c|c} +x & \log x & x - \log x \\ +\hline +7.897 & 0.89746 & 6.99954 \\ +7.898 & 0.89752 & 7.00048 +\end{array} +\] +By the final column the ratio of interpolation is~$46/94$. Hence $x = 7.8975$ to four +decimal places. +\end{Solution} + +%% -----File: 104.png---Folio 98------- + + +\begin{Example}[3.] +Solve $2x-\log x = 7$, the logarithm being to base~10. +\end{Example} + +\begin{Solution} +Evidently $x$ is a little less than~$4$. A table of common logarithms shows +at once that a fair approximation to $x$ is~$a=3.8$. Write +\[ +f(x) \equiv 2x - \log x - 7,\qquad +\log x = M \log_e x,\qquad +M = 0.4343. +\] +By calculus, the derivative of $\log_e x$ is~$1/x$. Hence +\begin{gather*} +f'(x) = 2 - \frac{M}{x},\qquad f'(a) = 2 - 0.1143 = 1.8857, \\ +f(a) = 0.6 - \log 3.8 = 0.6 - 0.57978 = 0.02022, \\ +-h = \frac{f(a)}{f'(a)} = 0.0107,\qquad a_1 = a + h = 3.7893, \\ +f(a_1) = 0.000041,\qquad f(3.7892) = -0.000148. \\ +\frac{148}{189} × 0.0001 = 0.000078,\qquad x = 3.789278. +\end{gather*} +All figures of~$x$ are correct as shown by Vega's table of logarithms to 10~places. +\end{Solution} + + +\begin{Exercises}{Page98} + +Find the angle~$x$ at the center of a circle subtended by a chord which cuts off a segment +whose ratio to the circle is +\begin{Problems}[2] + +\item[1.] $\frac{1}{4}$. + +\item[2.] $\frac{3}{8}$. +\end{Problems} + +When the logarithms are to base~10, +\begin{Problems}[2] +\item[3.] Solve $2x - \log x = 9$. + +\item[4.] Solve $3x - \log x = 9$. + +\ResetCols{1} + +\item[5.] Find the angle just $>15°$ for which +$\frac{1}{2} \sin x + \sin 2x = 0.64$. + +\item[6.] Find the angle just $>72°$ for which +$x - \frac{1}{2} \sin x = \frac{1}{4} \pi$. + +\item[7.] Find all solutions of Ex.~5 by replacing $\sin 2x$ by $2\sin x\cos x$, squaring, and +solving the quartic equation for~$\cos x$. + +\item[8.] Solve similarly $\sin x + \sin 2x = 1.2$. + +\item[9.] Find $x$ to 6~decimal places in $\sin x = x - 2$. + +\item[10.] Find $x$ to 5~decimal places in $x = 3\log_e x$. +\end{Problems} +\end{Exercises} +\index{Newton's!method of solution|)}% + + +\Section{79.}{Imaginary Roots} To find the imaginary roots $x+yi$ of an equation +$f(z)=0$ with real coefficients, expand $f(x+yi)$ by Taylor's theorem; +we get +\index{Imaginary!roots}% +\[ +f(x) + f'(x)yi + - f''(x) \frac{y^2}{1·2} + - f'''(x) \frac{y^3 i}{1·2·3} + \dotsb = 0. +\] + +%% -----File: 105.png---Folio 99------- + +Since $x$ and~$y$ are to be real, and $y \ne 0$, +\[ +\left\{ +\begin{aligned} +f(x) - f''(x) \frac{y^2}{1·2} + f''''(x) \frac{y^4}{1·2·3·4} - \dotsb = 0, \\ +f'(x) - f'''(x) \frac{y^2}{1·2·3} + f^{(5)}(x)\frac{y^4}{5!} - \dotsb = 0. +\end{aligned}\right. +\Tag{4} +\] + +In the Example and Exercises below, $f(z)$ is of degree~$4$ or less. Then +the second equation~\Eq{4} is linear in~$y^2$. Substituting the resulting value +of~$y^2$ in the first equation~\Eq{4}, we obtain an equation $E(x) = 0$, whose real +roots may be found by one of the preceding methods. If the degree of +$f(z)$ exceeds~$4$, we may find $E(x) = 0$ by eliminating $y^2$ between the two +equations~\Eq{4} by one of the methods to be explained in \ChapRef{X}. + + +\begin{Example} +For $f(z) = z^4 - z + 1$, equations~\Eq{4} are +\[ +x^4 - x + 1 - 6x^2 y^2 + y^4 = 0,\qquad +4x^3 - 1 - 4xy^2 = 0. +\] +Thus +\[ +y^2 = x^2 - \frac{1}{4x},\qquad +-4x^6 + x^2 + \frac{1}{16} = 0. +\] +The cubic equation in $x^2$ has the single real root +\[ +x^2 = 0.528727,\qquad x = ±0.72714. +\] +Then $y^2 = 0.184912$ or $0.87254$, and +\[ +z = x+yi + = 0.72714 ± 0.43001i,\qquad + -0.72714 ± 0.93409i. +\] +\end{Example} + + +\begin{Exercises}{Page99} + +Find the imaginary roots of +\begin{Problems}[2] +\item[1.] $z^3 - 2z - 5 = 0$. + +\item[2.] $28z^3 + 9z^2 - 1 = 0$. + +\ResetCols{2} + +\item[3.] $z^4 - 3z^2 - 6z = 2$. + +\item[4.] $z^4 - 4z^3 + 11z^2 - 14z + 10 = 0$. + +\ResetCols{1} + +\item[5.] $z^4 - 4z^3 + 9z^2 - 16z + 20 = 0$. Hint: +\[ +E(x) \equiv x(x - 2)(16x^4 - 64x^3 + 136x^2 - 144x + 65) = 0, +\] +and the last factor becomes $(w^2 + 1)(w^2 + 9)$ for $2x = w + 2$. +\end{Problems} +\end{Exercises} + +\begin{Note} +If we know a real root~$r$ of a cubic equation $f(z)=0$, we may remove the +factor $z-r$ and solve the resulting quadratic equation. When, as usual, $r$ involves +several decimal places, this method is laborious and unsatisfactory. But we may utilize +a device, explained in the author's \textit{Elementary Theory of Equations}, pp.~119--121, §§6,~7. +As there explained, a similar device may be used when we know two real roots of a +quartic equation. +\end{Note} + +%% -----File: 106.png---Folio 100------- + + +\begin{Exercises}[MISCELLANEOUS~]{Page100} + +(Give answers to 6~decimal places, unless the contrary is stated.) +\begin{Problems} + +\item[1.] What arc of a circle is double its chord? + +\item[2.] What arc of a circle is double the distance from the center of the circle to the +chord of the arc? + +\item[3.] If $A$ and~$B$ are the points of contact of two tangents to a circle of radius unity +from a point~$P$ without it, and if arc $AB$ is equal to~$PA$, find the length of the arc. + +\item[4.] Find the angle at the center of a circle of a sector which is bisected by its chord. + +\item[5.] Find the radius of the smallest hollow iron sphere, with air exhausted, which will +float in water if its shell is $1$~inch thick and the specific gravity of iron is~$7.5$. + +\item[6.] From one end of a diameter of a circle draw a chord which bisects the semicircle. + +\item[7.] The equation $x \tan x = c$ occurs in the theory of vibrating strings. Its approximate +solutions may be found from the graphs $y = \cot x$, $y = x/c$. Find $x$ when $c = 1$. + +\item[8.] The equation $\tan x = x$ occurs in the study of the vibrations of air in a spherical +cavity. From an approximate solution $x_1 = 1.5\pi$, we obtain successively better approximations +$x_2 = \tan^{-1} x_1 = 1.4334 \pi$, $x_3 = \tan^{-1} x_2, \dotsc$. Find the first three solutions to +4~decimal places. + +\item[9.] Find to 3~decimal places the first five solutions of +\[ +\tan x = \frac{2x}{2-x^2}, +\] +which occurs in the theory of vibrations in a conical pipe. + +\item[10.] $4 \tau x^3 - (3x - 1)^2 = 0$ arises in the study of the isothermals of a gas. Find its +roots when (i)~$\tau = 0.002$ and (ii)~$\tau = 0.99$. + +\item[11.] Solve $x^x = 100$. + +\item[12.] Solve $x = 10\log x$. + +\item[13.] Solve $x + \log x = x \log x$. + +\item[14.] Solve Kepler's equation $M = x - e \sin x$ when $M = 332° 28' 54.8''$, + $e = 14° 3' 20''$. + +\item[15.] In what time would a sum of money at 6\% interest compounded annually +amount to as much as the same sum at simple interest at~8\%? +\index{Compound interest}% + +\item[16.] In a semicircle of diameter~$x$ is inscribed a quadrilateral with sides $a$, $b$, $c$,~$x$; +then $x^3 - (a^2 + b^2 + c^2) x - 2abc = 0$ (I.~Newton). Given $a = 2$, $b = 3$, $c = 4$, find~$x$. + +\item[17.] What rate of interest is implied in an offer to sell a house for \$9000~cash, or +\$1000~down and \$3000 at the end of each year for three years? +\end{Problems} +\end{Exercises} +\index{Solution of numerical equations|)}% + +%% -----File: 107.png---Folio 101------- + + +\Chapter{VIII}{Determinants; Systems of Linear Equations} +\index{Determinants|(}% +\index{Linear equations!system|(}% + +% [** PP: ToC entry reads Solution of 2 or 3 Linear Equations by Determinants] +\Section[Solution of $2$~Linear Equations by Determinants] +{80.}{Solution of Two Linear Equations by Determinants of Order~$2$.} +Assume that there is a pair of numbers $x$ and~$y$ for which +\[ +\left\{ +\begin{aligned} +a_1 x + b_1 y &= k_1, \\ +a_2 x + b_2 y &= k_2. +\end{aligned} +\right. +\Tag{1} +\] +Multiply the members of the first equation by~$b_2$ and those of the second +equation by~$-b_1$, and add the resulting equations. We get +\[ +(a_1 b_2 - a_2 b_1)x = k_1 b_2 - k_2 b_1. +\] +Employing the respective multipliers $-a_2$ and~$a_1$, we get +\[ +(a_1 b_2 - a_2 b_1)y = a_1 k_2 - a_2 k_1. +\] +The common multiplier of $x$ and~$y$ is +\[ +a_1 b_2 - a_2 b_1, +\Tag{2} +\] +and is denoted by the symbol +\[ +\begin{vmatrix} + a_1 & b_1 \\ + a_2 & b_2 +\end{vmatrix}, +\Tag{2'} +\] +which is called a \emph{determinant of the second order}, and also called the determinant +of the coefficients of $x$ and~$y$ in equations~\Eq{1}. The results above +may now be written in the form +\[ +\begin{vmatrix} + a_1 & b_1 \\ + a_2 & b_2 +\end{vmatrix} x += +\begin{vmatrix} + k_1 & b_1 \\ + k_2 & b_2 +\end{vmatrix},\qquad +\begin{vmatrix} + a_1 & b_1 \\ + a_2 & b_2 +\end{vmatrix} y += +\begin{vmatrix} + a_1 & k_1 \\ + a_2 & k_2 +\end{vmatrix}. +\Tag{3} +\] +We shall call $k_1$ and~$k_2$ the known terms of our equations~\Eq{1}. Hence, +\emph{if $D$ is the determinant of the coefficients of the unknowns, the product of~$D$ by +any one of the unknowns is equal to the determinant obtained from~$D$ by +substituting the known terms in place of the coefficients of that unknown}. + +%% -----File: 108.png---Folio 102------- + +If $D \ne 0$, relations~\Eq{3} uniquely determine values of $x$ and~$y$: +\[ +x = \frac{k_1 b_2 - k_2 b_1}{D},\qquad +y = \frac{a_1 k_2 - a_2 k_1}{D}, +\] +and these values satisfy equations~\Eq{1}; for example, +\[ +a_1 x + b_1 y = \frac{(a_1 b_2 - a_2 b_1)k_1}{D} = k_1. +\] +Hence our equations~\Eq{1} have been solved by determinants when $D \ne 0$. +We shall treat in~§96 the more troublesome case in which $D = 0$. + + +\begin{Example} +For $2x - 3y = -4$, $6x - 2y = 2$, we have +\begin{align*} +&& +\begin{vmatrix} + 2 & -3 \\ + 6 & -2 +\end{vmatrix} x + &= +\begin{vmatrix} +-4 & -3 \\ +\Neg 2 & -2 +\end{vmatrix}, & 14x &= 14,\qquad x = 1, && \\ +&& +14y &= +\begin{vmatrix} +\Neg 2 & -4 \\ +\Neg 6 & \Neg 2 +\end{vmatrix} = 28, & y &= 2. +\end{align*} +\end{Example} + + +\begin{Exercises}{Page102} + +Solve by determinants the following systems of equations: +\begin{Problems}[3] +\item[1.] $\begin{System}[\,]{2} + 8x &-{}& y &= 34, \\ + x &+{}& 8y &= 53. +\end{System}$ + +\item[2.] $\begin{System}[\,]{2} + 3x &+{}& 4y &= 10, \\ + 4x &+{}& y &= 9. +\end{System}$ + +\item[3.] $\begin{System}[\,]{2} + ax &+{}& by &= a^2, \\ + bx &-{}& ay &= ab. +\end{System}$ +\end{Problems} +\end{Exercises} + + +% [** PP: No separate ToC entry] +\Section[Solution of $3$~Linear Equations by Determinants] +{81.}{Solution of Three Linear Equations by Determinants of Order~$3$.} + +Consider a system of three linear equations +\[ +\begin{aligned} +a_1 x + b_1 y + c_1 z &= k_1, \\ +a_2 x + b_2 y + c_2 z &= k_2, \\ +a_3 x + b_3 y + c_3 z &= k_3. +\end{aligned} +\Tag{4} +\] +Multiply the members of the first, second and third equations by +\[ +b_2 c_3 - b_3 c_2,\qquad +b_3 c_1 - b_1 c_3,\qquad +b_1 c_2 - b_2 c_1, +\Tag{5} +\] +respectively, and add the resulting equations. We obtain an equation +in which the coefficients of $y$ and~$z$ are found to be zero, while the coefficient +of~$x$ is +\[ +a_1 b_2 c_3 - +a_1 b_3 c_2 + +% +a_2 b_3 c_1 - +a_2 b_1 c_3 + +% +a_3 b_1 c_2 - +a_3 b_2 c_1. +\Tag{6} +\] +%% -----File: 109.png---Folio 103------- +Such an expression is called a \emph{determinant of the third order} and denoted +by the symbol +\[ +\begin{vmatrix} + a_1 & b_1 & c_1 \\ + a_2 & b_2 & c_2 \\ + a_3 & b_3 & c_3 +\end{vmatrix}. +\Tag{6'} +\] + +The nine numbers $a_1, \dotsc, c_3$ are called the \emph{elements} of the determinant. +In the symbol these elements lie in three (horizontal) \emph{rows}, and +also in three (vertical) \emph{columns}. Thus $a_2$, $b_2$, $c_2$ are the elements of the +second row, while the three $c$'s are the elements of the third column. +\index{Determinants!columns}% +\index{Determinants!elements}% +\index{Determinants!rows}% + +The equation (free of $y$ and~$z$), obtained above, may now be written +as +\[ +\begin{vmatrix} + a_1 & b_1 & c_1 \\ + a_2 & b_2 & c_2 \\ + a_3 & b_3 & c_3 +\end{vmatrix} x += +\begin{vmatrix} + k_1 & b_1 & c_1 \\ + k_2 & b_2 & c_2 \\ + k_3 & b_3 & c_3 +\end{vmatrix}, +\] +since the right member was the sum of the products of the expressions~\Eq{5} +by $k_1$, $k_2$, $k_3$, and hence may be derived from~\Eq{6} by replacing the +$a$'s by the~$k$'s. Thus the theorem of~§80 holds here as regards the +unknown~$x$. We shall later prove, without the laborious computations +just employed, that the theorem holds for all three unknowns. +\index{Linear equations!system|)}% + + +% [** PP: No ToC entry, matching running head] +\Section[Signs of the Terms of a Determinant] +{82.}{The Signs of the Terms of a Determinant of Order~$3$.} In the +\index{Determinants!signs of terms|(}% +six terms of our determinant~\Eq{6}, the letters $a$, $b$, $c$ were always written +in this sequence, while the subscripts are the six possible arrangements +of the numbers $1$, $2$, $3$. The first term $a_1 b_2 c_3$ shall be called the \emph{diagonal +term}, since it is the product of the elements in the main diagonal running +\index{Determinants!diagonal term}% +from the upper left-hand corner to the lower right-hand corner of the +symbol~\Eq{6'} for the determinant. The subscripts in the term $-a_1 b_3 c_2$ +are derived from those of the diagonal term by interchanging $2$ and~$3$, +and the minus sign is to be associated with the fact that an odd number +(here one) of interchanges of subscripts were used. To obtain the arrangement +$2$, $3$, $1$ of the subscripts in the term $+a_2 b_3 c_1$ from the natural order +$1$, $2$, $3$ (in the diagonal term), we may first interchange $1$ and~$2$, obtaining +the arrangement $2$, $1$, $3$, and then interchange $1$ and~$3$; an even number +(two) of interchanges of subscripts were used and the sign of the term +is plus. + +While the arrangement $1$, $3$, $2$ was obtained from $1$, $2$, $3$ by one interchange +$(2, 3)$, we may obtain it by applying in succession the three interchanges +%% -----File: 110.png---Folio 104------- +$(1, 2)$, $(1, 3)$, $(1, 2)$, and in many new ways. To show that the +number of interchanges which will produce the final arrangement $1$, $3$, $2$ +is odd in every case, note that each of the three possible interchanges, +viz., $(1, 2)$, $(1, 3)$, and~$(2, 3)$, changes the sign of the product +\[ +P = (x_1 - x_2) (x_1 - x_3) (x_2 - x_3), +\] +where the $x$'s are arbitrary variables. Thus a succession of $k$~interchanges +yields $P$ or~$-P$ according as $k$ is even or odd. Starting with the arrangement +$1$, $2$, $3$ and applying $k$~successive interchanges, suppose that we +obtain the final arrangement $1$, $3$, $2$. But if in $P$ we replace the subscripts +$1$, $2$, $3$ by $1$, $3$, $2$, respectively, i.e., if we interchange $2$ and~$3$, we obtain~$-P$. +Hence $k$ is odd. We have therefore proved the following rule +of signs: + +\begin{Thm}% +Although the arrangement $r$, $s$, $t$ of the subscripts in any term $±a_r b_s c_t$ of +the determinant may be obtained from the arrangement $1$, $2$, $3$ by various +successions of interchanges, the number of these interchanges is either always +an even number and then the sign of the term is plus or always an odd number +and then the sign of the term is minus. +\end{Thm} + + +\begin{Exercises}{} + +Apply the rule of signs to all terms of +\begin{Problems}[2] + +\item[1.] Determinant~\Eq{6}. + +\item[2.] Determinant $a_1 b_2 - a_2 b_1$. +\end{Problems} +\end{Exercises} + + +\Section[Even and Odd Arrangements] +{83.}{Number of Interchanges always Even or always Odd.} We now +extend the result in~§82 to the case of $n$~variables $x_1, \dotsc, x_n$. The +product of all of their differences $x_i - x_j$ ($i<j$) is +\begin{align*} +P = (x_1 - x_2)(x_1 - x_3) \dotsm &(x_1 - x_n) \\ + {} · (x_2 - x_3) \dotsm &(x_2 - x_n) \\ + \vdots & \\ + {} · &(x_{n-1} - x_n). +\end{align*} +Interchange any two subscripts $i$ and~$j$. The factors which involve neither +$i$ nor~$j$ are unaltered. The factor $(x_i - x_j)$ involving both is changed in +sign. The remaining factors may be paired to form the products +\[ +±(x_i - x_k)(x_j - x_k)\qquad (k = 1, \dotsc, n;\quad k \ne i,\ k \ne j). +\] +Such a product is unaltered. Hence $P$ is changed in sign. + +Suppose that an arrangement $i_1, i_2, \dotsc, i_n$ can be obtained from +%% -----File: 111.png---Folio 105------- +$1, 2, \dotsc$, $n$ by using $m$ successive interchanges and also by $t$~successive +interchanges. Make these interchanges on the subscripts in~$P$; the +resulting functions are equal to~$(-1)^m P$ and~$(-1)^t P$, respectively. But +the resulting functions are identical since either can be obtained at one +step from~$P$ by replacing the subscript~$1$ by~$i_1$, $2$ by~$i_2$, \ldots, $n$ by~$i_n$. Hence +\[ +(-1)^m P \equiv (-1)^t P, +\] +so that $m$ and~$t$ are both even or both odd. + +Thus \emph{if the same arrangement is derived from $1$, $2, \dotsc, n$ by $m$~successive +interchanges as by $t$~successive interchanges, then $m$ and~$t$ are both even or +both odd.} + + +\Section{84.}{Definition of a Determinant of Order~$n$} We define a determinant +of order~$4$ to be +\[ +\begin{vmatrix} + a_1 & b_1 & c_1 & d_1 \\ + a_2 & b_2 & c_2 & d_2 \\ + a_3 & b_3 & c_3 & d_3 \\ + a_4 & b_4 & c_4 & d_4 +\end{vmatrix} + = \sum_{(24)} ± a_q b_r c_s d_t, +\Tag{7} +\] +where $q, r, s, t$ is any one of the $24$~arrangements of $1, 2, 3, 4$, and the +sign of the corresponding term is $+$ or~$-$ according as an even or odd +number of interchanges are needed to derive this arrangement $q, r, s, t$ +from $1, 2, 3, 4$. Although different numbers of interchanges will produce +the same arrangement $q, r, s, t$ from $1, 2, 3, 4$, these numbers are all even +or all odd, as just proved, so that the sign is fully determined. + +We have seen that the analogous definitions of determinants of orders +$2$ and~$3$ lead to our earlier expressions~\Eq{2} and~\Eq{6}. + +We will have no difficulty in extending the definition to a determinant +of general order~$n$ as soon as we decide upon a proper notation for the $n^2$ +elements. The subscripts $1, 2, \dotsc, n$ may be used as before to specify +the rows. But the alphabet does not contain $n$~letters with which to +specify the columns. The use of $e', e'', \dotsc, e^{(n)}$ for this purpose would +conflict with the notation for derivatives and besides be very awkward +when exponents are used. It is customary in mathematical journals +and scientific books (a custom not always followed in introductory text +books, to the distinct disadvantage of the reader) to denote the $n$~letters +used to distinguish the $n$~columns by $e_1, e_2, \dotsc, e_n$ (or some other letter +with the same subscripts) and to prefix (but see~§85) such a subscript by +%% -----File: 112.png---Folio 106------- +the new subscript indicating the row. The symbol for the determinant +is therefore +\[ +D = \begin{vmatrix} +e_{11} & e_{12} & \cdots & e_{1n} \\ +e_{21} & e_{22} & \cdots & e_{2n} \\ +\Dots{4} \\ +e_{n1} & e_{n2} & \cdots & e_{nn} +\end{vmatrix}. +\Tag{8} +\] +By definition this shall mean the sum of the $n(n-1) \dotsm 2·1$ terms +\[ +(-1)^i e_{{i_1}1} e_{{i_2}2} \dotsm e_{{i_n}n} +\Tag{9} +\] +in which $i_1, i_2, \dotsc, i_n$ is an arrangement of $1, 2, \dotsc, n$, derived from +$1, 2, \dotsc, n$ by $i$~interchanges. Any term~\Eq{9} of the determinant~\Eq{8} is, +apart from sign, the product of $n$~factors, one and only one from each column, +and one and only one from each row. +\index{Determinants!interchanges}% + +For example, if we take $n=4$ and write $a_j, b_j, c_j, d_j$ for $e_{j1}, e_{j2}, e_{j3}, e_{j4}$, +the symbol~\Eq{8} becomes~\Eq{7} and the general term~\Eq{9} becomes the general +term $(-1)^i a_{i_1} b_{i_2} c_{i_3} d_{i_4}$ of the second member of~\Eq{7}. + + +\begin{Exercises}{Page106} + +\begin{Problems} + +\item[1.] Find the six terms involving $a_2$ in the determinant~\Eq{7}. + +\item[2.] What are the signs of $a_3b_5c_2d_1e_4$, $a_5b_4c_3d_2e_1$ in a determinant of order five? +\index{Determinants!signs of terms|)}% + +\item[3.] Show that the arrangement $4, 1, 3, 2$ may be obtained from $1, 2, 3, 4$ by use of +the two successive interchanges $(1, 4)$, $(1, 2)$, and also by use of the four successive +interchanges $(1, 4)$, $(1, 3)$, $(1, 2)$, $(2, 3)$. + +\item[4.] Write out the six terms of~\Eq{8} for $n = 3$, rearrange the factors of each term so that +the new first subscripts shall be in the order $1, 2, 3$, and verify that the resulting six +terms are those of the determinant~$D'$ in~§85 for $n = 3$. +\end{Problems} +\end{Exercises} + + +\Section{85.}{Interchange of Rows and Columns} +\begin{Thm} +Any determinant is not +altered in value if in its symbol we replace the elements of the first, second, +$\dotsc, n$th rows by the elements which formerly appeared in the same order +in the first, second, $\dotsc, n$th columns, or briefly if we interchange the corresponding +rows and columns. +\end{Thm}% No \par +For example, +\[ +\begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} += ad - bc = +\begin{vmatrix} +a & c\\ +b & d +\end{vmatrix}. +\] + +We are to prove that the determinant~$D$ given by~\Eq{8} is equal to +\[ +D' = +\begin{vmatrix} +e_{11} & e_{21} & \cdots & e_{n1} \\ +e_{12} & e_{22} & \cdots & e_{n2} \\ +\Dots{4} \\ +e_{1n} & e_{2n} & \cdots & e_{nn} +\end{vmatrix}. +\] +%% -----File: 113.png---Folio 107------- +If we give to $D'$ a more familiar aspect by writing $e_{ik} = a_{ki}$ for each element +so that, as in~\Eq{8}, the row subscript precedes instead of follows the column +subscript, the definition of the determinant in terms of the $a$'s gives $D'$ +in terms of the $e$'s as the sum of all expressions +\[ +(-1)^i e_{1{k_1}} e_{2{k_2}} \dotsm e_{n{k_n}}, +\] +% [** PP: Added comma after k_2] +in which $k_1, k_2, \dotsc, k_n$ is an arrangement of $1, 2, \dotsc, n$, derived from +the latter sequence by $i$~interchanges. + +As for the terms of~$D$, without altering~\Eq{9}, we may rearrange its factors +so that the first subscripts shall appear in the order $1, 2, \dotsc, n$, and +obtain +\[ +(-1)^i e_{1{k_1}} e_{2{k_2}} \dotsm e_{n{k_n}}. +\] +This can be done by performing in reverse order the $i$~successive interchanges +of the letters~$e$ corresponding to the $i$~successive interchanges +which were used to derive the arrangement $i_1, i_2, \dotsc, i_n$ of the first +subscripts from the arrangement $1, 2, \dotsc, n$. Thus the new second +subscripts $k_1, \dotsc, k_n$ are derived from the old second subscripts $1, \dotsc, n$ +by $i$~interchanges. The resulting signed product is therefore a term +of~$D'$. Hence $D = D'$. + + +% [** PP: Split combined ToC entry] +\Section{86.}{Interchange of Two Columns} +\begin{Thm} +A determinant is merely changed +in sign by the interchange of any two of its columns. +\end{Thm}% No \par +For example, +\[ +D = \begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} += ad-bc,\qquad +\Delta = \begin{vmatrix} +b & a \\ +d & c +\end{vmatrix} += bc - ad = -D. +\] + +Let $\Delta$ be the determinant derived from~\Eq{8} by the interchange of the +$r$th and $s$th columns. The terms of~$\Delta$ are therefore obtained from the +terms~\Eq{9} of~$D$ by interchanging $r$ and~$s$ in the series of second subscripts. +Interchange the $r$th and $s$th letters~$e$ to restore the second subscripts +to their natural order. Since the first subscripts have undergone an +interchange, the negative of any term of~$\Delta$ is a term of~$D$, and $\Delta = -D$. + + +% [** PP: No separate ToC entry] +\Section{87.}{Interchange of Two Rows} +\index{Determinants!interchanges}% +\begin{Thm} +A determinant~$D$ is merely changed +in sign by the interchange of any two rows. +\end{Thm} + +Let $\Delta$ be the determinant obtained from~$D$ by interchanging the $r$th +and $s$th rows. By interchanging the rows and columns in~$D$ and in~$\Delta$, +we get two determinants $D'$ and~$\Delta'$, either of which may be derived from +the other by the interchange of the $r$th and $s$th columns. Hence, by +§§85,~86, +\[ +\Delta = \Delta' = -D' = -D. +\] + +%% -----File: 114.png---Folio 108------- + + +% [** No ToC entry] +\Section{88.}{Two Rows or Two Columns Alike} +\begin{Thm} +A determinant is zero if any +two of its rows or any two of its columns are alike. +\end{Thm} + +For, by the interchange of the two like rows or two like columns, the +determinant is evidently unaltered, and yet must change in sign by §§86,~87. +Hence $D = -D$, $D = 0$. + + +\begin{Exercises}{} + +\begin{Problems} + +\item[1.] Prove that the equation of the straight line determined by the two distinct +points $(x_1, y_1)$ and $(x_2, y_2)$ is +\[ +\begin{vmatrix} +x & y & 1 \\ +x_1 & y_1 & 1 \\ +x_2 & y_2 & 1 +\end{vmatrix} = 0. +\] + +\item[2.] Show that +\[ +\begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} += +\begin{vmatrix} +a_2 & c_2 & b_2 \\ +a_1 & c_1 & b_1 \\ +a_3 & c_3 & b_3 +\end{vmatrix} += +\begin{vmatrix} +a_3 & a_1 & a_2 \\ +b_3 & b_1 & b_2 \\ +c_3 & c_1 & c_2 +\end{vmatrix}. +\] +\end{Problems} + +By use of the Factor Theorem~(§14) and the diagonal term, prove that +\begin{Problems} + +\item[3.] +\[ +\begin{vmatrix} +1 & 1 & 1 \\ +a & b & c \\ +a^2 & b^2 & c^2 +\end{vmatrix} = (b-a)(c-a)(c-b). +\] + +\item[4.] +\[ +\begin{vmatrix} +1 & 1 & \cdots & 1 \\ +x_1 & x_2 & \cdots & x_n \\ +x_1^2 & x_2^2 & \cdots & x_n^2 \\ +\Dots{4} \\ +x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} +\end{vmatrix} + = \prod^n_{\substack{i, j=1 \\ i>j}}(x_i-x_j). +\] + +This is known as the determinant of Vandermonde, who discussed it in~1770. The +symbol on the right means the product of all factors of the type indicated. +\index{Determinant!of Vandermonde}% + +\item[5.] Prove that a skew-symmetric determinant of odd order is zero: +\index{Determinants!skew symmetric}% +\[ +\begin{vmatrix} +\Neg 0 & \Neg a & \Neg b \\ + -a & \Neg 0 & \Neg c \\ + -b & -c & \Neg 0 +\end{vmatrix} = 0, +\qquad +\begin{vmatrix} +\Neg 0 & \Neg a & \Neg b & \Neg c & \Neg d \\ + -a & \Neg 0 & \Neg e & \Neg f & \Neg g \\ + -b & -e & \Neg 0 & \Neg h & \Neg j \\ + -c & -f & -h & \Neg 0 & \Neg k \\ + -d & -g & -j & -k & \Neg 0 +\end{vmatrix} = 0. +\] +\end{Problems} +\end{Exercises} + +%% -----File: 115.png---Folio 109------- + + +\Section{89.}{Minors} The determinant of order $n-1$ obtained by erasing +(or covering up) the row and column crossing at a given element of a +determinant of order~$n$ is called the \emph{minor} of that element. +\index{Determinants!minors}% + +\begin{Remark} +For example, in the determinant~\Eq{6'} of order~$3$, the minors of $b_1$, $b_2$, $b_3$ are respectively +\[ +B_1 = \begin{vmatrix} +a_2 & c_2 \\ +a_3 & c_3 +\end{vmatrix},\quad +% +B_2 = \begin{vmatrix} +a_1 & c_1 \\ +a_3 & c_3 +\end{vmatrix},\quad +% +B_3 = \begin{vmatrix} +a_1 & c_1 \\ +a_2 & c_2 +\end{vmatrix}. +\] +Again, \Eq{6'} is the minor of~$d_4$ in the determinant of order~$4$ given by~\Eq{7}. +\end{Remark} + + +\Section[Expansion] +{90.}{Expansion According to the Elements of a Row or Column.} In +\index{Determinants!expansion}% +\[ +D = \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix}, +\Tag{6'} +\] +denote the minor of any element by the corresponding capital letter, +so that $b_1$ has the minor~$B_1$, $b_3$ has the minor~$B_3$, etc., as in~§89. We +shall prove that +\begin{align*} +&& D &= \Neg a_1A_1 - b_1B_1 + c_1C_1, & + D &= \Neg a_1A_1 - a_2A_2 + a_3A_3, && \\ +% +&& D &= -a_2A_2 + b_2B_2 - c_2C_2, & + D &= -b_1B_1 + b_2B_2 - b_3B_3, && \\ +% +&& D &= \Neg a_3A_3 - b_3B_3 + c_3C_3, & + D &= \Neg c_1C_1 - c_2C_2 + c_3C_3. && +\end{align*} +The three relations at the left (or right) are expressed in words by saying +that a \emph{determinant~$D$ of the third order may be expanded according to the +elements of the first, second or third row} (\emph{or column}). To obtain the expansion, +we multiply each element of the row (or column) by the minor of +the element, prefix the proper sign to the product, and add the signed +products. The signs are alternately $+$ and~$-$, as in the diagram +\[ +\begin{matrix} ++ & - & + \\ +- & + & - \\ ++ & - & + +\end{matrix} +\] + +\begin{Remark} +For example, by expansion according to the second column, +\[ +\begin{vmatrix} +1 & 4 & 5 \\ +2 & 0 & 3 \\ +3 & 0 & 9 +\end{vmatrix} += -4 \begin{vmatrix} +2 & 3 \\ +3 & 9 +\end{vmatrix} += -4 × 9 = -36. +\] + +%% -----File: 116.png---Folio 110------- + +Similarly the value of the determinant~\Eq{7} of order~$4$ may be found by expansion +according to the elements of the fourth column: +\[ +-d_1 \begin{vmatrix} +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 \\ +a_4 & b_4 & c_4 +\end{vmatrix} +% ++ d_2 \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_3 & b_3 & c_3 \\ +a_4 & b_4 & c_4 +\end{vmatrix} +% +- d_3 \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_4 & b_4 & c_4 +\end{vmatrix} +% ++ d_4 \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix}. +\] +\end{Remark} + +We shall now prove that \emph{any determinant~$D$ of order~$n$ may be expanded +according to the elements of any row or any column}. + +Let $E_{ij}$ denote the minor of~$e_{ij}$ in~$D$, given by~\Eq{8}, so that $E_{ij}$ is +obtained by erasing the $i$th~row and $j$th~column of~$D$. + +(\emph{i}) We first prove that +\[ +D = e_{11}E_{11} - e_{21}E_{21} + e_{31}E_{31} - \dotsb + + (-1)^{n-1} e_{n1}E_{n1}, +\Tag{10} +\] +so that $D$ may be expanded according to the elements of its first column. +By~\Eq{9} the terms of~$D$ having the factor~$e_{11}$ are of the form +\[ +(-1)^i e_{11} e_{{i_2}2} \dotsm e_{{i_n}n}, +\] +where $1, i_2, \dotsc, i_n$ is an arrangement of $1, 2, \dotsc, n$, obtained from the +latter by $i$~interchanges, so that $i_2, \dotsc, i_n$ is an arrangement of $2, \dotsc, n$, +derived from the latter by $i$~interchanges. After removing from each +term the common factor~$e_{11}$ and adding the quotients, we obtain a sum +which, by definition, is the value of the determinant~$E_{11}$ of order~$n-1$. +Hence the terms of~$D$ having the factor~$e_{11}$ may all be combined into +$e_{11}E_{11}$, which is the first part of~\Eq{10}. + +We next prove that the terms of~$D$ having the factor~$e_{21}$ may be combined +into~$-e_{21}E_{21}$, which is the second part of~\Eq{10}. For, if $\Delta$ be the +determinant obtained from~$D$ by interchanging its first and second rows, +the result just proved shows that the terms of~$\Delta$ having the factor~$e_{21}$ +may be combined into the product of~$e_{21}$ by the minor +\[ +\begin{vmatrix} +e_{12} & e_{13} & \cdots & e_{1n} \\ +e_{32} & e_{33} & \cdots & e_{3n} \\ +\Dots{4} \\ +e_{n2} & e_{n3} & \cdots & e_{nn} +\end{vmatrix} +\] +of~$e_{21}$ in~$\Delta$. Now this minor is identical with the minor~$E_{21}$ of~$e_{21}$ in~$D$. +But $\Delta = -D$~(§87). Hence the terms of~$D$ having the factor~$e_{21}$ may be +%% -----File: 117.png---Folio 111------- +combined into $-e_{21} E_{21}$. Similarly, the terms of~$D$ having the factor~$e_{31}$ +may be combined into $e_{31} E_{31}$, etc., as in~\Eq{10}. + +(\emph{ii}) We next prove that~$D$ may be expanded according to the elements +of its $k$th~column ($k > 1$): +\[ +D = \sum_{j=1}^n (-1)^{j+k} e_{jk} E_{jk}. +\Tag{11} +\] +Consider the determinant~$\delta$ derived from~$D$ by moving the $k$th~column +over the earlier columns until it becomes the new first column. Since +this may be done by $k-1$ interchanges of adjacent columns, $\delta = (-1)^{k-1} D$. +The minors of the elements $e_{1k}, \dotsc, e_{nk}$ in the first column of~$\delta$ are evidently +the minors $E_{1k}, \dotsc, E_{nk}$ of $e_{1k}, \dotsc, e_{nk}$ in~$D$. Hence, by~\Eq{10}, +\[ +\delta = e_{1k} E_{1k} - e_{2k} E_{2k} + \dotsb + + (-1)^{n-1} e_{nk} E_{nk} + = \sum^n_{j=1} (-1)^{j+1} e_{jk} E_{jk}. +\] +Thus $D=(-1)^{k-1} \delta$ has the desired value~\Eq{11}. + +(\emph{iii}) Finally, $D$ may be expanded according to the elements of its +$k$th~row: +\[ +D = \sum^n_{j=1} (-1)^{j+k} e_{kj} E_{kj}. +\] +In fact, by Case~(\emph{ii}), the latter is the expansion of the equal determinant~$D'$ +in~§85 according to the elements of its $k$th~column. + + +\Section{91.}{Removal of Factors} +\index{Determinants!removal of factor}% +\begin{Thm} +A common factor of all of the elements of the +same row or same column of a determinant may be divided out of the elements +and placed as a factor before the new determinant. +\end{Thm} + +In other words, if all of the elements of a row or column are divided +by~$n$, the value of the determinant is divided by~$n$. For example, +\[ +\begin{vmatrix} +na_1 & nb_1 \\ + a_2 & b_2 +\end{vmatrix} += n \begin{vmatrix} +a_1 & b_1 \\ +a_2 & b_2 +\end{vmatrix},\qquad +\begin{vmatrix} +a_1 & nb_1 & c_1 \\ +a_2 & nb_2 & c_2 \\ +a_3 & nb_3 & c_3 +\end{vmatrix} += n \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix}. +\] + +Proof is made by expanding the determinants according to the elements +of the row or column in question and noting that the minors are the same +for the two determinants. Thus the second equation is equivalent to +\[ +-(nb_1)B_1 + (nb_2)B_2 - (nb_3)B_3 = n(-b_1B_1 + b_2B_2 - b_3B_3), +\] +where $B_i$ denotes the minor of~$b_i$ in the final determinant. + +%% -----File: 118.png---Folio 112------- + +% [** PP: Heading stubbornly refuses to start on a new page] +\clearpage +\begin{Exercises}{Page112} + +\begin{Problems}[2] + +\item[1.] $\ds +\begin{vmatrix} +3a & 3b & 3c \\ +5a & 5b & 5c \\ + d & e & f +\end{vmatrix} = 0$. + +\item[2.] $\ds +\begin{vmatrix} +2r & l & 3r \\ +2s & m & 3s \\ +2t & n & 3t +\end{vmatrix} = 0$. +\end{Problems} + +Expand by the shortest method and evaluate +\begin{Problems}[2] + +\item[3.] $\ds +\begin{vmatrix} +2 & 7 & 3 \\ +5 & 9 & 8 \\ +0 & 3 & 0 +\end{vmatrix}$. + +\item[4.] $\ds +\begin{vmatrix} +5 & 7 & 0 \\ +6 & 8 & 0 \\ +3 & 9 & 4 +\end{vmatrix}$. + +\ResetCols{1} + +\item[5.] $\ds +\begin{vmatrix} +a & b & c & d \\ +a^2 & b^2 & c^2 & d^2 \\ +a^3 & b^3 & c^3 & d^3 \\ +a^4 & b^4 & c^4 & d^4 +\end{vmatrix} += abcd(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$. +\end{Problems} +\end{Exercises} + + +\Section{92.}{Sum of Determinants} +\index{Determinants!sum of}% +\begin{Thm} +A determinant having $a_1+q_1$, $a_2+q_2, \dotsc$ as +the elements of a column is equal to the sum of the determinant having $a_1$, +$a_2, \dotsc$ as the elements of the corresponding column and the determinant +having $q_1$, $q_2, \dotsc$ as the elements of that column, while the elements of the +remaining columns of each determinant are the same as in the given determinant. +\end{Thm} + +For example, +\[ +\begin{vmatrix} +a_1 + q_1 & b_1 & c_1 \\ +a_2 + q_2 & b_2 & c_2 \\ +a_3 + q_3 & b_3 & c_3 +\end{vmatrix} += \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} ++ \begin{vmatrix} +q_1 & b_1 & c_1 \\ +q_2 & b_2 & c_2 \\ +q_3 & b_3 & c_3 +\end{vmatrix}. +\] + +To prove the theorem we have only to expand the three determinants +according to the elements of the column in question (the first column in +the example) and note that the minors are the same for all three determinants. +Hence $a_1 + q_1$ is multiplied by the same minor that $a_1$ and~$q_1$ +are multiplied by separately, and similarly for $a_2 + q_2$, etc. + +The similar theorem concerning the splitting of the elements of any +row into two parts is proved by expanding the three determinants according +to the elements of the row in question. For example, +\[ +\begin{vmatrix} +a+r & b+s \\ +c & d +\end{vmatrix} += \begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} ++ \begin{vmatrix} +r & s \\ +c & d +\end{vmatrix}. +\] + +%% -----File: 119.png---Folio 113------- + + +\Section{93.}{Addition of Columns or Rows} +\index{Determinants!addition of columns}% +\begin{Thm} +A determinant is not changed +in value if we add to the elements of any column the products of the corresponding +elements of another column by the same arbitrary number. +\end{Thm} + +Let $a_1$, $a_2, \dotsc$ be the elements to which we add the products of the +elements $b_1$, $b_2, \dotsc$ by~$n$. We apply~§92 with $q_1=nb_1$, $q_2=nb_2, \dotsc$. +Thus the modified determinant is equal to the sum of the initial determinant +and a determinant having $b_1$, $b_2, \dotsc$ in one column and $nb_1$, $nb_2, +\dotsc$ in another column. But~(§91) the latter determinant is equal to +the product of~$n$ by a determinant with two columns alike and hence +is zero~(§88). For example, +\[ +\begin{vmatrix} +a_1 + nb_1 & b_1 & c_1 \\ +a_2 + nb_2 & b_2 & c_2 \\ +a_3 + nb_3 & b_3 & c_3 +\end{vmatrix} += \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} ++ n\begin{vmatrix} +b_1 & b_1 & c_1 \\ +b_2 & b_2 & c_2 \\ +b_3 & b_3 & c_3 +\end{vmatrix}, +\] +and the last determinant is zero. + +Similarly, \emph{a determinant is not changed in value if we add to the elements +of any row the products of the corresponding elements of another row by the +same arbitrary number}. + +\begin{Remark} +For example, +\[ +\begin{vmatrix} +a + nc & b + nd \\ +c & d +\end{vmatrix} += \begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} ++ n \begin{vmatrix} +c & d \\ +c & d +\end{vmatrix} += \begin{vmatrix} +a & b \\ +c & d +\end{vmatrix}. +\] +\end{Remark} + + +\begin{Example} +Evaluate the first determinant below. +\[ +\begin{vmatrix} +1 & -2 & 1 \\ +1 & \Neg 2 & 3 \\ +6 & \Neg 4 & 3 +\end{vmatrix} += \begin{vmatrix} +1 & 0 & 1 \\ +1 & 8 & 3 \\ +6 & 10 & 3 +\end{vmatrix} += \begin{vmatrix} +\Neg 0 & 0 & 1 \\ + -2 & 8 & 3 \\ +\Neg 3 & 10 & 3 +\end{vmatrix} += \begin{vmatrix} + -2 & 8 \\ +\Neg 3 & 10 +\end{vmatrix} = -44. +\] +\end{Example} + +\begin{Solution} +First we add to the elements of the second column the products of the +elements of the last column by~$2$. In the resulting second determinant, we add to the +elements of the first column the products of the elements of the third column by~$-1$. +Finally, we expand the resulting third determinant according to the elements of its +first row. +\end{Solution} + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Prove that +\[ +\begin{vmatrix} +b + c & c + a & a + b \\ +b_1 + c_1 & c_1 + a_1 & a_1 + b_1 \\ +b_2 + c_2 & c_2 + a_2 & a_2 + b_2 +\end{vmatrix} += 2 \begin{vmatrix} +a & b & c \\ +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 +\end{vmatrix} +\] +\end{Problems} + +%% -----File: 120.png---Folio 114------- + +By reducing to a determinant of order~$3$, etc., prove that +\begin{Problems} +\item[2.] $\ds +\begin{vmatrix} +1 & 1 & 1 & 1 \\ +a & b & c & d \\ +a^2 & b^2 & c^2 & d^2 \\ +a^3 & b^3 & c^3 & d^3 +\end{vmatrix} += (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$. +\bigskip % [** PP: Explicit vertical space] + +\ResetCols{2} + +\item[3.] $\ds +\begin{vmatrix} +2 & -1 & \Neg 3 & -2 \\ +1 & \Neg 7 & \Neg 1 & -1 \\ +3 & \Neg 5 & -5 & \Neg 3 \\ +4 & -3 & \Neg 2 & -1 +\end{vmatrix} = -42$. + +\item[4.] $\ds +\begin{vmatrix} +1 & 1 & \phantom{1}1 & \phantom{1}1 \\ +1 & 2 & \phantom{1}3 & \phantom{1}4 \\ +1 & 3 & \phantom{1}6 & 10 \\ +1 & 4 & 10 & 20 +\end{vmatrix} = 1$. + +\end{Problems} +\end{Exercises} + + +\Section[System of $n$~Linear Equations in $n$~Unknowns] +{94.}{System of $n$~Linear Equations in $n$~Unknowns with $D \neq 0$.} In +\index{Linear equations!system|(}% +\[ +\begin{matrix} +a_{11} x_1 + a_{12} x_2 + \dotsb + a_{1n} x_n = k_1, \\ +\Dots{1} \\ +a_{n1} x_1 + a_{n2} x_2 + \dotsb + a_{nn} x_n = k_n, +\end{matrix} +\Tag{12} +\] +let $D$ denote the determinant of the coefficients of the $n$~unknowns: +\[ +D = \begin{vmatrix} +a_{11} & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +a_{n1} & a_{n2} & \cdots & a_{nn} +\end{vmatrix}. +\] +Then +{\small +\[ +Dx_1 = \begin{vmatrix} +a_{11}x_1 & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +a_{n1}x_1 & a_{n2} & \cdots & a_{nn} \\ +\end{vmatrix} += \begin{vmatrix} % [** PP: Removed commas after first column] +a_{11}x_1 + a_{12}x_2 + \dotsb + a_{1n}x_n & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +a_{n1}x_1 + a_{n2}x_2 + \dotsb + a_{nn}x_n & a_{n2} & \cdots & a_{nn} +\end{vmatrix}, +\]}% +where the second determinant was derived from the first by adding to +the elements of the first column the products of the corresponding elements +of the second column by~$x_2$, etc., and finally the products of the elements +of the last column by~$x_n$. The elements of the new first column are equal +to $k_1, \dotsc, k_n$ by~\Eq{12}. In this manner, we find that +\[ +Dx_1 = K_1,\qquad +Dx_2 = K_2,\qquad \dotsc,\qquad +Dx_n = K_n, +\Tag{13} +\] +in which $K_i$ is derived from~$D$ by substituting $k_1, \dotsc, k_n$ for the elements +$a_{1i}, \dotsc, a_{ni}$ of the $i$th~column of~$D$, whence +\[ +K_1 = \begin{vmatrix} +k_1 & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +k_n & a_{n2} & \cdots & a_{nn} +\end{vmatrix}, \dotsc\qquad +K_n = \begin{vmatrix} +a_{11} & \cdots & a_{1n-1} & k_1 \\ +\Dots{4} \\ +a_{n1} & \cdots & a_{nn-1} & k_n +\end{vmatrix}. +\] + +%% -----File: 121.png---Folio 115------- + +If $D \neq 0$, the unique values of $x_1, \dotsc, x_n$ determined by division +from~\Eq{13} actually satisfy equations~\Eq{12}. For instance, the first equation +is satisfied since +\[ +k_1D - a_{11} K_1 - a_{12} K_2 - \dotsb - a_{1n} K_n += \begin{vmatrix} +k_1 & a_{11} & a_{12} & \cdots & a_{1n} \\ +k_1 & a_{11} & a_{12} & \cdots & a_{1n} \\ +k_2 & a_{21} & a_{22} & \cdots & a_{2n} \\ +\Dots{5} \\ +k_n & a_{n1} & a_{n2} & \cdots & a_{nn} +\end{vmatrix}, +\] +as shown by expansion according to the elements of the first row; and +the determinant is zero, having two rows alike. + +\begin{Theorem} +If $D$ denotes the determinant of the coefficients of the +$n$~unknowns in a system of $n$~linear equations, the product of $D$ by any one +of the unknowns is equal to the determinant obtained from $D$ by substituting +the known terms in place of the coefficients of that unknown. If $D \neq 0$, we +obtain the unique values of the unknowns by division by~$D$. +\end{Theorem} + +We have therefore given a complete proof of the results stated and +illustrated in §80,~§81. % [** PP: Retaining instead of §§80,~81.] +Another proof is suggested in Ex.~7 below. +The theorem was discovered by induction in 1750 by G.~Cramer. + + +\begin{Exercises}{Page115} + +Solve by determinants the following systems of equations (reducing each determinant +to one having zero as the value of every element but one in a row or column, +as in the example in~§93). +\begin{Problems}[2] + +\item[1.] $\begin{System}{3} + x &+{}& y &+{}& z &= 11, \\ + 2x &-{}& 6y &-{}& z &= 0, \\ + 3x &+{}& 4y &+{}& 2z &= 0. +\end{System}$ + +\item[2.] $\begin{System}{3} + x &+{}& y &+{}& z &= 0, \\ + x &+{}& 2y &+{}& 3z &= -1, \\ + x &+{}& 3y &+{}& 6z &= 0. +\end{System}$ + +\ResetCols{2} + +\item[3.] $\begin{System}{3} + x &-{}& 2y &+{}& z &= 12, \\ + x &+{}& 2y &+{}& 3z &= 48, \\ +6x &+{}& 4y &+{}& 3z &= 84. +\end{System}$ + +% [** PP: Matching original alignment; not aligning like variables] +\item[4.] $\begin{System}{2} +3x &-{}& 2y &= 7, \\ +3y &-{}& 2z &= 6, \\ +3z &-{}& 2x &= -1. +\end{System}$ + +\ResetCols{2} + +\item[5.] $\begin{System}{4} +x &+{}& y &+{}& z &+{}& w &= 1, \\ +x &+{}& 2y &+{}& 3z &+{}& 4w &= 11, \\ +x &+{}& 3y &+{}& 6z &+{}& 10w &= 26, \\ +x &+{}& 4y &+{}& 10z &+{}& 20w &= 47. +\end{System}$ + +\item[6.] $\begin{System}{4} +2x &-{}& y &+{}& 3z &-{}& 2w &= 4, \\ + x &+{}& 7y &+{}& z &-{}& w &= 2, \\ +3x &+{}& 5y &-{}& 5z &+{}& 3w &= 0, \\ +4x &-{}& 3y &+{}& 2z &-{}& w &= 5. +\end{System}$ + +\ResetCols{1} + +\item[7.] Prove the first relation~\Eq{13} by multiplying the members of the first equation~\Eq{12} +by~$A_{11}$, those of the second equation by $-A_{21}, \dotsc$, those of the $n$th equation by +$(-1)^{n-1}A_{n1}$, and adding, where $A_{ij}$ by denotes the minor of~$a_{ij}$ in~$D$. Hint: The resulting +coefficient of~$x_2$ is the expansion, according to the elements of its first column, of a determinant +derived from $D$ by replacing $a_{11}$ by~$a_{12}$, $\dotsc$, $a_{n1}$ by~$a_{n2}$. +\end{Problems} +\end{Exercises} + +%% -----File: 122.png---Folio 116------- + + +\Section[Rank] +{95.}{Rank of a Determinant.} If we erase from a determinant~$D$ of +order~$n$ all but $r$~rows and all but $r$~columns, we obtain a determinant +of order~$r$ called an \emph{$r$-rowed minor of~$D$}. In particular, any element is +regarded as a one-rowed minor, and $D$ itself is regarded as an $n$-rowed +minor. +\index{Determinants!minors}% + +If a determinant~$D$ of order~$n$ is not zero, it is said to be of \emph{rank~$n$}. +If, for $0 < r < n$, some $r$-rowed minor of~$D$ is not zero, while every $(r+1)$-rowed +minor is zero, $D$ is said to be of \emph{rank~$r$}. It is said to be of rank +zero if every element is zero. +\index{Determinants!rank}% + +\begin{Remark} +For example, a determinant~$D$ of order~$3$ is of rank~$3$ if $D \neq 0$; of rank~$2$ if $D = 0$, +but some two-rowed minor is not zero; of rank~$1$ if every two-rowed minor is zero, +but some element is not zero. Again, every three-rowed minor of +\[ +\begin{vmatrix} +a & b & c & d \\ +e & f & g & h \\ +a & b & c & d \\ +e & f & g & h +\end{vmatrix} +\] +is zero since two pairs of its rows are alike. Hence it is of rank~$2$ if some two-rowed +minor is not zero. But it is of rank~$1$ if $a, b, c, d$ are not all zero and are proportional +to $e, f, g, h$, since all two-rowed minors are then zero. +\end{Remark} + + +\Section[System of $n$~Linear Equations in $n$~Unknowns] +{96.}{System of $n$~Linear Equations in $n$~Unknowns with $D=0$.} We +shall now discuss the equations~\Eq{12} for the troublesome case (previously +ignored) in which the determinant~$D$ of the coefficients of the unknowns +is zero. In view of~\Eq{13}, the given equations are evidently inconsistent +if any one of the determinants $K_1, \dotsc, K_n$ is not zero. But if $D$ and +these $K$'s are all zero, our former results~\Eq{13} give us no information +concerning the unknowns~$x_i$, and we resort to the following + +\begin{Theorem} +Let the determinant~$D$ of the coefficients of the unknowns +in equations~\Eq{12} be of rank~$r$, $r<n$. If the determinants~$K$ obtained from +the $(r+1)$-rowed minors of~$D$ by replacing the elements of any column by +the corresponding known terms~$k_i$ are not all zero, the equations are inconsistent. +But if these determinants $K$ are all zero, the $r$~equations involving +the elements of a non-vanishing $r$-rowed minor of $D$ determine uniquely $r$ +of the unknowns as linear functions of the remaining $n-r$ unknowns, which +are independent variables, and the expressions for these $r$~unknowns satisfy +also the remaining $n-r$ equations. +\end{Theorem} + +%% -----File: 123.png---Folio 117------- + +\begin{Remark} +Consider for example the three equations~\Eq{4} in the unknowns $x, y, z$. Five cases +arise: + +% [** PP: Not italicizing i.e.] +\begin{itemize} +\item[($\alpha$)] $D$ of rank~$3$, i.e., $D \neq 0$. + +\item[($\beta$)] $D$ of rank $2$ (i.e., $D=0$, but some two-rowed minor $\neq 0$), and +\[ +K_1 = \begin{vmatrix} +k_1 & b_1 & c_1 \\ +k_2 & b_2 & c_2 \\ +k_3 & b_3 & c_3 +\end{vmatrix},\quad +% +K_2 = \begin{vmatrix} +a_1 & k_1 & c_1 \\ +a_2 & k_2 & c_2 \\ +a_3 & k_3 & c_3 +\end{vmatrix},\quad +% +K_3 = \begin{vmatrix} +a_1 & b_1 & k_1 \\ +a_2 & b_2 & k_2 \\ +a_3 & b_3 & k_3 +\end{vmatrix} +\] +not all zero. + +\item[($\gamma$)] $D$ of rank~$2$ and $K_1$, $K_2$, $K_3$ all zero. + +\item[($\delta$)] $D$ of rank~$1$ (i.e., every two-rowed minor $= 0$, but some element $\neq 0$), and +\[ +\begin{vmatrix} +a_i & k_i \\ +a_j & k_j +\end{vmatrix},\qquad +% +\begin{vmatrix} +b_i & k_i \\ +b_j & k_j +\end{vmatrix},\qquad +\begin{vmatrix} +c_i & k_i \\ +c_j & k_j +\end{vmatrix}\qquad \text{($i$, $j$ chosen from $1$, $2$, $3$)} +\] +not all zero; there are nine such determinants~$K$. + +\item[($\epsilon$)] $D$ of rank~$1$, and all nine of the two-rowed determinants~$K$ zero. +\end{itemize} + +In case~($\alpha$) the equations have a single set of solutions~(§94). In cases ($\beta$) and~($\delta$) +there is no set of solutions. For ($\beta$) the proof follows from~\Eq{13}. In case~($\gamma$) one +of the equations is a linear combination of the other two; for example, if $a_1b_2 - a_2b_1 \neq 0$, +the first two equations determine $x$ and~$y$ as linear functions of~$z$ (as shown by transposing +the terms in~$z$ and solving the resulting equations for $x$ and~$y$), and the resulting +values of $x$ and~$y$ satisfy the third equation identically as to~$z$. Finally, in case~($\epsilon$), +two of the equations are obtained by multiplying the remaining one by constants. + +The reader acquainted with the elements of solid analytic geometry will see that +the planes represented by the three equations have the following relations: +\begin{itemize} + +\item[($\alpha$)] The three planes intersect in a single point. + +\item[($\beta$)] Two of the planes intersect in a line parallel to the third plane. + +\item[($\gamma$)] The three planes intersect in a common line. + +\item[($\delta$)] The three planes are parallel and not all coincident. + +\item[($\epsilon$)] The three planes coincide. + +\end{itemize} +\end{Remark} + +The remarks preceding our theorem furnish an illustration (the case +$r=n-1$) of the following + +\begin{Lemma}[1.] If every $(r+1)$-rowed minor~$M$ formed from certain $r+1$ +rows of~$D$ is zero, the corresponding $r+1$ equations~\Eq{12} are inconsistent +provided there is a non-vanishing determinant~$K$ formed from any~$M$ +by replacing the elements of any column by the corresponding known +terms~$k_i$. +\end{Lemma} + +For concreteness,\footnote + {All other cases may be reduced to this one by rearranging the $n$~equations and +relabelling the unknowns (replacing $x_3$ by the new~$x_1$, for example).} +let the rows in question be the first $r+1$ and let +%% -----File: 124.png---Folio 118------- +\[ +K = \begin{vmatrix} +a_{11} & \cdots & a_{1r} & k_1 \\ +\Dots{4} \\ +a_{r+11} & \cdots & a_{r+1r} & k_{r+1} +\end{vmatrix} \neq 0. +\] +Let $d_1, \dotsc, d_{r+1}$ be the minors of $k_1, \dotsc, k_{r+1}$ in~$K$. Multiply the +first $r+1$ equations~\Eq{12} by $d_1$, $-d_2, \dotsc, (-1)^rd_{r+1}$, respectively, and +add. The right member of the resulting equation is the expansion of~$±K$. +The coefficient of~$x_s$ is the expansion of +\[ +±\begin{vmatrix} +a_{11} & \cdots & a_{1r} & a_{1s} \\ +\Dots{4} \\ +a_{r+11} & \cdots & a_{r+1r} & a_{r+1s} +\end{vmatrix} +\] +and is zero, being an~$M$ if $s>r$, and having two columns identical if $s \leqq r$. +Hence $0 = ±K$. Thus if $K \neq 0$, the equations are inconsistent. + +\begin{Lemma}[2.] If all of the determinants $M$ and~$K$ in Lemma~1 are zero, +but an $r$-rowed minor of an~$M$ is not zero, one of the corresponding $r+1$ +equations is a linear combination of the remaining $r$ equations. +\end{Lemma} + +As before let the $r+1$ rows in question be the first $r+1$. Let the +non-vanishing $r$-rowed minor be +\[ +d_{r+1} = \begin{vmatrix} +a_{11} & \cdots & a_{1r} \\ +\Dots{3} \\ +a_{r1} & \cdots & a_{rr} +\end{vmatrix} \neq 0. +\Tag{14} +\] +Let the functions obtained by transposing the terms~$k_i$ in~\Eq{12} be +\[ +L_i \equiv a_{i1} x_1 + a_{i2} x_2 + \dotsb + a_{in} x_n - k_i. +\] +By the multiplication made in the proof of Lemma~1, +\[ +d_1L_1 - d_2L_2 + \dotsb + (-1)^rd_{r+1}L_{r+1} = \mp K = 0. +\] +Hence $L_{r+1}$ is a linear combination of $L_1, \dotsc, L_r$. + +The first part of the theorem is true by Lemma~1. The second part +is readily proved by means of Lemma~2. Let~\Eq{14} be the non-vanishing +$r$-rowed minor of~$D$. For $s>r$, the $s$th equation is a linear combination +of the first $r$~equations, and hence is satisfied by any set of solutions of +the latter. In the latter transpose the terms involving $x_{r+1}, \dotsc, x_n$. +Since the determinant of the coefficients of $x_1, \dotsc, x_r$ is not zero,~§94 +shows that $x_1, \dotsc, x_r$ are uniquely determined linear functions of +$x_{r+1}, \dotsc, x_n$ (which enter from the new right members). + +%% -----File: 125.png---Folio 119------- + + +\begin{Exercises}{Page119} + +Apply the theorem to the following four systems of equations and check the conclusions: +\begin{Problems}[2] + +\item[1.] $\begin{System}{3} +2x&+{}& y&+{}& 3z &= 1, \\ +4x&+{}& 2y&-{}& z &= -3, \\ +2x&+{}& y&-{}& 4z &= -4. +\end{System}$ + +\item[2.] $\begin{System}{3} +2x&+{}& y&+{}& 3z &= 1, \\ +4x&+{}& 2y&-{}& z &= 3, \\ +2x&+{}& y&-{}& 4z &= 4. +\end{System}$ + +\ResetCols{2} + +\item[3.] $\begin{System}{3} + x&-{}& 3y&+{}& 4z &= 1, \\ +4x&-{}& 12y&+{}& 16z &= 3, \\ +3x&-{}& 9y&+{}& 12z &= 3. +\end{System}$ + +\item[4.] $\begin{System}{3} + x&-{}& 3y&+{}& 4z &= 1, \\ +4x&-{}& 12y&+{}& 16z &= 4, \\ +3x&-{}& 9y&+{}& 12z &= 3. +\end{System}$ + +\ResetCols{1} + +\item[5.] Discuss the system +\[ +\begin{System}{3} +ax&+{}& y&+{}& z &= a-3, \\ + x&+{}& ay&+{}& z &= -2, \\ + x&+{}& y&+{}& az &= -2, +\end{System} +\] +when (\emph{i})~$a = 1$; (\emph{ii})~$a = -2$; (\emph{iii})~$a \neq 1$, $-2$, obtaining the simplest forms of the +unknowns. + +\item[6.] Discuss the system +\[ +\begin{System}{3} + x &+{}& y&+{}& z &= 1, \\ +ax &+{}& by&+{}& cz &= k, \\ +a^2x&+{}& b^2y&+{}& c^2z &= k^2, +\end{System} +\] +when (\emph{i})~$a$, $b$, $c$ are distinct; (\emph{ii}) $a = b \neq c$; (\emph{iii}) $a = b = c$. +\end{Problems} +\end{Exercises} + + +\Section[Homogeneous Equations]% [** PP: Allow line to break in ... list] +{97.}{Homogeneous Linear Equations.} When the known terms $k_1$, $\dotsc, +k_n$ in~\Eq{12} are all zero, the equations are called \emph{homogeneous}. The determinants~$K$ +are now all zero, so that the $n$~homogeneous equations are never +inconsistent. This is also evident from the fact that they have the set +of solutions $x_1 = 0, \dotsc, x_n = 0$. By~\Eq{13}, there is no further set of solutions +if $D \neq 0$. If $D = 0$, there are further sets of solutions. This is +shown by the theorem of~§96 which now takes the following simpler form. + +\begin{Thm}% +If the determinant~$D$ of the coefficients of $n$~linear homogeneous equations +in $n$~unknowns is of rank~$r$, $r<n$, the $r$~equations involving the elements of a +non-vanishing $r$-rowed minor of~$D$ determine uniquely $r$ of the unknowns as +linear functions of the remaining $n-r$ unknowns, which are independent variables, +and the expressions for these $r$~unknowns satisfy also the remaining +$n-r$ equations. +\end{Thm} + +The particular case mentioned is the much used theorem: + +\begin{Thm}% +\index{Linear equations!homogeneous}% +A necessary and sufficient condition that $n$~linear homogeneous equations +in $n$~unknowns shall have a set of solutions, other than the trivial one in which +each unknown is zero, is that the determinant of the coefficients be zero. +\end{Thm} + +%% -----File: 126.png---Folio 120------- + + +\begin{Exercises}{Page120} + +Discuss the following systems of equations: +\begin{Problems}[2] +\item[1.] $\begin{System}{3} +x &+{}& y &+{}& 3z &= 0,\\ +x &+{}& 2y &+{}& 2z &= 0,\\ +x &+{}& 5y &-{}& z &= 0. +\end{System}$ + +\item[2.] $\begin{System}{3} + 2x &-{}& y &+{}& 4z &= 0,\\ + x &+{}& 3y &-{}& 2z &= 0,\\ + x &-{}& 11y &+{}& 14z &= 0. +\end{System}$ + +\ResetCols{2} + +\item[3.] $\begin{System}{3} + x &-{}& 3y &+{}& 4z &= 0,\\ +4x &-{}& 12y &+{}& 16z &= 0,\\ +3x &-{}& 9y &+{}& 12z &= 0. +\end{System}$ + +\item[4.] $\begin{System}{4} +6x &+{}& 4y &+{}& 3z &-{}& 84w &= 0,\\ + x &+{}& 2y &+{}& 3z &-{}& 48w &= 0,\\ + x &-{}& 2y &+{}& z &-{}& 12w &= 0,\\ +4x &+{}& 4y &-{}& z &-{}& 24w &= 0. +\end{System}$ + +\ResetCols{1} + +\item[5.] $\begin{System}{4} +2x &+{}& 3y &-{}& 4z &+{}& 5w &= 0,\\ +3x &+{}& 5y &-{}& z &+{}& 2w &= 0,\\ +7x &+{}& 11y &-{}& 9z &+{}& 12w &= 0,\\ +3x &+{}& 4y &-{}& 11z &+{}& 13w &= 0. +\end{System}$ +\end{Problems} +\end{Exercises} + + +% [** PP: ToC entry reads System of $m$~Linear Equations in $n$~Unknowns, +% Matrix and Augmented Matrix] +\Section{98.}{System of $m$~Linear Equations in $n$~Unknowns} The case $m<n$ +may be treated by means of the lemmas in~§96. If $m>n$, we select +any $n$ of the equations and apply to them the theorems of §§94,~96. If +they are found to be inconsistent, the entire system is evidently inconsistent. +But if the $n$~equations are consistent, and if $r$ is the rank of the +determinant of their coefficients, we obtain $r$ of the unknowns expressed +as linear functions of the remaining $n-r$ unknowns. Substituting these +values of these $r$~unknowns in the remaining equations, we obtain a +system of $m-n$ linear equations in $n-r$ unknowns. Treating this system +in the same manner, we ultimately either find that the proposed +$m$~equations are consistent and obtain the general set of solutions, +or find that they are inconsistent. To decide in advance whether the +former or latter of these cases will arise, we have only to find the maximum +order~$r$ of a non-vanishing $r$-rowed determinant formed from the +coefficients of the unknowns, taken in the regular order in which they +occur in the equations, and ascertain whether or not the corresponding +($r+1$)-rowed determinants~$K$, formed as in~§96, are all zero. + +The last result may be expressed simply by employing the terminology +of matrices. The system of coefficients of the unknowns in any set of +linear equations +\[ +\begin{matrix} +a_{11} x_1 + \dotsb + a_{1n} x_n = k_1, \\ +\Dots{1} \\ +a_{m1} x_1 + \dotsb + a_{mn} x_n = k_m, +\end{matrix} +\Tag{15} +\] +arranged as they occur in the equations, is called the \emph{matrix} of the coefficients, +and is denoted by +\index{Matrix}% +\[ +A = \begin{pmatrix} +a_{11} & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +a_{m1} & a_{m2} & \cdots & a_{mn} +\end{pmatrix}. +\] +%% -----File: 127.png---Folio 121------- +By annexing the column composed of the known terms $k_i$ we obtain +the so-called \emph{augmented matrix} +\index{Matrix!augmented}% +\[ +B = \begin{pmatrix} +a_{11} & a_{12} & \cdots & a_{1n} & k_1\\ +\Dots{5}\\ +a_{m1} & a_{m2} & \cdots & a_{mn} & k_m +\end{pmatrix}. +\] + +The definitions of an $r$-rowed minor (determinant) of a matrix and of +the rank of a matrix are entirely analogous to the definitions in~§95. + +In view of Lemma~1 in~§96, our equations~\Eq{15} are inconsistent if +$B$ is of rank $r+1$ and $A$ is of rank $\leqq r$. By Lemma~2, if $A$ and~$B$ are both +of rank~$r$, all of our equations are linear combinations of~$r$ of them. Noting +also that the rank~$r$ of~$A$ cannot exceed the rank of~$B$, since every minor +of~$A$ is a minor of~$B$, and hence a non-vanishing $r$-rowed minor of~$A$ is a +minor of~$B$, so that the rank of~$B$ is not less than~$r$, we have the following + +\begin{Theorem} +A system of $m$~linear equations in $n$~unknowns is consistent +if and only if the rank of the matrix of the coefficients of the unknowns is +equal to the rank of the augmented matrix. If the rank of both matrices is~$r$, +certain $r$ of the equations determine uniquely $r$ of the unknowns as linear +functions of the remaining $n - r$ unknowns, which are independent variables, +and the expressions for these $r$~unknowns satisfy also the remaining +$m - r$ equations. +\index{Determinants!rank}% +\end{Theorem} + +When $m = n+1$, $B$ has an $m$-rowed minor called the \emph{determinant of +the square matrix~$B$}. If this determinant is not zero, $B$ is of rank~$m$. +Since $A$ has no $m$-rowed minor, its rank is less than~$m$. Hence we obtain +the + +\begin{Corollary} +Any system of $n+1$ linear equations in $n$~unknowns is +inconsistent if the determinant of the augmented matrix is not zero. +\end{Corollary} + + +\begin{Exercises}{Page121} + +Discuss the following systems of equations: +\begin{Problems}[2] +\item[1.] $\begin{System}{3} + 2x &+{}& y &+{}& 3z &= 1,\\ + 4x &+{}& 2y &-{}& z &= -3,\\ + 2x &+{}& y &-{}& 4z &= -4,\\ +10x &+{}& 5y &-{}& 6z &= -10. +\end{System}$ + +\item[2.] $\begin{System}{3} +2x &-{}& y &+{}& 3z &= 2,\\ + x &+{}& 7y &+{}& z &= 1,\\ +3x &+{}& 5y &-{}& 5z &= a,\\ +4x &-{}& 3y &+{}& 2z &= 1. +\end{System}$ + +\ResetCols{2} + +\item[3.] $\begin{System}{3} +4x &-{}& y &+{}& z &= 5,\\ +2x &-{}& 3y &+{}& 5z &= 1,\\ + x &+{}& y &-{}& 2z &= 2,\\ +5x & & &-{}& z &= 2. +\end{System}$ + +\item[4.] $\begin{System}{2} + 4x &-{}& 5y &= 2,\\ + 2x &+{}& 3y &= 12,\\ +10x &-{}& 7y &= 16. +\end{System}$ + +\ResetCols{1} + +\item[5.] Prove the Corollary by multiplying the known terms by $x_{n+1}=1$ and applying~§97 +with $n$ replaced by $n+1$. + +\item[6.] Prove that if the matrix of the coefficients of any system of linear homogeneous +\index{Linear equations!homogeneous}% +equations in $n$~unknowns is of rank~$r$, the values of certain $n-r$ of the unknowns may be +%% -----File: 128.png---Folio 122------- +assigned at pleasure and the others will then be uniquely determined and satisfy all of the equations. +\end{Problems} +\end{Exercises} +\index{Linear equations!system|)}% [** PP: Original page range is 114--121] + + +\Section{99.}{Complementary Minors} The determinant +\index{Determinants!complementary minors}% +\[ +D = \begin{vmatrix} +a_1 & b_1 & c_1 & d_1 \\ +a_2 & b_2 & c_2 & d_2 \\ +a_3 & b_3 & c_3 & d_3 \\ +a_4 & b_4 & c_4 & d_4 +\end{vmatrix} +\Tag{16} +\] +is said to have the \emph{two-rowed complementary minors} +\[ +M = \begin{vmatrix} +a_1 & b_1 \\ +a_3 & b_3 +\end{vmatrix},\qquad +M' = \begin{vmatrix} +c_2 & d_2 \\ +c_4 & d_4 +\end{vmatrix}, +\] +since either is obtained by erasing from~$D$ all the rows and columns having +an element which occurs in the other. + +In general, if we erase from a determinant~$D$ of order~$n$ all but $r$~rows +and all but $r$~columns, we obtain a determinant~$M$ of order~$r$ called an +$r$-rowed minor of~$D$. But if we had erased from~$D$ the $r$~rows and $r$~columns +previously kept, we would have obtained an ($n-r$)-rowed minor of~$D$ +called the \emph{minor complementary to~$M$}. In particular, any element is +regarded as a one-rowed minor and is complementary to its minor (of +order $n-1$). + + +% [** PP: Next two paragraphs have single ToC entry, Laplace's Development] +\Section{100.}{Laplace's Development by Columns} +\index{Determinants!Laplace's development|(}% +\begin{Thm} +Any determinant~$D$ is +equal to the sum of all the signed products $±MM'$, where $M$ is an $r$-rowed +minor having its elements in the first $r$~columns of~$D$, and $M'$ is the minor +complementary to~$M$, while the sign is $+$ or~$-$ according as an even or odd +number of interchanges of rows of~$D$ will bring $M$ into the position occupied +by the minor~$M_1$ whose elements lie in the first $r$~rows and first $r$~columns +of~$D$. +\end{Thm} + +\begin{Remark} +For $r = 1$, this development becomes the known expansion of $D$ according to the +elements of the first column~(§90); here $M_1 = e_{11}$. + +If $r=2$ and $D$ is the determinant~\Eq{16}, +\begin{align*} +D &= \begin{vmatrix} +a_1 & b_1 \\ +a_2 & b_2 +\end{vmatrix} · \begin{vmatrix} +c_3 & d_3 \\ +c_4 & d_4 +\end{vmatrix} - \begin{vmatrix} +a_1 & b_1 \\ +a_3 & b_3 +\end{vmatrix} · \begin{vmatrix} +c_2 & d_2 \\ +c_4 & d_4 +\end{vmatrix} + \begin{vmatrix} +a_1 & b_1 \\ +a_4 & b_4 +\end{vmatrix} · \begin{vmatrix} +c_2 & d_2 \\ +c_3 & d_3 +\end{vmatrix} \\[1ex] + &{}+ \begin{vmatrix} +a_2 & b_2 \\ +a_3 & b_3 +\end{vmatrix} · \begin{vmatrix} +c_1 & d_1 \\ +c_4 & d_4 +\end{vmatrix} - \begin{vmatrix} +a_2 & b_2 \\ +a_4 & b_4 +\end{vmatrix} · \begin{vmatrix} +c_1 & d_1 \\ +c_3 & d_3 +\end{vmatrix} + \begin{vmatrix} +a_3 & b_3 \\ +a_4 & b_4 +\end{vmatrix} · \begin{vmatrix} +c_1 & d_1 \\ +c_2 & d_2 +\end{vmatrix}. +\end{align*} +%% -----File: 129.png---Folio 123------- +The first product in the development is $M_1M_1'$; the second product is $-MM'$ (in the +notations of~§99), and the sign is minus since the interchange of the second and third +rows of $D$ brings this $M$ into the position of~$M_1$. The sign of the third product in +the development is plus since two interchanges of rows of~$D$ bring the first factor +into the position of~$M_1$. +\end{Remark} + +If $D$ is the determinant~\Eq{8}, then +\[ +M_1 = \begin{vmatrix} +e_{11} & \cdots & e_{1r} \\ +\Dots{3} \\ +e_{r1} & \cdots & e_{rr} +\end{vmatrix},\qquad +M_1' = \begin{vmatrix} +e_{r+1 r+1} & \cdots & e_{r+1 n}\\ +\Dots{3} \\ +e_{n r+1} & \cdots & e_{nn} +\end{vmatrix}. +\] + +Any term of the product $M_1M_1'$ is of the type +\[ + (-1)^i e_{{i_1}1} e_{{i_2}2} \dotsm e_{{i_r}r} +· (-1)^j e_{i_{r+1}{r+1}} \dotsm e_{{i_n}n}, +\Tag{17} +\] +where $i_1, \dotsc, i_r$ is an arrangement of $1, \dotsc, r$ derived from $1, \dotsc, r$ +by $i$~interchanges, while $i_{r+1}, \dotsc, i_n$ is an arrangement of $r+1, \dotsc, n$ +derived by $j$~interchanges. Hence $i_1, \dotsc, i_n$ is an arrangement of +$1, \dotsc, n$ derived by $i+j$ interchanges, so that the product~\Eq{17} is a term +of $D$ with the proper sign. + +It now follows from~§87 that any term of any of the products~$±MM'$ +mentioned in the theorem is a term of~$D$. Clearly we do not obtain twice +in this manner the same term of~$D$. + +Conversely, any term~$t$ of~$D$ occurs in one of the products~$±MM'$. +Indeed, $t$ contains as factors $r$~elements from the first $r$~columns of~$D$, +no two being in the same row, and the product of these is, except perhaps +as to sign, a term of some minor~$M$. Thus $t$ is a term of $MM'$ or +of~$-MM'$. In view of the earlier discussion, the sign of $t$ is that of the +corresponding term in $±MM'$, where the latter sign is given by the +theorem. + + +% [** PP: No separate ToC entry] +\Section{101.}{Laplace's Development by Rows} There is a Laplace development +of $D$ in which the $r$-rowed minors~$M$ have their elements in the first +$r$~rows of~$D$, instead of in the first $r$~columns as in~§100. To prove this, +we have only to apply~§100 to the equal determinant obtained by interchanging +the rows and columns of~$D$. + +There are more general (but less used) Laplace developments in which +the $r$-rowed minors~$M$ have their elements in any chosen $r$~columns (or +rows) of~$D$. It is simpler to apply the earlier developments to the determinant +$±D$ having the elements of the chosen $r$~columns (or rows) in the +new first $r$~columns (or rows). +\index{Determinants!Laplace's development|)}% + +%% -----File: 130.png---Folio 124------- + + +\begin{Exercises}{} +\begin{Problems} +\item[1.] Prove that +\[ +\begin{vmatrix} +a & b & c & d \\ +e & f & g & h \\ +0 & 0 & j & k \\ +0 & 0 & l & m +\end{vmatrix} = \begin{vmatrix} +a & b \\ +e & f +\end{vmatrix} · \begin{vmatrix} +j & k \\ +l & m +\end{vmatrix}. +\] + +\item[2.] By employing $2$-rowed minors from the first two rows, show that +\[ +\frac{1}{2}\begin{vmatrix} +a & b & c & d \\ +e & f & g & h \\ +a & b & c & d \\ +e & f & g & h +\end{vmatrix} = \begin{vmatrix} +a & b \\ +e & f +\end{vmatrix} · \begin{vmatrix} +c & d \\ +g & h +\end{vmatrix} - \begin{vmatrix} +a & c \\ +e & g +\end{vmatrix} · \begin{vmatrix} +b & d \\ +f & h +\end{vmatrix} + \begin{vmatrix} +a & d \\ +e & h +\end{vmatrix} · \begin{vmatrix} +b & c \\ +f & g +\end{vmatrix} = 0. +\] + +\item[3.] By employing $2$-rowed minors from the first two columns of the $4$-rowed determinant +in Ex.~2, show that the products in Laplace's development cancel. +\end{Problems} +\end{Exercises} + + +\Section{102.}{Product of Determinants} +\index{Determinants!product of}% +\begin{Thm} +The product of two determinants of +the same order is equal to a determinant of like order in which the element +of the $r$th row and $c$th column is the sum of the products of the elements of +the $r$th row of the first determinant by the corresponding elements of the $c$th +column of the second determinant. +\end{Thm} + +\begin{Remark} +For example, +\[ +\begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} · \begin{vmatrix} +e & f \\ +g & h +\end{vmatrix} = \begin{vmatrix} +ae + bg & af + bh \\ +ce + dg & cf + dh +\end{vmatrix}. +\Tag{18} +\] +\end{Remark} + +While for brevity we shall give the proof for determinants of order~$3$, +the method is seen to apply to determinants of any order. By Laplace's +development with $r = 3$~(§101), we have +\[ +\begin{vmatrix} +\Neg a_1 & \Neg b_1 & \Neg c_1 & 0 & 0 & 0 \\ +\Neg a_2 & \Neg b_2 & \Neg c_2 & 0 & 0 & 0 \\ +\Neg a_3 & \Neg b_3 & \Neg c_3 & 0 & 0 & 0 \\ + -1 & \Neg 0 & \Neg 0 & e_1 & f_1 & g_1 \\ +\Neg 0 & -1 & \Neg 0 & e_2 & f_2 & g_2 \\ +\Neg 0 & \Neg 0 & -1 & e_3 & f_3 & g_3 +\end{vmatrix} = \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} · \begin{vmatrix} +e_1 & f_1 & g_1 \\ +e_2 & f_2 & g_2 \\ +e_3 & f_3 & g_3 +\end{vmatrix}. +\Tag{19} +\] + +%% -----File: 131.png---Folio 125------- + +In the determinant of order~$6$, add to the elements of the fourth, fifth, +and sixth columns the products of the elements of the first column by +$e_1$, $f_1$, $g_1$, respectively (and hence introduce zeros in place of the former +elements $e_1$, $f_1$, $g_1$). Next, add to the elements of the fourth, fifth, and +sixth columns the products of the elements of the second column by +$e_2$, $f_2$, $g_2$, respectively. Finally, add to the elements of the fourth, fifth, +and sixth columns the products of the elements of the third column by +$e_3$, $f_3$, $g_3$, respectively. The new determinant is +\[ +\begin{vmatrix} +\Neg a_1 & \Neg b_1 & \Neg c_1 & + a_1e_1+b_1e_2+c_1e_3 & a_1f_1+b_1f_2+c_1f_3 & a_1g_1+b_1g_2+c_1g_3 \\ +% +\Neg a_2 & \Neg b_2 & \Neg c_2 & + a_2e_1+b_2e_2+c_2e_3 & a_2f_1+b_2f_2+c_2f_3 & a_2g_1+b_2g_2+c_2g_3 \\ +% +\Neg a_3 & \Neg b_3 & \Neg c_3 & + a_3e_1+b_3e_2+c_3e_3 & a_3f_1+b_3f_2+c_3f_3 & a_3g_1+b_3g_2+c_3g_3 \\ +% + -1 & \Neg 0 & \Neg 0 & 0 & 0 & 0 \\ +\Neg 0 & -1 & \Neg 0 & 0 & 0 & 0 \\ +\Neg 0 & \Neg 0 & -1 & 0 & 0 & 0 +\end{vmatrix}. +\] +By Laplace's development (or by expansion according to the elements of +the last row, etc.), this is equal to the $3$-rowed minor whose elements +are the long sums. Hence this minor is equal to the product in the right +member of~\Eq{19}. + + +\begin{Exercises}{} + +\begin{Problems} + +\item[1.] Prove~\Eq{18} by means of~§92. + +\item[2.] Prove that, if $s_i = \alpha^i + \beta^i + \gamma^i$, +\[ +\begin{vmatrix} + 1 & 1 & 1 \\ +\alpha & \beta & \gamma \\ +\alpha^2 & \beta^2 & \gamma^2 +\end{vmatrix} · \begin{vmatrix} +1 & \alpha & \alpha^2 \\ +1 & \beta & \beta^2 \\ +1 & \gamma & \gamma^2 +\end{vmatrix} = \begin{vmatrix} +3 & s_1 & s_2 \\ +s_1 & s_2 & s_3 \\ +s_2 & s_3 & s_4 +\end{vmatrix}. +\] + +\item[3.] If $A_i$, $B_i$, $C_i$ are the minors of $a_i$, $b_i$, $c_i$ in the determinant~$D$ defined by the second +factor below, prove that +\[ +\begin{vmatrix} +\Neg A_1 & -A_2 & \Neg A_3 \\ + -B_1 & \Neg B_2 & -B_3 \\ +\Neg C_1 & -C_2 & \Neg C_3 +\end{vmatrix} · \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} = \begin{vmatrix} +D & 0 & 0 \\ +0 & D & 0 \\ +0 & 0 & D +\end{vmatrix}. +\] +Hence the first factor is equal to $D^2$ if $D \neq 0$. + +%% -----File: 132.png---Folio 126------- + +\item[4.] Express $(a^2 + b^2 + c^2 + d^2)(e^2 + f^2 + g^2 + h^2)$ as a sum of four squares by writing +\index{Sum of!four squares}% +\[ +\begin{vmatrix} +\Neg a + bi & c + di \\ + -c + di & a - bi +\end{vmatrix} · \begin{vmatrix} +\Neg e + fi & g + hi \\ + -g + hi & e - fi +\end{vmatrix} +\] +as a determinant of order~$2$ similar to each factor. Hint: If $k'$ denotes the conjugate +of the complex number~$k$, each of the three determinants is of the form +\[ +\begin{vmatrix} +\Neg k & l \\ + -l' & k' +\end{vmatrix}. +\] +\end{Problems} +\end{Exercises} + + +\begin{Exercises}[MISCELLANEOUS~]{Page126} + +\begin{Problems} + +\item[1.] Solve +\[ +\begin{System}{3} + ax &+{}& by &+{}& cz &= k,\\ +a^2x &+{}& b^2y &+{}& c^2z &= k^2,\\ +a^4x &+{}& b^4y &+{}& c^4z &= k^4 +\end{System} +\] +by determinants for~$x$, treating all cases. + +\item[2.] In three linear homogeneous equations in four unknowns, prove that the values +of the unknowns are proportional to four determinants of order~$3$ formed from the +coefficients. +\end{Problems} + +Factor the following determinants: +\begin{Problems}[2] + +\item[3.] $\ds +\begin{vmatrix} +1 & a & bc \\ +1 & b & ca \\ +1 & c & ab +\end{vmatrix}$. + +\item[4.] $\ds +\begin{vmatrix} +x & x^2 & yz \\ +y & y^2 & xz \\ +z & z^2 & xy +\end{vmatrix} = \begin{vmatrix} +x^2 & x^3 & 1 \\ +y^2 & y^3 & 1 \\ +z^2 & z^3 & 1 +\end{vmatrix}$. + +\ResetCols{1} + +\item[5.] +\[ +\begin{vmatrix} +a & b & c \\ +c & a & b \\ +b & c & a +\end{vmatrix} = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega), +\] +where $\omega$ is an imaginary cube root of unity. + +\ResetCols{2} + +\item[6.] $\ds +\begin{vmatrix} +a & b & c & d \\ +b & a & d & c \\ +c & d & a & b \\ +d & c & b & a +\end{vmatrix}$. + +\item[7.] $\ds +\begin{vmatrix} +a & b & c & d \\ +d & a & b & c \\ +c & d & a & b \\ +b & c & d & a +\end{vmatrix}$. + +\ResetCols{1} + +% [** PP: Using array for better surrounding space in wide matrix] +\item[8.] If the points $(x_1, y_1), \dotsc, (x_4, y_4)$ lie on a circle, prove that +\[ +\left|\begin{array}{cccc} +x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ +\Dots{4} \\ +x_4^2 + y_4^2 & x_4 & y_4 & 1 +\end{array}\right| = 0. +\] + +%% -----File: 133.png---Folio 127------- + +\item[9.] Prove that +\begin{gather*} +\begin{vmatrix} +aa' + bb' + cc' & ea' + fb' + gc' \\ +ae' + bf' + cg' & ee' + ff' + gg' +\end{vmatrix} \\ +% +{} = \begin{vmatrix} +a & b \\ +e & f +\end{vmatrix} · \begin{vmatrix} +a' & b' \\ +e' & f' +\end{vmatrix} + \begin{vmatrix} +a & c \\ +e & g +\end{vmatrix} · \begin{vmatrix} +a' & c' \\ +e' & g' +\end{vmatrix} + \begin{vmatrix} +b & c \\ +f & g +\end{vmatrix} · \begin{vmatrix} +b' & c' \\ +f' & g' +\end{vmatrix}. +\end{gather*} + +\item[10.] Prove that the cubic equation +\index{Cubic equation}% +\[ +D(x) \equiv \begin{vmatrix} +a-x & b & c \\ +b & f-x & g \\ +c & g & h-x +\end{vmatrix} = 0 +\] +has only real roots. Hints: +\begin{gather*} +D(x) · D(-x) = \left|\begin{array}{lll} +a^2+b^2+c^2-x^2 & ab+bf+cg & ac+bg+ch \\ +ab+bf+cg & b^2+f^2+g^2-x^2 & bc+fg+gh \\ +ac+bg+ch & bc+fg+gh & c^2+g^2+h^2-x^2 +\end{array}\right| \\ +% +{} = -x^6+x^4(a^2+f^2+h^2+2b^2+2c^2+2g^2) - x^2(D_1+D_2+D_3)+ D^2(0), +\end{gather*} +where $D_3$ denotes the first determinant in Ex.~9 with all accents removed and with +$e = b$, while $D_1$ and~$D_2$ are analogous minors of elements in the main diagonal of the +present determinant of order~$3$ with $x = 0$. Hence the coefficient of~$-x^2$ is a sum of +squares. Since the function of degree~$6$ is not zero for a negative value of~$x^2$, $D(x)=0$ +has no purely imaginary root. If it had an imaginary root $r+si$, then $D(x+r)=0$ +would have a purely imaginary root~$si$. But $D(x+r)$ is of the form $D(x)$ with $a$, $f$, $h$ +replaced by $a-r$, $f-r$, $h-r$. Hence $D(x)=0$ has only real roots. The method is +applicable to such determinants of order~$n$. + +\item[11.] If $a_1, \dotsc, a_n$ are distinct, solve the system of equations +\[ +\frac{x_1}{k_i-a_1} + \frac{x_2}{k_i-a_2} + \dotsb + + \frac{x_n}{k_i - a_n} = 1\qquad (i=1, \dotsc, n). +\] + +Hint: Regard $k_1, \dotsc, k_n$ as the roots of an equation of degree~$n$ in $k$ formed from +the typical one above by substituting~$k$ for~$k_i$ and clearing of fractions; write $k = a_j-t$, +and consider the product of the roots of $t^n + \dotsb = 0$. Hence find~$x_j$. + +\item[12.] Solve the equation +\[ +\begin{vmatrix} +a+x & x & x \\ + x & b+x & x \\ + x & x & c+x +\end{vmatrix} = 0. +\] +\end{Problems} +\end{Exercises} +\index{Determinants|)}% + +%% -----File: 134.png---Folio 128------- + + +\Chapter{IX}{Symmetric Functions} + +\index{Sigma function|(}% +\index{Symmetric functions|(}% + +% [** PP: Unit has two separate ToC entries] +\Section{103.}{Sigma Functions, Elementary Symmetric Functions} A rational +function of the independent variables $x_1, x_2, \dotsc, x_n$ is said to be \emph{symmetric} +in them if it is unaltered by the interchange of any two of the variables. +For example, +\[ +x_1^2 + x_2^2 + x_3^2 +4x_1 + 4x_2 +4x_3 +\] +is a symmetric polynomial in $x_1$, $x_2$, $x_3$; the sum of the first three terms +is denoted by $\Sigma x_1^2$ and the sum of the last three by $4\Sigma x_1$. In general, +if $t$ is a rational function of $x_1, \dotsc, x_n, \Sigma t$ denotes the sum of $t$ and all +of the distinct functions obtained from $t$ by permutations of the variables; +such a $\Sigma$-function (read \emph{sigma function}) is symmetric in $x_1, \dotsc, x_n$. +\index{Symbol!f@{$\Sigma$\IndAdd{symmetric function}}}% [** PP: Manually alphabetized] + +For example, if there are three independent variables $\alpha$, $\beta$, $\gamma$, +\begin{gather*} +\begin{aligned} +\Sigma \alpha\beta + &= \alpha\beta + \alpha\gamma + \beta\gamma,\qquad & +\Sigma \alpha^2\beta + &= \alpha^2\beta + \alpha\beta^2 + + \alpha^2\gamma + \alpha\gamma^2 + + \beta^2 \gamma + \beta \gamma^2, \\ +% +\Sigma \frac{1}{\alpha} + &= \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}, & +\Sigma \frac{\beta}{\alpha} + &= \frac{\beta}{\alpha} + \frac{\alpha}{\beta} + + \frac{\beta}{\gamma} + \frac{\gamma}{\beta} + + \frac{\alpha}{\gamma} + \frac{\gamma}{\alpha}, +\end{aligned} \\ % Top-level break +% +\Sigma \frac{\alpha^2 + \beta^2}{\alpha\beta} + = \frac{\alpha^2 + \beta^2}{\alpha\beta} + + \frac{\alpha^2 + \gamma^2}{\alpha\gamma} + + \frac{\beta^2 + \gamma^2}{\beta\gamma}. +\end{gather*} + +In particular, $\Sigma \alpha = \alpha + \beta + \gamma$, $\Sigma \alpha\beta$, and $\alpha\beta\gamma$ are called the three \emph{elementary +symmetric functions} of $\alpha$, $\beta$, $\gamma$. In general, +\index{Elementary symmetric function}% +\[ +\Sigma \alpha_1,\quad +\Sigma \alpha_1\alpha_2,\quad +\Sigma \alpha_1\alpha_2\alpha_3, \dotsc,\quad +\Sigma \alpha_1\alpha_2 \dotsm \alpha_{n-1},\quad +\alpha_1 \alpha_2 \dotsc \alpha_n +\] +are the elementary symmetric functions of $\alpha_1$, $\alpha_2, \dotsc, \alpha_n$. In~§20 they +were written out more fully and proved to be equal to $-c_1$, $c_2$, $-c_3, \dotsc, +(-1)^n c_n$ if $\alpha_1, \dotsc, \alpha_n$ are the roots of the equation +\[ +x^n + c_1x^{n-1} + c_2x^{n-2} + \dotsb + c_n = 0 +\Tag{1} +\] +whose leading coefficient is unity. + +%% -----File: 135.png---Folio 129------- + + +\begin{Exercises}{} + +If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + r = 0$, +so that $\Sigma \alpha = -p$, $\Sigma \alpha\beta = q$, and $\alpha\beta\gamma = -r$, +prove that +\begin{Problems} + +\item[1.] $(\Sigma \alpha)^2 = \Sigma \alpha^2 + 2\Sigma \alpha\beta$, whence $\Sigma \alpha^2 = p^2 - 2q$. + +\item[2.] $\Sigma \alpha · \Sigma \alpha\beta = \Sigma \alpha^2\beta + 3\alpha\beta\gamma$, whence $\Sigma \alpha^2\beta = 3r-pq$. + +\item[3.] $\Sigma \alpha^2\beta\gamma = pr$. + +\item[4.] $\Sigma \alpha^2\beta^2 = (\Sigma \alpha\beta)^2 - 2\alpha\beta\gamma\Sigma \alpha = q^2 - 2pr$. +\end{Problems} + +If $\alpha$, $\beta$, $\gamma$, $\delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, prove that +\begin{Problems} +\item[5.]\qquad\qquad $\Sigma \dfrac{1}{\alpha} = \dfrac{-r}{s}$,\qquad + $\Sigma \dfrac{1}{\alpha\beta} = \dfrac{q}{s}$,\qquad + $\Sigma \dfrac{1}{\alpha^2} = \dfrac{r^2 - 2qs}{s^2}$. \\ +Hint: Compute the sum, sum of the products two at a time, and sum of the squares +of the roots of the equation +\[ +1 + py + qy^2 + ry^3 + sy^4 = 0, +\] +obtained by replacing~$x$ by~$1/y$ in the given quartic equation. + +\item[6.] $\Sigma \dfrac{\beta}{\alpha} = \Sigma \alpha · \Sigma \dfrac{1}{\alpha} - 4 = \dfrac{pr}{s} - 4$. + +\item[7.] $\Sigma \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \Sigma \dfrac{\beta}{\alpha}$. + +\item[8.] $\Sigma \dfrac{\beta\gamma}{\alpha^2} = \Sigma \alpha\beta · \Sigma \dfrac{1}{\alpha^2} - \Sigma \dfrac{\beta}{\alpha} = \dfrac{1}{s^2} (qr^2 - 2q^2s - prs + 4s^2)$. + +\item[9.] $\Sigma \dfrac{\gamma}{\alpha\beta} = \dfrac{3r - pq}{s}$. +\end{Problems} +\end{Exercises} + + +\Section[Fundamental Theorem] +{104.}{Fundamental Theorem on Symmetric Functions.} +\begin{Thm} +Any polynomial +symmetric in $x_1, \dotsc, x_n$ is equal to an integral rational function, +with integral coefficients, of the elementary symmetric functions +\[ +E_1 = \Sigma x_1,\qquad +E_2 = \Sigma x_1x_2,\qquad +E_3 = \Sigma x_1x_2x_3,\dotsc,\qquad +E_n = x_1x_2 \dotsm x_n +\Tag{2} +\] +and the coefficients of the given polynomial. In particular, any symmetric +polynomial with integral coefficients is equal to a polynomial in the elementary +symmetric functions with integral coefficients. +\end{Thm} + +\begin{Remark} +For example, if $n = 2$, +\[ +rx_1^2 + rx_2^2 + sx_1 + sx_2 \equiv r(E_1^2 - 2E_2) + sE_1. +\] +In case $r$ and~$s$ are integers, the resulting polynomial in $E_1$ and~$E_2$ has integral coefficients. +\end{Remark} + +The theorem is most frequently used in the equivalent form: + +%% -----File: 136.png---Folio 130------- + +\emph{Any polynomial symmetric in the roots of an equation, +\[ +x^n - E_1 x^{n-1} + E_2 x^{n-2} - \dotsb + (-1)^n E_n = 0, +\] +is equal to an integral rational function, with integral coefficients, of the coefficients +of the equation and the coefficients of the polynomial.} + +It is this precise theorem that is required in all parts of modern algebra +and the theory of numbers, where attention to the nature of the coefficients +is vital, rather than the inadequate, oft-quoted, theorem that any symmetric +function of the roots is expressible (rationally) in terms of the +coefficients. + +It suffices to prove the theorem for any homogeneous symmetric polynomial~$S$, +i.e., one expressible as a sum of terms % [** PP: Not italicizing i.e.] +\[ +h = ax_1^{k_1} x_2^{k_2} \dotsm x_n^{k_n} +\] +of constant total degree $k = k_1 + k_2 + \dotsb + k_n$ in the~$x$'s. Evidently +we may assume that no two terms of~$S$ have the same set of exponents +$k_1, \dotsc, k_n$ (since such terms may be combined into a single one). We +shall say that $h$ is \emph{higher} than the term $bx_1^{l_1} x_2^{l_2} \dotsm x_n^{l_n}$ if $k_1>l_1$, or if +$k_1 = l_1$, $k_2>l_2$, or if $k_1 = l_1$, $k_2 = l_2$, $k_3>l_3, \dotsc$, so that the first one of +the differences $k_1-l_1$, $k_2-l_2$, $k_3-l_3, \dotsc$ which is not zero is positive. + +We first prove that, if the above term $h$ is the highest term of~$S$, then +\[ +k_1 \geqq k_2 \geqq k_3 \dotsb \geqq k_n. +\] +For, if $k_1<k_2$, the symmetric polynomial~$S$ would contain the term +\[ +ax_1^{k_2} x_2^{k_1} x_3^{k_3} \dotsm x_n^{k_n}, +\] +which is higher than~$h$. If $k_2<k_3$, $S$ would contain the term +\[ +ax_1^{k_1} x_2^{k_3} x_3^{k_2} \dotsm x_n^{k_n}, +\] +which is higher than~$h$, etc. + +If the highest term in another homogeneous symmetric polynomial~$S'$ +is +\[ +h' = a'x_1^{k'_1} x_2^{k'_2} \dotsm x_n^{k'_n}, +\] +and that of $S$ is $h$, then the highest term in their product~$SS'$ is +\[ +hh' = aa'x_1^{k_1 + k'_1} \dotsm x_n^{k_n + k'_n}. +\] +%% -----File: 137.png---Folio 131------- +Indeed, suppose that $SS'$ has a term, higher than~$hh'$, +\[ +cx_1^{l_1 + l'_1} \dotsm x_n^{l_n + l'_n}, +\Tag{3} +\] +which is either a product of terms +\[ +t = b x_1^{l_1} \dotsm x_n^{l_n},\qquad +t' = b'x_1^{l'_1} \dotsm x_n^{l'_n} +\] +of $S$ and~$S'$ respectively, or is a sum of such products. Since~\Eq{3} is higher +than~$hh'$, the first one of the differences +\[ +l_1 + l'_1 - k_1 -k'_1, \dotsc, l_n + l'_n - k_n -k'_n +\] +which is not zero is positive. But, either all of the differences $l_1 - k_1, +\dotsc, l_n - k_n$ are zero or the first one which is not zero is negative, since +$h$ is either identical with $t$ or is higher than~$t$. Likewise for the differences +$l'_1 - k'_1, \dotsc, l'_n - k'_n$. We therefore have a contradiction. + +It follows at once that the highest term in a product of any number +of homogeneous symmetric polynomials is the product of their highest +terms. Now the highest terms in $E_1$, $E_2$, $E_3, \dotsc, E_n$, given by~\Eq{2}, are +\[ +x_1, \qquad x_1x_2,\qquad x_1x_2x_3,\quad\dotsc,\qquad +x_1x_2 \dotsm x_n, +\] +respectively. Hence the highest term in $E_1^{a_1} E_2^{a_2} \dotsm E_n^{a_n}$ is +\[ +x_1^{a_1 + a_2 + \dotsb + a_n} +x_2^{a_2 + \dotsb + a_n} \dotsm +x_n^{a_n}. +\] + +Thus the highest term in +\[ +\sigma = aE_1^{k_1 - k_2} E_2^{k_2 - k_3} \dotsm E_{n-1}^{k_{n-1} - k_n}E_n^{k_n} +\] +is~$h$. Hence $S_1 = S - \sigma$ is a homogeneous symmetric polynomial of the +same total degree~$k$ as~$S$ and having a highest term~$h_1$ not as high as~$h$. +As before, we form a product~$\sigma_1$ of the $E$'s whose highest term is this~$h_1$. +Then $S_2 = S_1 - \sigma_1$ is a homogeneous symmetric polynomial of total degree~$k$ +and with a highest term $h_2$ not as high as~$h_1$. We must finally reach +a difference $S_t - \sigma_t$ which is identically zero. Indeed, there is only a +finite number of products of powers of $x_1, \dotsc, x_n$ of total degree~$k$. +Among these are the parts $h'$, $h'_1$, $h'_2, \dotsc$ of~$h$, $h_1$, $h_2, \dotsc$ with the coefficients +suppressed. Since each $h_i$ is not as high as~$h_{i-1}$, the $h'$, $h'_1$, $h'_2, \dotsc$ are +all distinct. Hence there is only a finite number of~$h_i$. Since $S_t - \sigma_t \equiv 0$, +\[ +S = \sigma + S_1 + = \sigma + \sigma_1 + S_2 = \dotsb + = \sigma + \sigma_1 + \sigma_2 + \dotsb + \sigma_t. +\] +Hence $S$ is a polynomial in $E_1$, $E_2, \dotsc, E_n$ and $a$, $b, \dotsc$, with integral +coefficients. + +%% -----File: 138.png---Folio 132------- + +\begin{Example}[1.] +If $S = \Sigma x_1^2 x_2^2 x_3$ and $n>4$, we have +\begin{align*} +\sigma &= E_2E_3 + = S + 3 \Sigma x_1^2 x_2 x_3 x_4 + 10 \Sigma x_1x_2x_3x_4x_5, \\ +% +S_1 &= S - \sigma + = -3 \Sigma x_1^2 x_2 x_3 x_4 - 10 \Sigma x_1x_2x_3x_4x_5, \\ +% +\sigma_1 &= -3 E_1E_4 + = -3 (\Sigma x_1^2 x_2 x_3 x_4 + 5 \Sigma x_1x_2x_3x_4x_5), \\ +% +S_2 &= S_1 - \sigma_1 + = 5 \Sigma x_1x_2x_3x_4x_5 + = 5E_5, \\ +% +S &= \sigma + S_1 + = \sigma + \sigma_1 + S_2 + = E_2 E_3 - 3 E_1 E_4 + 5E_5. +\end{align*} +\end{Example} + + +\begin{Example}[2.] +If $S = \Sigma x_1^3 x_2 x_3$ and $n> 4$, +\begin{align*} +\sigma = E_1^2 E_3 + &= E_1(\Sigma x_1^2 x_2 x_3 + 4 \Sigma x_1 x_2 x_3 x_4) \\ +% + &= \Sigma x_1^3 x_2 x_3 + 2 \Sigma x_1^2 x_2^2 x_3 + + 3 \Sigma x_1^2 x_2 x_3 x_4 \\ +% + & \phantom{{}=\Sigma x_1^3x_2x_3 } + + 4 (\Sigma x_1^2 x_2 x_3 x_4 + 5 \Sigma x_1x_2x_3x_4x_5), +\end{align*} +\[ +S_1 = S - \sigma + = -2 \Sigma x_1^2 x_2^2 x_3 - 7 \Sigma x_1^2 x_2 x_3 x_4 + -20 \Sigma x_1x_2x_3x_4x_5. +\] +Take $\sigma_1 = -2 E_2E_3$ and proceed as in Ex.~1. +\end{Example} + +\begin{Example}[3.] +By examples 1 and~2, if $n>4$, +\[ +a \Sigma x_1^2 x_2^2 x_3 + +b \Sigma x_1^3 x_2 x_3 + = bE_1^2 E_3 - (3a + b)E_1E_4 + (a - 2b)E_2E_3 + 5(a + b)E_5. +\] +\end{Example} + + +% [** PP: ToC entry matches unit title, using running head] +\Section[Functions Symmetric in all but One Root] +{105.}{Rational Functions Symmetric in all but One of the Roots.} +\index{Symmetric functions!in all but one root|(}% +\begin{Thm} +If +$P$ is a rational function of the roots of an equation $f(x)= 0$ of degree~$n$ and +if $P$ is symmetric in $n-1$ of the roots, then $P$ is equal to a rational function, +with integral coefficients, of the remaining root and the coefficients of $f(x)$ and~$P$. +\end{Thm} + +\begin{Remark} +For example, $P = r \alpha_1 + \alpha_2^2 + \alpha_3^2 + \dotsb + \alpha_n^2$ is symmetric in $\alpha_2, \dotsc, \alpha_n$, and +\[ +P = r \alpha_1 + \Sigma \alpha_1^2 - \alpha_1^2 + = c_1^2 - 2c_2 + r \alpha_1 - \alpha_1^2, +\] +if $\alpha_1, \dotsc, \alpha_n$ are the roots of equation~\Eq{1}. +\end{Remark} + +Since\footnote + {If $N/D$ is symmetric in $\alpha_1$, $\alpha_2$, and the polynomials $N$ and~$D$ have no common + factor, while $N$ becomes~$N'$ and $D$ becomes~$D'$ when $\alpha_1$, $\alpha_2$ are interchanged, then + $ND'\equiv DN'$. Thus $N$ divides~$N'$ and both are of the same degree. Hence $N'=cN, + D' = cD$, where $c$ is a constant. By again interchanging $\alpha_1$, $\alpha_2$, we obtain $N$ from~$N'$, + whence $N = cN' = c^2N$, $c^2 = 1$. If $c = -1$, we take $\alpha_1 = \alpha_2$ and see that $N = N'= -N$, + $N = 0$, whence $N$ has the factor $\alpha_1 - \alpha_2$. Similarly, $D$ has the same factor, contrary to + hypothesis. Hence $c = +1$ and $N$ and~$D$ are each symmetric in $\alpha_1$, $\alpha_2$.} +any symmetric rational function is the quotient of two symmetric +polynomials, the above theorem will follow if proved for the case in which +the words rational function are in both places replaced by polynomial. + +If $\alpha_1$ is the remaining root, the polynomial~$P$ is symmetric in the roots +$\alpha_2, \dotsc, \alpha_n$ of $f(x)/(x - \alpha_1) = 0$, an equation of degree $n-1$ whose coefficients +are polynomials in $\alpha_1$, $c_1, \dotsc, c_n$ with integral coefficients. Hence~(§104), +$P$ is equal to a polynomial, with integral coefficients, in $\alpha_1$, $c_1, \dotsc, c_n$ +and the coefficients of~$P$. + +%% -----File: 139.png---Folio 133------- + +\begin{Example} +If $\alpha$, $\beta$, $\gamma$ are the roots of $f(x) \equiv x^3 + px^2 + qx + r = 0$, find +\[ +\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta} + = \frac{\alpha^2 + \beta^2}{\alpha + \beta} + + \frac{\alpha^2 + \gamma^2}{\alpha + \gamma} + + \frac{\beta^2 + \gamma^2}{\beta + \gamma}. +\] +\end{Example} + +\begin{Solution} +Since $\beta^2 + \gamma^2 = p^2 - 2q - \alpha^2$,\quad $\beta + \gamma = -p - \alpha$, +\[ +\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta} + = \Sigma \frac{p^2 - 2q - \alpha^2}{-p - \alpha} + = \Sigma \left(\alpha - p + \frac{2q}{\alpha + p}\right) + = - p - 3p + 2q \Sigma \frac{1}{\alpha + p}. +\] +But $\alpha + p$, $\beta + p$, $\gamma + p$ are the roots $y_1$, $y_2$, $y_3$ of the cubic equation obtained from +\index{Cubic equation|(}% [** PP: Using page range] +$f(x)=0$ by the substitution $x + p = y$, i.e., $x = y-p$. The resulting equation is % [** PP: Not italicizing i.e.] +\[ +y^3 - 2py^2 + (p^2 + q)y + r - pq = 0. +\] +Since we desire the sum of the reciprocals of $y_1$, $y_2$, $y_3$, we set $y = 1/z$ and find the sum +of the roots $z_1$, $z_2$, $z_3$ of +\[ +1 - 2pz + (p^2 + q)z^2 + (r - pq)z^3 = 0. +\] +Hence +\[ +\Sigma \frac{1}{\alpha + p} + = \Sigma \frac{1}{y_1} + = \Sigma z_1 = \frac{p^2 + q}{pq - r},\qquad +\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta} + = \frac{2q^2 - 2p^2q + 4pr}{pq - r}. +\] +\end{Solution} + + +\begin{Exercises}{Page133} + +[In Exs.~1--12, $\alpha$, $\beta$, $\gamma$ are the roots of $f(x) = x^3 + px^2 + qx + r = 0$.] + +Using $\beta\gamma + \alpha(\beta + \gamma) = q$, find +\begin{Problems}[2] +\item[1.] $\Sigma \dfrac{\beta\gamma + \alpha^2}{\beta + \gamma}$, % [** PP: Added ,] + +\item[2.] $\Sigma \dfrac{3\beta\gamma - 2\alpha^2}{\beta + \gamma - \alpha}$. + +\ResetCols{1} + +\item[3.] Why would the use of $\beta\gamma = -r/\alpha$ complicate Exs.\ 1,~2? Verify that +\[ +\beta\gamma + = \frac{-r}{\alpha} + = \frac{f(\alpha) - r}{\alpha} + = \alpha^2 + p \alpha + q. +\] + +\item[4.] Why would you use $\beta\gamma = -r/\alpha$ in finding + $\Sigma \dfrac{\beta^2 + \gamma^2}{\beta\gamma + c}$? + +\ResetCols{3} + +\item[5.] Find $\Sigma (\beta + \gamma)^2$. + +\item[6.] Find $\Sigma (\alpha + \beta - \gamma)^3$. + +\item[7.] Find $\smash{\Sigma \left(\dfrac{\beta - \gamma}{\beta + \gamma}\right)^2}$. + +\ResetCols{1} + +\item[8.] Find a necessary and sufficient condition on the coefficients that the roots, in +some order, shall be in harmonic progression. +Hint: If $\dfrac{1}{\alpha} + \dfrac{1}{\gamma} = \dfrac{2}{\beta}$, then $\dfrac{-3r}{q} - \beta = 0$, +and conversely. Hence the condition is +\[ +\left(\frac{-3r}{q} - \alpha\right) +\left(\frac{-3r}{q} - \beta\right) +\left(\frac{-3r}{q} - \gamma\right) + = f\left(\frac{-3r}{q}\right) + = 0. +\] + +%% -----File: 140.png---Folio 134------- + +\item[9.] Find the cubic equation with the roots +$\beta\gamma - \dfrac{1}{\alpha}$, +$\alpha\gamma - \dfrac{1}{\beta}$, +$\alpha\beta - \dfrac{1}{\gamma}$. +Hint: since these are $(-r - 1)/\alpha$, etc., make the substitution $(-r - 1)/x = y$. +\end{Problems} + +Find the substitution which replaces the given cubic equation by one with the roots +\begin{Problems} + +\item[10.] +$\alpha\beta + \alpha\gamma$, +$\alpha\beta + \beta\gamma$, +$\alpha\gamma + \beta\gamma$. + +\ResetCols{2} + +\item[11.] $\dfrac{2\alpha - 1}{\beta + \gamma - \alpha}$, etc. + +\item[12.] $\dfrac{\beta\gamma + 3\alpha^2}{\beta + \gamma - 2\alpha}$, etc. +\end{Problems} + +If $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, find +\begin{Problems}[2] + +\item[13.] $\Sigma\dfrac{\beta^2 + \gamma^2 + \delta^2}{\beta + \gamma + \delta}$. + +\item[14.] $\Sigma\dfrac{\beta\gamma + \beta\delta + \gamma\delta}{\beta + \gamma + \delta - 3}$. + +\ResetCols{1} + +\item[15.] Prove that if $y_1$, $y_2$, $y_3$ are the roots of $y^3 + py + q = 0$, the equation with the roots +$z_1 = (y_2 - y_3)^2$, $z_2 = (y_1 - y_3)^2$, $z_3 = (y_1 - y_2)^2$ is +\index{Equation for differences of roots!squares of differences}% +\[ +z^3 + 6pz^2 + 9p^2 z + 4p^3 + 27q^2 = 0. +\] +Hints: since $z_1 = \Sigma y_1^2 - 2y_2y_3 - y_1^2 = -2p + 2q/y_1 - y_1^2$, etc., we set $z = -2p + 2q/y - y^2$. +By the given equation, $y^2 + p + q/y = 0$. Thus the desired substitution is $z = -p + 3q/y$, +$y = 3q/(z + p)$. + +\item[16.] Hence find the discriminant of the reduced cubic equation. +\index{Cubic equation|)}% +\index{Discriminant!of cubic}% + +\item[17.] If $x_1, \dotsc, x_n$ are the roots of $f(x)=0$, show that +\[ +\Sigma \frac{1}{x_1 - c} = \frac{-f'(c)}{f(c)}. +\] +Hint: $x_1 - c = y_1, \dotsc, x_n - c = y_n$ are the roots of +\[ +f(c+y) = f(c) + yf'(c) + y^2(\quad)+ \dotsb = 0, +\] +as shown by Taylor's theorem. Or we may employ~\Eq{5} below % [** PP: Added `below'] +for $x = c$. +\end{Problems} +\end{Exercises} +\index{Symmetric functions!in all but one root|)}% + + +% [** PP: ToC entry appended with ``Newton's Identities''] +\Section{106.}{Sums of Like Powers of the Roots} If $\alpha_1, \dotsc, \alpha_n$ are the roots of +\index{Sum of!like powers of roots|(}% +\[ +f(x) \equiv x^n + c_1 x^{n-1} + c_2x^{n-2} + \dotsb + c_n = 0, +\Tag{1} +\] +we write $s_1$ for~$\Sigma \alpha_1$, $s_2$ for~$\Sigma \alpha_1^2$, and, in general, +\index{Symbol!g@{$s_k$\IndAdd{sum of $k$th powers}}}% [** PP: Manually alphabetized] +\[ +s_k = \Sigma \alpha_1^k + = \alpha_1^k + \alpha_2^k + \dotsb + \alpha_n^k. +\] + +The factored form of~\Eq{1} is +\[ +f(x) \equiv (x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n). +\Tag{4} +\] +%% -----File: 141.png---Folio 135------- +The derivative $f'(x)$ of this product is found by multiplying the derivative +(unity) of each factor by the product of the remaining factors and adding +the results. Hence +\index{Derivative}% +\begin{gather*} +f'(x) = (x - \alpha_2) \dotsm (x - \alpha_n) + + (x - \alpha_1)(x - \alpha_3) \dotsm (x - \alpha_n) + \dotsb, \\ +f'(x) \equiv \frac{f(x)}{x - \alpha_1} + \frac{f(x)}{x - \alpha_2} + \dotsb + + \frac{f(x)}{x - \alpha_n}. +\Tag{5} +\end{gather*} + +If $\alpha$ is any root of~\Eq{1}, $f(\alpha)= 0$ and +\begin{align*} +\frac{f(x)}{x - \alpha} + &= \frac{f(x) - f(\alpha)}{x - \alpha} + = \frac{x^n - \alpha^n}{x - \alpha} + + c_1\frac{x^{n-1} - \alpha^{n-1}}{x - \alpha} + \dotsb + + c_{n-1}\frac{x - \alpha}{x - \alpha} \\ +% + &= x^{n-1} + \alpha x^{n-2} + \alpha^2 x^{n-3} + \dotsb + + c_1(x^{n-2} + \alpha x^{n-3} + \dotsb) \\ + &\hphantom{{}=x^{n-1}} + c_2(x^{n-3} + \dotsb) + \dotsb, +\end{align*} +\[ +\begin{split} +\frac{f(x)}{x - \alpha} + &= x^{n-1} + (\alpha + c_1)x^{n-2} + + (\alpha^2 + c_1 \alpha + c_2)x^{n-3} + \dotsb \\ + &+ (\alpha^k + c_1 \alpha^{k-1} + c_2 \alpha^{k-2} + \dotsb + + c_{k-1} \alpha + c_k)x^{n-k-1} + \dotsb. +\end{split} +\Tag{6} +\] +Taking $\alpha$ to be $\alpha_1, \dotsc, \alpha_n$ in turn, adding the results, and applying~\Eq{5}, we +obtain +\begin{align*} +f'(x) = nx^{n-1} + &+ (s_1 + nc_1)x^{n-2} + (s_2 + c_1 s_1 + nc_2)x^{n-3} + \dotsb \\ + &+ (s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + + c_{k-1} s_1 + nc_k)x^{n-k-1} + \dotsb. +\end{align*} + +The derivative of~\Eq{1} is found at once by the rules of calculus (or by~§56) +to be +\[ +f'(x) = nx^{n-1} + (n-1)c_1 x^{n-2} + (n-2)c_2 x^{n-3} + \dotsb + + (n-k)c_k x^{n-k-1} + \dotsb. +\] +Since this expression is identical term by term with the preceding, we have +\[ +\begin{aligned} +&s_1 + c_1 = 0,\qquad s_2 + c_1 s_1 + 2c_2 = 0, \dotsc, \\ +&s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_{k-1} s_1 + kc_k = 0\quad + (k \leqq n - 1). +\end{aligned} +\Tag{7} +\] + +We may therefore find in turn $s_1$, $s_2, \dotsc, s_{n-1}$: +\[ +s_1 = -c_1,\qquad +s_2 = c_1^2 - 2c_2,\qquad +s_3 = -c_1^3 + 3c_1 c_2 - 3c_3, \dotsc. +\Tag{8} +\] + +To find $s_n$, replace $x$ in~\Eq{1} by $\alpha_1, \dotsc, \alpha_n$ in turn and add the resulting +equations. We get +\[ +s_n + c_1 s_{n-1} + c_2 s_{n-2} + \dotsb + c_{n-1} s_1 + nc_n = 0. +\Tag{9} +\] + +%% -----File: 142.png---Folio 136------- + +We may combine \Eq{7} and~\Eq{9} into +\[ +s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_{k-1} s_1 + kc_k = 0\qquad + (k=1, 2, \dotsc, n). +\Tag{10} +\] + +This set of formulas~\Eq{10} will be referred to as \emph{Newton's identities}. +\index{Newton's!identities}% +The student should be able to write them down from memory and, when +writing them, should always check the final one~\Eq{9} by deriving it as above. + +To derive a formula which shall enable us to compute the $s_k$ for +$k>n$, we multiply~\Eq{1} by~$x^{k-n}$, take $x = \alpha_1, \dotsc, x = \alpha_n$ in turn, and add +the resulting equations. We get +\[ +s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_n s_{k-n} = 0\qquad (k>n). +\Tag{11} +\] + +Instead of memorizing this formula, it is preferable to deduce it for the +particular equation for which it is needed, thus avoiding errors of substitution +as well as confusion with~\Eq{10}. + + +\begin{Example} +Find $s_k$ for $x^n - 1 = 0$. +\end{Example} +\index{Roots of unity}% + +\begin{Solution} +Comparing our equation with~\Eq{1}, we have $c_1 = 0, \dotsc, c_{n-1} = 0$, $c_n = -1$. +Hence in~\Eq{10} for $k<n$, each $c$ is zero and $s_k = 0$. But, for $k = n$, \Eq{10} becomes $s_n - n = 0$. +We may check the latter by substituting each root $\alpha_1, \dotsc, \alpha_n$ in our given equation +and adding. Finally, to find $s_l$ when $l>n$, multiply our equation by~$x^{l-n}$. In the +resulting equation $x^l - x^{l-n} = 0$ we substitute each root, add, and obtain $s_l - s_{l-n} = 0$. +Hence from $s_l$ we obtain an equal $s$ by subtracting $n$ from~$l$. After repeated subtractions, +we reach a value~$k$ for which $1 \leqq k \leqq n$. Since $s_k = 0$ or $n$ according as $k<n$ or +$k = n$, it follows that $s_l = 0$ or $n$ according as $l$ is not or is divisible by~$n$. +\end{Solution} + + +\begin{Exercises}{Page136} + +\begin{Problems} +\item[1.] For a cubic equation, $s_4 = c_1^4 - 4c_1^2 c_2 + 4c_1 c_3 + 2c_2^2$. + +\item[2.] For an equation of degree $\geqq 4$, $s_4 = c_1^4 - 4c_1^2 c_2 + 4c_1 c_3 + 2c_2^2- 4c_4$. + +\item[3.] Find $s_2$, $s_3$, $s_4$, $s_5$ for $x^2 - px + q = 0$. + +\item[4.] Find $s_k$ for $x^5 - 3 = 0$. + +\item[5.] Find $s_2$, $s_3$, $s_6$, $s_7$ for $x^5 - px + q = 0$. +\end{Problems} +\end{Exercises} + + +\Section[Waring's Formula] +{107.}{Waring's Formula for $s_k$ in Terms of the Coefficients.} While +\index{Waring's formula}% +we have learned how to find $s_1$, $s_2$, $s_3, \dotsc$ in turn by Newton's identities, +it is occasionally useful to know an explicit expression for~$s_k$, where $k$ +has an arbitrary value. The formula in question is applied ordinarily +only to a quadratic equation +\[ +x^2 + px + q = 0. +\] +Accordingly we shall treat this case in detail. If its roots are $\alpha$ and~$\beta$, +then +\[ +x^2 + px + q \equiv (x - \alpha)(x - \beta). +\] +%% -----File: 143.png---Folio 137------- +Replace $x$ by~$1/y$ and multiply by~$y^2$. We get +\[ +1 + py + qy^2 \equiv (1 - \alpha y)(1 - \beta y). +\Tag{12} +\] +Taking derivatives, we have +\[ +p + 2qy \equiv - \alpha (1 - \beta y) - \beta (1 - \alpha y). +\] +Change of signs and division by the members of~\Eq{12} gives +\[ +\frac{-p - 2qy}{1 + py + qy^2} + \equiv \frac{\alpha}{1 - \alpha y} + \frac{\beta}{1 - \beta y}. +\Tag{13} +\] + +The identity in Ex.~7, §14, with $n$ changed to~$k$, may be written in +the form +\[ +\frac{1}{1 - r} \equiv 1 + r + r^2 + \dotsb + r^{k-1} + \frac{r^k}{1-r}. +\Tag{14} +\] +Take $r = \alpha y$ and multiply the resulting terms by~$\alpha$; thus +\begin{align*} +\frac{\alpha}{1 - \alpha y} + &= \alpha + \alpha^2 y + \dotsb + \alpha^k y^{k-1} + + \frac{\alpha^{k+1} y^k}{1 - \alpha y}. \\ +\intertext{Similarly,} +\frac{\beta}{1 - \beta y} + &= \beta + \beta^2 y + \dotsb + \beta^k y^{k-1} + + \frac{\beta^{k+1} y^k}{1 - \beta y}. +\end{align*} +To show that on adding, and writing $s_k$ for $\alpha^k + \beta^k$, we obtain~\Eq{15}, we need +the sum of the final fractions, which by~\Eq{12} is +\[ +\frac{\phi y^k}{(1 - \alpha y)(1 - \beta y)} + = \frac{\phi y^k}{1 + py + qy^2},\qquad +\phi \equiv \alpha^{k+1}(1 - \beta y) + \beta^{k+1}(1 - \alpha y). +\] +Hence +\[ +\frac{\alpha}{1 - \alpha y} + \frac{\beta}{1 - \beta y} + = s_1 + s_2 y + \dotsb + s_k y^{k-1} + \frac{\phi y^k}{1 + py + qy^2}, +\Tag{15} +\] +where the exact expression for $\phi$ is immaterial. + +Next, we seek an expansion of the fraction in the left member of~\Eq{13}. +Its denominator will be identical with that in~\Eq{14} if we choose +$r = -py - qy^2$. Evidently~\Eq{14} may be written in the compact form +\[ +\frac{1}{1 - r} \equiv \sum_{t=0}^{k-1} r^t + \frac{r^k}{1 - r}. +\] +Hence it becomes +\[ +\frac{1}{1 + py + qy^2} + = \sum_{t=0}^{k-1} (-1)^t(py + qy^2)^t + \frac{\psi y^k}{1 + py + qy^2}, +\] +%% -----File: 144.png---Folio 138------- +where $\psi = (-p - qy)^k$, although no use will be made of the particular +form of the polynomial~$\psi$. By the binomial theorem, +\[ +(py + qy^2)^t = \sum \frac{(g+h)!}{g!h!} (py)^g(qy^2)^h, +\] +where the summation extends over all sets of integers $g$ and~$h$, each $\geqq 0$, +for which $g + h = t$, while $g!$~denotes the product of $1$, $2, \dotsc, g$ if $g \geqq 1$, +but denotes unity if $g = 0$. Hence +\begin{gather*} +\frac{-p - 2qy}{1 + py + qy^2} + = (p + 2qy) \sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^hy^{g+2h} + E, +\Tag{16} \\ +E \equiv \frac{(-p - 2qy) \psi y^k}{1 + py + qy^2}, +\end{gather*} +where the summation extends over all sets of integers $g$ and~$h$, each $\geqq 0$, +for which $g+h \leqq k-1$. + +Since the left members of \Eq{15} and~\Eq{16} are identically equal by~\Eq{13}, +their right members must be identical, so that the coefficients of $y^{k-1}$ +in them must be equal.\footnote + {In fact, the $(k-1)$th derivatives of the two right members are identical, and we + obtain the indicated result by substituting $y = 0$ in these two derivatives and equating + the results. Note that the final terms in both \Eq{15} and~\Eq{16} have~$y$ as a factor of their + $(k-1)$th derivatives.} % end footnote +Hence the coefficient~$s_k$ of~$y^{k-1}$ in~\Eq{15} is equal +to the coefficient of~$y^{k-1}$ in~\Eq{16}, which is made up of two parts, corresponding +to the two terms of the factor $p + 2qy$. When we use the constant +term~$p$, we must employ from~$\sum$ in~\Eq{16} the terms in which the exponent +of~$y$ is equal to~$k-1$. But when we use the other term~$2qy$, we must +employ from~$\sum$ the terms in which the exponent of~$y$ is equal to~$k-2$, in +order to obtain the combined exponent~$k-1$ of~$y$. Hence $s_k$ is equal to +the sum of the following two parts: +\begin{align*} + p\sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^h && (g+2h &= k-1),\\ +2q\sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^h && (g+2h &= k-2). +\end{align*} +In the upper sum, write $i$ for~$g+1$, and $j$ for~$h$. In the lower sum, write +$i$ for~$g$, and $j$ for~$h+1$. Hence +\[ +s_k = \sum (-1)^{i+j} \frac{(i+j-1)!}{(i-1)!j!} p^iq^j + + 2\sum (-1)^{i+j} \frac{(i+j-1)!}{i!(j-1)!} p^iq^j, +\] +%% -----File: 145.png---Folio 139------- +where now each summation extends over all sets of integers $i$ and~$j$, each +$\geqq 0$, for which +\[ +i + 2j = k. +\Tag{17} +\] + +Finally, we may combine our two sums. Multiply the numerator +and denominator of the first fraction by~$i$, and those of the second fraction +by~$j$. Thus +\[ +s_k = k\sum (-1)^{i+j} \frac{(i+j-1)!}{i!j!} p^iq^j, +\Tag{18} +\] +since the present fraction occurred first multiplied by~$i$ and second multiplied +by~$2j$, and, by~\Eq{17}, the sum of these multipliers is equal to~$k$. Our +final result is~\Eq{18}, where the summation extends over all sets of integers +$i$ and~$j$, each $\geqq 0$, satisfying~\Eq{17}. + +\begin{Remark} +If we replace $i$ by its value $k-2j$, and change the sign of~$p$, we obtain from~\Eq{18} +the result that \emph{the sum of the $k$th powers of the roots of $x^2 - px + q = 0$ is equal to} +\index{Quadratic equation!sum of powers of roots}% +\begin{align*} +s_k &= k\sum_{j=0}^K (-1)^j \frac{(k-j-1)!}{(k-2j)!j!} p^{k-2j} q^j +\Tag{19} \\ + &= p^k - kp^{k-2} q + \frac{k(k-3)}{1·2} p^{k-4} q^2 + - \frac{k(k-4)(k-5)}{1·2·3} p^{k-6} q^3 + \dotsb, % [** PP: . -> ,] +\end{align*} +where $K$ is the largest integer not exceeding~$k/2$. + +The product of the roots is equal to~$q$. Hence if $x$ denotes one root, the second +root is~$q/x$. Thus $s_k = x^k + (q/x)^k$. Again, the sum of the roots is $x + q/x = p$. Regard +$q$ as given and $p$ as unknown. Hence, if $c$ is an arbitrary constant, the equation +\[ +p^k - kqp^{k-2} + \frac{k(k-3)}{1·2} q^2p^{k-4} - \dotsb = c +\Tag{20} +\] +is transformed by the substitution $p = x + q/x$ into +\[ +x^k + \left(\frac{q}{x}\right)^k = c. +\] +Hence equation~\Eq{20} may be solved for $p$ by radicals by the method employed in~§43 +for a cubic equation. +\end{Remark} + +%% -----File: 146.png---Folio 140------- + +The above proof applies\footnote + {See the author's \textit{Elementary Theory of Equations}, pp.~72--74, where there is given + also a shorter proof by means of infinite series.} +without essential change to any equation +$x^n + c_1x^{n-1} + \dotsb + c_n = 0$ and leads to the following formula for the sum +of the $k$th~powers of its roots: +\[ +s_k = k\sum (-1)^{r_1 + \dotsb + r_n} + \frac{(r_1 + \dotsb + r_n-1)!}{r_1! \dotsm r_n!} + c_1^{r_1} \dotsm c_n^{r_n}, +\Tag{21} +\] +where the summation extends over all sets of integers $r_1, \dotsc, r_n$, each +$\geqq 0$, for which $r_1 + 2r_2 + 3r_3 + \dotsb + nr_n = k$. This result~\Eq{21} is known +as \emph{Waring's formula} and was published by him in 1762. + +\begin{Example} +Let $n=3$, $k=4$. Then $r_1 + 2r_2 + 3r_3 = 4$ and +\begin{align*} +(r_1, r_2, r_3) + &= (4, 0, 0),\qquad + (2, 1, 0),\qquad + (1, 0, 1),\qquad + (0, 2, 0), \\ +s_4 &= 4(\frac{3!}{4!} c_1^4 - \frac{2!}{2!1!} c_1^2c_2 + + \frac{1!}{1!1!} c_1c_3 + \frac{1!}{2!} c_2^2) \\ + &= c_1^4 - 4c_1^2c_2 + 4c_1c_3 + 2c_2^2. +\end{align*} +\end{Example} + + +\begin{Exercises}{Page140} + +\begin{Problems} +\item[1.] For the quadratic $x^2 - px + q = 0$ write out the expressions for $s_2$, $s_3$, $s_4$, $s_5$ given by~\Eq{19}, +and compare with those obtained from Newton's identities (Ex.~3, §106). + +\item[2.] Find $s_4$ for a quartic equation by Waring's formula. + +\item[3.] For $k=5$, \Eq{20} becomes De Moivre's quintic $p^5 - 5qp^3 + 5q^2p = c$. Solve it by +radicals for~$p$. +\index{De Moivre's!quintic}% [** PP: Using subitem] + +\item[4.] Solve \Eq{20} by radicals when $k=7$. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry, using running head] +\Section[Computation of Sigma Functions] +{108.}{$\Sigma$-functions Expressed in Terms of the Functions~$s_k$.} Since +we have learned two methods of expressing the $s_k$ in terms of the coefficients, +it is desirable to learn how to express any $\Sigma$-polynomial (and +hence any symmetric function) in terms of the~$s_k$. + +By performing the indicated multiplication, we find that +\[ +s_a s_b \equiv \Sigma \alpha_1^a · \Sigma \alpha_1^b + = \Sigma \alpha_1^{a+b} + m\Sigma \alpha_1^a \alpha_2^b, +\] +where $m = 1$ if $a\neq b$, $m=2$ if $a=b$. Transposing the first term, which +is equal to $s_{a+b}$, and dividing by~$m$, we obtain +\[ +\Sigma \alpha_1^a \alpha_2^b = \frac{1}{m} (s_a s_b - s_{a+b}). +\Tag{22} +\] + +%% -----File: 147.png---Folio 141------- + +In order to compute $\Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2$ in terms of the~$s_k$, we form the product +\[ +\Sigma \alpha_1^4 · \Sigma \alpha_1^3 \alpha_2^2 + = \Sigma \alpha_1^7 \alpha_2^2 + \Sigma \alpha_1^6 \alpha_2^3 + + \Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2. +\] +Making three applications of~\Eq{22}, we get +\[ +s_4(s_3s_2 - s_5) + = (s_7s_2 - s_9) + (s_6s_3 - s_9) + \Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2. +\] +Hence +\[ +\Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2 + = s_2s_3s_4 - s_2s_7 - s_3s_6 - s_4s_5 + 2s_9. +\] + + +\begin{Exercises}{Page141} + +For a quartic equation, express in terms of the $s_k$ and ultimately in terms of the +coefficients $c_1, \dotsc, c_4$: + +\begin{Problems}[4] + +\item[1.] $\Sigma \alpha_1^2 \alpha_2^2$. + +\item[2.] $\Sigma \alpha_1^3 \alpha_2$. + +\item[3.] $\Sigma \alpha_1^2 \alpha_2 \alpha_3$. + +\item[4.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3^2$. + +\ResetCols{1} + +\item[5.] If $a\geqq b > c > 0$, prove that +\[ +\Sigma \alpha_1^a \alpha_2^b \alpha_3^c + = \frac{1}{m} (s_a s_b s_c - s_a s_{b+c} + - s_b s_{a+c} - s_c s_{a+b} + 2s_{a+b+c}), +\] +where $m = 1$ if $a > b$, $m = 2$ if $a = b$. + +\item[6.] $\Sigma \alpha_1^a \alpha_2^b \alpha_3^b + = \frac{1}{2}(s_a s_b^2 - s_as_{2b} - 2s_b s_{a+b} + 2s_{a+2b})$,\qquad $a > b > 0$. + +\item[7.] $\Sigma \alpha_1^a \alpha_2^a \alpha_3^a + = \frac{1}{6}(s_a^3 - 3s_a s_{2a} + 2s_{3a})$,\qquad $a > 0$. +\end{Problems} +\end{Exercises} + + +\Section{109.}{Computation of Symmetric Functions} The method last explained +is practicable when a term of the $\Sigma$-function involves only a few distinct +roots, the largeness of the exponents not introducing a difficulty in the +initial work of expressing the $\Sigma$-function in terms of the~$s_k$. + +But when a term of the $\Sigma$-function involves a large number of roots +with small exponents, we resort to a method suggested by~§104, which +tells us which auxiliary simpler symmetric functions %[** PP: Typo fuctions] +should be multiplied +together to produce our $\Sigma$-function along with simpler ones. + +\begin{Remark} +For example, to find $\Sigma x_1^2x_2x_3x_4$, when $n > 4$, we employ +\begin{gather*} +E_1E_4 \equiv \Sigma x_1 · \Sigma x_1x_2x_3x_4 + = \Sigma x_1^2x_2x_3x_4 + 5 \Sigma x_1x_2x_3x_4x_5, \\ +\Sigma x_1^2x_2x_3x_4 = E_1E_4 - 5E_5. +\end{gather*} + +To find $\Sigma x_1^2x_2^2x_3^2x_4$, employ $E_3E_4 = \Sigma x_1x_2x_3 · \Sigma x_1x_2x_3x_4$. + +When many such products of $\Sigma$-functions are to be computed, it will save time +in the long run to learn and apply the ``method of leaders'' explained in the author's +\textit{Elementary Theory of Equations}, pp.~64--65. +\end{Remark} + +%% -----File: 148.png---Folio 142------- + + +\begin{Exercises}[MISCELLANEOUS~]{Page142} + +Express in terms of the coefficients $c_1, \dotsc, c_n$: +\begin{Problems}[4] + +\item[1.] $\Sigma \alpha_1^2 \alpha_2 \alpha_3$. + +\item[2.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3$. + +% [** PP: Force into four columns] +\item[3.] \makebox[0pt][l]{$\Sigma \alpha_1^2 \alpha_2^2 \alpha_3 \alpha_4$.} + +\item[\quad4.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3^2$. +\end{Problems} + +If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + r = 0$, +find a cubic equation with the roots +\begin{Problems}[3] +\item[5.] $\alpha^2$, $\beta^2$, $\gamma^2$. + +\item[6.] $\alpha\beta$, $\alpha\gamma$, $\beta\gamma$. + +\item[7.] $\dfrac{2}{\alpha}$, $\dfrac{2}{\beta}$, $\dfrac{2}{\gamma}$. + +\ResetCols{2} + +\item[8.] $\alpha^2 + \beta^2$, $\alpha^2 + \gamma^2$, $\beta^2 + \gamma^2$. + +\item[9.] $\alpha^2 + \alpha\beta + \beta^2$, etc. +\end{Problems} + +If $\alpha$, $\beta$, $\gamma$, $\delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, find +\begin{Problems} +\item[10.] $\Sigma \dfrac{\beta}{\alpha} + = \Sigma \dfrac{\beta + \gamma + \delta}{\alpha} + = \Sigma \dfrac{-p - \alpha}{\alpha} + = -4 - p \Sigma \frac{1}{\alpha}$. + +\item[11.] $\Sigma \dfrac{\beta}{\alpha^2}$. Use + $\Sigma \dfrac{1}{\alpha}·\Sigma \dfrac{\beta}{\alpha} + = \Sigma \dfrac{\beta}{\alpha^2} + + 3\Sigma \dfrac{1}{\alpha} + + 2\Sigma \dfrac{\gamma}{\alpha\beta}$. + +\item[12.] Express $\Sigma \alpha_1^a \alpha_2^b \alpha_3^c \alpha_4^d$ in terms of the $s_k$ when (\emph{i})~$a>b>c>d>0$, and (\emph{ii})~when +$a=b=c=d$. + +\item[13.] By solving the first $k$ of Newton's identities~\Eq{10} as a system of linear equations, +find an expression in the form of a determinant (\emph{i})~for $s_k$ in terms of +$c_1, \dotsc, c_k$, and +(\emph{ii})~for $c_k$ in terms of $s_1, \dotsc, s_k$. + +\item[14.] One set of $n$~numbers is a mere rearrangement of another set if $s_1, \dotsc, s_n$ +have the same values for each set. +\end{Problems} +\end{Exercises} +\index{Symmetric functions|)}% +\index{Sigma function|)}% +\index{Sum of!like powers of roots|)}% + +%% -----File: 149.png---Folio 143------- + + +% [** PP: Not matching running head] +\Chapter{X}{Elimination, Resultants And Discriminants} +\index{Elimination|(}% +\index{Resultant|(}% + +% [** PP: Next four units have combined ToC entry] +\Section{110.}{Elimination} If the two equations +\begin{flalign*} +&& ax+b &= 0,\qquad cx+d = 0 && \Rightmark{(a \ne 0,\ c \ne 0)} +\end{flalign*} +are simultaneous, i.e., if $x$ has the same value in each, then % [** PP: Not italicizing i.e.] +\[ +x = -\frac{b}{a} = -\frac{d}{c}, \qquad +R \equiv ad - bc = 0, +\] +and conversely. Hence a necessary and sufficient condition that the +equations have a common root is $R = 0$. We call $R$ the \emph{resultant} (or +\emph{eliminant}) of the two equations. + +The result of eliminating~$x$ between the two equations might equally +well have been written in the form $bc - ad = 0$. But the arbitrary selection +of~$R$ as the resultant, rather than the product of~$R$ by some constant, +as~$-1$, is a matter of more importance than is apparent at first sight. For, +we seek a \emph{definite} function of the coefficients $a$, $b$, $c$, $d$ of the \emph{functions} +$ax+b$, $cx+d$, and not merely a property $R = 0$ or $R\ne 0$ of the corresponding +\emph{equations}. Accordingly, we shall lay down the definition in~§111, +which, as the reader may verify, leads to~$R$ in our present example. + +Methods of elimination which seem plausible often yield not $R$ itself, +but the product of~$R$ by an extraneous function of the coefficients. This +point (illustrated in~§114) indicates that the subject demands a more +careful treatment than is often given. + + +% [** No separate ToC entry] +\Section[Resultant of Two Polynomials] +{111.}{Resultant of Two Polynomials in~$x$.} Let +\begin{flalign*} +&& +\left\{ +\begin{aligned} +f(x) &= a_0x^m + a_1x^{m-1} + \dotsb + a_m \\ +g(x) &= \;b_0x^n + \;b_1x^{n-1} + \dotsb + \, b_n +\end{aligned} +\right. +&& +\begin{aligned} +&\Rightmark{(a_0 \ne 0),} \\ +&\Rightmark{(b_0 \ne 0)} +\end{aligned} +\Tag{1} +\end{flalign*} +be two polynomials of degrees $m$ and~$n$. Let $\alpha_1, \dotsc, \alpha_m$ be the roots +of $f(x) = 0$. Since $\alpha_1$ is a root of $g(x)=0$ only when $g(\alpha_1)=0$, the two +equations have a root in common if and only if the product +\[ +g(\alpha_1)g(\alpha_2) \dotsm g(\alpha_m) +\] +%% -----File: 150.png---Folio 144------- +is zero. This symmetric function of the roots of $f(x)= 0$ is of degree~$n$ +in any one root and hence is expressible as a polynomial of degree~$n$ in the +elementary symmetric functions~(§104), which are equal to $-a_1/a_0$, +$a_2/a_0, \dotsc$. To be rid of the denominators~$a_0$, it therefore suffices to +multiply our polynomial by~$a_0^n$. We therefore define +\[ +R(f, g) = a_0^n g(\alpha_1)g(\alpha_2) \dotsm g(\alpha_m) +\Tag{2} +\] +to be the \emph{resultant} of $f$ and~$g$. It equals an integral rational function of +$a_0, \dotsc, a_m$, $b_0, \dotsc, b_n$ with integral coefficients. +\index{Symbol!h@{$R(f, g)$\IndAdd{resultant}}}% [** PP: Manually alphabetized] + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] If $m = 1$, $n = 2$,\quad + $R(f, g) = b_0 a_1^2 - b_1 a_0 a_1 + b_2 a_0^2$. + +\item[2.] If $m = 2$, $n = 1$,\quad + $R(f,g) = a_0(b_0\alpha_1 + b_1)(b_0\alpha_2 + b_1) + = a_0 b_1^2 - a_1 b_0 b_1 + a_2 b_0^2$, since +\[ + a_0(\alpha_1 + \alpha_2) = -a_1,\qquad a_0 \alpha_1 \alpha_2 = a_2. +\] + +\item[3.] If $\beta_1, \dotsc, \beta_n$ are the roots of $g(x)= 0$, so that +\[ +g(\alpha_i) + = b_0(\alpha_i - \beta_1)(\alpha_i - \beta_2) \dotsm (\alpha_i - \beta_n), +\] +then +\[ +\begin{array}{r@{}l} +R(f,g) = a_0^n b_0^m %[** PP: Added ')' after \beta_2 below] + & (\alpha_1 - \beta_1)(\alpha_1 - \beta_2) \dotsm (\alpha_1 - \beta_n) \\ + ·& (\alpha_2 - \beta_1)(\alpha_2 - \beta_2) \dotsm (\alpha_2 - \beta_n) \\ + & \Dots{1} \\ + ·& (\alpha_m - \beta_1)(\alpha_m - \beta_2) \dotsm (\alpha_m - \beta_n). +\end{array} +\] +Multiplying together the differences in each column, we see that +\[ +R(f, g) = (-1)^{mn} b_0^m f(\beta_1)f(\beta_2) \dotsm f(\beta_n) + = (-1)^{mn} R(g, f). +\] + +\item[4.] If $m=2$, $n=1$,\quad + $R(g,f) = b_0^2 f( -b_1/b_0) = a_0 b_1^2 - a_1 b_0 b_1 + a_2 b_0^2$, +which is equal to $R(f,g)$ +by Ex.~2. This illustrates the final result in Ex.~3. + +\item[5.] If $m=n=2$, %[*Equation split to new line] +\begin{align*} +R(f,g) &= a_0^2 b_0^2 \alpha_1^2 \alpha_2^2 + + a_0^2 b_0 b_1 \alpha_1 \alpha_2 (\alpha_1 + \alpha_2) \\ +% + &\quad + a_0^2 b_0 b_2(\alpha_1^2 + \alpha_2^2) + + a_0^2 b_1^2 \alpha_1 \alpha_2 + + a_0^2 b_1 b_2(\alpha_1 + \alpha_2) + a_0^2 b_2^2 \\ +% + &= b_0^2 a_2^2 - b_0 b_1 a_1 a_2 + b_0 b_2(a_1^2 - 2a_0 a_2) + + b_1^2 a_0 a_2 - b_1 b_2 a_0 a_1 + a_0^2 b_2^2. +\end{align*} +This equals $R(g,f)$, since it is unaltered when the $a$'s and~$b$'s are interchanged. + +\item[6.] Prove by~\Eq{2} that $R$ is homogeneous and of total degree~$m$ in $b_0, \dotsc, b_n$; and +by Ex.~3, that $R$ is homogeneous and of total degree~$n$ in $a_0, \dotsc, a_m$. Show that $R$ +has the terms $a_0^n b_n^m$ and $(-1)^{mn} b_0^m a_m^n$. + +\item[7.] $R(f, g_1 g_2) = R(f, g_1) · R(f, g_2)$. + +\item[8.] $R(f, x^n) = (-1)^{mn} R(x^n, f) = (-1)^{mn} a_m^n$. +\end{Problems} +\end{Exercises} + +%% -----File: 151.png---Folio 145------- + + +% [** PP: No separate ToC entry] +\Section[Sylvester's Method of Elimination] +{112.}{Sylvester's Dialytic Method of Elimination.\protect\footnotemark}% +\addtocounter{footnote}{1}% + \footnotetext{Given without proof by Sylvester, \textit{Philosophical Magazine}, 1840, p.~132.} +\addtocounter{footnote}{-1}% +Let the equations +\index{Sylvester's eliminant}% +\[ +f(x) \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad +g(x) \equiv b_0x^2 + b_1x + b_2 = 0 +\] +have a common root~$x$. Multiply the first equation by $x$ and the second +by $x^2$ and $x$ in turn. We now have five equations +\begin{alignat*}{2} +a_0x^4 + a_1x^3 &+ a_2x^2 + a_3x &&= 0, \\ + a_0x^3 &+ a_1x^2 + a_2x + a_3 &&= 0, \\ +b_0x^4 + b_1x^3 &+ b_2x^2 &&= 0, \\ + b_0x^3 &+ b_1x^2 + b_2x &&= 0, \\ + &\phantom{{}+{}} b_0x^2 + b_1x + b_2 &&= 0, +\end{alignat*} +which are linear and homogeneous in $x^4$, $x^3$, $x^2$, $x$,~$1$. Hence~(§97) +\[ +F = \begin{vmatrix} + a_0 & a_1 & a_2 & a_3 & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3 \\ + b_0 & b_1 & b_2 & 0 & 0 \\ + 0 & b_0 & b_1 & b_2 & 0 \\ + 0 & 0 & b_0 & b_1 & b_2 + \end{vmatrix} +\Tag{3} +\] +must be zero. Next, if $F=0$, there exist~(§97) values which, when +substituted for $x^4$, $x^3$, $x^2$, $x$ and~$1$, satisfy the five equations. But why is +the value for $x^4$ the fourth power of the value for~$x$, that for $x^3$ the cube of +the value for~$x$, etc.? Since the direct verification of these facts would +be very laborious, we resort to a device to show that, conversely, if $F=0$ +the two given equations have a root in common. + +In~\Eq{3} replace $a_3$ by $a_3 - z$ and consider the equation +\[ +\begin{vmatrix} + a_0 & a_1 & a_2 & a_3-z & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3-z \\ + b_0 & b_1 & b_2 & 0 & 0 \\ + 0 & b_0 & b_1 & b_2 & 0 \\ + 0 & 0 & b_0 & b_1 & b_2 +\end{vmatrix} += 0. +\Tag{4} +\] +To prove that it has the roots $f(\beta_1)$ and $f(\beta_2)$, where $\beta_1$ and~$\beta_2$ are the roots +of $g(x) = 0$, we take $z = f(\beta_i)$ and prove that the determinant is then equal +to zero. For, if we add to the last column the products of the elements +%% -----File: 152.png---Folio 146------- +of the first four columns by $\beta_i^4$, $\beta_i^3$, $\beta_i^2$, $\beta_i$, respectively, we find that all +of the elements of the new last column are zero. + +Since \Eq{4} reduces to~\Eq{3} for $z=0$, it is of the form +\[ +b_0^3z^2 + kz + F = 0, +\] +in which the value of $k$ is immaterial. By considering the product of +the roots of this quadratic equation, we see that +\[ +F = b_0^3 f(\beta_1) f(\beta_2). +\] +Hence the Sylvester determinant~$F$ is the resultant $R(g, f)$ and hence +is the resultant $R(f, g)$, since $mn$ is here even (Ex.~3,~§111). +\index{Determinants|(}% + +In general, if the equations are +\[ +f(x) \equiv a_0x^m + \dotsb + a_m = 0,\qquad +g(x) \equiv b_0x^n + \dotsb + b_n = 0, +\] +we multiply the first equation by $x^{n-1}$, $x^{n-2}, \dotsc, x$, $1$, in turn, and the +second by $x^{m-1}$, $x^{m-2}, \dotsc, x$, $1$, in turn. We obtain $n+m$ equations +which are linear and homogeneous in the $m+n$ quantities $x^{m+n-1}, \dotsc, +x$, $1$. Hence the determinant +\[ +F= +\left| +\begin{array}{ccccccccccc} + a_0 & a_1 & a_2 & \Dots{2} & a_m & 0 & \Dots{3} & 0 \\ + 0 & a_0 & a_1 & a_2 & \Dots{2} & a_m & 0 & \Dots{2} & 0 \\ + 0 & 0 & a_0 & a_1 & a_2 & \Dots{2} & a_m & 0 & \dots & 0 \\ + \Dots{11} \\ + 0 & \Dots{2} & 0 & a_0 & a_1 & a_2 & \Dots{3} & a_m \\ + b_0 & b_1 & \Dots{3} & b_n & 0 & \Dots{3} & 0 \\ + 0 & b_0 & b_1 & \Dots{4} & b_n & \Dots{2} & 0 \\ + \Dots{11} \\ + 0 & \dots & 0 & b_0 & b_1 & \Dots{5} & b_n +\end{array} +\right| +% Now set braces, using empty arrays to get correct vertical alignment +{\setlength{\arraycolsep}{0pt} +\begin{array}{c} +\left. +\begin{array}{c} +\\ +\\ +\\ +\\ +\\ +\end{array} +\right\} \text{$n$~rows} \\ +\left. +\begin{array}{c} +\\ +\\ +\\ +\\ +\end{array} +\right\} \text{$m$~rows} +\end{array}} +\Tag{5} +\] +is zero. It may be shown to be equal to the resultant $R(f, g)$, whether +$mn$ is even or odd, by the method employed in the above case $m = 3$, $n = 2$. + +We may also prove as follows that if $F = 0$ the equations $f = 0$ and +$g = 0$ have a common root. Since $F$ was obtained as the determinant +of the coefficients of +\[ +x^{n-1}f, \dotsc, xf, f,\qquad +x^{m-1}g, \dotsc, xg, g, +\] +$F = 0$ implies, by~§96, Lemma~2, the existence of a linear relation +\[ +B_0x^{n-1}f + \dotsb + B_{n-2}xf + B_{n-1}f + +A_0x^{m-1}g + \dotsb + A_{m-2}xg + A_{m-1}g \equiv 0, +\] +%% -----File: 153.png---Folio 147------- +identically in~$x$, with constant coefficients $B_0, \dotsc, A_{m-1}$ not all zero. +In other words, $\beta f + \alpha g \equiv 0$, where +\[ +\alpha\equiv A_0x^{m-1} + \dotsb + A_{m-2}x + A_{m-1},\quad +\beta \equiv B_0x^{n-1} + \dotsb + B_{n-2}x + B_{n-1}. +\Tag{6} +\] + +Neither $\alpha$ nor~$\beta$ is identically zero. For, if $\alpha\equiv 0$, for example, then +$\beta f\equiv 0$ and $\beta\equiv 0$, whereas the $A_i$ and~$B_i$ are not all zero. + +Consider the factored forms of $f$, $g$, $\alpha$, $\beta$. Suppose that $f$ and~$g$ have +no common linear factor. The highest power of each linear factor occurring +in $f$ divides $\alpha g \equiv -\beta f$ and hence divides~$\alpha$. Thus $f$ divides~$\alpha$, whereas +$f$ is of higher degree than~$\alpha$. Hence our assumption that $f = 0$ and $g = 0$ +have no common root has led to a contradiction. + +\begin{Remark} +A similar idea is involved in the method of elimination due to Euler (1707--1783). +If $f=0$ and $g=0$ have a common root~$c$, then $f\equiv (x-c)\alpha$, $-g\equiv (x-c)\beta$, identically in~$x$, +where $\alpha$ and~$\beta$ are polynomials in~$x$ of degrees $m-1$ and~$n-1$, respectively. Give +them the notations~\Eq{6}. In the identity $\beta f + \alpha g \equiv 0$, the coefficient of each power of~$x$ +is zero. Hence +\index{Euler's eliminant}% +\[ +\begin{array}{l@{}r@{}l@{}r@{}l} +a_0B_0 & &{}+b_0A_0 & &=0\\ +a_1B_0+a_0B_1 & &{}+b_1A_0+b_0A_1 & &=0\\ +\Dots{5}\\ + & a_mB_{n-2}+a_{m-1}B_{n-1} & & {}+b_nA_{m-2}+b_{n-1}A_{m-1}&=0\\ + & a_mB_{n-1} & & {}+b_nA_{m-1}&=0. +\end{array} +\] +Since these $m+n$ linear homogeneous equations in the unknowns $B_0, \dotsc, B_{n-1}$, $A_0, \dotsc, +A_{m-1}$ have a set of solutions not all zero, the determinant of the coefficients is zero. %[** PP: , -> .] +By interchanging the rows and columns, we obtain the determinant~\Eq{5}. +% [** PP: Added.] +\end{Remark} + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] For $m=n=2$, show that the resultant is +\[ +R= +\begin{vmatrix} + a_0 & a_1 & a_2 & 0 \\ + 0 & a_0 & a_1 & a_2 \\ + b_0 & b_1 & b_2 & 0 \\ + 0 & b_0 & b_1 & b_2 +\end{vmatrix} +\] +Interchange the second and third rows, apply Laplace's development, and prove that +\index{Determinants!Laplace's development|(}% +\[ +R= (a_0b_2)^2 - (a_0b_1)(a_1b_2), +\] +where $(a_0b_2)$ denotes $a_0b_2 - a_2b_0$, etc. + +%% -----File: 154.png---Folio 148------- + +\item[2.] For $m = n = 3$, write down the resultant~$R$ and, by interchanges of rows, derive +the second determinant in +\[ +R= +\begin{vmatrix} +a_0 & a_1 & a_2 & a_3 & 0 & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3 & 0 \\ + 0 & 0 & a_0 & a_1 & a_2 & a_3 \\ +b_0 & b_1 & b_2 & b_3 & 0 & 0 \\ + 0 & b_0 & b_1 & b_2 & b_3 & 0 \\ + 0 & 0 & b_0 & b_1 & b_2 & b_3 +\end{vmatrix} += - +\begin{vmatrix} +a_0 & a_1 & a_2 & a_3 & 0 & 0 \\ +b_0 & b_1 & b_2 & b_3 & 0 & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3 & 0 \\ + 0 & b_0 & b_1 & b_2 & b_3 & 0 \\ + 0 & 0 & a_0 & a_1 & a_2 & a_3 \\ + 0 & 0 & b_0 & b_1 & b_2 & b_3 +\end{vmatrix} +\] +To the second determinant apply Laplace's development, selecting minors from the +first two rows, and to the complementary minors apply a similar development. This +may be done by inspection and the following value of~$-R$ will be obtained: +\[ +\begin{split} + & (a_0b_1) \bigl\{(a_1b_2)(a_2b_3) - (a_1b_3)^2+(a_2b_3)(a_0b_3)\bigr\} \\ +-& (a_0b_2) \bigl\{(a_0b_2)(a_2b_3) - (a_0b_3)(a_1b_3)\bigr\} \\ ++& (a_0b_3) \bigl\{(a_0b_1)(a_2b_3) - (a_0b_3)^2\bigr\}. +\end{split} +\] +The third term of the first line and the first term of the last line are alike. Hence, +changing the signs, +\index{Determinants!Laplace's development|)}% [** PP: Original page range is 147--149] +\[ +\begin{split} +R = (a_0b_3)^3 + &- 2(a_0b_1)(a_0b_3)(a_2b_3) - (a_0b_2)(a_0b_3)(a_1b_3) \\ + &+ (a_0b_2)^2 (a_2b_3) + (a_0b_1)(a_1b_3)^2 - (a_0b_1)(a_1b_2)(a_2b_3). +\end{split} +\] +\end{Problems} + +Other methods of simplifying Sylvester's determinant~\Eq{5} are given in~§113. +\end{Exercises} + + +% [** PP: No separate ToC entry] +\Section{113.}{Bézout's Method of Elimination} When the two equations are +of the same degree, the method published by Bézout in~1764 will be clear +from the example +\index{Bézout's eliminant}% +\[ +f \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad +g \equiv b_0x^3 + b_1x^2 + b_2x + b_3 = 0. +\] +Then +\[ +\begin{gathered} +a_0g - b_0f, \\ +(a_0x + a_1)g - (b_0x + b_1)f, \\ +(a_0x^2 + a_1x + a_2)g - (b_0x^2 + b_1x + b_2)f +\end{gathered} +\Tag{7} +\] +are equal respectively to +\begin{alignat*}{4} +(a_0b_1)x^2 & & &+{}& (a_0b_2)\;x &+{}& (a_0b_3) &= 0,\\ +(a_0b_2)x^2 &{}+{}& \bigl\{(a_0b_3) &+{}& (a_1b_2)\bigr\}x &+{}& (a_1b_3) &= 0, +\Tag{8} \\ +(a_0b_3)x^2 & & &+{}& (a_1b_3)\;x &+{}& (a_2b_3) &= 0, +\end{alignat*} +%% -----File: 155.png---Folio 149------- +where $(a_0b_1) = a_0b_1 - a_1b_0$, etc. The determinant of the coefficients is the +negative of the resultant $R(f, g)$. Indeed, the negative of the determinant +is easily verified to have the expansion given at the end of Ex.~2 +just above. + +\begin{Remark} +To give a more instructive proof of the last fact, note that, by~\Eq{7}, equations~\Eq{8} +are linear combinations of +\[ +x^2f = 0,\qquad xf = 0,\qquad f = 0,\qquad +x^2g = 0,\qquad xg = 0,\qquad g = 0, +\] +the latter being the equations used in Sylvester's method of elimination. The determinant +\index{Sylvester's eliminant}% +of the coefficients in these six equations is the first determinant~$R$ in Ex.~2 just +above. The operations carried out to obtain equations~\Eq{8} are seen to correspond +step for step to the following operations on determinants. To the products of the elements +of the fourth row by $a_0$ add the products of the elements of the 1st, 2nd, 3rd, +5th, 6th rows by $-b_0$, $-b_1$, $-b_2$, $a_1$, $a_2$ respectively [corresponding to the formation +of the third function~\Eq{7}]. To the products of the elements of the fifth row by $a_0$ add +the products of the elements of the 2nd, 3rd, 6th rows by $-b_0$, $-b_1$, $a_1$ respectively [corresponding +to the second function~\Eq{7}]. Finally, to the products of the elements of the +sixth row by $a_0$ add the products of the elements of the third row by $-b_0$ [corresponding +to $a_0g - b_0f$]. Hence +\[ +a_0^3R = +\left| +\begin{array}{ccc;{2pt/2pt}ccc;{2pt/2pt}} +a_0 & a_1 & \Bare{a_2} & a_3 & 0 & \Bare{0} \\ + 0 & a_0 & \Bare{a_1} & a_2 & a_3 & \Bare{0} \\ + 0 & 0 & \Bare{a_0} & a_1 & a_2 & \Bare{a_3} \\ + 0 & 0 & 0 & (a_0b_3) & (a_1b_3) & (a_2b_3) \\ + 0 & 0 & 0 & (a_0b_2) & (a_0b_3)+(a_1b_2) & (a_1b_3) \\ + 0 & 0 & 0 & (a_0b_1) & (a_0b_2) & (a_0b_3) +\end{array}\; % Explicit space before right \vert +\right|, +\] +so that $R$ is equal to the $3$-rowed minor enclosed by the dots. The method of Bézout +therefore suggests a definite process for the reduction of Sylvester's determinant of +order~$2n$ (when $m = n$) to one of order~$n$. + +Next, for equations of different degrees, consider the example +\[ +f \equiv a_0x^4 + a_1x^3 + a_2x^2 + a_3x + a_4,\qquad +g \equiv b_0x^2 + b_1x + b_2. +\] +Then +\[ +a_0x^2g - b_0f,\qquad (a_0x + a_1)x^2g - (b_0x + b_1)f +\] +are equal respectively to +\[ +\begin{aligned} +&(a_0b_1)x^3 + (a_0b_2)x^2 - a_3b_0x - a_4b_0,\\ +&(a_0b_2)x^3 + \bigl\{(a_1b_2) - a_3b_0\bigr\}x^2 + - \bigl\{a_3b_1 + a_4b_0\bigr\}x - a_4b_1. +\end{aligned} +\] +The determinant of the coefficients of $x^3$, $x^2$, $x$, $1$ in these two functions and $xg$, $g$, after +the first and second rows are interchanged, is the determinant of order~$4$ enclosed by +dots in the second determinant below. Hence it is the resultant $R(f, g)$. + +%% -----File: 156.png---Folio 150------- + +As in the former example, we shall indicate the corresponding operations on Sylvester's +determinant +\index{Sylvester's eliminant}% +\[ +R= +\begin{vmatrix} + a_0 & a_1 & a_2 & a_3 & a_4 & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3 & a_4 \\ + b_0 & b_1 & b_2 & 0 & 0 & 0 \\ + 0 & b_0 & b_1 & b_2 & 0 & 0 \\ + 0 & 0 & b_0 & b_1 & b_2 & 0 \\ + 0 & 0 & 0 & b_0 & b_1 & b_2 +\end{vmatrix} +\] +Multiply the elements of the third and fourth rows by~$a_0$. In the resulting determinant +$a_0^2R$, add to the elements of the third row the products of the elements of the first, +second and fourth rows by $-b_0$, $-b_1$, $a_1/a_0$ respectively. Add to the elements of the +fourth row the products of those of the second by $-b_0$. We get +\[ +a_0^2R = +\left| +\begin{array}{cc;{2pt/2pt}cccc;{2pt/2pt}} + a_0 & \Bare{a_1} & a_2 & a_3 & a_4 & \Bare{0} \\ + 0 & \Bare{a_0} & a_1 & a_2 & a_3 & \Bare{a_4} \\ + 0 & 0 & (a_0b_2) & (a_1b_2)-a_3b_0 & -a_3b_1-a_4b_0 & -a_4b_1 \\ + 0 & 0 & (a_0b_1) & (a_0b_2) & -a_3b_0 & -a_4b_0 \\ + 0 & 0 & b_0 & b_1 & b_2 & 0 \\ + 0 & 0 & 0 & b_0 & b_1 & b_2 \\ + \end{array}\; +\right| +\] +Hence $R$ is equal to the minor enclosed by dots. +\end{Remark} + + +\begin{Exercises}{} + +\begin{Problems} + +\item[1.] For $m=3$, $n=2$, apply to Sylvester's determinant~$R$ exactly the same operations +as used in the last case in~§113 and obtain +\[ +R= +\begin{vmatrix} +(a_0b_2) & (a_1b_2)-a_3b_0 & -a_3b_1 \\ +(a_0b_1) & (a_0b_2) & -a_3b_0 \\ + b_0 & b_1 & b_2 +\end{vmatrix}. +\] + +\item[2.] For $m=n=4$, reduce Sylvester's~$R$ (as in the first case in~§113) to +\[ +\begin{vmatrix} +(a_0b_1) & (a_0b_2) & (a_0b_3) & (a_0b_4) \\ +(a_0b_2) & (a_0b_3) + (a_1b_2) & (a_0b_4)+(a_1b_3) & (a_1b_4) \\ +(a_0b_3) & (a_0b_4)+(a_1b_3) & (a_1b_4)+(a_2b_3) & (a_2b_4) \\ +(a_0b_4) & (a_1b_4) & (a_2b_4) & (a_3b_4) +\end{vmatrix}. +\] +\end{Problems} +\end{Exercises} + +%% -----File: 157.png---Folio 151------- + + +% [** PP: No ToC entry] +\Section{114.}{General Theorem on Elimination} +\begin{Thm} +If any method of eliminating~$x$ +between two equations in~$x$ leads to a relation $F = 0$, where $F$ is a polynomial +in the coefficients, then $F$ has as a factor the true resultant of the equations. +\end{Thm} + +Some of the preceding proofs become simpler if this theorem is applied. +For example, determinant~\Eq{3} is divisible by the resultant~$R$. Since the +diagonal term of~\Eq{3} is a term $a_0^2b_2^3$ of~$R$ (Ex.~6,~§111), $F$ is identical +with~$R$. + +The preceding general theorem is proved in the author's \textit{Elementary +Theory of Equations}, pp.~152--4. We shall here merely verify the theorem +in an instructive special case. Let +\[ +f \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad +g \equiv b_0x^3 + b_1x^2 + b_2x + b_3 = 0 +\] +have a common root $x\ne 0$. Then +\begin{align*} +-b_0f + a_0g &= (a_0b_1)x^2 + (a_0b_2)x + (a_0b_3), \\ +(b_3f - a_3g)/x &= (a_0b_3)x^2 + (a_1b_3)x + (a_2b_3). +\end{align*} +By Ex.~1 of~§112, the resultant of these two quadratic functions is +\[ +F= +\begin{vmatrix} +(a_0b_3) & (a_0b_1) \\ +(a_2b_3) & (a_0b_3) +\end{vmatrix}^2 +- +\begin{vmatrix} +(a_0b_3) & (a_0b_1) \\ +(a_1b_3) & (a_0b_2) +\end{vmatrix} +· +\begin{vmatrix} +(a_1b_3) & (a_0b_2) \\ +(a_2b_3) & (a_0b_3) +\end{vmatrix}. +\] +This is, however, not the resultant~$R$ of the cubic functions $f$,~$g$. To show +that $(a_0b_3)$ is an extraneous factor, note that the terms of $F$ not having +this factor explicitly are +\index{Elimination!extraneous factor}% +\[ +(a_0b_1) (a_2b_3) \bigl\{(a_0b_1)(a_2b_3) - (a_0b_2)(a_1b_3)\bigr\}. +\] +The quantity in brackets is equal to $-(a_0b_3)(a_1b_2)$, since, as in Ex.~2 +of~§101, +\[ +0 = \tfrac{1}{2} +\begin{vmatrix} + a_0 & a_1 & a_2 & a_3 \\ + b_0 & b_1 & b_2 & b_3 \\ + a_0 & a_1 & a_2 & a_3 \\ + b_0 & b_1 & b_2 & b_3 +\end{vmatrix} += (a_0b_1)(a_2b_3) - (a_0b_2)(a_1b_3) + (a_0b_3)(a_1b_2). +\] +We now see that $F=(a_0b_3)R$, where $R$ is given in Ex.~2 of~§112. This +method of elimination therefore introduces an extraneous factor $(a_0b_3)$. +The student should employ only methods of elimination (such as those +due to Sylvester, Euler, and Bézout) which have been proved to lead +to the true resultant. + +%% -----File: 158.png---Folio 152------- + + +\begin{Exercises}{Page152} + +Find the result of eliminating~$x$ and hence find all sets of common solutions of +\begin{Problems} +\item[1.] $x^2-y^2=9$, $xy = 5y$. + +\item[2.] $x^2 + y^2 = 25$, $x^2 + 3(c-1)x + c(y^2 - 25) = 0$. + +\item[3.] When $x^2 + ax + b = 0$ has a double root, what $3$-rowed determinant is zero? + +\item[4.] Find the roots of $x^6 + 3x^4 + 32x^3 + 67x^2 + 32x + 65 = 0$ by~§79. +\end{Problems} +\end{Exercises} + + +\Section{115.}{Discriminants} Let $\alpha_1, \dotsc, \alpha_m$ be the roots of +\index{Discriminant|(}% +\begin{flalign*} +&& f(x) &\equiv a_0x^m + a_1x^{m-1} + \dotsb + a_m = 0 + && \Rightmark{(a_0\ne 0),} +\Tag{9} \\ +\intertext{so that} +&& f(x) &\equiv a_0(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_m). +\Tag{10} +\end{flalign*} +As in~§44, we define the discriminant of~\Eq{9} to be +\[ +D = a_0^{2m-2}(\alpha_1 - \alpha_2)^2(\alpha_1 - \alpha_3)^2 \dotsm + (\alpha_1 - \alpha_m)^2(\alpha_2 - \alpha_3)^2 \dotsm + (\alpha_{m-1} - \alpha_m)^2. +\] +Evidently $D$ is unaltered by the interchange of any two roots. Since the +degree in any root is $2(m-1)$, the symmetric function~$D$ is equal to a +polynomial in $a_0, \dotsc, a_m$. Indeed, $a_0^{2m-2}$ is the lowest power of $a_0$ +sufficient to cancel the denominators introduced by replacing $\Sigma \alpha_1$ by +$-a_1/a_0, \dotsc, \alpha_1\alpha_2\dotsm \alpha_m$ by~$±a_m/a_0$. By differentiating~\Eq{10}, we see +that +\begin{align*} +f'(\alpha_1) + &= a_0(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3) \dotsm (\alpha_1 - \alpha_m),\\ +f'(\alpha_2) + &= a_0(\alpha_2 - \alpha_1)(\alpha_2 - \alpha_3) \dotsm (\alpha_2 - \alpha_m),\\ +f'(\alpha_3) + &= a_0(\alpha_3 - \alpha_1)(\alpha_3 - \alpha_2)(\alpha_3 - \alpha_4) \dotsm + (\alpha_3 - \alpha_m), +\end{align*} +etc. Hence +\begin{align*} + a_0^{m-1}f'(\alpha_1) \dotsm f'(\alpha_m) + &= a_0^{2m-1} (-1)^{1 + 2 + \dotsb + m-1} + (\alpha_1 - \alpha_2)^2\dotsm (\alpha_{m-1} - \alpha_m)^2 \\ + &= (-1)^{\frac{m(m-1)}{2}} a_0 D. +\end{align*} +By~\Eq{2}, the left member is the resultant of $f(x)$, $f'(x)$. Hence +\[ +D = (-1)^{\frac{m(m-1)}{2}} \frac{1}{a_0} R(f, f'). +\Tag{11} +\] + +%% -----File: 159.png---Folio 153------- + + +\begin{Exercises}{} +\index{Cubic equation|(}% + +\begin{Problems} + +\item[1.] Show that the discriminant of $f \equiv y^3 + py + q = 0$ is $-4p^3 - 27q^2$ by evaluating +the determinant of order five for $R(f, f')$. + +\item[2.] Prove that the discriminant of the product of two functions is equal to the product +of their discriminants multiplied by the square of their resultant. Hint: use +the expressions in terms of the differences of the roots. + +\item[3.] For $a_0 = 1$, show that the discriminant is equal to +\[ +\begin{vmatrix} +1 & \alpha_1 & \alpha_1^2 & \cdots & \alpha_1^{m-1} \\ +1 & \alpha_2 & \alpha_2^2 & \cdots & \alpha_2^{m-1} \\ +\Dots{5} \\ +1 & \alpha_m & \alpha_m^2 & \cdots & \alpha_m^{m-1} +\end{vmatrix}^2 += +\begin{vmatrix} + s_0 & s_1 & s_2 & \cdots & s_{m-1}\\ + s_1 & s_2 & s_3 & \cdots & s_m \\ +\Dots{5}\\ + s_{m-1} & s_m & s_{m+1} & \cdots & s_{2m-2} +\end{vmatrix} +\] +where $s_i = \alpha_1^i + \dotsb + \alpha_m^i$. See Ex.~4,~§88; Ex.~2,~§102. + +\item[4.] Hence verify that the discriminant of $x^3 + px + q = 0$ is equal to +\[ +\begin{vmatrix} + 3 & 0 & -2p \\ + 0 & -2p & -3q \\ +-2p & -3q & 2p^2 +\end{vmatrix} += -4p^3 - 27q^2. +\] + +\item[5.] By means of Ex.~1,~§113, show that the discriminant of $a_0x^3 + a_1x^2 + a_2x + a_3 = 0$ is +\[% [** PP: Not breaking] +-\begin{vmatrix} + 2a_0a_2 & a_1a_2 + 3a_0a_3 & 2a_1a_3 \\ + a_1 & 2a_2 & 3a_3 \\ + 3a_0 & 2a_1 & a_2 +\end{vmatrix} %\\ += 18a_0a_1a_2a_3 - 4a_0a_2^3 - 4a_1^3a_3 + a_1^2a_2^2 - 27a_0^2a_3^2. +\] +\end{Problems} +\end{Exercises} +\index{Discriminant|)}% + + +\begin{Exercises}[MISCELLANEOUS~]{Page153} + +\begin{Problems} +\item[1.] Find the equation whose roots are the abscissas of the points of intersection +of two general conics. + +\item[2.] Find a necessary and sufficient condition that +\[ + f(x) \equiv x^4 + px^3 + qx^2 + rx + s = 0 +\] +shall have one root the negative of another root. When this condition is satisfied, +what are the quadratic factors of~$f(x)$? Apply to Ex.~4,~§74. Hint: add and subtract +$f(x)$ and~$f(-x)$. + +\item[3.] Solve $f(x) \equiv x^4 - 6x^3 + 13x^2 - 14x + 6 = 0$, given that two roots $\alpha$ and~$\beta$ are such +that $2\alpha + \beta = 5$. Hint: $f(x)$ and $f(5-2x)$ have a common factor. + +\item[4.] Solve $x^3 + px + q = 0$ by eliminating $x$ between it and $x^2 + vx + w = y$ by the greatest +common divisor process, and choosing $v$ and~$w$ so that in the resulting cubic equation +for $y$ the coefficients of $y$ and~$y^2$ are zero. The next to the last step of the elimination +%% -----File: 160.png---Folio 154------- +gives $x$ as a rational function of~$y$. (Tschirnhausen, \textit{Acta Erudit.}, Lipsiae,~II, 1683, +p.~204.) + +\item[5.] Find the preceding $y$-cubic as follows. Multiply $x^2 + vx + w = y$ by~$x$ and replace +$x^3$ by~$-px-q$; then multiply the resulting quadratic equation in~$x$ by~$x$ and replace +$x^3$ by its value. The determinant of the coefficients of $x^2$, $x$, $1$ must vanish. + +\item[6.] Eliminate $y$ between $y^3 = v$, $x = ry + sy^2$, and get +\[ +x^3 - 3rsvx - (r^3v + s^3v^2) = 0. +\] +Take $s=1$ and choose %[** PP: Typo chose] +$r$ and~$v$ so that this equation shall be identical with $x^3 + px + q = 0$, +and hence solve the latter. (Euler,~1764.) + +\item[7.] Eliminate $y$ between $y^3 = v$, $x = f + ey + y^2$ and get +\[ +\begin{vmatrix} + 1 & e & f-x \\ + e & f-x & v \\ +f-x & v & ev +\end{vmatrix} +=0. +\] +This cubic equation in $x$ may be identified with the general cubic equation by choice +of $e$, $f$, $v$. % [** PP: , -> .] +Hence solve the latter. + +\item[8.] Determine $r$, $s$ and~$v$ so that the resultant of +\[ +y^3 = v,\qquad y = \frac{x+r}{y+s} +\] +shall be identical with $x^3 + px + q = 0$. (Bézout,~1762.) + +\item[9.] Show that the reduction of a cubic equation in~$x$ to the form $y^3 = v$ by the substitution +\[ +x = \frac{r + sy}{1 + y} +\] +is not essentially different from the method of Ex.~7. [Multiply the numerator and +denominator of~$x$ by $1 - y + y^2$.] + +\item[10.] Prove that the equation whose roots are the $n(n-1)$ differences $x_j-x_k$ of the +roots of $f(x)=0$ may be obtained by eliminating $x$ between the latter and $f(x+y)=0$ +and deleting from the eliminant the factor~$y^n$ (arising from $y = x_j - x_j = 0$). The +equation free of this factor may be obtained by eliminating~$x$ between $f(x)=0$ and +\index{Equation for differences of roots}% +\[ +\bigl\{f(x+y) - f(x)\bigr\}/y + = f'(x) + f''(x)\frac{y}{1·2} + \dotsb + + f^{(n)}(x)\frac{y^{n-1}}{1·2\dotsm n} = 0. +\] +This eliminant involves only even powers of~$y$, so that if we set $y^2 = z$ we obtain an +equation in~$z$ having as its roots the squares of the differences of the roots of $f(x)=0$. +\index{Equation for differences of roots!squares of differences}% +(Lagrange \textit{Résolution des équations}, 1798,~§8.) + +\item[11.] Compute by Ex.~10 the $z$-equation when $f(x) = x^3 + px + q$. +\end{Problems} +\end{Exercises} +\index{Cubic equation|)}% +\index{Determinants|)}% [** PP: Original entry range is 146--153] +\index{Elimination|)}% +\index{Resultant|)}% + +%% -----File: 161.png---Folio 155------- + + +\Appendix +\index{Fundamental theorem of algebra|(}% + +\begin{Theorem} +An equation of degree~$n$ with any complex coefficients +\[ +f(z) \equiv z^n + a_1 z^{n-1} + \dotsb + a_n = 0 +\] +has a complex \(real or imaginary\) root. +\end{Theorem} + +Write $z = x+iy$ where $x$ and~$y$ are real, and similarly $a_1 = c_1 + id_1$, etc. +By means of the binomial theorem, we may express any power of~$z$ in the +form $X+iY$. Hence +\[ +f(z) = \phi(x,y) + i\psi(x,y), +\Tag{1} +\] +where $\phi$ and~$\psi$ are polynomials with real coefficients. + +The first proof of the fundamental theorem was given by Gauss in +1799 and simplified by him in~1849. This simplified proof consists in +showing that the two curves represented by $\phi(x, y) = 0$ and $\psi(x, y) = 0$ +have at least one point $(x_1, y_1)$ in common, so that $z_1 = x_1 + iy_1$ is a root +of $f(z)= 0$. This proof is given in \ChapRef{V} of the author's \textit{Elementary +Theory of Equations}. + +We here give a shorter proof, the initial idea of which was suggested, +but not fully developed, by Cauchy.\footnote + {For a history of the fundamental theorem, see \textit{Encyclopédie des sciences mathématiques}, + tome~I, vol.~II, pp.~189--205.} + +\begin{Lemma}[1.] +$a_1 h + a_2 h^2 + \dotsb + a_n h^n$ is less in absolute value than any +assigned positive number~$p$ for all complex values of $h$ sufficiently small in +absolute value. +\end{Lemma} + +The proof differs from that of the auxiliary theorem in~§62 only in +reading ``in absolute value'' for ``numerically.'' + +We shall employ the notation $|z|$ for the absolute value $+\sqrt{x^2 + y^2}$ of +$z = x + iy$. +\index{Symbol!c@{$\lvert a\rvert$\IndAdd{modulus}}}% [** PP: Manually alphabetized] + +%% -----File: 162.png---Folio 156------- + +\begin{Lemma}[2.] +Given any positive number~$P$, we can find a positive number~$R$ +such that $|f(z)| > P$ if $|z| \geqq R$. +\end{Lemma} + +The proof is analogous to that in~§64. We have +\[ +f(z) = z^n(1+D),\qquad +D \equiv a_1\left(\frac{1}{z}\right) + \dotsb + + a_n\left(\frac{1}{z}\right)^n. +\] +Since (Ex.~5,~§8) the absolute value of a sum of two complex numbers +is equal to or greater than the difference of their absolute values, we have +\[ +|f(z)| \geqq |z|^n \bigl[1 - |D|\bigr]. +\] + +Let $p$ be any assigned positive number~$<1$. Applying Lemma~1 with +$h$ replaced by~$1/z$, we see that $|D| < p$ if $|1/z|$ is sufficiently small, i.e., % [** PP: Not italicizing i.e.] +if $\rho\equiv |z|$ is sufficiently large. Then +\[ +|f(z)| > \rho^n(1-p) \geqq P +\] +if $\rho^n \geqq P/(1-p)$, which is true if +\[ +\rho \geqq \sqrt[n]{\frac{P}{1-p}} \equiv R. +\] +This proves Lemma~2. + +\begin{Lemma}[3.] +Given a complex number a such that $f(a) \ne 0$, we can find +a complex number~$z$ for which $|f(z)| < |f(a)|$. +\end{Lemma} + +Write $z = a+h$. By Taylor's theorem~\Eq{8} of~§56, +\[ +f(a+h) = f(a) + f'(a)h + \dotsb + f^{(r)}(a)·\frac{h^r}{r!} + \dotsb + + f^{(n)}(a)·\frac{h^n}{n!}. +\] +Not all of the values $f'(a)$, $f''(a), \dotsc$ are zero since $f^{(n)}(a) = n!$. Let +$f^{(r)}(a)$ be the first one of these values which is not zero. Then +\[ +\frac{f(a+h)}{f(a)} + = 1 + \frac{f^{(r)}(a)}{f(a)}·\frac{h^r}{r!} + \dotsb + + \frac{f^{(n)}(a)}{f(a)}·\frac{h^n}{n!}. +\] +Writing the second member in the simpler notation +\[ +g(h) \equiv 1 + bh^r + ch^{r+1} + \dotsb + lh^n,\qquad b\ne 0, +\] +we shall prove that a complex value of~$h$ may be found such that $|g(h)| < 1$. +Then the absolute value of $f(z)/f(a)$ will be~$<1$ and Lemma~3 proved. +To find such a value of~$h$, write $h$ and~$b$ in their trigonometric forms~(§4) +\[ +h = \rho(\cos \theta + i \sin \theta),\qquad +b = |b|(\cos \beta + i \sin \beta). +\] +%% -----File: 163.png---Folio 157------- +Then by~§5,~§7,% [** PP: Retaining instead of §§5,~7] +\[ +bh^r = |b| \rho^r \bigl\{\cos(\beta+r\theta) + i\sin (\beta+r\theta)\bigr\}. +\] +Since $h$ is at our choice, $\rho$ and angle~$\theta$ are at our choice. We choose~$\theta$ +so that $b + r\theta = 180°$. Then the quantity in brackets reduces to~$-1$, +whence +\[ +g(h) = (1 - |b|\rho^r) + h^r(ch + \dotsb + lh^{n-r}). +\] +By Lemma~1, we may choose $\rho$ so small that +\[ +|ch + \dotsb + lh^{n-r}| < |b|. +\] +By taking $\rho$ still smaller if necessary, we may assume at the same time +that $|b| \rho^r < 1$. Then +\[ +|g(h)| < (1 - |b|\rho^r) + \rho^r|b|,\qquad |g(h)|<1. +\] + +\paragraph{\indent Minimum Value of a Continuous Function.} Let $F(x)$ be any polynomial +\index{Continuity}% +\index{Minimum}% +with real coefficients. Among the real values of~$x$ for which +$2\leqq x\leqq 3$, there is at least one value~$x_1$ for which $F(x)$ takes its minimum +value~$F(x_1)$, i.e., % [** PP: Not italicizing i.e.] +for which $F(x_1)\leqq F(x)$ for all real values of~$x$ such that +$2\leqq x\leqq 3$. This becomes intuitive geometrically. The portion of the +graph of $y = F(x)$ which extends from its point with the abscissa~$2$ to its +point with the abscissa~$3$ either has a lowest point or else has several +equally low points, each lower than all the remaining points. The arithmetic +proof depends upon the fact that $F(x)$ is continuous for each~$x$ +between $2$ and~$3$ inclusive~(§62). The proof is rather delicate and is +omitted since the theorem for functions of one variable~$x$ is mentioned +here only by way of introduction to our case of functions of two variables. + +We are interested in the analogous question for +\[ +G(x,y) = \phi^2(x, y) + \psi^2(x, y), +\] +which, by~\Eq{1}, is the square of~$|f(z)|$. As in the elements of solid analytic +geometry, consider the surface represented by $Z = G(x,y)$ and the right +circular cylinder $x^2 + y^2 = R^2$. Of the points on the first surface and on +or within their curve of intersection there is a lowest point or there are +several equally low lowest points, possibly an infinite number of them. +Expressed arithmetically, among all the pairs of real numbers $x$,~$y$ for +%% -----File: 164.png---Folio 158------- +which $x^2 + y^2\leqq R^2$, there is\footnote + {Harkness and Morley, \textit{Introduction to the Theory of Analytic Functions}, p.~79, + prove that a real function of two variables which is continuous throughout % [** PP: Typo thoroughout] + a closed + region has a minimum value at some point of the region.} +at least one pair $x_1$,~$y_1$ for which the +polynomial $G(x,y)$ takes a minimum value $G(x_1, y_1)$, i.e., for which % [** PP: Not italicizing i.e.] +$G(x_1, y_1) \leqq G(x, y)$ for all pairs of real numbers $x$,~$y$ for which $x^2 + y^2 \leqq R^2$. + +\paragraph{\indent Proof of the Fundamental Theorem.} Let $z'$ denote any complex +number for which $f(z')\ne 0$. Let $P$ denote any positive number exceeding +$|f(z')|$. Determine $R$ as in Lemma~2. In it the condition $|z|\geqq R$ may +be interpreted geometrically to imply that the point $(x,y)$ representing +$z = x + iy$ is outside or on the circle~$C$ having the equation $x^2 + y^2 = R^2$. +Lemma~2 thus states that, if $z$ is represented by any point outside or on +the circle~$C$, then $|f(z)|>P$. In other words, if $|f(z)|\leqq P$, the point +representing~$z$ is inside circle~$C$. In particular, the point representing~$z'$ +is inside circle~$C$. + +In view of the preceding section on minimum value, we have +\[ +G(x_1,y_1) \leqq G(x,y) +\] +for all pairs of real numbers $x$,~$y$ for which $x^2 + y^2\leqq R^2$, where $x_1$,~$y_1$ is one +such pair. Write $z_1$ for $x_1 + iy_1$. Since $|f(z)|^2 = G(x,y)$, we have +\[ +|f(z_1)|\leqq |f(z)| +\] +for all $z$'s represented by points on or within circle~$C$. Since $z'$ is represented +by such a point, +\[ +|f(z_1)|\leqq |f(z')| < P. +\Tag{2} +\] + +This number $z_1$ is a root of $f(z)=0$. For, if $f(z_1)\neq 0$, Lemma~3 shows +that there would exist a complex number $z$ for which +\[ +|f(z)| < |f(z_1)|. +\Tag{3} +\] +Then $|f(z)| < P$ by~\Eq{2}, so that the point representing~$z$ is inside circle~$C$, +as shown above. By the statement preceding~\Eq{2}, +\[ +|f(z_1)|\leqq |f(z)|. +\] +But this contradicts~\Eq{3}. Hence the fundamental theorem is proved. +\index{Fundamental theorem of algebra|)}% + +%% -----File: 165.png---Folio 159------- + +\PrintAnswers + +\begin{Answers}[5]{Page2}%, 3} + +\item[1.] $3i$. + +\item[2.] $2$. + +\item[3.] $-20 + 20i$. + +\item[4.] $-\frac{2}{3}$. + +\item[5.] $(8 + 2\sqrt{3})$. + +\ResetCols{2} + +\item[6.] $\frac{1}{5}(6 + \sqrt{5}) + \frac{1}{5}(2\sqrt{5} - 3)i$. + +\item[7.] $\dfrac{-9}{13} + \dfrac{19}{13} i$. + +\ResetCols{2} + +\item[8.] $\dfrac{a^2 - b^2}{a^2 + b^2} + \dfrac{2ab}{a^2 + b^2}i$. + +\item[10.] Yes. + +\ResetCols{2} + +\item[13.] $3$, $4$ and $-3$, $-4$. + +\item[14.] $±(5 + 6i)$. + +\ResetCols{2} + +\item[15.] $±(3 - 2i)$. + +\item[16.] $±\bigl[c + d + (c - d)i\bigr]$. +\end{Answers} + + +\begin{Answers}[1]{Page6}%, 7} + +\item[2.] $-3$, $-3\omega$, $-3\omega^2$;\quad $i$, $\omega i$;\quad $\omega^2 i$;\quad + $R = \cos 40° + i\sin 40°$, $\omega R$, $\omega^2 R$. + +\item[3.] $±(1 + i)/\sqrt{2}$;\quad $±(1 - i)/\sqrt{2}$;\quad $±\omega^2$. +\end{Answers} + + +\begin{Answers}[1]{Page9} + +\item[4.] $-1$, $\cos A + i \sin A$ ($A=36°$, $108°$, $252°$, $324°$). + +\item[6.] $R^3$, $R^6$, $R^9$. +\end{Answers} + + +\begin{Answers}{Page10} + +\item[5.] $p(p-1)$. + +\item[6.] $(p-1)(q-1)(r-1)$ if $n=pqr$. +\end{Answers} + + +\begin{Answers}{Page13} + +\item[1.] $51$. + +\item[2.] $13$. +\end{Answers} + + +\begin{Answers}{Page15} + +\item[1.] Rem.~$11$, quot.~$x^2 + 5x + 8$. + +\item[2.] $-61$, $2x^4 - 4x^3 + 7x^2 - 14x + 30$. + +\ResetCols{1} + +\item[3.] $-0.050671$, $x^2 + 6.09x + 10.5481$. + +\item[4.] $x^2 - x - 6$, $x+2$;\quad $4$, $3$, $-2$. + +\item[5.] $x^2 - x - 6 = 0$, $3$, $-2$. + +\ResetCols{3} + +\item[6.] $2±\sqrt{5}$. + +\item[7.] $2x^2 - x + 2$. + +\item[8.] $x^2 + 1$. +\end{Answers} + + +\begin{Answers}{Page17} + +\item[1.] $x^3 - 3x^2 + 2x = 0$. + +\item[2.] $x^4 - 5x^2 + 4 = 0$. + +\ResetCols{2} + +\item[3.] $x^4 - 18x^2 + 81 = 0$. + +\item[4.] $x^4 - 5x^3 + 9x^2 - 7x + 2 = 0$. + +\ResetCols{2} + +\item[5.] $b^2 = 4ac$. + +\item[7.] By theorem in~§18. +\end{Answers} + +%% -----File: 166.png---Folio 160------- + + +\begin{Answers}{Page19} + +\item[1.] $x^3 - 6x^2 + 11x - 6 = 0$. + +\item[2.] $x^4 - 8x^2 + 16 = 0$. + +\ResetCols{4} + +\item[3.] $1$, $2$. + +\item[5.] $4$, $\tfrac{3}{2}$, $-\tfrac{3}{2}$. + +\item[6.] $1$, $3$, $5$. + +\item[7.] $1$, $1$, $1$, $3$. + +\ResetCols{3} + +\item[8.] $2$, $-6$, $18$. + +\item[9.] $-3$, $1$, $5$. + +\item[10.] $5$, $2$, $-1$, $-4$. + +\ResetCols{2} + +\item[11.] $y^2 - (p^2 - 2q)y + q^2 = 0$. + +\item[12.] $y^2 - (p^3 - 3pq)y + q^3 = 0$. + +\ResetCols{1} + +\item[13.] (i) $y^2 - y(p^3 - 3pq)/q + q = 0$. \\ + (ii) $y^2 - q(p^2 - 2q)y + q^4 = 0$. \\ + (iii) $y^2 - (p + p/q)y + 2 + q + 1/q = 0$. + +\ResetCols{2} + +\item[14.] $p^3r = q^3$. + +\item[15.] $2$, $4$, $-6$. +\end{Answers} + + +\begin{Answers}{Page20} + +\item[1.] $5$, $-1±\sqrt{-3}$. + +\item[2.] $1±i$, $1±\sqrt{2}$. + +\ResetCols{2} + +\item[3.] $x^3 - 7x^2 + 19x - 13 = 0$. + +\item[4.] $4$, $1-\sqrt{-5}$, $x^3 - 6x^2 + 14x - 24 = 0$. + +\ResetCols{2} + +\item[6.] $± 1$, $2±\sqrt{3}$. + +\item[7.] $\sqrt{3}$, $2±i$. + +\ResetCols{2} + +\item[9.] $x^3 - \tfrac{3}{2}x^2 - \tfrac{5}{4}x + \tfrac{7}{8} = 0$. + +\item[10.] $2 + \sqrt{3}$, $x^2 + 2x + 2 = 0$. + +\ResetCols{2} + +\item[11,] \textbf{12.}~Not necessarily. + +\item[13.] No. +\end{Answers} + + +\begin{Answers}[5]{Page23} + +\item[1.] $19\tfrac{1}{4}$, $3$. + +\item[2.] $6$. + +\item[3.] $2$. + +\item[4.] $3$. + +\item[5.] $0, -7, -\tfrac{7}{3}$. +\end{Answers} + + +\begin{Answers}{Page25} + +\item[1.] $-1$, $-1$, $-6$. + +\item[2.] $-2$, $3$, $4$. + +\ResetCols{3} + +\item[3.] $1$, $3$, $6$. + +\item[4.] $-2$, $-4$. + +\item[5.] None. +\end{Answers} + + +\begin{Answers}{Page27} + +\item[1.] $2$, $-1$, $-4$, $5$. + +\item[2.] $9$. + +\ResetCols{3} + +\item[3.] $8$, $9$. + +\item[4.] $-12$, $-35$. + +\item[5.] $2$, $2$, $-3$. + +\end{Answers} + + +\begin{Answers}[4]{Page28} + +\item[1.] $1$, $3$, $9$, $\frac{1}{3}$. + +\item[2.] $1$, $\tfrac{1}{2}$, $\tfrac{1}{3}$. + +\item[3.] $-\tfrac{1}{6}$. + +\item[4.] $\tfrac{1}{2}$, $-\tfrac{1}{4}$, $-\tfrac{1}{4}$. + +\ResetCols{4} + +\item[5.] $\tfrac{1}{4}$, $-\tfrac{1}{4}$, $\tfrac{1}{6}$. + +\item[6.] $-\tfrac{1}{2}$, $\tfrac{1}{3}$, $\tfrac{1}{4}$. + +\item[7.] $\tfrac{1}{2}$. + +\item[8.] $\tfrac{2}{3}$. + +\ResetCols{2} + +\item[10.] $x^2 - 12x - 12 = 0$. + +\item[11.] $x^3 - 3x^2 - 12x + 54 = 0$. +\end{Answers} + + +\begin{Answers}[3]{Page30} + +\item[1.] $1$, $4$. + +\item[2.] $-1$, $-4$. + +\item[3.] $0.7$, $-5.7$. + +\ResetCols{3} + +\item[4.] $-0.7$, $5.7$. + +\item[5.] $2$, $2$. + +\item[6.] Imaginary. +\end{Answers} + + +\begin{Answers}[1]{Page40} + +\item[5.] $x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$. + +\item[6.] $-\tfrac{1}{2}(1±\sqrt{-3})$, $\tfrac{1}{2}(7±\sqrt{45})$. + +\item[10.] See~\Eq{11},~§32. + +\item[11.] Edges roots of $x^3 - 7x^2 + 12x - v = 0$, all real~(§45) and irrational. + +\item[14.] $\Delta = \text{area}$, $c = \text{hypotenuse}$, squares of legs $\tfrac{1}{2}(c^2 ± \sqrt{c^4 - 16\Delta^2})$. + +\item[15.] $\Delta$ area, $a$, $b$ given sides, square third side is $a^2 + b^2 ± 2\sqrt{a^2b^2 - 4\Delta^2}$. + +\item[16.] $y^4 - 2y^3 + (2 - g^2)y^2 - 2y + 1 = 0$, pos.\ roots $0.09125$, $10.95862$. +\end{Answers} + +%% -----File: 167.png---Folio 161------- + + +\begin{Answers}[1]{Page44} + +\item[3.] $g=2$, $R + R^8 + R^{12} + R^5$, etc., $z^3 + z^2 - 4z + 1 = 0$. + +\item[4.] $g=2$, $R+R^8$, $R^2+R^7$, $R^4+R^5$. + +\ResetCols{2} + +\item[5.] $\tfrac{1}{2}(1±\sqrt{-3})$, $\tfrac{1}{2}(-5±\sqrt{21})$. + +\item[6.] $-1$, $2±\sqrt{3}$, $\tfrac{1}{2} ± \tfrac{1}{2}\sqrt{-3}$. + +\ResetCols{2} + +\item[7.] $1$, $1$, $1$, $-1$, $\tfrac{1}{4}(1±\sqrt{-15})$. + +\item[8.] $-1$, $-2$, $-\tfrac{1}{2}$, $\tfrac{1}{6}(-5±\sqrt{-11})$. + +\end{Answers} + + +\begin{Answers}{Page46} + +\item[1.] $-5$, $\tfrac{1}{2}(5±\sqrt{-3})$. + +\item[2.] $-6$, $±\sqrt{-3}$. + +\ResetCols{2} + +\item[3.] $-2$, $1± i$. + +\item[4.] $\tfrac{1}{4}$, $\tfrac{1}{7}(-2±\sqrt{-3})$. + +\end{Answers} + + +\begin{Answers}{Page48} + +\item[1.] $\Delta = -400$, one. + +\item[2.] $\Delta = 4 · 27 · 121$, three. + +\ResetCols{2} + +\item[3.] $\Delta = 0$, two. + +\item[4.] $\Delta = 0$, two. + +\end{Answers} + + +\begin{Answers}{Page49} + +\item[1.] $-4$, $2±\sqrt{3}$. + +\item[2.] See Ex.~1,~§47. + +\ResetCols{2} + +\item[3.] $1.3569$, $1.6920$, $-3.0489$. %[** PP: Horizontal layout] + +\item[4.] $-1.201639$, $1.330058$, $-3.128419$. + +\ResetCols{2} + +\item[5.] $1.24698$, $-1.80194$, $-0.44504$. + +\item[6.] $1.1642$, $-1.7729$, $-3.3914$. + +\end{Answers} + + +\begin{Answers}[3]{Page51} + +\item[1.] $1$, $-1$, $4±\sqrt{6}$. + +\item[2.] $-1$, $-2$, $2$, $3$. + +\item[3.] $1± i$, $-1±\sqrt{2}$. + +\ResetCols{2} + +\item[4.] $1±\sqrt{2}$, $-1±\sqrt{-2}$. + +\item[5.] $4$, $-2$, $-1± i$. + +\end{Answers} + + +\begin{Answers}{Page54}% (bottom)} + +\item[1.] $(-3, 9)$. + +\item[2.] $\Delta=-250000$, $x=3$, $-2$, $±i$. + +\ResetCols{3} + +\item[3.] $(3,9)$, $(-2,4)$. + +\item[4.] $h=3$. + +\item[5.] $6.856$, $7$. + +\end{Answers} + + +\begin{Answers}[1]{Page59}%, 60} + +\item[2.] $2.1$. + +\item[3.] $(-0.845, 4.921)$, $(-3.155, 11.079)$;\quad between $-4$ and~$-5$. + +\item[4.] $1.1$, $-1.3$. %[** PP: Added .] + +\item[5.] Between $0$ and~$1$, $0$ and~$-1$, $2.5$ and~$3$, $-2.5$ and~$-3$. + +\item[9.] $120(x^3 + x)$, $120x^2 - 42$. + +\end{Answers} + + +\begin{Answers}[3]{Page62} + +\item[1.] $3$. + +\item[2.] $2$, $-2$. + +\item[3.] $-1$. + +\ResetCols{3} + +\item[4.] Double roots, $1$, $3$. + +\item[5.] None. + +\item[6.] $3$, $3$, $-3$, $6$. + +\end{Answers} + + +\begin{Answers}{Page64}%, 65} + +\item[3.] Use Ex.~3, p.~62, abscissas $-1$, $3$. + +\item[\qquad4.] Use Ex.~2, p.~62. + +\ResetCols{1} + +\item[6.] $y = -15x - 7$, $X^3 - 15X + 23 = 0$. + +\end{Answers} + +%% -----File: 168.png---Folio 162------- + + +\begin{Answers}{Page66} + +\item[1.] One real. + +\item[2.] $(±\sqrt{\frac{7}{3}}, 7\mp\frac{14}{3}\sqrt{\frac{7}{3}})$, three real. + +\ResetCols{2} + +\item[3.] $(±\sqrt\frac{2}{3}, -1\mp\frac{4}{3}\sqrt{\frac{2}{3}})$, three. + +\item[4.] $(-2±\sqrt{5}, 23\mp10\sqrt{5})$, one. +\end{Answers} + + +\begin{Answers}[1]{Page74}%, 75} + +\item[13.] $y^5 + 2y^4 + 5y^3 + 3y^2 - 2y - 9 = 0$. + +\item[14.] $y^3 + 15y^2 + 52y - 36 = 0$. + +\end{Answers} + + +\begin{Answers}{Page78} + +\item[1.] One, between $-2$ and~$-3$. + +\item[2.] One, between $1$ and~$2$. + +\end{Answers} + + +\begin{Answers}{Page79}%, 80} + +\item[1.] $(-4, -3)$, $(-2, -1)$, $(1, 2)$. + +\item[2.] $(-2, -1)$, $(0, 1)$. + +\ResetCols{2} + +\item[3.] $(-2, -1.5)$, $(-1.5, -1)$, $(3, 4)$. + +\item[4.] $(-2, -1)$, $(0, 1)$. + +\ResetCols{2} + +\item[5.] $(-7, -6)$, $(1, 2)$. + +\item[6.] $(0, 1)$, $(3, 4)$. + +\end{Answers} + + +\begin{Answers}[3]{Page83} + +\item[2.] $1$, $1$, $1$, $2$. + +\item[3.] $1$, $1$, $-2$, $-2$. + +\item[4.] $1$, $1$, two imaginary. + +\end{Answers} + + +\begin{Answers}{Page85} + +\item[1.] $(-2, -1)$, $(0, 1)$, $(1, 2)$. + +\item[2.] $(-4, -3)$, $(-2, -1)$, $(1, 2)$. + +\end{Answers} + + +\begin{Answers}{Page89}%, 90} + +\item[1.] Single, $-2.46955$. + +\item[2.] $-1.20164$, $1.33006$, $-3.12842$. + +\ResetCols{2} + +\item[3.] $1.24698$, $-1.80194$, $-0.44504$. + +\item[4.] $± 2.1213203$, $\Neg2.1231056$, $-6.1231056$. + +\ResetCols{2} + +\item[5.] $3.45592$, $21.43067$. + +\item[6.] $2.15443$. + +\ResetCols{1} + +\item[7.] $-1.7728656$, $\Neg1.1642479$, $-3.3913823$. + +\item[8.] $\Neg3.0489173$, $-1.3568958$, $-1.6920215$. + +\ResetCols{2} + +\item[9.] $2.24004099$. + +\item[10.] $1.997997997$. + +\ResetCols{2} + +\item[11.] $1.094551482$. + +\item[12.] $2.059$, $-1.228$. + +\ResetCols{2} + +\item[13.] $1.2261$. + +\item[14.] $0.6527 = \text{reciprocal of } 2 \cos 40°$. + +\ResetCols{3} + +\item[15.] $0.9397$. + +\item[16.] $1.3500$. + +\item[17.] $2.7138$, $3.3840$. + +\ResetCols{3} + +\item[18.] $5.46\%$. + +\item[19.] $5.57\%$. + +\item[20.] $9.70\%$. + +\end{Answers} + + +\begin{Answers}{Page94} + +\item[1.] $2$. + +\item[2.] $3$. + +\end{Answers} +%% -----File: 169.png---Folio 163------- + + +\begin{Answers}[3]{Page96} + +\item[1.] $2.24004099$. + +\item[2.] $2.3593041$. + +\item[3.] $1.997998$. + +\end{Answers} + + +\begin{Answers}[4]{Page98} + +\item[1.] $132° 20.7'$. + +\item[2.] $157° 12'$. + +\item[3.] $4.8425364$. + +\item[4.] $3.1668771$. + +\ResetCols{1} + +\item[5,] \textbf{7.}~$15° 16\tfrac{1}{2}'$, + $85° 56\tfrac{1}{2}'$, + $212° 49'$, + $225° 57'$. + +\ResetCols{4} + +\item[6.] $72° 17'$. + +\item[8.] $5° 56\tfrac{1}{2}'$, $25° 18'$. + +\item[9.] $2.5541949$. + +\item[10.] $1.85718$. + +\end{Answers} + + +\begin{Answers}{Page99} + +\item[1.] $-1.04727± 1.13594 i$. + +\item[2.] $-\frac{2}{7} ± \frac{1}{7}\sqrt{3}i$. + +\ResetCols{3} + +\item[3.] $-1±i$. + +\item[4.] $1±i$, $1±2i$. + +\item[5.] $2±i$, $±2i$. + +\end{Answers} + + +\begin{Answers}{Page100} + +\item[1.] $217° 12' 27.4'' = 3.790988$ radians. + +\item[2.] $42° 20' 47\tfrac{1}{4}''$ doubled. + +\ResetCols{3} + +\item[3.] $133° 33.8'$. + +\item[4.] $108° 36' 14''$. + +\item[5.] $21.468212$. + +\ResetCols{2} + +\item[6.] Angle at center $47° 39' 13''$. + +\item[7.] $49° 17' 36.5''$. + +\ResetCols{1} + +\item[8.] $1.4303\pi$, $2.4590\pi$, $3.4709\pi$;\quad $257° 27' 12.225''$ more exact than first. + +\item[9.] $x/\pi = 0.6625, 1.891, 2.930, 3.948, 4.959$. + +\item[10.] (i) $0.327739$, $0.339224$, $1124.333037$. \\ + (ii) $0.250279$, $0.894609$, $1.127839$. \hfill\break + % [** PP: If \\ above, LaTeX thinks next token is an optional argument] + [Set $x = 1 + y$, $y = 1/z$ and solve by trigonometry.] + +\ResetCols{2} + +\item[11.] $3.597285$. + +\item[12.] $10$, $1.371288$. + +\ResetCols{2} + +\item[13.] $0.326878$, $12.267305$. + +\item[14.] $324° 16' 29.55''$. + +\ResetCols{3} + +\item[15.] $10$~yr.\ $4$~mo.\ $0$~days. + +\item[\qquad16.] $6.074674$. + +\item[17.] $6.13$\%. + +\end{Answers} + + +\begin{Answers}[3]{Page102} + +\item[1.] $x = 5$, $y = 6$. + +\item[2.] $x = 2$, $y = 1$. + +\item[3.] $x = a$, $y = 0$. + +\end{Answers} + + +\begin{Answers}[1]{Page106} + +\item[1.] $-a_2b_1c_3d_4 + a_2b_1c_4d_3 + a_2b_3c_1d_4 + - a_2b_3c_4d_1 - a_2b_4c_1d_3 + a_2b_4c_3d_1$. + +\item[2.] $+$, $+$. +\end{Answers} + + +\begin{Answers}{Page112} + +\item[3.] $-3$. + +\item[4.] $-8$. + +\end{Answers} + + +\begin{Answers}{Page115} + +\item[1.] $x = -8$, $y = -7$, $z = 26$. + +\item[2.] $x = 3$, $y = -5$, $z = 2$. + +\ResetCols{2} + +\item[3.] $x = 6$, $y = 3$, $z = 12$. + +\item[4.] $x = 5$, $y = 4$, $z = 3$. + +\ResetCols{2} + +\item[5.] $x = -5$, $y = 3$, $z = 2$, $w = 1$. + +\item[6.] $x = 1$, $y = z = 0$, $w = -1$. + +\end{Answers} + +%% -----File: 170.png---Folio 164------- + + +\begin{Answers}[1]{Page119} + +\item[1.] Consistent: $y = -8/7 - 2x$, $z = 5/7$ (common line). + +\ResetCols{2} + +\item[2.] Inconsistent, case $(\beta)$. + +\item[3.] Inconsistent (two parallel planes). + +\ResetCols{1} + +\item[4.] Consistent (single plane). + +\item[5.] % [** PP: Changing ;s to .s] + (i) $z = -x-y-2$. \hfill + (ii) inconsistent. \hfill + (iii) $x = \dfrac{a - 1}{a + 2}$, $y = z =\dfrac{-3}{a + 2}$. + +\item[6.] + (i) $x = \dfrac{(k-b)(c-k)}{(a-b)(c-a)}$. + (ii) $y = \dfrac{k-c}{a-c}-x$, + $z = \dfrac{a-k}{a-c}$ if $k=a$ or $k=c$, but + inconsistent if $k$ is different from $a$ and~$c$. + (iii) $z = 1 - x - y$ if $k=a$, inconsistent if $k\ne a$. + +\end{Answers} + + +\begin{Answers}{Page120} + +\item[1.] $r = 2$, $x:y:z = -4:1:1$. + +\item[2.] $r = 2$, $x:y:z = -10:8:7$. + +\ResetCols{2} + +\item[3.] $r = 1$, two unknowns arbitrary. + +\item[4.] $r = 3$, $x:y:z:w = 6:3:12:1$. + +\ResetCols{1} + +\item[5.] $r = 2$, $z = -\frac{11}{3} x - \frac{19}{3} y$, + $w = -\frac{10}{3} x - \frac{17}{3} y$. + +\end{Answers} + + +\begin{Answers}[1]{Page121} + +\item[1.] Ranks of $A$ and~$B$ are~$2$;\quad $y = -8/7 - 2x, z = 5/7$. + +\item[2.] Consistent only when $a = -225/61$ and then $x = -\dfrac{5}{61}$, $y = \dfrac{3}{61}$, $z = \dfrac{45}{61}$. + +\item[3.] Rank of~$A$ is~$2$, rank of~$B$ is~$3$, inconsistent. + +\item[4.] $A$ and~$B$ of rank~$2$, $x = 3$, $y = 2$. + +\end{Answers} + + +\begin{Answers}[1]{Page126}%, 127} + +\item[1.] $x = \dfrac{k(b-k)(c-k)(k+b+c)}{a(b-a)(c-a)(a+b+c)}$, if $a$, $b$, $c$ are distinct and not zero and their +sum $\neq 0$. If $a = b \neq c$, $ac \ne 0$, equations are inconsistent unless $k = 0$, $a$, $c$, or $-a-c$, +and then $y = \dfrac{k(c-k)}{a(c-a)} - x$, $z = \dfrac{k(k-a)}{c(c-a)}$, $x$~arbitrary. + +\ResetCols{2} + +\item[3.] $(a-b)(b-c)(c-a)$. + +\item[4.] $(x-y)(y-z)(z-x)(xy + yz + zx)$. + +\ResetCols{1} + +\item[6.] $(a+b+c+d)(a+b-c-d)(a-b-c+d)(a-b+c-d)$. + +\item[7.] $(a+b+c+d)(a-b+c-d)(a+bi-c-di)(a-bi-c+di)$. + +\item[11.] $\ds x_j = (k_1-a_j)\dotsm(k_n-a_j) + \div \prod\limits^n_{\substack{s=1 \\ s\neq j}} (a_s-a_j)$. + +\item[12.] $x(ab + ac + bc) = -abc$. + +\end{Answers} + +%% -----File: 171.png---Folio 165------- + + +\begin{Answers}{Page133}%,134} + +\item[1.] $\dfrac{p^4 - 3p^2q + 5pr + q^2}{r - pq}$. + +\item[\qquad2.] $\dfrac{(5p^2-12q)(p^2-4q)}{4(p^3 - 4pq + 8r)} - \dfrac{13}{4}p$. + +\ResetCols{2} + +\item[5.] $2p^2-2q$. + +\item[\qquad6.] $24r-p^3$. + +\ResetCols{2} + +\item[7.] $\dfrac{3p^2q^2 - 4p^3r - 4q^3 - 2pqr - 9r^2}{(r - pq)^2}$. + +\item[\qquad8.] $27r^2 - 9pqr + 2q^3 = 0$. + +\ResetCols{2} + +\item[10.] $y = q+r/x$. + +\item[11.] $x = \dfrac{1-py}{2+2y}$. + +\ResetCols{2} + +\item[12.] $y = \dfrac{4x^2 + px + q}{-3x-p}$, see~§112. + +\item[13.] $\dfrac{2q(p^3 + 2pq - r)}{p^2q - pr + s} - 5p$, see Ex.~17. + +\end{Answers} + + +\begin{Answers}{Page136} + +\item[3.] $s_2 = p^2 - 2q$, \\ + $s_3 = p^3 - 3pq$, \\ + $s_4 = p^4 - 4p^2q + 2q^2$, \\ + $s_5 = p^5 - 5p^3q + 5pq^2$. + +\item[4.] $s_{5n} = 5·3^n$, \\ + $s_k = 0$ if $k$ is not divisible by~$5$. + +\item[5.] All zero. + +\end{Answers} + + +\begin{Answers}[1]{Page140} + +\item[2.] See Ex.~2, p.~136. + +\item[3.] $\epsilon^j \sqrt[5]{\frac{1}{2}c + \sqrt{Q}} + + \epsilon^{5-j} \sqrt[5]{\frac{1}{2}c - \sqrt{Q}}$,\quad + $Q = \frac{1}{4}c^2 - q^5$\hfill ($j=0$, $1$, $2$, $3$, $4$). + +\item[4.] $\epsilon^j \sqrt[7]{\frac{1}{2}c + \sqrt{Q}} + + \epsilon^{7-j} \sqrt[7]{\frac{1}{2}c - \sqrt{Q}}$,\quad + $Q = \frac{1}{4}c^2 - q^7$\hfill ($j=0$, $1,\dotsc, 6$). + +\end{Answers} + + +\begin{Answers}{Page141} + +\item[1.] $c_2^2 - 2c_1c_3 + 2c_4$. + +\item[2.] $c_1^2c_2 - 2c_2^2 - c_1c_3 + 4c_4$. + +\ResetCols{2} + +\item[3.] $c_1c_3 - 4c_4$. + +\item[4.] $c_3^2 - 2c_2c_4$. + +\end{Answers} + + +\begin{Answers}{Page142} + +\item[1.] $c_1c_3 - 4c_4$ if $n>3$, $c_1c_3$ if $n=3$. + +\item[\qquad2.] $3c_1c_4 - c_2c_3 - 5c_5$. + +\ResetCols{2} + +\item[3.] $c_2c_4 - 4c_1c_5 + 9c_6$. + +\item[\qquad4.] $c_3^2 - 2c_2c_4 + 2c_1c_5 - 2c_6$. + +\ResetCols{1} + +\item[5.] $y^3 - (p^2 - 2q)y^2 + (q^2 - 2pr)y - r^2 = 0$. + +\ResetCols{2} + +\item[6.] $y^3 - qy^2 + pry - r^2 = 0$. + +\item[7.] $ry^3 + 2qy^2 + 4py + 8 = 0$. + +\ResetCols{2} + +\item[8.] Eliminate $x$ by $y = s_2 - x^2$. + +\item[9.] Use $p^2 - q + px = y$. + +\ResetCols{2} + +\item[10.] $-4 + pr/s$. + +\item[11.] $(rs - pr^2 + 2pqs)/s^2$. + +\ResetCols{1} + +\item[12.] (i) $s_a s_b s_c s_d - \Sigma s_a s_b s_{c+d} + + 2\Sigma s_a s_{b+c+d} + \Sigma s_{a+b} s_{c+d} - 6s_{a+b+c+d}$. \\ + (ii) $\tfrac{1}{24}(s_a^4 - 6s_a^2s_{2a} + + 8s_as_{3a} + 3s_{2a}^2 - 6s_{4a})$. + +%% -----File: 172.png---Folio 166------- + +\item[13.] (i) + \[ + s_k = - \left| + \begin{array}{cccccc} + 1 & 0 & 0 & \ldots & 0 & c_1 \\ + c_1 & 1 & 0 & \ldots & 0 & 2c_2 \\ + c_2 & c_1 & 1 & \ldots & 0 & 3c_3 \\ + c_3 & c_2 & c_1 & \ldots & 0 & 4c_4 \\ + \Dots{6} \\ + c_{k-1} &c_{k-2} &c_{k-3} & \ldots & c_1 & kc_k + \end{array}\;\right|,\quad + s_3 = - + \begin{vmatrix} + 1 & 0 & c_1\\ + c_1 & 1 & 2c_2\\ + c_2 & c_1 & 3c_3 + \end{vmatrix}, + \] + where all but the last term in the main diagonal is~$1$, and all terms above the + diagonal are zero except those in the last column. If $k>n$, we must take + $c_j =0 \quad (j>n)$. + + (ii) + \[ + k!\,c_k = - \left| + \begin{array}{cccccc} + 1 & 0 & 0 & \ldots & 0 & s_1 \\ + s_1 & 2 & 0 & \ldots & 0 & s_2 \\ + s_2 & s_1 & 3 & \ldots & 0 & s_3 \\ + \Dots{6} \\ + \ s_{k-1} & s_{k-2} & s_{k-3} & \ldots & s_1 & s_k + \end{array}\;\right|,\quad + 3!\,c_3 = - + \begin{vmatrix} + 1 & 0 & s_1\\ + s_1 & 2 & s_2\\ + s_2 & s_1 & s_3 + \end{vmatrix}. + \] + +\end{Answers} + + +\begin{Answers}[1]{Page152} + +\item[1.] $y^2(16 - y^2)$;\quad $y=0$, $x=±3$;\quad $y=±4$, $x=+5$. + +\item[2.] $(c-1)^2(y^2 - 25)(y^2 - 16)$. If $c\neq 1$, $y=±5$, $x=0$;\quad $y=±4$, $x=+3$. + +\ResetCols{2} + +\item[3.] +$\begin{vmatrix} + 1 & a & b \\ + 2 & a & 0 \\ + 0 & 2 & a +\end{vmatrix} = 4b-a^2$. + +\item[4.] $2±3i$, $-2±i$, $±i$. + $\vphantom{\begin{vmatrix}1\\ 1\\ 1\end{vmatrix}}$ + +\end{Answers} + + +\begin{Answers}[1]{Page153}%, 154} + +\item[2.] $pqr - p^2s - r^2 = 0$, $x^2 + r/p$, $x^2 + px + ps/r$. + +\ResetCols{2} + +\item[3.] $1$, $3$, $1± i$. + +\item[11.] See Ex.~15, p.~134. + +\end{Answers} + +%% -----File: 173.png---Folio 167------- + +\PrintIndex + +\iffalse + +\title % INDEX + +Numbers refer to pages. + +\begin(theindex) + +\item Abscissa, 55 + +\item Absolute value, 3 + +\item Amplitude, 3 + +\item Argument, 3 + +\item Arithmetical progression, 19 + +\indexspace + +\item Bend point, 56, 64 + +\item Bézout's eliminant, 148 + +\item Budan's theorem, 83 + +\indexspace + +\item Cardan's formulas, 46, 48 + +\item Complex number, 1 + \subitem geometrical representation, 3, 6, 7 + \subitem trigonometric form, 3 + +\item Compound interest, 13, 90, 100 + +\item Conjugate, 1 + +\item Continuity, 66, 157 + +\item Cube root, 5, 48 + \subitem of unity, 3, 4 + +\item Cubic equation, 32, 40, 45, 127, 134, 153--4 + \subitem graph of, 65 + \subitem number real roots, 48, 65, 79 + \subitem reduced, 45, 64--5 + \subitem trigonometric solution, 49 + +\indexspace + +\item De Moivre's quintic, 140 + \subitem theorem, 5 + +\item Derivative, 57--60, 69, 83, 97, 135 + +\item Descartes' rule of signs, 71, 85 + +\item Determinants, 101--27, 146--53 + \subitem addition of columns, 113 + \subitem columns, 103 + \subitem complementary minors, 122 + \subitem diagonal term, 103 + \subitem elements, 103 + \subitem expansion, 109 + \subitem interchanges, 106, 107 + \subitem Laplace's development, 122--3, 147--9 + \subitem minors, 109, 116 + \subitem of Vandermonde, 108 + \subitem product of, 124 + \subitem rank, 116, 121 + +\item Determinants, removal of factor, 111 + \subitem rows, 103 + \subitem signs of terms, 103--6 + \subitem skew symmetric, 108 + \subitem sum of, 112 + +\item Discriminant, 152--3 + \subitem of cubic, 47, 65, 134 + \subitem of quadratic, 11, 12 + \subitem of quartic, 51, 81 + +\item Double root, 16 (see Discriminant) + +\item Duplication of cube, 35 + +\indexspace + +\item Elementary symmetric function, 128 + +\item Elimination, 143--154 + \subitem extraneous factor, 151 + +\item Equation for differences of roots, 154 + \subitem squares of differences, 134, 154 + +\item Euler's eliminant, 147 + +\indexspace + +\item Factor theorem, 12 + +\item Factored form, 11, 15 + +\item Fundamental theorem of algebra, 17, 155--8 + +\indexspace + +\item Geometrical construction, 29--44 + \subitem progression, 13, 19 + +\item Graphs, 55--70 + +\item Greatest common divisor, 61, 75 + +\indexspace + +\item Horner's method, 86 + +\indexspace + +\item Identical polynomials, 16 + +\item Identity, 11 + +\item Imaginary, 2 + \subitem roots, 19, 98 + +\item Inflexion, 62--64 + +\item Integral rational function, 12, 17, 20 + \subitem roots, 24--27 + +\item Interpolation, 93, 97 + +\item Interval, 78 + +\item Irreducible case, 48 + +\item Isolation of roots, 71 + +%% -----File: 174.png---Folio 168------- + +\item Linear equations, system, 101--3, 114--21 + \subitem homogeneous, 119, 121 + +\item Linear factors, 11, 17 + +\item Lower limit to roots, 23 + +\indexspace + +\item Matrix, 120 + \subitem augmented, 121 + +\item Maximum, 57 + +\item Minimum, 57, 157 + +\item Modulus, 3 + +\item Multiple roots, 16, 60, 82 + +\item Multiplicity of root, 16, 20, 61 + +\indexspace + +\item Newton's identities, 136 + \subitem method of solution, 90--98 + +\item Number of roots, 16, 17, 48, 52, 69, 72--85 + \subitem of negative roots, 74 + +\indexspace + +\item Order of radical, 32 + +\item Ordinate, 55 + +\indexspace + +\item Plotting, 55 + +\item Polynomial, 12, 66 + \subitem sign of, 68 + +\item Primitive root of unity, 9 + +\item Product of roots, 18 + +\item Pure imaginary, 2 + +\indexspace + +\item Quadratic equation, 11 + \subitem graphical solution, 29, 55 + \subitem sum of powers of roots, 139 + +\item Quadratic function a square, 12 + +\item Quartic equation, 50--54, 80--81 + +\item Quotient by synthetic division, 14 + +\indexspace + +\item Rational roots, 27 + +\item Real equation, 12, 20 + +\item Reciprocal equation, 37, 44 + +\item Regula falsi, 93 + +\item Regular polygon, 8 + \subitem 7 sides, 35--36 + \subitem 9 sides, 35, 39 + \subitem 17 sides, 41--44 + \subitem \textit(n) sides, 44 + +\item Regular decagon, 39 + \subitem pentagon, 39 + +\item Relations between roots and coefficients, 17 + +\item Relatively prime, 9, 10 + +\item Remainder theorem, 12 + +\item Resolvent cubic, 50, 51 + +\item Resultant, 143--154 + +\item Rolle's theorem, 69 + +\item Root between \textit(a) and \textit(b), 67 + +\item Roots of unity, 8, 36, 39, 44, 136 + \subitem periods of, 40 + +\item Roots, \textit(n)th, 7 + +\indexspace + +\item Sigma function, 128--142 + +\item Sign of polynomial, 68 + +\item Simple root, 16 + +\item Slope, 57, 59 + +\item Solution of numerical equations, 86--100 + +\item Specific gravity, 89 + +\item Square roots, 1, 30, 31, 96 + +\item Sturm's functions, 75--82 + +\item Sum of four squares, 126 + \subitem like powers of roots, 134--142 + \subitem products of roots, 18 + \subitem roots, 18 + +\item Surd roots in pairs, 20 + +\item Sylvester's eliminant, 145, 149, 150 + +\item Symbol \equiv, 11; \textit(f)(\textit(x)), 12; $$\textit(a)$$,23, 155; \textit(r)!, 59; +\textit(f^(k)(x)), 59; \Sigma, 128; \textit(s_k), 134; \textit(R)(\textit(j), \textit(g)), +144 + +\item Symmetric functions, 128--142 + in all but one root, 132--4 + +\item Synthetic division, 13, 86--95 + +\indexspace + +\item Tangents, 60, 62 + +\item Taylor's theorem, 59 + +\item Transformed equation, 28, 86 + +\item Triple root, 16 + +\item Trisection of angle, 34, 40 + +\indexspace + +\item Upper limit to roots, 21--23 + +\indexspace + +\item Variation of sign, 71 + +\indexspace + +\item Waring's formula, 136 + +\end(theindex) +\fi + + +% LICENSE + +\phantomsection +\pdfbookmark[0]{PG License}{PG License} +\fancyhead[C]{\Running{LICENSING}} + +\begin{PGtext} +End of the Project Gutenberg EBook of First Course in the Theory of Equations, by +Leonard Eugene Dickson + +*** END OF THIS PROJECT GUTENBERG EBOOK THEORY OF EQUATIONS *** + +***** This file should be named 29785-pdf.pdf or 29785-pdf.zip ***** +This and all associated files of various formats will be found in: + http://www.gutenberg.org/2/9/7/8/29785/ + +Produced by Peter Vachuska, Andrew D. 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{\makebox(0,0)[r]{\hbox{\color{rgb_000000}$A$}}} +\pgftext[at={\pgfpoint{0in}{0.597326in}}] {\makebox(0,0)[t]{\hbox{\color{rgb_000000}$\alpha$}}} +\pgftext[at={\pgfpoint{1.52767in}{0.1125in}}] {\makebox(0,0)[l]{\hbox{\color{rgb_000000}$B$}}} +\pgftext[at={\pgfpoint{1.5in}{0.652674in}}] {\makebox(0,0)[b]{\hbox{\color{rgb_000000}$\beta$}}} +\pgftext[at={\pgfpoint{1.18622in}{0.652674in}}] {\makebox(0,0)[b]{\hbox{\color{rgb_000000}$T$}}} +\pgftext[at={\pgfpoint{0.538116in}{0.599945in}}] {\makebox(0,0)[tr]{\hbox{\color{rgb_000000}$c$}}} +\pgftext[at={\pgfpoint{0.75in}{-0.166044in}}] {\makebox(0,0)[t]{\hbox{\color{rgb_000000}\textsc{Fig.~26}}}} +\end{tikzpicture} diff --git a/29785-t/images/sources/009a.xp b/29785-t/images/sources/009a.xp new file mode 100644 index 0000000..6a1b4e1 --- /dev/null +++ b/29785-t/images/sources/009a.xp @@ -0,0 +1,37 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double dp(4); + +int main() +{ + picture(P(0,0), P(2,1.25), "1.875 x 1.1in"); + // offset(P(0,0.1)); + begin(); + + P A(polar(2, M_PI/6)), A1(A&E_1), O(0,0), X(xmax(),0), Y(0,ymax()); + + bold(); + line(O, X); + line(O, Y); + + path Tr(O, A); + Tr.pt(A1).draw(); + + label(O, P(-dp,0), "$O$", l); + label(A, P( dp,0), "$A$", r); + label(X, P(0,-dp), "$X$", b); + label(Y, P(-dp,0), "$Y$", l); + + label(O, P(20,2), "$\\theta$", tr); + + label(0.5*A1, P(0,-dp), "$a$", b); + label(0.5*(A+A1), P(dp,0), "$b$", r); + label(0.5*A, P(-2,2), "$r$", tl); + + label(P(0.5*xmax(), ymin()), P(0,-18), "\\textsc{Fig.~1}",b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/010a.xp b/29785-t/images/sources/010a.xp new file mode 100644 index 0000000..1b11b74 --- /dev/null +++ b/29785-t/images/sources/010a.xp @@ -0,0 +1,52 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +int main() +{ + picture(P(-1,-1), P(1,1), "2 x 2in"); + + begin(); + arrow_inset(0.25); + arrow_ratio(1); + degrees(); + P O(0, 0), z0(1, 0), z1(cis(120)), z2(cis(240)); + + double rad1(0.25), rad2(0.66*rad1), dp(2), dt(5); + + bold(); + circle(); + + line(P(-0.5), z0); + line(O, z1); + line(O, z2); + + arc_arrow(O, rad1, 0, 120-dt); + arc_arrow(O, rad2, 0, 240-dt); + + plain(); + dashed(); + triangle(z0, z1, z2); + + label(z0, dp*z0, "$1$", r); + label(z1, dp*z1, "$\\omega$", tl); + label(z2, dp*z2, "$\\omega^2$", bl); + + label(O, P(0,-dp), "$O$", br); + + label(P(-0.375), P(0,dp), "$\\frac{1}{2}$", t); + label(P(-0.5, 0.25*sqrt(3)), P(-dp,0), "$\\frac{1}{2}\\sqrt{3}$", l); + label(0.5*z1, dp*J(z1), "$1$", bl); + + label(P(0, ymin()), P(0,-6), "\\textsc{Fig.~2}", b); + + font_size("scriptsize"); + label_angle(-30); + label(polar(rad1, 60), -4*z2, "$120°$", c); + + label_angle(60); + label(polar(rad2, 150), P(-4,4), "$240°$", c); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/012a.xp b/29785-t/images/sources/012a.xp new file mode 100644 index 0000000..497845b --- /dev/null +++ b/29785-t/images/sources/012a.xp @@ -0,0 +1,55 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +void myquad(const P& p1, const P& p2, const P& p3, const P& p4) +{ + line(p1, p2); + line(p2, p3); + line(p3, p4); + line(p4, p1); +} + +int main() +{ + picture(P(0,0), P(3,2), "2 x 1.33in"); + + begin(); + + P ptA(1.25, 0.5), ptC(0.6, 1.375), ptS(ptA+ptC), X(xmax()), Y(0,ymax()); + P O(0,0), ptE(ptA&E_1), ptF(ptC&E_1), ptH(ptS&E_1), ptG(ptS.x1(), ptA.x2()); + double dp(2); + + dashed(); + line(ptA, ptE); + line(ptC, ptF); + line(ptS, ptH); + line(ptA, ptG); + + solid(); + bold(); + line(O, X); + line(O, Y); + + myquad(O, ptA, ptS, ptC); + line(O, ptS); + + label(X, P(0,-dp), "$X$", b); + label(Y, P(-dp,0), "$Y$", l); + label(O, P(-dp,0), "$O$", l); + + label(ptE, P(0,-dp), "$E$", b); + label(ptF, P(0,-dp), "$F$", b); + label(ptH, P(0,-dp), "$H$", b); + + label(ptG, P(dp,0), "$G$", r); + label(ptS, P(dp,0), "$S$", r); + + label(ptA, P(-dp, dp), "$A$", tl); + label(ptC, P(-dp, dp), "$C$", tl); + + label(P(0.5*xmax(), ymin()), P(0,-18), "\\textsc{Fig.~3}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/012b.xp b/29785-t/images/sources/012b.xp new file mode 100644 index 0000000..b6ecd3d --- /dev/null +++ b/29785-t/images/sources/012b.xp @@ -0,0 +1,47 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +P times(const P& arg1, const P& arg2) +{ + return P(arg1.x1()*arg2.x1() - arg1.x2()*arg2.x1(), + arg1.x1()*arg2.x2() + arg1.x2()*arg2.x1()); +} + +void mytri(const P& p1, const P& p2, const P& p3) +{ + line(p1, p2); + line(p2, p3); + line(p3, p1); +} + +int main() +{ + picture(P(0,0), P(2,1.33), "2 x 1.33in"); + + begin(); + + P ptA(1.25, 0.6), ptC(0.5, 0.5), ptP(times(ptA, ptC)), X(xmax()); + P O(0,0), ptU(1); + double dp(2); + + bold(); + line(O, X); + + mytri(O, ptU, ptA); + mytri(O, ptC, ptP); + + label(X, P(0,-dp), "$X$", b); + label(ptU, P(0,-dp), "$U$", b); + label(O, P(-dp,0), "$O$", l); + + label(ptA, P(dp,0), "$A$", r); + label(ptC, P(dp,0), "$C$", r); + + label(ptP, P(-dp,dp), "$P$", tl); + + label(P(0.5*xmax(), ymin()), P(0,-18), "\\textsc{Fig.~4}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/014a.xp b/29785-t/images/sources/014a.xp new file mode 100644 index 0000000..aa95153 --- /dev/null +++ b/29785-t/images/sources/014a.xp @@ -0,0 +1,32 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +int main() +{ + picture(P(-1.25,-1.25), P(1.25,1.25), "1.5 x 1.5in"); + offset(P(0,0.125)); + + begin(); + degrees(); + double dp(2); + + bold(); + circle(); + P i0(1,0), i1(0,1), i2(-1,0), i3(0,-1); + + quad(i0, i1, i2, i3); + line(i0, i2); + line(i1, i3); + + label(P(0,0), P(-dp,-dp), "$O$", bl); + label(i0, dp*i0, "$1$", r); + label(i1, dp*i1, "$i$", t); + label(i2, dp*i2, "$-1$", l); + label(i3, dp*i3, "$-i$", b); + + label(P(0,ymin()), P(0,-6), "\\textsc{Fig.~5}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/035a.xp b/29785-t/images/sources/035a.xp new file mode 100644 index 0000000..3f6772a --- /dev/null +++ b/29785-t/images/sources/035a.xp @@ -0,0 +1,46 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +int main() +{ + picture(P(-1,-1), P(8,5), "2.4 x 1.6in"); + + begin(); + double ca(6), cb(4); + double dp(2); + P O(0,0), ptB(0,1), ptQ(ca, cb), X(xmax(),0), Y(0,ymax()); + Circle Crc(0.5*(ptB+ptQ), ptQ); + Segment Xax(O, X), Yax(O, Y), MN(Crc*Xax), OC(Crc*Yax); + + P ptC(OC.end1()), ptM(MN.end1()), ptN(MN.end2()), ptT(ptQ&E_1); + + Segment QT(ptQ, ptT), QD(Crc*QT); + P ptD(QD.end1()); + + bold(); + Crc.draw(); + Xax.draw(); + + line(O, ptC); + line(ptQ, ptT); + line(ptQ, ptB); + + label(O, P(-dp,0), "$O$", l); + label(ptC, P(-dp,0), "$C$", l); + label(ptB, P(-dp,-dp), "$B$", l); + label(ptQ, P( dp, dp), "$Q$", r); + + label(ptM, P(-2*dp,dp), "$M$", t); + label(ptN, P( 2*dp,dp), "$N$", t); + + label(ptD, P(dp,-dp), "$D$", r); + + label(ptT, P(0,-dp), "$T$", b); + label(X, P(0,-dp), "$X$", b); + + label(P(0.5*(xmin()+xmax()), ymin()), P(0,-6), "\\textsc{Fig.~6}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/037a.xp b/29785-t/images/sources/037a.xp new file mode 100644 index 0000000..701afe9 --- /dev/null +++ b/29785-t/images/sources/037a.xp @@ -0,0 +1,30 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +int main() +{ + picture(P(0,0), P(3,3), "1.5 x 1.5in"); + + begin(); + degrees(); + bold(); + double ca(1.5), cb(1.25), cp(ca*cb), th(30); + double dp(2); + P O(0,0), p1(1+cb,0), p2(polar(ca+cp, th)), p2b(polar(ca, th)); + + triangle(O, p1, p2); + triangle(O, P(1), p2b); + + label(P(0.5), P(0,-dp), "$1$", b); + label(P(1+0.5*cb), P(0,-dp), "$b$", b); + + label(polar(0.5*ca, th), P(-dp,dp), "$a$", tl); + label_angle(th); + label(polar(ca*(1+0.5*cb), th), 3*P(-dp,dp), "$p=a·b$", c); + + label_angle(0); + label(P(0.5*xmax()), P(0,-18), "\\textsc{Fig.~7}", b); + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/037b.xp b/29785-t/images/sources/037b.xp new file mode 100644 index 0000000..f46ef2a --- /dev/null +++ b/29785-t/images/sources/037b.xp @@ -0,0 +1,30 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +// illogical variables; modified labels from 037a.xp +int main() +{ + picture(P(0,0), P(2.5,3), "1.25 x 1.5in"); + + begin(); + degrees(); + bold(); + double ca(1.9), cb(1.25), cp(ca*cb), th(50); + double dp(2); + P O(0,0), p1(1+cb,0), p2(polar(ca+cp, th)), p2b(polar(ca, th)); + + triangle(O, p1, p2); + triangle(O, P(1), p2b); + + label(P(0.5), P(0,-dp), "$1$", b); + label(P(1+0.5*cb), P(0,-dp), "$q=a/b$", b); + + label(polar(0.5*ca, th), P(-dp,dp), "$b$", tl); + label(polar(ca*(1+0.5*cb), th), P(-dp,dp), "$a$", tl); + + label_angle(0); + label(P(0.5*xmax()), P(0,-18), "\\textsc{Fig.~8}", b); + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/037c.xp b/29785-t/images/sources/037c.xp new file mode 100644 index 0000000..fe2eeaa --- /dev/null +++ b/29785-t/images/sources/037c.xp @@ -0,0 +1,28 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +int main() +{ + picture(P(-3,0), P(3,6), "1.5 x 1.5in"); + + begin(); + degrees(); + bold(); + double rad(3), X(-2), ht(sqrt((rad-X)*(rad+X))); + double dp(2); + + arc(P(0,0), rad, 0, 180); + line(P(-rad), P(rad)); + line(P(X,0), P(X,ht)); + + label(P(0.5*(-rad+X)), P(0,-dp), "$1$", b); + label(P(0.5*( rad+X)), P(0,-dp), "$n$", b); + + label(P(X, 0.5*ht), P(dp,0), "$s$", r); + + label(P(0), P(0,-18), "\\textsc{Fig.~9}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/045a.xp b/29785-t/images/sources/045a.xp new file mode 100644 index 0000000..fb37295 --- /dev/null +++ b/29785-t/images/sources/045a.xp @@ -0,0 +1,40 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +int main() +{ + picture(P(-1,-1), P(1,1), "1.75 x 1.75in"); + offset(P(0,-0.5)); + + begin(); + degrees(); + bold(); + double dp(2); + + P O(0,0), A(-1,0), Ap(1,0), B(0,1), Bp(0,-1), M(0.5,0), C(0.5*(1-sqrt(5))); + P R(B-M); + + circle(); + line(A, Ap); + line(B, Bp); + + arc(M, norm(R), Atan2(R.x2(), R.x1()), 180); + line(M, B); + line(B, C); + + label(A, P(-dp,0), "$A$", l); + label(Ap, P( dp,0), "$A'$", r); + + label(B, P(0, dp), "$B$", t); + label(Bp, P(0,-dp), "$B'$", b); + + label(M, P(0,-dp), "$M$", b); + label(C, P(0,-dp), "$C$", b); + label(O, P(-dp,-dp), "$O$", bl); + + label(P(0,ymin()), P(0,-18), "\\textsc{Fig.~10}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/049a.xp b/29785-t/images/sources/049a.xp new file mode 100644 index 0000000..8227301 --- /dev/null +++ b/29785-t/images/sources/049a.xp @@ -0,0 +1,79 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +int main() +{ + picture(P(-1.25,-1), P(2.25,1), "3.0625 x 1.75in"); + // offset(P(0,-0.5)); + + begin(); + degrees(); + bold(); + double dp(2); + + P O(0,0), pD(-1,0), pC(1,0), pB(0,1), pA(0,-1), pS(-1,-1), pE(-0.25,-1); + P pF(pE+P(sqrt(1.0625))), pFp(pE-P(sqrt(1.0625))); // F and F' + P pH(pF+P(norm(pF))), pHp(pFp+P(norm(pFp))); // H and H' + P pT(pH.x1()), pQ(pT+P(0, pHp.x1() - pA.x1())); // T and Q + + P ctr(0.5*(pB+pQ)), dir(pB-pQ); // circ(BQ) parameters + Circle Crc(ctr, pB); + Segment Xax(O, pT), MN(Crc*Xax); + P pM(MN.end1()), pN(MN.end2()); // M and N + + double L(0.5*pM.x1()); + + P pL(L), pP(L, sqrt((1-L)*(1+L))); // L and P + + circle(); + line(pA, pB); + line(pD, pT); + + line(pFp, pH); + line(pH, pQ); + line(pQ, pB); + line(O, pE); + line(O, pF); + line(O, pFp); + line(pD, pS); + line(pL, pP); + + arc(pF, norm(pF), 0, Atan2(-pF.x2(), -pF.x1())); + arc(pFp, norm(pFp), 0, Atan2(-pFp.x2(), -pFp.x1())); + + double th(Atan2(dir.x2(), dir.x1())); + arc(ctr, 0.5*norm(dir), th, th+180); + + label(P(0.5*(xmin()+xmax()), ymin()), P(0,-18), "\\textsc{Fig.~11}", b); + + font_size("footnotesize"); + + label(pA, P(0,-dp), "$A$", b); + label(pB, P(0, dp), "$B$", t); + + label(pC, P( dp,-dp), "$C$", br); + label(pD, P(-dp,0), "$D$", l); + + label(pM, P(-2*dp,dp), "$M$", t); + label(pN, P( 14, -dp), "$N$", b); + + label(pL, P(-dp,-dp), "$L$", b); + label(pP, P( dp, 0), "$P$", r); + + label(O, P(-dp,dp), "$O$", tl); + + label(pFp, P(0,-dp), "$F'$", b); + label(pS, P(0,-dp), "$S$", b); + label(pE, P(-dp,-dp), "$E$", b); + label(pA, P(0,-dp), "$A$", b); + label(pHp, P(0,-dp), "$H'$", b); + label(pF, P(0,-dp), "$F$", b); + label(pH, P(0,-dp), "$H$", b); + + label(pQ, P(dp,0), "$Q$", r); + label(pT, P(dp,0), "$T$", r); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/061a.xp b/29785-t/images/sources/061a.xp new file mode 100644 index 0000000..dd8dc01 --- /dev/null +++ b/29785-t/images/sources/061a.xp @@ -0,0 +1,59 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double x) +{ + return -3 + x*(x-6); +} + +void label(double x, double f(double), const P& offset, epix_label_posn A) +{ + P loc(x, f(x)); + std::stringstream buf; + buf << "$(" << x << "," << f(x) << ")$"; + + marker(loc, TIMES); + label(loc, offset, buf.str(), A); +} + +int main() +{ + picture(P(-4,-12), P(8,4), "1.8 x 2.4in"); + + begin(); + double dp(4); + P Q(4), pP(4,f(4)); + + bold(); + line(P(-2,0), P(9,0)); + line(P(0,-7), P(0,ymax())); + + line(Q, pP); + + plot(f, -1, 7, 40); + label(-1, f, P(-dp, 0), l); + label( 0, f, P(-dp, 0), l); + label( 1, f, P(-dp, 0), l); + label( 2, f, P(-dp, 0), l); + label( 3, f, P( 0,-dp), b); + label( 4, f, P( dp, 0), r); + label( 5, f, P( dp, 0), r); + label( 6, f, P( dp, 0), r); + label( 7, f, P( dp, 0), r); + + label(P(2), P(0,-dp), "$4$", b); + label(P(4,-5.5), P(-dp,0), "$11$", l); + + label(P(0,0), P(2,2), "$O$", tr); + label(pP, P(-4,2), "$P$", l); + label(Q, P(0,2), "$Q$", t); + + label(P(0,ymax()), P(dp,0), "$Y$", r); + label(P(9,0), P(0,-dp), "$X$", b); + + label(P(2.5, ymin()), P(0,-24), "\\textsc{Fig.~12}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/062a.xp b/29785-t/images/sources/062a.xp new file mode 100644 index 0000000..4c1637d --- /dev/null +++ b/29785-t/images/sources/062a.xp @@ -0,0 +1,39 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double x) +{ + return -2 + x*(11 + x*(-9 + x*(-14 + 8*x))); +} + +void cross(double x, double f(double)) +{ + marker(P(x, f(x)), TIMES); +} + +int main() +{ + picture(P(-1.5,-13), P(3.5,2), "1.15 x 3.45in"); + + begin(); + bold(); + line(P(xmin(),0), P(3,0)); + line(P(0,-3), P(0,ymax())); + set_crop(); + + plot(f, xmin(), xmax(), 240); + + cross(-1, f); + cross( 0, f); + cross( 1, f); + cross( 2, f); + + label(P(0,0), P(-2,2), "$O$", tl); + label(P(-0.5,0), P(0,-2), "$1$", b); + + label(P(0.5*(xmin()+xmax()), ymin()), P(0,-12), "\\textsc{Fig.~13}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/062b.xp b/29785-t/images/sources/062b.xp new file mode 100644 index 0000000..f71f0d9 --- /dev/null +++ b/29785-t/images/sources/062b.xp @@ -0,0 +1,39 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double x) +{ + return -11 + x*x*(4 + x); +} + +double df(double x) +{ + return x*(8 + 3*x); +} + +int main() +{ + picture(P(-5,-15), P(5,5), "1.5 x 3in"); + offset(P(0,0.125)); + + begin(); + bold(); + line(P(xmin(),0), P(xmax(),0)); + line(P(0,ymin()), P(0,ymax())); + set_crop(); + + plot(f, xmin(), xmax(), 120); + + label(P(0,0), P(-2,2), "$O$", tl); + label(P(xmax(),0), P(0,-2), "$X$", b); + label(P(0,ymax()), P(-2,0), "$Y$", l); + + label(P(-8./3, f(-8./3)), P(0,2), "$M'$", t); + label(P(0,f(0)), P(2,-2), "$M$", br); + + label(P(0.5*(xmin()+xmax()), ymin()), P(0,-12), "\\textsc{Fig.~14}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/063a.xp b/29785-t/images/sources/063a.xp new file mode 100644 index 0000000..b9f8057 --- /dev/null +++ b/29785-t/images/sources/063a.xp @@ -0,0 +1,30 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double x) +{ + return x*x*x; +} + +int main() +{ + picture(P(-2,-2), P(2,2), "2.25 x 2.25in"); + + begin(); + bold(); + set_crop(); + line(P(xmin(),0), P(xmax(),0)); + line(P(0,ymin()), P(0,ymax())); + + plot(f, xmin(), xmax(), 120); + + label(P(xmax(),0), P(0,-2), "$X$", b); + label(P(0,ymax()), P(2, 0), "$Y$", r); + label(P(0,0), P(2, -2), "$O$", br); + + label(P(0,ymin()), P(0,-12), "\\textsc{Fig.~15}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/063b.xp b/29785-t/images/sources/063b.xp new file mode 100644 index 0000000..c25322c --- /dev/null +++ b/29785-t/images/sources/063b.xp @@ -0,0 +1,48 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double u) +{ + double x(0.5*u-1.25); + return 2 + 1.2*x*x*x; +} + +int main() +{ + picture(P(-0.5,-0.5), P(2.5,2), "2.7 x 2.25in"); + + begin(); + bold(); + double xp(0.375), xq(1.75); + P pP(xp, f(xp)), pQ(xq, f(xq)); + + line(P(xmin(),0), P(xmax(),0)); + line(P(0,ymin()), P(0,ymax())); + + line(pP, pP&E_1); + line(pQ, pQ&E_1); + line(pP, P(pQ.x1(), pP.x2())); + line(pP, pP + 1.2*(pQ - pP)); + + label(P(xmax(),0), P(0,-2), "$X$", b); + label(P(0,ymax()), P(-2,0), "$Y$", l); + label(P(0,0), P(-2, -2), "$O$", bl); + + label(P(0.5*xp), P(0,2), "$x$", t); + label(P(0.5*(xp+xq)), P(0,2), "$h$", t); + label(P(xp, 0.5*f(xp)), P(2,0), "$y$", r); + + label(P(xq, 0.5*(pP.x2() + pQ.x2())), P(2,0), "$Y-y$", r); + + label(pP, P(-2,0), "$P$", l); + label(pQ, P(-2,2), "$Q$", tl); + + set_crop(); + plot(f, 0.25, 2, 60); + + label(P(0.5*(xmin()+xmax()),ymin()), P(0,-12), "\\textsc{Fig.~16}", b); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/069a.xp b/29785-t/images/sources/069a.xp new file mode 100644 index 0000000..b54276f --- /dev/null +++ b/29785-t/images/sources/069a.xp @@ -0,0 +1,53 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +const std::string Ax("-{\\small Axis}"); + +int main() +{ + picture(P(0,0), P(2,2), "1.75 x 1.75in"); + // picture(P(0,0), P(2,2), "2 x 2in"); + + begin(); + degrees(); + bold(); + double x0(1.1), th(30); + P O(0,0), XY(P(x0, ymax())), dir(polar(0.5*x0*Sec(th), th)), Yax(XY-2*dir); + + line(P(xmin(),0), P(xmax(),0)); + line(P(0,ymin()), P(0,ymax())); + + line(O, 4*dir); + line(P(x0), XY); + + dashed(); + line(XY, Yax); + + label(P(0.5*x0,0), P(0,-2), "$X$", b); + label(P(0.5*(xmax()+x0),0), P(0,-2), "$X$"+Ax, b); + + label(0.5*Yax, P(2, 0), "$y$", r); + label(0.5*Yax + 2*dir, P(2, 0), "$y$", r); + label(P(x0, dir.x2()), P(2, 0), "$Y-y$", r); + + label(XY, P(-2, 2), "$(X,Y)$", tr); + label(XY, P(2,-2), "$(x,y)$", br); + + label(P(0,0), P(-2, 0), "$O$", l); + label(P(0,0), P(18, 2), "$\\theta$", tr); + + label(P(0.5*xmax(), ymin()), P(0,-18), "\\textsc{Fig.~17}", b); + + label_angle(th); + label(dir, P(0,2), "$x$", t); + label(dir+Yax, P(0,2), "$x$", t); + label(3*dir, P(0,0), "$x$"+Ax, r); + + label_angle(90); + label(Yax, P(-2,12), "$Y$"+Ax, tl); + label(Yax, P( 2,14), "$y$"+Ax, tr); + + tikz_format(); + end(); +} diff --git a/29785-t/images/sources/071a.xp b/29785-t/images/sources/071a.xp new file mode 100644 index 0000000..3242310 --- /dev/null +++ b/29785-t/images/sources/071a.xp @@ -0,0 +1,46 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +const double ht(0.375); +const double wd(sqrt(1.5)); +const double x0(wd*sqrt(1.0/3)); // critical point + +double g(double x) +{ + return -0.35*x*(wd-x)*(wd+x); +} + +double f(double x) +{ + return g(x) + ht; +} + +int main() +{ + picture(P(-1.5,-1), P(1.5,1), "2.4 x 1.25in"); + + begin(); + bold(); + P O(0,0); + + line(P(xmin(),0), P(xmax(),0)); + line(P(0,ymin()), P(0,ymax())); + + set_crop(); + plot(f, xmin(), xmax(), 60); + marker(P(x0, f(x0)), TIMES); + marker(P(-x0, f(-x0)), TIMES); + + label(O, P(-2,-2), "$O$", bl); + label(P(xmax(),0), P(0,-2), "$X$", b); + label(P(0,ymax()), P(2, 0), "$Y$", r); + + masklabel(P(0,ymin()), P(0,0), "$q\\geqq 2l\\sqrt{l}$", t); + + label(P(0,ymin()), P(0,-6), "\\textsc{Fig.~18}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/images/sources/071b.xp b/29785-t/images/sources/071b.xp new file mode 100644 index 0000000..20c64f5 --- /dev/null +++ b/29785-t/images/sources/071b.xp @@ -0,0 +1,46 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +const double ht(-0.375); +const double wd(sqrt(1.5)); +const double x0(wd*sqrt(1.0/3)); // critical point + +double g(double x) +{ + return -0.35*x*(wd-x)*(wd+x); +} + +double f(double x) +{ + return g(x) + ht; +} + +int main() +{ + picture(P(-1.5,-1), P(1.5,1), "2.4 x 1.25in"); + + begin(); + bold(); + P O(0,0); + + line(P(xmin(),0), P(xmax(),0)); + line(P(0,ymin()), P(0,ymax())); + + masklabel(P(0,ymin()), P(0,0), "$q\\leqq -2l\\sqrt{l}$", t); + + set_crop(); + plot(f, xmin(), xmax()-0.15, 60); + marker(P(x0, f(x0)), TIMES); + marker(P(-x0, f(-x0)), TIMES); + + label(O, P(-2,-2), "$O$", bl); + label(P(xmax(),0), P(0,-2), "$X$", b); + label(P(0,ymax()), P(2, 0), "$Y$", r); + + label(P(0,ymin()), P(0,-6), "\\textsc{Fig.~19}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/images/sources/071c.xp b/29785-t/images/sources/071c.xp new file mode 100644 index 0000000..6d49d9a --- /dev/null +++ b/29785-t/images/sources/071c.xp @@ -0,0 +1,46 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +const double ht(0.125); +const double wd(sqrt(1.5)); +const double x0(wd*sqrt(1.0/3)); // critical point + +double g(double x) +{ + return -0.35*x*(wd-x)*(wd+x); +} + +double f(double x) +{ + return g(x) + ht; +} + +int main() +{ + picture(P(-1.5,-1), P(1.5,1), "2.4 x 1.25in"); + + begin(); + bold(); + P O(0,0); + + line(P(xmin(),0), P(xmax(),0)); + line(P(0,ymin()), P(0,ymax())); + + set_crop(); + plot(f, xmin(), xmax()-0.15, 60); + marker(P(x0, f(x0)), TIMES); + marker(P(-x0, f(-x0)), TIMES); + + label(O, P(-2,-2), "$O$", bl); + label(P(xmax(),0), P(0,-2), "$X$", b); + label(P(0,ymax()), P(2, 0), "$Y$", r); + + masklabel(P(0,ymin()), P(0,0), "$-2l\\sqrt{l}<q<2l\\sqrt{l}$", t); + + label(P(0,ymin()), P(0,-6), "\\textsc{Fig.~20}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/images/sources/073a.xp b/29785-t/images/sources/073a.xp new file mode 100644 index 0000000..4620640 --- /dev/null +++ b/29785-t/images/sources/073a.xp @@ -0,0 +1,43 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +int main() +{ + picture(P(0,-M_PI), P(M_PI,M_PI), "1.25 x 2.5in"); + offset(P(0,-0.5)); + + begin(); + bold(); + double xa(M_PI_4), xb(2.25); + P O(0,0); + + line(P(xmin(),0), P(xmax(),0)); + dashed(); + line(P(M_PI_2,ymin()), P(M_PI_2,ymax())); + solid(); + + line(P(xa,0), P(xa, Tan(xa))); + line(P(xb,0), P(xb, Tan(xb))); + + set_crop(); + plot(Tan, xmin(), M_PI_2-0.001, 30); + plot(Tan, M_PI_2+0.001, xmax(), 30); + + marker(P(xa, Tan(xa)), TIMES); + marker(P(xb, Tan(xb)), TIMES); + + label(O, P(0,-2), "$O$", b); + label(P(M_PI,0), P(0, 2), "$\\pi$", t); + + label(P(xa,0), P(0,-2), "$a$", b); + label(P(xb,0), P(0, 2), "$b$", t); + + label(P(M_PI_2), P(-2,-2), "$\\frac{\\pi}{2}$", bl); + + label(P(M_PI_2, ymin()), P(0,-18), "\\textsc{Fig.~21}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/images/sources/097a.xp b/29785-t/images/sources/097a.xp new file mode 100644 index 0000000..b920d1b --- /dev/null +++ b/29785-t/images/sources/097a.xp @@ -0,0 +1,68 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double x) +{ + double u(x-7), v(u*u); + return 0.125*v*(0.001*v*v + 1) - 1; +} + +Deriv fp(f); + +double df(double x) +{ + return fp.eval(x); +} + +double iter(double x) +{ + return x - f(x)/df(x); +} + +int main() +{ + picture(P(0,-1), P(8,4), "2.5 x 1.5625in"); + + begin(); + bold(); + + double xQ(2), T(iter(xQ)), T1(iter(T)); + + P O(0,0); + + line(O, P(xmax(),0)); + line(O, P(0,3)); + + line(P(xQ,0), P(xQ, f(xQ))); + line(P(T,0), P(xQ, f(xQ))); + + plot(f, xQ, xmax(), 60); + + line_style("- - "); + line(P(T,0), P(T, f(T))); + line(P(T1,0), P(T, f(T))); + + label(O, P(0,-2), "$O$", b); + label(P(xQ), P(0,-2), "$Q$", b); + label(P(T), P(0,-2), "$T$", b); + label(P(T1), P(0,-2), "$T_1$", b); + + label(P(7,f(7)), P(0,-2), "$P_1$", b); + + label(P(0.5*xQ), P(0, 2), "$a$", t); + label(P(0.5*(xQ+T)), P(0, 2), "$h$", t); + label(P(newton(f, T1)), P(2, 2), "$S$", t); + label(P(xmax()), P(0, 2), "$X$", t); + + label(P(0,3), P(2,0), "$Y$", r); + + label(P(xQ,f(xQ)), P(2,0), "$P$", r); + label(P(xmax(),f(xmax())), P(2,0), "$P_2$", r); + + label(P(0.5*xmax(), ymin()), P(0,-18), "\\textsc{Fig.~22}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/images/sources/097b.xp b/29785-t/images/sources/097b.xp new file mode 100644 index 0000000..92eedff --- /dev/null +++ b/29785-t/images/sources/097b.xp @@ -0,0 +1,55 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double alpha(-1.25), beta(1.3); + +double g(double x) +{ + return 0.05*pow(fabs(x-alpha), 1.5)*pow(fabs(x-beta),1.25); +} + +double f(double x) +{ + return tanh(x) - 0.05 - g(x); +} + +Deriv fp(f); + +double df(double x) +{ + return fp.eval(x); +} + +double iter(double x) +{ + return x - f(x)/df(x); +} + +int main() +{ + picture(P(-2,-1), P(2,1), "2.5 x 1.25in"); + + begin(); + bold(); + + set_crop(); + line(P(xmin()), P(xmax())); + + line(P(alpha,f(alpha)), P(alpha)); + line(P(alpha,f(alpha)), P(iter(alpha))); + + line(P(beta, f(beta)), P(beta)); + line(P(beta, f(beta)), P(iter(beta))); + + plot(f, alpha, beta, 60); + + label(P(alpha), P(-2,-2), "$\\alpha$", bl); + label(P(beta), P(2, 2), "$\\beta$", tr); + + label(P(0, ymin()), P(0,-18), "\\textsc{Fig.~23}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/images/sources/098a.xp b/29785-t/images/sources/098a.xp new file mode 100644 index 0000000..c2bb0e5 --- /dev/null +++ b/29785-t/images/sources/098a.xp @@ -0,0 +1,58 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double x) +{ + return -1 + x*(0.5 + 0.125*x + 0.005*pow(x, 6)); +} + +Deriv fp(f); + +double df(double x) +{ + return fp.eval(x); +} + +double iter(double x) +{ + return x - f(x)/df(x); +} + +int main() +{ + picture(P(0,-1), P(2,1), "1.5 x 1.25in"); + + begin(); + bold(); + + double alpha(xmin()), beta(xmax()), T(iter(beta)); + + P xA(alpha), xB(beta), pA(P(alpha, f(alpha))), pB(beta, f(beta)); + + line(xA, xB); + + line(xA, pA); + line(xB, pB); + line(P(T), pB); + + plot(f, alpha, beta, 40); + + dashed(); + line(pA, pB); + + label(xA, P(0,2), "$\\alpha$", t); + label(pA, P(-2,0), "$A$", l); + + label(xB, P(0,-2), "$\\beta$", b); + label(pB, P(2, 0), "$B$", r); + + label(P(T), P(0,-2), "$T$", b); + label(Segment(pA, pB)*Segment(xA, xB), P(-2,2), "$c$", tl); + + label(P(0.5*xmax(), ymin()), P(0,-12), "\\textsc{Fig.~24}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/images/sources/098b.xp b/29785-t/images/sources/098b.xp new file mode 100644 index 0000000..6f5a37c --- /dev/null +++ b/29785-t/images/sources/098b.xp @@ -0,0 +1,52 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double x) +{ + return 0.5*(1 - 0.5*pow(2-x, 2) - 0.01*pow(2-x, 6)); +} + +Deriv fp(f); + +double df(double x) +{ + return fp.eval(x); +} + +double iter(double x) +{ + return x - f(x)/df(x); +} + +int main() +{ + picture(P(0,-1), P(2,1), "1.5 x 1.25in"); + + begin(); + bold(); + + double alpha(xmin()), beta(xmax()), T(iter(alpha)); + + P xA(alpha), xB(beta), pA(P(alpha, f(alpha))), pB(beta, f(beta)); + + line(xA, xB); + + line(xA, pA); + line(xB, pB); + line(P(T), pA); + + plot(f, alpha, beta, 40); + + label(xA, P(0,2), "$\\alpha$", t); + + label(xB, P(0,-2), "$\\beta$", b); + + label(P(T), P(0, 2), "$T$", t); + + label(P(0.5*xmax(), ymin()), P(0,-12), "\\textsc{Fig.~25}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/images/sources/098c.xp b/29785-t/images/sources/098c.xp new file mode 100644 index 0000000..fd39835 --- /dev/null +++ b/29785-t/images/sources/098c.xp @@ -0,0 +1,60 @@ +/* -*-ePiX-*- */ +#include "epix.h" +using namespace ePiX; + +double f(double x) +{ + return 0.5*(1 - 0.5*pow(x, 2) - 0.01*pow(x, 6)); +} + +Deriv fp(f); + +double df(double x) +{ + return fp.eval(x); +} + +double iter(double x) +{ + return x - f(x)/df(x); +} + +int main() +{ + picture(P(0,-1), P(2,1), "1.5 x 1.25in"); + + begin(); + bold(); + + double alpha(xmin()), beta(xmax()), T(iter(beta)); + + P xA(alpha), xB(beta), pA(P(alpha, f(alpha))), pB(beta, f(beta)); + + line(xA, xB); + + line(xA, pA); + line(xB, pB); + line(P(T), pB); + + dashed(); + line(pA, pB); + + solid(); + plot(f, alpha, beta, 40); + + label(pA, P(-2,0), "$A$", l); + label(xA, P(0,-2), "$\\alpha$", b); + + label(pB, P(2, 0), "$B$", r); + label(xB, P(0, 2), "$\\beta$", t); + + label(P(T), P(0, 2), "$T$", t); + + label(Segment(pA, pB)*Segment(xA, xB), P(-2,-2), "$c$", bl); + + label(P(0.5*xmax(), ymin()), P(0,-12), "\\textsc{Fig.~26}", b); + + tikz_format(); + + end(); +} diff --git a/29785-t/old/29785-t.tex b/29785-t/old/29785-t.tex new file mode 100644 index 0000000..1b12c2c --- /dev/null +++ b/29785-t/old/29785-t.tex @@ -0,0 +1,14422 @@ +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% The Project Gutenberg EBook of First Course in the Theory of Equations, by +% Leonard Eugene Dickson % +% % +% 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.org % +% % +% % +% Title: First Course in the Theory of Equations % +% % +% Author: Leonard Eugene Dickson % +% % +% Release Date: August 25, 2009 [EBook #29785] % +% % +% Language: English % +% % +% Character set encoding: ISO-8859-1 % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF EQUATIONS *** % +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{29785} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% %% +%% inputenc: Standard DP encoding. Required. %% +%% fontenc: Font encoding, for bold smallcaps. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. Required. %% +%% amssymb: AMS mathematics symbols. Required. %% +%% %% +%% calc: Infix arithmetic for length calculations. Required. %% +%% ifthen: Logical conditionals. Required. %% +%% %% +%% alltt: Fixed-width font environment for PG stanzas. Required. %% +%% %% +%% textcase: \MakeTextUppercase, for running heads. Required. %% +%% %% +%% array: Enhanced tabular environment. Required. %% +%% arydshln: Dashed lines in tabular/array environments. Required. %% +%% %% +%% indentfirst: Indent sectional units' first paragraph. Required. %% +%% %% +%% multicol: Multi-column environment. Required. %% +%% makeidx: Indexing capabilities. Required. %% +%% %% +%% wrapfig: Wrap text around figures. Required. %% +%% fancyhdr: Enhanced running headers and footers. Required. %% +%% %% +%% tikz: Replacement for picture environment. Required. %% +%% %% +%% geometry: Enhanced page layout package. Required. %% +%% hyperref: Hypertext embellishments for pdf output. Required. %% +%% %% +%% %% +%% Producer's Comments: %% +%% %% +%% The book comprises multiple sub-page units of varying font %% +%% size, figures wrapped by text, and tall, table-like %% +%% calculations. Vertical spacing is loose, so only a few %% +%% locations need visual checking. %% +%% %% +%% %% +%% Things to Check: %% +%% %% +%% * Table of contents chapter headers not separated from %% +%% first line of section headers. %% +%% * Wrapped images on page 64 are both positioned on the page. %% +%% * Wrapped images on page 73 are all positioned on the page. %% +%% * Tall tables on pages 98 ff. are not too badly spaced. %% +%% %% +%% %% +%% Notes on spelling: %% +%% %% +%% All changes to the text are [** PP: Noted] in this file. %% +%% The old spellings constructible and parallelopiped have been %% +%% retained, as have the variants coordinates and coördinates, %% +%% 2-rowed and two-rowed, etc. %% +%% %% +%% Spellcheck: OK %% +%% Smoothreading pool: Yes %% +%% lacheck: OK, about 100 false positives %% +%% %% +%% Lprep/gutcheck: OK %% +%% PDF pages: 207 %% +%% PDF page size: US Letter (8.5" x 11") %% +%% PDF bookmarks: created %% +%% PDF document info: filled in %% +%% ToC page numbers: OK %% +%% Images: 26 picture environments (written with tikz macros) %% +%% If compiling with tikz 2.00 of greater, may need to replace %% +%% the option [cap=round] with [line cap=round] in each illo. %% +%% %% +%% Summary of log file: %% +%% * Three underfull hboxes -- two from fixed-width boxes on title %% +%% page, one from illo on PDF page 64. %% +%% * One underfull vbox %% +%% %% +%% %% +%% Compile History: %% +%% %% +%% May, 2009: adhere (Andrew D. 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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.org + + +Title: First Course in the Theory of Equations + +Author: Leonard Eugene Dickson + +Release Date: August 25, 2009 [EBook #29785] + +Language: English + +Character set encoding: ISO-8859-1 + +*** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF EQUATIONS *** +\end{PGtext} +\end{minipage} +\end{center} +\vfill + +\clearpage + +% Credits and transcriber's note +\phantomsection +\pdfbookmark[0]{Transcriber's Note}{Transcriber's Note} + +\begin{center} +\begin{minipage}{\textwidth} +\begin{PGtext} +Produced by Peter Vachuska, Andrew D. Hwang, Dave Morgan, +and the Online Distributed Proofreading Team at +http://www.pgdp.net +\end{PGtext} +\end{minipage} +\end{center} +\vfill + +\begin{minipage}{0.85\textwidth} +\small +\subsection*{\centering\normalfont\Titling\TransNote} + +\raggedright +\TransNoteText +\end{minipage} + +%% -----File: 001.png---Folio xx------- + +\frontmatter +\pagenumbering{roman} +\pagestyle{empty} + +\settowidth{\TmpLen}{{\Large\Titling JOHN WILEY \& SONS, Inc.}} +\begin{center} +\Titling +{\Huge FIRST COURSE}\\[0.25in] +{\Large IN THE}\\[0.25in] +{\Huge THEORY OF EQUATIONS}\\[0.75in] +BY\TitleSkip +{\large LEONARD EUGENE DICKSON, Ph.D.} + +{\tiny CORRESPONDANT DE L'INSTITUT DE FRANCE}\\[-4pt] +{\tiny PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF CHICAGO} + +\vfill +NEW YORK\TitleSkip +% [** PP: Next line is underfull, but deliberately spaced out] +\makebox[\TmpLen][s]{\large JOHN WILEY \& SONS, Inc.}\\[0pt] +{\small London: CHAPMAN \& HALL, Limited} +\normalfont +\end{center} + +%% -----File: 002.png---Folio xx------- + +\clearpage + +\settowidth{\TmpLen}{\itshape This book or any part thereof must not} +\begin{center} +\null\vfill +\textsc{Copyright, 1922,}\TitleSkip +\textsc{by}\TitleSkip +{\Titling LEONARD EUGENE DICKSON}\TitleSkip +\rule{0.5in}{0.5pt}\TitleSkip +\textit{All Rights Reserved} + +\makebox[\TmpLen][s]{\itshape This book or any part thereof must not}\\ +% [** PP: Next line is underfull, but deliberately spaced out] +\makebox[\TmpLen][s]{\itshape be reproduced in any form without} \\ +\makebox[\TmpLen][s]{\itshape the written permission of the publisher.} +\vfill + +Printed in U. S. A.\\[0.25in] +{\normalfont\tiny\sffamily +PRESS OF \\[0.5ex] +BRAUNWORTH\quad \&\quad CO.,\quad INC. \\[0.5ex] % [** PP: Added .] +BOOK\quad MANUFACTURERS \\[0.5ex] +BROOKLYN, NEW YORK \\ +} +\end{center} + +\iffalse +11/30 % [** PP: Omitted printer's mark] +\fi + +%% -----File: 003.png---Folio xx------- + + +\Preface + +The theory of equations is not only a necessity in the subsequent +mathematical courses and their applications, but furnishes an illuminating +sequel to geometry, algebra and analytic geometry. Moreover, +it develops anew and in greater detail various fundamental ideas of +calculus for the simple, but important, case of polynomials. The +theory of equations therefore affords a useful supplement to differential +calculus whether taken subsequently or simultaneously. + +It was to meet the numerous needs of the student in regard to his +earlier and future mathematical courses that the present book was +planned with great care and after wide consultation. It differs essentially +from the author's \textit{Elementary Theory of Equations}, both in regard to +omissions and additions, and since it is addressed to younger students +and may be used parallel with a course in differential calculus. Simpler +and more detailed proofs are now employed. The exercises are simpler, +more numerous, of greater variety, and involve more practical applications. + +This book throws important light on various elementary topics. +For example, an alert student of geometry who has learned how to bisect +any angle is apt to ask if every angle can be trisected with ruler and +compasses and if not, why not. After learning how to construct regular +polygons of $3$, $4$, $5$, $6$, $8$ and~$10$ sides, he will be inquisitive about the +missing ones of~$7$ and~$9$ sides. The teacher will be in a comfortable position +if he knows the facts and what is involved in the simplest discussion to +date of these questions, as given in \ChapRef{III}. Other chapters throw +needed light on various topics of algebra. In particular, the theory +of graphs is presented in \ChapRef{V} in a more scientific and practical +manner than was possible in algebra and analytic geometry. + +There is developed a method of computing a real root of an equation +with minimum labor and with certainty as to the accuracy of all the +decimals obtained. We first find by Horner's method successive transformed +%% -----File: 004.png---Folio xx------- +equations whose number is half of the desired number of significant +figures of the root. The final equation is reduced to a linear equation +by applying to the constant term the correction computed from the +omitted terms of the second and higher degrees, and the work is completed +by abridged division. The method combines speed with control of +accuracy. + +Newton's method, which is presented from both the graphical and +the numerical standpoints, has the advantage of being applicable also to +equations which are not algebraic; it is applied in detail to various such +equations. + +In order to locate or isolate the real roots of an equation we may +employ a graph, provided it be constructed scientifically, or the theorems +of Descartes, Sturm, and Budan, which are usually neither stated, nor +proved, correctly. + +The long chapter on determinants is independent of the earlier chapters. +The theory of a general system of linear equations is here presented +also from the standpoint of matrices. + +For valuable suggestions made after reading the preliminary manuscript +of this book, the author is greatly indebted to Professor Bussey +of the University of Minnesota, Professor Roever of Washington University, +Professor Kempner of the University of Illinois, and Professor +Young of the University of Chicago. The revised manuscript was much +improved after it was read critically by Professor Curtiss of Northwestern +University. The author's thanks are due also to Professor Dresden of +the University of Wisconsin for various useful suggestions on the +proof-sheets. + +{\small\textsc{Chicago, 1921.}} + +%% -----File: 005.png---Folio xx------- + +\PrintContents + +\iffalse%%%% COMMENTED HARD-CODED TABLE OF CONTENTS %%%% +\begin{center} +\section*{CONTENTS} + +{\tiny Numbers refer to pages.} + +\rule{3em}{0.75pt} + +\subsection*{CHAPTER I} + +\textsc{Complex Numbers} +\end{center} + +Square roots, 1. Addition, multiplication and division of complex numbers, +2. Cube roots of unity, 3. Geometrical representation, 3. Product +and quotient, 4. De Moivre's theorem, 5. Cube roots, 5. $n$th roots, 7. +Roots of unity, 8. Primitive roots of unity, 9. + +\begin{center} +\subsection*{CHAPTER II} + +\textsc{Elementary Theorems on the Roots of an Equation} +\end{center} + +Quadratic equation, 11. Remainder theorem, 12. Synthetic division, 13. +Factored form of a polynomial, 15. Multiple roots, 16. Identical polynomials, +16. Relations between the roots and the coefficients, 17. Imaginary +roots occur in pairs, 19. Upper limit to the real roots, 21. Integral roots, +24. Rational roots, 27. + +\begin{center} +\subsection*{CHAPTER III} + +\textsc{Constructions with Ruler and Compasses} +\end{center} + +Graphical solution of a quadratic equation, 29. Analytic criterion for +constructibility, 30. Cubic equations with a constructible root, 32. Trisection +of an angle, 34. Duplication of a cube, 35. Regular polygons of +7, 9, 17, and $n$ sides, 35-44. Reciprocal equations, 37. + +\begin{center} +\subsection*{CHAPTER IV} + +\textsc{Cubic and Quartic Equations} +\end{center} + +Algebraic solution of a cubic, 45. Discriminant, 47. Number of real roots +of a cubic, 48. Trigonometric solution of a cubic, 49. Ferrari's and Descartes' +solutions of a quartic, 50. Resolvent cubic, 51. Discriminant of a quartic, 51. + +\begin{center} +\subsection*{CHAPTER V} + + +\textsc{The Graph of an Equation} +\end{center} + +Use of graphs, 55. Caution in plotting, 55. Bend points, 56. Derivatives, +58. Horizontal tangents, 60. Multiple roots, 60. Ordinary and inflexion +%% -----File: 006.png---Folio xx------- +* tangents, 62. Real roots of a cubic equation, 65. Continuity, 66. Condition +for a root between $a$ and $b$, 67. Sign of a polynomial at infinity, 68. +Rolle's theorem, 69. + +\begin{center} +\subsection*{CHAPTER VI} + +\textsc{Isolation of the Real Roots} +\end{center} + +Descartes' rule of signs, 71. Sturm's method, 75. Sturm's functions for +the general quartic equation, 80. Budan's theorem, 83. + +\begin{center} +\subsection*{CHAPTER VII} + +\textsc{Solution of Numerical Equations} +\end{center} + +Horner's method, 86. Newton's method, algebraic and graphical discussion, +systematic computation, also for functions not polynomials, 90. Imaginary +roots, 98. + +\begin{center} +\subsection*{CHAPTER VIII} + +\textsc{Determinants; Systems of Linear Equations} +\end{center} + +Solution of 2 or 3 linear equations by determinants, 101. Even and odd +arrangements, 103. Definition of a determinant of order $n$, 105. Interchange +of rows and columns, 106. Interchange of two columns or two rows, 107. +Minors, 109. Expansion, 109. Removal of factors, 111. Sum of determinants, +112. Addition of columns or rows, 113. Rank, 116. System +of $n$ linear equations in n unknowns, 114, 116. Homogeneous equations, 119. +System of $m$ linear equations in $n$ unknowns, matrix and augmented matrix, +120. Complementary minors, 122. Laplace's development, 122. Product of +determinants, 124. + + +\begin{center} +\subsection*{CHAPTER IX} + +\textsc{Symmetric Functions} +\end{center} + +Sigma functions, 128. Elementary symmetric functions, 128. Fundamental +theorem, 129. Rational functions symmetric in all but one of the +roots, 132. Sums of like powers of the roots, Newton's identities, 134. +Waring's formula, 136. Computation of symmetric functions, 141. + +\begin{center} +\subsection*{CHAPTER X} + +\textsc{Elimination, Resultants and Discriminants} +\end{center} + +Methods of Sylvester, Euler, and Bézout, 143. Discriminants, 152. + +\begin{center} +\subsection*{APPENDIX} + +\textsc{The Fundamental Theorem of Algebra} +\end{center} + +\textsc{Answers} \dotfill 159 + +\textsc{Index} \dotfill 167 +\fi%%%% END OF COMMENTED HARD-CODED TABLE OF CONTENTS %%%% + +%% -----File: 007.png---Folio 1------- + +\mainmatter +\pagenumbering{arabic} + +\ChapSkip +\begin{center} +\textbf{\LARGE First Course in}\\[0.25in] +\textbf{\LARGE The Theory of Equations} +\medskip + +\TB +\end{center} + +\Chapter{I}{Complex Numbers} + + +\Section{1.}{Square Roots} If $p$ is a positive real number, the symbol~$\sqrt{p}$ is +used to denote the positive square root of~$p$. It is most easily computed +by logarithms. +\index{Square roots}% + +We shall express the square roots of negative numbers in terms of the +symbol~$i$ such that the relation $i^2 = -1$ holds. Consequently we denote +the roots of $x^2 = -1$ by~$i$ and~$-i$. The roots of $x^2 = -4$ are written in the +form $± 2i$ in preference to $±\sqrt{-4}$. In general, if $p$ is positive, the roots +of $x^2=-p$ are written in the form $±\sqrt{p} i$ in preference to $±\sqrt{-p}$. + +\begin{Remark} +The square of either root is thus $(\sqrt{p})^2 i^2 = -p$. Had we used the less desirable +notation $±\sqrt{-p}$ for the roots of $x^2 = -p$, we might be tempted to find the square of +either root by multiplying together the values under the radical sign and conclude +erroneously that +\[ +\sqrt{-p}\,\sqrt{-p} = \sqrt{p^2} = +p. +\] +To prevent such errors we use $\sqrt{p}\,i$ and not~$\sqrt{-p}$. +\end{Remark} + + +% [** PP: ToC entry reads Addition, multiplication and division of complex numbers] +\Section{2.}{Complex Numbers} If $a$ and~$b$ are any two real numbers and +\index{Complex number}% +$i^2 = -1$, $a+bi$ is called a \emph{complex number}\footnote + {Complex numbers are essentially couples of real numbers. For a treatment from + this standpoint and a treatment based upon vectors, see the author's \textit{Elementary Theory + of Equations}, p.~21, p.~18.} +and $a-bi$ its \emph{conjugate}. Either +\index{Conjugate}% +is said to be \emph{zero} if $a = b = 0$. Two complex numbers $a+bi$ and $c+di$ +are said to be \emph{equal} if and only if $a = c$ and $b = d$. In particular, $a+bi=0$ +%% -----File: 008.png---Folio 2------- +if and only if $a = b = 0$. If $b \ne 0$, $a+bi$ is said to be \emph{imaginary}. In particular, +$bi$ is called a \emph{pure imaginary}. +\index{Pure imaginary}% +\index{Imaginary}% + +Addition of complex numbers is defined by +\[ +(a+bi) + (c+di) = (a+c) + (b+d)i. +\] +The inverse operation to addition is called subtraction, and consists in +finding a complex number~$z$ such that +\[ +(c+di) + z = a+bi. +\] +In notation and value, $z$ is +\[ +(a+bi) - (c+di) = (a-c) + (b-d)i. +\] + +Multiplication is defined by +\[ +(a+bi)(c+di) = ac-bd+(ad+bc)i, +\] +and hence is performed as in formal algebra with a subsequent reduction +by means of $i^2 = -1$. For example, +\[ +(a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2. +\] + +Division is defined as the operation which is inverse to multiplication, +and consists in finding a complex number~$q$ such that $(a+bi)q = e+fi$. +Multiplying each member by $a-bi$, we find that~$q$ is, in notation and +value, +\[ +\frac{e+fi}{a+bi} = \frac{(e+fi)(a-bi)}{a^2+b^2} + = \frac{ae+bf}{a^2+b^2} + \frac{af-be}{a^2+b^2} i. +\] +Since $a^2+b^2 = 0$ implies $a = b = 0$ when $a$ and~$b$ are real, we conclude that +division except by zero is possible and unique. + + +\begin{Exercises}{Page2} +Express as complex numbers +\begin{Problems}[2] + +\item[1.] $\sqrt{-9}$. + +\item[2.] $\sqrt{4}$. + +\ResetCols{2} + +\item[3.] $(\sqrt{25} + \sqrt{-25})\sqrt{-16}$. + +\item[4.] $-\frac{2}{3}$. + +\ResetCols{4} + +\item[5.] $8 + 2\sqrt{3}\vphantom{\dfrac{1}{1}}$. + +\item[6.] $\dfrac{3 + \sqrt{-5}}{2 + \sqrt{-1}}$. + +\item[7.] $\dfrac{3 + 5i}{2 - 3i}$. + +\item[8.] $\dfrac{a + bi}{a - bi}$. + +\ResetCols{1} + +\item[9.] Prove that the sum of two conjugate complex numbers is real and that their +difference is a pure imaginary. + +\item[10.] Prove that the conjugate of the sum of two complex numbers is equal to the +sum of their conjugates. Does the result hold true if each word sum is replaced by the +word difference? + +%% -----File: 009.png---Folio 3------- + +\item[11.] Prove that the conjugate of the product (or quotient) of two complex numbers +is equal to the product (or quotient) of their conjugates. + +\item[12.] Prove that, if the product of two complex numbers is zero, at least one of them +is zero. + +\item[13.] Find two pairs of real numbers $x$, $y$ for which +\[ +(x+yi)^2 = -7+24i. +\] +\end{Problems} + +As in Ex.~13, express as complex numbers the square roots of +\begin{Problems}[3] + +\item[14.] {$-11+60i$.} + +\item[15.] {$5-12i$.} + +\item[16.] {$4cd+(2c^2-2d^2)i$.} + +\end{Problems} + +\end{Exercises} + + +\Section{3.}{Cube Roots of Unity} Any complex number~$x$ whose cube is equal +to unity is called a \emph{cube root of unity}. Since +\index{Cube root!of unity}% +\[ +x^3-1 = (x-1) (x^2+x+1), +\] +the roots of $x^3=1$ are~$1$ and the two numbers~$x$ for which +\[ +x^2+x+1=0,\qquad +(x + \tfrac{1}{2})^2 = -\tfrac{3}{4}, \qquad + x + \tfrac{1}{2} = ±\tfrac{1}{2} \sqrt{3}i. +\] +Hence there are three cube roots of unity, viz., +\[ +1, \qquad +\omega = -\tfrac{1}{2} + \tfrac{1}{2}\sqrt{3}i,\qquad +\omega' = -\tfrac{1}{2} - \tfrac{1}{2}\sqrt{3}i. +\] + +In view of the origin of~$\omega$, we have the important relations +\[ +\omega^2 + \omega+1 = 0, \quad \omega^3 = 1. +\] +Since $\omega \omega' = 1$ and $\omega^3 = 1$, it follows that $\omega' = \omega^2$, $\omega = \omega'^2$. + + +\Section[Geometrical Representation] +{4.}{Geometrical Representation of Complex Numbers.} Using rectangular +axes of coördinates, $OX$ and~$OY$, we represent the complex number +$a+bi$ by the point~$A$ having the coördinates $a$, $b$ (Fig.~1). +\index{Complex number!geometrical representation}% + +%[Illustration: \textsc{Fig}. 1] +\begin{wrapfigure}{r}{2.25in} +\hfill\Input{009a} +\end{wrapfigure} +The positive number $r = \sqrt{a^2+b^2}$ giving +the length of~$OA$ is called the \emph{modulus} (or +\emph{absolute value}) of~$a+bi$. The angle $\theta = XOA$, +measured counter-clockwise from~$OX$ to~$OA$, +is called the \emph{amplitude} (or \emph{argument}) of~$a+bi$. +\index{Absolute value}% +\index{Amplitude}% +\index{Argument}% +\index{Modulus}% +Thus $\cos \theta = a/r$, $\sin \theta = b/r$, whence +\[ +a+bi = r(\cos\theta + i\sin\theta). +\Tag{1} +\] +The second member is called the \emph{trigonometric form} of~$a+bi$. +\index{Complex number!trigonometric form}% + +For the amplitude we may select, instead of~$\theta$, any of the angles $\theta±360°$, +$\theta±720°$, etc. + +%% -----File: 010.png---Folio 4------- + +Two complex numbers are equal if and only if their moduli are equal +and an amplitude of the one is equal to an amplitude of the other. + +%[Illustration: \textsc{Fig}. 2] +\begin{Remark} +\begin{wrapfigure}{l}{2.125in} +\Input{010a} +\end{wrapfigure} +For example, the cube roots of unity are~$1$ and +\index{Cube root!of unity}% +\begin{align*} +\omega &= -\tfrac{1}{2} + \tfrac{1}{2}\sqrt{3} i \\ + &= \cos 120° + i \sin 120°, \\ +\omega^2 &= -\tfrac{1}{2} - \tfrac{1}{2}\sqrt{3} i \\ + &= \cos 240° + i \sin 240°, +\end{align*} +and are represented by the points marked $1$, $\omega$, $\omega^2$ +at the vertices of an equilateral triangle inscribed +in a circle of radius unity and center at the origin~$O$ +(Fig.~2). The indicated amplitudes of~$\omega$ and~$\omega^2$ +are~$120°$ and~$240°$ respectively, while the modulus +of each is~$1$. + +The modulus of~$-3$ is $3$ and its amplitude is~$180°$ or~$180°$ plus or minus the product +of~$360°$ by any positive whole number. +\end{Remark} + + +% [** PP: ToC entry reads ``Product and quotient'', splitting] +\Section[Product] +{5.}{Product of Complex Numbers.} By actual multiplication, +\begin{align*} +&\quad +\bigl[r (\cos\theta + i \sin\theta)\bigr] +\bigl[r'(\cos\alpha + i \sin\alpha)\bigr] \\ +&= rr' +\bigl[ (\cos\theta \cos\alpha - \sin\theta \sin\alpha) + + i(\sin\theta \cos\alpha + \cos\theta \sin\alpha)\bigr] \\ +&= rr' +\bigl[ \cos(\theta + \alpha) + i \sin(\theta + \alpha)], \quad +\text{by trigonometry.} +\end{align*} +Hence \textit{the modulus of the product of two complex numbers is equal to the +product of their moduli, while the amplitude of the product is equal to the +sum of their amplitudes.} + +\begin{Remark} +For example, the square of $\omega = \cos 120°+ i \sin 120°$ has the modulus~$1$ and the amplitude +$120°+120°$ and hence is $\omega^2 = \cos 240°+i \sin 240°$. Again, the product of~$\omega$ and~$\omega^2$ +has the modulus~$1$ and the amplitude $120°+ 240°$ and hence is $\cos 360°+ i \sin 360°$, +which reduces to~$1$. This agrees with the known fact that $\omega^3 = 1$. +\end{Remark} + +Taking $r = r' = 1$ in the above relation, we obtain the useful formula +\[ +(\cos \theta + i \sin \theta) +(\cos \alpha + i \sin \alpha) + = \cos (\theta + \alpha) + i \sin (\theta + \alpha). +\Tag{2} +\] + +% [** PP: Split ToC entry, part 2] +\Section[Quotient] +{6.}{Quotient of Complex Numbers.} Taking $\alpha = \beta - \theta$ in~\Eq{2} and dividing +the members of the resulting equation by $\cos\theta + i \sin\theta$, we get +\[ +\frac{\cos \beta + i \sin \beta} + {\cos \theta + i \sin \theta} + = \cos(\beta - \theta) + i \sin(\beta - \theta). +\] +%% -----File: 011.png---Folio 5------- +Hence \textit{the amplitude of the quotient of $R(\cos \beta+i \sin \beta)$ by $r(\cos \theta+i \sin \theta)$ +is equal to the difference $\beta - \theta$ of their amplitudes, while the modulus of the +quotient is equal to the quotient~$R/r$ of their moduli.} + +The case $\beta = 0$ gives the useful formula +\[ +\frac{1}{\cos\theta + i \sin\theta} = \cos\theta - i \sin\theta. +\] + + +\Section{7.}{De Moivre's Theorem} +\index{De Moivre's!theorem}% +\begin{Thm} +If $n$ is any positive whole number, +\[ +(\cos\theta + i \sin\theta)^n = \cos n\theta + i \sin n\theta. +\Tag{3} +\] +\end{Thm} + +This relation is evidently true when $n = 1$, and when $n = 2$ it follows +from formula~\Eq{2} with $\alpha = \theta$. To proceed by mathematical induction, +suppose that our relation has been established for the values $1, 2, \dotsc, m$ +of~$n$. We can then prove that it holds also for the next value $m+1$ of~$n$. +For, by hypothesis, we have +\[ +(\cos\theta + i \sin\theta)^m = \cos m\theta + i \sin m\theta. +\] +Multiply each member by $\cos\theta + i \sin\theta$, and for the product on the right +substitute its value from~\Eq{2} with $\alpha = m \theta$. Thus +\begin{align*} +(\cos\theta + i \sin\theta)^{m+1} + &= (\cos\theta + i \sin\theta)(\cos m\theta + i \sin m\theta), \\ + &= \cos(\theta + m \theta) + i \sin(\theta + m \theta), +\end{align*} +which proves~\Eq{3} when $n = m+1$. Hence the induction is complete. + +\begin{Remark} +Examples are furnished by the results at the end of~§5: +\begin{align*} +(\cos 120° + i \sin 120°)^2 = \cos 240° + i \sin 240°, \\ +(\cos 120° + i \sin 120°)^3 = \cos 360° + i \sin 360°. +\end{align*} +\end{Remark} + + +\Section{8.}{Cube Roots} To find the cube roots of a complex number, we first +express the number in its trigonometric form. For example, +\index{Cube root}% +\[ +4\sqrt{2} + 4\sqrt{2} i = 8(\cos 45° + i \sin 45°). +\] +If it has a cube root which is a complex number, the latter is expressible +in the trigonometric form +\[ +r(\cos\theta + i \sin\theta). +\Tag{4} +\] +The cube of the latter, which is found by means of~\Eq{3}, must be equal +to the proposed number, so that +\[ +r^3(\cos 3\theta + i \sin 3\theta) = 8(\cos 45° + i \sin 45°). +\] +%% -----File: 012.png---Folio 6------- +The moduli $r^3$ and~$8$ must be equal, so that the positive real number~$r$ +is equal to~$2$. Furthermore, $3 \theta$ and~$45°$ have equal cosines and equal +sines, and hence differ by an integral multiple of~$360°$. Hence $3 \theta = 45°+ +k·360°$, or $\theta = 15°+k·120°$, where $k$ is an integer.\footnote + {Here, as elsewhere when the contrary is not specified, zero and negative as well + as positive whole numbers are included under the term ``integer.''} +Substituting this +value of~$\theta$ and the value~$2$ of~$r$ in~\Eq{4}, we get the desired cube roots. The +values $0$, $1$, $2$ of~$k$ give the distinct results +\begin{alignat*}{2} % [** PP: Re-breaking, aligning] +R_1 &= 2(\cos 15° &+{}& i \sin 15°), \\ +R_2 &= 2(\cos 135° &+{}& i \sin 135°),\\ +R_3 &= 2(\cos 255° &+{}& i \sin 255°). +\end{alignat*} + +Each new integral value of~$k$ leads to a result which is equal to~$R_1$, +$R_2$ or~$R_3$. In fact, from $k = 3$ we obtain~$R_1$, from $k = 4$ we obtain~$R_2$, from +$k = 5$ we obtain~$R_3$, from $k = 6$ we obtain~$R_1$ again, and so on periodically. + + +\begin{Exercises}{Page6} + +\begin{Problems} +\item[1.] Verify that $R_2 = \omega R_1$, $R_3 = \omega^2 R_1$. Verify that $R_1$ is a cube root of $8 (\cos 45°+ +i \sin 45°)$ by cubing~$R_1$ and applying De Moivre's theorem. Why are the new expressions +for~$R_2$ and~$R_3$ evidently also cube roots? + +\item[2.] Find the three cube roots of~$-27$; those of~$-i$; those of~$\omega$. + +\item[3.] Find the two square roots of~$i$; those of~$-i$; those of~$\omega$. + +\item[4.] Prove that the numbers $\cos\theta + i \sin\theta$ and no others are represented by points +on the circle of radius unity whose center is the origin. + +\item[5.] If $a+bi$ and $c+di$ are represented by the points~$A$ and~$C$ in Fig.~3, prove that +their sum is represented by the fourth vertex~$S$ of the parallelogram two of whose sides +are~$OA$ and~$OC$. Hence show that the modulus of the sum of two complex numbers +is equal to or less than the sum of their moduli, and is equal to or greater than the difference +of their moduli. +\index{Complex number!geometrical representation}% +%[Illustration: \textsc{Fig}. 3] +%[Illustration: \textsc{Fig}. 4] +\begin{figure*}[hbt] +\begin{center} +\Input{012a}\hfil +\Input{012b} +\end{center} +\end{figure*} + +%% -----File: 013.png---Folio 7------- + +\item[6.] Let $r$ and~$r'$ be the moduli and $\theta$ and~$\alpha$ the amplitudes of two complex numbers +represented by the points $A$ and~$C$ in Fig.~4. Let~$U$ be the point on the $x$-axis one +unit to the right of the origin~$O$. Construct triangle $OCP$ similar to triangle $OUA$ and +similarly placed, so that corresponding sides are $OC$ and~$OU, CP$ and~$UA$, $OP$ and~$OA$, +while the vertices $O$, $C$, $P$ are in the same order (clockwise or counter-clockwise) as +the corresponding vertices $O$, $U$, $A$. Prove that~$P$ represents the product~(§5) of the +complex numbers represented by $A$ and~$C$. + +\item[7.] If $a+bi$ and $e+fi$ are represented by the points $A$ and~$S$ in Fig.~3, prove that +the complex number obtained by subtracting $a+bi$ from $e+fi$ is represented by the point~$C$. +Hence show that the absolute value of the difference of two complex numbers is +equal to or less than the sum of their absolute values, and is equal to or greater than +the difference of their absolute values. + +\item[8.] By modifying Ex.~6, show how to construct geometrically the quotient of two +complex numbers. +\index{Complex number!geometrical representation}% +\end{Problems} +\end{Exercises} + + +% [** PP: ToC entry matches unit title, running head is as given] +\Section[Roots of Complex Numbers] +{9.}{$n$th~Roots.} As illustrated in~§8, it is evident that the $n$th roots +of any complex number $\rho(\cos A + i \sin A)$ are the products of the $n$th +roots of $\cos A + i \sin A$ by the positive real $n$th root of the positive real +number~$\rho$ (which may be found by logarithms). +\index{Rootz@{Roots, $n$th}}% [** PP: Rootz places entry after Roots] + +Let an $n$th root of $\cos A + i \sin A$ be of the form +\[ +r(\cos \theta+i \sin \theta). +\tag{4}%[ ** PP: [sic], equation repeated] +\] +Then, by De Moivre's theorem, +\[ +r^n(\cos n\theta + i \sin n\theta) = \cos A + i \sin A. +\] +The moduli $r^n$ and~$1$ must be equal, so that the positive real number~$r$ +is equal to~$1$. Since $n\theta$ and~$A$ have equal sines and equal cosines, they +differ by an integral multiple of~$360°$. Hence $n\theta = A + k·360°$, where~$k$ +is an integer. Substituting the resulting value of~$\theta$ and the value~$1$ of~$r$ +in~\Eq{4}, we get +\[ + \cos\left(\frac{A + k·360°}{n}\right) + + i \sin\left(\frac{A + k·360°}{n}\right). +\Tag{5} +\] + +For each integral value of~$k$, \Eq{5}~is an answer since its $n$th power reduces +to $\cos A + i \sin A$ by DeMoivre's theorem. Next, the value~$n$ of~$k$ gives +the same answer as the value~$0$ of~$k$; the value $n+1$ of~$k$ gives the same +answer as the value~$1$ of~$k$; and in general the value $n+m$ of~$k$ gives the +same answer as the value~$m$ of~$k$. Hence we may restrict attention to +the values $0, 1, \dotsc, n-1$ of~$k$. Finally, the answers~\Eq{5} given by these +%% -----File: 014.png---Folio 8------- +values $0, 1,\ldots, n-1$ of $k$ are all distinct, since they are represented by +points whose distance from the origin is the modulus~$1$ and whose amplitudes +are +\[ +\frac{A}{n},\qquad +\frac{A}{n} + \frac{360°}{n},\qquad +\frac{A}{n} + \frac{2·360°}{n},\dotsc, +\frac{A}{n} + \frac{(n-1)360°}{n}, +\] +so that these $n$ points are equally spaced points on a circle of radius unity. +Special cases are noted at the end of~§10. Hence +\begin{Thm}% +any complex number +different from zero has exactly $n$ distinct complex $n$th roots. +\end{Thm} + + +\Section{10.}{Roots of Unity} The trigonometric form of~$1$ is $\cos 0° + i \sin 0°$. +Hence by~§9 with $A=0$, the $n$ distinct $n$th roots of unity are +\index{Roots of unity}% +\[ +\cos\frac{2k \pi}{n} + i \sin\frac{2k \pi}{n}\quad +(k=0, 1, \dotsc, n-1), +\Tag{6} +\] +where now the angles are measured in radians (an angle of $180$~degrees +being equal to $\pi$~radians, where $\pi= 3.1416$, approximately). For $k = 0$, +\Eq{6} reduces to~$1$, which is an evident $n$th root of unity. For $k = 1$, \Eq{6}~is +\[ +R = \cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n}. +\Tag{7} +\] + +By De Moivre's theorem, the general number~\Eq{6} is equal to the +$k$th power of~$R$. Hence the $n$ distinct $n$th roots of unity are +\[ +R,\ R^2,\ R^3,\dotsc,\ R^{n-1},\ R^n = 1. +\Tag{8} +\] + +As a special case of the final remark in~§9, the $n$~complex numbers~\Eq{6}, +and therefore the numbers~\Eq{8}, are represented geometrically by the +vertices of a regular polygon of $n$~sides inscribed in the circle of radius +unity and center at the origin with one vertex on the positive $x$-axis. +\index{Regular polygon}% + +%[Illustraton: \textsc{Fig}. 5] +% [** PP: wrapfigure doesn't play nicely with Remark environment] +\noindent\raisebox{-12pt}{\Input{014a}}\hfill +\begin{minipage}[b]{\linewidth-1.75in} +\begin{Remark} +\hspace*{1.5em}% We're in a minipage +For $n =3$, the numbers~\Eq{8} are $\omega$, $\omega^2$, $1$, which are represented +in Fig.~2 by the vertices of an equilateral triangle. + +\hspace*{1.5em}% +For $n = 4$, $R = \cos\pi/2 + i \sin\pi/2 = i$. The four fourth roots +of unity~\Eq{8} are $i$, $i^2=-1$, $i^3=-i$, $i^4 = 1$, which are represented +by the vertices of a square inscribed in a circle of +radius unity and center at the origin~$O$ (Fig.~5). +\end{Remark} +\end{minipage} + +%% -----File: 015.png---Folio 9------- + +\begin{Exercises}{Page9} + +\begin{Problems} +\item[1.] Simplify the trigonometric forms~\Eq{6} of the four fourth roots of unity. Check +the result by factoring $x^4-1$. + +\item[2.] For $n=6$, show that $R = -\omega^2$. The sixth roots of unity are the three cube roots +of unity and their negatives. Check by factoring $x^6-1$. + +\item[3.] From the point representing $a+bi$, how do you obtain that representing $-(a+bi)$? +Hence derive from Fig.~2 and Ex.~2 the points representing the six sixth roots of unity. +Obtain this result another way. + +\item[4.] Find the five fifth roots of~$-1$. + +\item[5.] Obtain the trigonometric forms of the nine ninth roots of unity. Which of +them are cube roots of unity? + +\item[6.] Which powers of a ninth root~\Eq{7} of unity are cube roots of unity? + +\end{Problems} +\end{Exercises} + + +\Section[Primitive Roots of Unity] +{11.}{Primitive $n$th Roots of Unity.} An $n$th root of unity is called +\emph{primitive} if $n$ is the smallest positive integral exponent of a power of it +that is equal to unity. Thus $\rho$ is a primitive $n$th root of unity if and only +if $\rho^n=1$ and $\rho^l \neq 1$ for all positive integers $l<n$. +\index{Primitive root of unity}% + +Since only the last one of the numbers~\Eq{8} is equal to unity, the number~$R$, +defined by~\Eq{7}, is a primitive $n$th root of unity. We have shown that +the powers~\Eq{8} of~$R$ give all of the $n$th roots of unity. Which of these +powers of~$R$ are primitive $n$th roots of unity? + +\begin{Remark} +For $n=4$, the powers~\Eq{8} of $R=i$ were seen to be +\[ +i^1 = i,\ +i^2 = -1,\ +i^3 = -i,\ +i^4 = 1. +\] +The first and third are primitive fourth roots of unity, and their exponents $1$ and~$3$ +are relatively prime to~$4$, i.e., each has no divisor $>1$ in common with~$4$. But the +second and fourth are not primitive fourth roots of unity (since the square of $-1$ and the +first power of~$1$ are equal to unity), and their exponents $2$ and~$4$ have the divisor~$2$ in +common with $n=4$. These facts illustrate and prove the next theorem for the case +$n=4$. +\end{Remark} + +\begin{Theorem} +The primitive $n$th roots of unity are those of the numbers~\Eq{8} +whose exponents are relatively prime to $n$. +\end{Theorem} +\index{Relatively prime}% + +\begin{Proof} +If $k$ and~$n$ have a common divisor $d$ $(d>1)$, $R^k$ is not a primitive +$n$th root of unity, since +\[ +(R^k)^{\frac{n}{d}} = (R^n)^{\frac{k}{d}} = 1, +\] +and the exponent $n/d$ is a positive integer less than~$n$. + +%% -----File: 016.png---Folio 10------- + +But if $k$ and~$n$ are relatively prime, i.e., have no common divisor $>1$, +$R^k$ is a primitive $n$th root of unity. To prove this, we must show that +$(R^k)^{l} \ne 1$ if $l$~is a positive integer~$<n$. By De Moivre's theorem, +\index{Relatively prime}% +\[ +R^{kl} = \cos\frac{2kl\pi}{n} + + i \sin\frac{2kl\pi}{n}. +\] +If this were equal to unity, $2kl\pi/n$ would be a multiple of~$2\pi$, and hence +$kl$ a multiple of~$n$. Since $k$ is relatively prime to~$n$, the second factor~$l$ +would be a multiple of~$n$, whereas $0 < l < n$. +\end{Proof} + + +\begin{Exercises}{Page10} + +\begin{Problems} + +\item[1.] Show that the primitive cube roots of unity are $\omega$ and~$\omega^{2}$. + +\item[2.] For $R$ given by~\Eq{7}, prove that the primitive $n$th +roots of unity are (i)~for $n=6$, +$R$, $R^5$; (ii)~for $n=8$, $R$, $R^3$, $R^5$, $R^7$; (iii)~for $n=12$, $R$, $R^5$, $R^7$, $R^{11}$. + +\item[3.] When $n$ is a prime, prove that any $n$th root of unity, other than~$1$, is primitive. + +\item[4.] Let $R$ be a primitive $n$th root \Eq{7} of unity, where $n$ is a product of two different +primes $p$ and~$q$. Show that $R, \dotsc, R^n$ are primitive with the exception of $R^p$, $R^{2p}, \dotsc, +R^{qp}$, whose $q$th powers are unity, and $R^q$, $R^{2q}, \dotsc, R^{pq}$, whose $p$th powers are unity. +These two sets of exceptions have only $R^{pq}$ in common. Hence there are exactly +$pq - p - q + 1$ primitive $n$th roots of unity. + +\item[5.] Find the number of primitive $n$th roots of unity if $n$ is a square of a prime~$p$. + +\item[6.] Extend Ex.~4 to the case in which $n$ is a product of three distinct primes. + +\item[7.] If $R$ is a primitive $15$th root \Eq{7} of unity, verify that $R^3$, $R^6$, $R^9$, $R^{12}$ are the primitive +fifth roots of unity, and $R^5$ and~$R^{10}$ are the primitive cube roots of unity. Show +that their eight products by pairs give all the primitive $15$th roots of unity. + +\item[8.] If $\rho$ is any primitive $n$th root of unity, prove that $\rho$, $\rho^2, \dots, \rho^n$ are distinct and +give all the $n$th roots of unity. Of these show that $\rho^k$ is a primitive $n$th root of unity +if and only if $k$ is relatively prime to~$n$. + +\item[9.] Show that the six primitive $18$th roots of unity are the negatives of the primitive +ninth roots of unity. +\end{Problems} +\end{Exercises} + +%% -----File: 017.png---Folio 11------- + + +\Chapter[Theorems on Roots of Equations] +{II}{Elementary Theorems on the Roots of an Equation} + +\Section{12.}{Quadratic Equation} If $a$, $b$, $c$ are given numbers, $a \ne 0$, +\index{Quadratic equation}% +\[ +ax^2 + bx + c = 0 \quad (a \ne 0) +\Tag{1} +\] +is called a \emph{quadratic equation} or equation of the second degree. The +reader is familiar with the following method of solution by ``completing +the square.'' Multiply the terms of the equation by~$4a$, and transpose +the constant term; then +\[ +4a^{2}x^2 + 4abx = -4ac. +\] +Adding $b^2$ to complete the square, we get +\[ +(2ax + b)^2 = \Delta,\qquad \Delta = b^2 - 4ac, +\] +\[ +x_{1} = \frac{-b + \sqrt{\Delta}}{2a}\qquad +x_{2} = \frac{-b - \sqrt{\Delta}}{2a} +\Tag{2} +\] + +By addition and multiplication, we find that +\[ +x_{1} + x_{2} = \frac{-b}{a},\qquad +x_{1} x_{2} = \frac{ c}{a}. +\Tag{3} +\] +Hence for all values of the variable~$x$, +\[ +a(x - x_1)(x - x_2) + \equiv ax^2 - a(x_1 + x_2)x + ax_1 x_2 + \equiv ax^2 + bx + c, +\Tag{4} +\] +the sign $\equiv$ being used instead of~$=$ since these functions of~$x$ are \emph{identically +equal}, i.e., the coefficients of like powers of~$x$ are the same. We speak +of $a(x - x_1)(x - x_2)$ as the \emph{factored form} of the quadratic function $ax^2 + bx + c$, +and of $x - x_1$ and $x - x_2$ as its \emph{linear factors}. +\index{Factored form}% +\index{Identity}% +\index{Linear factors}% +\index{Symbol!a@{$\equiv$\IndAdd{identically equal to}}}% [** PP: Manually alphabetized] + +In \Eq{4} we assign to~$x$ the values $x_1$ and~$x_2$ in turn, and see that +\[ +0 = ax_1^2 + bx_1 + c,\qquad +0 = ax_2^2 + bx_2 + c. +\] + +Hence the values \Eq{2} are actually the roots of equation~\Eq{1}. + +We call $\Delta = b^2 - 4ac$ the \emph{discriminant} of the function $ax^2 + bx + c$ or +\index{Discriminant!of quadratic}% +of the corresponding equation~\Eq{1}. If $\Delta = 0$, the roots~\Eq{2} are evidently +equal, so that, by~\Eq{4}, $ax^2 + bx + c$ is the square of $\sqrt{a}(x - x_1)$, and conversely. +%% -----File: 018.png---Folio 12------- +We thus obtain the useful result that $ax^2 + bx + c$ \emph{is a perfect +square (of a linear function of $x$) if and only if $b^2 = 4ac$ (\emph{i.e.}, if its discriminant +is zero)}. +\index{Quadratic function a square}% + +Consider a \emph{real} quadratic equation, i.e., one whose coefficients $a$, $b$, $c$ +are all real numbers. Then if $\Delta$ is positive, the two roots~\Eq{2} are real. +But if $\Delta$ is negative, the roots are conjugate imaginaries~(§2). + +When the coefficients of a quadratic equation \Eq{1} are any complex +numbers, $\Delta$ has two complex square roots~(§9), so that the roots \Eq{2} of +\Eq{1} are complex numbers, which need not be conjugate. + +\begin{Remark} +For example, the discriminant of $x^2 - 2x + c$ is $\Delta = 4(1 - c)$. If $c = 1$, then $\Delta = 0$ and +$x^2 - 2x + 1 \equiv (x - 1)^2$ is a perfect square, and the roots $1$, $1$ of $x^2 - 2x + 1 = 0$ are equal. +If $c = 0$, $\Delta = 4$ is positive and the roots $0$ and~$2$ of $x^2 - 2x \equiv x(x - 2) = 0$ are real. If $c = 2$, +$\Delta = -4$ is negative and the roots $1 ± \sqrt{-1}$ of $x^2 - 2x + 2 = 0$ are conjugate complex +numbers. The roots of $x^2 - x + 1 + i = 0$ are $i$ and~$1 - i$, and are not conjugate. +\index{Discriminant!of quadratic}% +\end{Remark} + + +% [** PP: No ToC entry in original] +\Section[Polynomial] +{13.}{Integral Rational Function, Polynomial.} If $n$ is a positive integer +and $c_0$, $c_1, \dotsc, c_n$ are constants (real or imaginary), +\index{Integral!rational function}% +\index{Polynomial}% +\[ +f(x) \equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_{n-1} x + c_n +\] +is called a \emph{polynomial} in~$x$ of \emph{degree}~$n$, or also an \emph{integral rational function} +of~$x$ of degree~$n$. It is given the abbreviated notation~$f(x)$, just as the +logarithm of $x + 2$ is written $\log(x + 2)$. +\index{Symbol!b@{$f(x)$\IndAdd{polynomial}}}% [** PP: Manually alphabetized] + +If $c_0 \ne 0$, $f(x) = 0$ is an equation of degree~$n$. If $n = 3$, it is often called +a \emph{cubic equation}; and, if $n = 4$, a \emph{quartic equation}. For brevity, we often +speak of an equation all of whose coefficients are real as a \emph{real equation}. +\index{Real equation}% + + +\Section[Remainder Theorem] +{14.}{The Remainder Theorem.} +\index{Remainder theorem}% +\begin{Thm} +If a polynomial $f(x)$ be divided by +$x - c$ until a remainder independent of~$x$ is obtained, this remainder is equal +to~$f(c)$, which is the value of~$f(x)$ when $x = c$. +\end{Thm} + +Denote the remainder by~$r$ and the quotient by~$q(x)$. Since the +dividend is~$f(x)$ and the divisor is~$x - c$, we have +\[ +f(x) \equiv (x-c)q(x) + r, +\] +identically in~$x$. Taking $x = c$, we obtain $f(c) = r$. + +If $r = 0$, the division is exact. Hence we have proved also the following +useful theorem. + +\begin{Theorem}[The Factor Theorem.] +If $f(c)$ is zero, the polynomial $f(x)$ has the +factor $x - c$. In other words, if $c$ is a root of $f(x) = 0$, $x - c$ is a factor of~$f(x)$. +\end{Theorem} +\index{Factor theorem}% + +%% -----File: 019.png---Folio 13------- + +For example, $2$~is a root of $x^3 - 8 = 0$, so that $x - 2$ is a factor of $x^3 - 8$. Another +illustration is furnished by formula (4). % [** PP: Added period] + + +\begin{Exercises}{Page13} + +Without actual division find the remainder when +\begin{Problems} +\item[1.] $x^4 - 3x^2 - x - 6$ is divided by $x + 3$. + +\item[2.] $x^3 - 3x^2 + 6x - 5$ is divided by $x - 3$. +\end{Problems} + +Without actual division show that +\begin{Problems} +\item[3.] $18x^{10} + 19x^5 + 1$ is divisible by $x + 1$. + +\item[4.] $2x^4 - x^3 - 6x^2 + 4x - 8$ is divisible by $x - 2$ and $x + 2$. + +\item[5.] $x^4 - 3x^3 + 3x^2 - 3x + 2$ is divisible by $x - 1$ and $x - 2$. + +\item[6.] $r^3 - 1$, $r^4 - 1$, $r^5 - 1$ are divisible by $r - 1$. + +\item[7.] By performing the indicated multiplication, verify that +\[ +r^n - 1 \equiv (r - 1)(r^{n-1} + r^{n-2} + \dotsb + r + 1). +\] + +\item[8.] In the last identity replace $r$ by~$x/y$, multiply by~$y^n$, and derive +\[ +x^n - y^n \equiv (x-y)(x^{n-1} + x^{n-2}y + \dotsb + xy^{n-2} + y^{n-1}). +\] + +\item[9.] In the identity of Exercise~8 replace $y$ by $-y$, and derive +\begin{align*} +x^n + y^n + &\equiv (x+y)(x^{n-1} - x^{n-2} y + \dotsb - xy^{n-2} + y^{n-1}), + \quad \text{$n$~odd}; \\ +x^n - y^n + &\equiv (x+y)(x^{n-1} - x^{n-2} y + \dotsb + xy^{n-2} - y^{n-1}), + \quad\text{$n$~even}. +\end{align*} +\end{Problems} + +Verify by the Factor Theorem that $x + y$ is a factor. +\begin{Problems} +\item[10.] If $a$, $ar$, $ar^2, \dotsc, ar^{n-1}$ are $n$ numbers in \emph{geometrical progression} (the ratio of any +term to the preceding being a constant $r \ne 1$), prove by Exercise~7 that their sum is +equal to +\index{Geometrical!progression}% +\[ +\frac{a(r^n - 1)}{r - 1}. +\] + +\item[11.] At the end of each of $n$ years a man deposits in a savings bank $a$~dollars. With +annual compound interest at~4\%, show that his account at the end of $n$~years will be +\index{Compound interest}% +\[ +\frac{a}{.04} \bigl\{(1.04)^n - 1\bigr\} +\] +dollars. Hint: The final deposit draws no interest; the prior deposit will amount to +$a(1.04)$ dollars; the deposit preceding that will amount to $a(1.04)^2$ dollars, etc. Hence +apply Exercise~10 for $r = 1.04$. +\end{Problems} +\end{Exercises} + + +\Section{15.}{Synthetic Division} The labor of computing the value of a polynomial +in~$x$ for an assigned value of~$x$ may be shortened by a simple device. +To find the value of +\index{Synthetic division}% +\[ +x^4 + 3x^3 - 2x - 5 +\] +%% -----File: 020.png---Folio 14------- +for $x = 2$, note that $x^4 = x·x^3 = 2x^3$, so that the sum of the first two terms +of the polynomial is~$5x^3$. To $5x^3 = 5·2^2x$ we add the next term~$-2x$ and +obtain~$18x$ or~$36$. Combining~$36$ with the final term~$-5$, we obtain the +desired value~$31$. + +This computation may be arranged systematically as follows. After +supplying zero coefficients of missing powers of~$x$, we write the coefficients +in a line, ignoring the powers of~$x$. +\[ +\begin{array}{rRRRRPc} +1 & 3 & 0 & -2 & -5 && \Lcol{2} \\ +\cline{7-7} + & 2 & 10 & 20 & 36 && \\ +\cline{1-5} +1 & 5 & 10 & 18 & 31 && \Strut +\end{array} +\] +First we bring down the first coefficient~$1$. Then we multiply it by the +given value~$2$ and enter the product~$2$ directly under the second coefficient~$3$, +add and write the sum~$5$ below. Similarly, we enter the product of +$5$ by~$2$ under the third coefficient~$0$, add and write the sum~$10$ below; etc. +The final number~$31$ in the third line is the value of the polynomial when +$x = 2$. The remaining numbers in this third line are the coefficients, in +their proper order, of the quotient +\[ +x^3 + 5x^2 + 10x + 18, +\] +which would be obtained by the ordinary long division of the given polynomial +by $x - 2$. + +We shall now prove that this process, called \emph{synthetic division}, enables +us to find the quotient and remainder when any polynomial $f(x)$ is divided +by $x - c$. Write +\index{Quotient by synthetic division}% +\begin{align*} +f(x) &\equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_n, \\ +\intertext{and let the constant remainder be $r$ and the quotient be} +q(x) &\equiv b_0 x^{n-1} + b_1 x^{n-2} + \dotsb + b_{n-1}. +\end{align*} + +By comparing the coefficients of $f(x)$ with those in +\begin{multline*} +(x - c)q(x) + r + \equiv b_0 x^n + + (b_1 - cb_0) x^{n-1} \\ + + (b_2 - cb_1) x^{n-2} + \dotsb + + (b_{n-1} - cb_{n-2}) x + + r - cb_{n-1}, +\end{multline*} +we obtain relations which become, after transposition of terms, +\[ +b_0 = a_0,\ +b_1 = a_1 + cb_0,\ +b_2 = a_2 + cb_1, \dotsc,\ +b_{n-1} = a_{n-1} + cb_{n-2},\ +r = a_n + cb_{n-1}. +\] +%% -----File: 021.png---Folio 15------- +The steps in the work of computing the~$b$'s may be tabulated as follows: +\[ +\begin{array}{rRRRWWPc} +a_0 & a_1 & a_2 & \cdots & a_{n-1} & a_n && \Lcol{c} \\ +\cline{8-8} + & cb_0 & cb_1 & \cdots & cb_{n-2} & cb_{n-1} && \\ +\cline{1-6} +b_0 & b_1 & b_2 & \cdots & b_{n-1}, & r && \Strut +\end{array} +\] +In the second space below $a_0$ we write $b_0$ (which is equal to $a_0$). We +multiply $b_0$ by $c$ and enter the product directly under $a_1$, add and write +the sum $b_1$ below it. Next we multiply $b_1$ by $c$ and enter the product +directly under $a_2$, add and write the sum $b_2$ below it; etc. + + +\begin{Exercises}{Page15} + +Work each of the following exercises by synthetic division. +\begin{Problems} +\item[1.] Divide $x^3 + 3x^2 - 2x - 5$ by $x-2$. + +\item[2.] Divide $2x^5 - x^3 + 2x - 1$ by $x+2$. + +\item[3.] Divide $x^3 + 6x^2 + 10x - 1$ by $x - 0.09$. + +\item[4.] Find the quotient of $x^3 - 5x^2 - 2x + 24$ by $x-4$, and then divide the quotient by +$x-3$. What are the roots of $x^3 - 5x^2 - 2x + 24 = 0$? + +\item[5.] Given that $x^4 - 2x^3 - 7x^2 + 8x + 12 = 0$ has the roots $-1$ and~$2$, find the quadratic +equation whose roots are the remaining two roots of the given equation, and find these +roots. + +\item[6.] If $x^4 - 2x^3 - 12x^2 + 10x + 3 = 0$ has the roots $1$ and~$-3$, find the remaining two roots. + +\item[7.] Find the quotient of $2x^4 - x^3 - 6x^2 + 4x - 8$ by $x^2 - 4$. + +\item[8.] Find the quotient of $x^4 - 3x^3 + 3x^2 - 3x + 2$ by $x^2 - 3x + 2$. + +\item[9.] Solve Exercises 1, 2, 3, 6, 7 of~§14 by synthetic division. +\end{Problems} +\end{Exercises} + + +\Section{16.}{Factored Form of a Polynomial} Consider a polynomial %[** PP: Typo polynominal] +\index{Factored form}% +\[ +f(x) \equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_n \quad (c_0 \ne 0), +\] +whose leading coefficient $c_0$ is not zero. If $f(x) = 0$ has the root~$\alpha_1$, which +may be any complex number, the Factor Theorem shows that $f(x)$ has the +factor $x - \alpha_1$, so that +\[ +f(x) \equiv (x - \alpha_1)Q(x),\quad +Q(x) \equiv c_0 x^{n-1} + c_1' x^{n-2} + \dotsb + c_{\alpha-1}'. +\] +If $Q(x) = 0$ has the root~$\alpha_2$, then +\[ +Q(x) \equiv (x - \alpha_2)Q_1(x),\quad +f(x) \equiv (x - \alpha_1)(x - \alpha_2)Q_1(x). +\] +If $Q_1(x) = 0$ has the root~$\alpha_3$, etc., we finally get +\[ +f(x) \equiv c_0(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n). +\Tag{5} +\] + +We shall deduce several important conclusions from the preceding +discussion. First, suppose that the equation $f(x)=0$ of degree~$n$ is known +%% -----File: 022.png---Folio 16------- +to have $n$~distinct roots $\alpha_1, \dotsc, \alpha_n$. In $f(x) \equiv (x-\alpha_1)Q(x)$ take $x=\alpha_2$; +then $0=(\alpha_2-\alpha_1)Q(\alpha_2)$, whence $Q(\alpha_2)=0$ and $Q(x)=0$ has the root $\alpha_2$. +Similarly, $Q_1(x)=0$ has the root $\alpha_3$, etc. Thus all of the assumptions +(each introduced by an ``if'') made in the above discussion have been +justified and we have the conclusion~\Eq{5}. Hence \emph{if an equation $f(x)=0$ +of degree~$n$ has $n$~distinct roots $\alpha_1, \dotsc, \alpha_n$, $f(x)$ can be expressed in the +factored form~\Eq{5}}. +\index{Number!of roots}% + +It follows readily that the equation can not have a root~$\alpha$ different +from $\alpha_1, \dotsc, \alpha_n$. For, if it did, the left member of~\Eq{5} is zero when +$x=\alpha$ and hence one of the factors of the right member must then be zero, +say $\alpha-\alpha_j = 0$, whence the root~$\alpha$ is equal to~$\alpha_j$. We have now proved +the following important result. + +\begin{Theorem} +An equation of degree $n$ cannot have more than $n$ distinct roots. +\end{Theorem} + + +\Section[Multiple Roots] +{17.}{Multiple Roots.\protect\footnotemark}\addtocounter{footnote}{1}% + \footnotetext{Multiple roots are treated by calculus in~§58.}% +\addtocounter{footnote}{-1}% +\index{Multiple roots}% +Equalities may occur among the $\alpha$'s in~\Eq{5}. +Suppose that exactly $m_1$ of the $\alpha$'s (including~$\alpha_1$) are equal to~$\alpha_1$; that +$\alpha_2 \ne \alpha_1$, while exactly~$m_2$ of the~$\alpha$'s are equal to~$\alpha_2$; etc. Then \Eq{5} becomes +\[ +f(x) \equiv + c_0(x-\alpha_1)^{m_1} + (x-\alpha_2)^{m_2} \dotsm + (x-\alpha_k)^{m_k}, \quad m_1 + m_2 + \dotsb + m_k = n, +\Tag{6} +\] +where $\alpha_1, \dotsc, \alpha_k$ are distinct. We then call~$\alpha_1$ a \emph{root of multiplicity}~$m_1$ +of $f(x)= 0$, $\alpha_2$ a root of multiplicity~$m_2$, etc. In other words, $\alpha_1$ is a root +of multiplicity~$m_1$ of $f(x)=0$ if $f(x)$ is exactly divisible by $(x-\alpha_1)^{m_1}$, but is +not divisible by $(x-\alpha_1)^{m_1+1}$. We call $\alpha_1$ also an $m_1$-\emph{fold root}. In the +particular cases $m_1=1$, $2$, and~$3$, we also speak of $\alpha_1$ as a \emph{simple root}, \emph{double +root}, and \emph{triple root}, respectively. For example, $4$~is a simple root, $3$~a +\index{Double root|see{Discriminant}}% +\index{Multiplicity of root}% +\index{Simple root}% +\index{Triple root}% +double root, $-2$~a triple root, and $6$ a root of multiplicity~$4$ (or a $4$-fold +root) of the equation +\[ +7(x-4)(x-3)^2(x+2)^3(x-6)^4 = 0 +\] +of degree~$10$ which has no further root. This example illustrates the +next theorem, which follows from~\Eq{6} exactly as the theorem in~§16 +followed from~\Eq{5}. + +\begin{Theorem} +An equation of degree~$n$ cannot have more than $n$~roots, +a root of multiplicity~$m$ being counted as $m$~roots. +\end{Theorem} + + +\Section{18.}{Identical Polynomials} +\index{Identical polynomials}% +\begin{Thm} +If two polynomials in~$x$, +\[ +a_0 x^n + a_1 x^{n-1} + \dotsb + a_n,\qquad +b_0 x^n + b_1 x^{n-1} + \dotsb + b_n, +\] +%% -----File: 023.png---Folio 17------- +each of degree~$n$, are equal in value for more than $n$~distinct values of~$x$, they +are term by term identical, i.e., $a_0 = b_0$, $a_1 = b_1, \dotsc, a_n = b_n$. +\end{Thm} + +For, taking their difference and writing $c_0 = a_0 - b_0, \dotsc, c_n = a_n - b_n$, +we have +\index{Number!of roots}% +\[ +c_0 x^n + c_1 x^{n-1} + \dotsb + c_n = 0 +\] +for more than $n$~distinct values of~$x$. If $c_0 \ne 0$, we would have a contradiction +with the theorem in~§16. Hence $c_0 = 0$. If $c_1 \ne 0$, we would have +a contradiction with the same theorem with $n$~replaced by~$n-1$. Hence +$c_1 = 0$, etc. Thus $a_0 = b_0$, $a_1 = b_1$, etc. + + +\begin{Exercises}{Page17} + +\begin{Problems} +\item[1.] Find a cubic equation having the roots $0$, $1$, $2$. + +\item[2.] Find a quartic equation having the roots $±1$, $±2$. + +\item[3.] Find a quartic equation having the two double roots $3$ and~$-3$. + +\item[4.] Find a quartic equation having the root~$2$ and the triple root~$1$. + +\item[5.] What is the condition that $ax^2+bx+c=0$ shall have a double root? + +\item[6.] If $a_0 x^n + \dotsb + a_n = 0$ has more than $n$~distinct roots, each coefficient is zero. + +\item[7.] Why is there a single answer to each of Exercises 1--4, if the coefficient of the +highest power of the unknown be taken equal to unity? State and answer the corresponding +general question. +\end{Problems} +\end{Exercises} + + +\Section[Fundamental Theorem of Algebra] +{19.}{The Fundamental Theorem of Algebra.} +\index{Fundamental theorem of algebra}% +\begin{Thm} +Every algebraic equation +with complex coefficients has a complex \(real or imaginary\) root. +\end{Thm} + +This theorem, which is proved in the Appendix, implies that +\begin{Thm}% +every +equation of degree~$n$ has exactly $n$~roots if a root of multiplicity~$m$ be counted +as $m$~roots. +\end{Thm}% +In other words, +\begin{Thm}% +\index{Integral!rational function}% +\index{Linear factors}% +every integral rational function of degree~$n$ +is a product of $n$~linear factors. +\end{Thm}% +For, in~§16, equations $f(x)=0$, $Q(x)=0$, +$Q_1(x)=0, \dotsc$ each has a root, so that \Eq{5} and~\Eq{6} hold. + + +% [** PP: ToC entry matches unit title; using running head] +\Section[Relations between Roots and Coefficients] +{20.}{Relations between the Roots and the Coefficients.} In~§12 we +found the sum and the product of the two roots of any quadratic equation +and then deduced the factored form of the equation. We now apply +the reverse process to any equation +\index{Relations between roots and coefficients}% +\begin{align*} +f(x) &\equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_n = 0\qquad (c_0 \ne 0), +\Tag{7} \\ +\intertext{whose factored form is} +f(x) &\equiv c_0(x-\alpha_1)(x-\alpha_2) \dotsm (x-\alpha_n). +\Tag{8} +\end{align*} +Our next step is to find the expanded form of this product. The following +special products may be found by actual multiplication: +%% -----File: 024.png---Folio 18------- +\begin{align*}% [** PP: Aligning, breaking second line] +(x - \alpha_1)(x - \alpha_2) + &\equiv x^2 - (\alpha_1 + \alpha_2)x + \alpha_1\alpha_2, \\ +%[** PP: Typo (x_1 - \alpha_1)] +(x - \alpha_1)(x - \alpha_2)(x - \alpha_3) + &\equiv x^3 - (\alpha_1 + \alpha_2 + \alpha_3)x^2 \\ + &\qquad + (\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3)x + - \alpha_1\alpha_2\alpha_3. +\end{align*} +These identities are the cases $n = 2$ and $n = 3$ of the following general +formula: +\begin{multline*} +%[** PP: Typo (x_1 - \alpha_1)] +(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n) + \equiv x^n + - (\alpha_1 + \dotsb + \alpha_n)x^{n-1} \\ + + (\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3 + \dotsb + + \alpha_{n-1}\alpha_n)x^{n-2} \\ + - (\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \dotsb + + \alpha_{n-2}\alpha_{n-1}\alpha_n)x^{n-3} \\ + + \dotsb + (-1)^n \alpha_1\alpha_2 \dotsm \alpha_n, +\Tag{9} +\end{multline*} +the quantities in parentheses being described in the theorem below. If +we multiply each member of~\Eq{9} by $x - \alpha_{n+1}$, it is not much trouble to verify +that the resulting identity can be derived from~\Eq{9} by changing $n$ into +$n+1$, so that \Eq{9} is proved true by mathematical induction. Hence the +quotient of \Eq{7} by $c_0$ is term by term identical with~\Eq{9}, so that +\[ +\begin{aligned} +\alpha_1 + \alpha_2 + \dotsb + \alpha_n &= -c_1 / c_0, \\ +% +\alpha_1\alpha_2 + \alpha_1\alpha_3 + + \alpha_2\alpha_3 + \dotsb + + \alpha_{n-1}\alpha_n &= c_2 / c_0, \\ +% +\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \dotsb + + \alpha_{n-2}\alpha_{n-1}\alpha_n &= -c_3 / c_0, \\ +&\vdots\\ %<tb> +\alpha_1\alpha_2 \dotsm \alpha_{n-1}\alpha_n &= (-1)^n c_n / c_0. +\end{aligned} +\Tag{10} +\] +These results may be expressed in the following words: + +\begin{Theorem} +If $\alpha_1, \dotsc, \alpha_n$ are the roots of equation~\Eq{7}, the sum of the +roots is equal to~$-c_1 / c_0$, the sum of the products of the roots taken two at +a time is equal to~$c_2 / c_0$, the sum of the products of the roots taken three at a +time is equal to~$-c_3 / c_0$, etc.; finally, the product of all the roots is equal to~$(-1)^n c_n / c_0$. +\index{Product of roots}% +\index{Sum of!products of roots}% +\index{Sum of!roots}% +\end{Theorem} + +Since we may divide the terms of our equation~\Eq{7} by~$c_0$, the essential +part of our theorem is contained in the following simpler statement: + +\begin{Corollary} +In an equation in~$x$ of degree~$n$, in which the coefficient +of~$x^n$ is unity, the sum of the $n$~roots is equal to the negative of the coefficient +of~$x^{n-1}$, the sum of the products of the roots two at a time is equal to the coefficient +of~$x^{n-2}$, etc.; finally the product of all the roots is equal to the constant +term or its negative, according as $n$ is even or odd. +\end{Corollary} + +\begin{Remark} +For example, in a cubic equation having the roots $2$, $2$, $5$, and having unity as the +coefficient of~$x^3$, the coefficient of~$x$ is $2·2 + 2·5 + 2·5 = 24$. +\end{Remark} + +%% -----File: 025.png---Folio 19------- + + +\begin{Exercises}{Page19} + +\begin{Problems} +\item[1.] Find a cubic equation having the roots $1$, $2$, $3$. + +\item[2.] Find a quartic equation having the double roots $2$ and~$-2$. + +\item[3.] Solve $x^4 - 6x^3 + 13x^2 - 12x + 4 = 0$, which has two double roots. + +\item[4.] Prove that one root of $x^3 + px^2 + qx + r = 0$ is the negative of another root if and +only if $r = pq$. + +\item[5.] Solve $4x^3 - 16x^2 - 9x + 36 = 0$, given that one root is the negative of another. + +\item[6.] Solve $x^3 - 9x^2 + 23x - 15 = 0$, given that one root is the triple of another. + +\item[7.] Solve $x^4 - 6x^3 + 12x^2 - 10x + 3 = 0$, which has a triple root. + +\item[8.] Solve $x^3 - 14x^2 - 84x + 216 = 0$, whose roots are in geometrical progression, i.e., +with a common ratio $r$ [say $m/r$, $m$, $mr$]. +\index{Geometrical!progression}% + +\item[9.] Solve $x^3 - 3x^2 - 13x + 15 = 0$, whose roots are in arithmetical progression, i.e., +with a common difference $d$ [say $m-d$, $m$, $m+d$]. +\index{Arithmetical progression}% + +\item[10.] Solve $x^4 - 2x^3 - 21x^2 + 22x + 40 = 0$, whose roots are in arithmetical progression. +[Denote them by $c-3b$, $c-b$, $c+b$, $c+3b$, with the common difference $2b$]. % [** PP: Added period] + +\item[11.] Find a quadratic equation whose roots are the squares of the roots of +$x^2-px+q = 0$. + +\item[12.] Find a quadratic equation whose roots are the cubes of the roots of $x^2 - px + q = 0$. +Hint: $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$. + +\item[13.] If $\alpha$ and~$\beta$ are the roots of $x^2 - px + q = 0$, find an equation whose roots are (i)~$\alpha^2 / \beta$; +and~$\beta^2 / \alpha$; (ii)~$\alpha^3\beta$ and~$\alpha\beta^3$; (iii)~$\alpha+1 / \beta$ and~$\beta + 1 / \alpha$. + +\item[14.] Find a necessary and sufficient condition that the roots, taken in some order, +of $x^3 + px^2 + qx + r = 0$ shall be in geometrical progression. + +\item[15.] Solve $x^3 - 28x + 48 = 0$, given that two roots differ by~$2$. +\end{Problems} +\end{Exercises} + + +\Section{21.}{Imaginary Roots occur in Pairs} The two roots of a real quadratic +equation whose discriminant is negative are conjugate imaginaries~(§12). +This fact illustrates the following useful result. +\index{Imaginary!roots}% +\index{Surd roots in pairs}% [** PP: Original entry points to page 20] + +\begin{Theorem} +If an algebraic equation with real coefficients has the root $a+bi$, +where $a$ and~$b$ are real and $b \ne 0$, it has also the root~$a-bi$. +\end{Theorem} + +Let the equation be $f(x)= 0$ and divide $f(x)$ by +\[ +(x-a)^2 + b^2 \equiv (x - a-bi)(x - a+bi) +\Tag{11} +\] +until we reach a remainder $rx + s$ whose degree in~$x$ is less than the degree +of the divisor. Since the coefficients of the dividend and divisor are all +real, those of the quotient~$Q(x)$ and remainder are real. We have +\[ +f(x) \equiv Q(x)\bigl\{(x-a)^2 + b^2\bigr\} + rx + s, +\] +identically in~$x$. This identity is true in particular when $x = a+bi$, so +that +\[ +0 = r(a+bi)+s = ra+s+rbi. +\] +%% -----File: 026.png---Folio 20------- +Since all of the letters, other than~$i$, denote real numbers, we have~(§2) +$ra + s = 0$, $rb = 0$. But $b \ne 0$. Hence $r = 0$, and then $s = 0$. Hence $f(x)$ +is exactly divisible by the function~\Eq{11}, so that $f(x) = 0$ has the root $a - bi$. + +The theorem may be applied to the real quotient~$Q(x)$. We obtain +the + +\begin{Corollary} +If a real algebraic equation has an imaginary root of +multiplicity~$m$, the conjugate imaginary of this root is a root of multiplicity~$m$. +\index{Multiplicity of root}% +\end{Corollary} + +Counting a root of multiplicity~$m$ as $m$~roots, we see that a real equation +cannot have an odd number of imaginary roots. Hence by~§19, \emph{a real +equation of odd degree has at least one real root}. +\index{Real equation}% + +Of the $n$~linear factors of a real integral rational function of degree~$n$ +(§19), those having imaginary coefficients may be paired as in~\Eq{11}. +Hence +\begin{Thm}% +\index{Integral!rational function}% +every integral rational function with real coefficients can be expressed +as a product of real linear and real quadratic factors. +\end{Thm} + + +\begin{Exercises}{Page20} + +\begin{Problems} +\item[1.] Solve $x^3 - 3x^2 - 6x - 20 = 0$, one root being $-1 + \sqrt{-3}$. + +\item[2.] Solve $x^4 - 4x^3 + 5x^2 - 2x - 2 = 0$, one root being $1-i$. + +\item[3.] Find a cubic equation with real coefficients two of whose roots are $1$ and $3+2i$. + +\item[4.] If a real cubic equation $x^3 - 6x^2 + \dotsb = 0$ has the root $1 +\sqrt{-5}$, what are the +remaining roots? Find the complete equation. + +\item[5.] If an equation with \emph{rational} coefficients has a root $a + \sqrt{b}$, where $a$ and $b$ are +rational, but $\sqrt{b}$ is irrational, prove that it has the root $a - \sqrt{b}$. [Use the method of~§21.] + +\item[6.] Solve $x^4 - 4x^3 + 4x - 1 = 0$, one root being $2 + \sqrt{3}$. + +\item[7.] Solve $x^3 - (4 + \sqrt{3})x^2 + (5 + 4\sqrt{3})x - 5\sqrt{3} = 0$, having the root $\sqrt{3}$. + +\item[8.] Solve the equation in Ex.~7, given that it has the root $2+i$. + +\item[9.] Find a cubic equation with rational coefficients having the roots $\frac{1}{2}, \frac{1}{2} + \sqrt{2}$. + +\item[10.] Given that $x^4 - 2x^3 - 5x^2 - 6x + 2 = 0$ has the root $2 - \sqrt{3}$, find another root and +by means of the sum and the product of the four roots deduce, without division, the +quadratic equation satisfied by the remaining two roots. + +\item[11.] Granted that a certain cubic equation has the root~$2$ and no real root different +from~$2$, does it have two imaginary roots? + +\item[12.] Granted that a certain quartic equation has the roots $2 ± 3i$, and no imaginary +roots different from them, does it have two real roots? + +\item[13.] By means of the proof of Ex.~5, may we conclude as at the end of~§21 that +every integral rational function with rational coefficients can be expressed as a product +of linear and quadratic factors with rational coefficients? +\end{Problems} +\end{Exercises} + +%% -----File: 027.png---Folio 21------- + + +\Section{22.}{Upper Limit to the Real Roots} Any number which exceeds +\index{Upper limit to roots|(}% +all real roots of a real equation is called an \emph{upper limit to the real roots}. +We shall prove two theorems which enable us to find readily upper limits +to the real roots. For some equations Theorem~I gives a better (smaller) +upper limit than Theorem~II; for other equations, the reverse is true. +Evidently any positive number is an upper limit to the real roots of an +equation having no negative coefficients. + + +\begin{Theorem}[Theorem~I.] +If, in a real equation +\[ +f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_n = 0 \qquad (a_0 > 0), +\] +the first negative coefficient is preceded by $k$ coefficients which are positive or +zero, and if $G$~denotes the greatest of the numerical values of the negative +coefficients, then each real root is less than $1 + \sqrt[k]{G / a_0}$. +\end{Theorem} + +\begin{Remark} +For example, in $x^5 + 4x^4 - 7x^2 - 40x + 1 = 0$, $G = 40$ and $k = 3$ since we must supply +the coefficient zero to the missing power~$x^3$. Thus the theorem asserts that each root +is less than $1 + \sqrt[3]{40}$ and therefore less than~$4.42$. Hence $4.42$ is an upper limit to the +roots. +\end{Remark} + +\begin{Proof} +For positive values of~$x$, $f(x)$ will be reduced in value or remain +unchanged if we omit the terms $a_1 x^{n-1}, \dotsc, a_{k-1}x^{n-k+1}$ (which are +positive or zero), and if we change each later coefficient $a_k, \dotsc, a_n$ to~$-G$. +Hence +\[ +f(x) \geqq a_0 x^n - G(x^{n-k} + x^{n-k-1} + \dotsb + x + 1). +\] +But, by Ex.~7 of~§14, +\[ +x^{n-k} + \dotsb + x + 1 \equiv \frac{x^{n-k+1} - 1}{x - 1}, +\] +if $x \ne 1$. Furthermore, +\[ +a_0 x^n - G\left(\frac{x^{n-k+1} - 1}{x - 1}\right) + \equiv \frac{x^{n-k+1}\{a_0 x^{k-1}(x-1) - G\} + G}{x - 1}. +\] +Hence, if $x>1$, +\begin{align*} +f(x) &> \frac{x^{n-k+1}\bigl\{a_0 x^{k-1}(x-1) - G\bigr\}}{x - 1}, \\ +f(x) &> \frac{x^{n-k+1}\bigl\{a_0 (x-1)^k - G\bigr\}}{x - 1}. +\end{align*} +Thus, for $x>1$, $f(x)>0$ and $x$ is not a root if $a_0 (x-1)^k - G \geqq 0$, which +is true if $x \geqq 1 + \sqrt[k]{G / a_0}$. +\end{Proof} + +%% -----File: 028.png---Folio 22------- + + +% [** PP: No ToC entry] +\Section{23.}{Another Upper Limit to the Roots}\hfill\break +% [** PP: Format Theorem heading manually to avoid \BeforeSkip] +{\indent\normalfont\normalsize\scshape Theorem~II.}\quad +\begin{Thm}% +If, in a real algebraic equation in which the coefficient +of the highest power of the unknown is positive, the numerical value of each +negative coefficient be divided by the sum of all the positive coefficients which +precede it, the greatest quotient so obtained increased by unity is an upper +limit to the roots. +\end{Thm} + +\begin{Remark} +For the example in~§22, the quotients are $7/(1+4)$ and~$40/5$, so that Theorem~II +asserts that $1+8$ or~$9$ is an upper limit to the roots. Theorem~I gave the better upper +limit~$4.42$. But for $x^3 + 8x^2 - 9x + c^2 = 0$, Theorem~I gives the upper limit~$4$, while +Theorem~II gives the better upper limit~$2$. + +We first give the proof for the case of the equation +\[ +f(x) \equiv p_4 x^4 - p_3 x^3 + p_2 x^2 - p_1 x + p_0 = 0 +\] +in which each $p_i$ is positive. In view of the identities +\[ +x^4 \equiv (x-1) (x^3 + x^2 + x + 1) + 1,\qquad +x^2 \equiv (x-1) (x+1) + 1, +\] +$f(x)$ is equal to the sum of the terms +\begin{alignat*}{3} +p_4 (x-1) x^3 &+{}& p_4 (x-1)x^2 + p_4 (x-1)x &+{}& p_4 (x-1) &+ p_4, \\ + {} - p_3 x^3 & & + p_2 (x-1)x &+{}& p_2 (x-1) &+ p_2, \\ + & & {} - p_1 x & & &+ p_0. +\end{alignat*} +If $x>1$, negative terms occur only in the first and third columns, while the sum of the +terms in each of these two columns will be $\geqq 0$ if +\[ +p_4 (x-1) - p_3 \geqq 0,\quad (p_4 + p_2 )(x-1) - p_1 \geqq 0. +\] +Hence $f(x) > 0$ and $x$ is not a root if +\[ +x \geqq 1 + \frac{p_3}{p_4},\quad +x \geqq 1 + \frac{p_1}{p_4 + p_2}. +\] +This proves the theorem for the present equation. + +Next, let $f(x)$ be modified by changing its constant term to~$-p_0$. We modify the +above proof by employing the sum $(p_4 + p_2)x - p_0$ of all the terms in the corresponding +last two columns. This sum will be $>0$ if $x > p_0 / (p_4 + p_2)$, which is true if +\[ +x \geqq 1 + \frac{p_0}{p_4 + p_2}. +\] +\end{Remark} + +To extend this method of proof to the general case +\[ +f(x) \equiv a_n x^n + \dotsb + a_0\qquad (a_n > 0), +\] +we have only to employ suitable general notations. Let the negative +coefficients be $a_{k_1}, \dotsc, a_{k_t} $, where $k_1 > k_2 > \dots > k_t$. For each positive +%% -----File: 029.png---Folio 23------- +integer~$m$ which is $\leqq n$ and distinct from $k_1, \dotsc, k_t$, we replace~$x^m$ +by the equal value +\[ +d(x^{m-1} + x^{m-2} + \dotsb + x + 1) + 1 +\] +where $d \equiv x-1$. Let $F(x)$ denote the polynomial in~$x$, with coefficients +involving~$d$, which is obtained from~$f(x)$ by these replacements. Let +$x>1$, so that $d$ is positive. Thus the terms $a_{k_i} x^{k_i}$ are the only negative +quantities occurring in~$F(x)$. If $k_i > 0$, the terms of~$F(x)$ which involve +explicitly the power $x^{k_i}$ are $a_{k_i} x^{k_i}$ and the $a_m d x^{k_i}$ for the various positive +coefficients~$a_m$ which precede~$a_{k_i}$. The sum of these terms will be $\geqq 0$ +if $a_{k_i} + d \sum a_m \geqq 0$, i.e., if +\[ +x \geqq 1 + \frac{-a_{k_i}}{\sum a_m}. +\] + +There is an additional case if $k_t = 0$, i.e., if $a_0$ is negative. Then the +terms of~$F(x)$ not involving~$x$ explicitly are $a_0$ and the $a_m (d+1)$ for the +various positive coefficients~$a_m$. Their sum, $a_0 + x \sum a_m$, will be $> 0$ if +\[ +x > \frac{-a_0}{\sum a_m}, +\] +which is true if +\[ +x \geqq 1 + \frac{-a_0}{\sum a_m}. +\] +\index{Upper limit to roots|)}% + + +\begin{Exercises}{Page23} + +Apply the methods of both §22 and~§23 to find an upper limit to the roots~of +\begin{Problems} +\item[1.] $4x^5 - 8x^4 + 22x^3 + 98x^2 - 73x + 5 = 0$. + +\item[2.] $x^4 - 5x^3 + 7x^2 - 8x + 1 = 0$. + +\item[3.] $x^7 + 3x^6 - 4x^5 + 5x^4 - 6x^3 - 7x^2 - 8 = 0$. + +\item[4.] $x^7 + 2x^5 + 4x^4 - 8x^2 - 32 = 0$. + +\item[5.] A lower limit to the negative roots of $f(x) = 0$ may be found by applying our +theorems to $f(-x) = 0$, i.e., to the equation derived from $f(x) = 0$ by replacing~$x$ by~$-x$. +Find a lower limit to the negative roots in Exs.\ 2, 3,~4. +\index{Lower limit to roots}% + +\item[6.] Prove that every real root of a real equation $f(x) = 0$ is less than $1 + g / a_0$ if $a_0 > 0$, +where $g$ denotes the greatest of the numerical values of $a_1, \dotsc, a_n$. Hint: if $x>0$, +\[ +a_0 x^n + a_1 x^{n-1} + \dotsb \geqq a_0 x^n - g(x^{n-1} + \dotsb + x + 1). +\] +Proceed as in~§22 with $k = 1$. + +\item[7.] Prove that $1 + g \div |a_0|$ is an upper limit for the moduli of all complex roots of any +equation $f(x)=0$ with complex coefficients, where $g$ is the greatest of the values $|a_1|, +\dotsc, |a_n|$, and $|a|$ denotes the modulus of~$a$. Hint: use Ex.~5 of~§8. +\index{Symbol!c@{$\lvert a\rvert$\IndAdd{modulus}}}% [** PP: Manually alphabetized] +\end{Problems} +\end{Exercises} + +%% -----File: 030.png---Folio 24------- + + +\Section{24.}{Integral Roots} +\index{Integral!roots|(}% +\begin{Thm} +For an equation all of whose coefficients are integers, +any integral root is an exact divisor of the constant term. +\end{Thm} + +For, if~$x$ is an integer such that +\[ +a_0 x^n + \dotsb + a_{n-1}x + a_n = 0, +\Tag{12} +\] +where the $a$'s are all integers, then, by transposing terms, we obtain +\[ +x(-a_0x^{n-1} - \dotsb - a_{n-1}) = a_n. +\] +Thus $x$ is an exact divisor of~$a_n$ since the quotient is the integer given by +the quantity in parenthesis. + +\begin{Example}[1.] +Find all the integral roots of +\[ +x^3 + x^2 - 3x + 9 = 0. +\] +\end{Example} + +\begin{Solution} +The exact divisors of the constant term~$9$ are $±1$, $±3$,~$±9$. By trial, +no one of $±1$, $3$ is a root. Next, we find that~$-3$ is a root by synthetic division~(§15): +\[ +\begin{array}{rRRRPc} +1 & 1 & -3 & 9 && \Lcol{-3} \\ +\cline{6-6} + & -3 & 6 & -9 && \\ +\cline{1-4} +1 & -2 & 3 & 0 && \Strut +\end{array} +\] + +Hence the quotient is $x^2 - 2x + 3$, which is zero for $x = 1 ±\sqrt{-2}$. Thus~$-3$ is the +only integral root. +\end{Solution} + +When the constant term has numerous exact divisors, some device +may simplify the application of the theorem. + +\begin{Example}[2.\protect\footnotemark] Find all the integral roots of +\[ +y^3 + 12y^2 - 32y - 256 = 0. +\] + \footnotetext{This problem is needed for the solution~(§48) of a certain quartic equation.} +\end{Example} + +\begin{Solution} +Since all the terms except~$y^3$ are divisible by~$2$, an integral root~$y$ must +be divisible by~$2$. Since all the terms except~$y^3$ are now divisible by~$2^4$, we have $y = 4z$, +where $z$ is an integer. Removing the factor~$2^6$ from the equation in~$z$, we obtain +\[ +z^3 + 3z^2 - 2z - 4 = 0. +\] +An integral root must divide the constant term~$4$. Hence, if there are any integral +roots, they occur among the numbers $±1$, $±2$, $±4$. By trial, $-1$ is found to be a +root: +\[ +\begin{array}{rRRRPc} +1 & 3 & -2 & -4 && \Lcol{-1} \\ +\cline{6-6} + & -1 & -2 & 4 && \\ +\cline{1-4} +1 & 2 & -4 & 0 && \Strut +\end{array} +\] +%% -----File: 031.png---Folio 25------- +Hence the quotient is $z^2 + 2z - 4$, which is zero for $z = -1 ± \sqrt{5}$. Thus $y = 4z = -4$ +is the only integral root of the proposed equation. +\end{Solution} + + +\begin{Exercises}{Page25} + +Find all the integral roots of +\begin{Problems}[2] +\item[1.] $x^3 + 8x^2 + 13x + 6 = 0$. + +\item[2.] $x^3 - 5x^2 - 2x + 24 = 0$. + +\ResetCols{2} + +\item[3.] $x^3 - 10x^2 + 27x - 18 = 0$. + +\item[4.] $x^4 + 4x^3 + 8x + 32 = 0$. + +\ResetCols{2} + +\item[5.] The equation in Ex.~4 of~§23. + +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{25.}{Newton's Method for Integral Roots} In~§24 we proved that +an integral root~$x$ of equation~\Eq{12} having integral coefficients must be +an exact divisor of~$a_n$. Similarly, if we transpose all but the last two +terms of~\Eq{12}, we see that $a_{n-1}x + a_n$ must be divisible by~$x^2$, and hence +$a_{n-1} + a_n/x$ divisible by~$x$. By transposing all but the last three terms +of~\Eq{12}, we see that their sum must be divisible by~$x^3$, and hence $a_{n-2} + +(a_{n-1} + a_n/x)/x$ divisible by~$x$. We thus obtain a series of conditions +of divisibility which an integral root must satisfy. The final sum +$a_0 + a_{1}/x + \dotsb$ must not merely be divisible by~$x$, but be actually zero, +since it is the quotient of the function~\Eq{12} by~$x^n$. + +In practice, we must test in turn the various divisors~$x$ of~$a_n$. If a +chosen~$x$ is not a root, that fact will be disclosed by one of the conditions +mentioned. Newton's method is quicker than synthetic division since +it usually detects early and throws out wrong guesses as to a root, whereas +in synthetic division the decision comes only at the final step. + +\begin{Remark} +For example, the divisor~$-3$ of the constant term of +\[ +f(x) \equiv x^4 - 9x^3 + 24x^2 - 23x + 15 = 0 +\Tag{13} +\] +is not a root since $-23 + 15/(-3) = -28$ is not divisible by~$-3$. To show that none +of the tests fails for~$3$, so that $3$ is a root, we may arrange the work systematically as +follows: +\[ +\begin{array}{rRRRRPl} + 1 & -9 & 24 & -23 & 15 && \Lcol{\quad3} \\ +-1 & 6 & -6 & 5 & && \Lcol{\text{(divisor)}} \\ +\cline{1-4}\cline{7-7} + 0 & -3 & 18 & -18 & && \Strut +\end{array} +\Tag{14} +\] +First we divide the final coefficient~$15$ by~$3$, place the quotient~$5$ directly under the coefficient~$-23$, +and add. Next, we divide this sum~$-18$ by~$3$, place the quotient~$-6$ +directly under the coefficient~$24$, and add. After two more such steps we obtain the +sum zero, so that $3$ is a root. + +It is instructive to obtain the preceding process by suitably modifying synthetic +division. First, we replace~$x$ by~$1/y$ in~\Eq{13}, multiply each term by~$y^4$, and obtain +\[ +15y^4 - 23y^3 + 24y^2 - 9y + 1 = 0. +\] +%% -----File: 032.png---Folio 26------- +We may test this for the root $y = \frac{1}{3}$, which corresponds to the root $x = 3$ of~\Eq{13}, by +ordinary synthetic division: +\[ +\begin{array}{rRRRRPl} +15 & -23 & 24 & -9 & 1 && \Lcol{\quad\frac{1}{3}} \\ + & 5 & -6 & 6 & -1 && \Lcol{\text{(multiplier)}} \\ +\cline{1-5}\cline{7-7} +15 & -18 & 18 & -3 & 0 && \Strut +\end{array} +\] +The coefficients in the last two lines (after omitting~$15$) are the same as those of the last +two lines in~\Eq{14} read in reverse order. This should be the case since we have here +multiplied the same numbers by~$\frac{1}{3}$ that we divided by~$3$ in~\Eq{14}. The numbers in +the present third line are the coefficients of the quotient~(§15). Since we equate the +quotient to zero for the applications, we may replace these coefficients by the numbers +in the second line which are the products of the former numbers by~$\frac{1}{3}$. The numbers +in the second line of~\Eq{14} are the negatives of the coefficients of the quotient of~$f(x)$ +by $x-3$. +\end{Remark} + +\begin{Example} +Find all the integral roots of equation~\Eq{13}. +\end{Example} + +\begin{Solution} +For a negative value of~$x$, each term is positive. Hence all the real +roots are positive. By~§23, $10$ is an upper limit to the roots. By~§24, any integral +root is an exact divisor of the constant term~$15$. Hence the integral roots, if any, occur +among the numbers $1$, $3$,~$5$. Since $f(1) = 8$, $1$~is not a root. By~\Eq{14}, $3$~is a root. Proceeding +similarly with the quotient by~$x-3$, whose coefficients are the negatives of the +numbers in the second line of~\Eq{14}, we find that $5$ is a root. +\end{Solution} + + +\begin{Exercises}{}% [** PP: No answers] + +\begin{Problems} +\item[1.] Solve Exs.~1--4 of~§24 by Newton's method. + +\item[2.] Prove that, in extending the process~\Eq{14} to the general equation~\Eq{12}, we may +employ the final equations in~§15 with $r = 0$ and write +\[ +\begin{array}{rRRRWWNPl} + a_0 & a_1 & a_2 & \ldots & a_{n-2} & a_{n-1} & a_n && \Lcol{\quad c} \\ +-b_0 & -b_1 & -b_2 & \ldots & -b_{n-2} & -b_{n-1} & && \Lcol{\text{(divisor)}} \\ +\cline{1-6}\cline{9-9} + 0 & -c{b_0} & -cb_1 & \ldots & -cb_{n-3} & -cb_{n-2} & && \Strut +\end{array} +\] +Here the quotient, $-b_{n-1}$, of~$a_n$ by~$c$ is placed directly under $a_{n-1}$, and added to it to +yield the sum $-cb_{n-2}$, etc. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{26.}{Another Method for Integral Roots} An integral divisor~$d$ of the constant term is not a root if $d-m$ is not a divisor of~$f(m)$, where $m$ is +any chosen integer. For, if $d$ is a root of $f(x)=0$, then +\[ +f(x) \equiv (x-d)Q(x), +\] +where $Q(x)$ is a polynomial having integral coefficients~(§15). Hence +$f(m) = (m-d) q$, where $q$ is the integer~$Q(m)$. + +%% -----File: 033.png---Folio 27------- + +\begin{Remark} +In the example of~§25, take $d = 15$, $m=1$. Since $f(1)=8$ is not divisible by $15-1=14$, $15$ is not an integral root. + +Consider the more difficult example +\[ +f(x) \equiv x^3 - 20x^2 + 164x - 400 = 0, +\] +whose constant term has many divisors. There is evidently no negative root, while +$21$ is an upper limit to the roots. The positive divisors less than~$21$ of $400 = 2^4 5^2$ are +$d = 1$, $2$, $4$, $8$, $16$, $5$, $10$,~$20$. First, take $m=1$ and note that $f(1) = -255 = -3·5·17$. The +corresponding values of $d-1$ are $0$, $1$, $3$, $7$, $15$, $4$, $9$,~$19$; of these, $7$, $4$, $9$,~$19$ are not divisors +of~$f(1)$, so that $d = 8$, $5$, $10$ and~$20$ are not roots. Next, take $m=2$ and note that $f(2)= -144$ +is not divisible by $16-2 = 14$. Hence $16$ is not a root. Incidentally, $d=1$ and +$d=2$ were excluded since $f(d)\neq 0$. There remains only $d=4$, which is a root. +\end{Remark} + +In case there are numerous divisors within the limits to the roots, it +is usually a waste of time to list all these divisors. For, if a divisor is +found to be a root, it is preferable to employ henceforth the quotient, +as was done in the example in~§25. + + +\begin{Exercises}{Page27} + +Find all the integral roots of +\begin{Problems} +\item[1.] $x^4 - 2x^3 - 21x^2 + 22x + 40 = 0$. + +\item[2.] $y^3 - 9y^2 - 24y + 216 = 0$. + +\item[3.] $x^4 - 23x^3 + 187x^2 - 653x + 936 = 0$. + +\item[4.] $x^5 + 47x^4 + 423x^3 + 140x^2 + 1213x - 420 = 0$. + +\item[5.] $x^5 - 34x^3 + 29x^2 + 212x - 300 = 0$. +\end{Problems} +\end{Exercises} +\index{Integral!roots|)}% + + +\Section{27.}{Rational Roots} +\index{Rational roots}% +\begin{Thm} +If an equation with integral coefficients +\[ +c_0 x^n + c_1 x^{n-1} + \dotsb + c_{n-1} x + c_n = 0 +\Tag{15} +\] +has the rational root~$a/b$, where $a$ and~$b$ are integers without a common divisor +$>1$, then $a$ is an exact divisor of~$c_n$, and~$b$ is an exact divisor of~$c_0$. +\end{Thm} + +Insert the value $a/b$ of~$x$ and multiply all terms of the equation by~$b^n$. +We obtain +\[ +c_0 a^n + c_1 a^{n-1} b + \dotsb + c_{n-1} ab^{n-1} + c_n b^n = 0. +\] +Since $a$ divides all the terms preceding the last term, it divides that term. +But $a$ has no divisor in common with~$b^n$; hence $a$ divides~$c_n$. Similarly, +$b$~divides all the terms after the first term and hence divides~$c_0$. + +\begin{Example} +Find all the rational roots of +\[ +2x^3 - 7x^2 + 10x - 6 = 0. +\] +\end{Example} + +%% -----File: 034.png---Folio 28------- + +\begin{Solution} +By the theorem, the denominator of any rational root~$x$ is a divisor of~$2$. +Hence $y = 2x$ is an integer. Multiplying the terms of our equation by~$4$, we obtain +\index{Transformed equation}% +\[ +y^3 - 7y^2 + 20y - 24 = 0. +\] +There is evidently no negative root. By either of the tests in §§22,~23, an upper limit +to the positive roots of our equation in~$x$ is $1+7/2$, so that $y<9$. Hence the only +possible values of an integral root~$y$ are $1$, $2$, $3$, $4$, $6$,~$8$. Since~$1$ and~$2$ are not roots, we try~$3$: +\[ +\begin{array}{rRRRPc} + 1 & -7 & 20 & -24 && \Lcol{3} \\ +\cline{6-6} +-1 & 4 & -8 & && \\ +\cline{1-3} + 0 & -3 & 12 & && \Strut +\end{array} +\] +Hence $3$ is a root and the remaining roots satisfy the equation $y^2 - 4y + 8 = 0$ and are +$2±2i$. Thus the only rational root of the proposed equation is $x=3/2$. +\end{Solution} + +If $c_0=1$, then $b=±1$ and~$a/b$ is an integer. Hence we have the +\begin{Corollary} +Any rational root of an equation with integral coefficients, +that of the highest power of the unknown being unity, is an integer. +\end{Corollary} + +Given any equation with integral coefficients +\[ +a_0 y^n + a_1 y^{n-1} + \dotsb + a_n = 0, +\] +we multiply each term by~${a_0}^{n-1}$, write $a_0 y = x$, and obtain an equation~\Eq{15} +with integral coefficients, in which the coefficient~$c_0$ of~$x^n$ is now unity. +By the Corollary, each rational root~$x$ is an integer. Hence we need only +find all the integral roots~$x$ and divide them by~$a_0$ to obtain all the rational +roots~$y$ of the proposed equation. + +Frequently it is sufficient (and of course simpler) to set $ky = x$, where +$k$ is a suitably chosen integer less than~$a_0$. + + +\begin{Exercises}{Page28} + +Find all of the rational roots of +\begin{Problems}[2] + +\item[1.] $y^4 -\frac{40}{3}y^3 + \frac{130}{3}y^2 - 40y + 9 = 0$. + +\item[2.] $6y^3 - 11y^2 + 6y - 1 = 0$. + +\ResetCols{1} + +\item[3.] $108y^3 - 270y^2 - 42y + 1 = 0$. [Use $k = 6$.] + +\item[4.] $32y^3 - 6y - 1 = 0$. [Use the least~$k$.] + +\ResetCols{2} + +\item[5.] $96y^3 - 16y^2 - 6y + 1 = 0$. + +\item[6.] $24y^3 - 2y^2 - 5y + 1 = 0$. + +\ResetCols{2} + +\item[7.] $y^3 - \frac{1}{2}y^2 - 2y + 1 = 0$. + +\item[8.] $y^3 - \frac{2}{3}y^2 + 3y - 2 = 0$. + +\ResetCols{1} + +\item[9.] Solve Exs.~2--6 by replacing $y$ by~$1/x$. +\end{Problems} + +Find the equations whose roots are the products of 6 by the roots of +\begin{Problems}[2] + +\item[10.] $y^2 - 2y - \frac{1}{3} = 0$. + +\item[11.] $y^3 - \frac{1}{2}y^2 - \frac{1}{3}y + \frac{1}{4} = 0$. +\end{Problems} +\end{Exercises} + +%% -----File: 035.png---Folio 29------- + + +\Chapter{III}{Constructions with Ruler and Compasses} +\index{Geometrical!construction|(}% [** PP: Using index subitem] + +% [** PP: No ToC entry] +\Section{28.}{Impossible Constructions} We shall prove that it is not possible, +by the methods of Euclidean geometry, to trisect all angles, or to construct +a regular polygon of~$7$ or $9$~sides. The proof, which is beyond the +scope of elementary geometry, is based on principles of the theory of +equations. Moreover, the discussion will show that a regular polygon +of $17$~sides can be constructed with ruler and compasses, a fact not suspected +during the twenty centuries from Euclid to Gauss. + +\Section{29.}{Graphical Solution of a Quadratic Equation} If $a$ +and~$b$ are constructible, and +\index{Quadratic equation!graphical solution}% +\begin{flalign*}% [** PP: Hack to center over the narrowed text block] +\makebox[\linewidth-3.25in][c]{$x^2 - ax + b = 0$}&& & && +\Tag{1} +\end{flalign*} +%Illustration: \textsc{Fig}.~6 +\begin{wrapfigure}{r}{2.75in} +\hfil\Input{035a} +\end{wrapfigure} +has real coefficients and real roots, the roots +can be constructed with ruler and compasses +as follows. Draw a circle having as a diameter +the line~$BQ$ joining the points $B = (0, 1)$ +and $Q = (a, b)$ in Fig.~6. Then \emph{the +abscissas $ON$ and~$OM$ of the points of intersection +of this circle with the $x$-axis are the +roots of~\Eq{1}}. + +For, the center of the circle is $\bigl(a/2, (b+1)/2\bigr)$; the square of~$BQ$ is +$a^2 + (b-1)^2$; hence the equation of the circle is +\[ +\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b+1}{2}\right)^2 + = \frac{a^2 + (b-1)^2}{4}. +\] + +This is found to reduce to~\Eq{1} when $y = 0$, which proves the theorem. + +When the circle is tangent to the $x$-axis, so that $M$ and~$N$ coincide, +the two roots are equal. When the circle does not cut the $x$-axis, or +when $Q$ coincides with~$B$, the roots are imaginary. + +Another construction follows from~§30. + +%% -----File: 036.png---Folio 30------- + +\begin{Exercises}{Page30} + +Solve graphically: +\begin{Problems}[3] + +\item[1.] $x^2 - 5x + 4 = 0$. + +\item[2.] $x^2 + 5x + 4 = 0$. + +\item[3.] $x^2 + 5x - 4 = 0$. + +\ResetCols{3} + +\item[4.] $x^2 - 5x - 4 = 0$. + +\item[5.] $x^2 - 4x + 4 = 0$. + +\item[6.] $x^2 - 3x + 4 = 0$. + +\end{Problems} +\end{Exercises} + + +\Section{30.}{Analytic Criterion for Constructibility} The first step in our +consideration of a problem proposed for construction consists in formulating +the problem analytically. In some instances elementary algebra +suffices for this formulation. For example, in the ancient problem of +the duplication of a cube, we take as a unit of length a side of the given +cube, and seek the length~$x$ of a side of another cube whose volume is +double that of the given cube; hence +\[ +x^3 = 2. +\Tag{2} +\] + +But usually it is convenient to employ analytic geometry as in~§29; +a point is determined by its coordinates $x$ and~$y$ with reference to fixed +rectangular axes; a straight line is determined by an equation of the +first degree, a circle by one of the second degree, in the coordinates of the +general point on it. Hence we are concerned with certain numbers, +some being the coordinates of points, others being the coefficients of equations, +and still others expressing lengths, areas or volumes. These numbers +may be said to define analytically the various geometric elements +involved. + +\begin{Criterion} +A proposed construction is possible by ruler and compasses +if and only if the numbers which define analytically the desired geometric +elements can be derived from those defining the given elements by a +finite number of rational operations and extractions of real square +roots. +\end{Criterion} +\index{Square roots}% + +\begin{Remark} +In §29 we were given the numbers $a$ and~$b$, and constructed lines of lengths +\[ +\tfrac{1}{2}(a ± \sqrt{a^2 -4b)}. +\] +\end{Remark} + +\begin{Proof} +First, we grant the condition stated in the criterion and prove +that the construction is possible with ruler and compasses. For, a rational +function of given quantities is obtained from them by additions, subtractions, +multiplications, and divisions. The construction of the sum +or difference of two segments is obvious. The construction, by means +of parallel lines, of a segment whose length~$p$ is equal to the product $a·b$ +of the lengths of two given segments is shown in Fig.~7; that for the quotient +%% -----File: 037.png---Folio 31------- +$q = a/b$ in Fig.~8. Finally, a segment of length $s = \sqrt{n}$ may be constructed, +as in Fig.~9, by drawing a semicircle on a diameter composed +of two segments of lengths $1$ and~$n$, and then drawing a perpendicular +to the diameter at the point which separates the two segments. Or we +may construct a root of $x^2 - n = 0$ by~§29. +%[Illustration: Fig. 7] +%[Illustration: Fig. 8] +%[Illustration: Fig. 9] +\begin{figure*}[hbt] +\begin{center} +\Input{037a}\hfil +\Input{037b}\hfil\hfil +\Input{037c} +\end{center} +\end{figure*} + +Second, suppose that the proposed construction is possible with ruler +and compasses. The straight lines and circles drawn in making the construction +are located by means of points either initially given or obtained +as the intersections of two straight lines, a straight line and a circle, or +two circles. Since the axes of coordinates are at our choice, we may +assume that the $y$-axis is not parallel to any of the straight lines employed +in the construction. Then the equation of any one of our lines is +\[ +y = mx + b. +\Tag{3} +\] + +Let $y = m'x + b'$ be the equation of another of our lines which intersects~\Eq{3}. +The coordinates of their point of intersection are +\[ +x = \frac{b' - b}{m - m'},\qquad +y = \frac{mb' - m'b}{m - m'}, +\] +which are rational functions of the coefficients of the equations of the +two lines. + +Suppose that a line~\Eq{3} intersects the circle +\[ +(x - c)^2 + (y - d)^2 = r^2, +\] +with the center $(c, d)$ and radius~$r$. To find the coordinates of the points +of intersection, we eliminate~$y$ between the equations and obtain a quadratic +equation for~$x$. Thus $x$ (and hence also $mx + b$ or~$y$) involves no +%% -----File: 038.png---Folio 32------- +irrationality other than a real square root, besides real irrationalities +present in $m$, $b$, $c$, $d$,~$r$. +\index{Square roots}% + +Finally, the intersections of two circles are given by the intersections +of one of them with their common chord, so that this case reduces to the +preceding. +\end{Proof} + +\begin{Remark} +For example, a side of a regular pentagon inscribed in a circle of radius unity is +(Ex.~2 of~§37) +\[ +s = \tfrac{1}{2}\sqrt{10 - 2\sqrt{5}}, +\Tag{4} +\] +which is a number of the type mentioned in the criterion. Hence a regular pentagon +can be constructed by ruler and compasses (see the example above +quoted). +\end{Remark} + + +\Section{31.}{Cubic Equations with a Constructible Root} We saw that the +\index{Cubic equation}% +problem of the duplication of a cube led to a cubic equation~\Eq{2}. We +shall later show that each of the problems, to trisect an angle, and to construct +regular polygons of $7$ and~$9$ sides with ruler and compasses, leads +to a cubic equation. We shall be in a position to treat all of these problems +as soon as we have proved the following general result. + +\begin{Theorem} +It is not possible to construct with ruler and compasses a +line whose length is a root or the negative of a root of a cubic equation with +rational coefficients having no rational root. +\end{Theorem} + +Suppose that $x_1$ is a root of +\[ +x^3 + \alpha x^2 + \beta x + \gamma = 0 \qquad +\text{($\alpha$, $\beta$, $\gamma$ rational)} +\Tag{5} +\] +such that a line of length $x_1$ or~$-x_1$ can be constructed with ruler and compasses; +we shall prove that one of the roots of~\Eq{5} is rational. We have +only to discuss the case in which $x_1$ is irrational. + +By the criterion in~§30, since the given numbers in this problem are +$\alpha$, $\beta$, $\gamma$, all rational, $x_1$ can be obtained by a finite number of rational +operations and extractions of real square roots, performed upon rational +numbers or numbers derived from them by such operations. Thus $x_1$ +involves one or more real square roots, but no further irrationalities. + +As in the case of~\Eq{4}, there may be superimposed radicals. Such a +two-story radical which is not expressible as a rational function, with +rational coefficients, of a finite number of square roots of positive rational +numbers is said to be a radical of \emph{order}~2. In general, an $n$-story radical +is said to be of order~$n$ if it is not expressible as a rational function, with +\index{Order of radical}% +rational coefficients, of radicals each with fewer than $n$~superimposed +radicals, the innermost ones affecting positive rational numbers. + +%% -----File: 039.png---Folio 33------- + +We agree to simplify $x_1$ by making all possible replacements of certain +types that are sufficiently illustrated by the following numerical examples. + +If $x_1$ involves $\sqrt{3}$, $\sqrt{5}$, and~$\sqrt{15}$, we agree to replace $\sqrt{15}$ by~$\sqrt{3}·\sqrt{5}$. +If $x_1 = s - 7t$, where $s$ is given by~\Eq{4} and +\[ +t = \tfrac{1}{2} \sqrt{10 + 2\sqrt{5}}, +\] +so that $st = \sqrt{5}$, we agree to write $x_1$ in the form $s - 7\sqrt{5}/s$, which involves +a single radical of order~$2$ and no new radical of lower order. Finally, +we agree to replace $\sqrt{4 - 2\sqrt{3}}$ by its simpler form $\sqrt{3}-1$. + +After all possible simplifications of these types have been made, the +resulting expressions have the following properties (to be cited as our +agreements): no one of the radicals of highest order~$n$ in~$x_1$ is equal to +a rational function, with rational coefficients, of the remaining radicals +of order~$n$ and the radicals of lower orders, while no one of the radicals +of order $n-1$ is equal to a rational function of the remaining radicals of +order $n-1$ and the radicals of lower orders, etc. + +Let $\sqrt{k}$ be a radical of highest order~$n$ in~$x_1$. Then +\[ +x_1 = \frac{a + b \sqrt{k}}{c + d \sqrt{k}}, +\] +where $a$, $b$, $c$, $d$ do not involve $\sqrt{k}$, but may involve other radicals. If +$d = 0$, then $c \neq 0$ and we write $e$ for $a/c$, $f$ for $b/c$, and get +\begin{flalign*} +&& x_1 &= e + f\sqrt{k}, && \Rightmark{(f \neq 0)} +\Tag{6} +\end{flalign*} +where neither $e$ nor~$f$ involves~$\sqrt{k}$. If $d \neq 0$, we derive~\Eq{6} by multiplying +the numerator and denominator of the fraction for~$x_1$ by $c - d\sqrt{k}$, which +is not zero since $\sqrt{k} = c/d$ would contradict our above agreements. + +By hypothesis, \Eq{6} is a root of equation~\Eq{5}. After expanding the +powers and replacing the square of $\sqrt{k}$ by~$k$, we see that +\[ +(e + f \sqrt{k})^3 + + \alpha(e + f \sqrt{k})^2 + + \beta(e + f \sqrt{k}) + + \gamma = A + B\sqrt{k}, +\Tag{7} +\] +where $A$ and~$B$ are certain polynomials in $e$, $f$,~$k$ and the rational numbers +$\alpha$, $\beta$,~$\gamma$. Thus $A + B \sqrt{k} = 0$. If $B \neq 0$, $\sqrt{k} = -A/B$ is a rational function, +with rational coefficients, of the radicals, other than~$\sqrt{k}$, in~$x_1$, contrary +to our agreements. Hence $B = 0$ and therefore $A = 0$. + +When $e - f \sqrt{k}$ is substituted for~$x$ in the cubic function~\Eq{5}, the result +%% -----File: 040.png---Folio 34------- +is the left member of~\Eq{7} with $\sqrt{k}$ replaced by~$-\sqrt{k}$, and hence the result +is $A - B\sqrt{k}$. But $A = B = 0$. This shows that +\[ +x_2 = e - f \sqrt{k} +\Tag{8} +\] +is a new root of our cubic equation. Since the sum of the three roots +is equal to~$-\alpha$ by~§20, the third root is +\[ +x_3 = -\alpha - x_1 - x_2 = -\alpha - 2e. +\Tag{9} +\] + +Now $\alpha$ is rational. If also $e$ is rational, $x_3$ is a rational root and we have +reached our goal. We next make the assumption that $e$ is irrational +and show that it leads to a contradiction. Since $e$ is a component part +of the constructible root~\Eq{6}, its only irrationalities are square roots. +Let $\sqrt{s}$ be one of the radicals of highest order in~$e$. By the argument +which led to~\Eq{6}, we may write $e = e' + f'\sqrt{s}$, whence, by~\Eq{9}, +\begin{flalign*} +&& x_3 &= g + h \sqrt{s},&& \Rightmark{(h \neq 0)} +\Tag{9'} +\end{flalign*} +where neither $g$ nor~$h$ involves~$\sqrt{s}$. Then by the argument which led +to~\Eq{8}, $g - h \sqrt{s}$ is a root, different from~$x_3$, of our cubic equation, and hence +is equal to $x_1$ or~$x_2$ since there are only three roots~(§16). Thus +\[ +g - h \sqrt{s} = e ± f \sqrt{k}. +\] + +By definition, $\sqrt{s}$ is one of the radicals occurring in $e$. Also, by~\Eq{9'}, +every radical occurring in $g$ or~$h$ occurs in $x_3$ and hence in $e = \frac{1}{2}(-\alpha - x_3)$, +by~(9), $\alpha$ being rational. Hence $\sqrt{k}$ is expressible rationally in terms +of the remaining radicals occurring in $e$ and~$f$, and hence in~$x_1$, whose value +is given by~\Eq{6}. But this contradicts one of our agreements. + + +\Section{32.}{Trisection of an Angle} For a given angle~$A$, we can construct +\index{Trisection of angle}% +with ruler and compasses a line of length $\cos A$ or $-\cos A$, namely the +adjacent leg of a right triangle, with hypotenuse unity, formed by dropping +a perpendicular from a point in one side of $A$ to the other, produced if +necessary. If it were possible to trisect angle~$A$, i.e., construct the angle +$A/3$ with ruler and compasses, we could as before construct a line whose +length is $±\cos(A/3)$. Hence if we show that this last cannot be done +when the only given geometric elements are the angle~$A$ and a line of +unit length, we shall have proved that the angle~$A$ cannot be trisected. +We shall give the proof for $A = 120°$. + +We employ the trigonometric identity +\[ +\cos A = 4 \cos^3 \frac{A}{3} - 3 \cos \frac{A}{3}. +\] +%% -----File: 041.png---Folio 35------- +Multiply each term by~$2$ and write $x$ for $2\cos(A/3)$. Thus +\[ +x^3 - 3x = 2\cos A. +\Tag{10} +\] + +For $A = 120°$, $\cos A = -\frac{1}{2}$ and~\Eq{10} becomes +\[ +x^3 - 3x + 1 = 0. +\Tag{11} +\] + +Any rational root is an integer~(§27) which is an exact divisor of the +constant term~(§24). By trial, neither $+1$ nor~$-1$ is a root. Hence +\Eq{11} has no rational root. Hence~(§31) +\begin{Thm}% +it is not possible to trisect all +angles with ruler and compasses. +\end{Thm} + +\begin{Remark} +Certain angles, like $90°$, $180°$, can be trisected. When $A=180°$, the equation +\Eq{10} becomes $x^3 - 3x = -2$ and has the rational root $x = 1$. It is the rationality of a +root which accounts for the possibility of trisecting this special angle~$180°$. +\end{Remark} + + +\Section[Duplication of a Cube] +{33.}{Regular Polygon of $9$~Sides, Duplication of a Cube.} Since angle +\index{Duplication of cube}% +$120°$ cannot be trisected with ruler and compasses~(§32), angle~$40°$ cannot +be so constructed in terms of angle~$120°$ and the line of unit length +as the given geometric elements. Since the former of these elements +and its cosine are constructible when the latter is given, we may take +the line of unit length as the only given element. In a regular polygon +of $9$~sides, the angle subtended at the center by one side is $\frac{1}{9}·360° = 40°$. +Hence \emph{a regular polygon of $9$~sides cannot be constructed with ruler and compasses}. +\index{Regular polygon!09@{$9$~sides}}% +Here, as in similar subsequent statements where the given +elements are not specified, the only such element is the line of unit length. + +A rational root of $x^3 = 2$ is an integer~(§27) which is an exact divisor +of~$2$. The cubes of $±1$ and~$±2$ are distinct from~$2$. Hence there is no +rational root. Hence (§§30,~31) +\begin{Thm}% +it is not possible to duplicate a cube with +ruler and compasses. +\end{Thm} + + +% [** PP: ToC entry reads Regular Polygons of 7, 9, 17, and n Sides] +\Section{34.}{Regular Polygon of $7$~Sides} If we could construct with ruler +and compasses an angle~$B$ containing $360/7$ degrees, we could so construct +a line of length $x = 2 \cos B$. Since $7B = 360°$, $\cos 3B = \cos 4B$. +But +\index{Regular polygon!07@{$7$~sides}|(}% +\begin{align*} +2 \cos 3B &= 2(4 \cos^3 B - 3 \cos B) = x^3 - 3x, \\ +2 \cos 4B &= 2(2 \cos^2 2B - 1) = 4(2 \cos^2 B - 1)^2 - 2 = (x^2 - 2)^2 - 2. +\end{align*} +Hence +\[ +0 = x^4 - 4x^2 + 2 - (x^3 - 3x) = (x - 2)(x^3 + x^2 - 2x - 1). +\] +But $x = 2$ would give $\cos B = 1$, whereas $B$ is acute. Hence +\[ +x^3 + x^2 - 2x - 1 = 0. +\Tag{12} +\] + +%% -----File: 042.png---Folio 36------- + +Since this has no rational root, \emph{it is impossible to construct a regular +polygon of $7$~sides with ruler and compasses}. + + +% [** PP: No ToC entry] +\Section{35.}{Regular Polygon of $7$~Sides and Roots of Unity} If +\index{Roots of unity}% +\[ +R = \cos\frac{2\pi}{7} + i \sin\frac{2\pi}{7}, +\] +we saw in~§10 that $R$, $R^2$, $R^3$, $R^4$, $R^5$, $R^6$, $R^7 = 1$ give all the roots of $y^7 = 1$ +and are complex numbers represented by the vertices of a regular polygon +of $7$~sides inscribed in a circle of radius unity and center at the origin of +coordinates. By~§6, +\[ +\frac{1}{R} = \cos\frac{2\pi}{7} -i \sin\frac{2\pi}{7},\qquad +R + \frac{1}{R} = 2\cos\frac{2\pi}{7}. +\] + +We saw in~§34 that $2 \cos(2\pi/7)$ is one of the roots of the cubic +equation~\Eq{12}. This equation can be derived in a new manner by utilizing +the preceding remarks on $7$th roots of unity. Our purpose is not primarily +to derive~\Eq{12} again, but to illustrate some principles necessary in the +general theory of the construction of regular polygons. + +Removing from $y^7 - 1$ the factor $y-1$, we get +\[ +y^6 + y^5 + y^4 + y^3 + y^2 + y + 1 = 0, +\Tag{13} +\] +whose roots are $R$, $R^2, \dotsc, R^6$. Since we know that $R+1/R$ is one of +the roots of the cubic equation~\Eq{12}, it is a natural step to make the substitution +\[ +y+ \frac{1}{y} = x +\Tag{14} +\] +in~\Eq{13}. After dividing its terms by~$y^3$, we have +\[ +\left(y^3 + \frac{1}{y^3}\right) + +\left(y^2 + \frac{1}{y^2}\right) + +\left(y + \frac{1}{y }\right) + 1=0. +\Tag{13'} +\] +By squaring and cubing the members of~\Eq{14}, we see that +\[ +y^2 + \frac{1}{y^2} = x^2 - 2,\qquad +y^3 + \frac{1}{y^3} = x^3 - 3x. +\Tag{15} +\] +Substituting these values in~\Eq{13'}, we obtain +\[ +x^3 + x^2 - 2x - 1 = 0. +\tag{12}% [** PP: [sic], equation repeated] +\] +That is, the substitution~\Eq{14} converts equation~\Eq{13} into~\Eq{12}. +\index{Regular polygon!07@{$7$~sides}|)}% + +%% -----File: 043.png---Folio 37------- + +If in~\Eq{14} we assign to~$y$ the six values $R, \dotsc, R^6$, we obtain only +three distinct values of~$x$: +\[ +x_1 = R + \frac{1}{R} = R + R^6, \quad +x_2 = R^2 + \frac{1}{R^2} = R^2 + R^5, \quad +x_3 = R^3 + \frac{1}{R^3} = R^3 + R^4. +\Tag{16} +\] + +In order to illustrate a general method of the theory of regular polygons, +we start with the preceding sums of the six roots in pairs and find +the cubic equation having these sums as its roots. For this purpose we +need to calculate +\[ +x_1 + x_2 + x_3, \qquad +x_1 x_2 + x_1 x_3 + x_2 x_3, \qquad +x_1 x_2 x_3. +\] +First, by~\Eq{16}, +\[ +x_1 + x_2 + x_3 = R + R^2 + \dotsb + R^6 = -1, +\] +since $R, \dotsc, R^6$ are the roots of~\Eq{13}. Similarly, +\begin{gather*} +x_1 x_2 + x_1 x_3 + x_2 x_3 = 2(R + R^2 + \dotsb +R^6) = -2, \\ +x_1 x_2 x_3 = 2 + R + R^2 + \dotsb + R^6 = 1. +\end{gather*} + +Consequently~(§20), the cubic having $x_1$, $x_2$, $x_3$ as roots is~\Eq{12}. + + +\Section{36.}{Reciprocal Equations} Any algebraic equation such that the +reciprocal of each root is itself a root of the same multiplicity is called a +\emph{reciprocal equation}. +\index{Reciprocal equation}% + +\begin{Remark} +The equation $y^7 - 1 = 0$ is a reciprocal equation, since if $r$ is any root, $1/r$ is evidently +also a root. Since~\Eq{13} has the same roots as this equation, with the exception of unity +which is its own reciprocal, \Eq{13} is also a reciprocal equation. +\end{Remark} + +If $r$ is any root $\neq 0$ of any equation +\[ +f(y) \equiv y^n + \dotsb + c = 0, +\] +$1/r$ is a root of $f(1/y)=0$ and hence of +\[ +y^n f \left(\frac{1}{y}\right) \equiv 1 + \dotsb + cy^n = 0. +\] +If the former is a reciprocal equation, it has also the root~$1/r$, so that every +root of the former is a root of the latter equation. Hence, by~§18, the +left member of the latter is identical % [** PP: Typo indentical] +with $cf(y)$. Equating the constant +terms, we have $c^2 = 1$, $c= ±1$. Hence +\[ +y^n f\left(\frac{1}{y}\right) \equiv ±f(y). +\Tag{17} +\] +%% -----File: 044.png---Folio 38------- +Thus if $p_i y^{n-i}$ is a term of~$f(y)$, also $±p_i y^i$ is a term. Hence +\[ +f(y) \equiv y^n ± 1 + p_1(y^{n-1} ± y) + p_2 (y^{n-2} ± y^2) + \dotsb. +\Tag{18'} +\] + +If $n$ is \emph{odd}, $n = 2t+1$, the final term is $p_t(y^{t+1} ± y^t)$, and $y ± 1$ is a factor +of $f(y)$. In view of~\Eq{17}, the quotient +\[ +Q(y) \equiv \frac{f(y)}{y±1} +\] +has the property that +\[ +y^{n-1} Q \left(\frac{1}{y}\right) \equiv Q(y). +\] +Comparing this with~\Eq{17}, which implied~\Eq{18'}, we see that $Q(y)=0$ is +a reciprocal equation of the type +\[ +y^{2t} + 1 + + c_1 (y^{2t-1} + y) + + c_2 (y^{2t-2} + y^2) + \dotsb + + c_{t-1} (y^{t+1} + y^{t-1}) + + c_t y^t = 0. +\Tag{18} +\] + +If $n$ is \emph{even}, $n = 2t$, and if the upper sign holds in~\Eq{17}, then~\Eq{18'} is +of the form~\Eq{18}. Next, let the lower sign hold in~\Eq{17}. Since a term +$p_t y^t$ would imply a term~$-p_t y^t$, we have $p_t = 0$. The final term in~\Eq{18'} +is therefore $p_{t-1} (y^{t+1} - y^{t-1})$. Hence $f(y)$ has the factor $y^2-1$. The +quotient $q(y) \equiv f(y)/(y^2-1)$ has the property that +\[ +y^{n-2} q \left(\frac{1}{y}\right) \equiv q(y). +\] +Comparing this with~\Eq{17} as before, we see that $q(y)=0$ is of the form~\Eq{18} +where now $2t = n -2$. Hence, at least after removing one or both +of the factors $y±1$, \emph{any reciprocal equation may be given the form~\Eq{18}}. + +The method by which~\Eq{13} was reduced to a cubic equation may be +used to reduce any equation~\Eq{18} to an equation in~$x$ of half the degree. +First, we divide the terms of~\Eq{18} by~$y^t$ and obtain +\index{Cubic equation}% [** PP: Original entry points to Folio 40] +\[ +\left(y^t + \frac{1}{y^t}\right) + + c_1 \left(y^{t-1} + \frac{1}{y^{t-1}}\right) + \dotsb + + c_{t-1} \left(y + \frac{1}{y}\right) + c_t = 0. +\] +Next, we perform the substitution~\Eq{14} by either of the following methods: +We may make use of the relation +\[ +y^k + \frac{1}{y^k} + = x \left(y^{k-1} + \frac{1}{y^{k-1}}\right) + - \left(y^{k-2} + \frac{1}{y^{k-2}}\right) +\] +to compute the values of $y^k + 1/y^k$ in terms of~$x$, starting with the special +%% -----File: 045.png---Folio 39------- +cases \Eq{14} and~\Eq{15}. For example, +\[ +\begin{split} +y^4 + \frac{1}{y^4} + &= x \left(y^3 + \frac{1}{y^3}\right) + - \left(y^2 + \frac{1}{y^2}\right) \\ + &= x (x^3-3x) - (x^2-2) = x^4 - 4x^2 + 2. % [** PP: Added .] +\end{split} +\] +Or we may employ the explicit formula~\Eq{19} of~§107 for the sum $y^k + 1/y^k$ +of the $k$th~powers of the roots~$y$ and~$1/y$ of $y^2 - xy + 1 = 0$. + + +% [** PP: No separate ToC entry] +\Section[Regular Polygon of $9$~Sides] +{37.}{Regular Polygon of $9$~Sides and Roots of Unity.} If +\index{Regular polygon!09@{$9$~sides}}% +\index{Roots of unity}% +\[ +R = \cos\frac{2\pi}{9} + i \sin\frac{ 2\pi}{9}, +\] +the powers $R$, $R^2$, $R^4$, $R^5$, $R^7$, $R^8$, are the primitive ninth roots of unity~(§11). +They are therefore the roots of +\[ +\frac{y^9 - 1}{y^3 - 1} = y^6 + y^3 + 1 = 0. +\Tag{19} +\] +Dividing the terms of this reciprocal equation by~$y^3$ and applying the second +relation~\Eq{15}, we obtain our former cubic equation~\Eq{11}. + + +\begin{Exercises}{Page40} + +\begin{Problems} +\item[1.] Show by~\Eq{16} that the roots of~\Eq{12} are $2\cos 2\pi/7$, $2\cos 4\pi/7$, $2\cos 6\pi/7$. + +\begin{minipage}[b]{\linewidth-2.25in} +\item[2.] The imaginary fifth roots of unity satisfy +$y^4 + y^3 + y^2 + y + 1 = 0$, which by the substitution~\Eq{14} becomes +$x^2 + x - 1 = 0$. It has the root +\[ +R + \frac{1}{R} = 2 \cos\frac{2\pi}{5} = \frac{1}{2}(\sqrt{5}-1). +\] +In a circle of radius unity and center~$O$ draw two perpendicular +diameters $AOA'$, $BOB'$. With the middle +point~$M$ of~$OA'$ as center and radius~$MB$ draw a circle +cutting~$OA$ at~$C$ (Fig.~10). Show that $OC$ and~$BC$ +are the sides~$s_{10}$ and~$s_5$ of the inscribed regular decagon +and pentagon respectively. Hints: +\end{minipage}% +% Illustration: \textsc{Fig}. 10 +\hfill\Input{045a} +\index{Regular!decagon}% +\index{Regular!pentagon}% +\begin{align*} +MB &= \tfrac{1}{2}\sqrt{5},\qquad +OC = \tfrac{1}{2}(\sqrt{5} - 1),\qquad +BC = \sqrt{1+OC^2} = \tfrac{1}{2}\sqrt{10 - 2\sqrt{5}}, \\ +s_{10} &= 2 \sin 18° = 2 \cos\frac{2\pi}{5} = OC, \\ +{s_5}^2 &= (2 \sin 36°)^2 = 2\left(1 - \cos\frac{2\pi}{5}\right) + = \frac{1}{4}(10 - 2\sqrt{5}), \qquad s_5 = BC. +\end{align*} + +%% -----File: 046.png---Folio 40------- + +\item[3.] If $R$ is a root of~\Eq{19} verify as at the end of~§35 that $R+R^8$, $R^2+R^7$, and $R^4+R^5$ +are the roots of~\Eq{11}. + +\item[4.] Hence show that the roots of~\Eq{11} are $2\cos 2\pi/9$, $2\cos 4\pi/9$, $2\cos 8\pi/9$. + +\item[5.] Reduce $y^{11} = 1$ to an equation of degree~$5$ in~$x$. + +\item[6.] Solve $y^5 - 7y^4 + y^3 - y^2 + 7y - 1 = 0$ by radicals. [One root is~$1$.] + +\item[7.] After finding so easily in \ChapRef{I} the trigonometric forms of the complex roots +of unity, why do we now go to so much additional trouble to find them algebraically? + +\item[8.] Prove that every real root of $x^4 + ax^2 + b = 0$ can be constructed with ruler and +compasses, given lines of lengths $a$ and~$b$. + +\item[9.] Show that the real roots of $x^3 - px - q = 0$ are the abscissas of the intersections +of the parabola $y = x^2$ and the circle through the origin with the center +$(\frac{1}{2}q, \frac{1}{2} + \frac{1}{2}p)$. +\end{Problems} + +Prove that it is impossible, with ruler and compasses: +\begin{Problems} +\item[10.] To construct a straight line representing the distance from the circular base +of a hemisphere to the parallel plane which bisects the hemisphere. + +\item[11.] To construct lines representing the lengths of the edges of an existing rectangular +parallelopiped having a diagonal of length~$5$, surface area~$24$, and volume~$1$, $2$, $3$, or~$5$. + +\item[12.] To trisect an angle whose cosine is $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, $\frac{1}{8}$ or~$p/q$, where $p$ and~$q$ ($q>1$) are +integers without a common factor, and $q$ is not divisible by a cube. +\index{Trisection of angle}% +\end{Problems} + +Prove algebraically that it is possible, with ruler and compasses: +\begin{Problems} +\item[13.] To trisect an angle whose cosine is $(4a^3 - 3ab^2)/b^3$, where the integer~$a$ is numerically +less than the integer~$b$; for example, $\cos^{-1} 11/16$ if $a = -1$, $b = 4$. + +\item[14.] To construct the legs of a right triangle, given its area and hypotenuse. + +\item[15.] To construct the third side of a triangle, given two sides and its area. + +\item[16.] To locate the point~$P$ on the side $BC=1$ of a given square $ABCD$ such that +the straight line $AP$ cuts $DC$ produced at a point~$Q$ for which the +length of $PQ$ is a given +number~$g$. Show that $y=BP$ is a root of a reciprocal quartic equation, and solve it +when $g = 10$. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{38.}{The Periods of Roots of Unity} Before taking up the regular +\index{Roots of unity!periods of}% +polygon of $17$~sides, we first explain another method of finding the pairs +of imaginary seventh roots of unity $R$ and~$R^6$, $R^2$ and~$R^5$, $R^3$ and~$R^4$, +employed in~\Eq{16}. To this end we seek a positive integer~$g$ such that +the six roots can be arranged in the order +\[ +R,\quad R^g,\quad R^{g^2},\quad R^{g^3},\quad R^{g^4},\quad R^{g^5}, +\Tag{20} +\] +where each term is the $g$th power of its predecessor. Trying $g = 2$, we find +that the fourth term would then be $R^8 = R$. Hence $g\neq 2$. Trying $g = 3$, +we obtain +\[ +R,\quad R^3,\quad R^2,\quad R^6,\quad R^4,\quad R^5, +\Tag{21} +\] +where each term is the cube of its predecessor. + +%% -----File: 047.png---Folio 41------- + +To define three \emph{periods}, each of two terms, +\[ +R + R^6,\qquad +R^2 + R^5,\qquad +R^3 + R^4, +\Tag{16'} +\] +we select the first term~$R$ of~\Eq{21} and the third term $R^6$ after it and add +them, then the second term $R^3$ and the third term $R^4$ after it, and finally +$R^2$ and the third term $R^5$ after it. + +We may also define two periods, each of three terms, +\[ +z_1 = R + R^2 + R^4,\qquad +z_2 = R^3 + R^6 + R^5, +\] +by taking alternate terms in~\Eq{21}. + +\begin{Remark} +Since $z_1 + z_2 = -1$, $z_1 z_2 = 3 + R + \dotsb + R^6 = 2$, $z_1$ and $z_2$ are the roots of $z^2 + z + 2 = 0$. +Then $R$, $R^2$, $R^4$ are the roots of $w^3 - z_1w^2 + z_2w - 1 = 0$. +\end{Remark} + + +% [** PP: No separate ToC entry] +\Section{39.}{Regular Polygon of $17$~Sides} Let $R$ be a root $\neq 1$ of $x^{17} = 1$. +Then +\index{Regular polygon!17@{$17$~sides}|(}% +\[ +\frac{R^{17} - 1}{R - 1} = R^{16} + R^{15} + \dotsb + R + 1 = 0. +\] +As in §38, we may take $g=3$ and arrange the roots $R, \dotsc, R^{16}$ so that +each is the cube of its predecessor: +\[ +R,\ R^3,\ R^9,\ R^{10},\ +R^{13},\ R^5,\ R^{15},\ R^{11},\ +R^{16},\ R^{14},\ R^8,\ R^7,\ +R^4,\ R^{12},\ R^2,\ R^6. +\] + +Taking alternate terms, we get the two periods, each of eight terms, +\begin{align*} +y_1 &= R + R^9 + R^{13} + R^{15} + R^{16} + R^8 + R^4 + R^2, \\ +y_2 &= R^3 + R^{10} + R^5 + R^{11} + R^{14} + R^7 + R^{12} + R^6. +\end{align*} +Hence $y_1 + y_2 = -1$. We find that $y_1 y_2 = 4(R + \dotsb + R^{16}) = -4$. Thus +\[ +y_1,\ y_2 \quad\text{satisfy}\quad y^2 + y - 4 = 0. +\Tag{22} +\] + +Taking alternate terms in~$y_1$, we obtain the two periods +\[ +z_1 = R + R^{13} + R^{16} + R^4, \qquad +z_2 = R^9 + R^{15} + R^8 + R^2. +\] +Taking alternate terms in~$y_2$, we get the two periods +\[ +w_1 = R^3 + R^5 + R^{14} + R^{12}, \qquad +w_2 = R^{10} + R^{11} + R^7 + R^6. +\] +Thus $z_1 + z_2 = y_1$, $w_1 + w_2 = y_2$. We find that $z_1 z_2 = w_1 w_2 = -1$. Hence +\begin{align*} +z_1,\ z_2 &\quad\text{satisfy}\quad z^2 - y_1 z - 1 = 0, +\Tag{23} \\ +w_1,\ w_2 &\quad\text{satisfy}\quad w^2 - y_2 w - 1 = 0. +\Tag{24} +\end{align*} + +%% -----File: 048.png---Folio 42------- + +Taking alternate terms in~$z_1$, we obtain the periods +\[ +v_1 = R + R^{16}, \qquad v_2 = R^{13} + R^4. +\] +Now, $v_1 + v_2 = z_1$, $v_1v_2 = w_1$. Hence +\begin{align*} +v_1,\ v_2 &\quad\text{satisfy}\quad v^2 - z_1v + w_1 = 0, +\Tag{25} \\ +R,\ R^{16} &\quad\text{satisfy}\quad \rho^2 - v_1\rho + 1 = 0. +\Tag{26} +\end{align*} + +Hence we can find $R$ by solving a series of quadratic equations. Which +of the sixteen values of~$R$ we shall thus obtain depends upon which root +of~\Eq{22} is called $y_1$ and which~$y_2$, and similarly in \Eq{23}--\Eq{26}. We shall now +show what choice is to be made in each such case in order that we shall +finally get the value of the particular root +\[ +R = \cos\frac{2\pi}{17} + i \sin\frac{2\pi}{17}. +\] +Then +\begin{alignat*}{4} +\frac{1}{R} + &= \cos \frac{2\pi}{17} - i \sin \frac{2\pi}{17}, &\qquad +v_1 &= R &+{}& \frac{1}{R} &&= 2 \cos \frac{2\pi}{17}, \\ +% +R^4 &= \cos \frac{8\pi}{17} + i \sin \frac{8\pi}{17}, & +v_2 &= R^4 &+{}& \frac{1}{R^4} &&= 2 \cos \frac{8\pi}{17}. +\end{alignat*} +Hence $v_1 > v_2 > 0$, and therefore $z_1 = v_1 + v_2 > 0$. Similarly, +\begin{align*} +w_1 &= R^3 + \frac{1}{R^3} + R^5 + \frac{1}{R^5} + = 2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17} + = 2 \cos \frac{6\pi}{17} - 2 \cos \frac{7\pi}{17} > 0, \\ +% +y_2 &= 2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17} + + 2 \cos \frac{12\pi}{17} + 2 \cos \frac{14\pi}{17} < 0, +\end{align*} +since only the first cosine in $y_2$ is positive and it is numerically less than +the third. But $y_1 y_2 = -4$. Hence $y_1>0$. Thus \Eq{22}--\Eq{24} give +\begin{align*} +y_1 &= \tfrac{1}{2}( \sqrt{17}-1), & +y_2 &= \tfrac{1}{2}(-\sqrt{17}-1), \\ +% +z_1 &= \tfrac{1}{2}y_1 + \sqrt{1 + \tfrac{1}{4}y_1^2}, & +w_1 &= \tfrac{1}{2}y_2 + \sqrt{1 + \tfrac{1}{4}y_2^2}. +\end{align*} + +We may readily construct segments of these lengths. Evidently +$\sqrt{17}$ is the length of the hypotenuse of a right triangle whose legs are of +lengths $1$ and~$4$, while for the radical in~$z_1$ we employ legs of lengths $1$ +and~$\frac{1}{2}y_1$. We thus obtain segments representing the coefficients of the +%% -----File: 049.png---Folio 43------- +quadratic equation~\Eq{25}. Its roots may be constructed as in~§29. The +larger root is +\[ +v_1 = 2 \cos\frac{2\pi}{17}. +\] +Hence we can construct angle $2\pi/17$ with ruler and compasses, and therefore +a regular polygon of $17$~sides. + + +% [** PP: No ToC entry] +\Section{40.}{Construction of a Regular Polygon of $17$~Sides} In a circle of +radius unity, construct two perpendicular +diameters $AB$, $CD$, +and draw tangents at $A$, $D$, +which intersect at~$S$ (Fig.~11). +Find the point~$E$ in~$AS$ for which +$AE = \frac{1}{4} AS$, by means of two bisections. +Then +\[ +AE = \tfrac{1}{4}, \qquad +OE = \tfrac{1}{4} \sqrt{17}. +\] +%[Illustration: \textsc{Fig.} 11] +\begin{figure*}[b] +\begin{center} +\Input{049a} +\end{center} +\end{figure*} + +\noindent Let the circle with center~$E$ +and radius~$OE$ cut~$AS$ at~$F$ and~$F'$. +Then +\begin{align*} +AF &= EF - EA = OE - \tfrac{1}{4} = \tfrac{1}{2} y_1, \\ +AF' &= EF' + EA = OE + \tfrac{1}{4} = -\tfrac{1}{2} y_2, \\ +OF &= \sqrt{OA^2 + AF^2} = \sqrt{1 + \tfrac{1}{4} y_1^2}, \qquad +OF' = \sqrt{1 + \tfrac{1}{4} y_2^2}. +\end{align*} +Let the circle with center~$F$ and radius~$FO$ cut~$AS$ at~$H$, outside of~$F'F$; +that with center~$F'$ and radius~$F'O$ cut~$AS$ at~$H'$ between $F'$ and~$F$. Then +\begin{align*} +AH &= AF + FH = AF + OF + = \tfrac{1}{2} y_1 + \sqrt{1 + \tfrac{1}{4} y_1^2} = z_1, \\ +AH' &= F'H' - F'A = OF'- AF'= w_1. +\end{align*} + +It remains to construct the roots of equation~\Eq{25}. This will be done +as in~§29. Draw $HTQ$ parallel to $AO$ and intersecting $OC$ produced at~$T$. +Make $TQ = AH'$. Draw a circle having as diameter the line $BQ$ +joining $B = (0,1)$ with $Q = (z_1, w_1)$. The abscissas $ON$ and $OM$ of the intersections +of this circle with the $x$-axis $OT$ are the roots of~\Eq{25}. Hence +the larger root~$v_1$ is $OM = 2 \cos(2\pi/17)$. + +%% -----File: 050.png---Folio 44------- + +Let the perpendicular bisector $LP$ of $OM$ cut the initial circle of unit +radius at~$P$. Then +\[ +\cos LOP = OL = \cos\frac{2\pi}{17},\qquad LOP=\frac{2\pi}{17}. +\] + +Hence the chord~$CP$ is a side of the inscribed regular polygon of $17$~sides, +constructed with ruler and compasses. +\index{Regular polygon!17@{$17$~sides}|)}% + + +% [** PP: No separate ToC entry] +\Section{41.}{Regular Polygon of $n$ Sides} If $n$ be a prime such that $n-1$ is +\index{Regular polygon!n@{$n$~sides}}% +\index{Roots of unity}% +a power $2^h$ of~$2$ (as is the case when $n = 3$, $5$, $17$), the $n-1$ imaginary $n$th +roots of unity can be separated into 2~sets each of $2^{h-1}$ roots, each of these +sets subdivided into 2~sets each of $2^{h-2}$ roots, etc., until we reach the pairs +$R$, $1/R$ and $R^2$, $1/R^2$, etc., and in fact\footnote + {See the author's article ``Constructions with ruler and compasses; regular polygons,'' + in \textit{Monographs on Topics of Modern Mathematics}, Longmans, Green and Co., +1911, p.~374.} +in such a manner that we have a +series of quadratic equations, the coefficients of any one of which depend +only upon the roots of quadratic equations preceding it in the series. +Note that this was the case for $n = 17$ and for $n = 5$. It is in this manner +that it can be proved that the roots of $x^n = 1$ can be found in terms of +square roots, so that a regular polygon of $n$~sides can be inscribed by ruler +and compasses, provided $n$ be a prime of the form $2^h + 1$. + +If $n$ be a product of distinct primes of this form, or $2^k$ times such a +product (for example, $n = 15$, $30$ or~$6$), or if $n = 2^m$ ($m > 1$), it follows readily +(see Ex.~1 below) that we can inscribe with ruler and compasses a regular +polygon of $n$~sides. But this is impossible for all other values of~$n$. + + +\begin{Exercises}{Page44} + +\begin{Problems} +\item[1.] If $a$ and~$b$ are relatively prime numbers, so that their greatest common divisor +is unity, we can find integers $c$ and~$d$ such that $ac + bd = 1$. Show that, if regular polygons +of $a$ and~$b$ sides can be constructed and hence angles $2\pi/a$ and $2\pi/b$, a regular +polygon of $a·b$ sides can be derived. + +\item[2.] If $p = 2^h + 1$ is a prime, $h$ is a power of~$2$. For $h = 2^0$, $2^1$, $2^2$, $2^3$, the values of~$p$ +are $3$, $5$, $17$, $257$ and are primes. [Show that $h$ cannot have an odd factor other than +unity.] + +\item[3.] For $13$th roots of unity find the least~$g$~(§38), write out the three periods each +of four terms, and find the cubic equation having them as roots. +\index{Cubic equation}% + +\item[4.] For the primitive ninth roots of unity find the least~$g$ and write out the three +periods each of two terms. +\end{Problems} + +Solve the following reciprocal equations: +\index{Reciprocal equation}% +\begin{Problems}[2] +\item[5.] $y^4 + 4y^3 - 3y^2 + 4y + 1 = 0$. + +\item[6.] $y^5 - 4y^4 + y^3 + y^2 - 4y + 1 = 0$. + +\ResetCols{2} + +\item[7.] $2y^6 - 5y^5 + 4y^4 - 4y^2 + 5y - 2 = 0$. + +\item[\qquad8.] $y^5 + 1 = 31(y + 1)^5$. +\end{Problems} +\end{Exercises} +\index{Geometrical!construction|)}% [** PP: Using index subitem] + +%% -----File: 051.png---Folio 45------- + + +\Chapter[Cubic and Quartic Equations] +{IV}{Solution of Cubic and Quartic Equations; Their Discriminants} +\index{Cubic equation}% + +% [** PP: No ToC entry] +\Section{42.}{Reduced Cubic Equation} If, in the general cubic equation +\index{Cubic equation!reduced}% +\[ +x^3 + bx^2 + cx + d = 0, +\Tag{1} +\] +we set $x = y-b/3$, we obtain the \emph{reduced cubic equation} +\[ +y^3 + py + q = 0, +\Tag{2} +\] +lacking the square of the unknown~$y$, where +\[ +p = c - \frac{b^2}{3}, \qquad +q = d - \frac{bc}{3} + \frac{2b^3}{27}. +\Tag{3} +\] + +After finding the roots $y_1$, $y_2$, $y_3$ of~\Eq{2}, we shall know the roots of~\Eq{1}: +\[ +x_1 = y_1 - \frac{b}{3}, \qquad +x_2 = y_2 - \frac{b}{3}, \qquad +x_3 = y_3 - \frac{b}{3}. +\Tag{4} +\] + + +\Section[Algebraic Solution of a Cubic] +{43.}{Algebraic Solution of the Reduced Cubic Equation.} We shall +employ the method which is essentially the same as that given by Vieta +in~1591. We make the substitution +\[ +y = z - \frac{p}{3z} +\Tag{5} +\] +in~\Eq{2} and obtain +\[ +z^3 - \frac{p^3}{27z^3} + q = 0, +\] +since the terms in $z$ cancel, and likewise the terms in~$1/z$. Thus +\[ +z^6 + qz^3 - \frac{p^3}{27} = 0. +\Tag{6} +\] +Solving this as a quadratic equation for~$z^3$, we obtain +\[ +z^3 = -\frac{q}{2} ±\sqrt{R},\qquad +R = \left(\frac{p}{3}\right)^3 + \left(\frac{q}{2}\right)^2. +\Tag{7} +\] + +%% -----File: 052.png---Folio 46------- + +By §8, any number has three cube roots, two of which are the products +of the remaining one by the imaginary cube roots of unity: +\[ +\omega = -\tfrac{1}{2} + \tfrac{1}{2} \sqrt{3}i,\qquad +\omega^2 = -\tfrac{1}{2} - \tfrac{1}{2} \sqrt{3}i. +\Tag{8} +\] +We can choose particular cube roots +\[ +A = \sqrt[3]{-\frac{q}{2} + \sqrt{R}},\qquad +B = \sqrt[3]{-\frac{q}{2} - \sqrt{R}}, +\Tag{9} +\] +such that $AB = -p/3$, since the product of the numbers under the cube +root radicals is equal to~$(-p/3)^3$. Hence the six values of~$z$ are +\[ +A,\quad \omega A,\quad \omega^2 A,\quad +B,\quad \omega B,\quad \omega^2 B. +\] +These can be paired so that the product of the two in each pair is~$-p/3$: %[** PP: Punctuation indistinct] +\[ +AB = -\frac{p}{3},\qquad +\omega A·\omega^2 B = -\frac{p}{3},\qquad +\omega^2 A·\omega B = -\frac{p}{3}. +\] +Hence with any root~$z$ is paired a root equal to~$-p/(3z)$. By~\Eq{5}, the sum +of the two is a value of~$y$. Hence the \emph{three} values of~$y$ are +\[ +y_1 = A + B,\qquad +y_2 = \omega A + \omega^2 B,\qquad +y_3 = \omega^2 A + \omega B. +\Tag{10} +\] + +It is easy to verify that these numbers are actually roots of~\Eq{2}. For +example, since $\omega^3 = 1$, the cube of $y_2$ is +\[ +A^3 + B^3 + 3\omega A^2 B + 3\omega^2 AB^2 + = -q - p(\omega A + \omega^2 B) = -q - py_2, +\] +by~\Eq{9} and $AB = -p/3$. + +The numbers~\Eq{10} are known as \emph{Cardan's formulas} for the roots of a +\index{Cardan's formulas}% +reduced cubic equation~\Eq{2}. The expression $A + B$ for a root was first +published by Cardan in his \textit{Ars Magna} of~1545, although he had obtained +it from Tartaglia under promise of secrecy. + +\begin{Example} +Solve $y^3 - 15y - 126 = 0$. +\end{Example} + +\begin{Solution} +The substitution~\Eq{5} is here $y = z + 5/z$. We get +\[ +z^6 - 126z^3 + 125 = 0,\qquad +z^3 = 1 \text{ or } 125. +\] +The pairs of values of~$z$ whose product is $5$ are $1$ and~$5$, $\omega$ and $5\omega^2$, $\omega^2$ and $5\omega$. Their +sums $6$, $\omega + 5\omega^2$, and $\omega^2 + 5\omega$ give the three roots. +\end{Solution} + +\begin{Exercises}{Page46} + +Solve the equations: + +\begin{Problems}[2] + +\item[1.] $y^3 - 18y + 35 = 0$. + +\item[2.] $x^3 + 6x^2 + 3x + 18 = 0$. + +\ResetCols{2} + +\item[3.] $y^3 - 2y + 4 = 0$. + +\item[4.] $28x^3 + 9x^2 - 1 = 0$. +\end{Problems} +\end{Exercises} + +%% -----File: 053.png---Folio 47------- + +\Section{44.}{Discriminant} The product of the squares of the differences of +the roots of any equation in which the coefficient of the highest power of +the unknown is unity shall be called the \emph{discriminant} of the equation. +For the reduced cubic~\Eq{2}, the discriminant is +\index{Discriminant!of cubic}% +\[ +(y_1 - y_2)^2 (y_1 - y_3)^2 (y_2 - y_3)^2 = -4p^3 - 27q^2, +\Tag{11} +\] +a result which should be memorized in view of its important applications. +It is proved by means of~\Eq{10} and $\omega^3 = 1$, $\omega^2 + \omega + 1 = 0$, as follows: +\begin{gather*} +y_1 - y_2 = (1-\omega)(A-\omega^2 B), \qquad +y_1 - y_3 = (1-\omega^2)(A-\omega B), \\ +y_2 - y_3 = (\omega -\omega^2) (A-B), \\ +(1-\omega)(1-\omega^2) = 3, \quad +\omega - \omega^2 = \sqrt{3}i. +\end{gather*} +Since $1$, $\omega$, $\omega^2$ are the cube roots of unity, +\[ +(x-1)(x-\omega)(x-\omega^2) \equiv x^3 - 1, +\] +identically in~$x$. Taking $x = A/B$, we see that +\[ +(A-B)(A-\omega B)(A-\omega^2 B) = A^3 - B^3 = 2 \sqrt{R}, +\] +by~\Eq{9}. Hence +\[ +(y_1-y_2)(y_1-y_3)(y_2-y_3) = 6\sqrt{3}\sqrt{R}i. +\] +Squaring, we get~\Eq{11}, since $-108R = -4p^3 - 27q^2$ by~\Eq{7}. For later +use, we note that the discriminant of the reduced cubic is equal to $-108 R$. + +\emph{The discriminant $\Delta$ of the general cubic~\Eq{1} is equal to the discriminant +of the corresponding reduced cubic~\Eq{2}.} For, by~\Eq{4}, +\[ +x_1 - x_2 = y_1 - y_2, \qquad +x_1 - x_3 = y_1 - y_3, \qquad +x_2 - x_3 = y_2 - y_3. +\] + +Inserting in~\Eq{11} the values of $p$ and~$q$ given by~\Eq{3}, we get +\[ +\Delta = 18bcd - 4b^3 d + b^2 c^2 - 4c^3 - 27d^2. +\Tag{12} +\] + +\begin{Remark} +It is sometimes convenient to employ a cubic equation +\[ +ax^3 + bx^2 + cx +d = 0 \quad (a \neq 0), +\Tag{13} +\] +in which the coefficient of $x^3$ has not been made unity by division. The product~$P$ +of the squares of the differences of its roots is evidently derived from~\Eq{12} by replacing +$b$, $c$, $d$ by $b/a$, $c/a$, $d/a$. Hence +\[ +a^4 P = 18 abcd - 4b^3 d + b^2 c^2 - 4ac^3 - 27a^2 d^2. +\Tag{14} +\] +This expression (and not $P$ itself) is called the discriminant of~\Eq{13}. +\end{Remark} + +%% -----File: 054.png---Folio 48------- + + +\Section[Number of Real Roots of a Cubic] +{45.}{Number of Real Roots of a Cubic Equation.} +\index{Cubic equation!number of real roots}% [** PP: Added ``of''] +\index{Number!of roots}% +\begin{Thm} +A cubic equation +with real coefficients has three distinct real roots if its discriminant~$\Delta$ is positive, +a single real root and two conjugate imaginary roots if $\Delta$ is negative, and at +least two equal real roots if $\Delta$ is zero. +\end{Thm} + +If the roots $x_1$, $x_2$, $x_3$ are all real and distinct, the square of the difference +of any two is positive and hence $\Delta$ is positive. + +If $x_1$ and $x_2$ are conjugate imaginaries and hence $x_3$ is real~(§21), +$(x_1 -x_2)^2$ is negative. Since $x_1 - x_3$ and $x_2 - x_3$ are conjugate imaginaries, +their product is positive. Hence $\Delta$ is negative. + +If $x_1 = x_2$, $\Delta$ is zero. If $x_2$ were imaginary, its conjugate would be +equal to~$x_3$ by~§21, and $x_2$, $x_3$ would be the roots of a real quadratic +equation. The remaining factor $x - x_1$ of the cubic would have real +coefficients, whereas $x_1 = x_2$ is imaginary. Hence the equal roots must +be real. + +Our theorem now follows from these three results by formal logic. +For example, if $\Delta$ is positive, the roots are all real and distinct, since +otherwise either two would be imaginary and $\Delta$ would be negative, or two +would be equal and $\Delta$ would be zero. + + +\begin{Exercises}{Page48} + +Compute the discriminant and find the number of real roots of +\begin{Problems}[2] +\item[1.] $y^3 - 2y - 4 = 0$. + +\item[2.] $y^3 - 15y + 4 = 0$. + +\ResetCols{2} + +\item[3.] $y^3 - 27y + 54 = 0$. + +\item[4.] $x^3 + 4x^2 - 11x + 6 = 0$. + +\ResetCols{1} + +\item[5.] Show by means of~§21 that a double root of a real cubic is real. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{46.}{Irreducible Case} When the roots of a real cubic equation are +all real and distinct, the discriminant $\Delta$ is positive and $R = -\Delta/108$ is +negative, so that Cardan's formulas present the values of the roots in a +\index{Cardan's formulas}% +\index{Cube root}% +\index{Irreducible case}% +form involving cube roots of imaginaries. This is called the irreducible +case since it may be shown that a cube root of a general complex number +cannot be expressed in the form $a + bi$, where $a$ and $b$ involve only real +radicals.\footnote + {Author's \textit{Elementary Theory of Equations,} pp.~35, 36.} % [** PP: Added . after pp] +While we cannot always find these cube roots algebraically, +we have learned how to find them trigonometrically~(§8). + +\begin{Example} +Solve the cubic equation~\Eq{2} when $p = -12,\, q = -8\sqrt{2}$. +\end{Example} + +\begin{Solution} +By~\Eq{7}, $R = -32$ Hence formulas~\Eq{9} become +\[ +A = \sqrt[3]{4\sqrt{2} + 4\sqrt{2}i},\qquad +B = \sqrt[3]{4\sqrt{2} - 4\sqrt{2}i}. +\] +%% -----File: 055.png---Folio 49------- +The values of~$A$ were found in~§8. The values of~$B$ are evidently the conjugate imaginaries +of the values of~$A$. Hence the roots are +\[ +4\cos{15°},\quad +4\cos{135°},\quad +4\cos{255°}. +\] +\end{Solution} + + +\begin{Exercises}{Page49} +\begin{Problems}[2] + +\item[1.] Solve $y^3 -15y+4=0$. + +\item[2.] Solve $y^3 -2y-1=0$. + +\ResetCols{2} + +\item[3.] Solve $y^3 -7y+7=0$. + +\item[4.] Solve $x^3+ 3x^2 -2x-5=0$. + +\ResetCols{2} + +\item[5.] Solve $x^3 +x^2 -2x-1=0$. + +\item[6.] Solve $x^3 +4x^2 -7=0$. + +\end{Problems} +\end{Exercises} + + +\Section[Trigonometric Solution of a Cubic] +{47.}{Trigonometric Solution of a Cubic Equation with $\Delta>0$.} When +\index{Cubic equation!trigonometric solution}% +the roots of a real cubic equation are all real, i.e., if $R$ is negative, they +can be computed simultaneously by means of a table of cosines with much +less labor than required by Cardan's formulas. To this end we write +the trigonometric identity +\[ +\cos 3A = 4\cos^3 A - 3\cos A +\] +in the form +\[ +z^3 - \tfrac{3}{4}z - \tfrac{1}{4}\cos 3A = 0\qquad (z = \cos A). +\] +In the given cubic $y^3 + py + q = 0$ take $y=nz$; then +\[ +z^3 + \frac{p}{n^2}z + \frac{q}{n^3} = 0, +\] +which will be identical with the former equation in~$z$ if +\[ +n = \sqrt{-\tfrac{4}{3}p},\quad +\cos{3A} = -\tfrac{1}{2}q ÷ \sqrt{-p^{3}/27}. +\] +Since $R = p^3/27 + q^2/4$ is negative, $p$ must be negative, so that $n$ is real +and the value of $\cos{3A}$ is real and numerically less than unity. Hence +we can find $3A$ from a table of cosines. The three values of $z$ are then +\[ +\cos A,\qquad +\cos(A + 120°),\qquad +\cos(A+240°). +\] +Multiplying these by~$n$, we obtain the three roots~$y$ correct to a number +of decimal places which depends on the tables used. + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] For $y^3 - 2y - 1 =0$, show that $n^2 =8/3$, $\cos{3A} = \sqrt{27/32}$, $3A=23° 17' 0''$, +$\cos A = 0.99084$, $\cos (A+120°) = -0.61237$, $\cos (A+240°) = -0.37847$, and that the +roots~$y$ are $1.61804$, $-1$, $-0.61804$. + +\item[2.] Solve Exs.\ 1, 3, 4, 5, 6 of~§46 by trigonometry. +\end{Problems} +\end{Exercises} + +%% -----File: 056.png---Folio 50------- + + +% [** PP: ToC reads ``Ferrari's and Descartes' Solution of a Quartic''] +\Section{48.}{Ferrari's Solution of the Quartic Equation} The general quartic +equation +\index{Quartic equation|(}% +\[ +x^4 +bx^3 +cx^2 +dx+e=0, +\Tag{15} +\] +or equation of degree four, becomes after transposition of terms +\[ +x^4 + bx^3 = -cx^2 - dx - e. +\] +The left member contains two of the terms of the square of $x^2 +\tfrac{1}{2}bx$. +Hence by completing the square, we get +\[ +(x^2 + \tfrac{1}{2}bx)^2 = (\tfrac{1}{4}b^2 -c)x^2 - dx - e. +\] + +Adding $(x^2+ \frac{1}{2}bx)y+ \frac{1}{4}y^2$ to each member, we obtain +\[ +(x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y)^2 + = (\tfrac{1}{4}b^2 - c + y)x^2 + + (\tfrac{1}{2}by - d)x + + \tfrac{1}{4}y^2 - e. +\Tag{16} +\] +The second member is a perfect square of a linear function of $x$ if and +only if its discriminant is zero~(§12): +\[ +(\tfrac{1}{2}by - d)^2 - 4(\tfrac{1}{4}b^2 - c + y)(\tfrac{1}{4}y^2 - e) = 0, +\] +which may be written in the form +\[ +y^3 - cy^2 + (bd - 4e)y - b^{2}e + 4ce - d^2 = 0. +\Tag{17} +\] + +Choose any root~$y$ of this \emph{resolvent cubic equation~\Eq{17}}. Then the +right member of~\Eq{16} is the square of a linear function, say $mx+n$. Thus +\index{Resolvent cubic}% +\[ +x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y = mx+n \quad\text{or}\quad +x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y = -mx-n. +\Tag{18} +\] +The roots of these quadratic equations are the four roots of~\Eq{16} and +hence of the equivalent equation~\Eq{15}. This method of solution is due +to Ferrari (1522--1565). + +\begin{Example} +Solve $x^4 +2x^3 -12x^2 -10x+3 = 0$. +\end{Example} + +\begin{Solution} +Here $b=2$, $c=-12$, $d=-10$, $e = 3$. Hence~\Eq{17} becomes +\[ +y^3 + 12y^2 - 32y - 256 = 0, +\] +which by Ex.~2 of~§24 has the root $y=-4$. Our quartic may be written in the form +\[ +(x^2 + x)^2 = 13x^2 + 10x-3. +\] +Adding $(x^2 +x)(-4)+4$ to each member, we get +\begin{gather*} +(x^2 + x - 2)^2 = 9x^2 + 6x + 1 = (3x + 1)^2, \\ +x^2 + x - 2 = ±(3x + 1),\qquad +x^2 - 2x - 3 = 0 \text{ or } x^2 + 4x - 1 = 0, +\end{gather*} +whose roots are $3$, $-1$, $-2± \sqrt{5}$. As a check, note that the sum of the roots is~$-2$. +\end{Solution} + +%% -----File: 057.png---Folio 51------- + + +\begin{Exercises}{Page51} + +\begin{Problems} +\item[1.] Solve $x^4 - 8x^3 + 9x^2 + 8x - 10 = 0$. Note that~\Eq{17} is $(y - 9) (y^2 - 24) = 0$. + +\item[2.] Solve $x^4 - 2x^3 - 7x^2 + 8x + 12 = 0$. Since the right member of~\Eq{16} is +$(8 + y) (x^2 - x) + \frac{1}{4} y^2 - 12$, use $y = -8$. + +\item[3.] Solve $x^4 - 3x^2 + 6x - 2 = 0$. + +\item[4.] Solve $x^4 - 2x^2 - 8x - 3 = 0$. + +\item[5.] Solve $x^4 - 10x^2 - 20x - 16 = 0$. + +\end{Problems} +\end{Exercises} + + +\Section[Resolvent Cubic] +{49.}{Roots of the Resolvent Cubic Equation.} Let $y_1$ be the root~$y$ +which was employed in~§48. Let $x_1$ and~$x_2$ be the roots of the first +quadratic equation~\Eq{18}, and $x_3$ and~$x_4$ the roots of the second. Then +\index{Resolvent cubic}% +\[ +x_1 x_2 = \tfrac{1}{2} y_1 - n,\qquad +x_3 x_4 = \tfrac{1}{2} y_1 + n,\qquad +x_1 x_2 + x_3 x_4 = y_1. +\] +If, instead of $y_1$, another root $y_2$ or~$y_3$ of the resolvent cubic~\Eq{17} had been +employed in~§48, quadratic equations different from~\Eq{18} would have +been obtained, such, however, that their four roots are $x_1$, $x_2$, $x_3$, $x_4$, paired +in a new manner. The root which is paired with $x_1$ is $x_2$ or $x_3$ or~$x_4$. It +is now plausible that the values of the three $y$'s are +\[ +y_1 = x_1 x_2 + x_3 x_4,\qquad +y_2 = x_1 x_3 + x_2 x_4,\qquad +y_3 = x_1 x_4 + x_2 x_3. +\Tag{19} +\] + +To give a more formal proof that the $y$'s given by~\Eq{19} are the roots +of~\Eq{17}, we employ~(§20) +\begin{gather*} +x_1 + x_2 + x_3 + x_4 = -b,\qquad +x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 = -d, \\ +x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 = c,\qquad + x_1 x_2 x_3 x_4 = e. +\end{gather*} + +From these four relations we conclude that +\begin{gather*} +y_1 + y_2 + y_3 = c,\\ +\begin{split} +y_1 y_2 + y_1 y_3 + y_2 y_3 + &= (x_1 + x_2 + x_3 + x_4) (x_1 x_2 x_3 + \dotsb + x_2 x_3 x_4) + - 4x_1 x_2 x_3 x_4 \\ +&= bd - 4e, +\end{split} \\ +\begin{split} +y_1 y_2 y_3 + &= (x_1 x_2 x_3 + \dotsb )^2 + + x_1 x_2 x_3 x_4 \bigl\{(x_1 + \dotsb )^2 - 4( x_1 x_2 + \dotsb) \bigr\} \\ + &= d^2 + e ( b^2 - 4c ). +\end{split} +\end{gather*} +Hence (§20) $y_1$, $y_2$, $y_3$ are the roots of the cubic equation~\Eq{17}. + + +% [** PP: No ToC entry] +\Section{50.}{Discriminant} The discriminant $\Delta$ of the quartic equation~\Eq{15} +\index{Discriminant!of quartic}% +is defined to be the product of the squares of the differences of its roots: +\[ +\Delta = ( x_1 - x_2 )^2 ( x_1 - x_3 )^2 ( x_1 - x_4 )^2 + ( x_2 - x_3 )^2 ( x_2 - x_4 )^2 ( x_3 - x_4 )^2. +\] + +%% -----File: 058.png---Folio 52------- + +The fact that $\Delta$ is equal to the discriminant of the resolvent cubic +equation~\Eq{17} follows at once from~\Eq{19}, by which +\begin{align*} +y_1 - y_2 &= (x_1-x_4)(x_2-x_3),\qquad + y_1-y_3 = (x_1-x_3)(x_2-x_4), \\ +% +y_2 - y_3 &= (x_1-x_2)(x_3-x_4),\qquad + (y_1-y_2)^2 (y_1-y_3)^2 (y_2- y_3)^2 = \Delta. +\end{align*} +Hence (§44) $\Delta$ is equal to the discriminant $-4p^3 - 27q^2$ of the reduced +cubic $Y^3 + pY + q = 0$, obtained from~\Eq{17} by setting $y = Y + c/3$. Thus +\[ +p = bd - 4e - \tfrac{1}{3} c^2,\qquad +q = -b^2 e + \tfrac{1}{3} bcd + \tfrac{8}{3} ce - d^2 - \tfrac{2}{27} c^3. +\Tag{20} +\] + +\begin{Theorem} +The discriminant of any quartic equation~\Eq{15} is equal to +the discriminant of its resolvent cubic equation and therefore is equal to the +discriminant $-4p^3 - 27q^2$ of the corresponding reduced cubic $Y^3 + pY + q = 0$, +whose coefficients have the values~\Eq{20}. +\end{Theorem} + + +\begin{Exercises}{} +\index{Number!of roots}% + +\begin{Problems} +\item[1.] Find the discriminant of $x^4 - 3x^3 + x^2 + 3x - 2 = 0$ and show that the equation +has a multiple root. + +\item[2.] Show by its discriminant that $x^4 - 8x^3 + 22x^2 - 24x + 9 = 0$ has a multiple root. + +\item[3.] If a real quartic equation has two pairs of conjugate imaginary roots, show that +its discriminant~$\Delta$ is positive. Hence prove that, if $\Delta<0$, there are exactly two real +roots. + +\item[4.] Hence show that $x^4 - 3x^3 + 3x^2 - 3x + 2 = 0$ has two real and two imaginary roots. +\end{Problems} +\end{Exercises} + + +% [** PP: No separate ToC entry] +\Section{51.}{Descartes' Solution of the Quartic Equation} Replacing $x$ by +$z - b/4$ in the general quartic~\Eq{15}, we obtain the \emph{reduced} quartic equation +\[ +z^4 + qz^2 + rz + s = 0, +\Tag{21} +\] +lacking the term with~$z^3$. We shall prove that we can express the left +member of~\Eq{21} as the product of two quadratic factors +\[ +(z^2 + 2kz + l)(z^2 - 2kz + m) + = z^4 + (l + m - 4k^2)z^2 + 2k(m - l)z + lm. +\] +The conditions are +\[ +l + m - 4k^2 = q,\qquad +2k(m-l)=r,\qquad +lm = s. +\] +If $k\neq0$, the first two give +\[ +2l = q + 4k^2 - \frac{r}{2k},\qquad +2m = q + 4k^2 + \frac{r}{2k}. +\] +Inserting these values in $2l·2m = 4s$, we obtain +\[ +64k^6 + 32qk^4 + 4(q^2 - 4s)k^2 - r^2 = 0. +\Tag{22} +\] +%% -----File: 059.png---Folio 53------- +The latter may be solved as a cubic equation for~$k^2$. Any root $k^2 \neq 0$ +gives a pair of quadratic factors of~\Eq{21}: +\[ +z^2 ± 2kz + \tfrac{1}{2}q + 2k^2 \mp \frac{r}{4k}. +\Tag{23} +\] +The four roots of these two quadratic functions are the four roots of~\Eq{21}. +This method of Descartes (1596--1650) therefore succeeds unless every +root of~\Eq{22} is zero, whence $q = s = r = 0$, so that \Eq{12} is the trivial equation +$z^4 = 0$. + +\begin{Remark} +For example, consider $z^4 - 3z^2 + 6z - 2 = 0$. Then \Eq{22} becomes +\[ +64k^6 - 3·32k^4 + 4·17k^2 - 36 = 0. +\] +The value $k^2 = 1$ gives the factors $z^2 + 2z - 1$, $z^2 - 2z + 2$. Equating these to zero, we +find the four roots $-1 ± \sqrt{2}$, $1± \sqrt{-1}$. +\end{Remark} + + +% [** PP: No ToC entry] +\Section{52.}{Symmetrical Form of Descartes' Solution} To obtain this symmetrical +form, we use all three roots $k_1^2$, $k_2^2$, $k_3^2$ of~\Eq{22}. Then +\[ +k_1^2 + k_2^2 + k_3^2 = -\tfrac{1}{2}q,\qquad +k_1^2 k_2^2 k_3^2 = \frac{r^2}{64}. +\] +It is at our choice as to which square root of~$k_1^2$ is denoted by~$+k_1$ and +which by~$-k_1$, and likewise as to $±k_2$, $±k_3$. For our purposes any +choice of these signs is suitable provided the choice give +\[ +k_1 k_2 k_3 = -\frac{r}{8}. +\Tag{24} +\] + +Let $k_1 \neq 0$. The quadratic function~\Eq{23} is zero for $k = k_1$ if +\[ +(z ± k_1)^2 = -\frac{q}{2} - k_1^2 ± \frac{r}{4k_1} + = k_2^2 + k_3^2 \mp \frac{8k_1k_2k_3}{4k_1} + = (k_2 \mp k_3)^2. +\] +Hence the four roots of the quartic equation~\Eq{21} are +\[ + k_1 + k_2 + k_3,\qquad + k_1 - k_2 - k_3,\qquad +-k_1 + k_2 - k_3,\qquad +-k_1 - k_2 + k_3. +\Tag{25} +\] + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Solve Exs.~4, 5 of~§48 by the method of Descartes. + +\item[2.] By writing $y_1$, $y_2$, $y_3$ for the roots $k_1^2$, $k_2^2$, $k_3^2$ of +\[ +64y^3 + 32qy^2 + 4(q^2 - 4s)y - r^2 = 0, +\Tag{26} +\] +show that the four roots of~\Eq{21} are the values of +\[ +z = \sqrt{y_1} + \sqrt{y_2} + \sqrt{y_3} +\Tag{27} +\] +%% -----File: 060.png---Folio 54------- +for all combinations of the square roots for which +\[ +\sqrt{y_1}·\sqrt{y_2}·\sqrt{y_3} = -\frac{r}{8}. +\Tag{28} +\] + +\item[3.] Euler (1707--1783) solved~\Eq{21} by assuming that it has a root of the form~\Eq{27}. +Square~\Eq{27}, transpose the terms free of radicals, square again, replace the last factor +of $8\sqrt{y_1 y_2 y_3}\,(\sqrt{y_1} + \sqrt{y_2} + \sqrt{y_3})$ by~$z$, and identify the resulting quartic in~$z$ with~\Eq{21}. +Show that $y_1$, $y_2$, $y_3$ are the roots of~\Eq{26} and that relation~\Eq{28} holds. + +\item[4.] Find the six differences of the roots~\Eq{25} and verify that the discriminant $\Delta$ of~\Eq{21} +is equal to the quotient of the discriminant of~\Eq{26} by~$4^6$. + +\item[5.] In the theory of the inflexion points of a plane cubic curve there occurs the +equation +\[ +z^4 - Sz^2 - \tfrac{4}{3}Tz - \tfrac{1}{12}S^2 = 0. +\] +Show that~\Eq{26} now becomes +\[ +\left(y - \frac{S}{6}\right)^3 = C,\qquad +C \equiv \left(\frac{T}{6}\right)^2 - \left(\frac{S}{6}\right)^3, +\] +and that the roots of the quartic equation are +\[ +±\sqrt{\tfrac{1}{6}S + \sqrt[3]{C}} +±\sqrt{\tfrac{1}{6}S + \omega\sqrt[3]{C}} +±\sqrt{\tfrac{1}{6}S + \omega^{2}\sqrt[3]C}, +\] +where $\omega$ is an imaginary cube root of unity and the signs are to be chosen so that the product of the three summands is equal to~$+\tfrac{1}{6}T$. +\end{Problems} +\end{Exercises} + + +\begin{Exercises}[MISCELLANEOUS~]{Page54} + +\begin{Problems} +\item[1.] Find the coordinates of the single real point of intersection of the parabola +$y = x^2$ and the hyperbola $xy - 4x + y + 6 = 0$. + +\item[2.] Show that the abscissas of the points of intersection of $y=x^2$ and +$ax^2 - xy + y^2 - x - (a+5)y - 6 = 0$ +are the roots of $x^4 - x^3 - 5x^2 - x - 6 = 0$. Compute the discriminant +of the latter and show that only two of the four points of intersection are real. + +\item[3.] Find the coordinates of the two real points in Ex.~2. + +\item[4.] A right prism of height~$h$ has a square base whose side is~$b$ and whose diagonal +is therefore $b\sqrt{2}$. If $v$ denotes the volume and $d$ a diagonal of the prism, $v = hb^2$ and +$d^2 = h^2 + (b\sqrt{2})^2$. Multiply the last equation by~$h$ and replace $hb^2$ by~$v$. Hence +$h^3 - d^2h + 2v = 0$. +Its discriminant is zero if $d = 3\sqrt{3}$, $v = 27$; find~$h$. + +\item[5.] Find the admissible values of~$h$ in Ex.~4 when $d = 12$, $v = 332.5$. + +\item[6.] Find a necessary and sufficient condition that quartic equation~\Eq{15} shall have +one root the negative of another root. + +Hint: $(x_1 + x_2)(x_3 + x_4) = q - y_1$. Hence substitute $q$ for~$y$ in~\Eq{17}. +\index{Quartic equation|)}% + +\item[7.] In the study of parabolic orbits occurs the equation +\[% [** PP: Displayed for better line breaking.] +\tan\tfrac{1}{2}v + \tfrac{1}{3}\tan^3 \tfrac{1}{2}v = t. +\] +Prove that there is a single real root and that it has the same sign as~$t$. + +\item[8.] In the problem of three astronomical bodies occurs the equation $x^3 + ax + 2 = 0$. +Prove that it has three real roots if and only if $a\leqq{-3}$. +\end{Problems} +\end{Exercises} + +%% -----File: 061.png---Folio 55------- + + +% [** PP: Not matching running head] +\Chapter{V}{The Graph of an Equation} + +\index{Graphs|(}% +\Section[Use of Graphs] +{53.}{Use of Graphs in the Theory of Equations.} To find geometrically +the real roots of a real equation $f(x)=0$, we construct a graph of $y=f(x)$ +and measure the distances from the origin~$O$ to the intersections of the +graph and the $x$-axis, whose equation is $y=0$. + +%%{Illustration} \textsc{Fig. 12} +\begin{wrapfigure}[16]{r}{2.25in} +\quad\Input{061a} +\end{wrapfigure} +For example to find geometrically the real +roots of +\[ +x^2 - 6x - 3 = 0, +\Tag{1} +\] +we equate the left member to~$y$ and make a +graph of +\[ +y = x^2 - 6x - 3. +\Tag{1'} +\] +We obtain the parabola in Fig.~12. Of the +points shown, $P$ has the \emph{abscissa} +\index{Abscissa}% +$x = OQ = 4$ +and the \emph{ordinate} +\index{Ordinate}% +$y = -QP = -11$. From the +points of intersection of $y = 0$ (the $x$-axis $OX$) +with the parabola, we obtain the approximate +values $6.46$ and $-0.46$ of the roots of~\Eq{1}. + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Find graphically the real roots of $x^2 - 6x + 7 = 0$. +\index{Quadratic equation!graphical solution}% +\end{Problems} + +Hint: For each $x$, $y = x^2 - 6x + 7$ exceeds the $y$ in~\Eq{1'} by~$10$, so that the new graph +is obtained by shifting the parabola in Fig.~12 upward $10$~units, leaving the axes~$OX$ +and~$OY$ unchanged. What amounts to the same thing, but is simpler to do, we leave +the parabola and~$OY$ unchanged, and move the axis~$OX$ downward $10$~units. + +\begin{Problems} +\item[2.] Discuss graphically the reality of the roots of $x^2 - 6x + 12 = 0$. + +\item[3.] Find graphically the roots of $x^2 - 6x + 9 = 0$. +\end{Problems} +\end{Exercises} + +%%Illustration \textsc{Fig}.~13 +\begin{wrapfigure}[21]{l}{1.25in} +\Input{062a} +\end{wrapfigure} +\Section{54.}{Caution in Plotting} If the example set were +\index{Plotting}% +\[ +y = 8x^4 - 14x^3 - 9x^2 + 11x - 2, +\Tag{2} +\] +one might use successive integral values of~$x$, obtain the points $(-2, 180)$, +%% -----File: 062.png---Folio 56------- +$(-1, 0)$, $(0, -2)$, $(1, -6)$, $(2, 0)$, $(3, 220)$, all but the +first and last of which are shown (by crosses) in Fig.~13, +and be tempted to conclude that the graph is a +\Shape{U}-shaped curve approximately like that in Fig.~12 +and that there are just two real roots, $-1$ and~$2$, of +\[ +8x^4 - 14x^3 - 9x^2 + 11x - 2 = 0. +\Tag{2'} +\] +But both of these conclusions would be false. In fact, +the graph is a \Shape{W}-shaped curve (Fig.~13) and the +additional real roots are $\frac{1}{4}$ and~$\frac{1}{2}$. + +This example shows that it is often necessary to +employ also values of~$x$ which are not integers. The +purpose of the example was, however, not to point +out this obvious fact, but rather to emphasize the +chance of serious error in sketching a curve through a +number of points, however numerous. The true curve +between two points below the $x$-axis may not cross the +$x$-axis, or may have a peak and actually cross the $x$-axis +twice, or may be an \Shape{M}-shaped curve crossing it four +times, etc. + +%%Illustration \textsc{Fig}.~14 +\begin{wrapfigure}{r}{1.5in} +\Input{062b} +\end{wrapfigure} +For example, the graph (Fig.~14) of +\[ +y = x^3 + 4x^2 - 11 +\Tag{3} +\] +crosses the $x$-axis only once; but this fact cannot +be established by a graph located by a number +of points, however numerous, whose abscissas +are chosen at random. + +We shall find that correct conclusions regarding +the number of real roots may be deduced +from a graph whose bend points~(§55) have been +located. + + +\Section{55.}{Bend Points} A point (like $M$ or~$M'$ in +Fig.~14) is called a \emph{bend point} of the graph of +\index{Bend point}% +$y=f(x)$ if the tangent to the graph at that point +is horizontal and if all of the adjacent points of +the graph lie below the tangent or all above the +tangent. The first, but not the second, condition +%% -----File: 063.png---Folio 57------- +is satisfied by the point~$O$ of the graph of $y = x^3$ given in Fig.~15 (see~§57). +In the language of the calculus, $f(x)$ has a (relative) maximum or +minimum value at the abscissa of a bend point on the graph of $y=f(x)$. +\index{Maximum}% +\index{Minimum}% +%[Illustration: Fig. 15] +%[Illustration: Fig. 16] +\begin{figure*}[hbt] +\begin{center} +\Input{063a}\hfil +\Input{063b} +\end{center} +\end{figure*} + +Let $P = (x, y)$ and $Q = (x+h, Y)$ be two points on the graph, sketched +in Fig.~16, of $y=f(x)$. By the \emph{slope} of a straight line is meant the tangent +of the angle between the line and the $x$-axis, measured counter-clockwise +from the latter. In Fig.~16, the slope of the straight line~$PQ$ is +\index{Slope}% +\[ +\frac{Y - y}{h} = \frac{f(x+h) - f(x)}{h}. +\Tag{4} +\] + +For equation~\Eq{3}, $f(x) = x^3 + 4x^2 - 11$. Hence +\begin{align*} +f(x+h) &= (x+h)^3 + 4(x+h)^2 - 11 \\ + &= x^3 + 4x^2 - 11 + (3x^2 +8x)h + (3x+4)h^2 + h^3. +\end{align*} +The slope~\Eq{4} of the secant~$PQ$ is therefore here +\index{Derivative|(}% +\[ +3x^2 + 8x + (3x+4)h + h^2. +\] +Now let the point~$Q$ move along the graph toward~$P$. Then $h$ approaches +the value zero and the secant~$PQ$ approaches the tangent at~$P$. The +slope of the tangent at~$P$ is therefore the corresponding limit $3x^2 + 8x$ +of the preceding expression. We call $3x^2 + 8x$ the \emph{derivative} of $x^3 + 4x^2 - 11$. + +%% -----File: 064.png---Folio 58------- + +In particular, if $P$ is a bend point, the slope of the (horizontal) tangent +at~$P$ is zero, whence $3x^2 + 8x = 0$, $x = 0$ or $x = -\tfrac{8}{3}$. Equation~\Eq{3} gives +the corresponding values of~$y$. The resulting points +\[ +M = (0, -11),\qquad M' = (-\tfrac{8}{3}, -\tfrac{41}{27}) +\] +are easily shown to be bend points. Indeed, for $x>0$ and for $x$ between +$-4$ and~$0$, $x^2(x+4)$ is positive, and hence $f(x) > -11$ for such values of~$x$, +so that the function~\Eq{3} has a relative minimum at $x = 0$. Similarly, +there is a relative maximum at $x = -\tfrac{8}{3}$. We may also employ the general +method of~§59 to show that $M$ and~$M'$ are bend points. Since these bend +points are both below the $x$-axis we are now certain that the graph +crosses the $x$-axis only once. + +The use of the bend points insures greater accuracy to the graph than +the use of dozens of points whose abscissas are taken at random. + +\Section{56.}{Derivatives} We shall now find the slope of the tangent to the +graph of $y=f(x)$, where $f(x)$ is any polynomial +\[ +\Tag{5} +f(x) = a_0 x^n + a_1 x^{n-1} + \dotsb + a_{n-1} x + a_n. +\] +We need the expansion of $f(x+h)$ in powers of~$x$. By the binomial +theorem, +\begin{align*} +a_0 (x+h)^n + &= a_0 x^n + na_0 x^{n-1} h + \frac{n(n-1)}{2} a_0 x^{n-2} h^2 + \dotsb, \\ +a_1 (x+h)^{n-1} + &= a_1 x^{n-1} + (n-1)a_1 x^{n-2} h + + \frac{(n-1)(n-2)}{2} a_1 x^{n-3} h^2 + \dotsb, \\ +\multispan{2}{\dotfill} \\ +a_{n-2} (x+h)^2 + &= a_{n-2} x^2 +2a_{n-2} xh + a_{n-2} h^2, \\ +a_{n-1} (x+h) + &= a_{n-1} x + a_{n-1} h, \\ +a_n &= a_n. +\end{align*} +The sum of the left members is evidently $f(x+h)$. On the right, the +sum of the first terms (i.e., those free of~$h$) is~$f(x)$. The sum of the coefficients +of~$h$ is denoted by~$f'(x)$, the sum of the coefficients of $\tfrac{1}{2} h^2$ is denoted +by $f''(x), \dotsc$, the sum of the coefficients of +\[ +\frac{h^k}{1·2\dotsm k} +\] +%% -----File: 065.png---Folio 59------- +is denoted by~$f^{(k)}(x)$. Thus +\begin{align*} +f'(x) &= na_0 x^{n-1} + (n-1)a_1 x^{n-2} + \dotsb + 2a_{n-2} x + a_{n-1}, +\Tag{6} \\ +f''(x) &= n(n-1)a_0 x^{n-2} + (n-1)(n-2)a_1 x^{n-3} + \dotsb + 2a_{n-2}, +\Tag{7} +\intertext{etc. Hence we have} +f(x+h) &= f(x) + f'(x)h + f''(x) \frac{h^2}{1·2} + f'''(x) \frac{h^3}{1·2·3} +\Tag{8} \\ + & \qquad + \dotsb + f^{(r)}(x) \frac{h^r}{r!} + + \dotsb + f^{(n)}(x) \frac{h^n}{n!}, +\end{align*} +where $r!$ is the symbol, read $r$~\emph{factorial}, for the product $1·2·3\dotsm(r-1)r$. +Here $r$ is a positive integer, but we include the case $r = 0$ by the definition, +$0! = 1$. +\index{Symbol!d@{$r$"!\IndAdd{factorial}}}% [** PP: Manually alphabetized] + +This formula~\Eq{8} is known as \emph{Taylor's theorem} for the present case of +\index{Taylor's theorem}% +a polynomial~$f(x)$ of degree~$n$. We call $f'(x)$ the (\emph{first}) \emph{derivative of~$f(x)$}, +and $f''(x)$ the \emph{second derivative} of~$f(x)$, etc. +Concerning the fact that +$f''(x)$ is equal to the first derivative of $f'(x)$ and that, in general, the $k$th +derivative $f^{(k)}(x)$ of $f(x)$ is equal to the first derivative of $f^{(k-1)}(x)$, see +Exs.~6--9 of the next set. +\index{Symbol!e@{$f^{(k)}(x)$\IndAdd{$k$th derivative}}}% [** PP: Manually alphabetized] + +In view of~\Eq{8}, the limit of~\Eq{4} as $h$~approaches zero is $f'(x)$. Hence +\begin{Thm}% +$f'(x)$ is the slope of the tangent to the graph of $y=f(x)$ at the point~$(x, y)$. +\end{Thm} +\index{Slope}% + +In \Eq{5} and~\Eq{6}, let every $a$ be zero except~$a_0$. Thus the derivative of +$a_0 x^n$ is $na_0 x^{n-1}$, and hence is obtained by multiplying the given term by +its exponent~$n$ and then diminishing its exponent by unity. For example, +the derivative of~$2x^3$ is~$6x^2$. + +Moreover, the derivative of $f(x)$ is equal to the sum of the derivatives +of its separate terms. Thus the derivative of $x^3 + 4x^2 - 11$ is $3x^2 + 8x$, +as found also in~§55. + + +\begin{Exercises}{Page59} + +\begin{Problems} +\item[1.] Show that the slope of the tangent to $y = 8x^3 - 22x^2 + 13x - 2$ at $(x, y)$ is +$24x^2 - 44x + 13$, and that the bend points are $(0.37, 0.203)$, $(1.46, -5.03)$, approximately. +Draw the graph. + +\item[2.] Prove that the bend points of $y = x^3 - 2x - 5$ are $(.82, -6.09)$, $(-.82$, $-3.91)$, % [** PP: Allow line break between coordinates] +approximately. Draw the graph and locate the real roots. + +\item[3.] Find the bend points of $y = x^3 + 6x^2 + 8x + 8$. Locate the real roots. + +\item[4.] Locate the real roots of $f(x) = x^4 + x^3 - x - 2 = 0$. + +Hints: The abscissas of the bend points are the roots of $f'(x) = 4x^3 + 3x^2 - 1 = 0$. +The bend points of $y = f'(x)$ are $(0, -1)$ and $(-\frac{1}{2}, -\frac{3}{4})$, so that $f'(x)= 0$ has a single real +root (it is just less than $\frac{1}{2}$). The single bend point of $y=f(x)$ is $(\frac{1}{2}, -\frac{37}{16})$, approximately. + +%% -----File: 066.png---Folio 60------- + +\item[5.] Locate the real roots of $x^6 - 7x^4 - 3x^2 + 7 = 0$. + +\item[6.] Prove that $f''(x)$, given by~\Eq{7}, is equal to the first derivative of~$f'(x)$. + +\item[7.] If $f(x) = f_1(x) + f_2(x)$, prove that the $k$th derivative of~$f$ is equal to the sum of +the $k$th derivatives of $f_1$ and~$f_2$. Use~\Eq{8}. + +\item[8.] Prove that $f^{(k)}(x)$ is equal to the first derivative of $f^{(k-1)}(x)$. Hint: prove this +for $f = ax^m$; then prove that it is true for $f=f_1 + f_2$ if true for $f_1$ and~$f_2$. + +\item[9.] Find the third derivative of $x^6 + 5x^4$ by forming successive first derivatives; +also that of $2x^5 - 7x^3 + x$. + +\item[10.] Prove that if $g$ and~$k$ are polynomials in~$x$, the derivative of $gk$ is $g'k + gk'$. Hint: +multiply the members of $g(x+h) = g(x) + g'(x)h + \dotsb$ and $k(x+h) = k(x) + k'(x)h + \dotsb$ +and use~\Eq{8} for $f = gk$. + +\end{Problems} +\end{Exercises} + + +\Section{57.}{Horizontal Tangents} If $(x, y)$ is a bend point of the graph of +\index{Tangents}% +$y=f(x)$, then, by definition, the slope of the tangent at $(x, y)$ is zero. +Hence~(§56), the abscissa~$x$ is a root of $f'(x)=0$. In Exs.~1--5 of the +preceding set, it was true that, conversely, any real root of $f'(x)=0$ is +the abscissa of a bend point. However, this is not always the case. +We shall now consider in detail an example illustrating this fact. The +example is the one merely mentioned in~§55 to indicate the need of the +second requirement made in our definition of a bend point. + +The graph (Fig.~15) of $y = x^3$ has no bend point since $x^3$ increases when +$x$~increases. Nevertheless, the derivative $3x^2$ of~$x^3$ is zero for the real +value $x = 0$. The tangent to the curve at $(0, 0)$ is the horizontal line +$y=0$. It may be thought of as the limiting position of a secant through~$O$ +which meets the curve in two further points, seen to be equidistant +from~$O$. When one, and hence also the other, of the latter points approaches~$O$, +the secant approaches the position of tangency. In this sense the +tangent at~$O$ is said to meet the curve in three coincident points, their +abscissas being the three coinciding roots of $x^3 = 0$. In the language of~§17, +$x^3 = 0$ has the triple root $x = 0$. The subject of bend points, to which +we recur in~§59, has thus led us to a digression on the important subject +of multiple roots. +\index{Derivative|)}% + + +\Section{58.}{Multiple Roots} In~\Eq{8} replace $x$ by~$\alpha$, and $h$ by $x-\alpha$. Then +\index{Multiple roots}% +\begin{align*} +f(x) &= f(\alpha) + f'(\alpha) (x-\alpha) + + f''(\alpha) \frac{(x-\alpha)^2}{1·2} + + f'''(\alpha) \frac{(x-\alpha)^3}{1·2·3} + \dotsb +\Tag{9} \\ +&\phantom{{} = f(\alpha)} + {} + f^{(m-1)}(\alpha) \frac{(x-\alpha)^{m-1}}{(m-1)!} + + f^{(m)}(\alpha) \frac{(x-\alpha)^m}{m!} + \dotsb. +\end{align*} +By definition~(§17) $\alpha$ is a root of $f(x)=0$ of multiplicity~$m$ if $f(x)$ is exactly +%% -----File: 067.png---Folio 61------- +divisible by $(x-\alpha)^m$, but not by $(x-\alpha)^{m+1}$. Hence \emph{$\alpha$~is a root of multiplicity~$m$ +of $f(x) = 0$ if and only if} +\index{Multiplicity of root}% +\[ +f(\alpha) = 0,\quad +f'(\alpha) = 0,\quad +f''(\alpha) = 0, \dotsc,\quad +f^{(m-1)}(\alpha) = 0,\quad +f^{(m)}(\alpha) \ne 0. +\Tag{10} +\] + +\begin{Remark} +For example, $x^4 + 2x^3 =0$ has the triple root $x = 0$ since $0$ is a root, and since the +first and second derivatives $4x^3 +6x^2$ and $12x^2 +12x$ are zero for $x = 0$, while the third +derivative $24x + 12$ is not zero for $x = 0$. +\end{Remark} + +If in~\Eq{9} we replace $f$ by~$f'$ and hence $f^{(k)}$ by~$f^{(k+1)}$, or if we differentiate +every term with respect to~$x$, we see by either method that +\begin{multline*} +f'(x) = f'(\alpha) + f''(\alpha)(x-\alpha) + \dotsb + + f^{(m-1)} (\alpha)\frac{(x-\alpha)^{m-2}}{(m-2)!} \\ + + f^{(m)}(\alpha)\frac{(x-\alpha)^{m-1}}{(m-1)!} + \dotsb. +\Tag{11} +\end{multline*} + +Let $f(x)$ and $f'(x)$ have the common factor $(x-\alpha)^{m-1}$, but not the common +factor $(x-\alpha)^m$, where $m>1$. Since~\Eq{11} has the factor $(x-\alpha)^{m-1}$, we +have $f'(\alpha) = 0, \dotsc, f^{(m-1)}(\alpha) = 0$. Since also $f(x)$ has the factor $x-\alpha$, +evidently $f(\alpha)=0$. Then, by~\Eq{9}, $f(x)$ has the factor $(x-\alpha)^m$, which, +by hypothesis, is not also a factor of $f'(x)$. Hence, in~\Eq{11}, $f^{(m)}(\alpha)\ne 0$. +Thus, by~\Eq{10}, $\alpha$ is a root of $f(x)=0$ of multiplicity~$m$. + +Conversely, let $\alpha$ be a root of $f(x)=0$ of multiplicity~$m$. Then relations~\Eq{10} +hold, and hence, by~\Eq{11}, $f'(x)$ is divisible by $(x-\alpha)^{m-1}$, but +not by $(x-\alpha)^m$. Thus $f(x)$ and $f'(x)$ have the common factor $(x-\alpha)^{m-1}$, +but not the common factor $(x-\alpha)^m$. + +We have now proved the following useful result. + +\begin{Theorem} +If $f(x)$ and $f'(x)$ have a greatest common divisor $g(x)$ +\index{Greatest common divisor}% +involving~$x$, a root of $g(x)=0$ of multiplicity $m-1$ is a root of $f(x)=0$ of +multiplicity~$m$, and conversely any root of $f(x) =0$ of multiplicity~$m$ is a root +of $g(x)=0$ of multiplicity $m-1$. +\end{Theorem} + +In view of this theorem, the problem of finding all the multiple roots +of $f(x)=0$ and the multiplicity of each multiple root is reduced to the +problem of finding the roots of $g(x)=0$ and the multiplicity of each. + +\begin{Remark} +For example, let $f(x) = x^3 - 2x^2 - 4x + 8$. Then +\[ +f'(x) = 3x^2 - 4x - 4,\qquad +9f(x) = f'(x)(3x-2) - 32(x-2). +\] +Since $x - 2$ is a factor of $f'(x)$, it may be taken to be the greatest common divisor of~$f(x)$ +and~$f'(x)$, the choice of the constant factor~$c$ in $c(x-2)$ being here immaterial. Hence~$2$ +is a double root of $f(x)=0$, while the remaining root~$-2$ is a simple root. +\end{Remark} + +%% -----File: 068.png---Folio 62------- + +\begin{Exercises}{Page62} + +\begin{Problems} +\item[1.] Prove that $x^3 - 7x^2 + 15x - 9 = 0$ has a double root. + +\item[2.] Show that $x^4 - 8x^2 + 16 = 0$ has two double roots. + +\item[3.] Prove that $x^4 - 6x^2 - 8x - 3 = 0$ has a triple root. + +\item[4.] Test $x^4 - 8x^3 + 22x^2 - 24x + 9 = 0$ for multiple roots. + +\item[5.] Test $x^3 - 6x^2 + 11x - 6 = 0$ for multiple roots. + +\item[6.] Test $x^4 - 9x^3 + 9x^2 + 81x - 162 = 0$ for multiple roots. +\end{Problems} +\end{Exercises} + + +\Section{59.}{Ordinary and Inflexion Tangents} The equation of the straight +\index{Inflexion|(}% +\index{Tangents}% +line through the point $(\alpha, \beta)$ with the slope~$s$ is $y-\beta = s(x -\alpha)$. The slope +of the tangent to the graph of $y=f(x)$ at the point $(\alpha, \beta)$ on it is $s=f'(\alpha)$ +by~§56. Also, $\beta=f(\alpha)$. Hence the equation of the tangent is +\[ +y = f(\alpha) + f'(\alpha)(x-\alpha). +\Tag{12} +\] + +By subtracting the members of this equation from the corresponding +members of equation~\Eq{9}, we see that the abscissas~$x$ of the points of intersection +of the graph of $y=f(x)$ with its tangent satisfy the equation +\begin{multline*} +f''(\alpha)\frac{(x-\alpha)^2}{2!} + + f'''(\alpha)\frac{(x-\alpha)^3}{3!} + \dotsb + + f^{(m-1)}(\alpha)\frac{(x-\alpha)^{m-1}}{(m-1)!} \\ + + f^{(m)}(\alpha)\frac{(x-\alpha)^m}{m!} + \dotsb = 0. +\end{multline*} +Here the term containing $f^{(m-1)}(\alpha)$ must evidently be suppressed if $m = 2$, +since the term containing $f^{(m)}(\alpha)$ then coincides with the first term. + +If $\alpha$ is a root of multiplicity~$m$ of this equation, i.e., if the left member +is divisible by $(x -\alpha)^m$, but not by $(x-\alpha)^{m+1}$, the point $(\alpha, \beta)$ is counted +as $m$~coincident points of intersection of the curve with its tangent (just +as in the case of $y=x^3$ and its tangent $y = 0$ in~§57). This will be the case +if and only if +\[ +f''(\alpha)=0,\qquad +f'''(\alpha) =0, \dotsc,\qquad +f^{(m-1)}(\alpha)=0,\qquad +f^{(m)}(\alpha)\ne 0, +\Tag{13} +\] +in which $m>1$ and, as explained above, only the final relation $f''(\alpha)\ne 0$ +is retained if $m= 2$. If $m=3$, the conditions are $f''(\alpha)= 0$, $f^{(3)}(\alpha) \ne 0$. + +\begin{Remark} +For example, if $f(x) = x^4$ and $\alpha = 0$, then $f''(0) =f'''(0) =0$, $f^{(4)} (0) = 24\ne 0$, so that +$m=4$. The graph of $y = x^4$ is a \Shape{U}-shaped curve, whose intersection with the tangent +(the $x$-axis) at $(0, 0)$ is counted as four coincident points of intersection. +\end{Remark} + +Given $f(x)$ and $\alpha$, we can find, as in the preceding example, the value +of~$m$ for which relations~\Eq{13} hold. We then apply the +%% -----File: 069.png---Folio 63------- +\begin{Theorem} +If $m$ is even \($m>0$\), the points of the curve in the vicinity +of the point of tangency $(\alpha, \beta)$ are all on the same side of the tangent, which +is then called an \emph{\textbf{ordinary tangent}}. But if $m$ is odd \($m>1$\), the curve crosses +the tangent at the point of tangency $(\alpha, \beta)$, and this point is called an \emph{\textbf{inflexion +point}}, while the tangent is called an \emph{\textbf{inflexion tangent}}. +\end{Theorem} + +\begin{Remark} +For example, in Fig.~15, $OX$ is an inflexion tangent, while the tangent at any point +except~$O$ is an ordinary tangent. In Figs.~18, 19, 20, the tangents at the points marked +by crosses are ordinary tangents, but the tangent at the point midway between them +and on the $y$-axis is an inflexion tangent. +\end{Remark} + +To simplify the proof, we first take as new axes lines parallel to the +old axes and intersecting at $(\alpha, \beta)$. In other words, we set $x-\alpha = X$, +$y-\beta=Y$, where $X$, $Y$ are the coordinates of $(x, y)$ referred to the new +axes. Since $\beta = f(\alpha)$, the tangent~\Eq{12} becomes $Y = f'(\alpha)X$, while, by~\Eq{9}, +$y = f(x) = \beta + f'(\alpha)(x-\alpha) + \dotsb$ becomes +\[ +Y = f'(\alpha)X + f''(\alpha)\frac{X^2}{2} + \dotsb + = f'(\alpha)X + f^{(m)}(\alpha)\frac{X^m}{m!} + \dotsb, +\] +after omitting terms which are zero by~\Eq{13}. + +%Illustration: Fig. 17 +\begin{wrapfigure}[15]{r}{2.375in} +\hfil\Input{069a} +\end{wrapfigure} +To simplify further the algebraic work, +we pass to oblique axes,\footnote + {Since the earlier $x$, $y$ do not occur in~\Eq{14} and the new equation of the tangent, + we shall designate the final coordinates by $x$, $y$ without confusion.} +the new $y$-axis +coinciding with the $Y$-axis, while the new +$x$-axis is the tangent, the angle between +which and the $X$-axis is designated by $\theta$. +Then +\[ +\tan\theta=f'(\alpha). +\] +By Fig.~17, +\[ +X = x\cos\theta,\qquad +Y-y = f'(\alpha)X. +\] + +Hence when expressed in terms of the +new coordinates $x$, $y$, the tangent is $y = 0$, while the equation~\Eq{14} of the +curve becomes +\[ +y = cx^m + dx^{m+1} + \dotsb,\qquad +c = \frac{f^{(m)}(\alpha)\cos^m \theta}{m!} \ne 0. +\] + +For $x$ sufficiently small numerically, whether positive or negative, +the sum of the terms after $cx^m$ is insignificant in comparison with $cx^{m}$, +%% -----File: 070.png---Folio 64------- +so that $y$ has the same sign as $cx^m$~(§64). Hence, if $m$ is even, the points +of the curve in the vicinity of the origin and on both sides of it are all +on the same side of the $x$-axis, i.e., the tangent. But, if $m$ is odd, the points +with small positive abscissas~$x$ lie on one side of the $x$-axis and those with +numerically small negative abscissas lie on the opposite side. + +Our transformations of coordinates changed the equations of the +curve and of its tangent, but did not change the curve itself and its tangent. +Hence our theorem is proved. + +By our theorem, $\alpha$ is the abscissa of an inflexion point of the graph +of $y=f(x)$ if and only if conditions~\Eq{13} hold with $m$~odd ($m>1$). These +conditions include neither $f(\alpha) = 0$ nor $f'(\alpha)=0$, in contrast with~\Eq{10}. +In the theory of equations we are primarily interested in the abscissas +$\alpha$ of only those points of inflexion whose inflexion tangents are horizontal, +and are interested in them, because we must exclude such roots $\alpha$ of +$f'(x)=0$ when seeking the abscissas of bend points, which are the important +points for our purposes. A point on the graph at which the tangent is +both horizontal and an ordinary tangent is a bend point by the definition +in~§55. Hence if we apply our theorem to the special case $f'(\alpha)=0$, +we obtain the following + +\begin{Criterion} +Any root $\alpha$ of $f'(x) = 0$ is the abscissa of a bend point +of the graph of $y = f(x)$ or of a point with a horizontal inflexion tangent according +as the value of~$m$ for which relations~\Eq{13} hold is even or odd. +\index{Bend point}% +\end{Criterion} + +\begin{Remark} +For example, if $f(x) = x^4$, then $\alpha = 0$ and $m = 4$, so that $(0, 0)$ is a bend point of the +\Shape{U}-shaped graph of $y=x^4$. If $f(x)=x^3$, then $\alpha = 0$ and $m = 3$, so that $(0, 0)$ is a point +with a horizontal inflexion tangent $(OX$ in Fig.~15) of the graph of $y = x^3$. +\end{Remark} + + +\begin{Exercises}{Page64} + +\begin{Problems} + +\item[1.] If $f(x) = 3x^5 + 5x^3 + 4$, the only real root of $f'(x)=0$ is $x = 0$. Show that $(0, 4)$ +is an inflexion point, and thus that there is no bend point and hence that $f(x)=0$ has a +single real root. + +\item[2.] Prove that $x^3 - 3x^2 + 3x + c = 0$ has an inflexion point, but no bend point. + +\item[3.] Show that $x^5 - 10x^3 - 20x^2 - 15x + c = 0$ has two bend points and no horizontal +inflexion tangents. + +\item[4.] Prove that $3x^5 - 40x^3 + 240x + c = 0$ has no bend point, but has two horizontal +inflexion tangents. + +\item[5.] Prove that any function $x^3 - 3\alpha x^2 + \dotsb$ of the third degree can be written in +\index{Cubic equation!reduced}% [** PP: Not using page range] +the form $f(x) = (x-\alpha)^3 + ax + b$. The straight line having the equation $y = ax+b$ meets +the graph of $y=f(x)$ in three coincident points with the abscissa $\alpha$ and hence is an +inflexion tangent. If we take new axes of coordinates parallel to the old and intersecting +at the new origin $(\alpha, 0)$, i.e., if we make the transformation $x = X+\alpha$, $y = Y$, +%% -----File: 071.png---Folio 65------- +of coordinates, we see that the equation $f(x)=0$ becomes a reduced cubic equation +$X^3 + pX + q = 0$~(§42). + +\item[6.] Find the inflexion tangent to $y = x^3 + 6x^2 - 3x + 1$ and transform +$x^3 + 6x^2 - 3x + 1 = 0$ into a reduced cubic equation. +\end{Problems} +\end{Exercises} +\index{Inflexion|)}% [** PP: Original page range is 62--64] + + +\Section[Real Roots of a Cubic Equation] +{60.}{Real Roots of a Real Cubic Equation.} It suffices to consider +\begin{flalign*} +&& f(x) &= x^3 - 3lx + q && \Rightmark{(l \neq 0),} +\end{flalign*} +in view of Ex.~5 above. Then $f' = 3 (x^2 - l)$, $f'' = 6x$. If $l<0$, there is +no bend point and the cubic equation $f(x)=0$ has a single real root. +If $l>0$, there are two bend points +\[ +( \sqrt{l}, q - 2l\sqrt{l}),\qquad +(-\sqrt{l}, q + 2l\sqrt{l}), +\] +which are shown by crosses in Figs.~18--20 for the graph of $y=f(x)$ in the +\index{Cubic equation!graph of}% +\index{Cubic equation!number of real roots}% [** PP: Added ``of''] +three possible cases specified by the inequalities shown below the figures. +For a large positive~$x$, the term~$x^3$ in~$f(x)$ predominates, so that the graph +contains a point high up in the first quadrant, +thence extends downward to the +right-hand bend point, then ascends to +the left-hand bend point, and finally descends. +As a check, the graph contains +a point far down in the third quadrant, +since for $x$ negative, but sufficiently large +numerically, the term $x^3$ predominates and the sign of~$y$ is negative. +% [** Illustrations] +% \caption{$q \geq 2l\sqrt{l}$} FIG. 18 +% \caption{$q \leq -2l\sqrt{l}$} FIG. 19 +\begin{figure*}[htb] +\begin{center} +\Input{071a}\hfill +\Input{071b} +\end{center} +\end{figure*} + +% [** Illustration] +% \caption{$-2l\sqrt{l} < q < 2l\sqrt{l}$} Fig. 20 +\begin{wrapfigure}[9]{r}{2.6in} +\hfill\raisebox{\baselineskip}{\Input{071c}} +\end{wrapfigure} +If the equality sign holds in Fig.~18 or Fig.~19, a necessary and sufficient +condition for which is $q^2 = 4l^3$, one of the bend points is on the $x$-axis, and +the cubic equation has a double root. The inequalities in Fig.~20 hold +if and only if $q^2 < 4l^3$, which implies that $l>0$. Hence \emph{$x^3 - 3lx + q = 0$ +has three distinct real roots if and only if $q^2 < 4l^3$, a single real root if and +only if $q^2 > 4l^3$, a double root \(necessarily real\) if and only if $q^2 = 4l^3$ and $l\neq 0$, +and a triple root if $q^2 = 4l^3 = 0$}. +\index{Discriminant!of cubic}% + +%% -----File: 072.png---Folio 66------- + + +\begin{Exercises}{Page66} + +Find the bend points, sketch the graph, and find the number of real roots of +\begin{Problems}[2] +\item[1.] $x^3 + 2x - 4 = 0$. + +\item[2.] $x^3 - 7x + 7 = 0$. + +\item[3.] $x^3 - 2x - 1 = 0$. + +\item[4.] $x^3 + 6x^2 - 3x + 1 = 0$. +\end{Problems} +\begin{Problems} +\item[5.] Prove that the inflexion point of $y = x^3 - 3lx + q$ is $(0, q)$. + +\item[6.] Show that the theorem in the text is equivalent to that in~§45. + +\item[7.] Prove that, if $m$ and~$n$ are positive odd integers and $m>n$, $x^m + px^n + q = 0$ has +no bend point and hence has a single real root if $p>0$; but, if $p<0$, it has just two +bend points which are on the same side or opposite sides of the $x$-axis according as +\[ +\left(\frac{np}{m}\right)^m + \left(\frac{nq}{m-n}\right)^{m-n} +\] +is positive or negative, so that the number of real roots is $1$ or~$3$ in the respective cases. + +\item[8.] Draw the graph of $y = x^4 - x^2$. By finding its intersections with the line $y = mx + b$, solve $x^4 - x^2 - mx - b= 0$. + +\item[9.] Prove that, if $p$ and~$q$ are positive, $x^{2m} - px^{2n} + q = 0$ has four distinct real roots, +two pairs of equal roots, or no real root, according as +\[ +\left(\frac{np}{m}\right)^m - \left(\frac{nq}{m-n}\right)^{m-n} > 0, +\quad\text{${} = 0$,\quad or\quad ${} < 0$}. +\] + +\item[10.] Prove that no straight line crosses the graph of $y = f(x)$ in more than $n$~points if +the degree~$n$ of the real polynomial $f(x)$ exceeds unity. [Apply~§16.] This fact serves as a check on the accuracy of a graph. +\end{Problems} +\end{Exercises} + + +\Section[Continuity] +{61.}{Definition of Continuity of a Polynomial.} Hitherto we have +located certain points of the graph of $y=f(x)$, where $f(x)$ is a polynomial +in~$x$ with real coefficients, and taken the liberty to join them by a continuous +curve. +\index{Continuity}% + +A polynomial $f(x)$ with real coefficients shall be called \emph{continuous at} +$x = a$, where $a$ is a real constant, if the difference +\[ +D = f(a+h) - f(a) +\] +is numerically less than any assigned positive number~$p$ for all real values +of~$h$ sufficiently small numerically. + + +% [** PP: No ToC entry] +\Section[Continuity of Polynomials] +{62.}{Any Polynomial $f(x)$ with real Coefficients is continuous at $x = a$, +where $a$ is any real Constant.} Taylor's formula~\Eq{8} gives +\index{Polynomial}% +\[ +D = f'(a) h + \frac{f''(a)}{1·2} h^2 + \dotsb + + \frac{f^{(n)}(a)}{1·2\dotsm n} h^n. +\] +This polynomial is a special case of +\[ +F = a_1 h + a_2 h^2 + \dotsb + a_n h^n. +\] +%% -----File: 073.png---Folio 67------- +We shall prove that, \emph{if $a_1, \dotsc, a_n$ are all real, $F$ is numerically less than +any assigned positive number~$p$ for all real values of~$h$ sufficiently small +numerically}. Denote by~$g$ the greatest numerical value of $a_1, \dotsc, a_n$. +If $h$ is numerically less than~$k$, where $k<1$, we see that $F$ is numerically less than +\[ +g(k + k^2 + \dotsb + k^n) < g\frac{k}{1-k} < p,\qquad +\text{if } k < \frac{p}{p + g}. +\] +Hence a real polynomial $f(x)$ is continuous at every real value of~$x$. But +the function $\tan x$ is not continuous at $x=90°$~(§63). + + +\Section[Condition for a Root Between $a$ and~$b$] +{63.}{Root between $a$ and~$b$ if $f(a)$ and~$f(b)$ have opposite Signs.} +\begin{Thm} +If +the coefficients of a polynomial $f(x)$ are real and if $a$ and~$b$ are real numbers +such that $f(a)$ and~$f(b)$ have opposite signs, the equation $f(x) = 0$ has at least +one real root between $a$ and~$b$; in fact, an odd number of such roots, if an +$m$-fold root is counted as $m$~roots. +\end{Thm} + +%[Illustration: \textsc{Fig}. 21] [** PP: Move to top of paragraph] +\begin{wrapfigure}[15]{r}{1.3in} +\hfill\Input{073a} +\end{wrapfigure} +The only argument\footnote + {An arithmetical proof based upon a refined theory of irrational numbers is given + in Weber's \textit{Lehrbuch der Algebra}, ed.~2, vol.~1, p.~123.} +given here (other than that in Ex.~5 below) is +one based upon geometrical intuition. We are stating that, if the points +\[ +\bigl(a, f(a)\bigr),\qquad +\bigl(b, f(b)\bigr) +\] +lie on opposite sides of the $x$-axis, the graph of $y=f(x)$ crosses the $x$-axis +once, or an odd number of times, between the vertical lines through +these two points. Indeed, the part of the graph between +these verticals is a continuous curve having one +and only one point on each intermediate vertical line, +since the function has a single value for each value +of~$x$. + +This would not follow for the graph of $y^2 = x$, which +is a parabola with the $x$-axis as its axis. It may not +cross the $x$-axis between the two initial vertical lines, +but cross at a point to the left of each. + +A like theorem does not hold for $f(x) = \tan x$, when +$x$ is measured in radians and $0 < a < \pi/2 < b < \pi$, since +$\tan x$ is not continuous at $x=\pi/2$. When $t$ increases +from $a$ to $\pi/2$, $\tan x$ increases without limit. When +$x$ decreases from $b$ to~$\pi/2$, $\tan x$ decreases without +limit. There is no root between $a$ and~$b$ of $\tan x = 0$. + +%% -----File: 074.png---Folio 68------- + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Prove that $8x^3 - 4x^2 - 18x + 9 = 0$ has a root between $0$ and~$1$, one between $1$ and~$2$, +and one between $-2$ and~$-1$. + +\item[2.] Prove that $16x^4 - 24x^2 + 16x - 3 = 0$ has a triple root between $0$ and~$1$, and a +simple root between $-2$ and~$-1$. + +\item[3.] Prove that if $a < b < c \dotsb < l$, and $\alpha$, $\beta, \dotsc, \lambda$ are positive, these quantities +being all real, +\[ +\frac{\alpha}{x-a} + +\frac{\beta}{x-b} + +\frac{\gamma}{x-c} + \dotsb + +\frac{\lambda}{x-l} + t = 0 +\] +has a real root between $a$ and~$b$, one between $b$ and~$c, \dotsc$ one between $k$ and~$l$, and +if $t$ is negative one greater than~$l$, but if $t$ is positive one less than~$a$. + +\item[4.] Verify that the equation in Ex.~3 has no imaginary root by substituting $r+si$ +and $r-si$ in turn for~$x$, and subtracting the results. + +\item[5.] Admitting that an equation $f(x) \equiv x^n + \dotsb = 0$ with real coefficients has $n$~roots, +show algebraically that there is a real root between $a$ and~$b$ if $f(a)$ and~$f(b)$ have opposite +signs. Note that a pair of conjugate imaginary roots $c ± di$ are the roots of +\[ +(x-c)^2 + d^2 = 0 +\] +and that this quadratic function is positive if $x$ is real. Hence if $x_1, \dotsc, x_r$ are the +real roots and +\[ +\phi(x) \equiv (x-x_1) \dotsm (x-x_r), +\] +then $\phi(a)$ and~$\phi(b)$ have opposite signs. Thus $a-x_i$ and $b-x_i$ have opposite signs for +at least one real root~$x_i$. (Lagrange.) +\end{Problems} +\end{Exercises} + + +\Section[Sign of a Polynomial at Infinity] +{64.}{Sign of a Polynomial.} Given a polynomial +\index{Polynomial!sign of}% +\index{Sign of polynomial}% +\begin{flalign*} +&& f(x) &= a_0 x^n + a_1 x^{n-1} + \dotsb + a_n && \Rightmark{(a_0 \neq 0)} +\end{flalign*} +with real coefficients, we can find a positive number~$P$ such that $f(x)$ has +the same sign as $a_0 x^n$ when $x>P$. In fact, +\[ +f(x) = x^n (a_0 + \phi),\qquad +\phi = \frac{a_1}{x} + \frac{a_2}{x^2} + \dotsb + \frac{a_n}{x^n}. +\] +By the result in~§62, the numerical value of $\phi$ is less than that of~$a_0$ +when $1/x$ is positive and less than a sufficiently small positive number, +say $1/P$, and hence when $x>P$. Then $a_0 + \phi$ has the same sign as~$a_0$, +and hence $f(x)$ the same sign as~$a_0 x^n$. + +The last result holds also when $x$ is a negative number sufficiently large +numerically. For, if we set $x=-X$, the former case shows that $f(-X)$ +has the same sign as $(-1)^n a_0 X^n$ when $X$ is a sufficiently large positive +number. + +%% -----File: 075.png---Folio 69------- + +We shall therefore say briefly that, for $x = +\infty$, $f(x)$ has the same +sign as~$a_0$; while, for $x = -\infty$, $f(x)$ has the same sign as $a_0$ if $n$ is even, +but the sign opposite to $a_0$ if $n$ is odd. + + +\begin{Exercises}{} +\index{Number!of roots}% + +\begin{Problems} +\item[1.] Prove that $x^3 + ax^2 + bx - 4 = 0$ has a positive real root [use $x=0$ and $x=+\infty$]. + +\item[2.] Prove that $x^3 + ax^2 + bx + 4 = 0$ has a negative real root [use $x=0$ and $x=-\infty$]. + +\item[3.] Prove that if $a_0 > 0$ and $n$ is odd, $a_0x^n + \dotsb + a_n = 0$ has a real root of sign opposite +to the sign of $a_n$ [use $x = -\infty$, $0$, $+\infty$]. + +\item[4.] Prove that $x^4 + ax^3 + bx^2 + cx - 4 = 0$ has a positive and a negative root. + +\item[5.] Show that any equation of even degree $n$ in which the coefficient of $x^n$ and the +constant term are of opposite signs has a positive and a negative root. +\end{Problems} +\end{Exercises} + + +\Section{65.}{Rolle's Theorem} +\index{Rolle's theorem}% +\index{Root@{Root between $a$ and~$b$}}% +\begin{Thm} +Between two consecutive real roots $a$ and~$b$ of $f(x)=0$, +there is an odd number of real roots of $f'(x) = 0$, a root of multiplicity~$m$ +being counted as $m$~roots. +\end{Thm} + +Let +\begin{flalign*} +&& f(x) &\equiv (x-a)^r(x-b)^s Q(x), && \Rightmark{a<b,} +\end{flalign*} +where $Q(x)$ is a polynomial divisible by neither $x-a$ nor $x-b$. Then +by the rule for the derivative of a product (§56, Ex.~10), +\index{Derivative}% +\[ +\frac{(x-a)(x-b)f'(x)}{f(x)} + \equiv r(x-b) + s(x-a) + (x-a)(x-b) \frac{Q'(x)}{Q(x)}. +\] +The second member has the value $r(a-b) < 0$ for $x = a$ and the value +$s(b-a) > 0$ for $x=b$, and hence vanishes an odd number of times between +$a$ and $b$~(§63). But, in the left member, $(x-a)(x-b)$ and~$f(x)$ remain +of constant sign between $a$ and~$b$, since $f(x) = 0$ has no root between $a$ +and~$b$. Hence $f'(x)$ vanishes an odd number of times. + +\begin{Corollary} +Between two consecutive %[** PP: Typo consective] +real roots $\alpha$ and~$\beta$ of $f'(x) = 0$ +there occurs at most one real root of $f(x) = 0$. +\end{Corollary} + +For, if there were two such real roots $a$ and~$b$ of $f(x) = 0$, the theorem +shows that $f'(x) = 0$ would have a real root between $a$ and~$b$ and hence between +$\alpha$ and~$\beta$, contrary to hypothesis. + +Applying also~§63 we obtain the +\begin{Criterion} +If $\alpha$ and~$\beta$ are consecutive real roots of $f'(x) = 0$, then $f(x) = 0$ +has a single real root between $\alpha$ and~$\beta$ if $f(\alpha)$ and~$f(\beta)$ have opposite signs, +but no root if they have like signs. At most one real root of $f(x) = 0$ is greater +than the greatest real root of $f'(x) = 0$, and at most one real root of $f(x) = 0$ is +less than the least real root of $f'(x) = 0$. +\end{Criterion} + +%% -----File: 076.png---Folio 70------- + +If $f(\alpha) = 0$ for our root $\alpha$ of $f'(x) = 0, \alpha$ is a multiple root of $f(x) = 0$ and it would be removed before the criterion is applied. + +\begin{Example} +For $f(x) = 3x^5 - 25x^3 + 60x - 20$, +\[ +\tfrac{1}{15}f'(x) = x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4). +\] +Hence the roots of $f'(x)=0$ are $± 1,\, ± 2$. Now +{\footnotesize +\[ +f(-\infty) = -\infty,\ +f(-2) = -36,\ +f(-1) = -58,\ +f(1) = 18,\ +f(2) = -4,\ +f(+\infty) = +\infty. +\]}% +Hence there is a single real root in each of the intervals +\[ +(-1, 1),\quad (1, 2),\quad (2, +\infty), +\] +and two imaginary roots. The three real roots are positive. +\end{Example} + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Prove that $x^5 - 5x + 2 = 0$ has 1~negative, 2~positive and 2~imaginary roots. + +\item[2.] Prove that $x^6 + x - 1 = 0$ has 1~negative, 1~positive and 4~imaginary roots. + +\item[3.] Show that $x^5 - 3x^3 + 2x^2 - 5 = 0$ has two imaginary roots, and a real root in each +of the intervals $(-2, -1.5)$, $(-1.5, -1)$, $(1, 2)$. + +\item[4.] Prove that $4x^5 - 3x^4 - 2x^2 + 4x - 10 = 0$ has a single real root. + +\item[5.] Show that, if $f^{(k)}(x) = 0$ has imaginary roots, $f(x) = 0$ has imaginary roots. + +\item[6.] Derive Rolle's theorem from the fact that there is an odd number of bend points +between $a$ and~$b$, the abscissa of each being a root of $f'(x) = 0$ of odd multiplicity, while +the abscissa of an inflexion point with a horizontal tangent is a root of $f'(x) = 0$ of even +multiplicity. +\end{Problems} +\end{Exercises} +\index{Graphs|)}% + +%% -----File: 077.png---Folio 71------- + + +% [** PP: ToC entry reads Isolation of the Real Roots] +\Chapter[Isolation of Real Roots] +{VI}{Isolation of the Real Roots of a Real Equation} + +% [** PP: No ToC entry] +\Section{66.}{Purpose and Methods of Isolating the Real Roots} In the next +chapter we shall explain processes of computing the real roots of a given +real equation to any assigned number of decimal places. Each such +method requires some preliminary information concerning the root to +be computed. For example, it would be sufficient to know that the root +is between $4$ and~$5$, provided there be no other root between the same +limits. But in the contrary case, narrower limits are necessary, such +as $4$ and~$4.3$, with the further fact that only one root is between these new +limits. Then that root is said to be \emph{isolated}. +\index{Isolation of roots}% [** PP: Original has no page range] + +\begin{Remark} +If an equation has a single positive root and a single negative root, the real roots +are isolated, since there is a single root between $-\infty$ and~$0$, and a single one between +$0$ and~$+\infty$. However, for the practical purpose of their computation, we shall need +narrower limits, sufficient to fix the first significant figure of each root, for example +$-40$ and~$-30$, or $20$ and~$30$. +\end{Remark} + +We may isolate the real roots of $f(x)=0$ by means of the graph of +$y=f(x)$. But to obtain a reliable graph, we saw in \ChapRef{V} that we +must employ the bend points, whose abscissas occur among the roots of % [** PP: Added `of'] +$f'(x)=0$. Since the latter equation is of degree $n-1$ when $f(x)=0$ is of +degree~$n$, this method is usually impracticable when $n$ exceeds~$3$. The +method based on Rolle's theorem~(§65) is open to the same objection. + +The most effective method is that due to Sturm~(§68). We shall, +however, begin with Descartes' rule of signs since it is so easily applied. +Unfortunately it rarely tells us the exact number of real roots. + + +\Section{67.}{Descartes' Rule of Signs} Two consecutive terms of a real polynomial +\index{Descartes' rule of signs}% +or equation are said to present a \emph{variation of sign} if their coefficients +have unlike signs. By the variations of sign of a real polynomial or equation +we mean all the variations presented by consecutive terms. +\index{Variation of sign}% + +\begin{Remark} +Thus, in $x^5 - 2x^3 - 4x^2 + 3 = 0$, the first two terms present a variation of sign, and +likewise the last two terms. The number of variations of sign of the equation is two. +\end{Remark} + +%% -----File: 078.png---Folio 72------- + +\begin{Theorem}[Descartes' Rule] +\index{Number!of roots|(}% +The number of positive real roots of an equation +with real coefficients is either equal to the number of its variations of sign +or is less than that number by a positive even integer. A root of multiplicity~$m$ +is here counted as $m$~roots. +\end{Theorem} + +\begin{Remark} +For example, $x^6 - 3x^2 + x + 1 = 0$ has either two or no positive roots, the exact number +not being found. But $3x^3 - x - 1 = 0$ has exactly one positive root, which is a simple +root. +\end{Remark} + +Descartes' rule will be derived in §73 as a corollary to Budan's theorem. +The following elementary proof\footnote + {The proofs given in college algebras are mere verifications of special cases.} +was communicated to the author by +Professor D.~R. Curtiss. + +Consider any real polynomial +\[ +f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_l x^{n-l} +\qquad (a_0 \ne 0,\ a_l \ne 0). +\] + +Let $r$ be a positive real number. By actual multiplication, +\[ +F(x) \equiv (x-r)f(x) + \equiv A_0 x^{n+1} + A_1 x^n + \dotsb + A_{l+1}x^{n-l}, +\] +where +\[ +A_0 = a_0,\quad +A_1 = a_1 - ra_0,\quad +A_2 = a_2 - ra_1, \dotsc, +A_l = a_l - ra_{l-1},\quad +A_{l+1} = -r a_l. +\] +In $f(x)$ let $a_{k_1}$ be the first non-vanishing coefficient of different sign from~$a_0$, +let $a_{k_2}$ be the first non-vanishing coefficient following $a_{k_1}$ and of the +same sign as~$a_0$, etc., the last such term, $a_{k_v}$, being either $a_l$ or of the same +sign as~$a_l$. Evidently $v$ is the number of variations of sign of~$f(x)$. + +\begin{Remark} +For example, if $f(x) \equiv 2x^6 + 3x^5 - 4x^4 - 6x^3 + 7x$, we have $v=2$, $a_{k_1} = a_2 = -4$, $a_{k_2} = a_5 = 7$. +Note that $a_4 = 0$ since $x^2$ is absent. +\end{Remark} + +The numbers $A_0, A_{k_1}, \dotsc, A_{k_v}, A_{l+1}$ are all different from zero and +have the same signs as $a_0, a_{k_1}, \dotsc, a_{k_v}, -a_l$, respectively. This is +obviously true for $A_0 = a_0$ and $A_{l+1}= -ra_l$. Next, $A_{k_i}$ is the sum of the +non-vanishing number $a_{k_i}$ and the number $-ra_{k_i - 1}$, which is either zero +or else of the same sign as $a_{k_i}$ since $a_{k_i - 1}$ is either zero or of opposite sign +to~$a_{k_i}$. Hence the sum~$A_{k_i}$ is not zero and has the same sign as~$a_{k_i}$. + +By hypothesis, each of the numbers $a_0, a_{k_1}, \dotsc, a_{k_v}$ after the first +is of opposite sign to its predecessor, while $-a_l$ is of opposite sign to~$a_{k_v}$. +Hence each term after the first in the sequence $A_0, A_{k_1}, \dotsc, A_{k_v}, A_{l+1}$ +is of opposite sign to its predecessor. Thus these terms present $v+1$ +variations of sign. We conclude that $F(x)$ has at least one more variation +of sign than $f(x)$. But we may go further and prove the following + +%% -----File: 079.png---Folio 73------- + +\begin{Lemma} +The number of variations of sign of~$F(x)$ is equal to that of +$f(x)$ increased by some positive odd integer. +\end{Lemma} + +For, the sequence $A_0, A_1, \dotsc, A_{k_1}$ has an odd number of variations +of sign since its first and last terms are of opposite sign; and similarly +for the $v$~sequences +\[ +\begin{array}{c@{\quad}c@{\,}c@{\,}c} +A_{k_1}, & A_{k_1 + 1}, & \dotsc, & A_{k_2}; \\ +\Dots{4} \\ +A_{k_v}, & A_{k_v + 1}, & \dotsc, & A_{l+1}. +\end{array} +\] +The total number of variations of sign of the entire sequence $A_0, A_1,\dotsc, +A_{l+1}$ is evidently the sum of the numbers of variations of sign for the +$v+1$ partial sequences indicated above, and is thus the sum of $v+1$ positive +odd integers. Since each such odd integer may be expressed as $1$ +plus~$0$ or a positive even integer, the sum mentioned is equal to $v+1$ plus~$0$ +or a positive even integer, i.e., to $v$ plus a positive odd integer. + +To prove Descartes' rule of signs, consider first the case in which $f(x)= 0$ +has no positive real roots, i.e., no real root between $0$ and~$+\infty$. Then +$f(0)$ and $f(\infty$) are of the same sign~(§63), and hence the first and last +coefficients of $f(x)$ are of the same sign.\footnote + {In case $f(x)$ has a factor $x^{n-l}$, we use the polynomial $f(x)/x^{n-l}$ instead of $f(x)$ in + this argument.} +Thus $f(x)$ has either no variations +of sign or an even number of them, as Descartes' rule requires. + +Next, let $f(x)= 0$ have the positive real roots $r_1,\dotsc, r_k$ and no others. +A root of multiplicity~$m$ occurs here $m$~times, so that the $r$'s need not be +distinct. Then +\[ +f(x) \equiv (x - r_1)\dotsm (x - r_k)\phi(x), +\] +where $\phi(x)$ is a polynomial with real coefficients such that $\phi(x)=0$ has +no positive real roots. We saw in the preceding paragraph that $\phi(x)$ +has either no variations of sign or an even number of them. By the +Lemma, the product $(x - r_k)\phi(x)$ has as the number of its variations of +sign the number for $\phi(x)$ increased by a positive odd integer. Similarly +when we introduce each new factor $x - r_i$. Hence the number of variations +of sign of the final product $f(x)$ is equal to that of $\phi(x)$ increased +by $k$~positive odd integers, i.e., by $k$ plus $0$ or a positive even integer. +Since $\phi(x)$ has either no variations of sign or an even number of them, +the number of variations of sign of $f(x)$ is $k$ plus $0$ or a positive even integer, +a result equivalent to our statement of Descartes' rule. + +%% -----File: 080.png---Folio 74------- + +If $-p$ is a negative root of $f(x)= 0$, then $p$ is a positive root of $f(-x)= 0$. +Hence we obtain the + +\begin{Corollary} +\index{Number!of negative roots}% +The number of \emph{negative} roots of $f(x)=0$ is either equal +to the number of variations of sign of $f(-x)$ or is less than that number +by a positive even integer. +\end{Corollary} + +\begin{Remark} +For example, $x^4 + 3x^3 + x - 1 = 0$ has a single negative root, which is a simple root, +since $x^4 - 3x^3 - x - 1 = 0$ has a single positive root. +\end{Remark} + +As indicated in Exs.~10, 11 below, Descartes' rule may be used to isolate +the roots. + + +\begin{Exercises}{Page74} + +Prove by Descartes' rule the statements in Exs.~1--8, 12,~15. +\begin{Problems} + +\item[1.] An equation all of whose coefficients are of like sign has no positive root. Why +is this self-evident? + +\item[2.] There is no negative root of an equation, like $x^5 - 2x^4 - 3x^2 + 7x - 5 = 0$, in which +the coefficients of the odd powers of~$x$ are of like sign, and the coefficients of the even +powers (including the constant term) are of the opposite sign. Verify by taking $x= -p$, +where $p$ is positive. + +\item[3.] $x^3 + a^2 x + b^2 = 0$ has two imaginary roots if $b\ne 0$. + +\item[4.] For $n$~even, $x^n - 1 = 0$ has only two real roots. + +\item[5.] For $n$~odd, $x^n - 1 = 0$ has only one real root. + +\item[6.] For $n$~even, $x^n + 1 = 0$ has no real root; for $n$~odd, only one. + +\item[7.] $x^4 + 12x^2 + 5x - 9 = 0$ has just two imaginary roots. + +\item[8.] $x^4 + a^2 x^2 + b^2 x - c^2 = 0$ ($c\ne 0$) has just two imaginary roots. + +\item[9.] Descartes' rule enables us to find the exact number of positive roots only when +all the coefficients are of like sign or when +\[ +f(x) = x^n + p_1 x^{n-1} + \dotsb + p_{n-s} x^s + - p_{n-s+1} x^{s-1} - \dotsb - p_n = 0, +\] +each $p_i$ being $\geqq 0$. Without using that rule, show that the latter equation has one +and only one positive root~$r$. Hints: There is a positive root~$r$ by~§63 ($a=0$, $b=\infty$). +Denote by~$P(x)$ the quotient of the sum of the positive terms by~$x^s$, and by $-N(x)$ +that of the negative terms. Then $N(x)$ is a sum of powers of~$1/x$ with positive coefficients. +\begin{align*} +\text{If}\quad x>r,\qquad P(x)>P(r),\qquad N(x)<N(r),\qquad f(x)>0; \\ +\text{If}\quad x<r,\qquad P(x)<P(r),\qquad N(x)>N(r),\qquad f(x)<0. +\end{align*} + +\item[10.] Prove that we obtain an upper limit to the number of real roots of $f(x)=0$ +between $a$ and~$b$, if we set +\[ +x = \frac{a+by}{1+y}\qquad +\left(\therefore y=\frac{x-a}{b-x}\right), +\] +multiply by $(1+y)^n$, and apply Descartes' rule to the resulting equation in~$y$. + +%% -----File: 081.png---Folio 75------- + +\item[11.] Show by the method of Ex.~10 that there is a single root between $2$ and~$4$ of +$x^3 + x^2 - 17x + 15 = 0$. Here we have $27y^3 + 3y^2 - 23y - 7 = 0$. + +\item[12.] In the astronomical problem of three bodies occurs the equation +\[ +r^5 + (3 - \mu)r^4 + (3 - 2\mu )r^3 - \mu r^2 - 2\mu r - \mu = 0, +\] +where $0 < \mu < 1$. Why is there a single positive real root? + +\item[13.] Prove that $x^5 + x^3 - x^2 + 2x - 3 = 0$ has four imaginary roots by applying Descartes' +rule to the equation in~$y$ whose roots are the squares of the roots of the former. +Transpose the odd powers, square each new member, and replace $x^2$ by~$y$. + +\item[14.] As in Ex.~13 prove that $x^3 + x^2 + 8x + 6 = 0$ has imaginary roots. + +\item[15.] If a real equation $f(x)=0$ of degree~$n$ has $n$~real roots, the number of positive +roots is exactly equal to the number~$V$ of variations of sign. Hint: consider also +$f(-x)$. + +\item[16.] Show that $x^3 - x^2 + 2x + 1 = 0$ has no positive root. Hint: multiply by $x + 1$. +\end{Problems} +\end{Exercises} + + +\Section{68.}{Sturm's Method} Let $f(x) = 0$ be an equation with real coefficients, +\index{Sturm's functions|(}% +and $f'(x)$ the first derivative of~$f(x)$. The first step of the usual process +of finding the greatest common divisor of $f(x)$ and $f'(x)$, if it exists, consists +\index{Greatest common divisor}% +in dividing~$f$ by~$f'$ until we obtain a remainder $r(x)$, whose degree +is less than that of~$f'$. Then, if $q_1$ is the quotient, we have $f = q_1 f' + r$. +Instead of dividing~$f'$ by~$r$, as in the greatest common divisor process, and +proceeding further in that manner, we write $f_2 = -r$, divide~$f'$ by~$f_2$, and +denote by~$f_3$ the remainder with its sign changed. Thus +\[ +f = q_1 f' - f_2,\qquad +f' = q_2 f_2 - f_3,\qquad +f_2 = q_3 f_3 - f_4,\dotsc. +\] + +The latter equations, in which each remainder is exhibited as the negative +of a polynomial~$f_i$, yield a modified process, just as effective as the +usual process, of finding the greatest common divisor~$G$ of $f(x)$ and~$f'(x)$ if +it exists. + +Suppose that $-f_4$ is the first constant remainder. If $f_4 = 0$, then $f_3 = G$, +since $f_3$ divides~$f_2$ and hence also $f'$ and~$f$ (as shown by using our above +equations in reverse order); while, conversely, any common divisor of +$f$ and~$f'$ divides $f_2$ and hence also~$f_3$. + +But if $f_4$ is a constant $\ne 0$, $f$ and~$f'$ have no common divisor involving~$x$. +This case arises if and only if $f(x) = 0$ has no multiple root~(§58), +and is the only case considered in~§§69--71. + +Before stating Sturm's theorem in general, we shall state it for a +numerical case and illustrate its use. + +%% -----File: 082.png---Folio 76------- + + +\begin{Example} $f(x) =x^3 +4x^2 -7$. Then $f'=3x^2 +8x$, +\begin{alignat*}{3} +f &= (\tfrac{1}{3}x + \tfrac{4}{9})f'- f_2, &&\qquad +f_2 &\equiv{}& \tfrac{32}{9}x + 7, \\ +f' &= (\tfrac{27}{32}x + \tfrac{603}{1024})f_2 - f_3, &&\qquad +f_3 &={}& \tfrac{4221}{1024}. +\end{alignat*} + +For\footnote + {Before going further, check that the preceding relations hold when $x= 1$ by inserting +the computed values of $f$, $f'$, $f_2$ for $x =1$. Experience shows that most students make some +error in finding $f_2, f_3, \dotsc$, so that checking is essential.} +$x = 1$, the signs of $f$, $f'$, $f_2$, $f_3$, are ${}-{}+{}+{}+{}$, showing a single variation of +consecutive signs. For $x = 2$, the signs are ${}+{}+{}+{}+{}$, showing no variation of sign. +Sturm's theorem states that there is a \emph{single} real root between $1$ and~$2$. For $x= -\infty$, +the signs are ${}-{}+{}-{}+{}$, showing 3~variations of sign. The theorem states that there +are $3-1=2$ real roots between $-\infty$ and~$1$. Similarly, +\[ +\begin{array}{c@{\quad}|c|c} + x & \text{Signs} & \text{Variations} \\ +\hline +-1 & {}-{}-{}+{}+{} & 1 \\ +-2 & {}+{}-{}-{}+{} & 2 \\ +-3 & {}+{}+{}-{}+{} & 2 \\ +-4 & {}-{}+{}-{}+{} & 3 +\end{array} +\] +Hence there is a single real root between $-2$ and~$-1$, and a single one between $-4$ +and~$-3$. Each real root has now been \emph{isolated} since we have found two numbers +such that a single real root lies between these two numbers or is equal to one of them. + +Some of the preceding computation was unnecessary. After isolating a root between +$-2$ and~$-1$, we know that the remaining root is isolated between $-\infty$ and~$-2$. But +before we can compute it by Horner's method, we need closer limits for it. For that +purpose it is unnecessary to find the signs of all four functions, but merely the sign +of~$f$~(§63). +\end{Example} + + +\Section{69.}{Sturm's Theorem} +\begin{Thm} +Let $f(x) =0$ be an equation with real coefficients +and without multiple roots. Modify the usual process of seeking the greatest +common divisor of $f(x)$ and its first derivative\footnote + {The notation $f_1$ instead of the usual~$f'$, and similarly $f_0$ instead of~$f$, is used to regularize + the notation of all the~$f$'s, and enables us to write any one of the equations~\Eq{1} + in the single notation~\Eq{3}.} +$f_1(x)$ by exhibiting each +remainder as the negative of a polynomial~$f_i$: +\[ +f = q_1 f_1 - f_2,\ +f_1 = q_2 f_2 - f_3,\ +f_2 = q_3 f_3 - f_4, \dotsc,\ +f_{n-2} = q_{n-1}f_{n-1} - f_n, +\Tag{1} +\] +where\footnote + {If the division process did not yield ultimately a constant remainder $\ne 0$, $f$ and~$f_1$ + would have a common factor involving~$x$, and hence $f(x) =0$ a multiple root.} +$f_n$ is a constant $\ne 0$. If $a$ and~$b$ are real numbers, $a<b$, neither +%% -----File: 083.png---Folio 77------- +a root of $f(x) = 0$, the number of real roots of $f(x) = 0$ between $a$ and~$b$ is equal +to the excess of the number of variations of sign of +\[ +f(x),\quad f_1(x),\quad f_2(x), \dotsc, f_{n-1}(x),\quad f_n +\Tag{2} +\] +for $x = a$ over the number of variations of sign for $x = b$. Terms which vanish +are to be dropped out before counting the variations of sign. +\end{Thm} + +For brevity, let $V_x$ denote the number of variations of sign of the +numbers~\Eq{2} when $x$ is a particular real number not a root of $f(x)= 0$. + +First, if $x_1$ and~$x_2$ are real numbers such that no one of the continuous +functions~\Eq{2} vanishes for a value of~$x$ between $x_1$ and~$x_2$ or for $x = x_1$ or +$x = x_2$, the values of any one of these functions for $x = x_1$ and $x = x_2$ are +both positive or both negative~(§63), and therefore $V_{x_1} = V_{x_2}$. + +Second, let $\rho$ be a root of $f_i(x) = 0$, where $1 \leqq i < n$. Then +\[ +f_{i-1}(x) = q_i f_i(x) - f_{i+1}(x) +\Tag{3} +\] +and the equations~\Eq{1} following this one show that $f_{i-1}(x)$ and~$f_i(x)$ have +no common divisor involving~$x$ (since it would divide the constant~$f_n$). +By hypothesis, $f_i(x)$ has the factor $x-\rho$. Hence $f_{i-1}(x)$ does not have +this factor $x-\rho$. Thus, by~\Eq{3}, +\[ +f_{i-1}(\rho) = -f_{i+1}(\rho) \ne 0. +\] +Hence, if $p$ is a sufficiently small positive number, the values of +\[ +f_{i-1}(x),\quad f_i(x),\quad f_{i+1}(x) +\] +for $x = \rho - p$ show just one variation of sign, since the first and third +values are of opposite sign, and for $x = \rho + p$ show just one variation of +sign, and therefore show no change in the number of variations of sign +for the two values of~$x$. + +It follows from the first and second cases that $V_\alpha = V_\beta$ if $\alpha$ and~$\beta$ are +real numbers for neither of which any one of the functions~\Eq{2} vanishes +and such that no root of $f(x) = 0$ lies between $\alpha$ and~$\beta$. + +Third, let $r$ be a root of $f(x) = 0$. By Taylor's theorem~\Eq{8} of~§56, +\begin{align*} +f(r - p) &= -pf'(r) + \tfrac{1}{2} p^2 f''(r) - \dotsb, \\ +f(r + p) &= \Neg pf'(r) + \tfrac{1}{2} p^2 f''(r) + \dotsb. +\end{align*} +If $p$ is a sufficiently small positive number, each of these polynomials in~$p$ +has the same sign as its first term. For, after removing the factor~$p$, +%% -----File: 084.png---Folio 78------- +we obtain a quotient of the form $a_0 + s$, where $s = a_1 p + a_2 p^2 + \dotsb$ is +numerically less than~$a_0$ for all values of~$p$ sufficiently small~(§62). Hence +if $f'(r)$ is positive, $f(r-p)$ is negative and $f(r+p)$ is positive, so that the +terms $f(x)$, $f_1(x)\equiv f'(x)$ have the signs ${}-{}+{}$ for $x = r-p$ and the signs +${}+{}+{}$ for $x = r+p$. If $f'(r)$ is negative, these signs are ${}+{}-{}$ and~${}-{}-{}$ +respectively. In each case, $f(x)$, $f_1(x)$ show one more variation of sign +for $x = r-p$ than for $x = r+p$. Evidently $p$ may be chosen so small that +no one of the functions $f_1(x), \dotsc, f_n$ vanishes for either $x=r-p$ or $x=r+p$, +and such that $f_1(x)$ does not vanish for a value of~$x$ between $r-p$ and~$r+p$, +so that $f(x) = 0$ has the single real root~$r$ between these limits~(§65). +Hence by the first and second cases, $f_1, \dotsc, f_n$ show the same number +of variations of sign for $x = r-p$ as for $x = r+p$. Thus, for the entire +series of functions~\Eq{2}, we have +\[ +V_{r-p} - V_{r+p} = 1. +\Tag{4} +\] + +The real roots of $f(x)= 0$ within the main interval from $a$ to~$b$ (i.e., the % [** PP: Not italicizing i.e.] +aggregate of numbers between $a$ and~$b$) separate it into intervals. By +the earlier result, $V_x$ has the same value for all numbers in the same +interval. By the present result~\Eq{4}, the value~$V_x$ in any interval exceeds +the value for the next interval by unity. Hence $V_a$ exceeds~$V_b$ by the +number of real roots between $a$ and~$b$. +\index{Interval}% + +\begin{Corollary} +If $a<b$, then $V_a\geqq V_{b}$. +\end{Corollary} + +A violation of this Corollary usually indicates an error in the computation +of Sturm's functions~\Eq{2}. + + +\begin{Exercises}{Page78} + +Isolate by Sturm's theorem the real roots of +\begin{Problems}[2] +\item[1.] $x^3 +2x +20 = 0$. + +\item[2.] $x^3 +x-3 = 0$. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry] +\Section{70.}{Simplifications of Sturm's Functions} In order to avoid fractions, +we may first multiply $f(x)$ by a \emph{positive} constant before dividing it by +by $f_1(x)$, and similarly multiply~$f_1$ by a positive constant before dividing it +by~$f_2$, etc. Moreover, we may remove from any~$f_i$ any factor~$k_i$ which is +either a positive constant or a polynomial in~$x$ positive for\footnote + {Usually we would require that $k_i$ be positive for all values of~$x$, since we usually +wish to employ the limits $-\infty$ and~$+\infty$.} +$a\leqq x \leqq b$, +and use the remaining factor~$F_i$ as the next divisor. + +To prove that Sturm's theorem remains true when these modified +%% -----File: 085.png---Folio 79------- +functions $f$, $F_1, \dotsc, F_m$ are employed in place of functions~\Eq{2}, consider +the equations replacing~\Eq{1}: +\begin{align*} +f_1 = k_1 F_1,\qquad +c_2 f &= q_1 F_1 - k_2 F_2,\qquad c_3 F_1 = q_2 F_2 - k_3 F_3, \\ +c_4 F_2 &= q_3 F_3 - k_4 F_4, \dotsc, +c_m F_{m-2} = q_{m-1} F_{m-1} - k_m F_m, +\end{align*} +in which $c_2, c_3, \dotsc$ are positive constants and $F_m$ is a constant $\ne 0$. A +common divisor (involving~$x$) of~$F_{i-1}$ and~$F_i$ would divide $F_{i-2},\dotsc, +F_2, F_1$, $f$,~$f_1$, whereas $f(x)=0$ has no multiple roots. Hence if $\rho$ is a root +of $F_i(x)=0$, then $F_{i-1}(\rho)\ne 0$ and +\[ +c_{i+1} F_{i-1}(\rho) = -k_{i+1}(\rho) F_{i+1}(\rho),\qquad +c_{i+1}>0,\qquad k_{i+1}(\rho)>0. +\] +Thus $F_{i-1}$ and~$F_{i+1}$ have opposite signs for $x = \rho$. We proceed as in~§69. + + +\begin{Example}[1.] +If $f(x) = x^3 + 6x - 10$, $f_1 = 3(x^2 + 2)$ is always positive. Hence we may +employ $f$ and~$F_1 = 1$. For $x =-\infty$, there is one variation of sign; for $x =+\infty$, no +variation. Hence there is a single real root; it lies between $1$ and~$2$. +\end{Example} + +\begin{Example}[2.] +\index{Cubic equation!number of real roots}% [** PP: Added ``of''] +If $f(x) = 2x^4 - 13x^2 - 10x - 19$, we may take +\[ +f_1 = 4x^3 - 13x - 5. +\] +Then +\[ +2f = xf_1 - f_2,\qquad +f_2 = 13x^2 + 15x + 38 = 13(x + \tfrac{15}{26})^2 + \tfrac{1751}{52}. +\] +Since $f_2$ is always positive, we need go no further (we may take $F_2 = 1$). For $x =-\infty$, +the signs are ${}+{}-{}+{}$; for $x = +\infty$, ${}+{}+{}+{}$. Hence there are two real roots. The +signs for $x = 0$ are ${}-{}-{}+{}$. Hence one real root is positive and the other negative. +\end{Example} + + +\begin{Exercises}{Page79} + +Isolate by Sturm's theorem the real roots of +\begin{Problems}[2] + +\item[1.] $x^3 + 3x^2 - 2x - 5 = 0$. + +\item[2.] $x^4 + 12x^2 + 5x - 9 = 0$. + +\ResetCols{2} + +\item[3.] $x^3 - 7x - 7 = 0$. + +\item[4.] $3x^4 - 6x^2 + 8x - 3 = 0$. + +\ResetCols{1} + +\item[5.] $x^6 + 6x^5 - 30x^2 - 12x - 9 = 0$ [stop with~$f_2$]. + +\item[6.] $x^4 - 8x^3 + 25x^2 - 36x + 8 = 0$. + +\item[7.] For $f = x^3 + px + q$ ($p\ne 0$), show that $f_1 = 3x^2 + p$, $f_2 = -2px - 3q$, +\[ +4p^2 f_1 = (-6px + 9q)f_2 - f_3,\quad +f_3 = -4p^3 - 27q^2, +\] +so that $f_3$ is the discriminant~$\Delta$~(§44). Let $[p]$ denote the sign of~$p$. Then the signs +of $f$, $f_1$, $f_2$, $f_3$ are +\begin{align*} +&{}-{}+{}+ [p]\ [\Delta]\quad \text{for $x = -\infty$}, \\ +&{}+{}+{}- [p]\ [\Delta]\quad \text{for $x = +\infty$}. +\end{align*} +For $\Delta$ negative there is a single real root. For $\Delta$ positive and therefore $p$~negative, +there are three distinct real roots. For $\Delta = 0$, $f_2$~is a divisor of~$f_1$ and~$f$, so that +$x = -3q/(2p)$ is a double root. + +%% -----File: 086.png---Folio 80------- + +\item[8.] Prove that if one of Sturm's functions has $p$~imaginary roots, the initial equation +has at least $p$~imaginary roots. + +\item[9.] State Sturm's theorem so as to include the possibility of~$a$, or~$b$, or both $a$ and~$b$ being roots of $f(x)=0$. +\end{Problems} +\end{Exercises} + + +% [** PP: ToC reads Sturm's Functions for a General Quartic Equation] +\Section{71.}{Sturm's Functions for a Quartic Equation} For the reduced quartic +equation $f(z) =0$, +\index{Quartic equation|(}% +\[ +\left\{ +\begin{aligned} +f &= z^4 + qz^2 + rz + s, \\ +f_1 &= 4z^3 + 2qz + r, \\ +f_2 &= -2qz^2 - 3rz - 4s. +\end{aligned} +\right. +\Tag{5} +\] +Let $q\ne 0$ and divide $q^2 f_1$ by~$f_2$. The negative of the remainder is +\[ +f_3 = Lz - 12rs - rq^2,\qquad +L = 8qs - 2q^3 - 9r^2. +\Tag{6} +\] +Let $L\ne 0$. Then $f_4$ is a constant which is zero if and only if $f=0$ has +multiple roots, i.e., if its discriminant~$\Delta$ is zero. We therefore desire~$f_4$ % [** PP: Not italicizing i.e.] +expressed as a multiple of~$\Delta$. By~§50, +\[ +\Delta = -4P^3 - 27Q^2,\qquad +P = -4s - \frac{q^2}{3},\qquad +Q = \tfrac{8}{3}qs - r^2 - \tfrac{2}{27}q^3. +\Tag{7} +\] +We may employ $P$ and~$Q$ to eliminate +\[ +4s = -P - \frac{q^2}{3},\qquad +r^2 = -Q - \tfrac{2}{3}qP - \tfrac{8}{27}q^3. +\Tag{8} +\] +We divide $L^2 f_2$ by +\[ +f_3 = Lz + 3rP,\qquad +L = 9Q + 4qP. +\Tag{9} +\] +The negative of the remainder\footnote + {Found directly by the Remainder Theorem~(§14) by inserting the root $z = -3rP/L$ + of $f_3=0$ into $L^2 f_2$.} +is +\[ +18r^2 qP^2 - 9r^2 LP + 4sL^2 = q^2 \Delta. +\Tag{10} +\] +The left member is easily reduced to~$q^2\Delta$. Inserting the values~\Eq{8} and +replacing $L^2$ by $L(9Q + 4qP)$, we get +\[ +-18qQP^2 - 12q^2 P^3 - \tfrac{16}{3}q^4 P^2 + 2qP^2 L + + \tfrac{4}{3}q^3 PL - 3q^2 QL. +\] +Replacing $L$ by its value~\Eq{9}, we get~$q^2\Delta$. Hence we may take +\[ +f_4 = \Delta. +\Tag{11} +\] +Hence if $qL\Delta\ne 0$, we may take \Eq{5}, \Eq{9}, \Eq{11} as Sturm's functions. + +%% -----File: 087.png---Folio 81------- + +Denote the sign of~$q$ by~$[q]$. The signs of Sturm's functions are +\[ +\begin{array}{rrrrr@{\quad}l} ++ & - & -[q] & -[L] & [\Delta] & \text{for $x = -\infty$}, \\ ++ & + & -[q] & [L] & [\Delta] & \text{for $x = +\infty$}. +\end{array} +\] + +First, let $\Delta > 0$. If $q$ is negative and $L$ is positive, the signs are +${}+{}-{}+{}-{}+{}$ and~${}+{}+{}+{}+{}+{}$, so that there are four real roots. In each +of the remaining three cases for $q$ and~$L$, there are two variations of sign +in either of the two series and hence there is no real root. + +Next, let $\Delta < 0$. In each of the three cases in which $q$ and~$L$ are not +both positive, there are three variations of sign in the first series and one +variation in the second, and hence just two real roots. If $q$ and~$L$ are +both positive, the number of variations is~$1$ in the first series and~$3$ in the +second, so that this case is excluded by the Corollary to Sturm's theorem. +To give a direct proof, note that, by the value of~$L$ in~\Eq{6}, $L>0$, $q>0$ +imply $4s > q^2$, i.e., $s>0$, and hence, by~\Eq{7}, $P$~is negative, so that each term +of~\Eq{10} is $\geqq 0$, whence $\Delta > 0$. % [** PP: Added space] + +Hence, if $qL\Delta \ne 0$, there are four distinct real roots if and only if $\Delta$ +and~$L$ are positive, and $q$~negative; two distinct real and two imaginary +roots if and only if $\Delta$ is negative. + +Combining this result with that in Ex.~4 below, we obtain the + +\begin{Theorem} +If the discriminant~$\Delta$ of $z^4 + qz^2 + rz + s = 0$ is negative, there +are two distinct real roots and two imaginary roots; if $\Delta > 0$, $q<0$, $L>0$, +four distinct real roots; if $\Delta > 0$ and either $q\geqq 0$ or $L\leqq 0$, no real roots. +Here $L = 8qs - 2q^3 - 9r^2$. +\end{Theorem} +\index{Discriminant!of quartic}% +\index{Quartic equation|)}% + +Our discussion furnished also the series of Sturm functions, which +may be used in isolating the roots. + + +\begin{Exercises}{} + +\begin{Problems} + +\item[1.] If $q\Delta\ne 0$, $L = 0$, then $f_3 = 3rP$ is not zero (there being no multiple root) and its +sign is immaterial in determining the number of real roots. Prove that there are just +two real roots if $q<0$, and none if $q>0$. By~\Eq{10}, $q$ has the same sign as~$\Delta$. + +\item[2.] If $r\Delta\ne 0$, $q = 0$, obtain~$-f_3$ by substituting $z = -4s/(3r)$ in~$f_1$. Show that we may +take $f_3 = r\Delta$ and that there are just two real roots if $\Delta < 0$, and no real +roots if $\Delta >0$. + +\item[3.] If $\Delta \ne 0$, $q = r = 0$, prove that there are just two real roots if $\Delta<0$, and no real +roots if $\Delta > 0$. Since $\Delta = 256s^3$, check by solving $z^4 + s = 0$. + +\item[4.] If $\Delta \ne 0$, $qL = 0$, there are just two real roots if $\Delta < 0$, and no real roots if $\Delta > 0$. +[Combine the results in Exs.~1--3.] + +\item[5.] Apply the theorem to Exs.~2, 4, 6 of~§70. + +\item[6.] Isolate the real roots of Exs.~3, 4, 5 of~§48. +\end{Problems} +\end{Exercises} + +%% -----File: 088.png---Folio 82------- + + +% [** PP: No ToC entry] +\Section[Sturm's Theorem for Multiple Roots] +{72.}{Sturm's Theorem for the Case of Multiple Roots.} We might +\index{Multiple roots}% +remove the multiple roots by dividing $f(x)$ by\footnote + {The degree of~$f(x)$ is not~$n$, + nor was it necessarily~$n$ in~§69.} +$f_n(x)$, the greatest common +divisor of $f(x)$ and $f_1 = f'(x)$; but this would involve considerable +work, besides wasting the valuable information in hand. As before, we +suppose $f(a)$ and $f(b)$ different from zero. We have equations~\Eq{1} in +which $f_n$ is now not a constant. + +\begin{Thm}% +The difference $V_a - V_b$ is the number of real roots between $a$ and~$b$, each +multiple root being counted only once. +\end{Thm} + +If $\rho$ is a root of $f_i(x) = 0$, but not a multiple root of $f(x)= 0$, then $f_{i-1}(\rho)$ +is not zero. For, if it were zero, $x-\rho$ would by~\Eq{1} be a common factor +of $f$ and~$f_1$. We may now proceed as in the second case in~§69. + +The third case requires a modified proof only when $r$ is a multiple root. +Let $r$ be a root of multiplicity~$m$, $m\geqq 2$. Then $f(r)$, $f'(r), \dotsc, f^{(m-1)}(r)$ +are zero and, by Taylor's theorem, +\begin{align*} +f(r+p) &= \frac{p^m}{1·2\dotsm m} f^{(m)}(r) + \dotsb, \\ +f'(r+p) &= \frac{p^{m-1}}{1·2\dotsm (m-1)} f^{(m)}(r) + \dotsb. +\end{align*} +These have like signs if $p$ is a positive number so small that the signs of +the polynomials are those of their first terms. Similarly, $f(r-p)$ and +$f'(r-p)$ have opposite signs. Hence $f$ and~$f_1$ show one more variation +of sign for $x = r-p$ than for $x = r+p$. Now $(x-r)^{m-1}$ is a factor of~$f$ +and~$f_1$ and hence, by~\Eq{1}, of $f_2, \dotsc, f_n$. Let their quotients by this +factor be $\phi, \phi_1, \dotsc, \phi_n$. Then equations~\Eq{1} hold after the $f'$s are replaced +by the~$\phi$'s. Taking $p$ so small that $\phi_1(x) = 0$ has no root between $r-p$ +and $r+p$, we see by the first and second cases in~§69 that $\phi_1, \dotsc, \phi_n$ +show the same number of variations of sign for $x = r-p$ as for $x = r+p$. +The same is true for $f_1, \dotsc, f_n$ since the products of $\phi_1, \dotsc, \phi_n$ by +$(x-r)^{m-1}$ have for a given~$x$ the same signs as $\phi_1, \dotsc, \phi_n $ or the same +signs as $-\phi_1, \dotsc, -\phi_n$. But the latter series evidently shows the +same number of variations of sign as $\phi_1, \dotsc, \phi_n$. Hence~\Eq{4} is proved +and consequently the present theorem. +\index{Sturm's functions|)}% + +%% -----File: 089.png---Folio 83------- + + +\begin{Exercises}{Page83} + +\begin{Problems} + +\item[1.] For $f = x^4 - 8x^2 + 16$, prove that $F_1 = x^3 - 4x$, $F_2 = x^2 - 4$, $F_1 = xF_2$. Hence $n = 2$. +Verify that $V_{-\infty} = 2$, $V_{\infty} = 0$, and that there are just two real roots, each a double +root. +\end{Problems} + +Discuss similarly the following equations. +\begin{Problems}[3] +\item[2.] $x^4 - 5x^3 + 9x^2 - 7x + 2 = 0$. + +\item[3.] $x^4 + 2x^3 - 3x^2 - 4x + 4 = 0$. + +\item[4.] $x^4 - x^2 - 2x + 2 = 0$. +\end{Problems} +\end{Exercises} + + +\Section{73.}{Budan's Theorem} +\index{Budan's theorem}% +\index{Derivative}% +\begin{Thm}% +Let $a$ and~$b$ be real numbers, $a<b$, neither\footnote + {In case $a$ or~$b$ + is a root of $f(x)=0$, the theorem holds if we count the number of + roots $>a$ and $\leqq b$. This inclusive theorem has been proved, by means of Rolle's + theorem, by A.~Hurwitz, \textit{Mathematische Annalen}, Vol.~71, 1912, p.~584, who extended + Budan's theorem from the case of a polynomial to a function~$f(x)$ which is real and + regular for $a \leqq x < b$.} +a root of $f(x) = 0$, an equation of degree~$n$ with real coefficients. Let $V_a$ +denote the number of variations of sign of +\[ +f(x),\qquad f'(x),\qquad f''(x),\qquad \dotsc,\qquad f^{(n)}(x) +\Tag{12} +\] +for $x = a$, after vanishing terms have been deleted. Then $V_a - V_b$ is either +the number of real roots of $f(x) = 0$ between $a$ and~$b$ or exceeds the number +of those roots by a positive even integer. A root of multiplicity~$m$ is here +counted as $m$~roots. +\end{Thm} + +\begin{Remark} +For example, if $f(x) = x^3 - 7x - 7$, then $f' = 3x^2 - 7$, $f'' = 6x$, $f''' = 6$. Their values +for $x = 3$, $4$, $-2$, $-1$ are tabulated below. +\[ +\begin{array}{r|rrrr|c} + x & f & f' & f'' & f''' & \text{Variations} \\ +\hline + 3 & -1 & 20 & 18 & 6 & 1 \\ + 4 & 29 & 41 & 24 & 6 & 0 \\ +-2 & -1 & 5 & -12 & 6 & 3 \\ +-1 & -1 & -4 & -6 & 6 & 1 +\end{array} +\] +Hence the theorem shows that there is a single real root between $3$ and~$4$, and two +or no real roots between $-2$ and~$-1$. The theorem does not tell us the exact number +of roots between the latter limits. To decide this ambiguity, note that $f(-3/2) = +1/8$, +so that there is a single real root between $-2$ and~$-1.5$, and a single one between +$-1.5$ and~$-1$. +\end{Remark} + +The proof is quite simple if no term of the series~\Eq{12} vanishes for +$x = a$ or for $x = b$ and if no two consecutive terms vanish for the same +value of~$x$ between $a$ and~$b$. Indeed, if no one of the terms vanishes for +$x_1 \leqq x\leqq x_2$, then $V_{x_1} = V_{x_2}$, since any term has the same sign for $x = x_1$ +as for $x = x_2$. Next, let $r$ be a root of $f^{(i)}(x)=0$, $a<r<b$. By hypothesis, +%% -----File: 090.png---Folio 84------- +the first derivative $f^{(i+1)}(x)$ of $f^{(i)}(x)$ is not zero for $x=r$. As in the third +step (now actually the case $i = 0$) in~§69, $f^{(i)}(x)$ and $f^{(i+1)}(x)$ show one +more variation of sign for $x = r-p$ than for $x = r+p$, where $p$ is a sufficiently +small positive number. If $i>0$, $f^{(i)}$ is preceded by a term $f^{(i-1)}$ in~\Eq{12}. +By hypothesis, $f^{(i-1)}(x) \ne 0$ for $x=r$ and hence has the same sign for +$x = r - p$ and $x = r + p$ when $p$ is sufficiently small. For these values of~$x$, +$f^{(i)}(x)$ has opposite signs. Hence $f^{(i-1)}$ and $f^{(i)}$ show one more or one less +variation of sign for $x = r-p$ than for $x = r+p$, so that $f^{(i-1)}$, $f^{(i)}$, $f^{(i+1)}$ show +two more variations or the same number of variations of sign. + +Next, let no term of the series~\Eq{12} vanish for $x = a$ or for $x = b$, but +let several successive % [** PP: Typo succssive] +terms +\[ +f^{(i)}(x),\qquad f^{(i+1)}(x), \dotsc, f^{(i+j-1)}(x) +\Tag{13} +\] +all vanish for a value~$r$ of~$x$ between $a$ and~$b$, while $f^{(i+j)}(r)$ is not zero, +but is say positive.\footnote + {If negative, all signs in the table below are to be changed; but the conclusion holds.} +Let $I_1$ be the interval between $r-p$ and~$r$, and $I_2$ +the interval between $r$ and~$r+p$. Let the positive number~$p$ be so small +that no one of the functions~\Eq{13} or $f^{(i+j)}(x)$ is zero in these intervals, so +that the last function remains positive. Hence $f^{(i+j-1)}(x)$ increases with~$x$ +(since its derivative is positive) and is therefore negative in~$I_1$ +and positive in~$I_2$. Thus $f^{(i+j-2)}(x)$ decreases in~$I_1$ and increases in~$I_2$ +and hence is positive in each interval. In this manner we may verify the +signs in the following table: +\[ +\begin{array}{c|cccccccc} +\multicolumn{2}{r}{f^{(i)}} & f^{(i+1)} & f^{(i+2)} & \ldots & + f^{(i+j-3)} & f^{(i+j-2)} & f^{(i+j-1)} & f^{(i+j)} \\ +% +I_1 & (-)^{j} & (-)^{j-1} & (-)^{j-2} & \ldots & + - & + & - & + \\ +I_2 & + & + & + & \ldots & + & + & + & + +\end{array} +\] +Hence these functions show $j$~variations of sign in~$I_1$ and none in~$I_2$. + +If $i>0$, the first term of~\Eq{13} is preceded by a function $f^{(i-1)}(x)$ which +is not zero for $x = r$, and hence not zero in $I_1$ or $I_2$ if $p$ is sufficiently small. +If $j$ is even, the signs of $f^{(i-1)}$ and $f^{(i)}$ are ${}+{}+{}$ or ${}-{}+{}$ in both $I_1$ and~$I_2$, +showing no loss in the number of variations of sign. If $j$ is odd, their +signs are +\[ +\begin{array}{c|ccc} +I_1 & {}+{}-{} & \raisebox{-12pt}{\smash{\text{or}}} & {}-{}-{} \\ +I_2 & {}+{}+{} & & {}-{}+{} +\end{array} +\] +so that there is a loss or gain of a single variation of sign. Hence +\[ +f^{(i-1)}, \qquad +f^{(i)}, \qquad +f^{(i+1)} \quad \dotsc,\qquad +f^{(i+j)} +\] +%% -----File: 091.png---Folio 85------- +show a loss of $j$~variations of sign if $j$ is even, and a loss of $j±1$ if $j$ is odd, +and hence always a loss of an even number $\geqq 0$ of variations of sign. + +If $i=0$, $f^{(i)}\equiv f$ has $r$ as a $j$-fold root and the functions in the table show +$j$~more variations of sign for $x = r-p$ than for $x = r+p$. + +Thus, when no one of the functions~\Eq{12} vanishes for $x=a$ or for $x=b$, +the theorem follows as at the end of~§69 (with unity replaced by the +multiplicity of a root). + +Finally, let one of the functions~\Eq{12}, other than $f(x)$ itself, vanish for +$x= a$ or for $x = b$. If $\delta$ is a sufficiently small positive number, all of the +$N$~roots of $f(x)=0$ between $a$ and~$b$ lie between $a+\delta$ and $b-\delta$, and for +the latter values no one of the functions~\Eq{12} is zero. By the above +proof, +\begin{gather*} +V_{a+\delta} - V_{b-\delta} = N + 2t, \\ +V_a - V_{a + \delta} = 2j,\qquad +V_{b - \delta} - V_b = 2s, +\end{gather*} +where $t$, $j$, $s$ are integers $\geqq 0$. Hence $V_a - V_b = N+2(t+j+s)$. + +Descartes' rule of signs~(§67) is a corollary to Budan's theorem. Consider +any equation with real coefficients +\index{Descartes' rule of signs}% +\[ +f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_{n-1}x + a_n = 0, +\] +having $a_n \ne 0$. For $x = 0$ the functions~\Eq{12} have the same signs as +\[ +a_n,\qquad a_{n-1},\qquad \dotsc, \quad a_1,\qquad a_0. +\] +Hence $V_0$ is equal to the number~$V$ of variations of sign of~$f(x)$. + +For $x= +\infty$, the functions all have the same sign, which is that of~$a_0$. +Thus $V_0 - V_{\infty} = V$ is either the number of positive roots or exceeds that +number by a positive even integer. Finally, Descartes' rule holds if +$a_n = 0$, as shown by removing the factors~$x$. +\index{Number!of roots|)}% + + +\begin{Exercises}{Page85} + +Isolate by Budan's theorem the real roots of +\begin{Problems}[2] + +\item[1.] $x^3 -x^2 -2x+1=0$. + +\item[2.] $x^3 +3x^2 -2x-5=0$. + +\ResetCols{1} + +\item[3.] Prove that if $f{(a)}\ne 0$, $V_a$ equals the number of real roots $>a$ or exceeds that number by an even integer. + +\item[4.] Prove that there is no root greater than a number making each of the functions~\Eq{12} +positive, if the leading coefficient of $f(x)$ is positive. (Newton.) + +\item[5.] Hence verify that $x^4 -4x^3 - 3x + 23 = 0$ has no root~$>4$. + +\item[6.] Show that $x^4 - 4x^3 + x^2 + 6x + 2 = 0$ has no root~$>3$. +\end{Problems} +\end{Exercises} + +%% -----File: 092.png---Folio 86------- + + +\Chapter{VII}{Solution of Numerical Equations} +\index{Solution of numerical equations|(}% + + +\Section[Horner's Method] +{74.}{Horner's Method.\protect\footnotemark}\addtocounter{footnote}{1}% + \footnotetext{W.~G. Horner, London Philosophical Transactions, 1819. Earlier (1804) by + P.~Ruffini. See Bulletin American Math.\ Society, May, 1911.} +\addtocounter{footnote}{-1}% +\index{Horner's method}% +\index{Synthetic division|(}% +After we have isolated a real root of a real +equation by one of the methods in \ChapRef{VI}, we can compute the root +to any desired number of decimal places either by Horner's method, +which is available only for polynomial equations, or by Newton's method~(§75), +which is applicable also to logarithmic, trigonometric, and other +equations. + +To find the root between $2$ and~$3$ of +\[ +x^3 - 2x - 5 = 0, +\Tag{1} +\] +set $x = 2+p$. Direct substitution gives the \emph{transformed equation} for~$p$: +\index{Transformed equation}% +\[ +p^3 + 6p^2 + 10p - 1 = 0. +\Tag{2} +\] +The method just used is laborious especially for equations of high degree. +We next explain a simpler method. Since $p = x -2$, +\[ +x^3 - 2x - 5 \equiv (x-2)^3 + 6(x-2)^2 + 10(x-2) - 1, +\] +identically in~$x$. Hence $-1$ is the remainder obtained when the given +polynomial $x^3 - 2x - 5$ is divided by $x-2$. By inspection, the quotient~$Q$ +is equal to +\[ +(x-2)^2 + 6(x-2) + 10. +\] +Hence $10$ is the remainder obtained when $Q$ is divided by $x-2$. The +new quotient is equal to $(x-2) + 6$, and another division gives the +remainder~$6$. Hence to find the coefficients $6$, $10$, $-1$ of the terms following~$p^3$ +in the transformed equation~\Eq{2}, we have only to divide the given +polynomial $x^3 - 2x - 5$ by $x-2$, the quotient~$Q$ by $x-2$, etc., and take +the remainders in reverse order. However, when this work is performed +by synthetic division~(§15) as tabulated below, no reversal of order is +%% -----File: 093.png---Folio 87------- +necessary, since the coefficients then appear on the page in their desired +order. +\[ +\begin{array}{rRRRPc} +1 & 0 & -2 & -5 && \Lcol{2} \\ +\cline{6-6} + & 2 & 4 & 4 && \\ +\cline{1-4} +1 & 2 & \Rcol{2} & \mathbf{-1} && \\ + & 2 & \Rcol{8} & && \\ +\cline{1-3} +1 & \Rcol{4} & \mathbf{10} & && \\ + & \Rcol{2} & & && \\ +\cline{1-2} +1 & \mathbf{6} & & && +\end{array} +\] + +Thus $1$, $6$, $10$, $-1$ are the coefficients of the desired equation~\Eq{2}. + +To obtain an approximation to the decimal~$p$, we ignore for the moment +the terms involving $p^3$ and~$p^2$; then by $10p - 1 = 0$, $p=0.1$. But this +value is too large since the terms ignored are all positive. For $p=0.09$, +the polynomial in~\Eq{2} is found to be negative, while for $p=0.1$ it was just +seen to be positive. Hence $p = 0.09+h$, where $h$ is of the denomination +thousandths. The coefficients $1$, $6.27, \dotsc$ of the transformed equation +for~$h$ appear in heavy type just under the first zigzag line in the following +scheme: +\[ +\begin{array}{r<{\qquad}l Pr@{}l Pr@{}l r@{}l} +1 & 6 && 10 & && -1 & && \Lcol{0.09} \\ +\cline{10-10} + & 0.09 && 0 & .5481 && 0 & .949329 && \\ +\cline{1-8} +1 & 6.09 && 10 & .5481 &\Lcol{}& \mathbf{-0} & \mathbf{.050671} && \\ + & 0.09 && 0 & .5562 &\Lcol{}& & && \\ +\cline{1-5} +1 & 6.18 &\Lcol{}& \mathbf{11} & \mathbf{.1043} & & &\Rcol{}& & +\smash[b]{\raisebox{-1.5ex}{$\dfrac{0.05}{11.1}$}} \\ + & 0.09 &\Lcol{}& & & & &\Rcol{}& & \\ +\cline{1-2} +1 & \textbf{6.27} && & & & &\Rcol{}& $=\,$&$0.004$\\ +\cline{9-10} + & 0.004 && 0 & .025096 && 0 & .044517584 & & \\ +\cline{1-8} +1 & 6.274 &&11 & .129396 &\Lcol{}& \mathbf{-0} & \mathbf{.006153416} && \\ + & 0.004 && 0 & .025112 &\Lcol{}& & && \\ +\cline{1-5} +1 & 6.278 &\Lcol{}& \mathbf{11} & \mathbf{.154508} && & && \\ + & 0.004 &\Lcol{}& & && & && \\ +\cline{1-2} +1 & \textbf{6.282} && & && & && +\end{array} +\] +Hence $x=2.094+t$, where $t$ is a root of +\[ +t^3 + 6.282t^2 + 11.154508t - 0.006153416 = 0. +\] +By the last two terms, $t$ is between $0.0005$ and~$0.0006$. Then the value +%% -----File: 094.png---Folio 88------- +of $C\equiv t^3 +6.282t^2$ is found to lie between $0.00000157$ and~$0.00000227$. +Hence we may ignore~$C$ provided the constant term be reduced by an +amount between these limits. Whichever of the two limits we use, we +obtain the same dividend below correct to 6~decimal places. +\[ +\begin{array}{r>{\quad}r<{\quad}l@{}l} +\multicolumn{1}{r|}{1\xatop{1}.\xatop{1}\xatop{5}4508} +& 0.006151 & \multicolumn{1}{|l@{}}{0.000551} & =t \\ +\cline{1-1} \cline{3-3} +& 5577 & \\[-8pt] +& \settowidth{\TmpLen}{5577}\rule{\TmpLen}{0.5pt} & \\ +& 574 & \\ +& 558 & \\[-8pt] +& \settowidth{\TmpLen}{558}\rule{\TmpLen}{0.5pt} & \\ +& 16 & \\ +& 11 & \\[-8pt] +& \settowidth{\TmpLen}{11}\rule{\TmpLen}{0.5pt} & \\ +& 5 & +\end{array} +\] +Since the quotient is $0.0005+$, only two decimal places of the divisor are +used, except to see by inspection how much is to be carried when making +the first multiplication. Hence we mark a cross above the figure~$5$ in +the hundredths place of the divisor and use only~$11.15$. Before making +the multiplication by the second significant figure~$5$ of the quotient~$t$, +we mark a cross over the figure~$1$ in the tenths place of the divisor and +hence use only~$11.1$. Thus $x = 2.0945514+$, with doubt only as to whether +the last figure should be $4$ or~$5$. + +If we require a greater number of decimal places, it is not necessary +to go back and construct a new transformed equation from the equation +in~$t$. We have only to revise our preceding dividend on the basis of our +present better value of~$t$. We now know that $t$ is between $0.000551$ and +$0.000552$. To compute the new value of the correction~$C$, in which we +may evidently ignore~$t^3$, we use logarithms. +\[ +% [** PP: Alignment: log integer mantissa = integer mantissa, x 2] +\begin{array}{rr@{}l@{}>{{}}c<{{}}@{}r@{}l@{}>{\qquad}rr@{}l@{}>{{}}c<{{}}@{}r@{}l@{}} +\log & 5&.51 & = & &.74115 & \log & 5&.52 & = && .74194 \\ +\therefore +\log & 5&.51^2 & = & 1&.48230 &\therefore + \log & 5&.52^2 & = &1&.48388 \\ +\log & 6&.282 & = & &.79810 & \log & 6&.282 & = & &.79810 \\ +\cline{5-6} \cline{11-12} +\log & 190&.72 & = & 2&.28040 & \log &191&.42 & = &2&.28198 \Strut +\end{array} +\] +Hence $C$ is between $0.000001907$ and $0.000001915$. Whichever of the two +limits we use, we obtain the same new dividend below correct to 8~decimal +places. +%% -----File: 095.png---Folio 89------- +\[ +\begin{array}{r>{\quad}r<{\quad}l@{}l} +\multicolumn{1}{r|}{1\xatop{1}.\xatop{1}\xatop{5}\xatop{4}\xatop{5}08} +& 0.00615150 & \multicolumn{1}{|l@{}}{0.00055148} \\ +\cline{1-1} \cline{3-3} +& 557725 & \\[-8pt] +& \settowidth{\TmpLen}{557725}\rule{\TmpLen}{0.5pt} & \\ +& 57425 & \\ +& 55773 & \\[-8pt] +& \settowidth{\TmpLen}{55773}\rule{\TmpLen}{0.5pt} & \\ +& 1652 & \\ +& 1115 & \\[-8pt] +& \settowidth{\TmpLen}{1115}\rule{\TmpLen}{0.5pt} & \\ +& 537 & \\ +& 446 & \\[-8pt] +& \settowidth{\TmpLen}{446}\rule{\TmpLen}{0.5pt} & \\ +& 91 & \\ +& 89 & \\[-8pt] +& \settowidth{\TmpLen}{89}\rule{\TmpLen}{0.5pt} & \\ +& 2 & +\end{array} +\] +Hence, finally, $x=2.094551482$, with doubt only as to the last figure. + + +\begin{Exercises}{Page89} + +(The number of transformations made by synthetic division should be about half +the number of significant figures desired for a root.) + +By one of the methods in \ChapRef{VI}, isolate each real root of the following equations, +and compute each real root to 5~decimal places. +\begin{Problems}[2] + +\item[1.] $x^3 +2x+20=0$. + +\item[2.] $x^3 +3x^2 -2x-5=0$. + +\ResetCols{2} + +\item[3.] $x^3 +x^2 -2x-1=0$. + +\item[4.] $x^4 +4x^3 -17.5x^2 -18x+58.5=0$. + +\ResetCols{2} + +\item[5.] $x^4 -11,727x+40,385=0$. + +\item[6.] $x^3 =10$. +\end{Problems} + +Find to 7~decimal places all the real roots of +\begin{Problems}[2] +\item[7.] $x^3 +4x^2 -7=0$. + +\item[8.] $x^3 -7x-7=0$. +\end{Problems} + +Find to 8~decimal places +\begin{Problems} +\item[9.] The root between $2$ and~$3$ of $x^3 -x-9=0$ (make only 3~transformations). + +\item[10.] The real cube root of~$7.976$. + +\item[11.] The abscissa of the real point of intersection of the conics $y=x^2$, $xy+x+3y-6=0$. + +\item[12.] Find to 3~decimal places the abscissas of the points of intersection of $x^2+y^2=9$, +$y=x^2-x$. + +\item[13.] A sphere two feet in diameter is formed of a kind of wood a cubic foot of which +weighs two-thirds as much as a cubic foot of water (i.e., the specific gravity of the wood +is~$2/3$). Find to four significant figures the depth~$h$ to which the floating sphere +will sink in water. +\index{Specific gravity}% + +Hints: The volume of a sphere of radius~$r$ is~$\tfrac{4}{3}\pi r^3$. Hence our sphere whose radius +%% -----File: 096.png---Folio 90------- +is $1$~foot weighs as much as $\tfrac{4}{3}\pi·\tfrac{2}{3}$ cubic feet of water. The volume of the submerged +portion of the sphere is $\pi h^2 (r-\tfrac{1}{3}h)$ cubic feet. Since this is also the volume of the displaced +water, its value for $r=1$ must equal $\tfrac{4}{3}\pi·\tfrac{2}{3}$. Hence +$h^3 - 3h^2 + \tfrac{8}{3} = 0$. + +\item[14.] If the specific gravity of cork is~$1/4$, find to four significant figures how far a +cork sphere two feet in diameter will sink in water. + +\item[15.] Compute $\cos 20°$ to four decimal places by use of +\[ +\cos 3A = 4\cos^3 A - 3\cos A,\qquad +\cos 60° = \tfrac{1}{2}. +\] + +\item[16.] Three intersecting edges of a rectangular parallelopiped are of lengths $6$,~$8$, +and $10$~feet. If the volume is increased by $300$~cubic feet by equal elongations of the +edges, find the elongation to three decimal places. + +\item[17.] Given that the volume of a right circular cylinder is $\alpha\pi$ and the total area of +its surface is~$2\beta\pi$, prove that the radius~$r$ of its base is a root of $r^3 - \beta r + \alpha = 0$. If $\alpha = 56$, +$\beta = 28$, find to four decimal places the two positive roots~$r$. The corresponding altitude +is~$\alpha/r^2$. + +\item[18.] What rate of interest is implied in an offer to sell a house for \$2700 cash, or +in annual installments each of \$1000 payable 1,~2, and 3~years from date? +\index{Compound interest|(}% + +Hint: The amount of \$2700 with interest for 3~years should be equal to the sum +of the first payment with interest for 2~years, the amount of the second payment with +interest for 1~year, and the third payment. Hence if $r$ is the rate of interest and we +write~$x$ for~$1+r$, we have +\[ +2700x^3 = 1000x^2 + 1000x + 1000. +\] + +\item[19.] Find the rate of interest implied in an offer to sell a house for \$3500 cash, or in +annual installments each of \$1000 payable 1,~2, 3, and 4~years from date. + +\item[20.] Find the rate of interest implied in an offer to sell a house for \$3500 cash, or +\$4000 payable in annual installments each of \$1000, the first payable now. +\index{Compound interest|)}% +\end{Problems} +\end{Exercises} + + +% [** PP: Splitting combined ToC entries] +\Section{75.}{Newton's Method} Prior to 1676, Newton\footnote + {Isaac Newton, \textit{Opuscula},~I, 1794, p.~10, p.~37.} +\index{Newton's!method of solution|(}% +had already found +the root between $2$ and~$3$ of equation~\Eq{1}. He replaced~$x$ by~$2+p$ and +obtained~\Eq{2}. Since $p$ is a decimal, he neglected the terms in $p^3$ and~$p^2$, +and hence obtained $p=0.1$, approximately. Replacing $p$ by $0.1 + q$ in~\Eq{2}, +he obtained +\[ +q^3 + 6.3q^2 + 11.23q + 0.061 = 0. +\] +Dividing $-0.061$ by~$11.23$, he obtained $-0.0054$ as the approximate +value of~$q$. Neglecting $q^3$ and replacing $q$ by $-0.0054 + r$, he obtained +\[ +6.3r^2 + 11.16196r + 0.000541708 = 0. +\] +Dropping $6.3r^2$, he found $r$ and hence +\[ +x = 2 + 0.1 - 0.0054 - 0.00004853 = 2.09455147, +\] +%% -----File: 097.png---Folio 91------- +of which all figures but the last are correct~(§74). But the method will +not often lead so quickly to so accurate a value of the root. + +Newton used the close approximation $0.1$ to~$p$, in spite of the fact +that this value exceeds the root~$p$ and hence led to a negative correction +at the next step. This is in contrast with Horner's method in which each +correction is positive, so that each approximation must be chosen less +than the root, as $0.09$ for~$p$. + +Newton's method may be presented in the following general form, +which is applicable to any equation $f(x) = 0$, whether $f(x)$ is a polynomial +or not. Given an approximate value~$a$ of a real root, we can usually +find a closer approximation $a+h$ to the root by neglecting the powers +$h^2$, $h^3, \dotsc$ of the small number~$h$ in Taylor's formula~(§56) +\[ +f(a+h) = f(a) + f'(a)h + f''(a) \frac{h^2}{2} + \dotsb +\] +and hence by taking +\[ +f(a) + f'(a)h = 0,\qquad h = \frac{-f(a)}{f'(a)}. +\] +We then repeat the process with $a_1 = a+h$ in place of the former~$a$. + +Thus in Newton's example, $f(x) = x^3 - 2x - 5$, we have, for $a=2$, +\begin{align*} +h &= \frac{-f(2)}{f'(2)} + = \frac{1}{10},\qquad + a_1 = a+h = 2.1, \\ +% +h_1 &= \frac{-f(2.1)}{f'(2.1)} + = \frac{-0.061}{11.23} + = -0.0054\dotsc. +\end{align*} + + +% [** PP: Split ToC entry, part 2] +\Section[Algebraic and Graphical Discussion] +{76.}{Graphical Discussion of Newton's Method.} Using rectangular +coördinates, consider the graph of $y = f(x)$ and the point~$P$ on it with the +abscissa $OQ = a$ (Fig.~22). Let the tangent at~$P$ meet the $x$-axis at~$T$ +%[** Illustrations FIG. 22 FIG. 23] +\begin{figure*}[hbt] +\begin{center} +\Input{097a}\hfill +\Input{097b} +\end{center} +\end{figure*} +%% -----File: 098.png---Folio 92------- +and let the graph meet the $x$-axis at~$S$. Take $h = QT$, the subtangent. +Then +\begin{align*} +QP=f(a),\qquad f'(a) + &= \tan XTP = \frac{-f(a)}{h}, \\ +h &= \frac{-f(a)}{f'(a)}. +\end{align*} +In the graph in Fig.~22, $OT=a+h$ is a better approximation to the +root $OS$ than $OQ=a$. The next step (indicated by dotted lines) gives a +still better approximation~$OT_1$. + +If, however, we had begun with the abscissa~$a$ of a point~$P_1$ in Fig.~22 +near a bend point, the subtangent would be very large and the method +would probably fail to give a better approximation. Failure is certain +if we use a point~$P_2$ such that a single bend point lies between it and~$S$. + +We are concerned with the approximation to a root previously isolated +as the only real root between two given numbers $\alpha$ and~$\beta$. These should +be chosen so nearly equal that $f'(x)=0$ has no real root between $\alpha$ and~$\beta$, +and hence $f(x)=y$ has no bend point between $\alpha$ and~$\beta$. Further, if $f''(x)=0$ +has a root between our limits, our graph will have an inflexion point with +an abscissa between $\alpha$ and~$\beta$, and the method will likely fail (Fig.~23). + +Let, therefore, neither $f'(x)$ nor $f''(x)$ vanish between $\alpha$ and~$\beta$. Since +$f''$ preserves its sign in the interval from $\alpha$ to~$\beta$, while $f$ changes in sign, +$f''$ and~$f$ will have the same sign for one end point. According as the +abscissa of this point is $\alpha$ or~$\beta$, we take $a=\alpha$ or $a=\beta$ for the first step of +Newton's process. In fact, the tangent at one of the end points meets +the $x$-axis at a point~$T$ with an abscissa within the interval from $\alpha$ to~$\beta$. +If $f'(x)$ is positive in the interval, so that the tangent makes an acute +angle with the $x$-axis, we have Fig.~24 or Fig.~25; if $f'$ is negative, Fig.~26 or Fig.~22. +%[** Illustration \textsc{Fig.~24} \textsc{Fig.~25} \textsc{Fig.~26}] +\begin{figure*}[hbt] +\begin{center} +\Input{098a}\hfill +\Input{098b}\hfill +\Input{098c} +\end{center} +\end{figure*} + +%% -----File: 099.png---Folio 93------- + +\begin{Remark} +In Newton's example, the graph between the points with the abscissas $\alpha = 2$ and +$\beta = 3$ is of the type in Fig.~24, but more nearly like a vertical straight line. In view +of this feature of the graph, we may safely take $a=\alpha$, as did Newton, although our +general procedure would be to take $a = \beta$. The next step, however, accords with our +present process; we have $\alpha=2$, $\beta=2.1$ in Fig.~24 and hence we now take $a=\beta$, getting +\[ +\frac{0.061}{11.23} = 0.0054 +\] +as the subtangent, and hence $2.1 - 0.0054$ as the approximate root. +\end{Remark} + +If we have secured (as in Fig.~24 or Fig.~26) a better upper limit to the +root than~$\beta$, we may take the abscissa~$c$ of the intersection of the chord +$AB$ with the $x$-axis as a better lower limit than~$\alpha$. By similar triangles, +\[ +-f(\alpha) : c - \alpha = f(\beta) : \beta - c, +\] +whence +\[ +c = \frac{\alpha f(\beta) - \beta f(\alpha)}{f(\beta) - f(\alpha)}. +\Tag{3} +\] +This method of finding the value of~$c$ intermediate to $\alpha$ and~$\beta$ is called the +method of interpolation (\emph{regula falsi}). +\index{Interpolation}% +\index{Regula falsi}% + +\begin{Remark} +In Newton's example, $\alpha= 2$, $\beta=2.1$, +\[ +f(\alpha) = -1,\qquad +f(\beta) = 0.061,\qquad +c = 2.0942. +\] +\end{Remark} + +The advantage of having $c$ at each step is that we know a close limit +of the error made in the approximation to the root. + +We may combine the various possible cases discussed into one: + +\begin{Thm}% +If $f(x)=0$ has a single real root between $\alpha$ and~$\beta$, and $f'(x) = 0$, $f''(x) = 0$ +have no real root between $\alpha$ and~$\beta$, and if we designate by~$\beta$ that one of the +numbers $\alpha$ and~$\beta$ for which $f(\beta)$ and $f''(\beta)$ have the same sign, then the root +lies in the narrower interval from~$c$ to $\beta - f(\beta)/f'(\beta)$, where~$c$ is given by~\Eq{3}. +\end{Thm} + +It is possible to prove\footnote + {Weber's \textit{Algebra}, 2d~ed.,~I, pp.~380--382; \textit{Kleines Lehrbuch der Algebra}, 1912, p.~163.} +this theorem algebraically and to show that by +repeated applications of it we can obtain two limits $\alpha'$, $\beta'$ between which +the root lies, such that $\alpha' - \beta'$ is numerically less than any assigned positive +number. Hence the root can be found in this manner to any desired accuracy. + + +\begin{Example} +$f(x) = x^3 - 2x^2 - 2$,\quad $\alpha = 2\tfrac{1}{4}$,\quad $\beta = 2\tfrac{1}{2}$. Then +\[ +f(\alpha) = -\tfrac{47}{64},\qquad f(\beta) = \tfrac{9}{8}. +\] +%% -----File: 100.png---Folio 94------- +Neither of the roots $0$, $4/3$ of $f'(x)=0$ lies between $\alpha$ and~$\beta$, so that $f(x)=0$ has a single +real root between these limits~(§65). Nor is the root~$\tfrac{2}{3}$ of $f''(x)=0$ within these limits. +The conditions of the theorem are therefore satisfied. For $\alpha<x<\beta$, the graph is of +the type in Fig.~24. We find that approximately +\begin{gather*} +c = \tfrac{559}{238} = 2.3487,\qquad +\beta_1 = \beta - \frac{f(\beta)}{f'(\beta)} = 2.3714, \\ +\beta_1 - \frac{f(\beta_{1})}{f'(\beta_{1})} = 2.3597. +\end{gather*} +For $x = 2.3593$, $f(x) = -0.00003$. We therefore have the root to four decimal places. +For $a = 2.3593$, +\[ +f'(a) = 7.2620,\qquad +a - \frac{f(a)}{f'(a)} = 2.3593041, +\] +which is the value of the root correct to 7~decimal places. We at once verify that the +result is greater than the root in view of our work and Fig.~24, while if we change the +final digit from $1$ to~$0$, $f(x)$ is negative. +\end{Example} + + +\begin{Exercises}{Page94} + +\begin{Problems} +\item[1.] For $f(x) = x^4 + x^3 - 3x^2 - x - 4$, show by Descartes' rule of signs that $f'(x)=0$ +and $f''(x)=0$ each have a single positive root and that neither has a root between $1$ +and~$2$. Which of the values $1$ and~$2$ should be taken as~$\beta$? + +\item[2.] When seeking a root between $2$ and~$3$ of $x^3 - x - 9 = 0$, which value should be +taken as~$\beta$? +\end{Problems} +\end{Exercises} + + +% [** PP: Split ToC entry, part 3] +\Section[Systematic Computation] +{77.}{Systematic Computation of Roots by Newton's Method.} By way +of illustration we shall compute to 7~decimal places a positive root of +\[ +f(x) = x^4 + x^3 - 3x^2 - x - 4 = 0. +\] + +Since $f(1) = -6$, $f(2) = 6$, there is a real root between $1$ and~$2$. Since +\[ +f'(x) = 4x^3 + 3x^2 - 6x - 1,\qquad +f'(1) = 0,\qquad +f'(2) = 31, +\] +the graph of $y=f(x)$ is approximately horizontal near $(1, -6)$ and approximately +vertical near $(2, 6)$. Hence the root is much nearer to $2$ than to~$1$. +Thus in applying Newton's method we employ $a = 2$ as the first approximation +to the root. The correction~$h$ is then +\[ +h = \frac{-f(2)}{f'(2)} = \frac{-6}{31} = -0.2 \ldots . +\] +The work of performing the substitutions $x = 2+d$, $d = -0.2 + e, \dotsc$, +to find the transformed equations satisfied by $d$, $e, \dotsc$, is done by synthetic +%% -----File: 101.png---Folio 95------- +division, exactly as in Horner's method, except that some of the +multipliers are now negative: see \hyperref[table:1]{Table~1}, +page~\pageref*{table:1}. +% The structure below is so large it's almost certain to cause a +% horribly-underfull page. Instead, float it, setting up bidirectional +% anchors, and slightly change the wording of the previous sentence. +\phantomsection\label{tableanchor:1}% + +\begin{table*}[hp] +% ** PP: Dear future maintainer of this table code: +% The P column specifier has fixed width 1/8in and serves as +% padding. The idea is to set columns of numbers in stanzas rlP, +% with no inter-column space, and with the decimal points in the +% left-aligned column. \Lcol and \Rcol are wrappers for center-aligned +% columns with a vertical bar at the corresponding side. The floating +% sub-tables at the right side are rather terribly hacked; to get the +% ``underlines'' of correct width, entries are wrapped in left-aligned +% makebox commands. In short, this structure is a nest of hacks. +% +{\small% Coax width into text block +\begin{gather*}% Commit further abuse of semantics.... +\begin{array}{r<{\qquad} r@{}l@{}P<{\qquad} *{3}{>{\quad}r@{}l@{}P} r@{}lP} + 1 & 1 &&& -3 &&& -1 &&& -4 && &\Lcol{2} & \\ +\cline{14-14} + & 2 &&& 6 &&& 6 &&& 10 && &&\Strut \\ +\cline{1-11} + 1 & 3 &&& 3 &&& 5 &\Rcol{}&& \mathbf{6} && &&\Strut \\ + & 2 &&& 10 &&& 26&\Rcol{}&& && && \\ +\cline{1-9} + 1 & 5 &&& 13 &\Rcol{}&& \mathbf{31} &&& && &&\Strut \\ + & 2 &&& 14 &\Rcol{}&& &&& && && \\ +\cline{1-6} + 1 & 7 &\Rcol{}&& \mathbf{27} &&& &&& && &&\Strut \\ + & 2 &\Rcol{}&& &&& &&& && && \\ +\cline{1-3} +\mathbf{1} & \mathbf{9}\Strut + &&& &&& &&& && &\Lcol{\makebox[1em][l]{$-0.2$}} \\ +\cline{14-15} + & -0&.2&& -1&.76&& + -5&.048&& -5&.1904& &&\Strut \\ +\cline{2-12} + & 8&.8&& 25&.24&& + 25&.952&\Lcol{}& + \mathbf{0}&\mathbf{.8096}& &&\Strut \\ +% + & -0&.2 && -1&.72 && + -4&.704 &\Lcol{}& && && \\ +\cline{2-9} +% + & 8&.6 && 23&.52 &\Rcol{}& + \mathbf{21}&\mathbf{.248} && && && \Strut \\ +% + & -0&.2 && -1&.68 &\Rcol{}& + & && && && \\ +\cline{2-7} + & 8&.4 &\Rcol{}& + \mathbf{21}&\mathbf{.84}& &&& && && + \Lcol{\makebox[0pt][l]{\smash[b]{\raisebox{-1ex}{$\dfrac{-0.8096}{21.248}$}}}} \\ +% + &-0&.2 &\Rcol{}& + & & &&& && && \Lcol{} \\ +\cline{2-4} + \mathbf{1} & \mathbf{8}&\mathbf{.2}&& + & & &&& && && \Lcol{\makebox[0pt][l]{$= -0.04$}}&&\ \\ +\cline{14-16} + & -0&.04 && -0&.3264 & + & -0&.860544 & &-0&.81549824 && & \\ +\cline{2-12} + & 8&.16 && 21&.5136 & + & 20&.387456 &\Rcol{}& + \mathbf{-0}&\mathbf{.00589824}\Strut && & \\ +% + & -0&.04 && -0&.3248 & + & -0&.847552 &\Rcol{}& + & && & \\ +\cline{2-10} + & 8&.12 && 21&.1888 + &\Rcol{}& \mathbf{19}&\mathbf{.539904} &&&&&&\Strut \\ +% + & -0&.04 && -0&.3232 + &\Rcol{}& & &&&&&& \\ +\cline{2-7} + & 8&.08 &\Rcol{}& \mathbf{20}&\mathbf{.8656} + & & & &&&&&&\Strut \\ + & -0&.04 &\Rcol{}&& &&& &&& && \\ +\cline{2-4} + \mathbf{1} & \mathbf{8}&\mathbf{.04} && + & & & &&&\Strut&& +\smash{\makebox[0pt][r]{$g=\dfrac{0.005898}{19.54} = .000302$}} && +\end{array} \\[24pt] +\text{\hyperref[tableanchor:1]{\scshape Table~1.}}% Caption...! +\end{gather*}}% End of \small +\phantomsection\label{table:1}% +\end{table*} + +The root is $2 - 0.2 - 0.04 + 0.000302 = 1.760302$, in which the last +figure is in slight doubt. Indeed, it can be proved that \emph{if the final fraction~$g$, +when expressed as a decimal, has $k$~zeros between the decimal point and the +first significant figure, the division may be safely carried to $2k$~decimal places}. +In our example $k=3$, so that we retained 6~decimal places in~$g$. +\index{Synthetic division|)}% + +To proceed independently of this rule, we note that $g$ is obviously +between $0.00030$ and $0.00031$. Then the value of $g^4 + 8.04g^3 + 20.8656g^2$ +%% -----File: 102.png---Folio 96------- +is found to lie between $0.000001878$ and $0.000002006$. Whichever of these +limits we use as a correction by which to reduce the constant term, we +obtain the same dividend below correct to 6~decimal places. +\[ +\begin{array}{r>{\quad}r<{\quad}l@{}l} +\multicolumn{1}{r|}{\xatop{1}\xatop{9}.\xatop{5}\xatop{3}9904} +& 0.005896 & \multicolumn{1}{|l@{}}{0.0003017} \\ +\cline{1-1} \cline{3-3} + & 005862 & \\[-8pt] +& \settowidth{\TmpLen}{005862}\rule{\TmpLen}{0.5pt} & \\ + & 34 & \\ + & 20 & \\[-8pt] +& \settowidth{\TmpLen}{20}\rule{\TmpLen}{0.5pt} & \\ + & 14 & \\ + & 14 & \\[-8pt] +& \settowidth{\TmpLen}{14}\rule{\TmpLen}{0.5pt} & \\ +\end{array} +\] +Hence the root is $1.7603017$ to 7~decimal places. + + +\begin{Exercises}{Page96} + +\begin{Problems} +\item[1.] Find to 8~decimal places the root between $2$ and~$3$ of $x^3 - x - 9 = 0$. + +\item[2.] Find to 7~decimal places the root between $2$ and~$3$ of $x^3 - 2x^2 - 2 = 0$. + +\item[3.] Find the real cube root of $7.976$ to 6~decimal places. + +\item[4.] Explain by Taylor's expansion of $f(2+d)$ why the values of +\[ +f(2),\qquad f'(2),\qquad +\tfrac{1}{2}f''(2),\qquad +\frac{1}{2·3} f'''(2),\qquad +\frac{1}{2·3·4} f''''(2) +\] +are in reverse order the coefficients of the transformed equation +\[ +d^4 + 9d^3 + 27d^2 + 31d + 6 = 0, +\] +obtained in the Example in the text, and printed in heavy type. + +\item[5.] The method commonly used to find the positive square root of~$n$ by a computing +machine consists in dividing~$n$ by an assumed approximate value~$a$ of the square root +and taking half the sum of~$a$ and the quotient as a better approximation. Show that +the latter agrees with the value of $a+h$ given by applying Newton's method to +$f(x) = x^2-n$. +\index{Square roots}% +\end{Problems} +\end{Exercises} + + +% [** PP: Split ToC entry, part 4] +\Section[For Functions not Polynomials] +{78.}{Newton's Method for Functions not Polynomials.}\hfill\break +% [** PP: Format Example manually to avoid \BeforeSkip] +{\indent\normalfont\small\scshape Example~1.\quad\upshape +Find the angle~$x$ at the center of a circle subtended by a chord which +cuts off a segment whose area is one-eighth of that of the circle.} + +\begin{Solution} +If $x$ is measured in radians and if $r$ is the radius, the area of the segment +is equal to the left member of +\[ +\tfrac{1}{2} r^2(x - \sin x) = \tfrac{1}{8} \pi r^2, +\] +whence +\[ +x - \sin x = \tfrac{1}{4} \pi. +\] +%% -----File: 103.png---Folio 97------- +By means of a graph of $y = \sin x$ and the straight line represented by $y = x-\tfrac{1}{4}\pi$, we +see that the abscissa of their point of intersection is approximately $1.78$~radians or~$102°$. +Thus $a = 102°$ is a first approximation to the root of +\[ +f(x) \equiv x - \sin x - \tfrac{1}{4} \pi = 0. +\] +By Newton's method a better approximation is $a+h$, where\footnote + {The derivative of $\sin x$ is~$\cos x$. We need the limit of +\index{Derivative}% + \[ + \frac{\sin(x+2k) - \sin x}{2k} + \equiv \frac{2\cos \tfrac{1}{2}(2x + 2k) \sin \tfrac{1}{2}(2k)}{2k} + \equiv \frac{\cos(x+k) \sin k}{k} + \] + as $2k$ approaches zero. Since the ratio of $\sin k$ to~$k$ approaches~$1$, the limit is $\cos x$.} +\[ +h = \frac{-f(a)}{f'(a)} + = \frac{-a + \sin a + \tfrac{1}{4} \pi}{1 - \cos a}. +\] +\begin{gather*} +\begin{array}{@{}r@{}c@{}l@{}l@{}} + \sin 102° = & & 0.9781 & \\ +\tfrac{1}{4}(3.1416) = & & 0.7854 & \\ +\cline{3-3} + & & 1.7635 & \\ +& \\ + 102° &{}={} & 1.7802 & \text{ radians} \\ +\cline{2-3} + &{}-{} & 0.0167 & +\end{array} +\qquad +\begin{aligned} + \cos 102° &= -0.2079 \\ +1 - \cos 102° &= \Neg 1.2079 \\ +& \\ +h &= \frac{-0.0167}{1.2079} = -0.0138 \\ +a_1 &= a+h = 1.7664 +\end{aligned} \\ % Top-level end-of-line +h_1 = \frac{-f(a_1)}{f'(a_1)} + = \frac{-1.7664 + 0.9809 + 0.7854}{1.1944} + = -0.0001. +\end{gather*} +Hence $x = a_1 + h_1 = 1.7663$~radians, or $101° 12'$. % [** PP: Added period] +\end{Solution} + + +\begin{Example}[2.\protect\footnotemark] +\index{Interpolation}% + \footnotetext{This Ex.~2, which should be contrasted with Ex.~3, is solved by interpolation + since that method is simpler than Newton's method in this special case.} +Solve $x - \log x = 7$, the logarithm being to base~$10$. +\end{Example} + +\begin{Solution} +Evidently $x$ exceeds~$7$ by a positive decimal which is the value of~$\log x$. +Hence in a table of common logarithms, we seek a number~$x$ between $7$ and~$8$ whose +logarithm coincides approximately with the decimal part of~$x$. We read off the values +in the second column. +\[ +\begin{array}{c|c|c} +x & \log x & x - \log x \\ +\hline +7.897 & 0.89746 & 6.99954 \\ +7.898 & 0.89752 & 7.00048 +\end{array} +\] +By the final column the ratio of interpolation is~$46/94$. Hence $x = 7.8975$ to four +decimal places. +\end{Solution} + +%% -----File: 104.png---Folio 98------- + + +\begin{Example}[3.] +Solve $2x-\log x = 7$, the logarithm being to base~10. +\end{Example} + +\begin{Solution} +Evidently $x$ is a little less than~$4$. A table of common logarithms shows +at once that a fair approximation to $x$ is~$a=3.8$. Write +\[ +f(x) \equiv 2x - \log x - 7,\qquad +\log x = M \log_e x,\qquad +M = 0.4343. +\] +By calculus, the derivative of $\log_e x$ is~$1/x$. Hence +\begin{gather*} +f'(x) = 2 - \frac{M}{x},\qquad f'(a) = 2 - 0.1143 = 1.8857, \\ +f(a) = 0.6 - \log 3.8 = 0.6 - 0.57978 = 0.02022, \\ +-h = \frac{f(a)}{f'(a)} = 0.0107,\qquad a_1 = a + h = 3.7893, \\ +f(a_1) = 0.000041,\qquad f(3.7892) = -0.000148. \\ +\frac{148}{189} × 0.0001 = 0.000078,\qquad x = 3.789278. +\end{gather*} +All figures of~$x$ are correct as shown by Vega's table of logarithms to 10~places. +\end{Solution} + + +\begin{Exercises}{Page98} + +Find the angle~$x$ at the center of a circle subtended by a chord which cuts off a segment +whose ratio to the circle is +\begin{Problems}[2] + +\item[1.] $\frac{1}{4}$. + +\item[2.] $\frac{3}{8}$. +\end{Problems} + +When the logarithms are to base~10, +\begin{Problems}[2] +\item[3.] Solve $2x - \log x = 9$. + +\item[4.] Solve $3x - \log x = 9$. + +\ResetCols{1} + +\item[5.] Find the angle just $>15°$ for which +$\frac{1}{2} \sin x + \sin 2x = 0.64$. + +\item[6.] Find the angle just $>72°$ for which +$x - \frac{1}{2} \sin x = \frac{1}{4} \pi$. + +\item[7.] Find all solutions of Ex.~5 by replacing $\sin 2x$ by $2\sin x\cos x$, squaring, and +solving the quartic equation for~$\cos x$. + +\item[8.] Solve similarly $\sin x + \sin 2x = 1.2$. + +\item[9.] Find $x$ to 6~decimal places in $\sin x = x - 2$. + +\item[10.] Find $x$ to 5~decimal places in $x = 3\log_e x$. +\end{Problems} +\end{Exercises} +\index{Newton's!method of solution|)}% + + +\Section{79.}{Imaginary Roots} To find the imaginary roots $x+yi$ of an equation +$f(z)=0$ with real coefficients, expand $f(x+yi)$ by Taylor's theorem; +we get +\index{Imaginary!roots}% +\[ +f(x) + f'(x)yi + - f''(x) \frac{y^2}{1·2} + - f'''(x) \frac{y^3 i}{1·2·3} + \dotsb = 0. +\] + +%% -----File: 105.png---Folio 99------- + +Since $x$ and~$y$ are to be real, and $y \ne 0$, +\[ +\left\{ +\begin{aligned} +f(x) - f''(x) \frac{y^2}{1·2} + f''''(x) \frac{y^4}{1·2·3·4} - \dotsb = 0, \\ +f'(x) - f'''(x) \frac{y^2}{1·2·3} + f^{(5)}(x)\frac{y^4}{5!} - \dotsb = 0. +\end{aligned}\right. +\Tag{4} +\] + +In the Example and Exercises below, $f(z)$ is of degree~$4$ or less. Then +the second equation~\Eq{4} is linear in~$y^2$. Substituting the resulting value +of~$y^2$ in the first equation~\Eq{4}, we obtain an equation $E(x) = 0$, whose real +roots may be found by one of the preceding methods. If the degree of +$f(z)$ exceeds~$4$, we may find $E(x) = 0$ by eliminating $y^2$ between the two +equations~\Eq{4} by one of the methods to be explained in \ChapRef{X}. + + +\begin{Example} +For $f(z) = z^4 - z + 1$, equations~\Eq{4} are +\[ +x^4 - x + 1 - 6x^2 y^2 + y^4 = 0,\qquad +4x^3 - 1 - 4xy^2 = 0. +\] +Thus +\[ +y^2 = x^2 - \frac{1}{4x},\qquad +-4x^6 + x^2 + \frac{1}{16} = 0. +\] +The cubic equation in $x^2$ has the single real root +\[ +x^2 = 0.528727,\qquad x = ±0.72714. +\] +Then $y^2 = 0.184912$ or $0.87254$, and +\[ +z = x+yi + = 0.72714 ± 0.43001i,\qquad + -0.72714 ± 0.93409i. +\] +\end{Example} + + +\begin{Exercises}{Page99} + +Find the imaginary roots of +\begin{Problems}[2] +\item[1.] $z^3 - 2z - 5 = 0$. + +\item[2.] $28z^3 + 9z^2 - 1 = 0$. + +\ResetCols{2} + +\item[3.] $z^4 - 3z^2 - 6z = 2$. + +\item[4.] $z^4 - 4z^3 + 11z^2 - 14z + 10 = 0$. + +\ResetCols{1} + +\item[5.] $z^4 - 4z^3 + 9z^2 - 16z + 20 = 0$. Hint: +\[ +E(x) \equiv x(x - 2)(16x^4 - 64x^3 + 136x^2 - 144x + 65) = 0, +\] +and the last factor becomes $(w^2 + 1)(w^2 + 9)$ for $2x = w + 2$. +\end{Problems} +\end{Exercises} + +\begin{Note} +If we know a real root~$r$ of a cubic equation $f(z)=0$, we may remove the +factor $z-r$ and solve the resulting quadratic equation. When, as usual, $r$ involves +several decimal places, this method is laborious and unsatisfactory. But we may utilize +a device, explained in the author's \textit{Elementary Theory of Equations}, pp.~119--121, §§6,~7. +As there explained, a similar device may be used when we know two real roots of a +quartic equation. +\end{Note} + +%% -----File: 106.png---Folio 100------- + + +\begin{Exercises}[MISCELLANEOUS~]{Page100} + +(Give answers to 6~decimal places, unless the contrary is stated.) +\begin{Problems} + +\item[1.] What arc of a circle is double its chord? + +\item[2.] What arc of a circle is double the distance from the center of the circle to the +chord of the arc? + +\item[3.] If $A$ and~$B$ are the points of contact of two tangents to a circle of radius unity +from a point~$P$ without it, and if arc $AB$ is equal to~$PA$, find the length of the arc. + +\item[4.] Find the angle at the center of a circle of a sector which is bisected by its chord. + +\item[5.] Find the radius of the smallest hollow iron sphere, with air exhausted, which will +float in water if its shell is $1$~inch thick and the specific gravity of iron is~$7.5$. + +\item[6.] From one end of a diameter of a circle draw a chord which bisects the semicircle. + +\item[7.] The equation $x \tan x = c$ occurs in the theory of vibrating strings. Its approximate +solutions may be found from the graphs $y = \cot x$, $y = x/c$. Find $x$ when $c = 1$. + +\item[8.] The equation $\tan x = x$ occurs in the study of the vibrations of air in a spherical +cavity. From an approximate solution $x_1 = 1.5\pi$, we obtain successively better approximations +$x_2 = \tan^{-1} x_1 = 1.4334 \pi$, $x_3 = \tan^{-1} x_2, \dotsc$. Find the first three solutions to +4~decimal places. + +\item[9.] Find to 3~decimal places the first five solutions of +\[ +\tan x = \frac{2x}{2-x^2}, +\] +which occurs in the theory of vibrations in a conical pipe. + +\item[10.] $4 \tau x^3 - (3x - 1)^2 = 0$ arises in the study of the isothermals of a gas. Find its +roots when (i)~$\tau = 0.002$ and (ii)~$\tau = 0.99$. + +\item[11.] Solve $x^x = 100$. + +\item[12.] Solve $x = 10\log x$. + +\item[13.] Solve $x + \log x = x \log x$. + +\item[14.] Solve Kepler's equation $M = x - e \sin x$ when $M = 332° 28' 54.8''$, + $e = 14° 3' 20''$. + +\item[15.] In what time would a sum of money at 6\% interest compounded annually +amount to as much as the same sum at simple interest at~8\%? +\index{Compound interest}% + +\item[16.] In a semicircle of diameter~$x$ is inscribed a quadrilateral with sides $a$, $b$, $c$,~$x$; +then $x^3 - (a^2 + b^2 + c^2) x - 2abc = 0$ (I.~Newton). Given $a = 2$, $b = 3$, $c = 4$, find~$x$. + +\item[17.] What rate of interest is implied in an offer to sell a house for \$9000~cash, or +\$1000~down and \$3000 at the end of each year for three years? +\end{Problems} +\end{Exercises} +\index{Solution of numerical equations|)}% + +%% -----File: 107.png---Folio 101------- + + +\Chapter{VIII}{Determinants; Systems of Linear Equations} +\index{Determinants|(}% +\index{Linear equations!system|(}% + +% [** PP: ToC entry reads Solution of 2 or 3 Linear Equations by Determinants] +\Section[Solution of $2$~Linear Equations by Determinants] +{80.}{Solution of Two Linear Equations by Determinants of Order~$2$.} +Assume that there is a pair of numbers $x$ and~$y$ for which +\[ +\left\{ +\begin{aligned} +a_1 x + b_1 y &= k_1, \\ +a_2 x + b_2 y &= k_2. +\end{aligned} +\right. +\Tag{1} +\] +Multiply the members of the first equation by~$b_2$ and those of the second +equation by~$-b_1$, and add the resulting equations. We get +\[ +(a_1 b_2 - a_2 b_1)x = k_1 b_2 - k_2 b_1. +\] +Employing the respective multipliers $-a_2$ and~$a_1$, we get +\[ +(a_1 b_2 - a_2 b_1)y = a_1 k_2 - a_2 k_1. +\] +The common multiplier of $x$ and~$y$ is +\[ +a_1 b_2 - a_2 b_1, +\Tag{2} +\] +and is denoted by the symbol +\[ +\begin{vmatrix} + a_1 & b_1 \\ + a_2 & b_2 +\end{vmatrix}, +\Tag{2'} +\] +which is called a \emph{determinant of the second order}, and also called the determinant +of the coefficients of $x$ and~$y$ in equations~\Eq{1}. The results above +may now be written in the form +\[ +\begin{vmatrix} + a_1 & b_1 \\ + a_2 & b_2 +\end{vmatrix} x += +\begin{vmatrix} + k_1 & b_1 \\ + k_2 & b_2 +\end{vmatrix},\qquad +\begin{vmatrix} + a_1 & b_1 \\ + a_2 & b_2 +\end{vmatrix} y += +\begin{vmatrix} + a_1 & k_1 \\ + a_2 & k_2 +\end{vmatrix}. +\Tag{3} +\] +We shall call $k_1$ and~$k_2$ the known terms of our equations~\Eq{1}. Hence, +\emph{if $D$ is the determinant of the coefficients of the unknowns, the product of~$D$ by +any one of the unknowns is equal to the determinant obtained from~$D$ by +substituting the known terms in place of the coefficients of that unknown}. + +%% -----File: 108.png---Folio 102------- + +If $D \ne 0$, relations~\Eq{3} uniquely determine values of $x$ and~$y$: +\[ +x = \frac{k_1 b_2 - k_2 b_1}{D},\qquad +y = \frac{a_1 k_2 - a_2 k_1}{D}, +\] +and these values satisfy equations~\Eq{1}; for example, +\[ +a_1 x + b_1 y = \frac{(a_1 b_2 - a_2 b_1)k_1}{D} = k_1. +\] +Hence our equations~\Eq{1} have been solved by determinants when $D \ne 0$. +We shall treat in~§96 the more troublesome case in which $D = 0$. + + +\begin{Example} +For $2x - 3y = -4$, $6x - 2y = 2$, we have +\begin{align*} +&& +\begin{vmatrix} + 2 & -3 \\ + 6 & -2 +\end{vmatrix} x + &= +\begin{vmatrix} +-4 & -3 \\ +\Neg 2 & -2 +\end{vmatrix}, & 14x &= 14,\qquad x = 1, && \\ +&& +14y &= +\begin{vmatrix} +\Neg 2 & -4 \\ +\Neg 6 & \Neg 2 +\end{vmatrix} = 28, & y &= 2. +\end{align*} +\end{Example} + + +\begin{Exercises}{Page102} + +Solve by determinants the following systems of equations: +\begin{Problems}[3] +\item[1.] $\begin{System}[\,]{2} + 8x &-{}& y &= 34, \\ + x &+{}& 8y &= 53. +\end{System}$ + +\item[2.] $\begin{System}[\,]{2} + 3x &+{}& 4y &= 10, \\ + 4x &+{}& y &= 9. +\end{System}$ + +\item[3.] $\begin{System}[\,]{2} + ax &+{}& by &= a^2, \\ + bx &-{}& ay &= ab. +\end{System}$ +\end{Problems} +\end{Exercises} + + +% [** PP: No separate ToC entry] +\Section[Solution of $3$~Linear Equations by Determinants] +{81.}{Solution of Three Linear Equations by Determinants of Order~$3$.} + +Consider a system of three linear equations +\[ +\begin{aligned} +a_1 x + b_1 y + c_1 z &= k_1, \\ +a_2 x + b_2 y + c_2 z &= k_2, \\ +a_3 x + b_3 y + c_3 z &= k_3. +\end{aligned} +\Tag{4} +\] +Multiply the members of the first, second and third equations by +\[ +b_2 c_3 - b_3 c_2,\qquad +b_3 c_1 - b_1 c_3,\qquad +b_1 c_2 - b_2 c_1, +\Tag{5} +\] +respectively, and add the resulting equations. We obtain an equation +in which the coefficients of $y$ and~$z$ are found to be zero, while the coefficient +of~$x$ is +\[ +a_1 b_2 c_3 - +a_1 b_3 c_2 + +% +a_2 b_3 c_1 - +a_2 b_1 c_3 + +% +a_3 b_1 c_2 - +a_3 b_2 c_1. +\Tag{6} +\] +%% -----File: 109.png---Folio 103------- +Such an expression is called a \emph{determinant of the third order} and denoted +by the symbol +\[ +\begin{vmatrix} + a_1 & b_1 & c_1 \\ + a_2 & b_2 & c_2 \\ + a_3 & b_3 & c_3 +\end{vmatrix}. +\Tag{6'} +\] + +The nine numbers $a_1, \dotsc, c_3$ are called the \emph{elements} of the determinant. +In the symbol these elements lie in three (horizontal) \emph{rows}, and +also in three (vertical) \emph{columns}. Thus $a_2$, $b_2$, $c_2$ are the elements of the +second row, while the three $c$'s are the elements of the third column. +\index{Determinants!columns}% +\index{Determinants!elements}% +\index{Determinants!rows}% + +The equation (free of $y$ and~$z$), obtained above, may now be written +as +\[ +\begin{vmatrix} + a_1 & b_1 & c_1 \\ + a_2 & b_2 & c_2 \\ + a_3 & b_3 & c_3 +\end{vmatrix} x += +\begin{vmatrix} + k_1 & b_1 & c_1 \\ + k_2 & b_2 & c_2 \\ + k_3 & b_3 & c_3 +\end{vmatrix}, +\] +since the right member was the sum of the products of the expressions~\Eq{5} +by $k_1$, $k_2$, $k_3$, and hence may be derived from~\Eq{6} by replacing the +$a$'s by the~$k$'s. Thus the theorem of~§80 holds here as regards the +unknown~$x$. We shall later prove, without the laborious computations +just employed, that the theorem holds for all three unknowns. +\index{Linear equations!system|)}% + + +% [** PP: No ToC entry, matching running head] +\Section[Signs of the Terms of a Determinant] +{82.}{The Signs of the Terms of a Determinant of Order~$3$.} In the +\index{Determinants!signs of terms|(}% +six terms of our determinant~\Eq{6}, the letters $a$, $b$, $c$ were always written +in this sequence, while the subscripts are the six possible arrangements +of the numbers $1$, $2$, $3$. The first term $a_1 b_2 c_3$ shall be called the \emph{diagonal +term}, since it is the product of the elements in the main diagonal running +\index{Determinants!diagonal term}% +from the upper left-hand corner to the lower right-hand corner of the +symbol~\Eq{6'} for the determinant. The subscripts in the term $-a_1 b_3 c_2$ +are derived from those of the diagonal term by interchanging $2$ and~$3$, +and the minus sign is to be associated with the fact that an odd number +(here one) of interchanges of subscripts were used. To obtain the arrangement +$2$, $3$, $1$ of the subscripts in the term $+a_2 b_3 c_1$ from the natural order +$1$, $2$, $3$ (in the diagonal term), we may first interchange $1$ and~$2$, obtaining +the arrangement $2$, $1$, $3$, and then interchange $1$ and~$3$; an even number +(two) of interchanges of subscripts were used and the sign of the term +is plus. + +While the arrangement $1$, $3$, $2$ was obtained from $1$, $2$, $3$ by one interchange +$(2, 3)$, we may obtain it by applying in succession the three interchanges +%% -----File: 110.png---Folio 104------- +$(1, 2)$, $(1, 3)$, $(1, 2)$, and in many new ways. To show that the +number of interchanges which will produce the final arrangement $1$, $3$, $2$ +is odd in every case, note that each of the three possible interchanges, +viz., $(1, 2)$, $(1, 3)$, and~$(2, 3)$, changes the sign of the product +\[ +P = (x_1 - x_2) (x_1 - x_3) (x_2 - x_3), +\] +where the $x$'s are arbitrary variables. Thus a succession of $k$~interchanges +yields $P$ or~$-P$ according as $k$ is even or odd. Starting with the arrangement +$1$, $2$, $3$ and applying $k$~successive interchanges, suppose that we +obtain the final arrangement $1$, $3$, $2$. But if in $P$ we replace the subscripts +$1$, $2$, $3$ by $1$, $3$, $2$, respectively, i.e., if we interchange $2$ and~$3$, we obtain~$-P$. +Hence $k$ is odd. We have therefore proved the following rule +of signs: + +\begin{Thm}% +Although the arrangement $r$, $s$, $t$ of the subscripts in any term $±a_r b_s c_t$ of +the determinant may be obtained from the arrangement $1$, $2$, $3$ by various +successions of interchanges, the number of these interchanges is either always +an even number and then the sign of the term is plus or always an odd number +and then the sign of the term is minus. +\end{Thm} + + +\begin{Exercises}{} + +Apply the rule of signs to all terms of +\begin{Problems}[2] + +\item[1.] Determinant~\Eq{6}. + +\item[2.] Determinant $a_1 b_2 - a_2 b_1$. +\end{Problems} +\end{Exercises} + + +\Section[Even and Odd Arrangements] +{83.}{Number of Interchanges always Even or always Odd.} We now +extend the result in~§82 to the case of $n$~variables $x_1, \dotsc, x_n$. The +product of all of their differences $x_i - x_j$ ($i<j$) is +\begin{align*} +P = (x_1 - x_2)(x_1 - x_3) \dotsm &(x_1 - x_n) \\ + {} · (x_2 - x_3) \dotsm &(x_2 - x_n) \\ + \vdots & \\ + {} · &(x_{n-1} - x_n). +\end{align*} +Interchange any two subscripts $i$ and~$j$. The factors which involve neither +$i$ nor~$j$ are unaltered. The factor $(x_i - x_j)$ involving both is changed in +sign. The remaining factors may be paired to form the products +\[ +±(x_i - x_k)(x_j - x_k)\qquad (k = 1, \dotsc, n;\quad k \ne i,\ k \ne j). +\] +Such a product is unaltered. Hence $P$ is changed in sign. + +Suppose that an arrangement $i_1, i_2, \dotsc, i_n$ can be obtained from +%% -----File: 111.png---Folio 105------- +$1, 2, \dotsc$, $n$ by using $m$ successive interchanges and also by $t$~successive +interchanges. Make these interchanges on the subscripts in~$P$; the +resulting functions are equal to~$(-1)^m P$ and~$(-1)^t P$, respectively. But +the resulting functions are identical since either can be obtained at one +step from~$P$ by replacing the subscript~$1$ by~$i_1$, $2$ by~$i_2$, \ldots, $n$ by~$i_n$. Hence +\[ +(-1)^m P \equiv (-1)^t P, +\] +so that $m$ and~$t$ are both even or both odd. + +Thus \emph{if the same arrangement is derived from $1$, $2, \dotsc, n$ by $m$~successive +interchanges as by $t$~successive interchanges, then $m$ and~$t$ are both even or +both odd.} + + +\Section{84.}{Definition of a Determinant of Order~$n$} We define a determinant +of order~$4$ to be +\[ +\begin{vmatrix} + a_1 & b_1 & c_1 & d_1 \\ + a_2 & b_2 & c_2 & d_2 \\ + a_3 & b_3 & c_3 & d_3 \\ + a_4 & b_4 & c_4 & d_4 +\end{vmatrix} + = \sum_{(24)} ± a_q b_r c_s d_t, +\Tag{7} +\] +where $q, r, s, t$ is any one of the $24$~arrangements of $1, 2, 3, 4$, and the +sign of the corresponding term is $+$ or~$-$ according as an even or odd +number of interchanges are needed to derive this arrangement $q, r, s, t$ +from $1, 2, 3, 4$. Although different numbers of interchanges will produce +the same arrangement $q, r, s, t$ from $1, 2, 3, 4$, these numbers are all even +or all odd, as just proved, so that the sign is fully determined. + +We have seen that the analogous definitions of determinants of orders +$2$ and~$3$ lead to our earlier expressions~\Eq{2} and~\Eq{6}. + +We will have no difficulty in extending the definition to a determinant +of general order~$n$ as soon as we decide upon a proper notation for the $n^2$ +elements. The subscripts $1, 2, \dotsc, n$ may be used as before to specify +the rows. But the alphabet does not contain $n$~letters with which to +specify the columns. The use of $e', e'', \dotsc, e^{(n)}$ for this purpose would +conflict with the notation for derivatives and besides be very awkward +when exponents are used. It is customary in mathematical journals +and scientific books (a custom not always followed in introductory text +books, to the distinct disadvantage of the reader) to denote the $n$~letters +used to distinguish the $n$~columns by $e_1, e_2, \dotsc, e_n$ (or some other letter +with the same subscripts) and to prefix (but see~§85) such a subscript by +%% -----File: 112.png---Folio 106------- +the new subscript indicating the row. The symbol for the determinant +is therefore +\[ +D = \begin{vmatrix} +e_{11} & e_{12} & \cdots & e_{1n} \\ +e_{21} & e_{22} & \cdots & e_{2n} \\ +\Dots{4} \\ +e_{n1} & e_{n2} & \cdots & e_{nn} +\end{vmatrix}. +\Tag{8} +\] +By definition this shall mean the sum of the $n(n-1) \dotsm 2·1$ terms +\[ +(-1)^i e_{{i_1}1} e_{{i_2}2} \dotsm e_{{i_n}n} +\Tag{9} +\] +in which $i_1, i_2, \dotsc, i_n$ is an arrangement of $1, 2, \dotsc, n$, derived from +$1, 2, \dotsc, n$ by $i$~interchanges. Any term~\Eq{9} of the determinant~\Eq{8} is, +apart from sign, the product of $n$~factors, one and only one from each column, +and one and only one from each row. +\index{Determinants!interchanges}% + +For example, if we take $n=4$ and write $a_j, b_j, c_j, d_j$ for $e_{j1}, e_{j2}, e_{j3}, e_{j4}$, +the symbol~\Eq{8} becomes~\Eq{7} and the general term~\Eq{9} becomes the general +term $(-1)^i a_{i_1} b_{i_2} c_{i_3} d_{i_4}$ of the second member of~\Eq{7}. + + +\begin{Exercises}{Page106} + +\begin{Problems} + +\item[1.] Find the six terms involving $a_2$ in the determinant~\Eq{7}. + +\item[2.] What are the signs of $a_3b_5c_2d_1e_4$, $a_5b_4c_3d_2e_1$ in a determinant of order five? +\index{Determinants!signs of terms|)}% + +\item[3.] Show that the arrangement $4, 1, 3, 2$ may be obtained from $1, 2, 3, 4$ by use of +the two successive interchanges $(1, 4)$, $(1, 2)$, and also by use of the four successive +interchanges $(1, 4)$, $(1, 3)$, $(1, 2)$, $(2, 3)$. + +\item[4.] Write out the six terms of~\Eq{8} for $n = 3$, rearrange the factors of each term so that +the new first subscripts shall be in the order $1, 2, 3$, and verify that the resulting six +terms are those of the determinant~$D'$ in~§85 for $n = 3$. +\end{Problems} +\end{Exercises} + + +\Section{85.}{Interchange of Rows and Columns} +\begin{Thm} +Any determinant is not +altered in value if in its symbol we replace the elements of the first, second, +$\dotsc, n$th rows by the elements which formerly appeared in the same order +in the first, second, $\dotsc, n$th columns, or briefly if we interchange the corresponding +rows and columns. +\end{Thm}% No \par +For example, +\[ +\begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} += ad - bc = +\begin{vmatrix} +a & c\\ +b & d +\end{vmatrix}. +\] + +We are to prove that the determinant~$D$ given by~\Eq{8} is equal to +\[ +D' = +\begin{vmatrix} +e_{11} & e_{21} & \cdots & e_{n1} \\ +e_{12} & e_{22} & \cdots & e_{n2} \\ +\Dots{4} \\ +e_{1n} & e_{2n} & \cdots & e_{nn} +\end{vmatrix}. +\] +%% -----File: 113.png---Folio 107------- +If we give to $D'$ a more familiar aspect by writing $e_{ik} = a_{ki}$ for each element +so that, as in~\Eq{8}, the row subscript precedes instead of follows the column +subscript, the definition of the determinant in terms of the $a$'s gives $D'$ +in terms of the $e$'s as the sum of all expressions +\[ +(-1)^i e_{1{k_1}} e_{2{k_2}} \dotsm e_{n{k_n}}, +\] +% [** PP: Added comma after k_2] +in which $k_1, k_2, \dotsc, k_n$ is an arrangement of $1, 2, \dotsc, n$, derived from +the latter sequence by $i$~interchanges. + +As for the terms of~$D$, without altering~\Eq{9}, we may rearrange its factors +so that the first subscripts shall appear in the order $1, 2, \dotsc, n$, and +obtain +\[ +(-1)^i e_{1{k_1}} e_{2{k_2}} \dotsm e_{n{k_n}}. +\] +This can be done by performing in reverse order the $i$~successive interchanges +of the letters~$e$ corresponding to the $i$~successive interchanges +which were used to derive the arrangement $i_1, i_2, \dotsc, i_n$ of the first +subscripts from the arrangement $1, 2, \dotsc, n$. Thus the new second +subscripts $k_1, \dotsc, k_n$ are derived from the old second subscripts $1, \dotsc, n$ +by $i$~interchanges. The resulting signed product is therefore a term +of~$D'$. Hence $D = D'$. + + +% [** PP: Split combined ToC entry] +\Section{86.}{Interchange of Two Columns} +\begin{Thm} +A determinant is merely changed +in sign by the interchange of any two of its columns. +\end{Thm}% No \par +For example, +\[ +D = \begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} += ad-bc,\qquad +\Delta = \begin{vmatrix} +b & a \\ +d & c +\end{vmatrix} += bc - ad = -D. +\] + +Let $\Delta$ be the determinant derived from~\Eq{8} by the interchange of the +$r$th and $s$th columns. The terms of~$\Delta$ are therefore obtained from the +terms~\Eq{9} of~$D$ by interchanging $r$ and~$s$ in the series of second subscripts. +Interchange the $r$th and $s$th letters~$e$ to restore the second subscripts +to their natural order. Since the first subscripts have undergone an +interchange, the negative of any term of~$\Delta$ is a term of~$D$, and $\Delta = -D$. + + +% [** PP: No separate ToC entry] +\Section{87.}{Interchange of Two Rows} +\index{Determinants!interchanges}% +\begin{Thm} +A determinant~$D$ is merely changed +in sign by the interchange of any two rows. +\end{Thm} + +Let $\Delta$ be the determinant obtained from~$D$ by interchanging the $r$th +and $s$th rows. By interchanging the rows and columns in~$D$ and in~$\Delta$, +we get two determinants $D'$ and~$\Delta'$, either of which may be derived from +the other by the interchange of the $r$th and $s$th columns. Hence, by +§§85,~86, +\[ +\Delta = \Delta' = -D' = -D. +\] + +%% -----File: 114.png---Folio 108------- + + +% [** No ToC entry] +\Section{88.}{Two Rows or Two Columns Alike} +\begin{Thm} +A determinant is zero if any +two of its rows or any two of its columns are alike. +\end{Thm} + +For, by the interchange of the two like rows or two like columns, the +determinant is evidently unaltered, and yet must change in sign by §§86,~87. +Hence $D = -D$, $D = 0$. + + +\begin{Exercises}{} + +\begin{Problems} + +\item[1.] Prove that the equation of the straight line determined by the two distinct +points $(x_1, y_1)$ and $(x_2, y_2)$ is +\[ +\begin{vmatrix} +x & y & 1 \\ +x_1 & y_1 & 1 \\ +x_2 & y_2 & 1 +\end{vmatrix} = 0. +\] + +\item[2.] Show that +\[ +\begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} += +\begin{vmatrix} +a_2 & c_2 & b_2 \\ +a_1 & c_1 & b_1 \\ +a_3 & c_3 & b_3 +\end{vmatrix} += +\begin{vmatrix} +a_3 & a_1 & a_2 \\ +b_3 & b_1 & b_2 \\ +c_3 & c_1 & c_2 +\end{vmatrix}. +\] +\end{Problems} + +By use of the Factor Theorem~(§14) and the diagonal term, prove that +\begin{Problems} + +\item[3.] +\[ +\begin{vmatrix} +1 & 1 & 1 \\ +a & b & c \\ +a^2 & b^2 & c^2 +\end{vmatrix} = (b-a)(c-a)(c-b). +\] + +\item[4.] +\[ +\begin{vmatrix} +1 & 1 & \cdots & 1 \\ +x_1 & x_2 & \cdots & x_n \\ +x_1^2 & x_2^2 & \cdots & x_n^2 \\ +\Dots{4} \\ +x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} +\end{vmatrix} + = \prod^n_{\substack{i, j=1 \\ i>j}}(x_i-x_j). +\] + +This is known as the determinant of Vandermonde, who discussed it in~1770. The +symbol on the right means the product of all factors of the type indicated. +\index{Determinant!of Vandermonde}% + +\item[5.] Prove that a skew-symmetric determinant of odd order is zero: +\index{Determinants!skew symmetric}% +\[ +\begin{vmatrix} +\Neg 0 & \Neg a & \Neg b \\ + -a & \Neg 0 & \Neg c \\ + -b & -c & \Neg 0 +\end{vmatrix} = 0, +\qquad +\begin{vmatrix} +\Neg 0 & \Neg a & \Neg b & \Neg c & \Neg d \\ + -a & \Neg 0 & \Neg e & \Neg f & \Neg g \\ + -b & -e & \Neg 0 & \Neg h & \Neg j \\ + -c & -f & -h & \Neg 0 & \Neg k \\ + -d & -g & -j & -k & \Neg 0 +\end{vmatrix} = 0. +\] +\end{Problems} +\end{Exercises} + +%% -----File: 115.png---Folio 109------- + + +\Section{89.}{Minors} The determinant of order $n-1$ obtained by erasing +(or covering up) the row and column crossing at a given element of a +determinant of order~$n$ is called the \emph{minor} of that element. +\index{Determinants!minors}% + +\begin{Remark} +For example, in the determinant~\Eq{6'} of order~$3$, the minors of $b_1$, $b_2$, $b_3$ are respectively +\[ +B_1 = \begin{vmatrix} +a_2 & c_2 \\ +a_3 & c_3 +\end{vmatrix},\quad +% +B_2 = \begin{vmatrix} +a_1 & c_1 \\ +a_3 & c_3 +\end{vmatrix},\quad +% +B_3 = \begin{vmatrix} +a_1 & c_1 \\ +a_2 & c_2 +\end{vmatrix}. +\] +Again, \Eq{6'} is the minor of~$d_4$ in the determinant of order~$4$ given by~\Eq{7}. +\end{Remark} + + +\Section[Expansion] +{90.}{Expansion According to the Elements of a Row or Column.} In +\index{Determinants!expansion}% +\[ +D = \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix}, +\Tag{6'} +\] +denote the minor of any element by the corresponding capital letter, +so that $b_1$ has the minor~$B_1$, $b_3$ has the minor~$B_3$, etc., as in~§89. We +shall prove that +\begin{align*} +&& D &= \Neg a_1A_1 - b_1B_1 + c_1C_1, & + D &= \Neg a_1A_1 - a_2A_2 + a_3A_3, && \\ +% +&& D &= -a_2A_2 + b_2B_2 - c_2C_2, & + D &= -b_1B_1 + b_2B_2 - b_3B_3, && \\ +% +&& D &= \Neg a_3A_3 - b_3B_3 + c_3C_3, & + D &= \Neg c_1C_1 - c_2C_2 + c_3C_3. && +\end{align*} +The three relations at the left (or right) are expressed in words by saying +that a \emph{determinant~$D$ of the third order may be expanded according to the +elements of the first, second or third row} (\emph{or column}). To obtain the expansion, +we multiply each element of the row (or column) by the minor of +the element, prefix the proper sign to the product, and add the signed +products. The signs are alternately $+$ and~$-$, as in the diagram +\[ +\begin{matrix} ++ & - & + \\ +- & + & - \\ ++ & - & + +\end{matrix} +\] + +\begin{Remark} +For example, by expansion according to the second column, +\[ +\begin{vmatrix} +1 & 4 & 5 \\ +2 & 0 & 3 \\ +3 & 0 & 9 +\end{vmatrix} += -4 \begin{vmatrix} +2 & 3 \\ +3 & 9 +\end{vmatrix} += -4 × 9 = -36. +\] + +%% -----File: 116.png---Folio 110------- + +Similarly the value of the determinant~\Eq{7} of order~$4$ may be found by expansion +according to the elements of the fourth column: +\[ +-d_1 \begin{vmatrix} +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 \\ +a_4 & b_4 & c_4 +\end{vmatrix} +% ++ d_2 \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_3 & b_3 & c_3 \\ +a_4 & b_4 & c_4 +\end{vmatrix} +% +- d_3 \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_4 & b_4 & c_4 +\end{vmatrix} +% ++ d_4 \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix}. +\] +\end{Remark} + +We shall now prove that \emph{any determinant~$D$ of order~$n$ may be expanded +according to the elements of any row or any column}. + +Let $E_{ij}$ denote the minor of~$e_{ij}$ in~$D$, given by~\Eq{8}, so that $E_{ij}$ is +obtained by erasing the $i$th~row and $j$th~column of~$D$. + +(\emph{i}) We first prove that +\[ +D = e_{11}E_{11} - e_{21}E_{21} + e_{31}E_{31} - \dotsb + + (-1)^{n-1} e_{n1}E_{n1}, +\Tag{10} +\] +so that $D$ may be expanded according to the elements of its first column. +By~\Eq{9} the terms of~$D$ having the factor~$e_{11}$ are of the form +\[ +(-1)^i e_{11} e_{{i_2}2} \dotsm e_{{i_n}n}, +\] +where $1, i_2, \dotsc, i_n$ is an arrangement of $1, 2, \dotsc, n$, obtained from the +latter by $i$~interchanges, so that $i_2, \dotsc, i_n$ is an arrangement of $2, \dotsc, n$, +derived from the latter by $i$~interchanges. After removing from each +term the common factor~$e_{11}$ and adding the quotients, we obtain a sum +which, by definition, is the value of the determinant~$E_{11}$ of order~$n-1$. +Hence the terms of~$D$ having the factor~$e_{11}$ may all be combined into +$e_{11}E_{11}$, which is the first part of~\Eq{10}. + +We next prove that the terms of~$D$ having the factor~$e_{21}$ may be combined +into~$-e_{21}E_{21}$, which is the second part of~\Eq{10}. For, if $\Delta$ be the +determinant obtained from~$D$ by interchanging its first and second rows, +the result just proved shows that the terms of~$\Delta$ having the factor~$e_{21}$ +may be combined into the product of~$e_{21}$ by the minor +\[ +\begin{vmatrix} +e_{12} & e_{13} & \cdots & e_{1n} \\ +e_{32} & e_{33} & \cdots & e_{3n} \\ +\Dots{4} \\ +e_{n2} & e_{n3} & \cdots & e_{nn} +\end{vmatrix} +\] +of~$e_{21}$ in~$\Delta$. Now this minor is identical with the minor~$E_{21}$ of~$e_{21}$ in~$D$. +But $\Delta = -D$~(§87). Hence the terms of~$D$ having the factor~$e_{21}$ may be +%% -----File: 117.png---Folio 111------- +combined into $-e_{21} E_{21}$. Similarly, the terms of~$D$ having the factor~$e_{31}$ +may be combined into $e_{31} E_{31}$, etc., as in~\Eq{10}. + +(\emph{ii}) We next prove that~$D$ may be expanded according to the elements +of its $k$th~column ($k > 1$): +\[ +D = \sum_{j=1}^n (-1)^{j+k} e_{jk} E_{jk}. +\Tag{11} +\] +Consider the determinant~$\delta$ derived from~$D$ by moving the $k$th~column +over the earlier columns until it becomes the new first column. Since +this may be done by $k-1$ interchanges of adjacent columns, $\delta = (-1)^{k-1} D$. +The minors of the elements $e_{1k}, \dotsc, e_{nk}$ in the first column of~$\delta$ are evidently +the minors $E_{1k}, \dotsc, E_{nk}$ of $e_{1k}, \dotsc, e_{nk}$ in~$D$. Hence, by~\Eq{10}, +\[ +\delta = e_{1k} E_{1k} - e_{2k} E_{2k} + \dotsb + + (-1)^{n-1} e_{nk} E_{nk} + = \sum^n_{j=1} (-1)^{j+1} e_{jk} E_{jk}. +\] +Thus $D=(-1)^{k-1} \delta$ has the desired value~\Eq{11}. + +(\emph{iii}) Finally, $D$ may be expanded according to the elements of its +$k$th~row: +\[ +D = \sum^n_{j=1} (-1)^{j+k} e_{kj} E_{kj}. +\] +In fact, by Case~(\emph{ii}), the latter is the expansion of the equal determinant~$D'$ +in~§85 according to the elements of its $k$th~column. + + +\Section{91.}{Removal of Factors} +\index{Determinants!removal of factor}% +\begin{Thm} +A common factor of all of the elements of the +same row or same column of a determinant may be divided out of the elements +and placed as a factor before the new determinant. +\end{Thm} + +In other words, if all of the elements of a row or column are divided +by~$n$, the value of the determinant is divided by~$n$. For example, +\[ +\begin{vmatrix} +na_1 & nb_1 \\ + a_2 & b_2 +\end{vmatrix} += n \begin{vmatrix} +a_1 & b_1 \\ +a_2 & b_2 +\end{vmatrix},\qquad +\begin{vmatrix} +a_1 & nb_1 & c_1 \\ +a_2 & nb_2 & c_2 \\ +a_3 & nb_3 & c_3 +\end{vmatrix} += n \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix}. +\] + +Proof is made by expanding the determinants according to the elements +of the row or column in question and noting that the minors are the same +for the two determinants. Thus the second equation is equivalent to +\[ +-(nb_1)B_1 + (nb_2)B_2 - (nb_3)B_3 = n(-b_1B_1 + b_2B_2 - b_3B_3), +\] +where $B_i$ denotes the minor of~$b_i$ in the final determinant. + +%% -----File: 118.png---Folio 112------- + +% [** PP: Heading stubbornly refuses to start on a new page] +\clearpage +\begin{Exercises}{Page112} + +\begin{Problems}[2] + +\item[1.] $\ds +\begin{vmatrix} +3a & 3b & 3c \\ +5a & 5b & 5c \\ + d & e & f +\end{vmatrix} = 0$. + +\item[2.] $\ds +\begin{vmatrix} +2r & l & 3r \\ +2s & m & 3s \\ +2t & n & 3t +\end{vmatrix} = 0$. +\end{Problems} + +Expand by the shortest method and evaluate +\begin{Problems}[2] + +\item[3.] $\ds +\begin{vmatrix} +2 & 7 & 3 \\ +5 & 9 & 8 \\ +0 & 3 & 0 +\end{vmatrix}$. + +\item[4.] $\ds +\begin{vmatrix} +5 & 7 & 0 \\ +6 & 8 & 0 \\ +3 & 9 & 4 +\end{vmatrix}$. + +\ResetCols{1} + +\item[5.] $\ds +\begin{vmatrix} +a & b & c & d \\ +a^2 & b^2 & c^2 & d^2 \\ +a^3 & b^3 & c^3 & d^3 \\ +a^4 & b^4 & c^4 & d^4 +\end{vmatrix} += abcd(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$. +\end{Problems} +\end{Exercises} + + +\Section{92.}{Sum of Determinants} +\index{Determinants!sum of}% +\begin{Thm} +A determinant having $a_1+q_1$, $a_2+q_2, \dotsc$ as +the elements of a column is equal to the sum of the determinant having $a_1$, +$a_2, \dotsc$ as the elements of the corresponding column and the determinant +having $q_1$, $q_2, \dotsc$ as the elements of that column, while the elements of the +remaining columns of each determinant are the same as in the given determinant. +\end{Thm} + +For example, +\[ +\begin{vmatrix} +a_1 + q_1 & b_1 & c_1 \\ +a_2 + q_2 & b_2 & c_2 \\ +a_3 + q_3 & b_3 & c_3 +\end{vmatrix} += \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} ++ \begin{vmatrix} +q_1 & b_1 & c_1 \\ +q_2 & b_2 & c_2 \\ +q_3 & b_3 & c_3 +\end{vmatrix}. +\] + +To prove the theorem we have only to expand the three determinants +according to the elements of the column in question (the first column in +the example) and note that the minors are the same for all three determinants. +Hence $a_1 + q_1$ is multiplied by the same minor that $a_1$ and~$q_1$ +are multiplied by separately, and similarly for $a_2 + q_2$, etc. + +The similar theorem concerning the splitting of the elements of any +row into two parts is proved by expanding the three determinants according +to the elements of the row in question. For example, +\[ +\begin{vmatrix} +a+r & b+s \\ +c & d +\end{vmatrix} += \begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} ++ \begin{vmatrix} +r & s \\ +c & d +\end{vmatrix}. +\] + +%% -----File: 119.png---Folio 113------- + + +\Section{93.}{Addition of Columns or Rows} +\index{Determinants!addition of columns}% +\begin{Thm} +A determinant is not changed +in value if we add to the elements of any column the products of the corresponding +elements of another column by the same arbitrary number. +\end{Thm} + +Let $a_1$, $a_2, \dotsc$ be the elements to which we add the products of the +elements $b_1$, $b_2, \dotsc$ by~$n$. We apply~§92 with $q_1=nb_1$, $q_2=nb_2, \dotsc$. +Thus the modified determinant is equal to the sum of the initial determinant +and a determinant having $b_1$, $b_2, \dotsc$ in one column and $nb_1$, $nb_2, +\dotsc$ in another column. But~(§91) the latter determinant is equal to +the product of~$n$ by a determinant with two columns alike and hence +is zero~(§88). For example, +\[ +\begin{vmatrix} +a_1 + nb_1 & b_1 & c_1 \\ +a_2 + nb_2 & b_2 & c_2 \\ +a_3 + nb_3 & b_3 & c_3 +\end{vmatrix} += \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} ++ n\begin{vmatrix} +b_1 & b_1 & c_1 \\ +b_2 & b_2 & c_2 \\ +b_3 & b_3 & c_3 +\end{vmatrix}, +\] +and the last determinant is zero. + +Similarly, \emph{a determinant is not changed in value if we add to the elements +of any row the products of the corresponding elements of another row by the +same arbitrary number}. + +\begin{Remark} +For example, +\[ +\begin{vmatrix} +a + nc & b + nd \\ +c & d +\end{vmatrix} += \begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} ++ n \begin{vmatrix} +c & d \\ +c & d +\end{vmatrix} += \begin{vmatrix} +a & b \\ +c & d +\end{vmatrix}. +\] +\end{Remark} + + +\begin{Example} +Evaluate the first determinant below. +\[ +\begin{vmatrix} +1 & -2 & 1 \\ +1 & \Neg 2 & 3 \\ +6 & \Neg 4 & 3 +\end{vmatrix} += \begin{vmatrix} +1 & 0 & 1 \\ +1 & 8 & 3 \\ +6 & 10 & 3 +\end{vmatrix} += \begin{vmatrix} +\Neg 0 & 0 & 1 \\ + -2 & 8 & 3 \\ +\Neg 3 & 10 & 3 +\end{vmatrix} += \begin{vmatrix} + -2 & 8 \\ +\Neg 3 & 10 +\end{vmatrix} = -44. +\] +\end{Example} + +\begin{Solution} +First we add to the elements of the second column the products of the +elements of the last column by~$2$. In the resulting second determinant, we add to the +elements of the first column the products of the elements of the third column by~$-1$. +Finally, we expand the resulting third determinant according to the elements of its +first row. +\end{Solution} + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] Prove that +\[ +\begin{vmatrix} +b + c & c + a & a + b \\ +b_1 + c_1 & c_1 + a_1 & a_1 + b_1 \\ +b_2 + c_2 & c_2 + a_2 & a_2 + b_2 +\end{vmatrix} += 2 \begin{vmatrix} +a & b & c \\ +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 +\end{vmatrix} +\] +\end{Problems} + +%% -----File: 120.png---Folio 114------- + +By reducing to a determinant of order~$3$, etc., prove that +\begin{Problems} +\item[2.] $\ds +\begin{vmatrix} +1 & 1 & 1 & 1 \\ +a & b & c & d \\ +a^2 & b^2 & c^2 & d^2 \\ +a^3 & b^3 & c^3 & d^3 +\end{vmatrix} += (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$. +\bigskip % [** PP: Explicit vertical space] + +\ResetCols{2} + +\item[3.] $\ds +\begin{vmatrix} +2 & -1 & \Neg 3 & -2 \\ +1 & \Neg 7 & \Neg 1 & -1 \\ +3 & \Neg 5 & -5 & \Neg 3 \\ +4 & -3 & \Neg 2 & -1 +\end{vmatrix} = -42$. + +\item[4.] $\ds +\begin{vmatrix} +1 & 1 & \phantom{1}1 & \phantom{1}1 \\ +1 & 2 & \phantom{1}3 & \phantom{1}4 \\ +1 & 3 & \phantom{1}6 & 10 \\ +1 & 4 & 10 & 20 +\end{vmatrix} = 1$. + +\end{Problems} +\end{Exercises} + + +\Section[System of $n$~Linear Equations in $n$~Unknowns] +{94.}{System of $n$~Linear Equations in $n$~Unknowns with $D \neq 0$.} In +\index{Linear equations!system|(}% +\[ +\begin{matrix} +a_{11} x_1 + a_{12} x_2 + \dotsb + a_{1n} x_n = k_1, \\ +\Dots{1} \\ +a_{n1} x_1 + a_{n2} x_2 + \dotsb + a_{nn} x_n = k_n, +\end{matrix} +\Tag{12} +\] +let $D$ denote the determinant of the coefficients of the $n$~unknowns: +\[ +D = \begin{vmatrix} +a_{11} & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +a_{n1} & a_{n2} & \cdots & a_{nn} +\end{vmatrix}. +\] +Then +{\small +\[ +Dx_1 = \begin{vmatrix} +a_{11}x_1 & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +a_{n1}x_1 & a_{n2} & \cdots & a_{nn} \\ +\end{vmatrix} += \begin{vmatrix} % [** PP: Removed commas after first column] +a_{11}x_1 + a_{12}x_2 + \dotsb + a_{1n}x_n & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +a_{n1}x_1 + a_{n2}x_2 + \dotsb + a_{nn}x_n & a_{n2} & \cdots & a_{nn} +\end{vmatrix}, +\]}% +where the second determinant was derived from the first by adding to +the elements of the first column the products of the corresponding elements +of the second column by~$x_2$, etc., and finally the products of the elements +of the last column by~$x_n$. The elements of the new first column are equal +to $k_1, \dotsc, k_n$ by~\Eq{12}. In this manner, we find that +\[ +Dx_1 = K_1,\qquad +Dx_2 = K_2,\qquad \dotsc,\qquad +Dx_n = K_n, +\Tag{13} +\] +in which $K_i$ is derived from~$D$ by substituting $k_1, \dotsc, k_n$ for the elements +$a_{1i}, \dotsc, a_{ni}$ of the $i$th~column of~$D$, whence +\[ +K_1 = \begin{vmatrix} +k_1 & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +k_n & a_{n2} & \cdots & a_{nn} +\end{vmatrix}, \dotsc\qquad +K_n = \begin{vmatrix} +a_{11} & \cdots & a_{1n-1} & k_1 \\ +\Dots{4} \\ +a_{n1} & \cdots & a_{nn-1} & k_n +\end{vmatrix}. +\] + +%% -----File: 121.png---Folio 115------- + +If $D \neq 0$, the unique values of $x_1, \dotsc, x_n$ determined by division +from~\Eq{13} actually satisfy equations~\Eq{12}. For instance, the first equation +is satisfied since +\[ +k_1D - a_{11} K_1 - a_{12} K_2 - \dotsb - a_{1n} K_n += \begin{vmatrix} +k_1 & a_{11} & a_{12} & \cdots & a_{1n} \\ +k_1 & a_{11} & a_{12} & \cdots & a_{1n} \\ +k_2 & a_{21} & a_{22} & \cdots & a_{2n} \\ +\Dots{5} \\ +k_n & a_{n1} & a_{n2} & \cdots & a_{nn} +\end{vmatrix}, +\] +as shown by expansion according to the elements of the first row; and +the determinant is zero, having two rows alike. + +\begin{Theorem} +If $D$ denotes the determinant of the coefficients of the +$n$~unknowns in a system of $n$~linear equations, the product of $D$ by any one +of the unknowns is equal to the determinant obtained from $D$ by substituting +the known terms in place of the coefficients of that unknown. If $D \neq 0$, we +obtain the unique values of the unknowns by division by~$D$. +\end{Theorem} + +We have therefore given a complete proof of the results stated and +illustrated in §80,~§81. % [** PP: Retaining instead of §§80,~81.] +Another proof is suggested in Ex.~7 below. +The theorem was discovered by induction in 1750 by G.~Cramer. + + +\begin{Exercises}{Page115} + +Solve by determinants the following systems of equations (reducing each determinant +to one having zero as the value of every element but one in a row or column, +as in the example in~§93). +\begin{Problems}[2] + +\item[1.] $\begin{System}{3} + x &+{}& y &+{}& z &= 11, \\ + 2x &-{}& 6y &-{}& z &= 0, \\ + 3x &+{}& 4y &+{}& 2z &= 0. +\end{System}$ + +\item[2.] $\begin{System}{3} + x &+{}& y &+{}& z &= 0, \\ + x &+{}& 2y &+{}& 3z &= -1, \\ + x &+{}& 3y &+{}& 6z &= 0. +\end{System}$ + +\ResetCols{2} + +\item[3.] $\begin{System}{3} + x &-{}& 2y &+{}& z &= 12, \\ + x &+{}& 2y &+{}& 3z &= 48, \\ +6x &+{}& 4y &+{}& 3z &= 84. +\end{System}$ + +% [** PP: Matching original alignment; not aligning like variables] +\item[4.] $\begin{System}{2} +3x &-{}& 2y &= 7, \\ +3y &-{}& 2z &= 6, \\ +3z &-{}& 2x &= -1. +\end{System}$ + +\ResetCols{2} + +\item[5.] $\begin{System}{4} +x &+{}& y &+{}& z &+{}& w &= 1, \\ +x &+{}& 2y &+{}& 3z &+{}& 4w &= 11, \\ +x &+{}& 3y &+{}& 6z &+{}& 10w &= 26, \\ +x &+{}& 4y &+{}& 10z &+{}& 20w &= 47. +\end{System}$ + +\item[6.] $\begin{System}{4} +2x &-{}& y &+{}& 3z &-{}& 2w &= 4, \\ + x &+{}& 7y &+{}& z &-{}& w &= 2, \\ +3x &+{}& 5y &-{}& 5z &+{}& 3w &= 0, \\ +4x &-{}& 3y &+{}& 2z &-{}& w &= 5. +\end{System}$ + +\ResetCols{1} + +\item[7.] Prove the first relation~\Eq{13} by multiplying the members of the first equation~\Eq{12} +by~$A_{11}$, those of the second equation by $-A_{21}, \dotsc$, those of the $n$th equation by +$(-1)^{n-1}A_{n1}$, and adding, where $A_{ij}$ by denotes the minor of~$a_{ij}$ in~$D$. Hint: The resulting +coefficient of~$x_2$ is the expansion, according to the elements of its first column, of a determinant +derived from $D$ by replacing $a_{11}$ by~$a_{12}$, $\dotsc$, $a_{n1}$ by~$a_{n2}$. +\end{Problems} +\end{Exercises} + +%% -----File: 122.png---Folio 116------- + + +\Section[Rank] +{95.}{Rank of a Determinant.} If we erase from a determinant~$D$ of +order~$n$ all but $r$~rows and all but $r$~columns, we obtain a determinant +of order~$r$ called an \emph{$r$-rowed minor of~$D$}. In particular, any element is +regarded as a one-rowed minor, and $D$ itself is regarded as an $n$-rowed +minor. +\index{Determinants!minors}% + +If a determinant~$D$ of order~$n$ is not zero, it is said to be of \emph{rank~$n$}. +If, for $0 < r < n$, some $r$-rowed minor of~$D$ is not zero, while every $(r+1)$-rowed +minor is zero, $D$ is said to be of \emph{rank~$r$}. It is said to be of rank +zero if every element is zero. +\index{Determinants!rank}% + +\begin{Remark} +For example, a determinant~$D$ of order~$3$ is of rank~$3$ if $D \neq 0$; of rank~$2$ if $D = 0$, +but some two-rowed minor is not zero; of rank~$1$ if every two-rowed minor is zero, +but some element is not zero. Again, every three-rowed minor of +\[ +\begin{vmatrix} +a & b & c & d \\ +e & f & g & h \\ +a & b & c & d \\ +e & f & g & h +\end{vmatrix} +\] +is zero since two pairs of its rows are alike. Hence it is of rank~$2$ if some two-rowed +minor is not zero. But it is of rank~$1$ if $a, b, c, d$ are not all zero and are proportional +to $e, f, g, h$, since all two-rowed minors are then zero. +\end{Remark} + + +\Section[System of $n$~Linear Equations in $n$~Unknowns] +{96.}{System of $n$~Linear Equations in $n$~Unknowns with $D=0$.} We +shall now discuss the equations~\Eq{12} for the troublesome case (previously +ignored) in which the determinant~$D$ of the coefficients of the unknowns +is zero. In view of~\Eq{13}, the given equations are evidently inconsistent +if any one of the determinants $K_1, \dotsc, K_n$ is not zero. But if $D$ and +these $K$'s are all zero, our former results~\Eq{13} give us no information +concerning the unknowns~$x_i$, and we resort to the following + +\begin{Theorem} +Let the determinant~$D$ of the coefficients of the unknowns +in equations~\Eq{12} be of rank~$r$, $r<n$. If the determinants~$K$ obtained from +the $(r+1)$-rowed minors of~$D$ by replacing the elements of any column by +the corresponding known terms~$k_i$ are not all zero, the equations are inconsistent. +But if these determinants $K$ are all zero, the $r$~equations involving +the elements of a non-vanishing $r$-rowed minor of $D$ determine uniquely $r$ +of the unknowns as linear functions of the remaining $n-r$ unknowns, which +are independent variables, and the expressions for these $r$~unknowns satisfy +also the remaining $n-r$ equations. +\end{Theorem} + +%% -----File: 123.png---Folio 117------- + +\begin{Remark} +Consider for example the three equations~\Eq{4} in the unknowns $x, y, z$. Five cases +arise: + +% [** PP: Not italicizing i.e.] +\begin{itemize} +\item[($\alpha$)] $D$ of rank~$3$, i.e., $D \neq 0$. + +\item[($\beta$)] $D$ of rank $2$ (i.e., $D=0$, but some two-rowed minor $\neq 0$), and +\[ +K_1 = \begin{vmatrix} +k_1 & b_1 & c_1 \\ +k_2 & b_2 & c_2 \\ +k_3 & b_3 & c_3 +\end{vmatrix},\quad +% +K_2 = \begin{vmatrix} +a_1 & k_1 & c_1 \\ +a_2 & k_2 & c_2 \\ +a_3 & k_3 & c_3 +\end{vmatrix},\quad +% +K_3 = \begin{vmatrix} +a_1 & b_1 & k_1 \\ +a_2 & b_2 & k_2 \\ +a_3 & b_3 & k_3 +\end{vmatrix} +\] +not all zero. + +\item[($\gamma$)] $D$ of rank~$2$ and $K_1$, $K_2$, $K_3$ all zero. + +\item[($\delta$)] $D$ of rank~$1$ (i.e., every two-rowed minor $= 0$, but some element $\neq 0$), and +\[ +\begin{vmatrix} +a_i & k_i \\ +a_j & k_j +\end{vmatrix},\qquad +% +\begin{vmatrix} +b_i & k_i \\ +b_j & k_j +\end{vmatrix},\qquad +\begin{vmatrix} +c_i & k_i \\ +c_j & k_j +\end{vmatrix}\qquad \text{($i$, $j$ chosen from $1$, $2$, $3$)} +\] +not all zero; there are nine such determinants~$K$. + +\item[($\epsilon$)] $D$ of rank~$1$, and all nine of the two-rowed determinants~$K$ zero. +\end{itemize} + +In case~($\alpha$) the equations have a single set of solutions~(§94). In cases ($\beta$) and~($\delta$) +there is no set of solutions. For ($\beta$) the proof follows from~\Eq{13}. In case~($\gamma$) one +of the equations is a linear combination of the other two; for example, if $a_1b_2 - a_2b_1 \neq 0$, +the first two equations determine $x$ and~$y$ as linear functions of~$z$ (as shown by transposing +the terms in~$z$ and solving the resulting equations for $x$ and~$y$), and the resulting +values of $x$ and~$y$ satisfy the third equation identically as to~$z$. Finally, in case~($\epsilon$), +two of the equations are obtained by multiplying the remaining one by constants. + +The reader acquainted with the elements of solid analytic geometry will see that +the planes represented by the three equations have the following relations: +\begin{itemize} + +\item[($\alpha$)] The three planes intersect in a single point. + +\item[($\beta$)] Two of the planes intersect in a line parallel to the third plane. + +\item[($\gamma$)] The three planes intersect in a common line. + +\item[($\delta$)] The three planes are parallel and not all coincident. + +\item[($\epsilon$)] The three planes coincide. + +\end{itemize} +\end{Remark} + +The remarks preceding our theorem furnish an illustration (the case +$r=n-1$) of the following + +\begin{Lemma}[1.] If every $(r+1)$-rowed minor~$M$ formed from certain $r+1$ +rows of~$D$ is zero, the corresponding $r+1$ equations~\Eq{12} are inconsistent +provided there is a non-vanishing determinant~$K$ formed from any~$M$ +by replacing the elements of any column by the corresponding known +terms~$k_i$. +\end{Lemma} + +For concreteness,\footnote + {All other cases may be reduced to this one by rearranging the $n$~equations and +relabelling the unknowns (replacing $x_3$ by the new~$x_1$, for example).} +let the rows in question be the first $r+1$ and let +%% -----File: 124.png---Folio 118------- +\[ +K = \begin{vmatrix} +a_{11} & \cdots & a_{1r} & k_1 \\ +\Dots{4} \\ +a_{r+11} & \cdots & a_{r+1r} & k_{r+1} +\end{vmatrix} \neq 0. +\] +Let $d_1, \dotsc, d_{r+1}$ be the minors of $k_1, \dotsc, k_{r+1}$ in~$K$. Multiply the +first $r+1$ equations~\Eq{12} by $d_1$, $-d_2, \dotsc, (-1)^rd_{r+1}$, respectively, and +add. The right member of the resulting equation is the expansion of~$±K$. +The coefficient of~$x_s$ is the expansion of +\[ +±\begin{vmatrix} +a_{11} & \cdots & a_{1r} & a_{1s} \\ +\Dots{4} \\ +a_{r+11} & \cdots & a_{r+1r} & a_{r+1s} +\end{vmatrix} +\] +and is zero, being an~$M$ if $s>r$, and having two columns identical if $s \leqq r$. +Hence $0 = ±K$. Thus if $K \neq 0$, the equations are inconsistent. + +\begin{Lemma}[2.] If all of the determinants $M$ and~$K$ in Lemma~1 are zero, +but an $r$-rowed minor of an~$M$ is not zero, one of the corresponding $r+1$ +equations is a linear combination of the remaining $r$ equations. +\end{Lemma} + +As before let the $r+1$ rows in question be the first $r+1$. Let the +non-vanishing $r$-rowed minor be +\[ +d_{r+1} = \begin{vmatrix} +a_{11} & \cdots & a_{1r} \\ +\Dots{3} \\ +a_{r1} & \cdots & a_{rr} +\end{vmatrix} \neq 0. +\Tag{14} +\] +Let the functions obtained by transposing the terms~$k_i$ in~\Eq{12} be +\[ +L_i \equiv a_{i1} x_1 + a_{i2} x_2 + \dotsb + a_{in} x_n - k_i. +\] +By the multiplication made in the proof of Lemma~1, +\[ +d_1L_1 - d_2L_2 + \dotsb + (-1)^rd_{r+1}L_{r+1} = \mp K = 0. +\] +Hence $L_{r+1}$ is a linear combination of $L_1, \dotsc, L_r$. + +The first part of the theorem is true by Lemma~1. The second part +is readily proved by means of Lemma~2. Let~\Eq{14} be the non-vanishing +$r$-rowed minor of~$D$. For $s>r$, the $s$th equation is a linear combination +of the first $r$~equations, and hence is satisfied by any set of solutions of +the latter. In the latter transpose the terms involving $x_{r+1}, \dotsc, x_n$. +Since the determinant of the coefficients of $x_1, \dotsc, x_r$ is not zero,~§94 +shows that $x_1, \dotsc, x_r$ are uniquely determined linear functions of +$x_{r+1}, \dotsc, x_n$ (which enter from the new right members). + +%% -----File: 125.png---Folio 119------- + + +\begin{Exercises}{Page119} + +Apply the theorem to the following four systems of equations and check the conclusions: +\begin{Problems}[2] + +\item[1.] $\begin{System}{3} +2x&+{}& y&+{}& 3z &= 1, \\ +4x&+{}& 2y&-{}& z &= -3, \\ +2x&+{}& y&-{}& 4z &= -4. +\end{System}$ + +\item[2.] $\begin{System}{3} +2x&+{}& y&+{}& 3z &= 1, \\ +4x&+{}& 2y&-{}& z &= 3, \\ +2x&+{}& y&-{}& 4z &= 4. +\end{System}$ + +\ResetCols{2} + +\item[3.] $\begin{System}{3} + x&-{}& 3y&+{}& 4z &= 1, \\ +4x&-{}& 12y&+{}& 16z &= 3, \\ +3x&-{}& 9y&+{}& 12z &= 3. +\end{System}$ + +\item[4.] $\begin{System}{3} + x&-{}& 3y&+{}& 4z &= 1, \\ +4x&-{}& 12y&+{}& 16z &= 4, \\ +3x&-{}& 9y&+{}& 12z &= 3. +\end{System}$ + +\ResetCols{1} + +\item[5.] Discuss the system +\[ +\begin{System}{3} +ax&+{}& y&+{}& z &= a-3, \\ + x&+{}& ay&+{}& z &= -2, \\ + x&+{}& y&+{}& az &= -2, +\end{System} +\] +when (\emph{i})~$a = 1$; (\emph{ii})~$a = -2$; (\emph{iii})~$a \neq 1$, $-2$, obtaining the simplest forms of the +unknowns. + +\item[6.] Discuss the system +\[ +\begin{System}{3} + x &+{}& y&+{}& z &= 1, \\ +ax &+{}& by&+{}& cz &= k, \\ +a^2x&+{}& b^2y&+{}& c^2z &= k^2, +\end{System} +\] +when (\emph{i})~$a$, $b$, $c$ are distinct; (\emph{ii}) $a = b \neq c$; (\emph{iii}) $a = b = c$. +\end{Problems} +\end{Exercises} + + +\Section[Homogeneous Equations]% [** PP: Allow line to break in ... list] +{97.}{Homogeneous Linear Equations.} When the known terms $k_1$, $\dotsc, +k_n$ in~\Eq{12} are all zero, the equations are called \emph{homogeneous}. The determinants~$K$ +are now all zero, so that the $n$~homogeneous equations are never +inconsistent. This is also evident from the fact that they have the set +of solutions $x_1 = 0, \dotsc, x_n = 0$. By~\Eq{13}, there is no further set of solutions +if $D \neq 0$. If $D = 0$, there are further sets of solutions. This is +shown by the theorem of~§96 which now takes the following simpler form. + +\begin{Thm}% +If the determinant~$D$ of the coefficients of $n$~linear homogeneous equations +in $n$~unknowns is of rank~$r$, $r<n$, the $r$~equations involving the elements of a +non-vanishing $r$-rowed minor of~$D$ determine uniquely $r$ of the unknowns as +linear functions of the remaining $n-r$ unknowns, which are independent variables, +and the expressions for these $r$~unknowns satisfy also the remaining +$n-r$ equations. +\end{Thm} + +The particular case mentioned is the much used theorem: + +\begin{Thm}% +\index{Linear equations!homogeneous}% +A necessary and sufficient condition that $n$~linear homogeneous equations +in $n$~unknowns shall have a set of solutions, other than the trivial one in which +each unknown is zero, is that the determinant of the coefficients be zero. +\end{Thm} + +%% -----File: 126.png---Folio 120------- + + +\begin{Exercises}{Page120} + +Discuss the following systems of equations: +\begin{Problems}[2] +\item[1.] $\begin{System}{3} +x &+{}& y &+{}& 3z &= 0,\\ +x &+{}& 2y &+{}& 2z &= 0,\\ +x &+{}& 5y &-{}& z &= 0. +\end{System}$ + +\item[2.] $\begin{System}{3} + 2x &-{}& y &+{}& 4z &= 0,\\ + x &+{}& 3y &-{}& 2z &= 0,\\ + x &-{}& 11y &+{}& 14z &= 0. +\end{System}$ + +\ResetCols{2} + +\item[3.] $\begin{System}{3} + x &-{}& 3y &+{}& 4z &= 0,\\ +4x &-{}& 12y &+{}& 16z &= 0,\\ +3x &-{}& 9y &+{}& 12z &= 0. +\end{System}$ + +\item[4.] $\begin{System}{4} +6x &+{}& 4y &+{}& 3z &-{}& 84w &= 0,\\ + x &+{}& 2y &+{}& 3z &-{}& 48w &= 0,\\ + x &-{}& 2y &+{}& z &-{}& 12w &= 0,\\ +4x &+{}& 4y &-{}& z &-{}& 24w &= 0. +\end{System}$ + +\ResetCols{1} + +\item[5.] $\begin{System}{4} +2x &+{}& 3y &-{}& 4z &+{}& 5w &= 0,\\ +3x &+{}& 5y &-{}& z &+{}& 2w &= 0,\\ +7x &+{}& 11y &-{}& 9z &+{}& 12w &= 0,\\ +3x &+{}& 4y &-{}& 11z &+{}& 13w &= 0. +\end{System}$ +\end{Problems} +\end{Exercises} + + +% [** PP: ToC entry reads System of $m$~Linear Equations in $n$~Unknowns, +% Matrix and Augmented Matrix] +\Section{98.}{System of $m$~Linear Equations in $n$~Unknowns} The case $m<n$ +may be treated by means of the lemmas in~§96. If $m>n$, we select +any $n$ of the equations and apply to them the theorems of §§94,~96. If +they are found to be inconsistent, the entire system is evidently inconsistent. +But if the $n$~equations are consistent, and if $r$ is the rank of the +determinant of their coefficients, we obtain $r$ of the unknowns expressed +as linear functions of the remaining $n-r$ unknowns. Substituting these +values of these $r$~unknowns in the remaining equations, we obtain a +system of $m-n$ linear equations in $n-r$ unknowns. Treating this system +in the same manner, we ultimately either find that the proposed +$m$~equations are consistent and obtain the general set of solutions, +or find that they are inconsistent. To decide in advance whether the +former or latter of these cases will arise, we have only to find the maximum +order~$r$ of a non-vanishing $r$-rowed determinant formed from the +coefficients of the unknowns, taken in the regular order in which they +occur in the equations, and ascertain whether or not the corresponding +($r+1$)-rowed determinants~$K$, formed as in~§96, are all zero. + +The last result may be expressed simply by employing the terminology +of matrices. The system of coefficients of the unknowns in any set of +linear equations +\[ +\begin{matrix} +a_{11} x_1 + \dotsb + a_{1n} x_n = k_1, \\ +\Dots{1} \\ +a_{m1} x_1 + \dotsb + a_{mn} x_n = k_m, +\end{matrix} +\Tag{15} +\] +arranged as they occur in the equations, is called the \emph{matrix} of the coefficients, +and is denoted by +\index{Matrix}% +\[ +A = \begin{pmatrix} +a_{11} & a_{12} & \cdots & a_{1n} \\ +\Dots{4} \\ +a_{m1} & a_{m2} & \cdots & a_{mn} +\end{pmatrix}. +\] +%% -----File: 127.png---Folio 121------- +By annexing the column composed of the known terms $k_i$ we obtain +the so-called \emph{augmented matrix} +\index{Matrix!augmented}% +\[ +B = \begin{pmatrix} +a_{11} & a_{12} & \cdots & a_{1n} & k_1\\ +\Dots{5}\\ +a_{m1} & a_{m2} & \cdots & a_{mn} & k_m +\end{pmatrix}. +\] + +The definitions of an $r$-rowed minor (determinant) of a matrix and of +the rank of a matrix are entirely analogous to the definitions in~§95. + +In view of Lemma~1 in~§96, our equations~\Eq{15} are inconsistent if +$B$ is of rank $r+1$ and $A$ is of rank $\leqq r$. By Lemma~2, if $A$ and~$B$ are both +of rank~$r$, all of our equations are linear combinations of~$r$ of them. Noting +also that the rank~$r$ of~$A$ cannot exceed the rank of~$B$, since every minor +of~$A$ is a minor of~$B$, and hence a non-vanishing $r$-rowed minor of~$A$ is a +minor of~$B$, so that the rank of~$B$ is not less than~$r$, we have the following + +\begin{Theorem} +A system of $m$~linear equations in $n$~unknowns is consistent +if and only if the rank of the matrix of the coefficients of the unknowns is +equal to the rank of the augmented matrix. If the rank of both matrices is~$r$, +certain $r$ of the equations determine uniquely $r$ of the unknowns as linear +functions of the remaining $n - r$ unknowns, which are independent variables, +and the expressions for these $r$~unknowns satisfy also the remaining +$m - r$ equations. +\index{Determinants!rank}% +\end{Theorem} + +When $m = n+1$, $B$ has an $m$-rowed minor called the \emph{determinant of +the square matrix~$B$}. If this determinant is not zero, $B$ is of rank~$m$. +Since $A$ has no $m$-rowed minor, its rank is less than~$m$. Hence we obtain +the + +\begin{Corollary} +Any system of $n+1$ linear equations in $n$~unknowns is +inconsistent if the determinant of the augmented matrix is not zero. +\end{Corollary} + + +\begin{Exercises}{Page121} + +Discuss the following systems of equations: +\begin{Problems}[2] +\item[1.] $\begin{System}{3} + 2x &+{}& y &+{}& 3z &= 1,\\ + 4x &+{}& 2y &-{}& z &= -3,\\ + 2x &+{}& y &-{}& 4z &= -4,\\ +10x &+{}& 5y &-{}& 6z &= -10. +\end{System}$ + +\item[2.] $\begin{System}{3} +2x &-{}& y &+{}& 3z &= 2,\\ + x &+{}& 7y &+{}& z &= 1,\\ +3x &+{}& 5y &-{}& 5z &= a,\\ +4x &-{}& 3y &+{}& 2z &= 1. +\end{System}$ + +\ResetCols{2} + +\item[3.] $\begin{System}{3} +4x &-{}& y &+{}& z &= 5,\\ +2x &-{}& 3y &+{}& 5z &= 1,\\ + x &+{}& y &-{}& 2z &= 2,\\ +5x & & &-{}& z &= 2. +\end{System}$ + +\item[4.] $\begin{System}{2} + 4x &-{}& 5y &= 2,\\ + 2x &+{}& 3y &= 12,\\ +10x &-{}& 7y &= 16. +\end{System}$ + +\ResetCols{1} + +\item[5.] Prove the Corollary by multiplying the known terms by $x_{n+1}=1$ and applying~§97 +with $n$ replaced by $n+1$. + +\item[6.] Prove that if the matrix of the coefficients of any system of linear homogeneous +\index{Linear equations!homogeneous}% +equations in $n$~unknowns is of rank~$r$, the values of certain $n-r$ of the unknowns may be +%% -----File: 128.png---Folio 122------- +assigned at pleasure and the others will then be uniquely determined and satisfy all of the equations. +\end{Problems} +\end{Exercises} +\index{Linear equations!system|)}% [** PP: Original page range is 114--121] + + +\Section{99.}{Complementary Minors} The determinant +\index{Determinants!complementary minors}% +\[ +D = \begin{vmatrix} +a_1 & b_1 & c_1 & d_1 \\ +a_2 & b_2 & c_2 & d_2 \\ +a_3 & b_3 & c_3 & d_3 \\ +a_4 & b_4 & c_4 & d_4 +\end{vmatrix} +\Tag{16} +\] +is said to have the \emph{two-rowed complementary minors} +\[ +M = \begin{vmatrix} +a_1 & b_1 \\ +a_3 & b_3 +\end{vmatrix},\qquad +M' = \begin{vmatrix} +c_2 & d_2 \\ +c_4 & d_4 +\end{vmatrix}, +\] +since either is obtained by erasing from~$D$ all the rows and columns having +an element which occurs in the other. + +In general, if we erase from a determinant~$D$ of order~$n$ all but $r$~rows +and all but $r$~columns, we obtain a determinant~$M$ of order~$r$ called an +$r$-rowed minor of~$D$. But if we had erased from~$D$ the $r$~rows and $r$~columns +previously kept, we would have obtained an ($n-r$)-rowed minor of~$D$ +called the \emph{minor complementary to~$M$}. In particular, any element is +regarded as a one-rowed minor and is complementary to its minor (of +order $n-1$). + + +% [** PP: Next two paragraphs have single ToC entry, Laplace's Development] +\Section{100.}{Laplace's Development by Columns} +\index{Determinants!Laplace's development|(}% +\begin{Thm} +Any determinant~$D$ is +equal to the sum of all the signed products $±MM'$, where $M$ is an $r$-rowed +minor having its elements in the first $r$~columns of~$D$, and $M'$ is the minor +complementary to~$M$, while the sign is $+$ or~$-$ according as an even or odd +number of interchanges of rows of~$D$ will bring $M$ into the position occupied +by the minor~$M_1$ whose elements lie in the first $r$~rows and first $r$~columns +of~$D$. +\end{Thm} + +\begin{Remark} +For $r = 1$, this development becomes the known expansion of $D$ according to the +elements of the first column~(§90); here $M_1 = e_{11}$. + +If $r=2$ and $D$ is the determinant~\Eq{16}, +\begin{align*} +D &= \begin{vmatrix} +a_1 & b_1 \\ +a_2 & b_2 +\end{vmatrix} · \begin{vmatrix} +c_3 & d_3 \\ +c_4 & d_4 +\end{vmatrix} - \begin{vmatrix} +a_1 & b_1 \\ +a_3 & b_3 +\end{vmatrix} · \begin{vmatrix} +c_2 & d_2 \\ +c_4 & d_4 +\end{vmatrix} + \begin{vmatrix} +a_1 & b_1 \\ +a_4 & b_4 +\end{vmatrix} · \begin{vmatrix} +c_2 & d_2 \\ +c_3 & d_3 +\end{vmatrix} \\[1ex] + &{}+ \begin{vmatrix} +a_2 & b_2 \\ +a_3 & b_3 +\end{vmatrix} · \begin{vmatrix} +c_1 & d_1 \\ +c_4 & d_4 +\end{vmatrix} - \begin{vmatrix} +a_2 & b_2 \\ +a_4 & b_4 +\end{vmatrix} · \begin{vmatrix} +c_1 & d_1 \\ +c_3 & d_3 +\end{vmatrix} + \begin{vmatrix} +a_3 & b_3 \\ +a_4 & b_4 +\end{vmatrix} · \begin{vmatrix} +c_1 & d_1 \\ +c_2 & d_2 +\end{vmatrix}. +\end{align*} +%% -----File: 129.png---Folio 123------- +The first product in the development is $M_1M_1'$; the second product is $-MM'$ (in the +notations of~§99), and the sign is minus since the interchange of the second and third +rows of $D$ brings this $M$ into the position of~$M_1$. The sign of the third product in +the development is plus since two interchanges of rows of~$D$ bring the first factor +into the position of~$M_1$. +\end{Remark} + +If $D$ is the determinant~\Eq{8}, then +\[ +M_1 = \begin{vmatrix} +e_{11} & \cdots & e_{1r} \\ +\Dots{3} \\ +e_{r1} & \cdots & e_{rr} +\end{vmatrix},\qquad +M_1' = \begin{vmatrix} +e_{r+1 r+1} & \cdots & e_{r+1 n}\\ +\Dots{3} \\ +e_{n r+1} & \cdots & e_{nn} +\end{vmatrix}. +\] + +Any term of the product $M_1M_1'$ is of the type +\[ + (-1)^i e_{{i_1}1} e_{{i_2}2} \dotsm e_{{i_r}r} +· (-1)^j e_{i_{r+1}{r+1}} \dotsm e_{{i_n}n}, +\Tag{17} +\] +where $i_1, \dotsc, i_r$ is an arrangement of $1, \dotsc, r$ derived from $1, \dotsc, r$ +by $i$~interchanges, while $i_{r+1}, \dotsc, i_n$ is an arrangement of $r+1, \dotsc, n$ +derived by $j$~interchanges. Hence $i_1, \dotsc, i_n$ is an arrangement of +$1, \dotsc, n$ derived by $i+j$ interchanges, so that the product~\Eq{17} is a term +of $D$ with the proper sign. + +It now follows from~§87 that any term of any of the products~$±MM'$ +mentioned in the theorem is a term of~$D$. Clearly we do not obtain twice +in this manner the same term of~$D$. + +Conversely, any term~$t$ of~$D$ occurs in one of the products~$±MM'$. +Indeed, $t$ contains as factors $r$~elements from the first $r$~columns of~$D$, +no two being in the same row, and the product of these is, except perhaps +as to sign, a term of some minor~$M$. Thus $t$ is a term of $MM'$ or +of~$-MM'$. In view of the earlier discussion, the sign of $t$ is that of the +corresponding term in $±MM'$, where the latter sign is given by the +theorem. + + +% [** PP: No separate ToC entry] +\Section{101.}{Laplace's Development by Rows} There is a Laplace development +of $D$ in which the $r$-rowed minors~$M$ have their elements in the first +$r$~rows of~$D$, instead of in the first $r$~columns as in~§100. To prove this, +we have only to apply~§100 to the equal determinant obtained by interchanging +the rows and columns of~$D$. + +There are more general (but less used) Laplace developments in which +the $r$-rowed minors~$M$ have their elements in any chosen $r$~columns (or +rows) of~$D$. It is simpler to apply the earlier developments to the determinant +$±D$ having the elements of the chosen $r$~columns (or rows) in the +new first $r$~columns (or rows). +\index{Determinants!Laplace's development|)}% + +%% -----File: 130.png---Folio 124------- + + +\begin{Exercises}{} +\begin{Problems} +\item[1.] Prove that +\[ +\begin{vmatrix} +a & b & c & d \\ +e & f & g & h \\ +0 & 0 & j & k \\ +0 & 0 & l & m +\end{vmatrix} = \begin{vmatrix} +a & b \\ +e & f +\end{vmatrix} · \begin{vmatrix} +j & k \\ +l & m +\end{vmatrix}. +\] + +\item[2.] By employing $2$-rowed minors from the first two rows, show that +\[ +\frac{1}{2}\begin{vmatrix} +a & b & c & d \\ +e & f & g & h \\ +a & b & c & d \\ +e & f & g & h +\end{vmatrix} = \begin{vmatrix} +a & b \\ +e & f +\end{vmatrix} · \begin{vmatrix} +c & d \\ +g & h +\end{vmatrix} - \begin{vmatrix} +a & c \\ +e & g +\end{vmatrix} · \begin{vmatrix} +b & d \\ +f & h +\end{vmatrix} + \begin{vmatrix} +a & d \\ +e & h +\end{vmatrix} · \begin{vmatrix} +b & c \\ +f & g +\end{vmatrix} = 0. +\] + +\item[3.] By employing $2$-rowed minors from the first two columns of the $4$-rowed determinant +in Ex.~2, show that the products in Laplace's development cancel. +\end{Problems} +\end{Exercises} + + +\Section{102.}{Product of Determinants} +\index{Determinants!product of}% +\begin{Thm} +The product of two determinants of +the same order is equal to a determinant of like order in which the element +of the $r$th row and $c$th column is the sum of the products of the elements of +the $r$th row of the first determinant by the corresponding elements of the $c$th +column of the second determinant. +\end{Thm} + +\begin{Remark} +For example, +\[ +\begin{vmatrix} +a & b \\ +c & d +\end{vmatrix} · \begin{vmatrix} +e & f \\ +g & h +\end{vmatrix} = \begin{vmatrix} +ae + bg & af + bh \\ +ce + dg & cf + dh +\end{vmatrix}. +\Tag{18} +\] +\end{Remark} + +While for brevity we shall give the proof for determinants of order~$3$, +the method is seen to apply to determinants of any order. By Laplace's +development with $r = 3$~(§101), we have +\[ +\begin{vmatrix} +\Neg a_1 & \Neg b_1 & \Neg c_1 & 0 & 0 & 0 \\ +\Neg a_2 & \Neg b_2 & \Neg c_2 & 0 & 0 & 0 \\ +\Neg a_3 & \Neg b_3 & \Neg c_3 & 0 & 0 & 0 \\ + -1 & \Neg 0 & \Neg 0 & e_1 & f_1 & g_1 \\ +\Neg 0 & -1 & \Neg 0 & e_2 & f_2 & g_2 \\ +\Neg 0 & \Neg 0 & -1 & e_3 & f_3 & g_3 +\end{vmatrix} = \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} · \begin{vmatrix} +e_1 & f_1 & g_1 \\ +e_2 & f_2 & g_2 \\ +e_3 & f_3 & g_3 +\end{vmatrix}. +\Tag{19} +\] + +%% -----File: 131.png---Folio 125------- + +In the determinant of order~$6$, add to the elements of the fourth, fifth, +and sixth columns the products of the elements of the first column by +$e_1$, $f_1$, $g_1$, respectively (and hence introduce zeros in place of the former +elements $e_1$, $f_1$, $g_1$). Next, add to the elements of the fourth, fifth, and +sixth columns the products of the elements of the second column by +$e_2$, $f_2$, $g_2$, respectively. Finally, add to the elements of the fourth, fifth, +and sixth columns the products of the elements of the third column by +$e_3$, $f_3$, $g_3$, respectively. The new determinant is +\[ +\begin{vmatrix} +\Neg a_1 & \Neg b_1 & \Neg c_1 & + a_1e_1+b_1e_2+c_1e_3 & a_1f_1+b_1f_2+c_1f_3 & a_1g_1+b_1g_2+c_1g_3 \\ +% +\Neg a_2 & \Neg b_2 & \Neg c_2 & + a_2e_1+b_2e_2+c_2e_3 & a_2f_1+b_2f_2+c_2f_3 & a_2g_1+b_2g_2+c_2g_3 \\ +% +\Neg a_3 & \Neg b_3 & \Neg c_3 & + a_3e_1+b_3e_2+c_3e_3 & a_3f_1+b_3f_2+c_3f_3 & a_3g_1+b_3g_2+c_3g_3 \\ +% + -1 & \Neg 0 & \Neg 0 & 0 & 0 & 0 \\ +\Neg 0 & -1 & \Neg 0 & 0 & 0 & 0 \\ +\Neg 0 & \Neg 0 & -1 & 0 & 0 & 0 +\end{vmatrix}. +\] +By Laplace's development (or by expansion according to the elements of +the last row, etc.), this is equal to the $3$-rowed minor whose elements +are the long sums. Hence this minor is equal to the product in the right +member of~\Eq{19}. + + +\begin{Exercises}{} + +\begin{Problems} + +\item[1.] Prove~\Eq{18} by means of~§92. + +\item[2.] Prove that, if $s_i = \alpha^i + \beta^i + \gamma^i$, +\[ +\begin{vmatrix} + 1 & 1 & 1 \\ +\alpha & \beta & \gamma \\ +\alpha^2 & \beta^2 & \gamma^2 +\end{vmatrix} · \begin{vmatrix} +1 & \alpha & \alpha^2 \\ +1 & \beta & \beta^2 \\ +1 & \gamma & \gamma^2 +\end{vmatrix} = \begin{vmatrix} +3 & s_1 & s_2 \\ +s_1 & s_2 & s_3 \\ +s_2 & s_3 & s_4 +\end{vmatrix}. +\] + +\item[3.] If $A_i$, $B_i$, $C_i$ are the minors of $a_i$, $b_i$, $c_i$ in the determinant~$D$ defined by the second +factor below, prove that +\[ +\begin{vmatrix} +\Neg A_1 & -A_2 & \Neg A_3 \\ + -B_1 & \Neg B_2 & -B_3 \\ +\Neg C_1 & -C_2 & \Neg C_3 +\end{vmatrix} · \begin{vmatrix} +a_1 & b_1 & c_1 \\ +a_2 & b_2 & c_2 \\ +a_3 & b_3 & c_3 +\end{vmatrix} = \begin{vmatrix} +D & 0 & 0 \\ +0 & D & 0 \\ +0 & 0 & D +\end{vmatrix}. +\] +Hence the first factor is equal to $D^2$ if $D \neq 0$. + +%% -----File: 132.png---Folio 126------- + +\item[4.] Express $(a^2 + b^2 + c^2 + d^2)(e^2 + f^2 + g^2 + h^2)$ as a sum of four squares by writing +\index{Sum of!four squares}% +\[ +\begin{vmatrix} +\Neg a + bi & c + di \\ + -c + di & a - bi +\end{vmatrix} · \begin{vmatrix} +\Neg e + fi & g + hi \\ + -g + hi & e - fi +\end{vmatrix} +\] +as a determinant of order~$2$ similar to each factor. Hint: If $k'$ denotes the conjugate +of the complex number~$k$, each of the three determinants is of the form +\[ +\begin{vmatrix} +\Neg k & l \\ + -l' & k' +\end{vmatrix}. +\] +\end{Problems} +\end{Exercises} + + +\begin{Exercises}[MISCELLANEOUS~]{Page126} + +\begin{Problems} + +\item[1.] Solve +\[ +\begin{System}{3} + ax &+{}& by &+{}& cz &= k,\\ +a^2x &+{}& b^2y &+{}& c^2z &= k^2,\\ +a^4x &+{}& b^4y &+{}& c^4z &= k^4 +\end{System} +\] +by determinants for~$x$, treating all cases. + +\item[2.] In three linear homogeneous equations in four unknowns, prove that the values +of the unknowns are proportional to four determinants of order~$3$ formed from the +coefficients. +\end{Problems} + +Factor the following determinants: +\begin{Problems}[2] + +\item[3.] $\ds +\begin{vmatrix} +1 & a & bc \\ +1 & b & ca \\ +1 & c & ab +\end{vmatrix}$. + +\item[4.] $\ds +\begin{vmatrix} +x & x^2 & yz \\ +y & y^2 & xz \\ +z & z^2 & xy +\end{vmatrix} = \begin{vmatrix} +x^2 & x^3 & 1 \\ +y^2 & y^3 & 1 \\ +z^2 & z^3 & 1 +\end{vmatrix}$. + +\ResetCols{1} + +\item[5.] +\[ +\begin{vmatrix} +a & b & c \\ +c & a & b \\ +b & c & a +\end{vmatrix} = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega), +\] +where $\omega$ is an imaginary cube root of unity. + +\ResetCols{2} + +\item[6.] $\ds +\begin{vmatrix} +a & b & c & d \\ +b & a & d & c \\ +c & d & a & b \\ +d & c & b & a +\end{vmatrix}$. + +\item[7.] $\ds +\begin{vmatrix} +a & b & c & d \\ +d & a & b & c \\ +c & d & a & b \\ +b & c & d & a +\end{vmatrix}$. + +\ResetCols{1} + +% [** PP: Using array for better surrounding space in wide matrix] +\item[8.] If the points $(x_1, y_1), \dotsc, (x_4, y_4)$ lie on a circle, prove that +\[ +\left|\begin{array}{cccc} +x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ +\Dots{4} \\ +x_4^2 + y_4^2 & x_4 & y_4 & 1 +\end{array}\right| = 0. +\] + +%% -----File: 133.png---Folio 127------- + +\item[9.] Prove that +\begin{gather*} +\begin{vmatrix} +aa' + bb' + cc' & ea' + fb' + gc' \\ +ae' + bf' + cg' & ee' + ff' + gg' +\end{vmatrix} \\ +% +{} = \begin{vmatrix} +a & b \\ +e & f +\end{vmatrix} · \begin{vmatrix} +a' & b' \\ +e' & f' +\end{vmatrix} + \begin{vmatrix} +a & c \\ +e & g +\end{vmatrix} · \begin{vmatrix} +a' & c' \\ +e' & g' +\end{vmatrix} + \begin{vmatrix} +b & c \\ +f & g +\end{vmatrix} · \begin{vmatrix} +b' & c' \\ +f' & g' +\end{vmatrix}. +\end{gather*} + +\item[10.] Prove that the cubic equation +\index{Cubic equation}% +\[ +D(x) \equiv \begin{vmatrix} +a-x & b & c \\ +b & f-x & g \\ +c & g & h-x +\end{vmatrix} = 0 +\] +has only real roots. Hints: +\begin{gather*} +D(x) · D(-x) = \left|\begin{array}{lll} +a^2+b^2+c^2-x^2 & ab+bf+cg & ac+bg+ch \\ +ab+bf+cg & b^2+f^2+g^2-x^2 & bc+fg+gh \\ +ac+bg+ch & bc+fg+gh & c^2+g^2+h^2-x^2 +\end{array}\right| \\ +% +{} = -x^6+x^4(a^2+f^2+h^2+2b^2+2c^2+2g^2) - x^2(D_1+D_2+D_3)+ D^2(0), +\end{gather*} +where $D_3$ denotes the first determinant in Ex.~9 with all accents removed and with +$e = b$, while $D_1$ and~$D_2$ are analogous minors of elements in the main diagonal of the +present determinant of order~$3$ with $x = 0$. Hence the coefficient of~$-x^2$ is a sum of +squares. Since the function of degree~$6$ is not zero for a negative value of~$x^2$, $D(x)=0$ +has no purely imaginary root. If it had an imaginary root $r+si$, then $D(x+r)=0$ +would have a purely imaginary root~$si$. But $D(x+r)$ is of the form $D(x)$ with $a$, $f$, $h$ +replaced by $a-r$, $f-r$, $h-r$. Hence $D(x)=0$ has only real roots. The method is +applicable to such determinants of order~$n$. + +\item[11.] If $a_1, \dotsc, a_n$ are distinct, solve the system of equations +\[ +\frac{x_1}{k_i-a_1} + \frac{x_2}{k_i-a_2} + \dotsb + + \frac{x_n}{k_i - a_n} = 1\qquad (i=1, \dotsc, n). +\] + +Hint: Regard $k_1, \dotsc, k_n$ as the roots of an equation of degree~$n$ in $k$ formed from +the typical one above by substituting~$k$ for~$k_i$ and clearing of fractions; write $k = a_j-t$, +and consider the product of the roots of $t^n + \dotsb = 0$. Hence find~$x_j$. + +\item[12.] Solve the equation +\[ +\begin{vmatrix} +a+x & x & x \\ + x & b+x & x \\ + x & x & c+x +\end{vmatrix} = 0. +\] +\end{Problems} +\end{Exercises} +\index{Determinants|)}% + +%% -----File: 134.png---Folio 128------- + + +\Chapter{IX}{Symmetric Functions} + +\index{Sigma function|(}% +\index{Symmetric functions|(}% + +% [** PP: Unit has two separate ToC entries] +\Section{103.}{Sigma Functions, Elementary Symmetric Functions} A rational +function of the independent variables $x_1, x_2, \dotsc, x_n$ is said to be \emph{symmetric} +in them if it is unaltered by the interchange of any two of the variables. +For example, +\[ +x_1^2 + x_2^2 + x_3^2 +4x_1 + 4x_2 +4x_3 +\] +is a symmetric polynomial in $x_1$, $x_2$, $x_3$; the sum of the first three terms +is denoted by $\Sigma x_1^2$ and the sum of the last three by $4\Sigma x_1$. In general, +if $t$ is a rational function of $x_1, \dotsc, x_n, \Sigma t$ denotes the sum of $t$ and all +of the distinct functions obtained from $t$ by permutations of the variables; +such a $\Sigma$-function (read \emph{sigma function}) is symmetric in $x_1, \dotsc, x_n$. +\index{Symbol!f@{$\Sigma$\IndAdd{symmetric function}}}% [** PP: Manually alphabetized] + +For example, if there are three independent variables $\alpha$, $\beta$, $\gamma$, +\begin{gather*} +\begin{aligned} +\Sigma \alpha\beta + &= \alpha\beta + \alpha\gamma + \beta\gamma,\qquad & +\Sigma \alpha^2\beta + &= \alpha^2\beta + \alpha\beta^2 + + \alpha^2\gamma + \alpha\gamma^2 + + \beta^2 \gamma + \beta \gamma^2, \\ +% +\Sigma \frac{1}{\alpha} + &= \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}, & +\Sigma \frac{\beta}{\alpha} + &= \frac{\beta}{\alpha} + \frac{\alpha}{\beta} + + \frac{\beta}{\gamma} + \frac{\gamma}{\beta} + + \frac{\alpha}{\gamma} + \frac{\gamma}{\alpha}, +\end{aligned} \\ % Top-level break +% +\Sigma \frac{\alpha^2 + \beta^2}{\alpha\beta} + = \frac{\alpha^2 + \beta^2}{\alpha\beta} + + \frac{\alpha^2 + \gamma^2}{\alpha\gamma} + + \frac{\beta^2 + \gamma^2}{\beta\gamma}. +\end{gather*} + +In particular, $\Sigma \alpha = \alpha + \beta + \gamma$, $\Sigma \alpha\beta$, and $\alpha\beta\gamma$ are called the three \emph{elementary +symmetric functions} of $\alpha$, $\beta$, $\gamma$. In general, +\index{Elementary symmetric function}% +\[ +\Sigma \alpha_1,\quad +\Sigma \alpha_1\alpha_2,\quad +\Sigma \alpha_1\alpha_2\alpha_3, \dotsc,\quad +\Sigma \alpha_1\alpha_2 \dotsm \alpha_{n-1},\quad +\alpha_1 \alpha_2 \dotsc \alpha_n +\] +are the elementary symmetric functions of $\alpha_1$, $\alpha_2, \dotsc, \alpha_n$. In~§20 they +were written out more fully and proved to be equal to $-c_1$, $c_2$, $-c_3, \dotsc, +(-1)^n c_n$ if $\alpha_1, \dotsc, \alpha_n$ are the roots of the equation +\[ +x^n + c_1x^{n-1} + c_2x^{n-2} + \dotsb + c_n = 0 +\Tag{1} +\] +whose leading coefficient is unity. + +%% -----File: 135.png---Folio 129------- + + +\begin{Exercises}{} + +If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + r = 0$, +so that $\Sigma \alpha = -p$, $\Sigma \alpha\beta = q$, and $\alpha\beta\gamma = -r$, +prove that +\begin{Problems} + +\item[1.] $(\Sigma \alpha)^2 = \Sigma \alpha^2 + 2\Sigma \alpha\beta$, whence $\Sigma \alpha^2 = p^2 - 2q$. + +\item[2.] $\Sigma \alpha · \Sigma \alpha\beta = \Sigma \alpha^2\beta + 3\alpha\beta\gamma$, whence $\Sigma \alpha^2\beta = 3r-pq$. + +\item[3.] $\Sigma \alpha^2\beta\gamma = pr$. + +\item[4.] $\Sigma \alpha^2\beta^2 = (\Sigma \alpha\beta)^2 - 2\alpha\beta\gamma\Sigma \alpha = q^2 - 2pr$. +\end{Problems} + +If $\alpha$, $\beta$, $\gamma$, $\delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, prove that +\begin{Problems} +\item[5.]\qquad\qquad $\Sigma \dfrac{1}{\alpha} = \dfrac{-r}{s}$,\qquad + $\Sigma \dfrac{1}{\alpha\beta} = \dfrac{q}{s}$,\qquad + $\Sigma \dfrac{1}{\alpha^2} = \dfrac{r^2 - 2qs}{s^2}$. \\ +Hint: Compute the sum, sum of the products two at a time, and sum of the squares +of the roots of the equation +\[ +1 + py + qy^2 + ry^3 + sy^4 = 0, +\] +obtained by replacing~$x$ by~$1/y$ in the given quartic equation. + +\item[6.] $\Sigma \dfrac{\beta}{\alpha} = \Sigma \alpha · \Sigma \dfrac{1}{\alpha} - 4 = \dfrac{pr}{s} - 4$. + +\item[7.] $\Sigma \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \Sigma \dfrac{\beta}{\alpha}$. + +\item[8.] $\Sigma \dfrac{\beta\gamma}{\alpha^2} = \Sigma \alpha\beta · \Sigma \dfrac{1}{\alpha^2} - \Sigma \dfrac{\beta}{\alpha} = \dfrac{1}{s^2} (qr^2 - 2q^2s - prs + 4s^2)$. + +\item[9.] $\Sigma \dfrac{\gamma}{\alpha\beta} = \dfrac{3r - pq}{s}$. +\end{Problems} +\end{Exercises} + + +\Section[Fundamental Theorem] +{104.}{Fundamental Theorem on Symmetric Functions.} +\begin{Thm} +Any polynomial +symmetric in $x_1, \dotsc, x_n$ is equal to an integral rational function, +with integral coefficients, of the elementary symmetric functions +\[ +E_1 = \Sigma x_1,\qquad +E_2 = \Sigma x_1x_2,\qquad +E_3 = \Sigma x_1x_2x_3,\dotsc,\qquad +E_n = x_1x_2 \dotsm x_n +\Tag{2} +\] +and the coefficients of the given polynomial. In particular, any symmetric +polynomial with integral coefficients is equal to a polynomial in the elementary +symmetric functions with integral coefficients. +\end{Thm} + +\begin{Remark} +For example, if $n = 2$, +\[ +rx_1^2 + rx_2^2 + sx_1 + sx_2 \equiv r(E_1^2 - 2E_2) + sE_1. +\] +In case $r$ and~$s$ are integers, the resulting polynomial in $E_1$ and~$E_2$ has integral coefficients. +\end{Remark} + +The theorem is most frequently used in the equivalent form: + +%% -----File: 136.png---Folio 130------- + +\emph{Any polynomial symmetric in the roots of an equation, +\[ +x^n - E_1 x^{n-1} + E_2 x^{n-2} - \dotsb + (-1)^n E_n = 0, +\] +is equal to an integral rational function, with integral coefficients, of the coefficients +of the equation and the coefficients of the polynomial.} + +It is this precise theorem that is required in all parts of modern algebra +and the theory of numbers, where attention to the nature of the coefficients +is vital, rather than the inadequate, oft-quoted, theorem that any symmetric +function of the roots is expressible (rationally) in terms of the +coefficients. + +It suffices to prove the theorem for any homogeneous symmetric polynomial~$S$, +i.e., one expressible as a sum of terms % [** PP: Not italicizing i.e.] +\[ +h = ax_1^{k_1} x_2^{k_2} \dotsm x_n^{k_n} +\] +of constant total degree $k = k_1 + k_2 + \dotsb + k_n$ in the~$x$'s. Evidently +we may assume that no two terms of~$S$ have the same set of exponents +$k_1, \dotsc, k_n$ (since such terms may be combined into a single one). We +shall say that $h$ is \emph{higher} than the term $bx_1^{l_1} x_2^{l_2} \dotsm x_n^{l_n}$ if $k_1>l_1$, or if +$k_1 = l_1$, $k_2>l_2$, or if $k_1 = l_1$, $k_2 = l_2$, $k_3>l_3, \dotsc$, so that the first one of +the differences $k_1-l_1$, $k_2-l_2$, $k_3-l_3, \dotsc$ which is not zero is positive. + +We first prove that, if the above term $h$ is the highest term of~$S$, then +\[ +k_1 \geqq k_2 \geqq k_3 \dotsb \geqq k_n. +\] +For, if $k_1<k_2$, the symmetric polynomial~$S$ would contain the term +\[ +ax_1^{k_2} x_2^{k_1} x_3^{k_3} \dotsm x_n^{k_n}, +\] +which is higher than~$h$. If $k_2<k_3$, $S$ would contain the term +\[ +ax_1^{k_1} x_2^{k_3} x_3^{k_2} \dotsm x_n^{k_n}, +\] +which is higher than~$h$, etc. + +If the highest term in another homogeneous symmetric polynomial~$S'$ +is +\[ +h' = a'x_1^{k'_1} x_2^{k'_2} \dotsm x_n^{k'_n}, +\] +and that of $S$ is $h$, then the highest term in their product~$SS'$ is +\[ +hh' = aa'x_1^{k_1 + k'_1} \dotsm x_n^{k_n + k'_n}. +\] +%% -----File: 137.png---Folio 131------- +Indeed, suppose that $SS'$ has a term, higher than~$hh'$, +\[ +cx_1^{l_1 + l'_1} \dotsm x_n^{l_n + l'_n}, +\Tag{3} +\] +which is either a product of terms +\[ +t = b x_1^{l_1} \dotsm x_n^{l_n},\qquad +t' = b'x_1^{l'_1} \dotsm x_n^{l'_n} +\] +of $S$ and~$S'$ respectively, or is a sum of such products. Since~\Eq{3} is higher +than~$hh'$, the first one of the differences +\[ +l_1 + l'_1 - k_1 -k'_1, \dotsc, l_n + l'_n - k_n -k'_n +\] +which is not zero is positive. But, either all of the differences $l_1 - k_1, +\dotsc, l_n - k_n$ are zero or the first one which is not zero is negative, since +$h$ is either identical with $t$ or is higher than~$t$. Likewise for the differences +$l'_1 - k'_1, \dotsc, l'_n - k'_n$. We therefore have a contradiction. + +It follows at once that the highest term in a product of any number +of homogeneous symmetric polynomials is the product of their highest +terms. Now the highest terms in $E_1$, $E_2$, $E_3, \dotsc, E_n$, given by~\Eq{2}, are +\[ +x_1, \qquad x_1x_2,\qquad x_1x_2x_3,\quad\dotsc,\qquad +x_1x_2 \dotsm x_n, +\] +respectively. Hence the highest term in $E_1^{a_1} E_2^{a_2} \dotsm E_n^{a_n}$ is +\[ +x_1^{a_1 + a_2 + \dotsb + a_n} +x_2^{a_2 + \dotsb + a_n} \dotsm +x_n^{a_n}. +\] + +Thus the highest term in +\[ +\sigma = aE_1^{k_1 - k_2} E_2^{k_2 - k_3} \dotsm E_{n-1}^{k_{n-1} - k_n}E_n^{k_n} +\] +is~$h$. Hence $S_1 = S - \sigma$ is a homogeneous symmetric polynomial of the +same total degree~$k$ as~$S$ and having a highest term~$h_1$ not as high as~$h$. +As before, we form a product~$\sigma_1$ of the $E$'s whose highest term is this~$h_1$. +Then $S_2 = S_1 - \sigma_1$ is a homogeneous symmetric polynomial of total degree~$k$ +and with a highest term $h_2$ not as high as~$h_1$. We must finally reach +a difference $S_t - \sigma_t$ which is identically zero. Indeed, there is only a +finite number of products of powers of $x_1, \dotsc, x_n$ of total degree~$k$. +Among these are the parts $h'$, $h'_1$, $h'_2, \dotsc$ of~$h$, $h_1$, $h_2, \dotsc$ with the coefficients +suppressed. Since each $h_i$ is not as high as~$h_{i-1}$, the $h'$, $h'_1$, $h'_2, \dotsc$ are +all distinct. Hence there is only a finite number of~$h_i$. Since $S_t - \sigma_t \equiv 0$, +\[ +S = \sigma + S_1 + = \sigma + \sigma_1 + S_2 = \dotsb + = \sigma + \sigma_1 + \sigma_2 + \dotsb + \sigma_t. +\] +Hence $S$ is a polynomial in $E_1$, $E_2, \dotsc, E_n$ and $a$, $b, \dotsc$, with integral +coefficients. + +%% -----File: 138.png---Folio 132------- + +\begin{Example}[1.] +If $S = \Sigma x_1^2 x_2^2 x_3$ and $n>4$, we have +\begin{align*} +\sigma &= E_2E_3 + = S + 3 \Sigma x_1^2 x_2 x_3 x_4 + 10 \Sigma x_1x_2x_3x_4x_5, \\ +% +S_1 &= S - \sigma + = -3 \Sigma x_1^2 x_2 x_3 x_4 - 10 \Sigma x_1x_2x_3x_4x_5, \\ +% +\sigma_1 &= -3 E_1E_4 + = -3 (\Sigma x_1^2 x_2 x_3 x_4 + 5 \Sigma x_1x_2x_3x_4x_5), \\ +% +S_2 &= S_1 - \sigma_1 + = 5 \Sigma x_1x_2x_3x_4x_5 + = 5E_5, \\ +% +S &= \sigma + S_1 + = \sigma + \sigma_1 + S_2 + = E_2 E_3 - 3 E_1 E_4 + 5E_5. +\end{align*} +\end{Example} + + +\begin{Example}[2.] +If $S = \Sigma x_1^3 x_2 x_3$ and $n> 4$, +\begin{align*} +\sigma = E_1^2 E_3 + &= E_1(\Sigma x_1^2 x_2 x_3 + 4 \Sigma x_1 x_2 x_3 x_4) \\ +% + &= \Sigma x_1^3 x_2 x_3 + 2 \Sigma x_1^2 x_2^2 x_3 + + 3 \Sigma x_1^2 x_2 x_3 x_4 \\ +% + & \phantom{{}=\Sigma x_1^3x_2x_3 } + + 4 (\Sigma x_1^2 x_2 x_3 x_4 + 5 \Sigma x_1x_2x_3x_4x_5), +\end{align*} +\[ +S_1 = S - \sigma + = -2 \Sigma x_1^2 x_2^2 x_3 - 7 \Sigma x_1^2 x_2 x_3 x_4 + -20 \Sigma x_1x_2x_3x_4x_5. +\] +Take $\sigma_1 = -2 E_2E_3$ and proceed as in Ex.~1. +\end{Example} + +\begin{Example}[3.] +By examples 1 and~2, if $n>4$, +\[ +a \Sigma x_1^2 x_2^2 x_3 + +b \Sigma x_1^3 x_2 x_3 + = bE_1^2 E_3 - (3a + b)E_1E_4 + (a - 2b)E_2E_3 + 5(a + b)E_5. +\] +\end{Example} + + +% [** PP: ToC entry matches unit title, using running head] +\Section[Functions Symmetric in all but One Root] +{105.}{Rational Functions Symmetric in all but One of the Roots.} +\index{Symmetric functions!in all but one root|(}% +\begin{Thm} +If +$P$ is a rational function of the roots of an equation $f(x)= 0$ of degree~$n$ and +if $P$ is symmetric in $n-1$ of the roots, then $P$ is equal to a rational function, +with integral coefficients, of the remaining root and the coefficients of $f(x)$ and~$P$. +\end{Thm} + +\begin{Remark} +For example, $P = r \alpha_1 + \alpha_2^2 + \alpha_3^2 + \dotsb + \alpha_n^2$ is symmetric in $\alpha_2, \dotsc, \alpha_n$, and +\[ +P = r \alpha_1 + \Sigma \alpha_1^2 - \alpha_1^2 + = c_1^2 - 2c_2 + r \alpha_1 - \alpha_1^2, +\] +if $\alpha_1, \dotsc, \alpha_n$ are the roots of equation~\Eq{1}. +\end{Remark} + +Since\footnote + {If $N/D$ is symmetric in $\alpha_1$, $\alpha_2$, and the polynomials $N$ and~$D$ have no common + factor, while $N$ becomes~$N'$ and $D$ becomes~$D'$ when $\alpha_1$, $\alpha_2$ are interchanged, then + $ND'\equiv DN'$. Thus $N$ divides~$N'$ and both are of the same degree. Hence $N'=cN, + D' = cD$, where $c$ is a constant. By again interchanging $\alpha_1$, $\alpha_2$, we obtain $N$ from~$N'$, + whence $N = cN' = c^2N$, $c^2 = 1$. If $c = -1$, we take $\alpha_1 = \alpha_2$ and see that $N = N'= -N$, + $N = 0$, whence $N$ has the factor $\alpha_1 - \alpha_2$. Similarly, $D$ has the same factor, contrary to + hypothesis. Hence $c = +1$ and $N$ and~$D$ are each symmetric in $\alpha_1$, $\alpha_2$.} +any symmetric rational function is the quotient of two symmetric +polynomials, the above theorem will follow if proved for the case in which +the words rational function are in both places replaced by polynomial. + +If $\alpha_1$ is the remaining root, the polynomial~$P$ is symmetric in the roots +$\alpha_2, \dotsc, \alpha_n$ of $f(x)/(x - \alpha_1) = 0$, an equation of degree $n-1$ whose coefficients +are polynomials in $\alpha_1$, $c_1, \dotsc, c_n$ with integral coefficients. Hence~(§104), +$P$ is equal to a polynomial, with integral coefficients, in $\alpha_1$, $c_1, \dotsc, c_n$ +and the coefficients of~$P$. + +%% -----File: 139.png---Folio 133------- + +\begin{Example} +If $\alpha$, $\beta$, $\gamma$ are the roots of $f(x) \equiv x^3 + px^2 + qx + r = 0$, find +\[ +\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta} + = \frac{\alpha^2 + \beta^2}{\alpha + \beta} + + \frac{\alpha^2 + \gamma^2}{\alpha + \gamma} + + \frac{\beta^2 + \gamma^2}{\beta + \gamma}. +\] +\end{Example} + +\begin{Solution} +Since $\beta^2 + \gamma^2 = p^2 - 2q - \alpha^2$,\quad $\beta + \gamma = -p - \alpha$, +\[ +\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta} + = \Sigma \frac{p^2 - 2q - \alpha^2}{-p - \alpha} + = \Sigma \left(\alpha - p + \frac{2q}{\alpha + p}\right) + = - p - 3p + 2q \Sigma \frac{1}{\alpha + p}. +\] +But $\alpha + p$, $\beta + p$, $\gamma + p$ are the roots $y_1$, $y_2$, $y_3$ of the cubic equation obtained from +\index{Cubic equation|(}% [** PP: Using page range] +$f(x)=0$ by the substitution $x + p = y$, i.e., $x = y-p$. The resulting equation is % [** PP: Not italicizing i.e.] +\[ +y^3 - 2py^2 + (p^2 + q)y + r - pq = 0. +\] +Since we desire the sum of the reciprocals of $y_1$, $y_2$, $y_3$, we set $y = 1/z$ and find the sum +of the roots $z_1$, $z_2$, $z_3$ of +\[ +1 - 2pz + (p^2 + q)z^2 + (r - pq)z^3 = 0. +\] +Hence +\[ +\Sigma \frac{1}{\alpha + p} + = \Sigma \frac{1}{y_1} + = \Sigma z_1 = \frac{p^2 + q}{pq - r},\qquad +\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta} + = \frac{2q^2 - 2p^2q + 4pr}{pq - r}. +\] +\end{Solution} + + +\begin{Exercises}{Page133} + +[In Exs.~1--12, $\alpha$, $\beta$, $\gamma$ are the roots of $f(x) = x^3 + px^2 + qx + r = 0$.] + +Using $\beta\gamma + \alpha(\beta + \gamma) = q$, find +\begin{Problems}[2] +\item[1.] $\Sigma \dfrac{\beta\gamma + \alpha^2}{\beta + \gamma}$, % [** PP: Added ,] + +\item[2.] $\Sigma \dfrac{3\beta\gamma - 2\alpha^2}{\beta + \gamma - \alpha}$. + +\ResetCols{1} + +\item[3.] Why would the use of $\beta\gamma = -r/\alpha$ complicate Exs.\ 1,~2? Verify that +\[ +\beta\gamma + = \frac{-r}{\alpha} + = \frac{f(\alpha) - r}{\alpha} + = \alpha^2 + p \alpha + q. +\] + +\item[4.] Why would you use $\beta\gamma = -r/\alpha$ in finding + $\Sigma \dfrac{\beta^2 + \gamma^2}{\beta\gamma + c}$? + +\ResetCols{3} + +\item[5.] Find $\Sigma (\beta + \gamma)^2$. + +\item[6.] Find $\Sigma (\alpha + \beta - \gamma)^3$. + +\item[7.] Find $\smash{\Sigma \left(\dfrac{\beta - \gamma}{\beta + \gamma}\right)^2}$. + +\ResetCols{1} + +\item[8.] Find a necessary and sufficient condition on the coefficients that the roots, in +some order, shall be in harmonic progression. +Hint: If $\dfrac{1}{\alpha} + \dfrac{1}{\gamma} = \dfrac{2}{\beta}$, then $\dfrac{-3r}{q} - \beta = 0$, +and conversely. Hence the condition is +\[ +\left(\frac{-3r}{q} - \alpha\right) +\left(\frac{-3r}{q} - \beta\right) +\left(\frac{-3r}{q} - \gamma\right) + = f\left(\frac{-3r}{q}\right) + = 0. +\] + +%% -----File: 140.png---Folio 134------- + +\item[9.] Find the cubic equation with the roots +$\beta\gamma - \dfrac{1}{\alpha}$, +$\alpha\gamma - \dfrac{1}{\beta}$, +$\alpha\beta - \dfrac{1}{\gamma}$. +Hint: since these are $(-r - 1)/\alpha$, etc., make the substitution $(-r - 1)/x = y$. +\end{Problems} + +Find the substitution which replaces the given cubic equation by one with the roots +\begin{Problems} + +\item[10.] +$\alpha\beta + \alpha\gamma$, +$\alpha\beta + \beta\gamma$, +$\alpha\gamma + \beta\gamma$. + +\ResetCols{2} + +\item[11.] $\dfrac{2\alpha - 1}{\beta + \gamma - \alpha}$, etc. + +\item[12.] $\dfrac{\beta\gamma + 3\alpha^2}{\beta + \gamma - 2\alpha}$, etc. +\end{Problems} + +If $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, find +\begin{Problems}[2] + +\item[13.] $\Sigma\dfrac{\beta^2 + \gamma^2 + \delta^2}{\beta + \gamma + \delta}$. + +\item[14.] $\Sigma\dfrac{\beta\gamma + \beta\delta + \gamma\delta}{\beta + \gamma + \delta - 3}$. + +\ResetCols{1} + +\item[15.] Prove that if $y_1$, $y_2$, $y_3$ are the roots of $y^3 + py + q = 0$, the equation with the roots +$z_1 = (y_2 - y_3)^2$, $z_2 = (y_1 - y_3)^2$, $z_3 = (y_1 - y_2)^2$ is +\index{Equation for differences of roots!squares of differences}% +\[ +z^3 + 6pz^2 + 9p^2 z + 4p^3 + 27q^2 = 0. +\] +Hints: since $z_1 = \Sigma y_1^2 - 2y_2y_3 - y_1^2 = -2p + 2q/y_1 - y_1^2$, etc., we set $z = -2p + 2q/y - y^2$. +By the given equation, $y^2 + p + q/y = 0$. Thus the desired substitution is $z = -p + 3q/y$, +$y = 3q/(z + p)$. + +\item[16.] Hence find the discriminant of the reduced cubic equation. +\index{Cubic equation|)}% +\index{Discriminant!of cubic}% + +\item[17.] If $x_1, \dotsc, x_n$ are the roots of $f(x)=0$, show that +\[ +\Sigma \frac{1}{x_1 - c} = \frac{-f'(c)}{f(c)}. +\] +Hint: $x_1 - c = y_1, \dotsc, x_n - c = y_n$ are the roots of +\[ +f(c+y) = f(c) + yf'(c) + y^2(\quad)+ \dotsb = 0, +\] +as shown by Taylor's theorem. Or we may employ~\Eq{5} below % [** PP: Added `below'] +for $x = c$. +\end{Problems} +\end{Exercises} +\index{Symmetric functions!in all but one root|)}% + + +% [** PP: ToC entry appended with ``Newton's Identities''] +\Section{106.}{Sums of Like Powers of the Roots} If $\alpha_1, \dotsc, \alpha_n$ are the roots of +\index{Sum of!like powers of roots|(}% +\[ +f(x) \equiv x^n + c_1 x^{n-1} + c_2x^{n-2} + \dotsb + c_n = 0, +\Tag{1} +\] +we write $s_1$ for~$\Sigma \alpha_1$, $s_2$ for~$\Sigma \alpha_1^2$, and, in general, +\index{Symbol!g@{$s_k$\IndAdd{sum of $k$th powers}}}% [** PP: Manually alphabetized] +\[ +s_k = \Sigma \alpha_1^k + = \alpha_1^k + \alpha_2^k + \dotsb + \alpha_n^k. +\] + +The factored form of~\Eq{1} is +\[ +f(x) \equiv (x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n). +\Tag{4} +\] +%% -----File: 141.png---Folio 135------- +The derivative $f'(x)$ of this product is found by multiplying the derivative +(unity) of each factor by the product of the remaining factors and adding +the results. Hence +\index{Derivative}% +\begin{gather*} +f'(x) = (x - \alpha_2) \dotsm (x - \alpha_n) + + (x - \alpha_1)(x - \alpha_3) \dotsm (x - \alpha_n) + \dotsb, \\ +f'(x) \equiv \frac{f(x)}{x - \alpha_1} + \frac{f(x)}{x - \alpha_2} + \dotsb + + \frac{f(x)}{x - \alpha_n}. +\Tag{5} +\end{gather*} + +If $\alpha$ is any root of~\Eq{1}, $f(\alpha)= 0$ and +\begin{align*} +\frac{f(x)}{x - \alpha} + &= \frac{f(x) - f(\alpha)}{x - \alpha} + = \frac{x^n - \alpha^n}{x - \alpha} + + c_1\frac{x^{n-1} - \alpha^{n-1}}{x - \alpha} + \dotsb + + c_{n-1}\frac{x - \alpha}{x - \alpha} \\ +% + &= x^{n-1} + \alpha x^{n-2} + \alpha^2 x^{n-3} + \dotsb + + c_1(x^{n-2} + \alpha x^{n-3} + \dotsb) \\ + &\hphantom{{}=x^{n-1}} + c_2(x^{n-3} + \dotsb) + \dotsb, +\end{align*} +\[ +\begin{split} +\frac{f(x)}{x - \alpha} + &= x^{n-1} + (\alpha + c_1)x^{n-2} + + (\alpha^2 + c_1 \alpha + c_2)x^{n-3} + \dotsb \\ + &+ (\alpha^k + c_1 \alpha^{k-1} + c_2 \alpha^{k-2} + \dotsb + + c_{k-1} \alpha + c_k)x^{n-k-1} + \dotsb. +\end{split} +\Tag{6} +\] +Taking $\alpha$ to be $\alpha_1, \dotsc, \alpha_n$ in turn, adding the results, and applying~\Eq{5}, we +obtain +\begin{align*} +f'(x) = nx^{n-1} + &+ (s_1 + nc_1)x^{n-2} + (s_2 + c_1 s_1 + nc_2)x^{n-3} + \dotsb \\ + &+ (s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + + c_{k-1} s_1 + nc_k)x^{n-k-1} + \dotsb. +\end{align*} + +The derivative of~\Eq{1} is found at once by the rules of calculus (or by~§56) +to be +\[ +f'(x) = nx^{n-1} + (n-1)c_1 x^{n-2} + (n-2)c_2 x^{n-3} + \dotsb + + (n-k)c_k x^{n-k-1} + \dotsb. +\] +Since this expression is identical term by term with the preceding, we have +\[ +\begin{aligned} +&s_1 + c_1 = 0,\qquad s_2 + c_1 s_1 + 2c_2 = 0, \dotsc, \\ +&s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_{k-1} s_1 + kc_k = 0\quad + (k \leqq n - 1). +\end{aligned} +\Tag{7} +\] + +We may therefore find in turn $s_1$, $s_2, \dotsc, s_{n-1}$: +\[ +s_1 = -c_1,\qquad +s_2 = c_1^2 - 2c_2,\qquad +s_3 = -c_1^3 + 3c_1 c_2 - 3c_3, \dotsc. +\Tag{8} +\] + +To find $s_n$, replace $x$ in~\Eq{1} by $\alpha_1, \dotsc, \alpha_n$ in turn and add the resulting +equations. We get +\[ +s_n + c_1 s_{n-1} + c_2 s_{n-2} + \dotsb + c_{n-1} s_1 + nc_n = 0. +\Tag{9} +\] + +%% -----File: 142.png---Folio 136------- + +We may combine \Eq{7} and~\Eq{9} into +\[ +s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_{k-1} s_1 + kc_k = 0\qquad + (k=1, 2, \dotsc, n). +\Tag{10} +\] + +This set of formulas~\Eq{10} will be referred to as \emph{Newton's identities}. +\index{Newton's!identities}% +The student should be able to write them down from memory and, when +writing them, should always check the final one~\Eq{9} by deriving it as above. + +To derive a formula which shall enable us to compute the $s_k$ for +$k>n$, we multiply~\Eq{1} by~$x^{k-n}$, take $x = \alpha_1, \dotsc, x = \alpha_n$ in turn, and add +the resulting equations. We get +\[ +s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_n s_{k-n} = 0\qquad (k>n). +\Tag{11} +\] + +Instead of memorizing this formula, it is preferable to deduce it for the +particular equation for which it is needed, thus avoiding errors of substitution +as well as confusion with~\Eq{10}. + + +\begin{Example} +Find $s_k$ for $x^n - 1 = 0$. +\end{Example} +\index{Roots of unity}% + +\begin{Solution} +Comparing our equation with~\Eq{1}, we have $c_1 = 0, \dotsc, c_{n-1} = 0$, $c_n = -1$. +Hence in~\Eq{10} for $k<n$, each $c$ is zero and $s_k = 0$. But, for $k = n$, \Eq{10} becomes $s_n - n = 0$. +We may check the latter by substituting each root $\alpha_1, \dotsc, \alpha_n$ in our given equation +and adding. Finally, to find $s_l$ when $l>n$, multiply our equation by~$x^{l-n}$. In the +resulting equation $x^l - x^{l-n} = 0$ we substitute each root, add, and obtain $s_l - s_{l-n} = 0$. +Hence from $s_l$ we obtain an equal $s$ by subtracting $n$ from~$l$. After repeated subtractions, +we reach a value~$k$ for which $1 \leqq k \leqq n$. Since $s_k = 0$ or $n$ according as $k<n$ or +$k = n$, it follows that $s_l = 0$ or $n$ according as $l$ is not or is divisible by~$n$. +\end{Solution} + + +\begin{Exercises}{Page136} + +\begin{Problems} +\item[1.] For a cubic equation, $s_4 = c_1^4 - 4c_1^2 c_2 + 4c_1 c_3 + 2c_2^2$. + +\item[2.] For an equation of degree $\geqq 4$, $s_4 = c_1^4 - 4c_1^2 c_2 + 4c_1 c_3 + 2c_2^2- 4c_4$. + +\item[3.] Find $s_2$, $s_3$, $s_4$, $s_5$ for $x^2 - px + q = 0$. + +\item[4.] Find $s_k$ for $x^5 - 3 = 0$. + +\item[5.] Find $s_2$, $s_3$, $s_6$, $s_7$ for $x^5 - px + q = 0$. +\end{Problems} +\end{Exercises} + + +\Section[Waring's Formula] +{107.}{Waring's Formula for $s_k$ in Terms of the Coefficients.} While +\index{Waring's formula}% +we have learned how to find $s_1$, $s_2$, $s_3, \dotsc$ in turn by Newton's identities, +it is occasionally useful to know an explicit expression for~$s_k$, where $k$ +has an arbitrary value. The formula in question is applied ordinarily +only to a quadratic equation +\[ +x^2 + px + q = 0. +\] +Accordingly we shall treat this case in detail. If its roots are $\alpha$ and~$\beta$, +then +\[ +x^2 + px + q \equiv (x - \alpha)(x - \beta). +\] +%% -----File: 143.png---Folio 137------- +Replace $x$ by~$1/y$ and multiply by~$y^2$. We get +\[ +1 + py + qy^2 \equiv (1 - \alpha y)(1 - \beta y). +\Tag{12} +\] +Taking derivatives, we have +\[ +p + 2qy \equiv - \alpha (1 - \beta y) - \beta (1 - \alpha y). +\] +Change of signs and division by the members of~\Eq{12} gives +\[ +\frac{-p - 2qy}{1 + py + qy^2} + \equiv \frac{\alpha}{1 - \alpha y} + \frac{\beta}{1 - \beta y}. +\Tag{13} +\] + +The identity in Ex.~7, §14, with $n$ changed to~$k$, may be written in +the form +\[ +\frac{1}{1 - r} \equiv 1 + r + r^2 + \dotsb + r^{k-1} + \frac{r^k}{1-r}. +\Tag{14} +\] +Take $r = \alpha y$ and multiply the resulting terms by~$\alpha$; thus +\begin{align*} +\frac{\alpha}{1 - \alpha y} + &= \alpha + \alpha^2 y + \dotsb + \alpha^k y^{k-1} + + \frac{\alpha^{k+1} y^k}{1 - \alpha y}. \\ +\intertext{Similarly,} +\frac{\beta}{1 - \beta y} + &= \beta + \beta^2 y + \dotsb + \beta^k y^{k-1} + + \frac{\beta^{k+1} y^k}{1 - \beta y}. +\end{align*} +To show that on adding, and writing $s_k$ for $\alpha^k + \beta^k$, we obtain~\Eq{15}, we need +the sum of the final fractions, which by~\Eq{12} is +\[ +\frac{\phi y^k}{(1 - \alpha y)(1 - \beta y)} + = \frac{\phi y^k}{1 + py + qy^2},\qquad +\phi \equiv \alpha^{k+1}(1 - \beta y) + \beta^{k+1}(1 - \alpha y). +\] +Hence +\[ +\frac{\alpha}{1 - \alpha y} + \frac{\beta}{1 - \beta y} + = s_1 + s_2 y + \dotsb + s_k y^{k-1} + \frac{\phi y^k}{1 + py + qy^2}, +\Tag{15} +\] +where the exact expression for $\phi$ is immaterial. + +Next, we seek an expansion of the fraction in the left member of~\Eq{13}. +Its denominator will be identical with that in~\Eq{14} if we choose +$r = -py - qy^2$. Evidently~\Eq{14} may be written in the compact form +\[ +\frac{1}{1 - r} \equiv \sum_{t=0}^{k-1} r^t + \frac{r^k}{1 - r}. +\] +Hence it becomes +\[ +\frac{1}{1 + py + qy^2} + = \sum_{t=0}^{k-1} (-1)^t(py + qy^2)^t + \frac{\psi y^k}{1 + py + qy^2}, +\] +%% -----File: 144.png---Folio 138------- +where $\psi = (-p - qy)^k$, although no use will be made of the particular +form of the polynomial~$\psi$. By the binomial theorem, +\[ +(py + qy^2)^t = \sum \frac{(g+h)!}{g!h!} (py)^g(qy^2)^h, +\] +where the summation extends over all sets of integers $g$ and~$h$, each $\geqq 0$, +for which $g + h = t$, while $g!$~denotes the product of $1$, $2, \dotsc, g$ if $g \geqq 1$, +but denotes unity if $g = 0$. Hence +\begin{gather*} +\frac{-p - 2qy}{1 + py + qy^2} + = (p + 2qy) \sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^hy^{g+2h} + E, +\Tag{16} \\ +E \equiv \frac{(-p - 2qy) \psi y^k}{1 + py + qy^2}, +\end{gather*} +where the summation extends over all sets of integers $g$ and~$h$, each $\geqq 0$, +for which $g+h \leqq k-1$. + +Since the left members of \Eq{15} and~\Eq{16} are identically equal by~\Eq{13}, +their right members must be identical, so that the coefficients of $y^{k-1}$ +in them must be equal.\footnote + {In fact, the $(k-1)$th derivatives of the two right members are identical, and we + obtain the indicated result by substituting $y = 0$ in these two derivatives and equating + the results. Note that the final terms in both \Eq{15} and~\Eq{16} have~$y$ as a factor of their + $(k-1)$th derivatives.} % end footnote +Hence the coefficient~$s_k$ of~$y^{k-1}$ in~\Eq{15} is equal +to the coefficient of~$y^{k-1}$ in~\Eq{16}, which is made up of two parts, corresponding +to the two terms of the factor $p + 2qy$. When we use the constant +term~$p$, we must employ from~$\sum$ in~\Eq{16} the terms in which the exponent +of~$y$ is equal to~$k-1$. But when we use the other term~$2qy$, we must +employ from~$\sum$ the terms in which the exponent of~$y$ is equal to~$k-2$, in +order to obtain the combined exponent~$k-1$ of~$y$. Hence $s_k$ is equal to +the sum of the following two parts: +\begin{align*} + p\sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^h && (g+2h &= k-1),\\ +2q\sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^h && (g+2h &= k-2). +\end{align*} +In the upper sum, write $i$ for~$g+1$, and $j$ for~$h$. In the lower sum, write +$i$ for~$g$, and $j$ for~$h+1$. Hence +\[ +s_k = \sum (-1)^{i+j} \frac{(i+j-1)!}{(i-1)!j!} p^iq^j + + 2\sum (-1)^{i+j} \frac{(i+j-1)!}{i!(j-1)!} p^iq^j, +\] +%% -----File: 145.png---Folio 139------- +where now each summation extends over all sets of integers $i$ and~$j$, each +$\geqq 0$, for which +\[ +i + 2j = k. +\Tag{17} +\] + +Finally, we may combine our two sums. Multiply the numerator +and denominator of the first fraction by~$i$, and those of the second fraction +by~$j$. Thus +\[ +s_k = k\sum (-1)^{i+j} \frac{(i+j-1)!}{i!j!} p^iq^j, +\Tag{18} +\] +since the present fraction occurred first multiplied by~$i$ and second multiplied +by~$2j$, and, by~\Eq{17}, the sum of these multipliers is equal to~$k$. Our +final result is~\Eq{18}, where the summation extends over all sets of integers +$i$ and~$j$, each $\geqq 0$, satisfying~\Eq{17}. + +\begin{Remark} +If we replace $i$ by its value $k-2j$, and change the sign of~$p$, we obtain from~\Eq{18} +the result that \emph{the sum of the $k$th powers of the roots of $x^2 - px + q = 0$ is equal to} +\index{Quadratic equation!sum of powers of roots}% +\begin{align*} +s_k &= k\sum_{j=0}^K (-1)^j \frac{(k-j-1)!}{(k-2j)!j!} p^{k-2j} q^j +\Tag{19} \\ + &= p^k - kp^{k-2} q + \frac{k(k-3)}{1·2} p^{k-4} q^2 + - \frac{k(k-4)(k-5)}{1·2·3} p^{k-6} q^3 + \dotsb, % [** PP: . -> ,] +\end{align*} +where $K$ is the largest integer not exceeding~$k/2$. + +The product of the roots is equal to~$q$. Hence if $x$ denotes one root, the second +root is~$q/x$. Thus $s_k = x^k + (q/x)^k$. Again, the sum of the roots is $x + q/x = p$. Regard +$q$ as given and $p$ as unknown. Hence, if $c$ is an arbitrary constant, the equation +\[ +p^k - kqp^{k-2} + \frac{k(k-3)}{1·2} q^2p^{k-4} - \dotsb = c +\Tag{20} +\] +is transformed by the substitution $p = x + q/x$ into +\[ +x^k + \left(\frac{q}{x}\right)^k = c. +\] +Hence equation~\Eq{20} may be solved for $p$ by radicals by the method employed in~§43 +for a cubic equation. +\end{Remark} + +%% -----File: 146.png---Folio 140------- + +The above proof applies\footnote + {See the author's \textit{Elementary Theory of Equations}, pp.~72--74, where there is given + also a shorter proof by means of infinite series.} +without essential change to any equation +$x^n + c_1x^{n-1} + \dotsb + c_n = 0$ and leads to the following formula for the sum +of the $k$th~powers of its roots: +\[ +s_k = k\sum (-1)^{r_1 + \dotsb + r_n} + \frac{(r_1 + \dotsb + r_n-1)!}{r_1! \dotsm r_n!} + c_1^{r_1} \dotsm c_n^{r_n}, +\Tag{21} +\] +where the summation extends over all sets of integers $r_1, \dotsc, r_n$, each +$\geqq 0$, for which $r_1 + 2r_2 + 3r_3 + \dotsb + nr_n = k$. This result~\Eq{21} is known +as \emph{Waring's formula} and was published by him in 1762. + +\begin{Example} +Let $n=3$, $k=4$. Then $r_1 + 2r_2 + 3r_3 = 4$ and +\begin{align*} +(r_1, r_2, r_3) + &= (4, 0, 0),\qquad + (2, 1, 0),\qquad + (1, 0, 1),\qquad + (0, 2, 0), \\ +s_4 &= 4(\frac{3!}{4!} c_1^4 - \frac{2!}{2!1!} c_1^2c_2 + + \frac{1!}{1!1!} c_1c_3 + \frac{1!}{2!} c_2^2) \\ + &= c_1^4 - 4c_1^2c_2 + 4c_1c_3 + 2c_2^2. +\end{align*} +\end{Example} + + +\begin{Exercises}{Page140} + +\begin{Problems} +\item[1.] For the quadratic $x^2 - px + q = 0$ write out the expressions for $s_2$, $s_3$, $s_4$, $s_5$ given by~\Eq{19}, +and compare with those obtained from Newton's identities (Ex.~3, §106). + +\item[2.] Find $s_4$ for a quartic equation by Waring's formula. + +\item[3.] For $k=5$, \Eq{20} becomes De Moivre's quintic $p^5 - 5qp^3 + 5q^2p = c$. Solve it by +radicals for~$p$. +\index{De Moivre's!quintic}% [** PP: Using subitem] + +\item[4.] Solve \Eq{20} by radicals when $k=7$. +\end{Problems} +\end{Exercises} + + +% [** PP: No ToC entry, using running head] +\Section[Computation of Sigma Functions] +{108.}{$\Sigma$-functions Expressed in Terms of the Functions~$s_k$.} Since +we have learned two methods of expressing the $s_k$ in terms of the coefficients, +it is desirable to learn how to express any $\Sigma$-polynomial (and +hence any symmetric function) in terms of the~$s_k$. + +By performing the indicated multiplication, we find that +\[ +s_a s_b \equiv \Sigma \alpha_1^a · \Sigma \alpha_1^b + = \Sigma \alpha_1^{a+b} + m\Sigma \alpha_1^a \alpha_2^b, +\] +where $m = 1$ if $a\neq b$, $m=2$ if $a=b$. Transposing the first term, which +is equal to $s_{a+b}$, and dividing by~$m$, we obtain +\[ +\Sigma \alpha_1^a \alpha_2^b = \frac{1}{m} (s_a s_b - s_{a+b}). +\Tag{22} +\] + +%% -----File: 147.png---Folio 141------- + +In order to compute $\Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2$ in terms of the~$s_k$, we form the product +\[ +\Sigma \alpha_1^4 · \Sigma \alpha_1^3 \alpha_2^2 + = \Sigma \alpha_1^7 \alpha_2^2 + \Sigma \alpha_1^6 \alpha_2^3 + + \Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2. +\] +Making three applications of~\Eq{22}, we get +\[ +s_4(s_3s_2 - s_5) + = (s_7s_2 - s_9) + (s_6s_3 - s_9) + \Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2. +\] +Hence +\[ +\Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2 + = s_2s_3s_4 - s_2s_7 - s_3s_6 - s_4s_5 + 2s_9. +\] + + +\begin{Exercises}{Page141} + +For a quartic equation, express in terms of the $s_k$ and ultimately in terms of the +coefficients $c_1, \dotsc, c_4$: + +\begin{Problems}[4] + +\item[1.] $\Sigma \alpha_1^2 \alpha_2^2$. + +\item[2.] $\Sigma \alpha_1^3 \alpha_2$. + +\item[3.] $\Sigma \alpha_1^2 \alpha_2 \alpha_3$. + +\item[4.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3^2$. + +\ResetCols{1} + +\item[5.] If $a\geqq b > c > 0$, prove that +\[ +\Sigma \alpha_1^a \alpha_2^b \alpha_3^c + = \frac{1}{m} (s_a s_b s_c - s_a s_{b+c} + - s_b s_{a+c} - s_c s_{a+b} + 2s_{a+b+c}), +\] +where $m = 1$ if $a > b$, $m = 2$ if $a = b$. + +\item[6.] $\Sigma \alpha_1^a \alpha_2^b \alpha_3^b + = \frac{1}{2}(s_a s_b^2 - s_as_{2b} - 2s_b s_{a+b} + 2s_{a+2b})$,\qquad $a > b > 0$. + +\item[7.] $\Sigma \alpha_1^a \alpha_2^a \alpha_3^a + = \frac{1}{6}(s_a^3 - 3s_a s_{2a} + 2s_{3a})$,\qquad $a > 0$. +\end{Problems} +\end{Exercises} + + +\Section{109.}{Computation of Symmetric Functions} The method last explained +is practicable when a term of the $\Sigma$-function involves only a few distinct +roots, the largeness of the exponents not introducing a difficulty in the +initial work of expressing the $\Sigma$-function in terms of the~$s_k$. + +But when a term of the $\Sigma$-function involves a large number of roots +with small exponents, we resort to a method suggested by~§104, which +tells us which auxiliary simpler symmetric functions %[** PP: Typo fuctions] +should be multiplied +together to produce our $\Sigma$-function along with simpler ones. + +\begin{Remark} +For example, to find $\Sigma x_1^2x_2x_3x_4$, when $n > 4$, we employ +\begin{gather*} +E_1E_4 \equiv \Sigma x_1 · \Sigma x_1x_2x_3x_4 + = \Sigma x_1^2x_2x_3x_4 + 5 \Sigma x_1x_2x_3x_4x_5, \\ +\Sigma x_1^2x_2x_3x_4 = E_1E_4 - 5E_5. +\end{gather*} + +To find $\Sigma x_1^2x_2^2x_3^2x_4$, employ $E_3E_4 = \Sigma x_1x_2x_3 · \Sigma x_1x_2x_3x_4$. + +When many such products of $\Sigma$-functions are to be computed, it will save time +in the long run to learn and apply the ``method of leaders'' explained in the author's +\textit{Elementary Theory of Equations}, pp.~64--65. +\end{Remark} + +%% -----File: 148.png---Folio 142------- + + +\begin{Exercises}[MISCELLANEOUS~]{Page142} + +Express in terms of the coefficients $c_1, \dotsc, c_n$: +\begin{Problems}[4] + +\item[1.] $\Sigma \alpha_1^2 \alpha_2 \alpha_3$. + +\item[2.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3$. + +% [** PP: Force into four columns] +\item[3.] \makebox[0pt][l]{$\Sigma \alpha_1^2 \alpha_2^2 \alpha_3 \alpha_4$.} + +\item[\quad4.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3^2$. +\end{Problems} + +If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + r = 0$, +find a cubic equation with the roots +\begin{Problems}[3] +\item[5.] $\alpha^2$, $\beta^2$, $\gamma^2$. + +\item[6.] $\alpha\beta$, $\alpha\gamma$, $\beta\gamma$. + +\item[7.] $\dfrac{2}{\alpha}$, $\dfrac{2}{\beta}$, $\dfrac{2}{\gamma}$. + +\ResetCols{2} + +\item[8.] $\alpha^2 + \beta^2$, $\alpha^2 + \gamma^2$, $\beta^2 + \gamma^2$. + +\item[9.] $\alpha^2 + \alpha\beta + \beta^2$, etc. +\end{Problems} + +If $\alpha$, $\beta$, $\gamma$, $\delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, find +\begin{Problems} +\item[10.] $\Sigma \dfrac{\beta}{\alpha} + = \Sigma \dfrac{\beta + \gamma + \delta}{\alpha} + = \Sigma \dfrac{-p - \alpha}{\alpha} + = -4 - p \Sigma \frac{1}{\alpha}$. + +\item[11.] $\Sigma \dfrac{\beta}{\alpha^2}$. Use + $\Sigma \dfrac{1}{\alpha}·\Sigma \dfrac{\beta}{\alpha} + = \Sigma \dfrac{\beta}{\alpha^2} + + 3\Sigma \dfrac{1}{\alpha} + + 2\Sigma \dfrac{\gamma}{\alpha\beta}$. + +\item[12.] Express $\Sigma \alpha_1^a \alpha_2^b \alpha_3^c \alpha_4^d$ in terms of the $s_k$ when (\emph{i})~$a>b>c>d>0$, and (\emph{ii})~when +$a=b=c=d$. + +\item[13.] By solving the first $k$ of Newton's identities~\Eq{10} as a system of linear equations, +find an expression in the form of a determinant (\emph{i})~for $s_k$ in terms of +$c_1, \dotsc, c_k$, and +(\emph{ii})~for $c_k$ in terms of $s_1, \dotsc, s_k$. + +\item[14.] One set of $n$~numbers is a mere rearrangement of another set if $s_1, \dotsc, s_n$ +have the same values for each set. +\end{Problems} +\end{Exercises} +\index{Symmetric functions|)}% +\index{Sigma function|)}% +\index{Sum of!like powers of roots|)}% + +%% -----File: 149.png---Folio 143------- + + +% [** PP: Not matching running head] +\Chapter{X}{Elimination, Resultants And Discriminants} +\index{Elimination|(}% +\index{Resultant|(}% + +% [** PP: Next four units have combined ToC entry] +\Section{110.}{Elimination} If the two equations +\begin{flalign*} +&& ax+b &= 0,\qquad cx+d = 0 && \Rightmark{(a \ne 0,\ c \ne 0)} +\end{flalign*} +are simultaneous, i.e., if $x$ has the same value in each, then % [** PP: Not italicizing i.e.] +\[ +x = -\frac{b}{a} = -\frac{d}{c}, \qquad +R \equiv ad - bc = 0, +\] +and conversely. Hence a necessary and sufficient condition that the +equations have a common root is $R = 0$. We call $R$ the \emph{resultant} (or +\emph{eliminant}) of the two equations. + +The result of eliminating~$x$ between the two equations might equally +well have been written in the form $bc - ad = 0$. But the arbitrary selection +of~$R$ as the resultant, rather than the product of~$R$ by some constant, +as~$-1$, is a matter of more importance than is apparent at first sight. For, +we seek a \emph{definite} function of the coefficients $a$, $b$, $c$, $d$ of the \emph{functions} +$ax+b$, $cx+d$, and not merely a property $R = 0$ or $R\ne 0$ of the corresponding +\emph{equations}. Accordingly, we shall lay down the definition in~§111, +which, as the reader may verify, leads to~$R$ in our present example. + +Methods of elimination which seem plausible often yield not $R$ itself, +but the product of~$R$ by an extraneous function of the coefficients. This +point (illustrated in~§114) indicates that the subject demands a more +careful treatment than is often given. + + +% [** No separate ToC entry] +\Section[Resultant of Two Polynomials] +{111.}{Resultant of Two Polynomials in~$x$.} Let +\begin{flalign*} +&& +\left\{ +\begin{aligned} +f(x) &= a_0x^m + a_1x^{m-1} + \dotsb + a_m \\ +g(x) &= \;b_0x^n + \;b_1x^{n-1} + \dotsb + \, b_n +\end{aligned} +\right. +&& +\begin{aligned} +&\Rightmark{(a_0 \ne 0),} \\ +&\Rightmark{(b_0 \ne 0)} +\end{aligned} +\Tag{1} +\end{flalign*} +be two polynomials of degrees $m$ and~$n$. Let $\alpha_1, \dotsc, \alpha_m$ be the roots +of $f(x) = 0$. Since $\alpha_1$ is a root of $g(x)=0$ only when $g(\alpha_1)=0$, the two +equations have a root in common if and only if the product +\[ +g(\alpha_1)g(\alpha_2) \dotsm g(\alpha_m) +\] +%% -----File: 150.png---Folio 144------- +is zero. This symmetric function of the roots of $f(x)= 0$ is of degree~$n$ +in any one root and hence is expressible as a polynomial of degree~$n$ in the +elementary symmetric functions~(§104), which are equal to $-a_1/a_0$, +$a_2/a_0, \dotsc$. To be rid of the denominators~$a_0$, it therefore suffices to +multiply our polynomial by~$a_0^n$. We therefore define +\[ +R(f, g) = a_0^n g(\alpha_1)g(\alpha_2) \dotsm g(\alpha_m) +\Tag{2} +\] +to be the \emph{resultant} of $f$ and~$g$. It equals an integral rational function of +$a_0, \dotsc, a_m$, $b_0, \dotsc, b_n$ with integral coefficients. +\index{Symbol!h@{$R(f, g)$\IndAdd{resultant}}}% [** PP: Manually alphabetized] + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] If $m = 1$, $n = 2$,\quad + $R(f, g) = b_0 a_1^2 - b_1 a_0 a_1 + b_2 a_0^2$. + +\item[2.] If $m = 2$, $n = 1$,\quad + $R(f,g) = a_0(b_0\alpha_1 + b_1)(b_0\alpha_2 + b_1) + = a_0 b_1^2 - a_1 b_0 b_1 + a_2 b_0^2$, since +\[ + a_0(\alpha_1 + \alpha_2) = -a_1,\qquad a_0 \alpha_1 \alpha_2 = a_2. +\] + +\item[3.] If $\beta_1, \dotsc, \beta_n$ are the roots of $g(x)= 0$, so that +\[ +g(\alpha_i) + = b_0(\alpha_i - \beta_1)(\alpha_i - \beta_2) \dotsm (\alpha_i - \beta_n), +\] +then +\[ +\begin{array}{r@{}l} +R(f,g) = a_0^n b_0^m %[** PP: Added ')' after \beta_2 below] + & (\alpha_1 - \beta_1)(\alpha_1 - \beta_2) \dotsm (\alpha_1 - \beta_n) \\ + ·& (\alpha_2 - \beta_1)(\alpha_2 - \beta_2) \dotsm (\alpha_2 - \beta_n) \\ + & \Dots{1} \\ + ·& (\alpha_m - \beta_1)(\alpha_m - \beta_2) \dotsm (\alpha_m - \beta_n). +\end{array} +\] +Multiplying together the differences in each column, we see that +\[ +R(f, g) = (-1)^{mn} b_0^m f(\beta_1)f(\beta_2) \dotsm f(\beta_n) + = (-1)^{mn} R(g, f). +\] + +\item[4.] If $m=2$, $n=1$,\quad + $R(g,f) = b_0^2 f( -b_1/b_0) = a_0 b_1^2 - a_1 b_0 b_1 + a_2 b_0^2$, +which is equal to $R(f,g)$ +by Ex.~2. This illustrates the final result in Ex.~3. + +\item[5.] If $m=n=2$, %[*Equation split to new line] +\begin{align*} +R(f,g) &= a_0^2 b_0^2 \alpha_1^2 \alpha_2^2 + + a_0^2 b_0 b_1 \alpha_1 \alpha_2 (\alpha_1 + \alpha_2) \\ +% + &\quad + a_0^2 b_0 b_2(\alpha_1^2 + \alpha_2^2) + + a_0^2 b_1^2 \alpha_1 \alpha_2 + + a_0^2 b_1 b_2(\alpha_1 + \alpha_2) + a_0^2 b_2^2 \\ +% + &= b_0^2 a_2^2 - b_0 b_1 a_1 a_2 + b_0 b_2(a_1^2 - 2a_0 a_2) + + b_1^2 a_0 a_2 - b_1 b_2 a_0 a_1 + a_0^2 b_2^2. +\end{align*} +This equals $R(g,f)$, since it is unaltered when the $a$'s and~$b$'s are interchanged. + +\item[6.] Prove by~\Eq{2} that $R$ is homogeneous and of total degree~$m$ in $b_0, \dotsc, b_n$; and +by Ex.~3, that $R$ is homogeneous and of total degree~$n$ in $a_0, \dotsc, a_m$. Show that $R$ +has the terms $a_0^n b_n^m$ and $(-1)^{mn} b_0^m a_m^n$. + +\item[7.] $R(f, g_1 g_2) = R(f, g_1) · R(f, g_2)$. + +\item[8.] $R(f, x^n) = (-1)^{mn} R(x^n, f) = (-1)^{mn} a_m^n$. +\end{Problems} +\end{Exercises} + +%% -----File: 151.png---Folio 145------- + + +% [** PP: No separate ToC entry] +\Section[Sylvester's Method of Elimination] +{112.}{Sylvester's Dialytic Method of Elimination.\protect\footnotemark}% +\addtocounter{footnote}{1}% + \footnotetext{Given without proof by Sylvester, \textit{Philosophical Magazine}, 1840, p.~132.} +\addtocounter{footnote}{-1}% +Let the equations +\index{Sylvester's eliminant}% +\[ +f(x) \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad +g(x) \equiv b_0x^2 + b_1x + b_2 = 0 +\] +have a common root~$x$. Multiply the first equation by $x$ and the second +by $x^2$ and $x$ in turn. We now have five equations +\begin{alignat*}{2} +a_0x^4 + a_1x^3 &+ a_2x^2 + a_3x &&= 0, \\ + a_0x^3 &+ a_1x^2 + a_2x + a_3 &&= 0, \\ +b_0x^4 + b_1x^3 &+ b_2x^2 &&= 0, \\ + b_0x^3 &+ b_1x^2 + b_2x &&= 0, \\ + &\phantom{{}+{}} b_0x^2 + b_1x + b_2 &&= 0, +\end{alignat*} +which are linear and homogeneous in $x^4$, $x^3$, $x^2$, $x$,~$1$. Hence~(§97) +\[ +F = \begin{vmatrix} + a_0 & a_1 & a_2 & a_3 & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3 \\ + b_0 & b_1 & b_2 & 0 & 0 \\ + 0 & b_0 & b_1 & b_2 & 0 \\ + 0 & 0 & b_0 & b_1 & b_2 + \end{vmatrix} +\Tag{3} +\] +must be zero. Next, if $F=0$, there exist~(§97) values which, when +substituted for $x^4$, $x^3$, $x^2$, $x$ and~$1$, satisfy the five equations. But why is +the value for $x^4$ the fourth power of the value for~$x$, that for $x^3$ the cube of +the value for~$x$, etc.? Since the direct verification of these facts would +be very laborious, we resort to a device to show that, conversely, if $F=0$ +the two given equations have a root in common. + +In~\Eq{3} replace $a_3$ by $a_3 - z$ and consider the equation +\[ +\begin{vmatrix} + a_0 & a_1 & a_2 & a_3-z & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3-z \\ + b_0 & b_1 & b_2 & 0 & 0 \\ + 0 & b_0 & b_1 & b_2 & 0 \\ + 0 & 0 & b_0 & b_1 & b_2 +\end{vmatrix} += 0. +\Tag{4} +\] +To prove that it has the roots $f(\beta_1)$ and $f(\beta_2)$, where $\beta_1$ and~$\beta_2$ are the roots +of $g(x) = 0$, we take $z = f(\beta_i)$ and prove that the determinant is then equal +to zero. For, if we add to the last column the products of the elements +%% -----File: 152.png---Folio 146------- +of the first four columns by $\beta_i^4$, $\beta_i^3$, $\beta_i^2$, $\beta_i$, respectively, we find that all +of the elements of the new last column are zero. + +Since \Eq{4} reduces to~\Eq{3} for $z=0$, it is of the form +\[ +b_0^3z^2 + kz + F = 0, +\] +in which the value of $k$ is immaterial. By considering the product of +the roots of this quadratic equation, we see that +\[ +F = b_0^3 f(\beta_1) f(\beta_2). +\] +Hence the Sylvester determinant~$F$ is the resultant $R(g, f)$ and hence +is the resultant $R(f, g)$, since $mn$ is here even (Ex.~3,~§111). +\index{Determinants|(}% + +In general, if the equations are +\[ +f(x) \equiv a_0x^m + \dotsb + a_m = 0,\qquad +g(x) \equiv b_0x^n + \dotsb + b_n = 0, +\] +we multiply the first equation by $x^{n-1}$, $x^{n-2}, \dotsc, x$, $1$, in turn, and the +second by $x^{m-1}$, $x^{m-2}, \dotsc, x$, $1$, in turn. We obtain $n+m$ equations +which are linear and homogeneous in the $m+n$ quantities $x^{m+n-1}, \dotsc, +x$, $1$. Hence the determinant +\[ +F= +\left| +\begin{array}{ccccccccccc} + a_0 & a_1 & a_2 & \Dots{2} & a_m & 0 & \Dots{3} & 0 \\ + 0 & a_0 & a_1 & a_2 & \Dots{2} & a_m & 0 & \Dots{2} & 0 \\ + 0 & 0 & a_0 & a_1 & a_2 & \Dots{2} & a_m & 0 & \dots & 0 \\ + \Dots{11} \\ + 0 & \Dots{2} & 0 & a_0 & a_1 & a_2 & \Dots{3} & a_m \\ + b_0 & b_1 & \Dots{3} & b_n & 0 & \Dots{3} & 0 \\ + 0 & b_0 & b_1 & \Dots{4} & b_n & \Dots{2} & 0 \\ + \Dots{11} \\ + 0 & \dots & 0 & b_0 & b_1 & \Dots{5} & b_n +\end{array} +\right| +% Now set braces, using empty arrays to get correct vertical alignment +{\setlength{\arraycolsep}{0pt} +\begin{array}{c} +\left. +\begin{array}{c} +\\ +\\ +\\ +\\ +\\ +\end{array} +\right\} \text{$n$~rows} \\ +\left. +\begin{array}{c} +\\ +\\ +\\ +\\ +\end{array} +\right\} \text{$m$~rows} +\end{array}} +\Tag{5} +\] +is zero. It may be shown to be equal to the resultant $R(f, g)$, whether +$mn$ is even or odd, by the method employed in the above case $m = 3$, $n = 2$. + +We may also prove as follows that if $F = 0$ the equations $f = 0$ and +$g = 0$ have a common root. Since $F$ was obtained as the determinant +of the coefficients of +\[ +x^{n-1}f, \dotsc, xf, f,\qquad +x^{m-1}g, \dotsc, xg, g, +\] +$F = 0$ implies, by~§96, Lemma~2, the existence of a linear relation +\[ +B_0x^{n-1}f + \dotsb + B_{n-2}xf + B_{n-1}f + +A_0x^{m-1}g + \dotsb + A_{m-2}xg + A_{m-1}g \equiv 0, +\] +%% -----File: 153.png---Folio 147------- +identically in~$x$, with constant coefficients $B_0, \dotsc, A_{m-1}$ not all zero. +In other words, $\beta f + \alpha g \equiv 0$, where +\[ +\alpha\equiv A_0x^{m-1} + \dotsb + A_{m-2}x + A_{m-1},\quad +\beta \equiv B_0x^{n-1} + \dotsb + B_{n-2}x + B_{n-1}. +\Tag{6} +\] + +Neither $\alpha$ nor~$\beta$ is identically zero. For, if $\alpha\equiv 0$, for example, then +$\beta f\equiv 0$ and $\beta\equiv 0$, whereas the $A_i$ and~$B_i$ are not all zero. + +Consider the factored forms of $f$, $g$, $\alpha$, $\beta$. Suppose that $f$ and~$g$ have +no common linear factor. The highest power of each linear factor occurring +in $f$ divides $\alpha g \equiv -\beta f$ and hence divides~$\alpha$. Thus $f$ divides~$\alpha$, whereas +$f$ is of higher degree than~$\alpha$. Hence our assumption that $f = 0$ and $g = 0$ +have no common root has led to a contradiction. + +\begin{Remark} +A similar idea is involved in the method of elimination due to Euler (1707--1783). +If $f=0$ and $g=0$ have a common root~$c$, then $f\equiv (x-c)\alpha$, $-g\equiv (x-c)\beta$, identically in~$x$, +where $\alpha$ and~$\beta$ are polynomials in~$x$ of degrees $m-1$ and~$n-1$, respectively. Give +them the notations~\Eq{6}. In the identity $\beta f + \alpha g \equiv 0$, the coefficient of each power of~$x$ +is zero. Hence +\index{Euler's eliminant}% +\[ +\begin{array}{l@{}r@{}l@{}r@{}l} +a_0B_0 & &{}+b_0A_0 & &=0\\ +a_1B_0+a_0B_1 & &{}+b_1A_0+b_0A_1 & &=0\\ +\Dots{5}\\ + & a_mB_{n-2}+a_{m-1}B_{n-1} & & {}+b_nA_{m-2}+b_{n-1}A_{m-1}&=0\\ + & a_mB_{n-1} & & {}+b_nA_{m-1}&=0. +\end{array} +\] +Since these $m+n$ linear homogeneous equations in the unknowns $B_0, \dotsc, B_{n-1}$, $A_0, \dotsc, +A_{m-1}$ have a set of solutions not all zero, the determinant of the coefficients is zero. %[** PP: , -> .] +By interchanging the rows and columns, we obtain the determinant~\Eq{5}. +% [** PP: Added.] +\end{Remark} + + +\begin{Exercises}{} + +\begin{Problems} +\item[1.] For $m=n=2$, show that the resultant is +\[ +R= +\begin{vmatrix} + a_0 & a_1 & a_2 & 0 \\ + 0 & a_0 & a_1 & a_2 \\ + b_0 & b_1 & b_2 & 0 \\ + 0 & b_0 & b_1 & b_2 +\end{vmatrix} +\] +Interchange the second and third rows, apply Laplace's development, and prove that +\index{Determinants!Laplace's development|(}% +\[ +R= (a_0b_2)^2 - (a_0b_1)(a_1b_2), +\] +where $(a_0b_2)$ denotes $a_0b_2 - a_2b_0$, etc. + +%% -----File: 154.png---Folio 148------- + +\item[2.] For $m = n = 3$, write down the resultant~$R$ and, by interchanges of rows, derive +the second determinant in +\[ +R= +\begin{vmatrix} +a_0 & a_1 & a_2 & a_3 & 0 & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3 & 0 \\ + 0 & 0 & a_0 & a_1 & a_2 & a_3 \\ +b_0 & b_1 & b_2 & b_3 & 0 & 0 \\ + 0 & b_0 & b_1 & b_2 & b_3 & 0 \\ + 0 & 0 & b_0 & b_1 & b_2 & b_3 +\end{vmatrix} += - +\begin{vmatrix} +a_0 & a_1 & a_2 & a_3 & 0 & 0 \\ +b_0 & b_1 & b_2 & b_3 & 0 & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3 & 0 \\ + 0 & b_0 & b_1 & b_2 & b_3 & 0 \\ + 0 & 0 & a_0 & a_1 & a_2 & a_3 \\ + 0 & 0 & b_0 & b_1 & b_2 & b_3 +\end{vmatrix} +\] +To the second determinant apply Laplace's development, selecting minors from the +first two rows, and to the complementary minors apply a similar development. This +may be done by inspection and the following value of~$-R$ will be obtained: +\[ +\begin{split} + & (a_0b_1) \bigl\{(a_1b_2)(a_2b_3) - (a_1b_3)^2+(a_2b_3)(a_0b_3)\bigr\} \\ +-& (a_0b_2) \bigl\{(a_0b_2)(a_2b_3) - (a_0b_3)(a_1b_3)\bigr\} \\ ++& (a_0b_3) \bigl\{(a_0b_1)(a_2b_3) - (a_0b_3)^2\bigr\}. +\end{split} +\] +The third term of the first line and the first term of the last line are alike. Hence, +changing the signs, +\index{Determinants!Laplace's development|)}% [** PP: Original page range is 147--149] +\[ +\begin{split} +R = (a_0b_3)^3 + &- 2(a_0b_1)(a_0b_3)(a_2b_3) - (a_0b_2)(a_0b_3)(a_1b_3) \\ + &+ (a_0b_2)^2 (a_2b_3) + (a_0b_1)(a_1b_3)^2 - (a_0b_1)(a_1b_2)(a_2b_3). +\end{split} +\] +\end{Problems} + +Other methods of simplifying Sylvester's determinant~\Eq{5} are given in~§113. +\end{Exercises} + + +% [** PP: No separate ToC entry] +\Section{113.}{Bézout's Method of Elimination} When the two equations are +of the same degree, the method published by Bézout in~1764 will be clear +from the example +\index{Bézout's eliminant}% +\[ +f \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad +g \equiv b_0x^3 + b_1x^2 + b_2x + b_3 = 0. +\] +Then +\[ +\begin{gathered} +a_0g - b_0f, \\ +(a_0x + a_1)g - (b_0x + b_1)f, \\ +(a_0x^2 + a_1x + a_2)g - (b_0x^2 + b_1x + b_2)f +\end{gathered} +\Tag{7} +\] +are equal respectively to +\begin{alignat*}{4} +(a_0b_1)x^2 & & &+{}& (a_0b_2)\;x &+{}& (a_0b_3) &= 0,\\ +(a_0b_2)x^2 &{}+{}& \bigl\{(a_0b_3) &+{}& (a_1b_2)\bigr\}x &+{}& (a_1b_3) &= 0, +\Tag{8} \\ +(a_0b_3)x^2 & & &+{}& (a_1b_3)\;x &+{}& (a_2b_3) &= 0, +\end{alignat*} +%% -----File: 155.png---Folio 149------- +where $(a_0b_1) = a_0b_1 - a_1b_0$, etc. The determinant of the coefficients is the +negative of the resultant $R(f, g)$. Indeed, the negative of the determinant +is easily verified to have the expansion given at the end of Ex.~2 +just above. + +\begin{Remark} +To give a more instructive proof of the last fact, note that, by~\Eq{7}, equations~\Eq{8} +are linear combinations of +\[ +x^2f = 0,\qquad xf = 0,\qquad f = 0,\qquad +x^2g = 0,\qquad xg = 0,\qquad g = 0, +\] +the latter being the equations used in Sylvester's method of elimination. The determinant +\index{Sylvester's eliminant}% +of the coefficients in these six equations is the first determinant~$R$ in Ex.~2 just +above. The operations carried out to obtain equations~\Eq{8} are seen to correspond +step for step to the following operations on determinants. To the products of the elements +of the fourth row by $a_0$ add the products of the elements of the 1st, 2nd, 3rd, +5th, 6th rows by $-b_0$, $-b_1$, $-b_2$, $a_1$, $a_2$ respectively [corresponding to the formation +of the third function~\Eq{7}]. To the products of the elements of the fifth row by $a_0$ add +the products of the elements of the 2nd, 3rd, 6th rows by $-b_0$, $-b_1$, $a_1$ respectively [corresponding +to the second function~\Eq{7}]. Finally, to the products of the elements of the +sixth row by $a_0$ add the products of the elements of the third row by $-b_0$ [corresponding +to $a_0g - b_0f$]. Hence +\[ +a_0^3R = +\left| +\begin{array}{ccc;{2pt/2pt}ccc;{2pt/2pt}} +a_0 & a_1 & \Bare{a_2} & a_3 & 0 & \Bare{0} \\ + 0 & a_0 & \Bare{a_1} & a_2 & a_3 & \Bare{0} \\ + 0 & 0 & \Bare{a_0} & a_1 & a_2 & \Bare{a_3} \\ + 0 & 0 & 0 & (a_0b_3) & (a_1b_3) & (a_2b_3) \\ + 0 & 0 & 0 & (a_0b_2) & (a_0b_3)+(a_1b_2) & (a_1b_3) \\ + 0 & 0 & 0 & (a_0b_1) & (a_0b_2) & (a_0b_3) +\end{array}\; % Explicit space before right \vert +\right|, +\] +so that $R$ is equal to the $3$-rowed minor enclosed by the dots. The method of Bézout +therefore suggests a definite process for the reduction of Sylvester's determinant of +order~$2n$ (when $m = n$) to one of order~$n$. + +Next, for equations of different degrees, consider the example +\[ +f \equiv a_0x^4 + a_1x^3 + a_2x^2 + a_3x + a_4,\qquad +g \equiv b_0x^2 + b_1x + b_2. +\] +Then +\[ +a_0x^2g - b_0f,\qquad (a_0x + a_1)x^2g - (b_0x + b_1)f +\] +are equal respectively to +\[ +\begin{aligned} +&(a_0b_1)x^3 + (a_0b_2)x^2 - a_3b_0x - a_4b_0,\\ +&(a_0b_2)x^3 + \bigl\{(a_1b_2) - a_3b_0\bigr\}x^2 + - \bigl\{a_3b_1 + a_4b_0\bigr\}x - a_4b_1. +\end{aligned} +\] +The determinant of the coefficients of $x^3$, $x^2$, $x$, $1$ in these two functions and $xg$, $g$, after +the first and second rows are interchanged, is the determinant of order~$4$ enclosed by +dots in the second determinant below. Hence it is the resultant $R(f, g)$. + +%% -----File: 156.png---Folio 150------- + +As in the former example, we shall indicate the corresponding operations on Sylvester's +determinant +\index{Sylvester's eliminant}% +\[ +R= +\begin{vmatrix} + a_0 & a_1 & a_2 & a_3 & a_4 & 0 \\ + 0 & a_0 & a_1 & a_2 & a_3 & a_4 \\ + b_0 & b_1 & b_2 & 0 & 0 & 0 \\ + 0 & b_0 & b_1 & b_2 & 0 & 0 \\ + 0 & 0 & b_0 & b_1 & b_2 & 0 \\ + 0 & 0 & 0 & b_0 & b_1 & b_2 +\end{vmatrix} +\] +Multiply the elements of the third and fourth rows by~$a_0$. In the resulting determinant +$a_0^2R$, add to the elements of the third row the products of the elements of the first, +second and fourth rows by $-b_0$, $-b_1$, $a_1/a_0$ respectively. Add to the elements of the +fourth row the products of those of the second by $-b_0$. We get +\[ +a_0^2R = +\left| +\begin{array}{cc;{2pt/2pt}cccc;{2pt/2pt}} + a_0 & \Bare{a_1} & a_2 & a_3 & a_4 & \Bare{0} \\ + 0 & \Bare{a_0} & a_1 & a_2 & a_3 & \Bare{a_4} \\ + 0 & 0 & (a_0b_2) & (a_1b_2)-a_3b_0 & -a_3b_1-a_4b_0 & -a_4b_1 \\ + 0 & 0 & (a_0b_1) & (a_0b_2) & -a_3b_0 & -a_4b_0 \\ + 0 & 0 & b_0 & b_1 & b_2 & 0 \\ + 0 & 0 & 0 & b_0 & b_1 & b_2 \\ + \end{array}\; +\right| +\] +Hence $R$ is equal to the minor enclosed by dots. +\end{Remark} + + +\begin{Exercises}{} + +\begin{Problems} + +\item[1.] For $m=3$, $n=2$, apply to Sylvester's determinant~$R$ exactly the same operations +as used in the last case in~§113 and obtain +\[ +R= +\begin{vmatrix} +(a_0b_2) & (a_1b_2)-a_3b_0 & -a_3b_1 \\ +(a_0b_1) & (a_0b_2) & -a_3b_0 \\ + b_0 & b_1 & b_2 +\end{vmatrix}. +\] + +\item[2.] For $m=n=4$, reduce Sylvester's~$R$ (as in the first case in~§113) to +\[ +\begin{vmatrix} +(a_0b_1) & (a_0b_2) & (a_0b_3) & (a_0b_4) \\ +(a_0b_2) & (a_0b_3) + (a_1b_2) & (a_0b_4)+(a_1b_3) & (a_1b_4) \\ +(a_0b_3) & (a_0b_4)+(a_1b_3) & (a_1b_4)+(a_2b_3) & (a_2b_4) \\ +(a_0b_4) & (a_1b_4) & (a_2b_4) & (a_3b_4) +\end{vmatrix}. +\] +\end{Problems} +\end{Exercises} + +%% -----File: 157.png---Folio 151------- + + +% [** PP: No ToC entry] +\Section{114.}{General Theorem on Elimination} +\begin{Thm} +If any method of eliminating~$x$ +between two equations in~$x$ leads to a relation $F = 0$, where $F$ is a polynomial +in the coefficients, then $F$ has as a factor the true resultant of the equations. +\end{Thm} + +Some of the preceding proofs become simpler if this theorem is applied. +For example, determinant~\Eq{3} is divisible by the resultant~$R$. Since the +diagonal term of~\Eq{3} is a term $a_0^2b_2^3$ of~$R$ (Ex.~6,~§111), $F$ is identical +with~$R$. + +The preceding general theorem is proved in the author's \textit{Elementary +Theory of Equations}, pp.~152--4. We shall here merely verify the theorem +in an instructive special case. Let +\[ +f \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad +g \equiv b_0x^3 + b_1x^2 + b_2x + b_3 = 0 +\] +have a common root $x\ne 0$. Then +\begin{align*} +-b_0f + a_0g &= (a_0b_1)x^2 + (a_0b_2)x + (a_0b_3), \\ +(b_3f - a_3g)/x &= (a_0b_3)x^2 + (a_1b_3)x + (a_2b_3). +\end{align*} +By Ex.~1 of~§112, the resultant of these two quadratic functions is +\[ +F= +\begin{vmatrix} +(a_0b_3) & (a_0b_1) \\ +(a_2b_3) & (a_0b_3) +\end{vmatrix}^2 +- +\begin{vmatrix} +(a_0b_3) & (a_0b_1) \\ +(a_1b_3) & (a_0b_2) +\end{vmatrix} +· +\begin{vmatrix} +(a_1b_3) & (a_0b_2) \\ +(a_2b_3) & (a_0b_3) +\end{vmatrix}. +\] +This is, however, not the resultant~$R$ of the cubic functions $f$,~$g$. To show +that $(a_0b_3)$ is an extraneous factor, note that the terms of $F$ not having +this factor explicitly are +\index{Elimination!extraneous factor}% +\[ +(a_0b_1) (a_2b_3) \bigl\{(a_0b_1)(a_2b_3) - (a_0b_2)(a_1b_3)\bigr\}. +\] +The quantity in brackets is equal to $-(a_0b_3)(a_1b_2)$, since, as in Ex.~2 +of~§101, +\[ +0 = \tfrac{1}{2} +\begin{vmatrix} + a_0 & a_1 & a_2 & a_3 \\ + b_0 & b_1 & b_2 & b_3 \\ + a_0 & a_1 & a_2 & a_3 \\ + b_0 & b_1 & b_2 & b_3 +\end{vmatrix} += (a_0b_1)(a_2b_3) - (a_0b_2)(a_1b_3) + (a_0b_3)(a_1b_2). +\] +We now see that $F=(a_0b_3)R$, where $R$ is given in Ex.~2 of~§112. This +method of elimination therefore introduces an extraneous factor $(a_0b_3)$. +The student should employ only methods of elimination (such as those +due to Sylvester, Euler, and Bézout) which have been proved to lead +to the true resultant. + +%% -----File: 158.png---Folio 152------- + + +\begin{Exercises}{Page152} + +Find the result of eliminating~$x$ and hence find all sets of common solutions of +\begin{Problems} +\item[1.] $x^2-y^2=9$, $xy = 5y$. + +\item[2.] $x^2 + y^2 = 25$, $x^2 + 3(c-1)x + c(y^2 - 25) = 0$. + +\item[3.] When $x^2 + ax + b = 0$ has a double root, what $3$-rowed determinant is zero? + +\item[4.] Find the roots of $x^6 + 3x^4 + 32x^3 + 67x^2 + 32x + 65 = 0$ by~§79. +\end{Problems} +\end{Exercises} + + +\Section{115.}{Discriminants} Let $\alpha_1, \dotsc, \alpha_m$ be the roots of +\index{Discriminant|(}% +\begin{flalign*} +&& f(x) &\equiv a_0x^m + a_1x^{m-1} + \dotsb + a_m = 0 + && \Rightmark{(a_0\ne 0),} +\Tag{9} \\ +\intertext{so that} +&& f(x) &\equiv a_0(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_m). +\Tag{10} +\end{flalign*} +As in~§44, we define the discriminant of~\Eq{9} to be +\[ +D = a_0^{2m-2}(\alpha_1 - \alpha_2)^2(\alpha_1 - \alpha_3)^2 \dotsm + (\alpha_1 - \alpha_m)^2(\alpha_2 - \alpha_3)^2 \dotsm + (\alpha_{m-1} - \alpha_m)^2. +\] +Evidently $D$ is unaltered by the interchange of any two roots. Since the +degree in any root is $2(m-1)$, the symmetric function~$D$ is equal to a +polynomial in $a_0, \dotsc, a_m$. Indeed, $a_0^{2m-2}$ is the lowest power of $a_0$ +sufficient to cancel the denominators introduced by replacing $\Sigma \alpha_1$ by +$-a_1/a_0, \dotsc, \alpha_1\alpha_2\dotsm \alpha_m$ by~$±a_m/a_0$. By differentiating~\Eq{10}, we see +that +\begin{align*} +f'(\alpha_1) + &= a_0(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3) \dotsm (\alpha_1 - \alpha_m),\\ +f'(\alpha_2) + &= a_0(\alpha_2 - \alpha_1)(\alpha_2 - \alpha_3) \dotsm (\alpha_2 - \alpha_m),\\ +f'(\alpha_3) + &= a_0(\alpha_3 - \alpha_1)(\alpha_3 - \alpha_2)(\alpha_3 - \alpha_4) \dotsm + (\alpha_3 - \alpha_m), +\end{align*} +etc. Hence +\begin{align*} + a_0^{m-1}f'(\alpha_1) \dotsm f'(\alpha_m) + &= a_0^{2m-1} (-1)^{1 + 2 + \dotsb + m-1} + (\alpha_1 - \alpha_2)^2\dotsm (\alpha_{m-1} - \alpha_m)^2 \\ + &= (-1)^{\frac{m(m-1)}{2}} a_0 D. +\end{align*} +By~\Eq{2}, the left member is the resultant of $f(x)$, $f'(x)$. Hence +\[ +D = (-1)^{\frac{m(m-1)}{2}} \frac{1}{a_0} R(f, f'). +\Tag{11} +\] + +%% -----File: 159.png---Folio 153------- + + +\begin{Exercises}{} +\index{Cubic equation|(}% + +\begin{Problems} + +\item[1.] Show that the discriminant of $f \equiv y^3 + py + q = 0$ is $-4p^3 - 27q^2$ by evaluating +the determinant of order five for $R(f, f')$. + +\item[2.] Prove that the discriminant of the product of two functions is equal to the product +of their discriminants multiplied by the square of their resultant. Hint: use +the expressions in terms of the differences of the roots. + +\item[3.] For $a_0 = 1$, show that the discriminant is equal to +\[ +\begin{vmatrix} +1 & \alpha_1 & \alpha_1^2 & \cdots & \alpha_1^{m-1} \\ +1 & \alpha_2 & \alpha_2^2 & \cdots & \alpha_2^{m-1} \\ +\Dots{5} \\ +1 & \alpha_m & \alpha_m^2 & \cdots & \alpha_m^{m-1} +\end{vmatrix}^2 += +\begin{vmatrix} + s_0 & s_1 & s_2 & \cdots & s_{m-1}\\ + s_1 & s_2 & s_3 & \cdots & s_m \\ +\Dots{5}\\ + s_{m-1} & s_m & s_{m+1} & \cdots & s_{2m-2} +\end{vmatrix} +\] +where $s_i = \alpha_1^i + \dotsb + \alpha_m^i$. See Ex.~4,~§88; Ex.~2,~§102. + +\item[4.] Hence verify that the discriminant of $x^3 + px + q = 0$ is equal to +\[ +\begin{vmatrix} + 3 & 0 & -2p \\ + 0 & -2p & -3q \\ +-2p & -3q & 2p^2 +\end{vmatrix} += -4p^3 - 27q^2. +\] + +\item[5.] By means of Ex.~1,~§113, show that the discriminant of $a_0x^3 + a_1x^2 + a_2x + a_3 = 0$ is +\[% [** PP: Not breaking] +-\begin{vmatrix} + 2a_0a_2 & a_1a_2 + 3a_0a_3 & 2a_1a_3 \\ + a_1 & 2a_2 & 3a_3 \\ + 3a_0 & 2a_1 & a_2 +\end{vmatrix} %\\ += 18a_0a_1a_2a_3 - 4a_0a_2^3 - 4a_1^3a_3 + a_1^2a_2^2 - 27a_0^2a_3^2. +\] +\end{Problems} +\end{Exercises} +\index{Discriminant|)}% + + +\begin{Exercises}[MISCELLANEOUS~]{Page153} + +\begin{Problems} +\item[1.] Find the equation whose roots are the abscissas of the points of intersection +of two general conics. + +\item[2.] Find a necessary and sufficient condition that +\[ + f(x) \equiv x^4 + px^3 + qx^2 + rx + s = 0 +\] +shall have one root the negative of another root. When this condition is satisfied, +what are the quadratic factors of~$f(x)$? Apply to Ex.~4,~§74. Hint: add and subtract +$f(x)$ and~$f(-x)$. + +\item[3.] Solve $f(x) \equiv x^4 - 6x^3 + 13x^2 - 14x + 6 = 0$, given that two roots $\alpha$ and~$\beta$ are such +that $2\alpha + \beta = 5$. Hint: $f(x)$ and $f(5-2x)$ have a common factor. + +\item[4.] Solve $x^3 + px + q = 0$ by eliminating $x$ between it and $x^2 + vx + w = y$ by the greatest +common divisor process, and choosing $v$ and~$w$ so that in the resulting cubic equation +for $y$ the coefficients of $y$ and~$y^2$ are zero. The next to the last step of the elimination +%% -----File: 160.png---Folio 154------- +gives $x$ as a rational function of~$y$. (Tschirnhausen, \textit{Acta Erudit.}, Lipsiae,~II, 1683, +p.~204.) + +\item[5.] Find the preceding $y$-cubic as follows. Multiply $x^2 + vx + w = y$ by~$x$ and replace +$x^3$ by~$-px-q$; then multiply the resulting quadratic equation in~$x$ by~$x$ and replace +$x^3$ by its value. The determinant of the coefficients of $x^2$, $x$, $1$ must vanish. + +\item[6.] Eliminate $y$ between $y^3 = v$, $x = ry + sy^2$, and get +\[ +x^3 - 3rsvx - (r^3v + s^3v^2) = 0. +\] +Take $s=1$ and choose %[** PP: Typo chose] +$r$ and~$v$ so that this equation shall be identical with $x^3 + px + q = 0$, +and hence solve the latter. (Euler,~1764.) + +\item[7.] Eliminate $y$ between $y^3 = v$, $x = f + ey + y^2$ and get +\[ +\begin{vmatrix} + 1 & e & f-x \\ + e & f-x & v \\ +f-x & v & ev +\end{vmatrix} +=0. +\] +This cubic equation in $x$ may be identified with the general cubic equation by choice +of $e$, $f$, $v$. % [** PP: , -> .] +Hence solve the latter. + +\item[8.] Determine $r$, $s$ and~$v$ so that the resultant of +\[ +y^3 = v,\qquad y = \frac{x+r}{y+s} +\] +shall be identical with $x^3 + px + q = 0$. (Bézout,~1762.) + +\item[9.] Show that the reduction of a cubic equation in~$x$ to the form $y^3 = v$ by the substitution +\[ +x = \frac{r + sy}{1 + y} +\] +is not essentially different from the method of Ex.~7. [Multiply the numerator and +denominator of~$x$ by $1 - y + y^2$.] + +\item[10.] Prove that the equation whose roots are the $n(n-1)$ differences $x_j-x_k$ of the +roots of $f(x)=0$ may be obtained by eliminating $x$ between the latter and $f(x+y)=0$ +and deleting from the eliminant the factor~$y^n$ (arising from $y = x_j - x_j = 0$). The +equation free of this factor may be obtained by eliminating~$x$ between $f(x)=0$ and +\index{Equation for differences of roots}% +\[ +\bigl\{f(x+y) - f(x)\bigr\}/y + = f'(x) + f''(x)\frac{y}{1·2} + \dotsb + + f^{(n)}(x)\frac{y^{n-1}}{1·2\dotsm n} = 0. +\] +This eliminant involves only even powers of~$y$, so that if we set $y^2 = z$ we obtain an +equation in~$z$ having as its roots the squares of the differences of the roots of $f(x)=0$. +\index{Equation for differences of roots!squares of differences}% +(Lagrange \textit{Résolution des équations}, 1798,~§8.) + +\item[11.] Compute by Ex.~10 the $z$-equation when $f(x) = x^3 + px + q$. +\end{Problems} +\end{Exercises} +\index{Cubic equation|)}% +\index{Determinants|)}% [** PP: Original entry range is 146--153] +\index{Elimination|)}% +\index{Resultant|)}% + +%% -----File: 161.png---Folio 155------- + + +\Appendix +\index{Fundamental theorem of algebra|(}% + +\begin{Theorem} +An equation of degree~$n$ with any complex coefficients +\[ +f(z) \equiv z^n + a_1 z^{n-1} + \dotsb + a_n = 0 +\] +has a complex \(real or imaginary\) root. +\end{Theorem} + +Write $z = x+iy$ where $x$ and~$y$ are real, and similarly $a_1 = c_1 + id_1$, etc. +By means of the binomial theorem, we may express any power of~$z$ in the +form $X+iY$. Hence +\[ +f(z) = \phi(x,y) + i\psi(x,y), +\Tag{1} +\] +where $\phi$ and~$\psi$ are polynomials with real coefficients. + +The first proof of the fundamental theorem was given by Gauss in +1799 and simplified by him in~1849. This simplified proof consists in +showing that the two curves represented by $\phi(x, y) = 0$ and $\psi(x, y) = 0$ +have at least one point $(x_1, y_1)$ in common, so that $z_1 = x_1 + iy_1$ is a root +of $f(z)= 0$. This proof is given in \ChapRef{V} of the author's \textit{Elementary +Theory of Equations}. + +We here give a shorter proof, the initial idea of which was suggested, +but not fully developed, by Cauchy.\footnote + {For a history of the fundamental theorem, see \textit{Encyclopédie des sciences mathématiques}, + tome~I, vol.~II, pp.~189--205.} + +\begin{Lemma}[1.] +$a_1 h + a_2 h^2 + \dotsb + a_n h^n$ is less in absolute value than any +assigned positive number~$p$ for all complex values of $h$ sufficiently small in +absolute value. +\end{Lemma} + +The proof differs from that of the auxiliary theorem in~§62 only in +reading ``in absolute value'' for ``numerically.'' + +We shall employ the notation $|z|$ for the absolute value $+\sqrt{x^2 + y^2}$ of +$z = x + iy$. +\index{Symbol!c@{$\lvert a\rvert$\IndAdd{modulus}}}% [** PP: Manually alphabetized] + +%% -----File: 162.png---Folio 156------- + +\begin{Lemma}[2.] +Given any positive number~$P$, we can find a positive number~$R$ +such that $|f(z)| > P$ if $|z| \geqq R$. +\end{Lemma} + +The proof is analogous to that in~§64. We have +\[ +f(z) = z^n(1+D),\qquad +D \equiv a_1\left(\frac{1}{z}\right) + \dotsb + + a_n\left(\frac{1}{z}\right)^n. +\] +Since (Ex.~5,~§8) the absolute value of a sum of two complex numbers +is equal to or greater than the difference of their absolute values, we have +\[ +|f(z)| \geqq |z|^n \bigl[1 - |D|\bigr]. +\] + +Let $p$ be any assigned positive number~$<1$. Applying Lemma~1 with +$h$ replaced by~$1/z$, we see that $|D| < p$ if $|1/z|$ is sufficiently small, i.e., % [** PP: Not italicizing i.e.] +if $\rho\equiv |z|$ is sufficiently large. Then +\[ +|f(z)| > \rho^n(1-p) \geqq P +\] +if $\rho^n \geqq P/(1-p)$, which is true if +\[ +\rho \geqq \sqrt[n]{\frac{P}{1-p}} \equiv R. +\] +This proves Lemma~2. + +\begin{Lemma}[3.] +Given a complex number a such that $f(a) \ne 0$, we can find +a complex number~$z$ for which $|f(z)| < |f(a)|$. +\end{Lemma} + +Write $z = a+h$. By Taylor's theorem~\Eq{8} of~§56, +\[ +f(a+h) = f(a) + f'(a)h + \dotsb + f^{(r)}(a)·\frac{h^r}{r!} + \dotsb + + f^{(n)}(a)·\frac{h^n}{n!}. +\] +Not all of the values $f'(a)$, $f''(a), \dotsc$ are zero since $f^{(n)}(a) = n!$. Let +$f^{(r)}(a)$ be the first one of these values which is not zero. Then +\[ +\frac{f(a+h)}{f(a)} + = 1 + \frac{f^{(r)}(a)}{f(a)}·\frac{h^r}{r!} + \dotsb + + \frac{f^{(n)}(a)}{f(a)}·\frac{h^n}{n!}. +\] +Writing the second member in the simpler notation +\[ +g(h) \equiv 1 + bh^r + ch^{r+1} + \dotsb + lh^n,\qquad b\ne 0, +\] +we shall prove that a complex value of~$h$ may be found such that $|g(h)| < 1$. +Then the absolute value of $f(z)/f(a)$ will be~$<1$ and Lemma~3 proved. +To find such a value of~$h$, write $h$ and~$b$ in their trigonometric forms~(§4) +\[ +h = \rho(\cos \theta + i \sin \theta),\qquad +b = |b|(\cos \beta + i \sin \beta). +\] +%% -----File: 163.png---Folio 157------- +Then by~§5,~§7,% [** PP: Retaining instead of §§5,~7] +\[ +bh^r = |b| \rho^r \bigl\{\cos(\beta+r\theta) + i\sin (\beta+r\theta)\bigr\}. +\] +Since $h$ is at our choice, $\rho$ and angle~$\theta$ are at our choice. We choose~$\theta$ +so that $b + r\theta = 180°$. Then the quantity in brackets reduces to~$-1$, +whence +\[ +g(h) = (1 - |b|\rho^r) + h^r(ch + \dotsb + lh^{n-r}). +\] +By Lemma~1, we may choose $\rho$ so small that +\[ +|ch + \dotsb + lh^{n-r}| < |b|. +\] +By taking $\rho$ still smaller if necessary, we may assume at the same time +that $|b| \rho^r < 1$. Then +\[ +|g(h)| < (1 - |b|\rho^r) + \rho^r|b|,\qquad |g(h)|<1. +\] + +\paragraph{\indent Minimum Value of a Continuous Function.} Let $F(x)$ be any polynomial +\index{Continuity}% +\index{Minimum}% +with real coefficients. Among the real values of~$x$ for which +$2\leqq x\leqq 3$, there is at least one value~$x_1$ for which $F(x)$ takes its minimum +value~$F(x_1)$, i.e., % [** PP: Not italicizing i.e.] +for which $F(x_1)\leqq F(x)$ for all real values of~$x$ such that +$2\leqq x\leqq 3$. This becomes intuitive geometrically. The portion of the +graph of $y = F(x)$ which extends from its point with the abscissa~$2$ to its +point with the abscissa~$3$ either has a lowest point or else has several +equally low points, each lower than all the remaining points. The arithmetic +proof depends upon the fact that $F(x)$ is continuous for each~$x$ +between $2$ and~$3$ inclusive~(§62). The proof is rather delicate and is +omitted since the theorem for functions of one variable~$x$ is mentioned +here only by way of introduction to our case of functions of two variables. + +We are interested in the analogous question for +\[ +G(x,y) = \phi^2(x, y) + \psi^2(x, y), +\] +which, by~\Eq{1}, is the square of~$|f(z)|$. As in the elements of solid analytic +geometry, consider the surface represented by $Z = G(x,y)$ and the right +circular cylinder $x^2 + y^2 = R^2$. Of the points on the first surface and on +or within their curve of intersection there is a lowest point or there are +several equally low lowest points, possibly an infinite number of them. +Expressed arithmetically, among all the pairs of real numbers $x$,~$y$ for +%% -----File: 164.png---Folio 158------- +which $x^2 + y^2\leqq R^2$, there is\footnote + {Harkness and Morley, \textit{Introduction to the Theory of Analytic Functions}, p.~79, + prove that a real function of two variables which is continuous throughout % [** PP: Typo thoroughout] + a closed + region has a minimum value at some point of the region.} +at least one pair $x_1$,~$y_1$ for which the +polynomial $G(x,y)$ takes a minimum value $G(x_1, y_1)$, i.e., for which % [** PP: Not italicizing i.e.] +$G(x_1, y_1) \leqq G(x, y)$ for all pairs of real numbers $x$,~$y$ for which $x^2 + y^2 \leqq R^2$. + +\paragraph{\indent Proof of the Fundamental Theorem.} Let $z'$ denote any complex +number for which $f(z')\ne 0$. Let $P$ denote any positive number exceeding +$|f(z')|$. Determine $R$ as in Lemma~2. In it the condition $|z|\geqq R$ may +be interpreted geometrically to imply that the point $(x,y)$ representing +$z = x + iy$ is outside or on the circle~$C$ having the equation $x^2 + y^2 = R^2$. +Lemma~2 thus states that, if $z$ is represented by any point outside or on +the circle~$C$, then $|f(z)|>P$. In other words, if $|f(z)|\leqq P$, the point +representing~$z$ is inside circle~$C$. In particular, the point representing~$z'$ +is inside circle~$C$. + +In view of the preceding section on minimum value, we have +\[ +G(x_1,y_1) \leqq G(x,y) +\] +for all pairs of real numbers $x$,~$y$ for which $x^2 + y^2\leqq R^2$, where $x_1$,~$y_1$ is one +such pair. Write $z_1$ for $x_1 + iy_1$. Since $|f(z)|^2 = G(x,y)$, we have +\[ +|f(z_1)|\leqq |f(z)| +\] +for all $z$'s represented by points on or within circle~$C$. Since $z'$ is represented +by such a point, +\[ +|f(z_1)|\leqq |f(z')| < P. +\Tag{2} +\] + +This number $z_1$ is a root of $f(z)=0$. For, if $f(z_1)\neq 0$, Lemma~3 shows +that there would exist a complex number $z$ for which +\[ +|f(z)| < |f(z_1)|. +\Tag{3} +\] +Then $|f(z)| < P$ by~\Eq{2}, so that the point representing~$z$ is inside circle~$C$, +as shown above. By the statement preceding~\Eq{2}, +\[ +|f(z_1)|\leqq |f(z)|. +\] +But this contradicts~\Eq{3}. Hence the fundamental theorem is proved. +\index{Fundamental theorem of algebra|)}% + +%% -----File: 165.png---Folio 159------- + +\PrintAnswers + +\begin{Answers}[5]{Page2}%, 3} + +\item[1.] $3i$. + +\item[2.] $2$. + +\item[3.] $-20 + 20i$. + +\item[4.] $-\frac{2}{3}$. + +\item[5.] $(8 + 2\sqrt{3})$. + +\ResetCols{2} + +\item[6.] $\frac{1}{5}(6 + \sqrt{5}) + \frac{1}{5}(2\sqrt{5} - 3)i$. + +\item[7.] $\dfrac{-9}{13} + \dfrac{19}{13} i$. + +\ResetCols{2} + +\item[8.] $\dfrac{a^2 - b^2}{a^2 + b^2} + \dfrac{2ab}{a^2 + b^2}i$. + +\item[10.] Yes. + +\ResetCols{2} + +\item[13.] $3$, $4$ and $-3$, $-4$. + +\item[14.] $±(5 + 6i)$. + +\ResetCols{2} + +\item[15.] $±(3 - 2i)$. + +\item[16.] $±\bigl[c + d + (c - d)i\bigr]$. +\end{Answers} + + +\begin{Answers}[1]{Page6}%, 7} + +\item[2.] $-3$, $-3\omega$, $-3\omega^2$;\quad $i$, $\omega i$;\quad $\omega^2 i$;\quad + $R = \cos 40° + i\sin 40°$, $\omega R$, $\omega^2 R$. + +\item[3.] $±(1 + i)/\sqrt{2}$;\quad $±(1 - i)/\sqrt{2}$;\quad $±\omega^2$. +\end{Answers} + + +\begin{Answers}[1]{Page9} + +\item[4.] $-1$, $\cos A + i \sin A$ ($A=36°$, $108°$, $252°$, $324°$). + +\item[6.] $R^3$, $R^6$, $R^9$. +\end{Answers} + + +\begin{Answers}{Page10} + +\item[5.] $p(p-1)$. + +\item[6.] $(p-1)(q-1)(r-1)$ if $n=pqr$. +\end{Answers} + + +\begin{Answers}{Page13} + +\item[1.] $51$. + +\item[2.] $13$. +\end{Answers} + + +\begin{Answers}{Page15} + +\item[1.] Rem.~$11$, quot.~$x^2 + 5x + 8$. + +\item[2.] $-61$, $2x^4 - 4x^3 + 7x^2 - 14x + 30$. + +\ResetCols{1} + +\item[3.] $-0.050671$, $x^2 + 6.09x + 10.5481$. + +\item[4.] $x^2 - x - 6$, $x+2$;\quad $4$, $3$, $-2$. + +\item[5.] $x^2 - x - 6 = 0$, $3$, $-2$. + +\ResetCols{3} + +\item[6.] $2±\sqrt{5}$. + +\item[7.] $2x^2 - x + 2$. + +\item[8.] $x^2 + 1$. +\end{Answers} + + +\begin{Answers}{Page17} + +\item[1.] $x^3 - 3x^2 + 2x = 0$. + +\item[2.] $x^4 - 5x^2 + 4 = 0$. + +\ResetCols{2} + +\item[3.] $x^4 - 18x^2 + 81 = 0$. + +\item[4.] $x^4 - 5x^3 + 9x^2 - 7x + 2 = 0$. + +\ResetCols{2} + +\item[5.] $b^2 = 4ac$. + +\item[7.] By theorem in~§18. +\end{Answers} + +%% -----File: 166.png---Folio 160------- + + +\begin{Answers}{Page19} + +\item[1.] $x^3 - 6x^2 + 11x - 6 = 0$. + +\item[2.] $x^4 - 8x^2 + 16 = 0$. + +\ResetCols{4} + +\item[3.] $1$, $2$. + +\item[5.] $4$, $\tfrac{3}{2}$, $-\tfrac{3}{2}$. + +\item[6.] $1$, $3$, $5$. + +\item[7.] $1$, $1$, $1$, $3$. + +\ResetCols{3} + +\item[8.] $2$, $-6$, $18$. + +\item[9.] $-3$, $1$, $5$. + +\item[10.] $5$, $2$, $-1$, $-4$. + +\ResetCols{2} + +\item[11.] $y^2 - (p^2 - 2q)y + q^2 = 0$. + +\item[12.] $y^2 - (p^3 - 3pq)y + q^3 = 0$. + +\ResetCols{1} + +\item[13.] (i) $y^2 - y(p^3 - 3pq)/q + q = 0$. \\ + (ii) $y^2 - q(p^2 - 2q)y + q^4 = 0$. \\ + (iii) $y^2 - (p + p/q)y + 2 + q + 1/q = 0$. + +\ResetCols{2} + +\item[14.] $p^3r = q^3$. + +\item[15.] $2$, $4$, $-6$. +\end{Answers} + + +\begin{Answers}{Page20} + +\item[1.] $5$, $-1±\sqrt{-3}$. + +\item[2.] $1±i$, $1±\sqrt{2}$. + +\ResetCols{2} + +\item[3.] $x^3 - 7x^2 + 19x - 13 = 0$. + +\item[4.] $4$, $1-\sqrt{-5}$, $x^3 - 6x^2 + 14x - 24 = 0$. + +\ResetCols{2} + +\item[6.] $± 1$, $2±\sqrt{3}$. + +\item[7.] $\sqrt{3}$, $2±i$. + +\ResetCols{2} + +\item[9.] $x^3 - \tfrac{3}{2}x^2 - \tfrac{5}{4}x + \tfrac{7}{8} = 0$. + +\item[10.] $2 + \sqrt{3}$, $x^2 + 2x + 2 = 0$. + +\ResetCols{2} + +\item[11,] \textbf{12.}~Not necessarily. + +\item[13.] No. +\end{Answers} + + +\begin{Answers}[5]{Page23} + +\item[1.] $19\tfrac{1}{4}$, $3$. + +\item[2.] $6$. + +\item[3.] $2$. + +\item[4.] $3$. + +\item[5.] $0, -7, -\tfrac{7}{3}$. +\end{Answers} + + +\begin{Answers}{Page25} + +\item[1.] $-1$, $-1$, $-6$. + +\item[2.] $-2$, $3$, $4$. + +\ResetCols{3} + +\item[3.] $1$, $3$, $6$. + +\item[4.] $-2$, $-4$. + +\item[5.] None. +\end{Answers} + + +\begin{Answers}{Page27} + +\item[1.] $2$, $-1$, $-4$, $5$. + +\item[2.] $9$. + +\ResetCols{3} + +\item[3.] $8$, $9$. + +\item[4.] $-12$, $-35$. + +\item[5.] $2$, $2$, $-3$. + +\end{Answers} + + +\begin{Answers}[4]{Page28} + +\item[1.] $1$, $3$, $9$, $\frac{1}{3}$. + +\item[2.] $1$, $\tfrac{1}{2}$, $\tfrac{1}{3}$. + +\item[3.] $-\tfrac{1}{6}$. + +\item[4.] $\tfrac{1}{2}$, $-\tfrac{1}{4}$, $-\tfrac{1}{4}$. + +\ResetCols{4} + +\item[5.] $\tfrac{1}{4}$, $-\tfrac{1}{4}$, $\tfrac{1}{6}$. + +\item[6.] $-\tfrac{1}{2}$, $\tfrac{1}{3}$, $\tfrac{1}{4}$. + +\item[7.] $\tfrac{1}{2}$. + +\item[8.] $\tfrac{2}{3}$. + +\ResetCols{2} + +\item[10.] $x^2 - 12x - 12 = 0$. + +\item[11.] $x^3 - 3x^2 - 12x + 54 = 0$. +\end{Answers} + + +\begin{Answers}[3]{Page30} + +\item[1.] $1$, $4$. + +\item[2.] $-1$, $-4$. + +\item[3.] $0.7$, $-5.7$. + +\ResetCols{3} + +\item[4.] $-0.7$, $5.7$. + +\item[5.] $2$, $2$. + +\item[6.] Imaginary. +\end{Answers} + + +\begin{Answers}[1]{Page40} + +\item[5.] $x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$. + +\item[6.] $-\tfrac{1}{2}(1±\sqrt{-3})$, $\tfrac{1}{2}(7±\sqrt{45})$. + +\item[10.] See~\Eq{11},~§32. + +\item[11.] Edges roots of $x^3 - 7x^2 + 12x - v = 0$, all real~(§45) and irrational. + +\item[14.] $\Delta = \text{area}$, $c = \text{hypotenuse}$, squares of legs $\tfrac{1}{2}(c^2 ± \sqrt{c^4 - 16\Delta^2})$. + +\item[15.] $\Delta$ area, $a$, $b$ given sides, square third side is $a^2 + b^2 ± 2\sqrt{a^2b^2 - 4\Delta^2}$. + +\item[16.] $y^4 - 2y^3 + (2 - g^2)y^2 - 2y + 1 = 0$, pos.\ roots $0.09125$, $10.95862$. +\end{Answers} + +%% -----File: 167.png---Folio 161------- + + +\begin{Answers}[1]{Page44} + +\item[3.] $g=2$, $R + R^8 + R^{12} + R^5$, etc., $z^3 + z^2 - 4z + 1 = 0$. + +\item[4.] $g=2$, $R+R^8$, $R^2+R^7$, $R^4+R^5$. + +\ResetCols{2} + +\item[5.] $\tfrac{1}{2}(1±\sqrt{-3})$, $\tfrac{1}{2}(-5±\sqrt{21})$. + +\item[6.] $-1$, $2±\sqrt{3}$, $\tfrac{1}{2} ± \tfrac{1}{2}\sqrt{-3}$. + +\ResetCols{2} + +\item[7.] $1$, $1$, $1$, $-1$, $\tfrac{1}{4}(1±\sqrt{-15})$. + +\item[8.] $-1$, $-2$, $-\tfrac{1}{2}$, $\tfrac{1}{6}(-5±\sqrt{-11})$. + +\end{Answers} + + +\begin{Answers}{Page46} + +\item[1.] $-5$, $\tfrac{1}{2}(5±\sqrt{-3})$. + +\item[2.] $-6$, $±\sqrt{-3}$. + +\ResetCols{2} + +\item[3.] $-2$, $1± i$. + +\item[4.] $\tfrac{1}{4}$, $\tfrac{1}{7}(-2±\sqrt{-3})$. + +\end{Answers} + + +\begin{Answers}{Page48} + +\item[1.] $\Delta = -400$, one. + +\item[2.] $\Delta = 4 · 27 · 121$, three. + +\ResetCols{2} + +\item[3.] $\Delta = 0$, two. + +\item[4.] $\Delta = 0$, two. + +\end{Answers} + + +\begin{Answers}{Page49} + +\item[1.] $-4$, $2±\sqrt{3}$. + +\item[2.] See Ex.~1,~§47. + +\ResetCols{2} + +\item[3.] $1.3569$, $1.6920$, $-3.0489$. %[** PP: Horizontal layout] + +\item[4.] $-1.201639$, $1.330058$, $-3.128419$. + +\ResetCols{2} + +\item[5.] $1.24698$, $-1.80194$, $-0.44504$. + +\item[6.] $1.1642$, $-1.7729$, $-3.3914$. + +\end{Answers} + + +\begin{Answers}[3]{Page51} + +\item[1.] $1$, $-1$, $4±\sqrt{6}$. + +\item[2.] $-1$, $-2$, $2$, $3$. + +\item[3.] $1± i$, $-1±\sqrt{2}$. + +\ResetCols{2} + +\item[4.] $1±\sqrt{2}$, $-1±\sqrt{-2}$. + +\item[5.] $4$, $-2$, $-1± i$. + +\end{Answers} + + +\begin{Answers}{Page54}% (bottom)} + +\item[1.] $(-3, 9)$. + +\item[2.] $\Delta=-250000$, $x=3$, $-2$, $±i$. + +\ResetCols{3} + +\item[3.] $(3,9)$, $(-2,4)$. + +\item[4.] $h=3$. + +\item[5.] $6.856$, $7$. + +\end{Answers} + + +\begin{Answers}[1]{Page59}%, 60} + +\item[2.] $2.1$. + +\item[3.] $(-0.845, 4.921)$, $(-3.155, 11.079)$;\quad between $-4$ and~$-5$. + +\item[4.] $1.1$, $-1.3$. %[** PP: Added .] + +\item[5.] Between $0$ and~$1$, $0$ and~$-1$, $2.5$ and~$3$, $-2.5$ and~$-3$. + +\item[9.] $120(x^3 + x)$, $120x^2 - 42$. + +\end{Answers} + + +\begin{Answers}[3]{Page62} + +\item[1.] $3$. + +\item[2.] $2$, $-2$. + +\item[3.] $-1$. + +\ResetCols{3} + +\item[4.] Double roots, $1$, $3$. + +\item[5.] None. + +\item[6.] $3$, $3$, $-3$, $6$. + +\end{Answers} + + +\begin{Answers}{Page64}%, 65} + +\item[3.] Use Ex.~3, p.~62, abscissas $-1$, $3$. + +\item[\qquad4.] Use Ex.~2, p.~62. + +\ResetCols{1} + +\item[6.] $y = -15x - 7$, $X^3 - 15X + 23 = 0$. + +\end{Answers} + +%% -----File: 168.png---Folio 162------- + + +\begin{Answers}{Page66} + +\item[1.] One real. + +\item[2.] $(±\sqrt{\frac{7}{3}}, 7\mp\frac{14}{3}\sqrt{\frac{7}{3}})$, three real. + +\ResetCols{2} + +\item[3.] $(±\sqrt\frac{2}{3}, -1\mp\frac{4}{3}\sqrt{\frac{2}{3}})$, three. + +\item[4.] $(-2±\sqrt{5}, 23\mp10\sqrt{5})$, one. +\end{Answers} + + +\begin{Answers}[1]{Page74}%, 75} + +\item[13.] $y^5 + 2y^4 + 5y^3 + 3y^2 - 2y - 9 = 0$. + +\item[14.] $y^3 + 15y^2 + 52y - 36 = 0$. + +\end{Answers} + + +\begin{Answers}{Page78} + +\item[1.] One, between $-2$ and~$-3$. + +\item[2.] One, between $1$ and~$2$. + +\end{Answers} + + +\begin{Answers}{Page79}%, 80} + +\item[1.] $(-4, -3)$, $(-2, -1)$, $(1, 2)$. + +\item[2.] $(-2, -1)$, $(0, 1)$. + +\ResetCols{2} + +\item[3.] $(-2, -1.5)$, $(-1.5, -1)$, $(3, 4)$. + +\item[4.] $(-2, -1)$, $(0, 1)$. + +\ResetCols{2} + +\item[5.] $(-7, -6)$, $(1, 2)$. + +\item[6.] $(0, 1)$, $(3, 4)$. + +\end{Answers} + + +\begin{Answers}[3]{Page83} + +\item[2.] $1$, $1$, $1$, $2$. + +\item[3.] $1$, $1$, $-2$, $-2$. + +\item[4.] $1$, $1$, two imaginary. + +\end{Answers} + + +\begin{Answers}{Page85} + +\item[1.] $(-2, -1)$, $(0, 1)$, $(1, 2)$. + +\item[2.] $(-4, -3)$, $(-2, -1)$, $(1, 2)$. + +\end{Answers} + + +\begin{Answers}{Page89}%, 90} + +\item[1.] Single, $-2.46955$. + +\item[2.] $-1.20164$, $1.33006$, $-3.12842$. + +\ResetCols{2} + +\item[3.] $1.24698$, $-1.80194$, $-0.44504$. + +\item[4.] $± 2.1213203$, $\Neg2.1231056$, $-6.1231056$. + +\ResetCols{2} + +\item[5.] $3.45592$, $21.43067$. + +\item[6.] $2.15443$. + +\ResetCols{1} + +\item[7.] $-1.7728656$, $\Neg1.1642479$, $-3.3913823$. + +\item[8.] $\Neg3.0489173$, $-1.3568958$, $-1.6920215$. + +\ResetCols{2} + +\item[9.] $2.24004099$. + +\item[10.] $1.997997997$. + +\ResetCols{2} + +\item[11.] $1.094551482$. + +\item[12.] $2.059$, $-1.228$. + +\ResetCols{2} + +\item[13.] $1.2261$. + +\item[14.] $0.6527 = \text{reciprocal of } 2 \cos 40°$. + +\ResetCols{3} + +\item[15.] $0.9397$. + +\item[16.] $1.3500$. + +\item[17.] $2.7138$, $3.3840$. + +\ResetCols{3} + +\item[18.] $5.46\%$. + +\item[19.] $5.57\%$. + +\item[20.] $9.70\%$. + +\end{Answers} + + +\begin{Answers}{Page94} + +\item[1.] $2$. + +\item[2.] $3$. + +\end{Answers} +%% -----File: 169.png---Folio 163------- + + +\begin{Answers}[3]{Page96} + +\item[1.] $2.24004099$. + +\item[2.] $2.3593041$. + +\item[3.] $1.997998$. + +\end{Answers} + + +\begin{Answers}[4]{Page98} + +\item[1.] $132° 20.7'$. + +\item[2.] $157° 12'$. + +\item[3.] $4.8425364$. + +\item[4.] $3.1668771$. + +\ResetCols{1} + +\item[5,] \textbf{7.}~$15° 16\tfrac{1}{2}'$, + $85° 56\tfrac{1}{2}'$, + $212° 49'$, + $225° 57'$. + +\ResetCols{4} + +\item[6.] $72° 17'$. + +\item[8.] $5° 56\tfrac{1}{2}'$, $25° 18'$. + +\item[9.] $2.5541949$. + +\item[10.] $1.85718$. + +\end{Answers} + + +\begin{Answers}{Page99} + +\item[1.] $-1.04727± 1.13594 i$. + +\item[2.] $-\frac{2}{7} ± \frac{1}{7}\sqrt{3}i$. + +\ResetCols{3} + +\item[3.] $-1±i$. + +\item[4.] $1±i$, $1±2i$. + +\item[5.] $2±i$, $±2i$. + +\end{Answers} + + +\begin{Answers}{Page100} + +\item[1.] $217° 12' 27.4'' = 3.790988$ radians. + +\item[2.] $42° 20' 47\tfrac{1}{4}''$ doubled. + +\ResetCols{3} + +\item[3.] $133° 33.8'$. + +\item[4.] $108° 36' 14''$. + +\item[5.] $21.468212$. + +\ResetCols{2} + +\item[6.] Angle at center $47° 39' 13''$. + +\item[7.] $49° 17' 36.5''$. + +\ResetCols{1} + +\item[8.] $1.4303\pi$, $2.4590\pi$, $3.4709\pi$;\quad $257° 27' 12.225''$ more exact than first. + +\item[9.] $x/\pi = 0.6625, 1.891, 2.930, 3.948, 4.959$. + +\item[10.] (i) $0.327739$, $0.339224$, $1124.333037$. \\ + (ii) $0.250279$, $0.894609$, $1.127839$. \hfill\break + % [** PP: If \\ above, LaTeX thinks next token is an optional argument] + [Set $x = 1 + y$, $y = 1/z$ and solve by trigonometry.] + +\ResetCols{2} + +\item[11.] $3.597285$. + +\item[12.] $10$, $1.371288$. + +\ResetCols{2} + +\item[13.] $0.326878$, $12.267305$. + +\item[14.] $324° 16' 29.55''$. + +\ResetCols{3} + +\item[15.] $10$~yr.\ $4$~mo.\ $0$~days. + +\item[\qquad16.] $6.074674$. + +\item[17.] $6.13$\%. + +\end{Answers} + + +\begin{Answers}[3]{Page102} + +\item[1.] $x = 5$, $y = 6$. + +\item[2.] $x = 2$, $y = 1$. + +\item[3.] $x = a$, $y = 0$. + +\end{Answers} + + +\begin{Answers}[1]{Page106} + +\item[1.] $-a_2b_1c_3d_4 + a_2b_1c_4d_3 + a_2b_3c_1d_4 + - a_2b_3c_4d_1 - a_2b_4c_1d_3 + a_2b_4c_3d_1$. + +\item[2.] $+$, $+$. +\end{Answers} + + +\begin{Answers}{Page112} + +\item[3.] $-3$. + +\item[4.] $-8$. + +\end{Answers} + + +\begin{Answers}{Page115} + +\item[1.] $x = -8$, $y = -7$, $z = 26$. + +\item[2.] $x = 3$, $y = -5$, $z = 2$. + +\ResetCols{2} + +\item[3.] $x = 6$, $y = 3$, $z = 12$. + +\item[4.] $x = 5$, $y = 4$, $z = 3$. + +\ResetCols{2} + +\item[5.] $x = -5$, $y = 3$, $z = 2$, $w = 1$. + +\item[6.] $x = 1$, $y = z = 0$, $w = -1$. + +\end{Answers} + +%% -----File: 170.png---Folio 164------- + + +\begin{Answers}[1]{Page119} + +\item[1.] Consistent: $y = -8/7 - 2x$, $z = 5/7$ (common line). + +\ResetCols{2} + +\item[2.] Inconsistent, case $(\beta)$. + +\item[3.] Inconsistent (two parallel planes). + +\ResetCols{1} + +\item[4.] Consistent (single plane). + +\item[5.] % [** PP: Changing ;s to .s] + (i) $z = -x-y-2$. \hfill + (ii) inconsistent. \hfill + (iii) $x = \dfrac{a - 1}{a + 2}$, $y = z =\dfrac{-3}{a + 2}$. + +\item[6.] + (i) $x = \dfrac{(k-b)(c-k)}{(a-b)(c-a)}$. + (ii) $y = \dfrac{k-c}{a-c}-x$, + $z = \dfrac{a-k}{a-c}$ if $k=a$ or $k=c$, but + inconsistent if $k$ is different from $a$ and~$c$. + (iii) $z = 1 - x - y$ if $k=a$, inconsistent if $k\ne a$. + +\end{Answers} + + +\begin{Answers}{Page120} + +\item[1.] $r = 2$, $x:y:z = -4:1:1$. + +\item[2.] $r = 2$, $x:y:z = -10:8:7$. + +\ResetCols{2} + +\item[3.] $r = 1$, two unknowns arbitrary. + +\item[4.] $r = 3$, $x:y:z:w = 6:3:12:1$. + +\ResetCols{1} + +\item[5.] $r = 2$, $z = -\frac{11}{3} x - \frac{19}{3} y$, + $w = -\frac{10}{3} x - \frac{17}{3} y$. + +\end{Answers} + + +\begin{Answers}[1]{Page121} + +\item[1.] Ranks of $A$ and~$B$ are~$2$;\quad $y = -8/7 - 2x, z = 5/7$. + +\item[2.] Consistent only when $a = -225/61$ and then $x = -\dfrac{5}{61}$, $y = \dfrac{3}{61}$, $z = \dfrac{45}{61}$. + +\item[3.] Rank of~$A$ is~$2$, rank of~$B$ is~$3$, inconsistent. + +\item[4.] $A$ and~$B$ of rank~$2$, $x = 3$, $y = 2$. + +\end{Answers} + + +\begin{Answers}[1]{Page126}%, 127} + +\item[1.] $x = \dfrac{k(b-k)(c-k)(k+b+c)}{a(b-a)(c-a)(a+b+c)}$, if $a$, $b$, $c$ are distinct and not zero and their +sum $\neq 0$. If $a = b \neq c$, $ac \ne 0$, equations are inconsistent unless $k = 0$, $a$, $c$, or $-a-c$, +and then $y = \dfrac{k(c-k)}{a(c-a)} - x$, $z = \dfrac{k(k-a)}{c(c-a)}$, $x$~arbitrary. + +\ResetCols{2} + +\item[3.] $(a-b)(b-c)(c-a)$. + +\item[4.] $(x-y)(y-z)(z-x)(xy + yz + zx)$. + +\ResetCols{1} + +\item[6.] $(a+b+c+d)(a+b-c-d)(a-b-c+d)(a-b+c-d)$. + +\item[7.] $(a+b+c+d)(a-b+c-d)(a+bi-c-di)(a-bi-c+di)$. + +\item[11.] $\ds x_j = (k_1-a_j)\dotsm(k_n-a_j) + \div \prod\limits^n_{\substack{s=1 \\ s\neq j}} (a_s-a_j)$. + +\item[12.] $x(ab + ac + bc) = -abc$. + +\end{Answers} + +%% -----File: 171.png---Folio 165------- + + +\begin{Answers}{Page133}%,134} + +\item[1.] $\dfrac{p^4 - 3p^2q + 5pr + q^2}{r - pq}$. + +\item[\qquad2.] $\dfrac{(5p^2-12q)(p^2-4q)}{4(p^3 - 4pq + 8r)} - \dfrac{13}{4}p$. + +\ResetCols{2} + +\item[5.] $2p^2-2q$. + +\item[\qquad6.] $24r-p^3$. + +\ResetCols{2} + +\item[7.] $\dfrac{3p^2q^2 - 4p^3r - 4q^3 - 2pqr - 9r^2}{(r - pq)^2}$. + +\item[\qquad8.] $27r^2 - 9pqr + 2q^3 = 0$. + +\ResetCols{2} + +\item[10.] $y = q+r/x$. + +\item[11.] $x = \dfrac{1-py}{2+2y}$. + +\ResetCols{2} + +\item[12.] $y = \dfrac{4x^2 + px + q}{-3x-p}$, see~§112. + +\item[13.] $\dfrac{2q(p^3 + 2pq - r)}{p^2q - pr + s} - 5p$, see Ex.~17. + +\end{Answers} + + +\begin{Answers}{Page136} + +\item[3.] $s_2 = p^2 - 2q$, \\ + $s_3 = p^3 - 3pq$, \\ + $s_4 = p^4 - 4p^2q + 2q^2$, \\ + $s_5 = p^5 - 5p^3q + 5pq^2$. + +\item[4.] $s_{5n} = 5·3^n$, \\ + $s_k = 0$ if $k$ is not divisible by~$5$. + +\item[5.] All zero. + +\end{Answers} + + +\begin{Answers}[1]{Page140} + +\item[2.] See Ex.~2, p.~136. + +\item[3.] $\epsilon^j \sqrt[5]{\frac{1}{2}c + \sqrt{Q}} + + \epsilon^{5-j} \sqrt[5]{\frac{1}{2}c - \sqrt{Q}}$,\quad + $Q = \frac{1}{4}c^2 - q^5$\hfill ($j=0$, $1$, $2$, $3$, $4$). + +\item[4.] $\epsilon^j \sqrt[7]{\frac{1}{2}c + \sqrt{Q}} + + \epsilon^{7-j} \sqrt[7]{\frac{1}{2}c - \sqrt{Q}}$,\quad + $Q = \frac{1}{4}c^2 - q^7$\hfill ($j=0$, $1,\dotsc, 6$). + +\end{Answers} + + +\begin{Answers}{Page141} + +\item[1.] $c_2^2 - 2c_1c_3 + 2c_4$. + +\item[2.] $c_1^2c_2 - 2c_2^2 - c_1c_3 + 4c_4$. + +\ResetCols{2} + +\item[3.] $c_1c_3 - 4c_4$. + +\item[4.] $c_3^2 - 2c_2c_4$. + +\end{Answers} + + +\begin{Answers}{Page142} + +\item[1.] $c_1c_3 - 4c_4$ if $n>3$, $c_1c_3$ if $n=3$. + +\item[\qquad2.] $3c_1c_4 - c_2c_3 - 5c_5$. + +\ResetCols{2} + +\item[3.] $c_2c_4 - 4c_1c_5 + 9c_6$. + +\item[\qquad4.] $c_3^2 - 2c_2c_4 + 2c_1c_5 - 2c_6$. + +\ResetCols{1} + +\item[5.] $y^3 - (p^2 - 2q)y^2 + (q^2 - 2pr)y - r^2 = 0$. + +\ResetCols{2} + +\item[6.] $y^3 - qy^2 + pry - r^2 = 0$. + +\item[7.] $ry^3 + 2qy^2 + 4py + 8 = 0$. + +\ResetCols{2} + +\item[8.] Eliminate $x$ by $y = s_2 - x^2$. + +\item[9.] Use $p^2 - q + px = y$. + +\ResetCols{2} + +\item[10.] $-4 + pr/s$. + +\item[11.] $(rs - pr^2 + 2pqs)/s^2$. + +\ResetCols{1} + +\item[12.] (i) $s_a s_b s_c s_d - \Sigma s_a s_b s_{c+d} + + 2\Sigma s_a s_{b+c+d} + \Sigma s_{a+b} s_{c+d} - 6s_{a+b+c+d}$. \\ + (ii) $\tfrac{1}{24}(s_a^4 - 6s_a^2s_{2a} + + 8s_as_{3a} + 3s_{2a}^2 - 6s_{4a})$. + +%% -----File: 172.png---Folio 166------- + +\item[13.] (i) + \[ + s_k = - \left| + \begin{array}{cccccc} + 1 & 0 & 0 & \ldots & 0 & c_1 \\ + c_1 & 1 & 0 & \ldots & 0 & 2c_2 \\ + c_2 & c_1 & 1 & \ldots & 0 & 3c_3 \\ + c_3 & c_2 & c_1 & \ldots & 0 & 4c_4 \\ + \Dots{6} \\ + c_{k-1} &c_{k-2} &c_{k-3} & \ldots & c_1 & kc_k + \end{array}\;\right|,\quad + s_3 = - + \begin{vmatrix} + 1 & 0 & c_1\\ + c_1 & 1 & 2c_2\\ + c_2 & c_1 & 3c_3 + \end{vmatrix}, + \] + where all but the last term in the main diagonal is~$1$, and all terms above the + diagonal are zero except those in the last column. If $k>n$, we must take + $c_j =0 \quad (j>n)$. + + (ii) + \[ + k!\,c_k = - \left| + \begin{array}{cccccc} + 1 & 0 & 0 & \ldots & 0 & s_1 \\ + s_1 & 2 & 0 & \ldots & 0 & s_2 \\ + s_2 & s_1 & 3 & \ldots & 0 & s_3 \\ + \Dots{6} \\ + \ s_{k-1} & s_{k-2} & s_{k-3} & \ldots & s_1 & s_k + \end{array}\;\right|,\quad + 3!\,c_3 = - + \begin{vmatrix} + 1 & 0 & s_1\\ + s_1 & 2 & s_2\\ + s_2 & s_1 & s_3 + \end{vmatrix}. + \] + +\end{Answers} + + +\begin{Answers}[1]{Page152} + +\item[1.] $y^2(16 - y^2)$;\quad $y=0$, $x=±3$;\quad $y=±4$, $x=+5$. + +\item[2.] $(c-1)^2(y^2 - 25)(y^2 - 16)$. If $c\neq 1$, $y=±5$, $x=0$;\quad $y=±4$, $x=+3$. + +\ResetCols{2} + +\item[3.] +$\begin{vmatrix} + 1 & a & b \\ + 2 & a & 0 \\ + 0 & 2 & a +\end{vmatrix} = 4b-a^2$. + +\item[4.] $2±3i$, $-2±i$, $±i$. + $\vphantom{\begin{vmatrix}1\\ 1\\ 1\end{vmatrix}}$ + +\end{Answers} + + +\begin{Answers}[1]{Page153}%, 154} + +\item[2.] $pqr - p^2s - r^2 = 0$, $x^2 + r/p$, $x^2 + px + ps/r$. + +\ResetCols{2} + +\item[3.] $1$, $3$, $1± i$. + +\item[11.] See Ex.~15, p.~134. + +\end{Answers} + +%% -----File: 173.png---Folio 167------- + +\PrintIndex + +\iffalse + +\title % INDEX + +Numbers refer to pages. + +\begin(theindex) + +\item Abscissa, 55 + +\item Absolute value, 3 + +\item Amplitude, 3 + +\item Argument, 3 + +\item Arithmetical progression, 19 + +\indexspace + +\item Bend point, 56, 64 + +\item Bézout's eliminant, 148 + +\item Budan's theorem, 83 + +\indexspace + +\item Cardan's formulas, 46, 48 + +\item Complex number, 1 + \subitem geometrical representation, 3, 6, 7 + \subitem trigonometric form, 3 + +\item Compound interest, 13, 90, 100 + +\item Conjugate, 1 + +\item Continuity, 66, 157 + +\item Cube root, 5, 48 + \subitem of unity, 3, 4 + +\item Cubic equation, 32, 40, 45, 127, 134, 153--4 + \subitem graph of, 65 + \subitem number real roots, 48, 65, 79 + \subitem reduced, 45, 64--5 + \subitem trigonometric solution, 49 + +\indexspace + +\item De Moivre's quintic, 140 + \subitem theorem, 5 + +\item Derivative, 57--60, 69, 83, 97, 135 + +\item Descartes' rule of signs, 71, 85 + +\item Determinants, 101--27, 146--53 + \subitem addition of columns, 113 + \subitem columns, 103 + \subitem complementary minors, 122 + \subitem diagonal term, 103 + \subitem elements, 103 + \subitem expansion, 109 + \subitem interchanges, 106, 107 + \subitem Laplace's development, 122--3, 147--9 + \subitem minors, 109, 116 + \subitem of Vandermonde, 108 + \subitem product of, 124 + \subitem rank, 116, 121 + +\item Determinants, removal of factor, 111 + \subitem rows, 103 + \subitem signs of terms, 103--6 + \subitem skew symmetric, 108 + \subitem sum of, 112 + +\item Discriminant, 152--3 + \subitem of cubic, 47, 65, 134 + \subitem of quadratic, 11, 12 + \subitem of quartic, 51, 81 + +\item Double root, 16 (see Discriminant) + +\item Duplication of cube, 35 + +\indexspace + +\item Elementary symmetric function, 128 + +\item Elimination, 143--154 + \subitem extraneous factor, 151 + +\item Equation for differences of roots, 154 + \subitem squares of differences, 134, 154 + +\item Euler's eliminant, 147 + +\indexspace + +\item Factor theorem, 12 + +\item Factored form, 11, 15 + +\item Fundamental theorem of algebra, 17, 155--8 + +\indexspace + +\item Geometrical construction, 29--44 + \subitem progression, 13, 19 + +\item Graphs, 55--70 + +\item Greatest common divisor, 61, 75 + +\indexspace + +\item Horner's method, 86 + +\indexspace + +\item Identical polynomials, 16 + +\item Identity, 11 + +\item Imaginary, 2 + \subitem roots, 19, 98 + +\item Inflexion, 62--64 + +\item Integral rational function, 12, 17, 20 + \subitem roots, 24--27 + +\item Interpolation, 93, 97 + +\item Interval, 78 + +\item Irreducible case, 48 + +\item Isolation of roots, 71 + +%% -----File: 174.png---Folio 168------- + +\item Linear equations, system, 101--3, 114--21 + \subitem homogeneous, 119, 121 + +\item Linear factors, 11, 17 + +\item Lower limit to roots, 23 + +\indexspace + +\item Matrix, 120 + \subitem augmented, 121 + +\item Maximum, 57 + +\item Minimum, 57, 157 + +\item Modulus, 3 + +\item Multiple roots, 16, 60, 82 + +\item Multiplicity of root, 16, 20, 61 + +\indexspace + +\item Newton's identities, 136 + \subitem method of solution, 90--98 + +\item Number of roots, 16, 17, 48, 52, 69, 72--85 + \subitem of negative roots, 74 + +\indexspace + +\item Order of radical, 32 + +\item Ordinate, 55 + +\indexspace + +\item Plotting, 55 + +\item Polynomial, 12, 66 + \subitem sign of, 68 + +\item Primitive root of unity, 9 + +\item Product of roots, 18 + +\item Pure imaginary, 2 + +\indexspace + +\item Quadratic equation, 11 + \subitem graphical solution, 29, 55 + \subitem sum of powers of roots, 139 + +\item Quadratic function a square, 12 + +\item Quartic equation, 50--54, 80--81 + +\item Quotient by synthetic division, 14 + +\indexspace + +\item Rational roots, 27 + +\item Real equation, 12, 20 + +\item Reciprocal equation, 37, 44 + +\item Regula falsi, 93 + +\item Regular polygon, 8 + \subitem 7 sides, 35--36 + \subitem 9 sides, 35, 39 + \subitem 17 sides, 41--44 + \subitem \textit(n) sides, 44 + +\item Regular decagon, 39 + \subitem pentagon, 39 + +\item Relations between roots and coefficients, 17 + +\item Relatively prime, 9, 10 + +\item Remainder theorem, 12 + +\item Resolvent cubic, 50, 51 + +\item Resultant, 143--154 + +\item Rolle's theorem, 69 + +\item Root between \textit(a) and \textit(b), 67 + +\item Roots of unity, 8, 36, 39, 44, 136 + \subitem periods of, 40 + +\item Roots, \textit(n)th, 7 + +\indexspace + +\item Sigma function, 128--142 + +\item Sign of polynomial, 68 + +\item Simple root, 16 + +\item Slope, 57, 59 + +\item Solution of numerical equations, 86--100 + +\item Specific gravity, 89 + +\item Square roots, 1, 30, 31, 96 + +\item Sturm's functions, 75--82 + +\item Sum of four squares, 126 + \subitem like powers of roots, 134--142 + \subitem products of roots, 18 + \subitem roots, 18 + +\item Surd roots in pairs, 20 + +\item Sylvester's eliminant, 145, 149, 150 + +\item Symbol \equiv, 11; \textit(f)(\textit(x)), 12; $$\textit(a)$$,23, 155; \textit(r)!, 59; +\textit(f^(k)(x)), 59; \Sigma, 128; \textit(s_k), 134; \textit(R)(\textit(j), \textit(g)), +144 + +\item Symmetric functions, 128--142 + in all but one root, 132--4 + +\item Synthetic division, 13, 86--95 + +\indexspace + +\item Tangents, 60, 62 + +\item Taylor's theorem, 59 + +\item Transformed equation, 28, 86 + +\item Triple root, 16 + +\item Trisection of angle, 34, 40 + +\indexspace + +\item Upper limit to roots, 21--23 + +\indexspace + +\item Variation of sign, 71 + +\indexspace + +\item Waring's formula, 136 + +\end(theindex) +\fi + + +% LICENSE + +\phantomsection +\pdfbookmark[0]{PG License}{PG License} +\fancyhead[C]{\Running{LICENSING}} + +\begin{PGtext} +End of the Project Gutenberg EBook of First Course in the Theory of Equations, by +Leonard Eugene Dickson + +*** END OF THIS PROJECT GUTENBERG EBOOK THEORY OF EQUATIONS *** + +***** This file should be named 29785-pdf.pdf or 29785-pdf.zip ***** +This and all associated files of various formats will be found in: + http://www.gutenberg.org/2/9/7/8/29785/ + +Produced by Peter Vachuska, Andrew D. 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