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diff --git a/old/31076-pdf 2010-01-25.pdf b/old/31076-pdf 2010-01-25.pdf Binary files differnew file mode 100644 index 0000000..09fcd36 --- /dev/null +++ b/old/31076-pdf 2010-01-25.pdf diff --git a/old/31076-pdf 2010-01-25.zip b/old/31076-pdf 2010-01-25.zip Binary files differnew file mode 100644 index 0000000..f34b587 --- /dev/null +++ b/old/31076-pdf 2010-01-25.zip diff --git a/old/31076-t 2010-01-25.tex b/old/31076-t 2010-01-25.tex new file mode 100644 index 0000000..c731410 --- /dev/null +++ b/old/31076-t 2010-01-25.tex @@ -0,0 +1,7049 @@ +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% The Project Gutenberg EBook of Elliptic Functions, by Arthur L. Baker % +% % +% 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: Elliptic Functions % +% An Elementary Text-Book for Students of Mathematics % +% % +% Author: Arthur L. Baker % +% % +% Release Date: January 25, 2010 [EBook #31076] % +% % +% Language: English % +% % +% Character set encoding: ISO-8859-1 % +% % +% *** START OF THIS PROJECT GUTENBERG EBOOK ELLIPTIC FUNCTIONS *** % +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{31076} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% inputenc: Standard DP encoding. Required. %% +%% %% +%% fix-cm: For larger title page fonts. Optional. %% +%% ifthen: Logical conditionals. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. 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Required. %% +%% %% +%% %% +%% Producer's Comments: %% +%% %% +%% Changes are noted in this file in three ways. %% +%% 1. \DPnote{} for in-line `placeholder' notes. %% +%% 2. \DPtypo{}{} for typographical corrections, showing original %% +%% and replacement text side-by-side. %% +%% 3. [** PP: Note]s for lengthier or stylistic comments. %% +%% %% +%% %% +%% Compilation Flags: %% +%% %% +%% The following behavior may be controlled by boolean flags. %% +%% %% +%% ForPrinting (false by default): %% +%% Compile a screen-optimized PDF file. Set to false for print- %% +%% optimized file (pages cropped, one-sided, blue hyperlinks). %% +%% %% +%% %% +%% Things to Check: %% +%% %% +%% %% +%% Spellcheck: .................................. OK %% +%% Smoothreading pool: ......................... yes %% +%% %% +%% lacheck: ..................................... 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\addto@DPbody{\DPh@@kr&\DP@rint}\@xp\addto@DPbody\@xp{\@xp{\the\DP@lign@no}&\\}\DP@crtrue + \global\advance\DP@lign@no\@ne + \let\@next\parse@cr + \else\check@braces#1!Q!Q!Q!\ifbr@ce\addto@DPbody{{#1}}\else + \addto@DPbody{#1}\fi + \let\@next\parseb@dy + \fi\fi\fi\fi\fi\fi\fi\fi\@next} +\def\process@DPb@dy{\let\lintertext\@gobble\let\rintertext\@gobble + \@xp\start@align\@xp\tw@\@xp\st@rredtrue\@xp\m@ne\the\DP@lignb@dy} +\def\linter@text#1{\@xp\DPlint\@xp{\the\DP@lign@no}{#1}\parseb@dy} +\def\rinter@text#1{\@xp\DPrint\@xp{\the\DP@lign@no}{#1}\parseb@dy} +\def\DPlint#1#2{\@xp\def\csname DP@lint:#1\endcsname{\text{#2}}} +\def\DPrint#1#2{\@xp\def\csname DP@rint:#1\endcsname{\text{#2}}} +\def\DP@lint#1{\ifbalancedlrint\@xp\ifx\csname +DP@lint:#1\endcsname\relax\phantom + {\csname DP@rint:#1\endcsname}\else\csname DP@lint:#1\endcsname\fi + \else\csname DP@lint:#1\endcsname\fi} +\def\DP@rint#1{\ifbalancedlrint\@xp\ifx\csname +DP@rint:#1\endcsname\relax\phantom + {\csname DP@lint:#1\endcsname}\else\csname DP@rint:#1\endcsname\fi + \else\csname DP@rint:#1\endcsname\fi} +\def\eat@command#1#2{\ifcase#1\addto@DPbody{\intertext{#2}}\or + \addto@DPbody{\shortintertext{#2}}\fi\DP@crtrue + \global\advance\DP@lign@no\@ne\parseb@dy} +\def\parse@cr{\new@ifnextchar*{\parse@crst}{\parse@crst{}}} +\def\parse@crst#1{\addto@DPbody{#1}\new@ifnextchar[{\parse@crb}{\parseb@dy}} +\def\parse@crb[#1]{\addto@DPbody{[#1]}\parseb@dy} +\def\check@braces#1#2!Q!Q!Q!{\def\dp@lignt@stm@cro{#2}\ifx + \empty\dp@lignt@stm@cro\br@cefalse\else\br@cetrue\fi} +\def\eat@environment#1{\addto@DPbody{\begin{#1}}\begingroup + \def\@currenvir{#1}\let\@next\digest@env\@next} +\def\digest@env#1\end#2{% + \edef\begin@stack{\push@begins#1\begin\end \@xp\@gobble\begin@stack}% + \ifx\@empty\begin@stack + \@checkend{#2} + \endgroup\let\@next\parseb@dy\fi + \addto@DPbody{#1\end{#2}} + \@next} +\def\lintertext{lint}\def\rintertext{rint} +\newif\ifbalancedlrint +\let\DPh@@kl\empty\let\DPh@@kr\empty +\def\DPg@therl{&\omit\hfil$\displaystyle} +\def\DPg@therr{$\hfil} + +\newenvironment{DPalign*}[1][a]{% + \if m#1\balancedlrintfalse\else\balancedlrinttrue\fi + \global\DP@lign@no\z@\DP@crfalse + \DP@lignb@dy{&\DP@lint0&}\parseb@dy +}{% + \endalign +} +\newenvironment{DPgather*}[1][a]{% + \if m#1\balancedlrintfalse\else\balancedlrinttrue\fi + \global\DP@lign@no\z@\DP@crfalse + \let\DPh@@kl\DPg@therl + \let\DPh@@kr\DPg@therr + \DP@lignb@dy{&\DP@lint0&\DPh@@kl}\parseb@dy +}{% + \endalign +} +\makeatother + +%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%% + +\begin{document} + +\pagestyle{empty} +\pagenumbering{Alph} + +\phantomsection +\pdfbookmark[-1]{Front Matter}{Front Matter} + +%%%% PG BOILERPLATE %%%% +\Pagelabel{PGBoilerplate} +\phantomsection +\pdfbookmark[0]{PG Boilerplate}{Project Gutenberg Boilerplate} + +\begin{center} +\begin{minipage}{\textwidth} +\small +\begin{PGtext} +The Project Gutenberg EBook of Elliptic Functions, by Arthur L. Baker + +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: Elliptic Functions + An Elementary Text-Book for Students of Mathematics + +Author: Arthur L. Baker + +Release Date: January 25, 2010 [EBook #31076] + +Language: English + +Character set encoding: ISO-8859-1 + +*** START OF THIS PROJECT GUTENBERG EBOOK ELLIPTIC FUNCTIONS *** +\end{PGtext} +\end{minipage} +\end{center} + +\clearpage + + +%%%% Credits and transcriber's note %%%% +\begin{center} +\begin{minipage}{\textwidth} +\begin{PGtext} +Produced by Andrew D. Hwang, Brenda Lewis and the Online +Distributed Proofreading Team at http://www.pgdp.net (This +file was produced from images from the Cornell University +Library: Historical Mathematics Monographs collection.) +\end{PGtext} +\end{minipage} +\end{center} +\vfill + +\begin{minipage}{0.85\textwidth} +\small +\phantomsection +\pdfbookmark[0]{Transcriber's Note}{Transcriber's Note} +\subsection*{\centering\normalfont\scshape% +\normalsize\MakeLowercase{\TransNote}}% + +\raggedright +\TransNoteText +\end{minipage} + + +%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%% + +%% -----File: 001.png---Folio xx------- +%[** Title page] + +\frontmatter +\pagenumbering{roman} +\pagestyle{empty} + +\enlargethispage{36pt} + +% [** PP: Set line skip] +\ifthenelse{\boolean{ForPrinting}}{% + \setlength{\TmpLen}{0.2in}% +}{% + \setlength{\TmpLen}{0.125in}% +} + +\begin{center} +\makebox[0pt][c]{\centering\MyHuge\scshape Elliptic Functions.}\\[\Titleskip{4}] +{\Large\scshape An Elementary Text-Book for \\[\Titleskip{1}] +Students of Mathematics.}\\[\Titleskip{4}] +% +{\footnotesize BY}\\[\Titleskip{1}] +% +{\LARGE\scshape ARTHUR~L. BAKER, C.E., Ph.D.,}\\ +\makebox[0pt][c]{\centering\footnotesize\scshape +Professor of Mathematics in the Stevens School of the Stevens Institute of}\\ +\makebox[0pt][c]{\centering\footnotesize\scshape +Technology, Hoboken, N.~J.; formerly Professor in the Pardee} \\ +\makebox[0pt][c]{\centering\footnotesize\scshape +Scientific Department, Lafayette College, Easton, Pa.}\\[\Titleskip{4}] +% +\tb +\[ +\vphantom{\Bigg|}\sin \am u = \frac{1}{\sqrt{k}} · \frac{\Eta (u)}{\Theta (u)}. +\] +\tb\\[\Titleskip{4}] +% +NEW YORK: \\[2pt] +{\large \so{JOHN WILEY \& SONS},} \\ +{\scshape 53~East Tenth Street.} \\ +1890. +\end{center} +\clearpage + +%% -----File: 002.png---Folio xx------- +\null\vfill +\begin{center} +Copyright, 1890, \\ +{\footnotesize BY} \\ +Arthur L. Baker. +\end{center} +\vfill\vfill + +\settowidth{\TmpLen}{444 \& 446 Pearl Street,}% +\noindent\parbox[t]{\TmpLen}{\centering +\textsc{Robert Drummond}, \\ +\textit{Electrotyper}, \\ +444 \& 446 Pearl Street, \\ +New York.} +\hfill +\settowidth{\TmpLen}{326 Pearl Street,}% +\parbox[t]{\TmpLen}{\centering +\textsc{Ferris Bros}., \\ +\textit{Printers}, \\ +326 Pearl Street, \\ +New York.} + +\clearpage + +%% -----File: 003.png---Folio xx------- + +\begin{center} +{\Large PREFACE.}\\[8pt] +\tb\\[16pt] +\end{center} +\phantomsection +\pdfbookmark[0]{Preface}{Preface} + +\First{In} the works of Abel, Euler, Jacobi, Legendre, and others, +the student of Mathematics has a most abundant supply of +material for the study of the subject of Elliptic Functions. + +These works, however, are not accessible to the general +student, and, in addition to being very technical in their treatment +of the subject, are moreover in a foreign language. + +It is in the hope of smoothing the road to this interesting +and increasingly important branch of Mathematics, and of +putting within reach of the English student a tolerably complete +outline of the subject, clothed in simple mathematical +language and methods, that the present work has been compiled. + +New or original methods of treatment are not to be looked +for. The most that can be expected will be the simplifying of +methods and the reduction of them to such as will be intelligible +to the average student of Higher Mathematics. + +I have endeavored throughout to use only such methods as +are familiar to the ordinary student of Calculus, avoiding those +methods of discussion dependent upon the properties of double +periodicity, and also those depending upon Functions of Complex +Variables. For the same reason I have not carried the +discussion of the $\Theta$~and~$\Eta$ functions further. +%% -----File: 004.png---Folio xx------- + +Among the minor helps to simplicity is the use of zero +subscripts to indicate decreasing series in the Landen Transformation, +and of numerical subscripts to indicate increasing +series. I have adopted the notation of Gudermann, as being +more simple than that of Jacobi. + +{\stretchyspace +I have made free use of the following works: \textsc{Jacobi's} +Fundamenta Nova Theoriæ Func.\ Ellip.; \textsc{Houel's} Calcul +Infinitésimal; \textsc{Legendre's} Traité des Fonctions Elliptiques; +\textsc{Durege's} Theorie der Elliptischen Functionen; \textsc{Hermite's} +Théorie des Fonctions Elliptiques; \textsc{Verhulst's} Théorie des +Functions Elliptiques; \textsc{Bertrand's} Calcul Intégral; \textsc{Laurent's} +Théorie des Fonctions Elliptiques; \textsc{Cayley's} Elliptic +Functions; \textsc{Byerly's} Integral Calculus; \textsc{Schlomilch's} Die +Höheren Analysis; \textsc{Briot et Bouquet's} Fonctions Elliptiques.} + +I have refrained from any reference to the Gudermann or +Weierstrass functions as not within the scope of this work, +though the Gudermannians might have been interesting +examples of verification formulæ. The arithmetico-geometrical +mean, the march of the functions, and other interesting investigations +have been left out for want of room. + +%% -----File: 005.png---Folio xx------- + +\clearpage +\phantomsection +\pdfbookmark[0]{Contents}{Contents}% +\tableofcontents + +\iffalse + +CONTENTS. + +PAGE + +Introductory Chapter, 1 + +Chap. I. Elliptic Integrals, 4 + +II. Elliptic Functions, 16 + +III. Periodicity of the Functions, 22 + +IV. Landen's Transformation, 30 + +V. Complete Functions, 45 + +VI. Evaluation for \phi, 48 + +% [** PP: Next entry matches neither chapter title nor running head; +% using chapter title] +VII. Factorization of Elliptic Functions, 51 + +VIII. The \Theta Function, 66 + +IX. The \Theta and \Eta Functions, 69 + +X. Elliptic Integrals of the Second Order, 81 + +XI. Elliptic Integrals of the Third Order, 90 + +XII. Numerical Calculations, q, 94 + +XIII. Numerical Calculations, K, 98 + +XIV. Numerical Calculations, u, 102 + +XV. Numerical Calculations, \phi, 108 + +XVI. Numerical Calculations, E(k, \phi), 111 + +XVII. Applications, 115 + +\fi +%% -----File: 006.png---Folio xx------- +%[Blank Page] +%% -----File: 007.png---Folio 1------- + +\mainmatter +\phantomsection +\pdfbookmark[-1]{Main Matter}{Main Matter} +\pagenumbering{arabic} +\pagestyle{fancy} +\fancyfoot{} + +\IntroChapter{Introductory Chapter.\protect\footnotemark}{Elliptic Functions.} +\footnotetext{Condensed from an article by Rev.\ Henry Moseley, M.A., F.R.S., Prof.\ of Nat.\ Phil.\ and Ast., King's College, London.} + +\First{The} first step taken in the theory of Elliptic Functions +was the determination of a relation between the amplitudes of +three functions of either order, such that there should exist an +algebraic relation between the three functions themselves of +which these were the amplitudes. It is one of the most remarkable +discoveries which science owes to Euler. In 1761 he +gave to the world the complete integration of an equation of +two terms, each an elliptic function of the first or second order, +not separately integrable. + +This integration introduced an arbitrary constant in the +form of a third function, related to the first two by a given +equation between the amplitudes of the three. + +In 1775 Landen, an English mathematician, published his +celebrated theorem showing that any arc of a hyperbola may +be measured by two arcs of an ellipse, an important element +of the theory of Elliptic Functions, but \emph{then} an isolated result. +The great problem of comparison of Elliptic Functions of different +moduli remained unsolved, though Euler, in a measure, +exhausted the comparison of functions of the same modulus. +It was completed in 1784 by Lagrange, and for the computation +%% -----File: 008.png---Folio 2------- +of numerical results leaves little to be desired. The value of +a function may be determined by it, in terms of increasing or +diminishing moduli, until at length it depends upon a function +having a modulus of zero, or unity. + +For all practical purposes this was sufficient. The enormous +task of calculating tables was undertaken by Legendre. +His labors did not end here, however. There is none of the +discoveries of his predecessors which has not received some +perfection at his hands; and it was he who first supplied to +the whole that connection and arrangement which have made +it an independent science. + +The theory of Elliptic Integrals remained at a standstill +from 1786, the year when Legendre took it up, until the year +1827, when the second volume of his Traité des Fonctions +Elliptiques appeared. Scarcely so, however, when there appeared +the researches of Jacobi, a Professor of Mathematics in +Königsberg, in the 123d~number of the Journal of Schumacher, +and those of Abel, Professor of Mathematics at Christiania, in +the 3d~number of Crelle's Journal for 1827. + +These publications put the theory of Elliptic Functions +upon an entirely new basis. The researches of Jacobi have for +their principal object the development of that general relation +of functions of the first order having different moduli, of which +the scales of \DPtypo{Legrange}{Lagrange} and Legendre are particular cases. + +It was to Abel that the idea first occurred of treating the +Elliptic Integral as a function of its amplitude. Proceeding +from this new point of view, he embraced in his speculations +all the principal results of Jacobi. Having undertaken to develop +the principle upon which rests the fundamental proposition +of Euler establishing an algebraic relation between three +functions which have the same moduli, dependent upon a certain +relation of their amplitudes, he has extended it from three +to an indefinite number of functions; and from Elliptic Functions +to an infinite number of other functions embraced under +an indefinite number of classes, of which that of Elliptic Functions +%% -----File: 009.png---Folio 3------- +is but one; and each class having a division analogous to +that of Elliptic Functions into three orders having common +properties. + +The discovery of Abel is of infinite moment as presenting +the first step of approach towards a more complete theory of +the infinite class of ultra elliptic functions, destined probably +ere long to constitute one of the most important of the +branches of transcendental analysis, and to include among the +integrals of which it effects the solution some of those which at +present arrest the researches of the philosopher in the very +elements of physics. +%% -----File: 010.png---Folio 4------- + + +\Chapter{I}{Elliptic Integrals.} + +\First{The} integration of irrational expressions of the form +\begin{gather*} +X\, dx \sqrt{A + Bx + Cx^{2}},\\ +\intertext{or} +\frac{X\, dx}{\sqrt{A + Bx + Cx^{2}}}, +\end{gather*} +$X$ being a rational function of~$x$, is fully illustrated in most elementary +works on Integral Calculus, and shown to depend upon +the transcendentals known as logarithms and circular functions, +which can be calculated by the proper logarithmic and trigonometric +tables. + +When, however, we undertake to integrate irrational expressions +containing higher powers of~$x$ than the square, we meet +with insurmountable difficulties. This arises from the fact that +the integral sought depends upon a new set of transcendentals, +to which has been given the name of \emph{elliptic functions}, and +whose characteristics we will learn hereafter. + +The name of Elliptic Integrals has been given to the simple +integral forms to which can be reduced all integrals of the form +\[ +\Tag{(1)} +V = \int F(X, R)\, dx, +\] +where $F(X, R)$ designates a rational function of $x$~and~$R$, and +$R$~represents a radical of the form +\[ +R = \sqrt{Ax^{4} + Bx^{3} + Cx^{2} + Dx + E}, +\] +%% -----File: 011.png---Folio 5------- +where $A$,~$B$,~$C$, $D$,~$E$ indicate constant coefficients. + +We will show presently that all cases of \Eqref{Eq.}{}{(1)} can be +reduced to the three typical forms +\[ +\Tag{(2)} +\begin{aligned} +&\int_{0}^{x} \frac{dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}, \\ +&\int_{0}^{x} \frac{x^{2}\, dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}},\\ +&\int_{0}^{x} \frac{dx}{(x^{2} + a) \sqrt{(1 - x^{2})(1 - k^{2}x^{2})}}, +\end{aligned} +\] +which are called elliptic integrals of the first, second, and third +order. + +Why they are called \emph{Elliptic} Integrals we will learn further +on. The transcendental functions which depend upon these +integrals, and which will be discussed in \Chapref{Chapter}{IV}, are called \emph{Elliptic Functions}. + +The most general form of \Eqref{Eq.}{}{(1)} is +\[ +\Tag{(3)} +V = \int \frac{A + BR}{C + DR}\, dx; +\] +where $A$,~$B$,~$C$, and~$D$ stand for rational integral functions of~$x$. + +$\dfrac{A + BR}{C + DR}$ can be written +\begin{align*} +\frac{A + BR}{C + DR} + &= \frac{AC - BDR^{2}}{C^{2} - D^{2}R^{2}} + - \frac{(AD - CB)R^{2}}{C^{2} - D^{2}R^{2}} · \frac{1}{R}\\ + &= N - \frac{P}{R}; +\end{align*} +%% -----File: 012.png---Folio 6------- +$N$~and~$P$ being rational integral functions of~$x$. Whence \Eqref{Eq.}{}{(3)} +becomes +\[ +\Tag{(4)} +V = \int N\, dx - \int \frac{P\, dx}{R}. +\] + +\Eqref{Eq.}{}{(4)} shows that the most general form of~$V$ can be made +to depend upon the expressions +\[ +\Tag{(5)} +V' = \int \frac{P\, dx}{R}, +\] +and +\[ +\int N\, dx. +\] + +This last form is rational, and needs no discussion here. + +We can write +\begin{align*} +P &= \frac{G_{0} + G_{1}x + G_{2}x^{2} + \dotsb} + {H_{0} + H_{1}x + H_{2}x^{2} + \dotsb} \\ + &= \frac{G_{0} + G_{2}x^{2} + G_{4}x^{4} + \dotsb + (G_{1} + G_{3}x^{2} + \dotsb)x} + {H_{0} + H_{2}x^{2} + H_{4}x^{4} + \dotsb + (H_{1} + H_{3}x^{2} + \dotsb)x}. +\end{align*} +Multiplying both numerator and denominator by +\[ +H_{0} + H_{2}x^{2} + H_{4}x^{4} + \dotsb - (H_{1} + H_{3}x^{2} + H_{5}x^{4} + \dotsb)x, +\] +we have a new \DPtypo{numerator}{denominator} which contains only powers of~$x^{2}$. +The result takes the following form: +\begin{align*} +P &= \frac{M_{0} + M_{2}x^{2} + M_{4}x^{4} + \dotsb + + (M_{1} + M_{3}x^{2} + M_{5}x^{4} + \dotsb)x} + {N_{0} + N_{2}x^{2} + N_{4}x^{4} + N_{6}x^{6} + \dotsb} \\ + &= \Phi(x^{2}) + \Psi(x^{2})·x. +\end{align*} + +\Eqref{Equation}{}{(5)} thus becomes +\[ +\Tag{(6)} +V' = \int \frac{\Phi(x^{2})\, dx}{R} + \int \frac{\Psi(x^{2}) · x · dx}{R}. +\] +%% -----File: 013.png---Folio 7------- + +We shall see presently that $R$ can always be assumed to be +of the form +\[ +\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}. +\] + +Therefore, putting $x^{2} = z$, the second integral in \Eqref{Eq.}{}{(6)} +takes the form +\[ +\frac{1}{2} \int \frac{\Psi(z) · dz}{\sqrt{(1 - z)(1 - k^{2}z)}}, +\] +which can be integrated by the well-known methods of Integral +Calculus, resulting in logarithmic and circular transcendentals. + +There remains, therefore, only the form +\[ +\int \frac{\Phi(x^{2})\, dx}{R} +\] +to be determined. + +We will now show that $R$ can always be assumed to be in +the form +\[ +\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}. +\] +We have +\begin{align*} +R &= \sqrt{Ax^{4} + Bx^{3} + Cx^{2} + Dx + E} \\ + &= \sqrt{G(x - a)(x - b)(x - c)(x - d)}, +\end{align*} +$a$,~$b$,~$c$, and~$d$ being the roots of the polynomial of the fourth +degree, and $G$~any number, real or imaginary, depending upon +the coefficients in the given polynomial. + +Substituting in \Eqref{equation}{}{(1)} +\begin{align*} +x &= \frac{p + qy}{1 + y}, \\ +\intertext{we have} +\Tag{(7)} +V &= \int \phi(y, \rho)\, dy, +\end{align*} +%% -----File: 014.png---Folio 8------- +$\rho$~designating the radical +\[ +\rho = \sqrt{G[p - a + (q - a)y] + [p - b + (q - b)y] + [p - c + (q - c)y]\DPtypo{}{\dotsm}} \DPtypo{\dotsm}{\;}. +\] + +In order that the odd powers of~$y$ under the radical may +disappear we must have their coefficients equal to zero; i.e., +\begin{align*} +(p - a)(q - b) + (p - b)(q - a) &= 0, \\ +(p - c)(q - d) + (p - d)(q - c) &= 0; +\end{align*} +whence +\begin{align*} +2pq - (p + q)(a + b) + 2ab &= 0, \\ +2pq - (p + q)(c + d) + 2cd &= 0, +\end{align*} +and +\[ +\Tag{(8)} +\left\{ +\begin{aligned} +pq &= \frac{ab(c + d) - cd(a + b)}{a + b - (c + d)}, \\ +p + q &= \frac{2ab - 2cd}{a + b - (c + d)}. +\end{aligned} +\right. +\] +\Eqref{Equation}{}{(8)} shows that $p$ and~$q$ are real quantities, whether +the roots $a$,~$b$,~$c$, and~$d$ are real or imaginary; $a$,~$b$, and $c$,~$d$ being +the conjugate pairs. + +Hence \Eqref{equation}{}{(1)} can always be reduced to the form of +\Eqref{equation}{}{(7)}, which contains only the second and fourth powers +of the variable. + +This transformation seems to fail when $a + b - (c + d) = 0$; +but in that case we have +\[ +R = \sqrt{G[x^{2} - (a + b)x + ab][x^{2} - (a + b)x + cd]}, +\] +and substituting +\[ +x = y - \frac{a + b}{2} +\] +will cause the odd powers of~$y$ to disappear as before. + +If the radical should have the form +\[ +\sqrt{G(x - a)(x - b)(x - c)}, +\] +%% -----File: 015.png---Folio 9------- +placing $x = y^{2} + a$, we get +\begin{align*} +V &= \int \phi(y, \rho)\, dy, \\ +\rho &= \sqrt{G(y^{2} + a - b)(y^{2} + a - c)}, +\end{align*} +$\phi$~designating a rational function of $y$~and~$\rho$. + +Thus all integrals of the form contained in \Eqref{equation}{}{(1)}, in +which $R$~stands for a quadratic surd of the third or fourth +degree, can be reduced to the form +\[ +\Tag{(9)} +V = \int \phi(x, R)\, dx, +\] +$R$~designating a radical of the form +\[ +\sqrt{G(1 + mx^{2})(1 + nx^{2})}, +\] +$m$~and~$n$ designating constants. + +It is evident that if we put +\[ +x' = x\sqrt{-m},\quad k^{2} = -\frac{n}{m}, +\] +we can reduce the radical to the form +\[ +\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}. +\] + +We shall see later on that the quantity~$k^{2}$, to which has +been given the name \emph{modulus}, can always be considered real +and less than unity. + +Combining these results with \Eqref{equation}{}{(6)}, we see that the +integration of \Eqref{equation}{}{(1)} depends finally upon the integration +of the expression +\[ +\Tag{(10)} +V'' = \int \frac{\phi(x^{2})\, dx}{\sqrt{(1 - x^{2})(1 - k^{2}x^{2})}} + = \int \frac{\phi(x^{2})\, dx}{R}. +\] +%% -----File: 016.png---Folio 10------- + +The most general form of~$\phi(x^{2})$ is +\begin{align*} +\phi(x^{2}) + &= \frac{M_{0} + M_{2}x^{2} + M_{4}x^{4} + \dotsb} + {N_{0} + N_{2}x^{2} + N_{4}x^{4} + \dotsb} \\ + &= P_{0} + P_{2}x^{2} + P_{4}x^{4} + P_{6}x^{\DPtypo{2}{6}} + \dotsb \\ + &+ \sum \frac{L}{(x^{2} + a)^{n}}. %[** PP: Textstyle sums in original] +\end{align*} + +Hence +\[ +\Tag{(11)} +V'' = \sum P \int \frac{x^{2m}\, dx}{R} + + \sum L \int \frac{dx}{(x^{2} + a)^{n}\, R}. +\] + +But $\displaystyle\int \frac{x^{2m}\, dx}{R}$ depends upon $\displaystyle\int \frac{dx}{R}$ and $\displaystyle\int \frac{x^{2}\, dx}{R}$, which can +be shown as follows: + +Differentiating $Rx^{2m-3}$, we have +\begin{align*} +d[x^{2m-3}R] + &= d\left[x^{2m-3} \sqrt{\alpha + \beta x^{2} + \gamma x^{4}}\right] \\ + &= (2m - 3)x^{2m-4}\, dx \sqrt{\alpha + \beta x^{2} + \gamma x^{4}} +%\end{aligned} \\ + + \frac{x^{2m-3}(\beta x + 2\gamma x^{3})\, dx} + {\sqrt{\alpha + \beta x^{2} + \gamma x^{4}}}. +\end{align*} +Integrating and collecting, we get +\begin{align*} +Rx^{2m-3} + &= \begin{aligned}[t] + (2m - 3)\alpha \int \frac{x^{2m-4}\, dx}{R} + &+ (2m - 2)\beta \int \frac{x^{2m-2}\, dx}{R} \\ + &+ (2m - 1)\gamma \int \frac{x^{2m}\, dx}{R} + \end{aligned} \\ +% +\Tag{(12)} + &= \alpha' \int \frac{x^{2m-4}\, dx}{R} + + \beta' \int \frac{x^{2m-2}\, dx}{R} + + \gamma' \int \frac{x^{2m}\, dx}{R}. +\end{align*} +%% -----File: 017.png---Folio 11------- +Whence we get, by taking $m=2$, +\[ +\Tag{(13)} +Rx = \alpha \int \frac{dx}{R} + + \beta \int \frac{x^{2}\, dx}{R} + + \gamma \int \frac{x^{4}\, dx}{R}, +\] +which shows that the general expression $\displaystyle\int \frac{x^{2m}\, dx}{R}$ can be found +by successive calculations, when we are able to integrate the +expressions +\[ +\int \frac{dx}{R}\quad \text{and}\quad \int \frac{x^{2}\, dx}{R}, +\] +the first and second of \Eqref{equation}{}{(2)}. + +We will now consider the second class of terms in \Eqref{eq.}{}{(11)}, +viz., $\dfrac{L\, dx}{(x^{2} + a)^{n}\, R}$. + +This second term is as follows: +\begin{align*} +\Tag{(14)} +\sum \int \frac{L}{(x^{2} + a)^{n}\, R} + = \int \frac{A\, dx}{(x^{2} + a)^{n}\, R} + &+ \int \frac{B\, dx}{(x^{2} + a)^{n-1}\, R} \\ + &+ \int \frac{C\, dx}{(x^{2} + a)^{n-2}\, R} + \dotsb +\end{align*} + +Each of these terms can be shown to depend ultimately +upon terms of the form +\[ +\frac{x^{2}\, dx}{R},\quad \frac{dx}{R},\quad \text{and}\quad \frac{dx}{(x^{2} + a)\, R}. +\] + +The two former will be recognized as the two ultimate forms +already discussed, the first and second of \Eqref{equation}{}{(2)}. The +third is the third one of \Eqref{equation}{}{(2)}. + +This dependence of \Eqref{equation}{}{(14)} can be shown as follows: +%% -----File: 018.png---Folio 12------- + +We have +\begin{align*} +d\left[\frac{xR}{(x^{2} + a)^{n-1}}\right] + &= \frac{(x^{2} + a)^{n-1} (x\, dR + R\, dx) - 2x^{2} R(n + 1)(x^{2} + a)^{n-2}\, dx} + {(x^{2} + a)^{2n-2}} \\ + &= \frac{(x^{2} + a)(x\, dR + R\, dx) - 2x^{2} R(n - 1)\, dx}{(x^{2} + a)^{n}}. +\end{align*} + +Substituting the value of +\[ +R = \sqrt{\alpha + \beta x^{2} + \gamma x^{4}}\quad\text{and}\quad +dR = (\beta x + 2 \gamma x^{3})\, \frac{dx}{R}, +\] +we get +\Pagelabel{12}% +\begin{gather*} +\begin{aligned} +&d\left[\frac{xR}{(x^{2} + a)^{n-1}}\right] \\ %[** PP: Moving = to next line] +% +&= \frac{(x^{2} + a)(\beta x^{2} + 2 \gamma x^{4} + \alpha + \beta x^{2} + \gamma x^{4}) + - 2x^{2}(n-1)(\alpha + \beta x^{2} + \gamma x^{4})} + {(x^{2} + a)^{n}} · \frac{dx}{R} +\end{aligned} \\ +% +\begin{aligned} +&= \frac{\left\{\begin{aligned} + \bigl(3\gamma - 2(n-1)\gamma\bigr) x^6 + &+ \bigl(2\beta + 3a\gamma - 2(n-1)\beta\bigr) x^4 \\ + &+ \bigl(2a\beta + \alpha - 2(n-1)\alpha\bigr) x^2 + + a\alpha\end{aligned}\right\}} + {(x^{2} + a)^{n}} · \frac{dx}{R} \\ +% +&= \frac{-(2n - 5)\gamma x^6 + + \bigl(\DPnote{[**A]} - (2n - 4)\beta + 3a\gamma\bigr) x^4 + + \bigl(\DPnote{[**B]} - (2n - 3)\alpha + \DPnote{[**C]} 2a\beta\bigr) x^2 + + a\alpha} + {(x^{2} + a)^{n}} · \frac{dx}{R}; +\end{aligned} +\end{gather*} +%[** PP: This display has been completely re-set. Minor modifications +% were required to maintain algebraic correctness: +% **A & **B: (+ inserted, matching ) immediately precedes x^4, x^2 resp.; +% **C: - typo corrected to +] +or, by substituting in the numerator $x^{2} = z - a$, +\[ += \frac{\left\{\begin{aligned}&- (2n - 5)\gamma z^3 \\ + &+ \bigl((2n - 5) 3a\gamma - (2n - 4)\beta + 3a\gamma\bigr) z^2 \\ + &+ \bigl(\DPnote{[**D]} - (2n - 5) 3a^2\gamma + (2n - 4) 2a\beta - 6a^2\gamma + - (2n - 3)\alpha + 2a\beta\bigr) z \\ + &+ \bigl((2n - 5)a^3\gamma - (2n - 4) a^2\beta + 3a^3\gamma + + (2n - 3)a\alpha - 2a^2\beta + a\alpha\bigr)\end{aligned}\right\}}{(x^{2} + a)^{n}} · \frac{dx}{R}; +\] +%[**D: (+ inserted, matching ) immediately precedes z]} +%% -----File: 019.png---Folio 13------- +or, after resubstituting $z = x^2 + a$, and integrating, +\begin{align*} +\Tag{(15)} +\frac{xR}{(x^2 + a)^{n - 1}} + &= -(2n - 5)\gamma \int \frac{dx}{(x^2 + a)^{n - 3} R} \\ + &\quad -(2n - 4)(\beta - 3a \gamma) \int \frac{dx}{(x^2 + a)^{n - 2} R}\\ + &\quad -(2n - 3)(3 a^2 \gamma - 2a \beta + \alpha) \int \frac{dx}{(x^2 + a)^{n - 1} R}\\ + &\quad +(2n - 2)(a^3 \gamma - a^2 \beta + a \alpha) \int \frac{dx}{(x^2 + a)^n R}. +\end{align*} +\begin{align*} += \alpha_{1} \int \frac{dx}{(x^2 + a)^{n - 3}R} + + \beta_{1} \int \frac{dx}{(x^2 + a)^{n - 2}R} +&+ \gamma_{1} \int \frac{dx}{(x^2 + a)^{n - 1}R}\\ +&+ \delta_{1} \int \frac{dx}{(x^2 + a)^n R}. +\end{align*} +Making $n = 2$, we have +\begin{align*} +\Tag{(16)} +\frac{xR}{(x^2 + a)^{\DPtypo{-1}{1}}} + = \alpha_{1} \int \frac{(x^2 + a)\, dx}{R} + + \beta_{1} \int \frac{dx}{R} + &+ \gamma_{1} \int \frac{dx}{(x^2 + a)R} \\ + &+ \delta_{1} \int \frac{dx}{(x^2 + a)^2 R}. +\end{align*} + +\Eqref{Equation}{}{(16)} shows that +\[ +\int \frac{dx}{(x^2 + a)^2 R} +\] +depends upon the three forms +\[ +\int \frac{x^2\, dx}{R},\quad +\int \frac{dx}{R},\quad \text{and}\quad +\int \frac{dx}{(x^2 + a)R}, +\] +%% -----File: 020.png---Folio 14------- +the three types of \Eqref{equation}{}{(2)}, and \Eqref{equation}{}{(15)} shows that +the general form +\[ +\int \frac{dx}{(x^2 + a)^n R} +\] +depends ultimately upon the same three types. + +We have now discussed every form which the general \Eqref{equation}{}{(1)} +can assume, and shown that they all depend ultimately +upon one or more of the three types contained in \Eqref{equation}{}{(2)}. + +These three types are called the three Elliptic Integrals of +the first, second, and third kind, respectively. + +Legendre puts $x = \sin \phi$, and reduces the three integrals +to the following forms: +\begin{align*}%[** PP: Aligning next three lines] +\Tag{(17)} +F(k, \phi) &= \int_{0}^{\phi} \frac{d \phi}{\sqrt{1 - k^2 \sin^2 \phi}}; \\ +&\quad \llap{$\dfrac{1}{k^2}$} + \int_{0}^{\phi} \frac{d \phi}{\sqrt{1 - k^2 \sin^2 \phi}} + - \frac{1}{k^2} \int_{0}^{\phi} \sqrt{1 - k^2 \sin^{2} \phi} · d \phi; \\ +\Tag{(18)} +\varPi(n, k, \phi) + &= \int_{0}^{\phi} \frac{d \phi}{(1 - n \sin^2 \phi) \sqrt{1 - k^2 \sin^2 \phi}}; +\end{align*} +the first being Legendre's integral of the first kind; the form +\[ +\Tag{(19)} +E(k, \phi) = \int_{0}^{\phi} \sqrt{1 - k^2 \sin^2 \phi} · d \phi +\] +being the integral of the second kind; and the third one being +the integral of the third kind. + +The form of the integral of the second kind shows why they +are called Elliptic Integrals, the arc of an elliptic quadrant +being equal to +\[ +a \int_{0}^{\frac{\pi}{2}} \sqrt{1 - e^2 \sin^2 \phi} · d\phi, +\] +$\phi$ being the complement of the eccentric angle. +%% -----File: 021.png---Folio 15------- + +By easy substitutions, we get from \Eqref{Eqs.}{}{(17)},~\Eqref{}{}{(18)}, and~\Eqref{}{}{(19)} +the following solutions: +\setlength{\TmpLen}{1.5ex}% +\begin{align*} +\int_{0}^{\phi} \frac{\sin^2 \phi}{\Delta}\, d\phi + &= \frac{F - E}{k^2}; \\[\TmpLen] +% +\int_{0}^{\phi} \frac{\cos^2 \phi}{\Delta}\, d\phi + &= \frac{E - (1 - k^2)F}{k^2}; \\[\TmpLen] +% +\int_{0}^{\phi} \frac{\tan^2 \phi}{\Delta}\, d\phi + &= \frac{\Delta \tan \phi - E}{1 - k^2}; \\[\TmpLen] +% +\int_{0}^{\phi} \frac{\sec^2 \phi}{\Delta}\, d\phi + &= \frac{\Delta \tan \phi + (1 - k^2)F - E}{1 - k^2}; \\[\TmpLen] +% +\int_{0}^{\phi} \frac{1}{\Delta^3}\, d\phi + &= \frac{1}{1 - k^2} \left(E - \frac{k^2 \sin\phi \cos\phi}{\Delta} \right); \\[\TmpLen] +% +\int_{0}^{\phi} \frac{\sin^2 \phi}{\Delta^3}\, d\phi + &= \frac{1}{1 - k^2} \left( \frac{E - (1 - k^2)F}{k^2} + - \frac{\sin\phi \cos\phi}{\Delta} \right); \\[\TmpLen] +% +\int_{0}^{\phi} \frac{\cos^2 \phi}{\Delta^3}\, d\phi + &= \frac{F - E}{k^2} + \frac{\sin\phi \cos\phi}{\Delta}. +\end{align*} +%% -----File: 022.png---Folio 16------- + + +\Chapter{II}{Elliptic Functions.} + +\begin{DPgather*} +\lintertext{\indent\First{Let}} +u = \int_0^\phi \frac{d\phi}{\sqrt{1 - k^2 \sin^2 \phi}}. +\end{DPgather*} + +$\phi$\footnotemark~is called the \emph{amplitude} corresponding to the \emph{argument}~$u$, +and is written +\footnotetext{Legendre.} +\[ +\phi = \am (u, k) = \am u. +\] + +The quantity~$k$ is called the \emph{modulus}, and the expression +$\sqrt{1 - k^2 \sin^2 \phi}$ is written\footnotemark[1] +\[ +\sqrt{1 - k^2 \sin^2 \phi} = \Delta \am u = \Delta \phi, +\] +and is called the \emph{delta function} of the amplitude of~$u$, or \emph{delta +of~$\phi$}, or simply \emph{delta}~$\phi$. + +$u$~can be written +\[ +u = F(k, \phi). +\] + +The following abbreviations are used: +\begin{align*} +\sin \phi &= \sin \am u = \sn\footnotemark u; \\ +\cos \phi &= \cos \am u = \cn\footnotemark[2] u; \\ +\Delta \phi &= \Delta \am u = \dn\footnotemark[2] u = \Delta u; \\ +\tan \phi &= \tan \am u = \tn u. +\end{align*} +\footnotetext{\textit{Gudermann}, in his ``Theorie der Modularfunctionen'': Crelle's Journal, + Bd.~18.}% + +Let $\phi$ and~$\psi$ be any two arbitrary angles, and put +\begin{align*} +\phi &= \am u;\\ +\psi &= \am \nu.\\ +\end{align*} +%% -----File: 023.png---Folio 17------- + +%[Illustration] +\begin{wrapfigure}{r}{1.25in} + \Input[1in]{023a} +\end{wrapfigure} +In the spherical triangle~$ABC$ we have from +Trigonometry, $\DPtypo{c}{\mu}$~and~$C$ being constant, +\[ +\frac{d\phi}{\cos B} + \frac{d\psi}{\cos A} = 0. +\] + +Since $C$~and~$\DPtypo{c}{\mu}$ are constant, denoting by~$k$ an arbitrary constant, +we have +\[ +\Tag{(1)} +\frac{\sin C}{\sin \mu} = k. +\] + +But +\[ +\sin A + = \sin\psi \frac{\sin B}{\sin \phi} + = \sin\psi \frac{\sin C}{\sin \mu} + = k \sin\psi. +\] + +Whence +\[ +\cos A + = \sqrt{1 - \sin^{2}A} + = \sqrt{1 - k^{2} \sin^{2} \psi}. +\] + +In the same manner +\[ +\cos B + = \sqrt{1 - \sin^{2}B} + = \sqrt{1 - k^{2} \sin^{2} \phi}. +\] + +Substituting these values, we get +\[ +\Tag{(2)} +\frac{d\phi}{\sqrt{1 - k^{2} \sin^{2} \phi}} + +\frac{d\psi}{\sqrt{1 - k^{2} \sin^{2} \psi}} = 0. +\] + +Integrating this, there results +\[ +\Tag{(3)} +\int_0^\phi \frac{d\phi}{\sqrt{1 - \DPtypo{k_2}{k^2} \sin^{2}\phi}} + +\int_0^\psi \frac{d\psi}{\sqrt{1 - k^2 \sin^2\psi}} = \text{const}. +\] + +When $\phi = 0$, we have $\psi = \mu$, and therefore the constant +must be of the form +\[ +\int_0^\mu \frac{d\phi}{\sqrt{1 - k^2 \sin^2 \phi}}, +\] +%% -----File: 024.png---Folio 18------- +whence +\[ +\Tag{(4)} +\int_0^\phi \frac{d \phi}{\sqrt{1 - k^2 \sin^2 \phi}} + +\int_0^\psi \frac{d \psi}{\sqrt{1 - k^2 \sin^2 \psi}} = +\int_0^\mu \frac{d \phi}{\sqrt{1 - k^2 \sin^2 \phi}}, +\] +or +\[ +u + \nu = m; +\] +and evidently the amplitudes $\phi$,~$\psi$, and~$\mu$ can be considered as +the three sides of a spherical triangle, and the relations between +the sides of this spherical triangle will be the same as those +between $\phi$,~$\psi$, and~$\mu$. + +%[Illustration] +\begin{wrapfigure}{l}{1.125in} + \Input{024a} +\end{wrapfigure} +But the sides of this triangle have imposed upon them the +condition +\[ +\frac{\sin C}{\sin \mu} = k; +\] +and since $k < 1$, we must have $\mu > C$, which requires that one +of the angles of the triangle shall be obtuse and the other two +acute. + +In the figure, let $C$~be an acute angle of the triangle~$ABC$, +and $PQ$~the equatorial great circle of which $C$~is the pole. + +%[Illustration] +\ifthenelse{\boolean{ForPrinting}}{% + \begin{wrapfigure}[11]{r}{2.125in} + \vspace{-1.5\baselineskip} + \hfill\Input[2in]{025a} + \vspace{1.5\baselineskip} + \end{wrapfigure} +}{% + \begin{wrapfigure}{r}{2.125in} + \hfill\Input[2in]{025a} + \end{wrapfigure} +} +The arc~$PQ$ will be the measure of the angle~$C$. + +Let $AG$~and~$AH$ be the arcs of two great +circles perpendicular respectively to $CQ$ and~$CP$. +They will of course be shorter than~$PQ$. +Hence $AB = \mu$ must intersect~$CQ$ in points +between $CG$ and~$HQ$, since $\mu > (C=PQ)$. In +any case either $A$~or~$B$ will be obtuse according +as $B$~falls between $QH$ or~$CG$ respectively; and +the other angle will be acute. + +In the case where $C$~is an obtuse angle, it will be easily seen +that the angle at~$A$ must be acute, since the great circle~$AD$, +perpendicular to~$CP$, intersects~$PQ$ in~$D$, $PD$~being a quadrant. +The same remarks apply to the angle~$B$. Hence, in either +%% -----File: 025.png---Folio 19------- +case, one of the angles of the triangle is +obtuse and the other two are acute, as +a result of the condition +\[ +\frac{\sin C }{\sin \mu} = k < 1. +\] + +From Trigonometry we have +\[ +\cos \mu = \cos \phi \cos \psi + \sin \phi \sin \psi \cos C; +\] +and since the angle~$C$ is obtuse, +\[ +\cos C = - \sqrt{1- \sin^2C} = -\sqrt{1 - k^2 \sin^2 \mu}, +\] +and +\[ +\Tag{(5)} +\cos \mu = \cos \phi \cos \psi - \sin \phi \sin \psi \sqrt{1 - k^2 \sin^2 \mu}, +\] +the relation sought. + +The spherical triangle likewise gives the following relations +between the sides: +\[ +\Tag[5st]{(5)^*} +\left\{ +\begin{aligned} +\cos \phi &= \cos \mu \cos \psi + \sin \mu \sin \psi \sqrt{1 - k^2 \sin^2 \phi}; \\ +\cos \psi &= \cos \mu \cos \phi + \sin \mu \sin \phi \sqrt{1 - k^2 \sin^2 \psi}. +\end{aligned} +\right. +\] + +These give, by eliminating $\cos \mu$, +\[ +\sin \mu = \frac{\cos^2 \psi - \cos^2 \phi} + {\sin \phi \cos \psi \Delta \psi - \sin \psi \cos \phi \Delta \phi}; +\] +which, after multiplying by the sum of the terms in the denominator +and substituting $\cos^2 = 1 - \sin^2$, can be written +\[ +\sin \mu + = \frac{(\sin^2 \phi - \sin^2 \psi) + (\sin \phi \cos \psi \Delta \psi + \sin \psi \cos \phi \Delta \phi \DPtypo{}{)}} + { \sin^2 \phi \cos^2 \psi \Delta^2 \psi - \sin^2 \psi \cos^2 \phi \Delta^2 \phi}. +\] + +Since the denominator can be written +\begin{gather*} +(\sin^2 \phi - \sin^2 \psi)(1 - k^2 \sin^2 \phi \sin^2 \psi), \\ +\Tag{(6)} +\sin \mu = \frac{\sin \phi \cos \psi \Delta \psi + + \sin \psi \cos \phi \Delta \phi} + {1 - k^2 \sin^2 \phi \sin^2 \psi}. +\end{gather*} + +In a similar manner we get +\[ +\Tag[6st]{(6)^*} +\left\{ +\begin{aligned} +\cos \mu &= \frac{\cos \phi \cos \psi - \sin \phi \sin \psi \Delta \phi \Delta \psi} + {1 - k^2 \sin^2 \phi \sin^2 \psi}; \\ +% +\Delta \mu &= \frac{\Delta \phi \Delta \psi - k^2 \sin \phi \sin \psi \cos \phi \cos \psi} + {1 - k^2 \sin^2 \phi \sin^2 \psi}. +\end{aligned} +\right. +\] +%% -----File: 026.png---Folio 20------- + +These equations can also be written as follows: +\[ +\Tag{(7)} +\left\{ +\makebox[\linewidth-40pt][l]{$\begin{aligned} +\sin \am (u±\nu) + &= \frac{\sin \am u \cos \am \nu \Delta \am \nu + ± \sin \am \nu \cos \am u \Delta \am u} + {1 - k^{2} \sin^{2} \am u \sin^{2} \am \nu}; \\ +% +\cos \am (u±\nu) + &= \frac{\cos \am u \cos \am \nu + \mp \sin \am u \sin \am \nu \Delta \am u \Delta \am \nu} + {1 - k^{2} \sin^{2} \am u \sin^{2} \am \nu}; \\ +% +\Delta \am (u±\nu) + &= \frac{\Delta \am u \Delta \am \nu + \mp k^{2} \sin \am u \sin \am \nu \cos \am u \cos \am \nu} + {1 - k^{2} \sin^{2} \am u \sin^{2} \am \nu}; +\end{aligned}$} +\right. +\] +or +\[ +\Tag{(8)} +\left\{ +\begin{aligned} +\sn (u±\nu) + &= \frac{\sn u \cn \nu \dn \nu ± \sn \nu \cn u \dn u} + {1 - k^{2} \sn^{2} u \sn^{2} \nu}; \\ +% +\cn (u±\nu) + &= \frac{\cn u \cn \nu \mp \sn u \sn \nu \dn u \dn \nu} + {1 - k^{2} \sn^{2} u \sn ^{2} \nu}; \\ +% +\dn (u±\nu) + &= \frac{\dn u \dn \nu \mp k^{2} \sn u \sn \nu \cn u \cn \nu} + {1 - k^{2} \sn^{2} u \sn^{2} \nu}. +\end{aligned} +\right. +\] + +Making $u = \nu$, we get from the upper sign +\[ +\Tag{(9)} +\left\{ +\begin{aligned} +\sn 2u &= \frac{2 \sn u \cn u \dn u}{1 - k^{2} \sn^{4} u}; \\ +\cn 2u &= \frac{\cn^{2}u - \sn^{2}u \dn^{2}u}{1 - k^{2} \sn^{4}u} + = \frac{1 - 2\sn^{2}u + k^{2}\sn^{4}u}{1 - k^{2}\sn^{4}u}; \\ +\dn 2u &= \frac{\dn^{2}u - k^{2}\sn^{2}u \cn^{2}u}{1 - k^{2}\sn^{4}u} + = \frac{1 - 2k^{2}\sn^{2}u + k^{2}\sn^{4}u}{1 - k^{2}\sn^{4}u}. +\end{aligned} +\right. +\] + +From these +\[ +\Tag{(10)} +\left\{ +\begin{aligned} +1 - \cn 2u &= \frac{2 \cn^{2} u \dn^{2} u}{1 - k^{2} \sn^{4} u}; \\ +1 + \cn 2u &= \frac{2 \cn^{2} u}{1 - k^{2} \sn^{4} u}; \\ +1 - \dn \PadTo[l]{2u}{u} &= \frac{2k^{2} \sn^{2} u \cn^{2} u}{1 - k^{2} \sn^{4} u}; \\ +1 + \dn \PadTo[l]{2u}{u} &= \frac{2 \dn^{2} u}{1 - k^{2} \sn^{4} u}; +\end{aligned} +\right. +\] +%% -----File: 027.png---Folio 21------- +and therefore +\[ +\Tag{(11)} +\left\{ +\begin{aligned} +\sn^{2} u &= \frac{1 - \cn 2u}{1 + \dn 2u}; \\ +\cn^{2} u &= \frac{\dn 2u + \cn 2u}{1 + \dn 2u}; \\ +\dn^{2} u &= \frac{1 - k^{2} + \dn 2u + k^{2}\cn 2u}{1 +\dn 2u}; +\end{aligned} +\right. +\] +and by analogy +\[ +\Tag{(12)} +\left\{ +\begin{aligned} +\sn \dfrac{u}{2} &= \sqrt{\frac{1 - \cn u}{1 + \dn u}}; \\ +\cn \dfrac{u}{2} &= \sqrt{\frac{\cn u + \dn u}{1 + \dn u}}; \\ +\dn \dfrac{u}{2} &= \sqrt{\frac{1 - k^{2} + \dn u + k^{2} \cn u}{1 + \dn u}}. +\end{aligned} +\right. +\] + +In \Eqref{equations}{}{(7)} making $u = \nu$, and taking the lower sign, +we have +\[ +\Tag{(13)} +\left\{ +\begin{aligned} +\sn 0 &= 0; \\ +\cn 0 &= 1; \\ +\dn 0 &= 1. +\end{aligned} +\right. +\] + +Likewise, we get by making $u = 0$, +\[ +\Tag{(14)} +\left\{ +\begin{aligned} +\sn (-u) &= -\sn u; \\ +\cn (-u) &= +\cn u; \\ +\dn (-u) &= \dn u. +\end{aligned} +\right. +\] +%% -----File: 028.png---Folio 22------- + + +\Chapter{III}{Periodicity of the Functions.} + +\First{When} the elliptic integral +\[ +\int_{0}^{\phi} \frac{d \phi}{\sqrt{1 - k^{2} \sin^{2} \phi}} +\] +has for its amplitude~$\dfrac{\pi}{2}$, it is called, following the notation of +Legendre, the \emph{complete} function, and is indicated by~$K$, thus: +\[ +K = \int_{0}^{\frac{\pi}{2}} \frac{d \phi}{\sqrt{1-k^{2} \sin^{2} \phi}}. +\] + +When~$k$ becomes the complementary modulus,~$k'$, (see \Eqref{eq.}{IV}{\DPtypo{4}{(4)}}, +Chap.\ IV,) the corresponding complete function is indicated by~$K'$, +thus: +\[ +K' = \int_{0}^{\frac{\pi}{2}} \frac{d \phi}{\sqrt{1-k'^{2} \sin^{2} \phi}}. +\] + +From these, evidently, +\[ +\am (K, k) = \frac{\pi}{2},\qquad \am (K', k') = \frac{\pi}{2}. +\] +\[ +\Tag{(1)} +\left\{ +\begin{aligned} +\sn (K, k) &= 1; \\ +\cn (K, k) &= 0; \\ +\dn (K, k) &= k'. +\end{aligned} +\right. +\] +%% -----File: 029.png---Folio 23------- + +From \Eqref{eqs.}{II}{(7)},~\Eqref{}{II}{(8)}, and~\Eqref{}{II}{(9)}, Chap.~II, we have, by the substitution +of the values of~$\sn (K) = 1$, $\cn (K) = 0$, $\dn (K) = k'$, +\[ +\Tag{(2)} +\left\{ +\begin{aligned} +\sn 2K &= 0; \\ +\cn 2K &= -1; \\ +\dn 2K &= 1. +\end{aligned} +\right. +\] + +These equations, by means of \Eqref{}{II}{(1)},~\Eqref{}{II}{(2)}, and~\Eqref{}{II}{(3)} of Chap.~II, +give +\[ +\Tag{(3)} +\left\{ +\begin{aligned} +\sn (u+2K) &= - \sn u; \\ +\cn (u+2K) &= -\cn u; \\ +\dn (u+2K) &= \dn u; +\end{aligned} +\right. +\] +and these, by changing $u$ into~$u+2K$, give +\[ +\Tag{(4)} +\left\{ +\begin{aligned} +\sn (u+4K) &= \sn u; \\ +\cn (u+4K) &= \cn u; \\ +\dn (u+4K) &= \dn u. +\end{aligned} +\right. +\] + +From these equations it is seen that the elliptic functions +$\sn$,~$\cn$,~$\dn$, are periodic functions having for their period~$4K$. +Unlike the period of trigonometric functions, this period is not +a fixed one, but depends upon the value of~$k$, the modulus. + +From the Integral Calculus we have +\begin{align*} +\int_{0}^{n \frac{\pi}{2}} \frac{d \phi}{\Delta \phi} + &= \int_{0}^{\frac{\pi}{2}} \frac{d \phi}{\Delta \phi} + + \int_{\frac{\pi}{2}}^{\pi} \frac{d \phi}{\Delta \phi} + + \int_{\pi}^{\frac{3\pi}{2}} \frac{d \phi}{\Delta \phi} + \cdots + + \int_{(n-1)\frac{\pi}{2}}^{n \frac{\pi}{2}} \frac{d \phi}{\Delta \phi} \\ + &= n \int_{0}^{\frac{\pi}{2}} \frac{d \phi}{\Delta \phi} + = nK; +\end{align*} +from which we see that +\begin{DPalign*} +n \frac{\pi}{2} &= \am (nK); \\ +%% -----File: 030.png---Folio 24------- +\intertext{or, since $\dfrac{\pi}{2} = \am K$,} +\am (nK) &= n · \am K, \\ +\intertext{and} +n \pi &= \am (2nK), \\ +\lintertext{and also} +n \pi &= 2n \am K. +\end{DPalign*} +In the case of an Elliptic Integral with the arbitrary angle~$\alpha$, +we can put +\[ +\DPtypo{d}{\alpha} = n \pi ± \beta, +\] +where $\beta$ is an angle between $0$~and~$\dfrac{\pi}{2}$, the upper or the lower +sign being taken according as $\dfrac{\pi}{2}$ is contained in~$\alpha$ an even or +an uneven number of times. + +In the first case we have +\[ +\int_0^{n\pi+\beta} \frac{d \phi}{\Delta \phi} + = \int_0^{n\pi} \frac{d \phi}{\Delta \phi} + + \int_{n\pi}^{n\pi+\beta} \frac{d \phi}{\Delta \phi}; +\] +or, putting $\phi_{1} = \DPtypo{}{\phi} - n \pi$, +\[ +\int_0^{n\pi+\beta} \frac{d \phi}{\Delta \phi} + = 2nK + \int_0^\beta \frac{d \phi_{1}}{\Delta \phi_{1}}. +\] + +In the second case +\[ +\int_0^{n\pi-\beta} \frac{d \phi}{\Delta \phi} + = \int_0^{n\pi} \frac{d \phi}{\Delta \phi} + - \int_{n\pi-\beta}^{n\pi} \frac{d \phi}{\Delta \phi}; +\] +or, putting $\phi_{1} = n \pi - \phi$, +\[ +\int_0^{n\pi-\beta} \frac{d \phi}{\Delta \phi} + = 2nK - \int_0^\beta \frac{d \phi_{1}}{\Delta \phi_{1}}; +\] +%% -----File: 031.png---Folio 25------- +or in either case, +\[ +\int_0^{n\pi±\beta} \frac{d \phi}{ \Delta \phi} + = 2nK ± \int_0^\beta \frac{d \phi_{1} }{ \Delta \phi_{1}}. +\] + +Thus we see that the Integral with the general amplitude~$\alpha$ +can be made to depend upon the complete integral~$K$ and +an Integral whose amplitude lies between $0$~and~$\dfrac{\pi}{2}$. + +Put now +\[ +\int_0^\beta \frac{d \phi_{1} }{ \Delta \phi_1} = u,\qquad \beta = \am u. +\] + +This gives +\begin{DPalign*} +\int_0^{n\pi±\beta} \frac{d \phi }{ \Delta \phi} &= 2nK ± u,\\ +\lintertext{or} +\am (2nK ± u) + &= n \pi ± \beta\\ +\Tag{(5)} + &= n\pi ± \am u\\ +\Tag{(6)} + &= 2n · \am K ± \am u;\\ +\lintertext{or, since} +\am (-z) + &= -\am z,\\ +\am (u ± 2nK) + &= \am u ± n \pi\\ + &= \am u ± 2n · \am K.\\ +\end{DPalign*} + +Taking the sine and cosine of both sides, we have +\begin{align*} +\sn (u + 2nK) &= ± \sn u;\\ +\cn (u + 2nK) &= ± \cn u; +\end{align*} +the upper or the lower sign being taken according as $n$~is even +or odd. By giving the proper values to~$n$ we can get the same +results as in \Eqref{equations}{}{(3)} and~\Eqref{}{}{(4)}. + +Putting $n = 1$ in \Eqref{eq.}{}{(5)}, we have +\begin{align*} +\sn (2K - u) + &= \sin \pi \cn u - \cos \pi \sn u\\ +\Tag{(7)} + &= \sn u. +\end{align*} +%% -----File: 032.png---Folio 26------- + +Elliptic functions also have an imaginary period. In order +to show this we will, in the integral +\[ +\int_0^\phi \frac{d \phi}{\Delta \phi}, +\] +assume the amplitude as imaginary. Put +\[ +\sin \phi = i \tan \psi. +\] +From this we get +\[ +\Tag{(8)} +\left\{ +\begin{aligned} +\cos \phi + &= \frac{1}{\cos \psi}; \\ +\Delta \phi + &= \frac{\sqrt{1 - k'^{2} \sin^{2} \psi}}{\cos \psi} + = \frac{\Delta (\psi, k')}{\cos \psi}; \\ +d \phi + &= i \frac{d \psi}{\cos \psi}. +\end{aligned} +\right. +\] +From these, since $\phi$~and~$\psi$ vanish simultaneously, we easily get +\begin{align*} +\int_0^\phi \frac{d \phi}{\Delta \phi} + &= i \int_0^\psi \frac{d \psi}{\Delta (\psi, k')}. \\ +\intertext{Put} +\int_0^\psi \frac{d \psi}{\Delta (\psi, k')} &= u \quad \text{and} \quad +\psi = \am (u, k'), \\ +\intertext{whence} +\int_0^\phi \frac{d \phi}{\Delta \phi} &= iu \quad \text{and} \quad +\phi = \am (iu); +\end{align*} +and these substituted in \Eqref{Eq.}{}{(8)} give +\[ +\Tag{(9)} +\left\{ +\begin{aligned} +\sn iu &= i \tn (u, k'); \\ +\cn iu &= \frac{1}{\cn (u, k')}; \\ +\dn iu &= \frac{\dn (u, k')}{\cn (u, k')}. +\end{aligned} +\right. +\] +%% -----File: 033.png---Folio 27------- +By assuming +\[ +\int_0^\psi \frac{d \psi}{\Delta (\psi, k')} = iu \quad \text{and}\quad +\int_0^\phi \frac{d \phi}{\Delta \phi} = -u, +\] +we get +\begin{align*} +\sn (-u) &= i \tn (iu, k'), \\ +\cn (-u) &= \frac{1}{\cn(iu, k')}, \\ +\dn (-u) &= \frac{\dn (iu, k')}{\cn (iu, k')}; +\end{align*} +or, from \Eqref{eq.}{II}{(14)}, Chap.~II, +\[ +\Tag{(10)} +\left\{ +\begin{aligned} +\sn u &= -i \tn (iu, k'); \\ +\cn u &= \frac{1}{\cn (iu, k')}; \\ +\dn u &= \frac{\dn (iu, k')}{\cn (iu, k')}. +\end{aligned} +\right. +\] + +From \Eqref{eqs.}{II}{(7)}, Chap.~II, making $\nu = K$, we get, since +$\sn K = 1$, $\cn K = 0$, $\dn K = k'$, +\[ +\Tag{(11)} +\left\{ +\begin{aligned}%[** PP: Moved \mp out of numerator in second equation] +\sn (u ± K) &= ± \frac{\cn u \dn u}{1 - k^{2} \sn^{2} u} = ± \dfrac{\cn u}{\dn u}; \\ +\cn (u ± K) &= \mp \frac{\sn u \dn uk'}{\dn^{2} u} = \mp \frac{k' \sn u}{\dn u}; \\ +\dn (u ± K) &= + \frac{k'}{\dn u}. +\end{aligned} +\right. +\] + +In these equations, changing $u$ into~$iu$, we get, by means of +\Eqref{eqs.}{}{(9)}, +\[ +\Tag{(12)} +\left\{ +\begin{aligned} +\sn (iu ± K) &= ± \frac{1}{\dn (u, k')}; \\ +\cn (iu ± K) &= \mp \frac{ik' \sn (u, k')}{\dn (u, k')}; \\ +\dn (iu ± K) &= + \frac{k' \cn (u, k')}{\dn (u, k')}. +\end{aligned} +\right. +\] +%% -----File: 034.png---Folio 28------- + +Putting now in \Eqref{eqs.}{}{(9)} $u ± K'$ instead of~$u$, and making use +of \Eqref{eqs.}{}{(10)}, and interchanging $k$~and~$k'$, we have +\[ +\Tag{(13)} +\left\{ +\begin{aligned} +\sn (iu ± iK') &= - \frac{i \cn (u, k')}{k \sn (u, k')}; \\ +\cn (iu ± iK') &= \mp \frac{\dn (u, k')}{k \sn (u, k')}; \\ +\dn (iu ± iK') &= \mp \frac{1}{\sn (u, k')}. +\end{aligned} +\right. +\] + +Substituting in these~$-iu$ in place of~$u$, we get, by +means of \Eqref{eqs.}{II}{(9)} and \Eqref{eqs.}{II}{(14)} of Chap.~II, +\[ +\Tag{(14)} +\left\{ +\begin{aligned} +\sn (u ± iK') &= \frac{1}{k \sn u}; \\ +\cn (u ± iK') &= \mp \frac{i \dn u}{k \sn u}; \\ +\dn (u ± iK') &= \mp i \cot \am u. +\end{aligned} +\right. +\] + +In these equations, putting $u + K$ in place of~$u$, we get +\[ +\Tag{(15)} +\left\{ +\begin{aligned} +\sn (u + K ± iK') &= + \frac{\dn u}{k \cn u}; \\ +\cn (u + K ± iK') &= \mp \frac{ik'}{k \cn u}; \\ +\dn (u + K ± iK') &= ± ik' \tn u. +\end{aligned} +\right. +\] +Whence for $u = 0$ we get +\[ +\Tag{(16)} +\left\{ +\begin{aligned} +\sn (K ± iK') &= \frac{1}{k}; \\ +\cn (K ± iK') &= \mp \frac{ik'}{k}; \\ +\dn (K ± iK') &= 0. +\end{aligned} +\right. +\] +%% -----File: 035.png---Folio 29------- + +If in \Eqref{eqs.}{}{(14)} we put $u = 0$, we see that as $u$~approaches +zero, the expressions +\Pagelabel{29} +\[ +\sn (± iK'), \quad \cn (± iK'), \quad \dn (± iK') +\] +approach infinity. + +We see from what has preceded that Elliptic Functions +have two periods, one a real period, and one an imaginary +period. + +In the former characteristic they resemble Trigonometric +Functions, and in the latter Logarithmic Functions. + +On account of these two periods they are often called +Doubly Periodic Functions. Some authors make this double +periodicity the starting-point of their investigations. This +method of investigation gives some very beautiful results and +processes, but not of a kind adapted for an elementary work. + +It will be noticed that the Elliptic Functions $\sn u$,~$\cn u$, and~$\dn u$ +have a very close analogy to trigonometric functions, in +which, however, the independent variable~$u$ is not an angle, as +it is in the case of trigonometric functions. + +Like Trigonometric Functions, these Elliptic Functions can +be arranged in tables. These tables, however, require a double +argument, viz., $u$~and~$k$. In \Chapref{Chap.}{IX} these functions are developed +into series, from which their values may be computed +and tables formed. + +No complete tables have yet been published, though they +are in process of computation. +%% -----File: 036.png---Folio 30------- + + +\Chapter{IV}{Landen's Transformation} + +%[Illustration] +\begin{wrapfigure}[6]{l}{2in} + \vspace{-\baselineskip} + \Input[2in]{036a} + \vspace{\baselineskip} + \Pagelabel{30}% +\end{wrapfigure} +\First{Let} $AB$~be the diameter of a circle, +with the centre at~$O$, the radius $AO = r$, +and $C$~a fixed point situated upon~$OB$, +and $OC = k_{0}r$. Denote the angle~$PBC$ +by~$\phi$, and the angle~$PCO$ by~$\phi_{1}$. Let +$P'$~be a point indefinitely near to~$P$. + +Then +\[ +\frac{PP'}{PC} = \frac{\sin PCP'}{\sin PP'C} = \frac{\sin PCP'}{\cos OP'C}. +\] + +But $PP' = 2r\, d\phi$, and $\sin PCP' = PCP' = d \phi_{1}$; +therefore +\[ +\frac{2r\, d\phi}{PC} = \frac{d \phi_{1}}{\cos OP'C}. +\] + +But +\begin{align*} +\overline{PC}^{2} + &= r^{2} + r^{2}k_{0}^{2} + 2r^{2}k_{0} \cos 2 \phi \\ + &= (r + rk_{0})^{2} \cos^{2} \phi + (r - rk_{0})^{2} \sin^{2} \phi; +\end{align*} +\NegMathSkip +\begin{DPalign*} +\lintertext{also} +r^{2} \cos^{2} OP'C + &= r^{2} - r^{2} \sin^{2} OP'C \\ + &= r^{2} - r^{2}k_{0}^{2} \sin^{2} \phi_{1}. +\end{DPalign*} + +Therefore %[** PP: Next two equations aligned in original] +\[ +\frac{2\, d\phi}{\sqrt{(r + rk_{0})^{2} \cos^{2} \phi + \DPtypo{(r - rk_{0})}{(r - rk_{0})^{2}} \sin^{2} \phi}} + = \frac{d \phi_{1}}{\sqrt{r^{2} - r^{2}k_{0}^{2} \sin^{2} \phi_{1}}}, +\] +which can be written +\[ +\frac{2}{r + rk_{0}}\, + \frac{d \phi}{\sqrt{1 - \dfrac{4k_{0}r^{2}}{(r + rk_{0})^{2}} \sin^{2} \phi}} + = \frac{1}{r}\, \frac{d \phi_{1}}{\sqrt{1 - k_{0}^{2} \sin^{2} \phi_{1}}}, +\] +%% -----File: 037.png---Folio 31------- + +Putting +\[ +\Tag{(1)} +\frac{4k_{0}r^{2}}{(r + rk_{0})^{2}} = \frac{4k_{0}}{(1 + k_{0})^{2}} = k^{2}, +\] +we have +\[ +\Tag{(2)} +\int_{0}^{\phi} \frac{d \phi}{\sqrt{1 - k^{2} \sin^{2} \phi}} + = \frac{1 + k_{0}}{2} + \int_{0}^{\phi_{1}} \frac{d \phi_{1}}{\sqrt{1 - k_{0}^{2} \sin^{2} \phi_{1}}}; +\] +no constant being added because $\phi$ and~$\phi_{1}$ vanish simultaneously; +$\phi$ and~$\phi_{1}$ being connected by the equation +\[ +\Tag{(3)} +\frac{\sin OPC}{\sin OCP} + = \frac{\sin (2 \phi - \phi_{1})}{\sin \phi_{1}} + = \frac{rk_{0}}{r} + = k_{0}. +\] + +From the value of~$k^{2}$ we have +\[ +\Tag{(4)} +1 - k^{2} = k'^{2} = \frac{(1 - k_{0})^{2}}{(1 + k_{0})^{2}}, +\] +and therefore +\[ +\Tag{(5)} +k_{0} = \frac{1 - k'}{1 + k'}. +\] +$k'$~is called the \emph{complementary modulus}, and is evidently the +minimum value of~$\Delta \phi$, the value of~$\Delta \phi$ when~$\phi = 90°$: +\[ +\sqrt{1 - k^{2}} = k'. +\] + +From \Eqref{eq.}{}{(1)} we evidently have $k > k_{0}$, for, putting \Eqref{eq.}{}{(1)} +in the form +\[ +\frac{k^{2}}{k_{0}^{2}} = \frac{4}{k_{0} + 2k_{0}^{2} + k_{0}^{3}}, +\] +we see that if $k_{0} = 1$, then $k = k_{0}$, but as $k_{0} < 1$, always, as is +evident from the figure, $k$~must be greater than~$k_{0}$. + +It is also evident, from the figure, that~$\phi_{1} > \phi$. Or it may +be deduced directly from \Eqref{eq.}{}{(3)}. + +Since $k < 1$, we can write +\[ +k = \sin \theta, \qquad k' = \sqrt{1 - k^{2}} = \cos \theta. +\] +%% -----File: 038.png---Folio 32------- + +Substituting in \Eqref{eq.}{}{(5)}, we have +\[ +k_{0} = \frac{1 - k'}{1 + k'} = \tan^2 \tfrac{1}{2} \theta, +\] +and we can write +\[ +k_{0} = \sin \theta_{0}, \qquad k_{1}' = \sqrt{1 - k_{0}^2} = \cos \theta_{0}. +\] + +From \Eqref{eq.}{}{(5)} we have +\[ +1 + k_{0} = \frac{2}{1 + k'}. +\] + +Substituting the value of~$k_{0}$ in that for~$k_{1}'$, we get +\[ +k_{1}' = \frac{2\sqrt{k'}}{1 + k'}. +\] + +We also have +\begin{DPalign*} %[** aligning next group] +2 \phi - \phi_{1} & = \phi - (\phi_{1} - \phi) \\ +\phi_{1} & = \phi + (\phi_{1} - \phi), \\ +\lintertext{and, \Eqref{eq.}{}{(3)},} +\sn (2\phi - \phi_{1}) &= k_{0} \sin \phi_{1}, +\end{DPalign*} +becomes +\begin{multline*} %[** PP: Moving = to second line] +\sin \phi \cos (\phi_{1} - \phi) - \cos \phi \sin (\phi_{1} - \phi) \\ += k_{0} \sin \phi \cos (\phi_{1} - \phi) + k_{0} \cos \phi \sin (\phi_{1} - \phi), +\end{multline*} +or +\[ +\tan \phi - \tan (\phi_{1} - \phi) + = k_{0} \tan \phi + k_{0} \tan (\phi_{1} - \phi), +\] +or +\begin{align*} +\tan (\phi_{1} - \phi) + &= \frac{1 - k_{0}}{1 + k_{0}} \tan \phi \\ + &= k' \tan \phi. +\end{align*} + +Collecting these results, we have +\begin{align*} +\Tag{(6)} +k &= \frac{2\sqrt{k_{0}}}{1 + k_{0}} = \sin \theta; \\ +\Tag{(7)} +k_{0} + &= \frac{1 - k'}{1 + k'} + = \sin \theta_{0} + = \tan^{2} \tfrac{1}{2} \theta; +\end{align*} +%% -----File: 039.png---Folio 33------- +\begin{align*} +\Tag{(8)} +k_{1}' &= \frac{2\sqrt{k'}}{1 + k'} = \cos \theta_{0}; \\ +\Tag{(9)} +k' &= \frac{1 - k_{0}}{1 + k_{0}} = \cos \theta; \\ +\Tag{(10)} +1 + k_{0} &= \frac{2}{1 + k'} + = \frac{2\sqrt{k_{0}}}{k} + = \frac{k_{1}'}{\sqrt{k'}} + = \frac{1}{\cos^{2} \frac{1}{2} \theta}; +\end{align*} +\begin{align*} +\Tag{(11)} +\sin (2 \phi - \phi_{1}) &= k_{0} \sin \phi_{1}; \\ +\Tag{(12)} +\tan (\phi_{1} - \phi) &= k' \tan \phi; +\end{align*} +\begin{align*} +\Tag{(13)} +\int_{0}^{\phi} \frac{d \phi}{\Delta (k, \phi)} + &= \frac{1 + k_{0}}{2} \int_{0}^{\phi_{1}} \frac{d \phi_{1}}{\Delta (k_{0}, \phi_{1})}; \\ +\Tag{(14)} +k &= \sqrt{1 - k'^{2}}, \quad k' = \sqrt{1 - k^{2}}. +\end{align*} + +Upon examination it will easily appear that $k$ and~$k_{0}$, and $\theta$ +and~$\theta_{0}$, are the first two terms of a decreasing series of moduli +and angles; $k'$~and~$k_{1}'$, and $\phi$~and~$\phi_{1}$, of an increasing series; +the law connecting the different terms of the series being deduced +from \Eqref{eqs.}{}{(6)} to~\Eqref{}{}{(12)}. + +By repeated applications of these equations we would get +the following series of moduli and amplitudes: +\[ +\renewcommand{\arraystretch}{\SMSTR}% +\begin{array}{l*{2}{>{\qquad}l}} +k_{0n} = 0_{(n = \infty)} & k_{n}' = 1_{(n = \infty)} & \phi_{n} \\ +\PadTo{k_{00}}{\vdots} & \PadTo{k'_n}{\vdots} & \PadTo{\phi_n}{\vdots} \\ +k_{00} & k_{2}' & \phi_{2} \\ +k_{0} & k_{1}' & \phi_{1} \\ +k & k' & \phi +\end{array} +\] + +The upper limit of the one series of moduli is~$1$, and the +lower limit of the other series is~$0$, as is indicated. $k$~and~$k'$, +%% -----File: 040.png---Folio 34------- +which are bound by the relation $k^{2} + k'^{2} = 1$, are called the +\emph{primitives} of the series. + +\begin{Remark} +It will be noticed that the successive terms of a decreasing series +are indicated by the sub-accents $0, 00, 03, 04, \ldots 0n$; and the successive terms +of an increasing series by the sub-accents $1, 2, 3, \ldots n$. +\end{Remark} + +Again, by application of these equations, we can form a new +series running up from~$k$, viz., $k_{1}, k_{2}, k_{3}, \ldots k_{n} = 1_{(n = \infty)}$; and +also a new series running down from~$k'$, viz., $k_{0}', k'_{00}, \ldots \DPtypo{k_{0n}}{k_{0n}'} = +0_{(n = \infty)}$. So also with~$\phi$. + +Collecting these series, we have +\[ +\renewcommand{\arraystretch}{\SMSTR}% +\begin{array}{l@{}*{2}{p{1in}l}} +k_{0n} \rlap{$ = 0$} && k_{n}' \rlap{$ = 1$} && \phi_{n} \\ +\PadTo{k_{00}}{\vdots} && \PadTo{k'_n}{\vdots} && \PadTo{\phi_n}{\vdots} \\ +k_{02} && k_{2}' && \phi_{2} \\ +k_{0} && k_{1}' && \phi_{1} \\ +k & \Dots{1} & k' & \Dots{1} & \phi \\ +k_{1} && k_{0}' && \phi_{0} \\ +k_{2} && k'_{00}&& \phi_{00} \\ +\PadTo{k_{00}}{\vdots} && \PadTo{k'_n}{\vdots} && \PadTo{\phi_n}{\vdots} \\ +k_{n} \rlap{$ = 1$} && k'_{0n} \rlap{$ = 0$} && \phi_{0n}=0 +\end{array} +\] + +\begin{Remark} +In practice it will be found that generally $n$ will not need to be very +large in order to reach the limiting values of the terms, often only two or three +terms being needed. +\end{Remark} + +Applying \Eqref{eqs.}{}{(7)}, \Eqref{}{}{(12)},~\Eqref{}{}{(13)}, and~\Eqref{}{}{(14)} repeatedly, we get +\[ +\Tag[14sub1]{(14_{1})} +\left\{\begin{array}{ll} +k=\sin \theta, & k' = \cos \theta; \\ +k_{0} = \dfrac{1-k'}{1+k'} = \tan^{2} \frac{1}{2} \theta = \sin \theta_{0}, \qquad & k_{1}' = \cos \theta_{0}; \\ +k_{00} = \tan^{2} \frac{1}{2} \theta_{0} = \sin \theta_{00}, & k_{2}' = \cos \theta_{00}; \\ +k_{03} = \tan^{2} \frac{1}{2} \theta_{00} = \sin \theta_{03}, & k_{3}' = \cos \theta_{03}; \\ +\Dots{2} \\ +k_{0n} = \tan^{2} \frac{1}{2} \theta_{0(n-1)} = \sin \theta_{0n}, & k_{n}' = \cos \theta_{0n}. +\end{array}\right. +\] +%% -----File: 041.png---Folio 35------- +\begin{align*} +\Tag[14sub2]{(14_2)} +&\left\{ +\begin{array}{l} +\tan (\phi_{1} - \phi) = k' \tan \phi;\\ +\tan (\phi_{2} - \phi_{1}) = k_{1}' \tan \phi_{1};\\ +\tan (\phi_{3} - \phi_{2}) = k_{2}' \tan \phi_{2};\\ +\Dots{1} \\ +\tan (\phi_{n} - \phi_{n - 1}) = k'_{(n - 1)} \tan \phi_{n - 1}. +\end{array} \right. \\ +% +\Tag[14sub3]{(14_3)} +&\left\{ +\begin{array}{r@{}l} +F(k, \phi) &{}= \dfrac{1 + k_{0}}{2} F(k_{0}, \phi_{1});\\ +F(k_{0}, \phi_{1}) &{}= \dfrac{1 + k_{00}}{2} F(k_{00}, \phi_{2});\\ +F(k_{00}, \phi_{2}) &{}= \dfrac{1 + k_{03}}{2} F(k_{03}, \phi_{3});\\ +\Dots{2} \\ +F(k_{0(n - 1)}, \phi_{n - 1}) &{}= \dfrac{1 + k_{0n}}{2} F(k_{0n}, \phi_{n}). +\end{array} \right. +\end{align*} + +Multiplying these latter equations together, member by +member, we have +\[ +\Tag{(15)} +F(k, \phi) = (1 + k_{0})(1 + k_{00}) \dotsm (1 + k_{0n}) + \frac{F(k_{0n}, \phi_{n})}{2^{n}}; +\] +$k_{0}$,~$k_{00}$,~etc., and $\phi_{1}$,~$\phi_{2}$,~etc., being determined from the preceding +equations. + +From \Eqref{eqs.}{}{(9)} and~\Eqref{}{}{(10)} we get +\[ +1 + k_{0} = \frac{1}{\cos^{2} \frac{1}{2} \theta},\qquad +1 + k_{00} = \frac{1}{\cos^{2} \frac{1}{2} \theta_{0}},\quad \text{etc.} +\] + +Substituting these in \Eqref{eq.}{}{(15)}, we get +\[ +\Tag{(16)} +F(k, \phi) + = \frac{1}{\cos^{2} \dfrac{\theta}{2} + \cos^{2} \dfrac{\theta_{0}}{2} \dotsm + \cos^{2} \dfrac{\theta_{0n}}{2}} + · \frac{F(k_{0n}, \phi_{n})}{2^{n}}. +\] +%% -----File: 042.png---Folio 36------- + +From \Eqref{eqs.}{}{(15)} and~\Eqref{}{}{(10)} we get +\[ +F(k, \phi) + = \sqrt{\frac{k_{1}' k_{2}' k_{3}' \dotsm k_{n}'^{2}}{k'}} + · \frac{F(k_{0n}, \phi_{n})}{2^{n}}. +\] +And this with \Eqref{equations}{}{(8)} and~\Eqref{}{}{(9)} gives +\[ +\Tag{(17)} +F(k, \phi) + = \sqrt{\frac{\cos \theta_{0} \cos \theta_{00} \dotsm \cos^{2} \theta_{0n}}{\cos \theta}} + · \frac{F(k_{0n}, \phi_{n})}{2^{n}}. +\] + +Applying \Eqref{equation}{}{(13)} to $(k_{1}, \phi_{0})$, $(k_{2}, \phi_{00})$,~etc., we get +\[ +F(k_{1}, \phi_{0}) = \frac{1 + k}{2} F(k, \phi),\ \text{etc.}; +\] +but since, \Eqref{eq.}{}{(10)}, +\[ +\frac{1 + k}{2} = \frac{1}{1 + k_{0}'},\ \text{etc.}, +\] +these become +\[ +\begin{array}{r@{}l} +F(k, \phi) + &{} = (1 + k_{0}') F(k_{1}, \phi_{0}); \\ +F(k_{1}, \phi_{0}) + &{} = (1 + k'_{00}) F(k_{2}, \phi_{00}); \\ +\Dots{2} \\ +\llap{$F(k_{n-1}, \phi_{0(n-1)})$} + &{} = (1 + k'_{0n}) F(k_{\DPtypo{00}{n}}, \phi_{0n}); +\end{array} +\] +whence +\[ +\Tag{(18)} +F(k, \phi) = (1 + k_{0}')(1 + k'_{00}) \dotsm (1 + k'_{0n})F(k_{n}, \phi_{0n}), +\] +in which $k_{0}'$, $k_{00}'$,~etc., $k_{1}$, $k_{2}$,~etc., $\phi_{0}$, $\phi_{00}$,~etc., are determined +as follows: +\begin{DPalign*} +\lintertext{\indent Let} +k &= \sin \theta, \\ +k_{1} & = \sin \theta_{1}. +\end{DPalign*} +From \Eqref{eq.}{}{(10)}, +\[ +k_{1} = \frac{2\sqrt{k}}{1 + k} \quad \text{or}\quad +\sin \theta_{1} = \frac{2\sqrt{\sin \theta}}{1 + \sin \theta}. +\] +%% -----File: 043.png---Folio 37------- + +Solving this equation for~$\sin \theta$, we get +\[ +\sin \theta = \tan^{2} \tfrac{1}{2} \theta_{1}. +\] + +Hence we can write +\begin{align*} +\Tag[18sub1]{(18_{1})} +&\left\{\begin{array}{l} +k = \sin \theta = \tan^{2} \frac{1}{2} \theta_{1}; \\ +k_{1} = \sin \theta_{1} = \tan^{2} \frac{1}{2} \theta_{2}; \\ +\Dots{1} \\ +k_{n} = \sin \theta_{n}. +\end{array}\right. \\ +\intertext{\indent From \Eqref{equation}{}{(12)} we get} +\Tag[18sub2]{(18_{2})} +&\left\{\begin{array}{l} +\sin (2\phi_{0} - \phi) = k \sin \phi;\footnotemark \\ +\sin (2\phi_{00} - \phi_{0}) = k_{1} \sin \phi_{0}; \\ +\Dots{1} \\ +\rlap{$\sin (2\phi_{0n} - \phi_{0(n-1)}) = k_{n-1} \sin \phi_{0(n-1)}$.} +\end{array}\right. +\end{align*} +\footnotetext{When $\sin \phi = 1$ nearly, $\phi$~is best determined as follows: From \Eqref{eq.}{}{(12)} + we have + \begin{align*} + \tan (\phi - \phi_{0}) + &= k_{0}' \tan \phi_{0} \\ + &= k_{0}' \tan \phi\ \text{nearly}; + \intertext{whence} + \phi - \phi_{0} + &= Rk_{0}' \tan \phi\ \text{nearly}, + \end{align*} + $R$~being the radian in seconds, viz.\ $206264''.806$, and $\log R = 5.3144251$. + + Substituting the approximate value of~$\phi_{0}$, we can get a new approximation. + + \textit{Example}.\qquad\qquad $\phi_{0} = 82°\ 30'$\qquad\qquad\qquad $k'_{00} = \log^{-1} 5.8757219$ + \[ + \begin{array}{l<{\qquad\qquad}@{}r@{}>{\qquad}c} + \tan 82°\ 30' & 10.8805709 & \\ + k'_{00} & 5.8757219 & \\ + R & 5.3144251 & \\ + \cline{2-2} + & 2.0707179 & 117''.684 = 1'.9614 + \end{array} + \] + \begin{DPalign*} + \phi_{0} - \phi_{00} & = 1'.9614 \\ + \phi_{00} & = 82°\ 28'.0386 \rintertext{1st approximation.} + \end{DPalign*} + + This value gives + \begin{DPalign*} + \phi_{0} - \phi_{00} &= 117''.1675 = 1'.95279\\ + \therefore \phi_{00} &= 82°\ 28'.04721 + \rintertext{2d~approximation.}\\ + \intertext{\indent This value gives} + \phi_{0} - \phi_{00} &= 117''.1698 = 1'.95283\\ + \phi_{00} &= 82°\ 28'.04717 + \rintertext{3d~approximation.} + \end{DPalign*} +} +%% -----File: 044.png---Folio 38------- + +To determine $k'_{0}$, $k'_{00}$,~etc., we have +\[ +\Tag[18sub3]{(18_3)} +\left\{ +\begin{aligned} +k' &= \sin \eta, & k &= \cos \eta;\\ +k'_{0} &= \frac{1 - k}{1 + k} = \tan^{2} \tfrac{1}{2} \eta = \sin \eta_{0}, + & k_{1} &= \cos \eta_{0};\\ +k'_{00} &= \tan^{2} \tfrac{1}{2} \eta_{0} = \sin \eta_{00}, + & k_{2} &= \cos \eta_{00};\\ + & \quad \PadTo[l]{\tan^{2} \tfrac{1}{2} \eta_{0}}{\quad\text{etc.}} + \phantom{{}={}} + \PadTo{\sin \eta_{00}}{\quad\text{etc.}} & + & \PadTo{\cos \eta_{00}}{\text{etc.}} +\end{aligned} +\right. +\] + +Or, since $1 + k'_0 = \dfrac{1}{\cos^2 \frac{1}{2} \eta}$,\quad $1 + k'_{00} = \dfrac{1}{\cos^2 \frac{1}{2} \eta_0}$, \text{etc.}, +we can put \Eqref{eq.}{}{(18)} in the following form: +\[ +\Tag{(19)} +F(k, \phi) + = \frac{1}{\cos^{2} \frac{1}{2}\eta\, + \cos^{2} \frac{1}{2}\eta_{0} \dotsm + \cos^{2} \frac{1}{2}\eta_{0n}}\, F(k_{n}, \phi_{0n}). +\] + +From \Eqref{equation}{}{(13)} we have +\begin{align*} +\Tag[19st]{(19)^*} +F(k_1, \phi_0) &= \frac{1 + k}{2} F(k, \phi), \\ +\intertext{whence} +F(k, \phi) &= \frac{2}{1 + k} F(k_1, \phi_0). +\end{align*} + +By repeated applications this gives, after combining, +\begin{align*} +F(k, \phi) + &= \frac{2}{1 + k} + · \frac{2}{1 + k_{1}} \dotsm + \frac{2}{1 + k_{n - 1}} · F(k_{n}, \phi_{0n}) \\ +%% -----File: 045.png---Folio 39------- + &= \frac{k_{1}}{\sqrt{k}} · + \frac{k_{2}}{\sqrt{k_{1}}} \dotsm + \frac{k_{n}}{\sqrt{k_{n-1}}} · F(k_{n}, \phi_{0n}); +\end{align*} +\NegMathSkip +\[ +\Tag{(20)} +F(k, \phi) = \sqrt{\frac{k_{1} k_{2} \dotsm k_{n}^{2}}{k}} · F(k_{n}, \phi_{0n}); +\] +$k_{1}$,~$k_{2}$,~etc., being determined by repeated applications of +\[ +k_{1} = \frac{2\sqrt{k}}{1 + k}, +\] +or by \Eqref[18sub1]{equations}{}{(18_{1})}. + +In \Eqref[19st]{equation}{}{(19)^*} let us change $k_{1}$~and~$\phi_{0}$ into $k'$~and~$\phi$ respectively, +so that the first member may have for its complete +function +\[ +K' = F(k', \phi). +\] + +Upon examination of \Eqref[19st]{eq.}{}{(19)^*} we see that the modulus in +the second member must be the one next less than the one in +the first member, that is,~$k_{0}'$; and likewise that the amplitude +must be the one next greater than the amplitude in the first +member, viz.,~$\phi_{1}$; hence we get +\[ +F(k', \phi) = \frac{1 + k_{0}'}{2}\, F(k_{0}', \phi_{1}). +\] + +Indicating the complete functions by $K'$~and~$K_{0}'$, we have, +since $\phi = \dfrac{\pi}{2}$ when $\phi_{1} = \pi$ (see \Chapref{Chap.}{V}), +\[ +K' = (1 + k_{0}')K_{0}'; +\] +and in the same manner, +\[ +\begin{array}{r@{}l} +K_{0}' &{}= (1+k_{00}') K_{00}', \\ +K_{00}' &{}= (1+k_{03}') K_{03}', \\ +\Dots{2} \\ +\llap{$K_{0(n-1)}'$} &{}= (1 + k_{0n}')K_{0n}'; +\end{array} +\] +%% -----File: 046.png---Folio 40------- +whence +\[ +K' = (1 + k_{0}')(1 + k_{00}') \dotsm (1+k_{0n}')K_{0n}'. +\] + +Since +\begin{DPgather*} +K_{0n}' = \int_{0}^{\frac{\pi}{2}} d \phi = \frac{\pi}{2}, +\rintertext{($n =$ limit,)} +\end{DPgather*} +we have +\[ +\Tag[20st]{(20)^*} +(1 + k_{0}')(1 + k_{00}') \dotsm (1 + k_{0n}') = \frac{2K'}{\pi}. +\] + +From \Eqref[19st]{eq.}{}{(19)^*} we have, since [\Eqref{eq.}{IV}{(10)}, Chap\DPtypo{}{.}~IV] +\begin{gather*} +\frac{1 + k}{2} = \frac{1}{1 + k_{0}'}, \\ +(1 + k_{0}') \int_{0}^{\phi_{0}} \frac{d \phi_{0}}{\Delta (\phi_{0}, k_{1})} + = \int_{0}^{\phi} \frac{d \phi}{\Delta (\phi_{1}, k)}; +\end{gather*} +whence also, since for $\phi_{0} = \dfrac{\pi}{2}$, $\phi = \pi$, +\[ +\begin{array}{r@{}l} +(1 + k_{0}')K_{1} &{}= 2K, \\ +(1 + k_{00}')K_{2} &{}= 2K_{1}, \\ +\Dots{2} \\ +(1 + k_{0n}')K_{n} &{}= \rlap{$2K_{n-1}$,} +\end{array} +\] +and +\[ +(1 + k_{0}')(1 + k_{00}') \dotsm (1 + k_{0n}')K_{n} = 2^{n}K; +\] +or +\begin{DPalign*} +\frac{K_{n}}{2^{n}} + &= \frac{K}{(1 + k_{0}')(1 + k_{00}') \dotsm} +\rintertext{($n = \infty$)} \\ +\Tag{(21)} + & = \frac{\pi}{2K_{1}} K. +\end{DPalign*} +%% -----File: 047.png---Folio 41------- + +Let us find the limiting value of $F(k_{0n}, \phi_{n})$ in \Eqref{eq.}{}{(15)}. In +the equation $\tan (\phi_{n} - \phi_{n-1}) = k_{n-1} \tan \phi_{n-1}$, we see that when +$k_{n-1}$~reaches the limit~$1$, then $\phi_{n} - \phi_{n-1} = \phi_{n-1}$ or $\phi_{n} = 2\phi_{n-1}$. +Therefore +\begin{align*} +\frac{\phi_{n}}{2^{n}} + &= \frac{2\phi_{n - 1}}{2^{n}} + = \frac{\phi_{n-1}}{2^{n - 1}}; \\ +% +\frac{\phi_{n+1}}{2^{n+1}} + &= \frac{2\phi_{n}}{2^{n+1}} + = \frac{\phi_{n}}{2^{n}} + = \frac{\phi_{n-1}}{2^{n}}; \\ +% +\frac{\phi_{n+m}}{2^{n+m}} + &= \frac{\phi_{n - 1}}{2^{n}} + = \text{constant, whatever $m$ may be}. +\end{align*} +Therefore \Eqref{eq.}{}{(15)} becomes +\[ +\Tag[21st]{(21)^*} +F(k, \phi) = (1 + k_{0})(1 + k_{00}) \dotsm (1 + k_{0n}) \frac{\phi_{n}}{2^{n}}, +\] +$n$~being whatever number will carry $k_{0}$ and~$\dfrac{\phi_{1}}{2}$ to their limiting +values. + +In the same way, \Eqref{eqs.}{}{(16)} and~\Eqref{}{}{(17)} become +\begin{align*} +\Tag{(22)} +F(k, \phi) + &= \frac{1}{\cos^{2} \dfrac{\theta}{2} + \cos^{2} \dfrac{\theta_{0}}{2} \dotsm + \cos^{2} \dfrac{\theta_{0n}}{2}} · \frac{\phi_{n}}{2^{n}} \\ +\Tag{(23)} + &= \sqrt{\frac{\cos \theta_{0} \cos \theta_{00} \dotsm \cos^{2} \theta_{0n}}{\cos \theta}} + · \frac{\phi_{n}}{2^{n}}, +\end{align*} +$n - 1$ being the number which makes $k_{n-1}' = 1$. + +In these last three \DPtypo{equation}{equations} $k_{0}$,~$k_{00}$ are determined by \Eqref[14sub1]{eqs.}{}{(14_{1})}; +$\phi_{1}$,~$\phi_{2}$,~etc., by \Eqref[14sub2]{eqs.}{}{(14_{2})}\footnotemark; +\footnotetext{Taking for~$\phi_{1} - \phi$, etc., not always the least angle given by the tables, + but that which is nearest to~$\phi$.}% +$\theta$,~$\theta_{0}$,~etc., by \Eqref[14sub1]{eqs.}{}{(14_{1})}; and +$k'$,~$k_{1}'$, $\DPtypo{k_{2}}{k_{2}'}$,~etc., for use in \Eqref[14sub2]{eq.}{}{(14_{2})} by \Eqref[14sub1]{eqs.}{}{(14_{1})}. +%% -----File: 048.png---Folio 42------- + +\Section{BISECTED AMPLITUDES.} + +We have identically +\begin{DPalign*} +u &= 2 · \frac{u}{2} + = 2 \raisebox{-1ex}{\scalebox{2}{$\displaystyle\int$}} \frac{d \am \dfrac{u}{2}}{\sqrt{1 - k^{2} \sn^{2} \dfrac{u}{2}}}; \\ +\frac{u}{2} + &= 2 · \frac{u}{4} + = 2F\left(k, \am \frac{u}{4}\right); \\ + & \quad \text {etc.} \\ +\intertext{\indent Therefore} +u &= F(k, \am u) = 2^{n} F\left(k, \am \frac{u}{n}\right) \\ + &= 2^{n} · \am \frac{u}{n}, +\rintertext{($n =$ limit,)} +\end{DPalign*} +$\am \dfrac{u}{n}$ being determined by repeated applications of \Eqref{eq.}{II}{(12)} of +Chap.~II, as follows: +\begin{align*} +\sn^{2} \frac{u}{2} + &= \frac{1 - \cn u}{1 + \dn u} + = \frac{2 \sin^{2} \frac{1}{2} \am u}{1 + \dn u}; \\ +% +\Tag{(24)} +\sn \frac{u}{2} + &= \frac{\sin \frac{1}{2} \am u}{\sqrt{\dfrac{1 + \cos \beta}{2}}} + = \frac{\sin \dfrac{\am u}{2}}{\cos \frac{1}{2} \beta}; +\end{align*} +$\beta$ being an angle determined by the equation +\[ +\Tag{(25)} +\cos \beta = \dn u = \sqrt{1 - k^{2} \sn^{2} u}, +\] +and $n$ being the number which makes +\[ +2^{n} \am \frac{u}{n} = \text{constant}. +\] +$\am \dfrac{u}{n}$ is found by repeated applications of \Eqref{eq.}{}{(24)}. +%% -----File: 049.png---Folio 43------- + +Indicating the amplitudes as follows: +\[ +\begin{array}{r@{}l} +\am u &{}= \phi, \\ +\am \dfrac{u}{2} &{}= \phi_{02}, \\[2pt] +\am \dfrac{u}{4} &{}= \phi_{04}, \\[2pt] +\am \dfrac{u}{8} &{}= \phi_{08}, \\[2pt] +\Dots{2} \\ +%[** PP: Em-dash present in high-res scan, retaining] +\llap{$\am \dfrac{u}{2^{n}}$} &{}= \rlap{$\phi_{02^{n}},\text{---}$} \\ +\Tag{(26)} +\llap{$F(k, \phi)$} &{} = \rlap{$2^{n} \phi_{02^{n}}$;} +\end{array} +\] +$n$~being the limiting value. + +In \Eqref{eq.}{}{(18)}, when $k_{n}$~reaches its limit~$1$, we have +\[ +F(k_{n}, \phi_{0n}) + = \int_{0}^{\phi} \frac{d \phi_{0n}}{\cos \phi_{0n}} + = \log_{\epsilon} \tan (45^{\circ} + \tfrac{1}{2} \phi_{0n}), +\] +and \Eqref{eqs.}{}{(18)} and~\Eqref{}{}{(19)} become +\begin{align*} +\Tag{(27)} +F(k, \phi) + &= (1 + k_{0}')(1 + k_{00}') \dotsm (1 + k_{0n}') + \log_{\epsilon} \tan (45° + \tfrac{1}{2} \phi_{0n}) \\ + &= \frac{1}{\cos^{2} \frac{1}{2} \eta\, + \cos^{2} \frac{1}{2} \eta_{0} \dotsm + \cos^{2} \frac{1}{2} \eta_{0n}} + \log_{\epsilon} \tan (45° + \tfrac{1}{2} \phi_{0n}) \\ +\Tag{(28)} + &= \frac{1}{\cos^{2} \frac{1}{2} \eta\, + \cos^{2} \frac{1}{2} \eta_{0} \dotsm + \cos^{2} \frac{1}{2} \eta_{0n}} + · \frac{1}{M} \log \tan (45° + \tfrac{1}{2} \phi_{0n}); +\end{align*} +$n$~being the number which renders $k_{n} = 1$. + +\Eqref{Eq.}{}{(20)} becomes +%% -----File: 050.png---Folio 44------- +\begin{align*}%[** PP: Realigning first line on =] +\Tag{(29)} +F(k, \phi) + &= \sqrt{\frac{k_{1} k_{2} \dotsm k_{n}^{2}}{k}} + · \log_{\epsilon} \tan (45° + \tfrac{1}{2} \phi_{0n}) \\ +% + &= \sqrt{\frac{k_{1}k_{2} \dotsm k_{n}^{2}}{k}} + · \frac{1}{M} \log \tan (45° + \tfrac{1}{2} \phi_{0n}) \\ +% + &= \sqrt{\frac{\cos \eta_{0} \cos \eta_{00} \dotsm \cos^{2} \eta_{0n}}{\cos \eta}} + · \frac{1}{M} \log \tan (45° + \tfrac{1}{2} \phi_{0n}). +\end{align*} + +In these equations $k'_{0}$,~$k'_{00}$,~etc., are determined by \Eqref[18sub3]{eqs.}{}{(18_3)}; +$\eta$,~$\eta_{0}$,~etc., by \Eqref[18sub3]{eqs.}{}{(18_3)}; $\phi_{0}$,~$\phi_{00}$,~etc., by \Eqref[18sub2]{eqs.}{}{(18_2)}; $k_{1}$,~$k_{2}$,~etc., by \Eqref[18sub1]{eqs.}{}{(18_1)}. + +Substituting in \Eqref{eq.}{}{(27)} from \Eqref[20st]{eq.}{}{(20)^*}, we have +\begin{align*} +F(k, \phi) + &= \frac{2K'}{\pi} \log_{\epsilon} \tan (45° + \tfrac{1}{2} \phi_{0n})\\ +\Tag{(30)} + &= \frac{2K'}{\pi M} \log \tan (45° + \tfrac{1}{2} \phi_{0n}). +\end{align*} +%% -----File: 051.png---Folio 45------- + + +\Chapter{V}{Complete Functions} + +\First{Indicate} by~$K$ the complete integral +\[ +\Tag{(1)} +K = \int_{0}^{\frac{\pi}{2}} \frac{d\phi}{\sqrt{1 - k^{2} \sin^{2} \phi}}, +\] +and by~$K_{0}$ the complete integral +\[ +\Tag{(2)} +K_{0} = \int_{0}^{\frac{\pi}{2}} \frac{d\phi_{1}}{\sqrt{1 - k_{0}^{2} \sin^{2} \phi_{1}}}; +\] +and in a similar manner $K_{00}$,~$K_{03}$,~etc. + +From \Eqref{eq.}{IV}{(12)}, Chap.~IV, we have +\begin{align*} +\tan (\phi_{1} - \phi) + &= k' \tan \phi \\ + &= \frac{\tan \phi_{1} - \tan \phi}{1 + \tan \phi_{1} \tan \phi}, \\ +\intertext{whence} +\tan \phi_{1} + &= \frac{(1 + k') \tan \phi}{1 - k' \tan^{2} \phi} \\ + &= \frac{1 + k'}{\dfrac{1}{\tan \phi} - k' \tan \phi}. +\end{align*} +%% -----File: 052.png---Folio 46------- + +From this equation we see that when $\phi = \dfrac{\pi}{2}$, $\phi_{1} = \pi$. This +same result might also have been deduced from \hyperref[page:30]{Fig.~1}, Chap.~IV, +or from the equation +\[ +\Tag{(3)} +\phi_{1} = 2 \phi - k_{0} \sin 2\phi + \tfrac{1}{2} k_{0}^{2} \sin 4\phi - \text{etc.}, +\] +this last being the well-known trigonometrical formula +\begin{gather*} +\tan x = n \tan y, \\ +x = y - \frac{1 - n}{1 + n}\, \sin 2y + + \frac{1}{2} \left(\frac{1 - n}{1 + n}\right)^{2} \sin 4y + - \frac{1}{3} \left(\frac{1 - n}{1 + n}\right)^{3} \sin 6y + \text{etc.} +\end{gather*} +\begin{DPalign*} +\lintertext{Since} +\int_{0}^{\frac{\pi}{2}} \frac{d \phi_{1}}{\Delta (k_{0}\DPtypo{}{,} \phi_{1})} + &= K_{0},\ \text{we must have} \\ +\int_{0}^{\pi} \frac{d\phi_{1}}{\Delta (k_{0}\DPtypo{}{,} \phi_{1})} + &= 2K_{0}. +\end{DPalign*} + +These values substituted in \Eqref{eq.}{IV}{(13)}, Chap.~IV, give successively +\[ +\begin{array}{r@{}l} +\Tag{(4)} +K &{}= (1 + k_{0})K_{0}, \\ +K_{0} &{}= (1 + k_{00})K_{00}, \\ +\Dots{2} \\ +\llap{$K_{0(n - 1)}$} &{}= (1 + k_{0n})K_{0n}; +\end{array} +\] +whence +\[ +\Tag{(5)} +K = (1 + k_{0})(1 + k_{00}) \dotsm (1 + k_{0n})K_{0n}. +\] + +Since the limit of~$k_{0n}$ is~$0$, $K_{0n}$~becomes +\[ +K_{0n} = \int_{0}^{\frac{\pi}{2}} d \phi = \frac{\pi}{2}, +\] +%% -----File: 053.png---Folio 47------- +and we have +\begin{align*} +\Tag{(6)} +K &= \frac{\pi}{2} (1 + k_{0})(1 + k_{00}) \dotsm \\ +\Tag{(7)} + &= \frac{\tfrac{1}{2} \pi}{\cos^{2} \tfrac{1}{2}\theta + \cos^{2} \tfrac{1}{2}\theta_{0} \dotsm + \cos^{2} \tfrac{1}{2}\theta_{0n}}; +\end{align*} +$k_{1}$,~$k_{0}$,~etc., and $\theta_{1}$,~$\theta_{0}$,~etc., being found by \Eqref[14sub1]{eqs.}{IV}{(14_{1})} of Chap.~IV\@. + +From the formulæ in these two chapters we can compute +the values of~$u$ for all values of $\phi$~and~$k$ and arrange them in +tables. These are Legendre's Tables of Elliptic Integrals. +%% -----File: 054.png---Folio 48------- + + +\Chapter[Evaluation for phi.]{VI}{Evaluation for $\phi$.} + +\Section{TO FIND~$\phi$, $u$~AND~$k$ BEING GIVEN.} + +\DPtypo{From}{\First{From}} \Eqref{eqs.}{IV}{(21)} and~\Eqref{}{IV}{(23)}, Chap.~IV, we have ($n$~having the +value which makes $\cos \theta_{0n} = 1$) +\[ +\Tag{(1)} +\phi_{n} + = \frac{2^{n}u}{(1 + k_{0})(1 + k_{00}) \dotsm (1 + k_{0n})} + = \frac{2^{n}u \sqrt{\cos \theta}}{\sqrt{\cos \theta_{0} \dotsm \cos^{2} \theta_{0n}}}, +\] +from which $\phi_{n}$~can be calculated, $k_{0},~k_{00}$,~etc., being found by +means of \Eqref[14sub1]{equations}{IV}{(14_1)}, Chap.~IV\@. + +Then, having $\phi_{n}$, $k_{0}$,~$k_{00}$,~etc., we can find~$\phi$ by means of +the following equations: +\[ +\begin{array}{r@{}l} +\sin (2\phi_{n - 1} - \phi_{n}) &= k_{0n} \sin \phi_{n},\\ +\sin (2\phi_{n - 2} - \phi_{n - 1}) &= k_{0(n - 1)} \sin \phi_{n - 1},\\ +\Dots{2} \\ +\sin (2\phi - \phi_{1}) &= k_{0} \sin \phi_{1};\\ +\end{array} +\] +whence we can get the angle~$\phi$. + +When $k > \sqrt{\frac{1}{2}}$ the following formulæ will generally be found +to work more rapidly: + +From \Eqref{eq.}{IV}{(29)}, Chap.~IV, we have +\[ +\Tag{(2)} +\log \tan (45° + \tfrac{1}{2} \phi_{0n}) + = \frac{uM}{\sqrt{\dfrac{k_{1}k_{2} \dotsm k_{n}^{2}}{k}}}, +\] +%% -----File: 055.png---Folio 49------- +from which we can get~$\phi_{0n}$; $k_{1}$,~$k_{2}$,~etc., being calculated from +\Eqref[18sub1]{eqs.}{IV}{(18_{1})}, Chap.~IV, and $\phi$~being calculated from the following +equations: +\[ +\begin{array}{r@{}l} +\tan (\phi_{0(n - 1)} - \phi_{0n}) + &= k_{n} \tan \phi_{0n}, \\ +% +\Dots{2} \\ +% +\tan (\phi_{0} - \phi_{00}) + &= k_{2} \tan \phi_{02}, \\ +% +\tan (\phi - \phi_{0}) + &= k \tan \phi_{0}; +\end{array} +\] +whence we get~$\phi$. + +This gives a method of solving the equation +\[ +F\psi = n\, F\phi, +\] +where $n$~and~$\phi$ and the moduli are known, and $\psi$~is the +required quantity. $n$~and~$\phi$ give~$F\psi$, and then $\psi$~can be determined +by the foregoing methods. + +When $k = 1$ \emph{nearly}, \Eqref{equation}{}{(2)} takes a special form,--- + +\primo. When~$\tan \phi$ is very much less than~$\dfrac{1}{k'}$. In this case +\begin{align*} +F(k, \phi) + &= \int \frac{d\phi}{\sqrt{\cos^{2}\phi + k'^{2} \sin^{2} \phi}} + = \int \frac{d\phi}{\sqrt{(1 + k'^{2} \tan^{2} \phi) \cos^{2} \phi}} \\ + &= \int \frac{d\phi}{\cos \phi} + = \log \tan (45° + \tfrac{1}{2} \phi); +\end{align*} +whence we can find~$\phi$. + +\secundo. When~$\tan \phi$ and~$\dfrac{1}{k'}$ approach somewhat the same value, +and $k' \tan \phi$~cannot be neglected, $F(k, \phi)$~must be transposed +into another where $k'$~shall be much smaller, so that $k' \tan \phi$~can +be neglected. +%% -----File: 056.png---Folio 50------- + +These methods for finding~$\phi$ apply only when $\phi < \dfrac{\pi}{2}$, that +is, $u < K$. In the opposite case ($u > K$) put +\[ +u = 2nK ± \nu, +\] +the upper or the lower sign being taken according as $K$~is continued +in~$u$ an even or an odd number of times. In either case +$\nu < K$, and we can find~$\nu$ by the preceding methods. + +Having found~$\nu$, we have from \Eqref{eq.}{III}{(5)}, Chap.~III, +\begin{align*} +\am u + &= \am (2nK ± \nu)\\ + &= n \pi ± \am \nu. +\end{align*} +%% -----File: 057.png---Folio 51------- + + +\Chapter{VII}{Development of Elliptic Functions into Factors.} +\SetRunningHeads{Development into Factors} + +From \Eqref{eq.}{IV}{(12)}, Chap.~IV, we readily get +\begin{align*} +\sin (2\phi_{0} - \phi) + &= k \sin \phi;\\ +% +\sin \phi + &= \frac{\sin 2 \phi_{0}}{\sqrt{1 + k^{2} + 2k \cos 2\phi_0}}\\ + &= \frac{\sin 2 \phi_{0}}{\sqrt{(1 + k)\DPtypo{}{^2} - 4k \sin^{2} \phi_0}}\\ + &= \frac{1 + k_{0}'}{2} · \frac{\sin 2 \phi_{0}}{\sqrt{1 - k_{1}^{2} \sin^{2} \phi_{0}}}\\ +\end{align*} +$\Bigl(\text{since } \dfrac{4k}{1 + k} = k_{1} \text{ and } 1 + k = \dfrac{2}{1 + k_{0}'}, \text{\Eqref{eqs.}{IV}{(6)} and~\Eqref{}{IV}{(10)}, Chap.~IV}\Bigr)$; +and thence +\[ +\Tag{(1)} +\sin \phi = \frac{(1 + k_{0}') \sin \phi_{0} \cos \phi_{0}}{\Delta (\phi_{0}, k_{1})}. +\] + +From \Eqref{eq.}{IV}{(13)}, Chap.~IV, we have +\[ +\int_{0}^{\phi_{0}} \frac{d \phi_{0}}{\Delta (\phi_{0}, k_{1})} + = \frac{1 + k}{2} \int_{0}^{\phi} \frac{d\phi}{\Delta (\phi, k)}; +\] +and from \Eqref{eq.}{V}{(4)}, Chap.~V, passing up the scale of moduli +one step, +\[ +1 + k = \frac{K_{1}}{K}, +\] +%% -----File: 058.png---Folio 52------- +whence %[** PP: Next several displays aligned on = in original] +\[ +F(\phi_{0}, k_{1}) = \frac{K_{1}}{2K} F(\phi, k). +\] + +Put +\[ +F(\phi_{0}, k_{1}) = u_{1}\quad \text{and} \quad F(\phi, k) = u, +\] +whence +\[ +u_{1} = \frac{K_{1}}{2K} u. +\] + +Furthermore, +\begin{align*} +\phi &= \am (u, k);\\ +\phi_{1} = \am (u_{1}, k_{1}) &= \am \left( \frac{K_{1}}{2K} u, k_{1} \right). +\end{align*} + +Substituting these values in \Eqref{eq.}{}{(1)}, we have +\[ +\sn (u, k) + = (1 + k'_{0}) + \frac{\sn \left( \dfrac{K_{1}}{2K} u, k_{1} \right) + \cn \left( \dfrac{K_{1}}{2K} u, k_{1} \right)} + {\dn \left( \dfrac{K_{1}}{2K} u, k_{1} \right)}. +\] + +But from \Eqref{eq.}{III}{(11)}, Chap.~III, we have +\begin{align*} +\frac{\cn (v, k_{1})}{\dn (v, k_{1})} &= \sn (v + K_{1}, k_{1}), \\ +\intertext{or} +\frac{\cn \left( \dfrac{K_{1}}{2K} u, k_{1} \right)} + {\dn \left( \dfrac{K_{1}}{2K} u, k_{1} \right)} + &= \sn \left( \frac{K_{1}u}{2K} + K_{1}, k_{1} \right) \\ + &= \sn \left( \dfrac{K_{1}}{2K} (u + \DPtypo{K_{1}}{2K}), k_{1} \right); +\end{align*} +%% -----File: 059.png---Folio 53------- +whence +\begin{DPalign*} +\Tag{(2)} +\sn (u, k) + &= (1 + k_{0}') \sn \frac{K_{1}u}{2K} + \sn \left[\frac{K_{1}}{2K} (u + 2K)\right].\footnotemark \\ +& \rintertext{\llap{(Mod.${}= k_{1}$.)}} +\end{DPalign*} +\footnotetext{The analogous formula in Trigonometry is + \[ + \sin \phi = \DPtypo{\tfrac{1}{2}}{2} \sin \tfrac{1}{2} \phi\, + \sin \tfrac{1}{2} (\phi + \pi). + \]} + +From this equation, evidently, we have generally +\begin{DPgather*} +\Tag[2st]{(2)^*} +\sn (\nu, k_{n}) + = (1 + k'_{0(n + 1)}) \sn \frac{K_{n + 1}}{2K_{n}} \nu + \sn \left[\frac{K_{n + 1}}{2K_{n}} (\nu + 2K_{n})\right]. \\ +\rintertext{\llap{(Mod.${}= k_{n + 1}$.)}} +\end{DPgather*} + +Applying this general formula to the two factors of \Eqref{eq.}{}{(2)}, +we have +\begin{DPalign*} +\sn \left(\frac{K_{1}u}{2K}, k_{1}\right) + &= (1 + k'_{00}) \sn \frac{K_{2}}{2K_{1}} · \frac{K_{1}u}{2K} + · \sn \left[\frac{K_{2}}{2K_{1}} \left(\frac{K_{1}u}{2K} + 2K_{1}\right)\right] \\ +& \rintertext{\llap{(Mod.\ $k_{2}$)}} \\ + &= (1 + k'_{00}) \sn \frac{K_{2}u}{2^{2}K} \sn \frac{K_{2}}{2^{2}K} (u + 4K); +\rintertext{\llap{(Mod.\ $k_{2}$;)}} +\end{DPalign*} +\begin{DPgather*} +\Tag{(3)} +\sn \left[\frac{K_{1}}{2K} (u + 2K), k_{1}\right] + = (1 + k'_{00}) \sn \frac{K_{2}}{2^{2}K} (u + 2K) \\ + · \sn \frac{K_{2}}{2K_{1}} \left[\frac{K_{1}}{2K} (u + 2K) + 2K_{1}\right]. +\rintertext{(Mod.\ $k_{2}$.)} +\end{DPgather*} + +The last argument in this equation is equal to +\[ +\frac{K_{2}}{2^{2}K} (u + 6K); +\] +and since, \Eqref{eq.}{III}{(7)}, Chap.~III, +\[ +\sn (u, k_{2}) = \sn (2K_{2} - u, k_{2}), +\] +%% -----File: 060.png---Folio 54------- +we can put in place of this, +\[ +2K_{2} - \frac{K_{2}}{2^{2}K} (u + 6K\DPtypo{}{)} = \frac{K_{2}}{2^{2}K} (2K - u); +\] +whence \Eqref{eq.}{}{(3)} becomes +\begin{DPgather*}%[** PP: Re-breaking] +\sn \left[\frac{K_{1}}{2K} (u + 2K), k_{1}\right] + = (1 + k'_{00}) \sn \frac{K_{2}}{2^{2}K} (2K + u) + · \sn \frac{K_{2}}{2^{2}K} (2K - u). \\ +\rintertext{\llap{(Mod.\ $k_{2}$.)}} +\end{DPgather*} + +Substituting these values in \Eqref{eq.}{}{(2)}, we have +\begin{DPalign*}%[** PP: Aligning on equals sign] +\Tag{(4)} +\sn (u, k) + &= (1+k'_{0})(1 + k'_{00})^{2} \sn \frac{K_{2}u}{2^{2}K} \\ + &\quad + · \sn \frac{K_{2}}{2^{2}K} (2K ± u) \sn \frac{K_{2}}{2^{2}K} (4K+u), +\rintertext{(Mod.\ $k_{2}$,)} +\end{DPalign*} +in which the double sign indicates two separate factors which +are to be multiplied together. + +By the application of the general \Eqref[2st]{equation}{}{(2)^*} we find that +the arguments in the second member of \Eqref{eq.}{}{(4)} will each give +rise to two new arguments, as follows: +\[ +\frac{K_{2}u}{2^{2}K}\quad \text{gives}\quad \frac{K_{3}u}{2^{3}K}, +\] +and +\begin{gather*} +\frac{K_{3}}{2K_{2}} \left(\frac{K_{2}u}{2^{2}K} + 2K_{2}\right) + = \frac{K_{3}}{2^{3}K} (u + 8K); \\ +\frac{K_{2}}{2^{2}K} (2K ± u)\quad \text{gives}\quad +\frac{K_{3}}{2^{3}K} (2K ± u), +\end{gather*} +%% -----File: 061.png---Folio 55------- +and +\begin{gather*} +\Tag{(a)} +\frac{K_{3}}{2K_{2}} \left[\frac{K_{2}}{2^{2}K} (2K ± u) + 2K_{2}\right] + = \frac{K_{3}}{2^{3}K} (10K ± u), \\ +\frac{K_{2}}{2^{2}K} (4K + u) \quad \text{gives} \quad \frac{K_{3}}{2^{3}K} (4K + u), +\intertext{and} +\Tag{(b)} +\frac{K_{3}}{2K_{2}} \left[\frac{K_{2}}{2^{2}K} (4K + u) + 2K_{2}\right] + = \frac{K_{3}}{2^{3}K} (12K + u). +\end{gather*} + +Subtracting~\Eqno{(a)} and~\Eqno{(b)} from~$2K_{3}$, by which the sine of the +amplitudes will not be changed [\Eqref{eq.}{III}{(7)}, Chap.~III], and since +our new modulus is~$k_{3}$, we have for the expressions \Eqno{(a)}~and~\Eqno{(b)}, +\begin{gather*} +\Tag[apr]{(a')} +\frac{K_{3}}{2^{3}K} (6K \mp u); \\ +\Tag[bpr]{(b')} +\frac{K_{3}}{2^{3}K} (4K - u). +\end{gather*} + +Substituting these values in \Eqref{eq.}{}{(4)}, and remembering the +factor $(1 + k'_{03})$ introduced by each application of \Eqref[2st]{eq.}{}{(2)^*}, we +have +\begin{DPalign*}%[** PP: Align on equal signs, remove extraneous semicolons] +\sn (u, k) + &= (1 + k'_{0})(1 + k'_{00})^{2}(1 + k'_{03})^{4} + \sn \frac{K_{3}u}{2^{3}K}\DPtypo{;}{} \\ + &\quad· + \sn \frac{K_{3}}{2^{3}K} (2K ± u) + \sn \frac{K_{3}}{2^{3}K} (4K ± u)\DPtypo{;}{} \\ + &\quad· + \sn \frac{K_{3}}{2^{3}K} (6K ± u) + \sn \frac{K_{3}}{2^{3}K} (8K + u). +\rintertext{(Mod.\ $k_{3}$.)} +\end{DPalign*} +%% -----File: 062.png---Folio 56------- + +From this the law governing the arguments is clear, and we +can write for the general equation +\begin{DPalign*} +\Tag{(5)} +\sn (u, k) + &= (1 + k_{0}')(1 + k'_{00})^{2}(1 + k'_{03})^{4} \dotsm (1 + k'_{0n})^{2^{n-1}} \\ + &\quad· \sn \frac{K_{n}u}{2^{n}K} + \sn \frac{K_{n}}{2^{n}K} (2K ± u) \\ + &\quad· \sn \frac{K_{n}}{2^{n}K} (4K ± u) + \sn \frac{K_{n}}{2^{n}K} (6K ± u) \\ + &\quad \dotsm \sn \frac{K_{n}}{2^{n}K} \Bigl[(2^{n} - 2)K ± u\Bigr] \\ + &\quad· \sn \frac{K_{n}}{2^{n}K} (2^{n}K + u). +\rintertext{\llap{(Mod.\ $k_{n}$.)}} +\end{DPalign*} + +Indicate the continued product of the binomial factors by~$A'$, +and we have +\[ +A' = (1 + k_{0}')(1 + k'_{00})^{2}(1 + k'_{03})^{4}(1 + k'_{04})^{8} \dotsm. +\] + +Since the limit of~$k'_{0}$, $k'_{00}$,~etc., is zero, it is evident that +these factors converge toward the value unity. It can be shown +that the functional factors also converge toward the value +unity. Thus the argument of the last factor can be written +\[ +K_{n} + \frac{K_{n}u}{2^{n}K}. +\] + +From \Eqref{eq.}{III}{(11)}, Chap.~III, we get then +\begin{DPgather*} +\sn \left(K_{n} + \frac{K_{n}u}{2^{n}K}\right) + = \frac{\cn \dfrac{K_{n}u}{2^{n}K}} + {\dn \dfrac{K_{n}u}{2^{n}K}}. +\rintertext{(Mod.\ $k_{n}$.)} +\end{DPgather*} + +But since $k_{n}$ at its limit is equal to unity, $\cn = \dn$; whence +the last factor of \Eqref{eq.}{}{(5)} is unity. +%% -----File: 063.png---Folio 57------- + +From \Eqref{eq.}{IV}{(21)}, Chap.~IV, we have +\[ +\operatorname{limit} \frac{K_{n}}{2^{n}K} = \frac{2 \pi}{2 K'}. +\] + +Therefore for $n = \infty$, \Eqref{eq.}{}{(5)} becomes +\begin{DPalign*}%[** PP: Align on equals sign] +\sn (u, k) + &= A' \sn \frac{\pi u}{2K'} \sn \frac{\pi}{2K'} (2K ± u) \\ + &\quad· \sn \frac{\pi}{2K'}(4K ± u) \sn \frac{\pi}{2K'}(6K ± u), \ldots +\rintertext{(Mod.\ $1$,)} \\ +\intertext{or} +\Tag{(6)} +\sn (u, k) + &= A' \sn \frac{\pi u}{2K'} \left[{\textstyle\prod\limits_{1}^{\infty}} h\right] + \sn \frac{\pi}{2K'} (2hK ± u), +\rintertext{\llap{(Mod.\ $1$,)}} +\end{DPalign*} +where the sign~$\Prod$ indicates the continued product in the same +manner as $\sum$~indicates the continued sum. + +When~$k = 1$, $\displaystyle\int_{0}^{\phi} F(\phi, k)\DPtypo{}{\,d\phi}$ becomes +\[ +\nu = \int_{0}^{\phi} \frac{d \phi}{\cos \phi} + = \tfrac{1}{2} \DPtypo{\log^e}{\log_\epsilon} \frac{1 + \sin \phi}{1 - \sin \phi}; +\] +whence +\[ +e^{2\nu} = \frac{1 + \sin \phi}{1 - \sin \phi}, +\] +and +\[ +\sin \phi = \frac{e^{2\nu} - 1}{e^{2\nu} + 1} + = \frac{e^{\nu} - e^{-\nu}}{e^{\nu} + e^{-\nu}} + = \sn (\nu, 1). +\] +%% -----File: 064.png---Folio 58------- + +Hence in \Eqref{equation}{}{(6)} +\begin{align*} +\sin \frac{\pi u}{2K'} + &= \frac{e^{\frac{\pi u}{2K'}} - e^{-\frac{\pi u}{2K'}}} + {e^{\frac{\pi u}{2K'}} + e^{-\frac{\pi u}{2K'}}}; \\ +\sn \frac{\pi}{2K'}(2hK ± u) + &= \frac{e^{ \frac{h\pi K}{K'}} e^{±\frac{\pi u}{2K'}} + - e^{-\frac{h\pi K}{K'}} e^{\mp\frac{\pi u}{2K'}}} + {e^{ \frac{h\pi K}{K'}} e^{±\frac{\pi u}{2K'}} + + e^{-\frac{h\pi K}{K'}} e^{\mp\frac{\pi u}{2K'}}}. +\end{align*} + +Put +\[ +\Tag[6st]{(6)^*} +q = e^{-\frac{\pi K'}{K}}, \quad q' = e^{-\frac{\pi K}{K'}}, +\] +and the last expression becomes +\begin{align*} +\sn \frac{\pi}{2K'} & (2hK ± u) + = \frac{q'^{-h} e^{±\frac{\pi u}{2K'}} - q'^{h} e^{\mp\frac{\pi u}{2K'}}} + {q'^{-h} e^{±\frac{\pi u}{2K'}} + q'^{h} e^{\mp\frac{\pi u}{2K'}}}; \\ +% +\sn \frac{\pi}{2K'} & (2hK + u) \sn \frac{\pi}{2K'}(2hK - u) \\ + &= \frac{q'^{-h} e^{ \frac{\pi u}{2K'}} - q'^{h} e^{-\frac{\pi u}{2K'}}} + {q'^{-h} e^{ \frac{\pi u}{2K'}} + q'^{h} e^{-\frac{\pi u}{2K'}}} + · \frac{q'^{-h} e^{-\frac{\pi u}{2K'}} - q'^{h} e^{ \frac{\pi u}{2K'}}} + {q'^{-h} e^{-\frac{\pi u}{2K'}} + q'^{h} e^{ \frac{\pi u}{2K'}}} \\ +% + &= \frac{q'^{-2h} + q'^{2h} - \left(e^{\frac{\pi u}{K'}} + e^{-\frac{\pi u}{K'}}\right)} + {q'^{-2h} + q'^{2h} + \left(e^{\frac{\pi u}{K'}} + e^{-\frac{\pi u}{K'}}\right)}. +\end{align*} + +From plane trigonometry we have the equations +\[ +\frac{e^{x} - e^{-x}}{e^{x} + e^{-x}} = -i \tan ix, \quad e^{x} + e^{-x} = 2 \cos ix; +\] +%% -----File: 065.png---Folio 59------- +where $i = \sqrt{-1}$: which gives +\begin{DPalign*} +\sn \frac{\pi u}{2K'} + &= -i \tan \frac{\pi i u}{2K'}; +\rintertext{\llap{(Mod.\ $1$;)}} \\ +% +\sn \frac{\pi}{2K'} & (2hK + u) \sn \frac{\pi}{2K'}(2hK - u) \\ + &= \frac{q'^{-2h} + q'^{2h} - 2 \cos \dfrac{\pi iu}{K'}} + {q'^{-2h} + q'^{2h} + 2 \cos \dfrac{\pi iu}{K'}} \\ +% +%[** PP: No equation label in orig., but text refers specifically to (7).] +\DPtypo{}{\Tag{(7)}} + &= \frac{1 - 2q'^{2h} \cos \dfrac{\pi iu}{K'} + q'^{4h}} + {1 + 2q'^{2h} \cos \dfrac{\pi iu}{K'} + q'^{4h}}. +\end{DPalign*} + +From \Eqref{eq.}{III}{(10)}, Chap.~III, we have +\[ +\sn (u, k) = -i \tn (iu, k'). +\] + +Substituting these values in eq.~(6), we have +\[ +\tn (iu, k') = A' \tan \frac{\pi iu}{2K'} \Prod + \frac{1 - 2q'^{2h} \cos \dfrac{\pi iu}{K'} + q'^{4h}} + {1 + 2q'^{2h} \cos \dfrac{\pi iu}{K'} + q'^{4h}}. +\] + +Now in place of the series of moduli $k'$,~$k_{0}'$ and the corresponding +complete integral~$K'$, we are at liberty to substitute +the parallel series of moduli $k$,~$k_{0}$ and the corresponding complete +integral~$K$; calling the new integral~$u$, we have +\begin{align*} +\Tag{(8)} +\tn (u, k) + &= A \tan \frac{\pi u}{2K} \DPtypo{\{\textstyle\prod}{\Prod} + \frac{1 - 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 + 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} \\ +%% -----File: 066.png---Folio 60------- + &= A \tan \frac{\pi u}{2K} + \frac{1 - 2q^{2} \cos \dfrac{\pi u}{K} + q^{4}} + {1 + 2q^{2} \cos \dfrac{\pi u}{K} + q^{4}} \\ +%[** PP: Adding {} to the right of &s to coax · into math mode] + &\PadTo[r]{{}= A \tan \dfrac{\pi u}{2K}}{{}·{}} + \frac{1 - 2q^{4} \cos \dfrac{\pi u}{K} + q^{8}} + {1 + 2q^{4} \cos \dfrac{\pi u}{K} + q^{8}} \\ + &\PadTo[r]{{}= A \tan \dfrac{\pi u}{2K}}{{}·{}} + \frac{1 - 2q^{6} \cos \dfrac{\pi u}{K} + q^{12}} + {1 + 2q^{6} \cos \dfrac{\pi u}{K} + q^{12}} \dotsm, +\end{align*} +where +\[ +\Tag{(9)} +A = (1 + k_{0})(1 + k_{00})^{2}(1 + k_{03})^{4}(1 + k_{04})^{8} \dotsm +\] + +Now in \Eqref{equation}{}{(6)} put $u+K$~for~$u$, and we have, since +[\Eqref{eq.}{III}{(11)}, Chap.~III] $\sn (u+K)= \dfrac{\cn u}{\dn u}$, +\begin{DPalign*}%[** PP: Re-breaking] +\frac{\cn u}{\dn u} + &= A' \sn \frac{\pi (u+K)}{2K'} \\ + &\quad \Prod \sn \frac{\pi}{2K'} [(2h+1)K+u] + · \sn \frac{\pi}{2K'} [(2h-1)K-u]. \\ +&\rintertext{\llap{(Mod.\ $1$.)}} +\end{DPalign*} +Now from $2h-1$ and~$2h+1$ we have the following series of +numbers respectively: +\[ +\begin{array}{l@{\qquad}*{6}{@{\quad}r}} +2h-1: & 1, & 3, & 5, & 7, & 9, & \text{etc.} \\ +2h+1: & & 3, & 5, & 7, & 9, & \text{etc.} +\end{array} +\] + +It will be observed that the factor outside of the sign~$\Prod$, +viz., $\sin \am \dfrac{\pi (u + K)}{2K'}$, would, if placed under the sign~$\Prod$, supply +%% -----File: 067.png---Folio 61------- +the missing first term of the second series. Hence, placing this +factor within the sign, we have +\begin{DPalign*}%[** PP: Re-breaking] +\Tag{(10)} +\frac{\cn u}{\dn u} + &= A'\Prod \sn \frac{\pi}{2K'} \left[(2h-1)K + u\right] + · \sn \frac{\pi}{2K'} \left[(2h-1)K - u\right]. \\ + & \rintertext{\llap{(Mod.\ $1$.)}} +\end{DPalign*} + +Comparing this with \Eqref{equation}{}{(7)}, we see that the factors +herein differ from those in \Eqref{equation}{}{(7)} only in having $2h-1$ +in place of~$2h$; hence we have +\begin{DPgather*} +\sn \frac{\pi}{2K'} [(2h-1)K + u] +\sn \frac{\pi}{2K'} [(2h-1)K - u] \\ + = \frac{1 - 2q'^{2h-1} \cos \dfrac{\pi iu}{K'} + q'^{4h-2}} + {1 + 2q'^{2h-1} \cos \dfrac{\pi iu}{K'} + q'^{4h-2}}. +\rintertext{(Mod.\ $1$.)} +\end{DPgather*} + +From \Eqref{eqs.}{III}{(10)}, Chap.~III, we have +\[ +\frac{\cn (u, k)}{\dn (u, k)} = \frac{1}{\dn(iu, k')}; +\] +whence \Eqref{eq.}{}{(10)} becomes +\[ +\Tag{(11)} +\frac{1}{\dn(iu, k')} + = A'\Prod \frac{1 - 2q'^{2h-1} \cos \dfrac{\pi iu}{K'} + q'^{4h-2}} + {1 + 2q'^{2h-1} \cos \dfrac{\pi iu}{K'} + q'^{4h-2}}; +\] +and when in place of~$iu$, $k'$, $K'$, $q'$,~$A'$, we substitute~$u$, $k$, $K$, +$q$ and~$A$, and invert the equation, we have +\[ +\Tag{(12)} +\dn (u, k) + = \frac{1}{A} \Prod \frac{1 + 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}. +\] +%% -----File: 068.png---Folio 62------- + +Bearing in mind the remarkable property (Chap.~III, \Pageref{p.}{29}) +that the functions $\sn u$~and~$\dn u$ approach infinity for the same +value of~$u$, we see that both these functions, except as to the +factor independent of~$u$, must have the same denominator. +Furthermore, since $\sn u$~and~$\tn u$ disappear for the same value +of~$u$, they must, except for the independent factor, have the +same numerator. Hence, indicating by~$B$ a new quantity, +dependent upon~$k$ but independent of~$u$, we have +\[ +\Tag{(13)} +\sn (u, k) = B \sin \frac{\pi u}{2K} \Prod + \frac{1 - 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}; +\] +and since +\[ +\cn u = \frac{\sn u}{\tn u}, +\] +we also have, from \Eqref{eqs.}{}{(8)} and~\Eqref{}{}{(13)}, +\[ +\Tag{(14)} +\cn (u, k) = \frac{B}{A} \cos \frac{\pi u}{2K} \Prod + \frac{1 + 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}. +\] +Collecting these results, we have the following equations: +\begin{align*} +\Tag{(15)} +\sn (u, k) + &= B \sin \frac{\pi u}{2K} \Prod + \frac{1 - 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}, \\ +\Tag{(16)} +\cn (u, k) + &= \frac{B}{A} \cos \frac{\pi u}{2K} \Prod + \frac{1 + 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}, \\ +%% -----File: 069.png---Folio 63------- +\Tag{(17)} +\dn (u, k) &= \frac{1}{A} \Prod + \frac{1 + 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}. +\end{align*} + +To ascertain the values of $A$ and~$B$, we proceed as follows: + +In \Eqref{eq.}{}{(17)} we make~$u = 0$, whence, by \Eqref{eq.}{II}{(13)}, Chap.~II, +we have +\begin{align*} +1 &= \frac{1}{A} \Prod \left(\frac{1 + q^{2h-1}}{1 - q^{2h-1}}\right)^{2}; \\ +\intertext{whence} +\Tag{(18)} +\frac{1}{A} &= \Prod \left(\frac{1 - q^{2h-1}}{1 + q^{2h-1}}\right)^{2}. \\ +\intertext{\indent In \Eqref{equation}{}{(17)}, making~$u=K$, we get, by \Eqref{equation}{III}{(1)}, +Chap.~III,} +k' &= \frac{1}{A} \Prod \left(\frac{1 - q^{2h-1}}{1 + q^{2h-1}}\right)^{2} + = \frac{1}{A^{2}}; \\ +\Tag{(19)} +\therefore \frac{1}{A} &= \sqrt{k'}. +\end{align*} +We have identically +\begin{align*} +1 &= B \frac{1}{B} = B \frac{\dfrac{1}{A}}{\dfrac{B}{A}} + = B \frac{\sqrt{k'}}{\dfrac{B}{A}}; \\ +\intertext{whence} +\frac{B}{A} & = B \sqrt{k'}. +\end{align*} + +To calculate~$B$, put~$e^{\frac{i\pi u}{2K}} = \nu$; if we change $\dfrac{\pi u}{2K}$ into +%% -----File: 070.png---Folio 64------- +$\dfrac{\pi u}{2K} + \dfrac{i\pi K'}{2K}$, $\nu$~will change into~$\nu \sqrt{q}$, and $\sn u$ will become, by +\Eqref{eq.}{III}{(14)}, Chap.~III, +\[ +\sn (u+iK') = \frac{1}{k \sn u}. +\] + +Now, replacing $\sin \dfrac{\pi u}{2K}$ and $\cos \dfrac{\pi u}{K}$ by their exponential values, +and observing that +\[ +1 - 2q^{n} \cos \frac{\pi u}{K} + q^{2n} = (1 - q^{n}\nu^{2})(1 - q^{n}\nu^{-2}), +\] +we have +\[ +\sn u = \frac{B}{2} · \frac{\nu - \nu^{-1} }{\sqrt{-1}} + · \frac{\Prod (1 - q^{2h} \nu^{2})(1 - q^{2h} \nu^{-2})} + {\Prod (1 - q^{2h-1}\nu^{2})(1 - q^{2h-1}\nu^{-2})}. +\] + +Changing $u$ into~$u + iK'$, and consequently $\nu$~into~$\nu \sqrt{q}$, we +have +\[ +\frac{1}{k \sn u} + = \frac{B}{2} · \frac{\nu \sqrt{q} - \nu^{-1} \sqrt{q^{-1}}}{\sqrt{-1}} + · \frac{\Prod (1 - q^{2h+1}\nu^{2})(1 - q^{2h-1}\nu^{-2})} + {\Prod (1 - q^{2h} \nu^{2})(1 - q^{2h-2}\nu^{\DPtypo{2}{-2}})}. +\] + +Multiplying these equations together, member by member, +and observing that +\begin{align*} +\nu \sqrt{q} - \nu^{-1} \sqrt{q^{-1}} + &= \frac{1 - q\nu^{2}}{-\nu \sqrt{q}}, \\ +\nu - \nu^{-1} + &= \nu(1 - \nu^{-2}), +\end{align*} +we get +\begin{align*} +\frac{1}{k} + &= \frac{B^{2}}{4} + · \frac{1 - q\nu^{2}}{\nu \sqrt{q}} + · \nu(1 - \nu^{-2}) + · \frac{\Prod(1 - q^{2h+1}\nu^{2})(1 - q^{2h} \nu^{-2})} + {\Prod(1 - q^{2h-1}\nu^{2})(1 - q^{2h-2}\nu^{-2})} \\ +%% -----File: 071.png---Folio 65------- + &= \frac{B^{2}}{4 \sqrt{q}} (1 - q\nu^{2})\nu(1 - \nu^{-2}) + \frac{(1 - q^{3}\nu^{2})(1 - q^{5}\nu^{2}) \dotsm} + {(1 - q \nu^{2})(1 - q^{3}\nu^{2}) \dotsm} \\ + &\PadTo[r]{{}=\dfrac{B^{2}}{4 \sqrt{q}} (1 - q\nu^{2})\nu(1 - \nu^{-2})}{{}·{}} + \frac{(1 - q^{2}\nu^{-2})(1 - q^{4}\nu^{-2}) \dotsm} + {(1 - \nu^{-2})(1 - q^{2}\nu^{-2}) \dotsm} \\ + &= \frac{B^{2}}{4} · \frac{1}{\sqrt{q}}. +\end{align*} +\begin{align*} +\therefore B &= \frac{2 \sqrt[4]{q}}{\sqrt{k}}; \\ +\intertext{whence} +\frac{B}{A} &= 2 \sqrt[4]{q} \sqrt{\frac{k'}{k}}. +\end{align*} + +Substituting these values in \Eqref{eqs.}{}{(15)},~\Eqref{}{}{(16)}, and~\Eqref{}{}{(17)}, we +have +\begin{align*} +\Tag{(20)} +\sn (u, k) &= \frac{2 \sqrt[4]{q}}{\sqrt{k}} \sin \frac{\pi u}{2K} \Prod + \frac{1 - 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}; \\ +% +\Tag{(21)} +\cn (u, k) &= \frac{2 \sqrt{k'} \sqrt[4]{q}}{\sqrt{k}} \cos \frac{\pi u}{2K} \Prod + \frac{1 + 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}; \\ +% +\Tag{(22)} +\dn (u, k) &= \sqrt{k'} \Prod + \frac{1 + 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}. +\end{align*} +%% -----File: 072.png---Folio 66------- + + +\Chapter[The Theta Function.]{VIII}{The $\Theta$ Function.} + +\First{We} will indicate the denominator in \Eqref{eq.}{VII}{(20)}, Chap.~VII, by~$\phi (u)$, +thus: +\[ +\Tag{(1)} +\phi (u) = \Prod(1 - 2q^{2h - 1} \cos \frac{\pi u}{K} + q^{4h - 2}). +\] +We will now develop this into a series consisting of the cosines +of the multiples of $\dfrac{\pi u}{K}$. Put $\dfrac{\pi u}{2K} = x$, whence +\begin{align*} +2 \cos \frac{\pi u}{K} &= (e^{2ix} + e^{-2ix});\\ +\intertext{but} +1 - 2q^{2h - 1} \cos \frac{\pi u}{K} + q^{4h - 2} + &= (1 - q^{2h - 1}e^{2ix})(1 - q^{2h - 1}e^{-2ix}), +\end{align*} +and therefore +\begin{align*} +\Tag{(2)} +\phi (u) + &= (1 - qe^{2ix})(1 - q^{3}e^{2ix})(1 - q^{5}e^{2ix}) \dotsm \\ + &\PadTo{{}={}}{} + (1 - qe^{-2ix})(1 - q^{3}e^{-2ix})(1 - q^{5}e^{-2ix}) \dotsm +\end{align*} + +Putting now $u + 2iK'$ instead of~$u$, we have +\begin{align*} +x_{1} &= \frac{\pi(u + 2iK')}{2K} = x + \frac{\pi iK'}{K},\\ +2ix_{1} &= 2ix - \frac{2 \pi K'}{K}; +\end{align*} +%% -----File: 073.png---Folio 67------- +and +\begin{align*} +e^{2ix_{1}} &= q^{2}e^{2ix}, \\ +e^{-2ix_{1}} &= \frac{1}{q^{2}} e^{-2ix}. +\end{align*} +From these we have +\begin{align*} +\phi (u + 2iK') = - \frac{1}{q} e^{-2ix} + & (1 - qe^{2ix}) (1 - q^{3}e^{2ix}) \dotsm \\ + & (1 - qe^{-2ix})(1 - q^{3}e^{-2ix}) \dotsm; +\end{align*} +whence +\begin{align*} +\phi (u + 2iK') &= -\frac{1}{q} e^{-2ix} \phi (u), \\ +\intertext{or} +\Tag{(3)} +\phi (u + 2iK') &= -q^{-1} e^{-\frac{\pi iu}{K}} \phi (u). +\end{align*} + +Now put +\[ +\Tag{(4)} +\phi (u) = A + B \cos \frac{\pi u}{K} + + C \cos \frac{2\pi u}{K} + + D \cos \frac{3\pi u}{K} + \text{etc.} +\] + +Since +\[ +\cos \frac{\pi u}{K} = \tfrac{1}{2} \left(e^{2ix} + e^{-2ix}\right), +\] +this becomes +\begin{align*} +\Tag{(5)} +\phi (u) = A & + \tfrac{1}{2} Be^{2ix} + \tfrac{1}{2} Ce^{4ix} + \tfrac{1}{2} De^{6ix} + \dotsb \\ + & + \tfrac{1}{2} Be^{-2ix} + \tfrac{1}{2} Ce^{-4ix} + \tfrac{1}{2} De^{-6ix} + \dotsb; +\end{align*} +whence +\begin{align*} +\Tag{(6)} +-\frac{1}{q} e^{-2ix} \phi (u) = + & -\frac{A}{q} e^{-2ix} - \frac{B}{2q} - \frac{C}{2q} e^{2ix} - \frac{D}{2q} e^{4ix} - \dotsb \\ + & -\frac{B}{2q} e^{-4ix} - \frac{C}{2q} e^{-6ix} - \frac{D}{2q} e^{-8ix} - \dotsb\DPtypo{}{.} +\end{align*} +%% -----File: 074.png---Folio 68------- + +Now in \Eqref{equation}{}{(5)} put $u + 2iK'$ in place of~$u$, remembering +that $e^{2ix}$~and~$e^{-2ix}$ are thereby changed respectively into +$q^{2}e^{2ix}$~and~$q^{-2}e^{-2ix}$, and we have +\begin{align*} +\Tag{(7)} +\phi (u + 2iK') = A + & + \frac{Bq^{2}}{2} e^{2ix} + \frac{Cq^{4}}{2} e^{4ix} + \frac{Dq^{6}}{2} e^{6ix} + \dotsb \\ + & + \frac{B}{2q^{2}} e^{-2ix} + \frac{C}{2q^{4}} e^{-4ix} + \dotsb. +\end{align*} + +Since \Eqref{equations}{}{(6)} and~\Eqref{}{}{(7)} are equal, we have +\[ +\begin{array}{r@{}lcr@{}l} +-\dfrac{B}{2q} &{}= A, &\qquad\qquad& B &= -2qA; \\ +-\dfrac{C}{2q} &{}= \dfrac{Bq^{2}}{2}, && C &= +2q^{4}A; \\ +-\dfrac{D}{2q} &{}= \dfrac{Cq^{4}}{2}, && D &= -2q^{9}A; \\ +\Dots{2} && \Dots{2} \\ +\end{array} +\] +whence +\[ +\Tag{(8)} +\left\{ +\begin{aligned} %[** PP: Retain small parentheses] +&\Prod (1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}) \\ +&\begin{aligned} + {} = A(1 - 2q \cos \dfrac{\pi u}{K} + &+ 2q^{4} \cos \dfrac{2\pi u}{K} - 2q^{9} \cos \dfrac{3\pi u}{K} \\ + &+ 2q^{16} \cos \dfrac{4\pi u}{K} - \ldots). + \end{aligned} +\end{aligned} +\right. +\] + +The series in the second member has been designated by +Jacobi and subsequent writers by~$\Theta (u)$, thus: +\[ +\Tag{(9)} +\Theta (u) = 1 - 2q \cos \frac{\pi u}{K} + 2q^{4} \cos \frac{2\pi u}{K} - \dotsb +\] +%% -----File: 075.png---Folio 69------- + + +\Chapter[The Theta and Eta Functions.]{IX}{The $\Theta$ and $\Eta$ Functions.} + +\First{In} \Eqref{equation}{VII}{(20)}, Chap.~VII, viz., +\[ +\sn (u, k) = \frac{2 \sqrt[4]{q}}{\sqrt{k}} \sin \frac{\pi u}{2K} + \Prod \frac{1 - 2q^{2h} \cos \dfrac{\DPtypo{u \pi}{\pi u}}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}, +\] +the numerator and the denominator have been considered separately +by Jacobi, who gave them a special notation and developed +from them a theory second only in importance to the +elliptic functions themselves. + +Put [see \Eqref{equation}{VIII}{(8)}, Chap.~VIII] +\begin{gather*} +\Tag{(1)} +\Theta (u) = \frac{1}{A} \Prod (1 - 2q^{2h-1} \cos \frac{\pi u}{K} + q^{4h-2}). \\ +\Tag{(2)} +\Eta (u) = 2 \frac{1}{A} \sqrt[4]{q} \sin \frac{\pi u}{2K} + \Prod \left(1 - 2q^{2h} \cos \frac{\pi u}{K} + q^{4h}\right); +\end{gather*} +$A$ being a constant whose value is to be determined later. +From these we have +\[ +\Tag{(3)} +\sn (u, k) = \frac{1}{\sqrt{k}} · \frac{\Eta (u)}{\Theta (u)}. +\] +%% -----File: 076.png---Folio 70------- + +The functions $\sn u$~and~$\cn u$ can also be expressed in terms +of the new functions; thus we have +\[ +\Tag{(4)} +\cn (u, k) = \sqrt{\frac{k'}{k}} · 2 \sqrt[4]{q} \cos \frac{\pi u}{2K} + \Prod \frac{1 + 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}; +\] +or, since $\sin x = \cos \left(x + \dfrac{\pi}{2}\right)$ and $\cos x = -\DPtypo{\cos}{\sin} \left(x + \dfrac{\pi}{2}\right)$, +and putting $u = \dfrac{2Kx}{\pi}$, +\begin{align*} +\cn \left(\frac{2Kx}{\pi}, k\right) + &= \sqrt{\frac{k'}{k}} + \frac{\Eta \left[\dfrac{2K}{\pi} \left(x + \dfrac{\pi}{2}\right)\right]} + {\Theta \left(\dfrac{2Kx}{\pi}\right)} \\ + &= \sqrt{\frac{k'}{k}} + \frac{\Eta \left[\dfrac{2Kx}{\pi} + K\right]} + {\Theta \left(\dfrac{2Kx}{\pi}\right)}. +\end{align*} + +Replacing $\dfrac{2Kx}{\pi}$ by its value,~$u$, we have +\[ +\Tag{(5)} +\cn (u, k) = \sqrt{\frac{k'}{k}}\, \frac{\Eta (u + K)}{\Theta (u)}. +\] + +Furthermore, +\[ +\Tag{(6)} +\dn (u, k) = \sqrt{k'} \Prod + \frac{1 + 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}} +\] +%% -----File: 077.png---Folio 71------- +gives in the same manner +\begin{align*} +\dn \frac{2Kx}{\pi} + &= \sqrt{k'}\, + \frac{\Theta \left[\dfrac{2K}{\pi} \left(x + \dfrac{\pi}{2}\right)\right]} + {\Theta \left(\dfrac{2Kx}{\pi}\right)}, +\intertext{or} +\Tag{(7)} +\dn (u, k) + &= \sqrt{k'}\, \frac{\Theta (u + K)}{\Theta (u)}. +\end{align*} + +If we put +\begin{align*} +\Tag{(8)} +\Eta (u + K) &= \Eta_{1}(u), \\ +\Tag{(9)} +\Theta (u + K) &= \Theta_{1}(u), +\end{align*} +the three elliptic functions can be expressed by the following +formulas: +\begin{align*} +\Tag{(10)} +\sn (u, k) &= \frac{1}{\sqrt{k}} · \frac{\Eta (u)}{\Theta (u)}; \\ +\Tag{(11)} +\cn (u, k) &= \sqrt{\frac{k'}{k}} · \frac{\Eta_{1}(u)}{\Theta (u)}; \\ +\Tag{(12)} +\dn (u, k) &= \sqrt{k'} \frac{\Theta_{1} (u)}{\Theta (u)}. +\end{align*} + +These functions $\Theta$~and~$\Eta$ can be expressed in terms of +each other. By definition, +\[ +\Eta (u) = 2C \sqrt[4]{q} \sin \frac{\pi u}{2K} + \Prod \left(1 - 2q^{2h} \cos \frac{\pi u}{K} + q^{4h}\right); +\] +%% -----File: 078.png---Folio 72------- +but +\begin{align*} +1 - 2q^{h} \cos \frac{\pi u}{K} + q^{2h} + &= \Bigl(1 - q^{h}e^{ \frac{\pi u \sqrt{-1}}{K}}\Bigr) + \Bigl(1 - q^{h}e^{-\frac{\pi u \sqrt{-1}}{K}}\Bigr) \\ +\sin \frac{\pi u}{2K} + &= \frac{e^{\frac{\pi iu}{2K}} - e^{-\frac{\pi iu}{2K}}}{2 \sqrt{-1}} \\ + &= e^{\frac{-\pi iu}{2K}} \frac{1 - e^{\frac{\pi iu}{K}}}{2} \sqrt{-1}, +\end{align*} +and consequently +\[ +\Tag{(13)} +\Eta(u) = C \sqrt[4]{q} e^{-\frac{\pi iu}{2K}} + \sqrt{-1} \Bigl(1 - e^{\frac{\pi iu}{K}}\Bigr) + \Bigl(1 - q^{2}e^{-\frac{\pi iu}{K}}\Bigr) + \Bigl(1 - q^{2}e^{ \frac{\pi iu}{K}}\Bigr) \dotsm. +\] + +Now, changing $u$ into~$u + iK'$, and remembering that +$e^{-\frac{\pi K'}{K}} = q$, we have +\begin{multline*} %[** PP: Re-breaking] +\Tag{(14)} +\Eta(u + iK') + = Cq^{-\frac{1}{4}}e^{\DPtypo{\frac{-\pi iu}{2K}}{-\frac{\pi iu}{2K}}} + \sqrt{-1}\Bigl(1 - qe^{ \frac{\pi iu}{K}}\Bigr) + \Bigl(1 - qe^{-\frac{\pi iu}{K}}\Bigr) \\ + \Bigl(1 - q^{3}e^{ \frac{\pi iu}{K}}\Bigr) + \Bigl(1 - q^{3}e^{-\frac{\pi iu}{K}}\Bigr) \dotsm; +\end{multline*} +and reuniting the factors two by two, this becomes +\begin{multline*} %[** PP: Re-breaking] +\Tag{(15)} +\Eta(u + iK') + = C \sqrt{-1} q^{-\frac{1}{4}}e^{-\frac{\pi iu}{2K}} \\ + \left(1 - 2q \cos \frac{\pi u}{K} + q^{2}\right) + \left(1 - 2q^{3} \cos \frac{\pi u}{K} + q^{6}\right) \dotsm; +\end{multline*} +and finally, according to \Eqref{equation}{}{(1)}, +\[ +\Tag{(16)} +\Eta (u + iK') = \sqrt{-1} q^{-\frac{1}{4}} e^{-\frac{\pi iu}{2K}} \Theta (u). +\] +%% -----File: 079.png---Folio 73------- + +In the same manner, we can get +\[ +\Tag{(17)} +\Theta (u + iK') = \sqrt{-1} q^{-\frac{1}{4}}e^{-\frac{\pi iu}{2K}} \Eta (u). +\] + +Substituting $u+2K$ for~$u$ in \Eqref{equations}{}{(1)} and~\Eqref{}{}{(2)}, we get +\begin{align*} +\Tag{(18)} +\Theta (u + 2K) &= \Theta (u), \\ +\Tag{(19)} +\Eta (u + 2K) &= -\Eta (u), +\end{align*} +since $\cos \dfrac{\pi}{K} (u + 2K) = \cos \dfrac{\pi u}{K}$ and $\sin \dfrac{\pi}{2K} (u + 2K) = -\sin \dfrac{\pi u}{2K}$. + +The comparison of these four equations with \Eqref{equations}{}{(10)},~\Eqref{}{}{(11)}, +and~\Eqref{}{}{(12)} shows the periodicity of the elliptic functions. +For example, comparing \Eqref{eqs.}{}{(10)} and~\Eqref{}{}{(16)} and~\Eqref{}{}{(17)}, we +see that changing~$u$ into $u+iK'$ simply multiplies the numerator +and denominator of the second member of \Eqref{eq.}{}{(10)} by +the same number, and does not change their ratio. + +The addition of~$2K$ changes the sign of the function, but +not its value. + +We will define $\Theta_{1}$ and~$\Eta_{1}$ as follows: +\begin{align*} +\Tag{(20)} +\Theta_{1}(x) &= \Theta (x + K); \\ +\Tag{(21)} +\Eta_{1}(x) &= \Eta (x + K). +\end{align*} +Hence we get, from \Eqref{equation}{}{(17)}, +\begin{DPalign*} +\Theta_{1}(x + iK') + &= \Theta (x + iK' + K) = \Theta (x + K + iK') \\ + &= i\Eta (x + K)e^{-\frac{i\pi}{4K} (2x + 2K + iK')} \\ + &= i\Eta_{1}(x)e^{-\frac{i\pi}{4K} (2x + iK')} (-\sqrt{-1}), \\ +\lintertext{since} +e^{-\frac{i\pi}{2}} + &= \cos \frac{\pi}{2} - \sqrt{-1} \sin \frac{\pi}{2} = -\sqrt{-1}; +\end{DPalign*} +%% -----File: 080.png---Folio 74------- +whence +\[ +\Tag{(22)} +\Theta_{1}(x + iK') = \Eta_{1}(x)e^{-\frac{i\pi}{4K}(2x + iK')}. +\] +In a similar manner we get +\[ +\Tag[22st]{(22)^*} +\Eta_{1}(x + iK') = \Theta_{1}(x)e^{-\frac{i\pi}{4K}(2x + iK')}. +\] + +In \Eqref{eq.}{VIII}{(9)}, Chap.~VIII, put $u = \dfrac{2Kz}{\pi}$, and we get +\[ +\Tag{(23)} +\Theta \left(\frac{2Kz}{\pi}\right) = 1 - 2q \cos 2z + 2q^{4} \cos 4z - \dotsb. +\] + +Now, in this equation, changing~$z$ into $z + \dfrac{\pi}{2}$, and observing +\Eqref{eq.}{}{(20)}, we get +\[ +\Tag{(24)} +\Theta_{1} \left(\frac{2Kz}{\pi}\right) = 1 + 2q \cos 2z + 2q^{4} \cos 4z + \dotsb. +\] + +Applying \Eqref{eq.}{}{(22)} to this, we have +\begin{align*} +&\begin{aligned} +\Eta_{1} \left(\frac{2Kz}{\pi}\right) %[** Explicit sizing of () required?] + &= \Theta_{1}\left(\frac{2K}{\pi} \Bigl(z + \frac{\pi iK'}{2K}\Bigr)\right) e^{\frac{\pi i}{4K}\left(\frac{4Kz}{\pi} + iK'\right)} \\ + &= \Theta_{1}\left(\frac{2K}{\pi} \Bigl(z + \frac{\pi iK'}{2K}\Bigr)\right) e^{iz}q^{\frac{1}{4}} +\end{aligned} \\ +% + &= e^{iz}q^{\frac{1}{4}} + \left[1 + 2q \cos 2\Bigl(z + \frac{\pi iK'}{2K}\Bigr) + + 2q^{4} \cos 4\Bigl(z + \frac{\pi iK'}{2K}\Bigr) + \dotsb\right] \\ + &= e^{iz}q^{\frac{1}{4}} + \biggl[1 + q \Bigl(e^{2i\bigl(z + \frac{\pi iK'}{2K}\bigr)} + + e^{-2i\bigl(z + \frac{\pi iK'}{2K}\bigr)}\Bigr) \\ + & \PadTo{{}=e^{iz}q^{\frac{1}{4}} \biggl[1}{} + {} + q^{4} \Bigl(e^{4i\bigl(z + \frac{\pi iK'}{2K}\bigr)} + + e^{-4i\bigl(z + \frac{\pi iK'}{2K}\bigr)}\Bigr) + \dotsb\biggr] \\ +%% -----File: 081.png---Folio 75------- + &= e^{iz} q^{\frac{1}{4}} + \left[1 + q(qe^{2iz} + q^{-1}e^{-2iz}) + + q^{4}(q^{2}e^{4iz} + q^{-2}e^{-4iz}) + \dotsb \right] \\ +%[** PP: Not breaking next two lines] + &= e^{iz}q^{\frac{1}{4}} + \left[1 + q^{2}e^{2iz} + q^{6}e^{4iz} + \dotsb + + e^{-2iz} + q^{2}e^{-4iz} + \dotsb \right] \\ + &= q^{\frac{1}{4}} \left[e^{iz} + q^{2}e^{3iz} + q^{6}e^{5iz} + \dotsb + + e^{-iz} + q^{2}e^{-3iz} + q^{6}e^{-5iz} + \dotsb \right] \\ + &= 2q^{\frac{1}{4}} \left[\cos z + q^{2} \cos 3z + q^{6} \cos 5z + \dotsb\right]; +\end{align*} +whence +\[ +\Tag{(25)} +\Eta_{1} \left( \frac{2Kz}{\pi} \right) + = 2 \sqrt[4]{q} \cos z + + 2 \sqrt[4]{q^{9}} \cos 3z + + 2 \sqrt[4]{q^{25}} \cos 5z + \dotsb +\] + +In this equation, changing~$z$ into $z - \dfrac{\pi}{2}$, and applying \Eqref{eq.}{}{(21)}, +we get +\begin{align*} +\Tag{(26)} +\Eta \left( \frac{2Kz}{\pi} \right) + &= 2 \sqrt[4]{q} \sin z + - 2 \sqrt[4]{q^{9}} \sin 3z + + 2 \sqrt[4]{q^{25}} \sin 5z - \dotsb, \\ +\intertext{since} +\Eta_{1} \left( \frac{2Kz}{\pi} \right) + &= \Eta \left( \frac{2Kz}{\pi} + K \right). +\end{align*} + +We will now determine the constant~$A$ of \Eqref{eq.}{VIII}{(8)}, Chap.~VIII, +and \Eqref{eqs.}{}{(1)} and~\Eqref{}{}{(2)} of this chapter. Denote~$A$ by~$f(q)$, +and we have +\[ +\Tag[26st]{(26)^*} +\Prod(1 - 2q^{2h - 1} \cos \frac{\pi u}{K} + q^{4h - 2}) = f(q)\Theta (u). +\] + +Substituting herein $u = 0$ and $u = \dfrac{K}{2}$, we have +\begin{align*} +\Prod(1 - q^{2h - 1})^{2} &= f(q) \Theta (0);\\ +\Prod(1 + q^{4h - 2}) &= f(q) \Theta \left( \frac{K}{2} \right). +\end{align*} +%% -----File: 082.png---Folio 76------- + +From \Eqref{eq.}{VIII}{(9)}, Chap.~VIII, we get +\begin{align*} +\Tag{(27)} +\Theta (0) + &= 1 - 2q + 2q^{4} - 2q^{9} + 2q^{16} - \dotsb; \\ +\Tag{(28)} +\Theta \left( \frac{K}{2} \right) + &= 1 - 2q^{4} + 2q^{16} - 2q^{36} + 2q^{64} - \dotsb; +\end{align*} +from which we see that $\Theta (0)$ is changed into $\Theta \left( \dfrac{K}{2} \right)$ when we +put $q^{4}$~in place of~$q$. + +Whence +\[ +\Prod(1 - q^{8h - 4})^{2} = f(q^{4})\Theta \left( \frac{K}{2} \right); +\] +and therefore +\begin{align*} +\frac{f(q)}{f(q^{4})} + &= \Prod \frac{1 + q^{4h - 2}}{(1 - q^{8h - 4})^{2}}\\ +\Tag{(29)} + &= \Prod \frac{1}{(1 - q^{8h - 4})(1 - q^{4h - 2})}. +\end{align*} + +Now, the expressions $4h - 2$, $8h - 4$, and~$8h$ give the +following series of numbers: +\begin{center} +\small +\begin{tabular}{l<{\qquad}*{18}{@{\,}c@{\,}}} +$4h - 2$, &2,& &6,& &10,& &14,& &18,& &22,& &26,& &30,& &34;& \\ +$8h - 4$, & &4,& & & &12,& & & &20,& & & &28,& & & &36;\\ +$8h$, & & & &8,& & & &16,& & & &24,& & & &32.& & +\end{tabular} +\end{center} +Hence, the three expressions taken together contain all the +even numbers, and +\[ +\Prod(1 - q^{8h - 4})(1 - q^{4h - 2})(1 - q^{8h}) = \Prod(1 - q^{2h}). +\] +Therefore, multiplying \Eqref{eq.}{}{(29)} by +\begin{gather*} +\Prod \frac{1 - q^{8h}}{1 - q^{8h}},\\ +\intertext{we have} +\frac{f(q)}{f(q^{4})} = \Prod \frac{1 - q^{8h}}{1 - q^{2h}}. +\end{gather*} +%% -----File: 083.png---Folio 77------- + +Now in this equation, by successive substitutions of~$q^{4}$ for~$q$, +we get +\[ +\begin{array}{r@{}l} +\dfrac{f(q^{4})}{f(q^{16})} &{}= \Prod \dfrac{1 - q^{32h} }{1 - q^{8h}}; \\ +\dfrac{f(q^{16})}{f(q^{64})} &{}= \Prod \dfrac{1 - q^{128h}}{1 - q^{32h}}; \\ +\dfrac{f(q^{64})}{f(q^{256})} &{}= \Prod \dfrac{1 - q^{512h}}{1 - q^{128h}}; \\ +\Dots{2} \\ +\end{array} +\] + +Now $q$ being less than~$1$, $q^{n}$~tends towards the limit~$0$ as $n$~increases, +and consequently $1-q^{n}$ tends towards the limit~$1$. +Also, from \Eqref{eq.}{VIII}{(8)}, Chap.~VIII, we see that $f(0) = 1$. Hence, +multiplying the above equations together member by member, +we have +\begin{align*} +\Tag{(30)} +f(q) &= \Prod \frac{1}{1-q^{2h}}, \\ +\intertext{or} +\Tag{(31)} +A &= \frac{1}{(1 - q^{2})(1 - q^{4})(1 - q^{6}) \dotsm}. +\end{align*} + +Substituting this value in \Eqref{equation}{VIII}{(8)}, Chap.~VIII, we have, +after making $u = 0$, +\begin{align*} +(1 - q)^{2}(1 - q^{3})^{2}(1 - q^{5})^{2} \dotsm + &= \frac{1- 2q + 2q^{4} - 2q^{9} + \dotsb} + {(1 - q^{2})(1 - q^{4})(1 - q^{6}) \dotsm} \\ + &= \frac{\Theta (0)}{(1 - q^{2})(1 - q^{4})(1 - q^{6}) \dotsm}. +\end{align*} + +(See \Eqref{equation}{VIII}{(9)}, Chap.~VIII.) + +Transposing one of the series of products from the left-hand +member, we get +\[ +(1 - q)(1 - q^{3}) \dotsm + = \frac{\Theta (0)}{(1 - q)(1 - q^{2})(1 - q^{3})(1 - q^{4}) \dotsm}. +\] +%% -----File: 084.png---Folio 78------- + +Introducing on both sides of the equation the factors $1 - q^{2}$, +$1 - q^{4}$, $1 - q^{6}$,~etc., we get +\begin{align*} +(1 - q) &(1 - q^{2})(1 - q^{3})(1 - q^{4}) \dotsm \\ + &= \Theta (0) \frac{1 - q^{2}}{1 - q} + · \frac{1 - q^{4}}{1 - q^{2}} + · \frac{1 - q^{6}}{1 - q^{3}} + · \frac{1 - q^{8}}{1 - q^{4}} \dotsm \\ + &= \Theta (0) (1 + q)(1 + q^{2})(1 + q^{3}) \dotsm; +\intertext{whence} +\Tag{(32)} +\Theta (0) + &= \frac{(1 - q)(1 - q^{2})(1 - q^{3})\DPtypo{}{\dotsm}} + {(1 + q)(1 + q^{2})(1 + q^{3})\DPtypo{}{\dotsm}}. +\end{align*} + +Resuming \Eqref{equation}{VII}{(20)}, Chap.~VII, and dividing both +members of the equation by~$u$, we have +\[ +\frac{\sn u}{u} + = \frac{2 \sqrt[4]{q}}{\sqrt{k}}\, + \frac{\sin \dfrac{\pi u}{2K}}{u} + \Prod \frac{1 - 2q^{2h} \cos \dfrac{\pi u}{K} + q^{4h}} + {1 - 2q^{2h-1} \cos \dfrac{\pi u}{K} + q^{4h-2}}. +\] +This, for~$u = 0$, since the limiting value of $\dfrac{\sn u}{u}$ for~$u = 0$ is~$1$, +and of $\dfrac{\sin \dfrac{\pi u}{2K}}{u}$ for~$x=0$ is~$\dfrac{\pi}{2K}$, becomes +\begin{align*} +1 &= \frac{\sqrt[4]{q}}{\sqrt{k}} · \frac{\pi}{K} + · \frac{(1 - q^{2})^{2}(1 - q^{4})^{2}(1 - q^{6})^{2} \dotsm} + {(1 - q)^{2}(1 - q^{3})^{2}(1 - q^{5})^{2} \dotsm}, +\intertext{or} +\Tag{(33)} +\frac{\sqrt{k} K}{\pi \sqrt[4]{q}} + &= \left[\frac{(1 - q^{2})(1 - q^{4})(1 - q^{6}) \dotsm} + {(1 - q)(1 - q^{3})(1 - q^{5}) \dotsm}\right]^{2}. +\end{align*} + +Further, from \Eqref{equation}{VII}{(21)}, Chap.~VII, for~$u=0$, we have +\[ +\Tag{(34)} +\frac{\sqrt{k}}{2\sqrt{k'} \sqrt[4]{q}} + = \left[\frac{(1 + q^{2})(1 + q^{4})(1 + q^{6}) \dotsm} + {(1 - q)(1 - q^{3})(1 - q^{5}) \dotsm}\right]^{2}. +\] +%% -----File: 085.png---Folio 79------- + +The quotient of these two equations gives +\[ +\Tag{(35)} +\frac{2\sqrt{k'}K}{\pi} + = \left[\frac{(1 - q^{2})(1 - q^{4})(1 - q^{6}) \dotsm} + {(1 + q^{2})(1 + q^{4})(1 + q^{6}) \dotsm}\right]^{2}; +\] +or, substituting the value of~$\sqrt{k'}$ from \Eqref{eqs.}{VII}{(18)} and~\Eqref{}{VII}{(19)}, Chap.~VII, +\[ +\Tag{(36)} +\frac{2k'K}{\pi} + = \left[\frac{(1 - q)(1 - q^{2})(1 - q^{3}) \dotsm} + {(1 + q)(1 + q^{2})(1 + q^{3}) \dotsm}\right]^{2}. +\] + +Comparing this with \Eqref{equation}{}{(32)}, we easily get +\[ +\Tag{(37)} +\Theta (0) = \sqrt{\frac{2k'K}{\pi}}. +\] + +From \Eqref{equation}{VIII}{(9)}, Chap.~VIII, making~$u=K$, we get +\[ +\Tag{(38)} +\Theta (K) = 1 + 2q + 2q^{4} + 2q^{9} + 2q^{16} + \dotsb. +\] + +Making $z=0$ in \Eqref{equation}{IX}{(24)}, Chap.~IX, we have +\[ +\Tag{(39)} +\Theta_{1} (0) = 1 + 2q + 2q^{4} + 2q^{9} + \dotsb. +\] + +This might also have been derived from \Eqref{eq.}{}{(38)} by observing +that +\[ +\DPtypo{\Theta_{1}}{\Theta} (0 + K) = \Theta_{1}(0) = \Theta (K). +\] +Knowing $\Theta (0)$, it is easy to deduce $\Theta (K)$~and~$\Eta (K)$. + +From \Eqref{equation}{}{(7)} we have +\[ +\dn u = \sqrt{k'}\, \frac{\Theta (u + K)}{\Theta (u)}. +\] + +Making $u=0$, we have, since $\dn (0) = 1$, +\[ +\Tag{(40)} +\Theta (K) = \frac{\Theta (0)}{\sqrt{k'}}. +\] +%% -----File: 086.png---Folio 80------- + +From \Eqref{equation}{}{(5)} we get, in the same manner, +\[ +\Tag{(41)} +\Eta (K) = \sqrt{\frac{k'}{k}}\, \Theta (0). +\] + +From \Eqref{eq.}{IX}{(12)}, Chap.~IX, we have +\[ +\Tag[41st]{(41)^*} +\dn u + = \sqrt{1 - k^{2} \sin^{2} \phi} + = \sqrt{k'}\, \frac{\Theta_{1}(u)}{\Theta (u)}; +\] +and putting $x = \dfrac{\pi u}{2K}$, we have +\[ +\Tag{(42)} +\frac{\dn u}{\sqrt{k'}} + = \frac{1 + 2q \cos 2x + 2q^{4} \cos 4x + 2q^{9} \cos 6x + \dotsb} + {1 - 2q \cos 2x + 2q^{4} \cos 4x - 2q^{9} \cos 6x + \dotsb}. +\] + +Putting +\[ +\Tag[42st]{(42)^*} +\frac{\dn u}{\sqrt{k'}} = \cot \gamma, +\] +we have +\[ +\frac{\cot \gamma - 1}{\cot \gamma + 1} + = \tan (45° - \gamma) + = 2q \frac{\cos2x + q^{8}(4 \cos^{3} 2x - 3 \cos 2x) + \dotsb} + {1 + q^{4}(4 \cos^{2} 2x - 2)}; +\] +whence +\begin{multline*} +\Tag{(43)} +\cos 2x = \frac{\tan (45° - \gamma) [1 + q^{4}(4 \cos^{2}2x - 2)]}{2q} \\ + - q^{8}(4 \cos^{3}2x - 3\cos2x) - \dotsb, +\end{multline*} +and approximately, +\[ +\Tag{(44)} +\cos 2x = \frac{\tan (45° - \gamma)}{2q}. +\] + +From \Eqref{equations}{IX}{(37)} and~\Eqref{}{IX}{(40)}, Chap.~IX, we have +\begin{align*} +\Tag{(45)} +x &= \frac{u}{\Theta^{2}(K)}; \\ +\intertext{whence} +\Tag{(46)} +u &= x\Theta^{2}(K). +\end{align*} +%% -----File: 087.png---Folio 81------- + + +\Chapter{X}{Elliptic Integrals of the Second Order.} + +\First{From} Chap.~I, \Eqref{equation}{I}{(19)}, we have +\[ +E(k, \phi) = \int_{0}^{\phi} \sqrt{1 - k^{2} \sin^{2} \phi} · d\phi + = \int_{0}^{\phi} \Delta \phi · d\phi. +\] + +From this we have +\[ +E(\phi) + E(\psi) + = \int_{0}^{\phi} \Delta \phi · d\phi + \int_{0}^{\psi} \Delta \phi · d\phi. +\] + +Put +\[ +\Tag{(1)} +E\phi + E\psi = S. +\] + +Differentiating, we get +\[ +\Tag{(2)} +\Delta \phi · d\phi + \Delta \psi · d\psi = dS. +\] +But we have, Chap.~II, \Eqref{equation}{II}{(2)}, +\[ +\frac{d\phi}{\Delta \phi} + \frac{d\psi}{\Delta \psi} = 0, +\] +or +\[ +\Tag{(3)} +\Delta \psi · d\phi + \Delta \phi · d\psi = 0. +\] + +Adding equations \Eqref{}{}{(2)}~and~\Eqref{}{}{(3)}, we get +\[ +\Tag{(4)} +(\Delta \phi + \Delta \psi)(d\phi + d\psi) = dS. +\] +%% -----File: 088.png---Folio 82------- + +Substituting $\cos \mu$ from \Eqref{eq.}{II}{(5)}, in \Eqref[5st]{eq.}{II}{(5)^*}, Chap.~II, we get +\[ +\Tag{(5)} +\left\{ +\begin{aligned} +\Delta\phi + &= \frac{\sin\phi \cos\psi\, \Delta\mu + \cos\phi \sin\psi}{\sin\mu}, \\ +\Delta\psi + &= \frac{\sin\psi \cos\phi\, \Delta\mu + \cos\psi \sin\phi}{\sin\mu}; +\end{aligned} +\right. +\] +whence +\[ +\Tag{(6)} +\Delta \phi ± \Delta \psi = \frac{\Delta \mu ± 1}{\sin \mu} \sin (\phi ± \psi). +\] + +Substituting in \Eqref{equation}{}{(4)}, we have +\begin{align*} +dS &= \frac{\Delta \mu + 1}{\sin \mu} \sin (\phi + \psi)\,d(\phi + \psi) \\ +\Tag{(7)} + &= - \frac{\Delta \mu + 1}{\sin \mu}\,d \cos (\phi + \psi). +\end{align*} +Integrating equation~\Eqref{}{}{(7)}, we have +\[ +E\phi + E\psi = \frac{\Delta \mu + 1}{\sin \mu} \left[C - \cos (\phi + \psi)\right]. +\] + +The constant of integration,~$C$, is determined by making +$\phi = 0$; in this case $\psi = \mu$, $E\phi = 0$, $E\psi = E\mu$, and $S = E\mu$; +whence +\[ +E\mu = \frac{\Delta \mu + 1}{\sin \mu} (C - \cos \mu), +\] +and by subtraction, +\[ +E\phi + E\psi - E\mu + = \frac{\Delta \mu + 1}{\sin \mu} + (\cos \mu - \cos \phi \cos \psi + \sin \phi \sin \psi). +\] +But, Chap.~II, \Eqref{eq.}{II}{(5)}, +\[ +\cos\mu - \cos\phi \cos\psi = - \sin\phi \sin\psi\, \Delta\mu; +\] +%% -----File: 089.png---Folio 83------- +whence +\[ +E\phi + E\psi - E\mu = \frac{1 - \Delta^{2}\mu}{\sin \mu} \sin \phi \sin \psi +\] +whence +\[ +\Tag{(8)} +E\phi + E\psi = E\mu + k^{2} \sin\phi \sin\psi \sin\mu. +\] + +When $\phi = \psi$, we have +\begin{align*} +\Tag{(9)} +E\mu &= 2E\phi - k^{2} \sin^{2} \phi \sin\mu. \\ +\intertext{But in that case} +\Tag{(10)} +\cos \mu &= \cos^{2} \phi - \sin^{2} \phi\, \Delta \mu; \\ +\intertext{whence} +\Tag{(11)} +\sin \phi &= \sqrt{\frac{1 - \cos \mu}{1 + \Delta \mu}}. +\end{align*} + +Let $\phi$, $\phi_{\frac{1}{2}}$, $\phi_{\frac{1}{4}}$,~etc., be such values as will satisfy the equations +\begin{align*} +\Tag{(12)} +E\phi + &= 2E\phi_{\frac{1}{2}} - k^{2} \sin^{2} \phi_{\frac{1}{2}} \sin \phi,\\ +E\phi_{\frac{1}{2}} + &= 2E\phi_{\frac{1}{4}} - k^{2} \sin^{2} \phi_{\frac{1}{4}} \sin \phi_{\frac{1}{2}},\\ + &\PadTo{= 2E\phi_{\frac{1}{4}}}{\text{etc.}} + \PadTo{ - k^{2} \sin^{2} \phi_{\frac{1}{4}} \sin \phi_{\frac{1}{2}}}{\text{etc.}} +\end{align*} + +Assume an auxiliary angle~$\gamma$, such that +\[ +\Tag{(13)} +\sin\gamma = k \sin\phi; +\] +whence +\[ +\Delta\phi = \cos\gamma, +\] +and Chap.~IV, \Eqref{eq.}{IV}{(24)}, +\[ +\Tag{(14)} +\sin\phi_{\frac{1}{2}} + = \frac{\sin \frac{1}{2}\phi}{\cos \frac{1}{2}\gamma}. +\] +%% -----File: 090.png---Folio 84------- + +Applying eqs.\ \Eqref{}{}{(13)}~and~\Eqref{}{}{(14)} successively, we get +\[ +\Tag{(15)} +\left\{ +\begin{array}{r@{}l} +\sin \phi_{\frac{1}{2}} + &{}= \dfrac{\sin \frac{1}{2} \phi}{\cos \frac{1}{2} \gamma},\quad + \sin \gamma_{\frac{1}{2}} = k \sin \phi_{\frac{1}{2}}; \\ +\sin \phi_{\frac{1}{4}} + &{}= \dfrac{\sin \frac{1}{2} \phi_{\frac{1}{2}}} + {\cos \frac{1}{2} \gamma_{\frac{1}{2}}},\quad + \sin \gamma_{\frac{1}{4}} = k \sin \phi_{\frac{1}{4}}; \\ +\Dots{2} \\ +\sin \phi_{\frac{1}{2^{n}}} + &{}= \dfrac{\sin \frac{1}{2} \phi_{\frac{1}{2^{n - 1}}}} + {\cos \frac{1}{2} \gamma_{\frac{1}{2^{n - 1}}}}; +\end{array} +\right. +\] +whence +\begin{align*}%[** PP: Unbalanced parentheses in original] +\Tag{(16)} +E \phi &= 2^{n} E \phi_{\frac{1}{2^{n}}} + - \Bigl(\sin \phi \sin^{2} \gamma_{\frac{1}{2}} + + 2 \sin \phi_{\frac{1}{2}} \sin^{2} \gamma_{\frac{1}{4}} \\ + &+ 2^{2} \sin \phi_{\frac{1}{4}} \sin^{2} \gamma_{\frac{1}{8}} + \dotsb + 2^{n-1} \sin \phi_{\frac{1}{2^{n}}} \sin^{2} \gamma_{\frac{1}{2^{n - 1}}} \Bigr) +\end{align*} + +%[** PP: Modernizing factorial notation] +To find the limiting value, $E \phi_{\frac{1}{n}}$, we have, by the Binomial +Theorem, since $\sin \phi = 1 - \dfrac{\phi^{3}}{3!} + \dfrac{\phi^{5}}{5!} -{}$ etc., +\begin{align*} +\Delta \phi + &= (1 - k^{2} \sin^{2} \phi)^{\frac{1}{2}}\\ + &= 1 - \frac{k^{2}}{2} \left( \phi - \frac{\phi^{3}}{6} \right)^{2} +%[** PP: Fourth power in next line missing in original] + - \frac{k^{4}}{8} \left( \phi - \frac{\phi^{3}}{6} \right)^{4} + \dotsb\\ + &= 1 - \frac{k^{2}}{2} \phi^{2} + + \left( \frac{k^{2}}{6} - \frac{k^{4}}{8} \right) \phi^{4}. +%[** PP: Series truncated to polynomial, presumably 4th power approximation] +\end{align*} +Whence +\begin{align*} +E k \phi_{\frac{1}{2^{n}}} + &= \int_{0}^{\phi_{n}} \Delta \phi_{\frac{1}{2^{n}}}\, d\phi\\ +\Tag{(17)} + &= \phi_{\frac{1}{2^{n}}} - \frac{k^{2}}{6} \phi^{3}_{\frac{1}{2^{n}}} + + \frac{k^{2}(4 - 3k^{2})}{120} \phi^{5}_{\frac{1}{2^{n}}}. +\end{align*} +%% -----File: 091.png---Folio 85------- + +Substituting in \Eqref{eq.}{}{(16)} the numerical values derived from +\Eqref{equations}{}{(15)} and~\Eqref{}{}{(17)}, we are enabled to determine the value +of~$E\phi$. + +Landen's Transformation can also be applied to Elliptic +Integrals of this class. + +From \Eqref{eq.}{IV}{(11)}, Chap.~IV, we get, by easy transformation, +\[ +\Tag{(18)} +\sin^{2} 2\phi = \sin^{2} \phi_{1} (1 + k_{0} + 2k_{0} \cos 2\phi). +\] +From this we easily get +\begin{align*} +2k_{0} \cos 2\phi \sin^{2} \phi_{1} + &= \sin^{2} 2\phi - \sin^{2} \phi_{1} - k_{0}^{2} \sin^{2} \phi_{1} \\ + &= 1 - \cos^{2} 2\phi - \sin^{2} \phi_{1} - k_{0}^{2} \sin^{2} \phi_{1} \\ + &= \Delta^{2}k_{0}\phi_{1} - \sin^{2}\phi_{1} - \cos^{2}2\phi; +\end{align*} +whence +\[ +\cos^{2} 2\phi + 2k_{0} \sin^{2}\phi_{1} \cos 2\phi + = \Delta^{2}k_{0}\phi_{1} - \sin^{2}\phi_{1}; +\] +and from this, +\begin{align*} +\cos 2\phi + &= -k_{0} \sin^{2} \phi_{1} + ± \sqrt{\Delta^{2}k_{0}\phi_{1} - \sin^{2} \phi_{1} + + k_{0}^{2} \sin^{4} \phi_{1}} \\ +\Tag{(19)} + &= \cos \phi_{1} \Delta k_{0}\phi_{1} - k_{0} \sin^{2} \phi_{1}; +\end{align*} +whence, also, +\begin{align*} +1 - \cos^{2} 2\phi + &= 1 - \cos^{2} \phi_{1}\, \Delta^{2}\phi_{1} + + 2k \sin^{2}\phi_{1} \cos \phi_{1}\, \Delta k_{0}\phi_{1} + - k_{0}^{2} \sin^{4} \phi_{1} \\ + &= \sin^{2} \phi_{1} + (1 + k_{0}^{2} \cos^{2} \phi_{1} + + 2k_{0} \cos \phi_{1}\, \Delta k_{0}\phi_{1} + - k_{0}^{2} \sin^{2} \phi_{1}) +\end{align*} +and +\[ +\Tag{(20)} +\sin 2\phi = \sin \phi_{1} (\Delta k_{0}\phi_{1} + k_{0} \cos \phi_{1}). +\] + +Differentiating \Eqref{equation}{}{(19)}, we get +\[ +2 \sin 2\phi \frac{d\phi}{d\phi_{1}} + = \sin \phi_{1} + \frac{(k_{0} \cos \phi_{1} + \Delta k_{0}\phi_{1})^{2}} + {\Delta k_{0} \phi_{1}}. +\] +Dividing this by \Eqref{equation}{}{(20)}, we have +\[ +\frac{2d\phi}{d\phi_{1}} + = \frac{k_{0} \cos \phi_{1} + \Delta k_{0}\phi_{1}}{\Delta k_{0}\phi_{1}}. +\] +%% -----File: 092.png---Folio 86------- +But from~\Eqref{}{}{(19)}, and \Eqref{eq.}{IV}{(6)}, Chap.~IV, +\begin{align*} +k^{2} \sin^{2} \phi + &= \frac{k^{2}(1 - \cos 2\phi)}{2} \\ + &= \frac{2k_{0}}{(1 + k_{0})^{2}} + \{1 + k_{0} \sin^{2} \phi_{1} - \cos \phi_{1} \Delta k_{0} \phi_{1}\}; +\intertext{whence} +\Delta k\, \phi + &= \frac{\Delta k_{0}\phi_{1} + k_{0} \cos \phi_{1}}{1 + k_{0}}, +\intertext{and} +2\Delta k\, \phi · \frac{d\phi}{d\phi_{1}} + &= \frac{(k_{0} \cos \phi_{1} + \Delta k_{0}\, \phi_{1})^{2}} + {(1 + k_{0}) \Delta k_{0}\, \phi_{1}}, +\intertext{and} +d\phi\, \Delta k\, \phi + &= \frac{d\phi_{1}}{\Delta k_{0} \phi_{1}} + · \frac{(k_{0} \cos \phi_{1} + \Delta k_{0}\, \phi_{1})^{2}}{2(1 + k_{0})}. +\end{align*} +This gives immediately, by integration, +\begin{align*} +Ek\phi + &= \frac{1}{2(1+k_{0})} + \int \frac{d\phi_{1}}{\Delta k_{0}\, \phi_{1}} + \{k_{0} \cos \phi_{1} + \Delta k_{0} \phi_{1}\}^{2} \\ + &=\frac{1}{2(1+k_{0})} + \int \frac{d\phi_{1}}{\Delta k_{0}\, \phi_{1}} + \{2\Delta^{2}k_{0}\, \phi_{1} + + 2k_{0} \cos \phi_{1} \Delta k_{0}\phi_{1} + - k_{1}'^{2}\} \\ +\Tag{(21)} + &= \frac{Ek_{0}\phi_{1}}{1+k_{0}} + + \frac{k_{0} \sin \phi_{1}}{1+k_{0}} + - \tfrac{1}{2} (1 - k_{0})Fk_{0}\phi_{1}. +\end{align*} + +Thus the value of~$Ek\phi$ is made to depend upon~$Ek_{0}\phi_{1}$ +(containing a smaller modulus and a larger amplitude), and +upon the integral of the first class,~$Fk_{0}\phi_{1}$; $k_{0}$,~$\phi_{1}$,~etc., being +determined by the \Eqref{formulæ}{IV}{(6)} to~\Eqref{}{IV}{(12)} of Chap.~IV. + +By successive applications of \Eqref{equation}{}{(21)}, $Ek\phi$~may be +made to depend ultimately upon~$Ek_{0n}\phi_{n}$, where $k_{0n}$~approximates +to zero and $Ek_{0n}\phi_{n}$ to~$\phi_{n}$. + +Or, by reversing, it may be made to depend upon~$Ek_{n}\phi_{0n}$, +where $k_{n}$~approximates to unity and $Ek_{n}\phi_{0n}$ to~$-\cos \phi_{0n}$. +%% -----File: 093.png---Folio 87------- + +To facilitate this, assume +\[ +Gk\phi = Ek\phi - Fk\phi. +\] +Subtracting from \Eqref{equation}{}{(21)} the equation +\[ +Fk\phi = \frac{1 + k_{0}}{2} Fk_{0}\phi_{1} + \text{ (see \Eqref{eq.}{IV}{(13)}, Chap.~IV)}, +\] +we have +\[ +Gk\phi = \frac{1}{1 + k_{0}} + (Gk_{0}\phi_{1} + k_{0} \sin \phi_{1} - k_{0}\, Fk_{0}\, \phi_{1}). +\] +Repeated applications of this give +\[ +\begin{array}{r@{}l} +Gk_{0}\phi_{1} + &{}= \dfrac{1}{1 + k_{00}} + (Gk_{00}\phi_{2} + k_{00} \sin \phi_{2} - k_{00}\, Fk_{00}\, \phi_{2}),\\ +\Dots{2} \\ +\llap{$Gk_{0(n - 1)}\phi_{n - 1}$} + &{}= \dfrac{1}{1 + k_{0n}} \rlap{$(Gk_{0n}\phi_{n} + + k_{0n} \sin \phi_{n} - k_{0n}\, Fk_{0n}\, \phi_{n})$.} +\end{array} +\] +Whence +\[ +\Tag{(22)} +Gk\phi = \sum_{n}^{1} %[** PP: \textstyle sum in original] + \Biggl\{ \frac{k_{0n}(\sin \phi_{n} - Fk_{0n}\phi_{n})} + {\Prodlim(1 + k_{0n})} \Biggr\} + + \frac{Gk_{0n}\, \phi_{n}}{\Prodlim(1 + k_{0n})}. +\] +But since (compare \Eqref{eq.}{IV}{(13)}, Chap.~IV) +\[%[** PP: Next two displays aligned in original] +Fk\phi = \frac{Fk_{0n}\, \phi_{n}\Prodlim(1 + k_{0n})}{2^{n}}, +\] +or +\[ +\Tag{(23)} +\frac{Fk_{0n}\, \phi_{n}}{\Prodlim (1 + k_{0n})} + = \frac{2^{n} Fk\, \phi}{\Prodlim(1 + k_{0n})^{2}}; +\] +%% -----File: 094.png---Folio 88------- +and since, also, (compare \Eqref{eq.}{IV}{(6)}, Chap.~IV,) +\[ +\frac{k^{2}_{0(n-1)}}{k_{0n}} = \frac{2^{2}}{(1 + k_{0n})^{2}}, +\] +we have +\begin{align*} +\Tag{(24)} +\frac{2^{n}k_{0n}}{\Prodlim(1 + k_{0n})^{2}} + &= \frac{k_{0n}}{2^{n}} \Prodlim \frac{k^{2}_{0(n-1)}}{k_{0n}} \\ + &= \frac{k_{0n}}{2^{n}} \Prodlim \frac{k_{0(n-1)}}{k_{0n}} \Prodlim k_{0(n-1)} \\ + &= \frac{k_{0n}}{2^{n}} · \frac{k}{k_{0}} · \frac{k_{0}}{k_{00}} \dotsm \frac{k_{0(n-1)}}{k_{0n}} · k\Prodlim[2] k_{0(n-1)} \\ + &= \frac{k^{2}}{2^{n}} \Prodlim[2] k_{0(n-1)}. +\end{align*} + +Substituting these values in \Eqref{equation}{}{(22)}, and neglecting +the term containing $Gk_{0n}\phi_{n}$ since, carried to its limiting +value, +\begin{DPalign*} +Gk_{0n}\phi_{n} + &= Ek_{0n}\phi_{n} - Fk_{0n}\phi_{n} \\ + &= \phi_{n} - \phi_{n} = 0, +\rintertext{\llap{($n =$ limiting value,)}} +\end{DPalign*} +we have +\begin{gather*} +\Tag{(25)} +Gk\phi = \sum_{n}^{1} \Biggl\{ \frac{k\sqrt{k_{0n}} \sin \phi_{n} + \Prodlim[2] \sqrt{k_{0(n-1)}} - k^{2} \Prodlim[2] k_{0(n-1)}}{2^{n}} + \Biggr\} \\ + \begin{aligned} + &= k \left[\frac{\sqrt{k_{0}}}{2} \sin \phi_{1} + + \frac{\sqrt{k_{0}k_{00}}}{2^{2}} \sin \phi_{2} + + \frac{\sqrt{k_{0}k_{00}k_{03}}}{2^{3}} \sin \phi_{3} + \dotsb\right] \\ + & \quad - \frac{k^{2}}{2} \left[1 + \frac{k_{0}}{2} + \frac{k_{0}k_{00}}{2^{2}} + \frac{k_{0}k_{00}k_{03}}{2^{3}} + \dotsb\right]; +\end{aligned} +\end{gather*} +whence +\begin{gather*} +\Tag{(26)} +Ek\phi = Fk\phi \left[1 - \frac{k^{2}}{2} + \left(1 + \frac{k_{0}}{2} + \frac{k_{0}k_{00}}{2^{2}} + \dotsb\right)\right] \\ + + k \left[\frac{\sqrt{k_{0}}}{2} \sin \phi_{1} + + \frac{\sqrt{k_{0}k_{00}}}{2^{2}} \sin \phi_{2} + + \frac{\sqrt{k_{0}k_{00}k_{03}}}{2^{3}} \sin \phi_{3} + \dotsb\right]. +\end{gather*} +%% -----File: 095.png---Folio 89------- + +From \Eqref{eq.}{V}{(3)}, Chap.~V, we see that when $\phi = \dfrac{\pi}{2}$, +\[ +\phi_{n} = 2^{n-1} \pi. +\] + +Substituting these values in \Eqref{equation}{}{(26)}, we have for a +complete Elliptic Integral of the second class, +\begin{multline*} +\Tag{(27)} +E \left(k, \frac{\pi}{2}\right) = \\ +F \left(k, \frac{\pi}{2}\right) + \left[1 - \frac{k^{2}}{2} + \left(1 + \frac{k_{0}}{2} + + \frac{k_{0}k_{00}}{2^{2}} + + \frac{k_{0}k_{00}k_{03}}{2^{3}} + \dotsb\right)\right]. +\end{multline*} + +In a similar manner we could have found the formula for +$E (k, \phi)$ in terms of an increasing modulus, viz., +\begin{align*} +\Tag{(28)} +E (k, \phi) + &= F (k, \phi) + \biggl[1 + k\biggl(1 + \frac{2}{k_{1}} + \frac{2^{2}}{k_{1}k_{2}} + + \frac{2^{3}}{k_{1}k_{2}k_{3}} + \dotsb \\ + &{} + \frac{2^{n-2}}{k_{1}k_{2} \dotsm k_{n-2}} + - \frac{2^{n-1}}{k_{1}k_{2} \dotsm k_{n-1}}\biggr)\biggr] \\ + &{} - k \biggl[\sin \phi + + \frac{2}{\sqrt{k}} \sin \phi_{1} + + \frac{2^{2}}{\sqrt{kk_{1}}} \sin \phi_{2} + \dotsb \\ + &{} + \frac{2^{n-1}}{\sqrt{kk_{1} \dotsm k_{n-2}}} \sin \phi_{n-1} + - \frac{2^{n}}{\sqrt{kk_{1} \dotsm k_{n-1}}} \sin \phi_{n}\biggr]. +\end{align*} +%% -----File: 096.png---Folio 90------- + + +\Chapter{XI}{Elliptic Integrals of the Third Order.} + +\First{The} Elliptic Integral of the third order is +\[ +\Tag{(1)} +\Pi (n, k, \phi) = \int_{0}^{\phi} \frac{d\phi}{(1 + n \sin^{2} \phi)\, \Delta \phi}. +\] + +Put +\[ +\Tag{(2)} +\Pi (\phi) + \Pi (\psi) = S; +\] +whence we have immediately +\[ +\Tag{(3)} +dS = \frac{d\phi}{(1 + n \sin^{2} \phi)\, \Delta \phi} + + \frac{d\psi}{(1 + n \sin^{2} \psi)\, \Delta \psi}. +\] +But, \Eqref{eq.}{II}{(2)}, Chap.~II, +\[ +\Tag{(4)} +\frac{d\phi}{\Delta \phi} + \frac{d\psi}{\Delta \psi} = 0; +\] +whence +\begin{align*} +dS &= \left(\frac{1}{1 + n \sin^{2} \phi} - \frac{1}{1 + n \sin^{2} \psi}\right) \frac{d\phi}{\Delta \phi} \\ +\Tag{(5)} + &= \frac{n (\sin^{2} \psi - \sin^{2} \phi)} + {(1 + n \sin^{2} \phi)(1 + n \sin^{2} \psi)} · \frac{d\phi}{\Delta \phi}. +\end{align*} + +From \Eqref{equation}{X}{(8)}, Chap.~X, we get by differentiation, since +$\sigma$ (or~$\mu$) is constant, +\begin{align*} +\Delta \phi · d \phi + \Delta \psi · d\psi + &= k^{2} \sin \sigma\, d(\sin \phi \sin \psi), +\intertext{or, from \Eqref{equation}{}{(3)},} +(\sin^{2} \psi - \sin^{2} \phi)\, \frac{d\phi}{\Delta \phi} + &= \sin \sigma\, d(\sin \phi \sin \psi). +\end{align*} +%% -----File: 097.png---Folio 91------- +This, introduced into \Eqref{equation}{}{(5)}, gives +\[ +dS = \frac{n \sin \sigma\, d (\sin \phi \sin \psi)} + {1 + n (\sin^{2} \phi + \sin^{2} \psi) + n^{2} \sin^{2} \phi \sin^{2} \psi}. +\] + +Put +\[ +\sin \phi \sin \psi = q, \quad \sin^{2} \phi + \sin^{2} \psi = p; +\] +whence +\[ +\Tag{(6)} +dS = \frac{n \sin \sigma\, dq}{1 + np + n^{2}p^{2}}. +\] + +From \Eqref{equation}{II}{(5)}, Chap.~II, we have +\[ +\cos \sigma = \cos\phi \cos\psi - \sin\phi \sin\psi\, \Delta\sigma, +\] +from which we easily get +\begin{align*} +(\cos \sigma + q \Delta \sigma)^{2} + &= \cos^{2} \phi \cos^{2} \psi\DPtypo{)}{} \\ + &= (1 - \sin^{2} \phi)(1 - \sin^{2} \psi) \\ + &= 1 - p + q^{2}, +\end{align*} +and thence +\begin{align*} +p &= 1 + q^{2} - (\cos \sigma + q\, \Delta \sigma)^{2} \\ + &= \sin^{2} \sigma - 2 \cos \sigma \Delta \sigma q + k^{2} \sin^{2} \sigma · q^{2}. +\end{align*} + +This, substituted in \Eqref{eq.}{}{(6)}, gives +\begin{align*} +dS &= \frac{n \sin \sigma\, dq} + {1 + n \sin^{2} \sigma - 2n \cos \sigma \Delta \sigma q + n(n + k^{2} \sin^{2} \sigma) q^{2}} \\ + &= \frac{n \sin \sigma\, dq}{A - 2Bq +Cq^{2}}, +\end{align*} +where +\begin{align*} +A &= 1 + n \sin^{2} \sigma, \\ +B &= n \cos \sigma\, \Delta \sigma, \\ +C &= nk^{2} \sin^{2} \sigma + n^{2}. +\end{align*} +%% -----File: 098.png---Folio 92------- + +From this we get +\[ +S = n \sin \sigma \int \frac{dq}{A - 2Bq + Cq^{2}} + \text{Const.} +\] + +In order to determine the constant of integration we must +observe that for $\phi = 0$, $\psi = \sigma$ and~$q = 0$; whence +\begin{align*} +\Pi \sigma &= n \sin \sigma \int_{q=0} \frac{dq}{A - 2Bq + Cq^{2}} + \text{Const.}; +\intertext{whence} +S &= \Pi \sigma + n \sin \sigma \int_{0}^{q} \frac{dq}{A - 2Bq + Cq^{2}}, +\intertext{or} +\Tag{(7)} +\Pi \phi + \Pi \psi &= \Pi \sigma + n \sin \sigma \int_{0}^{q} \frac{dq}{A - 2Bq + Cq^{2}}. +\end{align*} + +But we have +\begin{align*} +dS &= \frac{CM\, dq}{AC - B^{2} + (Cq - B)^{2}} \\ + &= \frac{CM}{AC - B^{2}} · \frac{dq}{1 + \left(\dfrac{Cq - B}{\sqrt{AC - B^{2}}}\right)^{2}} \\ + &= \frac{M}{\sqrt{AC-B^{2}}} + · \frac{\dfrac{C\, dq}{\sqrt{AC - B^{2}}}} + {1 + \left(\dfrac{Cq - B}{\sqrt{AC - B^{2}}}\right)^{2}} +\end{align*} +where $M = n \sin \sigma$. + +The integral of the second member is +\[ +\frac{M}{\sqrt{AC - B^{2}}} \tan^{-1} \frac{Cq - B}{\sqrt{AC - B^{2}}}; +\] +%% -----File: 099.png---Folio 93------- +whence +\[ +\int_{0}^{q} dS = S_{1} + = \frac{M}{\sqrt{AC - B^{2}}} + \left[\tan^{-1} \frac{Cq - B}{\sqrt{AC - B^{2}}} + + \tan^{-1} \frac{B}{\sqrt{AC - B^{2}}}\right]; +\] +or, since +\begin{gather*} +\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x + y}{1 - xy}, \\ +S_{1} = \frac{M}{\sqrt{AC - B^{2}}} \tan^{-1} \frac{q\sqrt{AC - B^{2}}}{A - Bq}. +\end{gather*} + +Substituting the values of $A$,~$B$,~$C$ and~$M$, we have +\begin{align*} +AC - B^{2} + &= n(1 + n - \Delta^{2} \sigma)(1 + n \sin^{2} \sigma) - n^{2} \cos^{2} \sigma\, \Delta^{2} \sigma \\ + &= n(1 + n - \Delta^{2} \sigma + n(1 + n) \sin^{2} \sigma - n\, \Delta^{2} \sigma) \\ + &= n(1 + n)(1 - \Delta^{2} \sigma + n \sin^{2} \sigma) \\ + &= n(1 + n)(k^{2} + n) \sin^{2} \sigma; +\end{align*} +and putting +\[ +\frac{(1 + n)(k^{2} + n)}{n} = \Omega, +\] +we have +\[ +\sqrt{AC - B^{2}} = n \sqrt{\Omega} \sin \sigma. +\] +Substituting these values in \Eqref{eq.}{}{(7)}, we have +\begin{gather*} +\Pi (n, k, \phi) + \Pi (n, k, \psi) - \Pi (n, k, \sigma) = S_{1} \\ += \frac{1}{\sqrt{\Omega}} + \tan^{-1} \frac{n\sqrt{\Omega} \sin \phi \sin \psi \sin \sigma} + {1 + n \sin^{2} \sigma - n \sin \phi \sin \psi \cos \sigma\, \Delta \sigma}. +\end{gather*} +%% -----File: 100.png---Folio 94------- + + +\Chapter[Numerical Calculations. q.]{XII}{Numerical Calculations. $q$.} + +\Section{CALCULATION OF THE VALUE OF~$q$.} + +\First{From} \Eqref{eq.}{IX}{(7)}, Chap.~IX, we have +\[ +\dn u = \sqrt{k'}\, \frac{\Theta (u + K)}{\Theta (u)}; +\] +whence, \Eqref{eq.}{IV}{(9)}, Chap.~IV, \Eqref{eqs.}{IX}{(27)} and~\Eqref{}{IX}{(39)}, Chap.~IX, +\begin{align*} +\Tag{(1)} +\sqrt{\cos \theta} + &= \frac{1 - 2q + 2q^{4} - 2q^{9} + 2q^{16} - \dotsb} + {1 + 2q + 2q^{4} + 2q^{9} + 2q^{16} + \dotsb} \\ + &= 1 - 4q + 8q^{2} - 16q^{3} + 32q^{4} - 56q^{5} + \dotsb. +\end{align*} + +The first five terms of this series can be represented by +\[ +\sqrt{\cos \theta} = \frac{1 - 2q}{1 + 2q}. +\] +From this we get +\[ +\Tag{(2)} +q = \frac{1}{2} · \frac{1 - \sqrt{\cos \theta}}{1 + \sqrt{\cos \theta}}, +\] +which is exact up to the term containing~$q^{5}$. + +Or we can deduce a more exact formula as follows: From +\Eqref{eq.}{}{(1)}, +\begin{align*} +\frac{1 + \sqrt{\cos \theta}}{1 - \sqrt{\cos \theta}} + &= \frac{\sqrt{1 + \tan^{2} \frac{1}{2} \theta} + + \sqrt{1 - \tan^{2} \frac{1}{2} \theta}} + {\sqrt{1 + \tan^{2} \frac{1}{2} \theta} + - \sqrt{1 - \tan^{2} \frac{1}{2} \theta}} \\ + &= \frac{ 1 + 2q^{4} + 2q^{16} + \dotsb} + {2q + 2q^{9} + 2q^{25} + \dotsb}; +\end{align*} +%% -----File: 101.png---Folio 95------- +whence, by the method of indeterminate coefficients, +\begin{align*} +\Tag{(3)} +q &= \tfrac{1}{4} \tan^{2} \frac{\theta}{2} + + \tfrac{1}{16} \tan^{\DPtypo{2}{6}} \frac{\theta}{2} + + \tfrac{\DPtypo{57}{17}}{512} \tan^{10} \frac{\theta}{2} + + \tfrac{45}{2048} \tan^{14} \frac{\theta}{2} + \dotsb, \\ +\intertext{or} +\log q %[** PP: Re-breaking] + &= 2 \log \tan \frac{\theta}{2} - \log 4 \\ + &\qquad + + \log \Bigl(1 + \tfrac{1}{4} \tan^{4} \frac{\theta}{2} + + \tfrac{17}{128} \tan^{8} \frac{\theta}{2} + + \tfrac{45}{512} \tan^{12} \frac{\theta}{2} \dotsb\Bigr) \\ +\Tag{(4)} + &= 2 \log \tan \frac{\theta}{2} - \log 4 \\ + &\qquad + + M\Bigl(\tfrac{1}{4} \tan^{4} \frac{\theta}{2} + + \tfrac{13}{128} \tan^{8} \frac{\theta}{2} + + \tfrac{23}{384} \tan^{12} \frac{\theta}{2} + \dotsb\Bigr), +\end{align*} +$M$~being the modulus of the common system of logarithms. + +Put +\[ +\Tag{(5)}%[** PP: Re-breaking] +\log q = 2 \log \tan \frac{\theta}{2} + + 9.397940 + a \tan^{4} \frac{\theta}{2} + + b \tan^{8} \frac{\theta}{2} + + c \tan^{12} \frac{\theta}{2} + \dotsb, +\] +in which +\begin{align*} +\log a &= 9.0357243; \\ +\log b &= 8.64452; \\ +\log c &= 8.41518; \\ +\log d &= 8.25283. +\end{align*} + +\Example. Let $k' = \cos 10°\ 23'\ 46''$. To find~$q$. +\[ +\begin{array}[t]{r@{}c@{}l@{}} +4 \log \tan \dfrac{\theta}{2} + &{}={}& 5.835 \\ +\log a &{}={}& 9.036 \\ +\cline{3-3} + && 4.871 \\ +&& \\ +a \tan^{4} \dfrac{\theta}{2} + &{}={}& \rlap{$0.0000074$} +\end{array} +\qquad\qquad +\begin{array}[t]{r@{}c@{}l@{}} +2 \log \tan \dfrac{\theta}{2} + &{}={}& 7.9176842 \\ + & & 9.3979400 \\ + & & \PadTo[r]{9.3979400}{74} \\ +\cline{3-3} +\log q &{}={}& 7.3156316 +\end{array} +\] +%% -----File: 102.png---Folio 96------- + +When $\theta$ approaches~$90°$, $\tan \dfrac{\theta}{2}$~differs little from unity, and +the series in \Eqref{eq.}{}{(5)} is not very converging, but $q$~can be calculated +by means of \Eqref{eq.}{VII}{(6)}, Chap.~VII, viz., +\[ +q = e^{-\frac{\pi K'}{K}}, \qquad q' = e^{-\frac{\pi K}{K'}}. +\] + +By comparing these equations with \Eqref{eqs.}{IV}{(6)} and~\Eqref{}{IV}{(9)}, Chap.~IV, +we see that if +\begin{align*} +q &= f(k) = f(\theta), +\intertext{then} +q' &= f(k') = f(90° - \theta). +\end{align*} + +Therefore, having $\theta$, we can from its complement, $90° - \theta$, +find~$q'$ by \Eqref{eq.}{}{(5)}, and thence~$q$ by the following process. We +have +\[ +\frac{1}{q} = e^{\frac{\pi K'}{K}}, \qquad +\frac{1}{q'} = e^{\frac{\pi K}{K'}}; +\] +whence +\begin{align*} +\log \frac{1}{q} \log \frac{1}{q'} = M^{2}\pi^{2} &= 1.8615228, \\ +\Tag{(6)} +\log \log \frac{1}{q} + \log \log \frac{1}{q'} &= 0.2698684, +\end{align*} +by which we can deduce $q$ from~$q'$. + +\Example. Let $\theta = 79°\ 36'\ 14''$. To find~$q$. +\[ +90° - \theta = 10°\ 23'\ 46''. +\] + +By \Eqref{eq.}{}{(5)} we get +\begin{gather*} +\log q' = 7.3156316, \qquad \log \frac{1}{q'} = 2.6843684, \\ +\text{and} \quad \log \log \frac{1}{q'} = .4288421; +\end{gather*} +%% -----File: 103.png---Folio 97------- +and by \Eqref{eq.}{}{(6)}, +\begin{align*} +\log \log \frac{1}{q} &= 9.8410263; +\intertext{whence} +\log q &= 1.3065321. +\end{align*} + +When $k' = k = \cos 45° = \frac{1}{2} \sqrt{2}$, \Eqref{eq.}{}{(6)} becomes +\begin{DPalign*} +\Tag{(7)} +\log \frac{1}{q} + &= M \pi = 1.3643763; +\rintertext{($k = k'$;)} \\ +\intertext{whence} +\log q &= 2.6356237, \\ +q &= 0.0432138. +\rintertext{($k = k'$.)} +\end{DPalign*} + +\Example. Given $\theta = 10°\ 23'\ 46''$. Find~$q$. \\ +\null\hfill\textit{Ans.} $\log q = 7.3156316$. + +\Example. Given $\theta = 82°\ 45'$. Find~$q$. \\ +\null\hfill\textit{Ans.} $\log q = 9.37919$. +%% -----File: 104.png---Folio 98------- + + +\Chapter[Numerical Calculations. K.]{XIII}{Numerical Calculations. $K$.} + +\Section{CALCULATION OF THE VALUE OF $K$.} + +\First{We} have already found from \Eqref{eq.}{IX}{(37)}, Chap.~IX, +\[ +\Tag{(1)} +\Theta (0) = \sqrt{\frac{2k'K}{\pi}}, +\] +and from \Eqref{eq.}{IX}{(40)}, same chapter, +\[ +\Tag{(2)} +\Theta (K) = \frac{\Theta (0)}{\sqrt{k'}} = \sqrt{\frac{2K}{\pi}}. +\] + +But, \Eqref{eqs.}{IX}{(38)} and~\Eqref{}{IX}{(27)}, Chap.~IX, +\begin{align*} +\Theta (K) &= 1 + 2q + 2q^{4} + 2q^{9} + 2q^{16} + \dotsb, \\ +\Theta (0) &= 1 - 2q + 2q^{4} - 2q^{9} + 2q^{16} - \dotsb; +\end{align*} +whence, \Eqref{eq.}{}{(2)}, +\[ +\Tag{(3)} +K = \frac{\pi}{2} (1 + 2q + 2q^{4} + 2q^{9} + \dotsb)^{2}. +\] + +By adding \Eqref{eqs.}{}{(1)}~and~\Eqref{}{}{(2)} we get +\[ +\Theta (0) + \Theta (K) = \sqrt{\frac{2K}{\pi}} (1 + \sqrt{k'}); +\] +whence +\begin{align*} +K &= \frac{\pi}{2} \left(\frac{\Theta (0) + \Theta (K)}{1 + \sqrt{k'}}\right)^{2} \\ + &= \frac{\pi}{2} \left[\frac{2(1 + 2q^{4} + 2q^{16} + \dotsb)}{1 + \sqrt{k'}}\right]^{2} \\ +%% -----File: 105.png---Folio 99------- +\Tag{(4)} + &= \frac{\pi}{2} \left(\frac{2}{1 + \sqrt{k'}}\right)^{2} + (1 + 2q^{4} + 2q^{16} + \dotsb)^{2}. +\end{align*} + +\Example. Let $k = \sin \theta = \sin 19°\ 30'$. Required~$K$. + +\emph{First Method.}\quad By \Eqref{eq.}{}{(3)}. + +By \Eqref{eq.}{XII}{(5)}, Chap.~XII, we find $\log q = 8.6356236$. Applying +\Eqref{eq.}{}{(3)}, using only two terms of the series, we have +\[ +\begin{array}{r@{}l@{}} +1 + 2q &{} = 1.0147662 \\ +\PadTo[r]{1+2q}{\log (1 + 2q)} &{} = 0.0063660 \\ +\PadTo[r]{1+2q}{2 \log (1 + 2q)} &{} = 0.0127320 \\ +\log \dfrac{\pi}{2} &{} = 0.1961199 \\ +\cline{1-2} +\log K &{} = 0.2088519 \\ +K &{} = 1.615101 +\end{array} +\] + +\emph{Second Method.}\quad By \Eqref{eq.}{}{(4)}. + +\Eqref{Equation}{}{(4)} may be written, neglecting~$q^{4}$, +\[ +K = \frac{\pi}{2} \left(\frac{1 + \sqrt{\cos \theta}}{2}\right)^{-2}; +\] +whence +\begin{align*} +\log \cos \theta & = 9.9743466, \\ +\log \sqrt{\cos \theta} & = 9.9871733, \\ +1 + \sqrt{\cos \theta} & = 1.9708973, \\ +\frac{1 + \sqrt{\cos \theta}}{2} & = 0.98544865; +\intertext{and} +\log K & = 0.2088519, \\ +K & = 1.615101, +\end{align*} +the same result as above. +%% -----File: 106.png---Folio 100------- + +\emph{Third Method.}\quad By \Eqref{eq.}{V}{(7)}, Chap.~V. +\[ +\begin{array}{r@{}c@{}l<{\quad}|>{\quad}r@{}c@{}l} +\theta &{}={} & 19°\ 30' & + \theta_{0} &{}={}& 1°\ 41'\ 31''.1 \\ +\frac{1}{2} \theta &{}={}& 9°\ 45' & + \frac{1}{2} \theta_{0} &{}={}& 0°\ 50'\ 45''.5 \\ +\log \tan \frac{1}{2} \theta &{}={}& 9.235103 & & & \\ +\log \cos \frac{1}{2} \theta &{}={}& 9.993681 & + \log \cos \frac{1}{2} \theta_{0} &{}={}& 9.999953 \\ +\begin{array}{r} +\log \tan^{2} \frac{1}{2} \theta \\ \log \sin \theta_{0}\end{array}\biggr\} & = & 8.470206 & & & \\ +\theta_{0} & = & 1°\ 41'\ 31''.1 & & & +\end{array} +\] +\[ +\begin{array}{r@{}c@{}l@{}} +\log \cos^{2} \tfrac{1}{2} \theta &{}={}& 9.987362 \\ +\log \cos^{2} \tfrac{1}{2} \theta_{0} &{}={}& 9.999906 \\ +\cline{3-3} + & & 9.987268 \\ +\log \dfrac{\pi}{2} &{}={}& 0.196120 \\ +\cline{3-3} +\log K &{}={}& 0.208852 +\end{array} +\] + +$\theta_{00}$~is not calculated, as it is evident that its cosine will be~$1$. + +\Example. Given $k = \sin 75°$. Find~$K$. + +By \Eqref{eq.}{V}{(7)}, Chap.~V. + +From \Eqref[14sub1]{eqs.}{IV}{(14_{1})}, Chap.~IV, we find +\settowidth{\TmpLen}{$\tan^{2} \tfrac{1}{2} \theta_{00}$}% +\begin{align*} +k &= \sin\theta = \sin 75° & \log &= 9.9849438 \\ +\PadTo[l]{k_{00}}{k_{0}} &= \biggl\{ + \begin{aligned} + \makebox[\TmpLen][l]{$\tan^{2} \tfrac{1}{2} \theta$} + &= \tan^{2} 37°\ 30' \\ + \makebox[\TmpLen][l]{$\sin \theta_{0}$} + &= \sinP 36°\ \Z4'\ 16''.47\Z + \end{aligned}\biggr\} & & \PadTo{{}={}}{} 9.7699610 \\ +% +k_{00} &= \biggl\{ + \begin{aligned} + \makebox[\TmpLen][l]{$\tan^{2} \tfrac{1}{2} \theta_{0}$} + &= \tan^{2} 18°\ \Z2'\ \Z8''.235 \\ + \makebox[\TmpLen][l]{$\sin \theta_{00}$} + &= \sinP \Z6°\ \Z5'\ \Z9''.38 + \end{aligned}\biggr\} & & \PadTo{{}={}}{} 9.0253880 \\ +% +k_{03} &= \biggl\{ + \begin{aligned} + \makebox[\TmpLen][l]{$\tan^{2} \tfrac{1}{2} \theta_{00}$} + &= \tan^{2} \Z3°\ \Z2'\ 34''.69\Z \\ + \makebox[\TmpLen][l]{$\sin \theta_{03}$} + &= \PadTo[l]{\tan^2 18°}{\sin}\ \Z9'\ 42''.90 + \end{aligned}\biggr\} & & \PadTo{{}={}}{} 7.4511672 +\end{align*} +%% -----File: 107.png---Folio 101------- +\[ +\begin{array}{l@{\,}c@{\,}l@{\,}r@{\ }lcc<{\quad}@{}c@{}} + &&&&& \log & 2 \log & \ac 2 \log \\ +\cos \frac{1}{2} \theta + &=& \cos & 37° & 30' & 9.8994667 & 9.7989334 & 0.2010666 \\ +\cos \frac{1}{2} \theta_{0} + &=& \cos & 18° & 2'.13725 & 9.9781184 & 9.9562368 & 0.0437632 \\ +\cos \frac{1}{2} \theta_{02} + &=& \cos & 3° & 2'.57817 & 9.9993873 & 9.9987746 & 0.0012254 \\ +\cos \frac{1}{2} \theta_{03} + &=& \cos & & 4'.8575 & 9.9999995 & 9.9999990 & 0.0000010 \\ +\cline{8-8} +& & & & & & & 0.2460562 \\ +\multicolumn{1}{c}{\dfrac{\pi}{2}} & & & & & & \dfrac{\pi}{2} & \Z.1961199 \\ +\cline{8-8} +& & & & & \multicolumn{3}{r@{}}{% + \log K = \PadTo[r]{2.768064\quad \text{\textit{Ans.}}}{0.4421761}} \\ +& & & & & \multicolumn{3}{r@{}}{% + K = 2.768064\quad \text{\textit{Ans.}}} +\end{array} +\] + +\Example. Given $k = \sin 45°$. Find~$K$. + +Method of \Eqref{eq.}{V}{(7)}, Chap.~V. + +From \Eqref[14sub1]{eqs.}{IV}{(14_{1})}, Chap.~IV, we have +\settowidth{\TmpLen}{$\tan^{2} \tfrac{1}{2} \theta_{00}$}% +\begin{align*} +& & \PadTo{9.2344486}{\log} \\ +\PadTo[l]{k_{00}}{k_{0}} &= \biggl\{ + \begin{aligned} + \makebox[\TmpLen][l]{$\tan^{2} \frac{1}{2} \theta$} + &= \tan^{2} 22°\ 30' \\ + \makebox[\TmpLen][l]{$\sin \theta_{0}$} + &= \sinP \Z9°\ 52'.75683 + \end{aligned}\biggr\} & 9.2344486 \\ +% +k_{00} &= \biggl\{ + \begin{aligned} + \makebox[\TmpLen][l]{$\tan^{2} \tfrac{1}{2} \theta_{0}$} + &= \tan^{2} \Z4°\ 56'.37841 \\ + \makebox[\TmpLen][l]{$\sin \theta_{00}$} + &= \PadTo[l]{\tan^2 22°}{\sin}\ 25'.679 + \end{aligned}\biggr\} & 7.8733009 \\ +% +k_{03} &= \biggl\{ + \begin{aligned} + \tan^{2} \tfrac{1}{2} \theta_{00} + &= \PadTo[l]{\tan^2 22°}{\tan^{2}}\ 12'.3395\Z \\ + \makebox[\TmpLen][l]{$\sin \theta_{03}$} + &= \PadTo[l]{\tan^2 22°}{\sin}\ \Z0'.05 + \end{aligned}\biggr\} & 5.1445523 +\end{align*} +\[ +\begin{array}{l@{}r@{}l@{}} +\ac \log \cos^{2} \frac{1}{2} \theta && 0.0687694 \\ +\ac \log \cos^{2} \frac{1}{2} \theta_{0} && 0.0032320 \\ +\ac \log \cos^{2} \frac{1}{2} \theta_{00} && 0.0000060 \\ +\multicolumn{1}{c}{\log \dfrac{\pi}{2}} && 0.1961199 \\ +\cline{3-3} + & \log K = {}& 0.2681273 \\ + & K = {}& 1.8540747 \rlap{\quad\text{\textit{Ans.}}} +\end{array} +\] + +\Example. Given $\theta = 63°\ 30'$. Find~$K$. \\ +\null\hfill\textit{Ans.} $\log K = 0.3539686$. + +\Example. Given $\theta = 34°\ 30'$. Find~$K$. \\ +\null\hfill\textit{Ans.} $K = 1.72627$. +%% -----File: 108.png---Folio 102------- + + +\Chapter[Numerical Calculations. u.]{XIV}{Numerical Calculations. $u$} + +\Section{CALCULATION OF THE VALUE OF~$u$.} + +\First{When} $\theta° = \sin^{-1}k < 45°$. + +\Example. Let $\phi = 30°$, $k = \sin 45°$. Find~$u$. + +\emph{First Method.} \Eqref{Eq.}{IV}{(23)}, Chap.~IV, and \Eqref[14sub1]{eqs.}{IV}{(14_{1})}, \Eqref[14sub2]{}{IV}{(14_{2})},~\Eqref[14sub3]{}{IV}{(14_{3})}, +Chap.~IV\@. + +By \Eqref[14sub1]{equations}{IV}{(14_{1})}, +\begin{align*} +\frac{\theta}{2} &= 22°~30'; \\ +\log \tan \frac{\theta}{2} &= 9.6172243; \\ +\log \tan^{2} \frac{\theta}{2} &= 9.2344486 = \log k_{0} = \log \sin \theta_{0}; \\ +\theta_{0} &= 9°~52'~45''.41; \\ +\log \tan \frac{\theta_{0}}{2} &= 8.9366506; \\ +\log \tan^{2} \frac{\theta_{0}}{2} &= 7.8733012 = \log k_{00} = \log \sin \theta_{00}; \\ +\theta_{00} &= 0°~25'~40''.7; \\ +\log \tan^{2} \frac{\theta_{00}}{2} &= 5.144552 = \log k_{03}. +\end{align*} +%% -----File: 109.png---Folio 103------- + +By \Eqref[14sub2]{equations}{IV}{(14_{2})}, +\[ +\begin{array}{@{}r@{}l@{}} +\phi &{}= 30° \\ +\log \tan \phi &{}= 9.761439 \\ +\log \cos \theta &{}= 9.849485 \\ +\cline{1-2} +\llap{$\log \tan (\phi_{1} - \phi)$} &{}= 9.610924 \\ +\phi_{1} - \phi &{}= \rlap{$22°\ 12'\ 27''.56$} \\ +\phi_{1} &{}= \rlap{$52°\ 12'\ 27''.56$} \\ +\end{array} +\] +% +\[ +\begin{array}{@{}r@{}l@{}} +\log \tan \phi_{1} &{}= 0.110438 \\ +\log \cos \theta_{0} &{}= 9.993512 \\ +\cline{1-2} +\llap{$\log \tan (\phi_{2} - \phi_{1})$} &{}= 0.103949 \\ +\phi_{2} - \phi_{1} &{}= \rlap{$51°\ 47'\ 32''.59$} \\ +\end{array} +\] +% +\[ +\begin{array}{@{}r@{}l@{}} +\phi_{2} &{}= \rlap{$104°\ 0'\ 0''.15$} \\ +\log \tan \phi_{2} &{}= 0.603228 \\ +\log \cos \theta_{00} &{}= 9.999988 \\ +\cline{1-2} +% +\llap{$\log \tan (\phi_{3} - \phi_{2})$} &{}= 0.603216 \\ +\phi_{3} - \phi_{2} &{}= \rlap{$104°\ 0'\ 1''.5$} \\ +& \\ +% +\phi_{3} &{}= \rlap{$208°\ 0'\ 1''.65$} +\end{array} +\] + +Since $\dfrac{\phi_{2}}{4} = 26°\ 0'\ 0''.04$ and $\dfrac{\phi_{3}}{8} = 26°\ 0'\ 0''.21$, +we need not calculate~$\phi_{4}$. +\[ +\frac{\phi_{3}}{8} = 93600''.21. +\] + +Reducing this to radians, we have +\[ +\log \frac{\phi_{3}}{8} = 9.656852. +\] +%% -----File: 110.png---Folio 104------- + +Substituting in \Eqref{eq.}{IV}{(23)}, Chap.~IV, we have, since $\cos \theta_{03} = 1$, +\[ +\begin{array}{@{}r@{}c@{}l@{}} +\llap{$\ac$} \log \cos \theta &{}={}& 0.150515 \\ +\log \cos \theta_{0} &{}={}& 9.993512 \\ +\log \cos \theta_{00} &{}={}& 9.999988 \\ +\cline{1-3} +% +&& 0.144014 \\ +&& 0.072007 + \rlap{${} = \log\sqrt{\dfrac{\cos \theta_{0} \cos \theta_{00}}{\cos \theta}}$} \\ +\log \dfrac{\phi_{3}}{8} &{}={}& 9.656852 \\ +\cline{3-3} +% +\log u &{}={}& 9.728859 \\ +u &{}={}& 0.535623\rlap{,\quad\text{\textit{Ans.}}} +\end{array} +\] + +When $\theta = \sin^{-1} k > 45°$. %[** PP: Scan unclear] + +\Example. Given $k = \sin 75°$, $\tan \phi = \sqrt{\dfrac{2}{\sqrt{3}}}$. To find~$F(k, \phi)$. + +\emph{First Method. Bisected Amplitudes.} + +By \Eqref{equations}{IV}{(24)} and~\Eqref{}{IV}{(25)}, Chap.~IV, we get +\begin{align*} +\PadTo[l]{\phi_{\frac{1}{32}}}{\phi} + &= 47°\ \Z3'\ 30''.91, & & \\ +% +\PadTo[l]{\phi_{\frac{1}{32}}}{\phi_{\frac{1}{2}}} + &= 25°\ 36'\ \Z5''.64, & +\PadTo[l]{\beta_{04}}{\beta} &= 45°; \\ +% +\PadTo[l]{\phi_{\frac{1}{32}}}{\phi_{\frac{1}{4}}} + &= 13°\ \Z6'\ 30''.98, & +\PadTo[l]{\beta_{04}}{\beta_{0}} &= 24°\ 40'\ 10''.94; \\ +% +\PadTo[l]{\phi_{\frac{1}{32}}}{\phi_{\frac{1}{8}}} + &= \Z6°\ 35'\ 40''.74, & +\beta_{00} &= 12°\ 39'\ 15''.83; \\ +% +\PadTo[l]{\phi_{\frac{1}{32}}}{\phi_{\frac{1}{16}}} + &= \Z3°\ 18'\ \Z8''.75, & +\beta_{03} &= \Z6°\ 22'\ \Z8''.40; \\ +% +\phi_{\frac{1}{32}} &= \Z1°\ 39'\ \Z7''.43, & +\beta_{04} &= {} +\end{align*} + +Substituting in \Eqref{equation}{IV}{(26)}, Chap.~IV, we have +\begin{align*} +F(k, \phi) + &= 32 × 1°\ 39'\ 7''.43 \\ + &= 52°\ 51'\ 58''.03 \\ + &= 0.9226878.\quad \text{\textit{Ans.}} +\end{align*} +%% -----File: 111.png---Folio 105------- + +\emph{Second Method.} \Eqref{Equation}{IV}{(29)}, Chap.~IV. + +From \Eqref[18sub3]{equations}{IV}{(18_{3})}, Chap.~IV, we have +\[ +\begin{array}{r@{}l@{}l@{\;}c} +&&& \log \\ +k &{}= \cos \eta &\begin{aligned}{}= \cosP 15°\ \Z0'\ \Z0''.00\end{aligned} & 9.9849438 \\ +k' &{}= \sin \eta &\begin{aligned}{}= \sinP 15°\ \Z0'\ \Z0''.00\end{aligned} & 9.4129962 \\ +k_{0}' &{}= \biggl\{\begin{aligned} + &\tan^{2} \tfrac{1}{2} \eta \\ + &\sin \eta_{0} + \end{aligned} & + \begin{aligned} + {}= \tan^{2} \Z7°\ 30'\ \Z0''.00 \\ + {}= \sinP \Z0°\ 59'\ 35''.25 + \end{aligned}\;\biggr\} & + 8.2388582 \\ +k_{1} &{}= \cos \eta_{0} &\begin{aligned}{}= \cosP \Z0°\ 59'\ 35''.25\end{aligned} & 9.9999348 \\ +k'_{00} &{}= \biggl\{\begin{aligned} + &\tan^{2} \tfrac{1}{2} \eta_{0} \\ + &\sin \eta_{00} + \end{aligned}& + \begin{aligned} + {}= \tan^{2} \Z0°\ 29'\ 47''.62 \\ + {}= \sinP \Z0°\ \Z0'\ 15''.49 + \end{aligned}\;\biggr\} & + 5.8757219 \\ +k_{2} &{}= \cos \eta_{00} &\begin{aligned}{}= \cosP \Z0°\ \Z0'\ 15''.49\end{aligned} & 0.0000000 \\ +k'_{03} &{}= \left(\tfrac{1}{2} k'_{00}\right)^{2} && 1.1493838 +\end{array} +\] + +From \Eqref[18sub2]{equations}{IV}{(18_{2})}, Chap.~IV, we get +\begin{align*} +\phi &= 47°\ 3'\ 30''.95; \\ +2 \phi_{0} - \phi &= 45°; \\ +\phi_{0} &= 46°\ 1'\ 45''.475; \\ +\phi_{02} &= 46°\ 1'\ 29''.41; \\ +\phi_{03} &= 46°\ 1'\ 29''.41; \\ +45° + \tfrac{1}{2} \phi_{3} &= 68°\ 0'\ 44''.705. +\end{align*} + +Substituting these values in \Eqref{eq.}{IV}{(29)}, Chap.~IV, we get +\begin{align*} +F(k, \phi) + &= \sqrt{\frac{k_{1}}{k}} · \frac{1}{M} · \log \tan 68°\ 0'\ 44''.705 \\ + &= 0.9226877.\quad\text{\textit{Ans.}} +\end{align*} + +\emph{Third Method.} \Eqref{Equation}{IV}{\DPtypo{(23)^*}{(23)}}, Chap.~IV\@. +%% -----File: 112.png---Folio 106------- + +From \Eqref[14sub1]{equations}{IV}{(14_{1})}, Chap.~IV, we have +\[ +\begin{array}{r@{}l@{}l@{\;}c} +&&& \log \\ % [** Moving to prev. line, cf. prev. page] +k &{}= \sin \theta &\begin{aligned}{}= \sinP 75°\ \Z0'\ \Z0''\phantom{.00}\end{aligned} & 9.9849438 \\ +k' &{}= \cos \theta &\begin{aligned}{}= \cosP 75°\phantom{\ 00'\ 00''.00}\end{aligned} & 9.4129962 \\ +k_{0} &{}= \biggl\{\begin{aligned} + &\tan^{2} \tfrac{1}{2} \theta \\ + &\sin \theta_{0} + \end{aligned} & + \begin{aligned} + &{}= \tan^{2} 37°\ 30' \\ + &{}= \sinP 36°\ \Z4'\ 16''.47 + \end{aligned}\;\biggr\} & + 9.7699610 \\ +k_{1}' &{}= \cos \theta_{0} && 9.9075648 \\ +k_{02} &{}= \biggl\{\begin{aligned} + &\tan^{2} \tfrac{1}{2} \theta_{0} \\ + &\sin \theta_{00} + \end{aligned} & + \begin{aligned} + &{}= \tan^{2} 18°\ \Z2'\ 8''.235 \\ + &{}= \sinP \Z6°\ \Z5'\ 9''.38 + \end{aligned}\;\biggr\} & + 9.0253880 \\ +k_{2}' &{}= \cos \theta_{00} && 9.9975452 \\ +k_{03} &{}= \biggl\{\begin{aligned} + &\tan^{2} \tfrac{1}{2} \theta_{00} \\ + &\sin \theta_{03} + \end{aligned} & + \begin{aligned} + &{}= \tan^{2} \Z3°\ \Z2'\ 34''.69 \\ + &{}= \sinP \phantom{00°}\ \Z9'\ 42''.90 + \end{aligned}\;\biggr\} & + 7.4511672 \\ +k'_{3} &{}= \cos \theta_{03} && 9.9999982 \\ +k_{04} &{}= \left(\tfrac{1}{2} k_{03}\right)^{2} && 4.3002761 \\ +k_{4}' &{}={} && 0.0000000 +\end{array} +\] + +From \Eqref[14sub2]{equations}{IV}{(14_{2})}, Chap.~IV, we have +\begin{align*} +\phi &= \Z47°\ \Z3'\ 30''.94; \\ +\phi_{1} &= \Z62°\ 36'\ \Z3''.10; \\ +\phi_{2} &= 119°\ 55'\ 47''.67; \\ +\phi_{3} &= 240°\ \Z0'\ \Z0''.19; \\ +\phi_{4} &= 480°\ \Z0'\ \Z0''. +\end{align*} +Therefore the limit of $\phi$, $\dfrac{\phi_{1}}{2}$, $\dfrac{\phi_{2}}{4}$, or~$\dfrac{\phi_{n}}{2^{n}}$ is $30° = \dfrac{\pi}{6}$. + +Substituting these values in \Eqref{eq.}{IV}{\DPtypo{(23)^*}{(23)}}, Chap.~IV, we have +\begin{align*} +F(k, \phi) + &= \sqrt{\frac{k_{1}' k_{2}' k_{3}' k_{4}'}{k'}} · \frac{\pi}{6} \\ + &= 0.9226874.\quad\text{\textit{Ans.}} +\end{align*} + +\Example. Given $\phi = 30°$, $k = \sin 89°$. Find~$u$. + +Method of \Eqref{eq.}{IV}{(28)}, Chap.~IV\@. +%% -----File: 113.png---Folio 107------- + +From \Eqref[18sub1]{eqs.}{IV}{(18_{1})} we find +\[ +k_{1} = \sin \theta_{1}\quad \text{and}\quad +\tan^{2} \tfrac{1}{2} \theta_{1} = k = \sin \theta, +\] +from which we find that $k_{1} = 1$ as far as seven decimal places. + +From \Eqref[18sub2]{eqs.}{IV}{(18_{2})} we have +\[ +\begin{array}{r@{}c@{}l@{}} +\sin \phi &{}={}& 9.6989700 \\ +k &{}={}& 9.9999338 \\ +\cline{3-3} +% +\sin (2 \phi_{0} - \phi) &{}={}& 9.6989038 \\ +2 \phi_{0} - \phi &{}={}& \rlap{$29°\ 59'.69733$} \\ +2 \phi_{0} &{}={}& \rlap{$59°\ 59'.69733$} \\ +45° + \tfrac{1}{2} \phi_{0}\footnotemark &{}={}& \rlap{$59°\ 59'.92433$} \\ +\log \left(45° + \tfrac{1}{2} \phi_{0}\right) &{}={}& 0.2385385 +\end{array} +\] +\footnotetext{Since $k_{1} = 1$, $\phi_{00} = \phi_{0}$, and we need not carry the calculation further.}% + +From \Eqref[18sub3]{eqs.}{IV}{(18_{3})}, Chap.~IV, we have +\[ +k = \cos \eta = \cos 1°,\qquad \tfrac{1}{2} \eta = 30'. +\] + +Substituting in \Eqref{eq.}{IV}{(28)}, Chap.~IV, we have +\[ +\begin{array}{r@{}c@{}l@{}} +\ac \log \cos \tfrac{1}{2} \eta && 0.0000330 \\ +\log \log \left(45° + \tfrac{1}{2} \phi_{0}\right) && 9.3775585 \\ +\ac \log M && 0.3622157 \\ +\cline{3-3} +% +\log F(k, \phi) &{}={}& 9.7398072 \\ +F(k, \phi) &{}={}& 0.549297\PadTo[l]{0}{.}\rlap{\quad\text{\textit{Ans.}}} +\end{array} +\] + +\Example. Given $\phi = 79°$, $k = 0.25882$. Find~$u$. \\ +\null\hfill\textit{Ans.} $u = 0.39947$. + +\Example. Given $\phi = 37°$, $k = 0.86603$. Find~$u$. \\ +\null\hfill\textit{Ans.} $u = 0.68141$. +%% -----File: 114.png---Folio 108------- + + +\Chapter[Numerical Calculations. phi.]{XV}{Numerical Calculations. $\phi$.} + +\Example. Given $u = 1.368407$, $\theta = 38°$. Find~$\phi$. + +\emph{First Method.} From \Eqref{eqs.}{IX}{(46)} \Eqref[41st]{and}{IX}{(41)^*}, Chap.~IX, we have +\begin{align*} +u &= x \Theta^{2}(K), \\ +\Delta \phi &= \sqrt{k'} \frac{\Theta_{1}(x)}{\Theta (x)}. +\end{align*} + +From \Eqref{equations}{XII}{(5)}, Chap.~XII, and~\Eqref{}{IX}{(38)}, Chap.~IX, we +have +\[ +\begin{array}{r@{}c@{}l@{}} +\log q &{}={}& 8.4734187 \\ +\log \Theta^{2}(K) &{}={}& 0.0501955 \\ +\log u &{}={}& 0.1362153 \\ +\cline{3-3} +% +\log x &{}={}& 0.0860198 \\ +x &{}={}& \rlap{$69°\ 50'\ 46''.12$} +\end{array} +\] + +From \Eqref{equations}{IX}{(23)} and~\Eqref{}{IX}{(24)}, Chap.~IX, we get +\[ +\begin{array}{r@{}c@{}l@{}} +\log \Theta_{1}(x) &{}={}& 9.9798368 \\ +\log \Theta (x) &{}={}& 0.0192687 \\ +\cline{3-3} +% +&& 9.9605681 \\ +\log \sqrt{k'} &{}={}& 9.9482661 \\ +\cline{3-3} +% +\log \Delta \phi &{}={}& 9.9088342 \rlap{${} = \log \sin \lambda$} +\end{array} +\] + +But +\begin{align*} +k^{2} \sin^{2} \phi &= 1 - \Delta^{2} \phi, \\ +k \sin \phi &= \cos \lambda; +\end{align*} +%% -----File: 115.png---Folio 109------- +whence +\[ +\begin{array}{r@{}c@{}l@{}} +\log \cos \lambda &{}={}& 9.7675483 \\ +\log k &{}={}& 9.7893420 \\ +\cline{3-3} +% +\log \sin \phi &{}={}& 9.9782063 \\ +\phi &{}={}& \rlap{$72°$.\quad\text{\textit{Ans.}}} +\end{array} +\] + +\emph{Second Method.} From \Eqref{eq.}{VI}{(1)}, Chap.~VI\@. + +From \Eqref[14sub1]{eqs.}{IV}{(14_{1})} Chap.~IV, we find +\[ +\begin{array}{r@{}l@{}l@{\;}c} +&&& \log \\ %[** log on its own line, as on 113] +\PadTo[l]{k_{00}}{k_{0}} &{}= \biggl\{\begin{aligned} + &\tan^{2} \tfrac{1}{2} \theta\\ + &\sin \theta_{0} + \end{aligned} & + \begin{aligned} + {}= \tan^{2} 19°\phantom{\ 48'.54569} \\ + {}= \sinP \Z6°\ 48'.54569 + \end{aligned}\;\biggr\} & + 9.0739438 \\ + & \phantom{{}={}}\quad \begin{aligned}\cos \theta_{0}\end{aligned} && 9.9969260 \\ +% +k_{00} &{}= \biggl\{\begin{aligned} + &\tan^{2} \tfrac{1}{2} \theta_{0} \\ + &\sin \theta_{00} + \end{aligned} & + \begin{aligned} + {}= \tan^{2} \Z3°\ 24'.2784\Z \\ + {}= \DPtypo{\phantom{\sinP}}{\sinP} \phantom{00°}\ 12'.16659 + \end{aligned}\;\biggr\} & + 7.5488952 \\ + &\phantom{{}={}}\quad \begin{aligned}\cos \theta_{00}\end{aligned} &&9.9999974 \\ +% +k_{03} &{}= \biggl\{\begin{aligned} + &\tan^{2} \tfrac{1}{2} \theta_{00} \\ + &\sin \theta_{03} + \end{aligned} & + \begin{aligned} + {}= \tan^{2} \phantom{00°}\ \Z6'.08329 \\ + {} + \end{aligned}\;\biggr\} & + 4.4957316 \\ + &\phantom{{}={}}\quad \begin{aligned}\cos \theta_{03}\end{aligned} &&0.0000000 +\end{array} +\] + +Substituting these values in \Eqref{eq.}{VI}{(1)}, Chap.~VI, we have +\[ +\begin{array}{r<{\quad}@{}l@{}} +\log \cos \theta_{0} & 9.9969260 \\ +\log \cos \theta_{00} & 9.9999974 \\ +\cline{2-2} +% +& 9.9969234 \\ +\log \sqrt{\cos \theta_{0} \cos \theta_{00}} & 9.9984617 \\ +\ac \log \PadTo{\cos \theta_{0}}{\text{``}} + \PadTo{\cos \theta_{00}}{\text{``}} & 0.0015383 \\ +\log u & \Z.1362153 \\ +\log \sqrt{\cos \theta} & 9.9482660 \\ +\log 2^{3} & \Z.9030900\rlap{\footnotemark} \\ +\ac \log \sqrt{\cos \theta_{0} \cos \theta_{00}} & 0.0015383 \\ +\cline{2-2} +% +& 0.9891096 \\ +& 2.2418773 \\ +\cline{2-2} +% +\log \phi_{3}\addtocounter{footnote}{-1}\footnotemark & 2.7472323 \\ +\phi_{3} & \rlap{$558°\ 46'.140$} +\end{array} +\] +\footnotetext{$n$~is taken equal to~$3$, because $\cos\DPtypo{}{\theta}_{03} = 1$.} +%% -----File: 116.png---Folio 110------- + +Whence, by \Eqref{equations}{VI}{\DPtypo{(1)^*}{(1)}} of Chap.~VI, we get +\[ +\begin{array}{r@{}c@{}l@{}}%[** PP: Re-aligning first group] +k_{03} \log &{}={}& 4.4957316 \\ +\sin \phi_{3} && 9.5075232_{n} \\ +\cline{3-3} +% +\sin (2 \phi_{2} - \phi_{3}) && 4\DPtypo{\,}{.}0032548_{n} \\ +2 \phi_{2} - \phi_{3} &{}={}& -0'.00346 \\ +\phi_{2} &{}={}& \rlap{$279°\ 23'.06827$} +\end{array} +\] +% +\[ +\begin{array}{r@{}c@{}l@{}} +k_{00} \log &{}={}& 7.5488952 \\ +\sin \phi_{2} && 9.9941484_{n} \\ +\cline{3-3} +\sin (2 \phi_{1} - \phi_{2}) && 7.5430436_{n} \\ +2 \phi_{1} - \phi_{2} &{}={}& -12'.0039 \\ +\phi_{1} &{}={}& \rlap{$139°\ 35'.5321$} +\end{array} +\] +% +\[ +\begin{array}{r@{}c@{}l@{}} +k_{0} \log &{}={}& 9.0739438 \\ +\sin \phi_{1} && 9.8117249 \\ +\cline{3-3} +\sin (2 \phi - \phi_{1}) && 8.8856687 \\ +2 \phi - \phi_{1} &{}={}& \rlap{$\Z4°\ 24'.467$} \\ +\phi &{}={}& \rlap{$71°\ 59'.9999$} \\ + &{}={}& \rlap{$72°$.\quad\text{\textit{Ans.}}} +\end{array} +\] + +\Example. Given $u = 2.41569$, $\theta = 80°$. Find~$\phi$. \\ +\null\hfill\textit{Ans.} $\phi = 82°$. + +\Example. Given $u = 1.62530$, $k = \frac{1}{2}$. Find~$\phi$. \\ +\null\hfill\textit{Ans.} $\phi = 87°$. +%% -----File: 117.png---Folio 111------- + + +\Chapter[Numerical Calculations. E(k, phi).]{XVI} +{Numerical Calculations. $E(k, \phi)$.} + +\emph{First Method.} By Chap.~X, \Eqref{eqs.}{X}{(15)},~\Eqref{}{X}{(16)}, and~\Eqref{}{X}{(17)}. + +\Example. Given $k = 0.9327$, $\phi = 80°$. Find $E(k, \phi)$. + +By \Eqref{eq.}{X}{(15)}, Chap.~X, +\[ +\begin{array}{l@{}c@{}l<{\qquad\qquad}l@{}c@{}l} +\phi &{}={}& 80°; & + \gamma &{}={}& 67°\ 44'.\Z; \\ +\phi_{\frac{1}{2}} &{}={}& 50° 43'.6, & + \gamma_{\frac{1}{2}} &{}={}& 46°\ 40'.4; \\ +\phi_{\frac{1}{4}} &{}={}& 27° 48'.5, & + \gamma_{\frac{1}{4}} &{}={}& 26°\ \Z0'.1; \\ +\phi_{\frac{1}{8}} &{}={}& 14° 16'.7, & + \gamma_{\frac{1}{8}} &{}={}& 13°\ 24'.0; \\ +\phi_{\frac{1}{16}} &{}={}& \Z7° 11'.3, & + \gamma_{\frac{1}{16}} &{}={}& \Z6°\ 45'.2; \\ +\phi_{\frac{1}{32}} &{}={}& \Z3° 36'.0, & + \llap{$\log \sin{}$} \gamma_{\frac{1}{32}} &{}={}& 8.77094; \\ +\phi_{\frac{1}{32}} &{}={}& 0.062831. && \\ +\llap{$\therefore$ } \phi^{5}_{\frac{1}{32}} &{}<{}& 0.0000001. && +\end{array} +\] +Whence, by \Eqref{eq.}{X}{(17)}, +\[ +\begin{array}{r@{}c@{}l@{}} +E(k, \phi_{\frac{1}{32}}) &{}={}& 0.06279\rlap{$4$} \\ +\PadTo[l]{\sin \phi_{\frac{1}{16}}}{\sin \phi} + \PadTo[l]{\sin^{2} \gamma_{\frac{1}{32}}}{\sin^{2} \gamma_{\frac{1}{2}}} &{}={}& 0.52116 \\ +2\, \PadTo[l]{\sin \phi_{\frac{1}{16}}}{\sin \phi_{\frac{1}{2}}} + \PadTo[l]{\sin^{2} \gamma_{\frac{1}{32}}}{\sin^{2} \gamma_{\frac{1}{4}}} &{}={}& 0.29757 \\ +4\, \PadTo[l]{\sin \phi_{\frac{1}{16}}}{\sin \phi_{\frac{1}{4}}} + \PadTo[l]{\sin^{2} \gamma_{\frac{1}{32}}}{\sin^{2} \gamma_{\frac{1}{8}}} &{}={}& 0.10023 \\ +8\, \PadTo[l]{\sin \phi_{\frac{1}{16}}}{\sin \phi_{\frac{1}{8}}} + \sin^{2} \gamma_{\frac{1}{16}} &{}={}& 0.02728 \\ +16 \sin \phi_{\frac{1}{16}} \sin^{2} \gamma_{\frac{1}{32}} &{}={}& 0.00697 \\ +\cline{3-3} +&& 0.95321 +\end{array} +\] +Hence, by \Eqref{eq.}{X}{(16)}, +\begin{align*} +E(k, \phi) + &= 32E(k, \phi_{\frac{1}{32}}) - 0.95321 \\ + &= 2.0094 - 0.9532 = 1.0562. +\end{align*} +%% -----File: 118.png---Folio 112------- + +\emph{Second Method.} By Chap.~X, \Eqref{eq.}{X}{(26)}. + +\Example. Given $k = \sin 75°$, $\tan \phi = \sqrt{\dfrac{2}{\sqrt{3}}}$. Find~$E(k, \phi)$. + +From \Eqref[14sub1]{eqs.}{IV}{(14_{1})}, Chap.~IV, we have +\begin{align*}%XXXX +k &= \sin \theta = \sin 75°\ 0'\ 0'' & \log ={}& 9.9849438 \\ +k' &= \cos \theta = \cos 75° && 9.4129962 \\ +k_{0} + &= \biggl\{ +\begin{aligned} + \tan^{2} \tfrac{1}{2} \theta &= \tan^{2} 37°\ 30' \\ + \PadTo[l]{\tan^2 \tfrac{1}{2} \theta}{\sin \theta_{0}} &= \PadTo[l]{\tan^2}{\sin}\ 36°\ \Z4'\ 16''.47 +\end{aligned} +\biggr\} && 9.7699610 \\ +k_{1}' &= \cos \theta_{0} && 9.9075648 \\ +k_{02} &= \biggl\{ +\begin{aligned} + \tan^{2} \tfrac{1}{2} \theta_{0} &= \tan^{2} 18°\ \Z2'\ \Z8''.235 \\ + \PadTo[l]{\tan^2 \tfrac{1}{2} \theta_{0}}{\sin \theta_{00}} &= \PadTo[l]{\tan^2}{\sin}\ \Z6°\ \Z5'\ \Z9''.38 +\end{aligned} +\biggr\} && 9.0253880 \\ +k_{2}' &= \cos \theta_{00} && 9.9975452 \\ +k_{03} &= \biggl\{ +\begin{aligned} + \tan^{2} \tfrac{1}{2} \theta_{00} &= \tan^{2} \Z3°\ \Z2'\ 34''.69 \\ + \PadTo[l]{\tan^2 \tfrac{1}{2} \theta_{00}}{\sin \theta_{03}} &= \PadTo[l]{\tan^2 22°}{\sin}\ \Z9'\ 42''.90 +\end{aligned} +\biggr\} && 7.4511672 \\ +k_{3}' &= \cos \theta_{03} && 9.9999982 \\ +k_{04} &= \left(\tfrac{1}{2}k_{03}\right)^{2} && 4.3002761 \\ +k_{4}' &={} && 0.0000000 +\end{align*} + +From \Eqref[14sub2]{eqs.}{IV}{(14_{2})}, Chap.~IV, we have +\begin{align*} +\phi &= \Z47°\ \Z3'\ 30''.94; \\ +\phi_{1} &= \Z62°\ 36'\ \Z3''.10; \\ +\phi_{2} &= 119°\ 55'\ 47''.67; \\ +\phi_{3} &= 240°\ \Z0'\ \Z0''.19. +\end{align*} +%% -----File: 119.png---Folio 113------- + +Applying \Eqref{eq.}{X}{(26)}, Chap.~X, we have +\[ +\begin{array}{rr@{}l@{}c<{\qquad}@{}c@{}} +k^{2} & \log ={}& 9.9698876 && \\ +\ac 2 & & 9.6989700 && \\ +\cline{3-3} + & & 9.6688576 && .4665064 +\end{array} +\] +\[ +\begin{array}{rr@{}l@{}c<{\qquad}@{}c@{}} +k_{0} & \phantom{\log ={}} & 9.7699610 && \\ +\ac 2 & & 9.6989700 && \\ +\cline{3-3} + & & 9.1377886 && .1373373 +\end{array} +\] +\[ +\begin{array}{rr@{}l@{}c<{\qquad}@{}c@{}} +k_{00}& \phantom{\log ={}} & 9.0253880 && \\ +\ac 2 & & 9.6989700 && \\ +\cline{3-3} + & & 7.8621466 && .0072802 +\end{array} +\] +\[ +\begin{array}{rr@{}l@{}c<{\qquad}@{}c@{}} +k_{03}& \phantom{\log ={}} & 7.4511672 && \\ +\ac 2 & & 9.6989700 && \\ +\cline{3-3} + & & 5.0132838 && .0000103 \\ +\cline{5-5} + & & && .6111342 +\end{array} +\] +\[ +1 - .6111342 = 0.3888658. +\] + +From \Eqref{eq.}{IV}{\DPtypo{(23)^*}{(23)}}, Chap.~IV, we find $F(k, \phi) = 0.9226874$. + +Hence +\[ +\begin{array}{r@{}l@{}c@{}} %[** PP: Re-aligning first equation] +F(k, \phi) + \left[1 - \dfrac{k^{2}}{2} \left(1 + \dfrac{k_{0}}{2} + \dotsb\right)\right] + &{}=& 0.3588016 \\[4pt] +\dfrac{k \sqrt{k_{0}}}{2} \sin \phi_{1} + &{}=& 0.3290186 \\[4pt] +\dfrac{k \sqrt{k_{0}k_{00}}}{4} \sin \phi_{2} + &{}=& 0.0522872 \\[4pt] +\dfrac{k \sqrt{k_{0}k_{02}k_{03}}}{8} \sin \phi_{3} + &{}= -& 0.0013888 \\[4pt] +\dfrac{k \sqrt{k_{0} \dotsm k_{04}}}{16} \sin \phi_{4} + &{}=& 0.0000010 \\[4pt] +\cline{3-3} + && 0.3799180 +\end{array} +\] +%% -----File: 120.png---Folio 114------- +Whence +\[ +E(k, \phi) = 0.3588016 + 0.3799180 = 0.7387196.\quad\textit{Ans.} +\] + +\Example. Given $k = \sin 75°$. Find~$E\left(k, \dfrac{\pi}{2}\right)$. + +From Example~2, Chap.~XIII, we find +\[ +\begin{array}{r@{}c@{}l@{}} +\log F\left(k, \dfrac{\pi}{2}\right) &{}={}& 0.4421761 \\ +\log 0.3888658 &{}={}& 1.5897998 \\ +\cline{3-3} +\log E \left(k, \dfrac{\pi}{2}\right) &{}={}& 0.0319759 \\ +E\left(k, \frac{\pi}{2}\right) &{}={}& 1.076405\rlap{.\quad\text{\textit{Ans.}}} +\end{array} +\] + +\Example. Given $k = \sin 30°$, $\phi = 81°$. Find~$E(k, \phi)$. \\ +\null\hfill\textit{Ans.} $E(k, \phi) = 1.33124$. + +\Example. Find~$E(\sin 80°, 55°)$. \\ +\null\hfill\textit{Ans.} $0.82417$. + +\Example. Find~$E\left(\sin 27°, \dfrac{\pi}{2}\right)$. \\ +\null\hfill\textit{Ans.} $1.48642$. + +\Example. Find~$E(\sin 19°, 27°)$. \\ +\null\hfill\textit{Ans.} $0.46946$. +%% -----File: 121.png---Folio 115------- + + +\Chapter{XVII}{Applications.} + +\Section{RECTIFICATION OF THE LEMNISCATE.} + +\First{The} polar equation of the Lemniscate is $r = a \sqrt{\cos 2\theta}$, +referred to the centre as the origin. From this we get +\[ +\frac{dr}{d\theta} = -\frac{a\sin 2\theta}{\sqrt{\cos 2\theta}}; +\] +whence the length of the arc measured from the vertex to any +point whose co-ordinates are $r$~and~$\theta$ +\begin{align*} +s &= \int \biggl\{\left(\frac{dr}{d\theta}\right)^{2} + r^{2} \biggr\}^{\frac{1}{2}} d\theta + = a \int \biggl\{\frac{\sin^{2} 2\theta}{\cos 2\theta} + \cos 2\theta \biggr\}^{\frac{1}{2}} \DPtypo{}{d\theta} \\ + &= a \int \frac{d\theta}{\sqrt{\cos 2\theta}} + = a \int \frac{d\theta}{\sqrt{1 - 2 \sin^{2} \theta}}. +\end{align*} + +Let $\cos 2\theta = \cos^{2} \phi$, whence +\begin{align*} %[** PP: Aligning on equal signs] +s &= a \int \frac{\dfrac{d\theta}{d\phi}\, d\phi}{\cos \phi} + = a \int \frac{\sin \phi\, d\phi}{\sqrt{1 - \cos^{4} \phi}} \\ + &= a \int_{0}^{\phi} \frac{d\phi}{\sqrt{1 + \cos^{2} \phi}} + = \frac{a}{\sqrt{2}} \int_{0}^{\phi} \frac{d\phi}{\sqrt{1 - \frac{1}{2} \sin^{2} \phi}} \\ + &= \frac{a}{\sqrt{2}} F\left(\frac{1}{\sqrt{2}}, \phi\right). +\end{align*} +%% -----File: 122.png---Folio 116------- + +Since $r = a \sqrt{\cos 2\theta} = a \cos \phi$, the angle~$\phi$ can be easily +constructed by describing upon the axis~$a$ of the Lemniscate a +semicircle, and then revolving the radius vector until it cuts +this semicircle. In the right-angled triangle of which this is one +side, and the axis the hypotenuse, $\phi$~is evidently the angle between +the axis and the revolved position of the radius vector. + + +\Section{RECTIFICATION OF THE ELLIPSE.} + +Since the equation of the ellipse is $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1$, we can +assume $x = a\sin \phi$, $y = b\cos \phi$, so that $\phi$~is the complement +of the \emph{eccentric angle}. Hence +\begin{align*} +s &= \int \sqrt{dx^{2} + dy^{2}} + = a \int d\phi \sqrt{1 - e^{2} \sin^{2} \phi} \\ + &= aE(e, \phi), +\end{align*} +in which $e$, the eccentricity of the ellipse, is the modulus of the +Elliptic Integral. + +The length of the Elliptic Quadrant is +\[ +s' = aE\left(e, \frac{\pi}{2}\right). +\] + +\Example. The equation of an ellipse is +\[ %[** PP: Displaying +\frac{x^{2}}{16.81} + \dfrac{y^{2}}{16} = 1; +\] +required the length of an arc whose abscissas are $1.061162$ and +$4.100000$: of the quadrantal arc. \\ +\null\hfill\textit{Ans.} $5.18912$; $6.36189$. + + +\Section{RECTIFICATION OF THE HYPERBOLA.} + +On the curve of the hyperbola, construct a straight line +perpendicular to the axis~$x$, and at a distance from the centre +equal to the projection of~$b$, the transverse axis, upon the +asymptote, i.e.~equal to $\dfrac{b^{2}}{\sqrt{a^{2} + b^{2}}}$\DPtypo{}{.} Join the projection of the +%% -----File: 123.png---Folio 117------- +given point of the hyperbola on this line with the centre. The +angle which this joining line makes with the axis of~$x$ we will +call~$\phi$. If~$y$ is the ordinate of the point on the hyperbola, then +evidently +\[ +y = \frac{b^{2} \tan \phi}{\sqrt{a^{2} + b^{2}}}, +\] +and +\[ +x = \frac{a}{\cos \phi} \sqrt{1 - \frac{a^{2} \sin^{2} \phi}{a^{2} + b^{2}}} + = \frac{a}{\cos \phi} \sqrt{1 - \frac{1}{e^{2}} \sin^{2} \phi}; +\] +whence +\begin{align*} +s &= \int \sqrt{\DPtypo{ax}{dx}^{2} + dy^{2}} + = \frac{b^{2}}{c} \int_{0}^{\phi} \frac{d \phi}{\cos^{2} \phi \sqrt{1 - \dfrac{1}{e^{2}} \sin^{2} \phi}} \\ + &= \frac{b^{2}}{c} \int_{0}^{\phi} \frac{d \phi}{\cos^{2} \phi \sqrt{1 - k^{2} \sin^{2} \phi}}. +\end{align*} +But +\begin{gather*} +d(\tan \phi \sqrt{1 - k^{2} \sin^{2} \phi}) + = d \phi \sqrt{1 - k^{2} \sin^{2} \phi} + + d \phi \frac{1 - k^{2}}{\sqrt{1 - e^{2} \sin^{2} \phi}} \\ +% +- \frac{1 - k^{2}}{\cos^{2} \phi \sqrt{1 - e^{2} \sin^{2} \phi}}\, d \phi. +\end{gather*} +Consequently +\begin{align*} +s &= \frac{b^{2}}{c} \int_{0}^{\phi} \frac{d \phi}{\cos^{2} \phi \sqrt{1 - k^{2} \sin^{2} \phi}} \\ + &= \frac{b^{2}}{c} F(k, \phi) - cE(k, \phi) + c \tan \phi \Delta (k, \phi) \\ + &= \frac{b^{2}}{ae} F\left(\frac{1}{e}, \phi\right) - aeE\left(\frac{1}{e}, \phi\right) + ae \tan \phi \Delta \left(\frac{1}{e}, \phi\right). +\end{align*} +%% -----File: 124.png---Folio 118------- + +\Example. Find the length of the arc of the hyperbola +\[ %[** PP: Displaying here, below] +\frac{x^{2}}{20.25} - \frac{y^{2}}{400} = 1 +\] +from the vertex to the point whose ordinate +is~$\dfrac{40}{2.05} \tan 15°$. \\ +\null\hfill\textit{Ans.} $5.231184$. + +\Example. Find the length of the arc of the hyperbola +\[ +\frac{x^{2}}{144} - \frac{y^{2}}{81} = 100 +\] +from the vertex to the point whose ordinate +is~$0.6$. \\ +\null\hfill\textit{Ans.} $0.6582$. + + +%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%% + +\cleardoublepage + +\backmatter +\phantomsection +\pdfbookmark[-1]{Back Matter}{Back Matter} +\phantomsection +\pdfbookmark[0]{PG License}{Project Gutenberg License} +\fancyhead[C]{\textsc{LICENSING}} + +\begin{PGtext} +End of the Project Gutenberg EBook of Elliptic Functions, by Arthur L. 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