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% The Project Gutenberg EBook of An Elementary Treatise on Fourier's Series
% and Spherical, Cylindrical, and Ellipsoidal Harmonics, by William Elwood Byerly
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% Title: An Elementary Treatise on Fourier's Series and Spherical, Cylindrical, and Ellipsoidal Harmonics
%        With Applications to Problems in Mathematical Physics            %
%                                                                         %
% Author: William Elwood Byerly                                           %
%                                                                         %
% Release Date: August 19, 2009 [EBook #29779]                            %
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% Language: English                                                       %
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% *** START OF THIS PROJECT GUTENBERG EBOOK TREATISE ON FOURIER'S SERIES ***
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\begin{document}
\begin{verbatim}
The Project Gutenberg EBook of An Elementary Treatise on Fourier's Series
and Spherical, Cylindrical, and Ellipsoidal Harmonics, by William Elwood Byerly

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: An Elementary Treatise on Fourier's Series and Spherical, Cylindrical, and Ellipsoidal Harmonics
       With Applications to Problems in Mathematical Physics

Author: William Elwood Byerly

Release Date: August 19, 2009 [EBook #29779]

Language: English

Character set encoding: ISO-8859-1

*** START OF THIS PROJECT GUTENBERG EBOOK TREATISE ON FOURIER'S SERIES ***


Produced by Laura Wisewell, Carl Hudkins, Keith Edkins and
the Online Distributed Proofreading Team at
http://www.pgdp.net (The original copy of this book was
generously made available for scanning by the Department
of Mathematics at the University of Glasgow.)
\end{verbatim}

% -----File: 001.png
\newpage
% title page
\begin{center}

{\Large AN ELEMENTARY TREATISE}\\[3ex]

{\scriptsize ON}\\[3ex]

{\Huge FOURIER'S SERIES}\\[3ex]

{\scriptsize AND}\\[3ex]

{\Large SPHERICAL, CYLINDRICAL, AND ELLIPSOIDAL\\

HARMONICS,}\\[3ex]

{\scriptsize WITH}\\[3ex]

APPLICATIONS TO PROBLEMS IN MATHEMATICAL PHYSICS.\\[5ex]

{\scriptsize BY}\\

WILLIAM ELWOOD BYERLY, \textsc{Ph.D.},\\

{\scriptsize PROFESSOR OF MATHEMATICS IN HARVARD UNIVERSITY.}\\[4ex]

\hspace{1.5in} \hrulefill \hspace*{1.7in}

\vspace{5ex}

{\Large GINN \& COMPANY}\\

BOSTON · NEW YORK · CHICAGO · LONDON\\

\end{center}

% -----File: 002.png

\bigskip\bigskip

\begin{center}
{\footnotesize Copyright, 1893,\\
By WILLIAM ELWOOD BYERLY.}

\par\rule{5em}{0.5pt}\par

{\scriptsize ALL RIGHTS RESERVED.}
\end{center}

\bigskip\bigskip
\scriptsize \noindent \textsc{Transcriber's Note:} \emph{A few typographical errors have been corrected -
these are noted at the end of the text.} \normalsize

% -----File: 003.png

\frontmatter
\pagestyle{myheadings}
\newpage
\setcounter{footnote}{0}
\newpage
\medskip\begin{center}{\Large PREFACE.}\par\rule{5em}{0.5pt}\par\end{center}
\markright{}

About ten years ago I gave a course of lectures on Trigonometric Series,
following closely the treatment of that subject in Riemann's ``Partielle
Differentialgleichungen,'' to accompany a short course on The Potential
Function, given by Professor B.~O. Peirce.

My course has been gradually modified and extended until it has become an
introduction to Spherical Harmonics and Bessel's and Lamé's Functions.

Two years ago my lecture notes were lithographed by my class for their
own use and were found so convenient that I have prepared them for
publication, hoping that they may prove useful to others as well as to my
own students. Meanwhile, Professor Peirce has published his lectures on
``The Newtonian Potential Function'' (Boston, Ginn \& Co.), and the two
sets of lectures form a course (Math.~10) given regularly at Harvard, and
intended as a partial introduction to modern Mathematical Physics.

Students taking this course are supposed to be familiar with so much of the
infinitesimal calculus as is contained in my ``Differential Calculus'' (Boston,
Ginn \& Co.)\ and my ``Integral Calculus'' (second edition, same publishers),
to which I refer in the present book as ``Dif.\ Cal.''\ and ``Int.\ Cal.'' Here,
as in the ``Calculus,'' I speak of a ``derivative'' rather than a ``differential
coefficient,'' and use the notation $D_x$ instead of $\frac{\delta}{\delta x}$ for ``partial derivative with
respect to $x$.''

The course was at first, as I have said, an exposition of Riemann's ``Partielle
Differentialgleichungen.'' In extending it, I drew largely from Ferrer's
``Spherical Harmonics'' and Heine's ``Kugelfunctionen,'' and was somewhat
indebted to Todhunter (``Functions of Laplace, Bessel, and Lamé''), Lord
Rayleigh (``Theory of Sound''), and Forsyth (``Differential Equations'').

In preparing the notes for publication, I have been greatly aided by the
criticisms and suggestions of my colleagues, Professor B.~O. Peirce and Dr.\
Maxime Bôcher, and the latter has kindly contributed the brief historical
sketch contained in Chapter~IX.

\rightline{W.~E. BYERLY.\qquad}

{\small \textsc{Cambridge, Mass.}, Sept.\ 1893.}
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\setcounter{footnote}{0}
\newpage
\medskip\begin{center}{\Large ANALYTICAL TABLE OF CONTENTS.}\par\rule{5em}{0.5pt}\par\end{center}
\markright{}

\begin{center}CHAPTER I.\end{center}

\hfill \textsc{pages}

\textsc{Introduction} \hfill \pageref{ch1start}--\pageref{ch1end}\\

\textsc{Art.}\ 1. List of some important homogeneous linear partial differential equations
of Physics.---\textsc{Arts.}\ 2--4. Distinction between the general solution and a particular
solution of a differential equation. Need of additional data to make the solution
of a differential equation determinate. Definition of linear and of linear and
homogeneous.---\textsc{Arts.}\ 5--6. Particular solutions of homogeneous linear differential
equations may be combined into a more general solution. Need of development
in terms of normal forms.---\textsc{Art.}\ 7. Problem: Permanent state of temperatures
in a thin rectangular plate. Need of a development in sine series. Example.---\textsc{Art.}\
8. Problem: Transverse vibrations of a stretched elastic string. A development
in sine series suggested.---\textsc{Art.}\ 9. Problem: Potential function due to the
attraction of a circular ring of small cross-section. Surface Zonal Harmonics
(Legendre's Coefficients). Example.---\textsc{Art.}\ 10. Problem: Permanent state of
temperatures in a solid sphere. Development in terms of Surface Zonal Harmonics
suggested.---\textsc{Arts.}\ 11--12. Problem: Vibrations of a circular drumhead.
Cylindrical Harmonics (Bessel's Functions). Recapitulation.---\textsc{Art.}\ 13. Method
of making the solution of a linear partial differential equation depend upon solving
a set of ordinary differential equations by assuming the dependent variable equal
to a product of factors each of which involves but one of the independent variables.
\textsc{Arts.}\ 14--15 Method of solving ordinary homogeneous linear differential equations
by development in power series. Applications.---\textsc{Art.}\ 16. Application to
Legendre's Equation. Several forms of general solution obtained. Zonal
Harmonics of the second kind.---\textsc{Art.}\ 17. Application to Bessel's Equation.
General solution obtained for the case where $m$ is not an integer, and for the case
where $m$ is zero. Bessel's Function of the second kind and zeroth order.---\textsc{Art.}\
18. Method of obtaining the general solution of an ordinary linear differential
equation of the second order from a given particular solution. Application to
the equations considered in Arts.~14--17.\\[1ex]

\begin{center}CHAPTER II.\end{center}

\textsc{Development in Trigonometric Series} \hfill \pageref{ch2start}--\pageref{ch2end}\\

\textsc{Arts.}\ 19--22. Determination of the coefficients of $n$ terms of a sine series so that
the sum of the terms shall be equal to a given function of $x$ for $n$ given values
of $x$. Numerical example.---\textsc{Art.}\ 23. Problem of development in sine series
treated as a limiting case of the problem just solved.---\textsc{Arts.}\ 24--25. Shorter
method of solving the problem of development in series involving sines of whole
multiples of the variable. Working rule deduced. Recapitulation.---\textsc{Art.}\ 26. A
% -----File: 006.png
few important sine developments obtained. Examples.---\textsc{Arts.}\ 27--28. Development
in cosine series. Examples.---\textsc{Art.}\ 29. Sine series an odd function of the
variable, cosine series an even function, and both series periodic functions.---\textsc{Art.}\
30. Development in series involving both sines and cosines of whole
multiples of the variable. Fourier's series. Examples.---\textsc{Art.}\ 31. Extension of
the range within which the function and the series are equal. Examples.---\textsc{Art.}\
32. Fourier's Integral obtained.\\[1ex]

\markright{TABLE OF CONTENTS}
\begin{center}CHAPTER III.\end{center}

\textsc{Convergence of Fourier's Series} \hfill \pageref{ch3start}--\pageref{ch3end}\\

\textsc{Arts.}\ 33--36. The question of the convergence of the sine series for unity considered
at length.---\textsc{Arts.}\ 37--38. Statement of the conditions which are sufficient
to warrant the development of a function into a Fourier's series. Historical note.
\textsc{Art.}\ 39. Graphical representation of successive approximations to a sine series.
Properties of a Fourier's series inferred from the constructions.---\textsc{Arts.}\ 40--42.
Investigation of the conditions under which a Fourier's series can be differentiated
term by term.---\textsc{Art.}\ 43. Conditions under which a function can be expressed as
a Fourier's Integral.\\[1ex]

\begin{center}CHAPTER IV.\end{center}

\textsc{Solution of Problems in Physics by the Aid of Fourier's Integrals and
Fourier's Series } \hfill \pageref{ch4start}--\pageref{ch4end}\\

\textsc{Arts.}\ 44--48. Logarithmic Potential. Flow of electricity in an infinite plane,
where the value of the Potential Function is given along an infinite straight line;
along two mutually perpendicular straight lines; along two parallel straight lines.
Examples. Use of Conjugate Functions. Sources and Sinks. Equipotential
lines and lines of Flow. Examples.---\textsc{Arts.}\ 49--52. One-dimensional flow of heat.
Flow of heat in an infinite solid; in a solid with one plane face at the temperature
zero; in a solid with one plane face whose temperature is a function of the time
(Riemann's solution); in a bar of small cross section from whose surface heat
escapes into air at temperature zero. Limiting state approached when the temperature
of the origin is a periodic function of the time. Examples.---\textsc{Arts.}\ 53--54.
Temperatures due to instantaneous and to permanent heat sources and sinks,
and to heat doublets. Examples. Application to the case where there is
leakage.---\textsc{Arts.}\ 55--56. Transmission of a disturbance along an infinite stretched
elastic string. Examples.---\textsc{Arts.}\ 57--58. Stationary temperatures in a long
rectangular plate. Temperature of the base unity. Summation of a Trigonometric
series. Isothermal lines and lines of flow. Examples.---\textsc{Art.}\ 59. Potential
Function given along the perimeter of a rectangle. Examples.---\textsc{Arts.}\ 60--63.
One-dimensional flow of heat in a slab with parallel plane faces. Both faces at
temperature zero. Both faces adiathermanous. Temperature of one face a
function of the time. Examples.---\textsc{Art.}\ 64. Motion of a stretched elastic string
fastened at the ends. Steady vibration. Nodes. Examples.---\textsc{Art.}\ 65. Motion
of a string in a resisting medium.---\textsc{Art.}\ 66. Flow of heat in a sphere whose
surface is kept at a constant temperature.---\textsc{Arts.}\ 67--68. Cooling of a sphere in
air. Surface condition given by a differential equation. Development in a Trigonometric
series of which Fourier's Sine Series is a special case. Examples.---\textsc{Arts.}\
% -----File: 007.png
69--70. Flow of heat in an infinite solid with one plane face which is
exposed to air whose temperature is a function of the time. Solution for an
instantaneous heat source when the temperature of the air is zero. Examples.---\textsc{Arts.}\
71--73. Vibration of a rectangular drumhead. Development of a function
of two variables in a double Fourier's Series. Examples. Nodal lines in a
rectangular drumhead. Nodal lines in a square drumhead.\\[1ex]

\textsc{Miscellaneous Problems} \hfill \pageref{probstart}--\pageref{probend}\\

I. Logarithmic Potential. Polar Coördinates.---II. Potential Function in Space.
III. Conduction of heat in a plane.---IV. Conduction of heat in Space.\\[1ex]

\begin{center}CHAPTER V.\end{center}

\textsc{Zonal Harmonics} \hfill \pageref{ch5start}--\pageref{ch5end}\\

\textsc{Art.}\ 74. Recapitulation. Surface Zonal Harmonics (Legendrians). Zonal Harmonics
of the second kind.---\textsc{Arts.}\ 75--76. Legendrians as coefficients in a Power
Series. Special values.---\textsc{Art.}\ 77. Summary of the properties of a Legendrian.
List of the first eight Legendrians. Relation connecting any three successive
Legendrians.---\textsc{Arts.}\ 78--81. Problems in Potential. Potential Function due to
the attraction of a material circular ring of small cross section. Potential Function
due to a charge of electricity placed on a thin circular disc. Examples: Spheroidal
conductors. Potential Function due to the attraction of a material homogeneous
circular disc. Examples: Homogeneous hemisphere; Heterogeneous sphere;
Homogeneous spheroids. Generalisation.---\textsc{Art.}\ 82. Legendrian as a sum of
cosines.---\textsc{Arts.}\ 83--84. Legendrian as the $m$th derivative of the $m$th power of
$x^2--1$.---\textsc{Art.}\ 85. Equations derivable from Legendre's Equation.---\textsc{Art.}\ 86.
Legendrian as a Partial Derivative.---\textsc{Art.}\ 87. Legendrian as a Definite Integral.
\textsc{Arts.}\ 88--90. Development in Zonal Harmonic Series. Integral of the product of
two Legendrians of different degrees. Integral of the square of a Legendrian.
Formulas for the coefficients of the series.---\textsc{Arts.}\ 91--92. Integral of the product
of two Legendrians obtained by the aid of Legendre's Equation; by the aid of
Green's Theorem. Additional formulas for integration. Examples.---\textsc{Arts.}\ 93--94.
Problems in Potential where the value of the Potential Function is given on a
spherical surface and has circular symmetry about a diameter. Examples.---\textsc{Art.}\
95. Development of a power of $x$ in Zonal Harmonic Series.---\textsc{Art.}\ 96.
Useful formulas.---\textsc{Art.}\ 97. Development of $\sin n\theta$ and $\cos n\theta$ in Zonal Harmonic
Series. Examples. Graphical representation of the first seven Surface Zonal
Harmonics. Construction of successive approximations to Zonal Harmonic Series.
\textsc{Arts.}\ 98--99. Method of dealing with problems in Potential when the density is
given. Examples.---\textsc{Art.}\ 100. Surface Zonal Harmonics of the second kind.
Examples: Conal Harmonics.\\[1ex]

\begin{center}CHAPTER VI.\end{center}

\textsc{Spherical Harmonics} \hfill \pageref{ch6start}--\pageref{ch6end}\\

\textsc{Arts.}\ 101--102. Particular Solutions of Laplace's Equation obtained. Associated
Functions. Tesseral Harmonics. Surface Spherical Harmonics. Solid Spherical
Harmonics. Table of Associated Functions. Examples.---\textsc{Arts.}\ 103--108. Development
in Spherical Harmonic Series. The integral of the product of two
% -----File: 008.png
Surface Spherical Harmonics of different degrees taken over the surface of the
unit sphere is zero. Examples. The integral of the product of two Associated
Functions of the same order. Formulas for the coefficients of the series. Illustrative
example. Examples.---\textsc{Arts.}\ 109--110. Any homogeneous rational integral
Algebraic function of $x$, $y$, and $z$ which satisfies Laplace's Equation is a Solid
Spherical Harmonic. Examples.---\textsc{Art.}\ 111. A transformation of axes to a new
set having the same origin will change a Surface Spherical Harmonic into another
of the same degree.---\textsc{Arts.}\ 112--114. Laplacians. Integral of the product of a
Surface Spherical Harmonic by a Laplacian of the same degree. Development in
Spherical Harmonic Series by the aid of Laplacians. Table of Laplacians. Example.---\textsc{Art.}\
115. Solution of problems in Potential by direct integration.
Examples.---\textsc{Arts.}\ 116--118. Differentiation along an axis. Axes of a Spherical
Harmonic.---\textsc{Art.}\ 119. Roots of a Zonal Harmonic. Roots of a Tesseral Harmonic.
Nomenclature justified.\\[1ex]

\Needspace*{4\baselineskip}

\begin{center}CHAPTER VII.\end{center}

\textsc{Cylindrical Harmonics (Bessel's Functions)} \hfill \pageref{ch7start}--\pageref{ch7end}\\

\textsc{Art.}\ 120. Recapitulation. Cylindrical Harmonics (Bessel's Functions) of the
zeroth order; of the $n$th order; of the second kind. General solution of Bessel's
Equation.---\textsc{Art.}\ 121. Bessel's Functions as definite integrals. Examples.---\textsc{Art.}\
122. Properties of Bessel's Functions. Semi-convergent series for a Bessel's
Function. Examples.---\textsc{Art.}\ 123. Problem: Stationary temperatures in a
cylinder (\emph{a}) when the temperature of the convex surface is zero; (\emph{b}) when the
convex surface is adiathermanous; (\emph{c}) when the convex surface is exposed to air
at the temperature zero.---\textsc{Art.}\ 124. Roots of Bessel's functions.---\textsc{Art.}\ 125.
The integral of $r$ times the product of two Cylindrical Harmonics of the zeroth
order. Example.---\textsc{Art.}\ 126. Development in Cylindrical Harmonic Series.
Formulas for the coefficients. Examples.---\textsc{Art.}\ 127. Problem: Stationary
temperatures in a cylindrical shell. Bessel's Functions of the second kind
employed. Example: Vibration of a ring membrane.---\textsc{Art.}\ 128. Problem:
Stationary temperatures in a cylinder when the temperature of the convex surface
varies with the distance from the base. Bessel's Functions of a complex variable.
Examples.---\textsc{Art.}\ 129. Problem: Stationary temperatures in a cylinder when
the temperatures of the base are unsymmetrical. Bessel's Functions of the $n$th
order employed. Miscellaneous examples. Bessel's Functions of fractional
order.\\[1ex]

\begin{center}CHAPTER VIII.\end{center}

\textsc{\spreadout{Laplace's Equation in Curvilinear Coördinates.  Ellipsoidal}
}\\
\textsc{Harmonics} \hfill \pageref{ch8start}--\pageref{ch8end}\\

\textsc{Arts.}\ 130--131. Orthogonal Curvilinear Coördinates in general. Laplace's Equation
expressed in terms of orthogonal curvilinear coördinates by the aid of Green's
theorem.---\textsc{Arts.}\ 132--135. Spheroidal Coördinates. Laplace's Equation in
spheroidal coördinates, in normal spheroidal coördinates. Examples. Condition
that a set of curvilinear coördinates should be normal. Thermometric Parameters.
Particular solutions of Laplace's Equation in spheroidal coördinates. Spheroidal
Harmonics. Examples. The Potential Function due to the attraction of an
oblate spheroid. Solution for an external point. Examples.---\textsc{Arts.}\ 136--141.
% -----File: 009.png
\emph{Ellipsoidal Coördinates.} Laplace's Equation in ellipsoidal coördinates. Normal
ellipsoidal coördinates expressed as Elliptic Integrals. Particular solutions of
Laplace's Equation. Lamé's Equation. Ellipsoidal Harmonics (Lamé's Functions).
Tables of Ellipsoidal Harmonics of the degrees 1, 2, and 3. Lamé's
Functions of the second kind. Examples. Development in Ellipsoidal Harmonic
series. Value of the Potential Function at any point in space when its value is
given at all points on the surface of an ellipsoid.---\textsc{Art.}\ 142. \emph{Conical Coördinates.}
The product of two Ellipsoidal Harmonics a Spherical Harmonic.---\textsc{Art.}\ 143.
\emph{Toroidal Coördinates.} Laplace's Equation in toroidal coördinates. Particular
solutions. Toroidal Harmonics. Potential Function for an anchor ring.\\[1ex]

\begin{center}
CHAPTER IX.
\end{center}

\textsc{Historical Summary} \hfill \pageref{ch9start}--\pageref{ch9end}\\

\begin{center}APPENDIX.\end{center}

\textsc{Tables}                                           \hfill\pageref{tablestart}--\pageref{tableend}\\
Table I\@.  Surface Zonal Harmonics. Argument $\theta$      \hfill\pageref{tableI}\\
Table II\@.  Surface Zonal Harmonics. Argument $x$          \hfill\pageref{tableII}\\
Table III\@. Hyperbolic Functions                           \hfill\pageref{tableIII}\\
Table IV\@. Roots of Bessel's Functions                     \hfill\pageref{tableIV}\\
Table V\@.  Roots of Bessel's Functions                     \hfill\pageref{tableV}\\
Table VI\@. Bessel's Functions                              \hfill\pageref{tableVI}

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\mainmatter
%\pagestyle{myheadings}
\mychap{CHAPTER I.}{INTRODUCTION.}
\label{ch1start}

\mypara{1.} In many important problems in mathematical physics we are obliged
to deal with \emph{partial differential equations} of a comparatively simple form.

For example, in the Analytical Theory of Heat we have for the change of
temperature of any solid due to the flow of heat within the solid, the equation
\[
\tag*{\smallrom{I}}
D_t u = a^2(D_x^2 u + D_y^2 u + D_z^2 u),\footnote
  {For the sake of brevity we shall often use the symbol $\nabla^2$ for the operation $D_x^2 + D_y^2 + D_z^2$;
  and with this notation equation~\smallrom{I} would be written $D_t u = a^2\nabla^2 u$.}
\]
where $u$ represents the temperature at any point of the solid and $t$ the time.

In the simplest case, that of a slab of infinite extent with parallel plane
faces, where the temperature can be regarded as a function of one coördinate,
\smallrom{I} reduces to
\[
\tag*{\smallrom{II}}
D_t u = a^2 D_x^2 u,
\]
a form of considerable importance in the consideration of the problem of the
cooling of the earth's crust.

In the problem of the permanent state of temperatures in a thin rectangular
plate, the equation~\smallrom{I} becomes
\[
\tag*{\smallrom{III}}
D_x^2u + D_y^2u = 0.
\]

In \emph{polar} or \emph{spherical coördinates} \smallrom{I} is less simple, it is
\[
\tag*{\smallrom{IV}}
D_t u = \dfrac{a^2}{r^2}
        \left[ D_r(r^2D_r u)
             + \dfrac{1}{\sin\theta} D_\theta (\sin\theta D_\theta u)
             + \dfrac{1}{\sin^2\theta} D_\phi^2 u \right].
\]

In the case where the solid in question is a sphere and the temperature
at any point depends merely on the distance of the point from the centre
\smallrom{IV} reduces to
\[
\tag*{\smallrom{V}}
D_t(ru) = a^2D_r^2(ru).
\]

In \emph{cylindrical coördinates} \smallrom{I} becomes
\[
\tag*{\smallrom{VI}}
D_t u = a^2[D_r^2 u + \dfrac{1}{r} D_r u + \dfrac{1}{r^2} D_\phi^2 u + D_z^2 u].
\]

In considering the flow of heat in a cylinder when the temperature at
any point depends merely on the distance $r$ of the point from the axis
\smallrom{VI} becomes
\[
\tag*{\smallrom{VII}}
D_t u = a^2(D_r^2 u + \dfrac{1}{r} D_r u).
\]
% -----File: 012.png

In Acoustics in several problems we have the equation
\[
\tag*{\smallrom{VIII}}
  D_t^{2}y = a^{2}D_{x}^{2}y;
\]
for instance, in considering the transverse or the longitudinal vibrations of a
stretched elastic string, or the transmission of plane sound waves through
the air.

\markright{INTRODUCTION.}
If in considering the transverse vibrations of a stretched string we take
account of the resistance of the air \smallrom{VIII} is replaced by
\[\tag*{\smallrom{IX}}
D_t^2 y + 2kD_t y = a^2 D_x^2 y.
\]

In dealing with the vibrations of a stretched elastic membrane, we have the
equation
\[\tag*{\smallrom{X}}
D_t^{2}z = c^2(D_x^{2}z + D_y^{2}z),
\]
or in \emph{cylindrical coördinates}
\[\tag*{\smallrom{XI}}
D_{t}^2 z = c^2 (D_{r}^2 z + \dfrac{1}{r} D_{r}z + \dfrac{1}{r^2}D_{\phi}^2 z).
\]

In the theory of \emph{Potential} we constantly meet Laplace's Equation

\begin{flalign*}\tag*{\smallrom{XII}}
&\hfill & D_x^{2}V + &D_y^{2}V + D_z^{2}V=0&& \\
&\text{or}\hfill &       &\nabla^2 V = 0&&\hfill\phantom{or}
\end{flalign*}
which in \emph{spherical coördinates} becomes
\[\tag*{\smallrom{XIII}}
  \dfrac{1}{r^2} \left[
      rD_r^{2}(rV)
      + \dfrac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta}V)
      + \dfrac{1}{\sin^{2}\theta}D_{\phi}^{2}V \right] = 0,
\]
and in \emph{cylindrical coördinates}
\[\tag*{\smallrom{XIV}}
D_r^{2}V + \dfrac{1}{r} D_r V + \frac{1}{r^2} D_{\phi}^{2}V + D_z^{2} V = 0.
\]

In \emph{curvilinear coördinates} it is
\[\tag*{\smallrom{XV}}
  h_{1}h_{2}h_{3} \left[
    D_{\rho_1} \left( \frac{h_1}{h_2 h_3} D_{\rho_1} V \right)
  + D_{\rho_2} \left( \frac{h_2}{h_3 h_1} D_{\rho_2} V \right)
  + D_{\rho_3} \left( \frac{h_3}{h_1 h_2} D_{\rho_3} V \right) \right] = 0;\\
\]
where \hfill
$f_1 (x,y,z) = \rho_{1}$,
$f_2 (x,y,z) = \rho_{2}$,
$f_3 (x,y,z) = \rho_{3}$ \hfill\phantom{where}\\[1ex]
represent a set of surfaces which cut one another at right angles, no matter
what values are given to $\rho_1$, $\rho_2$, and $\rho_3$; and where
\begin{align*}
  h_1^2 &= (D_{x}\rho_1)^2 + (D_{y}\rho_{1})^2 + (D_{z}\rho_1)^2 \\
  h_2^2 &= (D_{x}\rho_2)^2 + (D_{y}\rho_{2})^2 + (D_{z}\rho_2)^2 \\
  h_3^2 &= (D_{x}\rho_3)^2 + (D_{y}\rho_{3})^2 + (D_{z}\rho_3)^2,
\end{align*}
and, of course, must be expressed in terms of $\rho_1$, $\rho_2$, and $\rho_3$.

If it happens that $\nabla^{2}\rho_1 = 0$, $\nabla^{2}\rho_{2}= 0$, and $\nabla^{2}\rho_3=0$, then Laplace's
Equation~\smallrom{XV} assumes the very simple form
\[\tag*{\smallrom{XVI}}
h_1^2 D_{\rho_1}^2 V + h_2^2 D_{\rho_2}^2 V + h_3^2 D_{\rho_3}^2 V= 0.
\]
% -----File: 013.png

\mypara{2.}  A \emph{differential equation} is an equation containing derivatives or differentials
with or without the primitive variables from which they are derived.

The \emph{general solution} of a differential equation is the equation expressing the
most general relation between the primitive variables which is consistent with
the given differential equation and which does not involve differentials or
derivatives. A general solution will always contain arbitrary (\emph{i.\,e.}, undetermined)
\emph{constants} or \emph{arbitrary functions}.

A \emph{particular solution} of a differential equation is a relation between the
primitive variables which is consistent with the given differential equation,
but which is less general than the general solution, although included in it.

Theoretically, every particular solution can be obtained from the general
solution by substituting in the general solution particular values for the arbitrary
constants or particular functions for the arbitrary functions; but in
practice it is often easy to obtain particular solutions directly from the differential
equation when it would be difficult or impossible to obtain the general
solution.

\mypara{3.}  If a problem requiring for its solution the solving of a differential equation
is \emph{determinate}, there must always be given in addition to the differential
equation enough outside conditions for the determination of all the arbitrary
constants or arbitrary functions that enter into the general solution of the
equation; and in dealing with such a problem, if the differential equation can
be readily solved the natural method of procedure is to obtain its general
solution, and then to determine the constants or functions by the aid of the
given conditions.

It often happens, however, that the general solution of the differential equation
in question cannot be obtained, and then, since the problem \emph{if determinate}
will be solved if by any means a solution of the equation can be found which
will also satisfy the given outside conditions, it is worth while to try to get
\emph{particular solutions} and so to combine them as to form a result which shall
satisfy the given conditions without ceasing to satisfy the differential equation.

\mypara{4.}  A differential equation is \emph{linear} when it would be of the first degree if
the dependent variable and all its derivatives were regarded as algebraic
unknown quantities. If it is linear and contains no term which does not
involve the dependent variable or one of its derivatives, it is said to be linear
and \emph{homogeneous}.

All the differential equations collected in Art.~1 are linear and homogeneous.

\mypara{5.}  \emph{If a value of the dependent variable has been found which satisfies a
given homogeneous, linear, differential equation, the product formed by multiplying
this value by any constant will also be a value of the dependent variable
which will satisfy the equation.}
% -----File: 014.png

For if all the terms of the given equation are transposed to the first member,
the substitution of the first-named value must reduce that member to
zero; substituting the second value is equivalent to multiplying each term of
the result of the first substitution by the same constant factor, which therefore
may be taken out as a factor of the whole first member. The remaining
factor being zero, the product is zero and the equation is satisfied.

\emph{If several values of the dependent variable have been found each of which
satisfies the given differential equation, their sum will satisfy the equation;} for
if the sum of the values in question is substituted in the equation each term
of the sum will give rise to a set of terms which must be equal to zero, and
therefore the sum of these sets must be zero.

\mypara{6.}  It is generally possible to get by some simple device \emph{particular solutions}
of such differential equations as those we have collected in Art.~1. The
object of the branch of mathematics with which we are about to deal is to
find methods of so combining these particular solutions as to satisfy any given
conditions which are consistent with the nature of the problem in question.

This often requires us to be able to develop any given function of the variables
which enter into the expression of these conditions in terms of \emph{normal
forms} suited to the problem with which we happen to be dealing, and suggested
by the form of particular solution that we are able to obtain for the
differential equation.

These normal forms are frequently sines and cosines, but they are often
much more complicated functions known as \emph{Legendre's Coefficients,} or \emph{Zonal
Harmonics; Laplace's Coefficients,} or \emph{Spherical Harmonics: Bessel's Functions,}
or \emph{Cylindrical Harmonics; Lamé's Functions,} or \emph{Ellipsoidal Harmonics,} \&c.

\mypara{7.} As an illustration, let us take Fourier's problem of the permanent state
of temperatures in a thin rectangular plate of breadth $\pi$ and of infinite length
whose faces are impervious to heat. We shall suppose that the two long
edges of the plate are kept at the constant temperature zero, that one of the
short edges, which we shall call the base of the plate, is kept at the temperature
unity, and that the temperatures of points in the plate decrease indefinitely
as we recede from the base; we shall attempt to find the temperature
at any point of the plate.

Let us take the base as the axis of $X$ and one end of the base as the origin.
Then to solve the problem we are to find the temperature $u$ of any point from
the equation
\[
  D_x^{2}u +D_y^{2}u= 0         \tag*{[III] Art.~1}
\]
subject to the conditions
\begin{alignat*}{3}
  u &= 0\quad &&\text{when}\quad & x &= 0    \tag{1} \\
  u &= 0    &&\quad\text{``}   & x &= \pi    \tag{2} \\
  u &= 0    &&\quad\text{``}   & y &= \infty \tag{3} \\
  u &= 1    &&\quad\text{``}   & y &= 0.     \tag{4}
\end{alignat*}
% -----File: 015.png

We shall begin by getting a particular solution of \smallrom{III}, and we shall use
a device which always succeeds when the equation is \textit{linear} and \textit{homogeneous}
and has \textit{constant coefficients}.

Assume\label{notep5}\footnote
  {This assumption must be regarded as purely tentative. It must be tested by substituting
  in the equation, and is justified if it leads to a solution.}
$u = e^{\alpha y + \beta x}$, where $\alpha$ and $\beta$ are constants, substitute in \smallrom{III} and
divide by $e^{\alpha y + \beta x}$, and we have $\alpha^2 + \beta^2 = 0$. If, then, this condition is satisfied
$u = e^{\alpha y + \beta x}$ is a solution.

Hence $u = e^{\alpha y \pm \alpha xi}$ \footnote
  {We shall regularly use the symbol $i$ for $\sqrt{-1}$.}
is a solution of \smallrom{III}, no matter what value may be
given to $\alpha$.

This form is objectionable, since it involves an imaginary. We can, however,
readily improve it.

Take $u=e^{\alpha y} e^{\alpha xi}$, a solution of \smallrom{III}, and $u = e^{\alpha y} e^{-\alpha xi}$, another solution
of \smallrom{III}; add these values of $u$ and divide the sum by 2 and we have
$e^{\alpha y} \cos\alpha x$. (v.~Int.\ Cal.\ Art.~35,\ [1].) Therefore by Art.~5
\[
u = e^{\alpha y} \cos \alpha x      \tag{5}
\]
is a solution of \smallrom{III}. Take $u=e^{\alpha y} e^{\alpha xi}$ and $u = e^{\alpha y} e^{-\alpha xi}$, subtract the
second value of $u$ from the first and divide by $2i$ and we have $e^{\alpha y} \sin\alpha x$.
(v.~Int.\ Cal.\ Art.~35,\ [2]). Therefore by Art.~5
\[
u = e^{\alpha y} \sin\alpha x     \tag{6}
\]
is a solution of \smallrom{III}.

Let us now see if out of these particular solutions we can build up a solution
which will satisfy the conditions (1), (2), (3), and (4).
\begin{flalign*}
&\text{\indent Consider }&
& u = e^{\alpha y} \sin\alpha x.  &&\phantom{\indent Consider }    \tag{6}
\end{flalign*}
It is zero when $x = 0$ for all values of $\alpha$. It is zero when $x = \pi$ if $\alpha$ is a
whole number. It is zero when $y = \infty$ if $\alpha$ is negative. If, then, we write
$u$ equal to a sum of terms of the form $Ae^{-my} \sin mx$, where $m$ is a positive
integer, we shall have a solution of \smallrom{III} which satisfies conditions (1), (2)
and (3). Let this solution be
\[
u = A_1e^{-y} \sin x + A_2e^{-2y} \sin 2x + A_3e^{-3y} \sin 3x + A_4e^{-4y} \sin 4x  + \cdots  \tag{7}
\]
$A_1$, $A_2$, $A_3$, $A_4$, \&c., being undetermined constants.

When $y = 0$ (7) reduces to
\[
u = A_1 \sin x + A_2 \sin 2x + A_3 \sin 3x + A_4 \sin 4x + \cdots . \tag{8}
\]

If now it is possible to develop unity into a series of the form (8), our
problem is solved; we have only to substitute the coefficients of that series for
$A_1$, $A_2$, $A_3$, \&c.\ in (7).
% -----File: 016.png

It will be proved later that
\[
1 = \frac{4}{\pi} \left( \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x
  + \frac{1}{7} \sin 7x + \cdots \right)
\]
for all values of $x$ between 0 and $\pi$; hence our required solution is
\[
  u = \frac{4}{\pi} \left[ e^{-y}\sin x + \frac{1}{3} e^{-3y} \sin 3x
       + \frac{1}{5} e^{-5y} \sin 5x + \frac{1}{7} e^{-7y} \sin 7x + \cdots \right]
\tag{9}
\]
for this satisfies the differential equation and all the given conditions.

If the given temperature of the base of the plate instead of being unity
is a function of $x$, we can solve the problem as before if we can express the
given function of $x$ as a sum of terms of the form $A \sin m x$, where $m$ is a
whole number.

The problem of finding the value of the \textit{potential function} at any point of
a long, thin, rectangular conducting sheet, of breadth $\pi$, through which an
electric current is flowing, when the two long edges are kept at potential zero,
and one short edge at potential unity, is mathematically identical with the
problem we have just solved.

\Example{}

Taking the temperature of the base of the plate described above as 100°
centigrade, and that of the sides of the plate as 0°, compute the temperatures
of the points
\begin{center}
(\textit{a}) $\displaystyle \left( \frac{\pi}{6}, 1 \right)$;
(\textit{b}) $\displaystyle \left( \frac{\pi}{3}, 2 \right)$;
(\textit{c}) $\displaystyle \left( \frac{\pi}{2}, 3 \right)$,
\end{center}
correct to the nearest degree. \hfill \emph{Ans.}\ (\textit{a}) 26°; (\textit{b}) 15°; (\textit{c}) 6°.

\mypara{8.} As another illustration, we shall take the problem of the transverse
vibrations of a stretched string fastened at the ends, initially distorted into
some given curve and then allowed to swing.

Let the length of the string be $l$. Take the position of equilibrium of the
string as the axis of $\text{X}$, and one of the ends as the origin, and suppose the
string initially distorted into a curve whose equation $y = f(x)$ is given.

We have then to find an expression for $y$ which will be a solution of the
equation
\[
D_t^2 y = a^2 D_x^2 y     \tag*{\smallrom{VIII}\ Art.~1,}
\]
while satisfying the conditions
\begin{alignat}{4}
      y &= 0     && \text{when}\quad& x &= 0     \tag{1} \\
      y &= 0     && \quad\text{``}  & x &= l     \tag{2} \\
      y &= f(x)  && \quad\text{``}  & t &= 0     \tag{3} \\
  D_t y &= 0     && \quad\text{``}  & t &= 0,    \tag{4}
\end{alignat}
the last condition meaning merely that the string starts from rest.
% -----File: 017.png

As in the last problem let\footnote{See note on page \pageref{notep5}.} $y = e^{\alpha x+\beta t}$ and substitute in \smallrom{VIII}. Divide
by $e^{\alpha x +\beta t}$  and we have $\beta^2=a^2\alpha^2$ as the condition that our assumed value of
$y$ shall satisfy the equation.
\[
y= e^{\alpha x \pm a\alpha t} \tag{5}
\]
is, then, a solution of \smallromr{VIII} whatever the value of $\alpha$.

It is more convenient to have a trigonometric than an exponential form to
deal with, and we can readily obtain one by using an imaginary value for $\alpha$ in (5).
Replace $\alpha$ by $\alpha i$ and (5) becomes $y=e^{(x\pm at)\alpha i}$, a solution of \smallrom{VIII}. Replace
$\alpha$ by $-\alpha i$ and (5) becomes $y=e^{-(x\pm at)\alpha i}$, another solution of \smallrom{VIII}. Add
these values of $y$ and divide by 2 and we have $\cos\alpha(x\pm at)$. Subtract the
second value of $y$ from the first and divide by $2i$ and we have $\sin\alpha(x\pm at)$.
\begin{align*}
& y = \cos \alpha(x + at)\\
& y = \cos \alpha(x - at)\\
& y = \sin \alpha(x + at)\\
& y = \sin \alpha(x - at)
\end{align*}
are, then, solutions of \smallrom{VIII}. Writing $y$ successively equal to half the sum
of the first pair of values, half their difference, half the sum of the last
pair of values, and half their difference, we get the very convenient particular
solutions of \smallrom{VIII}.
\begin{align*}
&y = \cos \alpha x \cos \alpha at\\
&y = \sin \alpha x \sin \alpha at\\
&y = \sin \alpha x \cos \alpha at\\
&y = \cos \alpha x \sin \alpha at.
\end{align*}

If we take the third form
\begin{align*}
&y = \sin \alpha x \cos \alpha at
\end{align*}
it will satisfy conditions (1) and (4), no matter what value may be given to
$\alpha$, and it will satisfy (2) if $\alpha = \dfrac{m\pi}{l}$
where $m$ is an integer.

If then we take
\[
y=A_1\sin\frac{\pi x}{l}\cos\frac{\pi at}{l} + A_2\sin\frac{2\pi x}{l}\cos\frac{2\pi at}{l} + A_3\sin\frac{3\pi x}{l}\cos\frac{3\pi at}{l} + \cdots \tag{6}
\]
where $A_{1}$, $A_{2}$, $A_{3}$ $\cdots$ are undetermined constants, we shall have a solution of
\smallrom{VIII} which satisfies (1), (2), and (4). When $t=0$ it reduces to
\[
y=A_1\sin\frac{\pi x}{l} + A_2\sin\frac{2\pi x}{l} + A_3\sin\frac{3\pi x}{l} + \cdots \tag{7}
\]

If now it is possible to develop $f(x)$ into a series of the form (7), we can
solve our problem completely. We have only to take the coefficients of this
series as values of $A_1$, $A_2$, $A_3$ $\cdots$ in (6), and we shall have a solution of
\smallrom{VIII} which satisfies all our given conditions.
% -----File: 018.png

In each of the preceding problems the \emph{normal function}, in terms of which a
given function has to be expressed, is the sine of a simple multiple of the
variable. It would be easy to modify the problem so that the \emph{normal form}
should be a cosine.

We shall now take a couple of problems which are much more complicated
and where the normal function is an unfamiliar one.

\mypara{9.} Let it be required to find the potential function due to a circular wire
ring of small cross section and of given radius $c$, supposing the matter of the
ring to attract according to the law of nature.

We can readily find, by direct integration, the value of the potential function
at any point of the axis of the ring. We get for it
\[
V=\frac{M}{\sqrt{c^2 + x^2}}  \tag{1}
\]
where $M$ is the mass of the ring, and $x$ the distance of the point from the
centre of the ring.

Let us use spherical coördinates, taking the centre of the ring as origin and
the axis of the ring as the polar axis.

To obtain the value of the potential function at any point in space, we must
satisfy the equation
\[
rD_r^2(rV) + \frac{1}{\sin\theta}D_\theta(\sin\theta D_\theta V) + \frac{1}{\sin^2\theta}D_\phi^2V=0,  \tag*{\smallrom{XIII} Art.~1,}
\]
subject to the condition
\[
V=\frac{M}{(c^2+r^2)^{\frac{1}{2}}} \quad \text{when} \quad \theta=0.     \tag{1}
\]

From the symmetry of the ring, it is clear that the value of the potential
function must be independent of $\phi$, so that \smallrom{XIII} will reduce to
\[
rD_r^2(rV) + \frac{1}{\sin\theta} D_\theta(\sin\theta D_\theta V)=0.    \tag{2}
\]

We must now try to get particular solutions of (2), and as the coefficients
are not constant, we are driven to a new device.

Let\footnote{See note on page \pageref{notep5}.} $V=r^m P$, where $P$ is a function of $\theta$ only, and $m$ is a positive integer,
and substitute in (2), which becomes
\[
m(m+1)r^mP + \frac{r^m}{\sin\theta}D_\theta(\sin\theta D_\theta P) =0.
\]
% -----File: 019.png
Divide by $r^m$ and use the notation of ordinary derivatives since $P$ depends
upon $\theta$ only, and we have the equation
\[
m(m+1)P + \frac{1}{\sin\theta}\frac{d\Big(\sin\theta\dfrac{dP}{d\theta}\Big)}{d\theta}=0,   \tag{3}
\]
from which to obtain $P$.

Equation (3) can be simplified by changing the independent variable. Let
$x=\cos\theta$ and (3) becomes
\[
\frac{d}{dx}\left [(1-x^2)\frac{dP}{dx}\right ] + m(m+1)P = 0.   \tag{4}
\]

Assume\footnote{See note on page \pageref{notep5}.} now that $P$ can be expressed as a sum or as a series of terms
involving whole powers of $x$ multiplied by constant coefficients.

Let $P=\sum a_nx^n$ and substitute this value of $P$ in (4). We get
\[
\textstyle\sum[n(n-1)a_n x^{n-2} - n(n+1) a_n x^n + m(m+1) a_nx^n] = 0,   \tag{5}
\]
where the symbol $\sum$ indicates that we are to form all the terms we can by
taking successive whole numbers for $n$.

As (5) must be true no matter what the value of $x$, the coefficient of any
given power of $x$, as for instance $x^k$, must vanish. Hence
\begin{flalign*}
&&(k+2)(k+1)a_{k+2}-k(k+1)a_k+m(m+ {}& 1)a_k = 0 &  \tag{6}\\
&\text{and} &  a_{k+2} = -\frac{m(m+1)-k(k+1)}{(k+1)(k+2)}a_k.& & \tag{7}
\end{flalign*}
If now any set of coefficients satisfying the relation (7) be taken, $P = \sum a_k x^k$
will be a solution of (4).
\begin{flalign*}
&\indent\text{If }&  &  k=m,\quad a_{k+2}=0,\quad a_{k+4}=0,\quad \text{\&c.}  &&\phantom{\indent If }
\end{flalign*}
Since it will answer our purpose if we pick out the simplest set of coefficients
that will obey the condition (7), we can take a set including $a_m$.

Let us rewrite (7) in the form
\[
a_k=-\frac{(k+2)(k+1)}{(m-k)(m+k+1)}a_{k+2}.  \tag{8}
\]
We get from (8), beginning with $k=m-2$,
\begin{align*}
a_{m-2}& = -\frac{m(m-1)}{2.(2m-1)}a_m\\
a_{m-4}& =  \frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3)}a_m\\
a_{m-6}& = -\frac{m(m-1)(m-2)(m-3)(m-4)(m-5)}{2.4.6.(2m-1)(2m-3)(2m-5)}a_m,  \text{\quad\&c.}
\end{align*}
% -----File: 020.png
If $m$ is even we see that the set will end with $a_0$, if $m$ is odd, with $a_1$.
\[
P = a_m \left[ x^m
             - \frac{m(m-1)}{2.(2m-1)} x^{m-2}
             + \frac{m(m-1)(m-2)(m-3)}
                    {2.4.(2m-1)(2m-3)} x^{m-4}
             - \cdots\,\right]
\]
where $a_m$ is entirely arbitrary, is, then, a solution of (4). It is found convenient
to take $a_m$ equal to
\[
\frac{(2m - 1)(2m - 3)\cdots 1}{m!}
\]
and it can be shown that with this value of $a_m$ $P = 1$ when $x = 1$.

$P$ is a function of $x$ and contains no higher powers of $x$ than $x^m$. It is
usual to write it as $P_m(x)$.

We proceed to compute a few values of $P_m(x)$ from the formula
\begin{align*}
P_m(x) &= \frac{(2m-1)(2m-3) \cdots 1}{m!} \left[x^m - \frac{m(m-1)}{2.(2m-1)}x^{m-2} \right. \\[1ex]
&\left. + \frac{m(m-1)(m-2)(m-3)}
       {2.4.(2m-1)(2m-3)} x^{m-4} - \cdots \,\right].  \tag{9}
\end{align*}

We have:
{
\[
\left.\begin{alignedat}{4}
P_0(x) &= 1				  && \text{or}\qquad	& P_0(\cos\theta) &= 1 \\
P_1(x) &= x				  && \:\text{``}	& P_1(\cos\theta) &= \cos\theta \\
P_2(x) &= \tfrac{1}{2}(3x^2 - 1)	  && \:\text{``}	& P_2(\cos\theta) &= \tfrac{1}{2}(3 \cos^2\theta - 1) \\
P_3(x) &= \tfrac{1}{2}(5x^3 - 3x)  \quad  && \:\text{``}	& P_3(\cos\theta) &= \tfrac{1}{2}(5 \cos^3\theta - 3 \cos\theta) \\
P_4(x) &= \rlap{$\tfrac{1}{8}(35x^4 - 30x^2 + 3)$ or} \\
     &&&& \llap{$P_4(\cos\theta) = \tfrac{1}{8}(35$} &\cos^4\theta - 30 \cos^2\theta + 3)  \\
P_5(x) &= \rlap{$\tfrac{1}{8}(63x^5 - 70x^3 + 15x)$ or} \\
     &&&& \llap{$P_5(\cos\theta) = \tfrac{1}{8}(63 \cos^5\theta $} &-70 \cos^3\theta + 15 \cos\theta).
\end{alignedat}
\tag{10}
\quad\right\}
\]}

We have obtained $P = P_m(x)$ as a particular solution of (4) and
$P = P_m(\cos\theta)$ as a particular solution of (3). $P_m(x)$ or $P_m(\cos\theta)$ is a
new function, known as a \textit{Legendre's Coefficient}, or as a \textit{Surface Zonal Harmonic},
and occurs as a normal form in many important problems.

$V = r^mP_m(\cos\theta)$ is a particular solution of (2) and $r^mP_m(\cos\theta)$ is sometimes
called a \textit{Solid Zonal Harmonic}.

We can now proceed to the solution of our original problem.
\[
V=A_0r^0P_0(\cos\theta) + A_1rP_1(\cos\theta) + A_2r^2P_2(\cos\theta) + A_3r^3P_3(\cos\theta) + \cdots \tag{11}
\]
where $A_0$, $A_1$, $A_2$, \&c., are entirely arbitrary, is a solution of (2) (v.\ Art.~5).
When $\theta = 0$ (11) reduces to
\[
V=A_0 + A_1 r + A_2 r^2 + A_3 r^3 + \cdots\,,
\]
since, as we have said, $P_m(x) = 1$ when $x = 1$, or $P_m(\cos\theta) = 1$ when $\theta = 0$.

By our condition (1)
\[
V=\frac{M}{(c^2 + r^2)^{\frac{1}{2}}}
\]
when $\theta = 0$.
% -----File: 021.png

By the Binomial Theorem
\[
\frac{M}{(c^2+r^2)^{\frac{1}{2}}}
= \frac{M}{c} \left[ 1 - \frac{1}{2} \frac{r^2}{c^2}
  + \frac{1.3}{2.4} \frac{r^4}{c^4}
  - \frac{1.3.5}{2.4.6} \frac{r^6}{c^6}
  + \cdots \,\right]
\]
provided $r < c$. Hence
% recast to fit line
\begin{align*}
V  = \frac{M}{c} \left[ P_0(\cos\theta)
 - \frac{1}{2} \frac{r^2}{c^2} P_2(\cos\theta)\right.
& + \frac{1.3}{2.4} \frac{r^4}{c^4} P_4(\cos\theta)  \\
& \left. - \frac{1.3.5}{2.4.6} \frac{r^6}{c^6} P_6(\cos\theta)
 + \cdots\, \right]  \tag{12}
\end{align*}
is our required solution if $r < c$; for it is a solution of equation~(2) and satisfies
condition~(1).

\Example{}

Taking the mass of the ring as one pound and the radius of the ring as one
foot, compute to two decimal places the value of the potential function due to
the ring at the points
{\def\bit#1#2#3{\text{(\textit{#1})\ }\left(r = #2, \, \theta = #3 \right);}
\begin{align*}
& \bit{a}{.2}{0} &
& \bit{d}{.6}{0} &
& \bit{f}{.6}{\frac{\pi}{3}} \\
%
& \bit{b}{.2}{\frac{\pi}{4}} &
& \bit{e}{.6}{\frac{\pi}{6}} &
& \bit{g}{.6}{\frac{\pi}{2}} \\
%
& \bit{c}{.2}{\frac{\pi}{2}}
\end{align*}
\hfill\smash{\raisebox{3ex}{\begin{tabular}
{l@{\hfill}}
\llap{\emph{Ans.} }(\emph{a}) .98;  (\emph{b}) .99; (\emph{c}) 1.01; (\emph{d}) .86;\\
    (\emph{e}) .90; (\emph{f}) 1.00; (\emph{g}) 1.10.
\end{tabular}}}}\break
The unit used is the potential due to a pound of mass concentrated at a point
and attracting a second pound of mass concentrated at a point, the two points
being a foot apart.

\mypara{10.} A slightly different problem calling for development in terms of Zonal
Harmonics is the following:

Required the permanent temperatures within a solid sphere of radius~1,
one half of the surface being kept at the constant temperature zero, and the
other half at the constant temperature unity.

Let us take the diameter perpendicular to the plane separating the unequally
heated surfaces as our axis and let us use spherical coördinates. As in the
last problem, we must solve the equation
\[
rD_r^2(ru) + \frac{1}{\sin\theta} D_\theta (\sin \theta D_\theta u) + \frac{1}{\sin^2\theta} D_\phi^2 u = 0  \tag*{\smallrom{XIII} Art.~1}
\]
which as before reduces to
\[
r D_r^2 (ru) + \frac{1}{\sin\theta} D_\theta (\sin\theta D_\theta u) = 0   \tag{1}
\]
from the consideration that the temperatures must be independent of $\phi$.

Our equation of condition is
\[
u = 1 \text{ from } \theta = 0 \text{ to } \theta = \frac{\pi}{2} \text{ and } u = 0 \text{ from } \theta = \frac{\pi}{2} \text{ to } \theta = \pi,  \tag{2}
\]
when $r = 1$.
% -----File: 022.png

As we have seen $u = r^mP_m(\cos \theta)$ is a particular solution of (1), $m$ being
any positive whole number, and
\[
  u = A_0r^0P_0 (\cos \theta)
    + A_1r  P_1 (\cos \theta)
    + A_2r^2P_2 (\cos \theta)
    + A_3r^3P_3 (\cos \theta) + \cdots
\tag{3}
\]
where $A_0$, $A_1$, $A_2$, $A_3 \cdots$ are undetermined constants, is a solution of (1).

When $r = 1$ (3) reduces to
\[
\tag{4}
u = A_0P_0 (\cos \theta) + A_1P_1 (\cos \theta) + A_2P_2(\cos \theta) + A_3P_3 (\cos \theta) + \cdots
\]
If then we can develop our function of $\theta$ which enters into equation~(2) in
a series of the form (4), we have only to take the coefficients of that series
as the values of $A_0$, $A_1$, $A_2$, \&c., in (3) and we shall have our required solution.

\mypara{11.} As a last example we shall take the problem of the vibration of a stretched
circular membrane fastened at the circumference, that is, of an ordinary drumhead.
We shall suppose the membrane initially distorted into any given form
which has circular symmetry\label{notep12}\footnote
  {A function of the coördinates of a point has \textit{circular symmetry} about an axis when its
  value is not affected by rotating the point through any angle about the axis. A surface has
  circular symmetry about an axis when it is a surface of revolution about the axis.}
about an axis through the centre perpendicular
to the plane of the boundary, and then allowed to vibrate.

Here we have to solve
\[
\tag*{\smallrom{XI} Art.~1}
D_t^2z=c^2\left(D_r^2z + \frac{1}{r}D_rz + \frac{1}{r^2} D_\phi^2z \right)
\]
subject to the conditions
\begin{alignat*}{3}
\tag{1}
  z &= f(r) \quad  && \text{when}\quad &  t &= 0  \\
\tag{2}
  D_t z &= 0  && \quad\text{``}   &  t &= 0  \\
\tag{3}
  z &= 0      && \quad\text{``}   &  r &= a.
\end{alignat*}

From the symmetry of the supposed initial distortion $z$ must be independent
of $\phi$, therefore \smallrom{XI} reduces to
\[
\tag{4}
D_t^2 z = c^2\left(D_r^2 z + \frac{1}{r} D_r z\right)
\]
and this is the equation for which we wish to find a particular solution.

We shall employ a device not unlike that used in Art.~9.

Assume\footnote
  {See note on page \pageref{notep5}.}
$z = R.T$ where $R$ is a function of $r$ alone and $T$ is a function of
$t$ alone.  Substitute this value of $z$ in (4) and we get
\[
R D_t^2 T = c^2 T\left(D_r^2 R + \frac{1}{r} D_r R\right)
\]
\begin{flalign*}
\tag{5}
&\text{or }&&
\frac{1}{c^2T} \frac{d^2T}{dt^2} = \frac{1}{R} \left(\frac{d^2R}{dr^2} + \frac{1}{r} \frac{dR}{dr}\right). &&
\end{flalign*}
The second member of (5) does not involve $t$, therefore its equal the first
member must be independent of $t$. The first member of (5) does not involve
% -----File: 023.png
$r$, and consequently since it contains neither $t$ nor $r$, it must be constant. Let
it equal $-\mu^2$, where $\mu$ of course is an undetermined constant.

Then (5) breaks up into the two differential equations
\begin{gather}
\frac{d^2T}{dt^2} + \mu^2c^2T=0   \tag{6} \\
\frac{d^2R}{dr^2} + \frac{1}{r}\frac{dR}{dr} + \mu^2R = 0. \tag{7}
\end{gather}
(6) can be solved by familiar methods, and we get $T = \cos \mu ct$ and $T = \sin \mu ct$
as simple particular solutions (v.\ Int.\ Cal.\ p.~319, §~21).

To solve (7) is not so easy. We shall first simplify it by a change of independent
variable. Let $r = \dfrac{x}{\mu}$.  (7) becomes
\[
\frac{d^2R}{dx^2} + \frac{1}{x}\frac{dR}{dx}+R=0.     \tag{8}
\]

Assume, as in Art.~9, that $R$ can be expressed in terms of whole powers of
$x$. Let $R = \sum a_n x^n$ and substitute in (8). We get
\[
  \textstyle\sum [n(n - 1)a_nx^{n-2} + na_nx^{n-2} + a_nx^n] = 0, \notag
\]
an equation which must be true no matter what the value of $x$. The coefficient
of any given power of $x$, as $x^{k-2}$, must, then, vanish, and
\begin{flalign*}
&&       k(k - 1)a_k + k&a_k + a_{k-2} = 0  &&\phantom{whence we obtain } \\
&\text{or }&  k^2a_k + a&_{k-2} = 0  && \\
&\text{whence we obtain }&
             a_{k-2} ={}& - k^2a_k  &&  \tag{9}
\end{flalign*}
as the only relation that need be satisfied by the coefficients in order that
$R = \sum a_k x^k$ shall be a solution of (8).
\[
\hbox to \displaywidth {\rlap{\indent If}\hfil $k = 0$,\quad $a_{k-2}= 0$,\quad $a_{k-4}=0$,\quad \&c.\hfil}
\]
We can then begin with $k = 0$ as our lowest subscript.
\begin{flalign*}
&\text{\indent From (9) }& a_k &=-\frac{a_{k-2}}{k^2}. &&\phantom{\indent From\ (9) }\\
&\text{Then }&   a_{2} &= -\frac{a_0}{2^{2}} \\
&&               a_{4} &= \frac{a_0}{2^2.4^2} \\
&&               a_{6} &= -\frac{a_0}{2^2.4^2.6^2}, \text{ \&c.} \\
&\text{Hence }&
  R = a_0\left[1-\frac{x^2}{2^2} \right.+{}& \frac{x^4}{2^2.4^2}- \left. \frac{x^6}{2^2.4^2.6^2} + \cdots\,\right]
\end{flalign*}
where $a_0$ may be taken at pleasure, is a solution of (8), provided the series is
convergent.
% -----File: 024.png

Take $a_0=1$, and then $R=J_0(x)$ where
\[
J_0(x)=1-\frac{x^2}{2^2} + \frac{x^4}{2^2.4^2} - \frac{x^6}{2^2.4^2.6^2} + \frac{x^8}{2^2.4^2.6^2.8^2} - \cdots \tag{10}
\]
is a solution of (8).

$J_0(x)$ is easily shown to be convergent for all values real or imaginary of $x$,
since the series made up of the moduli of the terms of $J_0(x)$ (v.\ Int.\ Cal.\
Art.~30)
\[
1 + \frac{r^2}{2^2} + \frac{r^4}{2^2.4^2} + \frac{r^6}{2^2.4^2.6^2} + \cdots,
\]
where $r$ is the modulus of $x$, is convergent for all values of $r$. For the ratio
of the $n + 1$st term of this series to the $n$th term is $\dfrac{r^2}{4n^2}$ and approaches
zero as its limit as $n$ is indefinitely increased, no matter what the value of $r$.
Therefore $J_0(x)$ is \textit{absolutely convergent}.

$J_0(x)$ is a new and important form. It is called a \textit{Bessel's Function} of the
zeroth order, or a \textit{Cylindrical Harmonic}.

Equation (8) was obtained from (7) by the substitution of $x=\mu r$, therefore
\[
R = J_0(\mu r) = 1 - \frac{(\mu r)^2}{2^2} + \frac{(\mu r)^4}{2^2.4^2} - \frac{(\mu r)^6}{2^2.4^2.6^2} + \cdots
\]
is a solution of (7), no matter what the value of $\mu$, and $z=J_0(\mu r)\cos\mu ct$
or $z=J_0(\mu r)\sin\mu ct$ is a solution of (4).

$z=J_0(\mu r)\cos\mu ct$ satisfies condition~(2) whatever the value of $\mu$. In
order that it should also satisfy condition~(3) $\mu$ must be so taken that
\[
J_0(\mu a)=0;   \tag{11}
\]
that is, $\mu$ must be a root of (11) regarded as an equation in $\mu$.

It can be shown that $J_0(x)=0$ has an infinite number of real positive
roots, any one of which can be obtained to any required degree of approximation
without serious difficulty. Let $x_1$, $x_2$, $x_3$, $\cdots$ be these roots. Then if
\begin{gather*}
\frac{x_1}{a}=\mu_1, \quad \frac{x_2}{a}=\mu_2, \quad \frac{x_3}{a}=\mu_3, \quad \text{\&c.\,}\\
z=A_1J_0(\mu_1r)\cos\mu_1ct + A_2J_0(\mu_2r)\cos\mu_2ct + A_3J_0(\mu_3r)\cos\mu_3ct + \cdots, \tag{12}
\end{gather*}
where $A_1$, $A_2$, $A_3$, \&c., are any constants, is a solution of (4) which satisfies
conditions (2) and (3).

When $t=0$ (12) reduces to
\[
z=A_1J_0(\mu_1 r) + A_2J_0(\mu_2 r) + A_3J_0(\mu_3 r) + \cdots.  \tag{13}
\]
If then $f(r)$ can be expressed as a series of the form just given, the solution
of our problem can be obtained by substituting the coefficients of that series
for $A_1$, $A_2$, $A_3$, \&c., in (12).
% -----File: 025.png

\Example{}

The temperature of a long cylinder is at first unity throughout. The
convex surface is then kept at the constant temperature zero. Show
that the temperature of any point in the cylinder at the expiration
of the time $t$ is
\[
u=A_{1}e^{-a^2 \mu_1^2t}J_{0}(\mu_1r) +
  A_{2}e^{-a^2 \mu_2^2t}J_{0}(\mu_2r) +
  A_{3}e^{-a^2 \mu_3^2t}J_{0}(\mu_3r) + \cdots
\]
where $\mu_{1}$, $\mu_{2}$, \&c., are the roots of
$J_{0}(\mu c) = 0$, and where
\[
1=A_{1}J_{0}(\mu_{1}r) + A_{2}J_{0}(\mu_{2}r) +A_{3}J_{0}(\mu_{3}r)
+ \cdots,
\]
$c$ being the radius of the cylinder.

\mypara{12.}  Each of the five problems which we have taken up forces upon us
the consideration of the development of a given function in terms of
some \emph{normal form}, and in two of them the normal form suggested is an
unfamiliar function. It is clear, then, that a complete treatment of
our subject will require the investigation of the properties and
relations of certain new and important functions, as well as the
consideration of methods of developing in terms of them.

\mypara{13.}  In each of the problems just taken up we have to deal with a
homogeneous linear partial differential equation involving two
independent variables, and we are content if we can obtain
particular solutions. In each case the assumption made in the last
problem, that there exists a solution of the equation in which the
dependent variable is the product of two factors each of which
involves but one of the independent variables, will reduce the
question to solving two ordinary differential equations which can be
treated separately.

If these equations are familiar ones their solutions can be written
down at once; if unfamiliar, the device used in problems 3 and 5 is
often serviceable, namely, that of assuming that the dependent
variable can be expressed as a sum or series of terms involving
whole powers of the independent variable, and then determining the
coefficients.

Let us consider again the equations used in the first, second and
third problems.
\begin{flalign}
&\text{\indent(\textit{a})}&&  D_x^2 u + D_y^2 u = 0 &&  \quad        \tag{1}
\end{flalign}
Assume $u = X.Y$ where $X$ involves $x$ but not $y$, and $Y$ involves $y$ but
not $x$.
\begin{flalign*}
&\text{Substitute in (1), }&&  YD_x^2 X + XD_y^2 Y = 0,  &&\phantom{Substitute in (1), }
\end{flalign*}
or, since
we are now dealing with functions of a single variable,
\begin{flalign}
&&             \frac{1}{X}\frac{d^2X}{dx^2} &+ \frac{1}{Y}\frac{d^2Y}{dy^2}=0, &&\notag\\
&\text{or}&    \frac{1}{Y}\frac{d^2Y}{dy^2} &= -\frac{1}{X}\frac{d^2X}{dx^2}.             \tag{2}
\end{flalign}
% -----File: 026.png

Since the first member of (2) does not contain $x$, and the second member
does not contain $y$, and the two members must be identically equal, neither of
them can contain either $x$ or $y$, and each must be equal to a constant, say $\alpha^2$.
\begin{flalign*}
&\text{Then}&          \frac{d^2Y}{dy^2} - \alpha^2 Y& = 0   &\tag{3}\\
&\text{and}&          \frac{d^2X}{dx^2} + \alpha^2 X &= 0;     &\tag{4}
\end{flalign*}
and if (3) and (4) can be solved, we can solve (1). They have for their complete
solutions
\begin{flalign*}
&                     &  Y &= Ae^{\alpha y} + Be^{-\alpha y}\\
&\text{and}&        X &= C \sin {\alpha x} + D \cos {\alpha x}. \quad\text{ (v.\ Int.\ Cal.\ p.~319, \S~21.)}
\end{flalign*}
Hence $Y = e^{\alpha y}$ and $Y=e^{-\alpha y}$ are particular solutions of (3), $X=\sin {\alpha x}$
and $X = \cos {\alpha x}$ are particular solutions of (1), and consequently
\[
u =e^{\alpha y} \sin{\alpha x},\quad u = e^{\alpha y} \cos {\alpha x}, \quad u = e^{-\alpha y} \sin {\alpha x}, \text{ and }u =e^{-\alpha y} \cos {\alpha x}
\]
are particular solutions of (1). These agree with the results of Art.~7.
\begin{flalign*}
&\text{\indent(\textit{b})} &           D_t^2 y &= a^2 D_x^2 y     &\quad\tag{1}
\end{flalign*}

Assume $y = T.X$ where $T$ is a function of $t$ only and $X$ a function of $x$
only; substitute in (1) and divide by $a^2 TX$. We get
\begin{flalign*}
&\phantom{(2)}&&    \frac{1}{a^2 T}\frac{d^2T}{dt^2} = \frac{1}{X}\frac{d^2X}{dx^2};  \tag{2}\\
\intertext{hence as in the last case $\dfrac{1}{X}\dfrac{d^2X}{dx^2}$ is a constant; call it $- \alpha^2$, and (2) breaks up into}
&&    &\frac{d^2X}{dx^2} + \alpha^2   X = 0 &\tag{3}\\
&\phantom{(4)}&    &\frac{d^2T}{dt^2} + \alpha^2a^2T = 0.&\tag{4}
\end{flalign*}
The complete solutions of (3) and (4) are
\begin{flalign*}
&&  &X = A \sin {\alpha x} + B \cos {\alpha x} \\
&\text{and }&      &T = C \sin {\alpha at} + D\cos {\alpha at}, \quad\text{ (v.\ Int.\ Cal.\ p.~319, \S~21).}
\end{flalign*}
\[
  y = \sin {\alpha x}\cos {\alpha at}, \
  y = \sin {\alpha x}\sin {\alpha at}, \
  y = \cos {\alpha x}\cos {\alpha at}, \
  y = \cos {\alpha x}\sin {\alpha at}
\]
are particular solutions of (1), and agree with the results of Art.~8.
\begin{flalign*}
&\text{\indent (\textit{c})}&   rD_r^2(rV) + \dfrac{1}{\sin{\theta}} D_\theta (\sin{\theta} D_{\theta} V)& = 0. &\quad\tag{1}
\end{flalign*}

Assume $V= R.\Theta$ where $R$ involves $r$ alone, and $\Theta$ involves $\theta$ alone; substitute
in (1), divide by $R.\Theta$, and transpose; we get
\[
\frac{r}{R} \frac{d^2(rR)}{dr^2} = - \frac{1}{\Theta \sin \theta}\frac{d\Big(\sin \theta \dfrac{d\Theta}{ d\theta}\Big)}{d\theta}.  \tag{2}
\]
% -----File: 027.png
Since by the reasoning used in (\textit{a}) and (\textit{b}) each member of (2) must be a constant,
say $\alpha^2$, we have
\begin{flalign*}
&&  r\frac{d^2(rR)}{dr^2} ={}& \alpha^2 R \tag{3} \\
&\text{and} &\frac{1}{\sin\theta}\frac{d\Big(\sin\theta \dfrac{d\Theta}{d\theta}\Big)}{d\theta} &+ \alpha^2\Theta = 0. &&\phantom{and}\tag{4}
\end{flalign*}
(3) can be expanded into
\[
  r^2 \frac{d^2 R}{dr^2} + 2r \frac{dR}{dr} - \alpha^2 R = 0. \tag{5}
\]
(5) can be solved (v.\ Int.\ Cal.\ p.~321, \S~23), and has for its complete solution
\[
  R = Ar^m + Br^n,
\]
where \hfill $m = -\frac{1}{2} + \sqrt{\alpha^2 + \frac{1}{4}}$ \quad and \quad $n = -\frac{1}{2} - \sqrt{\alpha^2 + \frac{1}{4}}$. \hfill\phantom{where }\\[1ex]
Hence $n = - m - 1$, and $\alpha^2$ may be written $m(m + 1)$, $m$ being wholly
arbitrary; and
\begin{gather*}
  R = Ar^m + Br^{-m-1}.
\\
  R = r^m,\quad \text{ and }\quad R = \frac{1}{r^{m + 1}}
\end{gather*}
are, then, particular solutions of
\[
  r^2 \frac{d^2R}{dr^2} + 2r\frac{dR}{dr} - m(m + 1) R = 0. \tag{6}
\]
With the new value of $\alpha^2$ (4) becomes
\[
  \frac{1}{\sin\theta}\frac{d\Big(\sin\theta \dfrac{d\Theta}{d\theta}\Big)}{d\theta} + m(m + 1) \Theta = 0. \tag{7}
\]
which has been treated in Art.~9 for the case where $m$ is a positive integer,
and the particular solution $\Theta = P_m (\cos\theta)$ has been obtained.
\begin{flalign*}
&\indent\text{Hence }&
  V &= r^m P_m (\cos \theta) &&\phantom{\indent Hence }\\
&\text{and }&
  V &= \frac{1}{r^{m + 1}} P_m (\cos \theta),  &&
\end{flalign*}
$m$ being a positive integer, are particular solutions of (1). The first of these
was obtained in Art.~9, but the second is new and exceedingly important.

\mypara{14.} The method of obtaining a particular solution of an ordinary linear
differential equation, which we have used in Articles 9 and 11, is of very
extensive application, and often leads to the general solution of the equation
in question.
% -----File: 028.png

As a very simple example, let us take the equation Art.~13 (\emph{a}) (4), which
we shall write
\[
\frac{d^2z}{dx^2}+\alpha^2z = 0.  \tag{1}
\]

Assume that there is a solution which can be expressed in terms of powers
of $x$; that is, let $z = \sum{a_nx^n}$, where the coefficients are to be determined.
Substitute this value for $z$ in (1) and we get
\[
\textstyle \sum{[n(n - 1)a_nx^{n-2} + \alpha^2a_nx^n]} = 0.
\]
Since this equation must be true from its form, without reference to the value
of $x$, that is, since it must be an identical equation, the coefficient of each
power of $x$ must equal zero, and we have
\begin{flalign*}
&&  (n + 1)(n + 2)a_{n+2} + \alpha^2a_n = 0; &  &&\phantom{whence }\\
&\text{whence }&     a_n = -\frac{(n+1)(n+2)}{\alpha^2}a_{n + 2}&  &&
\end{flalign*}
is the only relation that need hold between the coefficients in order that
$z = \sum{a_nx^n}$ should be a solution of (1).

If $n + 2 = 0$ or $n + 1 = 0$, $a_n$ will be zero and $a_{n-2}$, $a_{n-4}$, \&c., will be
zero. In the first case the series will begin with $a_0$, in the second with $a_1$.
\[
a_{n+2} = - \frac{\alpha^2}{(n+1)(n+2)}a_n.
\]
If we begin with $a_0$ we have
\begin{align*}
  a_2 &= -\frac{\alpha^2}{2!}a_0,  &
  a_4 &=  \frac{\alpha^4}{4!}a_0,  &
  a_6 &= -\frac{\alpha^6}{6!}a_0, \text{ \&c}.,  \:\cdots\\[-5ex]
\end{align*}
\begin{flalign*}
&\text{and }&
  z = a_0 \left( 1
               - \frac{\alpha^2x^2}{2!} \right.&
               + \frac{\alpha^4x^4 }{4!} - \left.
                 \frac{\alpha^6x^6}{6!} + \cdots \right) &&\tag{2}
\\
&\text{or }&    z = {}& a_0 \cos\alpha x &&\tag{3}
\end{flalign*}
is a particular solution of (1).

If we begin with $a_1$ we have
\begin{align*}
  a_3 &= -\frac{\alpha^2}{3!}a_1,  &
  a_5 &=  \frac{\alpha^4}{5!}a_1,  &
  a_7 &= -\frac{\alpha^6}{7!}a_1,  \text{ \&c}.,  \:\cdots\\[-5ex]
\end{align*}
\begin{flalign*}
&\text{and }&
  z &= a_1 \left( x
                - \frac{\alpha^2x^3}{3!}
                + \frac{\alpha^4x^5}{5!}
                - \frac{\alpha^6x^7}{7!} + \cdots \right) &&\tag{4}
\end{flalign*}
% -----File: 029.png
is a solution of (1); $a_1$ can be taken at pleasure. Let $a_1 = \alpha$, (4) becomes
\begin{flalign*}
&&z &= \alpha x - \frac{\alpha^3x^3}{3!} + \frac{\alpha^5x^5}{5!} - \frac{\alpha^7x^7}{7!} + \cdots  &&\phantom{or}\\
&\text{or}   & z &= \sin\alpha x&
\end{flalign*}
which, then, is a particular solution of (1).
\[
z = A\sin\alpha x + B\cos\alpha x   \tag{5}
\]
is, then, a solution of (1), and since it contains two arbitrary constants it is
the general solution.

\mypara{15.}  As another example we will take the equation
\[
x^2\frac{d^2z}{dx^2} + 2x\frac{dz}{dx} - m(m+1)z = 0,   \tag{1}
\]
which is in effect equation (6), Art.~13~(\emph{c}), and let $m$ be a positive integer.

Assume $z = \sum a_nx^n$ and substitute in (1). We get
\[
  \textstyle\sum [n(n+1) - m(m+1)]a_nx^n = 0.
\]
This is an identical equation, therefore
\[
[n(n+1) - m(m+1)]a_n = 0.
\]
Hence $a_n=0$ for all values of $n$ except those which make
\[
n(n+1) - m(m+1) = 0,
\]
that is, for all values of $n$ except $n=m$ and $n=-m-1$. Then
\[
z = Ax^m + Bx^{-m-1}      \tag{2}
\]
is the general solution of (1) and
\[
z=x^m \quad \text{and} \quad z=\frac{1}{x^{m+1}}
\]
are particular solutions. If $m$ is not a positive integer this method will not
lead to a result, and we are driven back to that employed in Art.~13~(\emph{c}).

\mypara{16.}  Let us now take the equation
\[
\frac{d}{dx}\left [(1-x^2)\frac{dz}{dx}\right ] + m(m+1)z = 0  \tag{1}
\]
which is in effect equation (4), Art.~9, and is known as \emph{Legendre's Equation}.
(1) may be written
\[
(1-x^2) \frac{d^2z}{dx^2}-2x\frac{dz}{dx} + m(m+1)z = 0.   \tag{2}
\]
% -----File: 030.png

Assume $z=\sum a_nx^n$ and substitute in (2). We get
\[
  \textstyle\sum \{n(n-1)a_nx^{n-2} + [m(m+1) - n(n+1)]a_nx^n\} = 0.
\]
\begin{flalign*}
&\text{Hence } & (&n+1)(n+2)a_{n+2} + [m(m+1)-n(n+1)]a_n = 0,  &&\phantom{Hence }
\end{flalign*}
\begin{flalign*}
&\text{or}  & a_n &= -\frac{(n+1)(n+2)}{m(m+1)-n(n+1)}a_{n+2}.  && \tag{3}
\intertext{If $a_n=0$, then $a_{n-2}=0$, $a_{n-4}=0$, \&c.; but $a_n=0$ if $n=-2$ or $n=-1$.
For the first case we have the sequence of coefficients}
&&a_2&=-\frac{m(m+1)}{2!}a_0\\
&&a_4&= \frac{m(m-2)(m+1)(m+3)}{4!}a_0\\
&&a_6&=-\frac{m(m-2)(m-4)(m+1)(m+3)(m+5)}{6!}a_0, \quad \text{\&c.}
\end{flalign*}
Let us take $a_0$, which is arbitrary, as 1. Then $z=p_m(x)$ where
\[
  p_m(x) = \left[ 1 - \frac{m(m+1)}{2!}x^2 + \frac{m(m-2)(m+1)(m+3)}{4!}x^4 - \cdots \right] \tag{4}
\]
is a solution of Legendre's Equation if $p_m(x)$ is a finite sum or a convergent
series.

For the second case we have the sequence of coefficients
\begin{align*}
&a_3=-\frac{(m-1)(m+2)}{3!}a_1\\
&a_5= \frac{(m-1)(m-3)(m+2)(m+4)}{5!}a_1\\
&a_7=-\frac{(m-1)(m-3)(m-5)(m+2)(m+4)(m+6)}{7!}a_1, && \text{\&c.}
\end{align*}
Let us take $a_1$, which is arbitrary, as 1. Then $z=q_m(x)$ where
\[
q_m(x) = \left[ x - \frac{(m-1)(m+2)}{3!}x^3 + \frac{(m-1)(m-3)(m+2)(m+4)}{5!}x^5 - \cdots \!\right]\!\text{(5)}
\]
is a solution of Legendre's Equation if $q_m(x)$ is a finite sum or a convergent
series.
% -----File: 031.png

If $m$ is a positive even whole number, $p_m(x)$ will terminate with the term
containing $x^m$, and is easily seen to be identical with
\[
  (-1)^\frac{m}{2}
  \frac{ 2^m\Big [\Gamma \Big(\dfrac{m}{2}+1\Big)\Big]^2 }
       { \Gamma(m+1) } P_m(x).  \tag*{[v.\ Art.~9~(9)]}
\]
For all other values of $m$, $p_m(x)$ is a series.

The ratio of the $(n+1)$st term of $p_m(x)$ to the $n$th, when $m$ is not a positive
even integer, is
\[
\frac{(2n-2-m)(2n-1+m)}{(2n-1)(2n)}x^2.
\]
Its limiting value, as $n$ is increased, is $x^2$, and the series is therefore convergent
if $-1 < x < 1$. It is divergent for all other values of $x$.

If $m$ is a positive odd whole number $q_m(x)$ will terminate with the term
containing $x^m$, and is easily seen to be identical with
\[
  (-1)^\frac{m-1}{2}
   \frac{ 2^{m-1} \Big[\Gamma\Big(\dfrac{m+1}{2} \Big) \Big]^2 }
        { \Gamma(m+1)} P_m(x).
\]
For all other values of $m$, $q_m(x)$ is a series, and can be shown to be convergent
if $-1 < x < 1$, and divergent for all other values of $x$.
\[
z=Ap_m(x)+Bq_m(x)  \tag{6}
\]
is the general solution of Legendre's Equation if $-1 < x < 1$, no matter
what the value of $m$. From Art.~13~(\emph{c}) it follows that
\[
\left.
\begin{aligned}
  V &= r^mp_m(\cos\theta)\\
  V &= \frac{1}{r^{m+1}}p_m(\cos\theta)\quad\\
  V &= r^mq_m(\cos\theta)\\
  V &= \frac{1}{r^{m+1}}q_m(\cos\theta)\\
\end{aligned}
\right\}  \tag{7}
\]
are particular solutions of
\[
rD_r^2(rV) + \frac{1}{\sin\theta}D_\theta(\sin\theta D_\theta V)=0,
\]
no matter what the value of $m$, provided $\cos\theta$ is neither one nor minus one.

In the work we shall have to do with Laplace's and Legendre's Equations,
it is generally possible to restrict $m$ to being a positive integer, and hereafter
we shall usually confine our attention to that case.
% -----File: 032.png

With this understanding let us return to (3), which may be rewritten
\[
a_{n+2}=-\frac{(m-n)(m+n+1)}{(n+1)(n+2)}a_n.\\[-3ex]
\]
\begin{flalign*}
&\text{If}    && \text{$a_{n+2}=0$, then $a_{n+4}=0$,\; $a_{n+6} = 0$, \&c.;} &&\qquad \\[1ex]
&\text{but}   && \text{$a_{n+2}=0$\phantom{, }if\; $n=m$,\; or\; $n=-m-1$.} &&
\end{flalign*}
If in (3) we begin with $n=m-2$, we get the sequence of coefficients already
obtained in Art.~9, and we have $z=P_m(x)$, where
\begin{align*}
  P_m(x) &= \frac{(2m-1)(2m-3)\cdots 1}{m!}
            \left[ x^m - \frac{m(m-1)}{2(2m-1)} x^{m-2} \right.
\\[1ex]
&+ \frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3)}
     x^{m-4}
\\[1ex]
&- \left.\frac{m(m-1)(m-2)(m-3)(m-4)(m-5)}
                {2.4.6.(2m-1)(2m-3)(2m-5)}
           x^{m-6} + \cdots \right], \tag{8}
\end{align*}
as a particular solution of Legendre's Equation.

If, however, we begin with $n=-m-3$, we have
\begin{align*}
a_{-m-3} &= \frac{(m+1)(m+2)}{2(2m+3)}a_{-m-1} \\[1ex]
a_{-m-5} &= \frac{(m+1)(m+2)(m+3)(m+4)}
                 {2.4.(2m+3)(2m+5)}a_{-m-1}
\\[1ex]
a_{-m-7}
&= \frac{(m+1)(m+2)(m+3)(m+4)(m+5)(m+6)}
        {2.4.6.(2m+3)(2m+5)(2m+7)} a_{-m-1}, \quad\text{\&c.}
\end{align*}
$a_{-m-1}$ may be taken at pleasure, and is usually taken as $\dfrac{m!}{1.3.5.\cdots(2m+1)}$,
and $z=Q_m(x)$ where
\begin{align*}
Q_m(x) &= \frac{m!}{(2m+1)(2m-1)\cdots1}
 \left[ \frac{1}{x^{m+1}}
      + \frac{(m+1)(m+2)}{2.(2m+3)} \frac{1}{x^{m+3}} \right.
\\[1ex]
&+ \left. \frac{(m+1)(m+2)(m+3)(m+4)}
               {2.4.(2m+3)(2m+5)}
          \frac{1}{x^{m+5}} + \cdots \right]  \tag{9}
\end{align*}
is a second particular solution of Legendre's Equation, provided the series is
convergent. $Q_m(x)$ is called a \emph{Surface Zonal Harmonic} of the \emph{second kind}.
% -----File: 033.png
It is easily seen to be convergent if $x < - 1$ or $x > 1$, and divergent if
$- 1 < x < 1$.

Hence if $m$ is a positive integer,
\[
z = A P_m(x) + B Q_m(x)           \tag{10}
\]
is the general solution of Legendre's Equation if $x < - 1$ or $x > 1$.

We have seen that for $- 1 < x < 1$
\begin{align*}
  P_m(x)
&= (-1)^\frac{m}{2}
   \frac{ \Gamma (m + 1) }
        { 2^m\left[\Gamma\left( \dfrac{m}{2} + 1 \right) \right]^2 }
   p_m(x)     \tag{11}
\intertext{if $m$ is an even integer, and }
  P_m(x)
&= (- 1)^\frac{m-1}{2}
   \frac{ \Gamma (m + 1) }
        { 2^{m-1}\Big[\Gamma\Big( \dfrac{m+1}{2} \Big) \Big]^2 }
   q_m(x)     \tag{12}
\end{align*}
if $m$ is an odd integer.

If now we define $Q_m(x)$ as follows when $- 1 < x < 1$
\begin{align*}
  Q_m(x)
&= (-1)^\frac{m+1}{2}
   \frac{ 2^{m-1} \Big[\Gamma\Big( \dfrac{m+1}{ 2} \Big)\Big]^2 }
        { \Gamma(m+1) } p_m (x)     \tag{13}
\intertext{if $m$ is an odd integer, and }
  Q_m(x)
&= (-1)^\frac{m}{2}
   \frac{ 2^m \Big[\Gamma\Big( \dfrac{m}{2}+1 \Big) \Big]^2 }
        { \Gamma(m+1) } q_m(x)     \tag{14}
\end{align*}
if $m$ is an even integer, then (10) will be the general solution of Legendre's
Equation if $m$ is a positive integer when $-1 < x < 1$, as well as when $x < -1$
or $x > 1$.

\mypara{17.} Let us last consider the equation
\[
\frac{d^2z}{dx^2} + \frac{1}{x} \frac{dz}{dx} + \left (1 - \frac{m^2}{ x^2} \right ) z = 0  \tag{1}
\]
which is known as Bessel's Equation, and which reduces to (8) Art.~11,
that is, to
\[
\frac{d^2 z}{ dx^2} + \frac{1}{x} \frac{dz}{dx} + z = 0
\]
when $m$ = 0;\footnote
  {This equation was first studied by Fourier in considering the cooling of a cylinder. We
  shall designate it as ``Fourier's Equation.''}
(1) can be simplified by a change of the dependent variable.
% -----File: 034.png
Let $z = x^m v$ and we get
\[
\frac{d^2 v}{ dx^2} + \frac{2m +1 }{ x}\frac{ dv}{dx} + v = 0          \tag{2}
\]
to determine $v$.

Assume $v = \sum a_n x^n$, and substitute in (2). We get
\[
  \textstyle\sum [n(2m + n)a_n x^{n-2} + a_n x^n] = 0;
\]
\begin{flalign*}
&\text{whence }&  &a_{n-2} = -n(2m + n)a_n.  &&\phantom{whence }
\end{flalign*}
If we begin with $n = 0$, then $a_{n-2} = 0$, $a_{n-4} = 0$, \&c., and we have the
set of values
\begin{align*}
a_2 &= - \frac{a_0}{ 2(2m+2)} = - \frac{a_0}{ 2^2 (m+1)}\\[1ex]
a_4 &= \frac{a_0}{ 2.4(2m + 2)(2m + 4)}
     = \frac{a_0}{ 2^4.2!(m + 1)(m + 2)}
\end{align*}\\[-3ex]
\[
a_6 = - \frac{a_0} {2.4.6(2m + 2)(2m + 4)(2m + 6)}
    = - \frac{a_0}{ 2^6.3!(m + 1)(m + 2)(m + 3)};
\]\\[-2ex]
\begin{flalign*}
&\text{whence }&  z = a_0 x^m \left[ 1 \vphantom{\dfrac{x^2}{(0)}}\right.
&- \frac{x^2}{ 2^2 (m+1)}  +  \frac{x^4}{ 2^4.2!(m + 1)(m + 2)}  &&\\[1ex]
&&  &- \left. \frac{x^6}{ 2^6.3!(m + 1)(m + 2)(m + 3)}  +  \cdots  \right ] &&  \tag{3}
\end{flalign*}
is a solution of Bessel's Equation. $a_0$ is usually taken as $\dfrac{1}{2^m m!}$ if $m$ is a positive
integer, or as $\dfrac{1}{ 2^m \Gamma (m + 1)}$ if $m$ is unrestricted in value, and the second
member of (3) is represented by $J_m(x)$ and is called a \textit{Bessel's Function} of the
$m$th order, or a \textit{Cylindrical Harmonic} of the $m$th order.

If $m = 0$, $J_m(x)$ becomes $J_0(x)$ and is the value of $z$ obtained in Art.~11
as the solution of equation~(8) of that article.

If in equation (1) we substitute $x^{-m }v$ in place of $x^m v$ for $z$, we get in place
of (2)  the equation
\[
\frac{d^2 v}{ dx^2} + \frac{1-2m}{x} \frac{dv}{dx} + v = 0
\]
and in place of (3)
\begin{align*}
z = a_0 x^{-m} \left[ 1 \vphantom{\dfrac{x^2}{(0)}}\right.
&- \frac{x^2}{ 2^2 (1-m)}  +  \frac{x^4}{ 2^4.2!(1-m)(2-m)} \\[1ex]
&- \left. \frac{x^6}{ 2^6.3!(1 - m)(2 - m)(3 - m)}  + \cdots \right ]    \tag{4}
\end{align*}
% -----File: 035.png

If $a_0$ is taken equal to $\dfrac{1}{2^{-m} \Gamma (1 - m)}$  the second member of (4) is the same
function of $-m$ and $x$ that $J_m (x)$ is of $+m$ and $x$ and may be written
$J_{-m}(x)$.
\begin{flalign*}
&\text{\indent Therefore }&
  z &= A J_m(x) + B J_{-m}(x)  &&\phantom{Therefore } \tag{5}
\end{flalign*}
is the general solution of (1) unless $J_m(x)$ and $J_{-m}(x)$ should prove not to be
independent.

It is easily seen that when $m = 0$, $J_{-m}(x)$ and $J_m(x)$ become identical
and (5) reduces to
\[
z = (A + B) J_0(x)
\]
and contains but a single arbitrary constant and is not the general solution of
Fourier's Equation~(8) Art.~(11).

It can be shown that $J_{-m} (x) = (-1)^m J_m(x)$ whenever $m$ is an integer,
and consequently that the solution~(5) is general only when $m$ if real is fractional
or incommensurable.

The general solution for the important case where $m = 0$ is, however, easily
obtained. Let $F(m,x)$ be the value which the second member of (3) assumes
when $a_0 = 1$; then the value which the second member of (4) assumes
when $a_0 = 1$ will be $F(-m,x)$, and it has been shown that $z = F(m,x)$ and
$z = F(-m,x)$ are solutions of Bessel's Equation; $z = F(m,x) - F(-m,x)$
is, then, a solution, as is also
\[
  z = \frac{F(m,x) - F(-m,x)}{ 2m},        \tag{6}
\]
but the limiting value which  $\dfrac{F(m,x) - F(-m,x)}{ 2m}$  approaches as $m$ approaches
zero is $[D_m F(m,x)]_{m=0}$ and consequently
\[
z=[D_m F(m,x)]_{m=0} \tag{7}
\]
is a solution of the equation
\[
  \frac{d^2 z}{dx^2} + \frac{1}{x} \frac{dz}{dx} + z = 0, \tag{8}
\]
and the general solution of (8) is
\begin{align*}
  z = {}& A J_0(x) + B[D_m F(m,x)]_{m=0}.
\\[1ex]
  F(m,x) = x^m \bigg[ 1
&- \frac{x^2}{ 2^2 (m + 1)}
 + \frac{x^4}{ 2^4.2! (m + 1)(m + 2)}
\\[1ex]
&- \frac{x^6}{2^6.3!(m+1)(m+2)(m+3)} + \cdots \bigg]\\
% -----File: 036.png
D_mF(m,x) = x^m\log x \bigg[ 1&-\frac{x^2}{2^2(m+1)} + \frac{x^4}{2^4.2!(m+1)(m+2)} - \cdots \bigg] & \\
           +x^mD_m  \bigg[ 1 &- \frac{x^2}{2^2(m+1)} + \frac{x^4}{2^4.2!(m+1)(m+2)} + \cdots \bigg].&
\end{align*}
The general term of the last parenthesis can be written
\[
(-1)^k\frac{x^{2k}}{2^{2k}.k!(m+1)(m+2)\cdots(m+k)},
\]
and its partial derivative with respect to $m$ is
\begin{gather*}
(-1)^k\frac{x^{2k}}{2^{2k}.k!} D_m\frac{1}{(m+1)(m+2)\cdots(m+k)}.\\
\log\frac{1}{(m+1)(m+2)\cdots(m+k)} =
\begin{aligned}[t]
  -[ &\log(m+1)+\log(m+2)+\cdots\\
     &+ \log(m+k)].
\end{aligned}
\end{gather*}

Take the $D_m$ of both members and we have
\begin{align*}
D_m & \frac{1}{(m+1)(m+2)\cdots(m+k)}\\
    &=-\frac{1}{(m+1)(m+2)\cdots(m+k)}
       \left[ \frac{1}{m+1} + \frac{1}{m+2} + \cdots\frac{1}{m+k} \right].
\\
D_m\left[1\vphantom{\frac{x^2}{(0)}}\right. & -\frac{x^2}{2^2(m+1)} + \frac{x^4}{2^4.2!(m+1)(m+2)} - \frac{x^6}{2^6.3!(m+1)(m+2)(m+3)}
\\
& +\cdots\left.\vphantom{\frac{x^2}{(0)}}\right] = \frac{x^2}{2^2}\frac{1}{(m+1)^2} - \frac{x^4}{2^4.2!}\frac{1}{(m+1)(m+2)} \left[ \frac{1}{m+1} + \frac{1}{m+2} \right]
\\
& +\frac{x^6}{2^6.3!}\frac{1}{(m+1)(m+2)(m+3)} \left[ \frac{1}{m+1} + \frac{1}{m+2} + \frac{1}{m+3}\right] + \cdots
\end{align*}
and we have
% recast to fit line
\begin{align*}
[D_mF(m,x)]_{m=0} &= J_0(x)\log x + \frac{x^2}{2^2(1!)^2}\frac{1}{1} - \frac{x^4}{2^4(2!)^2} \left(\frac{1}{1} + \frac{1}{2}\right)
\\ &+ \frac{x^6}{2^6(3!)^2} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \right)
-\frac{x^8}{2^8(4!)^2} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) + \cdots;
\end{align*}
\begin{flalign*}
&\text{and}  && \hspace{6em} z = AJ_0(x) + BK_0(x),   \tag{9}
\\
% recast to fit line
&\text{where} & K_0&(x) = J_0(x)\log x + \frac{x^2}{2^2} - \frac{x^4}{2^4(2!)^2} \left( \frac{1}{1}+\frac{1}{2}\right)
\\
&&&  + \frac{x^6}{2^6(3!)^2}\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3}\right)
-\frac{x^8}{2^8(4!)^2} \left( \frac{1}{1} + \frac{1}{2} +\frac{1}{3} + \frac{1}{4} \right) + \cdots \tag{10}
\end{flalign*}
is the general solution of Fourier's Equation (8).

$K_0(x)$ is known as a \emph{Bessel's Function} of the \emph{Second Kind}.
% -----File: 037.png

\mypara{18.} It is worth while to confirm the results of the last few articles by
getting the general solutions of the equations in question by a different and
familiar method.

The general solution of any ordinary linear differential equation of the
second order can be obtained when a particular solution of the equation has
been found [v.\ Int.\ Cal.\ p.~321, §~24~(\emph{a})].

The most general form of a homogeneous ordinary linear differential equation
of the second order is
\[
\frac{d^2y}{dx^2} + P \frac{dy}{dx} + Qy = 0 \tag{1}
\]
where $P$ and $Q$ are functions of $x$. Suppose that
\[
y = v       \tag{2}
\]
is a particular solution of (1). Substitute $y = vz$ in (1) and we get
\[
v\frac{d^2z}{dx^2}+\left(2\frac{dv}{dx} + Pv\right)\frac{dz}{dx} = 0. \tag{3}
\]
Call  $\dfrac{dz}{dx} = z'$. Then (3) becomes
\[
v\frac{dz'}{dx}+\left(2\frac{dv}{dx}+ Pv\right)z'=0,  \tag{4}
\]
a differential equation of the first order in which the variables can be separated.
Multiply by $dx$ and divide by $vz'$ and (4) reduces to
\[
\frac{dz'}{z'} + 2\frac{dv}{v} + Pdx = 0.
\]
Integrate and we have
\begin{flalign*}
&&\log{}&z' + \log v^2 + \int Pdx = C
\\
&\text{or } &   z'v^2 &= e^{C-\int\!Pdx}=Be^{-\int\!Pdx},  &&\phantom{or }\\
&&z'&=\frac{dz}{dx}=B\frac{e^{-\int\!Pdx}}{v^2},&\\
&&z &=A+B\int\frac{e^{-\int\!Pdx}}{v^2}dx;&\\
&\text{and }   & y=&\ v\bigg(A+B\int \frac{e^{-\int\!Pdx}}{v^2}dx\bigg) && \tag{5}
\end{flalign*}
is the general solution of (1), the only arbitrary constants in the second member
of (5) being those explicitly written, namely, $A$ and $B$.\\

(\emph{a})\quad Apply this formula to (1) Art.~14,
\[
\frac{d^2z}{dx^2} + \alpha^2z=0;        \tag{1}
\]
% -----File: 038.png
given: $z=\cos\alpha x$, as a particular solution. Substituting in (5) we have
since $P=0$
\begin{align*}
z &= \cos\alpha x\left(A+B\int\frac{dx}{\cos^2\alpha x}\right)  \\
  &= \cos\alpha x\left(A+\frac{B}{\alpha}\tan\alpha x\right)   \\
  &= A\cos\alpha x + B_1\sin\alpha x,  \tag{2}
\end{align*}
as the general solution of (1), and this agrees perfectly with (5) Art.~14.\\

(\emph{b})\quad  Take equation (1) Art.~15.
\[
x^2\frac{d^2z}{dx^2}+2x\frac{dz}{dx}-m(m+1)z=0;  \tag{1}
\]
given: $z=x^m$, as a particular solution.

\noindent Here $P=\dfrac{2}{x}$, $\displaystyle\int Pdx=2\log x =\log x^2$, and $e^{-\int\!Pdx}=\dfrac{1}{x^2}$. Hence by (5)
\[
z = x^m \left( A + B\int\frac{dx}{x^{2m+2}} \right)
  = x^m \left( A + \frac{B}{-2m-1} \frac{1}{x^{2m+1}} \right),\\[-3ex]
\]
\begin{flalign*}
&\text{that is} &  z= &\ Ax^m+\frac{B_1}{x^{m+1}}  &&\phantom{that is} \tag{2}
\end{flalign*}
is the general solution of (1), and agrees with (2) Art.~15.\\

(\emph{c})\quad Take Legendre's Equation, (2) Art.~16.
\[
(1-x^2)\frac{d^2z}{dx^2}-2x\frac{dz}{dx}+m(m+1)z=0;  \tag{1}
\]
given: $z=P_m(x)$, as a particular solution.

Here $P=\dfrac{-2x}{1-x^2}$, $\displaystyle\int Pdx=\log(1- x^2)$, and $e^{-\int\!Pdx}=\dfrac{1}{1-x^2}$.
\begin{flalign*}
&\text{Hence by (5)} &&
  z = P_m(x) \left( A + B\int\frac{dx}{(1-x^2)[P_m(x)]^2} \right)  &&\phantom{Hence by }  \tag{2}
\end{flalign*}
is the general solution of (1) and must agree with (10) Art.~16, if $m$ is an
integer, and therefore
\[
Q_m(x)=CP_m(x)\int\frac{dx}{(1-x^2)[P_m(x)]^2}    \tag{3}
\]
where $C$ is as yet undetermined, and no constant term is to be understood with
the integral in the second member.\\

(\emph{d})\quad  Take Bessel's Equation, (1) Art.~17.
\[
\frac{d^2z}{dx^2} + \frac{1}{x}\frac{dz}{dx} + \left(1-\frac{m^2}{x^2}\right)z = 0; \tag{1}
\]
given: $z=J_m(x)$, as a particular solution.
% -----File: 039.png

Here $P=\dfrac{1}{x}$, $\displaystyle\int Pdx=\log x$, and $e^{-\int\!Pdx}=\dfrac{1}{x}$. Hence by (5)
\[
z=J_m(x)\left(A+B\int\frac{dx}{x[J_m(x)]^2}\right)  \tag{2}
\]
is the general solution of Bessel's Equation.

If $m=0$ (2) becomes
\[
z=J_0(x)\left(A+B\int\frac{dx}{x[J_0(x)]^2}\right)  \tag{3}
\]
and must agree with (9) Art.~17. Therefore
\[
K_0(x)=CJ_0(x)\int\frac{dx}{x[J_0(x)]^2}, \tag{4}
\]
where $C$ is at present undetermined, and no constant term is to be taken
with the integral.

The first considerable subject suggested by the problems which we have
taken up in this introductory chapter is that of development in Trigonometric
Series (v.\ Arts.~7 and 8).
\label{ch1end}
% -----File: 040.png

\mychap{CHAPTER II.}{DEVELOPMENT IN TRIGONOMETRIC SERIES.}
\label{ch2start}

\mypara{19.} We have seen in Chapter I. that it is sometimes important to be able
to express a given function of a variable $x$, in terms of the sines or of the
cosines of multiples of $x$. The problem in its general form was first solved
by Fourier in his ``Analytic Theory of Heat'' (1822), and its solution plays a
very important part in most branches of modern Physics. Series involving
only sines and cosines of whole multiples of $x$, that is series of the form
\[
b_0 + b_1\cos x + b_2\cos 2x + \cdots + a_1\sin x + a_2\sin 2x + \cdots
\]
are generally known as Fourier's series.

Let us endeavor to develop a given function of $x$ in terms of $\sin x$, $\sin 2x$,
$\sin 3x$, \&c., in such a way that the function and the series shall be equal for
all values of $x$ between $x=0$ and $x=\pi$.

To fix our ideas let us suppose that we have a curve,
\[
y=f(x),
\]
given, and that we wish to form the equation,
\[
y = a_1\sin x + a_2\sin 2x + a_3\sin 3x + \cdots,
\]
of a curve which shall coincide with so much of the given curve as lies between
the points corresponding to $x=0$ and $x=\pi$.
It is clear that in the equation
\[
y = a_1 \sin x   \tag{1}
\]
$a_1$ may be determined so that the curve represented shall pass through any
given point. For if we substitute in (1) the coördinates of the point in question
we shall have an equation of the first degree in which $a_1$ is the only
unknown quantity and which will therefore give us one and only one value
for $a_1$.

In like manner the curve
\[
y = a_1 \sin x + a_2 \sin 2x
\]
may be made to pass through any two arbitrarily chosen points whose abscissas
lie between 0 and $\pi$ provided that the abscissas are not equal; and
\[
y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots + a_n \sin nx
\]
may be made to pass through any $n$ arbitrarily chosen points whose abscissas
lie between 0 and $\pi$ provided as before that their abscissas are all different.

If, then, the given function $f(x)$ is of such a character that for each value of $x$
between $x=0$ and $x=\pi$ it has one and only one value, and if between
$x=0$ and $x=\pi$ it is finite and continuous, or if discontinuous has only
\emph{finite discontinuities} (v.\ Int.\ Cal.\ Art.~83, p.~78), the coefficients in
\[
y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots + a_n \sin nx    \tag{2}
\]
% -----File: 041.png
can be determined so that the curve represented by (2) will pass through any
$n$ arbitrarily chosen points of the curve
\[
y = f(x)              \tag{3}
\]
whose abscissas lie between $0$ and $\pi$ and are all different, and these coefficients
will have but one set of values.

For the sake of simplicity suppose that the $n$ points are so chosen that their
projections on the axis of $X$ are equidistant.

\markright{DEVELOPMENT IN TRIGONOMETRIC SERIES.}
Call $\dfrac{\pi}{ n+1} = \Delta x$; then the coördinates of the $n$ points will be $[\Delta x,f(\Delta x)],$
$[2\Delta x,f(2\Delta x)]$, $[3\Delta x,f(3\Delta x)]$, $\cdots\ [n\Delta x,f(n\Delta x)]$. Substitute them in (2) and
we have
\begin{equation*}
\left.\begin{alignedat}{5}
f( \Delta x) &= {}& a_1 \sin  \Delta x &+{}& a_2 \sin 2 \Delta x &+{}
& a_3 \sin 3 \Delta x &+ \cdots +{}& a_n \sin n \Delta x\phantom{,} &
\\
f(2\Delta x) &= {}& a_1 \sin 2\Delta x &+{}& a_2 \sin 4 \Delta x &+{}
& a_3 \sin 6 \Delta x &+ \cdots +{}& a_n \sin 2n \Delta x\phantom{,} &
\\
f(3\Delta x) &= {}& a_1 \sin 3\Delta x &+{}& a_2 \sin 6 \Delta x &+{}
& a_3 \sin 9 \Delta x &+ \cdots +{}& a_n \sin 3n \Delta x\phantom{,} &
\\
  \vdots  \quad && \vdots \qquad && \vdots \qquad &
& \vdots \qquad && \vdots \qquad &
\\
f(n\Delta x) &= {}& a_1 \sin n\Delta x &+{}& a_2 \sin 2n \Delta x &+{}
& a_3 \sin 3n \Delta x &+ \cdots +{}& a_n \sin n^2 \Delta x, &
\end{alignedat}
\right\}
\tag{4}
\end{equation*}
$n$ equations of the first degree to determine the $n$ coefficients $a_1$, $a_2$, $a_3$, $\cdots$ $a_n$.

Not only can equations (4) be solved in theory, but they can be actually
solved in any given case by a very simple and ingenious method due to
Lagrange.

Let us take as an example the simple problem to determine the coefficients
$a_1$, $a_2$, $a_3$, $a_4$, and $a_5$, so that
\[
y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + a_4 \sin 4x + a_5 \sin 5x   \tag{5}
\]
shall pass through the five points of the line
\[
y = x
\]
which have the abscissas $\dfrac{\pi}{6}$, $\dfrac{2\pi}{6}$, $\dfrac{3\pi}{6}$, $\dfrac{4\pi}{6}$, and $\dfrac{5\pi}{6}$, $\dfrac{\pi}{6}$ here being $\Delta x$.

We must now solve the equations
\begin{equation*}
\left.\begin{alignedat}{6}
\frac{\pi}{6}\: &= {}& a_1 \sin \:\frac{ \pi}{6}\: &+{}
& a_2 \sin \:\frac{2\pi}{6}\: &+{}& a_3 \sin \:\frac{3\pi}{6}\: &+{}
& a_4 \sin \:\frac{4\pi}{6}\: &+{}& a_5 \sin \:\frac{5\pi}{6}\:\phantom{.}
\\
\frac{2\pi}{6} &= & a_1 \sin \frac{2\pi}{6} &+{}
& a_2 \sin \:\frac{4\pi}{6}\: &+{}& a_3 \sin \:\frac{6\pi}{6}\: &+{}
& a_4 \sin \:\frac{8\pi}{6}\: &+{}& a_5 \sin \frac{10\pi}{6}\phantom{.}
\\
\frac{3\pi}{6} &= & a_1 \sin \frac{3\pi}{6} &+{}
& a_2 \sin \:\frac{ 6\pi}{6}\: &+{}& a_3 \sin \:\frac{ 9\pi}{6}\: &+{}
& a_4 \sin \frac{12\pi}{6} &+{}& a_5 \sin \frac{15\pi}{6}\phantom{.}
\\
\frac{4\pi}{6} &= {}& a_1 \sin \frac{4\pi}{6} &+{}
& a_2 \sin \:\frac{ 8\pi}{6}\: &+{}& a_3 \sin \frac{12\pi}{6} &+{}
& a_4 \sin \frac{16\pi}{6} &+{}& a_5 \sin \frac{20\pi}{6}\phantom{.}
\\
\frac{5\pi}{6} &= {}& a_1 \sin \frac{5\pi}{6} &+{}
& a_2 \sin \frac{10\pi}{6} &+{}& a_3 \sin \frac{15\pi}{6} &+{}
& a_4 \sin \frac{20\pi}{6} &+{}& a_5 \sin \frac{25\pi}{6}.
\end{alignedat}
\right\}
\tag{6}
\end{equation*}
% -----File: 042.png

Multiply the first equation by $2 \sin \dfrac{\pi}{6}$, the second by $2 \sin \dfrac{2 \pi }{6}$, the third
by $2\sin\dfrac{3 \pi}{6}$, the fourth by $2\sin\dfrac{4\pi}{6}$, the fifth by $2 \sin \dfrac{5 \pi }{6}$ and add the
equations.

The coefficient of $a_2$ is
\begin{flalign*}
&&2 \sin \frac{\pi}{6} \sin \frac{2 \pi}{ 6} &+ 2 \sin \frac{2 \pi}{6} \sin \frac{4 \pi}{6} + 2 \sin \frac{3 \pi }{6} \sin \frac{6 \pi}{6} + 2 \sin \frac{4 \pi}{6} \sin \frac{8 \pi}{6} & \phantom{\text{but}}\\
&&&+ 2 \sin \frac{5 \pi}{6} \sin \frac{10 \pi}{6}; \\
&\text{but}  && \qquad 2 \sin \dfrac{\pi }{6} \sin \dfrac{2 \pi}{6} = \cos \dfrac{\pi}{6} - \cos \dfrac{3 \pi }{6}, \text{\&c.}
\end{flalign*}

Hence the coefficient of $a_2$ becomes
\begin{equation*}
\left.\begin{alignedat}{5}
&\cos \:\frac{\pi}{6} &+{}& \cos \frac{2 \pi}{6} &+{}& \cos \frac{3 \pi}{6} &+{}& \cos \:\frac{4 \pi}{6} &+{}& \cos \:\frac{5 \pi}{6}\\
-{}&\cos \frac{3 \pi}{6} &-{}& \cos \frac{6 \pi}{6} &-{}& \cos \frac{9 \pi}{6} &-{}& \cos \frac{12 \pi}{6} &-{}& \cos \frac{15 \pi}{6}
\end{alignedat}
\right\}
\tag{7}
\end{equation*}
and this may be reduced by the aid of an important Trigonometric formula
which we proceed to establish.

\mypara{20.} \textsc{Lemma.}
\begin{align*}
\cos \theta &+ \cos 2 \theta + \cos 3 \theta + \cdots + \cos n \theta = - \frac{1}{2} + \frac{1}{2} \cfrac{\sin(2n+1) \dfrac{\theta}{2}}{\sin \dfrac{\theta}{2}}.  \tag{1}
\end{align*}
For let $S=\cos \theta  + \cos 2 \theta + \cos 3 \theta + \cdots + \cos n \theta$ and multiply by $2 \cos \theta$.
\begin{align*}
2S \cos \theta &= 2 \cos^2 \theta + 2 \cos \theta \cos 2 \theta + 2 \cos \theta \cos 3 \theta + \cdots + 2 \cos \theta \cos n \theta&&\hspace{1em}\\
&= 1 + \cos \theta + \cos 2 \theta + \cdots + \cos (n - 1) \theta\\
&\hspace{1.85em}+ \cos 2\theta + \cos 3\theta + \cos 4\theta + \cdots + \cos (n + 1)\theta\\
&= 2S + 1 + \cos (n + 1) \theta - \cos \theta - \cos n \theta.  \qquad \text{Hence}\\[-5.5ex]
\end{align*}
\begin{flalign*}
&&  S &= - \frac{1}{2} + \frac{\cos n \theta - \cos (n + 1) \theta }{ 2(1- \cos \theta )} &&  \\
&\text{or }& S &= - \frac{1}{2} + \frac{1}{2} \cfrac{\sin (2n + 1) \dfrac{\theta}{2}}{ \sin \dfrac{\theta}{ 2 }}. && \tag*{\scriptsize Q.E.D.}
\end{flalign*}
% -----File: 043.png

\mypara{21.} Applying (1) Art.~20 to (7) Art.~19 the coefficient of $a_2$ reduces to
\[
  \cfrac{\sin\dfrac{11\pi}{12}}{2\sin\dfrac{ \pi}{12}}
- \cfrac{\sin\dfrac{33\pi}{12}}{2\sin\dfrac{3\pi}{12}};
\]
\begin{flalign*}
&\text{but}  && \frac{11\pi}{12} = \pi - \frac{\pi}{12},\text{ and }\frac{33\pi}{12} = 3\pi - \frac{3\pi}{12}; &&\phantom{\text{but}}
\end{flalign*}
\begin{flalign*}
&\text{therefore}  && \frac{\sin \Big( \pi- \dfrac{\pi}{12} \Big) }{2\sin\dfrac{\pi}{12}}
- \frac{\sin \Big(3\pi- \dfrac{3\pi}{12} \Big) }{2\sin\dfrac{3\pi}{12}} = \frac12 - \frac12 = 0, &&\phantom{\text{therefore}}
\end{flalign*}
and $a_2$ vanishes.

In like manner it may be shown that the coefficients of $a_3$, $a_4$, and $a_5$
vanish.

The coefficient of $a_1$ is
\begin{align*}
  &\ 2\sin^2 \frac{ \pi}{6} + 2\sin^2 \frac{2\pi}{6}
   + 2\sin^2 \frac{3\pi}{6} + 2\sin^2 \frac{4\pi}{6}
   + 2\sin^2 \frac{5\pi}{6}
\\[1ex]
= {}& \makebox[20.5em]{1 \hfill + \hfill 1 \hfill + \hfill 1 \hfill + \hfill 1 \hfill + \hfill 1}
\\
  &\!\!- \cos \frac{2\pi}{6} - \cos \frac{4\pi}{6}
   - \cos \frac{6\pi}{6} - \cos \frac{8\pi}{6}
   -\cos \frac{10\pi}{6}
\\[1ex]
= {}& 5 + \frac12
    - \cfrac{\sin \dfrac{11\pi}{6}}{2 \sin \cfrac{\pi}{6}}
= 5\tfrac12 - \cfrac{ \sin \left( 2\pi-\dfrac{\pi}{6} \right) }
                    { 2\sin \cfrac{\pi}{6} } = 6.
\end{align*}

The first member of the final equation is
\begin{gather*}
    \frac{2\pi}{6} \sin \frac{ \pi}{6}
+ 2 \frac{2\pi}{6} \sin \frac{2\pi}{6}
+ 2 \frac{3\pi}{6} \sin \frac{3\pi}{6}
+ 2 \frac{4\pi}{6} \sin \frac{4\pi}{6}
+ 2 \frac{5\pi}{6} \sin \frac{5\pi}{6}. \quad\text{Hence}\\
  a_1 = \frac26 \sum_{k=1}^{k=5} \frac{k\pi}{6} \sin \frac{k\pi}{6}
= \frac{\pi}{6} (2 + \surd3) = 2\text{\quad  approximately.}
\end{gather*}

If we multiply the first equation of (6) Art.~19 by $2\sin\dfrac{2\pi}{6}$, the second by
$2\sin\dfrac{4\pi}{6}$, the third by $2\sin\dfrac{6\pi}{6}$, the fourth by $2\sin\dfrac{8\pi}{6}$,  the fifth
by $2\sin\dfrac{10\pi}{6}$, add and reduce as before we shall find
\begin{align*}
  a_2 &= \frac26 \sum_{k=1}^{k=5} \frac{k\pi}{6} \sin \frac{2k\pi}{6}
= - \frac{\pi}{6} \surd3 = -0.9;
% -----File: 044.png
\intertext{and in like manner we get}
a_3 &= \frac{2}{6}\sum_{k=1}^{k=5} \frac{k\pi}{6}\sin\frac{3k\pi}{6}
= \frac{\pi}{6} = 0.5\\
a_4 &= \frac{2}{6}\sum_{k=1}^{k=5} \frac{k\pi}{6}\sin\frac{4k\pi}{6}
=-\frac{\pi\surd3}{18} = -0.3\\
a_5 &= \frac{2}{6}\sum_{k=1}^{k=5} \frac{k\pi}{6}\sin\frac{5k\pi}{6}
= \frac{\pi}{6}(2-\surd3) = 0.1.
\end{align*}

Therefore
\[
y=2\sin x - 0.9\sin 2x + 0.5\sin 3x - 0.3\sin 4x + 0.1\sin 5x  \tag{1}
\]
cuts the curve $y=x$ at the five points whose abscissas are $\dfrac{\pi}{6}$, $\dfrac{2\pi}{6}$, $\dfrac{3\pi}{6}$,
$\dfrac{4\pi}{6}$, and $\dfrac{5\pi}{6}$.

\mypara{22.} The equations (4) Art.~19 can be solved by exactly the same device.
To find any coefficient $a_m$ multiply the first equation by $2\sin m\Delta x$, the
second by $2\sin 2m\Delta x$, the third by $2\sin 3m\Delta x$, \&c.\ and add.

The coefficient of any other $a$ as $a_k$ in the resulting equation will be
\begin{align*}
  2&\sin k\Delta x\sin m\Delta x
+ 2\sin 2k\Delta x\sin 2m\Delta x
+ 2\sin 3k \Delta x\sin 3m\Delta x + \cdots\\
&\qquad+ 2\sin nk\Delta x\sin nm\Delta x\\
={}& \cos(m-k)\Delta x + \cos 2(m-k)\Delta x + \cos 3(m-k)\Delta x + \cdots + \cos n(m-k)\Delta x\\
-{}&\cos(m+k)\Delta x - \cos 2(m+k)\Delta x - \cos 3(m+k)\Delta x - \cdots - \cos n(m+k)\Delta x\\
={}& \cfrac{\sin\dfrac{2n+1}{2}(m-k)\Delta x}{2\sin\dfrac{(m-k)\Delta x}{2}}
-\cfrac{\sin\dfrac{2n+1}{2}(m+k)\Delta x}{2\sin\dfrac{(m+k)\Delta x}{2}};
\text{\qquad by (1) Art.~20.}
\end{align*}
\[
\frac{2n+1}{2}=n+1-\frac{1}{2} \quad \text{and} \quad (n + 1)\Delta x = \pi .
\]
Hence the coefficient of $a_k$ may be written
\[
 \cfrac{ \sin \Big[ (m-k)\pi - \dfrac{(m-k)\Delta x}{2} \Big] }
       {2\sin\dfrac{(m-k)\Delta x}{2} }
-\cfrac{ \sin \Big[ (m+k)\pi - \dfrac{(m+k)\Delta x}{2} \Big] }
       {2\sin\dfrac{(m+k)\Delta x}{2} }
\]
but this is equal to $\dfrac{1}{2}-\dfrac{1}{2}$ or $-\dfrac{1}{2}+\dfrac{1}{2}$ according as $m-k$ is odd or even
and so is zero in either case.
% -----File: 045.png

The coefficient of $a_m$ will be
\begin{align*}
&\;2\sin^2m\Delta x + 2\sin^22m\Delta x +
2\sin^23m\Delta x + \cdots + 2\sin^2nm\Delta x\\
&= \qquad 1 \qquad + \qquad 1 \qquad + \qquad 1 \qquad + \quad \cdots\quad  + \qquad 1\\
&\quad-\cos2m\Delta x - \cos4m\Delta x - \cos6m\Delta x - \;\cdots\; - \cos2nm\Delta x\\
&=n+\frac{1}{2}-\frac{\sin(2n+1)m\Delta x}{2\sin m\Delta x}, \text{ by (1) Art.~20.}
\end{align*}
\begin{flalign*}
&\text{\indent But } &
  (2n+1)m\Delta x = 2m(n & +1)\Delta x-m\Delta x = 2m\pi-m\Delta x,  &&\phantom{therefore }
\\
&\text{therefore } &
  \frac{\sin(2n+1)m\Delta x}{2\sin m\Delta x} &
= \frac{\sin(2m\pi-m\Delta x)}{2\sin m\Delta x} = -\frac{1}{2},  &&
\end{flalign*}
and the coefficient of $a_m$ is $n+1$.

The first member of our final equation will be
\[
2\sum_{k=1}^{k=n}f(k\Delta x)\sin km\Delta x.
\]

Hence
\[
a_m=\frac{2}{n+1}\sum_{k=1}^{k=n}f(k\Delta x)\sin km\Delta x,  \tag{1}
\]
and the curve
\[
y = a_1\sin x + a_2\sin 2x + \cdots + a_n\sin nx,     \tag{2}
\]
where the coefficients are given by (1) will pass through the $n$ points of the
curve $y =f(x)$ whose abscissas are $\Delta x$, $2\Delta x$, $3\Delta x$, $\cdots$ $n\Delta x$. $\Delta x$ being $\dfrac{\pi}{n+1}$.

It should be noted that since the $n$ equations (4) Art.~19 are all of the first
degree there will exist only one set of values for the $n$ quantities $a_1$, $a_2$, $a_3$,
$\cdots$ $a_n$ that can satisfy these equations. Consequently the solution which we
have obtained is the only solution possible.

\mypara{23.} The result just obtained obviously holds good no matter how great a
value of $n$ may be taken.

If now we suppose $n$ indefinitely increased the two curves (2) Art.~22 and
$y=f(x)$ will come nearer and nearer to coinciding throughout the whole of
their portions between $x=0$ and $x=\pi$, and consequently the limiting
form that equation (2) Art.~22 approaches as $n$ is indefinitely increased will
represent a curve absolutely coinciding between the values of $x$ in question
with $y=f(x)$.
% -----File: 046.png

Let us see what limiting value $a_m$ approaches as $n$ is indefinitely increased.
\begin{align*}
\phantom{(1) Art.~22.}a_m &=\frac{2}{n+1}\sum_{k=1}^{k=n}f(k\Delta x)\sin km\Delta x \tag*{(1) Art.~22.}\\
    &=\frac{2\Delta x}{\pi}\sum_{k=1}^{k=n}f(k\Delta x)\sin km\Delta x
\end{align*}
% recast to fit line
\begin{flalign*}
= \frac{2}{\pi} \left[
\begin{alignedat}{1}
 f( \Delta x)\sin  m\Delta x.\Delta x
 + f(2\Delta x)\sin 2m\Delta x.\Delta x & + \cdots  \\
 & + f(n\Delta x)\sin nm\Delta x.\Delta x
\end{alignedat}
\right]
\end{flalign*}
\begin{flalign*}
= \frac{2}{\pi} \left[
\begin{alignedat}{1}
f(    \Delta x)\sin m\Delta x.\Delta x
+ f(   2\Delta x)\sin2m\Delta x.\Delta &x  + \cdots\\
+ f(&\pi-\Delta x)\sin m(\pi-\Delta x).\Delta x
\end{alignedat}
\right]
\end{flalign*}
since $\Delta x=\dfrac{\pi}{n+1}$.

As $n$ is increased indefinitely $\Delta x$ approaches zero as a limit. Hence the
limiting value of $a_m$ as $n$ increases indefinitely is
\begin{flalign*}
& \frac{2}{\pi}\limit_{\Delta x\doteq 0}\left[
\begin{alignedat}{1}
f(    \Delta x)\sin  m\Delta x.\Delta x
+ f(   2\Delta x)\sin 2m\Delta x.\Delta x &+ \cdots\\
+ f(\pi-\Delta x)&\sin m(\pi-\Delta x).\Delta x
\end{alignedat}
\right]\footnotemark  &&\\[-8ex]
\end{flalign*}
\footnotetext{We shall use the sign $\doteq$ for \emph{approaches}. $\Delta x\doteq 0$ is read $\Delta x$ approaches zero.}
\begin{flalign*}
&&= \frac{2}{\pi}\int\limits_0^\pi f(x)\sin mx.dx.  \quad\text{[v.\ Int.\ Cal.\ Arts.~80, 81.]}\\[-5ex]
\end{flalign*}
\begin{flalign*}
&\text{\indent Hence} &  f(x)=a_1\sin x + a_2\sin 2x + a_3\sin 3x + \cdots,  &  \tag{2}&
\end{flalign*}
where any coefficient $a_m$ is given by the formula
\[
a_m=\frac{2}{\pi}\int\limits_0^\pi f(x)\sin mx.dx,   \tag{3}
\]
is a true development of $f(x)$ for all values of $x$ between $x=0$ and $x=\pi$
\emph{provided that the series} (2) \emph{is convergent}, for it is in that case only that we can
assume that the limiting value of the second member of (2) Art.~22 can be obtained
by adding the limiting values of the several terms.

When $x=0$ and when $x=\pi$ every term in the second member of (2)
is zero, and the second member is zero and will not be equal to $f(x)$ unless $f(x)$
is itself zero when $x=0$ and $x=\pi$; but even when $f(x)$ is not zero for
$x=0$ and $x=\pi$ the development given above holds good for any value
of $x$ between zero and $\pi$ no matter how near it may be taken to either of these
values.

\mypara{24.} Instead of actually performing the elimination in equations (4) Art.~19
and getting a formula for $a_m$ in terms of $n$, and then letting $n$ increase
indefinitely, we might have saved labor by the following method.

% -----File: 047.png

Return to equations (4) Art.~19 and multiply the first by $\Delta x\sin m\Delta x$,
the second by $\Delta x\sin 2m\Delta x$, and so on, that is multiply each equation by $\Delta x$
times the coefficient of $a_m$ in that equation, and then add the equations.

We get as the coefficient of $a_k$
\[
 \sin  k\Delta x\sin  m\Delta x.\Delta x
+\sin 2k\Delta x\sin 2m\Delta x.\Delta x + \cdots
+\sin nk\Delta x\sin nm\Delta x.\Delta x.
\]

Let us find its limiting value as $n$ is indefinitely increased. It may be
written, since $(n+1)\Delta x=\pi$,
\begin{flalign*}
\limit_{\Delta x\doteq 0} \left[
\begin{alignedat}{1}
\sin k\Delta x\sin m\Delta x.\Delta x
+ \sin2k\Delta x\sin2m&\Delta x.\Delta x +\cdots\\
+ \sin &k(\pi-\Delta x)\sin m(\pi-\Delta x).\Delta x
\end{alignedat}
\right]
\end{flalign*}
\[
      = \int\limits_0^\pi\sin kx\sin mx.dx;
\]
\begin{flalign*}
&\text{but }  & \int\limits_0^\pi\sin kx\sin mx.dx &=\tfrac{1}{2}\int\limits_0^\pi[\cos(m-k)x-\cos(m+k)x]dx  &&\qquad{}\\
&&&              = 0  \text{ if $m$ and $k$ are not equal.}  &&
\end{flalign*}

The coefficient of $a_m$ is
\[
\Delta x(\sin^2 m\Delta x + \sin^2 2m\Delta x + \sin^2 3m\Delta x + \cdots + \sin^2 nm\Delta x).
\]

Its limiting value
\begin{gather*}
\limit_{\Delta x\doteq 0} \bigg[
  \sin^2  m\Delta x.\Delta x
+ \sin^2 2m\Delta x.\Delta x + \cdots
+ \sin^2  m(\pi - \Delta x)\Delta x \bigg]
\\
= \int\limits_0^\pi\sin^2 mx.dx = \frac{\pi}{2}.
\end{gather*}

The first member is
\[
  f( \Delta x)\sin  m\Delta x.\Delta x
+ f(2\Delta x)\sin 2m\Delta x.\Delta x + \cdots
+ f(n\Delta x)\sin mn \Delta x.\Delta x
\]
and its limiting value is
\[
\int\limits_0^\pi f(x)\sin mx.dx.
\]

Hence the limiting form approached by the final equation as $n$ is increased is
\[
\int\limits_0^\pi f(x)\sin mx.dx = \frac{\pi}{2}a_m.
\]
\begin{flalign*}
&\text{\indent Whence } &  a_m = \frac{2}{\pi}\int\limits_0^\pi f(x)\sin mx.dx  && \text{ as before.}&
\end{flalign*}

This method is practically the same as \emph{multiplying the equation}
\[
f(x) = a_1\sin x + a_2\sin 2x + a_3\sin 3x + \cdots    \tag{1}
\]
\emph{by $\mathop\textit{sin} mx.dx$ and integrating both members from zero to $\pi$}.
% -----File: 048.png

It is exceedingly important to realize that the short method of determining
any coefficient $a_m$ of the series (1) which has just been described in the italicized
paragraph, is essentially the same as that of obtaining $a_m$ by actual
elimination from the equations (4) Art.~19, and then supposing $n$ to increase
indefinitely, thus making the curves (3) Art.~19 and (2) Art.~19 absolutely
coincide between the values of $x$ which are taken as the limits of the
definite integration.

\mypara{25.}  We see, then, that any function of $x$ which is single-valued, finite, and
continuous between $x = 0$ and $x = \pi$, or if discontinuous has only finite
discontinuities each of which is preceded and succeeded by continuous portions,
can probably be developed into a series of the form
\[
f(x) = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots      \tag{1}
\]
\begin{flalign*}
&\text{where }& a_m
= \frac{2}{\pi}\int\limits_{0}^{\pi} f(x) \sin mx.dx
= \frac{2}{\pi}\int\limits_{0}^{\pi} f(\alpha) \sin m\alpha.d\alpha;   &&\phantom{where } \tag{2}
\end{flalign*}
and the series and the function will be identical for all values of $x$ between
$x = 0$ and $x = \pi$, not including the values $x = 0$ and $x = \pi$ unless
the given function is equal to zero for those values.

An elaborate investigation of the question of the convergence of the series
(1), for which we have not space, entirely confirms the result formulated
above\label{notep38}\footnote
  {Provided the function has not an infinite number of maxima and minima in the neighborhood
  of a point. v.\ Arts.~37--38.}
and shows in addition that at a point of finite discontinuity the series
has a value equal to half the sum of the two values which the function
approaches as we approach the point in question from opposite sides.

The investigation which we have made in the preceding sections establishes
the fact that the curve represented by $y = f(x)$ need not follow the same
mathematical law throughout its length, but may be made up of portions of
entirely different curves. For example, a broken line or a locus consisting of
finite parts of several different and disconnected straight lines can be
represented perfectly well by $y =$ a sine series.

\mypara{26.}  Let us obtain a few sine developments.
\begin{flalign*}
&\text{\indent (\emph{a})\quad Let  }&f(x)=x.&        &&\phantom{\indent (a)\quad Let\ }         \tag{1}
\end{flalign*}
\begin{flalign*}
&\text{We have }&x = a_1 \sin x + {}& a_2 \sin 2x + a_3\sin 3x + \cdots   &&\phantom{We\ have } \tag{2}\\
&\text{where }&a_m = \frac{2}{ \pi} & \int\limits_{0}^{\pi} x \sin mx.dx&&                     \tag{3}
\end{flalign*}
% -----File: 049.png
\[
\int x\sin mx.dx = \frac{1}{m^2}(\sin mx - mx\cos mx),
\]
\[
\int\limits_0^\pi x\sin mx.dx = -\frac{(-1)^m \pi}{m},
\]
\begin{flalign*}
&\text{and } & x = 2 \left( \frac{\sin x}{1} - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \frac{\sin 4x}{4} + \cdots \right) & \tag{4}&&
\end{flalign*}
\begin{flalign*}
&\indent \text{(\emph{b})\quad Let } & f(x) = 1.  &   \tag{1}  &&\phantom{\indent(b)\quad Let\ }
\end{flalign*}
\[
a_m = \frac{2}{\pi}\int\limits_0^\pi\sin mx.dx;  \tag{2}
\]
\[
\int\sin mx.dx = -\frac{\cos mx}{m},
\]
\begin{align*}
\int\limits_0^\pi\sin mx.dx &= \frac{1}{m}(1-\cos m\pi) = \frac{1}{m}[1-(-1)^m]\\
&= 0 \text{ if $m$ is even}\\
&= \frac{2}{m} \text{ if $m$ is odd.}
\end{align*}
\begin{flalign*}
&\text{Hence } &  1 = \frac{4}{\pi} \left( \frac{\sin x}{1} + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \frac{\sin 7x}{7} + \cdots \right). &  \tag{3}  &&\phantom{Hence}
\end{flalign*}

It is to be noticed that (3) gives at once a sine development for any constant
$c$. It is,
\[
c = \frac{4c}{\pi} \left( \frac{\sin x}{1} + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \cdots \right).  \tag{4}
\]

If we substitute $x=\dfrac{\pi}{2}$ in (4)\,(\emph{a}) or (3)\,(\emph{b}) we get a familiar result, namely
\[
\frac{\pi}{4}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots,  \tag{5}
\]
a formula usually derived by substituting $x=1$ in the power series for
$\tan^{-1}x$. (v.\ Dif.\ Cal.\ Art.~135.)

(4)\,(\emph{a}) does not hold good when $x=\pi$, and (3)\,(\emph{b}) fails when $x=0$ and
when $x=\pi$, for in all these cases the series reduces to zero.\\

%[Illustration]
\pngright{001.png}{626}{390}{-24}

(\emph{c})\quad Let $f(x)=x$ from $x=0$ to $x=\dfrac{\pi}{2}$
and $f(x)=\pi-x$ from $x=\dfrac{\pi}{2}$ to $x=\pi$.
That is, let $y=f(x)$ represent the broken
line in the figure.

As the mathematical expression for
$f(x)$ is different in the two halves of the
curve we must break up
% -----File: 050.png
\[
\int\limits_0^\pi f(x)\sin mx.dx \quad \text{into}\quad \int\limits_0^\frac{\pi}{2}f(x)\sin mx.dx + \int\limits_\frac{\pi}{2}^\pi f(x)\sin mx.dx.
\]
We have, then,
\begin{align*}
a_m &= \frac{2}{\pi} \int\limits_0^\frac{\pi}{2}x\sin mx.dx + \frac{2}{\pi} \int\limits_\frac{\pi}{2}^\pi(\pi-x)\sin mx.dx  \tag{1}\\
&= \frac{4}{m^2\pi}\sin m \frac{\pi}{2}.\\[-6ex]
\end{align*}

\begin{flalign*}
&\text{\indent But}&
\begin{alignedat}[t]{5}
\sin m \dfrac{\pi}{2} &= 1      && \text{if}\quad && m=1\quad && \text{or}\quad && 4k+1 \\[-1ex]
                      &= 0      && ``        && m=2 && \,``        && 4k+2  \\
                      &= -1\;\; && ``        && m=3 && \,``        && 4k+3  \\
                      &= 0      && ``        && m=4 && \,``        && 4k.   \\
\end{alignedat}
&&\phantom{\indent But}
\end{flalign*}
Hence if $y=f(x)$ represents our broken line,
\[
f(x)=\frac{4}{\pi}\left(\frac{\sin x}{1^2}-\frac{\sin 3x}{3^2}+\frac{\sin 5x}{5^2}-\frac{\sin 7x}{7^2}+\cdots\right).  \tag{2}
\]

When $x=\dfrac{\pi}{2}$  $f(x)=\dfrac{\pi}{2}$ and we have
\[
\frac{\pi^2}{8} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \cdots  \tag{3}
\]

(\emph{d})\quad As a case where the function has a finite discontinuity, let
\begin{alignat*}{4}
f(x)=1 \quad& \text{from} \quad&& x=0\quad        && \text{to}\quad && x=\dfrac{\pi}{2} \quad \text{and}\\
f(x)=0 \quad& \;\;\text{``}     && x=\dfrac{\pi}{2} && \;\text{``}     && x=\pi.
\end{alignat*}
$y=f(x)$ will in this case represent the locus in the figure.

%[Illustration]
\pngrightl{002.png}{675}{3}{-24}

As before
\begin{align*}
\int\limits_0^\pi f(x)\sin mx.dx & = \int\limits_0^\frac{\pi}{2} f(x)\sin mx.dx \\
&+ \int\limits_\frac{\pi}{2}^\pi f(x)\sin mx.dx.
\end{align*}
\[
a_m = \frac{2}{\pi} \int\limits_0^\frac{\pi}{2}\sin mx.dx + \frac{2}{\pi}\int\limits_\frac{\pi}{2}^\pi 0
.\sin mx.dx. \tag{1}
\]
% -----File: 051.png
\[
a_m=\frac{2}{\pi} \int\limits_0^\frac{\pi}{2}\sin mx.dx
=\frac{2}{\pi}\frac{1}{m}\left(1-\cos m\frac{\pi}{2}\right).
\]

\begin{flalign*}
&\text{\indent But}&
\begin{alignedat}[t]{5}
\cos m \dfrac{\pi}{2} &= \phantom{-}0\quad    && \text{if}\quad && m=1\quad && \text{or}\quad && 4k+1 \\[-1ex]
                      &=-1      && ``        && m=2 && \,``        && 4k+2  \\
                      &= \phantom{-}0 && ``        && m=3 && \,``        && 4k+3  \\
                      &= \phantom{-}1      && ``        && m=4 && \,``        && 4k.   \\
\end{alignedat}
&&\phantom{\indent But}
\end{flalign*}

Hence
\[
f(x) = \frac{2}{\pi}
       \left(
   \frac{ \sin  x}{1} + \frac{2\sin 2x}{2} + \frac{ \sin 3x}{3}
 + \frac{ \sin 5x}{5} + \frac{2\sin 6x}{6} + \frac{ \sin 7x}{7}
 + \cdots \right). \tag{2}
\]

If $x=\dfrac{\pi}{2}$ the second member of (2) reduces to $\dfrac{1}{2}$, for
\[
  \frac{2}{\pi}
  \left( \frac{1}{1} - \frac{1}{3}
       + \frac{1}{5} - \frac{1}{7} + \cdots \right)
= \frac{1}{2}\qquad \text{by (5) ($b$)};
\]
and we see that the series represents the function completely for all values of
$x$ between $x=0$ and $x=\pi$ except for $x=\dfrac{\pi}{2}$ and there it has a
value which is the mean of the values approached by the function as $x$
approaches $\dfrac{\pi}{2}$ from opposite sides.

\EXAMPLE{S}

Obtain the following developments:---
\begin{flalign*}
&\indent\text{(1)}\quad x^2 = \dfrac{2}{\pi} \bigg[
  \bigg( \dfrac{\pi^2}{1} - \dfrac{4}{1^3} \bigg) \sin  x
       - \dfrac{\pi^2}{2}\sin 2x
+ \bigg( \dfrac{\pi^2}{3} - \dfrac{4}{3^3} \bigg) \sin 3x
       - \dfrac{\pi^2}{4}\sin 4x \\
& \hspace{8em} + \bigg( \dfrac{\pi^2}{5} - \dfrac{4}{5^3} \bigg) \sin 5x - \cdots \bigg].\\
%
&\indent\text{(2)}\quad x^3 = \dfrac{2}{\pi} \bigg[
  \bigg( \dfrac{\pi^3}{1} - \dfrac{6\pi}{1^3} \bigg) \sin  x
- \bigg( \dfrac{\pi^3}{2} - \dfrac{6\pi}{2^3} \bigg) \sin 2x
+ \bigg( \dfrac{\pi^3}{3} - \dfrac{6\pi}{3^3} \bigg) \sin 3x \\
&\hspace{8em} - \bigg( \dfrac{\pi^3}{4} - \dfrac{6\pi}{4^3} \bigg) \sin 4x + \cdots \bigg].  \\
%
&\indent\text{(3)}\quad f(x) = \dfrac{2}{\pi} \bigg[
  \dfrac{\sin  x}{1^2} + \dfrac{ \pi}{2^2}\sin 2x
- \dfrac{\sin 3x}{3^2} - \dfrac{2\pi}{4^2}\sin 4x
+ \dfrac{\sin 5x}{5^2}   \\
&\hspace{8em} + \dfrac{3\pi}{6^2}\sin 6x - \cdots \bigg],
\end{flalign*}
% -----File: 052.png
if $f(x)=x$ from $x=0$ to $x=\dfrac{\pi}{2}$ and $f(x)=0$ from $x=\dfrac{\pi}{2}$ to $x=\pi$.

\begin{flalign*}
&\indent\text{(4)}\quad \sin\mu x=\dfrac{2}{\pi}\sin\mu\pi \bigg[
  \dfrac{ \sin x}{1^2-\mu^2} - \dfrac{2\sin2x}{2^2-\mu^2}
+ \dfrac{3\sin3x}{3^2-\mu^2} - \dfrac{4\sin4x}{4^2-\mu^2} + \cdots \bigg]\\
&\text{if $\mu$ is a fraction.}\\
%
&\indent\text{(5)}\quad e^x = \dfrac{2}{\pi} \bigg[
  \dfrac{1}{ 2}(1+e^\pi)\sin  x + \dfrac{2}{5}(1-e^\pi)\sin 2x
+ \dfrac{3}{10}(1+e^\pi)\sin 3x \\
&\hspace{8em} + \dfrac{4}{17}(1-e^\pi)\sin 4x + \cdots \bigg].\\
%
&\indent\text{(6)}\quad \sinh x = \dfrac{2\sinh\pi}{\pi} \bigg[
  \dfrac{1}{ 2}\sin x - \dfrac{2}{5}\sin 2x
+ \dfrac{3}{10}\sin3x - \dfrac{4}{17}\sin4x + \cdots \bigg].\\
%
&\indent\text{(7)}\quad \cosh x = \dfrac{2}{\pi} \bigg[
  \dfrac{1}{2}(1+\cosh\pi)\sin  x
+ \dfrac{2}{5}(1-\cosh\pi)\sin 2x \\
&\hspace{8em} + \dfrac{3}{10}(1+\cosh\pi)\sin 3x + \cdots \bigg].
\end{flalign*}

\mypara{27.} Let us now try to develop a given function of $x$ in a series of cosines.

As before suppose that $f(x)$ has a single value for each value of $x$ between
$x=0$ and $x=\pi$, that it does not become infinite between $x=0$ and
 $x=\pi$, and that if discontinuous it has only finite discontinuities.

Assume
\[
f(x) = b_0 + b_1\cos x + b_2\cos 2x + b_3\cos 3x + \cdots   \tag{1}
\]

To determine any coefficient $b_m$
multiply (1) by $\cos mx.dx$ and integrate
each term from $0$ to $\pi$.
\begin{align*}
\int\limits_0^\pi &b_0\cos mx.dx = 0. \\
                \int\limits_0^\pi b_k\cos kx\cos mx.dx
&= \frac{b_k}{2}\int\limits_0^\pi[\cos(m-k)x+\cos(m+k)x]dx\\
&= 0 \text{ if $m$ and $k$ are not equal.}
\end{align*}
\[
\int b_m\cos^2 mx.dx = \frac{b_m}{2m}(mx+\cos mx\sin mx),
\]
\[
\phantom{if\ m\ is\ not\ zero.}
\int\limits_0^\pi b_m\cos^2 mx.dx = \frac{\pi}{2}b_m,    \tag*{if $m$ is not zero.}
\]
\begin{flalign*}
&\text{Hence} &  b_m
= \frac{2}{\pi} \int\limits_0^\pi f(x)\cos mx.dx
= \frac{2}{\pi} \int\limits_0^\pi f(\alpha)\cos m\alpha.d\alpha, &   \tag{2}&&
\end{flalign*}
if $m$ is not zero.
% -----File: 053.png

To get $b_0$ multiply (1) by $dx$ and integrate from zero to $\pi$.
\begin{flalign*}
&&&\int\limits_0^\pi b_0 dx = b_0 \pi,&&\\
&&&\int\limits_0^\pi b_k\cos kx.dx=0.&&\\
\label{err053}
&\text{\indent Hence} & b_0=\frac{1}{\pi} &\int\limits_0^\pi f(x)dx=\frac{1}{\pi}\int\limits_0^\pi f(\alpha)d\alpha, & \tag{3} &\phantom{Hence}
\end{flalign*}
which is just half the value that would be given by formula~(2) if zero were
substituted for $m$.

To save a separate formula (1) is usually written
\[
f(x) = \tfrac{1}{2} b_0 + b_1\cos x + b_2\cos 2x + b_3\cos 3x + \cdots   \tag{4}
\]
and then the formula
\[
b_m = \frac{2}{\pi} \int\limits_0^\pi f(x)\cos mx.dx
= \frac{2}{\pi}\int\limits_0^\pi f(\alpha)\cos m\alpha.d\alpha   \tag{2}
\]
will give $b_0$ as well as the other coefficients.

It is important to see clearly that what we have just done in determining
the coefficients of (1) is equivalent to taking $n+1$ terms of (4),
substituting in
\[
y = \tfrac{1}{2} b_0 + b_1\cos x + b_2\cos 2x + \cdots + b_n\cos nx   \tag{5}
\]
in turn the coördinates of the $n+1$ points of the curve
\[
y=f(x)
\]
whose projections on the axis of $X$ are equidistant, determining $b_0$, $b_1$, $b_2$, $\cdots$ $b_n$
by elimination from the $n+1$ resulting equations, and then taking the limiting
values they approach as $n$ is indefinitely increased.\quad (v.\ Art.~24.)

If $\Delta x = \dfrac{\pi}{n+1}$ the abscissas of the $n+1$ points used are 0, $\Delta x$, $2\Delta x$,
$3\Delta x$, $\cdots$ $n\Delta x$, so that we should expect our cosine development to hold for
 $x=0$ as well as for values of $x$ between zero and $\pi$.

\mypara{28.} Let us take one or two examples:
\begin{flalign*}
&\text{\indent($a$)\quad Let} &  f(x)=x. &  \tag{1}  &&\phantom{\indent\quad Let}
\end{flalign*}
\begin{gather*}
b_0 = \frac{2}{\pi} \int\limits_0^\pi x\,dx
    = \frac{2}{\pi} \frac{\pi^2}{2} = \pi.
\\
b_m = \frac{2}{\pi}\int\limits_0^\pi x\cos mx.dx
    = \frac{2}{m^2\pi} (\cos m\pi-1)
    = \frac{2}{m^2\pi} [(-1)^m-1].
\end{gather*}
% -----File: 054.png

\begin{flalign*}
&\text{\indent Hence} & x &= \frac{\pi}{2} - \frac{4}{\pi} \left(
    \cos x + \frac{\cos 3x}{3^2} + \frac{\cos 5x}{5^2}
  + \frac{\cos 7x}{7^2} + \cdots \right).   &&\phantom{Hence}  \tag{2}
\end{flalign*}

(2) holds good not only for values of $x$ between zero and $\pi$ but for $x = 0$
and $x = \pi$ as well, since for these values we have
\begin{flalign*}
&&  0 &= \frac{\pi}{ 2} - \frac{4}{\pi} \left( 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{ 1}{7^2} + \cdots \right)  &&\tag{3}\\
&\text{and }&
  \pi &= \frac{\pi}{2} + \frac{4}{\pi} \left(1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{ 1}{7^2} + \cdots \right)  &&  \tag{4}
\end{flalign*}
which are true by Art.~26 (\emph{c})(3).

\begin{flalign*}
&\text{\indent($b$)\quad Let}& f(x) &= x\sin x.  &&\phantom{\indent Let}  \tag{1}
\end{flalign*}
\begin{gather*}
b_0
= \frac{2}{\pi}\int\limits_{0}^{\pi} x\sin x.dx
= \frac{2}{\pi}\pi = 2,\\
b_1
= \frac{2}{\pi}\int\limits_{0}^{\pi} x\sin x \cos x.dx
= \frac{1}{\pi}\int\limits_{0}^{\pi} x\sin 2x.dx = -\frac{1}{2},\\
b_m
= \frac{2}{\pi}\int\limits_{0}^{\pi} x\sin x \cos mx.dx
= \frac{1}{\pi}\int\limits_{0}^{\pi} [x\sin (m+1)x - x \sin (m-1)x]dx\\
\begin{split}
&= \frac{2}{(m-1)(m+1)} \quad \text{if $m$ is odd}\\
&=-\frac{2}{(m-1)(m+1)} \quad \text{if $m$ is even.}\\
\end{split}
\end{gather*}

Hence
\[
x \sin x = 1 - \frac{\cos  x}{2}
             - \frac{2\cos 2x}{1.3}
             + \frac{2\cos 3x}{2.4}
             - \frac{2\cos 4x}{3.5} + \cdots   \tag{2}
\]
If $x = \dfrac{\pi}{2}$ we have
\[
\frac{\pi}{4} = \frac{1}{ 2}
              + \frac{1}{1.3}
              - \frac{1}{3.5}
              + \frac{1}{5.7} - \cdots.      \tag{3}
\]

\newpage
\EXAMPLE{S}

Obtain the following developments:
\begin{flalign*}
&\text{\indent(1)\quad}  f(x) = \frac{\pi}{ 4} - \frac{2}{\pi} \bigg[
    \frac{\cos  2x}{1^2} + \frac{\cos  6x}{3^2}
  + \frac{\cos 10x}{5^2} + \frac{\cos 14x}{7^2} + \cdots \bigg] &\\
&\text{if \quad $f(x) = x$ from $x = 0$ to $x = \dfrac{\pi}{ 2 }$ and $f(x) = \pi - x$ from $x = \dfrac{\pi}{ 2}$ to $x = \pi$.}\\
% -----File: 055.png
%
&\text{\indent(2)\quad} f(x) = \dfrac{1}{2} + \dfrac{2}{\pi}\bigg[\dfrac{\cos x}{1} - \dfrac{\cos 3x}{3} + \dfrac{\cos 5x}{5} - \dfrac{\cos 7x}{7} +\cdots\bigg], \\
&\text{if \quad $f(x) = 1$ from $x = 0$ to $x = \dfrac{\pi}{ 2}$ and $f(x) = 0$ from $x = \dfrac{\pi }{2}$ to $x = \pi$.}\\
%
&\text{\indent(3)\quad} x^2 = \dfrac{\pi^2}{ 3} - 4 \bigg[\dfrac{\cos x}{1^2} - \dfrac{\cos 2x}{2^2} + \dfrac{\cos 3x}{3^2} - \dfrac{\cos 4x}{4^2} +\cdots\bigg], \\
%
&\text{\indent(4)\quad} x^3 = \dfrac{\pi^3 }{4} - \dfrac{6}{\pi}\bigg[\bigg(\dfrac{\pi^2}{1^2} - \dfrac{4}{1^4}\bigg)\cos x - \dfrac{\pi^2}{2^2}\cos 2x + \bigg(\dfrac{\pi^2}{3^2} - \dfrac{4}{3^4}\bigg)\cos 3x \\
&\hspace{10em} - \dfrac{\pi^2}{4^2} \cos 4x + \bigg(\dfrac{\pi^2}{5^2} - \dfrac{4}{5^4}\bigg)\cos 5x - \cdots\bigg], \\
%
&\text{\indent(5)\quad} f(x) =  \dfrac{\pi }{ 8} +  \dfrac{2}{\pi}\bigg[\bigg( \dfrac{\pi}{2} - 1\bigg)\cos x -  \dfrac{2}{2^2} \cos 2x -  \dfrac{1}{3^2} \bigg( \dfrac{3\pi}{2}+1\bigg) \cos 3x \\
&\hspace{11em} +\dfrac{1}{5^2}\bigg(\dfrac{5\pi}{2} -1\bigg)\cos 5x -\dfrac{2 }{6^2} \cos 6x - \cdots\bigg], \\
&\text{if \quad $f(x) = x$ from $x = 0$ to $x = \dfrac{\pi }{2}$ and $f(x) = 0$ from $x = \dfrac{\pi}{ 2}$ to $x = \pi$.}\\
%
&\text{\indent(6)\quad} e^x = \dfrac{2}{\pi}\bigg[\dfrac{1}{2}(e^\pi-1)- \dfrac{1}{1+1^2} (e^\pi+1)\cos x + \dfrac{1}{1+2^2} (e^\pi-1)\cos 2x \\
&\hspace{7em} - \dfrac{1}{1+3^2} (e^\pi+1)\cos 3x + \cdots\bigg], \\
%
&\text{\indent(7)\quad} \cosh x = \dfrac{2 \sinh \pi}{\pi} \bigg[\dfrac{1}{2} - \dfrac{1}{2}\cos x + \dfrac{1}{5}\cos 2x - \dfrac{1}{10}\cos 3x \\
&\hspace{13em} + \dfrac{1}{17}\cos 4x - \cdots\bigg], \\
%
&\text{\indent(8)\quad} \sinh x =\dfrac{ 2}{\pi}\bigg[\dfrac{1}{2} (\cosh\pi-1) - \dfrac{1}{2}(\cosh\pi +1)\cos x \\
&\hspace{8em} + \dfrac{1}{5} (\cosh \pi-1)\cos 2x - \dfrac{1}{10}(\cosh\pi +1) \cos 3x + \cdots \bigg], \\
%
&\text{\indent(9)\quad} \cos \mu x = \dfrac{2\mu \sin \mu \pi}{\pi} \bigg[ \dfrac{1}{2\mu^2} - \dfrac{\cos x}{ \mu^2-1^2} + \dfrac{\cos 2x}{\mu^2-2^2} - \dfrac{\cos 3x}{\mu^2-3^2} \\
%
&\hspace{12.5em} + \dfrac{\cos 4x}{\mu^2-4^2} - \cdots \bigg], \\
&\indent\phantom{(9)}  \quad \text{if $\mu$ is a fraction.}
\end{flalign*}

\mypara{29.} Although any function can be expressed both as a sine series and as a
cosine series, and the function and either series will be equal for all values of
$x$ between zero and $\pi$, there is a decided difference in the two series for other
values of $x$.

Both series are periodic functions of $x$ having the period $2\pi$. If then we
let $y$ equal the series in question and construct the portion of the corresponding
% -----File: 056.png
curve which lies between the values $x = -\pi$ and $x = \pi$ the whole
curve will consist of repetitions of this portion.

Since $\sin mx = -\sin(- mx)$ the ordinate corresponding to any value of
$x$ between $-\pi$ and zero in the sine curve; will be the negative of the ordinate
corresponding to the same value of $x$ with the positive sign. In other words
the curve
\[
  y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots       \tag{1}
\]
is symmetrical with respect to the origin.

Since $\cos mx = \cos (-mx)$ the ordinate corresponding to any value of $x$
between $-\pi$ and zero in the cosine curve will be the same as the ordinate
belonging to the corresponding positive value of $x$. In other words the curve
\[
  y = \tfrac{1}{2}b_0 +b_1\cos x + b_2\cos 2x + b_3\cos 3x + \cdots   \tag{2}
\]
is symmetrical with respect to the axis of $Y$.

If then $f(x) = -f(-x)$, that is if $f(x)$ is an \textit{odd} function the sine series
corresponding to it will be equal to it for all values of $x$ between $-\pi$ and $\pi$,
except perhaps for the value $x = 0$ for which the series will necessarily be
zero.

If $f(x) =f(-x)$, that is if $f(x)$ is an \textit{even} function the cosine series corresponding
to it will be equal to it for all values of $x$ between $x = -\pi$ and
$x = \pi$, not excepting the value $x = 0$.

As an example of the difference between the sine and cosine developments
of the same function let us take the series for $x$.
\begin{align*}
  y &= 2\left[\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3}
                     - \frac{\sin 4x}{4} + \cdots \right]      \tag{3}
\\
  y &= \frac{\pi}{2} - \frac{4}{\pi} \left[
              \cos x + \frac{\cos 3x}{3^2} + \frac{\cos 5x}{5^2}
                     + \frac{\cos 7x}{7^2} + \cdots \right]   \tag{4}
\end{align*}
[v.\ Art.~26(\emph{a}) and Art.~28(\emph{a})].  (3) represents the curve
%[Illustration]
\pngcent{003.png}{1266}%{346}

\newpage\noindent and (4) the curve
%[Illustration]
\pngcent{004.png}{1266}%{236}
% -----File: 057.png

Both coincide with $y = x$ from $x = 0$ to $x = \pi$, (3) coincides with
 $y = x$ from $x = -\pi$ to $x = \pi$, and neither coincides with $y = x$ for
values of $x$ less than $-\pi$ or greater than $\pi$. Moreover (3), in addition to
the continuous portions of the locus represented in the figure, gives the isolated
points $(-\pi,0)$ $(\pi,0)$ $(3\pi,0)$ \&c.

\mypara{30.} We have seen that if $f(x)$ is an \textit{odd} function its development in sine
series holds for all values of $x$ from $-\pi$ to $\pi$, as does the development of
$f(x)$ in cosine series if $f(x)$ is an \textit{even} function.

Thus the developments of Art.~26(\emph{a}), Art.~26 Exs.~(2), (4), (6); Art.~28(\emph{b}),
Art.~28 Exs.~(3), (7), (9) are valid for all values of $x$ between $-\pi$ and $\pi$.

Any function of $x$ can be developed into a Trigonometric series to which it
is equal for all values of $x$ between $-\pi$ and $\pi$.

Let $f(x)$ be the given function of $x$. It can be expressed as the sum of an
even function of $x$ and an odd function of $x$ by the following device.
\[
f(x)=\frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}  \tag{1}
\]
identically; but $\dfrac{f(x)+f(-x)}{2}$ is not changed by reversing the sign of $x$ and
is therefore an \emph{even} function of $x$; and when we reverse the sign of $x$,
$\dfrac{f(x)-f(-x)}{2}$ is affected only to the extent of having its sign reversed and
is consequently an \textit{odd} function of $x$.

Therefore for all values of $x$ between $-\pi$ and $\pi$\\[-1ex]
\[
  \frac{f(x)+f(-x)}{2} = \frac{1}{2}b_0
                       + b_1\cos x + b_2\cos 2x + b_3\cos3x + \cdots \\[1ex]
\]
where \hfill$\displaystyle
  b_m = \frac{2}{\pi} \int\limits_{0}^{\pi} \frac{f(x) + f(-x)}{2}
                                     \cos mx.dx;
$ \hfill\quad and \\[1ex]
\[
  \frac{f(x)-f(-x)}{2} = a_1\sin x + a_2\sin 2x + a_3\sin 3x + \cdots \\[1ex]
\]
where \hfill$\displaystyle
  a_m = \frac{2}{\pi} \int\limits_{0}^{\pi} \frac{f(x) - f(-x)}{2}
                                     \sin mx.dx.
$ \hfill\phantom{where}\\[2ex]
\newpage\noindent $b_m$ and $a_m$ can be simplified a little. \\[-1ex]
\begin{align*}
  b_m &= \frac{2}{\pi} \int\limits_{0}^{\pi} \frac{f(x)+f(-x)}{2}
                                      \cos mx.dx
\\[1ex]
      &= \frac{1}{\pi}
         \bigg[ \int\limits_{0}^{\pi} f( x)\cos mx.dx
              + \int\limits_{0}^{\pi} f(-x)\cos mx.dx \bigg],
\end{align*}
% -----File: 058.png
but if we replace $x$ by $-x$, we get
\[
   \int\limits_{0}^{ \pi} f(-x) \cos mx.dx
= -\int\limits_{0}^{-\pi} f( x) \cos mx.dx
=  \int\limits_{-\pi}^{0} f( x) \cos mx.dx,\\[-4ex]
\]
\begin{flalign*}
&\text{and we have }&&
  b_m = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x) \cos mx.dx. &&\phantom{and\ we\ have }
\end{flalign*}

In the same way we can reduce the value of $a_m$ to
\[
  \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x) \sin mx.dx.
\]

Hence
\[
  \left\{ \begin{aligned}
 f(x) = \frac{1}{2} b_0 &+ b_1 \cos x + b_2 \cos 2x + b_3 \cos 3x + \cdots\\
                        &+ a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots
\end{aligned}
 \right \}\tag{2}\\[-3ex]
\]
\begin{flalign*}
&\text{where }&
  b_m &= \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x) \cos mx.dx
       = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(\alpha) \cos m\alpha.d\alpha.  &&\phantom{where} \tag{3}
\\
&\text{and }&
  a_m &= \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(x) \sin mx.dx
       = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(\alpha) \sin m\alpha.d\alpha.  && \tag{4}
\end{flalign*}
and this development holds for all values of $x$ between $-\pi$ and $ \pi$.

The second member of (2) is known as a Fourier's Series.

\EXAMPLE{S}

1.\quad Obtain the following developments, all of which are valid from $x = -\pi$
to $x=\pi$:---
\begin{flalign*}
&\indent\text{(1)}&
e^x
={}& \frac{2 \sinh \pi}{\pi}
     \left[ \frac{1}{2} - \frac{1}{2} \cos x + \frac{1}{5} \cos 2x
          - \frac{1}{10} \cos 3x + \frac{1}{17} \cos 4x + \cdots \right]&&
\\
&&&+ \frac{2 \sinh \pi}{ \pi}
      \left[ \frac{1}{2}\sin x - \frac{2}{5} \sin 2x
           + \frac{3}{10} \sin 3x - \frac{4}{17} \sin 4x + \cdots \right].
\end{flalign*}
\begin{flalign*}
&\indent\text{(2)}&
f(x)
={}& \frac{\pi}{ 4} - \frac{2}{ \pi}
     \left[ \cos x + \frac{\cos 3x }{ 3^2} + \frac{\cos 5x}{ 5^2}
     + \frac{\cos 7x }{ 7^2} + \cdots \right]&&\hspace{5.5em}
\\
&&&+ \frac{\sin x}{ 1} - \frac{\sin 2x}{ 2} + \frac{\sin 3x}{ 3}
 - \frac{\sin 4x}{ 4} + \cdots,
\end{flalign*}
where $f(x) = 0$ from $x = -\pi$ to $x=0$ and $f(x) = x$ from $x=0$ to $x=\pi$.
% -----File: 059.png
\begin{flalign*}
&\indent\text{(3)}  &f(x)={}&- \frac{3\pi}{16} + \frac{1}{\pi}\bigg[\frac{1}{1^2}\cos x + \frac{2}{2^2}\cos2x + \frac{1}{3^2}\cos3x + \frac{1}{5^2}\cos5x\\
&&&\hspace{5em} + \frac{2}{6^2}\cos 6x + \cdots\bigg]&&\hspace{2em}\\
&&&+ \frac{1}{\pi}\bigg[\biggl(\frac{3\pi}{2} - 1\biggr)\sin x - \frac{3\pi}{4}\sin2x + \biggl(\frac{3\pi}{6} + \frac{1}{3^2}\biggr)\sin3x\\
&&&\hspace{3em} - \frac{3\pi}{8}\sin4x + \biggl(\frac{3\pi}{10} - \frac{1}{5^2}\biggr)\sin5x - \cdots\bigg],\\[-6ex]
\end{flalign*}
\begin{flalign*}
&\text{where } &f(x)&=x \text{ from } x=-\pi \text{ to } x=0, \quad f(x)=0 \text{ from } x=0 \text{ to } x=\frac\pi2, &&\\
&\text{and }   &f(x)&=x - \frac{\pi}{2} \text{ from } x=\frac{\pi}{2} \text{ to } x=\pi.
\end{flalign*}

2.\quad Show that formula~(2) Art.~30 can be written
\[
f(x)=\frac{1}{2} c_0\cos \beta_0 + c_1\cos(x - \beta_1) + c_2\cos(2x - \beta_2) + c_3\cos(3x - \beta_3) + \cdots
\]
\vspace{-4ex}
\begin{flalign*}
&\text{where}
&c_m=(a_m^2 + b_m^2)^\frac{1}{2} \text{ and } \beta_m=\tan^{-1} \frac{a_m}{b_m}.&&\phantom{where}
\end{flalign*}

3.\quad Show that formula~(2) Art.~30 can be written
\[
f(x)=\frac{1}{2} c_0\sin \beta_0 + c_1\sin(x + \beta_1) + c_2\sin (2x + \beta_2) + c_3\sin (3x + \beta_3) + \cdots\\[-4ex]
\]
\begin{flalign*}
&\text{where}
&c_m=(a_m^2 + b_m^2)^\frac{1}{2} \text{ and } \beta_m=\tan^{-1} \frac{b_m}{a_m}.&&\phantom{where}
\end{flalign*}

\mypara{31.} In developing a function of $x$ into a Trigonometric series it is often
inconvenient to be held within the narrow boundaries $x=-\pi$ and $x=\pi$.
Let us see if we cannot widen them.

Let it be required to develop a function of $x$ into a Trigonometric series
which shall be equal to $f(x)$ for all values of $x$ between $x=-c$ and $x=c$.

Introduce a new variable
\[
z=\frac{\pi}{c} x,
\]
which is equal to $-\pi$ when $x=-c$ and to $\pi$ when $x=c$.

$f(x)=f\left(\dfrac{c}{\pi}z\right)$ can be developed in terms of $z$ by Art.~30 (2), (3), and (4).

We have
\begin{equation}
\left.
\begin{aligned}
f\left(\frac{c}{\pi}z\right)=\frac{1}{2} b_0 &+ b_1\cos z + b_2\cos 2z + b_3\cos 3z + \cdots\\
&+ a_1\sin z + a_2\sin 2z + a_3\sin 3z + \cdots
\end{aligned}\right\}\tag{1}
\end{equation}
\vspace{-3ex}
\begin{flalign*}
&\text{where}
&b_m&=\frac{1}{\pi} \int\limits_{-\pi}^{\pi} f\left(\frac{c}{\pi} z\right)\cos mz.dz.&\phantom{where}  \tag{2}\\
% -----File: 060.png
&\text{and}&
  a_m &= \frac{1}{\pi}
        \int\limits_{-\pi}^{\pi} f\left(\frac{c}{\pi} z\right)
                         \sin mz.dz.     &     \tag{3}
\end{flalign*}
and (1) holds good from $z = -\pi$ to $z = \pi$.

Replace $z$ by its value in terms of $x$ and (1) becomes
\[ \left . \begin{aligned}
f(x)=\frac{1}{2}b_0
&+ b_1 \cos\frac{\pi x}{c} + b_2 \cos \frac{2\pi x}{c}
 + b_3 \cos \frac{3\pi x}{c} + \cdots
\\
&+a _1 \sin \frac{\pi x}{c} + a_2 \sin \frac{2\pi x}{c} + a_3 \sin \frac{3\pi x}{c} + \cdots
\end{aligned}
\right \}
\tag{4}
\]

The coefficients in (4) are the same as in (1), and (4) holds good from
 $x = -c$ to $x = c$.

Formulas (2) and (3) can be put into more convenient shape.
\begin{flalign*}
&& b_m
&= \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f\left(\frac{c}{\pi} z\right) \cos mz.dz
 = \frac{1}{\pi} \int\limits_{-c}^{c} f(x) \cos \frac{m\pi x}{c} \frac{\pi}{c} dx&&
\\[1ex]
&\text{or}& b_m
&= \frac{1}{c} \int\limits_{-c}^{c} f(x) \cos \frac{m\pi x}{c} dx
 = \frac{1}{c} \int\limits_{-c}^{c} f(\lambda) \cos\frac{m\pi\lambda}{c} d\lambda. && \tag{5}
\\
\intertext{\indent In like manner we can transform (3) into}
&& a_m
&= \frac{1}{c} \int\limits_{-c}^{c} f(x) \sin \frac{m\pi x}{c} dx
 = \frac{1}{c} \int\limits_{-c}^{c} f(\lambda) \sin \frac{m\pi \lambda}{c} d\lambda.    &&  \tag{6}
\end{flalign*}

By treating in like fashion formulas (1) and (2) Art.~25 and formulas (4)
and (2) Art.~27 we get
\[
  f(x) = a_1 \sin \frac{\pi x}{c} + a_2 \sin \frac{2\pi x}{c}
       + a_3 \sin \frac{3\pi x}{c} + \cdots     \tag{7}
\]
\begin{flalign*}
&\text{where }& a_m
&= \frac{2}{c} \int\limits_{0}^{c} f(x) \sin \frac{m\pi x}{c} dx
 = \frac{2}{c} \int\limits_{0}^{c} f(\lambda) \sin \frac{m\pi \lambda}{c} d\lambda.   &&  \tag{8}
\\[1ex]
&\text{and}& f(x)
&= \frac{1}{2 } b_0 + b_1 \cos \frac{\pi x}{c}
 + b_2 \cos \frac{2\pi x}{c} + b_3 \cos \frac{3\pi x}{c} + \cdots &&    \tag{9}
\\[1ex]
&\text{where}& b_m
&= \frac{2}{c} \int\limits_{0}^{c} f(x) \cos \frac{m\pi x}{c} dx
 = \frac{2}{c} \int\limits_{0}^{c} f(\lambda) \cos \frac{m\pi \lambda}{c} d\lambda.  &&  \tag{10}
\end{flalign*}
and (7) and (9) hold good from $x = 0$ to $x = c$.
% -----File: 061.png

\EXAMPLE{S}

1.\quad Obtain the following developments:
\begin{flalign*}
&\rlap{\indent(1)\quad} &	 1={}&\dfrac{4}{\pi}\bigg[\sin\dfrac{\pi x}{c}+\dfrac{1}{3}\sin\dfrac{3\pi x}{c}+\dfrac{1}{5}\sin\dfrac{5\pi x}{c}+\cdots\bigg]\\
&&                 &\text{from } x=0 \text{ to } x=c.\\
&\rlap{\indent(2)\quad} &  x={}&\dfrac{2c}{\pi}\bigg[\sin\dfrac{\pi x}{c}-\dfrac{1}{2}\sin\dfrac{2\pi x}{c}+\dfrac{1}{3}\sin\dfrac{3\pi x}{c}-\dfrac{1}{4}\sin\dfrac{4\pi x}{c}+\cdots\bigg]\\
&&                 &\text{from } x=-c \text{ to } x=c.\\
&&	         x={}&\dfrac{c}{2}-\dfrac{4c}{\pi^2}\bigg[\cos\dfrac{\pi x}{c}+\dfrac{1}{3^2}\cos\dfrac{3\pi x}{c}+\dfrac{1}{5^2}\cos\dfrac{5\pi x}{c}+\dfrac{1}{7^2}\cos\dfrac{7\pi x}{c}+\cdots\bigg]\\
&&                 &\text{from } x=-c \text{ to } x=c.\\
&\rlap{\indent(3)\quad} &  x^2={}&\dfrac{2c^2}{\pi^3}\bigg[\bigg(\dfrac{\pi^2}{1}-\dfrac{4}{1^3}\bigg)\sin\dfrac{\pi x}{c}-\dfrac{\pi^2}{2}\sin\dfrac{2\pi x}{c}+\bigg(\dfrac{\pi^2}{3}-\dfrac{4}{3^2}\bigg)\sin\dfrac{3\pi x}{c}\\
&&                 &\hspace{3em}-\dfrac{\pi^2}{4}\sin\dfrac{4\pi x}{c}+\bigg(\dfrac{\pi^2}{5}-\dfrac{4}{5^3}\bigg)\sin\dfrac{5\pi x}{c}+\cdots\bigg]\\
&&                 &\text{from } x=0 \text{ to } x=c.\\
&&             x^2={}&\dfrac{c^2}{3}-\dfrac{4c^2}{\pi^2}\bigg[\!\cos\dfrac{\pi x}{c}-\dfrac{1}{2^2}\cos\dfrac{2\pi x}{c}+\dfrac{1}{3^2}\cos\dfrac{3\pi x}{c}-\dfrac{1}{4^2}\cos\dfrac{4\pi x}{c}+\cdots\!\bigg]\:\\
&&                 &\text{from } x=-c \text{ to } x=c.\\
&\rlap{\indent(4)\quad} &  e^x={}&2\pi\bigg[\dfrac{1+e^c}{c^2+\pi^2}\sin\dfrac{\pi x}{c}+\dfrac{2(1-e^c)}{c^2+4\pi^2}\sin\dfrac{2\pi x}{c}\\
&&                 &\hspace{3em}+\dfrac{3(1+e^c)}{c^2+9\pi^2}\sin\dfrac{3\pi x}{c}+\dfrac{4(1-e^c)}{c^2+16\pi^2}\sin\dfrac{4\pi x}{c}+\cdots\bigg],\\
&&             e^x={}&2c\bigg[\dfrac{1}{2}\dfrac{e^c-1}{c^2}-\dfrac{e^c+1}{c^2+\pi^2}\cos\dfrac{\pi x}{c}+\dfrac{e^c-1}{c^2+4\pi^2}\cos\dfrac{2\pi x}{c}\\
&&                 &\hspace{3em}-\dfrac{e^c+1}{c^2+9\pi^2}\cos\dfrac{3\pi x}{c}+\cdots\bigg]\\
&&                 &\text{from } x=0 \text{ to } x=c.
\end{flalign*}
\begin{flalign*}
&\rlap{\indent(5)\quad}   &  f(x)={}&\dfrac{4c}{\pi^2}\bigg[\sin\dfrac{\pi x}{c}-\dfrac{1}{3^2}\sin\dfrac{3\pi x}{c}+\dfrac{1}{5^2}\sin\dfrac{5\pi x}{c}+\cdots\bigg]\hspace{8em}\\
&&                 &\text{from } x=0 \text{ to } x=c,
\end{flalign*}
where $f(x)=x$ from $x=0$ to $x=\dfrac{c}{2}$ and $f(x)=c-x$ from $x=\dfrac{c}{2}$ to
$x=c$.\\
% -----File: 062.png

2.\quad Show that formula (4) Art.~31 can be written
\begin{multline*}
  f(x) = \frac{1}{2} c_0 \cos\beta_0
       + c_1 \cos \bigg( \frac{ \pi x}{c} - \beta_1 \bigg)
       + c_2 \cos \bigg( \frac{2\pi x}{c} - \beta_2 \bigg)
\\
       + c_3 \cos \bigg( \frac{3\pi x}{c} - \beta_3 \bigg) + \cdots
\end{multline*}
\vspace{-3ex}
\begin{flalign*}
&\text{where}&&
  c_m = (a_m^2 + b_m^2)^{\frac{1}{2}} \quad \text{and} \quad
  \beta_m = \tan^{-1}\frac{a_m}{b_m}.&&\phantom{where}
\end{flalign*}

3.\quad Show that formula (4) Art.~31 can be written
\begin{multline*}
  f(x) = \frac{1}{2} c_0 \sin\beta_0
       + c_1 \sin \bigg( \frac{ \pi x}{c} + \beta_1 \bigg)
       + c_2 \sin \bigg( \frac{2\pi x}{c} + \beta_2 \bigg)
\\
       + c_3 \sin \bigg( \frac{3\pi x}{c} + \beta_3 \bigg) + \cdots
\end{multline*}
\vspace{-3ex}
\begin{flalign*}
&\text{where}&&
  c_m = (a_m^2 + b_m^2)^{\frac{1}{2}} \quad \text{and} \quad
  \beta_m = \tan^{-1}\frac{a_m}{b_m}.&&\phantom{where}
\end{flalign*}

\mypara{32.} In the formulas of Art.~31 $c$ may have as great a value as we please,
so that we can obtain a Trigonometric Series for $f(x)$ that will represent the
given function through as great an interval as we may choose to take. If,
then, we can obtain the limiting form approached by the series (4) Art.~31 as
$c$ is indefinitely increased the expression in question ought to be equal to the
given function of $x$ for all values of $x$. Equation (4) Art.~31 can be written
as follows if we replace $b_0$, $b_1$, $b_2$, $\cdots$ $a_1$, $a_2$, $\cdots$ by their values given in
Art.~31 (5) and (6).
\begin{align*}
f(x) &= \frac{1}{c} \biggl[
\begin{aligned}[t]
& \frac{1}{2} \int\limits_{-c}^{c} f(\lambda) \,d\lambda
\\
&+ \int\limits_{-c}^{c} f(\lambda)
   \cos\frac{ \pi\lambda}{c} \cos\frac{ \pi x}{c} d\lambda
 + \int\limits_{-c}^{c} f(\lambda)
   \cos\frac{2\pi\lambda}{c} \cos\frac{2\pi x}{c} d\lambda + \cdots
\\
&+ \int\limits_{-c}^{c} f(\lambda)
   \sin\frac{ \pi\lambda}{c} \sin\frac{ \pi x}{c} d\lambda
 + \int\limits_{-c}^{c} f(\lambda)
   \sin\frac{2\pi\lambda}{c} \sin\frac{2\pi x}{c} d\lambda + \cdots \biggr]
\end{aligned}
\\
&= \frac{1}{c} \int\limits_{-c}^{c} f(\lambda) \,d\lambda \biggl[
\begin{aligned}[t]
&  \frac{1}{2}
 + \cos\frac{ \pi\lambda}{c} \cos\frac{\pi x}{c}
 + \sin\frac{ \pi\lambda}{c} \sin\frac{\pi x}{c}
\\[1ex]
&+ \cos\frac{2\pi\lambda}{c} \cos\frac{2\pi x}{c}
 + \sin\frac{2\pi\lambda}{c} \sin\frac{2\pi x}{c} + \cdots \biggr ]
\end{aligned}
\end{align*}
% -----File: 063.png
\begin{flalign*}
  f(x) &= \frac{1}{c} \int\limits_{-c}^c f(\lambda) \,d\lambda \biggl[
  \frac{1}{2} + \cos\frac{ \pi}{c} (\lambda-x)
              + \cos\frac{2\pi}{c} (\lambda-x) + \cdots \biggr]
\\
&= \frac{1}{2c} \int\limits_{-c}^c f(\lambda) \,d\lambda \biggl[
\begin{aligned}[t]
& 1 + \cos\frac{ \pi}{c} (\lambda-x)
    + \cos\frac{2\pi}{c} (\lambda-x) + \cdots
\\
&+ \cos \Bigl( -\frac{ \pi}{c} \Bigr) (\lambda-x)
 + \cos \Bigl( -\frac{2\pi}{c} \Bigr) (\lambda-x) + \cdots \biggr]
\end{aligned}
\end{flalign*}
since $\cos(-\phi)=\cos\phi$.
\begin{flalign*}
  f(x) = \frac{1}{2\pi} \int\limits_{-c}^c f(\lambda) \,d\lambda \biggl[ \cdots
&+ \frac{\pi}{c} \cos \Bigl( -\frac{2\pi}{c} \Bigr) (\lambda-x)
 + \frac{\pi}{c} \cos \Bigl( -\frac{ \pi}{c} \Bigr) (\lambda-x)
\\
&+ \frac{\pi}{c} \cos \frac{0\pi}{c} (\lambda-x)
 + \frac{\pi}{c} \cos \frac{ \pi}{c} (\lambda-x)
\\
&+ \frac{\pi}{c} \cos \frac{2\pi}{c} (\lambda-x) + \cdots \biggr] \tag{1}
\end{flalign*}

As $c$ is indefinitely increased the limiting value approached by the
parenthesis in (1) is
\[
\int\limits_{-\infty}^{\infty}\cos\alpha(\lambda-x).d\alpha.
\]

Hence the limiting form approached by (1) is
\[
  f(x)
= \frac{1}{2\pi}
  \int\limits_{-\infty}^{\infty} f(\lambda) \,d\lambda
  \int\limits_{-\infty}^{\infty} \cos\alpha(\lambda-x).d\alpha, \tag{2}
\]
and the second member of (2) must be equal to $f(x)$ for all values of $x$.

The double integral in (2) is known as \emph{Fourier's Integral}, and since it is a
limiting form of \emph{Fourier's Series} it is subject to the same limitations as the
series.

That is, in order that (2) should be true $f(x)$ must be finite, continuous, and
single valued for all values of $x$, or if discontinuous, must have only finite
discontinuities.\footnote{See note on page \pageref{notep38}.}

(2) is sometimes given in a slightly different form.

Since \quad $\displaystyle
  \int\limits_{-\infty}^{\infty} \cos\alpha(\lambda-x).d\alpha
= \int\limits_{-\infty}^0 \cos\alpha(\lambda-x).d\alpha
+ \int\limits_0^{\infty}  \cos\alpha(\lambda-x).d\alpha$\\
and
\begin{gather*}
  \int\limits_{-\infty}^0 \cos\alpha(\lambda-x).d\alpha
= \int\limits_{ \infty}^0 \cos(-\alpha)(\lambda-x).d(-\alpha)
=-\int\limits_{ \infty}^0 \cos\alpha(\lambda-x).d\alpha
\\
  \int\limits_{-\infty}^{\infty} \cos\alpha(\lambda-x).d\alpha
=2\int\limits_0^{\infty} \cos\alpha(\lambda-x).d\alpha
\end{gather*}
% -----File: 064.png
and (2) may be written
\[
f(x) = \frac{1}{\pi}
       \int\limits_{-\infty}^{\infty} f(\lambda) \,d\lambda
       \int\limits_{0}^{\infty} \cos \alpha(\lambda-x).d\alpha.      \tag{3}
\]

If $f(x)$ is an \textit{even} function or an \textit{odd} function (3) can be still further simplified.

Let \hfill $ f(x) = -f(-x)$. \hfill\phantom{\indent Let}

Since the limits of integration in (3) do not contain $\alpha$ or $\lambda$ the integrations
may be performed in whichever order we choose. That is
\[
  \int\limits_{-\infty}^{\infty} f(\lambda) \,d\lambda
  \int\limits_{0}^{\infty} \cos \alpha(\lambda-x).d\alpha
= \int\limits_{0}^{\infty} d\alpha
  \int\limits_{-\infty}^{\infty} f(\lambda) \cos\alpha(\lambda-x).d\lambda.
\]

Now
\begin{gather*}
  \int\limits_{-\infty}^{\infty} \!f(\lambda) \cos\alpha(\lambda-x).d\lambda
= \int\limits_{-\infty}^{0} \!f(\lambda) \cos\alpha(\lambda-x).d\lambda
+ \int\limits_{0}^{\infty} \!f(\lambda) \cos\alpha(\lambda-x).d\lambda.
\\
\begin{aligned}
   \int\limits_{-\infty}^{0} f(\lambda) \cos\alpha(\lambda-x).d\lambda
&= \int\limits_{\infty}^{0} f(-\lambda) \cos\alpha(-\lambda-x).d(-\lambda)
\\
&=-\int\limits_{0}^{\infty} f(\lambda) \cos\alpha(\lambda+x).d\lambda
\end{aligned}
\end{gather*}
and (3) becomes
\begin{align*}
f(x)
&= \frac{1}{\pi} \int\limits_{0}^{\infty} d\alpha \int\limits_{0}^{\infty}
   f(\lambda) [\cos\alpha(\lambda-x) - \cos\alpha(\lambda+x).d\lambda
\\
&= \frac{2}{\pi} \int\limits_{0}^{\infty} d\alpha \int\limits_{0}^{\infty}
   f(\lambda) \sin\alpha\lambda \sin\alpha x.d\lambda
\end{align*}
\begin{flalign*}
&\text{or}&&
  f(x) = \frac{2}{\pi} \int\limits_{0}^{\infty} f(\lambda)d\lambda \int\limits_{0}^{\infty} \sin\alpha\lambda \sin\alpha x.d\alpha. &&  \tag{4}
\end{flalign*}

If $f(x) = f(-x)$ (3) can be reduced in like manner to
\[
f(x) = \frac{2}{\pi} \int\limits_{0}^{\infty} f(\lambda) \,d\lambda \int\limits_{0}^{\infty} \cos\alpha\lambda \cos\alpha x.d\alpha.  \tag{5}
\]

Although (4) holds for all values of $x$ only in case $f(x)$ is an \textit{odd} function,
and (5) only in case $f(x)$ is an \textit{even}, function, both (4) and (5) hold for all
\textit{positive} values of $x$ in the case of any function.

\EXAMPLE{}

(1)\quad Obtain formulas (4) and (5) directly from (7) and (9) Art.~31.
\label{ch2end}

% -----File: 065.png

\mychap{CHAPTER III.}{CONVERGENCE OF FOURIER'S SERIES.}
\label{ch3start}

\mypara{33.} The question of the \textit{convergence} of a Fourier's Series is altogether too
large to be completely handled in an elementary treatise. We will, however,
consider at some length one of the most important of the series we have
obtained, namely
\[
  \frac{4}{\pi} \biggl[ \sin x + \frac{\sin 3x}{3}
  + \frac{\sin 5x}{5 } + \frac{\sin 7x}{7} + \cdots\biggr],
\tag*{[v.~(3) Art.~26(\emph{b}).]}
\]
and prove that for all values of $x$ between zero and $\pi$ its sum is absolutely
equal to unity; that is, that the limit approached by the sum of $n$ terms of the
series
\[
  \frac{2}{\pi} \biggl[
    \sin  x \int\limits_{0}^{\pi} \sin  \alpha.d\alpha
  + \sin 2x \int\limits_{0}^{\pi} \sin 2\alpha.d\alpha
  + \sin 3x \int\limits_{0}^{\pi} \sin 3\alpha.d\alpha + \cdots \biggr ],
\]
as $n$ is indefinitely increased, is $1$, provided that $x$ lies between zero and $\pi$.

Let
\begin{align*}
S_n = \frac{2}{\pi}\biggl[
      &  \sin  x \int\limits_0^{\pi} \sin  \alpha.d\alpha
       + \sin 2x \int\limits_0^{\pi} \sin 2\alpha.d\alpha
       + \sin 3x \int\limits_0^{\pi} \sin 3\alpha.d\alpha + \cdots
\\
      &+ \sin nx \int\limits_0^{\pi} \sin n\alpha.d\alpha \biggr ].  \tag{1}
\end{align*}

Then
\begin{align*}
  S_{n} &= \frac{2}{\pi} \int\limits_0^{\pi} [
     \sin\alpha\sin  x + \sin 2\alpha\sin 2x + \sin 3\alpha\sin 3x
   + \cdots + \sin n\alpha\sin nx ] d\alpha
\\
&=  \frac{1}{\pi} \int\limits_0^{\pi}
\begin{aligned}[t]
[ & \cos  (\alpha-x) - \cos  (\alpha+x)
  + \cos 2(\alpha-x) - \cos 2(\alpha+x) + \cdots
\\[3ex]
  &+\cos n(\alpha-x) - \cos n(\alpha+x) ] \,d\alpha
\end{aligned}
\\
&=  \frac{1}{\pi} \int\limits_0^{\pi}
    [ \cos  (\alpha-x) + \cos 2(\alpha-x) + \cos 3(\alpha-x) + \cdots
    + \cos n(\alpha-x) ] d\alpha
\\
&-\frac{1}{\pi} \int\limits_0^{\pi}
    [ \cos  (\alpha+x) + \cos 2(\alpha+x) + \cos 3(\alpha+x) + \cdots
    + \cos n(\alpha+x) ] d\alpha.
\end{align*}
% -----File: 066.png

Therefore by Art.~20 (1)
\begin{gather*}
\begin{aligned}
S_n ={}& \frac{1}{\pi} \int\limits_{0}^{\pi} \biggl[ -\frac{1}{2}
         + \frac{1}{2} \frac{ \sin (2n+1)\dfrac{\alpha-x}{2} }
                            { \sin       \dfrac{\alpha-x}{2} } \biggr] d\alpha
\\
&- \frac{1}{\pi} \int\limits_{0}^{\pi} \biggl[ -\frac{1}{2}
         + \frac{1}{2} \frac{ \sin (2n+1)\dfrac{\alpha+x}{2} }
                            { \sin       \dfrac{\alpha+x}{2} } \biggr] d\alpha.
\end{aligned}
\\
S_n = \frac{1}{2\pi} \int\limits_{0}^{\pi}
      \frac{ \sin (2n+1)\dfrac{\alpha-x}{2} }
           { \sin       \dfrac{\alpha-x}{2} } d\alpha
    - \frac{1}{2\pi} \int\limits_{0}^{\pi}
      \frac{ \sin (2n+1)\dfrac{\alpha+x}{2} }
           { \sin       \dfrac{\alpha+x}{2} } d\alpha.
\end{gather*}

In the first integral substitute $\beta$ for  $\dfrac{\alpha-x}{2}$, and in the second integral substitute
$\beta$ for $\dfrac{\alpha+x}{2}$.

We get
\[
S_n = \frac{1}{\pi} \int\limits_{-\frac{x}{2}}^{\frac{\pi}{2}-\frac{x}{2}}
      \frac{\sin(2n+1)\beta }{ \sin\beta } \,d\beta
    - \frac{1}{\pi} \int\limits_{ \frac{x}{2}}^{\frac{\pi}{2}+\frac{x}{2}}
      \frac{\sin(2n+1)\beta }{ \sin\beta } \,d\beta .   \tag{2}
\]

It remains to find the limit approached by $S_{n}$ as $n$ is indefinitely increased.

\markright{CONVERGENCE OF FOURIER'S SERIES.}
\mypara{34.}\vspace{-2ex}
\[
\int\limits_0^{\frac{\pi}{2}} \frac{ \sin(2n+1)\beta}{ \sin\beta} d\beta = \frac{\pi}{2}.  \tag{1}
\]

For
\begin{flalign*}
&\phantom{fr}& \frac{\sin(2n+1)\beta }{ 2 \sin\beta} = \tfrac{1}{2 }+ \cos 2\beta + \cos 4\beta &+ \cdots + \cos 2n\beta, && \text{by Art.~20.}
\\
&\text{and }& \int\limits_0^{\frac{\pi}{2}} \cos 2k\beta.d\beta &= 0.&&
\end{flalign*}

Let us construct the curve
\[
y = \frac{\sin(2n+1)x}{\sin x}.
\]

We have only to draw the curve $y = \sin (2n + 1)x$ and then to divide
the length of each ordinate by the value of the sine of the corresponding
abscissa.

In $y = \sin (2n + 1)x$ the successive arches into which the curve is
divided by the axis of $X$ are equal, and consequently their areas are equal.
% -----File: 067.png

%[Illustration]
\pngrightl{005.png}{476}{28}{-12}

Each arch has for its altitude unity and for its base
 $\dfrac{\pi}{2n+1}$ and is symmetrical with respect to the ordinate
of its highest or lowest point.

If now we form the curve $y = \dfrac{\sin(2n+1)x }{\sin x}$ from
the curve $y= \sin(2n + 1)x$, it is clear that, since
$\sin x$ increases as $x$ increases from $0$ to $\dfrac{\pi}{2}$, the ordinate
of any point of the new curve will be shorter
than the ordinate of the corresponding point in the
preceding arch, and that consequently the area of
each arch $y=\dfrac{\sin(2n + 1)x}{\sin x}$ will be less than
that of the arch before it.

If $a_{0}$, $a_{1}$, $a_{2}$, $\cdots a_{n-1}$ are the areas of the successive
arches and $a_n$ that of the incomplete arch terminated
by the ordinate corresponding to $x =\dfrac{\pi}{2}$
\[
\int\limits_0^{\frac{\pi}{2}} \frac{\sin(2n+1)x}{\sin x} \,dx = a_0-a_1+a_2-a_3+\cdots .
\]
But
\[
  \int\limits_0^{\frac{\pi}{2}} \frac{\sin(2n+1)x    }{\sin x    } \,dx
= \int\limits_0^{\frac{\pi}{2}} \frac{\sin(2n+1)\beta}{\sin \beta} \,d\beta
= \frac{\pi}{2}   \tag*{by (1).}
\]

Hence
\[
\frac{\pi}{2}= a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - \cdots + a_{n}  \quad \text{if $n$ is even,}
\]
or
\[
\frac{\pi}{2}= a_{0} - a_{1} + a_{2} - a_{3} + a_{4} - \cdots - a_{n}  \quad \text{if $n$ is odd.}
\]

These equations can be written
\begin{align*}
\frac{\pi}{2} ={}& a_{0} + (-a_{1} + a_{2}) + (-a_{3} + a_{4})\\
         &+ (-a_{5} + a_{6}) + \cdots +(-a_{n-1} + a_{n})
\end{align*}
if $n$ is even, and
\begin{align*}
\frac{\pi}{2} ={}& a_{0} + (-a_{1} + a_{2}) + (-a_{3} + a_{4})\\
         &+ (-a_{5} + a{_6}) + \cdots + (-a_{n-2}+a_{n-1})+(-a_{n})
\end{align*}
if $n$ is odd.
% -----File: 068.png

In either case each parenthesis is a negative quantity since
\[
  a_0 > a_1 > a_2 > a_3 \cdots > a_n,
\]
and it follows that $a_0$ is greater than $\dfrac{\pi}{2}$.

Again
\begin{align*}
&\frac{\pi}{2} = a_{0} - a_{1} + (a_2-a_3) + (a_4-a_5) + \cdots + (a_{n-2}-a_{n-1}) + a_n
\intertext{if $n$ is even and}
&\frac{\pi}{2} = a_{0} - a_{1} + (a_2-a_3) + (a_4-a_5) + \cdots + (a_{n-1}-a_{n})
\end{align*}
if $n$ is odd.

In either case each parenthesis is positive and it follows that $a_0-a_1$ is
is less than $\dfrac{\pi}{2}$.

Since
\[
  a_0 > \frac{\pi}{2} > a_0 - a_1,
\]
$a_{0}$ and $a_{0} - a_{1}$ differ from $\dfrac{\pi}{2}$  by less than they differ from each other, that
is, by less than $a_1$.

In like manner we can show that $a_0-a_1$ and $a_0-a_1+a_2$ differ from
$\dfrac{\pi}{2}$ by less than $a_2$; and in general that $a_0-a_1+a_2-a_3+\cdots \pm a_k$ differs
from $\dfrac{\pi}{2}$\vphantom{\fbox{$\dfrac00$}}
%[**F2: Spacer added to keep the dfrac from colliding with the line above.]
by less than $a_k$; or even that
\[
  a_0 - a_1 + a_2 - a_3 + \cdots \pm \frac{a_k}{p}
\]
differs from $\dfrac{\pi}{2}$ by less than $a_k$ no matter the value of $p$, provided $p$ is
greater than unity.

\mypara{35.} From what has been proved in the last article it follows that
\[
  \int\limits_0^b \frac{\sin(2n+1)x}{\sin x} \,dx,
\]
where $b$ is some value between $\dfrac{\pi}{2n+1}$ and $\dfrac{\pi}{2}$, differs from $\dfrac{\pi}{2}$ by less than
the area of the arch in which the ordinate of $y= \dfrac{\sin(2n+1)x}{\sin x}$ corresponding
to $x = b$ falls if this ordinate divides an arch, or by less than the area
of the arch next beyond the point $(b, 0)$ if the curve crosses the axis of $X$ at
that point.
% -----File: 069.png

The area of the arch in question is less than $\dfrac{\pi}{2n+1}$, its base, multiplied by
$\dfrac{1}{ \sin\Big( b-\dfrac{\pi}{ 2n+1} \Big)}$, a value greater than the length of its longest ordinate.

Therefore \hfill $\displaystyle
  \int\limits_0^b \frac{\sin(2n+1)x }{ \sin x } \,dx $\hfill\hfill\quad\\
differs from $\dfrac{\pi}{2}$ by less than $\dfrac{\pi}{2n+1} \dfrac{ 1}{\sin\Big( b-\dfrac{\pi}{ 2n+1} \Big)}$ .

If now $n$ is indefinitely increased $\dfrac{\pi}{2n+1} \dfrac{ 1}{ \sin\Big( b-\dfrac{\pi}{2n+1} \Big)}$  approaches
zero as its limit, and we get the very important result
\begin{flalign*}
&&  \limit_{n=\infty} \biggl[\int\limits_0^b \frac{\sin(2n+1)x }{ \sin x } \,dx \biggr]
= \frac{\pi}{2} && \phantom{\text{if } 0<b< \dfrac{\pi}{2}}\tag{1}
\\
&\text{if } 0<b< \dfrac{\pi}{2}.\\
\end{flalign*}

\mypara{36.}\vspace{-7.9ex}
\begin{flalign*}
&&S_n ={}
& \frac{1}{\pi} \int\limits_{-\frac{x}{2}}^{\frac{\pi}{2}-\frac{x}{2}}
  \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
- \frac{1}{\pi} \int\limits_{ \frac{x}{2}}^{\frac{\pi}{2}+\frac{x}{2}}
  \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta.
\hspace{6pt}\text{[Art.~33. (2)]}\\[1ex]
&&={}& \frac{1}{\pi} \int\limits_{-\frac{x}{2}}^0
        \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
      + \frac{1}{\pi} \int\limits_0^{\frac{\pi}{2}-\frac{x}{2}}
        \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
\\[1ex]
&&&- \frac{1}{\pi} \int\limits_0^{\frac{\pi}{2}}
      \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
    + \frac{1}{\pi} \int\limits_0^{\frac{x}{2}}
      \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
\\[1ex]
&&&- \frac{1}{\pi} \int\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}+\frac{x}{2}}
      \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta.
\end{flalign*}

This last value for $S_n$ can be somewhat simplified.

Substituting $\gamma = -\beta$ we get
\[
  \int\limits_{-\frac{x}{2}}^0 \frac{ \sin(2n+1)\beta }{\sin\beta } \,d\beta
=-\int\limits_{ \frac{x}{2}}^0 \frac{ \sin(2n+1)\gamma}{\sin\gamma} \,d\gamma
= \int\limits_0^{\frac{x}{2}} \frac{\sin(2n+1)\beta }{\sin\beta } \,d\beta.
\]
% -----File: 070.png

Substituting $\gamma = \pi - \beta$ in
\[
\int\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}+\frac{x}{2}}
    \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta  \tag*{we have}
\]
\begin{flalign*} &&
    \int\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}+\frac{x}{2}}
    \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
&= -\int\limits_{\frac{\pi}{2}}^{\frac{\pi}{2}-\frac{x}{2}}
    \frac{\sin(2n+1)\gamma}{ \sin\gamma} \,d\gamma
 =  \int\limits_{\frac{\pi}{2}-\frac{x}{2}}^{\frac{\pi}{2}}
    \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta  &&
\\[1ex]
&&&=  \int\limits_{0}^{\frac{\pi}{2}}
    \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
 -  \int\limits_{0}^{\frac{\pi}{2}-\frac{x}{2}}
    \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta.  &&
\end{flalign*}

Hence
\[
S_n = \frac{2}{\pi} \int\limits_{0}^{\frac{x}{2}}
      \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
    + \frac{2}{\pi} \int\limits_{0}^{\frac{\pi}{2}-\frac{x}{2}}
      \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta
    - \frac{2}{\pi} \int\limits_{0}^{\frac{\pi}{2}}
      \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta.
\]
\[
  \int\limits_{0}^{\frac{\pi}{2}}
  \frac{\sin(2n+1)\beta}{ \sin\beta} d\beta = \frac{\pi}{2 }
\tag*{by (1) Art.~34.}
\]
\begin{flalign*} &&
   \limit_{n=\infty} \biggl[ \int\limits_{0}^{\frac{x}{2}}
      \frac{\sin(2n+1)\beta}{ \sin\beta} \,d\beta \biggr]
&= \frac{\pi}{2 } \quad \text{if} \quad 0<x<\pi  &&\text{by (1) Art.~35}
\\
\intertext{and}
&&  \limit_{n=\infty} \biggl[ \int\limits_0^{\frac{\pi}{2}-\frac{x}{2}}
      \frac{ \sin(2n+1)\beta}{ \sin\beta} \,d\beta \biggr]
&= \frac{\pi}{2} \quad \text{if} \quad 0 < x < \pi   &&\text{by (1) Art.~35.}
\end{flalign*}
Therefore \hfill $\displaystyle\limit_{n=\infty} [S_n] = 1 + 1 - 1 = 1$ \quad if \quad $0 < x < \pi$ \hfill and
\[
  \frac{4}{\pi} \left[ \sin x + \frac{\sin 3x}{ 3}
+ \frac{\sin 5x}{ 5} + \frac{\sin 7x}{ 7} + \cdots \right] = 1
\]
for all values of $x$ between zero and $ \pi$.

\mypara{37.} By a somewhat long but not especially difficult extension of the reasoning
just given it can be shown that if $f(x)$ is \textit{single-valued} and \textit{finite}
between $x = -\pi$ and $x = \pi$, and has only a \textit{finite number} of \textit{discontinuities}
and of \textit{maxima} and \textit{minima} between $x = -\pi$ and $x = \pi$ the
\textit{Fourier's Series}
\begin{flalign*}
&&  \frac{1}{2} b_0 &+ b_1 \cos x + b_2 \cos 2x + b_3 \cos 3x + \cdots\\
&&                  &+ a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots\\
% -----File: 071.png
&\text{where}&  &\quad a_m = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(\alpha) \sin m\alpha.d\alpha  &&\phantom{where}\\
&\text{and}&    &\quad b_m = \frac{1}{\pi} \int\limits_{-\pi}^{\pi} f(\alpha) \cos m\alpha.d\alpha,  &&
\end{flalign*}
\textit{and that Fourier's Series only} is equal to $f(x)$ for all values of $x$ between
 $x = -\pi$ and $x = \pi$, \textit{excepting the values of $x$ corresponding to the
discontinuities} of $f(x)$, and the values $\pi$ and $-\pi$ if $f(\pi)$ is not equal to $f(-\pi)$;
and that if $c$ is a value of $x$ corresponding to a discontinuity of $f(x)$, the value
of the series when $x = c$ is
\[
\frac{1}{2} \limit_{\epsilon \doteq 0}[f(c-\epsilon) + f(c+\epsilon)];
\]
and that if $f(\pi)$ is not equal to $f(-\pi)$ the value of the series when $x = -\pi$
and when $x = \pi$ is
\[
   \frac{1}{2} [f(-\pi) + f(\pi)].
\]

If $f(x)$ while satisfying the conditions named in the preceding paragraph
except for a finite number of values of $x$, becomes infinite for those values, the
series is equal to the function except for the values of $x$ in question provided
that $\displaystyle\int\limits_{-\pi}^{\pi} f(x)\,dx$ is finite and determinate. \quad (v.\ Int.\ Cal.\ Arts.~83 and 84.)

\mypara{38.}  The question of the convergency of a Fourier's Series and the conditions
under which a function may be developed in such a series was first
attacked successfully by Dirichlet in 1829, and his conclusions have been
criticised and extended by later mathematicians, notably by Riemann, Heine,
Lipschitz, and du Bois Reymond. It may be noted that the criticisms relate
not to the sufficiency but to the necessity of Dirichlet's conditions.

An excellent résumé of the literature of the subject is given by Arnold
Sachse in a short dissertation published by Gauthier--Villars, Paris, 1880,
entitled ``Essai Historique sur la Représentation d'une Fonction Arbitraire
d'une seule variable par une Série Trigonométrique."

\mypara{39.}  A good deal of light is thrown on the peculiarities of trigonometric
series by the attempt to construct approximately the curves corresponding to them.

If we construct $y = a_1 \sin x$ and $y = a_2 \sin 2x$ and add the ordinates
of the points having the same abscissas we shall obtain points on the curve
% -----File: 072.png
\[
y = a_1 \sin x + a_2 \sin 2x.
\]

If now we construct $y = a_3 \sin 3x$ and add the ordinates to those of
 $y = a_1 \sin x + a_2 \sin 2x$ we shall get the curve
\[
y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x.
\]
By continuing this process we get successive approximations to
\[
y = a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + a_4 \sin 4x + \cdots
\]
Let us apply this method to a few of the series which we have obtained in
Chapter~II.

Take
\begin{align*}
  y &= \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots  \tag{1}
\\
    &= 0 \text{ when } x=0,\
       \frac{\pi}{4} \text{ from } x=0 \text{ to } x=\pi, \text{ and }
       0 \text{ when } x=\pi,
\\
\intertext{v.\ Art.~26 [\emph{b}](3).}
  y &= 2\left(\sin x - \frac{1}{2} \sin 2x + \frac{1}{3} \sin 3x
     - \frac{1}{4} \sin 4x + \cdots \right)        \tag{2}
\\
    &= x \text{ from } x=0 \text{ to } x=\pi, \text{ and }
       0 \text{ when } x=\pi,
\\
\intertext{Art.~26 [\emph{a}](4).}
  y &= \frac{4}{\pi}\left[\frac{1}{1^2} \sin x - \frac{1}{3^2}\sin 3x
     + \frac{1}{5^2}\sin 5x - \frac{1}{7^2} \sin 7x + \cdots \right]  \tag{3}
\\
    &= x \text{ from } x=0 \text{ to } x=\frac{\pi}{2}, \text{ and }
       \pi-x \text{ from } x=\frac{\pi}{2} \text{ to } x=\pi,
\intertext{Art.~26 [\emph{c}](2).}
  y &= \frac{1}{1} \sin  x + \frac{2}{2} \sin 2x + \frac{1}{3}\sin 3x
     + \frac{1}{5} \sin 5x - \frac{2}{6} \sin 6x + \frac{1}{7}\sin 7x + \cdots \tag{4}
\\
    &= 0 \text{ when } x=0, \
       \frac{\pi}{2} \text{ from } x=0 \text{ to } x=\frac{\pi}{2}, \text{ and }
       0 \text{ from } x=\frac{\pi}{2} \text{ to } x=\pi,
\end{align*}
v.\ Art.~26 [\emph{d}](2).

It must be borne in mind that each of these curves is periodic having the
period $2\pi$, and is symmetrical with respect to the origin.

The following figures I, II, III, and IV represent the first four approximations
to each of these curves.

In each figure the curve $y =$ the series, and the approximation in question
are drawn in continuous lines, and the preceding approximation and the
curve corresponding to the term to be added are drawn in dotted lines.

%[Illustration
\pngcent{006.png}{1466}%{2170}

\label{was63}

Figs.~I, II, III, and IV immediately suggest the following facts:

(\emph{a})  The curve representing each approximation is continuous even when
the curve representing the series is discontinuous.

(\emph{b})  When the curve representing the series is discontinuous the portion of
each successive approximate curve in the neighborhood of the point whose
abscissa is a value of $x$ for which the series curve is discontinuous approaches
more and more nearly a straight line perpendicular to the axis of $X$ and connecting
the separate portions of the series curve.

%[Illustration]
\pngcent{007.png}{1466}

\label{was64}

(\emph{c}) The curves representing successive approximations do not necessarily
tend to lose their wavy character, since each is obtained from the preceding
one by superposing upon it a wave line whose waves are shorter each time but
do not necessarily lose their sharpness of pitch. This is the case in Figures
I, II, and IV\@. In Fig.~III the waves of the superposed curves grow rapidly
flatter.

It follows from this that in such cases as those represented in Figures I, II,
and IV the direction of the approximate curve at a point having a given
abscissa does not in general approach the direction of the series curve at the
corresponding point, or indeed, approach any limiting value, as the approximation
is made closer and closer; and that the length of any portion of the
approximate curve will not in general approach the length of the corresponding
portion of the series curve.

% -----File: 073.png

Analytically this amounts to saying that the derivative of a function of $x$
cannot in general be obtained by differentiating term by term the Fourier's
Series which represents the function.

(\emph{d}) The area bounded by a given ordinate, the approximate curve, the axis of
$X$, and any second ordinate will approach as its limit the corresponding area of
the series curve if the series curve is continuous between the ordinates in
question; and will approach the area bounded by the given ordinate, the series
curve, the axis of $X$, any second ordinate, and a line perpendicular to the axis
of $X$, and joining the separate portions of the series curve if the latter has a
discontinuity between the ordinates in question.

% -----File: 074.png

% -----File: 075.png

Analytically this amounts to saying that the Fourier's Series corresponding
to any given function can be integrated term by term and the resulting series
will represent the integral of the function even when the function is
discontinuous (v.\ Int.\ Cal.\ Art.~83).

We may note in passing that if the function curve is continuous a curve
representing the integral of the function will be continuous and will not
change its direction abruptly at any point; while if the function curve is discontinuous
the curve representing the integral will still be continuous but will
change its direction abruptly at points corresponding to the discontinuities of
the given function.
% -----File: 076.png

\mypara{40.} The facts that the derivative of a Fourier's Series cannot in general be
obtained by differentiating the series term by term and that its integral can be
obtained by integrating the series term by term are so important that it is
worth while to look at the matter a little more closely. Let us consider the
differentiation of the series represented in Art.~39 Figure~I.

Let
\[
S_n = \sin x + \frac{1}{3} \sin 3x + \frac{1}{5} \sin 5x + \cdots + \frac{1}{2n+1} \sin (2n + 1)x.
\]
\begin{flalign*}
&\text{\indent Then}&\frac{dS_n}{dx} = \cos x + \cos 3x + \cos 5x + \cdots + \cos (2n + 1)x.&&\phantom{\indent Then}
\end{flalign*}

If $x = \dfrac{\pi}{2}$
\[
\frac{dS_n}{dx} = 0
\]
and the curve is parallel to the axis of $X$ for $x = \dfrac{\pi}{2}$ no matter what the
value of $n$.

If $x = 0$ or $x = \pi$
\[
\frac{dS_n}{dx} = 1 + 1 + 1 + 1 + \cdots + 1 = n + 1
\]
and the curve $y = S_n$ becomes more nearly perpendicular to the axis of $X$
at the origin and for $x = \pi$ as we increase $n$.

If $x = \dfrac{\pi}{3}$
\begin{flalign*}
&& \frac{dS_n}{dx}
&= \frac{1}{2} - 1 + \frac{1}{2} + \frac{1}{2} - 1 + \frac{1}{2} + \cdots \\
&\text{That is }& \frac{dS_n}{dx}& \!
\begin{alignedat}[t]{6}
&=  &&\frac{1}{2} &\ \ &\text{ if } &\ \ n&=0 &\ \ &\text{ or } &\ n&=3k    \\
&= -&&\frac{1}{2} &    &\text{ `` } &    n&=1 &    &\text{ `` } &  n&=3k+1  \\
&=  &&\,0         &    &\text{ `` } &    n&=2 &    &\text{ `` } &  n&=3k+2.
\end{alignedat} &&\phantom{That\ is}
\end{flalign*}
Consequently when $x = \dfrac{\pi}{3}$
$\dfrac{dS_n}{dx}$ does not approach any limiting value as $n$ is
indefinitely increased.  Indeed, in the successive approximations the point
whose abscissa is $\dfrac{\pi}{3}$ is successively on the rear, on the front, and on the crest
or in the trough of a wave, and although the waves are getting smaller they
do
not lose their sharpness of pitch.

If $x$ has any other value between $0$ and $\pi$ \ $\dfrac{dS_n}{dx}$ will change abruptly as $n$ is
changed and will not approach any limiting value as $n$ is increased.
% -----File: 077.png

\mypara{41.} In general if we differentiate a Fourier's Series
\begin{flalign*}
  S = \frac{1}{2} b_0
     &+ b_1 \cos x + b_2 \cos 2x + b_3 \cos 3x + \cdots  \\
   &+ a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots\\[-2ex]
\intertext{we get}\\[-6ex]
&\!
\begin{alignedat}{4}
   &- b_1 \sin x &&- 2b_2 \sin 2x &&- 3b_3 \sin 3x &&- \cdots   \\[1ex]
   &+ a_1 \cos x &&+ 2a_2 \cos 2x &&+ 3a_3 \cos 3x &&+ \cdots.
\end{alignedat}
\intertext{Differentiate again and we get}
   &- b_1 \cos x - 2^2b_2 \cos 2x - 3^2b_3 \cos 3x - \cdots  \\
   &- a_1 \sin x - 2^2a_2 \sin 2x - 3^2b_3 \sin 3x \,- \cdots .
\end{flalign*}

We see that each time we differentiate we multiply the coefficient of $\sin kx$
and of $\cos kx$ by $k$ while the term still involves $\cos kx$ or $\sin kx$.

Since the series
\begin{alignat*}{3}
  \cos x &+ \cos 2x &&+ \cos 3x &&+ \cdots  \\
+ \sin x &+ \sin 2x &&+ \sin 3x &&+ \cdots
\end{alignat*}
is not convergent, and a Fourier's Series converges only because its coefficients
decrease as we advance in the series, the differentiation of a Fourier's Series
must make its convergence less rapid if it does not actually destroy it, and
repetitions of the process will usually eventually make the derived series
diverge.

It is to be observed that the derived series are Fourier's Series, but of somewhat
special form, that is they lack the constant term. (v.\ Art.~30.)

If now we integrate a Fourier's Series
\begin{align*}
\frac{1}{2} b_0 &+ b_1 \cos x + b_2 \cos 2x + b_3 \cos 3x + \cdots  \\
                &+ a_1 \sin x + a_2 \sin 2x + a_3 \sin 3x + \cdots  \\[-6ex]
\end{align*}
\begin{flalign*}
&\text{we get } &&
\begin{alignedat}[t]{4}
  C + \frac{1}{2} b_0x
    &+ b_1 \sin x &&+ \frac{1}{2} b_2 \sin 2x &&+ \frac{1}{3} b_3 \sin 3x &&+ \cdots\\
    &- a_1 \cos x &&- \frac{1}{2} a_2 \cos 2x &&- \frac{1}{3} a_3 \cos 3x &&- \cdots,
\end{alignedat}
&&\phantom{we\ get}
\end{flalign*}
a Trigonometric Series which converges more rapidly than the given series.

It is to be observed that the series obtained by integrating a Fourier's
Series is not in general a Fourier's Series owing to the presence of the term
$\frac{1}{2} b_0x$.  (v.\ Art.~30.)

\mypara{42.} We are now ready to consider the conditions under which a function of
$x$ can be developed into a Fourier's Series whose term by term derivative shall
be equal to the derivative of the function.
% -----File: 078.png

Let the function $f(x)$ satisfy the conditions stated in Art.~37. Then there
is one Fourier's Series and but one which is equal to it. Call this series $S$.

Let the derivative $f'(x)$\footnote
  {We shall regularly use the notation $f'(x)$ for $\dfrac{df(x)}{dx}$. v.\ Dif.\ Cal.\ Art.~124.}
 of the given function also satisfy the conditions
stated in Art.~37. Then $f'(x)$ can be expressed as a Fourier's Series. By Art.~39
($d$) the integral of this latter series will be equal to the integral of $f'(x)$,
that is to $f(x)$ plus a constant, and one integral will be equal to $f(x)$.

If this integral which is necessarily a Trigonometric Series is a Fourier's
Series it must be identical with $S$. It will be a Fourier's Series only in case
the Fourier's Series for $f'(x)$ lacks the constant term $\frac{1}{2}b_0$.
\begin{flalign*}
&\indent\text{But}&  b_0 = \frac{1}{\pi} \int\limits_{-\pi}^\pi &  f'(x)dx  &\text{by (3) Art.~30.}&\\
&\text{Therefore }&  b_0 = \frac{1}{\pi}[f(\pi) & - f(-\pi)];  &&
\end{flalign*}
and will be zero if $f(\pi)=f(-\pi)$.

In order that $f'(x)$ shall satisfy the conditions stated in Art.~37 $f(x)$ while
satisfying the same conditions must in addition be finite and continuous
between $x=-\pi$ and $x=\pi$.

If, then, $f(x)$ is \emph{single-valued, finite, and continuous, and has only a finite
number of maxima and minima}, between $x=-\pi$ and $x=\pi$, (the values
 $x=-\pi$ and $x=\pi$ being included), \emph{and if} $f(\pi)=f(-\pi)$ $f(x)$ can be
developed into a Fourier's Series whose term by term derivative will be equal
to the derivative of the function.

It will be observed that in this case the periodic curve $y=S$ is continuous
throughout its whole extent.

\mypara{43.} Since a Fourier's Integral is a limiting case of a Fourier's Series the
conclusions stated in this chapter hold, \emph{mutatis mutandis} for a Fourier's
Integral.

For example if a function of $x$ is finite and single-valued for all values of $x$
and has not an infinite number of discontinuities or of maxima and minima in
the neighborhood of any value of $x$ it will be equal to the Fourier's Integral
\[
\frac{1}{\pi}\int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x).d\lambda
\]
and to that Fourier's Integral only, and the integral with respect to $x$ of this
Fourier's Integral will be equal to $\displaystyle\int f(x)dx$.

If in addition $f(x)$ is finite and continuous for all values of $x$ the derivative
of the Fourier's Integral with respect to $x$ will be equal to $\dfrac{df(x)}{dx}$.
\label{ch3end}

% -----File: 079.png

\mychap{CHAPTER IV.}{SOLUTION OF PROBLEMS IN PHYSICS BY THE AID OF FOURIER'S INTEGRALS AND FOURIER'S SERIES.}
\label{ch4start}

\mypara{44.} In Art.~7 we have already considered at some length a problem in
Heat Conduction which required the use of a Fourier's Series. We shall begin
the present chapter with a problem closely analogous in its treatment to that
of Art.~7, but calling for the use of a Fourier's Integral.

Suppose that electricity is flowing in a thin plane sheet of infinite extent
and that the value of the potential function is given for every point in some
straight line in the sheet, required the value of the potential function at any
point of the sheet.

Let us take the line as the axis of $X$ and consider at first only those points
for which $y$ is positive:

We have, then, to satisfy the equation
\[
  D_x^2 V + D_y^2 V = 0             \tag{1}
\]
subject to the conditions
\begin{flalign*}
&\quad&
\begin{aligned}
  V &= 0     &&\text{when }    &  y &= \infty  \\
  V &= f(x)  &&\text{\quad ``} &  y &= 0
\end{aligned}
&&
\begin{aligned} (2)& \\ (3)& \end{aligned}
\end{flalign*}
where $f(x)$ is a given function, and we are not concerned with negative
values of $y$.

As in Art.~7 we have $e^{-\alpha y} \sin \alpha x$ and $e^{-\alpha y} \cos \alpha x$ as particular values of $V$
which satisfy (1) and (2). We must multiply them by constant coefficients
and so combine them as to satisfy condition (3).

By (3) Art.~32
\[
f(x) = \frac{1}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty}
       f(\lambda) \cos \alpha(\lambda - x).d\lambda. \tag{4}
\]
We wish to build up a value of $V$ which will reduce to (4) when $y = 0$.
This requires a little care but not much ingenuity.
% -----File: 080.png

Take $e^{-\alpha y} \cos \alpha x$ and $e^{-\alpha y} \sin \alpha x$ and multiply the first by $\cos {\alpha\lambda}$, and
the second by $\sin {\alpha\lambda}$; they are still values of $V$ which satisfy (1). Add
these and we get
\[
e^{-\alpha y}\cos \alpha(\lambda- x),
\]
still a value of $V$ which satisfies (1), no matter what the values of $\alpha$ and $\lambda$.
Multiply by $f(\lambda)d\lambda$ and we have
\[
e^{-\alpha y}f(\lambda)\cos \alpha(\lambda - x).d\lambda   \tag{5}
\]
as a value of $V$ which satisfies (1).
\[
V= \int\limits_{-\infty}^{\infty} e^{-\alpha y}f(\lambda) \cos \alpha(\lambda - x).d\lambda           \tag{6}
\]
is still a solution of (1) since it is the limit of the sum of terms covered by
the form (5); and finally
\[
  V = \frac{1}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} e^{-\alpha y}f(\lambda) \cos \alpha(\lambda - x).d\lambda        \tag{7}
\]
is a solution of (1) as it is $\dfrac{1}{\pi}$ multiplied by the limit of the sum of terms
formed by multiplying the second member of (6) by $d\alpha$ and giving different
values to $\alpha$.

\markright{SOLUTION OF PROBLEMS IN PHYSICS.}
But (7) must be our required solution since while it satisfies (1) and (2), it
reduces to (4) when $y = 0$ and therefore satisfies condition (3).

If $f(x)$ is an \textit{even} function we can reduce (7) to the form
\[
  V = \frac{2}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_{0}^{\infty} e^{-\alpha y} f(\lambda) \cos \alpha x \cos \alpha\lambda.d\lambda      \tag{8}
\]
and if $f(x)$ is an \textit{odd} function to the form
\[
  V = \frac{2}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_{0}^{\infty} e^{-\alpha y} f(\lambda) \sin \alpha x \sin \alpha\lambda.d\lambda.      \tag{9}
\]

(7), (8), and (9) are valid only for positive values of $y$, but as the problem is
obviously symmetrical with respect to the axis of $X$, (7), (8), and (9) enable
us to get the value of the potential function at any point of the plane.

\EXAMPLE{S}

1.\quad  Obtain forms (8) and (9) directly by the aid of (5) and (4) Art.~32.\\

2.\quad  State a problem in statical electricity of which the solution given in
Art.~44 is the solution.
% -----File: 081.png

\mypara{45.} As a special case under Art.~44 let us consider the problem:---To find
the value of the potential function at any point of a thin plane sheet of infinite
extent where all points of a given line which lie to the left of the origin are
kept at potential zero, and all points which lie to the right of the origin are
kept at potential unity.

Here $f(x)=0$ if $x<0$ and $f(x)=1$ if $x>0$.

(7) Art.~44 gives us the required solution. It is
\[
V=\frac{1}{\pi}\int\limits_0^\infty d\alpha\int\limits_0^\infty e^{-\alpha y}\cos\alpha(\lambda-x).d\lambda; \tag{1}
\]
but this can be much simplified.

We have
\begin{flalign*}
&& V= {}& \frac{1}{\pi}\int\limits_0^\infty d\lambda\int\limits_0^\infty e^{-\alpha y}\cos\alpha(\lambda-x).d\alpha.  &&
\\
&\indent\text{Now}&  &\int\limits_0^\infty e^{-ax}\cos mx.dx = \frac{a}{a^2+m^2}  &&\phantom{\indent Now}
\end{flalign*}
if $a>0$. (Int.\ Cal.\ Art.~82, Ex.~8.)
\begin{flalign*}
&\indent\text{Hence} &  \int\limits_0^\infty e^{-\alpha y} \cos\alpha(\lambda-x).d\alpha = \frac{y}{y^2+(\lambda-x)^2},  &&\phantom{\indent Hence}  \\
&\text{and}
&  V = \frac{1}{\pi} \int\limits_0^\infty \frac{y\,d\lambda}{y^2+(\lambda-x)^2}
= \frac{1}{\pi} \left( \frac{\pi}{2} + \tan^{-1}\frac{x}{y} \right). &&
\end{flalign*}
\[
  \tan \left( \frac{\pi}{2} - \tan^{-1}\frac{x}{y} \right)
= \ctn \left( \tan^{-1}\frac{x}{y} \right) = \frac{y}{x};
\]
and  consequently
\[
V=\frac{1}{\pi}\left(\frac{\pi}{2}+\tan^{-1}\frac{x}{y}\right)
=1-\frac{1}{\pi}\tan^{-1}\frac{y}{x}.  \tag{2}
\]

Since \hfill $\log z=\log(x+yi)=\dfrac{1}{2}\log(x^2+y^2)+i\tan^{-1}\dfrac{y}{x}$, \hfill
{\,} \mbox{[Int.\ Cal.\ Art.~33 (2)],}
\[
  i - \frac{1}{\pi}\log z
= i - \frac{1}{\pi}\log(x+yi)
= -\frac{1}{2\pi}\log(x^2+y^2)
+ i\left( 1 - \frac{1}{\pi}\tan^{-1}\frac{y}{x} \right)
\]
and $1-\dfrac{1}{\pi}\tan^{-1}\dfrac{y}{x}$ and $-\dfrac{1}{2\pi}\log(x^2+y^2)$ are \emph{conjugate functions}. (v.\ Int.\
Cal.\ Arts.~209 and 210.) Hence
\[
V_1=-\frac{1}{2\pi}\log(x^2+y^2)  \tag{3}
\]
is a solution of the equation
\[
D_x^2V_1 + D_y^2V_1=0;  \tag{4}
\]
% -----File: 082.png
and the curves
\begin{flalign*}
&&&\frac{1}{\pi}\left(\frac{\pi}{2}+\tan^{-1}\frac{x}{y}\right)=a \tag{5}\\
&\text{and}&  &-\frac{1}{2\pi}\log(x^2+y^2)=b &&\phantom{and}  \tag{6}
\end{flalign*}
cut each other at right angles.

If we construct the curves obtained by giving different values to $a$ in (5) we
get a set of \emph{equipotential lines} for the conducting sheet described at the beginning
of this article, and the curves obtained by giving different values to $(b)$ in
(6) will be the \emph{lines of flow}.

Moreover since
\[
V_1=-\frac{1}{2\pi}\log(x^2+y^2)  \tag{3}
\]
is a solution of Laplace's Equation (4), the lines of flow just mentioned will be
equipotential lines for a certain distribution of potential, for which the equipotential
lines above mentioned will be lines of flow.

$V=a$, that is
\begin{flalign*}
&&\frac{1}{\pi}&\left(\frac{\pi}{2}+\tan^{-1}\frac{x}{y}\right)=a,  \tag{5}\\
&\text{reduces to} & y&=-x\tan a\pi. &&\phantom{reduces\ to}  \tag{7}
\end{flalign*}

If now we give to $a$ values differing by a constant amount we get a set of
straight lines radiating from the origin and at equal angular intervals.

$V_1=b$, that is
\begin{flalign*}
&&-\frac{1}{2\pi}&\log(x^2+y^2)=b,  &&\tag{6}\\[-1ex]
\intertext{reduces to}\\[-6ex]
&&x^2&+y^2=e^{-2\pi b}.  \tag{8}
\end{flalign*}

%[Illustration]
\pngright{008.png}{701}{390}{-24}

If we give to $b$ a set of values differing by a constant amount we get a set
of circles whose centres are at the origin and whose radii form a geometrical
progression. They are the equipotential lines for a thin plane sheet of infinite
extent where the potential function is kept equal to given different constant
values on the circumferences of two given concentric circles or where we have
a \emph{source} at the origin; and for this
system the lines (7) are lines of flow,
and (3) is the complete solution.

The figure gives the equipotential
lines and lines of flow for either system,
but only for positive values of $y$.
The complete figure has the axis of $X$
as an axis of symmetry.

% -----File: 083.png
\EXAMPLE{S}

1.\quad Solve the problem of Art.~44 for the case where
\[
  f(x)=-1 \quad\text{if}\quad x<0 \quad\text{and}\quad
  f(x)= 1 \quad\text{if}\quad x>0.
\]
\begin{flushright}
\textit{Ans.}, \quad $V = \dfrac{2}{\pi} \tan^{-1}\dfrac{x}{y}$.
\end{flushright}

2.\quad Solve the problem of Art.~44 for the case where
\[
  f(x)=a \quad\text{if}\quad x<0  \quad\text{and}\quad
  f(x)=b \quad\text{if}\quad x>0.
\]
\begin{flushright}
\textit{Ans.}, \quad $V = \dfrac{1}{2}(a+b) + \dfrac{1}{\pi}(b-a) \tan^{-1}\dfrac{x}{y}$.
\end{flushright}

3.\quad Reduce (7), (8), and (9) Art.~44 to the forms
\begin{align*}
  V &= \dfrac{1}{\pi} \int\limits_{-\infty}^\infty \dfrac{yf(\lambda)d\lambda}{y^2+(\lambda-x)^2},  \\
  V &= \dfrac{1}{\pi} \int\limits_0^\infty yf(\lambda)d\lambda \left[ \dfrac{1}{y^2+(\lambda-x)^2} + \dfrac{1}{y^2+(\lambda+x)^2} \right],  \\
  V &= \dfrac{1}{\pi} \int\limits_0^\infty yf(\lambda)d\lambda \left[ \dfrac{1}{y^2+(\lambda-x)^2} - \dfrac{1}{y^2+(\lambda+x)^2} \right],
\end{align*}
respectively.

\mypara{46.} An especially interesting case of Art.~44 is the following where
\[
  f(x)=0 \;\;\text{if}\;\;  x<-1, \quad
  f(x)=1 \;\;\text{if}\;\;  -1<x< 1, \quad \text{and} \quad
  f(x)=0 \;\;\text{if}\;\;  x> 1.
\]
\begin{flalign*}
&\indent\text{Here} & V = \frac{1}{\pi} \left[ \tan^{-1}\frac{1+x}{y} + \tan^{-1}\frac{1-x}{y} \right].  &&\phantom{\indent Here} \tag{1}
\end{flalign*}
\begin{flalign*}
&\text{Now} & \frac{1}{\pi} \log[(1-z)i]
&= \frac{1}{\pi} \log[(1-x-yi)i]
 = \frac{1}{\pi} \log[y+(1-x)i] &&\phantom{Now}
\\
&&&= \frac{1}{2\pi} \log[(1-x)^2+y^2] + \frac{i}{\pi} \tan^{-1}\frac{1-x}{y},  &&
\end{flalign*}
and
\begin{align*}
   -\frac{1}{\pi} \log[(-1-z)i]
&= -\frac{1}{\pi} \log[(-1-x-yi)i]
 = -\frac{1}{\pi} \log[y-(1+x)i]
\\
&= -\frac{1}{2\pi} \log[(1+x)^2+y^2]
 +  \frac{i}{ \pi} \tan^{-1}\frac{1+x}{y}.
\end{align*}
\[
  \frac{1}{ \pi} \log\frac{1-z}{-1-z}
= \frac{1}{2\pi} \log\frac{(1-x)^2+y^2}{(1+x)^2+y^2}
+ \frac{i}{\pi} \left[ \tan^{-1}\frac{1+x}{y} + \tan^{-1}\frac{1-x}{y} \right].
\]
% -----File: 084.png
Hence
\[
  \frac{1}{\pi} \left( \tan^{-1}\frac{1+x}{y} + \tan^{-1}\frac{1-x}{y} \right) \quad\text{and}\quad \frac{1}{2\pi} \log\frac{(1-x)^2+y^2}{(1+x)^2+y^2}
\]
are \emph{conjugate functions:}\footnote
  {The function conjugate to
\[
  \frac{1}{\pi} \left[ \tan^{-1}\frac{1+x}{y} + \tan^{-1}\frac{1-x}{y} \right]
\]
might have been found as follows. If $\phi$ is the required function and $\psi$ the given function we
have by Int.\ Cal.\ Arts.~211, 212, and 213 the relations
\[
D_x\phi = D_y\psi \quad \text{and} \quad D_y\phi = -D_x\psi.
\]
\begin{flalign*}
&\text{Here}&  D_y\psi
&= -\dfrac{1}{\pi} \left[ \dfrac{1+x}{(1+x)^2+y^2} +
\dfrac{1-x}{(1-x)^2+y^2} \right]  &&\phantom{Here}
\\
&\text{and} & -D_x\psi
&= -\dfrac{1}{\pi} \left[ \dfrac{y}{(1+x)^2+y^2}
 - \dfrac{y}{(1-x)^2+y^2} \right].  &&\phantom{and}
\end{flalign*}

If now we integrate $D_y\psi$ with respect to $x$ treating $y$ as a constant and add an arbitrary
function of $y$ we shall have $\phi$. So that
\begin{gather*}
  \phi = -\dfrac{1}{2\pi} \biggl\{ \log[(1+x)^2 + y^2\,] - \log[(1-x)^2 + y^2\,]\,\biggr\} + f(y).
\\
  D_y\phi = -\dfrac{1}{\pi} \biggl[ \dfrac{y}{(1+x)^2+y^2 } - \dfrac{y}{(1-x)^2+y^2} \biggr] + \dfrac{df(y)}{dy}
\end{gather*}
Comparing this with its equal $-D_x\psi$ above we find $\dfrac{df(y)}{dy}=0$  and $f(y) = C$ a constant
\begin{flalign*}
&\text{therefore} && \dfrac{1}{2\pi} \log{ \dfrac{(1-x)^2+y^2}{(1+x)^2+y^2} } + C,
&&\phantom{therefore}
\end{flalign*}
where $C$ may be taken at pleasure, is our required conjugate function.}\  %[endfootnotetext]
and
\[
\dfrac{1}{\pi} \biggl( \tan^{-1}\dfrac{1+x}{y} + \tan^{-1}\dfrac{1-x}{y} \biggl) = a   \tag{2}
\]
is any equipotential line, and
\[
\dfrac{1}{2\pi} \log \dfrac{(1-x)^2+y^2 }{(1+x)^2+y^2} = b      \tag{3}
\]
any line of flow for the system described at the beginning of this article; and
\[
V_1 = \dfrac{1}{2\pi} \log \dfrac{(1-x)^2+y^2}{(1+x)^2+y^2}   \tag{4}
\]
is the solution of a new problem for which (3) represents any equipotential
line and (2) any line of flow.
% -----File: 085.png

(2) reduces to
\[
\frac{2y}{x^2+y^2-1}=\tan a\pi
\]
\begin{flalign*}
&\text{or} & x^2+(y-\ctn a\pi)^2 = \csc^2 a\pi;&   \tag{5}&
\end{flalign*}
\begin{flalign*}
&\text{and (3) to} & x^2 + y^2 + 2\frac{e^{2b\pi}+1}{e^{2b\pi}-1}x + 1 = 0  &&\phantom{and\ (3)\ to}
\end{flalign*}
\begin{flalign*}
&\text{or}
& \left( x + \frac{e^{b\pi}+e^{-b\pi}}{e^{b\pi}-e^{-b\pi}} \right)^2 + y^2
= \left(     \frac{e^{b\pi}+e^{-b\pi}}{e^{b\pi}-e^{-b\pi}} \right)^2 - 1  &&\phantom{or}
\end{flalign*}
\begin{flalign*}
&\text{or} & (x+\ctnh b\pi)^2+y^2=\csch^2 b\pi.  &&\phantom{or} \tag{6}
\end{flalign*}
(5) and (6) are circles. The circles (5) have their centres in the axis of $Y$,
and pass through the points $(-1,0)$ and $(1,0)$; and the circles (6) have their
centres in the axis of $X$.

(4) is the complete solution, (6) is any equipotential line and (5) any line of
flow for a plane sheet in which the points in the circumferences of two given
circles whose centres are further apart than the sum of their radii are kept at
different constant potentials, or where a source and a sink of equal intensity
are placed at the points $(-1,0)$ and $(1,0)$. An important practical example
is where two wires connected with the poles of a battery are placed
with their free ends in contact with a thin plane sheet of conducting material.
The figure shows the equipotential lines and lines of flow of either system.

The complete figure would have the axis of $X$ for an axis of symmetry.

%[Illustration]
\pngcent{009.png}{1074}

{}\EXAMPLE{S}

1.\quad Show that if $f(x)=a_1$ when $x<-b$, $f(x)=a_2$ when $-b<x<b$,
$f(x)=a_3$ when $x > b$,
\[
V = \frac{a_1+a_3}{2} + \frac{1}{\pi} \left[ (a_2-a_1)\tan^{-1}\frac{b+x}{y} + (a_2-a_3)\tan^{-1}\frac{b-x}{y} \right].
\]
% -----File: 086.png

2.\quad Show that if $f(x) = 0$ if $x < 0$,
 $f(x) = a_1$ if $0 < x < b_1$,
 $f(x) = a_2$ if $b_1 < x < b_2$,
 $f(x) = a_3$ if $b_2 < x < b_3$, \&c.,
\begin{multline*}
V = \frac{1}{\pi}\left [a_1 \tan^{-1} \frac{x}{y }+ (a_1 - a_2) \tan^{-1} \frac{b_1 - x}{y }+ (a_2 - a_3) \tan^{-1} \frac{b_2 - x}{y} \right. \\
\left.{} + (a_3 - a_4) \tan^{-1} \frac{b_3 - x}{y }+ \cdots \right].
\end{multline*}

3.\quad Show that if $f(x) = -1$ if $x < -1$, $f(x) = x$ if $-1 < x < 1$,
$f(x) = 1$ if $x > 1$,
\[
V = \frac{1}{\pi}\left [(1 + x)\tan^{-1} \frac{1 + x}{y } - (1 - x) \tan^{-1}\frac {1 - x}{y }+ \frac{y}{2} \log \frac{(1 - x)^2 + y^2 }{ (1 + x)^2 + y^2}\right].
\]

4.\quad Show that if $f(x) = -1$ if $x < -1$, $f(x) = 0$ if $-1 < x < 1$,
$f(x) = 1$ if $x > 1$,
\[
V = \frac{1}{\pi}\left [\tan^{-1} \frac{1 + x }{ y }- \tan^{-1} \frac{1 - x}{y }\right ].
\]

Show that the equipotential lines are equilateral hyperbolas passing through
the points $(-1, 0)$ and $(1, 0)$, and that the lines of flow are Cassinian ovals
having $(-1, 0)$ and $(1, 0)$ as foci. The lines of flow are equipotential lines
and the equipotential lines are lines of flow for the case where the points
$(-1, 0)$ and $(1, 0)$ are kept at the same infinite potential, or where very small
ovals surrounding these points are kept at the same finite potential. The case
is approximately that of a pair of wires connected with the same pole of a
battery whose other pole is grounded, and then placed with their ends in contact
with a thin plane conducting sheet.\\

5.\quad Show that if $f(x) = 0$ if $x < 0$, $f(x) = -1$ if $0 < x < a$, $f(x) = 0$
 if $a < x < b$, and $f(x) = 1$ if $x > b$,
\[
V = \frac{1}{\pi}\left[\frac{\pi}{2} - \tan^{-1} \frac{a - x }{y }- \tan^{-1} \frac{b - x }{y} - \tan^{-1} \frac{x}{y}\right].
\]

The conjugate function
\[
V = \frac{1}{2\pi} \log \frac{x^2 + y^2 }{[(a-x)^2 + y^2] [(b-x)^2 + y^2]}
\]
is the solution for the case where a sink and two sources of equal intensity lie
on the axis of $X$, the sink at the origin and the sources at the distances $a$ and
$b$ to the right of the origin. One of the lines of flow is easily seen to be the
circle $ x^2 + y^2 = ab$.

\mypara{47.} If the plane conducting sheet has two straight edges at right angles
with each other and one is kept at potential zero while the value of the potential
% -----File: 087.png
function is given at each point of the second, that is if $V=0$ when
 $x=0$ and $V=f(x)$ when $y=0$, the solution is readily obtained. It is
\[
V = \frac{2}{\pi} \int\limits_0^\infty d\alpha
\int\limits_0^\infty e^{-\alpha y}
f(\lambda) \sin \alpha x \sin \alpha\lambda.d\lambda.   \tag{1}
\]
v.~(9) Art.~44.

\noindent This reduces to
\[
V = \frac{1}{\pi} \int\limits_0^\infty f(\lambda)d\lambda \left[ \frac{y}{y^2+(\lambda-x)^2} -\frac{y}{y^2+(\lambda+x)^2} \right].  \tag{2}
\]
v.\ Ex.~3 Art.~45.

\EXAMPLE{S}

1.\quad If $V=0$ when $y=0$ and $V=F(y)$ when $x=0$ show that
\begin{align*}
V &= \frac{2}{\pi} \int\limits_0^\infty d\alpha \int\limits_0^\infty e^{-\alpha x} F(\lambda) \sin\alpha y \sin\alpha\lambda.d\lambda\\
  &= \frac{1}{\pi} \int\limits_0^\infty F(\lambda) d\lambda \left[ \frac{x}{x^2+(\lambda-y)^2} - \frac{x}{x^2+(\lambda+y)^2} \right].
\end{align*}

2.\quad If $V=f(x)$ when $y=0$ and $V=F(y)$ when $x=0$ show that
\begin{align*}
V = \frac{1}{\pi} \int\limits_0^\infty & \left[ f(\lambda) \left( \frac{y}{y^2+(\lambda-x)^2} - \frac{y}{y^2+(\lambda+x)^2} \right) \right. \\
& \left. {}+ F(\lambda) \left( \frac{x}{x^2+(\lambda-y)^2} - \frac{x}{x^2+(\lambda+y)^2} \right) \right] d\lambda.
\end{align*}

3.\quad If $F(y)=b$ the result of Ex.~2 reduces to
\[
V = \frac{2b}{\pi} \tan^{-1}\frac{y}{x} + \frac{1}{\pi} \int\limits_0^\infty f(\lambda) d\lambda \left[ \frac{y}{y^2+(\lambda-x)^2} - \frac{y}{y^2+(\lambda+x)^2} \right].
\]

4.\quad If $F(y)=1$ for $0<y<1$ and $F(y)=0$ for $y>1$ while $f(x)=1$
 for $0<x<1$ and $f(x)=0$ for $x>1$
\begin{align*}
V = \frac{1}{\pi} & \left[ \tan^{-1}\frac{1-x}{y} - \tan^{-1}\frac{1+x}{y} + 2\tan^{-1}\frac{y}{x} \right.\\
& \left. {}+ \tan^{-1}\frac{1-y}{x} - \tan^{-1}\frac{1+y}{x} + 2\tan^{-1}\frac{x}{y} \right].
\end{align*}
% -----File: 088.png

5.\quad If one edge of the conducting sheet treated in Art.~47 is insulated, so that
 $D_{x}V=0$ if $x = 0$ and $V=f(x)$ when $y = 0$
\begin{align*}
  V &= \frac{2 }{ \pi } \int\limits_0^{\infty} d\alpha \int\limits_0^{\infty} e^{-\alpha y} f(\lambda) \cos\alpha x \cos\alpha\lambda.d\lambda\\
 &= \frac{1 }{ \pi } \int\limits_0^{\infty} f(\lambda) d\lambda \left [ \frac{y }{ y^2+(\lambda+x)^2 } + \frac{y }{ y^2+(\lambda-x)^2} \right].
\end{align*}

\mypara{48.} If the conducting sheet is a long strip with parallel edges one of which
is at potential zero while the value of the potential function is given at all
points of the other, that is if $V=0$ when $y = 0$ and $V=F(x)$ when
 $y = b$ the problem is not a very difficult one.

Since we are no longer concerned with the value of $V$ when $y = \infty$ \quad
$V= e^{\alpha y}\sin ax$ and $V=e^{\alpha y} \cos ax$ are available as particular solutions of the
equation
\[
D_x^2V+D_y^2V=0           \tag{1}
\]
as well as $V=e^{-\alpha y}\sin\alpha x$ and $V=e^{-\alpha y}\cos\alpha x$.
\begin{flalign*}
&\indent\text{Consequently }&&
  \frac{e^{\alpha y} + e^{-\alpha y}}{2} \sin  \alpha x = \cosh\alpha y \sin\alpha x  &  \text{[Int.\ Cal.\ Art.~43 (2)]}
\\
&\text{and}&&
  \frac{e^{\alpha y} - e^{-\alpha y} }{2} \sin  \alpha x = \sinh\alpha y \sin\alpha x &  \text{[Int.\ Cal.\ Art.~43 (1)]}\\[-5ex]
\end{flalign*}
\begin{flalign*}
&\text{and}&&
  \cosh \alpha y \cos \alpha x \quad \text{and}\quad  \sinh\alpha y \cos\alpha x  &&\phantom{and}
\end{flalign*}
are now available values of $V$ and can be used precisely as $e^{-\alpha y}\cos\alpha x$ and
 $e^{-\alpha y} \sin\alpha x$ are used in Art.~44.

Following the same course as in Art.~44 we get
\[
  V = \frac{1}{ \pi } \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} \frac{\sinh\alpha y}{\sinh\alpha b} F(\lambda) \cos\alpha(\lambda-x).d\lambda    \tag{2}
  \]
as a solution of (1) which will reduce to $V= F(x)$ when $y= b$
\begin{flalign*}
&\text{and to }&& V = 0 \quad\text{when}\quad y = 0\text{, since } \sinh 0 = \frac{1-1 }{2} = 0,  &&\phantom{and\ to}
\end{flalign*}
and (2) is therefore our required solution.

If $V$ is to be equal to zero when $y = b$ and to $f(x)$ when $y = 0$ we have
only to replace $y$ by $b-y $ and $F(x)$ by $f(x)$ in (2). We get
\[
  V =  \frac{1}{ \pi } \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty}\frac{\sinh \alpha(b-y) }{\sinh \alpha b} f(\lambda) \cos \alpha(\lambda-x).d\lambda.     \tag{3}
\]
% -----File: 089.png

If $V = f(x)$ when $y = 0$ and $V = F(x)$ when $y = b$ then
\begin{equation*} \begin{split}
V = {}& \frac{1}{\pi} \int\limits^\infty_{0} d\alpha \int\limits^\infty_{-\infty}
\dfrac{\sinh\alpha(b - y)}{\sinh\alpha b} f(\lambda)\cos {\alpha(\lambda - x)}.d\lambda \\
&+ \frac{1}{\pi} \int\limits^\infty_{0} d\alpha \int\limits^\infty_{-\infty} \frac{\sinh\alpha y}
{\sinh\alpha b} F(\lambda)\cos {\alpha(\lambda - x)}.d\lambda.
\end{split} \end{equation*}

This can be considerably simplified by the aid of the formula
\[
\int \limits^{\infty}_{0} \dfrac{\sinh px}{\sinh qx}\cos{rx.dx}
= \dfrac{\pi}{2q} \dfrac{\sin \dfrac{p \pi}{q}} {\cos\dfrac{p \pi}{q} + \cosh \dfrac{r \pi}{q}}
\]
if $p^2  <  q^2$. [Bierens de Haan, Tables of Def.\ Int.\ (7) 265] and becomes
\begin{flalign*} &&
V ={} & \dfrac{1}{2b} \sin \dfrac{\pi}{b} (b - y) \int \limits^\infty_{-\infty}f(\lambda)
\dfrac{d\lambda}{\cos \dfrac{\pi(b - y)}{b} + \cosh \dfrac{\pi}{b} (\lambda - x)}  &&
\\
&&\phantom{or}
& {}+ \dfrac{1}{2b} \sin \dfrac{\pi y}{b} \int \limits^\infty_{-\infty} F(\lambda) \dfrac{d\lambda}
{\cos \dfrac{\pi y}{b} + \cosh \dfrac{\pi}{b} (\lambda - x)}  &\text{or}&
\end{flalign*}
\[\tag{5}
V =  \dfrac{1}{2b}\sin \dfrac{\pi y}{b} \int\limits^\infty_{-\infty} \biggl
[\dfrac{f\lambda}{\cosh \dfrac{\pi}{b} (\lambda - x) - \cos\dfrac{\pi y}{b}}
+ \dfrac{F\lambda}{\cosh\dfrac{\pi}{b} (\lambda - x) + \cos\dfrac{\pi y}{b}}\biggr]
d\lambda.
\]

\EXAMPLE{S}

1.\quad Given the formula
\[
\int \dfrac{dx}{a+b \cosh x} = \dfrac{2}{\sqrt{b^2 - a^2}} \tan^{-1}\biggl(\sqrt{\dfrac{b-a}{b+a}}
\tanh \dfrac{x}{2}\biggr) \quad\text{ if } b >  a,
\]
show that if $V = 1$ when $y = 0$ and $V = 0$ when $y = b$\;\;$V = \dfrac{1}{b}(b - y).$\\

2.\quad Show that if $V = 0$ when $y = b$, $V = -1$ when $y = 0$ and
 $x < 0$, and $V = 1$ when $y = 0$ and $x > 0$
\[
V=\dfrac{2}{\pi}\tan^{-1}\Biggl[\dfrac{\tanh \dfrac{\pi x}{2b}} {\tan \dfrac{\pi y}{2b}}\Biggr]
\]
The solution for the conjugate system, that is, for a strip having a source at
$(0, 0)$ and an infinitely distant sink is
\[
V =  -\dfrac{1}{\pi}\log \biggl[\cosh^{2}\dfrac{\pi x}{2b} - \cos^{2}\dfrac{\pi y}{2b}\biggr].
\]
% -----File: 090.png

3.\quad Show that if $V= -1$ when $y = 0$ and $x<0$, $V=1$ when $y = 0$
and $x>0$, $V= -1$ when $y = b$ and $x < 0$, and $V=1$ when
 $y = b$ and $x > 0$,
\[
\begin{split}
V &
= \frac{2}{\pi}\tan^{-1}\bigg ( \tan \dfrac{\pi}{2b} (b -y) \tanh \dfrac{\pi x}{2b}\bigg )
+ \frac{2}{\pi}\tan^{-1} \bigg ( \tan \dfrac{\pi}{2b} y \tanh \dfrac{\pi x}{2b}\bigg )
\\
 &=\frac{2}{\pi}\tan^{-1} \Bigg[ \frac{\sinh \dfrac{\pi x}{b }} {\sin\dfrac{ \pi y}{b}} \Bigg].
 \end{split}
 \]
The solution for the conjugate system, that is, for a strip having a source and a
sink at the points $(0, 0)$ and $(0, b)$ is
 \[
V = \frac{1}{\pi}\log \Bigg[ \dfrac{\cosh\dfrac{ \pi x}{b} + \cos \dfrac{ \pi y}{b}}{\cosh\dfrac{ \pi x}{b} - \cos \dfrac{ \pi y}{b} } \Bigg].
\]

4.\quad If $V=0$ when $x = 0$, $V = f(x)$ when $y = 0$ and $x>0$, and $V=0$
when $y = b$ and $x > 0$.
\[
\begin{split}
V &= \frac{1}{\pi} \int\limits_0^{\infty} d\alpha \int\limits_0^{\infty} \frac{\sinh\alpha(b-y)}{\sinh\alpha b}
  [ \cos\alpha(\lambda-x) - \cos\alpha(\lambda+x) ] f(\lambda) d\lambda\\
&= \frac{1}{2b} \sin \dfrac{\pi y }{b} \int\limits_0^{\infty} \biggl[
   \frac{1} {\cosh \dfrac{\pi }{b} (\lambda-x) - \cos \dfrac{\pi y}{ b}}
 - \frac{1} {\cosh \dfrac{\pi }{b} (\lambda+x) - \cos \dfrac{\pi y}{ b}} \biggr] f(\lambda) d\lambda
\end{split}
\]
for positive values of $x$ and for values of $y$ between $0$ and $b$.\\

5.\quad If $V_1 = 0$ when $x = 0$, $V_1 = F(x)$ when $y = b$ and $x > 0$, and
 $V_1 = 0$ when $y = 0$ and $x>0$
\[
V_1 = \frac{1}{2b} \sin \frac{\pi y}{ b} \int\limits_0^{\infty} \biggl[
   \frac{1} {\cosh \dfrac{\pi }{b} (\lambda-x) + \cos \dfrac{\pi y}{ b}}
 - \frac{1} {\cosh \dfrac{\pi }{b} (\lambda+x) - \cos \dfrac{\pi y}{ b}} \biggr] F(\lambda) d\lambda
 \]
for positive values of $x$ and values of $y$ between $0$ and $b$.\\

6.\quad If $V_2=0$ when $x = 0$, $V_2 =f(x)$ when $y = 0$ and $x > 0$, and
 $V_2 = F(x)$ when $y = b$ and $x>0$
\[  \phantom{(v.\ Exs.\ 4 and 5)}
V_2 = V+ V_1 \quad \text{for} \quad x > 0\quad  \text{and} \quad 0 < y < b. \tag*{(v.\ Exs.\ 4 and 5)}
\]

7.\quad If one edge of the strip described in Art.~48 is insulated so that we have
 $V=f(x)$ when $y = 0$ and $D_y V=0$ when $y = b$ show that
\[
  V = \frac{1}{\pi} \int\limits_0^{\infty}d\alpha \int\limits_{-\infty}^{\infty} \frac{\cosh\alpha(b-y)}{\cosh\alpha b} f(\lambda) \cos\alpha(\lambda-x).d\lambda.
\]
% -----File: 091.png

By the aid of the formula
\begin{flalign*}
&\phantom{if\ p<q}&&
  \int\limits_0^\infty \frac{\cosh px}{\cosh qx} \cos rx.dx =
\frac{\pi}{q} \frac{\cosh\dfrac{r\pi}{2q}\cos\dfrac{p\pi}{2q}}
     {\cos\dfrac{p\pi}{q}+\cosh\dfrac{r\pi}{q}} && \text{if } p<q,&
\end{flalign*}
[Bierens de Haan, Def.\ Int.\ Tables (6) 265],\\
reduce this to
\[
V=\frac{1}{b}\sin\dfrac{\pi y}{2b}\int\limits_{-\infty}^\infty
\frac{f(\lambda)\cosh\dfrac{\pi}{2b}(\lambda-x)}
     {\cosh\dfrac{\pi}{b}(\lambda-x)-\cos\dfrac{\pi y}{b}}d\lambda.
\]

8.\quad If $V=0$ when $y=0 \text{\; or \;} b$ and $x<-a$, $V=1$ when $y=0 \text{\; or \;} b$
and $-a<x<a$, and $V=0$ when $y=0 \text{\; or \;} b$ and $x>a$
\[
V = \frac{1}{\pi} \Bigg[
 \tan^{-1}\frac{\sinh\dfrac{\pi(a-x)}{b}}{\sin\dfrac{\pi y}{b}}
+\tan^{-1}\frac{\sinh\dfrac{\pi(a+x)}{b}}{\sin\dfrac{\pi y}{b}} \Bigg].
\]

9.\quad If $V=0$ when $y=0 \text{\; or \;} b$ and $x<-a$, $V=1$ when $y=0$ and
 $-a<x<a$, $V=0$ when $y=0 \text{\; or \;} b$ and $x>a$, and $V=-1$ when
 $y=b$ and $-a<x<a$
\[
V=\frac{1}{\pi} \Bigg[
    \tan^{-1} \frac{\tanh\dfrac{\pi(a-x)}{b}}{\tan\dfrac{\pi y}{b}}
  + \tan^{-1} \frac{\tanh\dfrac{\pi(a+x)}{b}}{\tan\dfrac{\pi y}{b}} \Bigg].
\]

10.\quad A system conjugate to that of Ex.~9 is $V=+\infty$ when $y=0 \text{\; or \;} b$ \
and $x=-a$, $V=-\infty$ when $y=0 \text{\; or \;} b$ and  $x=a$. In this case
\[
V=\frac{1}{2\pi}\log
\frac{\sin^2\dfrac{\pi y}{b}+\sinh^2\dfrac{\pi(a-x)}{b}}
     {\sin^2\dfrac{\pi y}{b}+\sinh^2\dfrac{\pi(a+x)}{b}}.
\]

\mypara{49.} Let us take now a problem in the flow of heat. Suppose we have an
infinite solid in which heat flows only in one direction, and that at the start the
temperature of each point of the solid is given. Let it be required to find the
temperature of any point of the solid at the end of the time $t$.

Here we have to solve the equation
\[
D_t u = a^2D_x^2 u \tag{1}
\]
[v.\ Art.~1 \smallromr{II}] subject to the condition
\[
u=f(x)\quad \text{when}\quad t = 0.   \tag{2}
\]
% -----File: 092.png

As the equation (1) is linear with constant coefficients we can get a particular
solution by the device used in Arts.~7 and 8.

Let $u = e^{\beta t + \alpha x}$ and substitute in (1). We get
\[
  \beta = a^2 \alpha^2
\]
as the only relation which need hold between $\beta$ and $\alpha$.
\begin{flalign*}
&\indent\text{Hence}&&
  u = e^{\alpha x + a^2 \alpha^2 t} = e^{a^2 \alpha^2 t} e^{\alpha x}             &&\phantom{Hence}  \tag{3}
\end{flalign*}
is a solution of (1) no matter what value is given to $\alpha$.

To get a trigonometric form replace $\alpha$ by $\alpha i$.
\begin{flalign*}
&\indent\text{Then}&&  u = e^{-a^2 \alpha^2 t} e^{\alpha x i}.  &&\phantom{\indent Then}
\end{flalign*}

If in (3) we replace $\alpha$ by $-\alpha i$ we get
\[
    u = e^{-a^2 \alpha^2 t} e^{-\alpha x i}.
\]
As in Arts.~7 and 8 we get from these values
\[
  u = e^{-a^2 \alpha^2 t} \sin {\alpha x}  \quad \text{and} \quad
  u = e^{-a^2 \alpha^2 t} \cos {\alpha x}
\]
as particular solutions of (1), $\alpha$ being wholly unrestricted.

From these values we wish to build up a value of $u$ which shall reduce to
$f(x)$ when $t = 0$ and shall still be a solution of (1).
\begin{flalign*}
&\indent\text{We have }&&
  f(x) = \frac{1}{\pi } \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} f(\lambda) \cos \alpha (\lambda - x).d\lambda
&&\phantom{We\ have}   \tag{4}
\end{flalign*}
v.\ Art.~32 (3), and by proceeding as in Art.~44 we get
\[
  u = \frac{1}{\pi } \int\limits_0^{\infty} d\alpha \int\limits_{-\infty}^{\infty} e^{-a^2\alpha^2t} f(\lambda) \cos \alpha (\lambda - x).d\lambda         \tag{5}
\]
as our required value of $u$.

This can be considerably simplified.

Changing the order of integration
\begin{gather*}
  u = \frac{1}{\pi} \int\limits_{-\infty}^{\infty} f(\lambda) d\lambda \int\limits_0^{\infty} e^{-a^2\alpha^2t} \cos \alpha(\lambda - x).d\alpha.       \tag{6}
\\
  \int\limits_0^{\infty} e^{-a^2 \alpha^2 t} \cos \alpha (\lambda - x).d\alpha
= \frac{1}{2a } \sqrt{\frac{ \pi}{t }}.e^{-\frac{(\lambda - x)^2 }{4a^2 t}}  \tag{7}
\end{gather*}
by the formula
\begin{flalign*}
&&&  \int\limits_0^{\infty} e^{-a^2 x^2} \cos bx.dx
= \frac{\sqrt{ \pi}}{ 2a} e^{-\frac{b^2}{4a^2}}    \tag*{[Int.\ Cal.\ Art.~94 (2)]}\\
&\text{Hence}&&
  u = \frac{1}{2a \sqrt{ \pi t} } \int\limits_{-\infty}^{\infty} f(\lambda) e^{-\frac{(\lambda - x)^2 }{4a^2 t}} d\lambda.
&&\phantom{Hence}  \tag{8}\\[-5ex]
\end{flalign*}
% -----File: 093.png
\begin{flalign*}
&\indent\text{Let now} &&  \beta = \frac{\lambda-x}{2a \sqrt{t}},  &&\phantom{\indent Let\ now} \\[1ex]
&\text{then} &  \lambda &= x + 2a \sqrt{t}.\beta  &&\\
&\text{and } & u = \frac{1 }{\sqrt{\pi}}&\int\limits_{-\infty}^{\infty} f(x + 2a \sqrt{t}.\beta) e^{-\beta^2} d\beta. & \tag{9}&
\end{flalign*}

\EXAMPLE{S}

1.\quad Let the solid be of infinite extent and let the temperature be equal to a
constant $c$ at the time $t = 0$.
\begin{flalign*}
&\indent\text{Then }&&
  u = \frac{c }{\sqrt{\pi}} \int\limits_{-\infty}^{\infty} e^{-\beta^2} d\beta
= \frac{2c }{\sqrt{\pi}} \int\limits_0^{\infty} e^{-\beta^2} d\beta = c.  &&\phantom{\indent Then}
\end{flalign*}
v.\ Int.\ Cal.\ Art.~92 (2).\\

2.\quad Let $u = x$ when $t = 0$.
\begin{flalign*}
&\indent\text{Then}&&
  u =  \frac{1 }{\sqrt{\pi}} \int\limits_{-\infty}^{\infty} (x +  2a \sqrt{t}.\beta) e^{-\beta^2} d\beta = x.
&&\phantom{\indent Then}
\end{flalign*}

3.\quad Let $u = x^2$ when $t = 0$.
\begin{flalign*}
&\indent\text{Then}&&   u = x^2 + 2a^2t.  &&\phantom{\indent Then}
\end{flalign*}

4.\quad Let $u = 0$ if $x< -b$, $u = 1$ if $-b < x < b$, and $u = 0$ if $x > b$,
when $t = 0$.\\

Then
\[
u = \frac{1 }{\sqrt{\pi}} \!\!\int\limits_{-\frac{b+x}{2a\sqrt{t}}}^{\frac{b-x}{2a\sqrt{t}}}\!\! e^{-\beta^2} d\beta
= \frac{2 }{\sqrt{\pi}} \left [ \frac{b}{ 2a \sqrt{ t }} - \frac{b^3+3bx^2 }{ 3(2a \sqrt{ t })^3}
    + \frac{b^5 + 10b^3x^2 + 5bx^4 }{5.2!( 2a \sqrt{ t })^5} - \cdots \right].
\]

5.\quad Let $u = 0$ if $x < 0$ and $u = 1$ if $x > 0$ when $t = 0$.

Then
\begin{align*}
u &= \frac{1 }{\sqrt{\pi}} \!\!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} d\beta
= \frac{1 }{\sqrt{\pi}} \biggl[
    \int\limits_0^{\frac{x}{2a\sqrt{t}}}\!\! e^{-\beta^2} d\beta
  + \int\limits_0^{\infty} e^{-\beta^2 } d\beta \biggr]
= \frac{1 }{\sqrt{\pi}} \!\!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\!\! e^{-\beta^2 } d\beta + \frac{1 }{2}
\\
 &= \frac{1}{ 2} + \frac{1 }{\sqrt{\pi}} \left [
        \frac{ x  }{                2a \sqrt{ t }   }
      - \frac{ x^3}{ 3.(2a \sqrt{ t })^3}
      + \frac{ x^5}{ 5.2!(2a \sqrt{ t })^5}
      - \frac{ x^7}{ 7.3!(2a \sqrt{ t })^7} + \cdots \right].
\end{align*}

6.\quad An iron slab 10~c.\,m.\ thick is placed between and in contact with two
very thick iron slabs. The initial temperature of the middle slab is 100°, and
of each of the outer slabs 0°. Required the temperature of a point in the
middle of the inner slab fifteen minutes after the slabs have been put together.
Given \ $a^2 = 0.185$ \ in C.G.S. units.  \hfill   \emph{Ans.}, 21°.6.\\
% -----File: 094.png

7.\quad Two very thick iron slabs one of which is at the temperature 0° and the
other at the temperature 100° throughout are placed together face to face.
Find the temperature of each slab 10~c.\,m.\ from their common face fifteen
minutes after they have been placed together.     \hfill   \emph{Ans.}, 70°.8, 29°.2.\\

8.\quad Find a particular solution of $D_tu = a^2D_x^2u$ on the assumption that it
is of the form $u = T.X$ where $T$ is a function of  $t$ alone and $X$ is a function
of $x$ alone.

\mypara{50.} If our solid has one plane face which is kept at the constant temperature
zero, and we start with any given distribution of heat, the problem is
somewhat modified.

Take the origin of coördinates in the plane face. Then we have as before
the equation
\[
  D_t u = a^2D_x^2 u,               \tag{1}
\]
but our conditions are
\begin{alignat*}{3}
  u &= 0      && \text{when} \quad  &x = 0            \tag{2}\\
  u &= f(x)  && \text {\quad`` } \quad & t = 0            \tag{3}
\end{alignat*}
and we are concerned only with positive values of $x$.

We may then use the form (4) Art.~32
\[
f(x) = \frac{2}{\pi}\int\limits_0^{\infty} d\alpha \int\limits_0^{\infty} f(\lambda) \sin\alpha x \sin\alpha \lambda.d\lambda,  \tag{4}
\]
and proceeding as in the last section we get
\[
u = \frac{2}{\pi} \int\limits_0^{\infty}  d\alpha \int\limits_0^{\infty}  e^{-a^2 \alpha^2 t} f(\lambda) \sin\alpha x \sin\alpha\lambda.d\lambda  \tag{5}
\]
as our required solution. This may be reduced considerably.
\[
  u = \frac{1}{\pi} \int\limits_0^{\infty} f(\lambda) d\lambda \int\limits_0^{\infty} e^{-a^2 \alpha^2 t}[\cos \alpha (\lambda - x) - \cos \alpha (\lambda + x)]d\alpha,
\]
\[
\text{or}  \quad u = \frac{1}{2a\sqrt{\pi t}} \int\limits_0^{\infty} f(\lambda)(e^{-\frac{(\lambda-x)^2 }{ 4a^2 t}} - e^{-\frac{(\lambda+x)^2 }{ 4a^2 t}} ) d\lambda  \tag{6}
\]
by (7) Art.~49, and this may be reduced to the form
\[
u = \frac{1}{\sqrt{ \pi} }\biggl [\!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\!  e^{-\beta^2} f(x + 2a \sqrt{ t}.\beta) d\beta
 - \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} f(-x + 2a \sqrt{ t }.\beta) d\beta \biggr ] .  \tag{7}
\]

\EXAMPLE{S}

1.\quad Let the initial temperature be constant and equal to $c$.
% -----File: 095.png

Then
\[
\begin{split}
  u &= \frac{ c}{\sqrt{ \pi}} \biggl [ \!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} d\beta - \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} d\beta \biggr ]\\
    &= \frac{2c}{\sqrt{ \pi}} \!\int\limits_{0}^{\frac{x}{2a\sqrt{t}}}\!e^{-\beta^2} d\beta\\
    &= \frac{2c}{\sqrt{ \pi}} \left [ \frac{x }{ 2a \sqrt{ t}}
                  - \frac{x^3}{3.(2a    \sqrt{t})^3}
                  + \frac{x^5}{5.2!(2a  \sqrt{t})^5}
                  - \frac{x^7}{7.3!.(2a \sqrt{t})^7} + \cdots \right ].
\end{split}
\]

2.\quad Assuming that the earth was originally at the temperature 7000° Fahrenheit
throughout, and that the surface was kept at the constant temperature 0°,
find (1) the temperature 10 miles below the surface 10,000,000 years after the
cooling began; (2) the temperature 1 mile below the surface at the same
epoch; (3) the temperature 10 miles below the surface 100,000,000 years after
the cooling began; (4) the temperature 1 mile below the surface at the same
epoch; (5) the rate at which the temperature was increasing with the distance
from the surface at each point at each epoch.

Neglect the convexity of the earth's surface and take Sir Wm.\ Thomson's
value of $a^2 (400)$ the foot, the Fahrenheit degree, and the year being taken as
units.  (Thomson and Tait's Nat.\ Phil.\ Vol.\ II. Appendix.)

\emph{Ans.}, (1) 3114°; (2) 329°.5; (3) 1036°; (4) 103°; (5) 1° for every 20 feet, 3°
for every 50 feet, 1° for every 50 feet, 1° for every 50 feet.\\

3.\quad Let the initial temperature be constant and equal to $ -b$, then by Ex.~1
\[
  u = - \frac{2b}{\sqrt{\pi}} \!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2 }d\beta .
\]

4.\quad Let the temperature of the plane face be $b$ instead of zero, and let the
initial temperature be zero.

Then we have only to add $b$ to the second member of the solution in Ex.~3,
as we may since $u = b$ is a solution of (1) Art.~49, and we get
\[
  u = b \biggl ( 1 - \frac{2}{\sqrt{ \pi}} \!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2} d\beta \biggr ) .
\]

5.\quad Let $u = b$ when $x = 0$ and $u = f(x)$ when $t = 0$.

Then
\[
    u = b \biggl ( 1 - \frac{2}{\sqrt{ \pi}} \!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2} d\beta \biggr )
    + \frac{1}{ 2a \sqrt{\pi t }}\int\limits_0^{\infty} f(\lambda)
                         [ e^{-{\frac{(\lambda-x)^2 }{4a^2t}}}
                         - e^{-{\frac{(\lambda+x)^2 }{4a^2t }}}] d\lambda
\]
by (6) Art.~50.\\
% -----File: 096.png

6.\quad Let $u=b$ when $x=0$ and $u=c$ when $t=0$.
\begin{flalign*}
&\indent \text{Then}       && u = b + (c - b) \frac{2}{\sqrt{\pi}}\!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\! e^{-\beta^2} d\beta.&& \phantom{\indent Then}
\end{flalign*}

7.\quad If the earth has been cooling for 200,000,000 years from a uniform temperature,
prove that the rate of cooling is greatest at a depth of about 76
miles, and that at a depth of about 130 miles the rate of cooling has reached
its maximum value for all time.  Let $a^2 = 400$.\\

8.\quad Show that if the plane face of the solid considered in Art.~50 instead of
being kept at temperature zero is impervious to heat
\[\phantom{v.\ (6) Art.~50.}
u = \frac{1}{2a \sqrt{\pi t}} \int\limits_0^{\infty} f(\lambda)(e^{-\frac{(\lambda - x)^2 }{ 4a^2t}} + e^{-\frac{(\lambda + x)^2}{ 4a^2t}}) d\lambda.   \tag*{v.\ (6) Art.~50.}
\]

\mypara{51.} If the temperature of the plane face of the solid described in Art.~50
is a given function of the time and the initial temperature is zero, the solution
of  the problem can be obtained by a very ingenious method due to Riemann.

Here we have to solve the equation
\[
D_t u = a^2 D_x^2 u             \tag{1}
\]
subject to the conditions
\[
\left . \begin{aligned}
u& = F(t) &&\text{when} &x = 0\phantom{.}\\
u& = 0    &&\text{\quad``}        &t = 0.
\end{aligned}\right \}
\tag{2}
\]
We know that
\[
u= \frac{2}{\sqrt{ \pi}} \!\int\limits_0^{\frac{x}{2a\sqrt{t}}}\!  e^{-\beta^2} d\beta
\]
is a solution of (1), v.\ Ex.~1 Art.~50. It is easily shown that
\[
u=  \frac{2}{\sqrt{ \pi}} \!\!\int\limits_0^{\frac{x}{2a\sqrt{t-c}}}\!\!  e^{-\beta^2} d\beta, \tag{3}
\]
where $c$ is any constant, is a solution of (1).

For
\begin{align*}
D_t u &= -\frac{2}{\sqrt{ \pi}} \frac{ x}{2a}\frac{ 1}{2(t-c)^\frac{3}{2}} e^{-\frac{x^2}{4a^2(t-c)}} = -\frac{x}{2a \sqrt{ \pi} }(t - c)^{-\frac{3}{2}} e^{- \frac{x^2}{4a^2(t-c)}}\\
D_x u &= \frac{2}{\sqrt{ \pi}} \frac{ 1}{2a\sqrt{t-c}} e^{-\frac{x^2}{4a^2(t-c)}}\\
D_x^2u&= - \frac{2}{\sqrt{ \pi}} \frac{ 1}{2a\sqrt{t-c}}\frac{2x}{4a^2(t-c)}e^{-\frac{x^2}{ 4a^2(t-c)}}
       = -\frac{x}{2a^3 \sqrt{ \pi}}(t-c)^{-\frac{3}{2}} e^{-\frac{x^2}{4a^2(t-c)}}
\end{align*}
\begin{flalign*}
\text{and} && D_t u &= a^2 D_x^2 u.&&
\end{flalign*}
% -----File: 097.png

Let $\phi(x,t)$ be a function of $x$ and $t$ which shall be equal to zero if $t$ is
negative and shall be equal to
\[
1-\frac{2}{\sqrt\pi}\!\int\limits_0^\frac{x}{2a\sqrt t}\! e^{-\beta^2}d\beta
\]
if $t$ is equal to or greater than zero; so that if $x=0$\;\;$\phi(x,t)=1$ and if
 $t=0$\;\;$\phi(x,t)=0$.

We shall now attack the following problem, to solve equation (1) subject to
the conditions
\begin{alignat*}{5}
&u=0          && \text{if}\quad  && t=0      &&                    &&\\
&u=F(0)       && \text{``}       && x=0\quad && \text{and}\quad      && 0<t<\tau\\
&u=F(k\tau)\; && \text{``}       && x=0      && \phantom{a}\text{``} && k\tau<t<(k+1)\tau,
\end{alignat*}
where $k$ is any whole number and $\tau$ is any arbitrarily chosen interval of time.

If we form the value
\[
u=F(k\tau)[\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)] \tag{4}
\]
$u$ will satisfy equation (1) since zero, unity and
\[
\frac{2}{\sqrt\pi}\!\!\!\int\limits_0^\frac{x}{2a\sqrt{t-k\tau}}\!\!\! e^{-\beta^2}d\beta
\]
are values of $u$ which satisfy (1). $u$ will be zero if $t<k\tau$ by the definition
of the function $\phi(x,t)$; if $x=0$ $u=0$ if $t>(k+1)\tau$ and $u=F(k\tau)$ if
 $k\tau<t<(k+1)\tau$.

Therefore
\[
u=\sum_{k=0}^{k=\infty}F(k\tau)[\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)] \tag{5}
\]
is the solution of the problem stated above.

(5) can be simplified somewhat from the consideration that for a given value
of $t$\;\;$\phi(x,t-k\tau)=0$ if $k\tau>t$. If, then, $n\tau$ is the greatest whole multiple
of $\tau$ not exceeding $t$,
\[
u=\sum_{k=0}^{k=n}F(k\tau)[\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)].  \tag{6}
\]

If now we decrease $\tau$ indefinitely the limiting form of (6) will be the solution
of the problem stated at the beginning of this article.

(6) may be written
\[
u=\sum_{k=0}^{k=n}F(k\tau)\left[\frac{\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)}{\tau}\right]\tau \tag{7}
\]
% -----File: 098.png
and if $\tau$ is indefinitely decreased the limiting form of (7) is
\[
u=-\int\limits_0^t F(\lambda)D_\lambda\phi(x,t-\lambda)d\lambda.  \tag{8}
\]

Since $t-\lambda$ is positive between the limits of integration
\begin{flalign*}
&& \phi(x,t-\lambda)&=1-\frac{2}{\sqrt\pi}\!\!\int\limits_0^\frac{x}{2a\sqrt{t-\lambda}}\!\! e^{-\beta^2}d\beta, &&\\
&\text{and} &  D_\lambda\phi(x,t-\lambda)=&-\dfrac{x}{2a\sqrt\pi}e^{-\frac{x^2}{4a^2(t-\lambda)}}(t-\lambda)^{-\frac{3}{2}}; && \phantom{and}
\end{flalign*}
and (8) may be written
\[
u=\frac{x}{2a\sqrt\pi}\int\limits_0^t F(\lambda)e^{-\frac{x^2}{4a^2(t-\lambda)}}(t-\lambda)^{-\frac{3}{2}}d\lambda,  \tag{9}
\]
or if we let \qquad\qquad\qquad\qquad $\beta=\dfrac{x}{2a\sqrt{t-\lambda}}$
\[
u=\frac{2}{\sqrt\pi}\!\int\limits_{\frac{x}{2a\sqrt t}}^\infty\! e^{-\beta^2}F\left(t-\frac{x^2}{4a^2\beta^2}\right)d\beta.  \tag{10}
\]

\EXAMPLE{S}

1.\quad If $u=nt$ when $x=0$ and $u=0$ when $t=0$
\[
u=n\bigg(t+\frac{x^2}{2a^2}\bigg)\bigg[1-\frac{2}{\sqrt\pi}\!\int\limits_0^\frac{x}{2a\sqrt t}\!e^{-\beta^2}d\beta\bigg]-\frac{nx\sqrt t}{a\sqrt\pi}e^{-\frac{x^2}{4a^2t}}.
\]

2.\quad A thick iron slab is at the temperature zero throughout, one of its plane
faces is then kept at the temperature 100° Centigrade for 5 minutes, then at
the temperature zero for the next 5 minutes, then at the temperature 100° for
the next 5 minutes, and then at the temperature zero. Required the temperature
of a point in the slab 5 c.m.\ from the face at the expiration of 18
minutes.  Given; $a^2=.185$.  \hfill \textit{Ans.}, 20°.1.\\

3.\quad If $u=F(t)$ when $x=0$ and $u=f(x)$ when $t=0$, then
\[
u=\frac{2}{\sqrt\pi}\int\limits_\frac{x}{2a\sqrt t}^\infty e^{-\beta^2}F\bigg(t-\frac{x^2}{4a^2\beta^2}\bigg)d\beta
+\frac{1}{2a\sqrt{\pi t}}\int\limits_0^\infty(e^{-\frac{(\lambda-x)^2}{4a^2t}}-e^{-\frac{(\lambda+x)^2}{4a^2t}})f(\lambda)d(\lambda).
\]
v.~(6) Art.~50.\\

% -----File: 099.png
4.\quad If in Art.~(51) $F(t)$ is a periodic function of the time of period $T$ it can
be expressed by a Fourier's series of the form
\iffalse
\[
F(t) =\frac{1}{2}b_0 + \sum\limits_{m=1}^{m=\infty}[a_m\sin m\alpha t + b_m\cos m\alpha t],\qquad \text{where}\qquad \alpha=\frac{2\pi}{T},
\]
or \qquad\qquad $F(t)=\dfrac{1}{2b_0} + \sum\limits_{m=1}^{m=\infty}\rho_m\sin(m\alpha t + \lambda_m)$,
\begin{flalign*}
&\text{where} & \rho_m\cos\lambda_m=a_m \quad \text{and}\quad \rho_m\sin\lambda_m=b_m. && \text{v.\ Art.~31 Ex.~3.}&
\end{flalign*}
\fi
\begin{flalign*}
&&F(t) =\frac{1}{2}b_0 &+ \sum\limits_{m=1}^{m=\infty}[a_m\sin m\alpha t + b_m\cos m\alpha t],\quad \text{where} &&\alpha=\frac{2\pi}{T},\\
&\text{or} & F(t)=\dfrac{1}{2b_0} &+ \sum\limits_{m=1}^{m=\infty}\rho_m\sin(m\alpha t + \lambda_m),\\
&\rlap{where\hfill} && \rho_m\cos\lambda_m=a_m \quad \text{and}\quad \rho_m\sin\lambda_m=b_m. &&\text{v.\ Art.~31 Ex.~3.}
\end{flalign*}

Show that with this value of $F(t)$ (10) Art.~51.\
becomes
\begin{gather*}
\begin{split}
u=\frac{1}{\sqrt\pi}b_0\!\int\limits_\frac{x}{2a\sqrt t}^{\infty}\! e^{-\beta^2}d\beta + \frac{2}{\sqrt\pi}\sum\limits_{m=1}^{m=\infty}\rho_m\biggl[\sin(m \alpha t + \lambda_m)\int\limits_{\frac{x}{2a\sqrt t}}^{\infty}e^{-\beta^2}\cos\frac{m\alpha x^2}{4a^2\beta^2}d\beta\biggr.\\
\biggl.-\cos(m\alpha t + \lambda_m)\!\int\limits_\frac{x}{2a\sqrt t}^{\infty}\! e^{-\beta^2}\sin\frac{m\alpha x^2}{4a^2\beta^2}d\beta\biggr]
\end{split}
\end{gather*}
and that as $t$ increases $u$ approaches the value
\[
u= \frac{1}{2}b_0 + \sum\limits_{m=1}^{m=\infty}\rho_m e^{-\frac{x}{a}\sqrt\frac{m\alpha}{2}}\sin(m\alpha t - \frac{x}{a}\sqrt\frac{m\alpha}{2}+\lambda_m).
\]

Given that
\[
\int\limits_{0}^{\infty}e^{-x^2}\sin\frac{b^2}{x^2}dx=\frac{\sqrt\pi}{2}e^{-b\sqrt2}\sin b\sqrt2;\quad \int\limits_{0}^{\infty}e^{-x^2}\cos\frac{b^2}{x^2}dx=\frac{\sqrt\pi}{2}e^{-b\sqrt2}\cos b\sqrt2.
\]
v.\ \textit{Riemann, Lin.\ par.\ dif.\ gl.}\ §~54.\\

5.\quad If we are dealing with a bar of small cross-section where the heat not
only flows along the bar but at the same time escapes at the surface of the
bar into air at the temperature zero we have to solve the differential
equation
\[
\phantom{v.\ Fourier, Heat §~105.}
D_t u = a^2D_x^2 u - b^2u. \tag*{v.\ Fourier, Heat §~105.}
\]
Show that for this case
\[
u = e^{-(b^2+a^2\alpha^2)t}\sin\alpha x \qquad \text{and}\qquad u=e^{-(b^2+a^2\alpha^2)t}\cos\alpha x
\]
are particular solutions, and that if $u=f(x)$ when $t=0$
\[
u=\frac{e^{-b^2t}}{2a\sqrt{\pi t}}\int\limits_{-\infty}^{\infty}e^{-\frac{(\lambda-x)^2}{4a^2t}}f(\lambda)d\lambda
 =\frac{e^{-b^2t}}{\sqrt\pi}\int\limits_{-\infty}^{\infty} e^{-\beta^2}f(x+2a\sqrt t.\beta)d\beta.
\]
cf.~(8) and (9) Art.~49.
% -----File: 100.png

If $u = 0$ when $x = 0$ and $u = f(x)$ when $t = 0$

\[
u = \frac{e^{-b^2t}}{\sqrt{ \pi}}\biggl [\!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} f(x+2a \sqrt{ t }.\beta) d\beta - \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2} f(-x + 2a \sqrt{ t}.\beta) d\beta \biggr ].
\]
cf.~(7) Art.~50.

If $u = -e^{-\frac{bx}{a}}$ when $t = 0$ and $u = 0$ when $x = 0$
\[
u = \frac{1}{\sqrt{\pi}} \biggl [e^{\frac{bx}{a}} \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-(b \sqrt{ t} + \beta)^2} d\beta - e^{-\frac{bx}{a}}  \!\!\int\limits_{-\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-(b \sqrt{ t} + \beta)^2} d\beta \biggr ],
\]
and if $u = 1$ when $x = 0$ and $u = 0$ when $t = 0$ we have only to add
 $e^{-\frac{bx}{a}}$ to the second member of the last equation, since $u = e^{-\frac{bx}{a}}$ satisfies the
equation
\[
D_t u = a^2 D_x^2 u - b^2u.
\]

If $u = F(t)$ when $x = 0$ and $u = 0$ when $t = 0$ we can employ the
method of Art.~51.
\begin{gather*}
\phi(x,t- \lambda) = e^{-\frac{bx}{a}} + \frac{1}{\sqrt{ \pi}} \biggl  [e^{\frac{bx}{a}} \hspace{-10pt}\int\limits_{\frac{x}{2a\sqrt{t-\lambda}}}^{\infty}\hspace{-10pt} e^{-(b \sqrt{ t - \lambda}  + \beta)^2} d\beta - e^{-\frac{bx}{a}} \hspace{-10pt}\int\limits_{-\frac{x}{2a\sqrt{t-\lambda}}}^{\infty}\hspace{-10pt}  e^{-(b \sqrt{t-\lambda} + \beta)^2} d\beta \biggr ],\\
-D_{\lambda} \phi(x,t-\lambda) = \frac{x(t-\lambda)^{-\frac{3}{2}}} { 2a \sqrt{ \pi}} e^{-b^2(t-\lambda)- \frac{x^2}{4a^2 (t-\lambda)}};
\end{gather*}
\begin{flalign*}
&\text{and}  &&u = \frac{x}{2a\sqrt{\pi} }\int\limits_0^t  (t-\lambda)^{-\frac{3}{2}}  e^{-b^2(t-\lambda) -\frac{x^2}{4a^2(t-\lambda)}} F(\lambda)d\lambda,&&\phantom{and}
\end{flalign*}
cf.\ (9) Art.~51,
\begin{flalign*}
&\text{or}    &&u = \frac{2}{\sqrt{ \pi}} \!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\! e^{-\beta^2 -\frac{ b^2 x^2}{4a^2\beta^2} }F\biggl (t-\frac{x^2}{4a^2\beta^2}\biggr )d\beta,&&\phantom{or}
\end{flalign*}
cf.\ (10) Art.~51.

If $F(t)$ is periodic and has the value taken in Ex.~4, show that the value
approached by $u$ as $ t$ increases is
\[
u = \frac{1}{2}b_0 e^{-\frac{bx}{a}}+ \sum_{m=1}^{m=\infty} \rho_m e^{-\frac{x \sqrt{ 2}}{ 2a} p} \sin \biggl (m\alpha t - \frac{x \sqrt{2}}{2a }q + \lambda_m \biggr),
\]
\begin{flalign*}
&\text{where}  && p = (b^2 + \sqrt{b^4 + m^2\alpha^2})^{\frac{1}{2}} \quad \text{and} \quad q = (-b^2 + \sqrt{b^4 + m^2\alpha^2})^{\frac{1}{2}}. &&\phantom{where}
\end{flalign*}
% -----File: 101.png
\begin{flalign*}
&\indent\text{Given} & \int\limits_0^\infty e^{-x^2-\frac{a^2}{x^2}}dx&=\frac{\sqrt\pi}{2}e^{-2a}&\phantom{\indent Given}\\
&                    & \int\limits_0^\infty e^{-x^2-\frac{a^2}{x^2}}\sin\frac{b^2}{x^2}dx &=\frac{\sqrt\pi}{2}e^{-2c}\sin 2d&\\
&\text{and}          & \int\limits_0^\infty e^{-x^2-\frac{a^2}{x^2}}\cos\frac{b^2}{x^2}dx &=\frac{\sqrt\pi}{2}e^{-2c}\cos 2d,&
\end{flalign*}
where
\[
c=\frac{\sqrt2}{2}(a^2+\sqrt{a^4+b^4})^\frac{1}{2}\quad \text{and}\quad d=\frac{\sqrt2}{2}(-a^2+\sqrt{a^4+b^4})^\frac{1}{2}.
\]

Ängstrom's method of determining the conductivity of a metal is based on
the result just given (v.\ Phil.\ Mag.\ Feb.\ 1863), and is described by Sir Wm.~Thomson
(Encyc.\ Brit.\ Article ``Heat'') as by far the best that has yet been
devised.

\mypara{52.} If $u$ is a periodic function of the time when $x=0$ as in Art.~51 Ex.~4
and we are concerned with the limiting value approached by $u$ as $t$ increases
we can avoid evaluating a complicated definite integral if we take the following
course.

Since as we have seen in Art.~49 $u=e^{\beta t+\alpha x}$ is a solution of
\[
D_tu = a^2D_x^2u \tag{1}
\]
provided only that $\beta=a^2\alpha^2$
we have
\[
u=e^{\beta t±\frac{x}{a}\sqrt\beta}
\]
as a solution.

Replacing $\beta$ by $±\beta i$ this becomes
\begin{flalign*}
&          & u &=e^{±\beta ti±\frac{x}{a}\sqrt{\beta}\sqrt{±i}}&\\
&\text{or} & u &=e^{±\beta ti±\frac{x}{a}\sqrt{\frac{\beta}{2}}(1±i)}&\\
&\text{since} & \sqrt i   &=±\frac{1}{2}\sqrt2(1+i)&\\
&\text{and}   & \sqrt{-i} &=±\frac{1}{2}\sqrt2(1-i).&
\end{flalign*}

Hence
\begin{align*}
&& u&=e^{-\frac{x}{a}\sqrt\frac{\beta}{2}}\sin\biggl(\beta t-\frac{x}{a}\sqrt\frac{\beta}{2}\biggr),&
   u&=e^{-\frac{x}{a}\sqrt\frac{\beta}{2}}\cos\biggl(\beta t-\frac{x}{a}\sqrt\frac{\beta}{2}\biggr), \tag{2}&\\
&& u&=e^{ \frac{x}{a}\sqrt\frac{\beta}{2}}\sin\biggl(\beta t+\frac{x}{a}\sqrt\frac{\beta}{2}\biggr),&
   u&=e^{ \frac{x}{a}\sqrt\frac{\beta}{2}}\cos\biggl(\beta t+\frac{x}{a}\sqrt\frac{\beta}{2}\biggr), \tag{3}&
\end{align*}
are particular solutions of (1).
% -----File: 102.png

From these we get readily
\begin{flalign*}
&&u&=\rho_me^{-\frac{x}{a}\sqrt\frac{m\alpha}{2}}\sin\biggl(
m\alpha t-\frac{x}{a}\sqrt\frac{m\alpha}{2}+\lambda_m\biggr) &\tag{4}&
\intertext{as a solution. (4) reduces to}
&&u&=\rho_m\sin(m\alpha t+\lambda_m)\quad \text{when}\quad x=0&&\\
&\text{and to} &u&=\rho_me^{-\frac{x}{a}\sqrt\frac{m\alpha}{2}}\sin\biggl
(\lambda_m-\frac{x}{a}\sqrt\frac{m\alpha}{2}\biggr) \quad \text{when}\quad t=0.&&
\end{flalign*}

If we add a term which satisfies (1) and which is equal to zero when $x=0$
and to
$-\rho_me^{-\frac{x}{a}\sqrt\frac{m\alpha}{2}}\sin\biggl(\lambda_m-\dfrac{x}{a}\sqrt{\dfrac{m\alpha}{2}}\biggr)$
 when $t=0$ (v.\ Art.~50) we shall
have a solution of (1) which is zero when $t=0$ and which is
\[
\rho_m\sin(m\alpha t+\lambda_m)\quad \text{when}\quad x=0.
\]
The term in question approaches zero as $t$ increases [v.~(7) Art.~50] and we
have at once the solution given in Art.~51 Ex.~4, as our required result.

\EXAMPLE{}

Show that $u=e^{\beta t+\alpha x}$ is a solution of $D_tu=a^2D_x^2u-b^2u$ if $\beta=a^2\alpha^2-b^2$,
and hence that
\begin{align*}
u&=e^{\beta t \pm\frac{x}{a}\sqrt{b^2+\beta}},\quad
u =e^{ \pm\beta ti \pm\frac{x}{a}\sqrt{b^2 \pm\beta i}},\quad
u =e^{ \pm\beta ti \pm\frac{x}{a\sqrt2}(p \pm qi)},\\
u&=e^{ \pm\frac{px}{a\sqrt2}}\sin\biggl(\beta t \pm\frac{qx}{a\sqrt2}\biggr),\quad \text{and}\quad
u =e^{ \pm\frac{px}{a\sqrt2}}\cos\biggl(\beta t \pm\frac{qx}{a\sqrt2}\biggr),
\end{align*}
where
\[
p=[\sqrt{\beta^2+b^4}+b^2]^\frac{1}{2}\quad \text{and}\quad
q=[\sqrt{\beta^2+b^4}-b^2]^\frac{1}{2},
\]
are solutions.  Hence
\[
u=\rho_me^{-\frac{px}{a\sqrt2}}\sin\biggl(\beta t-\frac{qx}{a\sqrt2}+\lambda_m\biggr)
\]
is a solution.

If $\beta=m\alpha$ this last result reduces to $u=\rho_m\sin(m\alpha t+\lambda_m)$ when $x=0$
and by the reasoning of Art.~52 it must be the value $u$ approaches as $t$ increases
if we have the same conditions as in the last part of Art.~51 Ex.~5.

\mypara{53.} The whole problem of the flow of heat is treated by Sir William Thomson
(v.\ Math.\ and Phys.\ Papers, Vol.~II), and other recent writers from a different
and decidedly interesting point of view, which we shall briefly sketch
in connection with the problem of \textit{Linear Flow}.

Suppose we are dealing with a bar having a small cross-section and an adiathermanous
surface, and take as our unit of heat the amount required to raise by
a unit the temperature of a unit of length of the bar. If at a point of the bar a
% -----File: 103.png
quantity $Q$ of heat is suddenly generated the point is called an \emph{instantaneous
heat source} of strength $Q$.

If the heat instead of being suddenly generated is generated gradually and
at a rate that would give $Q$ units of heat per unit of time the point is called a
\emph{permanent heat source} of strength $Q$.

The temperature at any point of the bar at any time due to an instantaneous
source of strength $Q$ at the point $x=\lambda$ is easily found by the aid of formula
(8) Art.~49 as follows:---

If a quantity of heat $Q$ is suddenly generated along the portion of the bar
from $x=\lambda$ to $x=\lambda+\Delta\lambda$, where $\Delta\lambda$ is any arbitrary length, the temperature
of that portion will be suddenly raised to $\dfrac{Q}{\Delta\lambda}$, and we shall have by
(8) Art.~49
\[
u=\frac{Q}{2a\sqrt{\pi t}}\frac{1}{\Delta\lambda}\int\limits_\lambda^{\lambda+\Delta\lambda}e^{-\frac{(\lambda-x)^2}{4a^2t}}d\lambda  \tag{1}
\]
as the temperature of any point of the bar at any time $t$ thereafter.

If now we write $u$ equal to the limiting value approached by the second
member of (1) as $\Delta\lambda$ is made to approach zero we get
\[
u=\frac{Q}{2a\sqrt{\pi t}}e^{-\frac{(\lambda-x)^2}{4a^2t}} \tag{2}
\]
as the solution for the case where we have an instantaneous source at the
point $x=\lambda$.

It is to be observed that in (2) $u=0$ when $t=0$ and $u=\dfrac{Q}{2a\sqrt{\pi t}}$
when $x=\lambda$ and $t>0$.

If we have several sources we have only to add the temperatures due to the
separate sources.

Formula (8) Art.~49 may now be regarded as the solution for the case where
we start with an instantaneous heat source of strength $f(\lambda)d\lambda$ in every
element of length of the bar.

A source of strength $-Q$ is called a sink of strength $Q$; and (6) Art.~50
may be regarded as the solution for the case where we have at the start an
instantaneous source of strength $f(\lambda)d\lambda$ in every element of the bar whose distance
to the right of the origin is $\lambda$, and an instantaneous sink of strength
$f(\lambda)d\lambda$ in every element of the bar whose distance to the left of the origin is $\lambda$.

If we have an instantaneous source at the origin (2) reduces to
\[
u=\frac{Q}{2a\sqrt{\pi t}}e^{-\frac{x^2}{4a^2t}} \tag{3}
\]
% -----File: 104.png

For a permanent source of constant strength $Q$ at the origin (3) gives
\[
u=\frac{Q}{2a\sqrt\pi}\int\limits_0^t e^{-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{1}{2}}d\tau  \tag{4}
\]
and for a permanent source of variable strength $f(t)$
\[
u=\frac{1}{2a\sqrt\pi}\int\limits_0^t e^{-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{1}{2}}f(\tau)d\tau. \tag{5}
\]

In (4) and (5) $u$ obviously reduces to zero when $t=0$ and $x>0$, but its
value when $x=0$ is not easily determined. We can avoid the difficulty by
introducing the conception of a \emph{doublet}.

\mypara{54.} If a source and a sink of equal strength $Q$ are made to approach each
other while $Q$ multiplied by their distance apart is kept equal to a constant $P$
the limiting state of things is said to be due to a \emph{doublet} of strength $P$ whose
axis is tangent to the line of approach and points from sink to source. A
\emph{doublet} of strength $-P$ differs from a doublet of strength $P$ only in that its
axis has the opposite direction.

Let us find the temperature due to an instantaneous doublet of strength $P$
placed at the origin. For a source of strength $Q$ at $x=\eta$ and an equal sink
at $x=-\eta$ we have
\begin{align*}
u&=\frac{Q}{2a\sqrt{\pi t}}(e^{-\frac{(\eta-x)^2}{4a^2t}}-e^{-\frac{(\eta+x)^2}{4a^2t}}),\\
\intertext{or if $2\eta Q=P$,}
u&=\frac{P}{4a\eta\sqrt{\pi t}}e^{-\frac{(\eta^2+x^2)}{4a^2t}}
(e^{\frac{\eta x}{2a^2t}}-e^{-\frac{\eta x}{2a^2t}})\\
&=\frac{P}{2a\eta\sqrt{\pi t}}e^{-\frac{(\eta^2+x^2)}{4a^2t}}\sinh\frac{\eta x}{2a^2t}.
\end{align*}

If $\eta$ is made to approach zero
\[
\limit\left[\frac{1}{\eta}\sinh\frac{\eta x}{2a^2t}\right]=\frac{x}{2a^2t},
\]
\begin{flalign*}
&\text{and} & u=\frac{Px}{4a^3\sqrt{\pi t^3}}e^{-\frac{x^2}{4a^2t}} &\tag{1}&
\end{flalign*}
is the solution for the temperature at any time and place due to an instantaneous
doublet of strength $P$ placed at the origin. For a doublet at any other
point $x=\lambda$ we have
\[
u=\frac{P(x-\lambda)}{4a^3\sqrt{\pi t^3}}e^{-\frac{(x-\lambda)^2}{4a^2t}}. \tag{2}
\]
% -----File: 105.png

For a permanent doublet of constant strength $P$ placed at the origin we
have
\begin{flalign*}
&&u&=\frac{Px}{4a^3\sqrt\pi}\int\limits_0^t e^{-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{3}{2}}d\tau;& \tag{3}&\\
\intertext{and for a permanent doublet of variable strength $f(t)$}
&&u&=\frac{x}{4a^3\sqrt\pi}\int\limits_0^te^{-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{3}{2}}f(\tau)d\tau, & \tag{4}&\\
&\text{or} & u&=\frac{1}{a^2\sqrt\pi}\int\limits_\frac{x}{2a\sqrt t}^\infty e^{-\beta^2}f\left(t-\frac{x^2}{4a^2\beta^2}\right)d\beta & \tag{5}&\\
\intertext{if $x>0$, and}
&&u&=\frac{1}{a^2\sqrt\pi}\int\limits_\frac{x}{2a\sqrt t}^{-\infty}e^{-\beta^2}f\left(t-\frac{x^2}{4a^2\beta^2}\right)d\beta & \tag{6}&
\end{flalign*}
if $x<0$, if we let $\beta=\dfrac{x}{2a\sqrt{t-\tau}}$.

From (5) and (6) we see readily that $u=0$ when $t=0$ and that
 $u=\dfrac{f(t)}{2a^2}$ when $x=0$ if we approach the origin from the right and that
 $u=-\dfrac{f(t)}{2a^2}$ when $x=0$ if we approach the origin from the left.

If the point $x=0$ is kept at the constant temperature $b$ and we are concerned
only with positive values of $x$ we can get from (5) the solution given in
Art.~50 Ex.~4 by supposing a permanent doublet of strength $2a^2b$ placed at
the origin.

To solve the problem treated in Art.~51 we have only to suppose a permanent
doublet of strength $2a^2F(t)$ placed at $x=0$ and from (5) we get at once
(10) Art.~51.

\EXAMPLE{}

Show that if $D_tu=a^2D_x^2u-b^2u$ and an instantaneous source of strength
$Q$ is placed at $x=\lambda$
\[\phantom{v.\ Art.~51, Ex.~5.}
u=\frac{Q}{2a\sqrt{\pi t}}e^{-b^2t-\frac{(\lambda-x)^2}{4a^2t}}  \tag*{v.\ Art.~51, Ex.~5.}
\]

Show that if an instantaneous doublet of strength $P$ is placed at the point
 $x=0$
\[
u=\frac{Px}{4a^3\sqrt{\pi t^3}}e^{-b^2t-\frac{x^2}{4a^2t}}.
\]
% -----File: 106.png

If a permanent doublet of strength $f(t)$ is placed at $x=0$
\begin{align*}
u&=\frac{x}{4a^3\sqrt\pi}\int\limits_0^t e^{-b^2(t-\tau)-\frac{x^2}{4a^2(t-\tau)}}(t-\tau)^{-\frac{3}{2}}f(\tau)d\tau\\
 &=\frac{1}{a^2\sqrt\pi}\int\limits_{\frac{x}{2a\sqrt t}}^{±\infty} e^{-\beta^2-\frac{b^2x^2}{4a^2\beta^2}}f\left(t-\frac{x^2}{4a^2\beta^2}\right)d\beta,
\end{align*}
whence $u=0$ when $t=0$ and $x>0$ or $x<0$ and $u=±\dfrac{f(t)}{2a^2}$ when
 $x=0$.

Hence if we place at $x=0$ a permanent doublet of strength $2a^2F(t)$ we
get the solution given in Art.~51 Ex.~5 for the case where $u=F(t)$ when
 $x=0$ and $u=0$ when $t=0$ provided we are concerned only with positive
values of $x$.

If $F(t)=c$ this reduces to
\[
u=\frac{2c}{\sqrt\pi}\int\limits_{\frac{x}{2a\sqrt t}}^{\infty} e^{-\beta^2-\frac{b^2x^2}{4a^2\beta^2}}d\beta.
\]
\mypara{55.} As another example of the use of Fourier's Integral we shall consider
the transmission of a disturbance along a stretched elastic string.

Suppose we have a stretched elastic string so long that we need not consider
what happens at its ends, that is so long that we may treat its length as
infinite. Let the string be initially distorted into some given form and then
released; to investigate its subsequent motion.

Let us take the position of equilibrium of the string as the axis of $X$ and
any given point as origin.

We have, then, to solve the differential equation
\[
D_t^2y=a^2D_x^2 y \tag{1}
\]
[v.\ \smallromr{VIII} Art.~1] subject to the conditions
\begin{alignat*}{2}
   y&=f(x)   & \quad\text{when}\quad    & t=0 \tag{2}\\
D_ty&=0      & \quad\text{``}\quad\quad   & t=0. \tag{3}
\end{alignat*}

As in Art.~8 we find
\[
y=\cos\alpha(x±at)\quad \text{and}\quad y=\sin\alpha(x±at)
\]
as particular solutions of (1).

From these we must build up a value that will reduce to
\[
f(x)=\frac{1}{\pi} \int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x).d\lambda \tag{4}
\]
% -----File: 107.png
when $t=0$ and will at the same time satisfy (3).
\begin{flalign*}
&&y&=\cos\alpha\lambda\cos\alpha(x+at)+\sin\alpha\lambda\sin\alpha(x+at)&\phantom{or}\\
&\text{or}& y&=\cos\alpha(\lambda-x-at)&
\end{flalign*}
is a solution of (1).
\begin{flalign*}
&\indent\text{Hence}& y=\frac{1}{\pi}\int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x-at).d\lambda & \tag{5}&\hspace{2.5em}
\end{flalign*}
is also a solution of (1).

(5) reduces to $y=f(x)$ when $t=0$ but it gives
\[
D_ty=\frac{a}{\pi}\int\limits_0^\infty\alpha d\alpha
\int\limits_{-\infty}^\infty f(\lambda)\sin\alpha(\lambda - x).d\lambda
\]
when $t=0$ and consequently does not satisfy equation (3).

If in forming (5) we use $\cos\alpha(x-at)$ and $\sin\alpha(x-at)$ instead of
 $\cos\alpha(x+at)$ and $\sin\alpha(x+at)$ we get
\[
y=\frac{1}{\pi}\int\limits_0^\infty d\alpha
\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x+at).d\lambda \tag{6}
\]
which is a solution of (1), and reduces to $y=f(x)$ when $t=0$, but it gives
\[
D_ty=-\frac{a}{\pi}\int\limits_0^\infty\alpha d\alpha
\int\limits_{-\infty}^\infty f(\lambda)\sin\alpha(\lambda-x).d\lambda
\]
when $t=0$ and does not satisfy (3).

If, however, we take one-half the sum of the values of $y$ in (5) and (6) we
get
\begin{align*}
y=\frac{1}{2}\Bigg[\frac{1}{\pi}&\int\limits_0^\infty d\alpha
\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x-at).d\lambda\\
+&\frac{1}{\pi}\int\limits_0^\infty d\alpha
\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x+at).d\lambda\Bigg], \tag{7}
\end{align*}
a solution of (1) which satisfies both (2) and (3), and is, therefore, our required
solution.

This result can be very much simplified.

If we substitute $z=x+at$
\begin{align*}
 &\frac{1}{\pi}\int\limits_0^\infty d\alpha
\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x-at).d\lambda\\
=&\frac{1}{\pi}\int\limits_0^\infty d\alpha
\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-z).d\lambda
=f(z)=f(x+at);
\end{align*}
% -----File: 108.png
and in like manner we can show that
\[
\frac{1}{\pi}\int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty f(\lambda)\cos\alpha(\lambda-x+at).d\lambda=f(x-at).
\]

Hence our solution becomes
\[
y=\frac{1}{2}[f(x+at)+f(x-at)]. \tag{8}
\]

This result is of great importance in the theory of elastic strings and it
shows that the initial disturbance splits into two equal waves which run along
the string, one to the right and the other to the left, with a uniform velocity $a$,
and that there is nothing like a periodic motion or vibration of any sort unless
the ends of the string produce some effect.

\mypara{56.} If the string is not initially distorted but starts from its position of
equilibrium with a given initial velocity impressed upon each point we have to
solve the equation
\[
D_t^2y=a^2D_x^2y \tag{1}
\]
subject to the conditions
\begin{alignat}{3}
   y& = 0    &  \text{when\quad} & t& = 0   &\tag{2}\\
D_ty& = F(x) &  \text{``\qquad}   & t& = 0.  &\tag{3}
\end{alignat}

We get by the process used in Art.~55
\begin{align*}
y&=\frac{1}{2\pi a}\int\limits_0^\infty d\alpha\int\limits_{-\infty}^\infty F(\lambda)
\left[\frac{\sin\alpha(\lambda-x+at)}{\alpha}
     -\frac{\sin\alpha(\lambda-x-at)}{\alpha}\right]d\lambda\\
 &=\frac{1}{2\pi a}\int\limits_{-\infty}^\infty F(\lambda)d\lambda\int\limits_0^\infty
\left[\frac{\sin\alpha(\lambda-x+at)}{\alpha}
     -\frac{\sin\alpha(\lambda-x-at)}{\alpha}\right]d\alpha;
\end{align*}
but \qquad\qquad $\displaystyle\int\limits_0^\infty\frac{\sin\alpha(\lambda-x+at)}{\alpha}d\alpha
-\int\limits_0^\infty\frac{\sin\alpha(\lambda-x-at)}{\alpha}d\alpha=\pi$\\
if $x-at < \lambda < x+at$, and is equal to zero for all other values of $\lambda$; since
\begin{alignat*}{2}
\int\limits_0^\infty\frac{\sin mx}{x}dx
&= \phantom{-}\frac{\pi}{2} &\quad\text{if}\quad & m>0\\
&=          - \frac{\pi}{2} &\quad\text{if}\quad & m<0\\
&= \phantom{-}0             &\quad\text{if}\quad & m=0.
\end{alignat*}
v.\ Int.\ Cal.\ Art.~92~(3).
\begin{flalign*}
&\indent\text{Hence} & y=\frac{1}{2a}\int\limits_{x-at}^{x+at}F(\lambda)d\lambda &\tag{4}&\hspace{2.5em}
\end{flalign*}
is our required solution.
% -----File: 109.png

\EXAMPLE{S}

1.\quad If the string is initially distorted and starts with initial velocity so that
 $y=f(x)$ and $D_{t}y=F(x)$ when $t=0$
\[
y=\frac{1}{2}[f(x+at)+f(x-at)]+\frac{1}{2a}\int\limits_{x-at}^{x+at}F(\lambda)d\lambda.
\]

2.\quad If the initial disturbance is caused by a blow, as from the hammer in a
piano, which impresses upon all the points in a portion of the string of length
$c$ an equal transverse velocity $b$ show that the front of the wave which will be
seen to run to the left along the string will be a straight line having a slope
 equal to $\dfrac{b}{2a}$ and a length equal to $\displaystyle\frac{c}{2a}\sqrt{4a^2+b^2}$. Of course a wave having
a front of the same length with a slope equal to $-\dfrac{b}{2a}$ will be seen to run to
the right along the string, and the effect of the two waves will be to lift the
string bodily and permanently to a distance $\dfrac{bc}{2a}$ above its original position.

\mypara{57.}  We shall now take up a few examples of the use of \textit{Fourier's Series}.

In the problem of Art.~7 let the temperature of the base of the plate be a
given function of $x$, the other conditions remaining unchanged.
\begin{flalign*}
&\indent\text{Since} &    f(x)& =\sum_{m=1}^{m=\infty}(a_m\sin mx)&&\phantom{\indent Since}\\[-7ex]
\end{flalign*}
\begin{flalign*}
&\text{where}        &  a_m =& \frac{2}{\pi} \int\limits_0^\pi f(\alpha)\sin m\alpha.d\alpha&&\phantom{where}\\[-8ex]
\end{flalign*}
\begin{flalign*}
&\text{we have}      &u=\frac{2}{\pi}\sum_{m=1}^{m=\infty}\bigg[e^{-my}&\sin mx\int\limits_0^\pi f(\alpha)\sin m\alpha.d\alpha\bigg]. & \tag{1}&\hspace{1.5em}
\end{flalign*}

If the breadth of the plate is $a$ instead of $\pi$
\[
u=\frac{2}{a}\sum_{m=1}^{m=\infty}\bigg[e^{-\frac{m\pi y}{a}}\sin\frac{m\pi x}{a}\int\limits_0^a f(\lambda)\sin\frac{m\pi\lambda}{a}d\lambda\bigg]. \tag{2}
\]

\mypara{58.} If the temperature of the base is unity and the breadth of the plate is
$\pi$ the solution is, as we have seen in Art.~7,
\[
u=\frac{4}{\pi}\bigg[e^{-y}\sin x+\frac{1}{3}e^{-3y}\sin 3x+\frac{1}{5}e^{-5y}\sin 5x+\cdots\bigg].  \tag{1}
\]

This series can be summed without difficulty. We have the development
\[
\log(1+z)=\frac{z}{1}-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\cdots
\]
if the modulus of $z$ is less than 1. Int.\ Cal.\ Art.~221~(4).
% -----File: 110.png
\begin{flalign*}
& \text{Hence} & \log(1 - z) = -\frac{z}{1} -\frac{ z^2}{2} -\frac{ z^3}{3}  -\frac{ z^4}{4} - \cdots && \phantom{Hence}
\end{flalign*}
if  $\operatorname{mod.}  z<1$.
\begin{flalign*}
& \text{and}  & \frac{1}{2}[\log(1+z)-\log(1-z)]= \frac{ z}{1}  +\frac{ z^3}{3}  + \frac{ z^5}{5}  + \cdots  && \,\tag{2}
\end{flalign*}
if $\operatorname{mod.} z  <1$. \\

But
\begin{flalign*}
\log(1+z) &=\log[1+r(\cos \phi + i \sin \phi)]   \\
&=\frac{1}{2} \log[(1+r \cos \phi)^2 + (r \sin \phi)^2] + i \tan^{-1} \frac{r \sin \phi }{ 1+r \cos \phi} \\
&=\frac{1}{2} \log{(1 + 2r \cos \phi + r^2)} + i \tan^{-1}\frac{ r \sin \phi }{1+r \cos \phi},
\intertext{and}
\log(1-z) &= \frac{1}{2} \log(1-2r \cos\phi + r^2) - i \tan^{-1} \frac{r \sin \phi}{ 1-r \cos\phi},
\end{flalign*}
[Int.\ Cal.\ Art.~33~(2)],\\
and (2) becomes
\begin{multline*}
\frac{1}{2}\left[\frac{1}{2} \log\frac{ 1+2r \cos\phi+r^2 }{ 1-2r \cos\phi+r^2} + i \tan^{-1} \frac{2r \sin \phi }{1-r^2} \right]\\
= \frac{r(\cos\phi+i \sin\phi)}{1} +\frac{ r^3(\cos 3\phi + i\sin 3\phi) }{3} + \cdots  \tag{3}
\end{multline*}

From (3) we get two equations
\begin{gather*}
\frac{1}{4}\log {\frac{1+2r \cos \phi+r^2}{ 1-2r \cos \phi+r^2}} =  \frac{r\cos \phi}{1} + \frac{r^3 \cos 3 \phi}{3} + \frac{r^5 \cos 5 \phi}{5} + \cdots   \tag{4}\\
\frac{1}{2} \tan^{-1}\frac{ 2r \sin\phi}{ 1-r^2} = \frac{r \sin\phi}{1} + \frac{r^3 \sin3\phi}{3} + \frac{r^5 \sin5\phi}{5}+ \cdots  \tag{5}
\end{gather*}
both valid for all values of $\phi$ provided $r < 1$.

$e^{-y}$ is less than $1$ if  $y$ is positive.

Hence from (5)
\begin{align*}
\frac{e^{-y }\sin x}{1} + \frac{e^{-3y }\sin 3x}{3}+ \frac{e^{-5y }\sin 5x}{5} &+ \cdots = \frac{1}{2} \tan^{-1}\frac{ 2e^{-y }\sin x }{ 1-e^{-2y}}\\
                                         &= \frac{1}{2} \tan^{-1}\frac{ 2\sin x }{e^y-e^{-y} }= \frac{1}{2}\tan^{-1}\frac{ \sin x }{ \sinh y},
\end{align*}
and (1) may be written\\
\[
u = \frac{2}{\pi}\tan^{-1}\frac{ \sin x }{ \sinh y}.   \tag{6}
\]

% -----File: 111.png
If we replace $r$ by $e^{-y}$ and $\phi$ by $x$  in\\[-4ex]
\begin{flalign*}
&                  &     \log[1 + r(&\cos \phi + i \sin \phi)]&\\
&\text{it becomes} & \log[1 + e^{-y}&\cos x + i e^{-y} \sin x]\hfil\phantom{it becomes}&\\
&\text{or}         &       \log[1 + &\cos  z + i \sin z]&\\[-5ex]
\end{flalign*}
v.\ Int.\ Cal.\ Art.~35 (3) and (4)\\
a function of $z$ as a whole; and\\[-4ex]
\begin{flalign*}
&               & \log [1 - r(&\cos \phi + i \sin \phi)]&\phantom{becomes}\\
&\text{becomes} & \log (1 - &\cos z - i\sin z);&\\[-5ex]
\end{flalign*}
hence by Int.\ Cal.\ Arts.~212 and 213,\\[-4ex]
\begin{flalign*}
&&\frac{1}{4} \log &\frac{1 + 2e^{-y}\cos x + e^{-2y}}{1 - 2e^{-y}\cos x + e^{-2y}}\quad \text{and}\quad
 \frac{1}{2}\tan^{-1}\frac{2e^{-y}\sin x}{1 - e^{-2y}}&\\
&\text{or}&& \frac{1}{4}\log\dfrac{\cosh y + \cos x}{\cosh y -\cos x}\quad \text{and}\quad
  \frac{1}{2}\tan^{-1}\frac{\sin x}{\sinh y}&
\end{flalign*}
are conjugate functions, and
\[
u_1= \frac{1}{\pi}\log\dfrac{\cosh y + \cos x}{\cosh y - \cos x} \tag{7}
\]
is the solution for the problem where the isothermal lines are the lines of flow
of the present problem and the lines of flow are the isothermal lines of the
present problem.

For our problem, then, the isothermal lines are given by the equation
\begin{flalign*}
&          &   \frac{2}{\pi}\tan^{-1} &\frac{\sin x}{\sinh y} = a&&\\
&\text{or} & \dfrac{\sin x}{\sinh y}& = \tan\dfrac{a \pi}{2}& \tag{8}&
\end{flalign*}
and the lines of flow by
\begin{flalign*}
&& \frac{1}{\pi}&\log\dfrac{\cosh y + \cos x}{\cosh y - \cos x} = b,&&\\
&\text{or}&& \dfrac{\cosh y + \cos x}{\cosh y - \cos x}  = e^{\pi b}.&\tag{9}&
\end{flalign*}

\EXAMPLE{S}

1.\quad If $D_x^2u+D_y^2u=0$, and $u=1$ when $y=0$, and $u=0$ when
$x=0$ and when $x=a$,
\begin{align*}
u &= \frac{4}{\pi}\bigg[e^{-\frac{\pi y}{a}} \sin\frac{\pi x}{a} + \frac{1}{3}e^{-
\frac{3 \pi y}{a}} \sin\frac{3\pi x}{a} + \frac{1}{5}e^{-\frac{5 \pi y}{a}} \sin \frac{5\pi x}{a}
 +  \cdots \bigg] \\
\label{err111}
&= \frac{2}{\pi}\tan^{-1} \frac{\sin \dfrac{\pi x}{a}}{\sinh \dfrac{\pi y}{a}}.
\end{align*}

% -----File: 112.png

2.\quad If $u = \phi (x)$ when $y = 0$, $u =f(y)$ when $x = 0$, and $u = F(y)$
 when $x = a$
\begin{align*}
  &u= \frac{2}{a}
    \sum_{m=1}^{m=\infty} e^{-\frac{m\pi y}{a}} \sin \frac{m\pi x}{a}
    \int \limits_0^a \phi(\lambda) \sin \frac{m\pi\lambda}{a} d\lambda
\\
& + \frac{1}{2a} \sin \frac{\pi x}{a} \int \limits_0^\infty \ \biggl[ \
        \frac{1}{ \cosh \dfrac{\pi}{a} (\lambda-y)
                - \cos  \dfrac{\pi x}{a} }
      - \frac{1}{ \cosh \dfrac{\pi}{a} (\lambda+y)
                - \cos  \dfrac{\pi x}{a} }
    \ \biggr] \ f(\lambda)\, d\lambda
\\
& + \frac{1}{2a} \sin \frac{\pi x}{a} \int \limits_0^\infty  \ \biggl[ \
        \frac{1}{ \cosh \dfrac{\pi}{a} (\lambda-y)
                + \cos  \dfrac{\pi x}{a} }
      - \frac{1}{ \cosh \dfrac{\pi}{a} (\lambda+y)
                + \cos  \dfrac{\pi x}{a} }
    \ \biggr] \ F(\lambda)\, d\lambda
\end{align*}
v.\ Art.~48, Exs.~4, 5, and 6.

\mypara{59.} If three sides of a plane rectangular sheet of conducting material be
kept at potential zero and the value of the potential function at every point of
the fourth side be given; to find the value of this potential function at any
point of the sheet.

To formulate:---
\begin{alignat*}{3}
  &&\llap{$D_x^2V$} &\rlap{$+\;D_y^2V = 0.$} \tag{1}\\
  &V = 0    && \text{when}\quad && x = 0.   \tag{2}\\
  &V = 0    && \quad\text{``}   && x = a.   \tag{3}\\
  &V = 0    && \quad\text{``}   && y = b.   \tag{4}\\
  &V = f(x) && \quad\text{``}   && y = 0.   \tag{5}
\end{alignat*}

Working as in Art.~48 we get
\[
  \frac{ \sinh \dfrac{m\pi  }{a} (b-y) }
       { \sinh \dfrac{m\pi b}{a} }
  \sin \frac{m\pi x}{a}
\]
as a value of $V$ which satisfies equations (1), (2), (3), and (4) if $m$ is an integer.
Therefore
\[
  V = \frac{2}{a} \sum_{m=1}^{m=\infty} \ \biggl[ \
        \frac{ \sinh \dfrac{m\pi  }{a} (b-y) }
             { \sinh \dfrac{m\pi b}{a}       }
        \sin \frac{m\pi x}{a}
        \int \limits_0^a f(\lambda) \sin \frac{m\pi\lambda}{a} d\lambda
      \ \biggr]
\tag{6}
\]
is our required solution.

% -----File: 113.png
\EXAMPLE{S}

1.\quad If $f(x) = 1$ Eq.~(6) Art.~59 reduces to
\[
\begin{split}
V=\frac{4}{\pi}  \Biggl [ &\frac{\sinh \dfrac{\pi}{a} (b-y) }{ \sinh \dfrac{\pi b}{a }}\sin \frac{\pi x}{a}  + \frac{1}{3}\frac{\sinh \dfrac{3\pi}{a} (b-y) }{ \sinh \dfrac{3\pi b}{a} }\sin \frac{3\pi x}{a} \Biggr .\\
&+\Biggl .  \frac{1}{5}\frac{\sinh \dfrac{5\pi}{a} (b-y) } {\sinh \dfrac{5\pi b}{a}} \sin \frac{5\pi x}{a}+ \cdots \Biggr ]
\end{split}
\]

2.\quad If $V=0$ when $x = 0$, $V=0$ when $x = a$, $V= 0$ when $y = 0$,
 and $V=F(x)$ when $y = b$, then
\[
V = \frac{2}{a} \sum_{m=1}^{m=\infty}\Biggl  [\frac{\sinh \dfrac{m\pi y }{ a}} {\sinh \dfrac{m\pi b}{a}} \sin \frac{m\pi x}{a }\int \limits_0^a F(\lambda)\sin \frac{m\pi\lambda}{a} d\lambda \Biggr ].
\]

3.\quad If $F(x) = 1$ the answer of Ex.~2 reduces to
\[
V = \frac{4}{\pi}\Biggl [\frac{\sinh \dfrac{\pi y}{a }}{\sinh \dfrac{\pi b}{a}} \sin \frac{\pi x}{a }+ \frac{1}{3} \frac{\sinh \dfrac{3\pi y}{a}}{ \sinh \dfrac{3\pi b}{a}} \sin \frac{3\pi x}{a} + \frac{1}{5} \frac{\sinh \dfrac{5\pi y}{a}}{ \sinh \dfrac{5\pi b}{a}} \sin \frac{5\pi x}{a} + \cdots \Biggr ].
\]

4.\quad If $V=0$ when $x = 0$, $V=0$ when $x = a$, $V=f(x)$ when $y = 0$,
 and $V= F(x)$ when $y = b$, then
\[
\begin{split}
V = \frac{2}{a} \sum_{m=1}^{m=\infty} \Biggl [\sin \frac{m\pi x }{ a} \Biggl ( \frac{\sinh \dfrac{m\pi}{a} (b-y)}{ \sinh \dfrac{m\pi b}{a}} \int \limits_0^a f(\lambda)\sin \frac{m\pi\lambda}{a} d\lambda \Biggr . \Biggr .\\
+\Biggl .  \Biggl . \frac{\sinh\dfrac{ m\pi y }{ a} }{ \sinh \dfrac{m\pi b}{a }}\int \limits_0^a F(\lambda)\sin\frac{ m\pi\lambda}{a} d\lambda \Biggr ) \Biggr ].
\end{split}
\]

5.\quad If $f(x) = F(x)$ the answer of Ex.~4 reduces to
\[
V = \frac{2}{a }\sum_{m=1}^{m=\infty} \Biggl [  \frac{\cosh \dfrac{m\pi}{a } \Bigl (\dfrac{b}{2}-y \Bigr) }{\cosh \dfrac{m\pi b}{2a}} \sin \frac{m\pi x}{a} \int \limits_0^a f(\lambda)\sin \frac{m\pi \lambda}{a} d\lambda   \Biggr ].
\]
% -----File: 114.png

6.\quad If $f(x) = F(x) = 1$ the answer of Ex.~5 reduces to
\[\begin{split}
V = \frac{4}{\pi}\Biggl [ &\frac{\cosh\dfrac{ \pi}{a} \Bigl ( \dfrac{b}{2}-y \Bigr ) }{\cosh \dfrac{\pi b}{2a}} \sin\frac{ \pi x}{a} + \frac{1}{3}\dfrac{\cosh\dfrac{3 \pi}{a} \Bigl ( \dfrac{b}{2}-y \Bigr ) }{\cosh \dfrac{3\pi b}{2a}} \sin\frac{ 3\pi x}{a} \Biggr .\\
&+\Biggl . \frac{1}{5}\dfrac{\cosh\dfrac{5 \pi}{a} \Bigl ( \dfrac{b}{2}-y \Bigr ) }{\cosh \dfrac{5\pi b}{2a}} \sin\frac{ 5\pi x}{a} + \cdots \Biggr ].
\end{split}
\]

7.\quad If $V=f(x)$ when $y = 0$, $V=F(x)$ when $y = b$, $V=\phi (y)$ when
 $x = 0$, and $V=\chi (y)$ when $x=a$, then
\[
\begin{alignedat}{2}
V &= \frac{2}{a} \sum_{m=1}^{m=\infty}\Biggl  [&&\sin \frac{m\pi x}{a}  \Biggl (\dfrac{\sinh \dfrac{m\pi}{a}(b-y) }{ \sinh \dfrac{m\pi b}{a}} \int\limits_0^a f(\lambda)\sin \frac{m\pi\lambda}{a} d\lambda \Biggr . \Biggr .\\
&&&+ \Biggl . \Biggl. \dfrac{ \sinh \dfrac{m\pi y}{a }} {\sinh \dfrac{m\pi b}{a}} \int\limits_0^a
 F(\lambda)\sin \frac{m\pi\lambda}{a} d\lambda\Biggr) \Biggr ]\\
 &+ \frac{2}{b}\sum_{m=1}^{m=\infty}\Biggl [&&\sin \frac{m\pi y }{b} \Biggl (\frac{\sinh \dfrac{m\pi}{b}(a-x) }{ \sinh \dfrac{m\pi a}{b}} \int\limits_0^b \phi(\lambda)\sin \frac{m\pi\lambda}{b} d\lambda  \Biggr . \Biggr .\\
&&&+ \Biggl . \Biggl . \frac{\sinh \dfrac{m\pi x}{b} }{ \sinh \dfrac{m\pi a}{b}} \int\limits_0^b
 \chi(\lambda)\sin \frac{m\pi\lambda}{b} d\lambda \Biggr ) \Biggr ].
 \end{alignedat}
\]

8.\quad If $f(x)=\phi(y)=0$ and $F(x)=\chi(y)=1$ the answer of Ex.~7 may be
reduced to
\begin{align*}
V = \frac{2}{\pi} \Biggl [ &\frac{ \pi y}{2b} - \frac{\sinh \dfrac{\pi}{b} \left (\dfrac{a}{2}-x \right ) }{ \sinh \dfrac{\pi a}{2b }}\sin \frac{\pi y}{b} + \frac{1}{2}\frac{ \cosh \dfrac{2\pi}{b} \left (\dfrac{a}{2}-x \right ) }{\cosh \dfrac{2\pi a}{2b}} \sin \frac{2\pi y}{b} \Biggr .\\
 &- \Biggl . \frac{1}{3} \frac{\sinh \dfrac{3\pi}{b} \left (\dfrac{a}{2}-x \right ) }{ \sinh \dfrac{3\pi a}{2b}} \sin \frac{3\pi y}{b}  + \frac{1}{4} \frac{\cosh \dfrac{4\pi}{b} \left (\dfrac{a}{2}-x \right ) }{ \cosh \dfrac{4\pi a}{2b}} \sin \frac{4\pi y}{b} - \cdots \Biggr].
\end{align*}
% -----File: 115.png

9.\quad Find the temperature of the middle point of a thin square plate whose
faces are impervious to heat; 1st, when three edges are kept at the temperature
0° and the fourth edge at the temperature 100°; 2d, when two
opposite edges are kept at the temperature 0° and the other two at the temperature
100°; 3d, when two adjacent edges are kept at the temperature 0°
and the other edges at the temperature 100°. See examples 3, 6, and 8.\\
\phantom{kindly go to the right hand margin}\hfill
\textit{Ans.}, (1) 25°; (2) 50°; (3) 50°.

\mypara{60.} Let us pass on to the consideration of the flow of heat in one dimension.

Suppose that we have an infinite solid with two parallel plane faces whose
distance apart is $c$.

Take the origin in one face and the axis of $X$ perpendicular to the faces.
Let the initial temperature be any given function of $x$ and let the two faces be
kept at the constant temperature zero; to find the temperature at any point of
the slab at any time.

We have to solve the equation
\[
D_tu = a^2D_x^2u  \tag{1}
\]
subject to the conditions
\begin{alignat*}{3}
u& = 0  && \text{when}\quad & x & = 0  \tag{2}\\
u& = 0  && \quad\text{``}   & x & = c  \tag{3}\\
u& =f(x)&& \quad\text{``}   & t & = 0. \tag{4}
\end{alignat*}

In Art.~49 we have found
\begin{flalign*}
&&u &= e^{-a^2\alpha^2t}\sin\alpha x&\\
&\text{and} & u&=e^{-a^2\alpha^2t}\cos\alpha x&\\
\end{flalign*}
as particular solutions of (1).

$u=e^{-a^2\alpha^2t}\sin\alpha x$ satisfies (2) whatever value is given to $\alpha$. It satisfies (3)
if $\alpha=\dfrac{m\pi}{c}$ provided $m$ is an integer. Let us try to build a value of $u$ out of
terms of the form $Ae^{-\frac{a^2m^2\pi^2t}{c^2}}\sin\dfrac{m\pi x}{c}$ which shall satisfy (4).

We have
\begin{gather*}
f(x)=\frac{2}{c}\sum_{m=1}^{m=\infty}\Bigl[\sin\frac{m\pi x}{c}
\int\limits_0^c f(\lambda)\sin\frac{m\pi\lambda}{c}d\lambda\Bigr]. \tag{5}\\
u=\frac{2}{c}\sum_{m=1}^{m=\infty}\Bigl[e^{-\frac{m^2a^2\pi^2t}{c^2}}\sin\frac{m\pi x}{c}
\int\limits_0^c f(\lambda)\sin\frac{m\pi\lambda}{c}d\lambda\Bigr], \tag{6}
\end{gather*}
reduces to (5) when $t=0$ and is our required solution.
% -----File: 116.png

\EXAMPLE{S}

1.\quad If $f(\lambda)=b$, a constant, (6) Art.~60 reduces to
\[
u=\frac{4b}{\pi}\Bigl[e^{-\frac{  a^2\pi^2t}{c^2}}\sin\frac{ \pi x}{c}
          +\frac{1}{3}e^{-\frac{ 9a^2\pi^2t}{c^2}}\sin\frac{3\pi x}{c}
          +\frac{1}{5}e^{-\frac{25\pi^2a^2t}{c^2}}\sin\frac{5\pi x}{c}
+\cdots\Bigr].
\]

2.\quad An iron slab 10~cm.\ thick is placed between and in contact with two
other iron slabs each 10~cm.\ thick. The temperature of the middle slab is at
first 100° throughout, and of the outside slabs 0° throughout. The outer faces
of the outside slabs are kept at the temperature 0°. Required the temperature
of a point in the middle of the middle slab fifteen minutes after the slabs have
been placed in contact. Given $a^2=0.185$ in C.G.S. units. \hfill \textit{Ans.}, 10°.3.\\

3.\quad Two iron slabs each 20~cm.\ thick one of which is at the temperature 0°
and the other at the temperature 100° throughout, are placed together face to
face, and their outer faces are kept at the temperature 0°. Find the temperature
of a point in their common face and of points 10~cm.\ from the common
face fifteen minutes after the slabs have been put together.\\
\phantom{kindly go to the right hand margin}\hfill
\textit{Ans.}, 22°.8; 15°.1; 17°.2.\\

4.\quad One face of an iron slab 40~cm.\ thick is kept at the temperature 0° and
the other face at the temperature 100° until the permanent state of temperatures
is set up. Each face is then kept at the temperature 0°. Required
the temperature of a point in the middle of the slab, and of points 10~cm.\ from
the faces fifteen minutes after the cooling has begun.  \hfill \textit{Ans.}, 22°.8; 15°.6; 16°.7.

\mypara{61.} If the faces of the slab treated in Art.~60 instead of being kept at the
temperature zero are rendered impervious to heat, the solution of the problem
is easy.

In this case we have to solve the equation
\[
D_tu = a^2D_x^2u
\]
subject to the conditions
\begin{alignat*}{3}
D_xu& = 0   &&\text{when}\quad & x&=0\\
D_xu& = 0   &&\text{\quad``}        & x&=c\\
   u& = f(x)&&\text{\quad``}        & t&=0.
\end{alignat*}

We have only to use the particular solution
\begin{flalign*}
&                 & u& = e^{-a^2\alpha^2t}\cos\alpha x&\\
&\text{as we used}& u& = e^{-a^2\alpha^2t}\sin\alpha x&\phantom{as we used}
\end{flalign*}
in Art.~60. We get
\[
u=\frac{2}{c}\biggl[\frac{1}{2}\int\limits_0^c f(\lambda)d\lambda
+\sum_{m=1}^{m=\infty}
\Bigl(e^{-\frac{m^2a^2\pi^2t}{c^2}}\cos\frac{m\pi x}{c}
\int\limits_0^c f(\lambda)\cos\frac{m\pi\lambda}{c}d\lambda\Bigr)\biggr]. \tag{1}
\]
% -----File: 117.png

\EXAMPLE{S}

1.\quad Solve example  2  Art.~60 supposing that the outer surfaces are blanketed
after the slabs are placed together so that heat can neither enter nor escape.
Find in addition the temperature of the outer surfaces fifteen minutes after
the slabs are placed in contact. \hfill \textit{Ans.},  33°.3;   33°.3.\\

2.\quad Solve example  3  Art.~60 on the hypothesis just stated, getting in addition
the temperatures of points on the outer surfaces.\\
\phantom{kindly go to the right hand margin}\hfill
\textit{Ans.}, 50°;   33°.9;  66°.1;   27°.2;   72°.8.\\

3.\quad Solve example  4  Art.~60 supposing that heat neither enters nor escapes
at the outer surfaces after the permanent state of temperatures has been set
up. Find also the temperatures of points in the outer surfaces.\\
\phantom{kindly go to the right hand margin}\hfill
\textit{Ans.}, 50°;  39°.7;  60°.3;  35°.5;  64°.5.\\

4.\quad Show that if $u = 0$ when $x = 0$, $D_xu = 0$ when $x = c$, and $u=f(x)$
 when $t = 0$,
\[
u = \frac{2}{c}\sum_{m=0}^{m=\infty} \biggl ( e^{-\frac{(2m+1)^2a^2\pi^2t}{4c^2}} \sin\frac{(2m+1)\pi x}{2c} \int\limits_0^c f(\lambda) \sin \frac{(2m+1)\pi\lambda}{2c} d\lambda \biggr ).
\]

\textit{Suggestion}: Assume $u = 0$ when $x = 2c$ and $f(2c - x) =f(x)$, and see
(6) Art.~60.

\mypara{62.} If the temperature of the right-hand face of the slab considered in Art.~60
is a constant $\gamma$  instead of zero we have only to add to the second member
of  (6) Art.~60 a term  $u_1$  which shall satisfy the conditions
\begin{alignat*}{3}
&&\llap{$D_tu_1\:$} &\rlap{$= a^2D_x^2u_1$}\tag{1}\\
u_1 &= 0\quad  && \text{when} \quad & x &= 0       \tag{2}\\
u_1 &= 0       && \quad \text{``}   & t &= 0       \tag{3}\\
u_1 &= \gamma  && \quad \text{``}   & x &= c.      \tag{4}
\end{alignat*}

$u_1 = \dfrac{\gamma x}{c}$ obviously satisfies (1), (2), and (4); to make it satisfy (3) as well
we must add a term  $u_2$  which shall be equal to zero when $x = 0$ and when
 $x = c$ and to $-\dfrac{\gamma x}{c}$ when $t = 0$, while always satisfying (1).  It is given
immediately by (6) Art.~60 and is
\begin{gather*}
u_2=-\frac{2\gamma}{c^2} \sum_{m=1}^{m=\infty} \biggl (e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \int\limits_0^c \lambda \sin\frac{m\pi\lambda}{c} d\lambda \biggr ).      \tag{5}\\
\int\limits_0^c \lambda \sin\frac{m\pi\lambda}{c} d\lambda  =-\frac{c^2}{m\pi}\cos m\pi = (- 1)^{m+1} \frac{c^2}{m\pi},
\end{gather*}
% -----File: 118.png
\begin{flalign*}
&\text{and}       &&u_2=\frac{2\gamma}{\pi} \sum_{m=1}^{m=\infty} \left ( \frac{(-1)^m}{m }e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \right ).     &&  \tag{6}\\[-5ex]
\end{flalign*}
\begin{flalign*}
&\indent \text{Hence}  &&  u_1 = \gamma \biggl [ \frac{x}{c}  + \frac{2}{\pi} \sum_{m=1}^{m=\infty} \left ( \frac{(-1)^m}{m }e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \right ) \biggr ]. &&\hspace{2em} \tag{7}
\end{flalign*}

If the left-hand face of the slab considered in Art.~60 is to be kept at a
constant temperature  $\beta$  and the right-hand face at the temperature zero we
can get the term $u_3$ which must be added to the second member of (6)  Art.~60
by replacing  $\gamma$  by  $\beta$ and  $x$  by $c-x$  in (7). We then have
\[
u_3= \beta \biggl [\frac{c-x}{c} - \frac{2}{\pi} \sum_{m=1}^{m=\infty} \left ( \frac{1}{m}e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \right ) \biggr ].    \tag{8}
\]

\EXAMPLE{S}

1.\quad Show that if $u = \beta$ when $x = 0$, $u = \gamma$ when $x = c$, and $u=f(x)$
when $t = 0$
\[
\begin{split}
u=\beta + (\gamma - \beta) \biggl [ \frac{x}{c} + \frac{2}{\pi} \sum_{m=1}^{m=\infty} \left ( \frac{(-1)^m}{m }e^{-\frac{m^2\pi^2a^2t}{c^2}} \sin \frac{m\pi x}{c} \right ) \biggr ]\\
+\frac{2}{c} \sum_{m=1}^{m=\infty} \biggl ( e^{-\frac{m^2a^2\pi^2t}{c^2}} \sin \frac{m\pi x}{c} \int\limits_0^c [f(\lambda)-\beta] \sin \frac{m\pi\lambda}{c} d\lambda \biggr ).
\end{split}
\]

2.\quad Show that if $u = \beta$ when $x = 0$, $u = 0$ when $t = 0$, and $D_xu = 0$
 when $x = c$
\begin{align*}
&u =\beta \biggl [1 - \frac{4}{\pi}\sum_{m=0}^{m=\infty} \left (\frac{1}{2m+1} e^{-\frac{(2m+1)^2a^2\pi^2t}{4c^2}}\sin \frac{(2m+1)\pi x}{2c} \right ) \biggr ]\\
&=\beta \left [1 -\frac{4}{\pi} \left (e^{-\frac{  a^2\pi^2t}{4c^2}} \!\sin \frac{ \pi x}{2c} +
                            \frac{1}{3}e^{-\frac{ 9a^2\pi^2t}{4c^2}} \!\sin \frac{3\pi x}{2c} +
                            \frac{1}{5}e^{-\frac{25a^2\pi^2t}{4c^2}} \!\sin \frac{5\pi x}{2c} + \cdots \!\right ) \right ].
\end{align*}

\mypara{63.} If the temperature of the right-hand face of the slab just considered is
a function of the time instead of a constant and the temperature of the left-hand
face is zero the problem can be solved by a method nearly identical with
that of Art.~51.
% -----File: 119.png

Let $\phi(x,t)$ be a function of $x$ and $t$ which shall be zero if $t$ is less than zero
and shall be equal to
\[
\frac{x}{c}+\frac{2}{\pi}\sum_{m=1}^{m=\infty}
\left(\frac{(-1)^m}{m}e^{-\frac{m^2a^2\pi^2t}{c^2}}\sin\frac{m\pi x}{c}\right)
\]
[v.\ (7) Art.~62] if $t$ is equal to or greater than zero. So that
\begin{alignat*}{4}
\phi(x,t)& =0 \quad&&\text{if}\quad & t &< 0 &&\\
\phi(x,t)& =0      &&\text{``}      & t &= 0 & \quad\text{unless}\quad  x &= c\\
\phi(x,t)& =1      &&\text{``}      & t &= 0 &      \text{and}\quad     x &= c\\
\phi(x,t)& =1      &&\text{``}      & x &= c &&\\
\phi(x,t)& =0      &&\text{``}      & x &= 0.&&
\end{alignat*}

Precisely as in Art.~51 we get
\[
u=\limit_{\tau\doteq 0} \sum_{k=0}^{k=n}\left[F(k\tau)\frac{[\phi(x,t-k\tau)-\phi(x,t-(k+1)\tau)]\tau}{\tau}\right]  \tag{1}
\]
as the required solution of our problem, $n$ being as in Art.~51 the largest
integer in $\dfrac{t}{\tau}$ where $t$ is any given value of the time.

On our hypothesis the last term of (1), that is, $-F(n\tau)\phi[x,t-(n+1)\tau]=0$;
the next to the last term $F(n\tau)\phi(x,t-n\tau)$ has for its limiting value
\[
F(t)\phi(x,0)=F(t)\left[\frac{x}{c}+\frac{2}{\pi}\sum_{m=1}^{m=\infty}\left(\frac{(-1)^m}{m}\sin\frac{m\pi x}{c}\right)\right],
\]
while as in Art.~51 the limiting value of the rest of the sum is
\begin{gather*}
-\int\limits_0^t F(\lambda)D_\lambda\phi(x,t-\lambda)d\lambda.\\
D_\lambda\phi(x,t-\lambda)=\frac{2a^2\pi}{c^2}\sum_{m=1}^{m=\infty}\bigg[(-1)^m me^{-\frac{m^2a^2\pi^2}{c^2}(t-\lambda)}\sin\frac{m\pi x}{c}\bigg].
\end{gather*}

Hence
\begin{align*}
u=F(t)\biggl[\frac{x}{c}&+\frac{2}{\pi}\sum_{m=1}^{m=\infty}\biggl(\frac{(-1)^m}{m}\sin\frac{m\pi x}{c}\biggr)\biggr]\\
&-\frac{2a^2\pi}{c^2}\sum_{m=1}^{m=\infty}\biggl((-1)^m m\sin\frac{m\pi x}{c}\int\limits_0^t F(\lambda)e^{-\frac{m^2a^2\pi^2}{c^2}(t-\lambda)}d\lambda\biggr),
\end{align*}
% -----File: 120.png
\begin{multline*}
u=\frac{x}{c}F(t)+\frac{2}{\pi}\sum_{m=1}^{m=\infty}\biggl[
\frac{(-1)^m}{m}\sin\frac{m\pi x}{c}\biggl(F(t)\\
-\frac{m^2a^2\pi^2}{c^2}\int\limits_0^t F(\lambda)e^{-\frac{m^2a^2\pi^2}{c^2}(t-\lambda)}d\lambda\biggr)\biggr].
\tag{2}
\end{multline*}

If we substitute $\beta=\dfrac{m^2a^2\pi^2}{c^2}(t-\lambda)$ we get
\[
u=\frac{x}{c}F(t)+\frac{2}{\pi}\sum_{m=1}^{m=\infty}
\biggl[\frac{(-1)^m}{m}\sin\frac{m\pi x}{c}
\biggl(F(t)-\hspace{-10pt}\int\limits_0^{\frac{m^2a^2\pi^2t}{c^2}}\hspace{-10pt}e^{-\beta}
F\biggl(t-\frac{\beta c^2}{m^2a^2\pi^2}\biggr)d\beta\biggr)\biggr].  \tag{3}
\]

\EXAMPLE{S}

1.\quad If the temperature of the left-hand face is a function of $t$ and the temperature
of the right-hand face is zero and the initial temperature is zero
\[
u=\biggl(1-\frac{x}{c}\biggr)F(t)-\frac{2}{\pi}\sum_{m=1}^{m=\infty}
\biggl[\frac{1}{m}\sin\frac{m\pi x}{c}
\biggl(F(t)-\hspace{-10pt}\int\limits_0^{\frac{m^2a^2\pi^2t}{c^2}}\hspace{-10pt}e^{-\beta}
F\biggl(t-\frac{\beta c^2}{m^2a^2\pi^2}\biggr)d\beta\biggr)\biggr].
\]

2.\quad If the temperature of the left-hand face is a function of $t$, the initial
temperature is zero, and the right-hand face is impervious to heat
\begin{multline*}
u=F(t)-\frac{4}{\pi}\sum_{m=0}^{m=\infty}
\biggl[\frac{1}{2m+1} \sin\frac{(2m+1)\pi x}{2c}
\biggl(F(t)\\
-\frac{(2m+1)^2a^2\pi^2}{4c^2}\int\limits_0^t
F(\lambda)e^{-\frac{(2m+1)^2a^2\pi^2}{4c^2}(t-\lambda)}d\lambda\biggr)\biggr].
\end{multline*}

3.\quad If in Arts.~60--63 we are dealing with a bar of small cross-section and of
length $c$ and heat is radiating from the surface of the bar into air at the temperature
zero so that $D_tu=a^2D_x^2u-b^2u$, show that: (\textit{a}) the second members
of (6) Art.~60 and (1) Art.~61 must be multiplied by $e^{-b^2t}$; (\textit{b}) equation
(7) Art.~62 becomes
\[
u_1=\gamma\Biggl\{\frac{\sinh\dfrac{bx}{a}}{\sinh\dfrac{bc}{a}}+2a^2\pi e^{-b^2t}
\sum_{m=1}^{m=\infty}\left[(-1)^m\frac{m}{b^2c^2+m^2a^2\pi^2}
e^{-\frac{m^2a^2\pi^2t}{c^2}}\sin\frac{m\pi x}{c}\right]\Biggr\};
\]
% -----File: 121.png
(\textit{c}) equation (2) Art.~63 becomes
\begin{multline*}
u=\frac{\sinh \dfrac{bx}{a}} {\sinh \dfrac{bc}{a}} F(t)+2a^2\pi \sum_{m=1}^{m=\infty} \bigg \{ \frac{(-1)^m m}{b^2c^2+m^2a^2\pi^2} \sin\frac{ m\pi x}{c} \bigg [ F(t) \\
-  \frac{b^2c^2+m^2a^2\pi^2}{c^2} \int\limits_0^t e^{-\frac{b^2c^2+m^2a^2\pi^2}{c^2}(t-\lambda)}F(\lambda)d\lambda \bigg ] \bigg \}.
\end{multline*}

\mypara{64.} The problem of the motion of a finite stretched elastic string of length
$l$  fastened at the ends and distorted at first into some given curve $y=f(x)$,
 and then allowed to swing, has been treated and partially solved in Art.~8.

The complete solution is easily seen to be
\[
y=\frac{2}{l} \sum_{m=1}^{m=\infty} \sin \frac{m\pi x}{l} \cos \frac{m\pi at}{l }\int\limits_0^l f(\lambda) \sin \frac{m\pi\lambda}{l }d\lambda.    \tag{1}
\]

The second member of (1) is a periodic function of  $t$  having the period
$\dfrac{2l}{a}$.  The motion, then, unlike that in the case of an infinite string (Art.~55) is
a true vibration, a periodic motion. The period  $\dfrac{2l}{a}$  is the time it takes a disturbance
to travel twice the length of the string (v.\ Art.~55).

A careful examination of (1) will show that the actual motion is a good deal
like that in the case considered in Art.~55. The original disturbance breaks
up into two waves one of which runs to the right until it reaches the end of
the string and is then reflected, and runs back to the left or the under side of
the string, while the other wave runs to the left and is reflected at the left-hand
end of the string and runs back to the right under the string and is
again reflected, runs back to the left over the string and so on indefinitely.

If the curve into which the string is distorted at the start is of the form
 $y=b \sin \dfrac{m\pi x}{l}$ the solution is
\[
y=b \sin \frac{m\pi x}{l} \cos\frac{ m\pi at}{l}.    \tag{2}
\]

No matter what value  $t$  may have the curve is always of the form
\[
y=A \sin \frac{m\pi x}{l};
\]
that is, for different values of  $t$  we have a set of sine curves differing only in
the amplitude and not at all in the period of the curve. In this case either
the whole string if $m=1$, or each $m$th
of the string if  $m$  is not equal to
one, rises and falls, and there is no apparent onward motion. When this is
the case we are said to have a \textit{steady} vibration.
% -----File: 122.png

If $m = 1$ we get steady motion of the string as a whole and if the vibration
is rapid enough to give a musical note the note is said to be the pure fundamental
note of the string.  If $m = 2$ the vibration is twice as rapid as when
 $m = 1$, the middle point of the string does not move and is called a node, the
two halves of the string are in opposite phases of vibration at any instant, and
the note given is an octave higher than the fundamental note and is called its
pure \textit{first harmonic}.

If $m = 3$ the vibration is three times as rapid as in the first case, there are
two nodes $x=\dfrac{l}{3}$ and $x=\dfrac{2l}{3}$, and the note is the pure \textit{second harmonic} of
the fundamental note.

For any value of  $m$  the vibration is $m$ times as rapid as when $m = 1$, there
are ${m -1}$ nodes at the points $x=\dfrac{l}{m}, x=\dfrac{2l}{m}, \cdots  x=\dfrac{m-1}{m}l$, and we get the
${m - 1}$st harmonic of the fundamental note.

It is clear from (1) that no matter what the original form of the string the
resulting vibration can be regarded as a combination of steady vibrations each
of which alone would give the fundamental note of the string or one of its
harmonics, and that the complex note resulting is really a concord of the fundamental
note and some of its harmonics.

A finely trained ear can often recognize in a complex note the fundamental
note of the string and some of its harmonics and is capable of analyzing a
complex note into its component pure notes precisely as Fourier's Theorem
enables us to analyze the complex function representing the initial form of the
string into the simpler sine-functions which must be combined to form it.

\EXAMPLE{S}

1.\quad Show that if a point whose distance from the end of a harp string is
$\dfrac{1}{n}$th  the length of the string is drawn aside by the player's finger to a distance
$b$  from its position of equilibrium and then released, the form of the vibrating
string at any instant is given by the equation
\[
y=\frac{2bn^2}{(n-1)\pi^2} \sum_{m=1}^{m=\infty} \left ( \dfrac{1}{m^2} \sin \frac{m\pi}{n} \sin \frac{ m\pi x}{l} \cos \frac{m\pi at}{l} \right ).
\]
Show from this that all the harmonics of the fundamental note of the
string which correspond to forms of vibration having nodes at the point
drawn aside by the finger will be wanting in the complex note actually
sounded.\\
% -----File: 123.png

2.\quad If a stretched string starts from its position of equilibrium, each of its
points having a given initial velocity, so that we have
\begin{alignat*}{3}
   y&=0     & \text{when}\quad && t&=0\\
D_ty&=F(x)  & \text{``}\qquad  && t&=0\\
   y&=0     & \text{``}\qquad  && x&=0\\
   y&=0     & \text{``}\qquad  && x&=l,
\end{alignat*}
the solution of the problem of its vibration is easy and gives
\[
y=\frac{2}{a\pi}\sum_{m=1}^{m=\infty}\bigg(\frac{1}{m}\sin\frac{m\pi x}{l}\sin\frac{m\pi at}{l}
\int\limits_0^l F(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\bigg).
\]

3.\quad Write down the solution for the case where the string is initially distorted
and each point has a given initial velocity.

\mypara{65.} If we do not neglect the resistance of the air in the problem of the
vibration of a stretched string the differential equation is rather more complicated
and the solution is not so easily obtained. The equation is given as \smallromr{IX}
Art.~1.

Let us solve the problem for the case where there is no initial velocity.
\begin{flalign*}
&\indent\text{Here we have} & D_t^2y+2kD_ty=a^2D_x^2y.\qquad\qquad & \tag{1}&
\end{flalign*}
\vspace{-6.5ex}
\begin{alignat*}{3}
   y&=0     && \text{when}\quad  & x&=0  \tag{2}\\
   y&=0     && \quad\text{``}    & x&=l  \tag{3}\\
   y&=f(x)  && \quad\text{``}    & t&=0  \tag{4}\\
D_ty&=0     && \quad\text{``}    & t&=0. \tag{5}
\end{alignat*}

We get particular solutions of (1) in the usual way. Assume $y=e^{\alpha x+\beta t}$
 and substitute in (1). We have
\[
\beta^2+2k\beta=a^2\alpha^2
\]
as the only necessary relation between $\beta$ and $\alpha$. This gives
\begin{flalign*}
&&\beta=-k±\sqrt{a^2\alpha^2+k^2}.&&\\
&\indent\text{Hence}& y=e^{\alpha x-kt±t\sqrt{a^2\alpha^2+k^2}}& \tag{6}&
\end{flalign*}
is a solution of (1) no matter what the value of $\alpha$.

To throw it into Trigonometric form replace $\alpha$ by $\alpha i$, and since in actual
problems $k$, which is proportional to the resistance, is very small, take $-1$
out as a factor of the radical. We have
\[
y=e^{-kt}e^{(\alpha x±t\sqrt{a^2\alpha^2-k^2)}i}.
\]
% -----File: 124.png

Since $\alpha$
may be positive or negative we can get
\begin{flalign*}
&           &y = e^{-kt}\sin(\alpha x ± t\sqrt{a^2\alpha^2-k^2)}&&\\
&\text{and} &y = e^{-kt}\cos(\alpha x ± t\sqrt{a^2\alpha^2-k^2)}&&
\end{flalign*}
as solutions of (1), or by combining these
\begin{align*}
y& = e^{-kt}\sin \alpha x \cos t\sqrt{a^2\alpha^2-k^2} \tag{7}\\
y& = e^{-kt}\sin \alpha x \sin t\sqrt{a^2\alpha^2-k^2} \tag{8}\\
y& = e^{-kt}\cos \alpha x \cos t\sqrt{a^2\alpha^2-k^2} \tag{9}\\
y& = e^{-kt}\cos \alpha x \sin t\sqrt{a^2\alpha^2-k^2} \tag{10}
\end{align*}

(7) and (8) satisfy (1) and (2) for all values of $\alpha$. They satisfy (3) if
$\alpha=\dfrac{m\pi}{l}$. Let us see if out of them we cannot build up a value that will satisfy
(4) and (5) as well.
\begin{gather*}
f(x)=\frac{2}{l}\sum_{m=1}^{m=\infty}\biggl(\sin\frac{m\pi x}{l}\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr). \tag{11}\\
y=\frac{2}{l}e^{-kt}\sum_{m=1}^{m=\infty}\biggl(\sin\frac{m\pi x}{l}\cos t
\sqrt{\frac{m^2\pi^2 a^2}{l^2}-k^2}.\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr) \tag{12}
\end{gather*}
reduces to (11) when $t=0$ and therefore satisfies (4).
\begin{align*}
% recast to fit line
D_t y=&-\frac{2}{le^{kt}}\sum_{m=1}^{m=\infty}\biggl(\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}.
\sin\frac{m\pi x}{l}\sin t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}\\
&\hspace{18.5em}.\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr)\\
&-\frac{2k}{le^{kt}}\sum_{m=1}^{m=\infty}\biggl(
\sin\frac{m\pi x}{l}\cos t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}.\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr). \tag{13}
\end{align*}

When $t=0$ the first line of the second member of (13) vanishes but the
second line reduces to
\[
-\frac{2k}{l}\sum_{m=1}^{m=\infty}\biggl(\sin\frac{m\pi x}{l}\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr).
\]
We must, then, introduce into (12) an additional term which shall equal zero
when $t=0$ and whose derivative with respect to $t$ shall cancel the term above
when $t=0$.
% -----File: 125.png

This is easily seen to be
\[
\frac{2k}{l}e^{-kt}\sum_{m=1}^{m=\infty}\frac{1}{\sqrt{\dfrac{m^2\pi^2a^2}{l^2}-k^2}}
\sin\frac{m\pi x}{l}\sin t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}.\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda.
\]

Hence our complete solution is
\begin{multline*}
y=\frac{2}{l}e^{-kt}\sum_{m=1}^{m=\infty}\bigg[\bigg(\cos t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}\\
+\frac{k}{\sqrt{\dfrac{m^2\pi^2a^2}{l^2}-k^2}}\sin t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}\bigg)
 \sin\frac{m\pi x}{l}\int\limits_0^l f(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\bigg]. \tag{14}
\end{multline*}

Here the fact that $e^{-kt}$, which decreases rapidly as $t$ increases, is a factor of
the whole second member shows that the amplitude of the vibration rapidly
decreases.

Comparing this solution with that given in Art.~64 for the case where there
is no resistance we see that the period of any given term
\[
A\sin\frac{m\pi x}{l}\cos t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2},
\]
is greater than that of the corresponding term $A_1\sin\dfrac{m\pi x}{l}\cos\dfrac{m\pi at}{l}$ in Art.~64.
In other words the effect of the resistance of the air is to flatten somewhat
each component part of the note given by the string. More than this
since the periods of the different terms of (14) are no longer exact submultiples
of the period of the first term, the component notes are no longer in perfect
harmony with the fundamental note of the string, and the ideal perfect harmony
between the fundamental note and its harmonics is not quite realized in
any actual case.

When $k$ is very small, as in the case of a fine string, the departure from
perfect harmony is very slight; but in the case of a coarse string or worse still
of an elastic ribbon, where the resistance of the air is considerable, the
unmusical character of the sound is very noticeable.

\EXAMPLE{S}

1.\quad Solve Ex.~1 Art.~64 allowing for the resistance of the air.\\

2.\quad Solve Ex.~2 Art.~64 allowing for the resistance of the air;
\begin{multline*}
% recast to fit line
y=\frac{2}{l}e^{-kt}\sum_{m=1}^{m=\infty}\biggl(\frac{1}{\sqrt{\dfrac{m^2\pi^2a^2}{l^2}-k^2}}
\sin\frac{m\pi x}{l}\sin t\sqrt{\frac{m^2\pi^2a^2}{l^2}-k^2}\\[-5ex]
.
\int\limits_0^l F(\lambda)\sin\frac{m\pi\lambda}{l}d\lambda\biggr).
\end{multline*}
% -----File: 126.png

3.\quad Find a particular solution of (1) Art.~65 on the assumption that it is of
the form $y=T.X$, where $T$ is a function of $t$ alone and $X$ a function of $x$
alone.

\mypara{66.} We pass on now to a couple of problems that require the modification
and extension of Fourier's Theorem, \emph{the cooling of a sphere in air}, and the
\emph{vibration of a stretched rectangular membrane}, but as an introduction to the
former we shall first consider the following very simple problem; to find the
temperature of any point of a sphere whose initial temperature is any given
function of $r$ the distance of the point from the centre, and whose surface is
kept at the constant temperature $b$.

Here we are to solve
\[
D_t(ru) = a^2D_r^2(ru), \tag{1}
\]
see \smallrom{V} Art.~1, subject to the conditions
\begin{alignat*}{3}
u &=f(r)\ && \text{when}\quad &t&=0 \tag{2}\\
u &=b     && \quad\text{``}   &r&=c \tag{3}
\end{alignat*}
if $c$ is the radius.

Let $v=ru$, then our equations become
\begin{alignat*}{3}
&&\llap{$D_tv =\:$} &\rlap{$a^2D_r^2v$}   \tag{4}\\
v &=rf(r)\ && \text{when}\quad  &t&=0  \tag{5}\\
v &=bc     && \quad\text{``}    &r&=c  \tag{6}\\
v &=0      && \quad\text{``}    &r&=0. \tag{7}
\end{alignat*}
Our problem is now precisely that of Art.~62 and we have as our solution
\begin{align*}
ru=&\frac{2}{c}\sum_{m=1}^{m=\infty}\biggl(e^{-\frac{m^2a^2\pi^2}{c^2}t}\sin\frac{m\pi r}{c}
\int\limits_0^c\lambda f(\lambda)\sin\frac{m\pi\lambda}{c}d\lambda\biggr)\\
&+b\biggl[r+\frac{2c}{\pi}\sum_{m=1}^{m=\infty}\biggl(\frac{(-1)^m}{m}e^{-\frac{m^2a^2\pi^2}{c^2}t}\sin\frac{m\pi r}{c}\biggr)\biggr]. \tag{8}
\end{align*}

\EXAMPLE{S}

1.\quad If $f(r)=b$ (8) Art.~66 reduces to $u=b$ and there is no change of
temperature.

2.\quad If the initial temperature is constant and equal to $\beta$
\begin{multline*}
u= b + \frac{2c}{\pi r}(\beta-b)
       \bigg[e^{-\frac{ a^2\pi^2}{c^2}t}\sin\frac{ \pi r}{c}
-\frac{1}{2}e^{-\frac{4a^2\pi^2}{c^2}t}\sin\frac{2\pi r}{c}\\
+\frac{1}{3}e^{-\frac{9a^2\pi^2}{c^2}t}\sin\frac{3\pi r}{c}-\cdots\biggr].
\end{multline*}
% -----File: 127.png

3.\quad An iron sphere 40~cm.\ in diameter is heated to the temperature 100°
centigrade throughout; its surface is then kept at the constant temperature 0°.
Find the temperature of a point 10~cm.\ from the centre, and find the temperature
of the centre, 15 minutes after cooling has begun. Given $a^2 = 0.185$
 in C.G.S.\ units. \hfill \textit{Ans.}, 2°.1; 3°.3.

\mypara{67.} If instead of having the temperature of the surface of the sphere
constant, the sphere is placed in air which is kept at the constant temperature
zero, the problem is much more complicated. For in this case the
surface temperature can no longer be simply expressed but is given by a new
differential equation
\[
D_{r}u + hu = 0 \quad \text{when} \quad r = c,          \tag{1}
\]
where $h$ is an experimental constant depending upon what is called the surface
conductivity of the sphere.

Our equations, then, are
\begin{gather*}
D_{t}(ru) = a^{2}D_{r}^{2}(ru)            \tag{2}\\
u =f(r) \qquad \text{when}\qquad  t = 0                             \tag{3}\\
D_{r}u + hu = 0 \qquad \text{when} \qquad r = c.              \tag{4}
\end{gather*}
As in Art.~66 let  $v = ru$;  then we have
\begin{alignat*}{3}
&&\llap{$D_{t}v =\:$} & \rlap{$a^{2}D_{r}^{2}v$}         \tag{5}\\
v &= rf(r) \quad && \text{when} \quad & t &= 0           \tag{6}\\
v &= 0           &&\quad \text{``}    & r &= 0           \tag{7}\\[-5ex]
\end{alignat*}
\[
D_{r}v + \left ( h - \frac{1}{c} \right )v = 0 \quad \text{when} \quad r = c.        \tag{8}
\]

$v = e^{-a^2\alpha^{2}t} \cos \alpha r$ and $v = e^{-a^2\alpha^{2}t} \sin \alpha r$ have already been found as particular
solutions of (5) (see Art.~60).
\[
v = e^{-a^2\alpha^{2}t} \sin \alpha r           \tag{9}
\]
satisfies (7) for all values of $\alpha$.

Substitute this value of $v$ in (8) and we have
\[
\alpha c \cos \alpha c + (hc - 1) \sin \alpha c = 0.         \tag{10}
\]

If  $\alpha_{k}$ is a value of $\alpha$ which is a root of the transcendental equation (10)
\[
v = e^{-a^{2}\alpha_{k}^{2}t} \sin \alpha_{k}r            \tag{11}
\]
will satisfy (5), (7), and (8).

It remains to see whether out of terms of the form given in (11) we can
build up a value of $v$ which will satisfy (6).
% -----File: 128.png

When $t = 0$ the second member of (11) reduces to $\sin \alpha_kr$. If then we
can express  $rf(r)$  as a sum of terms of the form $b_k \sin \alpha_k r$ where $\alpha_k$ is a root
of (10)
\[
v = \sum b_k e^{-a^2\alpha_k^2t} \sin \alpha_k r       \tag{12}
\]
will satisfy all of the equations (5), (6), (7), and (8), and will be the required
solution.

Here, then, we have a new problem analogous to that of developing in a
Fourier's Series, but rather more complicated, namely, to develop any function
of  $x$ in a series of the form $\displaystyle\sum a_m \sin \alpha_m x$ where  $\alpha_m$  is a root of the equation
(10); or if we call $ac = \phi$ and $hc-1=p$, where $a_m = \dfrac{\phi_m}{c}$, $\phi_m$ being a root
of the equation
\begin{gather*}
\phi \cos \phi + p \sin \phi = 0   \tag{13}
\intertext{or more simply of}
\phi + p \tan \phi = 0;                  \tag{14}
\end{gather*}
remembering that the series and the function must be equal for all values of  $x$
between zero and  $c$.

If  $\phi_m$ is a root of (14)  $- \phi_m$  is also a root.

Since $\sin \dfrac{\phi_m}{c} x = - \sin \left (- \dfrac{\phi_m}{c} x \right )$ the terms of the required development
which correspond to negative roots may be combined with those corresponding
to positive roots, and therefore we need consider only positive roots.

$\phi = 0$ is a root of (14) but as $\sin 0 = 0$ there will be no corresponding
term in the development.
If we construct the curve
\begin{align*}
y &= - \frac{1}{p} x    \tag{15}
\intertext{and the curve}
y &= \tan x       \tag{16}
\end{align*}
the abscissas of their points of intersection are values of  $x$  which satisfy
 $\dfrac{x}{p} + \tan x = 0$, that is, are roots of equation (14). It is easy to see that
there will always be an infinite number of real positive roots, one for each of
the branches of the periodic curve $y = \tan x$ which lie to the right of the
origin. The numerical values of these roots can be obtained by an easy computation.
The construction suggested above shows that as  $m$  increases  $\phi_m$
will rapidly approach the value $(2m - 1) \dfrac{\pi}{2}$  if  $p$  is positive or if  $p$ is negative
and numerically less than unity, and  $(2m + 1) \dfrac{\pi}{2}\tablestrut$ if  $p$  is negative and numerically
greater than unity.
% -----File: 129.png

There exist, then, an infinite number of positive real roots of $\phi + p \tan \phi = 0$
 and consequently of
\[
\alpha c \cos \alpha c + (hc - 1) \sin \alpha c = 0.
\]

\mypara{68.} The development called for in the last article can be obtained very
easily from a simpler one which we shall now consider, namely, to develop  $f(x)$
into a series of the form
\[
f(x) = a_1 \sin\phi_1 x + a_2 \sin\phi_2 x + a_3 \sin\phi_3 x + \cdots  \tag{1}
\]
where $\phi_1$, $\phi_2$, $\phi_3 \cdots$ are roots of the equation
\[
\phi \cos\phi +p \sin\phi = 0,           \tag{2}
\]
the development to hold good for all values of  $x$ between $x = 0$ and $x = 1$.

Let us proceed as in Arts.~24 and 27. Call $\dfrac{1}{n+1} = \Delta x$ and form $n$ equations
by substituting for  $x$  in turn in the equation
\[
f(x) = a_1 \sin\phi_1 x + a_2 \sin\phi_2 x + a_3 \sin\phi_3 x + \cdots + a_n \sin\phi_n x   \tag{3}
\]
the values $\Delta x$,  $2\Delta x$,  $3\Delta x$,  $\cdots n\Delta x$; this being equivalent to making the values
of the sum and the function coincide for the n values of $x$ substituted.

To determine any coefficient  $a_m$  multiply the first equation by $\Delta x.\sin(\phi_m$ $\Delta x)$,
 the second by $\Delta x.\sin(2\phi_m\Delta x)$, the third by $\Delta x.\sin(3\phi_m \Delta x)$, and so on, the
$n$th equation by $\Delta x.\sin(n\phi_m\Delta x)$; add the equations and compute the limiting
values of the terms of the resulting equation as  $n$  is indefinitely increased.
This as in Art.~24 is seen to be equivalent to multiplying (3) by $\sin\phi_m x.dx$
and integrating between the limits $x = 0$ and $x = 1$.

The first member of the resulting equation is
\[
\int\limits_0^1 f(x) \sin \phi_m x.dx;
\]

The coefficient of $a_k$ is
\[
\int\limits_0^1 \sin\phi_k x \sin\phi_m x.dx,
\]
and of $a_m$ is
\[
\int\limits_0^1 \sin^2\phi_m x.dx.
\]
% -----File: 130.png
\begin{align*}
\int\limits_0^1 \sin\phi_k x \sin\phi_m x.dx &= \frac{1}{2}\int\limits_0^1 [\cos(\phi_k-\phi_m)x - \cos(\phi_k+\phi_m)x]dx\\
&=\frac{1}{2} \left [ \frac{\sin(\phi_k-\phi_m) }{ \phi_k-\phi_m }- \frac{\sin(\phi_k+\phi_m) }{ \phi_k+\phi_m} \right  ]\\
&= - \frac{\phi_k \cos\phi_k \sin\phi_m - \phi_m \sin\phi_k \cos\phi_m  }{  \phi_k^2 - \phi_m^2 }    \tag{4}
\end{align*}
\begin{flalign*}
& \indent \text{But} & \phi_k \cos\phi_k &+ p \sin\phi_k = 0&&\\
& \text{and} & \phi_m \cos\phi_m &+ p \sin\phi_m = 0\qquad\text{by (2)}.
\end{flalign*}

Hence the numerator of the second member of (4) is zero, and the coefficient
of  $a_k$  vanishes if $k$ is not equal to  $m$.

\[
\int\limits_0^1 \sin^2\phi_m x.dx = \frac{1}{2\phi_m }[\phi_m - \sin\phi_m \cos\phi_m]    = \frac{1}{2} \left [1 - \frac{\sin 2\phi_m }{ 2\phi_m} \right  ].   \tag{5}
\]
\begin{flalign*}
&\text{Therefore}& a_m &= \frac{2 }{1 - \dfrac{\sin 2\phi_m }{ 2\phi_m}} \int\limits_0^1 f(x) \sin\phi_m x.dx. &&\hspace{2em}\tag{6}
\end{flalign*}

The coefficient of the integral in (6) can be transformed as follows so as not
to involve trigonometric functions.
\begin{gather*}
\phantom{\qquad by (2)}\phi_m \cos\phi_m + p \sin\phi_m = 0,  \qquad \text{by (2)}\\
\phi_m \cos^2\phi_m + \frac{p}{2} \sin 2\phi_m = 0,\\
\frac{\sin 2\phi_m }{2\phi_m} = -\frac{\cos^2\phi_m }{p}.    \tag{7}\\
\phi_m^2 \cos^2\phi_m = p^2 \sin^2\phi_m,\\[1ex]
(\phi_m^2 + p^2) \cos^2\phi_m = p^2,\\[1ex]
\frac{\cos^2\phi_m }{p} = \frac{p }{ \phi_m^2 + p^2}.   \tag{8}
\end{gather*}

Hence by (7) and (8)
\begin{flalign*}
&  &&1-\frac{\sin 2\phi_m }{ 2\phi_m} = \frac{\phi_m^2+p(p+1) }{ \phi_m^2 + p^2},&&\\
&\text{and}   &a_m = &\frac{2(\phi_m^2+p^2) }{ \phi_m^2+p(p+1)} \int\limits_0^1 f(\alpha)\sin \phi_m \alpha.d\alpha.&&   \tag{9}
\end{flalign*}

Therefore our required development is
\[
f(x)=\sum_{m=1}^{m=\infty} \biggl ( \frac{2(\phi_m^2+p^2) }{ \phi_m^2+p(p+1)} \sin\phi_m x \int\limits_0^1 f(\alpha) \sin\phi_m \alpha.d\alpha \biggr ).  \tag{10}
\]
% -----File: 131.png

From (10) it easily follows that for values of  $x$  between $0$ and $c$
\begin{flalign*}
&&f(x) = &a_1 \sin a_1 x + a_2\sin a_2 x + a_3 \sin a_3 x + \cdots  \tag{11}\\
&\text{where} && a_m = \frac{2}{c}.\frac{\alpha_m^2 c^2 + p^2}{\alpha_m^2 c^2 + p(p+1)} \int\limits_0^c f(\lambda) \sin \alpha_m \lambda.d\lambda,  \hspace{2.5em}&&\tag{12}
\end{flalign*}
and  $\alpha_m$ is a root of the equation
\[
\alpha c \cos \alpha c + p \sin \alpha c =0.    \tag{13}
\]

It is to be observed that if  $p$  is infinite (13) reduces to $\sin \alpha c=0$, $\alpha_m$
becomes $\dfrac{m \pi}{c}$ and (11) and (12) give our regulation Fourier sine series (v.\ Art.\
31), and therefore the ordinary Fourier development in sine series is merely a
special case of the problem just solved.

Moreover since the Fourier method of determining the coefficients of such a
series requires that\\[-4ex]
\begin{flalign*}
&&\int\limits_0^c \sin \alpha_m x &\sin \alpha_n x.dx = 0,&&\\
&\text{that is that}      &\frac{\sin(\alpha_m-\alpha_n)c}{\alpha_m-\alpha_n} - &\frac{\sin(\alpha_m+\alpha_n)c}{\alpha_m+\alpha_n} = 0  &&\\[1ex]
&\text{or reducing, that} &\frac{\alpha_m c \cos \alpha_mc }{ \sin \alpha_mc}   &= \frac{\alpha_n c \cos \alpha_n c }{ \sin \alpha_n c}, &&\phantom{or reducing, that}
\end{flalign*}
or that $\alpha_m$ and $\alpha_n$ should be roots of the equation
\[
 \frac{\alpha c \cos \alpha c }{ \sin \alpha c}  = p
\]
where $p$ is some constant, it follows that we have obtained in (11) the most
general sine development that can be obtained by Fourier's method.

\EXAMPLE{S}

1.\quad Show that the solution of the problem of Art.~67 is
\begin{flalign*}
&  && ru = \sum_{m=1}^{m=\infty} b_m e^{-a^2\alpha_m^2 t}\sin \alpha_m r,  &&\\
&\text{where} &b_m = \frac{2}{c }.&\frac{\alpha_m^2c^2+(hc-1)^2 }{ \alpha_m^2c^2+hc(hc-1)} \int\limits_0^c \lambda f(\lambda) \sin \alpha_m \lambda.d\lambda  &&\phantom{where}
\end{flalign*}
and $\alpha_m$ is a root of
\[
\alpha c \cos \alpha c + (hc - 1) \sin \alpha c = 0.
\]

% -----File: 132.png
2.\quad If the initial temperature of the sphere is constant and equal to $\beta$
\begin{flalign*}
&&ru &= \sum_{m=1}^{m=\infty} b_m e^{-a^2\alpha_m^2t} \sin \alpha_m r &&\\
&\text{where} & b_m &= 2\beta h.\frac{ \alpha_m^2c^2+(hc-1)^2}{\alpha_m^2c^2+hc(hc-1)}.\frac{\sin \alpha_mc}{\alpha_m^2}    &&\phantom{where}\\
&&&  = \frac{2\beta hc}{\alpha_m}.\frac{ [\alpha_m^2c^2+(hc-1)^2]^{\frac{1}{2}}}{\alpha_m^2c^2+hc(hc-1)}. &&
\end{flalign*}

3.\quad If the temperature of the air is a constant $\gamma$ instead of zero the surface
equation of condition is
\[
D_r u + h (u - \gamma) = 0 \quad \text{when} \quad  r = c.
\]

The substitution of $u_1 = u - \gamma$, however, brings the problem under Ex.~1
and we get
\begin{flalign*}
&& r(u - \gamma) = \sum_{m=1}^{m=\infty} b_m e^{-a^2\alpha_m^2t} \sin \alpha_m &r   && \\
&\text{where} &  b_m = \frac{2}{c}.\frac{ \alpha_m^2c^2+(hc-1)^2}{\alpha_m^2c^2+hc(hc-1)} \int\limits_0^c \lambda[f(\lambda) - \gamma] &\sin \alpha_m \lambda.d\lambda.    &&\phantom{where}
\end{flalign*}

4.\quad An iron sphere 40 cm.\ in diameter is heated to the temperature 100°
centigrade throughout; it is then allowed to cool in air which is kept at the
constant temperature 0°. Find the temperature at the centre; at a point 10
cm.\ from the centre; and at the surface; 15 minutes after cooling has begun.
Given $a^2 = 0.185$ and $h = \dfrac{1}{800}$ in C.G.S. units.  (v.\ Ex.~3, Art.~66.)\\
\phantom{kindly go to the right hand margin}\hfill
\emph{Ans}., 97°.67;  97°.36;  96°.46.\\

5.\quad Show that if in the slab considered in Art.~60 one face is exposed to air
at the temperature zero, so that we have $D_t u = a^2D_x^2u$, $u = 0$ when $x = 0$,
 $u=f(x)$ when $t = 0$, and $D_x u + hu = 0$ when $x = c$, then
\begin{flalign*}
&&& u = \sum_{m=1}^{m=\infty} a_m e^{-a^2\alpha_m^2t} \sin \alpha_m x  &&\\
&\text{where} & a_m & = 2 \frac{\alpha_m^2+h^2}{\alpha_m^2c+h(hc+1)} \int\limits_0^c f(\lambda) \sin \alpha_m \lambda.d\lambda,  &&  \phantom{where}
\end{flalign*}
$\alpha_m$  being a root of $\alpha c \cos \alpha c + hc \sin \alpha c = 0$.\\

% -----File: 133.png
6.\quad If in the problem of Art.~57 heat escapes from one side of the plate into
air at the temperature zero so that we have $D_x^2u+D_y^2u = 0$, $u = 0$ when
$x = 0$, $u=f(x)$  when $y = 0$, and $D_{x}u + hu = 0$ when $x = a$, then
\begin{flalign*}
&&u = \sum_{m=1}^{m=\infty} a_m e^{-\alpha_m y} &\sin \alpha_m x &&\\
&\text{where} & a_m = 2 \frac{\alpha_m^2 + h^2}{\alpha_m^2 a + h(ha + 1)} &\int\limits_0^a f(\lambda) \sin \alpha_m\lambda.d\lambda,&& \phantom{where}
\end{flalign*}
$\alpha_m$  being a root of $\alpha a \cos \alpha a + ha \sin \alpha a = 0$.\\

7.\quad If in the problem of Art.~59 there is leakage at one side of the sheet so
that we have $D_x^2 V + D_y^2 V = 0$, $V = 0$ when $x = 0$, $V = 0$ when $y = b$,
 $V=f(x)$ when $y = 0$, and $D_xV+hV=0$ when $x = a$, then
\[
V=\sum_{m=1}^{m=\infty} a_m \frac{\sinh \alpha_m(b-y)}{\sinh \alpha_m b }\sin \alpha_m x,
\]
where  $a_m$  has the value given in Ex. 6.

\mypara{69.} If we have an infinite solid with one plane face which is exposed to air
at the temperatures $U = F(t)$ and heat can flow only at right angles to this
face, we can solve the problem readily for the case where the initial temperatures
are zero. We have
\[
D_tu=a^2D_x^2u\\
\]
subject to the conditions
\begin{flalign*}
&&\multispan{2}{\hfill $u = 0 \quad  \text{when} \quad t = 0$ \hfill}\\
&\text{and}  &D_xu+ h(U - &u) = 0  \quad \text{when} \quad x = 0. && \phantom{and}\\
&\text{Let}  & v ={}&u - \frac{1}{h}D_xu.            && \tag{1}
\end{flalign*}
Then $v$ will satisfy the equation
\[
D_tv = a^2D_x^2v,
\]
and we shall also have $v = U$ when $x = 0$.
\begin{flalign*}
&\text{\indent Since } U=F(t)  \qquad v = \frac{2}{\sqrt{\pi}}\!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\!e^{-\beta^2} F \left (t-\frac{x^2}{4a^2\beta^2} \right )d\beta        && \,\tag{2}
\end{flalign*}
by Art.~51 (10).
\begin{flalign*}
&      &&D_xu - hu = - hv   \qquad\text{by (1)}.&&\\
&\indent \text{Hence}   \quad      &&ue^{-hx} = - h \int e^{-hx} vdx+C; &&\phantom{\indent \text{Hence}}
\end{flalign*}
v.\ Int.\ Cal.\ \S~4, page 314.
% -----File: 134.png

Determining $C$ by the fact that $ue^{-hx} = 0$ when $x = \infty$ we have
\[
u=he^{hx} \int\limits_x^{\infty} e^{-hx} vdx.       \tag{3}
\]

Substituting the value of  $v$ from (2) we have
\[
u= \frac{2he^{hx}}{\sqrt{\pi}}\int\limits_x^{\infty} e^{-hx} dx \!\!\int\limits_{\frac{x}{2a\sqrt{t}}}^{\infty}\!\! e^{-\beta^2}F\left ( t- \frac{x^2}{4a^2\beta^2} \right ) d\beta,       \tag{4}
\]
as our required solution.

For an extension of this method to the flow of heat in two and three dimensions
and for the interpretation of the results by the aid of the theory of
\textit{Images}, see E.~W. Hobson, Proc.\ Lond.\ Math.\ Soc., Vol.\ XIX.

\EXAMPLE{S}

1.\quad If the temperature of the air is a periodic function of the time, say
 $\rho_{m}\sin (m\alpha t + \lambda_{m})$ and we care only for the limiting value of $u$ as $t$ increases,
show that this value is
\begin{align*}
\frac{h\rho_m e^{-\frac{x}{a}\frac{\sqrt{m\alpha}}{2}}}{\Big( h +\dfrac{1}{a} \sqrt{\dfrac{m\alpha}{2}}\Big)^2+\dfrac{m\alpha}{2a^2}} &\left [ \left ( h+\dfrac{1}{a} \sqrt{\frac{m\alpha}{2}}\right ) \sin \left (m\alpha t - \frac{x}{a} \sqrt{\frac{m\alpha}{2}}+\lambda_m \right ) \right .\\
& \quad-\left . \frac{1}{a} \sqrt{\frac{m\alpha}{2}}\cos \left (m\alpha t - \frac{x}{a} \sqrt{\frac{m\alpha}{2}}+\lambda_m \right ) \right ].
\end{align*}
v.\ Art.~52 and Art.~51  Ex.~4.
\begin{flalign*}
&\text{\indent Note that } & \int e^{ax}\sin bx.dx &= \frac{e^{ax}(a \sin bx - b \cos bx)}{a^2+b^2}   && \phantom{\indent Note that}\\
&\text{and } & \int e^{ax}\cos bx.dx &=\frac{e^{ax}(a \cos bx + b \sin bx)}{a^2+b^2}  &&
\end{flalign*}
v.\ Int.\ Cal.\ Table of Int.\ (235) and (236).\\

2.\quad If $D_{x}^{2}V+D_{y}^{2}V=0$, $V=0$ when $y = 0$ and $D_{x}V+ h[F(y) - V] = 0$
 when $x = 0$ show that
\[
V=\frac{he^{hx}}{\pi} \int\limits_x^{\infty}e^{-hx}dx \int\limits_0^{\infty}F(\lambda)d\lambda \left [ \frac{x}{x^2+(\lambda-y)^2}-\frac{x}{x^2+(\lambda+y)^2} \right ];\\
\]
v.\ Art.~47 Ex.~1.

\mypara{70.} The solution for an instantaneous heat source of strength   $Q$   at the
point $x = \lambda$ if heat escapes at the origin into air at the temperature zero, so
that $D_{x}u - hu = 0$ when $x = 0$, can be obtained by the aid of Art.~53.
% -----File: 135.png

Let $u = u_1  + u_2$ where $u_1$ is the temperature that would be due to the given
source if we had no boundary at the origin, so that
\[
u_1= \frac{Q}{2a\sqrt{\pi t}}e^{-\frac{(\lambda -x)^2}{4a^2t} }.     \tag*{ [Art.~53 (2)]}
\]
\vspace{-5ex}
\begin{flalign*}
&&D_{x}u - hu = D_{x}u_{1} - hu_{1} &+ D_{x}u_{2} - hu_{2} = 0  \quad \text{when }\quad x=0. && \\
&\text{Therefore} &        D_{x}u_{2} - hu_{2} &= -(D_{x}u_{1} - hu_{1})    &&    \tag{1}\\
\multispan{3}{when $x = 0$. \hfill}\\[-6ex]
\end{flalign*}
\begin{flalign*}
&\indent \text{But} &    -(D_x u_1-hu_1)&=- \frac{Q}{2a\sqrt{\pi t}} \left ( \frac{\lambda -x}{2a^2t} -h \right )e^{-\frac{(\lambda -x)^2}{4a^2t} }    && \phantom{But}\\
&&&     = -\frac{Q}{2a\sqrt{\pi t}} \left( \frac{\lambda}{2a^2t} -h \right ) e^{-\frac{\lambda^2}{4a^2t}} && \\
\multispan{3}{when $x = 0$.\hfill}
\end{flalign*}

This is easily seen to be the value to which
\[
-\frac{Q}{2a\sqrt{\pi t}} \left( \frac{\lambda+x}{2a^2t} -h \right ) e^{-\frac{(\lambda+x)^2}{4a^2t}}
\]
reduces when $x = 0$, and this last expression is
\[
(D_x+h)\frac{Q}{2a\sqrt{\pi t}}e^{-\frac{(\lambda+x)^2}{4a^2t}}
\]
and therefore satisfies the equation
\[
         D_{t}u=a^{2}D_{x}^{2}u;            \tag{2}
\]
since $\dfrac{Q}{2a\sqrt{\pi t}}e^{-\frac{(\lambda+x)^2}{4a^2t}}$ is the temperature due to a source at $x = - \lambda$.\\

If, then, we determine $u_2$ from the condition that
\[
D_x u_2-hu_2= -\frac{Q}{2a\sqrt{\pi t}} \left ( \frac{\lambda+x}{2a^2t}-h \right ) e^{-\frac{(\lambda+x)^2}{4a^2t}}     \tag{3}
\]
taking care not to introduce any arbitrary constant or arbitrary function of $t$
in our integration, $u_2$ will satisfy equation (2) and condition (1).

Integrating (3) [v.\ Int.\ Cal.\ \S~4, page 314] and determining the constants of
integration suitably we get
\[
u_2= \frac{Q}{2a\sqrt{\pi t}} \biggl [ e^{-\frac{(\lambda + x)^2}{4a^2t} } - 2he^{hx} \int\limits_x^{\infty} e^{-hx -\frac{(\lambda + x)^2}{4a^2t} } dx \biggr ].                \tag{4}
\]
Therefore the solution of our problem is
\[
u= \frac{Q}{2a\sqrt{\pi t}} \biggl [ e^{-\frac{(\lambda -x)^2}{4a^2t} } +e^{-\frac{(\lambda+x)^2}{4a^2t}}- 2he^{hx} \int\limits_x^{\infty} e^{-hx -\frac{(\lambda + x)^2}{4a^2t} } dx \biggr ].                                    \tag{5}
\]
% -----File: 136.png

If we replace $Q$ by $f(\lambda)d\lambda$ and integrate from $0$ to $\infty$ we get as the solution
for the case where $u=f(x)$ when $t = 0$ and $x > 0$, and $D_{x}u - hu = 0$
 when $x = 0$
\[
u=\frac{1}{2a\sqrt{\pi t}}\int\limits_0^{\infty}f(\lambda)d\lambda \biggl [ e^{-\frac{(\lambda-x)^2}{4a^2t}}+  e^{-\frac{(\lambda+x)^2}{4a^2t}} -2he^{hx} \int\limits_x^{\infty}  e^{- hx    -\frac{(\lambda+x)^2}{4a^2t}   }  dx \biggr ].                       \tag{6}
\]

For an interpretation of this result by the theory of Images and the
extension of the method to the conduction of heat in $n$ dimensions see G.~H.
Bryan, Proc.\ Lond.\ Math.\ Soc., Vol.\ XXII.

\EXAMPLE{}

Show that if $u=f(x)$ when $t = 0$ and $D_{x}u + h[F(t) - u] = 0$ when
 $x = 0$ we must take $u$ equal to the sum of the second members of (6) Art.~70
and of (4) Art.~69.

\mypara{71.} As another problem requiring a slight extension of Fourier's Theorem
let us consider the vibration of a rectangular stretched elastic membrane
fastened at the edges, that is of a rectangular drumhead.

If two of the sides are taken as axes and the plane of equilibrium of the
membrane as the plane of $XY$ the equation for the motion of the membrane is
\[
   D_{t}^{2}z= c^{2}(D_{x}^{2}z+D_{y}^{2}z)      \tag{1}
\]
see \smallrom{X} Art.~1.

Let the membrane be distorted at the start into some given form $z =f(x,y)$
 and then allowed to swing. Our equations of conditions are then
\begin{alignat*}{3}
  z &= 0     && \text{when}\qquad & x &= 0        \tag{2}\\
  z &= 0     &&\quad \text{``}    & x &= a        \tag{3}\\
  z &= 0     &&\quad \text{``}    & y &= 0        \tag{4}\\
  z &= 0     &&\quad \text{``}    & y &= b        \tag{5}\\
  z &=f(x,y) &&\quad \text{``}    & t &= 0        \tag{6}\\
  D_{t}z&=0  &&\quad \text{``}    & t &= 0.       \tag{7}
\end{alignat*}

We can get a particular solution of (1) by our usual device. Assume
\[
z=e^{\alpha x + \beta y + \gamma t}
\]
and substitute in (1). We get $\gamma^{2} = c^{2}(\alpha^{2} + \beta^{2})$ as the only
relation that
need hold between $\alpha$, $\beta$,  and $\gamma$, in order that $z=e^{\alpha x + \beta y + \gamma t}$ may be a
solution. This gives
\begin{flalign*}
&&                   \gamma = &\pm c \sqrt{ \alpha^2 + \beta^2}. &&\\
&\text{\indent Therefore}    &z = e&^{\alpha x + \beta y \pm ct \sqrt{ \alpha^2 + \beta^2}}&&\phantom{Therefore}
\end{flalign*}
is a solution of (1) no matter what values are given to $\alpha$ and $\beta$.
% -----File: 137.png

Replace $\alpha$ and $\beta$ by $ \alpha i$ and $\beta i$ and we have
\[
z=e^{(\alpha x + \beta y \pm ct \sqrt{\alpha^2+\beta^2 })i}
\]
as a solution, and from this we get
\begin{flalign*}
&  &z = \sin (\alpha x + \beta y \pm ct \sqrt{ \alpha^{2}+\beta^{2}})      &&  \tag{8}\\
&\text{and}\qquad  &z = \cos (\alpha x + \beta y \pm ct \sqrt{ \alpha^{2}+\beta^{2}})  &&        \tag{9}
\end{flalign*}
as particular solutions of (1), $\alpha$ and $\beta$ being unrestricted.

From (8) and (9) we can get solutions of the following forms
\[
\left . \begin{alignedat}{3}
     z &= \sin \alpha x &&\sin \beta y &&\sin ct \sqrt{\alpha^2 + \beta^2}\\
     z &= \sin \alpha x &&\sin \beta y &&\cos ct \sqrt{\alpha^2 + \beta^2}\\
     z &= \sin \alpha x &&\cos \beta y &&\sin ct \sqrt{\alpha^{2} + \beta^{2}}\\
     z &= \sin \alpha x &&\cos \beta y &&\cos ct \sqrt{\alpha^{2} + \beta^{2}}\\
     z &= \cos \alpha x &&\sin \beta y &&\sin ct \sqrt{\alpha^{2} + \beta^{2}}\\
     z &= \cos \alpha x &&\sin \beta y &&\cos ct \sqrt{\alpha^{2} + \beta^{2}}\\
     z &= \cos \alpha x &&\cos \beta y &&\sin ct \sqrt{\alpha^{2} + \beta^{2}}\\
     z &= \cos \alpha x &&\cos \beta y &&\cos ct \sqrt{\alpha^{2} + \beta^{2}},
\end{alignedat}
\right  \}
 \qquad \text{(10)}
\]
each of which will satisfy equation (1). The second of these will satisfy also
(2), (4) and (7) whatever values be taken for $\alpha$ and $\beta$. It will satisfy (3) and
(5) if $ \alpha$ and $\beta$ are equal $\dfrac{m\pi}{a}$ and $\dfrac{n\pi}{b}$ respectively.

If, then, we can so combine terms of the form
\[
\sin \frac{m\pi x}{a}\sin\frac{n\pi y}{b}\cos c\pi  t \sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}}
\]
as to satisfy (6) our problem will be completely solved.

This can be done if we can express  $f(x,y)$ as a sum of terms of the form
 $A \sin \dfrac{m\pi x}{a}\sin \dfrac{n\pi y}{b}$, the sum and the function being equal when $x$ lies
between $0$ and $a$ and $y$ between $0$ and  $b$.

$f(x,y)$ can be expressed in terms of $\sin \dfrac{m\pi x}{a}$ by Fourier's Theorem if we
regard $y$ as constant. We have
\begin{flalign*}
&&f(x,y)&= \sum_{m=1}^{m=\infty} a_m \sin \frac{m\pi x}{a}      &&     \tag{11}\\
% -----File: 138.png
&\text{where} &a_m=\frac{2}{a} &\int\limits_0^a f(\lambda,y)\sin \frac{m\pi\lambda}{a} d\lambda.         \tag{12}&&\;
\end{flalign*}
$f(\lambda,y)$ in (12) is a function of $y$ and may be developed by Fourier's Theorem.
\begin{flalign*}
&\text{We have} &   \quad f(\lambda,y)&= \sum_{n=1}^{n=\infty} b_n \sin \frac{n\pi y}{b }          \tag{13} &&\\
&\text{where} &  b_n=\frac{2}{b} &\int\limits_0^b f(\lambda,\mu)\sin \frac{n\pi\mu}{b} d\mu.    \hspace{3em}\tag{14}
\end{flalign*}

Substituting for $f(\lambda,y)$ in (12) the value just obtained we have
\[
a_m =\frac{2}{a}\frac{2}{b}\sum_{n=1}^{n=\infty}\biggl (\int\limits_0^a d\lambda \int\limits_0^b f(\lambda,\mu) \sin\frac{m\pi \lambda}{a}\sin\frac{n\pi \mu}{b} d\mu \biggr  ) \sin\frac{n\pi y}{b}
\]
and
\[
f(x,y) =\frac{4}{ab} \sum_{m=1}^{m=\infty}\sum_{n=1}^{n=\infty}\biggl ( \sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b} \int\limits_0^a d\lambda \int\limits_0^b f(\lambda,\mu) \sin\frac{m\pi \lambda}{a}\sin\frac{n\pi \mu}{b}d\mu\biggr ) .\tag{15}
\]
\begin{flalign*}
&\text{Hence} & z=&\sum_{m=1}^{m=\infty}\sum_{n=1}^{n=\infty}\biggl ( A_{m,n} \sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b} \cos c\pi t \sqrt{  \frac{m^2}{a^2}+\frac{n^2}{b^2} }\biggr ), && \tag{16}\\
&\text{where} &&\quad A_{m,n}=\frac{4}{ab} \int\limits_0^ad\lambda \int\limits_0^bf(\lambda,\mu)\sin\frac{m\pi \lambda}{a}\sin\frac{n\pi \mu}{b}d\mu.  &&\tag{17}
\end{flalign*}
is our required solution.

\EXAMPLE{S}

1.\quad Show that if the membrane starts from its position of equilibrium but
with a given initial velocity impressed upon each point so that $z = 0$ when
 $t = 0$ and $D_{t}z= F(x,y)$ when $t = 0$ the solution is
\[
z=\frac{1}{c\pi}\sum_{m=1}^{m=\infty}\sum_{n=1}^{n=\infty}\biggl ( A_{m,n} \dfrac{1}{\sqrt{  \dfrac{m^2}{a^2}+\dfrac{n^2}{b^2} }} \sin\frac{m\pi x}{a}\sin\frac{n\pi y}{b} \sin c\pi t \sqrt{  \frac{m^2}{a^2}+\frac{n^2}{b^2} }\biggr )\\[-2ex]
\]
\begin{flalign*}
&\text{where}  &A_{m,n}&=\frac{4}{ab} \int\limits_0^ad\lambda \int\limits_0^bF(\lambda,\mu)\sin\frac{m\pi \lambda}{a}\sin\frac{n\pi \mu}{b}d\mu. && \phantom{where}
\end{flalign*}

% -----File: 139.png
2.\quad If there is both initial distortion and initial velocity
\begin{align*}
% recast to fit line
z = \dfrac{4}{ab}\sum_{m=1}^{m=\infty}\,\sum_{n=1}^{n=\infty}\sin\dfrac{m\pi x}{a}
\sin\dfrac{n\pi y}{b}\biggl[&A_{m,n}\cos{c\pi t}\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}\\
 + &B_{m,n}\sin{c\pi t}\sqrt{\dfrac{m^2}{a^2} + \dfrac{n^2}{b^2}}\biggr]\\[-4ex]
\end{align*}
\begin{flalign*}
&\text{where}\hfil &&A_{m,n} = \int\limits_0^a d\lambda\int\limits_0^bf
(\lambda,\mu)\sin\dfrac{m\pi\lambda}{a}\sin\dfrac{n\pi\mu}{b}d\mu,\hfil\phantom{where}&&\\[-1ex]
&\text{and}\hfil &B_{m,n} =&  \dfrac{1}{c\pi\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}}
\int\limits_0^a d\lambda \int\limits_0^b  F(\lambda,\mu)\sin\dfrac{m\pi\lambda}{a}
\sin\dfrac{n\pi\mu}{b}d\mu.\hfil\phantom{and}
\end{flalign*}

3.\quad Obtain a particular solution of (1) Art.~71 by assuming $z = T.X.Y$
 where $T$ is a function of $t$ alone, $X$ of $x$ alone, and $Y$ of $y$ alone.

\mypara{72.} A number of interesting conclusions can be drawn from the results of
Art.~71 and Exs.~1 and 2.

(\emph{a}) No one of the three values of $z$ is in general a periodic function of $t$,
and consequently a vibrating rectangular membrane will not in general give a
musical note.

(\emph{b}) A stretched rectangular membrane can be made to give a musical note
by starting the vibration properly. For if the initial circumstances are such
that the solution reduces to a single term, as will be the case if the initial distortion
in the problem of Art.~71 be such that $f(x,y) = A_{m,n} \sin\dfrac{m\pi x}{a} \sin\dfrac{n\pi y}{b}$,
 or the initial velocity in Ex.~1 be such that $F(x,y) = B_{m,n} \sin\dfrac{m\pi x}{a}\tablestrut\sin\dfrac{n\pi y}{b}$,
 or the initial distortion and initial velocity in Ex.~2 be the values just given,
then the vibration will be periodic and will have the period

\[\tag{1}
T = \dfrac{2}{c\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}}.
\]

Since $T$ is a function of $m$ and $n$ and $m$ and $n$ are any whole numbers, the
same membrane is capable of giving a great variety of musical notes of different
pitches. If $m$ and $n$ are both unity we get the lowest note the membrane
can give, which is called its fundamental note. Its period
\[\tag{2}
T_1 = \dfrac{2}{c\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}}}=\dfrac{2ab}{c\sqrt{a^2+b^2}}
\]
If $m$ and $n$ are both equal to $k$ we get

\[\tag{3}
T_k = \dfrac{2ab}{kc\sqrt{a^2+b^2}};
\]
% -----File: 140.png
therefore the membrane can be made to give any harmonic of its fundamental
note.

More than this, since as we have seen
\[
T_{m,n}=\frac{2}{c\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}}
\]
is the period of any note the membrane can give, and since if  $m$  and $n$ are
replaced by $mk$ and $nk$ we get
\[
T_{mk,nk}= \frac{2}{ck\sqrt{\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}}}
\]
the membrane can sound all the harmonics of any note which it can give.

(\emph{c}) In the case considered above, where the solution reduces to the single
term
\[
z=\sin\frac {m\pi x}{a}\sin \frac {n\pi y}{b} \left  [A_{m,n}\cos c\pi t \sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}}+B_{m,n}\sin c\pi t \sqrt{\frac{m^2}{a^2}+\frac{n^2}{b^2}} \right ],
\]
if $x = \dfrac{a}{m}$, or  $\dfrac{2a}{m}$, or $\dfrac{3a}{m} \cdots$ or $ \dfrac{(m-1)a}{m}$, $z = 0$ for all values of  $t$, and
the lines $x = \dfrac{a}{m}$, $x = \dfrac{2a}{m}\tablestrut$, $\cdots$  $x = \dfrac{(m-1)a}{m}$ remain at rest during the whole
vibration and are nodes. The same thing is true of the lines
\[
y=\frac{b}{n},\; y=\frac{2b}{n},\; y=\frac{3b}{n},\; \cdots\;  y=\frac{(n-1)b}{n}.
\]

\mypara{73.} If the membrane is square it may have much more complicated nodes
than if the length and breadth are unequal, as in this case the period of any
term of the general solution reduces to
\[
T=\frac{2a}{c\sqrt{m^2+n^2}}   \tag{1}
\]
and there will in general be two terms having the same period, and a musical
note of the pitch corresponding to that period may be produced by initial circumstances
that bring in both terms. Thus
\begin{align*}
z=&\sin\frac {m\pi x}{a}\sin \frac{ n\pi y}{a} \left  [A_{m,n}\cos\frac{ c\pi t}{a} \sqrt{m^2+n^2}+B_{m,n}\sin \frac{c\pi t}{a} \sqrt{m^2+n^2} \right  ]\\
&+ \sin\frac {n\pi x}{a}\sin \frac{ m\pi y}{a} \left  [A_{n,m} \cos\frac{c\pi t}{a} \sqrt{m^2+n^2}+B_{n,m} \sin \frac{c\pi t}{a} \sqrt{m^2+n^2}\right ]
\end{align*}
% -----File: 141.png
is a form of vibration that will give a musical note. Let us write this
\begin{align*}
z = &\cos \frac{c\pi t}{a} \sqrt{m^2+n^2}\bigg [ A \sin \frac{m\pi x}{a} \sin \frac{n\pi y}{a} + B \sin\frac{n\pi x}{a} \sin \frac{m\pi y}{a}  \bigg ]\\
&+ \sin\frac{c\pi t}{a} \sqrt{m^2+n^2}\bigg [ C \sin \frac{m\pi x}{a} \sin \frac{n\pi y}{a} + D \sin\frac{n\pi x}{a} \sin \frac{m\pi y}{a}  \bigg ]    \tag{2}
\end{align*}
and in studying the forms of musical vibration of which the membrane is
capable we may take $A$, $B$, $C$, and $D$ at pleasure. Consider the simple case
where $A = C$ and $B = D$; then (2) reduces to
\begin{multline*}
z = \bigg ( A \sin\frac{m\pi x}{a} \sin \frac{n\pi y}{a} + B \sin\frac{n\pi x}{a} \sin\frac{m\pi y}{a} \bigg ) \biggl (\cos \frac{c\pi t}{a} \sqrt{m^2+n^2} \biggr .\\
\biggl .+ \sin \frac{c\pi t}{a} \sqrt{m^2+n^2} \biggr ). \tag{3}
\end{multline*}
Values of $x$ and $y$ that will reduce the first parenthesis in (3) to zero will correspond
to points of the membrane remaining motionless during the vibration.

Let us consider a few cases at length.

(\emph{a})\quad  If $m= 1$ and $n = 1$, the first parenthesis in (3) becomes
\[
(A + B) \sin\frac{\pi x}{a} \sin \frac{\pi y}{a},
\]
which is equal to zero only when $x = 0$ or $y = 0$, or $x = a$ or $y = a$,
 that is, for the four edges of the membrane. If, then, the membrane is sounding
its fundamental note it has no nodes.

(\emph{b})\quad  If $m = 1$ and $n = 2$, we have
\[
A \sin\frac{\pi x}{a} \sin \frac{2\pi y}{a} + B \sin\frac{2\pi x}{a} \sin \frac{\pi y}{a} = 0
\]
to give the nodes.

Let $B = 0$, then $\sin\dfrac{\pi x}{a} \sin \dfrac{2\pi y}{a} = 0$, which is satisfied by $y = \dfrac{a}{2}$; and
in addition to the edges the line $y = \dfrac{a}{2}$ is at rest and is a node.

If $A = 0$ \quad $x = \dfrac{a}{2}$ is a node.

If $A = B$
\begin{gather*}
\sin\frac{\pi x}{a} \sin \frac{2\pi y}{a}  + \sin\frac{2\pi x}{a} \sin \frac{\pi y}{a} = 0\\[1ex]
2 \sin\frac{\pi x}{a} \sin \frac{\pi y}{a} \cos \frac{\pi y}{a} + 2 \sin\frac{\pi x}{a} \cos\frac{\pi x}{a} \sin \frac{\pi y}{a} = 0\\[1ex]
\sin\frac{\pi x}{a} \sin \frac{\pi y}{a} \left(\cos \frac{\pi y}{a} + \cos \frac{\pi x}{a} \right) = 0.
\end{gather*}
% -----File: 142.png
The first factor gives the four edges of the membrane. The second written
equal to zero gives
\begin{gather*}
\cos \frac{\pi y}{a} = -\cos \frac{\pi x}{a} = \cos\left (\pi - \frac{\pi x}{a} \right )\\
\frac{\pi y}{a}=\pi-\frac{\pi x}{a}\\
x + y= a,
\end{gather*}
which is a diagonal of the square.

If $B = - A$
\begin{gather*}
\sin  \frac{\pi x}{a}\sin \frac{2\pi y}{a} - \sin \frac{2\pi x}{a} \sin\frac{\pi y}{a}=0\\
 \cos\frac{\pi y}{a} = \cos\frac{\pi x}{a}\\
x-y=0,
\end{gather*}
which is the other diagonal of the square.

Other relations between $A $ and $B$ will give Trigonometric curves of the form
\[
\cos \frac{\pi y}{a} =-\frac{B}{A} \cos \frac{\pi x}{a}
\]
which are easily constructed and which obviously all agree in passing through
the middle point of the square.

We give the figures for a few of the cases
%[Illustration]
\pngcent{010.png}{1288}

% -----File: 143.png

{}(\emph{c})\quad If $m = n = 2$ we have\\[-4ex]
\pngrightl{011.png}{344}{10}{-24}

\[
(A + B)\sin \frac{2\pi x}{a}\sin \frac{2\pi y}{a} = 0
\]
to give the nodes, which are merely the lines
\[
x = \frac{a}{2}, \quad \text{and} \quad y = \frac{a}{2}.
\]
This form gives the octave of the fundamental note.

(\emph{d})\quad  If $m = 1$ and $n = 3$ we have
\[
A \sin \frac{\pi x}{a} \sin \frac{3\pi y}{a} + B \sin \frac{3\pi x}{a} \sin \frac{\pi y}{a} = 0
\]
to give the nodes.
\begin{flalign*}
&\text{\indent If} \quad A = 0 \quad \text{we get}  &x &= \frac{a}{3}\quad  \text{and} \quad x = \frac{2a}{3} &&  \phantom{\indent If A = 0 we get}       \tag{1}   \\
&\text{\indent If} \quad B = 0 \quad \text{we get}  &y &= \frac{a}{3} \quad  \text{and}  \quad y = \frac{2a}{3} &&        \tag{2}
\end{flalign*}

If \quad $A = -B$ \quad we get
\begin{gather*}
  \sin \frac{\pi x}{a} \sin \frac{3\pi y}{a} - \sin \frac{3\pi x}{a} \sin \frac{\pi y}{a} = 0              \\[1ex]
  \sin \frac{\pi x}{a} \sin \frac{\pi y}{a} \left [4 \cos^2 \frac{\pi y}{a} - 1 - 4 \cos^2 \frac{\pi x}{a} + 1 \right ] = 0    \\[1ex]
  \cos^2 \frac{\pi y}{a} - \cos^2  \frac{\pi x}{a} = 0             \\[1ex]
  \left ( \cos \frac{\pi y}{a}  - \cos  \frac{\pi x}{a} \right ) \left ( \cos \frac{\pi y}{a}  +\cos  \frac{\pi x}{a} \right ) = 0 \\[-5ex]
\end{gather*}
\begin{flalign*}
&\text{or}\hspace{1em}   &      x - y = 0 \quad  &\text{and} \quad x + y = a.    \tag{3}     &&      \\[-5ex]
\end{flalign*}
\begin{flalign*}
&\text{\indent If} \quad A = B \quad \text{we get} & \cos^2 \frac{\pi y}{a} &+\cos^2 \frac{\pi x}{a} =  \frac{1}{2}   && \phantom{\indent If \quad A = B \quad we get}\\[-6ex]
\end{flalign*}
\begin{flalign*}
&\text{or}\hspace{1em}  & \cos \frac{2\pi y}{a} +{}& \cos \frac{2\pi x}{a} = -1,         \tag{4}         &&
\end{flalign*}
a Trigonometric curve easily constructed.

For other relations between $A$ and $B$ we get more complicated Trigonometric
curves coming under the general form
\[
A \cos \frac{2\pi y}{a} + B \cos \frac{2\pi x}{a} = - \frac{A + B}{ 2}   \tag{5}
\]
% -----File: 144.png
which all agree in containing the points
\[
\Bigl(\dfrac{a}{3}, \dfrac{a}{3}\Bigr),
\Bigl(\dfrac{a}{3}, \dfrac{2a}{3}\Bigr),
\Bigl(\dfrac{2a}{3}, \dfrac{a}{3}\Bigr),
\text{ and } \Bigl(\dfrac{2a}{3}, \dfrac{2a}{3}\Bigr).
\]
%[Illustration]
\pngcent{012.png}{1275}

\newpage
% -----File: 145.png

\label{ch4end}

\newpage
\begin{center}{MISCELLANEOUS PROBLEMS.}\end{center}
\markright{}
\label{probstart}

\begin{center}{I. \emph{Logarithmic Potential. Polar Coördinates.}}\end{center}

\vspace{\baselineskip}

1.\quad Show that $D_x^2V+D_y^2V=0$ becomes
\[
D_r^2V + \dfrac{1}{r}D_rV+ \dfrac{1}{r^2}D_\phi^2V = 0
\]
if we transform to Polar Coördinates.

\begin{flalign*}
&\text{\indent 2.\quad If in} && D_r^2V + \dfrac{1}{r}D_rV +  \dfrac{1}{r^2}D_\phi^2V = 0 && \phantom{\indent 2.\quad If in}(1)
\end{flalign*}
we let $V = R.\Phi$ we get
\begin{align*}
\begin{aligned}
\Phi &= A \cos \alpha\phi + B \sin \alpha\phi\ \\
    R&= A_1r^\alpha+B_1r^{-\alpha}
\end{aligned}
\bigg\}
\quad\text{or}\quad
\begin{aligned}
\Phi &= A e^{\alpha\phi} + B e^{-\alpha\phi}\\
    R& = A_1 \cos(\alpha \log r) + B_1 \sin(\alpha \log r);\
\end{aligned}
\bigg\}
\end{align*}
whence\rule{0em}{1.3em}
\[
 \begin{array}{l | l | l}
V = r^\alpha \cos \alpha\phi                     & V = e^{\alpha\phi} \cos(\alpha \log r)      & V = \cosh \alpha\phi \cos(\alpha \log r)\\[0.5em]
V = r^\alpha \sin \alpha\phi                      & V= e^{\alpha\phi} \sin(\alpha \log r)        &  V = \cosh \alpha\phi \sin(\alpha \log r)\\[0.5em]
V = \dfrac{1}{r^\alpha} \cos \alpha\phi & V = e^{-\alpha\phi} \cos(\alpha \log r) &  V = \sinh \alpha\phi \cos(\alpha \log r)\\[0.5em]
V = \dfrac{1}{r^\alpha} \sin \alpha\phi & V = e^{-\alpha\phi}\sin(\alpha \log r)  &  V = \sinh \alpha\phi \sin(\alpha \log r)
\end{array}
\]
are particular solutions of (1).\\

3.\quad Show that if $V$ satisfies (1) Ex. 2 and $V = f(\phi)$ when $r = a$
\begin{flalign*}
&                       &V      &= \dfrac{1}{2} b_0 + \sum^{m=\infty}_{m=1} \left(\dfrac{r}{a}\right)^m (b_m \cos m\phi + a_m \sin m\phi)\quad\text{for}\quad r < a\rule{3em}{0em}\\[0.5em]
&\text{and}    & V     &= \dfrac{1}{2} b_0 + \sum^{m=\infty}_{m=1}  \left(\dfrac{a}{r}\right)^m (b_m \cos m\phi + a_m \sin m\phi)\quad\text{for}\quad r > a,\\[0.5em]
&\text{where} &b_m& = \dfrac{1}{\pi} \int \limits ^\pi_{-\pi} f(\phi)\cos m\phi.d\phi\quad\text{and}\quad a_m =  \dfrac{1}{\pi} \int \limits ^\pi_{-\pi} f(\phi) \sin m\phi.d\phi &&
\end{flalign*}
% -----File: 146.png

4.\quad Show that if $V$ satisfies (1) Ex.~2 and $V=f(r)$ when $\phi=0$ and $r>0$
\begin{align*}
V&=\frac{1}{\pi}\int\limits_{-\infty}^\infty f(e^\lambda)d\lambda
                \int\limits_0^\infty\frac{\cosh\alpha(\pi-\phi)}{\cosh\alpha\pi}\cos\alpha(\lambda-\log r).d\alpha\\
 &=\frac{1}{\pi}\sin\frac{\phi}{2}\int\limits_{-\infty}^\infty f(e^\lambda)
\frac{\cosh\dfrac{1}{2}(\lambda-\log r)}
     {\cosh(\lambda-\log r)-\cos\phi}d\lambda.
\end{align*}

\markright{MISCELLANEOUS PROBLEMS.}
5.\quad If $V=1$ when $\phi=0$ and $0<r<1$, and $V=0$ when $\phi=0$ and
 $r>1$
\[
V=\frac{1}{\pi}\Bigg\{\frac{\pi}{2}-\tan^{-1}\Bigg[\frac{\sinh\dfrac{\log r}{2}}{\sin\dfrac{\phi}{2}}\Bigg]\Bigg\}
 =\frac{1}{\pi}\Bigg[\frac{\pi}{2}-\tan^{-1}\Bigg(\frac{r-1}{2\sqrt{r}.\sin\dfrac{\phi}{2}}\Bigg)\Bigg].
\]

6.\quad If $V=f(r)$ when $\phi=0$ and $V=0$ when $\phi=\beta$
\begin{align*}
V &=\frac{1}{\pi}\int\limits_{-\infty}^\infty f(e^\lambda) d\lambda\int\limits_0^\infty\frac{\sinh(\beta-\phi)\alpha}{\sinh\beta\alpha}\cos\alpha(\lambda-\log r).d\alpha\\
   &=\frac{1}{2\beta}\sin\frac{\pi\phi}{\beta}\int\limits_{-\infty}^\infty
\frac{f(e^\lambda)d\lambda}{\cosh\dfrac{\pi}{\beta}(\lambda-\log r)-\cos\dfrac{\pi}{\beta}\phi},
\end{align*}
if $0<\phi<\beta$.\\

7.\quad If $V=0$ when $\phi=0$ and $V=F(r)$ when $\phi=\beta$
\begin{align*}
V &=\frac{1}{\pi}\int\limits_{-\infty}^\infty F(e^\lambda)d\lambda\int\limits_0^\infty\frac{\sinh\phi\alpha}{\sinh\beta\alpha}\cos\alpha(\lambda-\log r).d\alpha\\
  &=\frac{1}{2\beta}\sin\frac{\pi\phi}{\beta}\int\limits_{-\infty}^\infty
\frac{F(e^\lambda)d\lambda}{\cosh\dfrac{\pi}{\beta}(\lambda-\log r)+\cos\dfrac{\pi}{\beta}\phi}.
\end{align*}

8.\quad If $V=\chi(r)$ when $\phi=0$ and $r<a$, $V=0$ when $\phi=\beta$, and
 $V=0$ when $r=a$
\begin{align*}
V=\frac{1}{2\beta}\sin\frac{\pi\phi}{\beta}\int\limits_{-\infty}^0\chi(ae^\lambda)
\Bigg[&\frac{d\lambda}{\cosh\dfrac{\pi}{\beta}\Big(\lambda-\log\dfrac{r}{a}\Big)-\cos\dfrac{\pi\phi}{\beta}}\\
    &-\frac{d\lambda}{\cosh\dfrac{\pi}{\beta}\Big(\lambda+\log\dfrac{r}{a}\Big)-\cos\dfrac{\pi\phi}{\beta}}\Bigg].
\end{align*}

% -----File: 147.png
9.\quad If $V=0$ when $r=1$, $V=1$ when $\phi=0$, $V=0$ when $\phi=\dfrac{\pi}{2}$
\[
V = \frac{2}{\pi}\tan^{-1}\biggl[\frac{1-r^2}{1+r^2}\ctn\phi\biggr].
\]

10.\quad If $V=0$ when $r=1$, $V=1$ when $\phi=0$, $V=1$ when $\phi=\dfrac{\pi}{2}$
\[
V = \frac{2}{\pi}\tan^{-1}\biggl[\frac{1-r^4}{2r^2\sin 2\phi}\biggr].
\]

11.\quad If $V=f(\phi)$ when $r=a$, $V=0$ when $\phi=0$, and $V=0$ when $\phi=\beta$
\begin{flalign*}
&&V&= \sum_{m=1}^{m=\infty} a_m\Bigl(\frac{r}{a}\Bigr)^\frac{m\pi}{\beta} \sin\frac{m\pi\phi}{\beta}\quad \text{if} \quad r<a &\\
&&V&= \sum_{m=1}^{m=\infty} a_m\Bigl(\frac{a}{r}\Bigr)^\frac{m\pi}{\beta} \sin\frac{m\pi\phi}{\beta}\quad \text{if} \quad r>a &\\
&\text{where} & a_m &= \frac{2}{\beta}\int\limits_0^\beta f(\phi)\sin\frac{m\pi\phi}{\beta}d\phi  \quad \text{and}\quad 0<\phi<\beta.&
\end{flalign*}

12.\quad If $V=f(\phi)$ when $r=a$, $V=0$ when $r=b$, $V=0$ when $\phi=0$,
 and $V=0$ when $\phi=\beta$, then if $a<r<b$ and $0<\phi<\beta$
\[
V = \sum_{m=1}^{m=\infty}\biggl\{\frac{a^{\frac{ m\pi}{\beta}} b^{\frac{ m\pi}{\beta}}}
                                      {a^{\frac{2m\pi}{\beta}}-b^{\frac{2m\pi}{\beta}}}
 \biggl[\Bigl(\frac{r}{b}\Bigr)^{\frac{m\pi}{\beta}}
       -\Bigl(\frac{b}{r}\Bigr)^{\frac{m\pi}{\beta}}\biggr]
a_m\sin\frac{m\pi\phi}{\beta}\biggr\}
\]
\begin{flalign*}
&\text{where} & a_m = \frac{2}{\beta}\int\limits_0^\beta f(\phi)
\sin\frac{m\pi\phi}{\beta}d\phi.&&
\end{flalign*}

13.\quad If $V=F(\phi)$ when $r=b$, $V=0$ when $r=a$, $V=0$ when $\phi=0$,
 and $V=0$ when $\phi=\beta$, then if $a<r<b$ and $0<\phi<\beta$
\[
V = \sum_{m=1}^{m=\infty}\biggl\{\frac{a^{\frac{ m\pi}{\beta}} b^{\frac{ m\pi}{\beta}}}
                                      {b^{\frac{2m\pi}{\beta}}-a^{\frac{2m\pi}{\beta}}}
 \biggl[\Bigl(\frac{r}{a}\Bigr)^\frac{m\pi}{\beta}
       -\Bigl(\frac{a}{r}\Bigr)^\frac{m\pi}{\beta}\biggr]
a_m\sin\frac{m\pi\phi}{\beta}\biggr\}
\]
\begin{flalign*}
&\text{ where }& a_m = \frac{2}{\beta}\int\limits_0^\beta F(\phi)
\sin\frac{m\pi\phi}{\beta} d\phi.&&
\end{flalign*}

14.\quad If $V=\chi(r)$ when $\phi=0$, $V=0$ when $\phi=\beta$, $V=0$ when $r=a$,
 and $V=0$ when $r=b$, then if $a<r<b$ and $0<\phi<\beta$
\[
V = \sum_{m=1}^{m=\infty}\Biggl\{ a_m\
\frac{\sinh\dfrac{m\pi(\beta-\phi)}{\log b - \log a}}
     {\sinh\dfrac{m\pi\beta}{\log b - \log a}}
\sin
\frac{m\pi(\log r - \log a)}
     {\log b - \log a}\Biggl\}
\]
% -----File: 148.png
\begin{flalign*}
&\text{where} & a_m = \frac{2}{\log b - \log a} \int\limits_0^{\log\frac{b}{a}} \chi(ae^x)
\sin\frac{m\pi x}{\log b - \log a} dx.&&
\end{flalign*}

15.\quad If $V=\psi(r)$ when $\phi=\beta$, $V=0$ when $\phi=0$, $V=0$ when $r=a$,
 and $V=0$ when $r=b$, then if $a<r<b$ and $0<\phi<\beta$
\[
V = \sum_{m = 1}^{m = \infty}\Biggl\{ a_m
\frac{\sinh\dfrac{m\pi\phi} {\log b - \log a}}
     {\sinh\dfrac{m\pi\beta}{\log b - \log a}}
\sin\frac{m\pi(\log r - \log a)}{\log b - \log a}\Biggr\}
\]
\begin{flalign*}
&\text{where} & a_m =\frac{2}{\log b - \log a} \int\limits_0^{\log\frac{b}{a}} \psi(ae^x)
\sin\frac{m\pi x}{\log b - \log a} dx.&&
\end{flalign*}

\vspace{1ex}

\begin{center}{II. \textit{Potential Function in Space.}}\end{center}

1.\quad  Show that
\[
f(x,y) = \frac{1}{\pi^2}\int\limits_0^\infty d\alpha
                        \int\limits_0^\infty d\beta
                        \int\limits_0^\infty d\lambda
                        \int\limits_0^\infty f(\lambda,\mu)\cos\alpha(\lambda-x)\cos\beta(\mu-y).d\mu,
\]
for all values of  $x$ and $y$.\\

2.\quad Find particular solutions of $D_x^2V+D_y^2V+D_z^2V=0$ in the forms
\begin{align*}
V& = e^{±z\sqrt{\alpha^2+\beta^2}}\cos(\alpha x ± \beta y)&\\
V& = e^{±z\sqrt{\alpha^2+\beta^2}}\sin(\alpha x ± \beta y)&\\
V& =\sinh z\sqrt{\alpha^2+\beta^2}.\sin(\alpha x ± \beta y)&\\
V& =\cosh z\sqrt{\alpha^2+\beta^2}.\sin(\alpha x ± \beta y)&\\
& \hspace{15ex}\text{\&c.}
\end{align*}

3.\quad Given $D_x^2V+D_y^2V+D_z^2V=0$, and $V=f(x,y)$ when $z=0$, solve for
positive values of $z$.
\begin{flalign*}
&\text{\emph{Result:}} & V=\frac{1}{2\pi}
\int\limits_{-\infty}^\infty d\lambda
\int\limits_{-\infty}^\infty \frac{zf(\lambda,\mu)d\mu}{[z^2+(\lambda-x)^2+(\mu-y)^2]^\frac{3}{2}}.&& \phantom{Result:}
\end{flalign*}

4.\quad Confirm the result of the last example by showing that if $f(x,y)$ is independent
of $y$
\[
\phantom{(v.\ Ex.~3 Art.~45)}V = \frac{1}{\pi}\int\limits_{-\infty}^\infty \frac{zf(\lambda,\mu)d\lambda}{z^2+(\lambda-x)^2} \tag*{\text{(v.\ Ex.~3 Art.~45)}.}
\]

% -----File: 149.png
5.\quad If $D_x^2V + D_y^2V + D_z^2V = 0$, and $V=1$ when $z=0$ for all points
within the rectangle bounded by the lines $x=a$, $x=-a$, $y=b$, and
 $y=-b$; and $V=0$ when $z=0$ for all points outside of this rectangle,
then
\begin{flalign*}
&\multispan{3}{$2\pi V = $\hfill} \\
 &&   \frac{b-y}{\sqrt{(b-y)^2}} \bigg\{\dfrac{\pi}{2} & + \dfrac{1}{2} \sin^{-1}\dfrac{(a-x)^2(b-y)^2 - z^2[(a-x)^2 + (b-y)^2 + z^2]}{(a-x)^2(b-y)^2 + z^2[(a-x)^2 + (b-y)^2 + z^2]}\\
 &&                                                    & + \dfrac{1}{2} \sin^{-1}\dfrac{(a+x)^2(b-y)^2 - z^2[(a+x)^2 + (b-y)^2 + z^2]}{(a+x)^2(b-y)^2 + z^2[(a+x)^2 + (b-y)^2 + z^2]}\bigg\}\\
 && + \frac{b+y}{\sqrt{(b+y)^2}} \bigg\{\dfrac{\pi}{2} & + \dfrac{1}{2} \sin^{-1}\dfrac{(a-x)^2(b+y)^2 - z^2[(a-x)^2+(b+y)^2+z^2]}{(a-x)^2(b+y)^2 + z^2[(a-x)^2 + (b+y)^2 + z^2]}\\
 &&                                                    & + \dfrac{1}{2} \sin^{-1}\dfrac{(a+x)^2(b+y)^2 - z^2[(a+x)^2+(b+y)^2+z^2]}{(a+x)^2(b+y)^2 + z^2[(a+x)^2 + (b+y)^2 + z^2]}\bigg\}
\end{flalign*}
if $-a<x<a$, and
\begin{alignat*}{2}
4\pi V &= \frac{b-y}{\sqrt{(b-y)^2}} \bigg\{&\phantom{- }&\sin^{-1}\dfrac{(a-x)^2(b-y)^2 - z^2[(a-x)^2 + (b-y)^2 + z^2]}{(a-x)^2(b-y)^2 + z^2[(a-x)^2 + (b-y)^2 + z^2]}      \\
       &                                    &         -{}&\sin^{-1}\dfrac{(a+x)^2(b-y)^2 - z^2[(a+x)^2 + (b-y)^2 + z^2]}{(a+x)^2(b-y)^2 + z^2[(a+x)^2 + (b-y)^2 + z^2]}\bigg\}\\
       &+ \frac{b+y}{\sqrt{(b+y)^2}} \bigg\{&            &\sin^{-1}\dfrac{(a-x)^2(b+y)^2 - z^2[(a-x)^2 + (b+y)^2 + z^2]}{(a-x)^2(b+y)^2 + z^2[(a-x)^2 + (b+y)^2 + z^2]}      \\
       &                                    &         -{}&\sin^{-1}\dfrac{(a+x)^2(b+y)^2 - z^2[(a+x)^2 + (b+y)^2 + z^2]}{(a+x)^2(b+y)^2 + z^2[(a+x)^2 + (b+y)^2 + z^2]}\bigg\};
\end{alignat*}
if $x<-a$ or $x>a$.\\

6.\quad If the value of the potential function $V$ is given at every point of the base
of an infinite rectangular prism and if the sides of the prism are at potential
zero the value of $V$ at any point within the prism is\label{err149}
\begin{multline*}
% recast to fit line
V = \dfrac{4}{ab}      \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} e^{-\pi z \sqrt{ \frac{m^2}{a^2} + \frac{n^2}{b^2}}} \sin \dfrac{m\pi x}{a}\sin \dfrac{n\pi y}{b}\\
\int\limits^a_0 d\lambda \int\limits^b_0 f(\lambda,\,\mu) \sin \dfrac{m\pi\lambda}{a}\sin \dfrac{n\pi\mu}{b} d\mu.
\end{multline*}
If $V=1$ on the base of the prism this reduces to
\[
V = \dfrac{16}{ \pi^2}\sum\limits^{m=\infty}_{m=0} \sum\limits^{n=\infty}_{n=0} e^{-\pi z  \sqrt{ \frac{(2m+1)^2 }{a^2} + \frac{(2n+1)^2}{b^2}}}\:\dfrac{\sin\dfrac{ (2m+1)\pi x}{a} \sin \dfrac{(2n+1)\pi y}{ b}}{ (2m+1)(2n+1)}.
\]

7.\quad If the value of the potential function on five faces of a rectangular
parallelopiped, whose length, breadth, and height are $a$, $b$, and $c$, is zero, and
% -----File: 150.png
if the value of $V$ is given for every point of the sixth face, then for any
point within the parallelopiped
\[
V =\sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} A_{m,n} \dfrac{\sinh \pi (c - z) \sqrt{\dfrac{ m^2}{a^2} + \dfrac{n^2}{ b^2}}}{\sinh \pi c \sqrt{\dfrac{ m^2 }{a^2} + \dfrac{n^2 }{ b^2}}} \sin \dfrac{m \pi x}{a} \sin \dfrac{n \pi y}{ b}
\]
\begin{flalign*}
&\text{where}  &A_{m,n}&= \dfrac{4}{ab} \int\limits^a_0 d\lambda \int\limits^b_0 f(\lambda,\;\mu) \sin \dfrac{m\pi\lambda}{a} \sin \dfrac{n\pi\mu}{b} d\mu.&
\end{flalign*}

8.\quad If the value of the potential function is given on two opposite faces of a
rectangular parallelopiped and is zero on the four remaining faces, then within
the parallelopiped
\begin{align*}
V =&   \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} A_{m,n} \dfrac{\sinh \pi (c - z) \sqrt{\dfrac{ m^2}{a^2} + \dfrac{n^2}{ b^2}}}{\sinh \pi c \sqrt{\dfrac{ m^2 }{a^2} + \dfrac{n^2 }{ b^2}}} \sin \dfrac{m \pi x}{a} \sin \dfrac{n \pi y}{ b}\\
      &+ \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} B_{m,n} \dfrac{\sinh \pi z         \sqrt{\dfrac{ m^2}{a^2} + \dfrac{n^2}{ b^2}}}{\sinh \pi c \sqrt{\dfrac{ m^2 }{a^2} + \dfrac{n^2 }{ b^2}}} \sin \dfrac{m \pi x}{a} \sin \dfrac{n \pi y}{ b}
\end{align*}
\begin{flalign*}
&\text{where}   &A_{m,n}&=\dfrac{4}{ab}\int\limits^a_0 d\lambda \int\limits^b_0 f (\lambda,\,\mu) \sin \dfrac{m\pi\lambda}{a}\sin \dfrac{n\pi\mu}{b} d\mu&\\
&\text{and}       &B_{m,n}&=\dfrac{4}{ab}\int\limits^a_0 d\lambda \int\limits^b_0 F(\lambda,\,\mu) \sin \dfrac{m\pi\lambda}{a}\sin \dfrac{n\pi\mu }{b} d\mu.&
\end{flalign*}

9.\quad If the value of the potential function is given at every point on the surface
of a rectangular parallelopiped, what is its value at any point within the
parallelopiped?

\vspace{1ex}
\begin{center}{III. \textit{Conduction of Heat in a Plane.}}\end{center}

1.\quad Find particular solutions of $D_tu = a^2(D_x^2 u + D_y^2 u)$ of the forms
\begin{align*}
&u = e^{-a^2(\alpha^2 + \beta^2)t} \sin(\alpha x\pm \beta y)\\
&u = e^{-a^2(\alpha^2 + \beta^2)t} \cos(\alpha x\pm \beta y).
\end{align*}

2.\quad Given the initial temperature of every point in a thin plane plate, find
the temperature of any point at any time,
% -----File: 151.png
\begin{align*}
u &= \dfrac{1}{4a^2 \pi t}\int\limits^{\infty}_{-\infty} d\lambda \int\limits^{\infty}_{-\infty} e^{-\frac{(\lambda-x)^2+(\mu -y)^2}{4a^2t}} f(\lambda,\, \mu)d\mu\\
    &= \dfrac{1}{\pi}\int\limits^{\infty}_{-\infty}e^{-\beta^2}d\beta\int\limits^{\infty}_{-\infty}e^{-\gamma^2}f(x+2a\sqrt{t}.\beta,\,y+2a\sqrt{t}.\gamma)d\gamma.
\end{align*}

3.\quad For an instantaneous \emph{source} of strength $Q$ at $(\lambda, \mu)$
\begin{flalign*}
&\phantom{v.\ Art.~53.}&u &= \dfrac{Q}{ 4\pi a^2t}e^{-\frac{(\lambda-x)^2+(\mu -y)^2}{4a^2t}} & \text{v.\ Art.~53}.
\end{flalign*}

For an instantaneous \emph{doublet} of strength $P$ at $(0, \mu)$ with its axis perpendicular
to the axis of $Y$
\begin{flalign*}
&\phantom{v.\ Art.~54.}&u &= \dfrac{Px}{ 8\pi a^4t^2}e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} & \text{v.\ Art.~54}.
\end{flalign*}

For a permanent doublet of strength $P$ at $(0, \mu)$ with its axis perpendicular
to the axis of $Y$
\[
u = \dfrac{P}{2\pi a^2}  \frac{x}{x^2+(\mu-y)^2}e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} .
\]

If the strength of the doublet were $Pd\mu$ and the heat were uniformly
generated and absorbed along the element $d\mu$ of the axis of $Y$ beginning at
(0, $\mu$) we should have
\[
u = \dfrac{P }{2\pi a^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} \dfrac{xd\mu}{ x^2+(\mu-y)^2} = \dfrac{P }{ 2\pi a^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2t}}  d \tan^{-1}\dfrac{ \mu-y}{x},
\]
and since $d \tan^{-1}\dfrac{ \mu-y}{x}$ is the angle $ARA'$, where $A$ and $A'$ are the points (0, $\mu$)
 and (0, $\mu + d\mu$) and $R$ is the point $(x, y)$, $u = 0$ when $x = 0$ unless
 $\mu < y < \mu + d\mu$, in which case $u = \dfrac{P}{2a^2}$ if $x$ approaches zero from the
positive side; and $u = 0$ when $t = 0$ except in the element $d\mu$. If then
 $u = 0$ when $t = 0$ and $u=f(y)$ when $x=0$ we have only to suppose a
doublet of strength $2a^2f(x)dx$ placed in each element of the axis of $Y$ and
then to integrate; we get
\[
u = \dfrac{1}{ \pi} \int\limits^{\infty}_{-\infty}e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} \dfrac{xf(\mu)}{x^2+(\mu -y)^2}{}d\mu.
\]

For a permanent doublet of strength $F(t)$ at (0, $\mu$) we have
\begin{align*}
u &= \dfrac{x}{ 8\pi a^4} \int\limits^t_0 e^{-\frac{x^2+(\mu-y)^2}{4a^2(t-\tau)}}   (t-\tau)^{-2}F(\tau)d\tau.\\
   &= \dfrac{1}{2\pi a^2}\bigg[\dfrac{xF(0)}{x^2+(\mu-y)^2}e^{-\frac{x^2+(\mu-y)^2}{4a^2t}}+\int\limits^t_0\dfrac{xF'(\tau)}{x^2+(\mu-y)^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2(t-\tau)}}  d\tau\bigg].
\end{align*}

% -----File: 152.png
From the reasoning above this must be zero when $t = 0$ except at the point
(0, $\mu$), must be $2a^2F(t)$ at the point (0, $\mu$), and 0 at every other point of the
axis of $Y$ when $t$ is not zero.

Hence if $u = 0$ when $t = 0$ and $u = F(y,t)$ when $x = 0$
\[
u = \dfrac{1}{\pi}  \!\int\limits^{\infty}_{-\infty}\! \dfrac{xF(\mu,0)}{x^2 + (\mu - y)^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2t}} d\mu + \dfrac{1}{\pi}  \!\int\limits^{\infty}_{-\infty}\!  d\mu \int\limits^t_0 \dfrac{xD_\tau F(\mu,\tau)}{x^2 + (\mu - y)^2} e^{-\frac{x^2+(\mu-y)^2}{4a^2(t-\tau)}}d\tau.
\]

For an extension of this solution by the method of images to the case where
there are other rectilinear boundaries and for its application to the corresponding
problems in the flow of heat in three dimensions see E.~W. Hobson in Vol.\
XIX. Proc.\ Lond.\ Math.\ Soc.\\

4.\quad If the perimeter of a thin plane rectangular plate is kept at the temperature
zero and the initial temperatures of all points of the plate are given,
then for any point of the plate
\begin{multline*}
% recast to fit line
u = \dfrac{4 }{ bc} \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1}e^{-a
^2\pi^2\left(\frac{m^2}{b^2} +\frac{ n^2}{c^2}\right)t} \sin \dfrac{m\pi x}{b} \sin \dfrac{n\pi y}{c}\\
\int\limits^b_0d\lambda \int\limits^c_0  f(\lambda,\mu) \sin \dfrac{m\pi\lambda}{b} \sin \dfrac{n\pi\mu}{c} d\mu.
\end{multline*}
if $b$ is the length and $c$ the breadth of the plate.\\

5.\quad A large mass of iron at the temperature 0° contains an iron core in the
shape of a long prism 40 cm.\ square. The core is removed and heated to the
temperature of 100° throughout and then replaced. Find the temperature of a
point in the axis of the core fifteen minutes afterward. Given $a^2 = .185$ in
C.G.S.
units.                       \hfill \emph{Ans}., 52°.9.

6.\quad If the prism described in Ex.~5 after being heated to 100° has its lateral
faces kept for 15 minutes at the temperature 0° find the temperature of a point
in its axis.                 \hfill \emph{Ans}., 20°.8.

\vspace{1ex}
\begin{center}IV.~ \textit{Conduction of Heat in Space.}\end{center}

1.\quad Show that
\begin{multline*}
% recast to fit line
\dfrac{1}{\pi^3}\int\limits^\infty_0  d\alpha \int\limits^\infty_0 d\beta \int\limits^\infty_0 d\gamma \int\limits^\infty_{-\infty} d\lambda \int \limits^\infty_{-\infty} d\mu \int\limits^\infty_{-\infty} f(\lambda,\mu,\nu) \\
\cos \alpha(\lambda - x) \cos \beta(\mu - y) \cos \gamma(\nu - z).d\nu
= f(x,y,z)
\end{multline*}
for all values of $x$, $y$, and $z$.\\

2.\quad Show that
\begin{flalign*}
&&f(x,y,z) = \sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1}\sum\limits^{p=\infty}_{p=1}A_{m,n,p} &\sin \dfrac{m\pi x}{a} \sin \dfrac{n\pi y}{b} \sin \dfrac{p\pi z}{c}&&\\
&\text{where}\quad & A_{m,n,p} = \dfrac{8}{abc}\displaystyle \int \limits^a_0 d\lambda \int \limits^b_0 d\mu \int \limits^c_0 f(\lambda,\mu,\nu) &\sin \dfrac{m\pi \lambda}{a} \sin \dfrac{n\pi \mu}{b} \sin \dfrac{p\pi \nu}{c} d\nu,
\end{flalign*}
for $0 < x < a$, $0 < y < b$, $0 < z < c$.\\

% -----File: 153.png
3.\quad Obtain particular solutions of $D_tu = a^2(D_x^2u + D_y^2u + D_z^2u)$ of the
forms
\begin{align*}
u &= e^{-a^2(\alpha^2 + \beta^2 + \gamma^2)t} \sin(\alpha x \pm\beta y \pm \gamma z).\\
u &= e^{-a^2(\alpha^2 + \beta^2 + \gamma^2)t} \cos(\alpha x \pm \beta y \pm \gamma z).
\end{align*}

4.\quad Given the initial temperature of every point in an infinite homogeneous
solid find the temperature of any point at any time.
\begin{align*}
&u=\dfrac{1}{8a^3(\pi t)^{\frac{3}{2}}} \int\limits^{\infty}_{-\infty} d\lambda                   \int\limits^{\infty}_{-\infty} d\mu                                     \int\limits^{\infty}_{-\infty} e^{-\frac{(\lambda-x)^2+(\mu-y)^2+(\nu-z)}{4a^2t}} f(\lambda,\mu,\nu)d\nu\\
  &=\dfrac{1}{\pi^{\frac{3}{2}}}               \!\int\limits^{\infty}_{-\infty}\! e^{-\beta^2} d\beta \!\int\limits^{\infty}_{-\infty}\! e^{-\gamma^2} d\gamma \!\int\limits^{\infty}_{-\infty}\! e^{-\delta^2} f(x+2a \sqrt{t}.\beta,\,y+2a \sqrt{t}.\gamma,\,z+2a \sqrt{t}.\delta)d\delta.
\end{align*}

5.\quad If the surface of a rectangular parallelopiped is kept at the temperature
zero and the initial temperatures of all points of the parallelopiped are given,
then for any point of the parallelopiped
\[
u=\sum\limits^{m=\infty}_{m=1} \sum\limits^{n=\infty}_{n=1} \sum\limits^{p=\infty}_{p=1} A_{m,n,p}e^{-a^2\pi^2\left(\frac{m^2}{b^2}+\frac{n^2}{c^2}+\frac{p^2}{d^2}\right)t} \sin \dfrac{m\pi x}{b} \sin \dfrac{n\pi y}{c} \sin \dfrac{p\pi z}{d}\\[-5ex]
\]
\begin{flalign*}
&\text{where}&A_{m,n,p}&=\dfrac{8}{bcd}\int \limits^b_0 d \lambda \int \limits^c_0 d\mu \int \limits^d_0 f(\lambda,\,\mu,\,\nu)\sin \dfrac{m\pi\lambda}{b} \sin \dfrac{n\pi\mu}{c} \sin \dfrac{p\pi\nu}{d} d\nu.&
\end{flalign*}

6.\quad An iron cube 40 cm.\ on an edge is heated to the uniform temperature of
100° Centigrade and then tightly enclosed in a large iron mass which is at the
uniform temperature of 0°. Find the temperature of the centre of the cube
fifteen minutes afterwards.                         \hfill \emph{Ans}., 38°.4.\\

7.\quad An iron cube 40 cm.\ on an edge is heated to the uniform temperature of
100° and then its surface is kept for fifteen minutes at the temperature 0°.
Required the temperature of its centre.             \hfill \emph{Ans.}, 9°.5.

\label{probend}

% -----File: 154.png
\label{ch5start}
\mychap{CHAPTER V.\footnotemark}{ZONAL HARMONICS.}

\footnotetext{ Before reading this chapter the student is advised to re-read carefully articles 9, 10, 13(\emph{c}),
15, 16, and 18(\emph{c}).}

\mypara{74.} In Art.~16 we obtained
\[
z = Ap_m(x) + Bq_m(x) \tag{1}
\]
[v.\ (6) Art.~16] as the general solution of Legendre's Equation
\[
(1-x^2)\dfrac{d^2z}{dx^2}-2x\dfrac{dz}{dx}+m(m+1)z = 0,     \tag{2}
\]
$m$ being wholly unrestricted in value and $x$ lying between $-1$ and 1; where
\begin{multline*}
p_m(x)=1-\dfrac{m(m +1)}{2!} x^2 + \dfrac{m(m -2)(m + 1)(m + 3)}{4!} x^4\\
-\dfrac{m(m - 2)(m - 4)(m + 1)(m + 3)(m + 5)}{6!} x^6+ \cdots  \tag{3}
\end{multline*}
and
\begin{multline*}
q_m(x) = x-\dfrac{(m-1)(m + 2)}{3!} x^3 + \dfrac{(m-1)(m-3)(m + 2)(m + 4)}{5!} x^5\\
-\dfrac{(m - 1)(m - 3)(m - 5)(m + 2)(m + 4)(m + 6)}{7!} x^7 +\cdots;  \tag{4}
\end{multline*}
\begin{flalign*}
\left.\!\!
\begin{aligned}
&\text{and we found}\hspace{5em}& V&=r^mp_m(\cos\theta)\\
&&             V&=\dfrac{1}{r^{m+1}} p_m(\cos\theta)\\
&&             V&=r^mq_m(\cos\theta)\\
&&             V&=\dfrac{1}{r^{m+1}} q_m(\cos\theta),
\end{aligned}
\right\}&&\hfill\tag{5}
\end{flalign*}
$m$ being unrestricted in value, as particular solutions of the special form
assumed by Laplace's Equation in spherical coördinates when $V$ is independent
of $\phi$; that is, of the equation
\[
rD_r^2(rV) + \dfrac{1}{\sin\theta} D_\theta(\sin\theta D_\theta V) = 0.  \tag{6}
\]

% -----File: 155.png
For the important case where $m$ is a positive integer we found
\[
z = AP_m(x) + BQ_m(x)                  \tag{7}
\]
[v.\ (10) Art.~16] as the general solution of Legendre's Equation (2), whence
\[\left .
\begin{aligned}
V&= r^mP_m(\cos \theta)\\
V&=\frac{1}{r^{m+1}} P_m(\cos\theta)\\
V&=r^mQ_m(\cos \theta)\\
V&=\frac{1}{r^{m+1}} Q_m(\cos\theta)
\end{aligned}
\right \}
\tag{8}
\]
are particular solutions of (6) if m is a positive integer.
\begin{multline*}
P_m(x)=\frac{(2m-1)(2m-3) \cdots 1}{m!} \left  [x^m-\frac{m(m-1)}{2(2m-1)}x^{m-2} \right . \\
+\left . \frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3) }x^{m-4} - \cdots \right ]  \tag{9}
\end{multline*}
[v.\ (8) Art.~16] and is a finite sum terminating with the term which involves
$x$ if $m$ is odd and with the term involving $x^0$ if $m$ is even.

\markright{ZONAL HARMONICS.}

It is called a \textit{Surface Zonal Harmonic}, or a \textit{Legendre's Coefficient}, or more
briefly a \textit{Legendrian}.
\begin{multline*}
Q_m(x)=\frac{m!}{(2m+1)(2m-1)\cdots 1} \biggl [\frac{1}{x^{m+1}} + \frac{(m+1)(m+2)}{2.(2m+3)} \frac{1}{x^{m+3}} \biggr . \\
+ \biggl . \frac{(m+1)(m+2)(m+3)(m+4)}{2.4.(2m+3)(2m+5)} \frac{1}{x^{m+5}} + \cdots \biggr ]  \tag{10}
\end{multline*}
if $x < -1$ or $x > 1.$ [v.\ (9) Art.~16.]

It is called a \textit{Surface Zonal Harmonic} of the \textit{second kind}.
 \begin{align*}
Q_m(x)&=(-1)^\frac{m+1}{2}  \frac{2^{m-1}\Big[\Gamma\Big(\dfrac{m+1}{2}\Big)\Big]^2}{\Gamma(m+1)} p_m(x)   \\
   &=(-1)^\frac{m+1}{2} \frac{2.4.6.\cdots (m-1)}{3.5.7.\cdots m} p_m(x) \tag{11}
\end{align*}
[v.\ (13) Art.~16]  if $m$ is odd and $-1 < x < 1$.
 \begin{align*}
Q_m(x)&=(-1)^\frac{m}{2}  \frac{2^m\Big[\Gamma\Big(\dfrac{m+1}{2}\Big)\Big]^2}{\Gamma(m+1)} q_m(x)   \\
   &=(-1)^\frac{m}{2} \frac{2.4.6.\cdots m}{1.3.5.\cdots (m-1)} q_m(x) \tag{12}
\end{align*}
[v.\ (14) Art.~16] if  $m$ is even and $-1 < x < 1$.

% -----File: 156.png
In most of the work that immediately follows we shall regard $x$ in $P_m(x) $as
equal to $\cos\theta$ and therefore as lying between $-1$ and 1.\footnote{English writers on Spherical Harmonics generally use $\mu$ in place of $x$ for $\cos \theta$. We
shall follow them, however, only when we should thereby avoid confusion.}

\mypara{75.} In Article 9 the undetermined coefficient $a_m$ of $x^m$ in $P_m(x)$ was
arbitrarily written in the form $\dfrac{(2m-1)(2m-3)\cdots 1}{m!}$ for reasons which shall
now be given.

In Articles 9 and 16 $z = P_m(x)$ was obtained as a particular solution of
Legendre's Equation
\[
(1-x^2)\dfrac{ d^2z}{dx^2} - 2x\dfrac{ dz}{dx} + m(m+1)z = 0  \tag{1}
\]
by the device of assuming that z could be expressed as a sum or a series of
terms of the form $a_nx^n$ and then determining the coefficients. We can, however,
obtain a particular solution of Legendre's Equation by an entirely different
method.

The potential function due to a unit of mass concentrated at a given point
$(x_1, y_1, z_1)$ is
\[
V= \dfrac{1}{\sqrt{ (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2}}\tag{2}
\]
and this must be a particular solution of Laplace's Equation
\[
D_x^2 V+D_y^2V+D_z^2 V= 0, \tag{3}
\]
as is easily verified by direct substitution.

If we transform (2) to spherical coördinates using the formulas of
transformation
\begin{flalign*}
&&x&=r\cos\theta & \\
&&y&=r\sin\theta\cos\phi&\\
&&z&=r\sin\theta\sin\phi& \text{we get}
\end{flalign*}
\[
V = \dfrac{1 }{\sqrt{ r^2 - 2rr_1[\cos\theta \cos\theta_1 + \sin\theta \sin\theta_1 \cos(\phi-\phi_1)] + r_1^2}} \tag{4}
\]
as a solution of Laplace's Equation in Spherical Coördinates
\begin{flalign*}
&&&&rD_r^2(rV)+\dfrac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta} V)+\dfrac{1}{\sin^2\theta} {D_\phi}^2V=0  \quad \text{ [{\scriptsize XIII}] Art.~1}.
\end{flalign*}

If the given point $(x_1, y_1, z_1)$  is taken on the axis of $X$, as it must be that
(4) may be independent of $\phi$, $\theta_1 = 0$, and
\[
V = \dfrac{1}{ \sqrt{ r^2-2rr_1 \cos\theta + r_1^2}}\tag{5}
\]
% -----File: 157.png
is a solution of
\[
rD_{r}^{2}(rV) + \dfrac{1}{\sin\theta} D_\theta(\sin\theta D_{\theta}V) = 0.      \tag{6}
\]
Equation (5) may be written
\begin{flalign*}
&   &V&=\frac{1}{r}\frac{1}{\sqrt {1-2 \dfrac{r_1}{r} \cos\theta + \dfrac{r_1^2}{r^2}} }     \tag{7}  && \\[1ex]
&\text{or}  &V&=\frac{1}{r_1}\frac{ 1}{\sqrt{ 1-2 \dfrac{r}{r_1} \cos\theta + \dfrac{r^2}{r_1^2}}}.    \tag{8}   &&
\end{flalign*}

$\sqrt{ 1- 2z \cos \theta + z^{2}}$ is finite and continuous for all values real or complex of
$z$.  It is double-valued but the two branches of the function are distinct except
for the values of $z$ which make $1 - 2z \cos \theta + z^{2} = 0$ namely $z = \cos \theta + i \sin \theta$
 and $z = \cos \theta - i\sin \theta$, both of which have the modulus unity and which are
\emph{critical} values.\\

$\dfrac{1}{\sqrt{ 1- 2z\cos\theta + z^2}}$ is finite and continuous except for the values of
 $z = \cos \theta -i\sin \theta$ and $z=\cos\theta + i\sin\theta$ for which it becomes infinite; it is
double-valued but has as critical values only these values of $z$. It is then
\textit{holomorphic} within a circle described with the origin as centre and the radius
unity, and can be developed into a power series which will be convergent for
all values of $ z$ having moduli less than one. (Int.\ Cal.\ Arts.~207, 212, 214,
220.)

If then $r>r_1$ $\displaystyle\frac{1}{\sqrt{1-\dfrac{2r_1}{r}\cos \theta + \dfrac{r_1^2}{r^2}}}$ can be developed into a convergent series
involving whole powers of $\dfrac{r_1}{r}$.

Let $\displaystyle\sum p_m \dfrac{r_1^m}{r^m}$ be this series, $p_{m}$, of course, being a function of  $\cos\theta$.
Then
\[
V = \frac{1}{r }\sum p_m \frac{r_1^m}{r^m}
\]
[v.\ (7)] is a solution of (6). Substitute this value of $V$ in (6) and we get
\[
\sum \left [\frac{r_1^m}{r^{m+1}} m(m+1)p_m + \frac{r_1^m}{r^{m+1}} \frac{1}{\sin\theta} \frac{d}{d\theta} \left ( \sin\theta \frac{dp_m}{d\theta} \right ) \right ] = 0.
\]
As this must hold whatever the value of r provided $ r > r_1$ the coefficient of
each power of  $r$ must be zero, and hence the equation
\[
\frac{1}{\sin \theta} \frac{d}{d\theta} \left (\sin \theta \frac{dp_{m}}{d\theta} \right )+m(m+1)p_{m}=0  \tag{9}
\]
must be true.

% -----File: 158.png
But as we have seen in Art.~9 the substitution of $x = \cos \theta$ in (9) reduces
it to
\[
(1 - x^2) \dfrac{d^2p_m}{ dx^2} - 2x \dfrac{dp_m }{ dx} + m(m + 1)p_m = 0,
\]
\begin{flalign*}
&\text{and therefore}  &        z &= p_m & \phantom{and therefore}
\end{flalign*}
is a solution of Legendre's Equation (1).

If $r < r_1$ $\dfrac{1 }{\sqrt{ 1 - \dfrac{2r}{r_1}\cos \theta + \dfrac{r^2}{r_1^2}} }$ can be developed into a convergent series
involving whole powers of $\dfrac{r^2}{r_1^2}$.

Let $\displaystyle \sum p_m \dfrac{r^m}{r_1^m}$ be this series. Then
\[
V = \dfrac{1}{r_1}\sum p_m\dfrac{r^m}{r_1^m}
\]
(v.\ 8) is a solution of (6); substituting in (6) we get
\[
  \sum\left[\dfrac{r^m}{r_1^{m+1}} m(m + 1) p_m + \dfrac{r^m}{r_1^{m+1}}\dfrac{ 1}{\sin \theta} \dfrac{d}{d \theta} \left(\sin \theta \dfrac{dp_m}{d \theta}\right)\right] = 0,
\]
whence it follows as before that
\[
z = p_m
\]
is a solution of Legendre's Equation.

But $p_m$ is the coefficient of the $m$th power of $\dfrac{r}{r_1}$ in the development of
$\left(1 - 2 \dfrac{r}{r_1} \cos \theta + \dfrac{r^2}{r_1^2}\right)^{-\frac{1}{2}}$ according to powers of $\dfrac{r}{r_1}$, or of the $m$th power of $\dfrac{r_1}{r }$ in
the development of $\left(1 - 2 \dfrac{r_1}{r} \cos \theta + \dfrac{r_1^2}{r^2}\right)^{-\frac{1}{2}}$ according to powers of $\dfrac{r_1}{r }$, or
more briefly it is the coefficient of the $m$th power of z in the development of
$(1 - 2xz + z^2)^{-\frac{1}{2}}$ according to powers of $z$, $x$ standing for $\cos \theta$.
\[
(1 - 2xz + z^2)^{-\frac{1}{2}} = [1 - z(2x - z)]^{-\frac{1}{2}}
\]
and can be developed by the Binomial Theorem; the coefficient of $z^m$ is easily
picked out and is
\begin{multline*}
\dfrac{(2m - 1)(2m - 3)\cdots 1 }{ m!}\left[x^m - \dfrac{m(m - 1)}{ 2(2m - 1)} x^{m-2}\right.\\
\left.+\dfrac{m(m - 1)(m - 2)(m - 3)}{2.4.(2m - 1)(2m - 3)} x^{m-4} - \cdots\right].
\end{multline*}
But this is precisely $P_m(x)$. [v.\ Art.~74 (9)]

Hence $P_m(x)$ is equal to the coefficient of the $m$th power of $z$ in
the development of $[1 - 2xz + z^2]^{-\frac{1}{2}}$ into a power series, the modulus of $z$
being less than unity.
% -----File: 159.png

\mypara{76.}  If $x=1$\;\;$P_m(x)=1$. For if $x=1$\;\;$(1-2xz+z^2)^{-\frac{1}{2}}$ reduces to
$(1-2z+z^2)^{-\frac{1}{2}}$ that is to $(1-z)^{-1}$, which develops into
\[
1 + z + z^2 + z^3 + z^4 + \cdots,
\]
and the coefficient of each power of $z$ is unity. Therefore
\[
P_m(1) = 1. \tag{1}
\]

We have seen that if $m$ is even $P_m(x)$ contains only even powers of $x$ and
terminates with the term involving $x^0$, that is with the constant term.

The value of this constant term can be picked out from the formula for
$P_m(x)$ [v.\ Art.~74 (9)]. It is
$(-1)^{\frac{m}{2}}\dfrac{1.3.5.\cdots(m-1)}
                         {2.4.6.\cdots m}$; or it can be found as
follows:---It is clearly the value $P_m(x)$ assumes when $x=0$; it is, then, the
coefficient of $z^m$ in the development of $(1+z^2)^{-\frac{1}{2}}$; but
\[
(1+z^2)^{-\frac{1}{2}}=1
 -\frac{1}{2}z^2
 +\frac{1.3}{2.4}z^4
 -\frac{1.3.5}{2.4.6}z^6
 +\frac{1.3.5.7}{2.4.6.8}z^8 - \cdots
\]
and the coefficient of $z^m$, $m$ being an even number, is
$(-1)^{\frac{m}{2}}\dfrac{1.3.5\cdots(m-1)}
                        {2.4.6\cdots m}$.

If $m$ is odd $P_m(x)$ contains only odd powers of $x$ and terminates with the
term involving $x$ to the first power.  The coefficient of this term can be
picked out from (9) Art.~74 and is
$(-1)^{\frac{m-1}{2}}\dfrac{3.5.7.\cdots m}
                           {2.4.6.\cdots(m-1)}$; or it can be
found as follows:---It is clearly the value assumed by $\dfrac{dP_m(x)}{dx}$ when $x=0$.

It is, then, the coefficient of $z^m$ in the development of $\dfrac{z}{(1+z^2)^{\frac{3}{2}}}$.
\[
\frac{z}{(1+z^2)^{\frac{3}{2}}} = z
 -\frac{3}{2}z^3
 +\frac{3.5}{2.4}z^5
 -\frac{3.5.7}{2.4.6}z^7
 +\cdots
\]
and the coefficient of $z^m$ in this development is
$(-1)^{\frac{m-1}{2}}\dfrac{3.5.7\cdots m}
                           {2.4.6\cdots(m-1)}$,
 $m$ being an odd number.

\mypara{77.} To recapitulate:
\begin{align*}
P_m(x) ={}&\frac{1.3.5\cdots(2m-1)}{m!}\bigg[
x^m-\frac{m(m-1)}{2(2m-1)}x^{m-2}\\
  &+\frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3)}x^{m-4}\\
  &-\frac{m(m-1)(m-2)(m-3)(m-4)(m-5)}{2.4.6.(2m-1)(2m-3)(2m-5)}x^{m-6}
   +\cdots\bigg], \tag{1}
\end{align*}
% -----File: 160.png
$m$ being a positive integer, is a \textit{Surface Zonal Harmonic} or \textit{Legendrian} of the
$m$th order. It is a finite sum terminating with the first power of $x$ if $m$ is
odd, and with the zeroth power of $x$ if $m$ is even.

$P_m(x)$ is the coefficient of the $m$th power of $z$ in the development of
$(1 - 2xz + z^2)^{-\frac{1}{2}}$ into a power series. Hence if  $z < 1$
\begin{multline*}
(1 - 2xz + z^2)^{-\frac{1}{2}} = P_0(x) + P_1(x).z + P_2(x).z^2 + P_3(x).z^3\\
+ P_4(x).z^4 + P_5(x).z^5 + \cdots + P_m(x).z^m + \cdots.  \tag{2}
\end{multline*}
Whence
\begin{flalign*}
 \left . \begin{aligned}
\frac{1}{\sqrt{ r^2 - 2rr_1 \cos\theta + r_1^2}} = \frac{1}{ r } \left [ P_0 (\cos \theta) + \frac{r_1 }{ r }P_1 (\cos \theta) + \frac{r_1^2 }{r^2 }P_2 (\cos \theta) + \cdots \right .  \\
+ \left . \frac{r_1^m }{ r^m} P_m (\cos \theta) + \cdots  \right  ] \quad \text{if} \quad r > r_1  \\
     = \frac{1}{ r_1} \left  [P_0(\cos \theta) +\frac{ r }{ r_1} P_1(\cos \theta) +\frac{ r^2 }{ r_1^2} P_2(\cos \theta) + \cdots \right .  \\
+\left . \frac{ r^m }{r_1^m} P_m(\cos \theta) + \cdots \right ]\;\text{ if } \;r < r_1.
\end{aligned} \right \} (3)
\end{flalign*}
\[
z = P_m(x)
\]
is a solution of Legendre's Equation
\[
(1 - x^2) \frac{d^2z }{ dx^2} - 2x \frac{dz}{ dx} + m(m + 1)z = 0
\]
when $m$ is a positive integer.
\begin{flalign*}
 &                         &  V&= r^mP_m(\cos \theta)  &&  \\
&\text{and}       &     V&= \frac{1}{ r^{m+1}} P_m (\cos \theta)  && \phantom{and}
\end{flalign*}
are solutions of the form of Laplace's Equation in Spherical Coördinates
which is independent of $\phi$, namely
\begin{gather*}
rD^2_r (rV) + \frac{1}{ \sin(\theta) }D_\theta (\sin \theta D_\theta V) = 0. \tag{4}\\
P_m(1) = 1.              \tag{5}\\[1ex]
P_{2m}(-x) = P_{2m}(x).             \tag{6}\\[1ex]
P_{2m+1} (-x) =-P _{2m+1}(x).              \tag{7}\\[1ex]
P_{2m+1}(0) = 0.             \tag{8}\\
P_{2m}(0) = (-1)^m \frac{1.3.5. \cdots  (2m - 1) }{2.4.6. \cdots 2m}.        \tag{9}\\[1ex]
% -----File: 161.png
\left[\frac{dP_{2m+1}(x)}{dx}\right]_{x=0}
 =(-1)^m\frac{3.5.7.\cdots(2m+1)}
             {2.4.6.\cdots 2m}. \tag{10}
\end{gather*}

For convenience of reference we write out a few Zonal Harmonics. They
are obtained by substituting successive integers for $m$ in formula (1).
\[\left.
\begin{aligned}
P_0(x) &= 1\\[1ex]
P_1(x) &= x\\
P_2(x) &= \frac{1}{2}(3x^2 - 1)\\
P_3(x) &= \frac{1}{2}(5x^3 - 3x)\\
P_4(x) &= \frac{1}{8}(35x^4 - 30x^2 + 3)\\
P_5(x) &= \frac{1}{8}(63x^5 - 70x^3 + 15x)\\
P_6(x) &= \frac{1}{16}(231x^6 - 315x^4 + 105x^2 - 5)\\
P_7(x) &= \frac{1}{16}(429x^7 - 693x^5 + 315x^3 - 35x)\\
P_8(x) &= \frac{1}{128}(6435x^8 - 12012x^6 + 6930x^4 - 1260x^2 + 35).
\end{aligned}\right\} \tag{11}
\]

Any Surface Zonal Harmonic may be obtained from the two of next lower
orders by the aid of the formula
\[
(n+1)P_{n+1}(x)-(2n+1)xP_n(x)+nP_{n-1}(x)=0 \tag{12}
\]
which is easily obtained and is convenient when the numerical value of $x$ is
given.

Differentiate (2) with respect to $z$ and we get
\[
\frac{-(z-x)}{(1-2xz+z^2)^{\frac{3}{2}}}
 =P_1(x) + 2P_2(x).z + 3P_3(x).z^2 + \cdots
\]
whence
\[
\frac{-(z-x)}{(1- 2xz + z^2)^{\frac{1}{2}}}
 =(1-2xz+z^2)(P_1(x) + 2P_2(x).z + 3P_3(x).z^2 + \cdots).
\]
Hence by (2)
\begin{align*}
(1-2xz+z^2)(P_1(x) &
+2P_2(x).z + 3P_3(x).z^2 + \cdots)\\
&+(z-x)(P_0(x)+P_1(x).z
 +      P_2(x).z^2 + \cdots) = 0 \tag{13}
\end{align*}
% -----File: 162.png

(13) is identically true, hence the coefficient of each power of $z$ must vanish.
Picking out the coefficient of $z^n$ and writing it equal to zero we have formula
(12) above.\footnote
{For tables of Surface Zonal Harmonics v.\ Appendix Tables I and II.}

\mypara{78.} We are now able to solve completely the problem considered in Art.~9.

We were to find a solution of the differential equation
\[
rD_r^2(rV) + \frac{1}{\sin\theta}D_\theta(\sin\theta D_\theta V) = 0 \tag{1}
\]
subject to the condition
\[
V = \frac{M}{(c^2 + r^2)^\frac{1}{2}}\quad \text{when}\quad \theta = 0. \tag{2}
\]

We know (v.\ Art.~77) that
\begin{flalign*}
&&V = r^mP_m(\cos\theta)&&\\
&\text{and} & V=\frac{1}{r^{m+1}}P_m(\cos\theta)&&&\phantom{and}
\end{flalign*}
are solutions of (1).

For values of $r<c$
\[
\frac{M}{(c^2+r^2)^\frac{1}{2}} = \frac{M}{c}
\Bigl[1-\frac{1}{2}\frac{r^2}{c^2}
       +\frac{1.3}{2.4}\frac{r^4}{c^4}
       -\frac{1.3.5}{2.4.6}\frac{r^6}{c^6}
       +\cdots\Bigr]. \tag{3}
\]

Therefore for values of $r<c$
\begin{align*}
V = \frac{M}{c}\Big[P_0(\cos\theta)-\frac{1}{2}\frac{r^2}{c^2}&P_2(\cos\theta)\\
&+\frac{1.3}{2.4}\frac{r^4}{c^4}P_4(\cos\theta)
-\frac{1.3.5}{2.4.6}\frac{r^6}{c^6}P_6(\cos\theta)+\cdots\Big]\tag{4}
\end{align*}
is our required solution; because each term satisfies equation (1), and therefore
the whole value satisfies (1), and when $\theta=0$
\[
P_m(\cos\theta) = P_m(1) = 1
\]
[v.\ (5) Art.~77], and hence (4) reduces to (3) and (2) is satisfied.

For values of $r>c$
\begin{align*}
\frac{M}{(c^2+r^2)^\frac{1}{2}} &= \frac{M}{r}
  \Bigl[1-\frac{1}{2}\frac{c^2}{r^2}
         +\frac{1.3}{2.4}\frac{c^4}{r^4}
         -\frac{1.3.5}{2.4.6}\frac{c^6}{r^6}
         +\cdots\Bigr] \tag{5}\\
&=M\Bigl[\frac{1}{r}-\frac{1}{2}\frac{c^2}{r^3}
        +\frac{1.3}{2.4}\frac{c^4}{r^5}
        -\frac{1.3.5}{2.4.6}\frac{c^6}{r^7}
        +\cdots\Bigr].
\end{align*}

% -----File: 163.png
Therefore for values of $r>c$
\begin{align*}
V=\frac{M}{c}\Big[\frac{c}{r}P_0(\cos\theta)-\frac{1}{2}&\frac{c^3}{r^3}P_2(\cos\theta)\\
&+\frac{1.3}{2.4}\frac{c^5}{r^5}P_4(\cos\theta)
 -\frac{1.3.5}{2.4.6}\frac{c^7}{r^7}P_6(\cos\theta)
 +\cdots\Big] \tag{6}
\end{align*}
is our required solution. For it satisfies (1) and reduces to (2) when $\theta=0$.

\mypara{79.} As another example let us suppose a conductor in the form of a thin
circular disc charged with electricity, and let it be required to find the value
of the potential function at any point in space.

If the magnitude of the charge is $M$ and the radius of the plate is $a$ the
surface density at a point of the plate at a distance $r$ from the centre is
\[
\sigma = \frac{M}{4a\pi\sqrt{a^2-r^2}}
\]
and all points of the conductor are at the potential $\dfrac{\pi M}{2a}$. (v.\ Peirce's Newtonian
Potential Function, §~61.)

The value of the potential function at a point in the axis of the plate at the
distance $x$ from the plate is easily seen to be
\begin{align*}
V &= \frac{M}{a}\int\limits_0^a\frac{r dr}{\sqrt{(a^2-r^2)(x^2+r^2)}}\\
  &= \frac{M}{2a}\cos^{-1}\frac{x^2-a^2}{x^2+a^2}.
\end{align*}
\begin{alignat*}{2}
&\frac{d}{dx}\Bigl(\frac{M}{2a}\cos^{-1}\frac{x^2-a^2}{x^2+a^2}\Bigr) &&= -\frac{M}{a^2+x^2}\\
&&&= -\frac{M}{a^2}\Bigl[1- \frac{x^2}{a^2}+\frac{x^4}{a^4}-\frac{x^6}{a^6}+\cdots\Bigr]\\
&\text{if} \quad x<a,\\
&&&= -\frac{M}{x^2}\Bigl[1- \frac{a^2}{x^2}+\frac{a^4}{x^4}-\frac{a^6}{x^6}+\cdots\Bigr]\\
&\text{if} \quad x>a.
\end{alignat*}

Integrating and then determining the arbitrary constant we have
\begin{alignat*}{2}
&\frac{M}{2a}\cos^{-1}\frac{x^2-a^2}{x^2+a^2}
&&= \frac{M}{a}\Bigl[\frac{\pi}{2}-\frac{x}{a}
  +\frac{x^3}{3a^3}-\frac{x^5}{5a^5}+\frac{x^7}{7a^7}-\cdots\Big]\\[1ex]
&\quad\text{if} \quad x<a,\\
&&&= \frac{M}{a}\Big[\frac{a}{x}-\frac{a^3}{3x^3}
  +\frac{a^5}{5x^5}-\frac{a^7}{7x^7}+\cdots\Bigr]\\
&\quad\text{if} \quad x>a.
\end{alignat*}
% -----File: 164.png

We have, then, to solve the equation
\[
rD^2_r(rV)+\frac{1}{\sin\theta}D_\theta(\sin\theta D_\theta V)=0
\]
subject to the conditions
\begin{flalign*}
&&V&=\frac{M}{a}\bigg[\frac{\pi}{2}-\frac{r}{a}
                    +\frac{r^3}{3a^3}-\frac{r^5}{5a^5}
                    +\frac{r^7}{7a^7}-\cdots\bigg]&\\
&&&\text{when}\quad \theta=0\quad \text{and}\quad r<a&\\
&\text{and} & V &=\frac{M}{a}\bigg[\frac{a}{r}-\frac{a^3}{3r^3}
                    +\frac{a^5}{5r^5}-\frac{a^7}{7r^7}+\cdots\bigg]&\phantom{and}\\
&&&\text{when}\quad \theta=0\quad \text{and}\quad r>a.&
\end{flalign*}

The required solution is easily seen to be
\begin{flalign*}
&&V&=\frac{M}{a}\bigg[
\frac{\pi}{2}-\frac{r}{a}P_1(\cos\theta)
  +\frac{1}{3}\frac{r^3}{a^3}P_3(\cos\theta)
  -\frac{1}{5}\frac{r^5}{a^5}P_5(\cos\theta)+\cdots\bigg]
\intertext{if $r<a$ and $\theta<\dfrac{\pi}{2}$,}
&\text{and}& V&=\frac{M}{a}\bigg[\frac{a}{r}
-\frac{1}{3}\frac{a^3}{r^3}P_2(\cos\theta)
+\frac{1}{5}\frac{a^5}{r^5}P_4(\cos\theta)
-\frac{1}{7}\frac{a^7}{r^7}P_6(\cos\theta)+\cdots\bigg]
\intertext{if $r>a$.}
\end{flalign*}

\EXAMPLE{S}

1.\quad Given that if a charge $M$ of electricity is placed on an ellipsoidal conductor
the surface density at any point $P$ of the conductor is equal to $\dfrac{Mp}{4\pi abc}$,
 where $p$ is the distance from the centre of the conductor to the tangent plane at
$P$ (v.\ Peirce, New.\ Pot.\ Func.\ §~61); find the value of the potential function at
any external point when the conductor is the oblate spheroid generated by the
rotation of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ about its minor axis.

\textit{Ans.} (1) If the point is on the axis of revolution
\[
V=\frac{M}{2\sqrt{a^2-b^2}}\bigg[\sin^{-1}\bigg(\frac{bx+a^2-b^2}{a\sqrt{x^2+a^2-b^2}}\bigg)
                               -\sin^{-1}\bigg(\frac{bx-a^2+b^2}{a\sqrt{x^2+a^2-b^2}}\bigg)\bigg]
\]
$x$ being the distance from the centre.

(2) If the point is on the surface of the spheroid
\[
V=\frac{M}{2\sqrt{a^2-b^2}}\bigg[\frac{\pi}{2}-\sin^{-1}\bigg(\frac{2b^2-a^2}{a^2}\bigg)\bigg]
 =\frac{M}{ \sqrt{a^2-b^2}}\bigg[\frac{\pi}{2}-\tan^{-1}\bigg(\frac{b}{\sqrt{a^2-b^2}}\bigg)\bigg].
\]

% -----File: 165.png
(3) If the distance $r$ of the point from the centre is less than $\sqrt{ a^2 - b^2}$ and
$\theta < \dfrac{\pi}{2}\tablestrut$
\begin{multline*}
V = \frac{M }{\sqrt{ a^2 - b^2}} \left  [ \frac{\pi }{2} - \frac{r }{(a^2-b^2)^{\frac{1}{2}}} P_1(\cos \theta) \right . \\
\left . + \frac{r^3}{ 3(a^2-b^2)^{\frac{3}{2}} }P_3(\cos \theta) - \frac{r^5 }{5(a^2-b^2)^{\frac{5}{2}}} P_5(\cos \theta)+  \cdots \right ].
\end{multline*}

(4)  If the distance $ r$ of the point from the centre is greater than $\sqrt{ a^2 - b^2}$
\begin{multline*}
V = \frac{M }{\sqrt{ a^2 - b^2}} \biggl  [ \frac{(a^2-b^2)^{\frac{1}{2}}}{r} - \frac{(a^2-b^2)^{\frac{3}{2}}}{3r^3} P_2(\cos \theta) \biggr .\\
\biggl . + \frac{(a^2-b^2)^{\frac{5}{2}}}{5r^5} P_4(\cos \theta) - \frac{(a^2-b^2)^{\frac{7}{2}}}{7r^7}P_6(\cos \theta) + \cdots \biggr ].
\end{multline*}

2.\quad If the conductor is the prolate spheroid generated by the rotation of the
ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ about its major axis, show that if the point is an external
point and is on the axis at a distance $x$ from the centre,
\[
V = \frac{M }{2\sqrt{ a^2 - b^2}} \log\frac{ x + \sqrt{ a^2 - b^2}} {x - \sqrt {a^2 - b^2}}.
\]

If the point is not on the axis and $r > \sqrt{a^2 - b^2}$
\begin{multline*}
V = \frac{M }{\sqrt{ a^2 - b^2}} \biggl  [ \frac{(a^2-b^2)^{\frac{1}{2}}}{r} + \frac{(a^2-b^2)^{\frac{3}{2}}}{3r^3} P_2(\cos \theta) \biggr .\\
\biggl . + \frac{(a^2-b^2)^{\frac{5}{2}}}{5r^5} P_4(\cos \theta) + \frac{(a^2-b^2)^{\frac{7}{2}}}{7r^7}P_6(\cos \theta) + \cdots \biggr ].
\end{multline*}

\mypara{80.} As a third example we will find the value of the potential function due
to a thin homogeneous circular disc, of density $\rho$, thickness $k$, and radius $a$.

The value of  $V$ at a point in the axis of the disc at a distance $x$ from its
centre is readily found and proves to be
\[
V_0 = 2 \pi \rho k(\sqrt{x^2 + a^2} - x) = \frac{2M}{ a^2} [\sqrt{x^2 + a^2} - x].
\]

If $x>a$
\begin{flalign*}
% recast to fit line
&&\sqrt{x^2+a^2} &= x \left (1+\frac{a^2}{x^2} \right )^{\frac{1}{2}} \\
&&               &= x \left [1+\frac{1}{2} \frac{a^2}{x^2} - \frac{1.1}{2.4} \frac{a^4}{x^4}+ \frac{1.1.3}{2.4.6} \frac{a^6}{x^6} - \frac{1.1.3.5}{2.4.6.8} \frac{a^8}{x^8} + \cdots \right ] &&\\
&\text{and} &V_0 &=\frac{ 2M}{a} \left  [ \frac{1}{2} \frac{a}{x} - \frac{1.1}{2.4} \frac{a^3}{x^3} + \frac{1.1.3}{2.4.6} \frac{a^5}{x^5} - \frac{1.1.3.5}{2.4.6.8} \frac{a^7}{x^7} + \cdots \right  ].  &&
\end{flalign*}

% -----File: 166.png
If $x < a$
\begin{flalign*}
% recast to fit line
&&\sqrt{x^2+a^2} &= a\left(1+\dfrac{x^2}{a^2}\right)^{\frac{1}{2}}\\
&&&               = a\left[1+\dfrac{1}{2} \dfrac{x^2}{a^2} - \dfrac{1.1}{2.4}\dfrac{ x^4}{a^4} + \dfrac{1.1.3}{2.4.6} \dfrac{x^6}{a^6} +\cdots\right]&\\
&\text{and}& V_0 = \dfrac{2M}{a}{}&\left[1 -\dfrac{ x}{a} + \dfrac{1}{2} \dfrac{x^2}{a^2} - \dfrac{1.1}{2.4}\dfrac{ x^4}{a^4} + \dfrac{1.1.3}{2.4.6} \dfrac{x^6}{a^6} - \dfrac{1.1.3.5}{2.4.6.8}\dfrac{ x^8}{a^8} + \cdots \right].&
\end{flalign*}

Hence the solution for any external point is
\begin{multline*}
V = \dfrac{2M}{a} \left[\dfrac{1}{2} \dfrac{a}{r} - \dfrac{1.1}{2.4} \dfrac{a^3}{r^3} P_2(\cos\theta)\right.\\
                +\left. \dfrac{1.1.3}{2.4.6} \dfrac{a^5}{r^5}P_4(\cos\theta) - \dfrac{1.1.3.5}{2.4.6.8}\dfrac{ a^7}{r^7}P_6(\cos\theta) + \cdots \right]
\end{multline*}
if $r > a$, and
\begin{multline*}
V = \dfrac{2M}{a}\left[1-\dfrac{r}{a}P_1(\cos\theta)\right.\\
+\left.\dfrac{1}{2} \dfrac{r^2}{a^2}P_2(\cos\theta) - \dfrac{1.1}{2.4}\dfrac{ r^4}{a^4} P_4(\cos\theta) + \dfrac{1.1.3}{2.4.6} \dfrac{r^6}{a^6}P_6(\cos\theta) -\cdots\right]
\end{multline*}
if $r < a$ and $\theta < \dfrac{\pi}{2}$.

\EXAMPLE{S}

1.\quad The potential function due to a homogeneous hemisphere whose axis is
taken as the polar axis, is
\begin{multline*}
V = \dfrac{M}{a} \left[\dfrac{a}{r} + \dfrac{3.1}{2.4} \dfrac{a^2}{r^2}P_1(\cos\theta) \right.\\
   -\left. \dfrac{3.1.1}{2.4.6} \dfrac{a^4}{r^4} P_3(\cos\theta) + \dfrac{3.1.1.3}{2.4.6.8} \dfrac{a^6}{r^6}P_5(\cos\theta) -\cdots \right]
\end{multline*}
if $r > a$, and is
\begin{multline*}
V = \dfrac{M}{a} \left[\dfrac{3}{2} + \dfrac{3}{2} \dfrac{r}{a}P_1(\cos\theta) + \dfrac{r^2}{a^2}P_2(\cos\theta)\right.\\
          +\left. \dfrac{3.1}{2.4}\dfrac{ r^3}{a^3} P_3(\cos\theta) - \dfrac{3.1.1}{2.4.6} \dfrac{r^5}{a^5}P_5(\cos\theta)+ \cdots \right]
\end{multline*}
if $r < a$ and $\theta > \dfrac{\pi}{2}$.\\

2.\quad The potential function due to a solid sphere whose density is proportional
to the distance from a diametral plane is, at an external point,
\begin{multline*}
V = \dfrac{8}{15}\dfrac{ M}{a} \left[\dfrac{5.3}{2.4}\dfrac{ a}{r} + \dfrac{5.3.1}{2.4.6} \dfrac{a^3}{r^3}P_2(\cos\theta)\right.\\
 -\left. \dfrac{5.3.1.1}{2.4.6.8} \dfrac{a^5}{r^5} P_4(\cos\theta) + \dfrac{5.3.1.1.3}{2.4.6.8.10}\dfrac{ a^7}{r^7}P_6(\cos\theta) - \cdots \right].
\end{multline*}

% -----File: 167.png
3.\quad The potential function due to the homogeneous oblate spheroid generated
by the rotation of $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ about its minor axis is, at an external point,
\begin{multline*}
V = \frac{3}{2}\frac{ M}{(a^2-b^2)} \left  [ \frac{x^2+a^2-b^2 }{ 2(a^2-b^2)^{\frac{1}{2}}} \left ( \sin^{-1} \frac{(a^2-b^2+bx)}{a \sqrt{ x^2+a^2-b^2}} \right . \right . \\
\left .  \left . + \sin^{-1} \frac{(a^2-b^2-bx)}{a \sqrt{ x^2+a^2-b^2}} \right  ) - x \right ]
\end{multline*}
if the point is on the axis of the spheroid at a distance $x$ from its centre.
\begin{multline*}
V = \frac{ 3M}{(a^2-b^2)^{\frac{1}{2} }} \left [\frac{1}{1.3}\frac{ (a^2-b^2)^{\frac{1}{2}}}{ r}  - \frac{1}{3.5}\frac{ (a^2-b^2)^{\frac{3}{2}}} {r^3 }P_2(\cos \theta) \right . \\
\left  . + \frac{1}{5.7}\frac{(a^2-b^2)^{\frac{5}{2}}} {r^5} P_4(\cos \theta) - \cdots \right ]
\end{multline*}
if $r > (a^2 - b^2)^{\frac{1}{2}}$, and
\begin{multline*}
V = \frac{3M}{(a^2-b^2)^{\frac{1}{2}}} \left  [ \frac{\pi}{4} -\frac{ r}{(a^2-b^2)^{\frac{1}{2}}}  P_1(\cos \theta) + \frac{\pi}{4} \frac{r^2}{(a^2-b^2)} P_2(\cos \theta) \right .\\
\left . - \frac{1}{1.3} \frac{r^3}{(a^2-b^2)^{\frac{3}{2}}} P_3(\cos \theta) + \frac{1}{3.5}\frac{ r^5}{(a^2-b^2)^{\frac{5}{2}}} P_5(\cos \theta) - \cdots \right  ]
\end{multline*}
if $r<(a^2-b^2)^{\frac{1}{2}}$ and $\theta < \dfrac{\pi }{2}$.\\

4.\quad  The potential function due to the homogeneous prolate spheroid
generated by the rotation of $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ about its major axis is, at an
external point,
\begin{gather*}
V = \frac{3M}{(a^2-b^2)^{\frac{1}{2}}} \left  [\frac{1}{1.3}\frac{ (a^2-b^2)^{\frac{1}{2}}}{r } + \frac{1}{3.5}\frac{ (a^2-b^2)^{\frac{3}{2}}}{r^3} P_2(\cos \theta) \right . \\
+\left . \frac{1}{5.7}\frac{ (a^2-b^2)^{\frac{5}{2}}}{ r^5} P_4(\cos \theta) + \cdots \right ]
\end{gather*}
if $r>(a^2 -b^2)^{\frac{1}{2}}$.

\mypara{81.} The method employed in the last three articles may be stated in
general as follows:---Whenever in a problem involving the solving of the
special form of Laplace's Equation
\[
rD^2_r (rV) + \frac{1}{\sin\theta }D_\theta (\sin \theta D_\theta V) = 0,
\]
the value of  $V$ is given or can be found for all points on the axis of $X$ and
this value can be expressed as a sum or a series involving only whole powers
positive or negative of the radius vector of the point, the solution for a point
% -----File: 168.png
not on the axis can be obtained by multiplying each term by the appropriate
Zonal Harmonic, subject only to the condition that the result if a series must
be convergent.

It will be shown in the next article that $P_m(\cos\theta)$ is never greater than
one nor less than minus one. Hence the series in question will be convergent
for all values of $r$ for which the original series was \textit{absolutely convergent}.\

\mypara{82.} In addition to the form given in (1) Art.~77 for $P_m(x)$ other forms
are often useful.

It ought to be possible to develop $P_m(\cos\theta)$, which may be regarded as a
function of $\theta$, into a Fourier's Series, and such a development may be obtained,
though with much labor, by the methods of Chapter II.

The development in terms of cosines of multiples of $\theta$ may be obtained
much more easily by the following device.

We have seen in Art.~75 that $P_m(\cos\theta)$ is the coefficient of the $m$th power
of $z$ in the development of $(1 - 2z \cos\theta + z^2)^{-\frac{1}{2}}$ in a power series, and that
if mod $z < 1$\;\;$(1 - 2z~\cos\theta + z^2)^{-\frac{1}{2}}$ can be developed into such a series. We
know by the Theory of Functions that only one such series exists, so that the
method by which we may choose to obtain the development will not affect the
result.
\begin{align*}
(1 - 2z\cos\theta + z^2)^{-\frac{1}{2}} &= (1-z(e^{\theta i} + e^{-\theta i}) + z^2)^{-\frac{1}{2}}\\
                            &= (1-ze^{\theta i})^{-\frac{1}{2}}(1-ze^{-\theta i})^{-\frac{1}{2}}.
\end{align*}
$(1-ze^{\theta i})^{-\frac{1}{2}}$  may be developed into an absolutely convergent series if
 mod $z <1$, by the Binomial Theorem. We have
\[
(1-ze^{\theta i})^{-\frac{1}{2} }= 1 + \frac{1}{2} ze^{\theta i} + \frac{1.3}{2.4} z^2e^{2\theta i }
+ \frac{1.3.5}{2.4.6} z^3e^{3\theta i} + \frac{1.3.5.7}{2.4.6.8 }z^4e^{4\theta i }+ \cdots\\[-4ex]
\]
\begin{multline*}
(1-ze^{-\theta i})^{-\frac{1}{2} }= 1 + \frac{1}{2} ze^{-\theta i} + \frac{1.3}{2.4} z^2e^{-2\theta i } \\
+ \frac{1.3.5}{2.4.6} z^3e^{-3\theta i} + \frac{1.3.5.7}{2.4.6.8 }z^4e^{-4\theta i }+ \cdots
\end{multline*}

The product of these series will give a development for $(1-2z\cos\theta+z^2)^{-\frac{1}{2}}$
 in power series. The coefficient of $z^m$ is easily picked out, and must be equal
to $P_m(\cos\theta)$. We thus get
\begin{align*}
% recast to fit line
P_m(\cos\theta)={}&\frac{ 1.3.5. \cdots (2m-1) }{ 2.4.6.\cdots 2m} \bigg[e^{m\theta i} + e^{-m\theta i} \\
&\quad + \frac{1}{2}.\frac{2m}{2m-1}(e^{(m-2)\theta i} + e^{-(m-2)\theta i}) \\
&\quad + \frac{1.3}{2.4}.\frac{  2m(2m-2) }{ (2m-1)(2m-3)}  (e^{(m-4)\theta i} + e^{-(m-4)\theta i}) + \cdots \bigg]\\[1ex]
% -----File: 169.png
P_m(\cos\theta)={}&\frac{1.3.5\cdots(2m-1)}{2.4.6.\cdots 2m}
\bigg[2\cos m\theta + 2\frac{1.m}{1.(2m-1)}\cos(m-2)\theta\\
&\quad +2\frac{1.3\,m(m-1)}{1.2(2m-1)(2m-3)}\cos(m-4)\theta\\
&\quad +2\frac{1.3.5}{1.2.3}\frac{m(m-1)(m-2)}{(2m-1)(2m-3)(2m-5)}\cos(m-6)\theta +\cdots\bigg]. \tag{1}
\end{align*}

If $m$ is odd the development runs down to $\cos\theta$; if $m$ is even to $\cos(0)$, but
in that case the coefficient of $\cos(0)$, that is, the constant term, will not contain
the factor 2 which is common to all the other terms, but will be simply
$\bigg[\dfrac{1.3.5\cdots(m-1)}{2.4.6.\cdots m}\bigg]^2$.

We write out the values of $P_m(\cos\theta)$ for a few values of $m$
\[\left.
\begin{aligned}
P_0(\cos\theta) &=1\\[1ex]
P_1(\cos\theta) &=\cos\theta\\
P_2(\cos\theta) &=\frac{1}{4}(3\cos 2\theta+1)\\
P_3(\cos\theta) &=\frac{1}{8}(5\cos 3\theta+3\cos\theta)\\
P_4(\cos\theta) &=\frac{1}{64}(35\cos 4\theta+20\cos 2\theta+9)\\
P_5(\cos\theta) &=\frac{1}{128}[63\cos 5\theta+35\cos 3\theta+30\cos\theta]\\
P_6(\cos\theta) &=\frac{1}{512}[231\cos 6\theta+126\cos 4\theta+105\cos 2\theta+50]\\
P_7(\cos\theta) &=\frac{1}{1024}[429\cos 7\theta+231\cos 5\theta+189\cos 3\theta+175\cos\theta]\\
P_8(\cos\theta) &=\frac{1}{16384}[6435\cos 8\theta+3432\cos 6\theta+2772\cos 4\theta\\
                &\qquad\qquad +2520\cos 2\theta+1225].
\end{aligned}\right\} \tag{2}
\]

Since all the coefficients in the second member of (1) are positive, and since
each cosine has unity for its maximum value it is clear that $P_m(\cos\theta)$ has
its maximum value when $\theta=0$; but we have shown in Art.~76 that $P_m(1)=1$.
 Therefore $P_m(\cos\theta)$ is never greater than unity if $\theta$ is real. It is also easily
seen from (1) that $P_m(\cos\theta)$ can never be less than $-1$.

% -----File: 170.png
\mypara{83.} $P_m(x)$ can be very simply expressed as a derivative. We have
\begin{multline*}
\hspace{12pt}P_m(x)=\frac{(2m-1)(2m-3)\cdots 1}{m!}\bigg[x^m-\frac{m(m-1)}{2.(2m-1)}x^{m-2}\hspace{8em}\\
\shoveright{+\frac{m(m-1)(m-2)(m-3)}{2.4.(2m-1)(2m-3)}x^{m-4}-\cdots\bigg]\phantom{.}\quad}\\
%
\shoveleft{\int\limits_0^x P_m(x)dx=\frac{(2m-1)(2m-3)\cdots 1}{(m+1)!}\bigg[x^{m+1}-\frac{(m+1)m}{2.(2m-1)}x^{m-1}}\\
\shoveright{+\frac{(m+1)m(m-1)(m-2)}{2.4.(2m-1)(2m-3)}x^{m-3}-\cdots\bigg]}\\
%
\shoveleft{\int\limits_0^x{}^2 P_m(x)dx^2= \int\limits_0^x dx \int\limits_0^x P_m(x)dx}\\
\shoveleft{\hspace{62pt}=\frac{(2m-1)(2m-3)\cdots 1}{(m+2)!}\bigg[x^{m+2}-\frac{(m+2)(m+1)}{2.(2m-1)}x^m}\\
\shoveright{+\frac{(m+2)(m+1)m(m-1)}{2.4.(2m-1)(2m-3)}x^{m-2}-\cdots\bigg]}\\
%
\shoveleft{\int\limits_0^x{}^m P_m(x)dx^m =\frac{(2m-1)(2m-3)\cdots 1}{(2m)!}\bigg[x^{2m}-\frac{2m(2m-1)}{2(2m-1)}x^{2m-2}}\\
\shoveright{+\frac{2m(2m-1)(2m-2)(2m-3)}{2.4.(2m-1)(2m-3)}x^{2m-4}-\cdots\bigg]}\\
\shoveleft{\hspace{68pt}=\frac{(2m-1)(2m-3)\cdots 1}{(2m)!}\bigg[x^{2m}-mx^{2m-2}+\frac{m(m-1)}{2!}x^{2m-4}}\\
\shoveright{-\frac{m(m-1)(m-2)}{3!}x^{2m-6}+\cdots\bigg].\quad}\\[-4ex]
\end{multline*}
\indent The quantity in brackets obviously differs from $(x^2-1)^m$ by terms involving
lower powers of $x$ than the $m$th.
\begin{flalign*}
&\text{Hence}
& P_m(x)&=\frac{1.3.5\cdots(2m-1)}{(2m)!}\frac{d^m}{dx^m}(x^2-1)^m,&&\phantom{Hence}\\
&\text{or}
&& P_m(x)=\frac{1}{2^mm!}\frac{d^m}{dx^m}(x^2-1)^m. & \tag{1}&
\end{flalign*}

This important formula is entirely general and holds not merely when
 $x=\cos\theta$, but for all values of $x$.
% -----File: 171.png

\mypara{84.} The last result is so important that it is worth while to confirm it by
obtaining it directly from Legendre's Equation
\[
(1-x^2)\frac{d^2z}{dx^2}-2x\frac{dz}{dx}+m(m+1)z=0 \tag{1}
\]
v.\ (1) Art.~75.

Let us differentiate (1) with respect to $x$ a few times representing
$\dfrac{dz}{dx}$ by $z'$, $\dfrac{d^2z}{dx^2}$ by $z''$, $\dfrac{d^3z}{dz^3}$ by $z'''$, \&c. We get
\begin{gather*}
(1-x^2)\frac{d^2z'}{dx^2}-2.2x\frac{dz'}{dx}+[m(m+1)-2]z'=0,\\
(1-x^2)\frac{d^2z''}{dx^2}-2.3x\frac{dz''}{dx}+[m(m+1)-2(1+2)]z''=0,\\
(1-x^2)\frac{d^2z'''}{dx^2}-2.4x\frac{dz'''}{dx}+[m(m+1)-2(1+2+3)]z'''=0,
\end{gather*}
and in general
\[
(1-x^2)\frac{d^2z^{(n)}}{dx^2}-2(n+1)x\frac{dz^{(n)}}{dx}+[m(m+1)-2(1+2+3+\cdots+n)]z^{(n)}=0\\[-3ex]
\]
\begin{flalign*}
&\text{or} & (1-x^2)\frac{d^2z^{(n)}}{dx^2}-2(n+1)x\frac{dz^{(n)}}{dx}+m(m+1)-n(n+1)]z^{(n)}=0. \hspace{15pt}(2)
\end{flalign*}

Following the analogy of these steps it is easy to write equations that will
differentiate into (1).

Let $\dfrac{dz_1}{dx}=z$, $\dfrac{d^2z_2}{dx^2}=z$, $\dfrac{d^3z_3}{dx^3}=z$, \&c. Then
\[
(1-x^2)\frac{d^2z_1}{dx^2}+m(m+1)z_1=0
\]
will differentiate into (1),
\[
(1-x^2)\frac{d^2z_2}{dx^2}+2.1x\frac{dz_2}{dx}+[m(m+1)-2.1]z_2=0
\]
if differentiated twice will give (1),
\[
(1-x^2)\frac{d^2z_3}{dx^2}+2.2x\frac{dz_3}{dx}+[m(m+1)-2(1+2)]z_3=0
\]
if differentiated three times will give (1), and in general
\[
(1-x^2)\frac{d^2z_n}{dx^2}+2(n-1)x\frac{dz_n}{dx}+[m(m+1)-n(n-1)]z_n=0 \tag{3}
\]
if differentiated $n$ times with respect to $x$ will give (1).

If $n=m+1$ (3) reduces to
\[
(1-x^2)\frac{d^2z_{m+1}}{dx^2}+2mx\frac{dz_{m+1}}{dx}=0, \tag{4}
\]
% -----File: 172.png
and the $(m+1)$st derivative with respect to $x$ of any function of $x$ which
satisfies (4) will be a solution of (1). (4) can be written
\[
(1-x^2)\frac{dz_m}{dx}+2mxz_m=0
\]
and can be readily solved by separating the variables and integrating. v.\ Int.\
Cal.~(1) page 314. It gives
\begin{flalign*}
&&&\quad z_m = C(x^2-1)^m.&\phantom{Hence}\\
&\text{Hence} & z=&\frac{d^mz_m}{dx^m}=C\frac{d^m(x^2-1)^m}{dx^m} &\tag{5}
\end{flalign*}
is a solution of Legendre's Equation (1) and agrees with the value of $P_m(x)$
obtained in Art.~83.

\mypara{85.} The equations obtained in Art.~84 are so curious and so simply related
that it is worth while to consider them a little more fully.

We have seen that
\[
(1-x^2)\frac{d^2z}{dx^2}+2mx\frac{dz}{dx}=0 \tag{1}
\]
differentiates into
\[
(1-x^2)\frac{d^2z}{dx^2}+2(m-1)x\frac{dz}{dx}+2mz=0; \tag{2}
\]
that if we differentiate (2) $m$ times we get Legendre's Equation
\[
(1-x^2)\frac{d^2z}{dx^2}-2x\frac{dz}{dx}+m(m+1)z=0; \tag{3}
\]
that if we differentiate (2) $2m$ times we get
\[
(1-x^2)\frac{d^2z}{dx^2}-2(m+1)x\frac{dz}{dx}=0; \tag{4}
\]
that if we differentiate (2) $m-n$ times we have
\[
(1-x^2)\frac{d^2z}{dx^2}+2(n-1)x\frac{dz}{dx}+[m(m+1)-n(n-1)]z=0; \tag{5}
\]
and that if we differentiate (2) $m+n$ times we have
\[
(1-x^2)\frac{d^2z}{dx^2}-2(n+1)x\frac{dz}{dx}+[m(m+1)-n(n+1)]z=0. \tag{6}
\]
By the aid of (1) we found in the last article a particular solution of (2),
namely
\[
z=(x^2-1)^m.
\]
% -----File: 173.png

If we substitute in (2) $z=u(x^2-1)^m$ following the method illustrated
fully in Art.~18, we get as the general solution of (2)
\[
z=A(x^2-1)^m+B(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}, \tag{7}
\]
$A$ and $B$ being arbitrary constants.

$\displaystyle\int\frac{dx}{(x^2-1)^{m+1}}$ is easily written out [v.\ formula (42) page 6. Table of Integrals.
Int.\ Cal.\ Appendix]. If $x<1$ it vanishes when $x=0$. If $x>1$ it
vanishes when $x=\infty$. If then $x<1$ (7) can be written
\[
z=A(x^2-1)^m+B(x^2-1)^m\int\limits_0^x\frac{dx}{(x^2-1)^{m+1}} \tag{8}
\]
and if $x>1$
\[
z=A(x^2-1)^m+B(x^2-1)^m\int\limits_x^\infty\frac{dx}{(x^2-1)^{m+1}} \tag{9}
\]
and in these forms unnecessary arbitrary constants are avoided.

From (7) we can get the general solutions of (3), (4), (5), and (6).
\[
z=A\frac{d^m(x^2-1)^m}{dx^m}+B\frac{d^m}{dx^m}\bigg[(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}\bigg] \tag{10}
\]
is the general solution of (3).
\[
z=A\frac{d^{2m}(x^2-1)^m}{dx^{2m}}+B\frac{d^{2m}}{dx^{2m}}\bigg[(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}\bigg] \tag{11}
\]
is the general solution of (4).
\[
z=A\frac{d^{m-n}(x^2-1)^m}{dx^{m-n}}+B\frac{d^{m-n}}{dx^{m-n}}\bigg[(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}\bigg] \tag{12}
\]
is the general solution of (5).
\[
z=A\frac{d^{m+n}(x^2-1)^m}{dx^{m+n}}+B\frac{d^{m+n}}{dx^{m+n}}\bigg[(x^2-1)^m\int\frac{dx}{(x^2-1)^{m+1}}\bigg] \tag{13}
\]
is the general solution of (6).

In each of these forms $A$ and $B$ are arbitrary constants and the integral is
to be taken from $0$ to $x$ if $x<1$ and from $x$ to $\infty$ if $x>1$.

Of course (10) must be identical with the forms already obtained in Arts.~16
and 18 as general solutions of Legendre's Equation.

Equation (4) is so simple that it can be solved directly, and we get its
solution in the form
\[
z=A_1+B_1\int\frac{dx}{(x^2-1)^{m+1}}  \tag{14}
\]
which must be equivalent to (11).
% -----File: 174.png

Comparing (14) with (7), the solution of (2), we see that every solution of (4)
can be obtained from a solution of (2) by dividing the latter by $(x^2 - 1)^m$, or
in other words that if we write (2)
\begin{flalign*}
&&(1 - x^{2})&\dfrac{d^2 z}{dx^2} + 2(m-1)x\dfrac{dz}{dx} + 2mz = 0,\tag{2}\\
&\text{and (4) as} &  (1 -{}&x^2)\dfrac{d^2 z_1}{dx^2} - 2(m + 1)
x\dfrac{dz_1}{dx} = 0
\qquad\qquad\hfill\phantom{and (4) as}\tag{4}
\end{flalign*}

$z = z_1 (x^2 - 1)^m$; and the substitution of this value in (2) will give (4), and
the substitution of $z_1 = \dfrac{z}{(x^2 - 1)^m}$ in (4) will give (2).

We have, then, two ways of obtaining (4) from (2); we may differentiate (2)
$2m$ times with respect to $x$, or we may replace $z$ in (2) by $z_1 (x^2 - 1)^m$.

If we use the first method we have seen that Legendre's Equation (3) is
midway between (2) and (4). That is if we differentiate (2) $m$ times we get
(3) and if we then differentiate (3) $m$ times we get (4). Let us see if the
half-way equation in our second process is Legendre's Equation.
\begin{flalign*}
&\indent\text{If}        &&z  = y (x^2 - 1)^{\tfrac{m}{2}}&&\\
&\text{and}               &&y  = z_1 (x^2 - 1)^{\tfrac{m}{2}}\hfill\phantom{and}&&\\
&                               &&z  = z_1 (x^2 - 1)^{m}.&&
\end{flalign*}

So that if in (2) we replace $z$ by $y(x^2 - 1)^{\tfrac{m}{2}}$ and then repeat the operation
on the resulting equation we shall get (4). Making the first substitution we
find,
\[\tag{15}
(1 - x^2)\dfrac{d^2 y}{dx^2} - 2x\dfrac{dy}{dx} + \biggl[m(m + 1) -
\dfrac{m^2}{1 - x^2}\biggr] y = 0,
\]
not Legendre's Equation but a somewhat more general form. Of course its solution is
\[\tag{16}
y = A(x^2 - 1)^{\tfrac{m}{2}} + B(x^2 - 1)^{\tfrac{m}{2}}\int\dfrac{dx}{(x^2 - 1)^{m+1}}.
\]

(2) and (4) are special forms of (5) and (6). Let us try the experiment of
substituting in (5) $z = y(1 - x^2)^{\tfrac{n}{2}}$ and in (6)
 $z = \dfrac{y}{(1 - x^2)^{\tfrac{n}{2}}}$.
We find that both substitutions give the same equation
\[\tag{17}
(1 - x^{2})\dfrac{d^2 y}{dx^{2}} - 2x \dfrac{dy}{dx} + \biggl[m(m + 1) -
\dfrac{n^{2}}{1 - x^{2}}\biggr] y = 0.
\]
% -----File: 175.png

The solution of (17) can be obtained from either (12) or (13) and is
\[\tag{18}
y = \dfrac{1}{(1 - x^2)^{\tfrac{n}{2}}} \biggl\lbrace A \dfrac{d^{m - n}(x^2 - 1)^m}{dx^{m-n}} +
B \dfrac{d^{m - n}} {dx^{m - n}} \biggl[(x^2 - 1)^{m} \int\dfrac{dx}{(x^2  - 1)^{m + 1}}\biggr]\biggr\rbrace
\]
or
\[\tag{19}
y = (1 - x^2)^{\tfrac{n}{2}} \biggl\lbrace A_1 \dfrac{d^{m+n} (x^2 - 1)^m}{dx^{m + n}} + B_1
\dfrac{d^{m+n}}{dx^{m + n}}\biggl[(x^2 -1)^{m} \int\dfrac{dx}{(x^2 -1)^{m+1}}\biggr]\biggr\rbrace
\]
which of course must be equivalent.

\mypara{86.}  In addition to the value of $P_{m}(x)$ given in (1) Art.~83 there is another
important derivative form which we shall proceed to obtain. It is
\[\tag{1}
P_m (\cos \theta) = \dfrac{(-1)^m}{m!} r^{m+1} D_x^m \biggl(\dfrac{1}{r}\biggr).
\]

We have seen in Art.~75 that $\dfrac{1}{r}\dfrac{1}{\sqrt{1 - 2\dfrac{r_1}{r}\cos\theta +
\dfrac{r_1^2}{r^2}}}$ can be developed into a convergent series if $r_1 <  r$ and that the
$(m + 1)$st term of that series is $\dfrac{P_m (\cos\theta) r_1^m} {r^{m+1}}$.
Let us obtain this term by Taylor's Theorem.

\begin{align*}
\dfrac{1}{r}\dfrac{1}{\sqrt{1-2\dfrac{r_1}{r} \cos\theta + \dfrac{r_1^2}{r^2}}}
&= \dfrac{1}{\sqrt{r^2-2r_1 r\cos\theta +r_1^2}} = \dfrac{1}{\sqrt{x^2+y^2+z^2-2xr_1+r_1^2}}\\
&=\dfrac{1}{\sqrt{(x-r_1)^2+y^2+z^2}}
\end{align*}

Regarding this as a function of $(x - r_1)$ and developing according to powers
of $r_1$ by Taylor's Theorem we get as the $(m + 1)$st term \\
\begin{flalign*}
&&\dfrac{(-1)^m}{m!} r_1^m D_x^m&\biggl[\dfrac{1}{\sqrt{x^2 + y^2 + z^2}}
\biggr]\quad\text{or}\quad \dfrac{(-1)^m}{m!}{ r_1^{m}} D_x^{m} \biggl(\dfrac{1}{r}\biggr).&& \\[1ex]
&\text{Hence}&&\dfrac{P_m (\cos \theta)}{r^{m + 1}} = \dfrac{(-1)^m}{m!}
D_x^m \biggl(\dfrac{1}{r}\biggr).&&\hfil\phantom{Hence}
\end{flalign*}

\mypara{87.}  We have now obtained four different forms for our \emph{zonal harmonic},
a polynomial in $x$, an expression involving cosines of multiples of $\theta$, a form
involving an ordinary $m$th derivative with respect to $x$, and a form involving
a partial $m$th derivative with respect to $x$. We shall now get a form due
to Laplace, involving a definite integral.

\[\tag{1}
\int\limits_0^\pi
\dfrac{d\phi}{a-b\cos\phi} = \dfrac{\pi}{(a^2-b^2)^{\frac{1}{2}}}
\]
if $a^2 > b^2$ [v.\ Int.\ Cal.\ page 68].
% -----File: 176.png

$\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}}$ can be expressed in the form $\dfrac{1}{(a^2-b^2)^{\frac{1}{2}}}$ by taking $a = 1 - zx$
and $b = z \sqrt{ x^2 - 1}$ and no matter what value $x$ may have $z$ can be taken so small
that $a^2$ will be greater than $b^2$. Then by (1)
\begin{align*}
% recast to fit line
\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}}&= \frac{1}{\pi} \int\limits_0^{\pi} \frac{d\phi}{1-zx-z\sqrt{  x^2 -1}.\cos \phi} \\
&= \frac{1}{\pi} \int\limits_0^{\pi} \frac{d\phi}{1-z(x+\sqrt{ x^2 -1}.\cos \phi)}\\
&= \frac{1}{\pi} \int\limits_0^{\pi} [1+(x+ \sqrt{ x^2 -1}.\cos \phi)z+(x+ \sqrt{ x^2 -1}.\cos \phi)^2 z^2\\
&\qquad\qquad+ (x+ \sqrt{ x^2 -1}.\cos \phi)^3 z^3 + \cdots ]d\phi
\end{align*}
if $z$ is taken so small that the modulus of $z(x + \sqrt{ x^2 -1}.\cos \phi)$ is less than 1. But
by Art.~77 (2) $P_m(x)$ is the coefficient of $z^m$ in the development of $\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}}$,
\begin{flalign*}
&\text{hence} &  & P_m(x)=\frac{1}{\pi} \int\limits_0^{\pi} [x+ \sqrt{ x^2 -1}.\cos\phi]^m d\phi.           \tag{2}&&\hspace{1em}
\end{flalign*}
By replacing $\phi$  by $\pi - \phi$ in (2) we get
\[
P_m(x)=\frac{1}{\pi} \int\limits_0^{\pi} [x- \sqrt{ x^2 -1}.\cos \phi]^m d\phi.        \tag{3}
\]

$\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}} = \dfrac{1}{z}\dfrac{1}{\Big(1-2x \dfrac{1}{z }+ \dfrac{1}{z^2}\Big)^{\frac{1}{2} }}$ and if mod $\dfrac{1}{z}<1$ or in other words if
mod $z>1$\;\;$\dfrac{1}{\Big(1-2x \dfrac{1}{z} + \dfrac{1}{z^2} \Big)^{\frac{1}{2}}}$ can be developed into a convergent series involving
powers of $\dfrac{1}{z}$, and the coefficient of $\Big(\dfrac{1}{z} \Big)^m$ will be $P_m(x)$; but this will be
the coefficient of $z^{-m-1}$ in the development of $\dfrac{1}{(1-2xz+z^2)^{\frac{1}{2}}}$ according to
descending powers of $z$, mod $z$ being greater than 1.

If now we let $a = zx - 1$ and $b=z\sqrt{x^2 -1}$, $a^2 - b^2 = 1 - 2xz +z^2$ and
$z$ may be taken so great that $a^2 - b^2 > 0$. Then by (1)

\begin{align*}
\frac{1}{(1-2xz+z^2)^{\frac{1}{2}}} &= \frac{1}{\pi} \int\limits_0^{\pi}\frac{ d\phi}{zx-1-z \sqrt{ x^2 -1}.\cos \phi}&\\
  ={}&\frac{1}{\pi}\int\limits_0^{\pi}\frac{ d\phi}{z(x-\sqrt{x^2-1}.\cos \phi) \Big[1- \dfrac{1}{ z(x-\sqrt{x^2-1}.\cos \phi)} \Big]}&\\
  ={}&\frac{1}{\pi}\int\limits_0^{\pi}\frac{ 1}{(x- \sqrt{ x^2 -1}.\cos \phi) }\left [z^{-1}+ \frac{1}{(x- \sqrt{ x^2 -1}.\cos \phi) } z^{-2} \right .&\\
   &  \left .  \qquad   \qquad \qquad \qquad + \frac{1}{(x- \sqrt{ x^2 -1}.\cos \phi)^2} z^{-3} + \cdots \right ] d\phi &
\end{align*}
% -----File: 177.png
and the coefficient of $z^{-m-1}$ is
$\displaystyle\frac{1}{\pi}\int\limits_0^\pi\frac{d\phi}{[x-\sqrt{x^2-1}.\cos\phi]^{m+1}}$.
\begin{flalign*}
&\text{Hence} & P_m(x)=\frac{1}{\pi}\int\limits_0^\pi\frac{d\phi}{[x-\sqrt{x^2-1}.\cos\phi]^{m+1}}. & \tag{4}&
\end{flalign*}

Replace $\phi$ by $\pi-\phi$ and we get
\[
P_m(x) =\frac{1}{\pi}\int\limits_0^\pi\frac{d\phi}{[x+\sqrt{x^2-1}.\cos\phi]^{m+1}}. \tag{5}
\]

\mypara{88.} In the problems in which we have already used \emph{Zonal Harmonics}
(v.\ Arts.~78--81) we have been able to start with the value of the Potential
Function at any point on the axis of $X$, and it has been necessary to develop
the expression for $V$ on that axis in terms of ascending or descending powers
of $x$. If, however, we start with the value of $V$ in terms of $\theta$ for some given
value of $r$, that is on the surface of some sphere, we must develop the function
of $\theta$ in terms of \emph{zonal harmonics} of $\cos\theta$ (v.\ Art.~10), and our problem becomes
the following:---To develop a given function of $\cos\theta$ in terms of zonal harmonics
of $\cos\theta$, or to develop a given function of $x$ in terms of the functions
$P_m(x)$, $x$ lying between 1 and $-1$.

The problem resembles closely that of developing in a Fourier's series,
which we have already considered at such length.
\begin{flalign*}
&\indent\text{Let} & f(x) = A_0P_0(x) + A_1P_1(x) + A_2P_2(x) + A_3P_3(x) + \cdots & \tag{1}&
\end{flalign*}
for all values of $x$ from $-1$ to 1 and let it be required to determine the
coefficients.

If $f(x)$ is single-valued and has only finite discontinuities between $x=-1$
 and $x=1$ we may proceed as in Art.~19.

Let us take $n+1$ terms of (1) and attempt to determine the coefficients.
Take $n+1$ values of $x$ at equal intervals $\Delta x$ between $x=-1$ and $x=1$
 so that $(n+2)\Delta x=2$; $f(-1+\Delta x)$, $f(-1+2\Delta x)$, $f(-1+3\Delta x)$, $\cdots$
$f[-1+(n+1)\Delta x]$ will be the corresponding values of $f(x)$. Substitute
these values in (1) and we have
\begin{equation*}\left.
\begin{aligned}
f&(-1+\Delta x) = A_0 P_0(-1+\Delta x)+A_1 P_1(-1+\Delta x)\\
&\hspace{22ex}+ A_2 P_2(-1+\Delta x)+ \cdots + A_n P_n(-1+\Delta x)\\
f&(-1+2\Delta x) = A_0 P_0(-1+2\Delta x)+A_1 P_1(-1+2\Delta x)\\
&\hspace{20ex}+ A_2 P_2(-1+2\Delta x)+ \cdots + A_n P_n(-1+2\Delta x)\\
&\vdots\hspace{8ex}\vdots\hspace{8ex}\vdots\hspace{8ex}\vdots\hspace{8ex}\vdots
\hspace{8ex}\vdots\hspace{8ex}\vdots\hspace{8ex}\vdots\hspace{8ex}\\
f&(1-\Delta x) = A_0 P_0(1-\Delta x) + A_1 P_1(1-\Delta x) + A_2 P_2(1-\Delta x) + \cdots\\
&\hspace{48ex}+ A_n P_n(1-\Delta x),
\end{aligned}\right\}\tag{2}
\end{equation*}
that is, $n+1$ equations from which in theory the $n+1$ coefficients
$A_0$, $A_1$, $\cdots$ $A_n$ can be determined.

% -----File: 178.png
Following the analogy of Art.~24 let us multiply the first equation by
$P_{m}(- 1 + \Delta x).\Delta x$, the second by $P_{m}(-1 + 2\Delta x).\Delta x$, the third by
$P_{m}(- 1 +3\Delta x).\Delta x$, \&c., and add the equations. The first member of the
resulting equation is
\[
\sum_{k=1}^{k=n+1} f(-1+k\Delta x)P_m(-1+k\Delta x).\Delta x,       \tag{3}
\]
and the coefficient of any $A$ as $A_l$ in the second member is
\[
 \sum_{k=1}^{k=n+1} P_m(-1+k\Delta x)P_l(-1+k\Delta x).\Delta x.    \tag{4}
\]
If now $n$ is indefinitely increased (3) approaches as its limiting value
\begin{flalign*}
&&&\int\limits_{-1}^1 f(x) P_m(x) dx   \tag{5}\\
&\text{and (4) approaches}&& \int\limits_{-1}^1 P_m(x) P_l(x)dx. && \phantom{and (4) approaches}\tag{6}
\end{flalign*}\\
\indent We have now to find the value of the integral (6) or as we shall write
it for the sake of greater convenience
\[
 \int\limits_{-1}^1 P_m(x) P_n(x) dx.
 \]

\mypara{89.} $\displaystyle\int\limits_{-1}^1 P_m(x) P_n(x) dx = \frac{1}{2^{m+n} m!n!} \int\limits_{-1}^1 \frac{d^m(x^2-1)^m}{dx^m}.\frac{d^n(x^2-1)^n}{dx^n}dx$
by (1) Art.~83.
\begin{flalign*}
\int\limits_{-1}^1 \frac{d^m(x^2-1)^m}{dx^m}.\frac{d^n(x^2-1)^n}{dx^n}dx &= \bigg[\frac{d^m(x^2-1)^m}{dx^m}.\frac{d^{n-1}(x^2-1)^n}{dx^{n-1}} \mathop{\bigg]}_{-1}^1 \\
  &-\int\limits_{-1}^1 \frac{d^{m+1}(x^2-1)^m}{dx^{m+1}}.\frac{d^{n-1}(x^2-1)^n}{dx^{n-1}}dx   \tag{1}
\end{flalign*}
by \textit{integration by parts}.

Now if $z = X(x^2 - 1)^n$
\begin{align*}
% recast to fit line
\frac{dz}{dx} &= 2nxX(x^2-1)^{n-1} + (x^2-1)^n \frac{dX}{dx} \\
              &= (x^2-1)^{n-1}  \left [2nxX+(x^2-1)\frac{dX}{dx} \right ].    \tag{2}
\end{align*}

Hence the $p$th derivative with respect to $x$ of any function of $x$ containing
 $(x^2- 1)^n$ as a factor will contain $(x^2- 1)^{n-p}$ as a factor if $p < n$.

% -----File: 179.png
$\dfrac{d^{n-1} (x^2 -1)^n }{ dx^{n-1}}$, then, contains $(x^2 - 1)$ as a factor and is zero when $x = 1$
 and when $x = -1$, so that (1) reduces to
\[
\int\limits^1_{-1}\dfrac{d^m (x^2 -1)^m }{ dx^m}.\dfrac{d^n (x^2 -1)^n}{dx^n} dx = - \int\limits^1_{-1} \dfrac{d^{m+1} (x^2 -1)^m }{dx^{m+1}}.\dfrac{ d^{n-1} (x^2 -1)^n }{ dx^{n-1}} dx.
\]
It follows that
\begin{align*}
% recast to fit line
\int\limits^1_{-1} \dfrac{d^m (x^2 -1)^m}{dx^m}.\dfrac{d^n (x^2 -1)^n }{ dx^n} &dx \\
={}&(-1)^p \int \limits^1_{-1} \dfrac{d^{m+p} (x^2 -1)^m }{ dx^{m+p}}.\dfrac{ d^{n-p} (x^2 -1)^n }{ dx^{n-p} }dx\\
={}&(-1)^p \int \limits^1_{-1}          \dfrac{d^{m-p} (x^2 -1)^m }{dx^{m-p}}.\dfrac{d^{n+p}(x^2 -1)^n}{dx^{n+p}} dx. \tag{3}
\end{align*}

If $m < n$ we get from (3)
\begin{align*}
\int\limits^1_{-1} \dfrac{d^m (x^2 -1)^m }{ dx^m}.\dfrac{d^n (x^2 -1)^n }{dx^n} dx &= (-1)^m \int\limits^1_{-1} \dfrac{d^{2m} (x^2 -1)^m }{ dx^{2m} }.\dfrac{d^{n-m} (x^2 -1)^n }{ dx^{n-m}} dx\\
                &= (-1)^m (2m)! \bigg[ \dfrac{d^{n-m-1} (x^2 -1)^n}{ dx^{n-m-1} }\mathop{\bigg]}^1_{-1} = 0,\\[-6ex]
\end{align*}
\begin{flalign*}
&\text{since}&& \dfrac{d^{2m} (x^2 -1)^m }{ dx^{2m}} = (2m)!.&\phantom{since}
\end{flalign*}

If $m > n$
\begin{align*}
\int \limits^1_{-1} \dfrac{d^m (x^2 -1)^m }{dx^m }.\dfrac{d^n (x^2 -1)^n }{ dx^n}dx &= (-1)^n \int \limits^1_{-1} \dfrac{d^{m-n} (x^2 -1)^m }{ dx^{m-n}}.\dfrac{d^{2n} (x^2 -1)^n }{ dx^{2n}} dx\\
                      &= (-1)^n (2n)! \bigg[ \dfrac{d^{m-n-1} (x^2 -1)^m }{dx^{m-n-1}} \mathop{\bigg]}^1_{-1} = 0.
\end{align*}

If, then, $m$ is not equal to $n$
\[
\int\limits^1_{-1} P_m (x) P_n (x) dx = 0.       \tag{4}
\]
If $m = n$ we have to find $\displaystyle \int\limits^1_{-1} [P_m (x)]^2 dx$.
\[
\int\limits^1_{-1} [P_m (x)]^2 dx = \dfrac{1}{2^{2m} (m!)^2} \int\limits^1_{-1} \dfrac{ d^m (x^2 -1)^m }{ dx^m}.\dfrac{ d^m (x^2 -1)^m }{ dx^m} dx.
\]
\begin{flalign*}
&&\int\limits^1_{-1} \dfrac{d^m (x^2 -1)^m }{dx^m }.\dfrac{d^m (x^2 -1)^m }{ dx^m} dx &= (-1)^m \int\limits^1_{-1} \dfrac{ d^{2m} (x^2 -1)^m }{ dx^{2m}}.(x^2 - 1)^m dx&&\\
&\multispan{2}{by (3),\hfill}&= (-1)^m (2m)! \int\limits^1_{-1}  (x^2 - 1)^m dx.&&
\end{flalign*}

% -----File: 180.png
\begin{align*}
\int\limits_{-1}^1(x^2 -1)^m dx &= \int\limits_{-1}^1 (x-1)^m (x+1)^m dx = -\frac{m}{m+1} \int\limits_{-1}^1 (x-1)^{m-1} (x+1)^{m+1} dx\\
&= (-1)^m \frac{m!}{(m+1)(m+2) \cdots 2m} \int\limits_{-1}^1 (x+1)^{2m} dx\\
&= (-1)^m \frac{2^{2m+1} m!}{(m+1)(m+2) \cdots (2m+1)}.
\end{align*}
\begin{flalign*}
&\text{Hence} &\int\limits_{-1}^1 [P_m (x)]^2 dx &= \frac{1}{2^{2m} (m!)^2}\frac{ (-1)^m (2m)!(-1)^m m!2^{2m+1}}{(m+1)(m+2) \cdots (2m+1)} &&\phantom{Hence}\\
&\text{or}  && \int\limits_{-1}^1 [P_m(x)]^2 dx = \frac{2}{2m+1}.        \tag{5}&&
\end{flalign*}

\mypara{90.} The solution of the problem in Art.~88 is now readily obtained, and
we have
\begin{flalign*}
&&f(x) = &A_0P_0(x) + A_1P_1(x) + A_2P_2(x) + \cdots     \tag{1} &&\\
&\text{where}&&A_m = \frac{2m+1}{2 }\int\limits_{-1}^1 f(x)P_m(x)dx.          \tag{2} &&\hspace{1em}
\end{flalign*}

The function and the series are equal for all values of $x$ from $x = -1$ to
 $x = 1$, and $f(x)$ is subject to no conditions save those which would enable us
to develop it in a Fourier's Series. [v.\ Chapter III.]

Of course (1) can be written
\begin{flalign*}
&&f(\cos \theta) &= A_0P_0(\cos \theta) +A_1P_1(\cos \theta) + A_2P_2(\cos \theta) + \cdots &&\\
&\text{where} && A_m = \frac{2m+1}{2} \int\limits_{-1}^1 f(\cos \theta)P_m(\cos \theta)d(\cos \theta)     && \phantom{where}
\intertext{or if $f(\cos \theta) = F(\theta)$}
&&F(\theta) = &A_0P_0(\cos \theta) + A_1P_1(\cos \theta) + A_2P_2(\cos \theta) + \cdots    \tag{3} \\
&\text{where}  &&  A_m = \frac{2m+1}{2} \int\limits_{0}^{\pi} F(\theta)P_m(\cos \theta)\sin\theta.d\theta  &&      \tag{4}
\end{flalign*}
and the development holds good from $ \theta = 0$ to $\theta = \pi$.

If $f(x)$ is an even function, that is, if $f(-x) =f(x)$ (1) and (2) can be
somewhat simplified.  For in that case it can be easily shown (v.\ Art.~77) that
\begin{flalign*}
&&\int\limits_{-1}^1 f(x)&P_{2k}(x)dx = 2 \int\limits_0^1 f(x)P_{2k}(x)dx,\\
% -----File: 181.png
&\text{and that}&  &\int\limits_{-1}^1 f(x)P_{2k+1} (x)dx = 0; && \phantom{and that}
\intertext{so that if $f(-x) =f(x)$}
&&f(x) = A_0P_0(x) &+A_2P_2(x) + A_4P_4(x) + A_6P_6(x) + \cdots  \tag{5} \\
&\text{where}&  A_{2k} &= (4k+1)\int\limits_0^1  f(x)P_{2k}(x)dx.   &&    \tag{6}
\end{flalign*}

If $f(x)$ is an odd function, that is, if $f(-x) = -f(x)$ it can be shown in
like manner that
\begin{flalign*}
&&f(x) = A_1P_1&(x) +A_3P_3(x) + A_5P_5(x) + A_7P_7(x) + \cdots  &&\tag{7}\\
&\text{where}&& A_{2k+1} = (4k+3) \int\limits_0^1 f(x)P_{2k+1}(x)dx.     && \phantom{where}   \tag{8}
\end{flalign*}

If it is only necessary that the development should hold for $0 < x < 1$ any
function may be expressed in form (5) or (7) at pleasure.

\mypara{91.}  We can establish the fact that $\displaystyle\int\limits_{-1}^1 P_m(x)P_n(x)dx = 0$ by a more general
method than that used in Art.~89.

Let $X_m$ be any solution of Legendre's Equation
\begin{flalign*}
&\hspace{6em}&&\frac{d}{dx} \left [ (1-x^2)\frac{ dz}{dx} \right ] + m(m+1)z = 0 && \tag*{[v.\ (1) Art.~16].}
\intertext{which with its first derivative with respect to $x$ is finite, continuous, and
single-valued for values of $x$ between $-1$ and $1$, $-1$ and $1$ being included.}
&\text{\indent Then}&&\frac{d}{dx} \left [ (1-x^2) \frac{dX_m }{ dx} \right  ] + m(m+1)X_m = 0  \tag{1}&&\\
&\text{and}&&\frac{d}{dx} \left [ (1-x^2) \frac{dX_n }{ dx} \right  ] + n(n+1)X_n = 0  \tag{2}&&\phantom{and}
\end{flalign*}

Multiply (1) by $X_n$ and (2) by $X_m$ and subtract and integrate and we get
\begin{multline*}
[m(m+1)-n(n+1)] \int\limits_{-1}^1 X_mX_n dx = \int\limits_{-1}^1 X_m \frac{d}{dx} \left  [ (1-x^2) \frac{dX_n }{ dx} \right ] dx\\
-	\int\limits_{-1}^1 X_n \frac{d}{dx} \left [ (1-x^2)\frac{ dX_m }{ dx} \right  ] dx.
\end{multline*}
% -----File: 182.png

Integrate by parts,
\begin{multline*}
[m(m + 1) - n(n + 1)] \int\limits_{-1}^1 X_m X_n dx =  \bigg[X_m(1-x^2)\frac{dX_n}{dx} - X_n(1-x^2)\frac{dX_m}{dx} \!\!\!\!\!\mathop{\bigg]}_{x=-1}^{x=1}  \\
 -\int\limits_{-1}^1 (1-x^2)\frac{dX_n}{dx}\frac{ dX_m}{dx} dx + \int\limits_{-1}^1 (1-x^2)\frac{dX_m}{dx} \frac{dX_n}{dx} dx. \tag{3}\\[-5ex]
\end{multline*}
\begin{flalign*}
&\text{Whence}&&\int\limits_{-1}^1 X_m X_n dx = 0 && \phantom{Whence}     \tag{4}\\[-5ex]
\end{flalign*}
unless $m = n$.

(3) gives at once the important formula
\[
\int\limits_x^1X_mX_ndx =\frac{ (1-x^2) \Big [X_n \dfrac{dX_m}{dx} - X_m\dfrac{dX_n}{dx} \Big ]}{m(m + 1) - n(n + 1) }   \tag{5} \\
\]
from which come as special cases
\[
\int\limits_x^1P_m(x)P_n(x)dx = \frac{(1-x^2) \Big [P_n(x)\dfrac{dP_m(x)}{dx} - P_m(x)\dfrac{dP_n(x)}{dx} \Big ] }{ m(m + 1) - n(n + 1)}   \tag{6}
\]
and since $P_0(x) = 1$
\[
\int\limits_x^1P_m(x)dx = \frac{(1-x^2)\dfrac{dP_m(x)}{dx}}{ m(m + 1)},    \tag{7}\\
\]
unless $m = 0$.

\EXAMPLE{S}

1.\quad Show that
\begin{align*}
\int\limits_0^1 P_m(x)dx &= 0 \quad\text{ if $m$ is even and is not zero.}\\
&=(-1)^{\frac{m-1}{2}}\frac{ 1}{m(m+1)}\frac{ 3.5.7.\cdots m }{ 2.4.6.\cdots (m-1) } \quad\text{if  $m$ is odd. }
\end{align*}
v.\ Art.~91 (7) and Art.~77 (10).\\

2.\quad  Show that
\begin{multline*}
\int\limits_0^1P_m(x)P_n(x)dx = 0 \quad \text{if $m$ and $n$ are both even or both odd.}\\
 = (-1)^{\frac{m+n+1}{2}}  \frac{m!\;n!}{ 2^{m+n-1} (m-n)(m+n+1) \left (\dfrac{m}{2} ! \right )^2 \Big (\dfrac{n-1}{2} ! \Big  )^2}
 \end{multline*}
 if $m$ is even and $n$ odd. v.\ Art.~91 (6) and Art.~77 (8), (9), and (10). cf.\ J.~W. Strutt
(Lord Rayleigh) Lond.\ Phil.\ Trans.\ 1870, page 579.\\

3.\quad  Show that \quad $\displaystyle\int\limits_0^1[P_m(x)]^2dx = \dfrac{1}{2m+1}$ \quad v.\ Art.~89 (5).

% -----File: 183.png
\mypara{92.}  Formula (4) Art.~91 can be obtained directly from Laplace's Equation
by the aid of \emph{Green's Theorem} (v.\ Peirce's Newt.\ Pot.\ Func.\ \S~48).

Take the special form of Green's Theorem, [(148) \S~48 Peirce's Newt.\ Pot.\
Func.]
\[
\iiint(U\nabla^2V - V\nabla^2U)dxdydz = \int (UD_n V - VD_n U)ds \tag{1}
\]
where $\nabla^2$ stands for $(D_x^2 + D_y^2 + D_2^2)$, $D_n$ is the partial derivative along the
external normal, and the left-hand member is the space-integral through the
space bounded by any closed surface, and the right-hand member is the surface
integral taken over the same surface.  (v.\ Int.\ Cal.\ Chapter XIV.)

If $U$ and $V$ are solutions of Laplace's Equation $\nabla^2 V = \nabla^2 U = 0$ and (1)
reduces to
\[
\int(UD_n V - V D_n U) ds = 0.  \tag{2}
\]

Now $r^mX_m$ and $r^nX_n$ are solutions of Laplace's Equation if $x = \cos \theta$
 (v.~Art.~16).

If the unit sphere is taken as the bounding surface and $U = r^mX_m$ and
 $V = r^nX_n$ (1) and (2) will hold good.
\begin{gather*}
D_nU = D_r(r^mX_m) = mr^{m-1}X_m,\\
D_nV = nr^{n-1}X_n,\\
ds = \sin \theta.d\theta d\phi,
\end{gather*}
and (2) becomes  $\displaystyle \int\limits^{2\pi}_0 d\phi \int\limits^{\pi}_0 (n X_m X_n - m X_m X_n) \sin \theta.d \theta = 0$
\begin{flalign*}
&\text{or}&  2\pi(n - m) \int\limits^{\pi}_0  X_m X_n \sin \theta.d \theta &= 0. &      \tag{3}
\end{flalign*}
Since $x = \cos\theta$, $\sin\theta.d\theta = - dx$ and (3) reduces to
\[
\int\limits^1_{-1} X_m X_n dx = 0\tag{4}\footnotemark
\]
unless $m = n$.

\footnotetext{It should be noted that this proof is no more general than that of the last article, for, in
order that Green's Theorem should apply to $r^mX_m$ this function and its first derivatives must
be finite continuous and single-valued within and on the surface of the unit sphere. (v.\ Peirce,
Newt.\ Pot.\ Func.\ \S~48.) }

\mypara{93.}  We can now solve completely the problem of Art.~10 which was in
that article carried to the point where it was only necessary to develop a
certain function of $\theta$ in the form
\[
A_0P_0(\cos\theta) + A_1P_1(\cos\theta) + A_2P_2(\cos\theta) + \cdots\\[-2ex]
\]
% -----File: 184.png
\begin{flalign*}
&\text{given that}& f(\theta) = 1 \quad\text{from}\quad\theta = 0\quad\text{to}\quad\theta = \dfrac{\pi}{2} &&\phantom{given that}\\
&\text{and}       & f(\theta) = 0 \quad\text{from}\quad\theta = \dfrac{\pi}{2}\quad\text{to}\quad\theta = \pi. &&
\end{flalign*}

This amounts to the same thing as developing $F(x)$ into the series
\[
F(x) = A_0 P_0(x) + A_1 P_1(x) + A_2 P_2(x) + A_3 P_3(x) + \cdots\\[-2ex]
\]
\begin{flalign*}
&\text{where} &&F(x) = 0\quad\text{from}\quad x = -1\quad\text{to}\quad x = 0&&\phantom{where}\\
&\text{and}   &&F(x) = 1\quad\text{from}\quad x =  0\phantom{-}\quad\text{to}\quad x = 1.&&
\end{flalign*}
By Art.~90 (1) and (2)
\[
A_0 = \dfrac{1}{2} \int\limits^1_0  P_0(x)dx = \dfrac{1}{2} \int\limits^1_0 dx = \dfrac{1}{2} ,
\]
\begin{flalign*}
&\text{and any coefficient}&&A_m = \dfrac{(2m + 1)}{2} \int\limits^1_0 P_m(x)dx. &&\phantom{and any coefficient}
\end{flalign*}
By Art.~91, Ex.~1
\[
\int\limits^1_0 P_m(x)dx  = 0\quad  \text{ if $m$ is even}\rule{14.7em}{0em}\\[-5ex]
\]
\begin{flalign*}
&&                                             &= (-1)^{\tfrac{m-1}{2}} \dfrac{1}{m(m+1)}\dfrac{ 3.5.7.\cdots m}{2.4.6.\cdots (m-1)} \quad \text{if $m$ is odd.}&\\
&\text{\indent Hence}& A_m &= 0\quad \text{ if $m$ is even}&\\
&&                                              &= (-1)^{\tfrac{m-1}{2}} \dfrac{2m+1}{2m+2}.\dfrac{ 1.3.5.\cdots(m-2)}{2.4.6.\cdots (m-1)}\quad \text{ if $m$ is odd}.&\\
&\text{\indent Then}&  F(x)    &= \dfrac{1}{2} + \dfrac{3}{4} P_1 (x) - \dfrac{7}{8}.\dfrac{1}{2}  P_3(x) + \dfrac{11}{12}.\dfrac{1.3}{2.4} P_5(x) -\cdots &\tag{1}
\end{flalign*}
and \quad $u = \dfrac{1}{2} + \dfrac{3}{4} r P_1(\cos\theta) - \dfrac{7}{8}.\dfrac{1}{2} r^3 P_3 (\cos\theta) + \dfrac{11}{12}.\dfrac{1.3}{2.4} r^5 P_5(\cos \theta) + \cdots$\hfill(2)\\[1ex]
for any point within the sphere.

\mypara{94.} If in a problem on the Potential Function the value of $V$ is given at
every point of a spherical surface and has circular symmetry\footnote{See note on page \pageref{notep12}.} about a diameter
of that surface the value of $V$ at any point in space can be obtained.

We have to solve Laplace's Equation in the form
\[
r D_r^2(rV) + \dfrac{1}{\sin\theta} D_\theta (\sin\theta D_\theta V) = 0     \tag{1}
\]
% -----File: 185.png
subject to the conditions
\begin{alignat*}{3}
&V=f(\theta)\quad && \text{when}\quad &&r=a\\
&V=0 &&\quad  \text{``}  &&r=\infty.
\end{alignat*}
\begin{flalign*}
&\text{We have}& f(\theta) &= A_0 P_0(\cos\theta) + A_1 P_1(\cos\theta) + A_2 P_2(\cos\theta) +\cdots&&\\
&\text{where}&& A_m = \frac{(2m+1)}{2} \int\limits_0^{\pi} f(\theta) P_m(\cos\theta) \sin\theta.d\theta. &&  \tag*{v.\ Art.~90 (4).}
\end{flalign*}
Hence
\[
V=A_0 + A_1 \left (\frac{r}{a} \right )P_1(\cos\theta) + A_2 \left (\frac{r}{a} \right )^2 P_2(\cos\theta) + A_3 \left (\frac{r}{a} \right )^3 P_3(\cos\theta) + \cdots  (2)
\]
is the required solution for a point within the sphere, and
\[
V=A_0\left (\frac{a}{r} \right ) + A_1 \left (\frac{a}{r} \right ) ^2P_1(\cos\theta) + A_2 \left (\frac{a}{r} \right ) ^3 P_2(\cos\theta) + A_3 \left (\frac{a}{r} \right ) ^4 P_3(\cos\theta) + \cdots   \tag{3}
\]
is the required solution for an external point.

\EXAMPLE{S}

1.\quad If on the surface of a sphere of radius $c$ $V$ is constant and equal to a
show that $V=a$ for any point within the sphere and $V=\dfrac{ac}{r}$ for any
external point.\\

2.\quad Two equal thin hemispherical shells of radius $c$ placed together to form
a spherical surface are separated by a thin non-conducting layer. Charges of
statical electricity are placed on the two hemispheres one of which is then
found to be at potential $a$ and the other at potential $b$. Find the value of the
potential function at any point.
\begin{multline*}
V= \frac{a+b}{2} + (b-a) \left [\frac{3}{4} \frac{r}{c} P_1(\cos\theta)- \frac{7}{8}.\frac{ 1}{2}\frac{ r^3}{c^3} P_3(\cos\theta) \right.\\
        \left .   + \frac{11}{12}.\frac{1.3}{2.4}\frac{ r^5}{c^5} P_5(\cos\theta)- \cdots \right ]
\end{multline*}
for an internal point
\begin{multline*}
V= \frac{a+b}{2}.\frac{c}{r} + (b-a) \left [\frac{3}{4} \frac{c^2}{r^2} P_1(\cos\theta)- \frac{7}{8}.\frac{ 1}{2}\frac{ c^4}{r^4} P_3(\cos\theta) \right.\\
        \left .   + \frac{11}{12}.\frac{1.3}{2.4}\frac{ c^6}{r^6} P_5(\cos\theta)- \cdots \right ]
\end{multline*}
for an external point.\\

% -----File: 186.png
3.\quad If $V_1 =f(\cos\theta)$ when $r = a$ and $V_1 = 0$ when $r = b$ show that for
 $a < r < b$
\begin{flalign*}
&&V_1 = \sum_{m=0}^{m=\infty} &A_m \left  ( \frac{b^{m+1}}{r^{m+1}} - \frac{r^m}{b^m} \right ) \left (\frac{b^{m+1}}{a^{m+1}} -\frac{ a^m}{b^m} \right )^{-1} P_m(\cos\theta)\\
&\text{where} &&A_m = \frac{2m+1}{2} \int\limits_{-1}^1 f(x)P_m(x)dx.&& \phantom{where}
\end{flalign*}

4.\quad If $V_2 =F(\cos\theta)$ when $r = b$ and $V_2 = 0$ when $r = a$ show that for
 $a < r < b$
\begin{flalign*}
&&V_2 = \sum_{m=0}^{m=\infty} &B_m \left  ( \frac{r^{m}}{a^{m}} - \frac{a^{m+1}}{r^{m+1}} \right ) \left (\frac{b^{m}}{a^{m}} -\frac{ a^{m+1}}{b^{m+1}} \right )^{-1} P_m(\cos\theta)\\
&\text{where} &&B_m = \frac{2m+1}{2} \int\limits_{-1}^1 F(x)P_m(x)dx.&& \phantom{where}
\end{flalign*}

5.\quad If the value of the potential function is given arbitrarily on the surfaces
of a spherical shell but has circular symmetry\footnote{See note on page 12} about a diameter $V= V_1 + V_2$
 (v.\ Exs.~3 and 4).\\

6.\quad Two concentric hollow spherical conductors are insulated and charged.
The inner one of radius $a$ is at potential $p$, and the outer one of radius $b$ is at
potential $q$. Find $V$ for any point in space.
\begin{align*}
V&=p \quad\text{if}\quad  r < a,\\
V&= \frac{pa}{b-a}\bigg  ( \frac{b}{r}-1 \bigg ) +\frac{ qb}{b-a} \bigg (1- \frac{a}{r} \bigg ) \quad\text{if} \quad  a<r<b,\\
V&=\frac{qb}{r} \quad\text{if} \quad r>b.
\end{align*}

7.\quad If $V=0$ on the base of a hemisphere and $V=f(\cos\theta)$ on the convex
surface, show that for a point within the hemisphere
\begin{flalign*}
&&V={}&\sum_{k=0}^{k=\infty} A_{2k+1}\bigg  (\frac{r}{a} \bigg )^{2k+1} P_{2k+1} (\cos\theta)\\
&\text{where}& A_{2k+1} &= (4k + 3) \int\limits_0^1 f(x)P_{2k+1}(x)dx  \tag*{ [v.\ Art.~90 (8)].}&&
\end{flalign*}

8.\quad If the convex surface of a solid hemisphere of radius $a$ is kept at the
constant temperature unity and the base at the constant temperature zero
show that after the permanent state of temperatures is set up the temperature
of any internal point is
\[
u = \frac{3}{2} \frac{ r}{a} P_1(\cos\theta) - \frac{ 7}{4}.\frac{1}{2} \frac{ r^3}{a^3} P_3(\cos\theta) +  \frac{11}{6}.\frac{ 1.3}{2.4 } \frac{r^5}{a^5} P_5(\cos\theta) - \cdots
\]

% -----File: 187.png
9.\quad A sphere of radius $a$ and with blackened surface is exposed to the direct
rays of the sun in air at the temperature zero. Find the \textit{stationary temperature}
of any internal point.\\

\textit{Suggestion:}\quad $D_ru + hu - Mf(\theta) = 0$ \quad when \quad $r = a$.
\begin{flalign*}
&\text{Let} &u= \sum A_m\frac{ r^m }{a^m} P_m(\cos\theta), \quad \text{and}\quad  f(\theta) = \sum  B_m P_m(\cos\theta).&&
\end{flalign*}
Then we have
\begin{flalign*}
&&\sum m \frac{A_m}{a} P_m(\cos\theta) + h \sum &A_m P_m (\cos\theta) - M \sum B_m P_m (\cos\theta) = 0,\\
&\multispan{2}{whence\hfill}&A_m = \frac{MB_m }{ h+ \dfrac{m}{a}}. &&\phantom{whence}
\end{flalign*}

Here $f(\theta) = \cos\theta$ if $0 < \theta <\dfrac{\pi}{2}$ and $f(\theta) = 0$ if $\dfrac{\pi}{2} < \theta < \pi$.
\begin{multline*}
f(\theta) = \frac{1}{4} + \frac{1}{2} P_1(\cos\theta) + \frac{5}{16} P_2(\cos\theta) - \frac{3}{32} P_4(\cos\theta) + \cdots\\
+ (-1)^{k+1}\frac{ (4k+1)(2k)!}{(4k+4)(2k-1)2^{2k} (k!)^2} P_{2k} (\cos \theta) + \cdots
\end{multline*}
v.\ Art.~91 Exs.~(2) and (3). cf. J.~W. Strutt (Lord Rayleigh), Lond.\ Phil.\
Trans.\ vol.\ 160, page 587.

\mypara{95.} The formulas of Art.~90 enable us to develop a given function of $x$ in
terms of \textit{Zonal Surface Harmonics}, the development holding true for values of
$x$ between $ -1$ and $+1$. If, however, we can show by outside considerations
that a given function of $x$ can be expressed in Zonal Surface Harmonics, the
development holding true for all values of $ x$, the formulas of Art.~90 will give
us the development in question.

For example if $n$ is a positive integer $x^n$ can be expressed in terms of Zonal
Surface Harmonics no matter what the value of $x$, and no Harmonic of higher
order than $n$ will enter. For the formulas giving the values of $P_1(x)$, $P_2(x), \cdots
P_n(x)$ (v.\ Art.~77) may be regarded as $n$ algebraic equations of the first degree
in terms of $x$, $x^2$, $x^3, \cdots  x^n$ and $P_1(x)$, $P_2(x), \cdots P_n(x)$.

From these equations the $n -1$ quantities $x$, $x^2$, $x^3, .\cdots x^{n-1}$, can be eliminated,
and there will result an equation of the first degree in $x^n$ and $P_1(x)$,
$P_2(x), \cdots P_n(x)$, which will enable us to express $x^n$ in the form
\[
A_0 +A_1 P_1(x) + A_2 P_2(x) + \cdots + A_n P_n(x),
\]
no matter what the value of $x$, and we shall have the same formula when
 $-1 < x < 1$ as when $x > 1$ or $x < -1$.

% -----File: 188.png
Let us obtain this development. By Art.~90 (1) and (2)
\begin{flalign*}
&&x^n &= A_0 P_0(x) + A_1 P_1(x) + A_2 P_2(x) + \cdots\tag{1}\\
&\text{where}&   &\qquad A_m = \dfrac{2m + 1}{2}\int\limits^1_{-1} x^n P_m(x)dx. &\phantom{where}\tag{2}\\
&\text{Then}\hspace{2em}&    A_m &= \dfrac{2m + 1}{2}\dfrac{ 1}{2^m m!} \int\limits^1_{-1}  x^n \dfrac{d^m(x^2-1)^m }{dx^m} dx &   \tag*{by (1) Art.~83.}
\end{flalign*}

By \emph{integration by parts} we get
\begin{alignat*}{2}
\int\limits^1_{-1} x^n \dfrac{d^m(x^2-1)^m}{dx^m} dx &\multispan{3}{$\displaystyle{}= n(n-1)(n-2) \cdots(n-m+1) \int\limits^1_{-1} x^{n-m} (1-x^2)^m dx,$} \\[-2ex]
    &&& \text{if}\quad m < n + 1,\tag{3}\\
&=0  \quad &&\text{if}\quad m > n.
\end{alignat*}

By \emph{integration by parts} we readily obtain the reduction formula
\begin{flalign*}
&&\int\limits^1_{-1}x^p(1 - x^2)^q dx &= \dfrac{2q}{p+1} \int\limits^1_{-1} x^{p+2} (1 -x^2)^{q-1} dx  &   \text{whence}\\[-5ex]
\end{flalign*}
\[
\int\limits^1_{-1} x^{n-m} (1- x^2)^m dx = \dfrac{2^m m!}{(n - m + 1)(n - m + 3)\cdots (n + m - 1)} \int\limits^1_{-1} x^{n+m} dx.\\[-5ex]
\]
\begin{flalign*}
\int\limits^1_{-1} x^{n+m} dx &=  \dfrac{2}{(n + m + 1)}\quad \text{if}\quad n + m \quad\text{is even,}\\
                                                   &= 0\quad\text{if}\quad n + m \quad\text{is odd.}
\end{flalign*}
\begin{flalign*}
&\text{Hence} &A_m &= \dfrac{(2m + 1)n(n - 1)(n - 2)\cdots (n - m + 1) }{(n - m + 1)(n - m + 3) (n - m + 5) \cdots (n + m + 1)}&\phantom{hence}\\[1ex]
&&& \hspace{6em}\text{if } m < n + 1 \text{ and } m + n \text{ is even,}\\[1ex]
&&&= 0 \quad \text{if } m > n \text{ or if } m + n \text{ is odd.}
\end{flalign*}

Therefore
\begin{align*}
&\rlap{$\displaystyle x^n = \dfrac{n! }{1.3.5 \cdots (2n+1)}\left [(2n+1) P_n(x) + (2n-3) \dfrac{(2n+1)}{2} P_{n-2} (x)\right.$}\\
&&&+ (2n-7) \dfrac{(2n+1)(2n-1)}{2.4 } P_{n-4}(x)\\
&\hspace{7em}&&+ \left.(2n-11) \dfrac{(2n+1)(2n-1)(2n-3)}{2.4.6 } P_{n-6}(x) + \cdots \right]  && \tag{4}
\end{align*}
the second member ending with the term $\dfrac{1}{n+1} P_0(x)$ if $n$ is even and with
the term $\dfrac{3}{n+2} P_1(x)$ if $n$ is odd.
% -----File: 189.png

For convenience of reference we write out a few powers of $x$.
\[ \left . \begin{aligned}
x^0 &= 1 = P_0(x)\\
x &= P_1(x)\\
x^2&=\frac{2}{3} P_2(x)+ \frac{1}{3}P_0(x)\\
x^3&=\frac{2}{5} P_3(x) + \frac{3}{5}  P_1(x)\\
x^4 &= \frac{8}{35} P_4(x) + \frac{4}{7 }P_2(x)+ \frac{1}{5} P_0(x)\\
x^5 &= \frac{8}{63} P_5(x) + \frac{4}{9} P_3(x) + \frac{3}{7}P_1(x)\\
x^6 &= \frac{16}{231} P_6(x) +\frac{24}{77} P_4(x) +\frac{10}{21 }P_2(x) +\frac{1}{7} P_0(x)\\
x^7 &= \frac{16}{429} P_7(x) +\frac{8}{39} P_5(x) +\frac{14}{33} P_3(x) +\frac{1}{3 }P_1 (x)\\
x^8 &= \frac{128}{6435} P_8(x) + \frac{64}{495} P_6(x) +\frac{48}{143} P_4(x) + \frac{40}{99 }P_2 (x) + \frac{1}{9} P_0(x).\\
\end{aligned}
\right  \}
\tag{ 5}
\]

If a given function of $x$ can be expressed as a \textit{terminating power series} it can
be developed into a Zonal Harmonic Series by the aid of (4). Given that
\begin{flalign*}
&&f(x) &= a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots,\\[1ex]
&\text{let} &f(x) = B_0 &+ B_1P_1(x) + B_2P_2(x) + B_3P_3(x) + \cdots ;&&\phantom{let}
\end{flalign*}
then picking out carefully the coefficient of $P_m(x)$ we have
\begin{multline*}
B_m = \frac{m! }{ 1.3.5.\cdots (2m - 1)} \left  [ a_m + \frac{(m+1)(m+2)}{ 2.(2m+3)} a_{m+2} \right .\\
\left . + \frac{(m + 1)(m + 2)(m + 3)(m + 4) }{2.4.(2m + 3)(2m + 5)}  a_{m+4} + \cdots \right ].  \tag{6}
\end{multline*}

\mypara{96.} The development of $\dfrac{dP_n(x) }{dx}$ is useful and is easily obtained.
\begin{flalign*}
&\text{\indent Let}&\frac{dP_n(x) }{dx} = &A_0P_0(x) + A_1P_1(x) + A_2P_2(x) + \cdots &&\phantom{\indent Let}\\
&\text{Then}& A_m &= \frac{2m+1 }{2} \int\limits_{-1}^1 P_m(x)\frac{ dP_n(x) }{dx} dx  &&      \tag{1}
\end{flalign*}
by Art.~90 (2);
\[
\int\limits_{-1}^1 P_m(x) \frac{dP_n(x)}{dx } dx = \Big[P_m(x)P_n(x)\Big]_{x=-1}^{x=1}-\int\limits_{-1}^1 P_n(x) \frac{dP_m(x) }{dx} dx.    \tag{2}
\]
% -----File: 190.png
\begin{align*}
\Big[P_m(x)P_n(x)\Big]_{x=-1}^{x=1}& =0 \quad \text{if $m + n$ is even}\\
                 & =2 \quad \text{if $m + n$ is odd.}
\end{align*}
Since $P_n(x)$ is an algebraic polynomial of the $n$th degree in $x$, $\dfrac{dP_n(x)}{dx}$ is an
algebraic polynomial of the $n-1$st degree in $x$. Therefore in (1) $m$ is less
than $n$; consequently $\dfrac{dP_m(x)}{dx}$ is an algebraic polynomial in $x$ of lower degree
than $n$ and
\begin{flalign*}
&\phantom{by Art. 95 (3).} &&\int\limits_{-1}^1 P_n(x) \frac{dP_m(x)}{dx} dx = 0   &&  \tag*{by Art.~95 (3).}
\end{flalign*}
\begin{flalign*}
&\text{We get then}&A_m & = 2m+1 \quad \text{if $m+n$ is odd and $m<n$,}&&\phantom{We get then}\\[1ex]
                      &&& = 0 \quad \text{if $m+n$ is even or $m > n-1$;} &&\tag*{and}
\end{flalign*}
\[
\frac{dP_n(x)}{dx }= (2n-1)P_{n-1}(x) + (2n-5)P_{n-3}(x) + (2n-9)P_{n-5}(x) + \cdots \tag{3}
\]
the second member ending with the term $3P_1(x)$ if $n$ is even and with the
term $P_0(x)$ if $n$ is odd.

From (3) a number of simple formulas are readily obtained. For example
\begin{gather*}
\frac{dP_{n+1}(x)}{dx} - \frac{dP_{n-1}(x)}{dx} =(2n+1)P_n(x)     \tag{4}\\
\int\limits_x^1P_n(x) dx = \frac{1}{2n+1} [P_{n-1}(x)-P_{n+1}(x)].   \tag{5}\\
(2n+1)x \frac{dP_n(x)}{dx} = n \frac{dP_{n+1}(x)}{dx} + (n+1) \frac{dP_{n-1}(x)}{dx }   \tag{6}
\intertext{[v.\ (4) and Article 77 (12)].}
(x^2-1) \frac{dP_n(x)}{dx} = nxP_n(x)-nP_{n-1}(x)          \tag{7}
\end{gather*}
[v.\ (5) and Article 91 (7).]

\mypara{97.} By the aid of the formulas of Art.~96 a number of valuable developments
can be obtained.

Let us get $\cos n\theta$ and $\sin n\theta$, $n$ being any positive real.

$z=\cos n\theta$ and $z=\sin n\theta$ are solutions of the equation
\[
\frac{d^2z}{d\theta^2} +n^2z =0
\]
% -----File: 191.png
or if we let $x = \cos \theta$, of the equation
\[
(1-x^2)\dfrac{d^2z}{dx^2} -x\dfrac{dz}{dx} +n^2z=0. \tag{1}
\]
\begin{flalign*}
&\text{\indent Let}&&        a_0P_0(x) + a_1P_1(x)+a_2P_2(x) + \cdots&&\phantom{\indent Let}
\end{flalign*}
be the required development of $\cos n\theta$ or of $\sin n\theta$.
\begin{flalign*}
&\text{\indent Then}  &&\sum\limits^{m=\infty}_{m=0} a_m\left[(1-x^2)\dfrac{d^2P_m(x)}{dx^2} - x\dfrac{ dP_m(x)}{dx} + n^2P_m(x)\right] = 0 \;&&\text{ by (1).}
\end{flalign*}
$z = P_m(x)$ is a solution of Legendre's Equation (v.\ Art.~77). Hence
\[
(1-x^2)\dfrac{d^2P_m(x)}{dx^2}-x\dfrac{dP_m(x)}{dx}=x\dfrac{dP_m(x)}{dx}-m(m+1)P_m(x),
\]
and (1) becomes
\[
\sum\limits^{m=\infty}_{m=0} a_m\left[x\dfrac{dP_m(x)}{dx}+[n^2-m(m+1)]P_m(x)\right]=0. \tag{2}
\]

Formulas (4) and (6) of Art.~96 enable us to throw (2) into the form
\[
\sum\limits^{m=\infty}_{m=0}a_m\left[\dfrac{n^2-m^2}{2m+1}\dfrac{dP_{m+1}(x)}{dx} -\dfrac{ n^2-(m+1)^2}{2m+1}
\dfrac{dP_{m-1}(x)}{dx}\right]=0.\tag{3}
\]

(3) must be identically true. Therefore the coefficient of $\dfrac{dP_{m+1}(x)}{dx}$ must
equal zero, and we have
\[
a_{m+2}=\dfrac{2m+5}{2m+1}.\dfrac{n^2-m^2}{n^2-(m+3)^2} a_m. \tag{4}
\]

If we are developing $\cos n\theta$
\begin{flalign*}
&&a_0&=\dfrac{1}{2}\int\limits^{\pi}_0  \cos n\theta \sin\theta.d\theta &&\tag*{  by Art.~90 (4),}\\
&&       &=\dfrac{1}{4}\int\limits^{\pi}_0 [\sin(n+1)\theta-\sin(n-1)\theta]d\theta,&
\end{flalign*}
\begin{flalign*}
&&a_0&=-\dfrac{1}{2}.\dfrac{1+\cos n\pi}{n^2-1};  && \tag{5}\\
&\text{and}\hspace{6em}&      a_1={}&\dfrac{3}{2}\int\limits^{\pi}_0 \cos n\theta\cos\theta\sin\theta.d\theta  &&\tag*{by Art.~90 (4),}\\
&&a_1&=-\dfrac{3}{2}.\dfrac{1-\cos n\pi }{n^2-4}. && \tag{6}
\end{flalign*}
% -----File: 192.png

(4), (5), and (6) give us
\begin{align*}
\cos n\theta=-\frac{1+\cos n\pi}{2(n^2-1)}
\Big[ & P_0(\cos\theta)+5\frac{n^2}{n^2-3^2}P_2(\cos\theta)\\
      & +9\frac{n^2(n^2-2^2)}{(n^2-3^2)(n^2-5^2)}P_4(\cos\theta)+\cdots\Big]\\
-\frac{1-\cos n\pi}{2(n^2-2^2)}
\Big[ & 3P_1(\cos\theta)+7\frac{n^2-1^2}{n^2-4^2}P_3(\cos\theta)\\
      & +11\frac{(n^2-1^2)(n^2-3^2)}{(n^2-4^2)(n^2-6^2)}P_5(\cos\theta)+\cdots\Big]. \tag{7}
\end{align*}
If $n$ is a whole number $1+\cos n\pi$ or $1-\cos n\pi$ will vanish and the series
will end with the term involving $P_n(\cos\theta)$. For this case (7) may be rewritten
\begin{align*}
\cos n\theta={}&\frac{1}{2}.\frac{2.4.6.\cdots 2n}{3.5.7.\cdots(2n+1)}
\Big[(2n+1)P_n(\cos\theta)\\
& +(2n-3)\frac{n^2-(n+1)^2}{n^2-(n-2)^2}P_{n-2}(\cos\theta)\\
& +(2n-7)\frac{[n^2-(n+1)^2][n^2-(n-1)^2]}{[n^2-(n-2)^2][n^2-(n-4)^2]}P_{n-4}(\cos\theta)+\cdots\Big]. &&\tag{8}
\end{align*}

If we are developing $\sin n\theta$
\begin{align*}
a_0=\frac{1}{2}\int\limits_0^\pi\sin n\theta\sin\theta.d\theta
=-\frac{1}{2}.\frac{\sin n\pi}{n^2-1},\\
a_1=\frac{3}{2}\int\limits_0^\pi\sin n\theta\cos\theta\sin\theta.d\theta
=\frac{3}{2}.\frac{\sin n\pi}{n^2-2^2} \tag*{and}\\[-5ex]
\end{align*}
\begin{align*}
\sin n\theta=-\frac{1}{2}.\frac{\sin n\pi}{n^2-1}
\Big[ & P_0(\cos\theta)+5\frac{n^2}{n^2-3^2}P_2(\cos\theta)\\
      & + 9\frac{n^2(n^2-2^2)}{(n^2-3^2)(n^2-5^2)}P_4(\cos\theta)+\cdots\Big]\\
             +\frac{1}{2}.\frac{\sin n\pi}{n^2-2^2}
\Big[ &3P_1(\cos\theta)+7\frac{n^2-1^2}{n^2-4^2}P_3(\cos\theta)\\
      & + 11\frac{(n^2-1^2)(n^2-3^2)}{(n^2-4^2)(n^2-6^2)}P_5(\cos\theta)+\cdots\Big]. \tag{9}
\end{align*}

If $n$ is a whole number $\sin n\pi=0$, and all the terms of (9) vanish except
those involving $P_{n-1}(\cos\theta)$, $P_{n+1}(\cos\theta)$, $P_{n+3}(\cos\theta)$ \&c., which become indeterminate.
For this case it is necessary to compute $a_{n-1}$
independently.

% -----File: 193.png
We have
\begin{align*}
a_{n-1} & = \frac{2n-1}{2}\int\limits_0^\pi\sin n\theta P_{n-1}(\cos\theta)\sin\theta.d\theta\\
        & = \frac{2n-1}{4}\int\limits_0^\pi[\cos(n-1)\theta-\cos(n+1)\theta]P_{n-1}(\cos\theta)d\theta.
\end{align*}
\begin{flalign*}
&\text{Hence} & a_{n-1}=\frac{2n-1}{4}.\frac{1.3.5.\cdots(2n-3)}{2.4.6.\cdots(2n-2)}\pi& \tag*{\text{[v.\ Art.~82 (1)],}}&
\end{flalign*}
and
\begin{align*}
\sin n\theta= & \frac{\pi}{4}.\frac{1.3.\cdots(2n-3)}{2.4.\cdots(2n-2)}\Big[(2n-1)P_{n-1}(\cos\theta)\\
              &\quad\quad +(2n+3)\frac{n^2-(n-1)^2}{n^2-(n+2)^2}P_{n+1}(\cos\theta)\\
              &\quad\quad +(2n+7)\frac{[n^2-(n-1)^2][n^2-(n+1)^2]}{[n^2-(n+2)^2][n^2-(n+4)^2]}P_{n+3}(\cos\theta)+\cdots\Big]. \tag{10}
\end{align*}

\EXAMPLE{S}

1.\quad Show that
\[
\csc\theta=\frac{\pi}{2}\Big[1
+5 \Big(\frac{1}{2}\Big)^2P_2(\cos\theta)
+9 \Big(\frac{1.3}{2.4}\Big)^2P_4(\cos\theta)
+13\Big(\frac{1.3.5}{2.4.6}\Big)^2P_6(\cos\theta)+\cdots\Big]
\]
whence
\[
\frac{1}{\sqrt{1-x^2}}=\frac{\pi}{2}\Big[1
+5 \Big(\frac{1}{2}\Big)^2P_2(x)
+9 \Big(\frac{1.3}{2.4}\Big)^2P_4(x)
+13\Big(\frac{1.3.5}{2.4.6}\Big)^2P_6(x)+\cdots\Big]
\]
[v.\ Art.~90 (4) and Art.~82].\\

2.\quad Show that
\[
\ctn\theta=\frac{\pi}{2}\Big[3\Big(\frac{1}{2}\Big)P_1(\cos\theta)
 +7 \Big(\frac{3}{4}\Big)\Big(\frac{1}{2}\Big)^2P_3(\cos\theta)
 +11\Big(\frac{5}{6}\Big)\Big(\frac{1.3}{2.4}\Big)^2P_5(\cos\theta)+\cdots\Big]
\]
whence
\[
\frac{1}{\sqrt{1-x^2}}=\frac{\pi}{2}\Big[3\Big(\frac{1}{2}\Big)P_1(x)
 +7 \Big(\frac{3}{4}\Big)\Big(\frac{1}{2}\Big)^2P_3(x)
 +11\Big(\frac{5}{6}\Big)\Big(\frac{1.3}{2.4}\Big)^2P_5(x)+\cdots\Big]
\]
[v.\ Art.~90 (4) and Art.~82].\\

3.\quad By integrating the result of Ex.~1 and simplifying by the aid of Art.~96
(5), obtain the development

\begin{align*}
\sin^{-1}x=\frac{\pi}{2}\Big[3 &
   \Big(\frac{1}{2}\Big)^2P_1(x)
+7 \Big(\frac{1}{2.4}\Big)^2P_3(x)\\
&+11\Big(\frac{1.3}{2.4.6}\Big)^2P_5(x)
+15\Big(\frac{1.3.5}{2.4.6.8}\Big)^2P_7(x)+\cdots\Big]
\end{align*}
% -----File: 194.png
whence \qquad
$\displaystyle\theta=\frac{\pi}{2}\Big[P_0(\cos\theta)
 -3\Big(\frac{1}{2}\Big)^2P_1(\cos\theta)
 -7\Big(\frac{1}{2.4}\Big)^2P_3(\cos\theta)$
\[
-11\Big(\frac{1.3}{2.4.6}\Big)^2P_5(\cos\theta)-\cdots\Big].
\]

4.\quad By integrating the result of Ex.~2 and simplifying by the aid of Art.~96
(5) obtain
\begin{gather*}
\sqrt{1-x^2}=\frac{\pi}{2}\Big[\frac{1}{2}
 -5\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)^2P_2(x)
 -9\Big(\frac{3}{6}\Big)\Big(\frac{1}{2.4}\Big)^2P_4(x)
\\
 -13\Big(\frac{5}{8}\Big)\Big(\frac{1.3}{2.4.6}\Big)^2P_6(x)+\cdots\Big]
\end{gather*}
whence
\[
\sin\theta=\frac{\pi}{2}\Big[\frac{1}{2}P_0(\cos\theta)
 -5\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)^2P_2(\cos\theta)
 -9\Big(\frac{3}{6}\Big)\Big(\frac{1}{2.4}\Big)^2P_4(\cos\theta)
 -\cdots\Big].
\]

To make clearer the analogy of development in Zonal Harmonic Series with
development in Fourier's Series we give on page \label{was184}\pageref{was185} a cut representing the
first seven Surface Zonal Harmonics $P_1(\cos\theta)$, $P_2(\cos\theta)$, $\cdots$  $P_7(\cos\theta)$, which
are of course somewhat complicated Trigonometric curves resembling roughly
$\cos\theta$, $\cos2\theta$, $\cdots$ $\cos7\theta$; and on page \pageref{was186}, the first four successive approximations
to the Zonal Harmonic Series
\[
 \frac{1}{2}+\frac{3}{4}P_1(\cos\theta)
-\frac{7}{8}.\frac{1}{2}P_3(\cos\theta)
+\frac{11}{12}.\frac{1.3}{2.4}P_5(\cos\theta)
-\cdots \tag*{\smallrom{I}}
\]
[v.\ (1) Art.~93], and
\begin{multline}
\frac{\pi}{2}\Big[P_0(\cos\theta)
 -3\Big(\frac{1}{2}\Big)^2P_1(\cos\theta)
 -7\Big(\frac{1}{2.4}\Big)^2P_3(\cos\theta)
\\
-11\Big(\frac{1.3}{2.4.6}\Big)^2P_5(\cos\theta)
-\cdots\Big] \tag*{\smallrom{II}}
\end{multline}
(v.\ Ex.~3 Art.~97).

\smallrom{I} is equal to 1 from $\theta=0$ to $\theta=\dfrac{\pi}{2}$, and to 0 from $\theta=\dfrac{\pi}{2}$ to $\theta=\pi$; and
\smallrom{II} is equal to $\theta$ from $\theta=0$ to $\theta=\pi$.

The figures on page \pageref{was186} are constructed on precisely the same principle as
those on pages \pageref{was63} and \pageref{was64}, with which they should be carefully compared.

\pngcent{013.png}{1720}

\label{was185}

\begin{center}
{\scriptsize The curves $y=P_0(\cos\theta)$, $y=P_1(\cos\theta)$, \ldots $y=P_7(\cos\theta)$. \qquad (v.\ page \pageref{was184}.)}
\end{center}

\mypara{98.} By applying \emph{Gauss's Theorem} (B.\ O.\ Peirce, Newt.\ Pot.\ Func.\ §~31) or
the special Form of \emph{Green's Theorem},
\[
\iiint\nabla^2Vdxdydz=\int D_nVds=-4\pi\iiint\rho dxdydz,
\]
% -----File: 195.png
% -----File: 196.png
% -----File: 197.png
\noindent[Peirce, N.~P.~F.\ §~49 (149)] to a box cut from an infinitely thin shell of
attracting matter by a tube of force whose end is an element of the surface of
the shell we readily obtain the important result
\[
4\pi\rho\kappa = D_nV_1-D_nV_2. \tag{1}
\]
where $\rho$ is the density and $\kappa$ the thickness of the shell, $V_1$ the value of the
potential function due to the shell at an internal point and $V_2$ its value at an
external point, and where $D_n$ is the partial derivative along the external normal
to the outer surface of the shell.

If we have to deal with a surface distribution of matter we have only to
replace $\rho\kappa$ in (1) by $\sigma$ where $\sigma$ is the surface density, whence
\[
4\pi\sigma = D_nV_1-D_nV_2 \tag{2}
\]
(v.\ Peirce, N.~P.~F.\ §§~45, 46, and 47).

\pngcent{014.png}{1321}

\label{was186}
%[Illustration: I.]
%[Illustration: II.]

Formulas (1) and (2) enable us to solve problems in attraction when we
know the density of the attracting mass, and problems in Statical Electricity
when we know the distribution of the charge, by methods analogous to that of
Art.~94.

For example let us find the value of the potential function due to a thin
material spherical shell of density $\rho$ and radius $a$.

Since $V$ must be a solution of Laplace's Equation and must be finite both
when $r=0$ and $r=\infty$ we have
\begin{align*}
V_1&=\sum A_mr^mP_m(\cos\theta)\\
V_2&=\sum B_m \frac{1}{r^{m+1}}P_m(\cos\theta).
\end{align*}
$V_1$ and $V_2$ must approach the same limiting values as $r$ approaches $a$. Hence
\begin{flalign*}
&&\frac{B_m}{a^{m+1}}&=A_ma^m&\\
&\text{or} &  B_m=A&_ma^{2m+1}.&\\
&&D_nV_1=D_rV_1=\sum &mr^{m-1}A_mP_m(\cos\theta),&\\
&&D_nV_2=D_rV_2=-\sum(&m+1)\frac{A_ma^{2m+1}}{r^{m+2}}P_m(\cos\theta).&
\end{flalign*}
Therefore by (1)
\[
4\pi\rho\kappa=\sum(2m+1)A_ma^{m-1}P_m(\cos\theta)
\]
if $\kappa$ is the thickness of the shell.
% -----File: 198.png
\begin{flalign*}
&\indent\text{Let} & \rho=f(\cos\theta)=&\sum C_mP_m(\cos\theta) && \phantom{\indent Let}\\
&\text{where}      & C_m=\frac{2m+1}{2}&\int\limits_{-1}^1 f(x)P_m(x)dx && \tag*{by Art.~90 (2).}\\
&\indent\text{Then}       & 4\pi\kappa C_m=(2&m+1)A_ma^{m-1}, && \tag*{and}  \\
&&A_m=\frac{4\pi\kappa C_m}{(2m+1)a^{m-1}},\quad &\text{and}\quad B_m=\frac{4\pi\kappa}{2m+1}C_ma^{m+2}, \\
&\text{and} &  V_1=4\pi a\kappa\sum&\frac{C_m}{2m+1}\frac{r_m}{a_m}P_m(\cos\theta), & \tag{3}\\
&\text{and} &  V_2=4\pi a\kappa\sum&\frac{C_m}{2m+1}\frac{a^{m+1}}{r^{m+1}}P_m(\cos\theta). & \tag{4}
\end{flalign*}

\mypara{99.} We can now get the value of the potential function due to a spherical
shell of finite thickness, provided that its density can be expressed as a sum of
terms of the form $Cr^kP_m(\cos\theta)$.

Let $a$ be the radius of the outer surface and $b$ be the radius of the inner
surface of the shell.

1st.---Let $\rho=Cr^kP_m(\cos\theta)$. Then for the shell of radius $s$ and thickness $ds$
\begin{flalign*}
&&V_1={}&4\pi sds\frac{Cs^k}{2m+1}\frac{r^m}{s^m}P_m(\cos\theta) && \tag*{by (3) Art.~98,}\\
&\text{and}\hspace{6em} &V_2=4&\pi sds\frac{Cs^k}{2m+1}\frac{s^{m+1}}{r^{m+1}}P_m(\cos\theta) && \tag*{by (4) Art.~98.}
\end{flalign*}

Then if $r<b$
\[
V=\int\limits_b^a V_1=\frac{4\pi C}{(2m+1)}\frac{(a^{k-m+2}-b^{k-m+2})}{(k-m+2)}r^mP_m(\cos\theta), \tag{1}
\]
if $r>a$
\[
V=\int\limits_b^a V_2=\frac{4\pi C}{(2m+1)}\frac{(a^{k+m+3}-b^{k+m+3})}{(k+m+3)}\frac{P_m(\cos\theta)}{r^{m+1}}, \tag{2}
\]
and if $b<r<a$
\begin{multline}
V=\int\limits_b^r V_2+\int\limits_r^a V_1=\frac{4\pi C}{2m+1}\bigg[\frac{r^{k+m+3}-b^{k+m+3}}{(k+m+3)r^{m+1}}\\
+\frac{a^{k-m+2}-r^{k-m+2}}{(k-m+2)}r^m\bigg]P_m(\cos\theta). \tag{3}
\end{multline}

2d.---If $\displaystyle\rho=\sum C_mr^kP_m(\cos\theta)$ the solutions will consist of sums of terms of
the forms given in (1), (2), and (3).
% -----File: 199.png

\EXAMPLE{S}

1.\quad If the shell is homogeneous
\begin{align*}
V&=2\pi\rho(a^2-b^2)\quad \text{if}\quad r<b,\\
V&=\frac{4}{3}\pi\rho(a^3-b^3)\frac{1}{r}=\frac{M}{r}\quad \text{if}\quad r>a,\\
V&=2\pi\rho\bigg[a^2-\frac{2b^3}{3r}-\frac{r^2}{3}\bigg]\quad \text{if}\quad b<r<a.
\end{align*}

2.\quad If the density is any given function of the distance from the centre
 $V=\dfrac{M}{r}$ if $r>a$, and $V=$ a constant if $r<b$.\\

3.\quad If the density at any point of a solid sphere is proportional to the square
of the distance from a diametral plane
\[
V=\frac{M}{a}\bigg[\frac{a}{r}+\frac{2}{7}\frac{a^3}{r^3}P_2(\cos\theta)\bigg]\quad \text{if}\quad r>a.
\]

4.\quad If the density at any point of a solid sphere is proportional to its distance
from a diametral plane
\[
V=\frac{M}{a}\bigg[\frac{a}{r}
 +\frac{1}{6}         \frac{a^3}{r^3}P_2(\cos\theta)
 -\frac{1.1}{6.8}     \frac{a^5}{r^5}P_4(\cos\theta)
 +\frac{1.1.3}{6.8.10}\frac{a^7}{r^7}P_6(\cos\theta)
 -\cdots\bigg]
\]
if $r>a$. Compare Ex.~2 Art.~80.

\mypara{100.} We have seen in Art.~18 (\emph{c}) (3) that
\[
Q_m(x)=CP_m(x)\int\frac{dx}{(1-x^2)[P_m(x)]^2}, \tag{1}
\]
no constant term being understood with $\displaystyle\int\frac{dx}{(1-x^2)[P_m(x)]^2}$.

$\dfrac{1}{(1-x^2)[P_m(x)]^2}$ is a rational fraction and becomes infinite only for $x=1$,
 $x=-1$, and for the roots of $P_m(x)=0$, all of which are real and lie
between $-1$ and 1, as can be proved by the aid of the relation
 $P_m(x)=\dfrac{1}{2^mm!}\dfrac{d^m(x^2-1)^m}{dx^m}$.

If $x^2>1$ $\displaystyle\int\limits_{x}^\infty\frac{dx}{(1-x^2)[P_m(x)]^2}$ is finite and determinate and contains no
constant term. Hence if $x^2>1$
\[
Q_m(x)=-P_m(x)\int\limits_{x}^\infty\frac{dx}{(1-x^2)[P_m(x)]^2}=P_m(x)\int\limits_{x}^\infty\frac{dx}{(x^2-1)[P_m(x)]^2} \tag{2}
\]
for the constant factor of $Q_m(x)$ has been chosen so that $C=-1$.

% -----File: 200.png
If $x^2 < 1$ the second member of (2) is not finite and determinate, and we
are thrown back to the form (1), and $C$ proves to be unity.

(1) gives us readily
\begin{flalign*}
&& &\quad Q_0(x) = \dfrac{1}{2}\log \dfrac{1+x}{1-x} &&\tag{3}\\
&&&Q_1(x) = -1+ \dfrac{x}{2} \log \dfrac{1+x}{ 1-x}    &&\tag{4}
\intertext{if $x^2<1$.}\\[-7ex]
&\text{\indent (2) gives us}&&\quad Q_0(x) = \dfrac{1}{2}\log \dfrac{x+1}{x-1}&&\tag{5}\hspace{4em}\\
&&&Q_1(x) = -1 + \dfrac{x}{2}\log \dfrac{x+1}{ x-1}&    &\tag{6}
\end{flalign*}
if $x^2>1$.

From Art.~85 (10) it follows that
\begin{align*}
Q_m(x) &= C\dfrac{d^m}{dx^m}\bigg[(x^2-1)^m \int\limits^x_0\dfrac{dx}{(x^2-1)^{m+1}}\bigg]\quad\text{if}\quad x^2 < 1,\\
    &= C\dfrac{d^m}{dx^m}\bigg[(x^2-1)^m \int\limits^{\infty}_x\dfrac{dx}{(x^2-1)^{m+1}}\bigg]\quad\text{if}\quad x^2  > 1.
\end{align*}
$C$ can be determined and is equal to $\dfrac{(-1)^{m+1}2^mm!}{(2m)!}$ if $x^2 < 1$, and is equal
to $\dfrac{(-1)^m2^m m!}{(2m)!}$ if $x^2 > 1$.
\begin{flalign*}
&\text{Hence}&  Q_m(x) &= \dfrac{(-1)^{m+1}2^mm!}{(2m)!}\dfrac{d^m}{dx^m}\bigg[(x^2-1)^m \int\limits^x_0\dfrac{dx}{(x^2-1)^{m+1}}\bigg]& (7)\\[-2ex]
\intertext{if $x^2<1$,}\\[-6ex]
&\text{and}     &  Q_m(x) &= \dfrac{(-1)^m2^m m!}{(2m)!}\dfrac{d^m}{dx^m}\bigg[(x^2-1)^m \int\limits^{\infty}_x\dfrac{dx}{(x^2-1)^{m+1}}\bigg]&(8)\\[-5ex]
\end{flalign*}
if $x^2>1$.

(7) and (8) give us for $Q_0(x)$ and $Q_1(x)$ the values already written in (3),
(4), (5), and (6).

By the repeated application of the formula
\[
(m+1)Q_{m+1}(x) - (2m+1)xQ_m(x) + mQ_{m-1}(x)=0,   \tag{9}
\]
which may be obtained for the case where $x^2 < 1$ from Art.~16 (13) and (14),
and for the case where $x^2 > 1$ from Art.~16 (9), any Surface Zonal Harmonic
of the Second Kind can be obtained from $Q_0(x)$ and $Q_1(x)$ as given in (3), (4),
(5), and (6).

% -----File: 201.png
Analogous formulas for $p_m(x)$ and $q_m(x)$ can be obtained without difficulty
from Art.~16 (4) and (5). They are
\begin{flalign*}
&&(m + 1)^2q_{m+1}(x) &- (2m + 1)xp_m(x) - m^2 q_{m-1}(x) = 0 \tag{10}\\
&\text{and}\hspace{1em}  &     p_{m+1}(x) + &(2m + 1)xq_m(x) - p_{m-1}(x) = 0 &\,\tag{11}
\end{flalign*}
and they hold good for any value of $m$.

\EXAMPLE{S}

1.\quad Confirm the values of $Q_0(x)$ and $Q_1(x)$ given in Art.~100 (3), (4), (5), and
(6) by expanding them and comparing them with Art.~16 (13), (14), and (9).\\

2.\quad If the value of $V$ on the surface of a cone of revolution can be expressed
in terms of whole powers positive or negative of $r$, $V$ can be found for any
point in space, cf.\ Art.~81.

If $V=\displaystyle \sum\left(A_mr^m + \dfrac{B_m }{r^{m+1}}\right)$ when $\theta = \alpha$ then
\[
V=\sum\left(A_mr^m + \dfrac{B_m }{r^{m+1}}\right)\dfrac{P_m(\cos \theta)}{P_m(\cos \alpha)}.
\]

3.\quad If $V=\displaystyle \sum\left(A_mr^m + \dfrac{B_m }{r^{m+1}}\right)$ when $\theta = \alpha$, and $V = 0$ when $\theta = \beta$,
\[
V=\sum\left(A_mr^m + \dfrac{B_m }{r^{m+1}}\right)\left[\dfrac{Q_m(\cos\beta)P_m(\cos\theta)-P_m(\cos\beta)Q_m(\cos\theta)}{P_m(\cos \alpha)Q_m(\cos\beta)-P_m(\cos\beta)Q_m(\cos \alpha)}\right].
\]

4.\quad Find $V$ for points corresponding to values of $\theta$ between $\alpha$ and $\beta$ when
$V$ can be given in terms of whole powers of $r$ for $\theta = \alpha$ and for $\theta = \beta$.\\

5.\quad Find by the method of Art.~16 solutions of Legendre's Equation of the
form
\begin{align*}
z = {}_1P_m(x) &= 1 + \dfrac{m(m+1) }{ 2} (x - 1) + \dfrac{(m-1)m(m+1)(m+2) }{ 2^2 (2!)^2} (x - 1)^2\\
             &\quad+ \dfrac{(m-2)(m-1)m(m+1)(m+2)(m+3) }{2^3 (3!)^2} (x-1)^3 + \cdots,\\
z = {}_{-1}P_m(x) &= 1 - \dfrac{m(m+1)}{2}(x + 1) + \dfrac{(m-1)m(m+1)(m+2) }{2^2(2!)^2} (x + 1)^2\\
             &\quad+ \dfrac{(m-2)(m-1)m(m+1)(m+2)(m+3)}{2^3(3!)^2} (x + 1)^3 + \cdots.
\end{align*}
If $m$ is a whole number, ${}_1P_m(x) = P_m(x)$ and ${}_{-1}P _m(x) = (-1)^m P_m (x)$. No
matter what the value of $m$, ${}_1P_m (x)$ is absolutely convergent for $-1 < x < 3$,
and ${}_{-1}P_m(x)$ is absolutely convergent for  $ -3 < x < 1$.\\
% -----File: 202.png

6.\quad By the aid of (7) Art.~16 show that
\[
\arraycolsep=0.3em
\begin{array}{c@{\quad} |@{\quad} c}
V= \dfrac{1}{\sqrt{r}} \sin (n \log r)k_n(\cos \theta),  & V= \dfrac{1}{\sqrt{r}} \sin (n \log r)l_n(\cos \theta),\\[1.3em]
V= \dfrac{1}{\sqrt{r}} \cos (n \log r)k_n(\cos \theta),  & V= \dfrac{1}{\sqrt{r}} \cos (n \log r)l_n(\cos \theta),
\end{array}
\]
are solutions of Laplace's Equation
\begin{flalign*}
&&r&D_r^2(rV) + \dfrac{1}{\sin \theta} D_\theta (\sin \theta D_\theta V) = 0,     &\text{if}\\
&&k_n(x) = p_{-\frac{1}{2}+ni} (x) &= 1 + \dfrac{n^2 + \Big(\dfrac{1}{2}\Big)^2}{ 2!} x^2 + \dfrac{\Big[n^2 + \Big(\dfrac{1}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{5}{2}\Big)^2\Big]}{ 4!} x^4\\[1ex]
&&          &+ \dfrac{\Big[n^2 + \Big(\dfrac{1}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{5}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{9}{2}\Big)^2\Big]}{ 6!} x^6 + \cdots
\end{flalign*}
and\\[-5ex]
\begin{flalign*}
&&l_n(x) = -q_{-\frac{1}{2}+ni} (x) &= x + \dfrac{n^2 + \Big(\dfrac{3}{2}\Big)^2}{ 3!} x^3 + \dfrac{\Big[n^2 + \Big(\dfrac{3}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{7}{2}\Big)^2\Big]}{ 5!} x^5&&\\[1ex]
&&          &+ \dfrac{\Big[n^2 + \Big(\dfrac{3}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{7}{2}\Big)^2\Big]\Big[n^2 + \Big(\dfrac{11}{2}\Big)^2\Big]}{ 7!} x^7 + \cdots&&
\end{flalign*}
$k_n (x)$ and $l_n (x)$ are convergent if $x^2 < 1$, but are divergent if $x^2 = 1$.\\

7.\quad Show by the aid of Example 5 that
\[
\arraycolsep=0.3em
\begin{array}{c@{\quad} |@{\quad} c}
V= \dfrac{1}{\sqrt{r}} \sin (n \log r)K_n(\cos \theta),  & V= \dfrac{1}{\sqrt{r}} \sin (n \log r)K_n(-\cos \theta),\\[1em]
V= \dfrac{1}{\sqrt{r}} \cos (n \log r)K_n(\cos \theta),  & V= \dfrac{1}{\sqrt{r}} \cos (n \log r)K_n(-\cos \theta),
\end{array}
\]
\begin{flalign*}
&\text{are solutions of}&   &rD_r^2(rV) + \dfrac{1}{\sin \theta} D_\theta (\sin \theta D_\theta V) = 0&\phantom{are solutions of}\\[-5ex]
\end{flalign*}
\begin{flalign*}
&\text{if }&  K_n(x) =& {}_1P_{-\frac{1}{2}+ni}(x)=1-\dfrac{n^2+\Big(\dfrac{1}{2}\Big)^2}{2}(x-1)&\\[0.5em]
&                &                   &+ \dfrac{\Big[n^2+\Big(\dfrac{1}{2}\Big)^2\Big]\Big[n^2+\Big(\dfrac{3}{2}\Big)^2\Big]}{2^2(2!)^2}(x-1)^2                &\\[0.5em]
&                &                   & - \dfrac{\Big[n^2+\Big(\dfrac{1}{2}\Big)^2\Big]\Big[n^2+\Big(\dfrac{3}{2}\Big)^2\Big]\Big[n^2+\Big(\dfrac{5}{2}\Big)^2\Big]}{2^2(3!)^2}(x-1)^3+\cdots&
\end{flalign*}
% -----File: 203.png
and\\[-5ex]
\begin{align*}
K_n(-x) = &{}_{-1}P_{-\frac{1}{2} + ni}(x) = 1 + \dfrac{n^2+\Big(\dfrac{1}{2}\Big)^2}{2} (x + 1)\\
&\phantom{{}_-}+\dfrac{
          \Big[n^2+\Big(\dfrac{1}{2}\Big)^2\Big]
          \Big[n^2+\Big(\dfrac{3}{2}\Big)^2\Big]}{2^2(2!)^2}(x+1)^2\\
&\phantom{{}_-}+\dfrac{
          \Big[n^2+\Big(\dfrac{1}{2}\Big)^2\Big]
          \Big[n^2+\Big(\dfrac{3}{2}\Big)^2\Big]
          \Big[n^2+\Big(\dfrac{5}{2}\Big)^2\Big]}{{2^3(3!)^2}} (x+1)^3 +\cdots.
\end{align*}
$K_n(\cos \theta)$ is convergent except for $\theta =\pi$, and $K_n(-\cos\theta)$ is convergent
except for $\theta = 0$.

$k_n(x)$, $l_n(x)$, $K_n(x)$, and $K_n(-x)$ are sometimes called \emph{Conal Harmonics}.
They are particular values of $z$ which satisfy Legendre's Equation written in
the form
\[
(1-x^2)\dfrac{d^2z}{dx^2}- 2x
       \dfrac{dz}{dx}-\Big(n^2+
       \dfrac{1}{4}\Big)z=0.
\]
For an elaborate treatment of them see E.~W. Hobson on ``A Class of Spherical
Harmonics of Complex Degree.'' Trans.\ Camb.\ Phil.\ Soc., Vol.\ XIV.\\

8.\quad If $V=f(r)$ when $\theta=\beta,$
\[
 V = \dfrac{1}{\pi\sqrt{r}}
     \int\limits_{-\infty}^\infty d\lambda
     \int\limits_0^\infty e^{\tfrac{\lambda}{2}} f(e^\lambda)
     \dfrac{K_\alpha(\cos\theta)}{K_\alpha(\cos\beta)}
     \cos[\alpha(\lambda-\log r)]d\alpha;
     \quad\text{if}\quad\theta<\beta.
\]

9.\quad If $V=f(r)$ when $\theta = \beta$ and $r<a$, and $V=0$ when $r=a$,
\[
 V = \dfrac{2}{\pi}
     \sqrt{\frac{a}{r}}
     \int\limits_{-\infty}^0d\lambda
     \int\limits_0^{\infty}e^{\tfrac{\lambda}{2}} f(ae^\lambda)
     \dfrac{K_\alpha(\cos\theta)}{K_\alpha(\cos\beta)}
     \sin \alpha\lambda \sin\left(\alpha \log \dfrac{r}{a}\right)d\alpha;
     \quad\text{if}\quad\theta<\beta.
\]

10.\quad If $V = f(r)$ when $\theta =\beta$ and $a < r < b$, and $V=0$ when $r = a$
 and when $r = b$,
 \iffalse
\[
 V = \sum\limits_{m=1}^{m=\infty}A_m
     \dfrac{K_{m^\prime}(\cos\theta)}{K_{m^\prime}(\cos\beta)}
     \sin\left[\dfrac{m\pi(\log r - \log a)}{\log b - \log a}\right]\\[-3ex]
\]
\begin{flalign*}
  &\text{where}
  &m^\prime = \dfrac{m\pi}{\log b - \log a}
  &&\text{and}
\end{flalign*}
\[
 A_m = \dfrac{2}{\log b-\log a}
       \sqrt{\dfrac{a}{r}}
       \int\limits_{0}^{\log{\tfrac{b}{a}}}
       e^{\tfrac{x}{2}} f(ae^x)
       \sin \dfrac{m\pi x}{\log b-\log a}dx;
       \quad\text{if}\quad\theta<\beta.
\]
\fi
\begin{flalign*}
 &&V = \sum\limits_{m=1}^{m=\infty}A_m
     \dfrac{K_{m^\prime}(\cos\theta)}{K_{m^\prime}(\cos\beta)}
     &\sin\left[\dfrac{m\pi(\log r - \log a)}{\log b - \log a}\right]\\
 &\text{where} &  m^\prime ={}&\dfrac{m\pi}{\log b - \log a}   &&\text{and}\\
 &&\llap{\hfill $\displaystyle A_m = \dfrac{2}{\log b-\log a}
       \sqrt{\dfrac{a}{r}}
       \int\limits_{0}^{\log{\tfrac{b}{a}}}
       e^{\tfrac{x}{2}} f(ae^x$}
       &)\rlap{$\displaystyle \sin \dfrac{m\pi x}{\log b-\log a}dx;
       \quad\text{if}\quad\theta<\beta.$\hfill}
\end{flalign*}
% -----File: 204.png

11.\quad If $\theta > \beta$\;\;$\cos \theta$ must be replaced by $(- \cos \theta)$ in examples 8, 9, and 10.\\

12.\quad If $V=f(r)$ when $\theta=\beta$, and $V=0$ when $\theta=\gamma$,
\begin{multline*}
% recast to fit line
 V = \dfrac{1}{\pi\sqrt{r}}
     \int\limits^{\infty}_{-\infty}d\lambda
     \int\limits^{\infty}_0 e^{\tfrac{\lambda}{2}} f(e^{\lambda})\\
     \dfrac{k_\alpha(\cos \theta)l_\alpha(\cos \gamma) - k_\alpha(\cos \gamma)l_\alpha(\cos \theta) }
           {k_\alpha(\cos \beta)l_\alpha(\cos \gamma) - k_\alpha(\cos \gamma) l_\alpha(\cos \beta) }
     \cos[\alpha(\lambda - \log r)] d\alpha;
\end{multline*}
if $\beta<\theta<\gamma$.\\

13.\quad If $V=f(r)$ when $\theta = \beta$ and $a<r<b$, $V=0$ when $\theta = \gamma$ and
 $a<r<b$, and $V=0$ when $r = a$ and when $r = b$,
\iffalse
\[
V = \sum^{m=\infty}_{m=1} A_m
    \dfrac{k_{m'}(\cos \theta) l_{m'}(\cos \gamma) - k_{m'}(\cos \gamma)l_{m'}(\cos \theta)}
          {k_{m'}(\cos \beta)  l_{m'}(\cos \gamma) - k_{m'}(\cos \gamma)l_{m'}(\cos \beta)}
    \sin\dfrac{m \pi (\log r - \log a)}{\log b - \log a},
\]
\begin{flalign*}
&\text{where}& m' &= \dfrac{m \pi }{ \log b - \log a}&\text{and}
\end{flalign*}
\[
  A_m = \dfrac{2}{\log b-\log a}
        \sqrt{\dfrac{a}{r}}
        \int\limits^{\log\tfrac{b}{a}}_0 e^{\tfrac{x}{2}}f(ae^x)
        \sin\dfrac{m\pi x}{\log b-\log a}dx;
\]
\fi
\begin{flalign*}
&&&&\llap{$\displaystyle V = \sum^{m=\infty}_{m=1} A_m
    \dfrac{k_{m'}(\cos \theta) l_{m'}(\cos \gamma) - k_{m'}(\cos \gamma)l_{m'}(\cos \theta)}
          {k_{m'}(\cos \beta)  l_{m'}(\cos \gamma) - k_{m'}(\cos \gamma)l_{m'}(\cos \beta)}
    \sin\dfrac{m \pi (\log r - \log a)}{\log b - \log a},\:$}\\
&\text{where}& m' &= \dfrac{m \pi }{ \log b - \log a}&\text{and}\\
&&  A_m = \dfrac{2}{\log b-\log a}
        \sqrt{\dfrac{a}{r}}
        &\int\limits^{\log\tfrac{b}{a}}_0 e^{\tfrac{x}{2}}f(ae^x)
        \sin\dfrac{m\pi x}{\log b-\log a}dx;
\end{flalign*}
if $\beta<\theta<\gamma$ and $a<r<b$.\\

14.\quad If $V=f(r)$ when $\theta=\beta$ and $a<r<b$, and $V=0$ when $r= a$
 and $D_rV+hV=0$ when $r = b$,
\begin{flalign*}
  &\phantom{where}&V&=\sum\limits^{m=\infty}_{m=1}A_m
  \dfrac{{K_{\alpha_m}(\cos \theta)}}{{K_{\alpha_m}(\cos \beta)}}
  \sin \left(\alpha_m\log \dfrac{r}{a}\right), &\text{where}
\end{flalign*}
\[
 A_m = \dfrac{2(\alpha_m^2 + h^2b^2)}{\alpha_m^2(\log b -\log a)+hb[hb(\log b - \log a)+1]}
       \int\limits^{\log\tfrac{b}{a}}_0
       e^{\tfrac{x}{2}}f(ae^x)\sin \alpha_mx.dx
\]
and $\alpha_m$ is a root of the equation
\begin{flalign*}
&\phantom{v.\ Art.~68 Ex.~5.}&& \alpha\cos\left(\alpha\log \dfrac{b}{a}\right)+
    hb\sin \left(\alpha\log\dfrac{b}{a}\right)=0&\text{v.\ Art.~68 Ex.~5.}
\end{flalign*}
\label{ch5end}

% -----File: 205.png

\mychap{CHAPTER VI.}{SPHERICAL HARMONICS.}
\label{ch6start}

\mypara{101.} When we are dealing with problems in finding the \textit{potential function}
due to forces which have not circular symmetry\footnote{See note, page 12.} about an axis and are using
Spherical Coördinates, we have to solve Laplace's Equation in the form

\[
rD_r^2(rV) + \frac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta}V) + \frac{1}{\sin^2\theta }D_{\phi}^2V = 0   \tag{1}
\]
[v.\ \smallromr{XIII} Art.~1].

To get a particular solution of (1) we shall assume as usual that $V$ is a
product of functions each of which involves but a single variable.

Let $V=R.\Theta.\Phi$; where $R$ involves $r$ only, $\Theta$ involves $\theta $ only, and $ \Phi$ $ \phi$ only.
Substitute in (1) and we get
\begin{flalign*}
  &&\frac{r}{R}\frac{ d^2(rR)}{dr^2 }&+
  \frac{1}{\Theta \sin\theta }
  \frac{d\Big (\sin\theta \dfrac{d\Theta}{d\theta}\Big )}{d\theta} +
  \frac{1}{\Phi \sin^2\theta} \frac{d^2\Phi}{d\phi^2}=0  && \tag{2}
\\
&\text{or}&
  \frac{ r \sin^2\theta}{R} &\frac{ d^2(rR)}{dr^2} +
  \frac{ \sin\theta}{\Theta } \frac{ d\Big (\sin\theta \dfrac{d\Theta}{d\theta}\Big )}{d\theta} = -
  \frac{ 1}{\Phi } \frac{ d^2\Phi}{d\phi^2} .&&\phantom{or}
\end{flalign*}

As the first member does not contain $\phi$ the second member cannot contain
$\phi$, and as it contains no other variable it must be constant; call it $n^2$. Equation
(2) is then equivalent to the two equations
\begin{flalign*}
&&\frac{d^2\Phi}{d\phi^2} &+ n^2\Phi = 0     &&     \tag{3}
\\
&\text{and}&
  \frac{r}{R}\frac{ d^2(rR)}{dr^2 }+
  \frac{1}{\Theta \sin\theta }
  &\frac{d\Big [\sin\theta \dfrac{d\Theta}{d\theta}\Big ]}{d\theta} -
  \frac{n^2}{\sin^2\theta} = 0   \tag{4} &&\phantom{\,}
\end{flalign*}

(3) has been solved before and gives us
\[
\Phi = A \cos n\phi + B \sin n\phi          \tag{5}
\]
[v.\ Art.~13(\emph{a})].

The first term of (4) does not involve $\theta$ and the second and third terms do\\
not involve $r$.\\
% -----File: 206.png

$\dfrac{r}{R}\dfrac{ d^2(rR)}{dr^2}$ must, then, be a constant; we shall call it $m(m + 1)$ as in Art.~13(\emph{c}). Then (4) breaks up into
\[
r\frac{ d^2(rR)}{dr^2}  = m(m+1) R       \tag{6}
\]
\begin{flalign*}
&\text{and}&\frac{ 1}{\sin \theta }\frac{d \Big [\sin \theta \dfrac{d \Theta}{ d \theta} \Big ]}{d \theta} + \left  [m(m+1) -  \frac{n^2}{\sin^2 \theta} \right ] \Theta = 0.   \tag{7} &&\phantom{\,}
\end{flalign*}

\markright{SPHERICAL HARMONICS.}
(6) was solved in Art.~13(\emph{c}) and gives
\[
R = A_1 r^m + B_1 r^{-m-1}.   \tag{8}
\]

If in (7) we replace $\cos \theta$ by $\mu$ we get
\[
\frac{d}{d \mu} \left  [(1- \mu^2) \frac{d \Theta}{d \mu} \right ] + \left [m(m+1) - \frac{n^2}{1- \mu^2} \right ] \Theta = 0,   \tag{9}\\
\]
the equivalent of
\[
(1- x^2) \frac{d^2z}{dx^2} - 2x \frac{dz}{dx} +  \left [m(m+1) - \frac{n^2}{1- x^2} \right ] z = 0,   \tag{10}
\]
[v.\ (17) Art.~85], which was solved in Art.~85 for the case where $m$ and $n$ are
positive integers and $n < m + 1$. v.\ (18) and (19) Art.~85.

From (19) Art.~85 we get as a particular solution of (9)
\[
\Theta = (1- \mu^2)^{\tfrac{n}{2}} \frac{d^n P_m(\mu)}{d \mu^n} = \sin^n \theta \frac{d^n P_m(\mu)}{d \mu^n},            \tag{11}
\]
if we restrict ourselves to whole positive values of $m$ and $n$, as we shall do
hereafter unless the contrary is explicitly stated, and suppose $m$ not less
than $n$.

A second but less useful particular solution of (9) is
\[
\Theta = (1- \mu^2)^{\tfrac{n}{2}} \frac{d^n Q_m(\mu)}{d \mu^n}.
\]

Combining our results we have as important particular solutions of (1)
\begin{flalign*}
&&&\quad V = r^m(A \cos n \phi + B \sin n \phi) \sin^n \theta \frac{d^n P_m (\mu)}{d \mu^n},   \tag{12}
\\
&\text{and}&&
V = \frac{1}{r^{m+1}} (A \cos n \phi + B \sin n \phi)
    \sin^n \theta \frac{d^n P_m (\mu)}{d \mu^n},    \tag{13} &&
\end{flalign*}
where $m$ and $n$ are positive integers and $n< m + 1$.
% -----File: 207.png

\mypara{102.} $\sin^n\theta \dfrac{d^nP_m(\mu)}{d\mu^n}$ or $(1-\mu^2)^{\tfrac{n}{2}} \dfrac{d^nP_m(\mu)}{d\mu^n}$ is a new function of $\mu$, that is
of $\cos \theta$, and we shall represent it by $P_m^n(\mu)$\footnote{Most of the English writers represent this function by $T_m^n(\mu)$.}  and shall call it an \emph{associated function} of the $n$th order and $m$th degree. It is a value of $ \Theta$ satisfying
equation (9) Art.~101.

By differentiating the value of $P_m(x)$ given in (9) Art.~74 we get the formula
\begin{multline*}
P_m^n(\mu) =
    \frac{(2m)! \sin^n\theta}{2^m m!(m-n)!}
    \left [\mu^{m-n} - \frac{(m-n)(m-n-1)}{2.(2m-1)} \mu^{m-n-2}  \right .\\
    \left . + \frac{(m-n)(m-n-1)(m-n-2)(m-n-3)}{2.4.(2m-1)(2m-3)} \mu^{m-n-4} -\cdots \right ]   \tag{1}
\end{multline*}
the expression in the parenthesis ending with the term involving $\mu^0$ if $m-n$ is
even and with the term involving $\mu$ if $m-n$ is odd.

For convenience of reference we give on the next page a table from which
$P_m^n(\mu)$ can be readily obtained for values of $m$ and $n$ from 1 to 8.

$\cos n\phi P_m^n(\mu)$ and $\sin n\phi P_m^n(\mu)$, that is,
\[
\cos n\phi  \sin^n\theta\frac{ d^nP_m(\mu)}{d\mu^n}\quad \text{and} \quad  \sin n\phi \sin^n\theta\frac{ d^nP_m(\mu)}{d\mu^n}
\]
are called \textit{Tesseral Harmonics} of the $m$th degree and $n$th order, and are
values of $V$ which satisfy the equation
\begin{gather*}
m(m+1)V + \frac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta} V) + \frac{1}{\sin^2\theta} D_{\phi}^2 V = 0    \tag{2}
\intertext{or its equivalent}
m(m+1)V + D_{\mu} [(1-\mu^2) D_{\mu} V] + \frac{1}{1-\mu^2} D_{\phi}^2 V = 0.   \tag{3}
\end{gather*}
There are obviously $2m+1$ Tesseral Harmonics of the $m$th degree, namely
\begin{align*}
P_m(\mu), \qquad\qquad \cos\phi \sin\theta \frac{dP_m(\mu)}{d\mu},  \qquad &  \sin\phi \sin\theta \frac{dP_m(\mu)}{d\mu}\\
   \cos 2\phi \sin^2\theta \frac{d^2P_m(\mu)}{d\mu^2},  \qquad &  \sin 2\phi \sin^2\theta \frac{d^2P_m(\mu)}{d\mu^2}\\
   \cos 3\phi \sin^3\theta \frac{d^3P_m(\mu)}{d\mu^3},  \qquad &  \sin 3\phi \sin^3\theta \frac{d^3P_m(\mu)}{d\mu^3}\\[-1ex]
\intertext{\phantom{$P_m(\mu),$}\qquad\qquad\qquad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad . \quad}
   \cos m\phi \sin^m\theta \frac{d^mP_m(\mu)}{d\mu^m},  \qquad &  \sin m\phi \sin^m\theta \frac{d^mP_m(\mu)}{d\mu^m}
\end{align*}

If each of these is multiplied by a constant and their sum taken, this sum
is called a \textit{Surface Spherical Harmonic} of the $m$th degree, and is a solution of
equations (2) and (3). We shall represent it by $Y_m(\mu,\phi)$ or by $Y_m(\theta,\phi)$.

\newpage

% -----File: 208.png

\renewcommand{\arraystretch}{0.2}
\begin{center}
\textit{Table for} $\csc^n\theta P_m^n(\mu)=\dfrac{d^nP_m(\mu)}{d\mu^n}.$

\vspace{\baselineskip}

% recast to fit page width
\begin{footnotesize}
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{}\\
\hline
\tablestrut{$m$} & $n=1.$
               & $n=2.$\\
\hline
&&\\
\hline
\tablestrut{1} & 1 &\\
\hline
\tablestrut{2} & $3\mu$
               & 3 \\
\hline
\tablestrut{3} & $\dfrac{3}{2}(5\mu^2-1)$
               & $15\mu$\\
\hline
\tablestrut{4} & $\dfrac{5}{2}(7\mu^3-3\mu)$
               & $\dfrac{15}{2}(7\mu^2-1)$\\
\hline
\tablestrut{5} & $\dfrac{15}{8}(21\mu^4-14\mu^2+1)$
               & $\dfrac{105}{2}(3\mu^3-\mu)$\\
\hline
\tablestrut{6} & $\dfrac{21}{8}(33\mu^5-30\mu^3+5\mu)$
               & $\dfrac{105}{8}(33\mu^4-18\mu^2+1$\\
\hline
\tablestrut{7} & $\dfrac{7}{16}(429\mu^6-495\mu^4+135\mu^2-5)$
               & $\dfrac{63}{8}(143\mu^5-110\mu^3+15\mu)$\\
\hline
\tablestrut{8} & $\dfrac{9}{16}(715\mu^7-1001\mu^5+385\mu^3-35\mu)$
               & $\dfrac{315}{16}(143\mu^6-143\mu^4+33\mu^2-1)$\\
\hline
\multicolumn{3}{|c|}{}\\
\hline
\end{tabular}
\end{footnotesize}

\vspace{\baselineskip}

\begin{footnotesize}
\begin{tabular}{|c|c|c|}
\hline
\multicolumn{3}{|c|}{}\\
\hline
\tablestrut{$m$} & $n=3.$
               & $n=4.$\\
\hline
&&\\
\hline
\tablestrut{1} & &\\
\hline
\tablestrut{2} & &\\
\hline
\tablestrut{3} & 15 &\\
\hline
\tablestrut{4} & $105\mu$
& 105\\
\hline
\tablestrut{5} & $\dfrac{105}{2}(9\mu^2-1)$
& $945\mu$\\
\hline
\tablestrut{6} & $\dfrac{315}{2}(11\mu^3-3\mu)$
& $\dfrac{945}{2}(11\mu^2-1)$\\
\hline
\tablestrut{7} & $\dfrac{315}{8}(143\mu^4-66\mu^2+3)$
& $\dfrac{3465}{2}(13\mu^3-3\mu)$\\
\hline
\tablestrut{8} & $\dfrac{3465}{8}(39\mu^5-26\mu^3+3\mu)$
& $\dfrac{10395}{8}(65\mu^4-26\mu^2+1)$\\
\hline
\multicolumn{3}{|c|}{}\\
\hline
\end{tabular}
\end{footnotesize}

\vspace{\baselineskip}

\begin{footnotesize}
\begin{tabular}{|c|c|c|c|c|}
\hline
\multicolumn{5}{|c|}{}\\
\hline
 \tablestrut{$m$}
&\tablestrut{$n=5.$}
&\tablestrut{$n=6.$}
&\tablestrut{$n=7.$}
&\tablestrut{$n=8.$}\\
\hline
&&&&\\
\hline
1& \tablestrut{ }&&&\\
\hline
2& \tablestrut{ }&&&\\
\hline
3& \tablestrut{ }&&&\\
\hline
4& \tablestrut{ }&&&\\
\hline
5& \tablestrut{945}&&&\\
\hline
6& \tablestrut{$10395\mu$}
&             10395&&\\
\hline
7& \tablestrut{$\dfrac{10395}{2}(13\mu^2-1)$}
&            $135135\mu$
&             135135&\\
\hline
8& \tablestrut{$\dfrac{135135}{2}(5\mu^3-\mu)$}
&            $\dfrac{135135}{2}(15\mu^2-1)$
&            $2027025\mu$
&             2027025\\
\hline
\multicolumn{5}{|c|}{}\\
\hline
\end{tabular}
\end{footnotesize}
\end{center}
\renewcommand{\arraystretch}{1}

\vspace{\baselineskip}

$r^mY_m(\mu,\phi)$ and $\dfrac{1}{r^{m+1}}Y_m(\mu,\phi)$ are called \emph{Solid Spherical Harmonics} of the
$m$th degree, and are solutions of Laplace's Equation (1) Art.~101.

To formulate:---
\[
Y_m(\mu,\phi) = \sum_{n=0}^{n=m}\Big[
   A_n\cos n\phi\sin^n\theta\frac{d^nP_m(\mu)}{d\mu^n}
 + B_n\sin n\phi\sin^n\theta\frac{d^nP_m(\mu)}{d\mu^n}\Big] \tag{4}\\[-5ex]
\]
\begin{flalign*}
&\text{or} & Y_m(\mu,\phi) = A_0P_m(\mu) + \sum_{n=1}^{n=m}[
  A_n\cos n\phi P_m^n(\mu)
+ B_n\sin n\phi P_m^n(\mu)] & \tag{5} &\\[-5ex]
\end{flalign*}
is a Surface Spherical Harmonic of the $m$th degree.

A Tesseral Harmonic is a special case of a Surface Spherical Harmonic, and
a Zonal Harmonic a special case of a Tesseral Harmonic; $P_m(\mu)$ being the
Tesseral Harmonic of the zeroth order and the $m$th degree; it might be
written $P_m^0(\mu)$.

\EXAMPLE{S}

1.\quad Show that
\[
(1-x^2)\frac{d^2z}{dx^2}-2x\frac{dz}{dx}+\bigg[m(m+1)-\frac{n^2}{1-x^2}\bigg]z = 0
\]
reduces to
\[
(1-x^2)\frac{d^2y}{dx^2}-2(n+1)x\frac{dy}{dx}+[m(m+1)-n(n+1)]y=0
\]
if we substitute $(1-x^2)^\frac{n}{2}y$ for $z$, even when $m$ and $n$ are unrestricted.\\
% -----File: 209.png

2.\quad Show that if in the second equation of Ex.~1 we let $y=\sum a_kx^k$ we get
\[
a_{k+2}=-\frac{(m-n-k)(m+n+1+k)}{(k+1)(k+2)}a_k  \tag*{\text{(v.\ Art.~16)}}
\]
whence $z=p_m^n(x)$ and $z=q_m^n(x)$ are solutions of the first equation of Ex.~1,
no matter what the values of $m$ and $n$, if
\begin{multline*}
p_m^n(x)=(1-x^2)^\frac{n}{2}\bigg[1-\frac{(m-n)(m+n+1)}{2!}x^2\\
+\frac{(m-n)(m-n-2)(m+n+1)(m+n+2)}{4!}x^4-\cdots\bigg]
\end{multline*}
and
\begin{multline*}
q_m^n(x)=(1-x^2)^\frac{n}{2}\bigg[x-\frac{(m-n-1)(m+n+2)}{3!}x^3\\
+\frac{(m-n-1)(m-n-3)(m+n+2)(m+n+4)}{5!}x^5-\cdots\bigg].
\end{multline*}

If $m-n$ is a positive integer, $p_m^n(x)$ or $q_m^n(x)$ will terminate with the term
involving $x^{m-n}$, and in that case
\begin{multline*}
z=(1-x^2)^\frac{n}{2}\bigg[x^{m-n}-\frac{(m-n)(m-n-1)}{2.(2m-1)}x^{m-n-2}\\
+\frac{(m-n)(m-n-1)(m-n-2)(m-n-3)}{2.4.(2m-1)(2m-3)}x^{m-n-4}-\cdots\bigg].
\end{multline*}
% -----File: 210.png
the parenthesis ending with a term involving $x^0$ if $m-n$ is even and $x$ if
$m-n$ is odd, is a solution of the first equation of Ex.~1. If $m$ and $n$ are
integers this value of $z$ is $\dfrac{2^mm!(m-n)!}{(2m)!}P_m^n(x)$.

\mypara{103.} We have seen in the last chapter that in many problems it is important
to be able to express a given function of $\cos\theta$, that is of $\mu$, in terms of
Zonal Harmonics of $\mu$. So it is often desirable to express a given function of
$\mu$ and $\phi$ in terms of Tesseral Harmonics of $\mu$ and $\phi$.

If, for example, we are trying to find the \emph{Potential Function} due to certain
forces and have the value of the function given for some given value of $r$,
that is, on the surface of some given sphere whose centre is at the origin of
coördinates, of course the given value will be a function of $\theta$ and $\phi$ and if we
can express it in terms of Spherical Harmonics of $\theta$ and $\phi$ we have only to
multiply each term by the proper power of $r$ to get the required solution of
the problem. For we shall then have a value of $V$ satisfying Laplace's
Equation and reducing to the given function of $\theta$ and $\phi$ on the surface of the
given sphere.

\mypara{104.} Suppose that we have a function of $\mu$ and $\phi$ given for all points on
the unit sphere, that is, for all values of $\mu$ from $-1$ to 1 and for all values of
$\phi$ from 0 to $2\pi$, $\mu$ and $\phi$ being independent variables, and that we wish to
express it in terms of Surface Spherical Harmonics.

Assume that\\[1ex]
$\displaystyle f(\mu,\phi)\!=\!\sum_{m=0}^{m=\infty}\!\Big[A_{0,m}P_m(\mu)
\!+\!\sum_{n=1}^{n=m}\!\Big(A_{n,m}\cos n\phi P_m^n(\mu)
+B_{n,m}\sin n\phi P_m^n(\mu)\Big)\Big].\hfill (1)$\\

Let us consider first a finite case, and attempt to determine the coefficients
so that\\[1ex]
$\displaystyle f(\mu,\phi)\!=\!\sum_{m=0}^{m=p}\!\Big[A_{0,m}P_m(\mu)
\!+\!\sum_{n=1}^{n=m}\!\Big(A_{n,m}\cos n\phi P_m^n(\mu)
+B_{n,m}\sin n\phi P_m^n(\mu)\Big)\Big] \hfill (2)$\\[1ex]
shall hold good at as many points of the sphere as possible. The expression
in brackets in the second member of (2) is a Surface Spherical Harmonic of
the $m$th degree and contains $2m+1$ constant coefficients. The whole number
of coefficients to be determined is then the sum of an Arithmetical Progression
of $p+1$ terms the first term of which is 1 and the last is $2p+1$, and is
therefore equal to $(p+1)^2$.

Let the interval from $\mu=-1$ to $\mu=1$ be divided into $p+2$ parts each of
which is $\Delta\mu$ so that $(p+2)\Delta\mu=2$, and let the interval from $\phi=0$ to $\phi=2\pi$
be divided into $p+2$ parts each of which is $\Delta\phi$ so that $(p+2)\Delta\phi=2\pi$.
% -----File: 211.png

Then if we substitute in equation (2) in turn the values $(-1 + \Delta\mu$, $\Delta\phi)$,
$(-1 + 2\Delta\mu$, $\Delta\phi)$, $\cdots [-1 + (p + 1)\Delta\mu$, $\Delta\phi]$; $(-1 + \Delta\mu$, $2\Delta\phi)$,
$(-1 + 2\Delta\mu$, $2\Delta\phi)$, $\cdots[-1 + (p + 1)\Delta\mu$, $2\Delta\phi]$; $\cdots [-1 + \Delta\mu$, $(p + 1)\Delta\phi]$,
$[-1 + 2\Delta\mu$, $(p + 1)\Delta\phi]$, $\cdots [-1 + (p + 1)\Delta\mu, (p + 1)\Delta\phi]$; since the first
member in each case will be known we shall have $(p + 1)^2$ equations of the
first degree containing no unknown except the $(p + 1)^2$ coefficients, and from
them the coefficients can be determined. When they are substituted in equation
(2) it will hold good at the $(p + 1)^2$ points of the unit sphere where $p + 1$
circles of latitude whose planes are equidistant intersect $p + 1$ meridians
which divide the equator into equal arcs. If now $p$ is indefinitely increased
the limiting values of the coefficients will be the coefficients in equation (1),
and (1) will hold good all over the surface of the unit sphere.

To determine any particular constant we multiply each of our $(p + 1)^2$
equations by $\Delta\mu \Delta\phi$ times the coefficient of the constant in question in that
equation and add the equations and then investigate the limiting form
approached by the resulting equation as $p$ is indefinitely increased.

As $p$ is indefinitely increased the summation in question will approach an
integration; and since $d\mu d\phi = -\sin\theta.d\theta d\phi$ is the element of surface of the
unit sphere, and as the limits $-1$ and 1 of $\mu$ correspond to $\pi$ and 0 of $\theta$ the
integration is a \emph{surface integration} over the surface of the unit sphere.

In determining any coefficient as $A_{n,m}$ in (1) the first member of the limiting
form of our resulting equation will be
\[
\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} f(\mu,\phi)\cos n\phi P_m^n(\mu)d\mu.
\]

In the second member we shall come across terms of the forms
\begin{gather*}
\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1}  \sin l\phi \cos n\phi P_m^{\,l}(\mu)P_m^n(\mu)d\mu,\quad
\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1}  \cos l\phi \cos n\phi P_m^{\,l}(\mu)P_m^n(\mu)d\mu,\\
\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1}  \sin n\phi \cos n\phi[P_m^n(\mu)]^2d\mu,\quad
\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1}  \cos^2 n\phi[P_m^n(\mu)]^2d\mu,
\end{gather*}
and other terms all of which come under the form
\[
\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} Y_l(\mu,\phi) Y_m(\mu,\phi)d\mu,
\]
where $Y_m(\mu,\phi)$ and $Y_l(\mu,\phi)$ are Surface Spherical Harmonics of different
degrees.

If we are determining a coefficient $B_{n,m}$ the only difference is that $\sin n\phi$
and $\cos n\phi$ will be interchanged in the forms just specified.
% -----File: 212.png

\mypara{105.}  \textit{The integral over the surface of the unit sphere of the product of two
Surface Spherical Harmonics of different degrees is zero}.
\begin{flalign*}
&\text{\indent That is}&
  \int\limits_0^{2\pi} d\phi
  \int\limits_{-1}^1 Y_l(\mu,\phi)Y_m(\mu,\phi)d\mu=0.\tag{1}&&\hspace{3em}
\end{flalign*}

For as we have seen $U=r^lY_l(\mu,\phi)$ and $V=r^mY_m(\mu,\phi)$ are solutions of
Laplace's Equation. Hence by \textit{Green's Theorem}
\begin{flalign*}
           \int (U&D_nV-VD_nU)ds=0      \tag*{v.\ Art.~92.}\\
           D_n V =& D_r V = m r^{m-1} Y_m (\mu,\phi),\\
            D_n U &= D_r U = l r^{l-1} Y_l (\mu,\phi);\\
U D_n V - V D_n U &= (m-l) r^{l + m - 1} Y_l (\mu,\phi) Y_m (\mu,\phi),\\
                  &= (m - l) Y_l (\mu,\phi) Y_m (\mu,\phi)
\end{flalign*}
on the surface of the unit sphere; and
\[
 (m-l)
 \int Y_l (\mu,\phi) Y_m (\mu,\phi)ds = (m-l)
 \int\limits_0^{2\pi} d\phi
 \int\limits_{1}^1 Y_l (\mu,\phi) Y_m (\mu,\phi) d\mu = 0.
\]

Hence unless $l = m$
\[
 \int\limits_0^{2\pi} d \phi
 \int\limits_{-1}^1 Y_l (\mu,\phi) Y_m (\mu, \phi) d \mu = 0.
\]

\EXAMPLE{S}

1.\quad  Obtain (1) Art.~105 directly from the equation
\[
 m(m+1) Y_m(\mu,\phi) + D_\mu [(1-\mu^2)D_\mu Y_m (\mu,\phi)] +
 \dfrac{1}{1- \mu^2} D_\phi^2 Y_m (\mu,\phi) = 0
\]
v.\ (3) Art.~102, and Art.~91.\\

2.\quad Show that the integral over the surface of the unit sphere of the product
of two Tesseral Harmonics of the same degree but of different orders is zero.\\

\textit{Suggestion:}
\[
 \int\limits_0^{2\pi} \sin k \phi \cos l \phi.d\phi =
 \int\limits_0^{2\pi} \sin k\phi \sin l \phi.d \phi =
 \int\limits_0^{2\pi} \cos k \phi \cos l \phi.d \phi = 0.
\]

\mypara{106.}
\vspace{-7.5ex}
\begin{align*}
  &&\int\limits_{-1}^1 P_l^n (\mu)P_m^n (\mu) d \mu &= 0 \quad\text{unless}\quad  l = m&\\[-1ex]
  &&&=\frac{2}{2m+1}\frac{ (m+n)!}{(m-n)!}\quad  \text{if} \quad l = m.
\end{align*}
% -----File: 213.png

For
\begin{align*}
  \int\limits^1_{-1}P_l^n(\mu)P_m^n(\mu)d\mu &=
  \int\limits^1_{-1}(1-\mu^2)^n \dfrac{d^nP_l(\mu)}{d\mu^n}.\dfrac{d^nP_m(\mu)}{d\mu^n} d\mu\\
                             &=(1-\mu^2)^n \dfrac{d^nP_m(\mu)}{d\mu^n}.\dfrac{ d^{n-1} P_l(\mu)}{d\mu^{n-1}}\mathop{\Big]}^1_{-1}\\
                             &\phantom{=}-\int\limits^1_{-1}\dfrac{d^{n-1} P_l(\mu)}{d\mu^{n-1}}.\dfrac{ d}{d\mu}\left[(1-\mu^2)^n \dfrac{d^n P_m(\mu)}{d\mu^n}\right]d\mu,\\
                             &= - \int\limits^1_{-1}\dfrac{ d^{n-1} P_l(\mu)}{d\mu^{n-1}}.\dfrac{d}{d\mu}\left[(1-\mu^2)^n\dfrac{ d^nP_m(\mu)}{d\mu^n}\right]d\mu;
\end{align*}
by \emph{integration by parts}.

Replacing \hfill $n$ \hfill by \hfill $n - 1$ \hfill in \hfill equation \hfill (2) \hfill Art.~84 \hfill and \hfill remembering \hfill that\\
$\dfrac{d^{n-1} P_m(x)}{dx^{n-1}}$
is a possible value of $z^{(n - 1)}$ we get
\[
(1-\mu^2)\dfrac{d^{n+1} P_m(\mu)}{d\mu^{n+1}}-2n\mu \dfrac{d^n P_m(\mu)}{d\mu^n}+[m(m+1)-n(n-1)]\dfrac{d^{n-1} P_m(\mu)}{d\mu^{n-1}}=0,
\]
or if we multiply by $(1-\mu^2)^{n-1}$
\begin{multline*}
(1-\mu^2)^n \dfrac{d^{n+1} P_m(\mu)}{d\mu^{n+1}}-2n\mu(1-\mu^2)^{n-1}\dfrac{d^n P_m(\mu)}{d\mu^n}\\
                      +(m+n)(m-n+1)(1-\mu^2)^{n-1}\dfrac{ d^{n-1} P_m(\mu)}{d\mu^{n-1}}=0,
\end{multline*}
or
\[
\dfrac{d}{d\mu}\left[(1-\mu^2)^n \dfrac{d^nP_m(\mu)}{d\mu^n}\right]=-(m+n)(m-n+1)(1-\mu^2)^{n-1} \dfrac{d^{n-1} P_m(\mu)}{d\mu^{n-1}}.
\]

Hence follows the \emph{reduction formula}
\begin{multline*}
\int\limits^1_{-1}(1-\mu^2)^n \dfrac{d^nP_l(\mu)}{d\mu^n}.\dfrac{d^n P_m(\mu)}{d\mu^n} d\mu\\
=(m+n)(m-n+1) \int\limits^1_{-1} (1-\mu^2)^{n-1}\dfrac{d^{n-1}P_l(\mu)}{d\mu^{n-1}}.\dfrac{d^{n-1} P_m(\mu)}{d\mu^{n-1}} d\mu.
\end{multline*}
Using this formula $n$ times we get
\begin{align*}
\int\limits^1_{-1} P_l^n(\mu)P_m^n(\mu)d\mu&=\dfrac{(m+n)!}{ (m-n)!} \int\limits^1_{-1} P_l(\mu)P_m(\mu)d\mu\\
                         &= 0\quad \text{unless}\quad l = m\\
                         &=\dfrac{2}{2m+1}\dfrac{ (m+n)!}{(m-n)!}\quad \text{if}\quad l = m
\end{align*}
v.\ Art.~89 (4) and (5).
% -----File: 214.png

\mypara{107.} We are now able to complete the solution of the problem in Art.~104
and since $\displaystyle\int\limits_0^{2\pi}\cos^2 n\phi.d\phi= \displaystyle\int\limits_0^{2\pi} \sin^2 n\phi.d\phi=\pi$ and $\displaystyle\int\limits_0^{2\pi}d\phi=2\pi$ we get as the
coefficients in (1) Art.~104
\begin{gather*}
A_{0,m}=\frac{2m+1}{4\pi}
        \int\limits_0^{2\pi} d\phi
        \int\limits_{-1}^1 f(\mu,\phi)P_m(\mu)d\mu,              \tag{1}\\
A_{n,m}=\frac{2m+1}{2\pi}.\frac{(m-n)!}{(m+n)!}
        \int\limits_0^{2\pi} d\phi
        \int\limits_{-1}^1 f(\mu,\phi)\cos n\phi P_m^n(\mu)d\mu, \tag{2}\\
B_{n,m}=\frac{2m+1}{2\pi}.\frac{(m-n)!}{(m+n)!}
        \int\limits_0^{2\pi} d\phi
        \int\limits_{-1}^1 f(\mu,\phi)\sin n\phi P_m^n(\mu)d\mu, \tag{3}
\intertext{whence}
f(\mu,\phi)=\sum_{m=0}^{m=\infty}\bigg[ A_{0,m}P_m(\mu) + \sum_{n=1}^{n=m}(A_{n,m}\cos n\phi + B_{n,m} \sin n\phi)P_m^n(\mu) \bigg] \tag{4}
\end{gather*}
and the development holds good for all values of $\mu$ and $\phi$ corresponding to
points on the unit sphere, provided only that the given function satisfies the
conditions that would have to be satisfied if it were to be developed into a
\textit{Fourier's Series}.

If we use $\mu_1$ and $\phi_1$ in place of $\mu$ and $\phi$ in (1), (2), and (3), we can write (4)
in the form
\begin{multline*}
f(\mu,\phi) =
   \frac{1}{2\pi}
   \sum_{m=0}^{m=\infty} (2m+1) \bigg [ \frac{1}{2}
   \int\limits_0^{2\pi} d\phi_1
   \int\limits_{-1}^1f(\mu_1,\phi_1) P_m(\mu)P_m(\mu_1)d\mu_1 \\
    +\sum_{n=1}^{n=m}\frac{(m-n)!}{(m+n)!}
   \int\limits_0^{2\pi}d\phi_1
   \int\limits_{-1}^1f(\mu_1,\phi_1)P_m^n(\mu)P_m^n(\mu_1)\cos n(\phi-\phi_1)d\mu_1\bigg ]. \tag{5}
\end{multline*}

Formulas (1), (2), (3), and (4) are convenient for actual work; (5) is rather
more compactly written.

\mypara{108.} As an example let us express $\sin^2 \theta \cos^2 \theta \sin \phi \cos \phi$ in terms of
Surface Spherical Harmonics.
\begin{flalign*}
&\text{\indent Here}& f(\mu,\phi)=\frac{1}{2}\mu^2(1-\mu^2)\sin2\phi.&&\phantom{\indent Here}\\[-5ex]
\end{flalign*}
\begin{gather*}
\quad A_{0,m}=\frac{2m+1}{8\pi}
   \int\limits_{-1}^1 \mu^2(1-\mu^2)P_m(\mu)d\mu
   \int\limits_0^{2\pi}\sin2\phi.d\phi=0,\\
% -----File: 215.png
A_{n,m} ={} \dfrac{2m+1}{4\pi}.
           \dfrac{ (m-n)!}{(m+n)!}
           \int\limits_{-1}^1 \mu^2(1-\mu^2)P_m^n(\mu)d\mu
           \int\limits^{2\pi}_0 \sin2\phi \cos n\phi.d\phi=0,\\
\end{gather*}
\begin{align*}
B_{n,m}& = \dfrac{2m+1}{4\pi}.
           \dfrac{(m-n)!}{(m+n)!}
           \int\limits^1_0 \mu^2(1-\mu^2)P_m^n(\mu)d\mu
           \int\limits^{2\pi}_0 \sin2\phi \sin n\phi.d\phi,\\
       & = 0\quad\text{unless}\quad n = 2.
\end{align*}
\begin{flalign*}
&\text{If $n = 2$}  &
    \int\limits^{2\pi}_0 \sin 2\phi \sin n\phi.d\phi& =
    \int\limits^{2\pi}_0 \sin^2 2\phi.d\phi=\pi, &\text{and}\\[-7ex]
\end{flalign*}
\begin{align*}
B_{2,m}& = \dfrac{2m+1}{4}.
           \dfrac{(m-2)!}{(m+2)!}
           \int\limits_{-1}^1 \mu^2(1-\mu^2)^2
           \dfrac{d^2P_m(\mu)}{d\mu^2} d\mu\\
       & = \dfrac{1}{2^m m!}\dfrac{ 2m+1}{4}\dfrac{ (m-2)!}{(m+2)!}
           \int\limits_{-1}^1 \mu^2(1-\mu^2)^2
           \dfrac{d^{m+2}(\mu^2-1)^m}{ d\mu^{m+2}} d\mu.
\end{align*}
\[
    \int\limits_{-1}^1 \mu^2(1-\mu^2)^2 \dfrac{d^{m+2}(\mu^2-1)^m }{ d\mu^{m+2}} d\mu=720
    \int\limits_{-1}^1 \dfrac{d^{m-4}(\mu^2-1)^m }{ d\mu^{m-4}} d\mu
\]
by repeated integration by parts,
\begin{flalign*}
&&& = 0\quad\text{if}\quad m>4,\\
&&& = 720 \int\limits_{-1}^1 (\mu^2-1)^4 d\mu=\dfrac{4096}{7}\quad\text{if}\quad m=4,\qquad
\\
    &\text{and} &B_{2,4} =
     \dfrac{1}{2^44!}&.\dfrac{9}{4}.
     \dfrac{2!}{6!}.\dfrac{4096}{7}=\dfrac{1}{105}.
\end{flalign*}

By a like process we find
\begin{flalign*}
   &&&\qquad\qquad B_{2,3} = 0\quad\text{and}\quad
     B_{2,2}  = \dfrac{1}{42}. &&\tag*{Hence}
\\[1ex]
&&    \sin^2\theta \cos^2\theta \sin\phi &\cos\phi =
     \dfrac{1}{42} P_2^2(\mu)\sin 2\phi +
     \dfrac{1}{105} P_4^2(\mu)\sin 2\phi,  \tag{1}
\\[1ex]
&&   &=\dfrac{1}{42}\sin 2\phi \sin^2\theta
     \dfrac{d^2P_2(\mu)}{d\mu^2} +
     \dfrac{1}{105}\sin 2\phi \sin^2\theta
     \dfrac{d^2P_4(\mu)}{d\mu^2},&\tag{2}\\[1ex]
&&   &=\dfrac{1}{14}\sin^2\theta \sin 2\phi +
     \dfrac{1}{14} \sin^2\theta (7\mu^2-1)\sin 2\phi.  \tag{3}
\end{flalign*}

The required expression might have been obtained without using the
formulas of Art.~107, by a very simple device, as follows:
\[
   \sin^2\theta \cos^2\theta \sin\phi \cos\phi =
   \dfrac{1}{2}\mu^2\sin^2\theta \sin 2\phi.\tag{4}
\]
% -----File: 216.png

If now we can express $\mu^2$ in the form $\displaystyle\sum\dfrac{ d^2P_m(\mu)}{d\mu^2}$ the work will be done.
\begin{flalign*}
&&                          \mu^2&=\dfrac{1}{4.3}\dfrac{ d^2(\mu^4)}{d\mu^2}, &\\[0.5em]
&&                          \mu^4&=\dfrac{8}{35}P_4(\mu)+\dfrac{4}{7}P_2(\mu)+\dfrac{1}{5}P_0(\mu),  &\text{(5) Art.~95.}\\[0.5em]
&&  \dfrac{d^2(\mu^4)}{d\mu^2}&=\dfrac{8}{35}\dfrac{d^2P_4(\mu)}{d\mu^2}+\dfrac{4}{7}\dfrac{d^2P_2(\mu)}{d\mu^2}; &\\[0.5em]
&\text{whence}& \mu^2&=\dfrac{2}{105}\dfrac{d^2P_4(\mu)}{d\mu^2}+\dfrac{1}{21}\dfrac{d^2P_2(\mu)}{d\mu^2},&
\end{flalign*}
and substituting this value in (4) we get (2).

\EXAMPLE{S}

1.\quad Show that
\begin{multline*}
\cos^3\theta \sin^3\theta \sin\phi \cos^2\phi=\left[\dfrac{1}{6930}P_6^3(\mu)+\dfrac{1}{1540}P_4^3(\mu)\right]\sin 3\phi\\
   -\left[\dfrac{2}{693}P_6^1(\mu)-\dfrac{1}{770}P_4^1(\mu)-\dfrac{1}{63}P_2^1(\mu)\right]\sin\phi.
\end{multline*}

2.\quad Show that
\[
\cos 2\phi=2 \cos 2\phi\left[\dfrac{5}{4!}P_2^2(\mu)+\dfrac{9.2!}{6!}P_4^2(\mu)+\dfrac{13.4!}{8!}P_6^2(\mu)+\cdots\right].
\]

3.\quad If in a problem on the Potential Function $V=f(\mu,\phi)$ when $r = a$, we
shall obviously have
\[
V= \sum\limits^{m=\infty}_{m=0}\dfrac{ r^m}{a^m}\bigg[A_{0,m} P_m(\mu) + \sum\limits^{n=m}_{n=1} (A_{n,m} \cos n\phi+B_{n,m} \sin n\phi)P_m^n(\mu)\bigg]
\]
at an internal point and
\[
V= \sum\limits^{m=\infty}_{m=0}\dfrac{a^{m+1}}{r^{m+1}}\bigg[A_{0,m} P_m(\mu) + \sum\limits^{n=m}_{n=1}  (A_{n,m} \cos n\phi+B_{n,m} \sin n\phi)P_m^n(\mu)\bigg]
\]
at an external point, where $A_{0,m}$, $A_{n,m}$, and $B_{n,m}$ have the values given in (1),
(2), and (3) Art.~107.\\

4.\quad Solve problems (3), (4), and (5) of Art.~94 for the case where $V$ is not
symmetrical with respect to an axis.

\mypara{109.} Any Solid Spherical Harmonic $r^mY_m(\mu,\phi)$ being a value of $V$ that
satisfies Laplace's Equation in Spherical Coördinates will transform into a
function of $x$, $y$, and $z$ satisfying $\nabla^2 V = 0$ if we change to a set of rectangular
% -----File: 217.png
axes having the same origin and the same axis of $X$ as the polar system.
Moreover the new function will be a homogeneous rational integral Algebraic
function of $x$, $y$, $z$, of the $m$th degree.

For each term of $r^m \cos n\phi P_m^n(\mu)$ is of the form
\begin{flalign*}
&&&Cr^m \cos^{n-2k} \phi \sin^{2k} \phi \sin^n \theta \cos^{m-2l-n} \theta &&\\[1ex]
&\text{where}&&2k < n + 1\quad\text{and}\quad 2l < m - n + 1.&&\phantom{where}
\end{flalign*}

This may be written\\[-5ex]
\begin{flalign*}
\intertext{\qquad$Cr^{2l}.r^{m-2l-n}\cos^{m-2l-n} \theta.r^{n-2k}\sin^{n-2k} \theta \cos^{n-2k} \phi.r^{2k}\sin^{2k} \theta \sin^{2k} \phi$}
&\text{which becomes}&& C(x^2 + y^2 + z^2)^l x^{m-2l-n} y^{n-2k} z^{2k},&&\phantom{which becomes}
\end{flalign*}
and is a homogeneous rational integral Algebraic function of $x$, $y$, and $z$ of the
$m$th degree. The same thing may be shown of each term of $r^m \sin n\phi P_m^n(\mu)$.
 Consequently $r^mY_m(\mu,\phi)$ is a homogeneous rational integral Algebraic function
of the $m$th degree in $x$, $y$, and $z$.

\mypara{110.} Any homogeneous rational integral Algebraic function $S_m(x, y, z)$ of
the $m$th degree in $x$, $y$, and $z$, which is a value of $V$ satisfying $\nabla^2 V = 0$ contains
$2m + 1$ arbitrary constant coefficients.

For $S_m(x, y, z)$ will in general consist of $\dfrac{(m+1)(m+2)}{2}$ terms and will
therefore contain $\dfrac{(m+1)(m+2)}{2}$ coefficients.

$\nabla^2 S_m(x, y, z)$ will be homogeneous of the $(m - 2)$d degree and will contain
$\dfrac{m(m-1)}{2}$ coefficients, which, of course, will be functions of the coefficients in
$S_m(x, y, z)$.  Since $\nabla^2 S_m(x, y, z) = 0$ independently of the numerical values
of $x$, $y$, and $z$ the $\dfrac{m(m-1)}{2}$ coefficients in $\nabla^2 S_m(x, y, z)$ must be separately
zero, and that fact will give us $\dfrac{m(m-1)}{2}$ equations of condition between the
$\dfrac{(m+1)(m+2)}{2}$ original coefficients and will leave $\dfrac{(m+1)(m+2)}{2} -\dfrac{ m(m-1)}{2}$
or $2m + 1$ of them undetermined. $S_m(x, y, z)$ contains, then, the same number
of arbitrary coefficients as $r^m Y_m(\mu,\phi)$.

We can then choose the coefficients in $r^m Y_m(\mu,\phi)$ so that it will transform
into any given $S_m(x, y, z)$.

Consequently a Solid Spherical Harmonic of the $m$th degree might be
defined as \textit{a homogeneous rational integral Algebraic function of x, y, and z,}
$S_m(x, y, z)$, \textit{of the $m$th degree satisfying the equation} $\nabla^2 S_m(x, y, z) = 0$; and a
Surface Spherical Harmonic of the $m$th degree as such a function divided by
$(x^2 + y^2 + z^2)^{\frac{m}{2}}$, that is by $r^m$.

% -----File: 218.png

\EXAMPLE{S}

1.\quad Show that if $S_m(x,y,z)$ is a Solid Spherical Harmonic of the $m$th degree
\[
\nabla^2[r^nS_m(x,y,z)]=n(2m+n+1)r^{n-2}S_m(x,y,z).
\]

\emph{Suggestion:}
\[
  \nabla^2S_m=0. \quad \nabla^2r=\dfrac{2}{r}.\quad
  D_rS_m=\dfrac{mS_m}{r}.\quad
  (D_xr)^2+(D_yr)^2+(D_zr)^2=1.
\]

2.\quad Show that if $f_n(x,y,z)$ is a rational integral homogeneous function of $x$,
$y$, and $z$ of the $n$th degree it can be expressed in the form
\[
f_n(x,y,z)=S_n(x,y,z) +r^2S_{n-2}(x,y,z) + r^4S_{n-4}(x,y,z) + \cdots,\tag{1}
\]
terminating with $r^{n-1}S_1(x,y,z)$ if $n$ is odd, and with $r^nS_0(x,y,z)$ if $n$ is even.\\

\emph{Suggestion}: If a term $rS_{n-1}$ were present in the second member of (1), and
we were to operate with $\nabla^2$ on both members we should by Ex.~1 have a term
$\dfrac{2n}{r }S_{n-1}$ which would be irrational when all the other terms of the resulting
equation were rational. No such term, then, could occur. In the same way
it may be shown by operating twice on (1) with $\nabla^2$ that there can be no term
$r^3S_{n-3}$ in (1); and thus step by step we can reach the result formulated in (1).\\

3.\quad Express $x^2yz$ in the form $S_4+r^2S_2+r^4S_0$.
\begin{flalign*}
&\text{\indent \emph{Suggestion}: let}&    x^2yz&=S_4+r^2S_2+r^4S_0 \rule{6em}{0em}&\
\intertext{and take $\nabla^2$ of both members we get}\\[-4ex]
&& 2yz&=14S_2+20r^2S_0.&\\[1ex]
&\text{Operate again with $\nabla^2$.}&   0&=120S_0.     &\text{Whence}
\end{flalign*}
\[
S_0=0, \quad S_2=\dfrac{1}{7}yz,\quad \text{and}\quad S_4=\dfrac{1}{7}(6x^2-y^2-z^2)yz.
\]

4.\quad Express $\sin^2\theta \cos^2\theta \sin\phi \cos\phi$ in terms of Surface Spherical Harmonics.
\begin{flalign*}
&\text{\indent \emph{Suggestion:}}&    &\sin^2\theta \cos^2\theta \sin\phi \cos\phi =\dfrac{x^2yz}{r^4}.\rule{6em}{0em}&
\end{flalign*}
For result v.\ Art.~108 (3).

\mypara{111.} A transformation of coördinates to a new set of axes having the same
origin as the old set will change a given Surface Spherical Harmonic into
another of the same degree. For such a transformation does not change the
form of Laplace's Equation $\nabla^2V=0$ if both sets of axes are rectangular,
and it is effected by replacing $x$, $y$, and $z$ in the Solid Harmonic corresponding
to the given Surface Harmonic by $x\cos\alpha_1+y\cos\alpha_2+z\cos\alpha_3$,
$x\cos\beta_1+y\cos\beta_2+z\cos\beta_3$ and $x\cos\gamma_1+y\cos\gamma_2+z\cos\gamma_3$ respectively
where the cosines are the \emph{direction cosines} of the new axes, and it will leave
% -----File: 219.png
the function a homogeneous function of the $m$th degree in the new variables,
and on dividing this by the $m$th power of the unchanged radius vector we shall
have a Surface Spherical Harmonic of the $m$th degree.

\mypara{112.} We have seen in Art.~75 that if $(x_1, y_1, z_1)$ are the coördinates of a
given point
\[
V = \dfrac{1}{\sqrt{ (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2}}\tag{1}
\]
is a solution of Laplace's Equation $\nabla^2V=0$, and transforming to spherical
coördinates that
\[
   V=\dfrac{1}{\sqrt{r^2-2rr_1[\cos\theta \cos\theta_1 +
   \sin\theta \sin\theta_1 \cos(\phi-\phi_1)] + r_1^2}}\tag{2}
\]
is a solution of
\[
   rD_r^2(rV) + \dfrac{1}{\sin\theta}
   D_\theta (\sin\theta D_\theta V) + \dfrac{1}{\sin^2 \theta}
   D_\phi^2 V = 0.   \tag{3}
\]

If $\gamma$ is the angle between the radii vectores $r$ and $r_1$ of the points $(x, y, z)$
and $(x_1, y_1, z_1)$ (1) can be written
\[
V = \dfrac{1}{\sqrt{ r^2-2rr_1 \cos\gamma + r_1^2}}\tag{4}
\]
which must be equivalent to (2), and hence
\[
\cos\gamma = \cos\theta \cos\theta_1 + \sin\theta \sin\theta_1 \cos(\phi - \phi_1).
\]
(4) which is a solution of (3) is of the same form as (5) Art.~75 and by
developing it as we developed (5) Art.~75 we find that
\[
V=P_m(\cos\gamma)
\]
is a solution of the equation
\begin{flalign*}
&&m(&m+1)V + \dfrac{1}{\sin\theta} D_\theta(\sin\theta D_\theta V) +
   \dfrac{1 }{ \sin^2\theta} D_\phi^2 V = 0 &&\tag{5}
\\
&\text{and that}&& V= r^m P_m(\cos\gamma)\quad\text{and}\quad  V=\dfrac{1}{r^{m+1} }P_m(\cos\gamma)&&\phantom{and that}
\end{flalign*}
are solutions of (3).

If we transform our coördinates keeping the origin unchanged and taking as
our new polar axis the radius vector of $(x_1, y_1, z_1)$ $\gamma$ becomes our new $\theta$ and
$P_m(\cos\gamma)$ reduces to $P_m(\cos\theta)$, a Surface Zonal Harmonic, or a \emph{Legendrian},\footnote{v.\ Art.~74.} of
the $m$th degree. It is then a Legendrian having for its axis not the original
polar axis but the radius vector of $(x_1, y_1, z_1)$. Since a Legendrian is a Surface
Spherical Harmonic,
\[
P_m(\cos\gamma) = P_m[\cos\theta \cos\theta_1 + \sin\theta \sin\theta_1 \cos(\phi-\phi_1)]
\]
is a Surface Spherical Harmonic of the $m$th degree.

% -----File: 220.png
It is, however, of very special form, since being a determinate function of
$\mu$, $\phi$, $\mu_1$, and $\phi_1$ it contains but two arbitrary constants if we regard it as a
function of $\mu$ and $\phi$, instead of containing $2m + 1$.

It is known as a \textit{Laplace's Coefficient}, or briefly as a Laplacian, of the $m$th
degree.

We shall soon express it in the regulation form of a Surface Spherical
Harmonic.

The radius vector of $(x_1, y_1, z_1)$ is called the axis of the Laplacian and the
point where the axis cuts the surface of the unit sphere is the \textit{pole} of the
Laplacian.

We shall represent the Laplacian $P_m(\cos\gamma)$ by $L_m(\mu, \phi, \mu_1, \phi_1)$. Of course
$L_m(\mu, \phi, 1, \phi_1) = P_m(\mu) = P_m(\cos\theta)$, and is really independent of $\phi$.

\mypara{113.} \emph{If the product of a Surface Spherical Harmonic of the $m$th degree by a
Laplacian of the same degree is integrated over the surface of the unit sphere, the
result is equal to $\dfrac{4\pi }{ 2m+1}$  multiplied by the value of the Spherical Harmonic at
the pole of the Laplacian.}

That is,
\[
\int\limits^{2\pi}_0 d\phi \int\limits^1_{-1} Y_m(\mu, \phi) L_m(\mu, \phi, \mu_1, \phi_1) d\mu = \dfrac{4\pi }{ 2m+1} Y_m(\mu_1, \phi_1).\tag{1}
\]

Transform to the axis of the Laplacian as a new polar axis, and let $Z_m(\mu, \phi)$
be the transformed Spherical Harmonic. $L_m(\mu, \phi, \mu_1, \phi_1)$ will become $P_m(\mu)$,
and (1) will be proved if we can show that
\begin{gather*}
   \int \limits^{2\pi}_0 d\phi
   \int \limits^1_{-1} Z_m(\mu,\phi) P_m(\mu)d\mu =
   \dfrac{4\pi }{ 2m+1} Z_m\text{(1, 0)}.\tag{2}\\
   Z_m(\mu, \phi) P_m(\mu) = A_0[P_m(\mu)]^2 +
   \sum \limits^{n=m}_{n=1} (A_n \cos n\phi + B_n \sin n\phi) P_m^n(\mu) P_m(\mu)
\end{gather*}
\indent(v.\ (5) Art.~102).
\begin{flalign*}
   &\phantom{(v.\ (5) Art.~89).}&&\int\limits^{2\pi}_0 Z_m(\mu,\phi)
   P_m(\mu)d\phi = 2\pi A_0[P_m(\mu)]^2   &\text{and}\\
   &&&\int\limits^1_{-1} d\mu \int\limits^{2\pi}_0
   Z_m(\mu,\phi) P_m(\mu)d\phi = \dfrac{4\pi}{2m+1} A_0& \text{(v.\ (5) Art.~89).}
\end{flalign*}

But $Z_m(1,0) = A_0$, since $P_m(1) = 1$ and $P_m^n(1)$ contains $(1 - 1)^{\frac{n}{2}}$ as a factor
and is equal to zero.

Hence (2) is proved.
% -----File: 221.png
\mypara{114.} We can now express a Laplacian in the regulation form as a Spherical
Harmonic, by the formulas of Art.~107.
\begin{flalign*}
   L_m(\mu,\phi,\mu_1,\phi_1)
   =& P_m(\cos \gamma) =
   P_m[\cos \theta \cos \theta_1 +
   \sin \theta \sin \theta_1 \cos(\phi - \phi_1)]&\\
   =&\sum\limits_{k = 0}^{k = \infty}\bigg[A_{0,k} P_k(\mu) +
   \sum\limits_{n = 1}^{n = k}(A_{n,k} \cos n\phi + B_{n,k} \sin n\phi) P_k^n (\mu)\bigg]&
\end{flalign*}
\begin{flalign*}
&\text{where}&
   A_{0,m}
   &= \dfrac{2m + 1}{4\pi}
      \int\limits_{0}^{2\pi}d\phi
      \int\limits_{-1}^{1} L_m(\mu,\phi,\mu_1,\phi_1)P_m(\mu)d\mu.&&\phantom{where}\\
   &&&= \dfrac{2m + 1}{4\pi}
      \dfrac{4\pi}{2m + 1}P_m(\mu_1)=P_m(\mu_1) &\tag*{by (1) Art.~113,}
\end{flalign*}
\begin{flalign*}
&&  A_{n,m} &= \dfrac{2m + 1}{2\pi}\dfrac{(m - n)!}{(m + n)!}
      \int\limits_{0}^{2\pi}d\phi
      \int\limits_{-1}^{1}L_m(\mu,\phi,\mu_1,\phi_1) \cos n\phi P_m^n(\mu)d\mu\\
&&   &= \dfrac{2(m-n)!}{(m+n)!}\cos n\phi_1 P_m^n(\mu_1)  &\tag*{by (1) Art.~113, and}\\
&&   B_{n,m} &=
      \dfrac{2m+1}{2\pi}
      \dfrac{(m-n)!}{(m+n)!}
      \int\limits_{0}^{2\pi}d\phi
      \int\limits_{-1}^{1}L_m(\mu,\phi,\mu_1,\phi_1) \sin n\phi P_m^n(\mu)d\mu\\
&&   &= \dfrac{2(m-n)!}{(m+n)!}\sin n\phi_1 P_m^n(\mu_1)      &\tag*{by (1) Art.~113,}
\end{flalign*}
and $A_{0,k} = A_{n,k} = B_{n,k} = 0$ by Art.~105 unless $k = m$. Hence
\begin{multline*}
   L_m(\mu,\phi,\mu_1,\phi_1) =\\
   P_m(\mu)P_m(\mu_1) + 2\sum\limits_{n=1}^{n=m}
      \left[\dfrac{(m-n)!}{(m+n)!}P_m^n(\mu)P_m^n(\mu_1)\cos n(\phi - \phi_1)\right]. \tag{1}
\end{multline*}
Each term of a Laplacian involves a numerical coefficient, a factor which is
a function of $\mu$, a second factor which is the same function of $\mu_1$, and a third
factor which is of the form $\cos k(\phi - \phi_1)$. We give below a table
of the first few Laplacians, taken from Minchin's Statics, omitting in each term
for the sake of brevity the function of $\mu_1$.

By the aid of (1) we can write (5) Art.~107 more compactly. It becomes
\begin{flalign*}
&&f(\mu,\phi) &= \dfrac{1}{4\pi}
   \sum\limits_{m=0}^{m=\infty} (2m + 1)
   \int\limits_{0}^{2\pi}       d\phi_1
   \int\limits_{-1}^{1}         f(\mu_1,\phi_1)L_m(\mu,\phi,\mu_1,\phi_1)d\mu_1   \tag{2}\\
&\text{or}&
   F(\theta,\phi) &=
   \dfrac{1}{4\pi}
   \sum\limits_{m=0}^{m=\infty}(2m + 1)
   \int\limits_{0}^{2\pi}d\phi_1
   \int\limits_{0}^{\pi}F(\theta_1,\phi_1)P_m(\cos \gamma) \sin \theta_1 d \theta_1. &&\tag{3}
\end{flalign*}

% -----File: 222.png

\begin{center}
LAPLACIANS.
\end{center}
\[
\arraycolsep=0.3em
\begin{array}{|c|c|c|c|}
\hline
 \rule{3.5em}{0em}
      &\text{coef.\ of }\cos 0(\phi-\phi_1)
      &\text{coef.\ of}\cos(\phi-\phi_1)
      &\text{coef.\ of}\cos 2(\phi-\phi_1)\rule{0em}{1.5em}\\[0.5em]
\hline
L_0& 1&&\rule{0em}{1.5em}\\[0.5em]
L_1& \mu&(1 - \mu^2)^{\frac{1}{2}}&\rule{0em}{1.5em}\\[0.5em]
L_2& \dfrac{1}{4}(3\mu^2 - 1)
   & 3\mu(1-\mu^2)^{\frac{1}{2}}
   & \dfrac{3}{4}(1 - \mu^2)\\[1em]
L_3& \dfrac{1}{4}(5\mu^3 - 3\mu)
   & \dfrac{3}{8}(1 - \mu^2)^{\frac{1}{2}}(5\mu^2 - 1)
   & \dfrac{15}{4} \mu(1 -\mu^2)\\[1em]
L_4& \dfrac{1}{64}(35\mu^4 - 30\mu^2 + 3)
   & \dfrac{5}{8}(1- \mu^2)^{\frac{1}{2}}(7\mu^3 - 3\mu)
   & \dfrac{5}{16}(1 - \mu^2)(7\mu^2 - 1)\\[1em]
\hline
\end{array}
\]

\vspace{\baselineskip}

\[
\arraycolsep=0.3em
\begin{array}{|c|c|c|}
\hline
 \rule{3.5em}{0em}
      &\text{coef.\ of}\cos 3(\phi-\phi_1)
      &\text{coef.\ of}\cos 4(\phi-\phi_1)\rule{0em}{1.5em}\\[0.5em]
\hline
L_0&&\rule{0em}{1.5em}\\[0.5em]
L_1&&\rule{0em}{1.5em}\\[0.5em]
L_2&&\rule{0em}{1.5em}\\[1em]
L_3& \dfrac{5}{8}(1 - \mu^2)^{\frac{3}{2}}
   & \rule{0em}{1.5em}\\[1em]
L_4& \dfrac{35}{8} \mu(1 - \mu^2)^{\frac{3}{2}}
   & \dfrac{35}{64} (1 - \mu^2)^2\rule{0em}{1.5em}\\[1em]
\hline
\end{array}
\]

\EXAMPLE{}

Work the problems of Art.~108 and Art.~108 Exs.~1 and 2 by the aid of (3)
Art.~114.

% -----File: 223.png
\mypara{115.} Such problems as we have handled in Arts.~98 and 99, and also problems
differing from them in not having circular symmetry about an axis, can
now be solved by direct integration.

For instance let it be required to find the value at an external point of the
potential function due to the attraction of a solid sphere whose density at any
point is proportional to the product of any power of the radius vector by a
Surface Spherical Harmonic.
\begin{flalign*}
&\text{\indent Let}& \rho = Cr_1^kY_m(\mu_1,\phi_1).  &&\phantom{\indent Let}
\end{flalign*}
Then using our ordinary notation we have
\[
  V = \int\limits_0^a dr_1
      \int\limits_0^{2\pi} d\phi_1
      \int\limits_{-1}^1
      \dfrac{Cr^k_1 Y_m(\mu_1,\phi_1)r_1^2d\mu_1 }{\sqrt{r^2 - 2rr_1 \cos\gamma + r_1^2}}.
\]
But by (3) Art.~77
\begin{flalign*}
   \dfrac{1 }{\sqrt{ r^2 - 2rr_1 \cos \gamma + r_1^2}} =
   \dfrac{1}{r} &\bigg [P_0(\cos\gamma) +
   \dfrac{r_1}{r }P_1(\cos\gamma) \\
   & + \dfrac{r_1^2}{r^2}P_2(\cos\gamma) + \cdots +
   \dfrac{r_1^m }{r^m }P_m(\cos\gamma) + \cdots \bigg]\\[1ex]
\text{if}  \quad &r > r_1.
\end{flalign*}
Consequently since
\[
   \int\limits_0^{2\pi} d\phi_1
   \int\limits_{-1}^1 Y_m(\mu_1,\phi_1)Y_n(\mu_1,\phi_1)d\mu_1 = 0,
\]
$V$ reduces to the single term
\begin{align*}
   V &= \frac{C}{r^{m+1}}
        \int\limits_0^a r_1^{m+k+2} dr_1
        \int\limits_0^{2\pi} d\phi_1
        \int\limits_{-1}^1Y_m(\mu_1, \phi_1)P_m(\cos\gamma)d\mu_1\\
     &= \frac{C}{r^{m+1}}
        \int\limits_0^a r_1^{m+k+2}
        \left (\frac{4\pi }{ 2m+1} Y_m(\mu,\phi)  \right )dr_1    \tag*{by Art.~113.}
\end{align*}
\[
   \therefore \:V =
      \frac{4\pi C }{ 2m+1}.
      \frac{a^{m + k + 3} }{m + k + 3}.
      \frac{Y_m(\mu,\phi) }{r^{m+1}}.
\]
% -----File: 224.png

\EXAMPLE{S}

1.\quad Solve by direct integration the problems worked in Arts.~98 and 99 and
Examples 1, 2, 3, and 4 of Art.~99.\\

2.\quad The density of a solid sphere is proportional to the product of the
squares of the distances from two mutually perpendicular diametral planes;
find the value of the potential function at an external point.
\begin{alignat*}{3}
&\textit{\indent Ans.}\qquad &\rho&= kr_1^4 \cos^2\theta_1 \sin^2\theta_1 \cos^2\phi_1 &&
\\[1ex]
&&&= kr_1^4 \bigg [\frac{1}{15} P_0(\mu_1) + \frac{1}{21} P_2(\mu_1) &&+ \frac{1}{42} \cos 2\phi_1 P_2^2(\mu_1)   \\
&&&&& - \frac{4}{35} P_4(\mu_1) + \frac{1}{105} \cos 2\phi_1 P_4^2(\mu_1) \bigg  ].\\[-8ex]
\end{alignat*}
\begin{multline*}
V = \frac{M}{a} \left [ \frac{a}{r} + \frac{a^3}{r^3} \left ( \frac{1}{9}P_2(\mu) + \frac{1}{18} \cos 2\phi P_2^2(\mu) \right ) \right .\\
\left . - \frac{a^5}{r^5} \left ( \frac{4}{33} P_4(\mu) - \frac{1}{99} \cos 2\phi P_4^2(\mu) \right ) \right ].
\end{multline*}

3.\quad Solve Example 2 by an extension of the method of Arts.~98 and 99.\\

4.\quad A conducting sphere of radius $a$ connected with the ground by a wire is
placed in the field of force due to an electrified point at which $m$ units of
electricity are concentrated. Find the value of the potential function due to
the induced charge.\\

\textit{Suggestion:} Let $V_1$ be the potential function due to the point, and $V_2$ that
due to the induced charge, and let $b$ be the distance of the point from the
centre of the sphere. Then
\begin{align*}
V_1 &= \frac{m}{\sqrt{\strxx b^2-2br \cos\theta + r^2} } &&\\
&=\frac{m}{b} \left [ P_0(\cos\theta) +\frac{ r}{b}P_1(\cos\theta) + \frac{r^2}{b^2}P_2(\cos\theta) + \cdots \right ]  \quad \text{if} \quad  r<b.&&\\
&=\frac{m}{r} \left [ P_0(\cos\theta) +\frac{ b}{r}P_1(\cos\theta) + \frac{b^2}{r^2}P_2(\cos\theta) + \cdots \right ]  \quad \text{if} \quad  r>b.&&\\
V_2 &= A_0P_0(\cos\theta) + A_1 \frac{r}{a }P_1(\cos\theta) + A_2 \frac{r^2}{a^2 }P_2(\cos\theta) + \cdots   \quad \text{if} \quad  r<a. &&\\
&= A_0\frac{a}{r}P_0(\cos\theta) + A_1 \frac{a^2}{r^2}P_1(\cos\theta) + A_2 \frac{a^3}{r^3 }P_2(\cos\theta) + \cdots   \quad \text{if} \quad  r>a.&&
\end{align*}
When $r = a$\;\;$V_1 + V_2 = 0$. Hence
\[
A_0 = -\dfrac{m}{b}, \quad A_1 = -\dfrac{ma}{b^2}, \quad A_2 = -\dfrac{ma^2}{b^3}, \cdots
\]
% -----File: 225.png
and
\begin{align*}
   V_2&=-\dfrac{m}{b}\left[P_0(\cos\theta)+
         \dfrac{r}{b} P_1(\cos\theta) +
         \dfrac{ r^2}{b^2} P_2(\cos\theta)+
         \cdots\right] \quad \text{if}\quad r<a\\
      &=-\dfrac{ma}{br}\left[P_0(\cos\theta)+
         \dfrac{a^2}{br} P_1(\cos\theta) +
         \dfrac{a^4}{b^2r^2} P_2(\cos\theta) +
         \cdots\right] \quad \text{if}\quad r>a.
\end{align*}

Hence the effect of the induced charge is precisely the same at an external
point as if the sphere were replaced by $\dfrac{ma}{b}$ units of negative electricity concentrated
at the point $r=\dfrac{a^2}{b}$, $\theta=0$. v.\ Peirce, Newt.\ Pot.\ Func., \S~66.

\mypara{116.}  If the two points $P$ and $P'$ are taken on the line $OH$ whose direction
cosines are $\lambda$, $\mu$, and $\nu$, and if $u$ and $u'$ are the values at $P$ and $P'$ of any continuous
function of the space coördinates, then $\displaystyle\limit_{PP'\doteq 0} \left[\dfrac{u'-u}{PP'}\right]$  is called the
\emph{partial derivative} of $u$ along the line $OH$ and will be represented by $D_hu$.

Let $x$, $y$, $z$ be the coördinates of $P$ and $x + \Delta x$, $y + \Delta y$, $z + \Delta z$ the coördinates
of $P'$; then
\[
u'-u=D_xu.\Delta x+D_yu.\Delta y+D_zu.\Delta z+\epsilon
\]
where $\epsilon$ is an infinitesimal of higher order than the first if $\Delta x$, $\Delta y$, and $\Delta z$ are
infinitesimal (v.\ Dif.\ Cal.\ Art.~198).
\begin{flalign*}
&\text{\indent Hence}& \dfrac{u'-u}{PP'} &=D_x u.\dfrac{\Delta x}{PP'} + D_y u.\dfrac{\Delta y}{PP'} + D_z u.\dfrac{\Delta z}{PP'} + \dfrac{\epsilon}{PP'}.&\phantom{\indent Henc}\\
&\text{But}&& \dfrac{\Delta x}{PP'} = \lambda,\quad \dfrac{\Delta y}{PP'}=\mu,\quad\text{and}\quad\dfrac{ \Delta z}{PP'}=\nu.&\\
&\text{Therefore}&  &\qquad D_hu=\lambda D_xu+\mu D_yu+ \nu D_zu.  & \tag{1}
\end{flalign*}

If $\nabla^2u=0$, $D_x^pD_y^qD_z^r u$ is a solution of Laplace's Equation.
\begin{flalign*}
   &\text{\indent For}&
       \nabla^2(D_x^pD_y^qD_z^ru) &=
       D_x^pD_y^qD_z^r(\nabla^2u) = 0.&\phantom{\indent For}
\end{flalign*}

Hence if $\nabla^2u=0$ $D_hu$ is a solution of Laplace's Equation, and if $OH_1$,
$OH_2$, $OH_3$, $\cdots$ are a set of lines through the origin $D_{h_1}D_{h_2}D_{h_3}\cdots u$ is a solution
of Laplace's Equation.

\mypara{117.} If $H_k$ is a rational integral homogeneous Algebraic function of $x$, $y$,
and $z$ of the $k$th degree
\begin{align*}
D_x\left(\dfrac{H_k}{r^l}\right) &= D_r\left(\dfrac{H_k}{r^l}\right)D_xr+\dfrac{1}{r^l}D_x(H_k)\\
                                                           &=-\dfrac{lxH_k}{r^{l+2}} + \dfrac{H_{k-1}}{r^l} = -\dfrac{lxH_k}{r^{l+2}} + \dfrac{r^2H_{k-1}}{r^{l+2}},
\end{align*}
and is of the form $\dfrac{H_{k+1}}{r^{l+2}}$.
% -----File: 226.png

The same thing can be proved of $D_y\Big(\dfrac{H_k}{r^l}\Big)$ and
$D_z\Big(\dfrac{H_k}{r^l}\Big)$ and therefore holds
good of $D_h\Big(\dfrac{H_k}{r^l}\Big)$.

If $u$ is a homogeneous function of $x$, $y$, and $z$ of the degree $-m-1$ and
 $\nabla^2u=0$ then $\nabla^2(r^{2m+1}u)=0$.
\begin{flalign*}
&&\llap{$\displaystyle \nabla^2(r^{2m+1}u)
    ={}$}&(2m+1)(2m+2)r^{2m-1}u\\
&&    &+2(2m+1)r^{2m-1}(xD_xu + yD_yu + zD_zu) + r^{2m+1}\nabla^2u\hspace{1.5em}\\
&&    ={}&0,\\
&\text{since} &&\;xD_xu+yD_yu+zD_zu=-(m+1)u
\end{flalign*}
by Euler's Theorem (v.\ Dif.\ Cal.\ Art.~220).

\mypara{118.} $\dfrac{M}{r}=\dfrac{M}{ \sqrt{x^2+y^2+z^2 }}$ is a solution of Laplace's Equation (v.\ Art.~75)
and is of the form $\dfrac{H_0}{r}$.

$D_{h_1}D_{h_2}D_{h_3} \cdots D_{h_m} \Big (\dfrac{M}{r} \Big )$ is then a solution of Laplace's Equation by Art. 116;
it is of the form $\dfrac{H_m}{r^{2m+1}}$ by Art.~117 and is a homogeneous function of the
degree $ -m-1$.

Therefore $r^{2m+1}D_{h_1}D_{h_2}D_{h_3} \cdots D_{h_m} \Big (\dfrac{M}{r} \Big )$ is a solution of Laplace's Equation,
and is a rational integral homogeneous Algebraic function of $x$, $y$, and $z$ of the
$m$th degree, and is consequently a Solid Spherical Harmonic of the $m$th
degree (v.\ Art.~110); and $r^{m+1}D_{h_1}D_{h_2}D_{h_3} \cdots D_{h_m} \Big (\dfrac{M}{r} \Big )$ is a Surface Spherical Harmonic of the $m$th degree.

Moreover since the direction of each of the lines $OH_1$, $OH_2$, $\cdots$ $OH_m$ depends
upon two angles which may be taken at pleasure, these angles and $M$ are
$2m+1$ arbitrary constants and may be so chosen that $r^{m+1}D_{h_1}D_{h_2}D_{h_3} \cdots$
$D_{h_m} \Big (\dfrac{M}{r} \Big )$ may be any given Surface Spherical Harmonic.

Consequently any given Surface Spherical Harmonic may be regarded as
formed by differentiating $\dfrac{M}{r}$ successively along $m$ determinate lines $OH_1$, $OH_2$ $\cdots$
$OH_m$, and is given except for the undetermined factor $M$ when these lines are given.

The lines $OH_1$, $OH_2$, $OH_3, \cdots OH_m$ are called the \textit{axes} of the Harmonic, and
the points where they meet the surface of the unit sphere the \textit{poles} of the
Harmonic. The $m$ axes of a Zonal Harmonic coincide with the axis of coördinates
(v.\ Art.~86) and consequently the $m$ axes of a Laplacian coincide with
what we have called the axis of the Laplacian (v.\ Art.~112).

% -----File: 227.png
\mypara{119.} Any \emph{Surface Zonal Harmonic} $P_m(\mu)$ is equal to zero for $m$ real and
distinct values of $\mu$ which lie between $-1$ and 1; and any \emph{Associated Function}
$P_m^n(\mu)$ is equal to zero for $m-n$ real and distinct values of $\mu$, which lie
between $-1$ and 1.
\begin{flalign*}
&\phantom{v.\ Art.~83 (1).}&P_m(\mu) &= \dfrac{1}{ 2^m m!}.\dfrac{ d^m (\mu^2 - 1)^m }{ d \mu^m}.  & \text{v.\ Art.~83 (1).}
\end{flalign*}
$\dfrac{d^k (\mu^2 - 1)^m }{ d \mu^k}$ contains $(\mu^2 - 1)^{m-k}$ as a factor.  v.\ Art.~89.

From Rolle's Theorem, ``If $f(x)$ is continuous and single-valued and is equal
to zero for the real values $a$ and $b$ of $x$, $\dfrac{df(x)}{dx}$ is equal to zero for at least one
real value of $x$ between $a$ and $b$,'' (v.\ Dif.\ Cal.\ Art.~126) it follows that since
 $(\mu^2 - 1)^m = 0$ when $\mu = -1$ and when $\mu = 1$ $\dfrac{d(\mu^2 - 1)^m }{ d\mu} = 0$ for at least
one value of $\mu$ between $-1$ and 1. $\dfrac{d(\mu^2 - 1)^m }{ d\mu}$ cannot be equal to zero for
more than one value of $\mu$ between $-1$ and 1, for it contains $(\mu^2 - 1)^{m-1}$ as a
factor and is a rational Algebraic polynomial of the $2m - 1$st degree.

In like manner we can show that $\dfrac{d^2 (\mu^2 - 1)^m }{d\mu^2} = 0$ has $m-2$ roots equal to
$-1$, $m-2$ roots equal to 1 and two real roots between $-1$ and 1 which
separate the three distinct roots of $\dfrac{d(\mu^2 - 1)^m }{ d\mu} = 0$; and in general if $k < m+1$
that $\dfrac{d^k(\mu^2 - 1)^m }{ d\mu^k} = 0$ has $m-k$ roots equal to $-1$, $m-k$ roots equal to 1,
and $k$ real roots separating the $k+1$ distinct roots of $\dfrac{d^{k-1}(\mu^2 - 1)^m}{d\mu^{k-1}} = 0$.

Hence $P_m(\mu) = 0$ or $\dfrac{1}{2^m m!}.\dfrac{d^m (\mu^2 - 1)^m }{ d\mu^m} = 0$ has $m$ real and distinct roots
between $-1$ and 1, and it has no more since it is of the $m$th degree.

The same reasoning shows that $\dfrac{d^{m+n}(\mu^2 - 1)^m }{ d\mu^{m+n}} = 0$ has $m-n$ distinct real
roots between $-1$ and 1, and therefore that $P_m^n(\mu)$ is equal to zero for $m-n$
distinct real values of $\mu$ between $-1$ and 1. Since $P_m^n(\mu)$ contains $\sin^n \theta$ as a
factor it is also equal to zero when $\mu = -1$ and when $\mu = 1$.

$\cos n\phi$ is equal to zero for $2n$ equidistant values of $\phi$, and $\sin n\phi$ is equal to
zero for $2n$ values of $\phi$. Hence any \emph{Tesseral Harmonic} $\sin n\phi P_m^n(\mu)$ or
$\cos n\phi P_m^n(\mu)$ is equal to zero for $2n$ equidistant values of $\phi$, for $\mu = 1$, for
$\mu = -1$, and for $m-n$ real and different values of $\mu$ between $-1$ and 1.

It follows that the value of any Surface Zonal Harmonic $P_m(\mu)$ at a point
on the surface of the unit sphere will have the same sign so long as the point
remains on one of the \textit{zones} into which the surface of the sphere is divided by
% -----File: 228.png
the $m$ circles of latitude corresponding to the $m$ roots of $P_m(\mu) = 0$, and will
change sign whenever the point passes from one of these zones into an adjoining
one; and that the value of any Tesseral Harmonic $\sin n\phi P^n_m(\mu)$ at a point
on the surface of the unit sphere will have the same sign so long as the point
remains on any one of the \textit{tesserae} into which the surface of the sphere is
divided by the $m-n$ circles of latitude corresponding to the roots of $P_m^n(\mu)=0$
and the $2n$ meridians corresponding to the roots of $\sin n\phi = 0$, and will change
sign whenever the point passes from one of these tesserae into an adjoining
one.
\label{ch6end}
% -----File: 229.png
\mychap{CHAPTER VII.\footnotemark}{CYLINDRICAL HARMONICS (BESSEL'S FUNCTIONS).}
\label{ch7start}

\footnotetext{The student should re-read carefully Arts.~11, 17,
and 18(\emph{d}) before beginning this chapter.}

\mypara{120.} In Arts.~11 and 17 we obtained
\[
   z = AJ_0(x) + BK_0(x) \tag{1}
\]
as the general solution of \textit{Fourier's Equation}
\begin{flalign*}
   &&&\frac{d^2z}{dx^2} +
   \frac{1}{x} \frac{dz}{dx}+ z = 0, \tag{2}
\\
&\text{where}& J_0(x) ={}&1 -
   \frac{x^2}{2^2} +
   \frac{x^4}{2^2.4^2 } -
   \frac{x^6}{2^2.4^2.6^2} + \cdots \tag{3}
&&\phantom{where}
\end{flalign*}
and is called a \textit{Cylindrical Harmonic} or \textit{Bessel's Function} of the zeroth order;
and where
\[
   K_0(x) = J_0(x)\log x +
   \frac{x^2}{2^2} -
   \frac{x^4}{2^2.4^2}\left  (
   \frac{1}{1} +
   \frac{1}{2} \right ) +
   \frac{x^6}{2^2.4^2.6^2 } \left (
   \frac{1}{1} +
   \frac{1}{2} +
   \frac{1}{3 }\right )  - \cdots \tag{4}
\]
and is called a \textit{Cylindrical Harmonic} or \textit{Bessel's Function} of the Second Kind,
and of the zeroth order.

In Art.~17 we found that\qquad\quad $z = J_n(x)$

\noindent is a particular solution of \textit{Bessel's Equation}
\[
   \frac{d^2z}{dx^2} +
   \frac{1}{x}\frac{ dz}{dx} +\left (1 -
   \frac{n^2}{x^2} \right ) z = 0, \tag{5}
\]
where if $n$ is unrestricted in value
\begin{multline*}
   J_n(x) = \frac{x^n}{2^n \Gamma(n+1)} \left [1 -
   \frac{x^2}{2^2(n+1)} +
   \frac{x^4}{2^4.2!(n+1)(n+2)} \right . \\ \left . -
   \frac{x^6}{2^6.3!(n+1)(n+2)(n+3)} + \cdots \right ]\tag{6}
\end{multline*}
and is called a \textit{Cylindrical Harmonic} or \textit{Bessel's Function} of the $n$th order;
and that unless $n$ is an integer
\[
   z = AJ_n(x)+ BJ_{-n}(x)
\]
is the general solution of Bessel's Equation.

% -----File: 230.png
If $n$ is an integer it can be shown that
\[
J_n (x) = (-1)^n J_{-n} (x),
\]
(v.\ Forsyth's Diff.\ Eq.\ Art.~102), and then
\[
z = AJ_n(x) + B\{K_n(x)\}
\]
is the general solution of Bessel's Equation and
\begin{multline*}
   \{K_n(x)\}= J_n(x)\log x -
      \dfrac{1}{2}\left(
      \dfrac{x}{2}\right)^{-n}
      \sum\limits^{k=n-1}_{k=0}
      \dfrac{(n-k-1)!}{k!} \left(\dfrac{x}{2}\right)^{2k}\\
      - \dfrac{1}{2}\left(\dfrac{x}{2}\right)^n
      \sum\limits^{k=\infty}_{k=0}
      \dfrac{(-1)^k}{(n+k)!k!}\left[1 +
      \dfrac{1}{2}+\dfrac{1}{3}+\cdots +
      \dfrac{1}{k}\right.\\
      \left.+1+\dfrac{1}{2} +
      \dfrac{1}{3}+\cdots +
      \dfrac{1}{n+k}\right]\left(
      \dfrac{x}{2}\right)^{2k}\tag{7}
\end{multline*}
v.\ M.\ Bôcher, Ann.\ Math.\ Vol.\ VI, No.\ 4.

\markright{CYLINDRICAL HARMONICS.}
\mypara{121.} A useful expression for $J_n (x)$ as a definite integral can be obtained
without difficulty from Bessel's Equation [(5) Art.~120] by a slight modification
of the method given by Forsyth (Diff.\ Eq.\ Art.~136).

It was shown in Art.~17 that $z = x^nv$ is a solution of Bessel's Equation if
$v$ satisfies the equation
\begin{flalign*}
   &&\dfrac{d^2v}{dx^2} &+\dfrac{2n+1}{x}
   \dfrac{dv}{dx} + v = 0. \tag{1}
\\
   &\text{\indent Assume} &&v =
    \int\limits^b_a T\cos (xt)dt &\phantom{\indent Assume}\tag{2}
\end{flalign*}
where $x$ and $t$ are independent, $T$ is an unknown function of $t$, and $a$ and $b$
are at present undetermined.
\begin{flalign*}
   &\text{\indent Then}&
     \dfrac{dv}{dx} &= -
     \int\limits^b_a tT\sin(xt)dt&\phantom{\indent Then}\\
   &\text{and}&
     \dfrac{d^2v}{dx^2}& = -
     \int\limits^b_at^2 T \cos(xt)dt.
\end{flalign*}
Substituting in (1) after multiplying through by $x$, we have
\[
   \int\limits^b_a(1 - t^2) Tx \cos(xt)dt -
   \int\limits^b_a(2n + 1)tT\sin(xt)dt = 0. \tag{3}
\]
% -----File: 231.png
By \emph{integration by parts} we find that
\begin{multline*}
   \int\limits^b_a(1 - t^2)Tx \cos(xt)dt = \Big[(1 - t^2) T\sin (xt)\mathop{\Big]}\limits^b_a\\
  -\int\limits^b_a\Big[(1 - t^2)\dfrac{dT}{dt} - 2tT\Big]\sin(xt)dt,
\end{multline*}
and (3) reduces to
\[
\Big[\vphantom{\dfrac{X}{Y}}(1 - t^2)T\sin(xt)\mathop{\Big]}\limits^b_a - \int\limits^b_a\Big[(1 - t^2)\dfrac{dT}{dt} + (2n - 1)tT\Big]\sin(xt)dt = 0.  \tag{4}
\]

If we determine $T$ so that
\begin{flalign*}
&&&(1-t^2) \dfrac{dT}{dt} +(2n-1)tT=0,  \tag{5}
\\
&\text{and $a$ and $b$ so that}&&   \Big[(1 - t^2)T\sin(xt)\mathop{\Big]}\limits^b_a =0 &\phantom{and a and b so that} \tag{6}
\end{flalign*}
(4) will be satisfied and our problem will be solved. (5) gives
\[
T=C(1-t^2)^{n- \frac{1}{2}}, \tag{7}
\]
and (6) will obviously be satisfied if $a = -1$ and $b = 1$.
\begin{flalign*}
&\text{\indent Hence}\qquad\qquad\quad & v&=C \int\limits^1_{-1}\dfrac{ (1-t^2)^n \cos(xt)dt }{ \sqrt{1-t^2}}&\text{is a solution of (1),}\\
&\text{and}  &   z&=Cx^n \int\limits^1_{-1}\dfrac{(1-t^2)^n \cos(xt)dt}{ \sqrt{1-t^2}}     &    (8)
\end{flalign*}
is a solution of Bessel's Equation.

If we let $t = \cos\phi$ in (8) we get
\[
z=Cx^n \int\limits^{\pi}_0 \sin^{2n} \phi \cos(x\cos\phi)d\phi.
\]

Expand $\cos(x\cos\phi)$ into a series involving powers of $x\cos\phi$, integrate
term by term by the aid of the formulas
\begin{flalign*}
&\hspace{8.5em}&\int\limits^{\tfrac{\pi}{2}}_0 \sin^n x.dx &= \dfrac{\sqrt{\pi}}{2}.\dfrac{ \Gamma\Big(\dfrac{n+1}{2}\Big)}{\Gamma\Big(\dfrac{n}{2}+1\Big)} &\text{ [Int.\ Cal.\ (1) Art.~99],}
\end{flalign*}
% -----File: 232.png
\[
   \int\limits_0^{\tfrac{\pi}{2}} \sin^n x \cos^m x.dx =
   \frac{\Gamma \Big (\dfrac{m+1}{2}\Big ) \Gamma \Big (
   \dfrac{n+1}{2} \Big ) }{2\Gamma \Big (\dfrac{m+n}{2} + 1 \Big )}
\]
(Int.\ Cal.\ Art.~99 Ex.~2), and compare with (6) Art.~120, and we get
\[
   J_n(x) = \frac{x^n }{2^n \sqrt{\pi}\
   \Gamma\Big(n + \dfrac{1}{2}\Big)}
   \int\limits_0^{\pi}\sin^{2n} \phi \cos(x \cos\phi)d\phi.     \tag{9}
\]

If $n$ is a positive integer (9) reduces to
\[
   J_n(x) = \frac{1}{\pi}.
   \frac{x^n }{1.3.5.\cdots(2n-1)}
   \int\limits_0^{\pi} \sin^{2n} \phi \cos(x \cos\phi)d\phi.    \tag{10}
\]

Let $n = 0$ in (9) or (10) and we get
\[
   J_0(x) = \frac{1}{\pi}
   \int\limits_0^{\pi} \cos(x \cos\phi)d\phi.       \tag{11}
\]

\EXAMPLE{S}

1.\quad Obtain Formula (11) directly from Fourier's Equation, (2) Art.~120.\\

2.\quad Prove by \textit{integration by parts} that if $n > -\dfrac{1}{2}$
\[
   \int\limits_0^{\pi} \sin^{2n} \phi \cos\phi \sin(x \cos\phi)d\phi =
   \frac{x}{ 2n+1}
   \int\limits_0^{\pi}\sin^{2n+2} \phi \cos(x \cos\phi)d\phi.
\]

3.\quad Prove by \textit{integration by parts} that if  $n > \dfrac{1}{2}$
\begin{align*}
   \int\limits_0^{\pi} \sin^{2n} \phi \cos\phi &\sin(x \cos\phi)d\phi\\
   &= \frac{1}{x} \int\limits_0^{\pi} [2n \sin^{2n} \phi - (2n-1)\sin^{2n-2} \phi] \cos(x \cos\phi)d\phi.
\end{align*}

\mypara{122.} We can now readily obtain a number of useful formulas.

\noindent Differentiate (11) Art.~121 with respect to $x$ and we get
\begin{flalign*}
\hspace{6em}   \dfrac{dJ_0(x) }{dx}& = -\dfrac{1}{\pi}
   \int\limits_0^{\pi} \cos\phi \sin(x \cos\phi)d\phi\\
   &= -\dfrac{x}{\pi}
   \int\limits_0^{\pi} \sin^2 \phi \cos(x \cos\phi)d\phi      \tag*{by Ex.~2 Art.~121.}\\[-6ex]
\end{flalign*}
% -----File: 233.png
\begin{flalign*}
   &\text{\indent Hence by (10) Art.~121} \qquad
   \dfrac{ dJ_0(x)}{dx} = -J_1(x).\tag{1}&&
\end{flalign*}

In like manner by the aid of Exs.~3 and 2, Art.~121, we can obtain the
relations
\[
   \frac{d[x^n J_n(x)] }{ dx} = x^n J_{n-1}(x)\tag{2}
\]
if $n > \dfrac{1}{2}$,
\[
   \frac{d[x^{-n} J_n (x)] }{ dx }= -x^{-n} J_{n+1} (x)\tag{3}
\]
if $n > -\dfrac{1}{2}$.\\

(2) can be written
\[
   \int\limits_0^x x^nJ_{n-1}(x)dx=x^nJ_n(x)\tag{4}\\[-4ex]
\]
if $n > \dfrac{1}{2}$.\\

(2) and (3) can be written
\begin{flalign*}
 &&x^n \frac{dJ_n(x)}{dx} +{}&nx^{n-1}J_n(x) = x^nJ_{n-1}(x)
\\[2ex]
   &\text{and}  &x^{-n}
      \frac{dJ_n(x)}{dx} - n&x^{-n-1}J_n(x) = -x^{-n}J_{n+1}(x), &&
\\[2ex]
    &\text{or} &
      \frac{dJ_n(x)}{dx} &= J_{n-1}(x) - \frac{n}{x}J_n(x)\tag{5}&&
\\[2ex]
   &\text{and} &
      \frac{dJ_n(x)}{dx} ={}&- J_{n+1}(x) + \frac{n}{x}J_n(x);\tag{6}&&
\\[2ex]
   &\text{whence}&
      2\frac{dJ_n(x)}{dx} &= J_{n-1}(x) - J_{n+1}(x)\tag{7}&&
\\[2ex]
   &\text{and}&
      \frac{2n}{x}J_n(x) &= J_{n-1}(x) + J_{n+1}(x).&&\tag{8}
\end{flalign*}

The repeated use of formula (8) will enable us to get from $J_0(x)$ and $J_1(x)$
any of Bessel's Functions whose order is a positive integer. For example, we
have
\begin{gather*}
   J_2(x) = \frac{2}{x}J_1(x) - J_0(x)\\
   J_3(x) = \Big ( \frac{8}{x^2}-1 \Big )J_1(x) - \frac{4}{x}J_0(x).
\end{gather*}
% -----File: 234.png

From a table giving the values of $J_0(x)$ and $J_1(x)$, then, tables for the
functions of higher order are readily constructed. Such a table taken from
Rayleigh's Sound (Vol.~I., page 265) will be found in the Appendix (Table VI.).

By the aid of (5) and (6) any derivative of  $J_n(x)$ can be expressed in terms
of $J_n(x)$ and $J_{n+1}(x)$. For example
\[
   \dfrac{d^2J_n(x)}{dx^2} = \bigg [
   \dfrac{n(n-1)}{x^2} -1\bigg ] J_n(x) +
   \dfrac{1}{x } J_{n+1}(x).
\]

If we write $J_0(x)$ for $z$ in Fourier's Equation [(2) Art.~120], then multiply
through by $xdx$ and integrate from zero to $x$, simplifying the resulting equation
by \textit{integration by parts}, we get
\begin{flalign*}
   &&x &\dfrac{dJ_0(x)}{dx} +
      \int\limits_0^x x J_0(x)dx = 0;
\\
   &\text{whence by (1)}&&
      \int\limits_0^x x J_0(x)dx = xJ_1(x).   \tag{9} &&\phantom{whence by}
\end{flalign*}

If we write $J_0(x)$ for $z$ in Fourier's Equation, then multiply through by
$x^2 \dfrac{dJ_0(x)}{dx}\,dx$ and integrate from zero to $x$, simplifying by \textit{integration by parts}
we get
\begin{flalign*}
   &&\dfrac{x^2}{2} &\bigg  [ \left (
   \dfrac{dJ_0(x)}{dx} \right )^2+(J_0(x))^2 \bigg ] -
   \int\limits_0^x x(J_0(x))^2 dx = 0;&&\qquad
\\
   &\text{whence by (1)}&
      &\int\limits_0^x x(J_0(x))^2 dx =
      \dfrac{x^2}{2} [(J_0(x))^2 + (J_1(x))^2].        \tag{10}&&
\end{flalign*}

In like manner we can get from Bessel's Equation [(5) Art.~120] the formula
\[
   \int\limits_0^x x(J_n(x))^2 dx=
      \dfrac{1}{2} \bigg [x^2 \left (
      \dfrac{dJ_n(x)}{dx} \right )^2 +
      (x^2 - n^2)(J_n(x))^2 \bigg ]   \tag{11}
\]
which (6) enables us to reduce to the form
\[
   \int\limits_0^x x(J_n(x))^2 dx=
   \dfrac{x^2}{2} [(J_n(x))^2 +
   (J_{n+1}(x))^2] - nxJ_n(x)J_{n+1}(x).\tag{12}
\]

Formulas (9), (10), (11), and (12) will prove useful when we attempt to
develop in terms of \textit{Cylindrical Harmonics}.
% -----File: 235.png

Values of $J_n(x)$ for larger values of $x$ than those given in Table VI.,
Appendix, may be computed very easily from the formula
\begin{flalign*}
   J_n(x)& = \sqrt{\frac{ 2}{\pi x} }
      \left [1-\frac{(1^2-4n^2)(3^2-4n^2)}{2!(8x)^2} \right .\\
      & \quad  \left . +
      \frac{(1^2-4n^2)(3^2-4n^2)(5^2-4n^2)(7^2-4n^2)}{4!(8x)^4} -
      \cdots \right ]\cos \left (x-\frac{\pi}{4}-n\frac{\pi}{2} \right )     \\
   & + \sqrt{ \frac{2}{\pi x}} \left [\frac{1^2-4n^2}{1!8x} \right .\\
     & \quad \left . - \frac{ (1^2-4n^2)(3^2-4n^2)(5^2-4n^2)}{3!(8x)^3} +
        \cdots \right  ]\sin \left (x-\frac{\pi}{4}-n\frac{\pi}{2} \right ).\tag{13}
\end{flalign*}
v.\ Lommel, Studien über die Bessel'schen Functionen, page 59.\\

The series terminates if $2n$ is an odd integer, but otherwise it is divergent.
It can be proved, however, that in any case the sum of $m$ terms differs from $J_n(x)$
by less than the last term included, and consequently the formula can safely
be used for numerical computation.

\EXAMPLE{S}

1.\quad Confirm (1), (2), and (3), Art.~122, by obtaining them from (3) and (6),
Art.~120.\\

2.\quad Confirm (1), Art.~122, by showing that Fourier's Equation will differentiate
into the special form assumed by Bessel's Equation when $n=1$.\\

3.\quad Show that (9), Art.~122, is a special case of (4), Art.~122.\\

4.\quad Show that the limit approached by $J_n(x)$ as $ n$ increases indefinitely is
zero, and by the aid of this fact and of (8), Art.~122, prove that
\[
   J_{n-1}(x) =
      \frac{2}{x} [nJ_n(x) -
      (n+2)J_{n+2}(x) +
      (n+4)J_{n+4}(x) + \cdots ].
\]

5.\quad Prove that
\[
   \frac{dJ_n(x)}{dx} =
   \frac{2}{x} [
   \tfrac{1}{2}nJ_n(x) -
      (n+2)J_{n+2}(x) +
      (n+4)J_{n+4}(x) - \cdots ].
\]

6.\quad Show that the substitution of $\left (1-\dfrac{y^2}{n^2}\right )^{\frac{1}{2}}$ for $x$ in Legendre's Equation
will reduce it to the form
\[
   \left (1-\frac{y^2}{n^2}\right )
   \frac{d^2z}{dy^2} +\left  (
   \frac{1}{y} -\frac{2y}{n^2} \right )
   \frac{dz}{dy }+  \left (1+
   \frac{1}{n} \right )z=0,
\]
and that the limiting form approached by this equation as $n$ is indefinitely
increased is Fourier's Equation, and hence that $J_0(x)$ can be regarded as some
constant factor multiplied by the limiting value approached by $P_n \bigg (1-\dfrac{x^2}{n^2} \bigg )^{\frac{1}{2}}$
as $n$ is indefinitely increased.
% -----File: 236.png

\mypara{123.} To complete the solution of the drumhead problem taken up in
Art. 11, we found that it would be necessary to develop a given function of $r$
in the form
\[
   f(r) = A_1 J_0(\mu_1 r) +
      A_2 J_0(\mu_2 r)  +
      A_3 J_0(\mu_3 r) + \cdots
\]
where $\mu_1$, $\mu_2$, $\mu_3$, \&c., are the roots of the transcendental equation $J_0(\mu a) = 0$;
and in Art.~11, Ex.\ the development of unity in a series of precisely the
same form was needed.\\

(\emph{a})\quad Let us consider another problem.

The convex surface and one base of a cylinder of radius $a$ and length $b$ are
kept at the constant temperature zero, the temperature at each point of the
other base is a given function of the distance of the point from the centre of
the base; required the temperature of any point of the cylinder after the
permanent temperatures have been established.

Here we have to solve Laplace's Equation in Cylindrical Coördinates
(\smallrom{XIV} Art.~1).
\[
   D_r^2 u + \frac{1}{r }D_r u +
   \frac{1}{r^2}D_{\phi}^2 u + D_z^2 u = 0  \tag{1}
\]
subject to the conditions
\begin{alignat*}{3}
      &u = 0   &&\text{when}\quad   &&z = 0\\
      &u = 0   &&\quad\text{``}     &&r = a\\
      &u =f(r) &&\quad\text{``}     &&z = b,
\end{alignat*}
and from the symmetry of the problem we know that $D_{\phi}^2u = 0$.

Assuming as usual $u = R.Z$ we break (1) up into the equations
\begin{flalign*}
&&&   \frac{d^2Z}{dz^2} -
      \mu^2 Z = 0\\[1ex]
&&   \frac{d^2R}{dr^2} &+
      \frac{1}{r}
      \frac{dR}{dr} + \mu^2 R =0,
\\[1ex]
   &\text{whence}& u &=
      \sinh (\mu z)J_0(\mu r)\tag{2}&&\phantom{whence}\\[1ex]
   &\text{and}& u &=
      \cosh (\mu z)J_0(\mu r)\tag{3}&&\\[-1ex]
\intertext{are particular solutions of (1).}
\end{flalign*}
\begin{flalign*}
   &\text{\indent If $\mu_k$ is a root of} & J_0(\mu a) = 0 &&\phantom{\indent If mu is a root of}\tag{4}
\end{flalign*}
\[
   u = \sinh(\mu_k z)J_0(\mu_k r)
\]
satisfies (1) and two of the three equations of condition.
\begin{flalign*}
   & \text{\indent If then}& f(r) =
      A_1 J_0(\mu_1 r) +
      A_2 J_0(\mu_2 r) +
      A_3 J_0(\mu_3 r) + \cdots &&\tag{5}
\end{flalign*}
$\mu_1$, $\mu_2$, $\mu_3$, \&c., being roots of (4),
\[
   u = A_1 \frac{ \sinh(\mu_1z)}{\sinh(\mu_1b)} J_0(\mu_1 r)  +
       A_2 \frac{\sinh(\mu_2z)}{\sinh(\mu_2b)} J_0(\mu_2 r) +
       A_3\frac{ \sinh(\mu_3z)}{\sinh(\mu_3b) }J_0(\mu_3 r) + \cdots  \tag{6}
\]
satisfies (1) and all of the equations of condition, and is the required solution.\\
% -----File: 237.png

(\emph{b})\quad If instead of keeping the convex surface of the cylinder at the temperature
zero we surround it by a jacket impervious to heat, the equation of
condition $u = 0$ when $r = a$ will be replaced by $D_ru = 0$ when $r = a$, or if
\begin{flalign*}
   &\qquad  & u =
      \sinh(\mu z&)J_0(\mu r), &&\\
   &\text{by}&
      \frac{dJ_0(\mu r)}{dr} &= 0&&\tag*{when $r = a$,}\\
   &\text{that is by}&
      \mu J'_0(\mu a) &= 0\footnotemark&&\tag*{or (v.\ (1) Art.~122)}\\[1ex]
   &\text{by} & J_1(\mu a)&=0.        &&      \tag{7}
\end{flalign*}

\footnotetext{We shall find it convenient to use the familiar notation of $f'(x) = \dfrac{df(x)}{dx}$ (v.\ Dif. Cal., p.\ 119).}

If now in (5) and (6) $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of (7), (6) will be the solution
of our new problem.\\

(\emph{c})\quad If instead of keeping the convex surface of the cylinder at the temperature
zero we allow it to cool in air at the temperature zero, the condition $u = 0$
when $r = a$ will be replaced by $D_ru + hu = 0$ when $r = a$, or if
\begin{flalign*}
&&u = \sinh (\mu z)J_0(&\mu r)&&\\[1ex]
&\text{by}& \mu J'_0(\mu r) + hJ_0(\mu r) &= 0    && \tag*{when \quad $r = a$}\\[1ex]
&\text{that is by}& \mu a J'_0(\mu a) + ahJ_0(\mu a) &=0   &&  \tag*{or (v.\ (1) Art.~122)}\\[1ex]
&\text{by}& \mu a J_1(\mu a) - ahJ_0(\mu a) &= 0.       &&    \tag{8}
\end{flalign*}

If now in (5) and (6) $\mu_1$, $\mu_2$, $\mu_3$,  \&c., are roots of (8), (6) will be the
solution of our present problem.

\mypara{124.} It can be shown that\\[-7.1ex]
\begin{flalign*}
   && J_0(x) &= 0 &&\tag{1}\\
   && J_1(x) &= 0 && \tag{2}\\
   &\text{and} & xJ'_0(x) + \lambda &J_0(x) = 0&& \tag{3}\\[-4ex]
\end{flalign*}
have each an infinite number of real positive roots (v.\ Riemann, Par.\ Dif.\ Gl.,
\S~97). The earlier roots of these equations can be computed without serious
difficulty from the table for the values of $J_0(x)$ (Table VI., Appendix).

The first twelve roots of $J_0(x) = 0$ and $J_1(x) =0$ are given in Table IV.,
Appendix, a table due to Stokes. Large roots of $J_0(x) = 0$ and of  $J_1(x) = 0$
may be very easily computed from the formulas
\begin{gather*}
   \frac{x_{\phantom{(}0}^{(s)}}{\pi} = s-.25 +
      \frac{.050661}{4s-1} -
      \frac{.053041}{(4s-1)^3 } +
      \frac{.262051}{(4s-1)^5 }- \cdots \tag{4}\\
   \frac{x_{\phantom{(}1}^{(s)}}{\pi} = s+.25 -
      \frac{.151982}{4s+1} +
      \frac{.015399}{(4s+1)^3} -
      \frac{.245270}{(4s+1)^5} + \cdots \tag{5}
\end{gather*}
given by Stokes in Camb.\ Phil.\ Trans., Vol.\ IX., $x_{\phantom{(}0}^{(s)}$ representing the $s$th root
of $J_0(x) = 0$, and $x_{\phantom{(}1}^{(s)}$ the $s$th root of $J_1(x) = 0$.

% -----File: 238.png
\mypara{125.} We have seen in Art.~123 that
 $U=\sinh(\mu_k z)J_0(\mu_k r)$ and $V=\sinh(\mu_l z)J_0(\mu_l r)$ are solutions of $\nabla^2U=0$
 and $\nabla^2V=0$ if we express Laplace's Equation in terms of Cylindrical
Coördinates (v.\ (1) Art.~123).

Hence, if $\int dS$ represents the surface integral over any closed surface, we
have
\[
\int(UD_nV-VD_nU)dS=0
\]
by Green's Theorem (v.\ Art.~92).

If we take the cylinder of Art.~123 as our surface, and perform the
integrations and simplify the resulting equation, we find
\begin{multline*}
\int\limits^a_0rJ_0(\mu_k r)J_0(\mu_l r)dr=\dfrac{-1}{{\mu_k}^2-{\mu_l}^2} [\mu_kaJ_0(\mu_l a)J'_0(\mu_k a)
-\mu_laJ_0(\mu_k a)J'_0(\mu_l a)]\\
=\dfrac{-1}{{\mu_l}^2-{\mu_k}^2} [\mu_kaJ_0(\mu_l a)J_1(\mu_k a)
-\mu_laJ_0(\mu_k a)J_1(\mu_l a)]. \tag{1}
\end{multline*}

Hence if $\mu_k$ and $\mu_l$ are different roots of
\begin{flalign*}
&              &                                  J_0(\mu a)&=0, &&\\
&\text{or of}  &                                  J_1(\mu a)&=0, &&\phantom{or of}\\
&\text{or of}&           \mu aJ_1(\mu a)-\lambda &J_0(\mu a)=0,  &&\\
&\text{then} & \int\limits^a_0 rJ_0(\mu_k r)J_0(&\mu_l r)dr=0. &&\tag{2}
\end{flalign*}

\EXAMPLE{}

Obtain (1) Art.~125 directly from Fourier's Equation
\[
\dfrac{d^2 J_0(\mu r)}{dr^2} + \dfrac{1}{r}\dfrac{ dJ_0(\mu r)}{dr} + \mu^2 J_0 (\mu r)=0.
\]

\mypara{126.}  We are now able to obtain the developments called for in Art.~123.
\begin{flalign*}
&\text{\indent Let}&   f(r)&=A_1J_0(\mu_1 r)+A_2J_0(\mu_2 r)+A_3J_0(\mu_3 r)+\cdots&\phantom{\indent Let} \tag{1}
\end{flalign*}
$\mu_1$, $\mu_2$, $\mu_3$, \&c., being roots of $J_0(\mu a)=0$, or of $J_1(\mu a)=0$, or of
\[
\mu aJ_1(\mu a)-\lambda J_0(\mu a)=0.
\]

To determine any coefficient $A_k$ multiply (1) by $rJ_0(\mu_k r)dr$ and integrate
from zero to $a$. The first member will become
\[
\int\limits^a_0 rf(r)J_0(\mu_k r)dr.
\]
% -----File: 239.png

Every term of the second member will vanish by (2) Art.~125 except the
term
\begin{gather*}
   A_k \int\limits_0^a r(J_0(\mu_k r))^2dr.\\
   \int\limits_0^a r(J_0(\mu_k r))^2dr =
   \frac{1}{\mu_k^2}
   \int\limits_0^{\mu_ka} x(J_0(x))^2dx =
   \frac{a^2}{2}[(J_0(\mu_k a))^2+(J_1(\mu_k a))^2]
\end{gather*}
by (10) Art.~122.
\begin{flalign*}
   &\text{\indent Hence}& A_k =
      \frac{2}{a^2[(J_0(\mu_k a))^2 +
      (J_1(\mu_k a))^2] }
      \int\limits_0^a rf(r)J_0(\mu_k r)dr. && \tag{2}
\end{flalign*}
The development (1) holds good from $r = 0$ to $r = a$ (v.\ Arts.~24, 25, and 88).

If $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of $J_0(\mu a)=0$, (2) reduces to
\[
   A_k =
      \frac{2}{a^2(J_1(\mu_k a))^2}
      \int\limits_0^a rf(r)J_0(\mu_k r)dr.  \tag{3}
\]

If $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of $J_1(\mu a)=0$, (2) reduces to
\[
   A_k = \frac{2}{a^2(J_0(\mu_k a))^2}
      \int\limits_0^a rf(r)J_0(\mu_k r)dr.  \tag{4}
\]

If $\mu_1$, $\mu_2$, $\mu_3$, \&c., are roots of $\mu aJ_1(\mu a)-\lambda J_0(\mu a)=0$, (2) reduces to\\
\[
   A_k = \frac{2\mu_k^2}{(\lambda^2+\mu_k^2a^2)(J_0(\mu_k a))^2}
      \int\limits_0^a rf(r)J_0(\mu_k r)dr.  \tag{5}
\]

For the important case where $ f(r) = 1$
\[
  \int\limits_0^a r f(r) J_0(\mu_k r) dr =
     \int\limits_0^a r J_0(\mu_k r) dr =
     \frac{1}{\mu_k^2}
     \int\limits_0^{\mu_k a}xJ_0(x) dx =
     \frac{a}{\mu_k} J_1(\mu_k a)   \tag{6}
\]
by (9) Art.~122, and (3) reduces to\\
\begin{flalign*}
&&   A_k&=\frac{2}{\mu_k aJ_1(\mu_k a)},\hspace{1.6em}  && \tag{7}
\end{flalign*}

(4) reduces to $A_k = 0$ except for $k = 1$ when $\mu_k = 0$ and we have $A_1 = 1$,

\begin{flalign*}
&\indent\text{(5) reduces to} &A_k &=
      \frac{2\lambda }{(\lambda^2 + \mu_k^2 a^2)J_0(\mu_k a)}.   && \phantom{(5) reduces to} \tag{8}
\end{flalign*}
% -----File: 240.png

\EXAMPLE{S}

1.\quad Show that in (12) Art.~11 any coefficient $A_k$ has the value given in (3)
Art.~126; and in the answer to Art.~11, Ex.~the value given in (7) Art.~126.\\

2.\quad Show that if a drumhead be initially distorted so that it has circular
symmetry, it will not in general give a musical note; that it may be initially
distorted so as to give a musical note; that in this case the vibration will be
a \textit{steady} vibration; that the frequencies of the various musical notes that can be
given when the distortion has circular symmetry are proportional to the roots
of $J_0(x) = 0$; that the possible nodes for such vibrations are concentric circles
whose radii are proportional to the roots of $J_0(x) = 0$.\\

3.\quad A cylinder of radius one meter and altitude one meter has its upper
surface kept at the temperature 100°, and its base and convex surface at the
temperature 15°, until the \textit{stationary temperature} is set up. Find the temperature
at points on the axis 25 cm., 50 cm., and 75 cm.\ from the base, and also
at a point 25 cm.\ from the base and 50 cm.\ from the axis.\\
\phantom{kindly go to the right hand margin}\hfill
\emph{Ans.}, 29°.6; 47°.6; 71°.2; 25°.8.\\

4.\quad An iron cylinder one meter long and twenty centimeters in diameter has its
convex surface covered with a so-called non-conducting cement one centimeter
thick. One end and the convex surface of the cylinder thus coated are kept at the
temperature zero, the other end at the temperature of 100°. Find to the nearest
tenth of a degree the temperature of the middle point of the axis, and of the
points of the axis twenty centimeters from each end after the temperatures
have ceased to change. Given that the conductivity of iron is 0.185 and of
cement 0.000162 in C. G. S. units. Find also the temperature of a point on
the surface midway between the ends, and of points on the surface twenty
centimeters from each end. Find the temperatures of the three points of the
axis, supposing the coating a perfect non-conductor, and again, supposing the
coating absent. Neglect the curvature of the coating.\\
\phantom{right margin}\hfill
\emph{Ans.}, 15°.4;  40°.85;  72°.8;  15°.3;  40°.7;  72°.5;  0°.0;  0°.0;  1°.3.

\mypara{127.} If instead of considering the cooling of a cylinder as in Art.~123 we
have to deal with a cylindrical shell whose curved surfaces are co-axial
cylinders, we are obliged to use the Bessel's Functions of the second kind.
Let our equations of condition be
\begin{alignat*}{6}
&u = 0    &&\text{when}\quad &&z = 0, \qquad\qquad && u = 0\quad  &&\text{when}\quad &&r = a,\\
&u = f(r) &&\quad\text{`` }  &&z = b,              && u = 0       &&\quad\text{`` }  &&r = c.
\end{alignat*}

Then (v.\ Art.~123)
\[
    u = \sinh(\mu_k z)
        \left [J_0(\mu_k r) -
        \frac{J_0(\mu_kc)}{K_0(\mu_kc)}K_0(\mu_k r)
        \right ]
\]
% -----File: 241.png
where $\mu_k$ is a root of the equation
\[
J_0(\mu a)-\dfrac{J_0(\mu c)}{K_0(\mu c)} K_0(\mu a)=0 \tag{1}
\]
will satisfy Laplace's Equation [(1) Art.~123] and all of the equations of
condition except the second.
\begin{flalign*}
&\text{\indent Hence} & u&= \sum\limits^{k=\infty}_{k=1} A_k \dfrac{\sinh(\mu_k z)}{\sinh(\mu_k b)}\left[J_0(\mu_k r)-\dfrac{J_0(\mu_kc)}{K_0(\mu_kc)} K_0(\mu_k r)\right]&\phantom{Hence}\tag{2}
\intertext{is the required solution if}
&&&f(r) = \sum\limits^{k=\infty}_{k=1} A_k \bigg[J_0(\mu_k r)-\dfrac{J_0(\mu_kc)}{K_0(\mu_kc)} K_0(\mu_k r)\bigg].    \tag{3}
\end{flalign*}

The development (3) is easily obtained.

Call the parenthesis for the sake of brevity $B_0(\mu_kr)$. Then by the method
of Art.~125 we get if we integrate over our cylindrical shell
\[
\int\limits^c_a rB_0(\mu_kr)B_0(\mu_lr)dr=0 \tag{4}
\]
if $\mu_k$ and $\mu_l$ are roots of (1); and by an easy extension of (10) Art.~122
\[
   \int\limits^c_ar[B_0(\mu_kr]^2dr =
   \tfrac{1}{2}\lbrace c^2[B'_0(\mu_kc)]^2-a^2[B'_0(\mu_ka)]^2\rbrace. \tag{5}
\]

Determining the coefficients in (3) as in Art.~124 and simplifying by the
aid of (4) we have

\noindent
\[
   A_k = \dfrac{\displaystyle 2\int\limits^c_a  rf(r)
      \Big[J_0(\mu_k r) -
      \dfrac{J_0(\mu_kc)}{K_0(\mu_kc)}K_0(\mu_k r)\Big]dr}
   {c^2\Big[J'_0(\mu_k c) -
      \dfrac{J_0(\mu_kc)}{K_0(\mu_kc)} K'_0
         (\mu_k c)\Big]^2-a^2\Big[{J_0}'(\mu_k a) -
      \dfrac{J_0(\mu_kc)}{K_0(\mu_kc)} K'_0(\mu_k a)\Big]^2}.\tag{6}
\]

\EXAMPLE{}

If a membrane bounded by concentric circles of radius $a$ and radius $b$, and
fastened at the edges, is initially distorted into a form symmetrical with respect
to the centre, and then allowed to vibrate
\[
y= \sum\limits^{k=\infty}_{k=1}  A_k \cos(\mu_kct)\left[J_0(\mu_k r)-\dfrac{J_0(\mu_kb)}{K_0(\mu_kb)} K_0(\mu_k r)\right]
\]
where $A_k$ is obtained from (6) Art.~127 by replacing $c$ by $b$.

% -----File: 242.png
\mypara{128.} If in the cooling of a cylinder $u = 0$ when $z = 0$, $u = 0$ when $z = b$,
and $u = f(z)$ when $r = a$, the problem is easily solved.

If in (2) and (3) Art.~123 $\mu$ is replaced by $\mu i$ we can readily obtain
\begin{flalign*}
&                    &z &= \sin (\mu z) J_0(\mu ri)&\\
&\text{and} & z& = \cos (\mu z) J_0(\mu ri)&\phantom{and}
\end{flalign*}
as particular solutions of Laplace's Equation [(1) Art.~123]; and
\[
           J_0(xi)=1 + \dfrac{x^2}{2^2} + \dfrac{x^4}{2^2.4^2} + \dfrac{x^6}{2^2.4^2.6^2} +\cdots \tag{1}
\]
and is real.
\begin{flalign*}
&&               &f(z)= \sum \limits^{k=\infty}_{k=1} A_k \sin \dfrac{k\pi z}{b} &&\\
&\text{where}    &&A_k = \dfrac{2}{b}\int \limits^b_a f(z) \sin \dfrac{k\pi z}{b} dz &&\tag{2}
\intertext{by Art.~31 (7) and (8).}
&\text{\indent Hence}&u&=\sum \limits^{k=\infty}_{k=1} A_k \sin \dfrac{k\pi z}{b}
\dfrac{ J_0\Big(\dfrac{k\pi ri}{b}\Big)}
          {J_0\Big(\dfrac{k\pi ai}{b}\Big)} && \tag{3}
&\intertext{is our required solution.}
\end{flalign*}

\EXAMPLE{S}

1.\quad If the cylinder is hollow and we have $u = 0$ when $z = 0$, $u = 0$ when
$z = b$, $u = 0$ when $r = c$, and $u =f(z)$ when $r = a$; then
% recast to fit page width
\begin{multline*}
u=\sum\limits^{k=\infty}_{k=1} A_k \sin \dfrac{k\pi z}{b}
\left[
\dfrac{
J_0\Big(\dfrac{k\pi ri}{b}\Big)}{
J_0\Big(\dfrac{k\pi ci}{b}\Big)}
-\dfrac{
\overline{K_0}\Big(\dfrac{k\pi ri}{b}\Big)}{
\overline{K_0}\Big(\dfrac{k\pi ci}{b}\Big)}
\right]\\
\div
\left [
\dfrac{
J_0\Big(\dfrac{k\pi ai}{b}\Big)}{
J_0\Big(\dfrac{k\pi ci}{b}\Big)}
-\dfrac{
\overline{K_0}\Big(\dfrac{k\pi ai}{b}\Big)}{
\overline{K_0}\Big(\dfrac{k\pi ci}{b}\Big)}
\right]
\end{multline*}
where $A_k$ has the value given in (2) Art.~128, and
\begin{align*}
\overline{K_0}(xi) &= K_0(xi) - J_0(xi) \log i\\
            &= J_0(xi) \log x - \dfrac{x^2}{2^2} - \dfrac{x^4}{2^2.4^2}(\tfrac{1}{1}+\tfrac{1}{2}) - \dfrac{x^6}{2^2.4^2.6^2}(\tfrac{1}{1}+\tfrac{1}{2}+\tfrac{1}{3}) -\cdots
\end{align*}
[v.\ (4) Art.~120], and is real.\\

2.\quad A hollow cylinder 6 feet long whose inner surface has the radius 3 inches,
and whose outer surface has the radius 1 foot, has its bases and outer surface
kept at the temperature 0°, and its inner surface at the temperature 100°, until
% -----File: 243.png
the permanent state of temperatures is established; find the temperatures of
two points in a plane parallel to the bases and half-way between them, one of
which is 6 inches and the other 9 inches from the axis.   \hfill \emph{Ans.}, 49°.6; 20°.2.

\mypara{129.} If in the problem of Art.~123 the temperatures of the points of the
upper base of the cylinder are unsymmetrical so that $u=f(r,\theta)$ when $z = b$,
 we have to get particular solutions of Laplace's Equation [(1) Art.~123] for
the case where $D_\phi^2u$ is not equal to zero. We readily find that
\begin{flalign*}
&&                   u&=\sinh\;(\mu z)[A \cos n\phi +B \sin n\phi]J_n(\mu r)&\\[1ex]
&\text{and}&u&=\cosh\;(\mu z)[A \cos n\phi +B \sin n\phi]J_n(\mu r)&\phantom{and}
\end{flalign*}
are such solutions, and that
\[
u= \sum\limits^{n=\infty}_{n=0} \sum\limits^{k=\infty}_{k=1}
\dfrac{\sinh \mu_{k} z}{\sinh \mu_k b}
 [A_{n,k} \cos n\phi +B_{n,k} \sin n\phi]J_n(\mu_k r) \tag{1}
\]
is the solution of the given problem if
\[
f(r,\phi) = \sum\limits^{n=\infty}_{n=0} \sum\limits^{k=\infty}_{k=1}
 (A_{n,k} \cos n\phi +B_{n,k} \sin n\phi)J_n(\mu_k r)  \tag{2}
\]
where $\mu_k$ is a root of the equation
\[
\dfrac{J_n(\mu a)}{\mu^n a^n} = 0. \tag{3}
\]

\EXAMPLE{S}

1.\quad Show that
% recast to fit page width
\begin{align*}
\int\limits^a_0
rJ_n(\mu_k r)J_n(\mu_l r)&dr\\[-2ex]
&=\dfrac{a}{{\mu_k}^2-{\mu_l}^2}
[\mu_l J_n(\mu_k a)J'_n(\mu_l a)-\mu_k J_n(\mu_l a)J'_n(\mu_k a)]\\
&=\dfrac{a}{{\mu_k}^2-{\mu_l}^2}
[\mu_k J_n(\mu_l a)J_{n+1}(\mu_k a)-\mu_l J_n(\mu_k a)J_{n+1}(\mu_l a)].
\end{align*}

2.\quad Show that
% recast to fit page width
\begin{align*}
\int\limits^a_0 r [J_n(\mu_k r)]^2&dr\\[-2ex]
&=\dfrac{1}{2}   \Big[a^2(J'_n(\mu_k a))^2+\Big(a^2 - \dfrac{n^2}{{\mu_k}^2}\Big) (J_n(\mu_k a))^2\Big]\\
&=\dfrac{a^2}{2}[(J_n (\mu_k a))^2 + (J_{n+1} (\mu_k a))^2]- \dfrac{na}{\mu_k}  J_n (\mu_k a)J_{n+1} (\mu_k a).
\end{align*}

% -----File: 244.png
3.\quad Show that in Art.~129
\begin{align*}
A_{0,k}&=\dfrac{1}{\pi}
\dfrac{\displaystyle \int\limits^{2\pi}_0d\phi\int\limits^a_0
rf(r,\phi)J_0(\mu_kr)dr}
{a^2[J_1(\mu_ka)]^2},\\[0.5em]
B_{0,k}&=0,\\
A_{n,k}&=\dfrac{2}{\pi}
\dfrac{\displaystyle\int\limits^{2\pi}_0 d\phi \int\limits^a_0
rf(r,\phi)\cos n\phi J_n(\mu_k r)dr}
{a^2[J_{n+1}(\mu_ka)]^2},\\
B_{n,k}&=\dfrac{2}{\pi}
\dfrac{\displaystyle\int\limits^{2\pi}_0 d\phi \int\limits^a_0
rf(r,\phi)\sin n\phi J_n(\mu_kr)dr}
{a^2[J_{n+1}(\mu_ka)]^2}.
\end{align*}

4.\quad Obtain the coefficients for the case where the convex surface of the
cylinder is impervious to heat.\\

5.\quad Obtain the coefficients for the case where the convex surface of the
cylinder is exposed to air at the temperature zero.\\

6.\quad Show that if in a drumhead problem of Art.~11 the initial distortion is
unsymmetrical, so that we have to solve the equation \smallrom{XI} Art.~1 subject to
the conditions $ z = f(r,\phi)$ when $t = 0$, $D_t z = 0$ when $t = 0$, $z = 0$ when $r = a$,
the solution is
\[
z=\sum\limits^{n=\infty}_{n=0}\sum\limits^{k=\infty}_{k=1}\cos(\mu_k ct)(A_{n,k}\cos n\phi + B_{n,k}\sin n\phi) J_n(\mu_k r)
\]
where $A_{0,k}$, $B_{0,k}$, $A_{n,k}$, and $B_{n,k}$ have the values given in Ex.~3.\\

7.\quad What modifications do the statements made in Ex.~2, Art.~126, need to
make them apply to the unsymmetrical case treated in Ex.~6?

Show that any possible nodal system in Ex.~6 is composed of concentric
circles and of radii whose outer extremities are equidistant. v.\ Rayleigh's
Sound, Vol.\ I., Arts.~(202-207).\\

8.\quad Solve the problem of Art.~127 and of Art.~127.\ Ex.~for the unsymmetrical
case. \quad \emph{Suggestion:} $AJ_n(x) + BK_n(x)$  is a solution of Bessel's
Equation.\\

9.\quad Solve the problem of Art.~128 and of Art.~128.\ Ex.~1.\ for the case where
$u = f(z,\phi)$ when $r = a$. \quad \emph{Suggestion:} $u = \sin \mu z (A \cos n\phi + B \sin n\phi) J_n(\mu ri)$
is a solution of Laplace's Equation, and $f(z,\phi)$ can be developed into a double
Fourier's Series [v.\ (15) Art.~71].\\
% -----File: 245.png

10.\quad Show that in dealing with a wedge cut from a cylinder by planes
passed through the axis, or with a membrane in the form of a circular sector,
it may be necessary to use Bessel's Functions of fractional or incommensurable
orders.\\

11.\quad \textit{Bernouilli's Problem} (v.\ Chapter IX). In considering small transverse
vibrations of a uniform, heavy, flexible, inelastic string fastened at one end
and initially distorted into some given curve, we have to solve the equation
 $D_t^2y=c^2(xD_x^2y+D_xy)$, subject to the conditions $D_ty=0$ when $t = 0$,
 $y =f(x)$ when $t = 0$, $y = 0$ when $x = a$; the origin being taken at the
distance $a$ below the point of suspension and the axis of $X$ taken vertical.

\begin{flalign*}
&\text{Show that} &&y=\sum_{k=1}^{k=\infty}A_k \cos\mu_k ct\,B_0(\mu_k^2x),&&\\
&\text{where} & B_0(x)&=1-\frac{x}{1^2}+\frac{x^2}{1^2.2^2}-\frac{x^3}{1^2.2^2.3^2}+\cdots&&\phantom{where}\\[1ex]
&&&=J_0(2\sqrt{x})&&
\end{flalign*}
and $\mu_k$ is a root of the equation
\[B_0(\mu^2 a)=J_0(2\mu \sqrt{a})=0,\]
\begin{flalign*}
&\text{and}& A_k = \frac{\displaystyle\int\limits_0^af(x)B_0(\mu_k^2 x)dx}{\mu^2a^2[B'_0(\mu^2_ka)]^2}=\frac{\displaystyle\int\limits_0^af(x)J_0(2\mu_k\sqrt{x})dx}{a[J_1(2\mu_k\sqrt{a})]^2}.&&\\.
\end{flalign*}

12.\quad As a simple case under Example 10 consider the vibrations of a circular
membrane fastened at the perimeter and also along a radius and then initially
distorted (v.\ Rayleigh's Sound, Art.~207). In this case we must modify the
formula given in Ex.~6 by dropping out the terms involving $\cos n\phi$ and by
taking $n = \dfrac{m}{2}$. The required solution is
\[
z=\sum_{m=1}^{m=\infty}\sum_{k=1}^{k=\infty}B_{m,k}\cos\mu_kct\sin\frac{m\phi}{2}J_{\frac{m}{2}}(\mu_kr)
\]
\begin{flalign*}
&\text{where $\mu_k$ is a root of} &\frac{J_{\frac{m}{2}}(\mu a) }{\mu^{\frac{m}{2}}a^{\frac{m}{2}}}&=0&&\phantom{where mu is a root of}
\end{flalign*}
\begin{flalign*}
&\text{and} &B_{m,k}&=\frac{2}{\pi}\frac{\displaystyle\int\limits_0^{2\pi}d\phi\int\limits_0^a rf(r,\phi)\sin\frac{m\phi}{2}J_{\frac{m}{2}}(\mu_k r)dr}{a^2[{J_{\frac{m}{2}}}'(\mu_ka)]^2}.&&\phantom{and}
\end{flalign*}
% -----File: 246.png

For the terms in which $m$ is odd, $J_{\frac{m}{2}}(x)$ can be readily obtained from (13)
Art.~122, which will become a finite sum.

For example, (13) Art.~122 gives the values
\begin{gather*}
J_{\frac{1}{2}}(x)=\sqrt{\frac{2}{\pi x }}\sin x;  \quad J_{\frac{3}{2}}(x) =\sqrt{\frac{2}{\pi x } }\bigg(\frac{1}{x} \sin x-\cos x \bigg);\\
J_{\frac{5}{2}}(x)=-\sqrt{\frac{2}{\pi x }}\bigg[\bigg(1+\frac{3}{x^2}\bigg)\sin x+\frac{3}{x} \cos x \bigg]; \quad \text{\&c}.
\end{gather*}

13.\quad The question of the flow of heat in three dimensions involves a problem
not unlike the last.

Suppose the initial temperatures of all points in a sphere of radius $c$ given,
and let the surface be kept at the temperature zero. Then we have to solve
the equation
\[
D_t u=\frac{a^2}{r^2}\biggr [D_r(r^2 D_r u) + \frac{1}{\sin\theta} D_{\theta}(\sin\theta D_{\theta} u) + \frac{1}{\sin^2\theta} D_{\phi}^2 u\biggr]    \tag{1}
\]
(\smallrom{IV} Art.~1) subject to the conditions
\begin{gather*}
   u = 0 \quad \text{when}\quad  r = c,\\
   u =f(r, \theta, \phi)\quad \text{when}\quad  t = 0.
\end{gather*}
If we assume $u = T.R.V$  where $T$ is a function of $t$ only, $R$ of $ r$ only, and $V$
of $\theta$ and $\phi$ only, (1) can be broken up into
\begin{gather*}
\frac{dT}{dt} + a^2 \alpha^2 T = 0    \tag{2}\\
 m(m+1)V + \frac{1}{\sin\theta }D_{\theta} (\sin\theta D_{\theta} V) + \frac{1}{\sin^2 \theta} D_{\phi}^2 V = 0    \tag{3}
\end{gather*}
\begin{flalign*}
&\text{and}&\frac{ d^2R}{dr^2} + \frac{2}{r }\frac{dR}{dr }\ + \biggr[\alpha^2 - \frac{m(m+1)}{r^2} \biggr ]R=0.  && \tag{4}
\end{flalign*}
Hence $T = e^{-a^2 \alpha^2 t}$, $V=Y_m (\mu, \phi)$ [v.\ Art.~102 (2)], and $R$ is still to be found.
If in (4) we let $x=\alpha r$ and $z=R\sqrt{\alpha r}$ it becomes
\[
\frac{d^2z}{dx^2} + \frac{1}{x} \frac{dz}{dx} + \biggr [ 1 - \frac{(m+\frac{1}{2})^2 }{x^2} \biggr ] z=0
\]
which is satisfied by $z=J_{m+\frac{1}{2}} (x)$. (v.\ Art.~17.)
\begin{flalign*}
&\text{\indent Therefore}&  R= \frac{1}{\sqrt{\alpha r }}J_{m+\frac{1}{2}} (\alpha r).&&\phantom{\indent Therefore}
\end{flalign*}
% -----File: 247.png
\begin{flalign*}
&&f(r,\theta,\phi)&=\dfrac{1}{4\pi}\sum\limits^{m=\infty}_{m=0} (2m+1) \int\limits^{2\pi}_0 d\phi_1  \int\limits^{\pi}_0 f(r,\theta_1,\phi_1)P_m(\cos \gamma)\sin\theta_1 d\theta_1&\tag*{by (3) Art.~114,}\\
&&&=\sum\limits^{m=\infty}_{m=0} \sum\limits^{n=m}_{n=0} [A_{m,n}f_{m,n}(r)\cos n\phi + B_{m,n}F_{m,n}(r)\sin n\phi]P_m^n(\mu).&
\end{flalign*}
\[
\sqrt{r}f_{m,n}(r)=\sum\limits^{k=\infty}_{k=0}  C_{m,n,k}J_{m+\frac{1}{2}}(\alpha_kr)
\]
where $\alpha_k$ is a root of the equation
\[
\dfrac{J_{m+\frac{1}{2}}(\alpha c)}{(\alpha c)^{m+\frac{1}{2}}}=0
\]
\begin{flalign*}
&\text{and}   &&C_{m,n,k}= \dfrac{2\displaystyle \int\limits^c_0 r^{\frac{3}{2}}f_{m,n}(r)J_{m+\frac{1}{2}}(\alpha_kr)dr}{c^2[J'_{m+\frac{1}{2}}(\alpha_kc)]^2}.&&\\
&                       &&\sqrt{r}F_{m,n}(r)=\sum\limits^{m=\infty}_{m=0}  D_{m,n,k}J_{m+\frac{1}{2}}(\alpha_kr)\\
&\text{where }&&      D_{m,n,k}= \dfrac{2\displaystyle\int\limits^c_0 r^{\frac{3}{2}}F_{m,n}(r)J_{m+\frac{1}{2}}(\alpha_kr)dr}{c^2[J'_{m+\frac{1}{2}}(\alpha_kc)]^2}.&&\phantom{where}
\end{flalign*}

The final solution is
\begin{multline*}
u=\dfrac{1}{\sqrt{r}}\sum\limits^{m=\infty}_{m=0} \sum\limits^{n=m}_{n=0} \bigg[P_m^n(\mu)\sum\limits^{k=\infty}_{k=1} (A_{m,n}C_{m,n,k}\cos n\phi\\
                         \vphantom{ \sum\limits^{n=m}_{n=0}}
                         +B_{m,n}D_{m,n,k} \sin n\phi)e^{-a^2\alpha_k^2 t} J_{m+\frac{1}{2}}(\alpha_kr)\bigg]
\end{multline*}
cf.\ Riemann, Par.\ Dif.\ Gl., \S\S~72 and 73.
\label{ch7end}

% -----File: 248.png
\mychap{CHAPTER VIII.}{LAPLACE'S EQUATION IN CURVILINEAR COÖRDINATES. ELLIPSOIDAL HARMONICS.}
\label{ch8start}

\mypara{130.} \textit{Orthogonal Curvilinear Coördinates.}

\begin{flalign*}
&\raisebox{3.5ex}{If}&
\left.
\begin{gathered}
F_1(x,y,z) =\rho_1\\
F_2(x,y,z) = \rho_2\\
F_3(x,y,z) = \rho_3
\end{gathered}
\right\}&&\tag{1}
\end{flalign*}
are the equations in rectangular coördinates of three surfaces that are mutually
perpendicular no matter what the values of $\rho_1$, $\rho_2$, and $\rho_3$, the parameters $\rho_1$,
$\rho_2$, and $\rho_3$, may be regarded as a set of coördinates for a point of intersection
of the three surfaces, in the sense that when $\rho_1$, $\rho_2$, $\rho_3$ are given the point in
question is determined, and when the point is given the corresponding values
of $\rho_1$, $\rho_2$, $\rho_3$, can be found.

From equations (1) $x$, $y$, and $z$ can be expressed in terms of $\rho_1$, $\rho_2$, and $\rho_3$.
Suppose this done. If now $x$, $y$, $z$ are the rectangular coördinates of the
point $\rho_1=a$, $\rho_2=b$, $\rho_3=c$, the rectangular coördinates of the points
$\rho_1=a+d\rho_1$, $\rho_2=b$, $\rho_3=c$, are obviously $x+D_{\rho_1}x.d\rho_1+\epsilon_1$, $y+D_{\rho_1} y.d\rho_1+\epsilon_2$,
$z+D_{\rho_1}z.d\rho_1+\epsilon_3$, where $\epsilon_1$, $\epsilon_2$, and $\epsilon_3$ are infinitesimals of higher order than
$d\rho_1$. Hence the square of the distance between the points will differ by an
infinitesimal of higher order than that of $d\rho_1^2$ from $dn_1^2$ where
\[
dn_1^2=[(D_{\rho_1}x)^2+(D_{\rho_1}y)^2+(D_{\rho_1}z)^2]d\rho_1 ^2.
\]
\begin{flalign*}
&\raisebox{6.5ex}{\indent Let}&
\left.
\begin{aligned}
\frac{1}{h_1^2}&=(D_{\rho_1}x)^2+(D_{\rho_1}y)^2+(D_{\rho_1}z)^2\\
\frac{1}{h_2^2}&=(D_{\rho_2}x)^2+(D_{\rho_2}y)^2+(D_{\rho_2}z)^2\\
\frac{1}{h_3^2}&=(D_{\rho_3}x)^2+(D_{\rho_3}y)^2+(D_{\rho_3}z)^2.
\end{aligned}
\right\}&&\tag{2}
\end{flalign*}

Then if $dn_1$ is the element of length normal to the surface $\rho_1=a$, $dn_2$
normal to $\rho_2=b$, and $dn_3$ normal to $\rho_3=c$
\[
dn_1=\frac{d\rho_1}{h_1},\quad dn_2=\frac{d\rho_2}{h_2},\quad dn_3=\frac{d\rho_3}{h_3}. \tag{3}
\]
% -----File: 249.png
The element of surface $dS_1$ on the surface $\rho_1=a$ is easily seen to be
\[
dS_1=\frac{d\rho_2d\rho_3}{h_2h_3}; \tag{4}
\]
and the element of volume $dv$ is
\[
dv=\frac{d\rho_1d\rho_2d\rho_3}{h_1h_2h_3}. \tag{5}
\]

\EXAMPLE{}

\begin{flalign*}
&\text{\indent Show that} &
  h_1^2&=(D_x\rho_1)^2+(D_y\rho_1)^2+(D_z\rho_1)^2&\phantom{\indent Show that}\\
&&h_2^2&=(D_x\rho_2)^2+(D_y\rho_2)^2+(D_z\rho_2)^2&\\
&&h_3^2&=(D_x\rho_3)^2+(D_y\rho_3)^2+(D_z\rho_3)^2.&
\end{flalign*}

\textit{Suggestion:} If $h_1$ has the value just given $\dfrac{D_x\rho_1}{h_1}$, $\dfrac{D_y\rho_1}{h_1}$, $\dfrac{D_z\rho_1}{h_1}$ are the
direction cosines of the normal at any given point of $\rho_1=a$. (v.\ Int.\ Cal.\
page 161.) Then
\[
dn_1 = \frac{D_x\rho_1}{h_1}dx
     + \frac{D_y\rho_1}{h_1}dy
     + \frac{D_z\rho_1}{h_1}dz
= \frac{1}{h_1}d\rho_1.
\]

\markright{ELLIPSOIDAL HARMONICS.}
\mypara{131.} \textit{Laplace's Equation in orthogonal curvilinear coördinates.}

If we apply the special form of Green's Theorem
\[
\iiint\nabla^2Vdxdydz=\int D_nVdS \tag{v.\ Art.~98}
\]
to the space bounded by the surfaces $\rho_1=a$, $\rho_2=b$, $\rho_3=c$, $\rho_1=a+d\rho_1$,
 $\rho_2=b+d\rho_2$, $\rho_3=c+d\rho_3$, we have
% recast to fit page width
\begin{align*}
\frac{\nabla^2Vd\rho_1d\rho_2d\rho_3}{h_1h_2h_3}&=\\
-h_1D_{\rho_1}&V\frac{d\rho_2d\rho_3}{h_2h_3}+h_1D_{\rho_1}V\frac{d\rho_2d\rho_3}{h_2h_3}+D_{\rho_1}\left(\frac{h_1}{h_2h_3}D_{\rho_1}V\right)d\rho_1d\rho_2d\rho_3\\
-h_2D_{\rho_2}&V\frac{d\rho_3d\rho_1}{h_3h_1}+h_2D_{\rho_2}V\frac{d\rho_3d\rho_1}{h_3h_1}+D_{\rho_2}\left(\frac{h_2}{h_3h_1}D_{\rho_2}V\right)d\rho_1d\rho_2d\rho_3\\
-h_3D_{\rho_3}&V\frac{d\rho_1d\rho_2}{h_1h_2}+h_3D_{\rho_3}V\frac{d\rho_1d\rho_2}{h_1h_2}+D_{\rho_3}\left(\frac{h_3}{h_1h_2}D_{\rho_3}V\right)d\rho_1d\rho_2d\rho_3;
\end{align*}
whence
\[
\nabla^2V=h_1h_2h_3\bigg[
  D_{\rho_1}\bigg(\frac{h_1}{h_2h_3}D_{\rho_1}V\bigg)
+ D_{\rho_2}\bigg(\frac{h_2}{h_3h_1}D_{\rho_2}V\bigg)
+ D_{\rho_3}\bigg(\frac{h_3}{h_1h_2}D_{\rho_3}V\bigg)\bigg], \tag{6}
\]
and Laplace's Equation in our curvilinear system is
\[
h_1h_2h_3\bigg[
  D_{\rho_1}\bigg(\frac{h_1}{h_2h_3}D_{\rho_1}V\bigg)
+ D_{\rho_2}\bigg(\frac{h_2}{h_3h_1}D_{\rho_2}V\bigg)
+ D_{\rho_3}\bigg(\frac{h_3}{h_1h_2}D_{\rho_3}V\bigg)\bigg]=0. \tag{7}
\]

% -----File: 250.png
If it happens that $\nabla^2\rho_1=0$, $V=\rho_1$ will satisfy (7) and we shall have
 $h_1h_2h_3D_{\rho_1}\bigg(\dfrac{h_1}{h_2h_3}\bigg)=0$. In like manner if $\nabla^2\rho_2=0$
 we have $D_{\rho_2}\bigg(\dfrac{h_2}{h_3h_1}\bigg)=0$,
 and if $\nabla^2\rho_3=0$ we have $D_{\rho_3}\bigg(\dfrac{h_3}{h_1h_2}\bigg)=0$; and therefore (7) reduces to
\[
h_1^2 D_{\rho_1}^2V + h_2^2 D_{\rho_2}^2V + h_3^2 D_{\rho_3}^2V=0\tag{8}
\label{err250}
\]
when $\nabla^2\rho_1=0$, $\nabla^2\rho_2=0$, and $\nabla^2\rho_3=0$.

\mypara{132.} If instead of having the value of the Potential Function $V$ given on
the surface of a sphere as in our Spherical Harmonic problem, we have it
given at all the points on the surface of an \emph{oblate spheroid}, and are required to
find its value at any internal or external point, we can easily get a solution by
methods in no essential respect different from those already employed, if only
we rightly choose our system of coördinates.

If we take an ellipse and an hyperbola having the same foci, and revolve
them about the minor axis of the ellipse, we shall get a pair of surfaces which
are mutually perpendicular; a plane through the axis of revolution will cut
both the \emph{spheroid} and the \emph{hyperboloid} orthogonally.

The equations of the three surfaces can be written:---
\begin{gather*}
F_1(x,y,z,\lambda)=\frac{x^2}{\lambda^2}+\frac{y^2}{\lambda^2-b^2}+\frac{z^2}{\lambda^2}-1=0 \tag{1}\\
F_2(x,y,z,\mu)    =\frac{x^2}{\mu^2}    +\frac{y^2}{\mu^2     -b^2}+\frac{z^2}{\mu^2}   -1=0 \tag{2}\\
F_3(x,y,z,\nu)    =z - \nu x = 0, \tag{3}
\end{gather*}
where $\lambda^2>b^2>\mu^2$, $2b$ being the distance between the foci.

For all values of $\lambda$, $\mu$, and $\nu$ consistent with the inequality above written
the surfaces (1), (2), (3) intersect in real points and cut orthogonally.

$\lambda$, $\mu$, and $\nu$ can be so chosen that the surfaces will intersect in any given
point, and therefore can be taken as a set of curvilinear coördinates, and
Laplace's Equation can be expressed in terms of them by the aid of Formula
\smallrom{XV} Art.~1.

From (1), (2), and (3) we readily get
\begin{gather*}\left.
\begin{split}
x^2 &= \frac{\lambda^2\mu^2}{b^2(1+\nu^2)}\\
y^2 &= \frac{(\lambda^2-b^2)(b^2-\mu^2)}{b^2}\\
z^2 &= \frac{\lambda^2\mu^2\nu^2}{b^2(1+\nu^2)};
\end{split}\right\}\tag{4}
\end{gather*}
% -----File: 251.png
whence \quad $D_\lambda x=\dfrac{\mu}{b\sqrt{1+\nu^2}}$,
       \quad $D_\lambda y=\dfrac{\lambda}{b}\sqrt{\dfrac{b^2-\mu^2}{\lambda^2-b^2}}$,
       \quad $D_\lambda z=\dfrac{\mu\nu}{b\sqrt{1+\nu^2}}$;
\begin{flalign*}
      &\text{and} & \frac{1}{h_1^2} =
       \frac{\lambda^2-\mu^2}{\lambda^2-b^2}& \tag{5}&
\end{flalign*}
[v.~130~(2)]. In like manner we get
\[
   \frac{1}{h_2^2} =
   \frac{\lambda^2-\mu^2}{b^2-\mu^2} \tag{6}
\]
\begin{flalign*}
   &\text{and} & \frac{1}{h_3^2} =
    	\frac{\lambda^2\mu^2}{b^2(1+\nu^2)^2}, &\tag{7}&
\end{flalign*}
and \smallrom{XV} Art.~1 becomes
% recast to fit page width
\begin{align*}
   &\frac{\mu}{b(1+\nu^2)\sqrt{b^2-\mu^2}}D_\lambda[\lambda\sqrt{\lambda^2-b^2}.D_\lambda V]\\
 + &\frac{\lambda}{b(1+\nu^2)\sqrt{\lambda^2-b^2}}D_\mu[\mu\sqrt{b^2-\mu^2}.D_\mu V]\\
 + &\frac{b(\lambda^2-\mu^2)}{\lambda\mu\sqrt{(\lambda^2-b^2)(b^2-\mu^2)}}D_\nu[(1+\nu^2)D_\nu V]=0,
\tag{8}
\end{align*}
which is Laplace's Equation in terms of our \emph{Spheroidal Coördinates} $\lambda$, $\mu$, and $\nu$.

If now in place of $\lambda$, $\mu$, and $\nu$ we can introduce some function of $\lambda$, some
function of $\mu$, and some function of $\nu$ which, therefore, will represent the
same set of orthogonal surfaces, and if we can choose these functions $\alpha$, $\beta$,
and $\gamma$, which of course are functions of $x$, $y$, and $z$, so that $\nabla^2\alpha=0$,
 $\nabla^2\beta=0$, and $\nabla^2\gamma=0$, equation (8) must reduce to the simple and symmetrical
form given in \smallrom{XVI} Art.~1.

These functions $\alpha$, $\beta$, and $\gamma$ are easily found. Equation (8) is $\nabla^2V=0$
 expressed in terms of $\lambda$, $\mu$, and $\nu$. Assume that $V$ is a function of $\lambda$ only;
then $D_\mu V=0$, and $D_\nu V=0$, and (8) reduces to
\begin{flalign*}
&&D_\lambda[&\lambda\sqrt{\lambda^2-b^2}.D_\lambda V]=0&&\\
&\text{whence} && \lambda\sqrt{\lambda^2-b^2}\ \frac{dV}{d\lambda}=c_1,&&\phantom{whence}\\
&&&\hspace{0.5em}dV=\frac{c_1d\lambda}{\lambda\sqrt{\lambda^2-b^2}},&&\\
&\text{and} & &\hspace{1em}V=\frac{c_1}{b}\sec^{-1}\frac{\lambda}{b},&&
\end{flalign*}
and is a function of $\lambda$ which satisfies Laplace's Equation.

% -----File: 252.png
Take this as $\alpha$ leaving $c_1$ at present undetermined, so that
\[
d\alpha=\frac{c_1d\lambda}{\lambda\sqrt{\lambda^2-b^2}}\quad \text{and}\quad \alpha=\frac{c_1}{b}\sec^{-1}\frac{\lambda}{b}.
\]

In the same way we get
\[
d\beta=\frac{c_2d\mu}{\mu\sqrt{b^2-\mu^2}}\quad \text{and}\quad \beta=\frac{c_2}{b}\sech^{-1}\frac{\mu}{b},
\]
(v.\ Int.\ Cal.\ Art.~46, Ex.)
\[
d\gamma=\frac{c_3d\nu}{1+\nu^2},\quad \text{and}\quad \gamma=c_3\tan^{-1}\nu.
\]
Substituting these values in (8) and taking $c_1=-c_2=b$, and $c_3=1$,
 (8) reduces at once to
\[
\frac{D_\alpha^2V}{\lambda^2}+\frac{D_\beta^2V}{\mu^2}+\frac{\lambda^2-\mu^2}{\lambda^2\mu^2}D_\gamma^2V=0, \tag{9}\\[-4ex]
\]
\begin{flalign*}
&\text{or since} & \lambda&=b\sec\alpha,\quad \mu=b\sech\beta,\quad \text{and}\quad \nu=\tan\gamma, & \tag{10}&\\[1ex]
&\text{to} & \cos^2\alpha &D_\alpha^2V+\cosh^2\beta D_\beta^2V+(\cosh^2\beta-\cos^2\alpha)D_\gamma^2V=0 & \tag{11}&
\end{flalign*}
which is Laplace's Equation in terms of what we may call \emph{Normal Oblate
Spheroidal Coördinates}.

In using (11) it is to be noted that the point whose coördinates are $(\alpha, \beta, \gamma)$
is the point of intersection of an oblate spheroid whose semi-axes are $b\sec\alpha$
and $b\tan\alpha$, an unparted hyperboloid of revolution whose semi-axes are
$b\sech\beta$ and $b\tanh\beta$, and a plane containing the axis of the system and
making the angle $\gamma$ with a fixed plane; and that if the axis of revolution is
the axis of $Y$ and the fixed plane is the plane of $XY$, the rectangular coördinates
of $(\alpha, \beta, \gamma)$ are
\[
x=b\sec\alpha\sech\beta\cos\gamma,\quad y=b\tan\alpha\tanh\beta,\quad z=b\sec\alpha\sech\beta\sin\gamma \tag{12}
\]
[v.~(4)].

If now we let $\alpha$ range from 0 to $\dfrac{\pi}{2}$, $\beta$ from $-\infty$ to $\infty$, and $\gamma$ from 0 to $2\pi$,
we shall be able to represent all points in space; and if we agree that negative
values of $\beta$ shall belong to points below a plane through the origin and
perpendicular to the axis of revolution and positive values of $\beta$ to points
above that plane, not only shall we have no ambiguity, but also the rectangular
coördinates of any point as given in (12) will have their proper signs.
% -----File: 253.png

\EXAMPLE{S}

1.\quad If the spheroid is a \emph{prolate} spheroid, the ellipse and confocal hyperbola
must be revolved about the major axis of the ellipse, and the plane must contain
that axis. In place of equations (1), (2), and (3) of Art.~132 we have,
then,
\begin{align*}
\frac{x^2}{\lambda^2}+\frac{y^2}{\lambda^2-b^2}+\frac{z^2}{\lambda^2-b^2}-1&=0\\
\frac{x^2}{\mu^2}+\frac{y^2}{\mu^2-b^2}+\frac{z^2}{\mu^2-b^2}-1&=0
\end{align*}
\[
z-\nu y=0
\]
\begin{flalign*}
&\text{where} & \lambda^2>b^2>\mu^2. &&\phantom{where}
\end{flalign*}
\[
h_1^2=\frac{\lambda^2-b^2}{\lambda^2-\mu^2},\quad h_2^2=\frac{b^2-\mu^2}{\lambda^2-\mu^2},\quad h_3^2=\frac{b^2(1+\nu^2)^2}{(\lambda^2-b^2)(b^2-\mu^2)}.
\]
Laplace's Equation becomes
\begin{align*}
 \frac{1}{b^2(1+\nu^2)}D_\lambda[(\lambda^2-b^2)D_\lambda V]
&+\frac{1}{b^2(1+\nu^2)}D_\mu[(b^2-\mu^2)D_\mu V]\\
&+\frac{\lambda^2-\mu^2}{(\lambda^2-b^2)(b^2-\mu^2)}D_\nu[(1+\nu^2)D_\nu V]=0. \tag{1}
\end{align*}
\begin{flalign*}
&\text{(1) reduces to} \quad \frac{D_\alpha^2V}{\lambda^2-b^2}
                       + \frac{D_\beta^2V}{b^2-\mu^2}
                       + \frac{\lambda^2-\mu^2}{(\lambda^2-b^2)(b^2-\mu^2)}D_\gamma^2V=0, &\hfill\tag{2}\\[-5ex]
\end{flalign*}
\begin{flalign*}
&\text{where} && d\alpha=-\frac{bd\lambda}{\lambda^2-b^2},\quad
                d\beta=\frac{bd\mu}{b^2-\mu^2},\quad
                d\gamma=\frac{d\nu}{1+\nu^2},&&\phantom{where}
\end{flalign*}
\[
   \alpha=\ctnh^{-1}\frac{\lambda}{b},\quad
   \beta =\tanh^{-1}\frac{\mu}{b},\quad \text{and}\quad
   \gamma=\tan^{-1}\nu.
\]
\begin{flalign*}
&\indent\text{Since} & \lambda=b\ctnh\alpha,\quad
                       \mu    =b\tanh\beta,\quad \text{and}\quad
                       \nu    =  \tan\gamma&&\phantom{\indent Since}
\end{flalign*}
(2) can be reduced to
\[
\sinh^2\alpha D_\alpha^2V +\cosh^2\beta D_\beta^2 V + (\sinh^2\alpha + \cosh^2\beta)D_\gamma^2 V=0. \tag{3}
\]

In using (3) it is to be noted that the point $(\alpha, \beta, \gamma)$ is the point of intersection
of a prolate spheroid whose semi-axes are $b\ctnh\alpha$ and $b\csch\alpha$, a
biparted hyperboloid of revolution whose semi-axes are $b\tanh\beta$ and $b\sech\beta$,
 and a plane containing the axis of revolution and making the angle $\gamma$ with a
fixed plane.
% -----File: 254.png

If the fixed plane is that of ($XY$) the rectangular coördinates of any point
$(\alpha, \beta, \gamma)$  are
\[
x = b \ctnh \alpha \tanh \beta, \quad y = b \csch \alpha \sech \beta \cos \gamma, \quad z = b \csch \alpha \sech \beta \sin\gamma,
\]
and $\alpha$ may range from $\infty$ to 0, $\beta$ from $-\infty$ to $\infty$, and $\gamma$ from 0 to $2\pi$.
Negative values of $\beta$ are to be taken for points lying to the left of a plane
through the origin perpendicular to the axis of revolution.\\

2.\quad Transform Laplace's Equation in Spherical Coördinates \smallrom{XIII} Art.~1
to the symmetrical form
\[
\alpha^2D_\alpha^2V + \cosh^2\beta D_\beta^2V + \cosh^2\beta D_\gamma^2V = 0
\]
\begin{flalign*}
&\text{where}&&  \alpha = \dfrac{1}{r},\quad \beta =\log\tan \dfrac{\theta}{2},\quad \text{and}\quad \gamma = \phi.&&\phantom{where}
\end{flalign*}

3.\quad Transform Laplace's Equation in Cylindrical Coördinates \smallrom{XIV} Art.~1
to the symmetrical form
\[
D_\alpha^2V + D_\beta^2V + e^{2\alpha}D_\gamma^2V = 0
\]
\begin{flalign*}
&\text{where}& &\alpha = \log r, \quad \beta = \phi, \quad \text{and}\quad \gamma = z.&&\phantom{where}
\end{flalign*}

\mypara{133.} In each of the cases we have considered, it has been easy to pass
from Laplace's Equation in terms of the chosen coördinates representing an
orthogonal system of surfaces to the symmetrical form \smallrom{XVI} Art.~1; and it is
evident that our new coördinate $\alpha$ is a value of $V$ corresponding to such a
distribution that the surfaces obtained by giving particular values to $\rho_1$ are
\emph{equipotential} surfaces; that $\beta$ is a value of $V$ corresponding to such a
distribution that the surfaces obtained by giving particular values to $\rho_2$ are
equipotential surfaces; and that $\gamma$ is a value of $V$ corresponding to such a
distribution that the surfaces obtained by giving particular values to $\rho_3$ are
equipotential surfaces. $\alpha$, $\beta$, and $\gamma$ are called by Lamé ``\emph{thermometric
parameters}.''

The condition that these values should exist, for a given system of surfaces,
that is, that the distribution described above should be possible, is readily
obtained. We shall work it out for $\alpha$. It is merely the condition that $V$ in
Laplace's Equation may be a function of $\rho_1$ alone.

If $V$ is a function of $\rho_1$ alone
\[
D_xV = \dfrac{dV}{d\rho_1}D_x\rho_1, \quad D_yV = \dfrac{dV}{d\rho_1}D_y\rho_1,\quad D_zV = \dfrac{dV}{d\rho_1}D_z\rho_1,\\[-2ex]
\]
\begin{align*}
D_x^2 V &= \dfrac{d^2 V}{{d \rho_1}^2} (D_x \rho_1)^2 + \dfrac{dV}{d \rho_1} D_x^2 \rho_1\\
% -----File: 255.png
D_y^2 V &= \dfrac{d^2 V}{{d \rho_1}^2} (D_y \rho_1)^2 + \dfrac{dV}{d \rho_1} D_y^2 \rho_1\\
D_z^2 V &= \dfrac{d^2 V}{{d \rho_1}^2} (D_z \rho_1)^2 + \dfrac{dV}{d \rho_1} D_z^2 \rho_1.
\end{align*}
Therefore $[(D_x \rho_1)^2 + (D_y \rho_1)^2 + (D_z \rho_1)^2]\dfrac{d^2 V}{d {\rho_1}^2} + [D_x^2 \rho_1 + D_y^2 \rho_1 + D_z^2 \rho_1]\dfrac{ dV}{d \rho_1} = 0$
\begin{flalign*}
&\text{whence}&   &\dfrac{D_x^2 \rho_1 + D_y^2 \rho_1 + D_z^2 \rho_1}{(D_x \rho_1)^2 + (D_y \rho_1)^2 + (D_z \rho_1)^2} = - \dfrac{d^2V}{d{\rho_1}^2}\div \dfrac{ dV}{d \rho_1}.&&\phantom{whence}
\end{flalign*}
\begin{flalign*}
&\text{or}          &   &\dfrac{\nabla^2\rho_1}{h_1^2} = F_1 (\rho_1)&&\phantom{or}
\end{flalign*}
where $F_1(\rho_1)$ may be any function of $\rho_1$ alone. Our required conditions are
then
\[
\left.
\begin{aligned}
\dfrac{\nabla^2 \rho_1}{h_1^2} &= F_1 (\rho_1)\\
\dfrac{\nabla^2 \rho_2}{h_2^2} &= F_2 (\rho_2)\\
\dfrac{\nabla^2 \rho_3}{h_3^2} &= F_3 (\rho_3)
\end{aligned}
\right\}
\tag{1}
\]
and when they are fulfilled the original curvilinear coördinates $\rho_1$, $\rho_2$, $\rho_3$,
correspond to possible \emph{equipotential} or \emph{isothermal} surfaces, \emph{thermometric
parameters} $\alpha$, $\beta$, and $\gamma$ exist, and the reduction of Laplace's Equation to the
symmetrical form \smallrom{XVI} Art.~1 is possible.

\mypara{134.} Returning to our Oblate Spheroid problem of Art.~132 we can proceed
as usual to break up our equation (\textbf{11}) Art.~132.

Assume that $V = L.M.N$, where $L$ is a function of $\alpha$ only, $M$ of $\beta$ only,
and $N$ of $\gamma$ only.  (11) Art.~132 becomes
\begin{flalign*}
&&\dfrac{\cos^2 \alpha}{L}\dfrac{ d^2L}{d \alpha^2} +  \dfrac{\cosh^2 \beta}{M}\dfrac{ d^2 M}{ d \beta^2} +\dfrac{[\cosh^2 \beta - \cos^2 \alpha]}{N}\dfrac{ d^2N}{d\gamma^2} &= 0\\
&\text{or }& \dfrac{1}{L}\dfrac{\cos^2 \alpha}{ \cosh^2 \beta - \cos^2 \alpha} \dfrac{d^2L}{d \alpha^2} + \dfrac{1}{M}\dfrac{\cosh^2 \beta}{ \cosh^2 \beta - \cos^2 \alpha} \dfrac{d^2M}{d \beta^2} = -&\dfrac{ 1}{N}\dfrac{ d^2 N}{d \gamma^2}.&&\phantom{or}
\end{flalign*}

The first member is independent of $\gamma$, and the second member is independent
of $\alpha$ and $\beta$, and the two members are identically equal. The second member
is then independent of $\alpha$, $\beta$, and $\gamma$ and must be constant; call it $n^2$. We have,
then,
\begin{flalign*}
&\rule{14em}{0em} \dfrac{d^2N}{d \gamma^2} + n^2 N = 0 \tag{1}&\\
% -----File: 256.png
&\text{and} \rule{4em}{0em} \dfrac{ \cos^2\alpha}{L}\dfrac{ d^2L}{d\alpha^2} + \dfrac{\cosh^2\beta}{M}\dfrac{ d^2M}{d\beta^2} - n^2(\cosh^2\beta - \cos^2\alpha) = 0.&&&(2)\\
&\text{(1) gives us} \rule{6.5em}{0em} N = A \cos n\gamma + B \sin n\gamma. && &\tag{3}
\end{flalign*}
(2) can be written
\begin{flalign*}
&&\dfrac{\cos^2\alpha}{L}\dfrac{ d^2L}{d\alpha^2} + n^2\cos^2&\alpha = n^2\cosh^2\beta - \dfrac{\cosh^2\beta}{M}\dfrac{d^2M}{d\beta^2} = m(m+1), &\quad\\
&\text{whence}&\phantom{c}\cos^2\alpha \dfrac{d^2L}{d\alpha^2}& + [n^2\cos^2\alpha - m(m+1)]L = 0&\tag{4}\\
&\text{and}       &\cosh^2\beta \dfrac{d^2M}{d\beta^2} &+ [m(m+1) - n^2\cosh^2\beta]M = 0.  &\tag{5}
\end{flalign*}

If we introduce $x = \tanh\beta$ in (5) it becomes
\[
(1-x^2)\dfrac{d^2M}{dx^2} - 2x \dfrac{dM}{dx} + \left[m(m+1) - \dfrac{n^2}{1-x^2}\right]M = 0  \tag{6}
\]
where since $x = \tanh\beta$ and $\beta$ may have any value from $-\infty$ to $\infty$, $x$ may
have any value between $-1$ and 1.  (6) is a familiar equation having for a
particular solution
\[
M = (1-x^2)^{\frac{n}{2}}\dfrac{ d^nP_m(x)}{dx^n} = P_m^n(x) = P_m^n(\tanh\beta).\tag{7}
\]
(v.\ Arts.~101 and 102).

If we introduce in (4) $x = \tan\alpha$ it reduces to
\[
(1+x^2)\dfrac{d^2L}{dx^2} + 2x \dfrac{dL}{dx} + \left[\dfrac{n^2}{1+x^2} - m(m+1)\right]L = 0.\tag{8}
\]
(8) is an unfamiliar equation, but it can be treated as (6) was treated if we
take the pains to go back to the beginning and follow the steps of the treatment
of Legendre's Equation.

This labor can be saved, however, by noting that if we let $x = \dfrac{y}{i}$ (8) becomes
\[
(1-y^2)\dfrac{ d^2L}{dy^2} - 2y\dfrac{ dL}{dy} + \left[m(m+1) - \dfrac{n^2}{1-y^2}\right]L = 0
\]
and is identical in form with (6).  Hence
\begin{flalign*}
&&\rule{4em}{0em}L &= P_m^n(y)\quad\text{and}\quad  L = (1-y^2)^{\frac{n}{2}}\dfrac{d^nQ_m(y)}{dy^n}&\text{(v.\ Art.~101)},
\end{flalign*}
where $y = i \tan\alpha$, are particular solutions of (4).

We can avoid imaginaries if we use the values
\[
L = (-i)^{m-n} P_m^n(y)\quad \text{and}\quad  L = i^{m+n+1}(1-y^2)^{\frac{n}{2}}\dfrac{ d^nQ_m(y)}{dy^n}.\qquad \tag{9}
\]
% -----File: 257.png
Since we assumed $V= L.M.N$ we have
\begin{flalign*}
&\raisebox{-1.7ex}{and}&
\left.
\begin{aligned}
V &= (A \cos n \gamma + B \sin n \gamma) P_m^n(\tanh \beta)(-i)^{m-n} P_m^n (i \tan \alpha)\\
V &= (A \cos n \gamma + B \sin n \gamma) P_m^n(\tanh \beta) i^{m+n+1} \sec^n \alpha \dfrac{d^n Q_m (i \tan \alpha)}{ (d(i \tan \alpha))^n}
\end{aligned}
\right\} (10)\\[-6ex]
\end{flalign*}
as particular solutions of (\textbf{11}) Art.~132.

If the problem is symmetrical with respect to the axis of the spheroid
 $D_\gamma^2 V = 0$, $n^2 = 0$ and our particular solutions (10) reduce to
\begin{flalign*}
&\raisebox{-2ex}{and}&
\left.
\begin{aligned}
V& = (-i)^m P_m (i \tan \alpha) P_m (\tanh \beta)&\\
V&= i^{m+1} Q_m (i \tan \alpha) P_m (\tanh \beta).
\end{aligned}
\right\}&&\tag{11}
\end{flalign*}

If, then, $V$ is given on the surface of a spheroid as a function of $\beta$ and $\gamma$,
we must express it as a function of $\tanh \beta$ and $\gamma$, and shall be obliged to
develop it in terms of \emph{Spherical Harmonics} of $\tanh \beta$ and $\gamma$ by the formulas of
Chapter VII, using the first equation in (10) for the value of $V$ at an internal
point, and the second for the value of $V$ at an external point. If the problem
is symmetrical, we must develop in Zonal Harmonics of $\tanh \beta$ by the formulas
of Chapter VI.

A convenient form for $Q_m (i \tan \alpha)$ is obtained from (2) Art.~100; it is
\begin{flalign*}
&&Q_m (i \tan \alpha) = -i P_m &(i \tan \alpha) \int\limits^{\infty}_{\tan\alpha}\dfrac{ dx}{ (1 + x^2)[P_m(xi)]^2}.& \tag{12} \\
&\text{Hence} &Q_0 (i \tan \alpha) = -&i \int\limits^{\infty}_{\tan\alpha}\dfrac{dx}{ 1 + x^2} = -i \left(\dfrac{\pi}{ 2} - \alpha\right).& \tag{13}
\end{flalign*}

\EXAMPLE{S}

1.\quad A conductor in the form of an oblate spheroid whose semi-axes are
$b \sec \alpha_0$ and $b \tan \alpha_0$ is charged with electricity and is found to be at potential
$V_0$; find the value of the potential function at any internal or external point.

Here $V_0 = V_0P_0(\tanh \beta)$. Hence at an internal point
\[
V = V_0 \dfrac{P_0 (i \tan \alpha) }{ P_0 (i \tan \alpha_0)} P_0 (\tanh \beta) = V_0, \tag{1}
\]
and at an external point
\[
V = V_0 \dfrac{Q_0 (i \tan \alpha) }{Q_0 (i \tan \alpha_0)} P_0 (\tanh \beta) =V_0 \dfrac{ \left(\dfrac{\pi}{2} - \alpha\right) }{\left(\dfrac{ \pi }{ 2} - \alpha_0\right)}. \tag{2}
\]
Since $V$ in (2) involves $\alpha$ only, the equipotential surfaces are all spheroids
confocal with the conductor.\\

% -----File: 258.png
2.\quad The upper half of an oblate spheroid whose semi-axes are $b \sec \alpha_0$ and
$b \tan \alpha_0$ is kept at the temperature unity, and the lower half at the temperature
zero. Find the permanent temperature at any internal point.
\begin{flushright}
\textit{Ans.} \quad $u = \dfrac{1}{2} + \dfrac{3}{4} \dfrac{P_1(i \tan\alpha)}{P_1(i \tan\alpha_0)}P_1(\tanh\beta) - \dfrac{7}{8}.\dfrac{1}{2 }\dfrac{P_3(i \tan\alpha)}{P_3(i \tan\alpha_0)}P_3(\tanh\beta) + \cdots$
\end{flushright}
(v.\ Art.~93). $u$ may be expressed in terms of $x$, $y$, and $z$ without serious
difficulty [v.\ (12) Art.~132].
\[
u = \frac{1}{2 }+ \frac{3}{4} \frac{y}{c} - \frac{7}{8}.\frac{1}{2}.\frac{1}{2}\frac{ [25y^3-15y(x^2+y^2+z^2-b^2)-9b^2y]}{5c^3+3b^2c} + \cdots
\]
if $2c = 2b \tan \alpha_0 = $ minor axis of spheroid.

\mypara{135.} Let us now find the potential function at an external point due to
the attraction of a solid homogeneous oblate spheroid, using the method employed
in Arts.~98 and 99.

Consider first the potential function due to a shell bounded by the spheroids
for which $\alpha = \phi$ and $\alpha = \phi + d\phi$.

By (1) Art.~98 we have
\[
4\pi\rho\kappa = [D_nV_1 - D_nV_2]_{\alpha=\phi},       \tag{1}
\]
where $\rho$ is the density and $\kappa$ the thickness of the shell, $V_1$ the value of the
potential function at an internal point, and $V_2$ the value of the potential
function at an external point.
\begin{flalign*}
&\indent \text{Let }&V_1 &= \sum A_m(-i)^mP_m(i \tan\alpha)P_m(\tanh\beta)&\\
&\text{and}&V_2 &= \sum B_m i^{m+1}Q_m(i \tan\alpha)P_m(\tanh\beta)     \tag*{[v.\ (11) Art.~134].}&
\end{flalign*}

Since $V_1$ and $V_2$ must have the same value when $\alpha = \phi$
\[
A_m = B_m i^{2m+1}\frac{ Q_m(i \tan\phi)}{P_m(i \tan\phi)} = (-1)^m B_m \int\limits_{\tan\phi}^{\infty}\frac{dx}{(1+x^2)[P_m(xi)]^2}    \tag{2}
\]
[v.\ (12) Art.~134].
\begin{flalign*}
\left.\!\!
\begin{alignedat}{2}
&\indent \text{Hence\quad} & V_1 &= \sum i^mB_mP_m(\tanh\beta)P_m(i \tan\alpha) \int\limits_{\tan\phi}^{\infty}\frac{dx}{(1+x^2)[P_m(xi)]^2}\\
&\text{and}& V_2 &= \sum i^mB_mP_m(\tanh\beta)P_m(i \tan\alpha) \int\limits_{\tan\alpha}^{\infty}\frac{dx}{(1+x^2)[P_m(xi)]^2} .
\end{alignedat}
\right \}\hspace{2.8pt}(3)
\end{flalign*}
\[
D_nV_1 = D_{\alpha}V_1.D_n\alpha. \qquad\qquad D_nV_2 = D_{\alpha}V_2.D_n\alpha
\]
% -----File: 259.png
\begin{align*}
[D_nV_1-D_nV_2]_{\alpha=\phi}&=[D_\alpha V_1-D_\alpha V_2]_{\alpha=\phi}(D_n\alpha)_{\alpha=\phi}\\
                             &\qquad =[D_\alpha(V_1-V_2)]_{\alpha=\phi}[D_n\alpha]_{\alpha=\phi}.
\end{align*}
\[
V_1-V_2= \sum  i^mB_mP_m(\tanh\beta)P_m(i \tan\alpha) \int\limits^{\tan\alpha}_{\tan\phi}\dfrac{ dx}{(1+x^2)[P_m(xi)]^2}.
\]
\begin{multline*}
D_\alpha(V_1-V_2)= \sum i^mB_mP_m(\tanh\beta)\bigg[P_m(i \tan\alpha) \dfrac{\sec^2\alpha}{(1+\tan^2\alpha)[P_m(i \tan\alpha)]^2}\\
     + \dfrac{dP_m(i \tan\alpha)}{d\alpha} \int\limits^{\tan\alpha}_{\tan\phi}\dfrac{ dx}{(1+x^2)[P_m(xi)]^2}\bigg].
\end{multline*}
\begin{gather*}
D_\alpha[V_1-V_2]_{\alpha=\phi} = \sum i^mB_m \dfrac{P_m(\tanh\beta)}{P_m(i \tan\phi)}.\\
               D_n\alpha = \dfrac{d\alpha}{dn}
\end{gather*}
\[
dn=\dfrac{d\rho_1}{h_1}=\dfrac{d\lambda}{h_1}= \dfrac{\sqrt{ \lambda^2-\mu^2}}{\sqrt{\lambda^2-b^2}}d\lambda=b \sec\alpha \sqrt{\tan^2\alpha+\tanh^2\beta}.d\alpha  \tag{4}
\]
v.\ Art.~130 (3), and Art.~132 (5) and (10).
\begin{flalign*}
&                      &[D_n\alpha]_{\alpha=\phi}                          &=\dfrac{1}{b \sec\phi\sqrt{\tan^2\phi+\tanh^2\beta}}.&
% recast to fit page width
&\intertext{\indent Hence}
&&  [D_nV_1-D_nV_2]_{\alpha=\phi}& = \dfrac{1}{b \sec\phi\sqrt{\tan^2\phi+\tanh^2\beta}}\sum  i^mB_m \dfrac{P_m(\tanh\beta)}{P_m(i \tan\phi)}.&\\
&                      &        \kappa = [dn]_{\alpha=\phi} &= b\sec\phi \sqrt{\tan^2\phi+\tanh^2\beta}.d\phi&
\end{flalign*}
by (4), and (1) may be written
\[
4\pi\rho b^2 \sec^2\phi(\tan^2\phi+\tanh^2\beta)d\phi = \sum  i^mB_m \dfrac{P_m(\tanh\beta)}{P_m(i \tan\phi)}.\tag{5}
\]
\begin{flalign*}
&\indent\text{Since}&      \tanh^2\beta &= \tfrac{1}{3}P_0(\tanh\beta)+\tfrac{2}{3}P_2(\tanh\beta)&\phantom{\indent Since}
\end{flalign*}
by (5) Art.~95, to satisfy (5) we must give $m$ the values 0 and 2 and
\begin{flalign*}
&                      &B_0&=\tfrac{4}{3}\pi\rho b^2 \sec^2\phi(3\tan^2\phi+1)d\phi&\\
&\text{and}& B_2&=\tfrac{4}{3}\pi\rho b^2 \sec^2\phi(3\tan^2\phi+1)d\phi.&
\end{flalign*}

% -----File: 260.png
So that by (3)
\begin{flalign*}
&&V_1&=\tfrac{4}{3} \pi \rho b^2 \sec^2 \phi (3 \tan^2 \phi + 1)d \phi \bigg[\int\limits^{\infty}_{\tan\phi}\dfrac{ dx}{1+x^2}\\
&&&\hspace{7em}-P_2(\tanh \beta)P_2(i \tan \alpha) \int\limits^{\infty}_{\tan\phi} \dfrac{dx}{(1+x^2)[P_2(xi)]^2}\bigg] \tag{6} \\
&\text{and}&  V_2&=\tfrac{4}{3} \pi \rho b^2 \sec^2 \phi (3 \tan^2 \phi + 1)d \phi [i Q_0 (i \tan \alpha)&\\
&&&\hspace{16em}+ i^3 P_2 (\tanh \beta) Q_2 (i \tan \alpha)]. &&\tag{7}
\end{flalign*}
The potential function at an external point due to the solid spheroid for
which $\alpha = \alpha_0$ is
\[
V=\!\!\int\limits^{\phi=\alpha_0}_{\phi=0}\!\!V_2 = \tfrac{4}{3} \pi \rho b^2 \sec^2 \alpha_0 \tan \alpha_0 [i Q_0 (i \tan \alpha) + i^3 P_2 (\tanh \beta) Q_2 (i \tan \alpha)].\tag{8}
\]

If $2a$ is the major axis and $2c$ the minor axis of the spheroid
\[
\tfrac{4}{3} \pi\rho b^2 \sec^2\alpha_0 \tan\alpha_0 =\tfrac{4}{3} \dfrac{ \pi\rho a^2c}{b}=\dfrac{M}{b}
\]
where $M$ is the mass of the spheroid. Therefore
\[
V = \dfrac{M}{b} [i Q_0 (i \tan \alpha) + i^3 P_2 (\tanh \beta)Q_2 (i \tan \alpha)] \tag{9}
\]
is the required value.  (9) can be reduced to\\[2ex]
\phantom{.}$\displaystyle\;V = \dfrac{M}{b} \left\{\dfrac{\pi}{2} - \alpha + \dfrac{1}{4}\left[\left(\dfrac{\pi}{2} - \alpha\right)(3 \tan^2 \alpha+1) - 3 \tan \alpha\right][3 \tanh^2 \beta - 1]\right\}.$\hfill(10)

\EXAMPLE{S}

1.\quad Break up the equation (3) Ex.~1, Art.~132, for the prolate spheroid, and
obtain particular solutions of the term
\begin{gather*}
V= (A \cos n\gamma + B \sin n\gamma)P_m^n(\tanh\beta)P_m^n(\ctnh\alpha),\\
V= (A \cos n\gamma + B \sin n\gamma)P_m^n(\tanh\beta)(-1)^{\frac{n}{2}} \csch^n \alpha \dfrac{d^n Q_m (\ctnh\alpha)}{(d \ctnh \alpha)^n}.
\end{gather*}

2.\quad Break up and solve the equations of Exs.\ 2 and 3, Art.~132, and show
that they lead to familiar forms.\\

3.\quad If in Ex.~1, Art.~132, the conductor is a prolate spheroid whose semi-axes
are $b \ctnh\alpha_0$ and $b \csch\alpha_0$ show that
\[
V= V_0\text{ at an internal point.}\qquad V = V_0\dfrac{ \alpha}{\alpha_0}\text{  at an external point}.
\]

% -----File: 261.png
4.\quad Show that the potential function at an external point due to the attraction
of a homogeneous solid prolate spheroid is
\[
V = \dfrac{M}{b} [Q_0(\ctnh \alpha) - P_2(\tanh \beta)Q_2(\ctnh \alpha)].
\]

\vspace{1ex}\begin{center}
\emph{Ellipsoidal Harmonics.}
\end{center}\vspace{-1ex}

\mypara{136.} If we are dealing with an \emph{ellipsoid} instead of a spheroid, we can take
as our orthogonal system of surfaces a set of \emph{confocal quadrics};
\[
\left.
\begin{aligned}
\dfrac{x^2}{\lambda^2} + \dfrac{y^2}{\lambda^2 -b^2} + \dfrac{z^2}{\lambda^2 - c^2} - 1& = 0\;\\
\dfrac{x^2}{\mu^2} + \dfrac{y^2}{\mu^2 - b^2} + \dfrac{z^2}{\mu^2 - c^2} - 1 &= 0\\
\dfrac{x^2}{\nu^2} + \dfrac{y^2}{\nu^2 - b^2} + \dfrac{z^2}{\nu^2 - c^2} - 1 &= 0
\end{aligned}
\right\}
\tag{1}
\]
where $\lambda^2 > c^2 > \mu^2 > b^2 > \nu^2$. Here the first surface is an ellipsoid, the
second an unparted hyperboloid, and the third a biparted hyperboloid. Each
of the three principal sections of the system consists of confocal conics, and it
is well known and is easily shown that the surfaces cut orthogonally.  $\lambda$, $\mu$,
and $\nu$ will be our curvilinear coördinates, and are known as Ellipsoidal
Coördinates.

We find without difficulty that
% recast to fit page width
\begin{multline*}
x^2 = \dfrac{\lambda^2 \mu^2 \nu^2}{b^2 c^2},\quad y^2=\dfrac{(\lambda^2 - b^2)(\mu^2 - b^2)(b^2 - \nu^2)}{ b^2(c^2 - b^2)}, \\ z^2 = \dfrac{(\lambda^2 - c^2)(c^2 - \mu^2)(c^2 - \nu^2)}{ c^2(c^2 - b^2)},\tag{2}
\end{multline*}
% recast to fit page width
\begin{multline*}
h_1^2 = \dfrac{(\lambda^2 - b^2)(\lambda^2 - c^2)}{(\lambda^2 - \mu^2)(\lambda^2 - \nu^2)}, \quad h_2^2 = \dfrac{(\mu^2 - b^2)(c^2 - \mu^2)}{(\mu^2 -\nu^2)(\lambda^2 - \mu^2)},\\  h_3^2 = \dfrac{(b^2 - \nu^2)(c^2 - \nu^2)}{(\lambda^2 - \nu^2)(\mu^2 - \nu^2)}.\tag{3}
\end{multline*}

To avoid ambiguity, we shall suppose that of the nine semi-axes in (1)
$\sqrt{\strxx c^2 - \mu^2}$ is to be taken with the positive sign for a point on the half of the
unparted hyperboloid on which $z$ is positive, and with the negative sign for a
point on the half on which $z$ is negative; $\sqrt{\strxx b^2 - \nu^2}$ is to be taken with the
positive sign for a point on the half of the biparted hyperboloid on which $y$ is
positive, and with the negative sign for a point on the half on which $y$ is
negative; $\nu$ is to be taken positive for a point on the half of the biparted
hyperboloid on which $x$ is positive, and negative for a point on the half on
which $x$ is negative, and that the remaining six are to be always positive. It
follows that our Ellipsoidal Coördinates have the disadvantage that to fully
fix a point we need to know not merely the values of its coördinates $\lambda$, $\mu$, and
$\nu$, but the signs of $\sqrt{\strxx c^2 - \mu^2}$, and $\sqrt{\strxx b^2 - \nu^2}$ as well.

% -----File: 262.png
We shall see later, Art.~139, when we come to introduce what we may call
the \emph{Normal Ellipsoidal Coördinates} $\alpha$, $\beta$, and $\gamma$ that they are free from this
disadvantage.

It is to be observed that $\lambda$ may range from $c$ to $\infty$, $\mu$ from $b$ to $c$, and $\nu$ from
$-b$ to $b$.

The element of length perpendicular to the Ellipsoid is
\[
dn = \dfrac{d \lambda}{h_1} = \sqrt{\dfrac{(\lambda^2 - \mu^2)(\lambda^2 - \nu^2)}{(\lambda^2 - b^2)(\lambda^2 - c^2)}}.d \lambda. \tag{4}
\]

The element of Ellipsoidal surface is
\[
dS = \dfrac{d \mu d \nu }{ h_2 h_3} = (\mu^2 - \nu^2) \sqrt{\dfrac{ (\lambda^2 - \mu^2)(\lambda^2 - \nu^2)}{(\mu^2 - b^2)(c^2 - \mu^2)(b^2 - \nu^2)(c^2 - \nu^2)}}.d \mu d\nu, \tag{5}
\]
and the element of volume is
% recast to fit page width
\begin{align*}
dv &=\dfrac{ d \lambda d \mu d \nu}{ h_1h_2h_3}\\
   &=\dfrac{ (\lambda^2 - \mu^2)(\lambda^2 - \nu^2)(\mu^2 - \nu^2)}{\sqrt{(\lambda^2 - b^2)(\lambda^2 - c^2)(\mu^2 - b^2)(c^2 - \mu^2)(b^2 - \nu^2)(c^2 - \nu^2)}} d \lambda d \mu d \nu.  \tag{6}
\end{align*}

The surface integral of any given function of $\mu$ and $\nu$ taken over the
ellipsoid is
\begin{multline*}
\int  f(\mu, \nu)dS =\int\limits^{b}_{-b} d \nu \int\limits^{c}_{b} [f_1 (\mu, \nu) + f_2 (\mu, \nu) + f_3 (\mu, \nu)\\
          +f_4 (\mu, \nu)](\mu^2 - \nu^2) \sqrt{\dfrac{(\lambda^2 - \mu^2)(\lambda^2 - \nu^2)}{(\mu^2 - b^2)(c^2 - \mu^2)(b^2 - \nu^2)(c^2 - \nu^2)}}.d \mu, \tag{7}
\end{multline*}
where $f_1(\mu, \nu)$, $f_2(\mu, \nu)$, $f_3(\mu, \nu)$ and $f_4(\mu, \nu)$ are the values of the given function
on the four quarters of the ellipsoid into which it is divided by the planes of
($XY$) and ($XZ$).

Laplace's Equation proves reducible to
\[
(\mu^2 - \nu^2) D_{\alpha}^2 V + (\lambda^2 - \nu^2) D_{\beta}^2 V + (\lambda^2 - \mu^2) D_{\gamma}^2 V = 0 \tag{8}
\]
\begin{flalign*}
&\text{where}& \alpha = c \int\limits^{\lambda}_c \dfrac{d\lambda}{\sqrt{(\lambda^2 - b^2)(\lambda^2 - c^2)}},\quad \beta &= c \int\limits^{\mu}_b \dfrac{d \mu}{\sqrt{(c^2 - \mu^2)(\mu^2 - b^2)}},&\\
&            & \gamma &= c \int \limits^{\nu}_0 \dfrac{d \nu }{\sqrt{(b^2 - \nu^2)(c^2 - \nu^2)}}. &\tag{9}
\end{flalign*}

% -----File: 263.png
$\alpha$, $\beta$, and $\gamma$ can be expressed as Elliptic Integrals of the first class and are
\begin{multline*}
\alpha=F\Big(\dfrac{b}{c}, \dfrac{\pi}{2}\Big)-F\Big(\dfrac{b}{c}, \sin^{-1} \dfrac{c}{\lambda}\Big),\quad \beta=F\Big(\sqrt{ 1-\dfrac{ b^2}{c^2}},\, \sin^{-1} \sqrt{\dfrac{1- \dfrac{b^2}{\mu^2}}{1- \dfrac{b^2}{c^2}}}\Big),\\
                                                     \gamma=F\Big(\dfrac{b}{c}, \sin^{-1} \dfrac{\nu}{b}\Big); \tag{10}
\end{multline*}
\begin{flalign*}
&\text{whence}
  &\lambda = \frac{c}{\sn(K-\alpha)} &\Big(\bmod \frac{b}{c}\Big)
= c\,\frac{ \dn \alpha}{\cn \alpha}   \Big(\bmod \frac{b}{c}\Big),&&
\\
&&  \mu = \frac{b}{\dn \beta }
        &\Big( \bmod\Big( 1-\dfrac{b^2}{c^2} \Big)^{\frac{1}{2}} \Big),\quad
  \nu = b \sn \gamma\Big( \bmod \frac{b}{c} \Big)&&
\tag{11}
\end{flalign*}
(v.\ Int.\ Cal.\ Arts.~179, 192, and 196).

\mypara{137.} If in (8) Art.~136 we assume $V=L.M.N$ where $L$ involves $\alpha$ only,
$M$ involves $\beta$ only, and $N$ involves $\gamma$ only, (8) can be written
\[
\dfrac{\mu^2-\nu^2}{L}\dfrac{ d^2 L}{d\alpha^2} + \dfrac{\lambda^2 - \nu^2}{M} \dfrac{d^2 M}{d\beta^2} + \dfrac{\lambda^2 - \mu^2}{N}\dfrac{ d^2 N}{d\gamma^2} =0. \tag{1}
\]
(1) is too complicated to be broken up by our usual method.

If, however, we let
\[
\dfrac{1}{L}\dfrac{d^2L}{d\alpha^2} ={\textstyle\sum} a_k \lambda^k,\quad\dfrac{1}{M}\dfrac{ d^2M}{d\beta^2} ={\textstyle\sum}  b_k \mu^k,\quad \dfrac{1}{N}\dfrac{ d^2N}{d\gamma^2} ={\textstyle\sum} c_k \nu^k,
\]
substitute in (1) and make use of the fact that the result must be identically
zero, we find that the coefficients are zero for all values of $k$ except $k = 0$ and
$k = 2$, and that $a_0 = -b_0 = c_0$, and $a_2 = -b_2 = c_2$.

Therefore (1) can be broken up into the three equations
\begin{align*}
\dfrac{d^2L}{d\alpha^2}&= (a_0 + a_2 \lambda^2)L \\
\dfrac{d^2M}{d\beta^2} &= -(a_0 + a_2 \mu^2)M \\
\dfrac{d^2N}{d\gamma^2}& = (a_0 + a_2 \nu^2)N.
\end{align*}

% -----File: 264.png
We shall find it convenient to take $a_2$ as $m(m + 1)$ and $a_0$ as $-(b^2 + c^2)p$;
whence
\[\left .\begin{aligned}
\frac{d^2 L }{d \alpha^2} &- [m(m + 1) \lambda^2 - (b^2 + c^2)p]L = 0\\[1ex]
\frac{d^2 M }{ d \beta^2} &+ [m(m + 1) \mu^2 - (b^2 + c^2)p]M = 0\;  \\[1ex]
\frac{d^2 N }{d \gamma^2} &- [m(m + 1) \nu^2 - (b^2 + c^2)p]N = 0 .
\end{aligned}
\right \}
\tag{2}\]

If now in (2) we replace $\alpha$, $\beta$, and $\gamma$ by their values in terms of $\lambda$, $\mu$, and
$\nu$, we get
\begin{equation}\left .
\begin{aligned}
(\lambda^2 - b^2) (\lambda^2 - c^2) \frac{d^2 L }{d\lambda^2}
+ \lambda (\lambda^2 - b^2 + \lambda^2 - c^2) \frac{dL }{d\lambda}\hspace{8em}\\
- [m(m + 1) \lambda^2 - (b^2 + c^2)p] L = 0\phantom{.}\\
(\mu^2 - b^2) (\mu^2 - c^2) \frac{d^2 M }{ d\mu^2}
+ \mu (\mu^2 - b^2 + \mu^2 - c^2) \frac{dM }{d\mu}\hspace{7em}\\
- [m(m + 1) \mu^2 - (b^2 + c^2)p] M = 0\phantom{.}\\
(\nu^2 - b^2)(\nu^2 - c^2)\frac{ d^2 N }{d\nu^2}
+ \nu (\nu^2 - b^2 + \nu^2 - c^2) \frac{dN}{ d\nu}\hspace{7.5em}\\
- [m(m + 1) \nu^2 - (b^2 + c^2)p] N = 0 .
\end{aligned}
\right \}
\tag{3}\end{equation}
Whence if $L = E_m^p (\lambda)$, it follows that $M =  E_m^p (\mu)$ and $N =  E_m^p (\nu)$, and that
\[
V =  E_m^p (\lambda)  E_m^p (\mu)  E_m^p (\nu)           \tag{4}
\]
is a solution of Laplace's Equation, (8) Art.~136.

The equation
\begin{multline*}
  (x^2 - b^2) (x^2 - c^2)\frac{ d^2 z }{ dx^2}
+ x(x^2 - b^2 + x^2 - c^2) \frac{dz }{dx}\\
- [m(m + 1) x^2 - (b^2 + c^2)p] z = 0  \tag{5}
\end{multline*}
is known as Lamé's Equation, and  $E_m^p (x)$ as a \textit{Lamé's Function} or an \textit{Ellipsoidal}
\textit{Harmonic}. We shall suppose $m$ a positive integer.

To get a particular solution of (5) let $z = \sum a_k x^k$.  Substitute in (5) and
reduce and we get
\begin{multline*}
  [k(k + 1) - m(m + 1)] a_k - (b^2 + c^2) [(k + 2)^2 - p] a_{k+2}\\
+ b^2 c^2 (k + 3)(k + 4) a_{k+4} = 0.  \tag{6}
\end{multline*}
We have now only to choose a sequence of coefficients satisfying (6), and we
may take any two consecutive coefficients arbitrarily.

% -----File: 265.png
(6) which is ordinarily a relation connecting three consecutive coefficients
reduces to a relation between two when $k = m$, when $k = -3$, and when
 $k = -4$. If we take $a_{m + 2} = 0$, $a_{m+4}$, $a_{m+6}$, \&c., will vanish. Let $a_m = 1$.
 If $m$ is even the coefficient of $a_0$ in (6) will be zero; if $p$ has such a value
that $a_{-2}$ is zero, $a_{-4}$, $a_{-6}$, \&c., will be zero, and there will be no terms in
the solution involving negative powers of $x$.

If we write the values of $a_{m-2}$, $a_{m-4}$, \&c., by the aid of (6) we see that
$a_{m-2}$ is of the first degree in $p$, $a_{m-4}$ of the second degree in $p$, \&c., and $a_{-2}$
of the degree $\dfrac{m}{2} +1$  in $p$. There are then $\dfrac{m}{2} +1$  values of $p$ which we shall
call $p_1$, $p_2$, $p_3$, \&c., for which $a_{-2}$ will vanish, and for which our solutions will
be of the form
\[
 E_m^p(x)=x^m+a_{m-2}x^{m-2}+a_{m-4}x^{m-4}+ \cdots+a_0
\]
if $m$ is even.

If $m$ is odd, the coefficient of $a_1$ in (6) will vanish and we can choose $p$ so
that $a_{-1}$ shall be zero, and then all coefficients of lower order will vanish.
$a_{-1}$ is of the degree $\dfrac{m+1}{2}$  in $p$, and there will be $\dfrac{m+1}{2}$ values $p_1$, $p_2$, $p_3$,
\&c., of $p$ for which
\[
 E_m^p(x)=x^m+a_{m-2}x^{m-2}+a_{m-4}x^{m-4}+ \cdots+a_1x.
\]

Following Heine we shall call the solution just obtained $K_m^p(x)$ so that
\[
 K_m^p(x)=x^m+a_{m-2}x^{m-2}+a_{m-4}x^{m-4}+ \cdots  \tag{7}
\]
terminating with $ a_0$ if $m$ is even, and with $a_1x$ if $m$ is odd. If $m$ is even,
there are $\dfrac{m}{2} +1$ of these functions  $K_m^{p_1}(x)$,  $K_m^{p_2}(x)$, \&c., and there are $\dfrac{m+1}{2}$
of them if $m$ is odd. The coefficients can be computed by the aid of (6).

If in Lamé's Equation (5) we let $z=v\sqrt{\strxx x^2-b^2}$ we get the equation
\begin{multline*}
  (x^2-b^2) (x^2-c^2)\frac{ d^2v}{dx^2}
+ x[x^2-b^2+3(x^2-c^2)]\frac{ dv}{dx}\\
-[(m+2)(m-1)x^2 + c^2 - (b^2+c^2)p] v = 0.    \tag{8}
\end{multline*}

Letting $v=\sum a_kx^k$ we obtain the relation
\begin{multline*}
  [k(k+3)-(m+2) (m-1)]a_k
- \{(b^2+c^2) [(k+2)^2-p] + c^2(2k+5)\} a_{k+2}\\
+ b^2c^2 (k+3) (k+4) a_{k+4} = 0.      \tag{9}
\end{multline*}
% -----File: 266.png

Proceeding exactly as before, we find that there are $\dfrac{m}{2}$ values $q_1$, $q_2$, $q_3$, \&c.,
of $ p$ for which $v=x^{m-1}+a_{m-3}x^{m-3} + \cdots + a_{1}x$ if $m$ is even, and $\dfrac{m+1}{2}$
values for which $v=x^{m-1}+a_{m-3}x^{m-3} + \cdots + a_{0}$ if $m$ is odd.

Calling $v\sqrt{\strxx x^2-b^2}$ $L_{m}^{p}(x)$ so that
\[
L_{m}^{p}(x) = \sqrt{ x^2-b^2}[x^{m-1}+a_{m-3}x^{m-3} + a_{m-5}x^{m-5} + \cdots],   \tag{10}
\]
terminating with $a_1x$ if $m$ is even and with $a_0$ if $m$ is odd, we have $\dfrac{m}{2}$
values of $E_m^p(x)$, namely $L_m^{q_1}(x)$, $L_m^{q_2}(x)$, \&c., of the form (10) if $m$ is even
and $\dfrac{m+1}{2}$ values if $m$ is odd.

By interchanging $b$ and $c$ in (8), (9), and (10) we may show that if
\[
M_{m}^{p}(x) =  \sqrt{x^2-c^2}[x^{m-1}+a_{m-3}x^{m-3} + a_{m-5}x^{m-5} + \cdots ]  \tag{11}
\]
there are $\dfrac{m}{2}$ values of $E_m^p(x)$, namely $M_m^{r_1}(x)$, $M_m^{r_2}(x)$, $M_m^{r_3}(x)$, \&c., of the form
(11) if $m$ is even and $\dfrac{m+1}{ 2}$ values if $m$ is odd.

Finally if in Lamé's Equation (5) we let $z=v \sqrt{\strxx(x^2-b^2)(x^2-c^2)}$ we get
\begin{multline*}
(x^2-b^2)(x^2-c^2)\frac{d^2v}{dx^2}+3x(x^2-b^2+x^2-c^2)\frac{dv}{dx}\\
 -[(m+3)(m-2)x^2-(b^2+c^2)(p-1)]v=0.    \tag{12}
\end{multline*}
If now we let $v= \sum a_kx^k$ we obtain the relation
\begin{multline*}
[k(k+5)-(m-2)(m+3)]a_k\\
-(b^2+c^2)[(k+2)(k+4)+1-p]a_{k+2}+b^2c^2(k+3)(k+4)a_{k+4}=0.   \tag{13}
\end{multline*}

Proceeding as before we find that there are $\dfrac{m}{2}$ values $s_1$, $s_2$, $s_3$, \&c., of  $ p$
for which $v=x^{m-2} + a_{m-4}x^{m-4} + a_{m-6}x^{m-6} + \cdots + a_0$ if $m$ is even, and
$\dfrac{m+1}{2}$ values for which $ v=x^{m-2}+a_{m-4}x^{m-4} + \cdots + a_1x$ if $m$ is odd.

Calling $v\sqrt{\strxx(x^2-b^2)(x^2-c^2) }$ $N_m^p(x)$ so that
\[
N_m^p(x) = \sqrt{(x^2-b^2)(x^2-c^2) }[x^{m-2} + a_{m-4}x^{m-4} + a_{m-6}x^{m-6} + \cdots]   \tag{14}
\]
terminating with $a_0$ if $m$ is even and with $a_1x$ if $m$ is odd, we have $\dfrac{m}{2 }$ values
of  $ E_m^p(x) $, namely $N_m^{s_1}(x)$, $N_m^{s_2}(x)$, $N_m^{s_3}(x)$, \&c., of the form (14) if $m$ is even and
$\dfrac{m-1}{2}$ values if $m$ is odd.

% -----File: 267.png
Summing up our results we see that there are $2m+1$ Ellipsoidal Harmonics
$E_m^p(x)$ each of which is a finite sum of the $m$th degree in $x$, or in $x$ and $\sqrt{\strxx x^2-b^2}$,
or in $x$ and $\sqrt{\strxx x^2-c^2}$, or in $x$ and $\sqrt{\strxx x^2-b^2}$ and $\sqrt{\strxx x^2-c^2}$.

It was proved by Lamé that the $2m + 1$ values of $p$, namely $p_1$, $p_2$, $p_3$, \&c.,
$q_1$, $q_2$, $q_3$, \&c., $r_1$, $r_2$, $r_3$, \&c., $s_1$, $s_2$, $s_3$, \&c., were all real, and by Liouville that
they were all different.

We give tables of the Ellipsoidal Harmonics for $m = 0$, $m = 1$, $m = 2$, and
$m = 3$. The coefficients were obtained by the aid of formulas (6), (9),
and (13).

\rule{3em}{0em} \begin{minipage}{0.25\linewidth}{
\[
\arraycolsep=0.7em
\begin{array}{|l@{}l|}
\multicolumn{2}{c}{E_0(x)}\\[0.5em]
\hline
K_0(x)&=1\rule{0em}{1.3em}\\[0.5em]
L_0(x)&=0\\[0.5em]
M_0(x)&=0\\[0.5em]
N_0(x)&=0\\[0.5em]
\hline
\end{array}
\]
}
\end{minipage}
\begin{minipage}{0.6\linewidth}{
\[
\begin{array}{|l@{}l|}
\multicolumn{2}{c}{E_1(x)}\\[0.5em]
\hline
K_1(x)&=x\rule{0em}{1.3em}\\[0.5em]
L_1(x)&=\sqrt{\strxx x^2-b^2}\\[0.5em]
M_1(x)&=\sqrt{\strxx x^2-c^2}\\[0.5em]
N_1(x)&=0\\[0.5em]
\hline
\end{array}\rule{3.7em}{0em}
\]
}
\end{minipage}
\[
\begin{array}{|l@{}l|}
\multicolumn{2}{c}{E_2(x)}\\[0.5em]
\hline
K_2^{p_1}(x)&=x^2 - \frac{1}{3} [b^2 + c^2 - \sqrt{(b^2 + c^2)^2-3b^2c^2}]\rule{0em}{1.3em}\\[0.5em]
K_2^{p_2}(x)&=x^2 - \frac{1}{3} [b^2 + c^2 + \sqrt{(b^2 + c^2)^2-3b^2c^2}]\\[0.5em]
L_2(x)&=x \sqrt{\strxx x^2-b^2}\\[0.5em]
M_2(x)&=x \sqrt{\strxx x^2-c^2}\\[0.5em]
N_2(x)&=\sqrt{(x^2-b^2)(x^2-c^2)}\\[0.5em]
\hline
\end{array}
\]
\[
\begin{array}{|l@{}l|}
\multicolumn{2}{c}{E_3(x)}\\[0.5em]
\hline
K_3^{p_1}(x)&=x^3 - \dfrac{x}{5} [2(b^2+c^2) - \sqrt{4(b^2+c^2)^2-15b^2c^2}]
\rule{0em}{1.3em}\\[0.9em]

K_3^{p_2}(x)&=x^3 -\dfrac{x}{5}  [2(b^2+c^2) + \sqrt{4(b^2+c^2)^2-15b^2c^2}]\\[0.9em]
L_3^{q_1}(x)&=\sqrt{\strxx x^2-b^2 }[x^2-\frac{1}{5} (b^2+2c^2 - \sqrt{(b^2+2c^2)^2-5b^2c^2})]\\[0.7em]
L_3^{q_2}(x)&=\sqrt{\strxx x^2-b^2} [x^2-\frac{1}{5} (b^2+2c^2 + \sqrt{(b^2+2c^2)^2-5b^2c^2})]\\[0.7em]
M_3^{r_1}(x)&=\sqrt{\strxx x^2-c^2} [x^2-\frac{1}{5} (2b^2+c^2 - \sqrt{(2b^2+c^2)^2-5b^2c^2})]\\[0.7em]
M_3^{r_2}(x)&=\sqrt{\strxx x^2-c^2} [x^2-\frac{1}{5} (2b^2+c^2 + \sqrt{(2b^2+c^2)^2-5b^2c^2})]\\[0.7em]
N_3(x)&=x \sqrt{(x^2-b^2)(x^2-c^2)}\\[0.5em]
\hline
\end{array}
\]
\vspace{\baselineskip}

% -----File: 268.png
It is to be noted that since in the solution (4) of Laplace's Equation,
\[
V = E_m^p(\lambda) E_m^p(\mu) E_m^p(\nu),
\]
we have the same $m$ and $p$ in each of the three factors, we shall have to deal
merely with products made up of factors of the same form, for example,
\[
K_m^{p_k}(\lambda)K_m^{p_k}(\mu)K_m^{p_k}(\nu),\quad L_m^{q_k}(\lambda)L_m^{q_k}(\mu)L_m^{q_k}(\nu),\quad \text{\&c.;}
\]
and that in a solution of the form
\[
V=\sum A_{m,p} E_m^p(\lambda) E_m^p(\mu) E_m^p(\nu)
\]
we shall have for a given $m$ just $2m+1$ terms.

\mypara{138.} From the particular solution of Lamé's Equation [(5) Art.~137]
$z=E_m^p(x)$, we can get by formula (5), Art.~18, the general solution.
\begin{flalign*}
&\indent\text{It is} & z=AE_m^p(x)+BE_m^p(x)\int\limits_x^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}[E_m^p(x)]^2}. & \tag{1}&
\end{flalign*}
Making $A=0$ and $B=2m+1$ we get a second form of particular solution of
Lamé's Equation, $z=F_m^p(x)$ where
\[
F_m^p(x)=(2m+1)E_m^p(x)\int\limits_x^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}[E_m^p(x)]^2}. \tag{2}
\]
We shall call $F_m^p(x)$ a \emph{Lamé's Function of the second kind}.

It is easily seen to approach the value zero as $x$ is indefinitely increased.

\EXAMPLE{S}

1.\quad If an ellipsoidal conductor is charged with electricity, and is found to
be at potential $V_0$, show that since $V_0=V_0K_0(\lambda)$,
\[
V=V_0K_0(\lambda)K_0(\mu)K_0(\nu)=V_0
\]
at an internal point, and
\begin{multline*}
V=V_0K_0(\mu)K_0(\nu)\bigg[K_0(\lambda)\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}[K_0(x)]^2}\\
÷ K_0(\lambda_0)\int\limits_{\lambda_0}^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}[K_0(x)]^2}\bigg]
\end{multline*}
\[
=V_0\bigg[\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}}
   ÷ \int\limits_{\lambda_0}^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}}\bigg]
=V_0\frac{F\Big(\dfrac{b}{c},\sin^{-1}\dfrac{c}{\lambda}\Big)}{F\Big(\dfrac{b}{c},\sin^{-1}\dfrac{c}{\lambda_0}\Big)},
\]
% -----File: 269.png
\begin{flalign*}
&\text{whence} & V=V_0\dfrac{\Big(F\Big(\dfrac{b}{c},\dfrac{\pi}{2}\Big)-\alpha\Big)}{F\Big(\dfrac{b}{c},\dfrac{\pi}{2}\Big)-\alpha_0} & \tag*{\text{v.~(10) Art.~136.}}&
\end{flalign*}

2.\quad Find the value of the potential function at an external point due to the
attraction of a solid homogeneous ellipsoid (v.\ Art.~135).

Observe that
\begin{align*}
(l^2-\mu^2)(l^2&-\nu^2)=\tfrac{1}{3}[3l^4-2(b^2+c^2)l^2+b^2c^2]K_0(\mu)K_0(\nu)\\
               &+\tfrac{1}{2}\bigg[1+\frac{b^2+c^2-3l^2}{\sqrt{(b^2+c^2)^2-3b^2c^2}}\bigg]K_2^{p_1}(\mu)K_2^{p_1}(\nu)\\
               &+\tfrac{1}{2}\bigg[1-\frac{b^2+c^2-3l^2}{\sqrt{(b^2+c^2)^2-3b^2c^2}}\bigg]K_2^{p_2}(\mu)K_2^{p_2}(\nu);
\end{align*}
and that
\[
\int\limits_0^{\lambda_0}\tfrac{4}{3}\pi\rho\frac{3l^4-2(b^2+c^2)l^2+b^2c^2}{\sqrt{(l^2-b^2)(l^2-c^2)}}dl
=\tfrac{4}{3}\pi\rho\lambda_0\sqrt{(\lambda_0^2-b^2)(\lambda_0^2-c^2)}=M
\]
where $M$ is the mass of the ellipsoid.
% recast to fit page width
\begin{flalign*}
&\textit{Ans.} & V=M\bigg\{&\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}}-\frac{3}{2\sqrt{(b^2+c^2)^2-3b^2c^2}}\\
&&&\bigg[K_2^{p_1}(\mu)K_2^{p_1}(\nu)K_2^{p_1}(\lambda)
\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}.(K_2^{p_1}(x))^2}\\
&&&-K_2^{p_2}(\mu)K_2^{p_2}(\nu)K_2^{p_2}(\lambda)
\int\limits_\lambda^\infty\frac{dx}{\sqrt{(x^2-b^2)(x^2-c^2)}.(K_2^{p_2}(x))^2}\bigg]\bigg\}.
\end{flalign*}

\mypara{139.} If for the sake of brevity we represent $\dfrac{b}{c}$ by $k$, and $\Big(1-\dfrac{b^2}{c^2}\Big)^\frac{1}{2}$ by $k'$ in
the formulas (11) Art.~136 we have
\[
\lambda=c\,\frac{\dn\alpha}{\cn\alpha}(\bmod k),\quad \mu=\frac{b}{\dn\beta(\bmod k')},\quad \nu=b\sn\gamma(\bmod k) \tag{1}
\]
and from these we get without difficulty (v.\ Int.\ Cal.\ Art.~192)
\begin{equation*}\left.
\begin{aligned}
\sqrt{\lambda^2-b^2}&=\frac{ck'}{\cn\alpha(\bmod k)},         &\sqrt{\mu^2-b^2}    &=\frac{bk'\sn\beta}{\dn\beta}(\bmod k'),\;\\
\sqrt{b^2-\nu^2}    &=b\cn\gamma(\bmod k),                    &\sqrt{\lambda^2-c^2}&=\frac{ck'\sn\alpha}{\cn\alpha}(\bmod k),\\
\sqrt{c^2-\mu^2}    &=\frac{ck'\cn\beta}{\dn\beta}(\bmod k'), &\sqrt{c^2-\nu^2}    &=c\dn\gamma(\bmod k).
\end{aligned}\right\} \tag{2}
\end{equation*}

% -----File: 270.png
If we let $\alpha$ range from 0 to $K$, and $\beta$ from 0 to $2K'$, and $\gamma$ from 0 to $4K$,
where $K$ and $K'$ are the complete Elliptic Integrals $F\Big(k,\dfrac{\pi}{2}\Big)$ and $F\Big(k',\dfrac{\pi}{2}\Big)$
respectively, $(\alpha, \beta, \gamma)$  may represent any point in space, and there will be no
ambiguity in sign (v.\ Art.~136).

We may note that if $0<\beta<K'$, $z$ is positive; if $K'<\beta<2K'$, $z$ is
negative; if $0<\gamma<K$, $x$ and $y$ are both positive; if $K<\gamma<2K$, $x$ is
positive and $y$ negative; if $2K<\gamma<3K$, $x$ and $y$ are both negative; and if
$3K<\gamma<4K$, $x$ is negative and $y$ positive (v.\ Art.~136).

We can write the values in (4), (5), (6), and (7), Art.~136, more neatly by
bringing in $\alpha$, $\beta$, and $\gamma$. We get
\begin{gather*}
dn=\frac{1}{c}\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}d\alpha, \tag{3}\\
dS=\frac{1}{c^2}(\mu^2-\nu^2)\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}d\beta d\gamma, \tag{4}\\
dv=\frac{1}{c^3}(\lambda^2-\mu^2)(\lambda^2-\nu^2)(\mu^2-\nu^2)d\alpha d\beta d\gamma. \tag{5}
\end{gather*}

For the integral of any function of $\alpha$, $\beta$, and $\gamma$ over the ellipsoid $\alpha=\alpha_0$, we
shall have\\
$\displaystyle\int F(\alpha,\beta,\gamma)dS
=\frac{1}{c^2}\int\limits_0^{2K'}d\beta\int\limits_0^{4K}
F(\alpha_0,\beta,\gamma)(\mu^2-\nu^2)\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}d\gamma.\hfill(6)$

\mypara{140.} If we make use of the formula (2) Art.~92
\[
\int(UD_n V - VD_n U)dS=0 \tag{1}
\]
and take as our closed surface any given ellipsoid, we can get a very important
result.
\begin{flalign*}
&\indent\text{If} & U &=E_m^p(\lambda)E_m^p(\mu)E_m^p(\nu)\quad \text{and}\quad
                    V  =E_n^q(\lambda)E_n^q(\mu)E_n^q(\nu)&&\\
\intertext{then\hspace*{\stretch{1}} $\nabla^2U=\nabla^2V=0$.\hspace*{\stretch{1}}}
&&D_nU&=D_\alpha U D_n \alpha
=E_m^p(\mu)E_m^p(\nu)\frac{dE_m^p(\lambda)}{d\alpha}\frac{c}{\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}},&&\\
&\text{and}& D_nV &=D_\alpha V D_n \alpha
=E_n^q(\mu)E_n^q(\nu)\frac{dE_n^q(\lambda)}{d\alpha}\frac{c}{\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}},&&
\end{flalign*}
% recast to fit page width
\begin{multline*}
U D_n V - V D_n U
=E_m^p(\mu) E_m^p(\nu) E_n^q(\mu) E_n^q(\nu)\\
\bigg(E_m^p(\lambda) \frac{dE_n^q(\lambda)}{d\alpha}
    - E_n^q(\lambda)\frac{dE_m^p(\lambda)}{d\alpha}\bigg)
\frac{c}{\sqrt{(\lambda^2-\mu^2)(\lambda^2-\nu^2)}}.
\end{multline*}

% -----File: 271.png
Integrating $UD_nV-VD_nU$ over the whole ellipsoid, and writing the result
equal to zero, we have
% recast to fit page width
\begin{multline*}
\frac{1}{c}\int\limits_0^{2K'}d\beta\int\limits_0^{4K} E_m^p(\mu)E_m^p(\nu)E_n^q(\mu)E_n^q(\nu)\\
\bigg[E_m^p(\lambda)\frac{dE_n^q(\lambda)}{d\alpha}-E_n^q(\lambda)\frac{dE_m^p(\lambda)}{d\alpha}\bigg]
(\mu^2-\nu^2)d\gamma=0.
\end{multline*}
\begin{flalign*}
&\indent\text{Hence} & \int\limits_0^{2K'}d\beta\int\limits_0^{4K} E_m^p(\mu)E_m^p(\nu)&E_n^q(\mu)E_n^q(\nu)(\mu^2-\nu^2)d\gamma=0 && \tag{2}&\\
&\text{unless} & E_m^p(\lambda)\frac{dE_n^q(\lambda)}{d\alpha}&-E_n^q(\lambda)\frac{dE_m^p(\lambda)}{d\alpha}=0. && \tag{3}&
\end{flalign*}

But as our ellipsoid may be taken at pleasure, $\lambda$ and $\alpha$ are unrestricted, and
if (3) is true it must be true identically.

If we divide (3) by $[E_m^p(\lambda)]^2$ it becomes
\[
\frac{d}{d\alpha}\bigg[\frac{E_n^q(\lambda)}{E_m^p(\lambda)}\bigg]=0\quad \text{and}\quad \frac{E_n^q(\lambda)}{E_m^p(\lambda)}=\text{a constant;}
\]
and this obviously cannot be true unless $n=m$ and $q=p$.

\EXAMPLE{S}

1.\quad Show that it follows from (2) Art.~140 that
\[
\int\limits_{-K'}^{K'}d\beta\int\limits_{-K}^{K}
E_m^p(\mu)E_m^p(\nu)E_n^q(\mu)E_n^q(\nu)(\mu^2-\nu^2)d\gamma=0.
\]
\textit{Suggestion:}
\begin{align*}
 \int\limits_0^{2K'}    E_m^p(\mu)E_n^q(\mu)(\mu^2-\nu^2)d\beta
=\int\limits_0^{K'}     E_m^p(\mu)&E_n^q(\mu)(\mu^2-\nu^2)d\beta\\
&+\int\limits_{K'}^{2K'} E_m^p(\mu)E_n^q(\mu)(\mu^2-\nu^2)d\beta.
\end{align*}

If in the last integral we replace $\beta$ by $\beta+2K'$ it becomes
\[
 ±\int\limits_{-K'}^0 E_m^p(\mu)E_n^q(\mu)(\mu^2-\nu^2)d\beta
\]
v.\ Arts.~136 and 139 and Int.\ Cal.\ Art.~196.\\

2.\quad Show that
\[
  \int\limits_0^{2K'} d\beta \int\limits_0^{4K} [E_m^p(\mu)E_m^p(\nu)]^2(\mu^2-\nu^2)d\gamma
=8\int\limits_0^{K'} d\beta \int\limits_0^{K} [E_m^p(\mu)E_m^p(\nu)]^2(\mu^2-\nu^2)d\gamma.
\]

% -----File: 272.png
\mypara{141.}  We can now solve the problem of finding the value of $V$ at any point
in space when it is given at all the points on the surface of the ellipsoid
$\alpha=\alpha_0$.

We have first to develop in Ellipsoidal Harmonics a function of $\mu$ and $\nu$ or
rather of $\alpha$ and $\beta$ given at all points on the surface of the ellipsoid in question;
and this is now easily accomplished by our usual method, which leads us to
the result
\[
f(\alpha_0,\beta,\gamma)=\sum_{m=0}^{m=\infty}\sum_{k=1}^{k=2m+1}\hspace{-0.7em}A_{m,p_k} E^{p_k}_m(\mu)E^{p_k}_m(\nu), \tag{1}
\]
\begin{flalign*}
&\text{where} & A_{m,p_k} =
\frac{\displaystyle\int\limits_0^{2K'}d\beta \int\limits_0^{4K}f(\alpha_0,\beta,\gamma)
E_m^{p_k}(\mu)E_m^{p_k}(\nu)(\mu^2-\nu^2)d\gamma}
    {\displaystyle 8\int\limits_0^{K'}d\beta\int\limits_0^{K}[E_m^{p_k}(\mu)E_m^{p_k}(\nu)]^2(\mu^2-\nu^2)d\gamma}. & \tag{2}&
\end{flalign*}

Our final solution is
\[
V=\sum_{m=0}^{m=\infty}\sum_{k=1}^{k=2m+1}\hspace{-0.7em}A_{m,p_k}
\frac{E^{p_k}_m(\lambda)}{E^{p_k}_m(\lambda_0)}E^{p_k}_m(\mu)E^{p_k}_m(\nu) \tag{3}
\]
at an internal point;
\[
V=\sum_{m=0}^{m=\infty}\sum_{k=1}^{k=2m+1}\hspace{-0.7em}A_{m,p_k}\frac{F^{p_k}_m(\lambda)}{F^{p_k}_m(\lambda_0)}E^{p_k}_m(\mu)E^{p_k}_m(\nu) \tag{4}
\]
at an external point.

Lamé has proved rather ingeniously that
\[
\int\limits_0^{K'}d\beta \int\limits_0^{K}[E^{p_k}_m(\mu)E^{p_k}_m(\nu)]^2(\mu^2-\nu^2)d\gamma
\]
can always be found and that it is equal to $\dfrac{\pi}{2}$ multiplied by a rational integral
function of the coefficients of $E^{p_k}_m(x)$ and of $c^2$ and $\Big(\dfrac{b}{c}\Big)^2$.

Of course the labor of obtaining even a few terms of the development of a
function that is in the least complicated is enormous.

\mypara{142.} If in Laplace's Equation (8) Art.~136 we let $V=E^p_m(\lambda)U$ supposing
$U$ to be a function of $\beta$ and $\gamma$ only, we get after replacing $\dfrac{1}{E_m^p(\lambda)}\dfrac{d^2E_m^p(\lambda)}{d\alpha^2}$
by its value $m(m+1)\lambda^2-(b^2+c^2)p$\, [v.~(2) Art.~137]\\
\[
  (\lambda^2-\nu^2)D^2_\beta U
+ (\lambda^2-\mu^2)D^2_\gamma U
+ (\mu^2-\nu^2)[m(m+1)\lambda^2-(b^2+c^2)p] U=0; \tag{1}
\]
% -----File: 273.png
and since by hypothesis $U$ is independent of $\lambda$, the coefficient of $\lambda^2$ in (1)
must vanish. Hence
\[
D^2_\beta U + D^2_\gamma U + (\mu^2 - \nu^2) m (m + 1) U = 0. \tag{2}
\]
Of course $U= E^p_m(\mu)E^p_m(\nu)$ will satisfy (2).

\EXAMPLE{S}

1.\quad Substitute $U=E^p_m(\mu)E^p_m(\nu)$ in (2) Art.~142 and by the aid of (2) Art.~137
show that the equation (2) Art.~142 is satisfied.\\

2.\quad Obtain (2) Art.~140 directly from (2) Art.~142.\\

3.\quad \textit{Conical Coördinates.} Consider the system of coördinates defined by the
equations
\begin{equation*}\left.
\begin{aligned}
x^2+y^2&+z^2=r^2\\
\dfrac{x^2}{\mu^2}+\dfrac{y^2}{\mu^2-b^2}&+\dfrac{z^2}{\mu^2-c^2}=0\\
\dfrac{x^2}{\nu^2}+\dfrac{y^2}{\nu^2-b^2}&+\dfrac{z^2}{\nu^2-c^2}=0\;
\end{aligned} \right\}\tag{1}
\end{equation*}
where $c^2>\mu^2>b^2>\nu^2$.

Show that
\[
x^2=\frac{r^2\mu^2\nu^2}{b^2c^2},\quad
y^2=\frac{r^2(\mu^2-b^2)(\nu^2-b^2)}{b^2(b^2-c^2)},\quad
z^2=\frac{r^2(\mu^2-c^2)(\nu^2-c^2)}{c^2(c^2-b^2)};
\]
\[
h_1^2=\frac{(\mu^2-b^2)(c^2-\mu^2)}{r^2(\mu^2-\nu^2)},\quad
h_2^2=\frac{(b^2-\nu^2)(c^2-\nu^2)}{r^2(\mu^2-\nu^2)},\quad
h_3^2=1.
\]

Laplace's Equation is
\[
D^2_\alpha V + D^2_\beta V +(\mu^2-\nu^2)D_r(r^2 D_r V) = 0 \tag{2}
\]
where  \hspace{\stretch{1}}$\displaystyle\alpha=\int\limits_b^\mu\frac{d\mu}{\sqrt{(\mu^2-b^2)(c^2-\mu^2)}}\quad \text{and}\quad
\beta = \int\limits_0^\nu\frac{d\nu}{\sqrt{(b^2-\nu^2)(c^2-\nu^2)}},$\hspace{\stretch{1}}

If $V=U.R$ (2) breaks up into
\begin{flalign*}
&&\frac{d}{dr}\bigg(r^2&\frac{dR}{dr}\bigg)=m(m+1)R, \tag{3}\\
&&D^2_\alpha U + D^2_\beta U &+ m(m+1)(\mu^2-\nu^2)U=0. \tag{4}\\[1ex]
&\text{(3) gives} & R&=Ar^m+Br^{-m-1}. &&\\[1ex]
&\text{(4) gives} && U=E^p_m(\mu)E^p_m(\nu) & \tag*{(v.\ Art.~142).}&
\end{flalign*}

So that a solution of (2) is
\[
V=Ar^m E^p_m(\mu)E^p_m(\nu).
\]

But since (2) is Laplace's Equation, $V=Ar^mY_m(\mu,\phi)$, if expressed in
Conical Coördinates, must satisfy it, consequently $E^p_m(\mu)E^p_m(\nu)$ must be simply
a Spherical Harmonic of the $m$th degree.
% -----File: 274.png

\vspace{1ex}\Needspace*{5\baselineskip}
\begin{center}
\emph{Toroidal Coördinates.}
\end{center}\vspace{-1ex}

\mypara{143.} Any pair of circles belonging to the orthogonal system obtained and
figured in Art.~46 can be represented by the equations
\[
\left.
\begin{aligned}
\dfrac{2ax}{\sinh \alpha} &= \dfrac{x^2 + y^2 + a^2}{ \cosh \alpha}\;\\
\dfrac{2ay }{\sin \beta}  &= \dfrac{x^2 + y^2 - a^2 }{\cos \beta}
\end{aligned}
\right\}
\tag{1}
\]
if we take $2a$ instead of 2 as the distance between the points common to the
second set of circles.

If we rotate the system about the axis of $y$ we get a set of spheres and a
set of anchor rings which cut orthogonally.  These and a set of planes through
the axis of revolution will form an orthogonal system of surfaces, and the
parameters corresponding to them may be taken as a set of curvilinear
coördinates and may be called \emph{Toroidal Coördinates}.

If we take the axis of the system as the axis of $Z$, the equations of a set of
the surfaces may be written
\[
\left.
\begin{aligned}
\dfrac{4a^2(x^2+y^2) }{\sinh^2 \alpha} &= \dfrac{[x^2+y^2+z^2+a^2]^2}{\cosh^2 \alpha}\;\\
\dfrac{2az}{\sin \beta} &= \dfrac{x^2+y^2+z^2-a^2}{\cos \beta}\\
y &= x \tan\gamma
\end{aligned}
\right\}
\tag{2}
\]

$\alpha$, $\beta$, and $\gamma$ being regarded as the coördinates of a point of intersection of the
three surfaces.

Finding Laplace's Equation in the usual manner we get
\begin{gather*}
x= \dfrac{a \sinh\alpha \cos\gamma}{\cosh\alpha \mp \cos\beta}, \quad y=\dfrac{a \sinh\alpha \sin\gamma}{\cosh\alpha \mp \cos\beta},\quad z=\dfrac{a \sin\beta}{\cosh\alpha \mp \cos\beta},\\[0.5em]
r=\sqrt{x^2+y^2}=\dfrac{a \sinh\alpha}{\cosh\alpha \mp \cos\beta},\quad a+z \ctn\beta=\dfrac{a \cosh\alpha}{\cosh\alpha \mp \cos\beta};\\[0.5em]
h_1=\dfrac{\cosh\alpha \mp \cos\beta}{a},\quad  h_2=\dfrac{\cosh\alpha \mp \cos\beta}{a},\quad h_3=\dfrac{\cosh\alpha \mp \cos\beta}{a \sinh\alpha};
\end{gather*}
and Laplace's Equation becomes
\begin{align*}
D_\alpha\left[\dfrac{a \sinh\alpha}{\cosh\alpha \mp \cos\beta} D_\alpha V\right]&+D_\beta\left[\dfrac{a \sinh\alpha}{\cosh\alpha \mp \cos\beta} D_\beta V\right]\\[0.5em]
&+D_\gamma\left[\dfrac{a}{\sinh\alpha(\cosh\alpha \mp \cos\beta)} D_\gamma V\right] = 0, \tag{1}
\end{align*}
% -----File: 275.png
\begin{flalign*}
&\text{or}& D_\alpha(rD_\alpha V) + D_\beta(rD_\beta V) + \dfrac{1}{\sinh^2\alpha} rD_\gamma^2 V &= 0.& \tag{2}
\end{flalign*}

We cannot proceed further by our usual method, for the assumption that $V$
is a function of $\alpha$ alone, or that $V$ is a function of $\beta$ alone, proves to be
inadmissible. Indeed, not only are $\alpha$, $\beta$, and $\gamma$ not \textit{thermometric parameters}
(v.\ Art.~133), but no thermometric parameters exist, and no possible distribution
can make our anchor rings or our spheres a set of equipotential surfaces.

We can, however, simplify (2). It can be written
\[
D_\alpha^2(V\sqrt{r}) + D_\beta^2(V\sqrt{r})+ \dfrac{1}{\sinh^2\alpha} D_\gamma^2(V\sqrt{r})-V(D_\alpha^2 \sqrt{r}+D_\beta^2\sqrt{r})=0. \tag{3}
\]

$D_\alpha^2\sqrt{r}+D_\beta^2\sqrt{r}$ proves equal to $-\dfrac{\sqrt{r}}{4\sinh^2\alpha}$; hence if $U = V\sqrt{r}$ (3) becomes
\[
\sinh^2\alpha(D_\alpha^2U+D_\beta^2U) + D_\gamma^2 U  + \tfrac{1}{4} U=0, \tag{4}
\]
for which particular solutions can readily be found by our usual process.

(4) can be broken up into the three equations
\begin{gather*}
\dfrac{d^2N}{d\gamma^2} + (m+\tfrac{1}{2})^2N = 0 \tag{5}\\
\dfrac{d^2M}{d\beta^2} + n^2M = 0 \tag{6}\\
\sinh^2\alpha \dfrac{d^2L}{d\alpha^2} - [m(m+1)+n^2\sinh^2\alpha]L=0.  \tag{7}\\
N = A \cos(m+\tfrac{1}{2})\gamma + B \sin(m+\tfrac{1}{2})\gamma\\[1ex]
M = A_1 \cos n\beta + B_1 \sin n\beta.
\end{gather*}

If we introduce into (7) $x = \ctnh \alpha$ it becomes
\[
(1-x^2) \dfrac{d^2L}{dx^2} - 2x\dfrac{ dL}{dx} + \left[m(m+1) - \dfrac{n^2}{1-x^2}\right]L=0,
\]
a solution of which is
\begin{flalign*}
&&L &= P_m^n(x) = (1-x^2)^{\frac{n}{2}} \dfrac{d^nP_m(x)}{dx^n} &\text{(v.\ Art.~102).}
\end{flalign*}
It is to be noted that since $\ctnh \alpha$ is greater than 1
\[
P_m^n(\ctnh \alpha) = i^{\frac{n}{2}}\csch^n \alpha \dfrac{d^n P_m(\ctnh \alpha)}{(d \ctnh \alpha)^n}.
\]
% -----File: 276.png
The constant coefficient $i^{\frac{n}{2}}$ can be rejected and we get
\[
U=[A \cos(m+\tfrac{1}{2})\gamma + B \sin(m+\tfrac{1}{2})\gamma](A_1 \cos n\beta + B_1 \sin n\beta)\csch^n \alpha \dfrac{d^nP_m(\ctnh \alpha)}{(d \ctnh \alpha)^n}
\]
as a particular solution of (4).
\[
\dfrac{1}{i^{\frac{n}{2}}} P_m^n(\ctnh \alpha)=\csch^n \alpha \dfrac{d^nP_m(\ctnh \alpha)}{(d \ctnh\alpha)^n}
\]
has been called a Toroidal Harmonic.

\EXAMPLE{S}

1.\quad Given the value of the potential function at all points on the surface of
an anchor ring; find its value at any point within the ring.

\emph{Suggestion}: If $V = f(\beta,\gamma)$ when $\alpha = \alpha_0$, the function to be developed is
\[
\sqrt{r}.f(\beta,\gamma)\quad \text{\textit{i.e}.}\quad \left[\dfrac{ a \sinh \alpha_0}{ \cosh \alpha_0\mp \cos \beta} \right]^{\frac{1}{2}} f(\beta,\gamma)
\]
and the development will be in a double Fourier's Series (v.\ Art.~71).\\

2.\quad Show that if we let $\alpha$ range from 0 to $\infty$, $\beta$ from $-\pi$ to $\pi$, and $\gamma$ from
0 to $2\pi$, each of the double signs on page 264 may be replaced by the minus
sign without loss of generality.
\label{ch8end}

% -----File: 277.png
\mychap{CHAPTER IX.\footnotemark}{HISTORICAL SUMMARY.}
\label{ch9start}

\footnotetext{See preface.}The method of development in series which has enabled us in the preceding
chapters to solve problems in various branches of mathematical physics, had
its origin, as might have been expected, in the theory of the musical vibrations
of a stretched string. It was in the year 1753\footnote{See two articles by Bernoulli and one by Euler in the Memoirs of the Academy of
Berlin for this year.} that Daniel Bernoulli
enunciated the principle of the coexistence of small oscillations, which, in
connection with Taylor's and John Bernoulli's theory of the vibrating string,
led him to believe that the general solution of this problem could be put in
the form of a trigonometric series. This principle also led him and Euler to
treat in a similar manner the problems of the vibration of a column of air and
of an elastic rod. The problem of the vibration of a heavy string suspended
from one end was also treated in the same manner by these mathematicians
and deserves special mention here as in it Bessel's functions of the zeroth
order appear for the first time.\footnote{See the Transactions of the Academy of St.\ Petersburg for 1732-33, 1734 and 1781.} In none of these cases, however, was any
method given for determining the coefficients of the series.

This last remark also applies to the more complicated problems of the
vibration of rectangular and circular membranes, which were discussed by
Euler\footnote{Transactions of the Academy of St.\ Petersburg.} in 1764, and in the last of which the general Bessel's functions of
integral orders occur.

It is in problems connected with astronomy that the first completely
successful application of the method here considered occurs. Legendre in a
paper published in the Mémoires des Savants Étrangers for 1785, first
introduced the zonal harmonics $P_m$ and applied them to the determination of
the attraction of solids of revolution. He was followed by Laplace, who in
one of the most remarkable memoirs ever written\footnote{``Théorie des attractions des sphéroïdes et de la figure des Planètes''  Mémoires de
l'académie des sciences 1782. This article, although bearing an earlier date than that of
Legendre, was really inspired by it. It is here that ``Laplace's equation'' first appears,
occurring, however, only in polar coördinates.} determined the potential
of a solid differing but little from a sphere by means of the development
according to the spherical harmonics $Y_m$.

% -----File: 278.png
Very closely related to this problem is Gauss's celebrated treatment of the
theory of terrestrial magnetism,\footnote{Resultate aus den Beobachtungen des magnetischen Vereins im Jahre 1838. Leipzig,
1839. Reprinted in Gauss's collected works, Vol.\ V., p.\ 121.} which we will for that reason mention here,
although it was not published until more than half a century later. This
paper is particularly noteworthy as it contains a numerical application of the
method on a larger scale than has ever been attempted before or since.

After the researches of Legendre and Laplace there was a pause of a
quarter of a century until in 1812 Fourier's extensive memoir: \textit{Théorie du
mouvement de la chaleur dans les corps solides} was crowned by the French
Academy. Although not printed until the years 1824-26,\footnote{Mémoires de l'académie des sciences for 1819-20 and 1821-22.} the manuscript of
this work was in the meantime accessible to the other French mathematicians
presently to be mentioned. The first part of this memoir, which was reproduced
with but few alterations in the \textit{Théorie analytique de la chaleur} (1822),
contains a treatment of the following problems and of practically all of their
special cases:

(\emph{a}) The one dimensional flow of heat. (\emph{b}) The two dimensional flow of
heat in a rectangle. (\emph{c}) The three dimensional flow of heat in a rectangular
parallelopiped. (\emph{d}) The flow of heat in a sphere when the temperature
depends only on the distance from the centre. (\emph{e}) The flow of heat in a
right circular cylinder when the temperature depends only on the distance
from the axis. In these problems not merely the simpler boundary conditions
are considered but also the question of radiation into an atmosphere. In
special cases of the first three problems just mentioned (when one or more
dimensions become infinite) the series degenerate into ``Fourier's integrals."

\markright{HISTORICAL SUMMARY.}
More important even than any of these special problems is the great
advance which Fourier caused the theory of trigonometric series to make.
In a posthumous paper Euler had given the formulae for determining the
coefficients,\footnote{Lagrange had practically determined these coefficients long before but failed to notice
what he had got.} but Fourier was the first to assert and to attempt to prove that
any function, even though for different values of the argument it is expressed
by different analytical formulae, can be developed in such a series. The fact
that the real importance of trigonometric series was thus for the first time
shown justifies us in associating Fourier's name with them, although, as we
have seen, they were known long before his day.

Fourier's results were extended by Laplace in 1820\footnote{Connaissance des Temps pour l'an 1823.} to the general (unsymmetrical)
case of the flow of heat in a sphere, and by Poisson\footnote{Journal de l'École Polytechnique, 19\textsuperscript{e} Cahier. Although the final forms to which Poisson
reduces his results are similar to Fourier's, his methods are very different.} (1821) to the
unsymmetrical flow of heat in a cylinder.

% -----File: 279.png
In 1835 Green published a paper\footnote{``On the determination of the exterior and interior attraction of ellipsoids of variable
densities.'' Transactions of the Cambridge Philosophical Society.} in which the method we are considering
is employed to determine the potential of a heterogeneous ellipsoid. This
paper, in which the analysis is performed at once for space of $n$ dimensions,
anticipates much that was subsequently done by others, but has failed to
exert an influence proportional to its importance.

At about this time Lamé began a series of publications which have connected
his name inseparably with the problem of the permanent state of
temperature of an ellipsoid. In the first of these\footnote{Mémoires des Savants Étrangers, Vol.\ V\@. Although the volume is dated 1838 this paper
(which was reprinted in Liouville's Journal, 1837) must have appeared at least as early as 1835.} the equation $\nabla^2V=0$ is
transformed to ellipsoidal coördinates and is then broken up into three
ordinary differential equations. The rest of the solution, however, is hardly
touched upon. Lamé's most important work on this subject\footnote{
``Sur l'équilibre des Températures dans un ellipsoïde à trois axes inégaux.'' An article
by the same author on the two dimensional potential will be found in Vol.\ I. of this Journal.} was published
in Liouville's Journal in 1839, and in it the complete solution of the problem
is given. Lamé clearly shows in this paper how he arrived at his solution, by
considering first the simpler case of a sphere where, instead of the polar
coördinates $\theta$ and $\phi$, the parameters of two families of confocal cones of the
second degree are used as coördinates. This system of curvilinear coördinates,
which, when applied to the complete sphere, merely gives the old results of
Laplace in a new form, is barely mentioned in Lamé's later publications. In
the same volume of Liouville's Journal Lamé published a second paper in
which he applies his results to the special cases of ellipsoids of revolution.

These two papers form the starting-point for a series of articles on the
same subject by Heine and Liouville. Heine in his doctor dissertation\footnote{Reprinted in Crelle's Journal, Vol.\ 26 (1843).\\
In the same Journal for 1847 F.~Neumann discussed the related problem of the magnetisation
of a soft iron ellipsoid of revolution.} (1842)
determined the potential not merely for the interior of an ellipsoid of
revolution when the value of the potential is given on the surface, but also
for the exterior of such an ellipsoid and for the shell between two confocal
ellipsoids of revolution. Even in the first of these problems, which is
equivalent to that of Lamé, he simplified Lamé's solution materially by
showing that the functions used may be reduced to spherical harmonics,
while in the other two problems he introduced spherical harmonics of the
second kind, which were then new. Shortly afterwards\footnote{\textit{Heine:} Crelle's Journal, Vol.\ 29, 1845. \textit{Liouville:} Liouville's Journal, Vol.\ X.,
1845, and Vol.\ XI., 1846. For a treatment of the problem of the potential of an ellipsoidal
shell by means of a development of $\dfrac{1}{r}$ in terms of Lamé's functions, see a paper by Heine
in Crelle's Journal, Vol.\ 42, 1851.} Heine and Liouville
% -----File: 280.png
published simultaneously two papers in which they arrived independently of
each other at about the same results. In each of these papers attention is
called to the fact that the product of two Lamé's functions is a spherical
harmonic, and this fact is made use of to throw Lamé's solution of the
problem of the permanent state of temperatures of an ellipsoid into a more
elementary form. Besides this the second solution of Lamé's equation is
introduced for the sake of solving the potential problem for the \emph{exterior} of
the ellipsoid.

In thus following up the theory of heat and the related potential problems,
we have lost sight of the question of small vibrations, to which during the
early part of the century a great deal of attention had been devoted by
Poisson, who frequently made use of the method of development in series.
In his memoirs\footnote{See especially the one in the Mémoires de l'académie des sciences, Vol.\ VIII., 1829.} most of the problems left unfinished by Bernoulli and Euler
are thoroughly treated, as well as various slight modifications of them.
When, however, he attacked the problem of the vibration of an elastic plate
he was unable to make much progress, owing in part to the erroneous form of
his boundary conditions. He was, nevertheless, able to solve the problem of
the \emph{symmetrical} vibration of a free circular plate. The complete theory of the
vibration of a free circular plate was first given by Kirchhoff.\footnote{Crelle's Journal, Vol.\ 40, 1850.}

Passing now to a new subject, the theory of the equilibrium of an elastic
spherical shell, we find a solution by Lamé in Liouville's Journal for 1854,
and by Sir William Thomson (1862) in the Philosophical Transactions for
1863. Both of these papers consist of an application of the spherical harmonic
analysis to this rather complicated problem. Thomson, however, considers
besides Lamé's problem certain related questions and the form of his analysis
is very different from Lamé's, being of the same nature as that used in the
Appendix B of his Natural Philosophy of which we shall have to speak
presently. These investigations form the starting point for a number of
recent memoirs among which those of G.~H. Darwin on cosmographical
questions deserve special mention.

Closely related to this last mentioned problem is the theory of the small
vibrations of an elastic sphere. While the simplest case of this problem was
treated by Poisson in the memoir referred to above, the general solution has
been only recently obtained by Jaerisch (1879)\footnote{Crelle's Journal, Vol.\ 88.} and Lamb (1882).\footnote{Proc.\ Lond.\ Math.\ Soc.} The
functions involved are the same as those which occur in the problem of the
non-stationary flow of heat in a sphere as solved by Laplace.

The Appendix B of Thomson and Tait's Natural Philosophy,\footnote{First edition, 1867.  This appendix
was evidently written as early as 1862, as Thomson refers to it in the memoir quoted above.} to which we
have already referred, deserves to be regarded as one of the most important
% -----File: 281.png
contributions to the general theory. The way in which spherical harmonics
are introduced (as homogeneous functions of the rectangular coördinates) was
then new\footnote{The same method was used at about the same time by Clebsch.} and the solution of the potential problem for a variety of new
solids was indicated; viz., for solids whose boundaries consist of concentric
spheres, cones of revolution, and planes. We shall have more to say presently
concerning the method employed for the solution of these problems.

Although connected only indirectly with the theory we are discussing, it
will be well to mention at this point the method of electrical images which is
also due to Sir William Thomson (1845). This method enables us to solve
many potential problems for the inverse of any solid when once we have
solved it for the solid itself. By means of this method most of the solutions
of potential problems obtained by our method may be applied at once with
very little modification to systems of curvilinear coördinates derived by
inversion from those we have used. It will not be necessary to mention
separately problems of this sort, as it is clearly immaterial whether they be
solved directly or by means of the method of inversion.\footnote{A case in point would be the potential problem for the shell between two non-intersecting
eccentric spheres, since these spheres can be inverted into concentric spheres. This problem
was treated directly by C. Neumann in a monograph published in Halle in 1862.}

Returning now to the Continent, we find as the next important question
taken up the problem of the potential of an anchor ring. The first publication
on this subject is a monograph by C. Neumann\footnote{``Theorie der Elektricitäts- und Wärme-Vertheilung in einem Ringe.'' Halle.} (1864), but in Riemann's
posthumous papers which were not published until 1876, ten years after his
death, will be found a short fragment on this subject, which (\emph{cf.}\ the last page
of Hattendorf's edition of Riemann's lectures: ``Partielle Differentialgleichungen'')
would appear to date back to the winter 1860-61. This fragment
is of peculiar interest, as the opening paragraphs clearly show that Riemann
had in mind an extended article on the fundamental principles of our subject.

We will next mention two papers by Mehler in which the functions known
as ``conal harmonics,'' which had already been introduced by Thomson in the
Appendix B above mentioned, were applied to the solution of two problems in
electrostatics. The first of these papers\footnote{Crelle's Journal, Vol.\ 68, 1868.} (1868) deals with the solid bounded
by two intersecting spheres, while in the second\footnote{Jahresbericht des Gymnasiums zu Elbing.} (1870) the infinite cone of
revolution is treated. Both of these problems are essentially different from
those discussed in the ``Appendix B,'' inasmuch as the infinite series which
we usually have degenerate in these cases into definite integrals, just as they
do in some simpler cases treated by Fourier. The later of the two papers
just quoted also contains valuable information concerning the nature of the
% -----File: 282.png
solution of similar problems for the hyperboloids and paraboloids of revolution.
The solutions of these problems are not, however, given.

It remains, in order to close the history of this part of the subject, to mention
a number of memoirs which although treating entirely new problems are of far
less importance than most of those considered up to this point, partly because
the solution is not brought to a point where it can be of much immediate use,
and partly because most of the methods employed are such as could not fail
to present themselves to any one attacking these problems.

Of these the first is a paper by Mathieu\footnote{Liouville's Journal, Vol.\ XIII.} on the vibration of an elliptic
membrane (1868), in which the functions of the elliptic cylinder occur for the
first time.

This was followed in the same year by a paper on closely allied subjects by
H. Weber,\footnote{``Ueber die Integration der partiellen Differentialgleichung $\frac{\delta^2 u}{\delta x^2} + \frac{\delta^2 u}{\delta y^2} + k^2 u=0.$''
Math.\ Ann., Vol.\ I\@.  No \emph{physical} problem is mentioned in this paper.} in which not merely the case of the complete ellipse is briefly
considered, but also that in which the boundary consists of two arcs of
confocal ellipses and two arcs of hyperbolas confocal with them. The special
case in which the ellipses and hyperbolas become confocal parabolas is also
considered, whereby the functions of the parabolic cylinder are for the first
time introduced.

In Mathieu's ``Cours de physique mathématique'' (1873) the problem of
the non-stationary flow of heat in an ellipsoid is touched upon, and an
elaborate though not very satisfactory treatment of the special cases where
we have ellipsoids of revolution is given. New functions appear in all of
these problems.

Of late years C. Baer has supplied a number of missing links in the chain
of problems here considered by treating in succession the potential problem
for the paraboloid of revolution,\footnote{``Ueber das Gleichgewicht und die Bewegung der Wärme in einem Rotationsparaboloid.''
Dissertation, Halle, 1881.} the parabolic cylinder\footnote{``Die Funktion des parabolischen Cylinders,'' Gymnasialprogramm Cüstrin, 1883.} and the general
paraboloid.\footnote{``Parabolische Coordinaten,'' Frankfurt, 1888. See also a paper by Greenhill in the
Proc.\ Lond.\ Math.\ Soc., Vol.\ XIX., 1889 (read Dec.\ 8, 1887). Also a posthumous paper by
Lamé in Liouville's Journal for 1874, Vol.\ XIX.} In the first of these problems Bessel's functions occur, as had
already been stated by Mehler, while in the last we find the functions of the
elliptic cylinder. For each of the three systems of coördinates employed the
same author also touches upon the more general problem of the non-stationary
flow of heat, in which new functions occur.

Except in the case of the anchor ring we have found so far only such solids
treated by our method as are bounded by surfaces of the first or second
% -----File: 283.png
degree. Wangerin\footnote{Preisschriften der Jablanowski'schen Gesellschaft, No.\ XVIII., and Crelle's Journal,
Vol.\ 82. See also, concerning a still further extension, the Berliner Monatsberichten
for 1878.} (1875-76) considered in connection with the theory of
the potential, more general systems of curvilinear coördinates than had
previously been used in physical questions, namely, \textit{cyclidic} coördinates.\footnote{Cyclids are a kind of surface of the fourth order (see Salmon's Geom.\ of three Dimensions,
p.\ 527). In his first memoir Wangerin considers only cyclids of revolution.} He
showed, however, merely how to break up Laplace's equation into three
ordinary differential equations.\footnote{See also a paper by this author in Grünert's Archiv for 1873, where the problem of the
equilibrium of elastic solids of revolution is treated.}

An important branch of our theory which we have not yet touched upon
dates back to the year 1836, when Sturm published a series of fundamentally
important papers in the first two volumes of Liouville's Journal. The
physical question which lies at the basis of these papers is the problem of the
flow of heat in a heterogeneous bar.\footnote{The similar problem of the vibration of a heterogeneous string under the action of an
external force was treated by Maggi (Giornale di Matematiche, 1880). Several special cases
are also considered here in detail.} The method here employed depends
upon the fact that the functions which occur are characterized by the number
of times they vanish in a certain interval. This same idea reappears in
Thomson and Tait's Appendix B already referred to, but first finds its full
expression in this more general field of the three dimensional potential in an
article by Klein: ``Ueber Körper welche von confocalen Flächen zweiten
Grades begrenzt sind''\footnote{Math.\ Ann., 18.} (1881). Still more recently (1889-90) Klein has in
his lectures extended this theory to the treatment of solids bounded by six
confocal cyclids, and has indicated how all the potential problems heretofore
treated by our method are special cases of this one.\footnote{For an exposition of this theory see the treatise: Ueber die Reihenentwickelungen der
Potentialtheorie, Leipsic, Teubner, 1894, by the writer of the present chapter.}

Of late years, especially since the year 1880, the younger English mathematicians
have done a vast amount of work in the theory we are here
considering. Although much of this work is of great value, hardly any of it
can be regarded as being a real \emph{development} of the method; it is rather an
application of it to a great variety of problems. We must therefore content
ourselves with giving a mere list of a few of the more important of these
papers.

\textit{Niven:} On the Conduction of Heat in Ellipsoids of Revolution. Phil.\
Trans., 1880.

\textit{Niven:} On the Induction of Electric Currents in Infinite Plates and
Spherical Shells.\ Phil.\ Trans., 1881.

% -----File: 284.png
\emph{Hicks:} On Toroidal Functions. Phil.\ Trans., 1881.

\emph{Hicks:} On the Steady Motion and Small Vibrations of a Hollow Vortex.
Phil.\ Trans., 1884, 1885.

\emph{Lamb:} On Ellipsoidal Current Sheets. Phil.\ Trans., 1887.

\emph{Chree:} The Equations of an Isotropic Elastic Solid in Polar and Cylindrical
Coördinates, their Solution and Application. Camb.\ Phil.\ Soc.\ Trans.,
XIV., 1889.

\emph{Hobson:} On a Class of Spherical Harmonics of Complex Degree with
Applications to Physical Problems. Camb.\ Phil.\ Soc.\ Trans., XIV., 1889.

\emph{Chree:} On some Compound Vibrating Systems. Camb.\ Phil.\ Soc.\ Trans.,
XV., 1891.

\emph{Niven:} On Ellipsoidal Harmonics. Phil.\ Trans., 1892.

The historical sketch we have just given would naturally require as a
supplement some account of the work that has been done on the question of
the convergence of the various series which occur. This, however, would
carry us too far, and we will content ourselves with mentioning the two
fundamental memoirs by Dirichlet in Crelle's Journal, one in 1829 on
Fourier's series, and one, which has been criticised to some extent by subsequent
mathematicians, in 1837 on Laplace's spherical harmonic development.

Another subject which naturally presents itself here is the theory of the
various new functions we have met. Those properties of these functions,
however, which the physicist needs have usually been investigated by the
physicists themselves in the papers mentioned above; while any thorough
account of the development of the theory of these functions would lead us
into the vast region of the modern theory of linear differential equations.

We will therefore close by merely giving a list of books which will be
found useful by those wishing to continue their study of the subject further.

We begin with the books relating directly to physical questions:

\emph{Fourier:} Théorie Analytique de la Chaleur, 1822.

\emph{Lamé:} Leçons sur les Functions inverses des Transcendantes et les Surfaces
isothermes, 1857.

\emph{Lamé:} Leçons sur les Coordonnées Curvilignes et leurs diverses Applications,
1859.

\emph{Mathieu:} Cours de Physique Mathématique, 1873.

\emph{Riemann:} Partielle Differentialgleichungen, und deren Anwendung auf
phys\-ikalische Fragen (edited by Hattendorf), third edition. 1882.

\emph{F.~Neumann:} Theorie des Potentials und der Kugelfunktionen (edited by
C. Neumann), 1887.

\emph{Thomson and Tait:} Natural Philosophy, second edition, 1879.

\emph{Rayleigh:} Theory of Sound, 1877.

\emph{Basset:} Hydrodynamics, 1888.

\emph{Love:} Theory of Elasticity, 1892.

% -----File: 285.png
\emph{Heine:} Handbuch der Kugelfunktionen (second edition), 1878-81.

\emph{Ferrers:} Spherical Harmonics, 1881.

\emph{Haentzschel:} Reduction der Potentialgleichung auf gewöhnliche Differentialgleichungen,
1893.

These last three books would also belong in the following list of books
relating to the theory of the various functions we use:

\emph{Todhunter:} The Functions of Laplace, Lamé and Bessel, 1875.

\emph{Lommel:} Studien über die Bessel'schen Funktionen, 1868.

\emph{F.~Neumann:} Beiträge zur Theorie der Kugelfunktionen, 1878.

And finally concerning the question of convergence:

\emph{C.~Neumann:} Über die nach Kreis-, Kugel- und Cylinder-Functionen
fort\-schreitenden Entwickelungen, 1881.
\label{ch9end}
% -----File: 286.png
%[Blank Page]
% -----File: 287.png
\label{tablestart}
\setcounter{footnote}{0}
\newpage
\medskip\begin{center}{\Large APPENDIX.}\par\rule{5em}{0.5pt}\par\bigskip{TABLES.}\end{center}
\markright{}

Table I., a table of Surface Zonal Harmonics (Legendrians), gives the values
of the first seven Harmonics $P_1(\cos \theta)$, $P_2(\cos \theta)$, $\cdots P_7(\cos \theta)$ for the argument
$\theta$ in degrees. It is taken from the Philosophical Magazine for December,
1891, and was computed by Messrs.\ C.~E. Holland, P.~R. James, and C.~G. Lamb, under the direction of Professor John Perry.

Table II., a table of Surface Zonal Harmonics (Legendrians), gives the
values of the first seven Harmonics $P_1(x)$, $P_2(x)$, $\cdots P_7(x)$ for the argument $x$.
It is reduced from the Tables of Legendrian Functions computed under the
direction of Dr.\ J.~W.~L. Glaisher, and published in the Report of the British
Association for the Advancement of Science for the year 1879.

Table III., the table of Hyperbolic Functions, gives the values of $e^x$, $e^{-x}$,
$\sinh x$, $\cosh x$, and $\gd x$ (Gudermannian of $x$) for values of $x$ from 0.00 to 1.00;
and the values of $\log \sinh x$ and $\log \cosh x$ for values of $x$ from 1.00 to 10.0.
The values of $\gd x$, $\log \sinh x$, and $\log \cosh x$ are taken from the Mathematical
Tables prepared by Professor J.~M. Peirce (Boston: Ginn \& Co.).

The $\log \sinh x$ and $\log \cosh x$ for values of $x$ between 0.00 and 1.00 can be
obtained from the values given for the Gudermannian of $x$ in the table by the
aid of the relations
\[
\log \sinh x = \log \tan (\gd x)\qquad
\log \cosh x = \log \sec (\gd x).
\]

Table IV.\ gives the first twelve roots of $J_0(x) = 0$ and $J_1(x) = 0$ each
divided by $\pi$. The table is taken from Lord Rayleigh's Sound, Vol.\ I.,
page 274, and is due to Professor Stokes, Camb.\ Phil.\ Trans., Vol.\ IX.,
page 186.

Table V.\ gives the first nine roots of $J_0(x) = 0$, $J_1(x) = 0$, $\cdots  J_5(x)=0$.
The table is taken from Rayleigh's Sound, Vol.\ I., page 274, and is due to
Professor J. Bourget, Ann.\ de l'Ecole Normale, T. III., 1866, page 82.

Table VI., the table of Bessel's Functions, gives the values of the Bessel's
Functions $J_0(x)$ and $J_1(x)$ for the argument $x$ from $x = 0$ to $x = 15$. It is
taken from Rayleigh's Sound, Vol.\ I., page 265, and from Lommel's Bessel'sche
Functionen.

% -----File: 288.png
\newpage
\markright{APPENDIX}
\label{tableI}
\begin{center}
TABLE I. --- \textsc{Surface Zonal Harmonics}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|c|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|}{\tablestrut\theta}&
\multicolumn{1}{c|}{P_1(\cos\theta)}&
\multicolumn{1}{c|}{P_2(\cos\theta)}&
\multicolumn{1}{c|}{P_3(\cos\theta)}&
\multicolumn{1}{c|}{P_4(\cos\theta)}&
\multicolumn{1}{c|}{P_5(\cos\theta)}&
\multicolumn{1}{c|}{P_6(\cos\theta)}&
\multicolumn{1}{c|}{P_7(\cos\theta)}\\
\hline
\vrule width0ptdepth0ptheight3.5ex
\phantom{\text{°}0}0\text{°}	&	1.0000	&	1.0000	&	1.0000	&	1.0000	&	1.0000	&	1.0000	&	1.0000	\\
\phantom{0}1    &	.9998	&	.9995   &	.9991   &	.9985	&	.9977	&	.9967	&	.9955	\\
\phantom{0}2    &	.9994	&	.9982	&	.9963	&	.9939	&	.9909	&	.9872	&	.9829	\\
\phantom{0}3    &	.9986	&	.9959	&	.9918	&	.9863	&	.9795	&	.9713	&	.9617	\\
\phantom{0}4    &	.9976	&	.9927	&	.9854	&	.9758	&	.9638	&	.9495	&	.9329	\\[2ex]
\phantom{0}5    &	.9962	&	.9886	&	.9773	&	.9623	&	.9437	&	.9216	&	.8961	\\
\phantom{0}6    &	.9945	&	.9836	&	.9674	&	.9459	&	.9194	&	.8881	&	.8522	\\
\phantom{0}7    &	.9925	&	.9777	&	.9557	&	.9267	&	.8911	&	.8476	&	.7986	\\
\phantom{0}8    &	.9903	&	.9709	&	.9423	&	.9048	&	.8589	&	.8053	&	.7448	\\
\phantom{0}9    &	.9877	&	.9633	&	.9273	&	.8803	&	.8232	&	.7571	&	.6831	\\[2ex]
10      &	.9848	&	.9548	&	.9106	&	.8532	&	.7840	&	.7045	&	.6164	\\
11	&	.9816	&	.9454	&	.8923	&	.8238	&	.7417	&	.6483	&	.5461	\\
12	&	.9781	&	.9352	&	.8724	&	.7920	&	.6966	&	.5892	&	.4732	\\
13	&	.9744	&	.9241	&	.8511	&	.7582	&	.6489	&	.5273	&	.3940	\\
14	&	.9703	&	.9122	&	.8283	&	.7224	&	.5990	&	.4635	&	.3219	\\[2ex]
15	&	.9659	&	.8995	&	.8042	&	.6847	&	.5471	&	.3982	&	.2454	\\
16	&	.9613	&	.8860	&	.7787	&	.6454	&	.4937	&	.3322	&	.1699	\\
17	&	.9563	&	.8718	&	.7519	&	.6046	&	.4391	&	.2660	&	.0961	\\
18	&	.9511	&	.8568	&	.7240	&	.5624	&	.3836	&	.2002	&	.0289	\\
19	&	.9455	&	.8410	&	.6950	&	.5192	&	.3276	&	.1347	&	-.0443	\\[2ex]
20	&	.9397	&	.8245	&	.6649	&	.4750	&	.2715	&	.0719	&	-.1072	\\
21	&	.9336	&	.8074	&	.6338	&	.4300	&	.2156	&	.0107	&	-.1662	\\
22	&	.9272	&	.7895	&	.6019	&	.3845	&	.1602	&	-.0481	&	-.2201	\\
23	&	.9205	&	.7710	&	.5692	&	.3386	&	.1057	&	-.1038	&	-.2681	\\
24	&	.9135	&	.7518	&	.5357	&	.2926	&	.0525	&	-.1559	&	-.3095	\\[2ex]
25	&	.9063	&	.7321	&	.5016	&	.2465	&	.0009	&	-.2053	&	-.3463	\\
26	&	.8988	&	.7117	&	.4670	&	.2007	&	-.0489	&	-.2478	&	-.3717	\\
27	&	.8910	&	.6908	&	.4319	&	.1553	&	-.0964	&	-.2869	&	-.3921	\\
28	&	.8829	&	.6694	&	.3964	&	.1105	&	-.1415	&	-.3211	&	-.4052	\\
29	&	.8746	&	.6474	&	.3607	&	.0665	&	-.1839	&	-.3503	&	-.4114	\\[2ex]
30	&	.8660	&	.6250	&	.3248	&	.0234	&	-.2233	&	-.3740	&       -.4101	\\
31	&	.8572	&	.6021	&	.2887	&	-.0185	&	-.2595	&	-.3924	&	-.4022	\\
32	&	.8480	&	.5788	&	.2527	&	-.0591	&	-.2923	&	-.4052	&	-.3876	\\
33	&	.8387	&	.5551	&	.2167	&	-.0982	&	-.3216	&	-.4126	&	-.3670	\\
34	&	.8290	&	.5310	&	.1809	&	-.1357	&	-.3473	&	-.4148	&	-.3409	\\[2ex]
35	&	.8192	&	.5065	&	.1454	&	-.1714	&	-.3691	&	-.4115	&	-.3096	\\
36	&	.8090	&	.4818	&	.1102	&	-.2052	&	-.3871	&	-.4031	&	-.2738	\\
37	&	.7986	&	.4567	&	.0755	&	-.2370	&	-.4011	&	-.3898	&	-.2343	\\
38	&	.7880	&	.4314	&	.0413	&	-.2666	&	-.4112	&	-.3719	&	-.1918	\\
39	&	.7771	&	.4059	&	.0077	&	-.2940	&	-.4174	&	-.3497	&	-.1469	\\[2ex]
40	&	.7660	&	.3802	&	-.0252	&	-.3190	&	-.4197	&	-.3234	&	-.1003	\\
41	&	.7547	&	.3544	&	-.0574	&	-.3416	&	-.4181	&	-.2938	&	-.0534	\\
42	&	.7431	&	.3284	&	-.0887	&	-.3616	&	-.4128	&	-.2611	&	-.0065	\\
43	&	.7314	&	.3023	&	-.1191	&	-.3791	&	-.4038	&	-.2255	&	.0398	\\
44	&	.7193	&	.2762	&	-.1485	&	-.3940	&	-.3914	&	-.1878	&	.0846	\\[2ex]
\phantom{\text{°}}45\text{°}
	&	.7071	&	.2500	&	-.1768	&	-.4062	&	-.3757	&	-.1485	&	.1270	\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 289.png
\begin{center}
TABLE I. --- \textsc{Surface Zonal Harmonics}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|c|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|}{\tablestrut\theta}&
\multicolumn{1}{c|}{P_1(\cos\theta)}&
\multicolumn{1}{c|}{P_2(\cos\theta)}&
\multicolumn{1}{c|}{P_3(\cos\theta)}&
\multicolumn{1}{c|}{P_4(\cos\theta)}&
\multicolumn{1}{c|}{P_5(\cos\theta)}&
\multicolumn{1}{c|}{P_6(\cos\theta)}&
\multicolumn{1}{c|}{P_7(\cos\theta)}\\
\hline
\vrule width0ptdepth0ptheight3.5ex
\phantom{\text{°}}45\text{°}
	&	.7071	&	.2500	&	-.1768	&	-.4062	&	-.3757	&	-.1485	&	.1270\\
46	&	.6947	&	.2238	&	-.2040	&	-.4158	&	-.3568	&	-.1079	&	.1666\\
47	&	.6820	&	.1977	&	-.2300	&	-.4252	&	-.3350	&	-.0645	&	.2054\\
48	&	.6691	&	.1716	&	-.2547	&	-.4270	&	-.3105	&	-.0251	&	.2349\\
49	&	.6561	&	.1456	&	-.2781	&	-.4286	&	-.2836	&	.0161	&	.2627\\[2ex]
50	&	.6428	&	.1198	&	-.3002	&	-.4275	&	-.2545	&	.0563	&	.2854\\
51	&	.6293	&	.0941	&	-.3209	&	-.4239	&	-.2235	&	.0954	&	.3031\\
52	&	.6157	&	.0686	&	-.3401	&	-.4178	&	-.1910	&	.1326	&	.3153\\
53	&	.6018	&	.0433	&	-.3578	&	-.4093	&	-.1571	&	.1677	&	.3221\\
54	&	.5878	&	.0182	&	-.3740	&	-.3984	&	-.1223	&	.2002	&	.3234\\[2ex]
55	&	.5736	&	-.0065	&	-.3886	&	-.3852	&	-.0868	&	.2297	&	.3191\\
56	&	.5592	&	-.0310	&	-.4016	&	-.3698	&	-.0510	&	.2559	&	.3095\\
57	&	.5446	&	-.0551	&	-.4131	&	-.3524	&	-.0150	&	.2787	&	.2949\\
58	&	.5299	&	-.0788	&	-.4229	&	-.3331	&	.0206	&	.2976	&	.2752\\
59	&	.5150	&	-.1021	&	-.4310	&	-.3119	&	.0557	&	.3125	&	.2511\\[2ex]
60	&	.5000	&	-.1250	&	-.4375	&	-.2891	&	.0898	&	.3232	&	.2231\\
61	&	.4848	&	-.1474	&	-.4423	&	-.2647	&	.1229	&	.3298	&	.1916\\
62	&	.4695	&	-.1694	&	-.4455	&	-.2390	&	.1545	&	.3321	&	.1571\\
63	&	.4540	&	-.1908	&	-.4471	&	-.2121	&	.1844	&	.3302	&	.1203\\
64	&	.4384	&	-.2117	&	-.4470	&	-.1841	&	.2123	&	.3240	&	.0818\\[2ex]
65	&	.4226	&	-.2321	&	-.4452	&	-.1552	&	.2381	&	.3138	&	.0422\\
66	&	.4067	&	-.2518	&	-.4419	&	-.1256	&	.2615	&	.2996	&	.0021\\
67	&	.3907	&	-.2710	&	-.4370	&	-.0955	&	.2824	&	.2819	&	-.0375\\
68	&	.3746	&	-.2896	&	-.4305	&	-.0650	&	.3005	&	.2605	&	-.0763\\
69	&	.3584	&	-.3074	&	-.4225	&	-.0344	&	.3158	&	.2361	&	-.1135\\[2ex]
70	&	.3420	&	-.3245	&	-.4130	&	-.0038	&	.3281	&	.2089	&	-.1485\\
71	&	.3256	&	-.3410	&	-.4021	&	.0267	&	.3373	&	.1786	&	-.1811\\
72	&	.3090	&	-.3568	&	-.3898	&	.0568	&	.3434	&	.1472	&	-.2099\\
73	&	.2924	&	-.3718	&	-.3761	&	.0864	&	.3463	&	.1144	&	-.2347\\
74	&	.2756	&	-.3860	&	-.3611	&	.1153	&	.3461	&	.0795	&	-.2559\\[2ex]
75	&	.2588	&	-.3995	&	-.3449	&	.1434	&	.3427	&	.0431	&	-.2730\\
76	&	.2419	&	-.4112	&	-.3275	&	.1705	&	.3362	&	.0076	&	-.2848\\
77	&	.2250	&	-.4241	&	-.3090	&	.1964	&	.3267	&	-.0284	&	-.2919\\
78	&	.2079	&	-.4352	&	-.2894	&	.2211	&	.3143	&	-.0644	&	-.2943\\
79	&	.1908	&	-.4454	&	-.2688	&	.2443	&	.2990	&	-.0989	&	-.2913\\[2ex]
80	&	.1736	&	-.4548	&	-.2474	&	.2659	&	.2810	&	-.1321	&	-.2835\\
81	&	.1564	&	-.4633	&	-.2251	&	.2859	&	.2606	&	-.1635	&	-.2709\\
82	&	.1392	&	-.4709	&	-.2020	&	.3040	&	.2378	&	-.1926	&	-.2536\\
83	&	.1219	&	-.4777	&	-.1783	&	.3203	&	.2129	&	-.2193	&	-.2321\\
84	&	.1045	&	-.4836	&	-.1539	&	.3345	&	.1861	&	-.2431	&	-.2067\\[2ex]
85	&	.0872	&	-.4886	&	-.1291	&	.3468	&	.1577	&	-.2638	&	-.1779\\
86	&	.0698	&	-.4927	&	-.1038	&	.3569	&	.1278	&	-.2811	&	-.1460\\
87	&	.0523	&	-.4959	&	-.0781	&	.3648	&	.0969	&	-.2947	&	-.1117\\
88	&	.0349	&	-.4982	&	-.0522	&	.3704	&	.0651	&	-.3045	&	-.0735\\
89	&	.0175	&	-.4995	&	-.0262	&	.3739	&	.0327	&	-.3105	&	-.0381\\[2ex]
\phantom{\text{°}}90\text{°}
	&	.0000	&	-.5000	&	.0000	&	.3750	&	.0000	&	-.3125	&	.0000\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 290.png
\label{tableII}
\begin{center}
TABLE II.---\textsc{Surface Zonal Harmonics.}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|}{\tablestrut x}&
\multicolumn{1}{c|}{P_1(x)}&
\multicolumn{1}{c|}{P_2(x)}&
\multicolumn{1}{c|}{P_3(x)}&
\multicolumn{1}{c|}{P_4(x)}&
\multicolumn{1}{c|}{P_5(x)}&
\multicolumn{1}{c|}{P_6(x)}&
\multicolumn{1}{c|}{P_7(x)}\\
\hline
\vrule width0ptdepth0ptheight3.5ex
0.00	&	0.0000	&	-.5000	&	0.0000	&	0.3750	&	0.0000	&	-.3125	&	0.0000\\
  .01	&	.0100	&	-.4998	&	-.0150	&	.3746	&	.0187	&	-.3118	&	-.0219\\
  .02	&	.0200	&	-.4994	&	-.0300	&	.3735	&	.0374	&	-.3099	&	-.0436\\
  .03	&	.0300	&	-.4986	&	-.0449	&	.3716	&	.0560	&	-.3066	&	-.0651\\
  .04	&	.0400	&	-.4976	&	-.0598	&	.3690	&	.0744	&	-.3021	&	-.0862\\[2ex]
  .05	&	.0500	&	-.4962	&	-.0747	&	.3657	&	.0927	&	-.2962	&	-.1069\\
  .06	&	.0600	&	-.4946	&	-.0895	&	.3616	&	.1106	&	-.2891	&	-.1270\\
  .07	&	.0700	&	-.4926	&	-.1041	&	.3567	&	.1283	&	-.2808	&	-.1464\\
  .08	&	.0800	&	-.4904	&	-.1187	&	.3512	&	.1455	&	-.2713	&	-.1651\\
  .09	&	.0900	&	-.4878	&	-.1332	&	.3449	&	.1624	&	-.2606	&	-.1828\\[2ex]
  .10	&	.1000	&	-.4850	&	-.1475	&	.3379	&	.1788	&	-.2488	&	-.1995\\
  .11	&	.1100	&	-.4818	&	-.1617	&	.3303	&	.1947	&	-.2360	&	-.2151\\
  .12	&	.1200	&	-.4784	&	-.1757	&	.3219	&	.2101	&	-.2220	&	-.2295\\
  .13	&	.1300	&	-.4746	&	-.1895	&	.3129	&	.2248	&	-.2071	&	-.2427\\
  .14	&	.1400	&	-.4706	&	-.2031	&	.3032	&	.2389	&	-.1913	&	-.2545\\[2ex]
  .15	&	.1500	&	-.4662	&	-.2166	&	.2928	&	.2523	&	-.1746	&	-.2649\\
  .16	&	.1600	&	-.4616	&	-.2298	&	.2819	&	.2650	&	-.1572	&	-.2738\\
  .17	&	.1700	&	-.4566	&	-.2427	&	.2703	&	.2769	&	-.1389	&	-.2812\\
  .18	&	.1800	&	-.4514	&	-.2554	&	.2581	&	.2880	&	-.1201	&	-.2870\\
  .19	&	.1900	&	-.4458	&	-.2679	&	.2453	&	.2982	&	-.1006	&	-.2911\\[2ex]
  .20	&	.2000	&	-.4400	&	-.2800	&	.2320	&	.3075	&	-.0806	&	-.2935\\
  .21	&	.2100	&	-.4338	&	-.2918	&	.2181	&	.3159	&	-.0601	&	-.2943\\
  .22	&	.2200	&	-.4274	&	-.3034	&	.2037	&	.3234	&	-.0394	&	-.2933\\
  .23	&	.2300	&	-.4206	&	-.3146	&	.1889	&	.3299	&	-.0183	&	-.2906\\
  .24	&	.2400	&	-.4136	&	-.3254	&	.1735	&	.3353	&	.0029	&	-.2861\\[2ex]
  .25	&	.2500	&	-.4062	&	-.3359	&	.1577	&	.3397	&	.0243	&	-.2799\\
  .26	&	.2600	&	-.3986	&	-.3461	&	.1415	&	.3431	&	.0456	&	-.2720\\
  .27	&	.2700	&	-.3906	&	-.3558	&	.1249	&	.3453	&	.0669	&	-.2625\\
  .28	&	.2800	&	-.3824	&	-.3651	&	.1079	&	.3465	&	.0879	&	-.2512\\
  .29	&	.2900	&	-.3738	&	-.3740	&	.0906	&	.3465	&	.1087	&	-.2384\\[2ex]
  .30	&	.3000	&	-.3650	&	-.3825	&	.0729	&	.3454	&	.1292	&	-.2241\\
  .31	&	.3100	&	-.3558	&	-.3905	&	.0550	&	.3431	&	.1492	&	-.2082\\
  .32	&	.3200	&	-.3464	&	-.3981	&	.0369	&	.3397	&	.1686	&	-.1910\\
  .33	&	.3300	&	-.3366	&	-.4052	&	.0185	&	.3351	&	.1873	&	-.1724\\
  .34	&	.3400	&	-.3266	&	-.4117	&	-.0000	&	.3294	&	.2053	&	-.1527\\[2ex]
  .35	&	.3500	&	-.3162	&	-.4178	&	-.0187	&	.3225	&	.2225	&	-.1318\\
  .36	&	.3600	&	-.3056	&	-.4234	&	-.0375	&	.3144	&	.2388	&	-.1098\\
  .37	&	.3700	&	-.2946	&	-.4284	&	-.0564	&	.3051	&	.2540	&	-.0870\\
  .38	&	.3800	&	-.2834	&	-.4328	&	-.0753	&	.2948	&	.2681	&	-.0635\\
  .39	&	.3900	&	-.2718	&	-.4367	&	-.0942	&	.2833	&	.2810	&	-.0393\\[2ex]
  .40	&	.4000	&	-.2600	&	-.4400	&	-.1130	&	.2706	&	.2926	&	-.0146\\
  .41	&	.4100	&	-.2478	&	-.4427	&	-.1317	&	.2569	&	.3029	&	.0104\\
  .42	&	.4200	&	-.2354	&	-.4448	&	-.1504	&	.2421	&	.3118	&	.0356\\
  .43	&	.4300	&	-.2226	&	-.4462	&	-.1688	&	.2263	&	.3191	&	.0608\\
  .44	&	.4400	&	-.2096	&	-.4470	&	-.1870	&	.2095	&	.3249	&	.0859\\[2ex]
  .45	&	.4500	&	-.1962	&	-.4472	&	-.2050	&	.1917	&	.3290	&	.1106\\
  .46	&	.4600	&	-.1826	&	-.4467	&	-.2226	&	.1730	&	.3314	&	.1348\\
  .47	&	.4700	&	-.1686	&	-.4454	&	-.2399	&	.1534	&	.3321	&	.1584\\
  .48	&	.4800	&	-.1544	&	-.4435	&	-.2568	&	.1330	&	.3310	&	.1811\\
  .49	&	.4900	&	-.1398	&	-.4409	&	-.2732	&	.1118	&	.3280	&	.2027\\[2ex]
  .50	&	.5000	&	-.1250	&	-.4375	&	-.2891	&	.0898	&	.3232	&	.2231\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 291.png
\begin{center}
TABLE II.---\textsc{Surface Zonal Harmonics.}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|}{\tablestrut x}&
\multicolumn{1}{c|}{P_1(x)}&
\multicolumn{1}{c|}{P_2(x)}&
\multicolumn{1}{c|}{P_3(x)}&
\multicolumn{1}{c|}{P_4(x)}&
\multicolumn{1}{c|}{P_5(x)}&
\multicolumn{1}{c|}{P_6(x)}&
\multicolumn{1}{c|}{P_7(x)}\\
\hline
\vrule width0ptdepth0ptheight3.5ex
.50	&	.5000	&	-.1250	&	-.4375	&	-.2891	&	.0898	&	.3232	&	.2231\\
.51	&	.5100	&	-.1098	&	-.4334	&	-.3044	&	.0673	&	.3166	&	.2422\\
.52	&	.5200	&	-.0944	&	-.4285	&	-.3191	&	.0441	&	.3080	&	.2596\\
.53	&	.5300	&	-.0786	&	-.4228	&	-.3332	&	.0204	&	.2975	&	.2753\\
.54	&	.5400	&	-.0626	&	-.4163	&	-.3465	&	-.0037	&	.2851	&	.2891\\[2ex]
.55	&	.5500	&	-.0462	&	-.4091	&	-.3590	&	-.0282	&	.2708	&	.3007\\
.56	&	.5600	&	-.0296	&	-.4010	&	-.3707	&	-.0529	&	.2546	&	.3102\\
.57	&	.5700	&	-.0126	&	-.3920	&	-.3815	&	-.0779	&	.2366	&	.3172\\
.58	&	.5800	&	.0046	&	-.3822	&	-.3914	&	-.1028	&	.2168	&	.3217\\
.59	&	.5900	&	.0222	&	-.3716	&	-.4002	&	-.1278	&	.1953	&	.3235\\[2ex]
.60	&	.6000	&	.0400	&	-.3600	&	-.4080	&	-.1526	&	.1721	&	.3226\\
.61	&	.6100	&	.0582	&	-.3475	&	-.4146	&	-.1772	&	.1473	&	.3188\\
.62	&	.6200	&	.0766	&	-.3342	&	-.4200	&	-.2014	&	.1211	&	.3121\\
.63	&	.6300	&	.0954	&	-.3199	&	-.4242	&	-.2251	&	.0935	&	.3023\\
.64	&	.6400	&	.1144	&	-.3046	&	-.4270	&	-.2482	&	.0646	&	.2895\\[2ex]
.65	&	.6500	&	.1338	&	-.2884	&	-.4284	&	-.2705	&	.0347	&	.2737\\
.66	&	.6600	&	.1534	&	-.2713	&	-.4284	&	-.2919	&	.0038	&	.2548\\
.67	&	.6700	&	.1734	&	-.2531	&	-.4268	&	-.3122	&	-.0278	&	.2329\\
.68	&	.6800	&	.1936	&	-.2339	&	-.4236	&	-.3313	&	-.0601	&	.2081\\
.69	&	.6900	&	.2142	&	-.2137	&	-.4187	&	-.3490	&	-.0926	&	.1805\\[2ex]
.70	&	.7000	&	.2350	&	-.1925	&	-.4121	&	-.3652	&	-.1253	&	.1502\\
.71	&	.7100	&	.2562	&	-.1702	&	-.4036	&	-.3796	&	-.1578	&	.1173\\
.72	&	.7200	&	.2776	&	-.1469	&	-.3933	&	-.3922	&	-.1899	&	.0822\\
.73	&	.7300	&	.2994	&	-.1225	&	-.3810	&	-.4026	&	-.2214	&	.0450\\
.74	&	.7400	&	.3214	&	-.0969	&	-.3666	&	-.4107	&	-.2518	&	.0061\\[2ex]
.75	&	.7500	&	.3438	&	-.0703	&	-.3501	&	-.4164	&	-.2808	&	-.0342\\
.76	&	.7600	&	.3664	&	-.0426	&	-.3314	&	-.4193	&	-.3081	&	-.0754\\
.77	&	.7700	&	.3894	&	-.0137	&	-.3104	&	-.4193	&	-.3333	&	-.1171\\
.78	&	.7800	&	.4126	&	.0164	&	-.2871	&	-.4162	&	-.3559	&	-.1588\\
.79	&	.7900	&	.4362	&	.0476	&	-.2613	&	-.4097	&	-.3756	&	-.1999\\[2ex]
.80	&	.8000	&	.4600	&	.0800	&	-.2330	&	-.3995	&	-.3918	&	-.2397\\
.81	&	.8100	&	.4842	&	.1136	&	-.2021	&	-.3855	&	-.4041	&	-.2774\\
.82	&	.8200	&	.5086	&	.1484	&	-.1685	&	-.3674	&	-.4119	&	-.3124\\
.83	&	.8300	&	.5334	&	.1845	&	-.1321	&	-.3449	&	-.4147	&	-.3437\\
.84	&	.8400	&	.5584	&	.2218	&	-.0928	&	-.3177	&	-.4120	&	-.3703\\[2ex]
.85	&	.8500	&	.5838	&	.2603	&	-.0506	&	-.2857	&	-.4030	&	-.3913\\
.86	&	.8600	&	.6094	&	.3001	&	-.0053	&	-.2484	&	-.3872	&	-.4055\\
.87	&	.8700	&	.6354	&	.3413	&	.0431	&	-.2056	&	-.3638	&	-.4116\\
.88	&	.8800	&	.6616	&	.3837	&	.0947	&	-.1570	&	-.3322	&	-.4083\\
.89	&	.8900	&	.6882	&	.4274	&	.1496	&	-.1023	&	-.2916	&	-.3942\\[2ex]
.90	&	.9000	&	.7150	&	.4725	&	.2079	&	-.0411	&	-.2412	&	-.3678\\
.91	&	.9100	&	.7422	&	.5189	&	.2698	&	.0268	&	-.1802	&	-.3274\\
.92	&	.9200	&	.7696	&	.5667	&	.3352	&	.1017	&	-.1077	&	-.2713\\
.93	&	.9300	&	.7974	&	.6159	&	.4044	&	.1842	&	-.0229	&	-.1975\\
.94	&	.9400	&	.8254	&	.6665	&	.4773	&	.2744	&	.0751	&	-.1040\\[2ex]
.95	&	.9500	&	.8538	&	.7184	&	.5541	&	.3727	&	.1875	&	.0112\\
.96	&	.9600	&	.8824	&	.7718	&	.6349	&	.4796	&	.3151	&	.1506\\
.97	&	.9700	&	.9114	&	.8267	&	.7198	&	.5954	&	.4590	&	.3165\\
.98	&	.9800	&	.9406	&	.8830	&	.8089	&	.7204	&	.6204	&	.5115\\
.99	&	.9900	&	.9702	&	.9407	&	.9022	&	.8552	&	.8003	&	.7384\\[2ex]
1.00	&	1.0000	&	1.0000	&	1.0000	&	1.0000	&	1.0000	&	1.0000	&	1.0000\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 292.png
\label{tableIII}
\begin{center}
TABLE III.---\textsc{Hyperbolic Functions.}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|}{\tablestrut x}&
\multicolumn{1}{c|}{e^x}&
\multicolumn{1}{c|}{e^{-x}}&
\multicolumn{1}{c|}{\sinh x}&
\multicolumn{1}{c|}{\cosh x}&
\multicolumn{1}{c|}{\gd x}\\
\hline
\vrule width0ptdepth0ptheight3.5ex
\;0.00 & \;1.0000 & \;1.0000 & \;0.0000 & \;1.0000 & \;0\raisebox{0.8ex}{\text{°}}\!\!\!.0000\\
.01 & 1.0100 & 0.9900 & .0100 & 1.0000 & 0.5729\\
.02 & 1.0202 & .9802 & .0200 & 1.0002 & 1.1458\\
.03 & 1.0305 & .9704 & .0300 & 1.0004 & 1.7186\\
.04 & 1.0408 & .9608 & .0400 & 1.0008 & 2.2912\\[2ex]
.05 & 1.0513 & .9512 & .0500 & 1.0012 & 2.8636\\
.06 & 1.0618 & .9418 & .0600 & 1.0018 & 3.4357\\
.07 & 1.0725 & .9324 & .0701 & 1.0025 & 4.0074\\
.08 & 1.0833 & .9231 & .0801 & 1.0032 & 4.5788\\
.09 & 1.0942 & .9139 & .0901 & 1.0040 & 5.1497\\[2ex]
.10 & 1.1052 & .9048 & .1002 & 1.0050 & 5.720\phantom{0}\\
.11 & 1.1163 & .8958 & .1102 & 1.0061 & 6.290\phantom{0}\\
.12 & 1.1275 & .8869 & .1203 & 1.0072 & 6.859\phantom{0}\\
.13 & 1.1388 & .8781 & .1304 & 1.0085 & 7.428\phantom{0}\\
.14 & 1.1503 & .8694 & .1405 & 1.0098 & 7.995\phantom{0}\\[2ex]
.15 & 1.1618 & .8607 & .1506 & 1.0113 & 8.562\phantom{0}\\
.16 & 1.1735 & .8521 & .1607 & 1.0128 & 9.128\phantom{0}\\
.17 & 1.1853 & .8437 & .1708 & 1.0145 & 9.694\phantom{0}\\
.18 & 1.1972 & .8353 & .1810 & 1.0162 & 10.258\phantom{0}\\
.19 & 1.2092 & .8270 & .1911 & 1.0181 & 10.821\phantom{0}\\[2ex]
.20 & 1.2214 & .8187 & .2013 & 1.0201 & 11.384\phantom{0}\\
.21 & 1.2337 & .8106 & .2115 & 1.0221 & 11.945\phantom{0}\\
.22 & 1.2461 & .8025 & .2218 & 1.0243 & 12.505\phantom{0}\\
.23 & 1.2586 & .7945 & .2320 & 1.0266 & 13.063\phantom{0}\\
.24 & 1.2712 & .7866 & .2423 & 1.0289 & 13.621\phantom{0}\\[2ex]
.25 & 1.2840 & .7788 & .2526 & 1.0314 & 14.177\phantom{0}\\
.26 & 1.2969 & .7711 & .2629 & 1.0340 & 14.732\phantom{0}\\
.27 & 1.3100 & .7634 & .2733 & 1.0367 & 15.285\phantom{0}\\
.28 & 1.3231 & .7558 & .2837 & 1.0395 & 15.837\phantom{0}\\
.29 & 1.3364 & .7483 & .2941 & 1.0423 & 16.388\phantom{0}\\[2ex]
.30 & 1.3499 & .7408 & .3045 & 1.0453 & 16.937\phantom{0}\\
.31 & 1.3634 & .7334 & .3150 & 1.0484 & 17.484\phantom{0}\\
.32 & 1.3771 & .7261 & .3255 & 1.0516 & 18.030\phantom{0}\\
.33 & 1.3910 & .7189 & .3360 & 1.0549 & 18.573\phantom{0}\\
.34 & 1.4049 & .7118 & .3466 & 1.0584 & 19.116\phantom{0}\\[2ex]
.35 & 1.4191 & .7047 & .3572 & 1.0619 & 19.656\phantom{0}\\
.36 & 1.4333 & .6977 & .3678 & 1.0655 & 20.195\phantom{0}\\
.37 & 1.4477 & .6907 & .3785 & 1.0692 & 20.732\phantom{0}\\
.38 & 1.4623 & .6839 & .3892 & 1.0731 & 21.267\phantom{0}\\
.39 & 1.4770 & .6771 & .4000 & 1.0770 & 21.800\phantom{0}\\[2ex]
.40 & 1.4918 & .6703 & .4108 & 1.0811 & 22.331\phantom{0}\\
.41 & 1.5068 & .6636 & .4216 & 1.0852 & 22.859\phantom{0}\\
.42 & 1.5220 & .6570 & .4325 & 1.0895 & 23.386\phantom{0}\\
.43 & 1.5373 & .6505 & .4434 & 1.0939 & 23.911\phantom{0}\\
.44 & 1.5527 & .6440 & .4543 & 1.0984 & 24.434\phantom{0}\\[2ex]
.45 & 1.5683 & .6376 & .4653 & 1.1030 & 24.955\phantom{0}\\
.46 & 1.5841 & .6313 & .4764 & 1.1077 & 25.473\phantom{0}\\
.47 & 1.6000 & .6250 & .4875 & 1.1125 & 25.989\phantom{0}\\
.48 & 1.6161 & .6188 & .4986 & 1.1174 & 26.503\phantom{0}\\
.49 & 1.6323 & .6126 & .5098 & 1.1225 & 27.015\phantom{0}\\[2ex]
0.50 & 1.6487 & 0.6065 & 0.5211 & 1.1276 & 27\raisebox{0.8ex}{\text{°}}\!\!\!.524\phantom{0}\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 293.png
\begin{center}
TABLE III.---\textsc{Hyperbolic Functions.}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|}{\tablestrut x}&
\multicolumn{1}{c|}{e^x}&
\multicolumn{1}{c|}{e^{-x}}&
\multicolumn{1}{c|}{\sinh x}&
\multicolumn{1}{c|}{\cosh x}&
\multicolumn{1}{c|}{\gd x}\\
\hline
\vrule width0ptdepth0ptheight3.5ex
\;0.50 & \;1.6487 & \;0.6065 & \;0.5211 & \;1.1276 & \;27\raisebox{0.8ex}{\text{°}}\!\!\!.524\\
.51 & 1.6653 & .6005 & .5324 & 1.1329 & 28.031\\
.52 & 1.6820 & .5945 & .5438 & 1.1383 & 28.535\\
.53 & 1.6989 & .5886 & .5552 & 1.1438 & 29.037\\
.54 & 1.7160 & .5827 & .5666 & 1.1494 & 29.537\\[2ex]
.55 & 1.7333 & .5770 & .5782 & 1.1551 & 30.034\\
.56 & 1.7507 & .5712 & .5897 & 1.1609 & 30.529\\
.57 & 1.7683 & .5655 & .6014 & 1.1669 & 31.021\\
.58 & 1.7860 & .5599 & .6131 & 1.1730 & 31.511\\
.59 & 1.8040 & .5543 & .6248 & 1.1792 & 31.998\\[2ex]
.60 & 1.8221 & .5488 & .6367 & 1.1855 & 32.483\\
.61 & 1.8404 & .5433 & .6485 & 1.1919 & 32.965\\
.62 & 1.8589 & .5379 & .6605 & 1.1984 & 33.444\\
.63 & 1.8776 & .5326 & .6725 & 1.2051 & 33.921\\
.64 & 1.8965 & .5273 & .6846 & 1.2119 & 34.395\\[2ex]
.65 & 1.9155 & .5220 & .6967 & 1.2188 & 34.867\\
.66 & 1.9348 & .5169 & .7090 & 1.2258 & 35.336\\
.67 & 1.9542 & .5117 & .7213 & 1.2330 & 35.802\\
.68 & 1.9739 & .5066 & .7336 & 1.2402 & 36.265\\
.69 & 1.9937 & .5016 & .7461 & 1.2476 & 36.726\\[2ex]
.70 & 2.0138 & .4966 & .7586 & 1.2552 & 37.183\\
.71 & 2.0340 & .4916 & .7712 & 1.2628 & 37.638\\
.72 & 2.0544 & .4867 & .7838 & 1.2706 & 38.091\\
.73 & 2.0751 & .4819 & .7966 & 1.2785 & 38.540\\
.74 & 2.0959 & .4771 & .8094 & 1.2865 & 38.987\\[2ex]
.75 & 2.1170 & .4724 & .8223 & 1.2947 & 39.431\\
.76 & 2.1383 & .4677 & .8353 & 1.3030 & 39.872\\
.77 & 2.1598 & .4630 & .8484 & 1.3114 & 40.310\\
.78 & 2.1815 & .4584 & .8615 & 1.3199 & 40.746\\
.79 & 2.2034 & .4538 & .8748 & 1.3286 & 41.179\\[2ex]
.80 & 2.2255 & .4493 & .8881 & 1.3374 & 41.608\\
.81 & 2.2479 & .4449 & .9015 & 1.3464 & 42.035\\
.82 & 2.2705 & .4404 & .9150 & 1.3555 & 42.460\\
.83 & 2.2933 & .4360 & .9286 & 1.3647 & 42.881\\
.84 & 2.3164 & .4317 & .9423 & 1.3740 & 43.299\\[2ex]
.85 & 2.3396 & .4274 & .9561 & 1.3835 & 43.715\\
.86 & 2.3632 & .4232 & .9700 & 1.3932 & 44.128\\
.87 & 2.3869 & .4190 & .9840 & 1.4029 & 44.537\\
.88 & 2.4109 & .4148 & .9981 & 1.4128 & 44.944\\
.89 & 2.4351 & .4107 & 1.0122 & 1.4229 & 45.348\\[2ex]
.90 & 2.4596 & .4066 & 1.0265 & 1.4331 & 45.750\\
.91 & 2.4843 & .4025 & 1.0409 & 1.4434 & 46.148\\
.92 & 2.5093 & .3985 & 1.0554 & 1.4539 & 46.544\\
.93 & 2.5345 & .3946 & 1.0700 & 1.4645 & 46.936\\
.94 & 2.5600 & .3906 & 1.0847 & 1.4753 & 47.326\\[2ex]
.95 & 2.5857 & .3867 & 1.0995 & 1.4862 & 47.713\\
.96 & 2.6117 & .3829 & 1.1144 & 1.4973 & 48.097\\
.97 & 2.6379 & .3791 & 1.1294 & 1.5085 & 48.478\\
.98 & 2.6645 & .3753 & 1.1446 & 1.5199 & 48.857\\
.99 & 2.6912 & .3716 & 1.1598 & 1.5314 & 49.232\\[2ex]
1.00 & 2.7183 & 0.3679 & 1.1752 & 1.5431 & 49\raisebox{0.8ex}{\text{°}}\!\!\!.605\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 294.png
\begin{center}
TABLE III.---\textsc{Hyperbolic Functions.}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|}{\tablestrut x}&
l\sinh x&
l\cosh x&
\multicolumn{1}{c|}{x}&
l\sinh x&
l\cosh x&
\multicolumn{1}{c|}{x}&
l\sinh x&
l\cosh x\\
\hline
\vrule width0ptdepth0ptheight3.5ex
\;1.00 & 0.0701 & 0.1884 & \;1.50 & 0.3282 & 0.3715 & \;2.00 & 0.5595 & 0.5754\\
1.01 & .0758 & .1917 & 1.51 & .3330 & .3754 & 2.01 & .5640 & .5796\\
1.02 & .0815 & .1950 & 1.52 & .3378 & .3794 & 2.02 & .5685 & .5838\\
1.03 & .0871 & .1984 & 1.53 & .3426 & .3833 & 2.03 & .5730 & .5880\\
1.04 & .0927 & .2018 & 1.54 & .3474 & .3873 & 2.04 & .5775 & .5922\\[2ex]
1.05 & .0982 & .2051 & 1.55 & .3521 & .3913 & 2.05 & .5820 & .5964\\
1.06 & .1038 & .2086 & 1.56 & .3569 & .3952 & 2.06 & .5865 & .6006\\
1.07 & .1093 & .2120 & 1.57 & .3616 & .3992 & 2.07 & .5910 & .6048\\
1.08 & .1148 & .2154 & 1.58 & .3663 & .4032 & 2.08 & .5955 & .6090\\
1.09 & .1203 & .2189 & 1.59 & .3711 & .4072 & 2.09 & .6000 & .6132\\[2ex]
1.10 & .1257 & .2223 & 1.60 & .3758 & .4112 & 2.10 & .6044 & .6175\\
1.11 & .1311 & .2258 & 1.61 & .3805 & .4152 & 2.11 & .6089 & .6217\\
1.12 & .1365 & .2293 & 1.62 & .3852 & .4192 & 2.12 & .6134 & .6259\\
1.13 & .1419 & .2328 & 1.63 & .3899 & .4232 & 2.13 & .6178 & .6301\\
1.14 & .1472 & .2364 & 1.64 & .3946 & .4273 & 2.14 & .6223 & .6343\\[2ex]
1.15 & .1525 & .2399 & 1.65 & .3992 & .4313 & 2.15 & .6268 & .6386\\
1.16 & .1578 & .2435 & 1.66 & .4039 & .4353 & 2.16 & .6312 & .6428\\
1.17 & .1631 & .2470 & 1.67 & .4086 & .4394 & 2.17 & .6357 & .6470\\
1.18 & .1684 & .2506 & 1.68 & .4132 & .4434 & 2.18 & .6401 & .6512\\
1.19 & .1736 & .2542 & 1.69 & .4179 & .4475 & 2.19 & .6446 & .6555\\[2ex]
1.20 & .1788 & .2578 & 1.70 & .4225 & .4515 & 2.20 & .6491 & .6597\\
1.21 & .1840 & .2615 & 1.71 & .4272 & .4556 & 2.21 & .6535 & .6640\\
1.22 & .1892 & .2651 & 1.72 & .4318 & .4597 & 2.22 & .6580 & .6682\\
1.23 & .1944 & .2688 & 1.73 & .4364 & .4637 & 2.23 & .6624 & .6724\\
1.24 & .1995 & .2724 & 1.74 & .4411 & .4678 & 2.24 & .6668 & .6767\\[2ex]
1.25 & .2046 & .2761 & 1.75 & .4457 & .4719 & 2.25 & .6713 & .6809\\
1.26 & .2098 & .2798 & 1.76 & .4503 & .4760 & 2.26 & .6757 & .6852\\
1.27 & .2148 & .2835 & 1.77 & .4549 & .4801 & 2.27 & .6802 & .6894\\
1.28 & .2199 & .2872 & 1.78 & .4595 & .4842 & 2.28 & .6846 & .6937\\
1.29 & .2250 & .2909 & 1.79 & .4641 & .4883 & 2.29 & .6890 & .6979\\[2ex]
1.30 & .2300 & .2947 & 1.80 & .4687 & .4924 & 2.30 & .6935 & .7022\\
1.31 & .2351 & .2984 & 1.81 & .4733 & .4965 & 2.31 & .6979 & .7064\\
1.32 & .2401 & .3022 & 1.82 & .4778 & .5006 & 2.32 & .7023 & .7107\\
1.33 & .2451 & .3059 & 1.83 & .4824 & .5048 & 2.33 & .7067 & .7150\\
1.34 & .2501 & .3097 & 1.84 & .4870 & .5089 & 2.34 & .7112 & .7192\\[2ex]
1.35 & .2551 & .3135 & 1.85 & .4915 & .5130 & 2.35 & .7156 & .7235\\
1.36 & .2600 & .3173 & 1.86 & .4961 & .5172 & 2.36 & .7200 & .7278\\
1.37 & .2650 & .3211 & 1.87 & .5007 & .5213 & 2.37 & .7244 & .7320\\
1.38 & .2699 & .3249 & 1.88 & .5052 & .5254 & 2.38 & .7289 & .7363\\
1.39 & .2748 & .3288 & 1.89 & .5098 & .5296 & 2.39 & .7333 & .7406\\[2ex]
1.40 & .2797 & .3326 & 1.90 & .5143 & .5337 & 2.40 & .7377 & .7448\\
1.41 & .2846 & .3365 & 1.91 & .5188 & .5379 & 2.41 & .7421 & .7491\\
1.42 & .2895 & .3403 & 1.92 & .5234 & .5421 & 2.42 & .7465 & .7534\\
1.43 & .2944 & .3442 & 1.93 & .5279 & .5462 & 2.43 & .7509 & .7577\\
1.44 & .2993 & .3481 & 1.94 & .5324 & .5504 & 2.44 & .7553 & .7619\\[2ex]
1.45 & .3041 & .3520 & 1.95 & .5370 & .5545 & 2.45 & .7597 & .7662\\
1.46 & .3090 & .3559 & 1.96 & .5415 & .5587 & 2.46 & .7642 & .7705\\
1.47 & .3138 & .3598 & 1.97 & .5460 & .5629 & 2.47 & .7686 & .7748\\
1.48 & .3186 & .3637 & 1.98 & .5505 & .5671 & 2.48 & .7730 & .7791\\
1.49 & .3234 & .3676 & 1.99 & .5550 & .5713 & 2.49 & .7774 & .7833\\[2ex]
1.50 & 0.3282 & 0.3715 & 2.00 & 0.5595 & 0.5754 & 2.50 & 0.7818 & 0.7876\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 295.png
\begin{center}
TABLE III.---\textsc{Hyperbolic Functions.}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|}{\tablestrut x}&
l\sinh x&
l\cosh x&
\multicolumn{1}{c|}{x}&
l\sinh x&
l\cosh x&
\multicolumn{1}{c|}{x}&
l\sinh x&
l\cosh x\\
\hline
\vrule width0ptdepth0ptheight3.5ex
\;2.50 & 0.7818 & 0.7876 & \;2.75 & 0.8915 & 0.8951 & 3.0 & 1.0008 & 1.0029\\
2.51 & .7862 & .7919 & 2.76 & .8959 & .8994 & 3.1 & 1.0444 & 1.0462\\
2.52 & .7906 & .7962 & 2.77 & .9003 & .9037 & 3.2 & 1.0880 & 1.0894\\
2.53 & .7950 & .8005 & 2.78 & .9046 & .9080 & 3.3 & 1.1316 & 1.1327\\
2.54 & .7994 & .8048 & 2.79 & .9090 & .9123 & 3.4 & 1.1751 & 1.1761\\[2ex]
2.55 & .8038 & .8091 & 2.80 & .9134 & .9166 & 3.5 & 1.2186 & 1.2194\\
2.56 & .8082 & .8134 & 2.81 & .9178 & .9209 & 3.6 & 1.2621 & 1.2628\\
2.57 & .8126 & .8176 & 2.82 & .9221 & .9252 & 3.7 & 1.3056 & 1.3061\\
2.58 & .8169 & .8219 & 2.83 & .9265 & .9295 & 3.8 & 1.3491 & 1.3495\\
2.59 & .8213 & .8262 & 2.84 & .9309 & .9338 & 3.9 & 1.3925 & 1.3929\\[2ex]
2.60 & .8257 & .8305 & 2.85 & .9353 & .9382 & 4.0 & 1.4360 & 1.4363\\
2.61 & .8301 & .8348 & 2.86 & .9396 & .9425 & 4.1 & 1.4795 & 1.4797\\
2.62 & .8345 & .8391 & 2.87 & .9440 & .9468 & 4.2 & 1.5229 & 1.5231\\
2.63 & .8389 & .8434 & 2.88 & .9484 & .9511 & 4.3 & 1.5664 & 1.5665\\
2.64 & .8433 & .8477 & 2.89 & .9527 & .9554 & 4.4 & 1.6098 & 1.6099\\[2ex]
2.65 & .8477 & .8520 & 2.90 & .9571 & .9597 & 4.5 & 1.6532 & 1.6533\\
2.66 & .8521 & .8563 & 2.91 & .9615 & .9641 & 4.6 & 1.6967 & 1.6968\\
2.67 & .8564 & .8606 & 2.92 & .9658 & .9684 & 4.7 & 1.7401 & 1.7402\\
2.68 & .8608 & .8649 & 2.93 & .9702 & .9727 & 4.8 & 1.7836 & 1.7836\\
2.69 & .8652 & .8692 & 2.94 & .9746 & .9770 & 4.9 & 1.8270 & 1.8270\\[2ex]
2.70 & .8696 & .8735 & 2.95 & .9789 & .9813 & 5.0 & 1.8704 & 1.8705\\
2.71 & .8740 & .8778 & 2.96 & .9833 & .9856 & 6.0 & 2.3047 & 2.3047\\
2.72 & .8784 & .8821 & 2.97 & .9877 & .9900 & 7.0 & 2.7390 & 2.7390\\
2.73 & .8827 & .8864 & 2.98 & .9920 & .9943 & 8.0 & 3.1733 & 3.1733\\
2.74 & .8871 & .8907 & 2.99 & .9964 & .9986 & 9.0 & 3.6076 & 3.6076\\[2ex]
2.75 & 0.8915 & 0.8951 & 3.00 & 1.0008 & 1.0029 & 10.0 & 4.0419 & 4.0419\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 296.png
\begin{center}
\label{tableIV}TABLE IV.---\textsc{Roots of Bessel's Functions.}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|c|c|r@{\quad}|c|c|}
\hline
\tablestrut{}&
\dfrac{x}{\pi}$ for $J_0(x)=0&
\dfrac{x}{\pi}$ for $J_1(x)=0&
\qquad&
\dfrac{x}{\pi}$ for $J_0(x)=0&
\dfrac{x}{\pi}$ for $J_1(x)=0\\
\hline
\vrule width0ptdepth0ptheight3.5ex
1 & 0.7655 & 1.2197 &  7 & \phantom{0}6.7519 & \phantom{0}7.2448\\
2 & 1.7571 & 2.2330 &  8 & \phantom{0}7.7516 & \phantom{0}8.2454\\
3 & 2.7546 & 3.2383 &  9 & \phantom{0}8.7514 & \phantom{0}9.2459\\
4 & 3.7534 & 4.2411 & 10 & \phantom{0}9.7513 & 10.2463\\
5 & 4.7527 & 5.2428 & 11 & 10.7512 & 11.2466\\
\;6 & 5.7522 & 6.2439 & \;12 & 11.7511 & 12.2469\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}

\bigskip
\begin{center}
\label{tableV}TABLE V.---\textsc{Roots of $J_n(x)=0$.}
\end{center}
\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|c|c|c|c|c|c|}
\hline
\tablestrut{}&
\;n=0\;&
\;n=1\;&
\;n=2\;&
\;n=3\;&
\;n=4\;&
\;n=5\;\\
\hline
\vrule width0ptdepth0ptheight3.5ex
1 & \phantom{0}2.405 & \phantom{0}3.832 & \phantom{0}5.135 & \phantom{0}6.379 & \phantom{0}7.586 & \phantom{0}8.780\\
2 & \phantom{0}5.520 & \phantom{0}7.016 & \phantom{0}8.417 & \phantom{0}9.760 & 11.064 & 12.339\\
3 & \phantom{0}8.654 & 10.173 & 11.620 & 13.017 & 14.373 & 15.700\\
4 & 11.792 & 13.323 & 14.796 & 16.224 & 17.616 & 18.982\\
5 & 14.931 & 16.470 & 17.960 & 19.410 & 20.827 & 22.220\\
6 & 18.071 & 19.616 & 21.117 & 22.583 & 24.018 & 25.431\\
7 & 21.212 & 22.760 & 24.270 & 25.749 & 27.200 & 28.628\\
8 & 24.353 & 25.903 & 27.421 & 28.909 & 30.371 & 31.813\\
\;9 & 27.494 & 29.047 & 30.571 & 32.050 & 33.512 & 34.983\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\newpage

% -----File: 297.png
\label{tableVI}
\begin{center}
TABLE VI.--- \textsc{Bessel's Functions.}
\end{center}

\begin{scriptsize}
\[
\arraycolsep=0.2em
\begin{array}{|c|}
\hline\\[-2.1ex]
\arraycolsep=0.5em
\begin{array}{|r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|||r@{\quad}|r@{\quad}|r@{\quad}|}
\hline
\multicolumn{1}{|c|} {\tablestrut x}      &
\multicolumn{1}{c|}  {J_0(x)} &
\multicolumn{1}{c|||}{J_1(x)} &
\multicolumn{1}{c|}  {x }     &
\multicolumn{1}{c|}  {J_0(x)} &
\multicolumn{1}{c|||}{J_1(x)} &
\multicolumn{1}{c|}  {x}      &
\multicolumn{1}{c|}  {J_0(x)} &
\multicolumn{1}{c|}  {J_1(x)}\\
\hline
\vrule width0ptdepth0ptheight3.5ex
\;0.0	&	1.0000	&	0.0000	&	5.0	&	-.1776	&	-.3276	&	\;10.0	&	-.2459	&	.0435\\
0.1	&	.9975	&	.0499	&	5.1	&	-.1443	&	-.3371	&	10.1	&	-.2490	&	.0184\\
0.2	&	.9900	&	.0995	&	5.2	&	-.1103	&	-.3432	&	10.2	&	-.2496	&	-.0066\\
0.3	&	.9776	&	.1483	&	5.3	&	-.0758	&	-.3460	&	10.3	&	-.2477	&	-.0313\\
0.4	&	.9604	&	.1960	&	5.4	&	-.0412	&	-.3453	&	10.4	&	-.2434	&	-.0555\\[2ex]
0.5	&	.9385	&	.2423	&	5.5	&	-.0068	&	-.3414	&	10.5	&	-.2366	&	-.0789\\
0.6	&	.9120	&	.2867	&	5.6	&	.0270	&	-.3343	&	10.6	&	-.2276	&	-.1012\\
0.7	&	.8812	&	.3290	&	5.7	&	.0599	&	-.3241	&	10.7	&	-.2164	&	-.1224\\
0.8	&	.8463	&	.3688	&	5.8	&	.0917	&	-.3110	&	10.8	&	-.2032	&	-.1422\\
0.9	&	.8075	&	.4060	&	5.9	&	.1220	&	-.2951	&	10.9	&	-.1881	&	-.1604\\[2ex]
1.0	&	.7652	&	.4401	&	6.0	&	.1506	&	-.2767	&	11.0	&	-.1712	&	-.1768\\
1.1	&	.7196	&	.4709	&	6.1	&	.1773	&	-.2559	&	11.1	&	-.1528	&	-.1913\\
1.2	&	.6711	&	.4983	&	6.2	&	.2017	&	-.2329	&	11.2	&	-.1330	&	-.2039\\
1.3	&	.6201	&	.5220	&	6.3	&	.2238	&	-.2081	&	11.3	&	-.1121	&	-.2143\\
1.4	&	.5669	&	.5419	&	6.4	&	.2433	&	-.1816	&	11.4	&	-.0902	&	-.2225\\[2ex]
1.5	&	.5118	&	.5579	&	6.5	&	.2601	&	-.1538	&	11.5	&	-.0677	&	-.2284\\
1.6	&	.4554	&	.5699	&	6.6	&	.2740	&	-.1250	&	11.6	&	-.0446	&	-.2320\\
1.7	&	.3980	&	.5778	&	6.7	&	.2851	&	-.0953	&	11.7	&	-.0213	&	-.2333\\
1.8	&	.3400	&	.5815	&	6.8	&	.2931	&	-.0652	&	11.8	&	.0020	&	-.2323\\
1.9	&	.2818	&	.5812	&	6.9	&	.2981	&	-.0349	&	11.9	&	.0250	&	-.2290\\[2ex]
2.0	&	.2239	&	.5767	&	7.0	&	.3001	&	-.0047	&	12.0	&	.0477	&	-.2234\\
2.1	&	.1666	&	.5683	&	7.1	&	.2991	&	.0252	&	12.1	&	.0697	&	-.2157\\
2.2	&	.1104	&	.5560	&	7.2	&	.2951	&	.0543	&	12.2	&	.0908	&	-.2060\\
2.3	&	.0555	&	.5399	&	7.3	&	.2882	&	.0826	&	12.3	&	.1108	&	-.1943\\
2.4	&	.0025	&	.5202	&	7.4	&	.2786	&	.1096	&	12.4	&	.1296	&	-.1807\\[2ex]
2.5	&	-.0484	&	.4971	&	7.5	&	.2663	&	.1352	&	12.5	&	.1469	&	-.1655\\
2.6	&	-.0968	&	.4708	&	7.6	&	.2516	&	.1592	&	12.6	&	.1626	&	-.1487\\
2.7	&	-.1424	&	.4416	&	7.7	&	.2346	&	.1813	&	12.7	&	.1766	&	-.1307\\
2.8	&	-.1850	&	.4097	&	7.8	&	.2154	&	.2014	&	12.8	&	.1887	&	-.1114\\
2.9	&	-.2243	&	.3754	&	7.9	&	.1944	&	.2192	&	12.9	&	.1988	&	-.0912\\[2ex]
3.0	&	-.2601	&	.3391	&	8.0	&	.1717	&	.2346	&	13.0	&	.2069	&	-.0703\\
3.1	&	-.2921	&	.3009	&	8.1	&	.1475	&	.2476	&	13.1	&	.2129	&	-.0489\\
3.2	&	-.3202	&	.2613	&	8.2	&	.1222	&	.2580	&	13.2	&	.2167	&	-.0271\\
3.3	&	-.3443	&	.2207	&	8.3	&	.0960	&	.2657	&	13.3	&	.2183	&	-.0052\\
3.4	&	-.3643	&	.1792	&	8.4	&	.0692	&	.2708	&	13.4	&	.2177	&	.0166\\[2ex]
3.5	&	-.3801	&	.1374	&	8.5	&	.0419	&	.2731	&	13.5	&	.2150	&	.0380\\
3.6	&	-.3918	&	.0955	&	8.6	&	.0146	&	.2728	&	13.6	&	.2101	&	.0590\\
3.7	&	-.3992	&	.0538	&	8.7	&	-.0125	&	.2697	&	13.7	&	.2032	&	.0791\\
3.8	&	-.4026	&	.0128	&	8.8	&	-.0392	&	.2641	&	13.8	&	.1943	&	.0984\\
3.9	&	-.4018	&	-.0272	&	8.9	&	-.0653	&	.2559	&	13.9	&	.1836	&	.1166\\[2ex]
4.0	&	-.3972	&	-.0660	&	9.0	&	-.0903	&	.2453	&	14.0	&	.1711	&	.1334\\
4.1	&	-.3887	&	-.1033	&	9.1	&	-.1142	&	.2324	&	14.1	&	.1570	&	.1488\\
4.2	&	-.3766	&	-.1386	&	9.2	&	-.1367	&	.2174	&	14.2	&	.1414	&	.1626\\
4.3	&	-.3610	&	-.1719	&	9.3	&	-.1577	&	.2004	&	14.3	&	.1245	&	.1747\\
4.4	&	-.3423	&	-.2028	&	9.4	&	-.1768	&	.1816	&	14.4	&	.1065	&	.1850\\[2ex]
4.5	&	-.3205	&	-.2311	&	9.5	&	-.1939	&	.1613	&	14.5	&	.0875	&	.1934\\
4.6	&	-.2961	&	-.2566	&	9.6	&	-.2090	&	.1395	&	14.6	&	.0679	&	.1999\\
4.7	&	-.2693	&	-.2791	&	9.7	&	-.2218	&	.1166	&	14.7	&	.0476	&	.2043\\
4.8	&	-.2404	&	-.2985	&	9.8	&	-.2323	&	.0928	&	14.8	&	.0271	&	.2066\\
4.9	&	-.2097	&	-.3147	&	9.9	&	-.2403	&	.0684	&	14.9	&	.0064	&	.2069\\[2ex]
5.0	&	-.1776	&	-.3276	&	10.0	&	-.2459	&	.0435	&	15.0	&	-.0142	&	.2051\\[1ex]
\hline
\end{array}\\[-2.1ex]
{}\\
\hline
\end{array}
\]
\end{scriptsize}
\label{tableend}
\newpage
\pagestyle{empty}

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\begin{LARGE}
APPLICATIONS OF THE CALCULUS\\
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By \textsc{E. R. Hedrick}, Professor of Mathematics in the University of Missouri,\\
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\noindent{\Huge T}HIS book presents a completed summary of those parts
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\begin{center}\textsc{Typographical Errors corrected in Project Gutenberg edition}\end{center}

p.~\pageref{err053}~Equation (3), the second $\displaystyle \int\limits_0^\pi$ was not printed.\\

p.~\pageref{err111}~Example~1, the = sign on the second line (beginning $\dfrac{2}{\pi}$) was not printed.\\

p.~\pageref{err149}~Example~6, second sin term $\sin \dfrac{m\pi y}{b}$ in original, amended to $\sin \dfrac{n\pi y}{b}$.\\

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[292] [1] [2] [3] [4] [5] [6] [7] [8] (./29779-t.aux)

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