% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
%                                                                         %
% The Project Gutenberg EBook of First Course in the Theory of Equations, by 
% Leonard Eugene Dickson                                                  %
%                                                                         %
% This eBook is for the use of anyone anywhere at no cost and with        %
% almost no restrictions whatsoever.  You may copy it, give it away or    %
% re-use it under the terms of the Project Gutenberg License included     %
% with this eBook or online at www.gutenberg.org                          %
%                                                                         %
%                                                                         %
% Title: First Course in the Theory of Equations                          %
%                                                                         %
% Author: Leonard Eugene Dickson                                          %
%                                                                         %
% Release Date: August 25, 2009 [EBook #29785]                            %
%                                                                         %
% Language: English                                                       %
%                                                                         %
% Character set encoding: ISO-8859-1                                      %
%                                                                         %
% *** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF EQUATIONS ***       %
%                                                                         %
% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %

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\begin{PGtext}
The Project Gutenberg EBook of First Course in the Theory of Equations, by 
Leonard Eugene Dickson

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org


Title: First Course in the Theory of Equations

Author: Leonard Eugene Dickson

Release Date: August 25, 2009 [EBook #29785]

Language: English

Character set encoding: ISO-8859-1

*** START OF THIS PROJECT GUTENBERG EBOOK THEORY OF EQUATIONS ***
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\begin{PGtext}
Produced by Peter Vachuska, Andrew D. Hwang, Dave Morgan,
and the Online Distributed Proofreading Team at
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\small
\subsection*{\centering\normalfont\Titling\TransNote}

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\frontmatter
\pagenumbering{roman}
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\settowidth{\TmpLen}{{\Large\Titling JOHN WILEY \& SONS, Inc.}}
\begin{center}
\Titling
{\Huge FIRST COURSE}\\[0.25in]
{\Large IN THE}\\[0.25in]
{\Huge THEORY OF EQUATIONS}\\[0.75in]
BY\TitleSkip
{\large LEONARD EUGENE DICKSON, Ph.D.}

{\tiny CORRESPONDANT DE L'INSTITUT DE FRANCE}\\[-4pt]
{\tiny PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF CHICAGO}

\vfill
NEW YORK\TitleSkip
% [** PP: Next line is underfull, but deliberately spaced out]
\makebox[\TmpLen][s]{\large JOHN WILEY \& SONS, Inc.}\\[0pt]
{\small London: CHAPMAN \& HALL, Limited}
\normalfont
\end{center}

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\clearpage

\settowidth{\TmpLen}{\itshape This book or any part thereof must not}
\begin{center}
\null\vfill
\textsc{Copyright, 1922,}\TitleSkip
\textsc{by}\TitleSkip
{\Titling LEONARD EUGENE DICKSON}\TitleSkip
\rule{0.5in}{0.5pt}\TitleSkip
\textit{All Rights Reserved}

\makebox[\TmpLen][s]{\itshape This book or any part thereof must not}\\
% [** PP: Next line is underfull, but deliberately spaced out]
\makebox[\TmpLen][s]{\itshape be reproduced in any form without} \\
\makebox[\TmpLen][s]{\itshape the written permission of the publisher.}
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Printed in U. S. A.\\[0.25in]
{\normalfont\tiny\sffamily
PRESS OF \\[0.5ex]
BRAUNWORTH\quad \&\quad CO.,\quad INC. \\[0.5ex] % [** PP: Added .]
BOOK\quad MANUFACTURERS \\[0.5ex]
BROOKLYN, NEW YORK \\
}
\end{center}

\iffalse
11/30 % [** PP: Omitted printer's mark]
\fi

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\Preface

The theory of equations is not only a necessity in the subsequent
mathematical courses and their applications, but furnishes an illuminating
sequel to geometry, algebra and analytic geometry. Moreover,
it develops anew and in greater detail various fundamental ideas of
calculus for the simple, but important, case of polynomials. The
theory of equations therefore affords a useful supplement to differential
calculus whether taken subsequently or simultaneously.

It was to meet the numerous needs of the student in regard to his
earlier and future mathematical courses that the present book was
planned with great care and after wide consultation. It differs essentially
from the author's \textit{Elementary Theory of Equations}, both in regard to
omissions and additions, and since it is addressed to younger students
and may be used parallel with a course in differential calculus. Simpler
and more detailed proofs are now employed. The exercises are simpler,
more numerous, of greater variety, and involve more practical applications.

This book throws important light on various elementary topics.
For example, an alert student of geometry who has learned how to bisect
any angle is apt to ask if every angle can be trisected with ruler and
compasses and if not, why not. After learning how to construct regular
polygons of $3$, $4$, $5$, $6$, $8$ and~$10$ sides, he will be inquisitive about the
missing ones of~$7$ and~$9$ sides. The teacher will be in a comfortable position
if he knows the facts and what is involved in the simplest discussion to
date of these questions, as given in \ChapRef{III}. Other chapters throw
needed light on various topics of algebra. In particular, the theory
of graphs is presented in \ChapRef{V} in a more scientific and practical
manner than was possible in algebra and analytic geometry.

There is developed a method of computing a real root of an equation
with minimum labor and with certainty as to the accuracy of all the
decimals obtained.  We first find by Horner's method successive transformed
%% -----File: 004.png---Folio xx-------
equations whose number is half of the desired number of significant
figures of the root. The final equation is reduced to a linear equation
by applying to the constant term the correction computed from the
omitted terms of the second and higher degrees, and the work is completed
by abridged division. The method combines speed with control of
accuracy.

Newton's method, which is presented from both the graphical and
the numerical standpoints, has the advantage of being applicable also to
equations which are not algebraic; it is applied in detail to various such
equations.

In order to locate or isolate the real roots of an equation we may
employ a graph, provided it be constructed scientifically, or the theorems
of Descartes, Sturm, and Budan, which are usually neither stated, nor
proved, correctly.

The long chapter on determinants is independent of the earlier chapters.
The theory of a general system of linear equations is here presented
also from the standpoint of matrices.

For valuable suggestions made after reading the preliminary manuscript
of this book, the author is greatly indebted to Professor Bussey
of the University of Minnesota, Professor Roever of Washington University,
Professor Kempner of the University of Illinois, and Professor
Young of the University of Chicago. The revised manuscript was much
improved after it was read critically by Professor Curtiss of Northwestern
University. The author's thanks are due also to Professor Dresden of
the University of Wisconsin for various useful suggestions on the
proof-sheets.

{\small\textsc{Chicago, 1921.}}

%% -----File: 005.png---Folio xx-------

\PrintContents

\iffalse%%%% COMMENTED HARD-CODED TABLE OF CONTENTS %%%%
\begin{center}
\section*{CONTENTS}

{\tiny Numbers refer to pages.}

\rule{3em}{0.75pt}

\subsection*{CHAPTER I}

\textsc{Complex Numbers}
\end{center}

Square roots, 1. Addition, multiplication and division  of complex numbers,
2. Cube roots of unity, 3. Geometrical representation, 3. Product
and quotient, 4. De Moivre's theorem, 5. Cube roots, 5. $n$th roots, 7.
Roots of unity, 8. Primitive roots of unity, 9.

\begin{center}
\subsection*{CHAPTER II}

\textsc{Elementary Theorems on the Roots of an Equation}
\end{center}

Quadratic equation, 11. Remainder theorem, 12. Synthetic division, 13.
Factored form of a polynomial, 15. Multiple roots, 16. Identical polynomials,
16. Relations between the roots and the coefficients, 17. Imaginary
roots occur in pairs, 19. Upper limit to the real roots, 21. Integral roots,
24. Rational roots, 27.

\begin{center}
\subsection*{CHAPTER III}

\textsc{Constructions with Ruler and Compasses}
\end{center}

Graphical solution of a  quadratic equation, 29. Analytic criterion for
constructibility, 30. Cubic equations with a constructible root, 32. Trisection
of an angle, 34. Duplication of a cube, 35. Regular polygons of
7, 9, 17, and $n$ sides, 35-44. Reciprocal equations, 37.

\begin{center}
\subsection*{CHAPTER IV}

\textsc{Cubic and Quartic Equations}
\end{center}

Algebraic solution of a cubic, 45. Discriminant, 47. Number of real roots
of a cubic, 48. Trigonometric solution of a cubic, 49. Ferrari's and Descartes'
solutions of a quartic, 50. Resolvent cubic, 51. Discriminant of a quartic, 51.

\begin{center}
\subsection*{CHAPTER V}


\textsc{The Graph of an Equation}
\end{center}

Use of graphs, 55. Caution in plotting, 55. Bend points, 56. Derivatives,
58. Horizontal tangents, 60. Multiple roots, 60. Ordinary and inflexion
%% -----File: 006.png---Folio xx-------
* tangents, 62. Real roots of a cubic equation, 65. Continuity, 66. Condition
for a root between $a$ and $b$, 67. Sign of a polynomial at infinity, 68.
Rolle's theorem, 69.

\begin{center}
\subsection*{CHAPTER VI}

\textsc{Isolation of the Real Roots}
\end{center}

Descartes' rule of signs, 71. Sturm's method, 75. Sturm's functions for
the general quartic equation, 80. Budan's theorem, 83.

\begin{center}
\subsection*{CHAPTER VII}

\textsc{Solution of Numerical Equations}
\end{center}

Horner's method, 86. Newton's method, algebraic and graphical discussion,
systematic computation, also for functions not polynomials, 90. Imaginary
roots, 98.

\begin{center}
\subsection*{CHAPTER VIII}

\textsc{Determinants; Systems of Linear Equations}
\end{center}

Solution of 2 or 3 linear equations by determinants, 101. Even and odd
arrangements, 103. Definition of a determinant of order $n$, 105. Interchange
of rows and columns, 106. Interchange of two columns or two rows, 107.
Minors, 109. Expansion, 109. Removal of factors, 111. Sum of determinants,
112. Addition of columns or rows, 113. Rank, 116. System
of $n$ linear equations in n unknowns, 114, 116. Homogeneous equations, 119.
System of $m$ linear equations in $n$ unknowns, matrix and augmented matrix,
120. Complementary minors, 122. Laplace's development, 122. Product of
determinants, 124.


\begin{center}
\subsection*{CHAPTER IX}

\textsc{Symmetric Functions}
\end{center}

Sigma functions, 128. Elementary symmetric functions, 128. Fundamental
theorem, 129. Rational functions symmetric in all but one of the
roots, 132. Sums of like powers of the roots, Newton's identities, 134.
Waring's formula, 136. Computation of symmetric functions, 141.

\begin{center}
\subsection*{CHAPTER X}

\textsc{Elimination, Resultants and Discriminants}
\end{center}

Methods of Sylvester, Euler, and Bézout, 143. Discriminants, 152.

\begin{center}
\subsection*{APPENDIX}

\textsc{The Fundamental Theorem of Algebra}
\end{center}

\textsc{Answers} \dotfill 159

\textsc{Index} \dotfill 167
\fi%%%% END OF COMMENTED HARD-CODED TABLE OF CONTENTS %%%%

%% -----File: 007.png---Folio 1-------

\mainmatter
\pagenumbering{arabic}

\ChapSkip
\begin{center}
\textbf{\LARGE First Course in}\\[0.25in]
\textbf{\LARGE The Theory of Equations}
\medskip

\TB
\end{center}

\Chapter{I}{Complex Numbers}


\Section{1.}{Square Roots} If $p$ is a positive real number, the symbol~$\sqrt{p}$ is
used to denote the positive square root of~$p$. It is most easily computed
by logarithms.
\index{Square roots}%

We shall express the square roots of negative numbers in terms of the
symbol~$i$ such that the relation $i^2 = -1$ holds. Consequently we denote
the roots of $x^2 = -1$ by~$i$ and~$-i$. The roots of $x^2 = -4$ are written in the
form $± 2i$ in preference to $±\sqrt{-4}$. In general, if $p$ is positive, the roots
of $x^2=-p$ are written in the form $±\sqrt{p} i$ in preference to $±\sqrt{-p}$.

\begin{Remark}
The square of either root is thus $(\sqrt{p})^2 i^2 = -p$. Had we used the less desirable
notation $±\sqrt{-p}$ for the roots of $x^2 = -p$, we might be tempted to find the square of
either root by multiplying together the values under the radical sign and conclude
erroneously that
\[
\sqrt{-p}\,\sqrt{-p}  = \sqrt{p^2} = +p.
\]
To prevent such errors we use $\sqrt{p}\,i$ and not~$\sqrt{-p}$.
\end{Remark}


% [** PP: ToC entry reads Addition, multiplication and division of complex numbers]
\Section{2.}{Complex Numbers} If $a$ and~$b$ are any two real numbers and
\index{Complex number}%
$i^2 = -1$, $a+bi$ is called a \emph{complex number}\footnote
  {Complex numbers are essentially couples of real numbers. For a treatment from
  this standpoint and a treatment based upon vectors, see the author's \textit{Elementary Theory
  of Equations}, p.~21, p.~18.}
and $a-bi$ its \emph{conjugate}. Either
\index{Conjugate}%
is said to be \emph{zero} if $a = b = 0$. Two complex numbers $a+bi$ and $c+di$
are said to be \emph{equal} if and only if $a = c$ and $b = d$. In particular, $a+bi=0$
%% -----File: 008.png---Folio 2-------
if and only if $a = b = 0$. If $b \ne 0$, $a+bi$ is said to be \emph{imaginary}.  In particular,
$bi$ is called a \emph{pure imaginary}.
\index{Pure imaginary}%
\index{Imaginary}%

Addition of complex numbers is defined by
\[
(a+bi) + (c+di) = (a+c) + (b+d)i.
\]
The inverse operation to addition is called subtraction, and consists in
finding a complex number~$z$ such that
\[
(c+di) + z = a+bi.
\]
In notation and value, $z$ is
\[
(a+bi) - (c+di) = (a-c) + (b-d)i.
\]

Multiplication is defined by
\[
(a+bi)(c+di) = ac-bd+(ad+bc)i,
\]
and hence is performed as in formal algebra with a subsequent reduction
by means of $i^2 = -1$.  For example,
\[
(a+bi)(a-bi) = a^2-b^2i^2 = a^2+b^2.
\]

Division is defined as the operation which is inverse to multiplication,
and consists in finding a complex number~$q$ such that $(a+bi)q = e+fi$.
Multiplying each member by $a-bi$, we find that~$q$ is, in notation and
value,
\[
\frac{e+fi}{a+bi} = \frac{(e+fi)(a-bi)}{a^2+b^2}
  = \frac{ae+bf}{a^2+b^2} + \frac{af-be}{a^2+b^2} i.
\]
Since $a^2+b^2 = 0$ implies $a = b = 0$ when $a$ and~$b$ are real, we conclude that
division except by zero is possible and unique.


\begin{Exercises}{Page2}
Express as complex numbers
\begin{Problems}[2]

\item[1.] $\sqrt{-9}$.

\item[2.] $\sqrt{4}$.

\ResetCols{2}

\item[3.] $(\sqrt{25} + \sqrt{-25})\sqrt{-16}$.

\item[4.] $-\frac{2}{3}$.

\ResetCols{4}

\item[5.] $8 + 2\sqrt{3}\vphantom{\dfrac{1}{1}}$.

\item[6.] $\dfrac{3 + \sqrt{-5}}{2 + \sqrt{-1}}$.

\item[7.] $\dfrac{3 + 5i}{2 - 3i}$.

\item[8.] $\dfrac{a + bi}{a - bi}$.

\ResetCols{1}

\item[9.] Prove that the sum of two conjugate complex numbers is real and that their
difference is a pure imaginary.

\item[10.] Prove that the conjugate of the sum of two complex numbers is equal to the
sum of their conjugates. Does the result hold true if each word sum is replaced by the
word difference?

%% -----File: 009.png---Folio 3-------

\item[11.]  Prove that the conjugate of the product (or quotient) of two complex numbers
is equal to the product (or quotient) of their conjugates.

\item[12.]  Prove that, if the product of two complex numbers is zero, at least one of them
is zero.

\item[13.]  Find two pairs of real numbers $x$, $y$ for which
\[
(x+yi)^2 = -7+24i.
\]
\end{Problems}

As in Ex.~13, express as complex numbers the square roots of
\begin{Problems}[3]

\item[14.]  {$-11+60i$.}

\item[15.] {$5-12i$.}

\item[16.] {$4cd+(2c^2-2d^2)i$.}

\end{Problems}

\end{Exercises}


\Section{3.}{Cube Roots of Unity} Any complex number~$x$ whose cube is equal
to unity is called a \emph{cube root of unity}. Since
\index{Cube root!of unity}%
\[
x^3-1 = (x-1) (x^2+x+1),
\]
the roots of $x^3=1$ are~$1$ and the two numbers~$x$ for which
\[
x^2+x+1=0,\qquad
(x + \tfrac{1}{2})^2 = -\tfrac{3}{4}, \qquad
 x + \tfrac{1}{2}    = ±\tfrac{1}{2} \sqrt{3}i.
\]
Hence there are three cube roots of unity, viz.,
\[
1, \qquad
\omega  = -\tfrac{1}{2} + \tfrac{1}{2}\sqrt{3}i,\qquad
\omega' = -\tfrac{1}{2} - \tfrac{1}{2}\sqrt{3}i.
\]

In view of the origin of~$\omega$, we have the important relations
\[
\omega^2 + \omega+1 = 0, \quad \omega^3 = 1.
\]
Since $\omega \omega' = 1$ and $\omega^3 = 1$, it follows that $\omega' = \omega^2$, $\omega = \omega'^2$.


\Section[Geometrical Representation]
{4.}{Geometrical Representation of Complex Numbers.} Using rectangular
axes of coördinates, $OX$ and~$OY$, we represent the complex number
$a+bi$ by the point~$A$ having the coördinates $a$, $b$ (Fig.~1).
\index{Complex number!geometrical representation}%

%[Illustration: \textsc{Fig}. 1]
\begin{wrapfigure}{r}{2.25in}
\hfill\Input{009a}
\end{wrapfigure}
The positive number $r = \sqrt{a^2+b^2}$ giving
the length of~$OA$ is called the \emph{modulus} (or
\emph{absolute value}) of~$a+bi$. The angle $\theta = XOA$,
measured counter-clockwise from~$OX$ to~$OA$,
is called the \emph{amplitude} (or \emph{argument}) of~$a+bi$.
\index{Absolute value}%
\index{Amplitude}%
\index{Argument}%
\index{Modulus}%
Thus $\cos \theta = a/r$, $\sin \theta = b/r$, whence
\[
a+bi = r(\cos\theta + i\sin\theta).
\Tag{1}
\]
The second member is called the \emph{trigonometric form} of~$a+bi$.
\index{Complex number!trigonometric form}%

For the amplitude we may select, instead of~$\theta$, any of the angles $\theta±360°$,
$\theta±720°$, etc.

%% -----File: 010.png---Folio 4-------

Two complex numbers are equal if and only if their moduli are equal
and an amplitude of the one is equal to an amplitude of the other.

%[Illustration: \textsc{Fig}. 2]
\begin{Remark}
\begin{wrapfigure}{l}{2.125in}
\Input{010a}
\end{wrapfigure}
For example, the cube roots of unity are~$1$ and
\index{Cube root!of unity}%
\begin{align*}
\omega   &= -\tfrac{1}{2} + \tfrac{1}{2}\sqrt{3} i \\
         &= \cos 120° + i \sin 120°, \\
\omega^2 &= -\tfrac{1}{2} - \tfrac{1}{2}\sqrt{3} i \\
         &= \cos 240° + i \sin 240°,
\end{align*}
and are represented by the points marked $1$, $\omega$, $\omega^2$
at the vertices of an equilateral triangle inscribed
in a circle of radius unity and center at the origin~$O$
(Fig.~2). The indicated amplitudes of~$\omega$ and~$\omega^2$
are~$120°$ and~$240°$ respectively, while the modulus
of each is~$1$.

The modulus of~$-3$ is $3$ and its amplitude is~$180°$ or~$180°$ plus or minus the product
of~$360°$ by any positive whole number.
\end{Remark}


% [** PP: ToC entry reads ``Product and quotient'', splitting]
\Section[Product]
{5.}{Product of Complex Numbers.} By actual multiplication,
\begin{align*}
&\quad
\bigl[r (\cos\theta + i \sin\theta)\bigr]
\bigl[r'(\cos\alpha + i \sin\alpha)\bigr] \\
&= rr'
\bigl[  (\cos\theta \cos\alpha - \sin\theta \sin\alpha)
     + i(\sin\theta \cos\alpha + \cos\theta \sin\alpha)\bigr] \\
&= rr'
\bigl[  \cos(\theta + \alpha) + i \sin(\theta + \alpha)], \quad
\text{by trigonometry.}
\end{align*}
Hence \textit{the modulus of the product of two complex numbers is equal to the
product of their moduli, while the amplitude of the product is equal to the
sum of their amplitudes.}

\begin{Remark}
For example, the square of $\omega = \cos 120°+ i \sin 120°$ has the modulus~$1$ and the amplitude
$120°+120°$ and hence is $\omega^2 = \cos 240°+i \sin 240°$. Again, the product of~$\omega$ and~$\omega^2$
has the modulus~$1$ and the amplitude $120°+ 240°$ and hence is $\cos 360°+ i \sin 360°$,
which reduces to~$1$. This agrees with the known fact that $\omega^3 = 1$.
\end{Remark}

Taking $r = r' = 1$ in the above relation, we obtain the useful formula
\[
(\cos \theta + i \sin \theta)
(\cos \alpha + i \sin \alpha)
  = \cos (\theta + \alpha) + i \sin (\theta + \alpha).
\Tag{2}
\]

% [** PP: Split ToC entry, part 2]
\Section[Quotient]
{6.}{Quotient of Complex Numbers.} Taking $\alpha = \beta - \theta$ in~\Eq{2} and dividing
the members of the resulting equation by $\cos\theta + i \sin\theta$, we get
\[
\frac{\cos \beta  + i \sin \beta}
     {\cos \theta + i \sin \theta}
  = \cos(\beta - \theta) + i \sin(\beta - \theta).
\]
%% -----File: 011.png---Folio 5-------
Hence \textit{the amplitude of the quotient of $R(\cos \beta+i \sin \beta)$ by $r(\cos \theta+i \sin \theta)$
is equal to the difference $\beta - \theta$ of their amplitudes, while the modulus of the
quotient is equal to the quotient~$R/r$ of their moduli.}

The case $\beta = 0$ gives the useful formula
\[
\frac{1}{\cos\theta + i \sin\theta} = \cos\theta - i \sin\theta.
\]


\Section{7.}{De Moivre's Theorem}
\index{De Moivre's!theorem}%
\begin{Thm}
If $n$ is any positive whole number,
\[
(\cos\theta + i \sin\theta)^n = \cos n\theta + i \sin n\theta.
\Tag{3}
\]
\end{Thm}

This relation is evidently true when $n = 1$, and when $n = 2$ it follows
from formula~\Eq{2} with $\alpha = \theta$. To proceed by mathematical induction,
suppose that our relation has been established for the values $1, 2, \dotsc, m$
of~$n$. We can then prove that it holds also for the next value $m+1$ of~$n$.
For, by hypothesis, we have
\[
(\cos\theta + i \sin\theta)^m = \cos m\theta + i \sin m\theta.
\]
Multiply each member by $\cos\theta + i \sin\theta$, and for the product on the right
substitute its value from~\Eq{2} with $\alpha = m \theta$. Thus
\begin{align*}
(\cos\theta + i \sin\theta)^{m+1}
  &= (\cos\theta + i \sin\theta)(\cos m\theta + i \sin m\theta), \\
  &= \cos(\theta + m \theta) + i \sin(\theta + m \theta),
\end{align*}
which proves~\Eq{3} when $n = m+1$. Hence the induction is complete.

\begin{Remark}
Examples are furnished by the results at the end of~§5:
\begin{align*}
(\cos 120° + i \sin 120°)^2 = \cos 240° + i \sin 240°, \\
(\cos 120° + i \sin 120°)^3 = \cos 360° + i \sin 360°.
\end{align*}
\end{Remark}


\Section{8.}{Cube Roots} To find the cube roots of a complex number, we first
express the number in its trigonometric form. For example,
\index{Cube root}%
\[
4\sqrt{2} + 4\sqrt{2} i = 8(\cos 45° + i \sin 45°).
\]
If it has a cube root which is a complex number, the latter is expressible
in the trigonometric form
\[
r(\cos\theta + i \sin\theta).
\Tag{4}
\]
The cube of the latter, which is found by means of~\Eq{3}, must be equal
to the proposed number, so that
\[
r^3(\cos 3\theta + i \sin 3\theta) = 8(\cos 45° + i \sin 45°).
\]
%% -----File: 012.png---Folio 6-------
The moduli $r^3$ and~$8$ must be equal, so that the positive real number~$r$
is equal to~$2$. Furthermore, $3 \theta$ and~$45°$ have equal cosines and equal
sines, and hence differ by an integral multiple of~$360°$. Hence $3 \theta = 45°+
k·360°$, or $\theta = 15°+k·120°$, where $k$ is an integer.\footnote
  {Here, as elsewhere when the contrary is not specified, zero and negative as well
  as positive whole numbers are included under the term ``integer.''}
Substituting this
value of~$\theta$ and the value~$2$ of~$r$ in~\Eq{4}, we get the desired cube roots. The
values $0$, $1$, $2$ of~$k$ give the distinct results
\begin{alignat*}{2} % [** PP: Re-breaking, aligning]
R_1 &= 2(\cos 15°  &+{}& i \sin 15°), \\
R_2 &= 2(\cos 135° &+{}& i \sin 135°),\\
R_3 &= 2(\cos 255° &+{}& i \sin 255°).
\end{alignat*}

Each new integral value of~$k$ leads to a result which is equal to~$R_1$,
$R_2$ or~$R_3$. In fact, from $k = 3$ we obtain~$R_1$, from $k = 4$ we obtain~$R_2$, from
$k = 5$ we obtain~$R_3$, from $k = 6$ we obtain~$R_1$ again, and so on periodically.


\begin{Exercises}{Page6}

\begin{Problems}
\item[1.] Verify that $R_2 = \omega R_1$,  $R_3 = \omega^2 R_1$. Verify that $R_1$ is a cube root of $8 (\cos 45°+
i \sin 45°)$ by cubing~$R_1$ and applying De Moivre's theorem. Why are the new expressions
for~$R_2$ and~$R_3$ evidently also cube roots?

\item[2.] Find the three cube roots of~$-27$; those of~$-i$; those of~$\omega$.

\item[3.] Find the two square roots of~$i$; those of~$-i$; those of~$\omega$.

\item[4.] Prove that the numbers $\cos\theta + i \sin\theta$ and no others are represented by points
on the circle of radius unity whose center is the origin.

\item[5.] If $a+bi$ and $c+di$ are represented by the points~$A$ and~$C$ in Fig.~3, prove that
their sum is represented by the fourth vertex~$S$ of the parallelogram two of whose sides
are~$OA$ and~$OC$. Hence show that the modulus of the sum of two complex numbers
is equal to or less than the sum of their moduli, and is equal to or greater than the difference
of their moduli.
\index{Complex number!geometrical representation}%
%[Illustration: \textsc{Fig}. 3]
%[Illustration: \textsc{Fig}. 4]
\begin{figure*}[hbt]
\begin{center}
\Input{012a}\hfil
\Input{012b}
\end{center}
\end{figure*}

%% -----File: 013.png---Folio 7-------

\item[6.] Let $r$ and~$r'$ be the moduli and $\theta$ and~$\alpha$ the amplitudes of two complex numbers
represented by the points $A$ and~$C$ in Fig.~4. Let~$U$ be the point on the $x$-axis one
unit to the right of the origin~$O$. Construct triangle $OCP$ similar to triangle $OUA$ and
similarly placed, so that corresponding sides are $OC$ and~$OU, CP$ and~$UA$, $OP$ and~$OA$,
while the vertices $O$, $C$, $P$ are in the same order (clockwise or counter-clockwise) as
the corresponding vertices $O$, $U$, $A$. Prove that~$P$ represents the product~(§5) of the
complex numbers represented by $A$ and~$C$.

\item[7.] If $a+bi$ and $e+fi$ are represented by the points $A$ and~$S$ in Fig.~3, prove that
the complex number obtained by subtracting $a+bi$ from $e+fi$ is represented by the point~$C$.
Hence show that the absolute value of the difference of two complex numbers is
equal to or less than the sum of their absolute values, and is equal to or greater than
the difference of their absolute values.

\item[8.] By modifying Ex.~6, show how to construct geometrically the quotient of two
complex numbers.
\index{Complex number!geometrical representation}%
\end{Problems}
\end{Exercises}


% [** PP: ToC entry matches unit title, running head is as given]
\Section[Roots of Complex Numbers]
{9.}{$n$th~Roots.} As illustrated in~§8, it is evident that the $n$th roots
of any complex number $\rho(\cos A + i \sin A)$ are the products of the $n$th
roots of $\cos A + i \sin A$ by the positive real $n$th root of the positive real
number~$\rho$ (which may be found by logarithms).
\index{Rootz@{Roots, $n$th}}% [** PP: Rootz places entry after Roots]

Let an $n$th root of $\cos A + i \sin A$ be of the form
\[
r(\cos \theta+i \sin \theta).
\tag{4}%[ ** PP: [sic], equation repeated]
\]
Then, by De Moivre's theorem,
\[
r^n(\cos n\theta + i \sin n\theta) = \cos A + i \sin A.
\]
The moduli $r^n$ and~$1$ must be equal, so that the positive real number~$r$
is equal to~$1$. Since $n\theta$ and~$A$ have equal sines and equal cosines, they
differ by an integral multiple of~$360°$. Hence $n\theta = A + k·360°$, where~$k$
is an integer. Substituting the resulting value of~$\theta$ and the value~$1$ of~$r$
in~\Eq{4}, we get
\[
      \cos\left(\frac{A + k·360°}{n}\right)
  + i \sin\left(\frac{A + k·360°}{n}\right).
\Tag{5}
\]

For each integral value of~$k$, \Eq{5}~is an answer since its $n$th power reduces
to $\cos A + i \sin A$ by DeMoivre's theorem. Next, the value~$n$ of~$k$ gives
the same answer as the value~$0$ of~$k$; the value $n+1$ of~$k$ gives the same
answer as the value~$1$ of~$k$; and in general the value $n+m$ of~$k$ gives the
same answer as the value~$m$ of~$k$. Hence we may restrict attention to
the values $0, 1, \dotsc, n-1$ of~$k$.   Finally, the answers~\Eq{5} given by these
%% -----File: 014.png---Folio 8-------
values $0, 1,\ldots, n-1$ of $k$ are all distinct, since they are represented by
points whose distance from the origin is the modulus~$1$ and whose amplitudes
are
\[
\frac{A}{n},\qquad
\frac{A}{n} + \frac{360°}{n},\qquad
\frac{A}{n} + \frac{2·360°}{n},\dotsc,
\frac{A}{n} + \frac{(n-1)360°}{n},
\]
so that these $n$ points are equally spaced points on a circle of radius unity.
Special cases are noted at the end of~§10. Hence
\begin{Thm}%
any complex number
different from zero has exactly $n$ distinct complex $n$th roots.
\end{Thm}


\Section{10.}{Roots of Unity} The trigonometric form of~$1$ is $\cos 0° + i \sin 0°$.
Hence by~§9 with $A=0$, the $n$ distinct $n$th roots of unity are
\index{Roots of unity}%
\[
\cos\frac{2k \pi}{n} + i \sin\frac{2k \pi}{n}\quad
(k=0, 1, \dotsc, n-1),
\Tag{6}
\]
where now the angles are measured in radians (an angle of $180$~degrees
being  equal to $\pi$~radians, where $\pi= 3.1416$, approximately). For $k = 0$,
\Eq{6} reduces to~$1$, which is an evident $n$th root of unity. For $k = 1$, \Eq{6}~is
\[
R = \cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n}.
\Tag{7}
\]

By De Moivre's theorem, the general number~\Eq{6} is equal to the
$k$th power of~$R$.   Hence the $n$ distinct $n$th roots of unity are
\[
R,\ R^2,\ R^3,\dotsc,\ R^{n-1},\ R^n = 1.
\Tag{8}
\]

As a special case of the final remark in~§9, the $n$~complex numbers~\Eq{6},
and therefore the numbers~\Eq{8}, are represented geometrically by the
vertices of a regular polygon of $n$~sides inscribed in the circle of radius
unity and center at the origin with one vertex on the positive $x$-axis.
\index{Regular polygon}%

%[Illustraton: \textsc{Fig}. 5]
% [** PP: wrapfigure doesn't play nicely with Remark environment]
\noindent\raisebox{-12pt}{\Input{014a}}\hfill
\begin{minipage}[b]{\linewidth-1.75in}
\begin{Remark}
\hspace*{1.5em}% We're in a minipage
For $n =3$, the numbers~\Eq{8} are $\omega$, $\omega^2$, $1$, which are represented
in Fig.~2 by the vertices of an equilateral triangle.

\hspace*{1.5em}%
For $n = 4$, $R = \cos\pi/2 + i \sin\pi/2 = i$. The four fourth roots
of unity~\Eq{8} are $i$, $i^2=-1$, $i^3=-i$, $i^4 = 1$, which are represented
by the vertices of a square inscribed in a circle of
radius unity and center at the origin~$O$ (Fig.~5).
\end{Remark}
\end{minipage}

%% -----File: 015.png---Folio 9-------

\begin{Exercises}{Page9}

\begin{Problems}
\item[1.] Simplify the trigonometric forms~\Eq{6} of the four fourth roots of unity. Check
the result by factoring $x^4-1$.

\item[2.] For $n=6$, show that $R = -\omega^2$. The sixth roots of unity are the three cube roots
of unity and their negatives. Check by factoring $x^6-1$.

\item[3.] From the point representing $a+bi$, how do you obtain that representing $-(a+bi)$?
Hence derive from Fig.~2 and Ex.~2 the points representing the six sixth roots of unity.
Obtain this result another way.

\item[4.] Find the five fifth roots of~$-1$.

\item[5.] Obtain the trigonometric forms of the nine ninth roots of unity. Which of
them are cube roots of unity?

\item[6.] Which powers of a ninth root~\Eq{7} of unity are cube roots of unity?

\end{Problems}
\end{Exercises}


\Section[Primitive Roots of Unity]
{11.}{Primitive $n$th Roots of Unity.} An $n$th root of unity is called
\emph{primitive} if $n$ is the smallest positive integral exponent of a power of it
that is equal to unity. Thus $\rho$ is a primitive $n$th root of unity if and only
if $\rho^n=1$ and $\rho^l \neq 1$ for all positive integers $l<n$.
\index{Primitive root of unity}%

Since only the last one of the numbers~\Eq{8} is equal to unity, the number~$R$,
defined by~\Eq{7}, is a primitive $n$th root of unity. We have shown that
the powers~\Eq{8} of~$R$ give all of the $n$th roots of unity. Which of these
powers of~$R$ are primitive $n$th roots of unity?

\begin{Remark}
For $n=4$, the powers~\Eq{8} of $R=i$ were seen to be
\[
i^1 =  i,\
i^2 = -1,\
i^3 = -i,\
i^4 =  1.
\]
The first and third are primitive fourth roots of unity, and their exponents $1$ and~$3$
are relatively prime to~$4$, i.e., each has no divisor $>1$ in common with~$4$. But the
second and fourth are not primitive fourth roots of unity (since the square of $-1$ and the
first power of~$1$ are equal to unity), and their exponents $2$ and~$4$ have the divisor~$2$ in
common with $n=4$. These facts illustrate and prove the next theorem for the case
$n=4$.
\end{Remark}

\begin{Theorem}
The primitive $n$th roots of unity are those of the numbers~\Eq{8}
whose exponents are relatively prime to $n$.
\end{Theorem}
\index{Relatively prime}%

\begin{Proof}
If $k$ and~$n$ have a common divisor $d$ $(d>1)$, $R^k$ is not a primitive
$n$th root of unity, since
\[
(R^k)^{\frac{n}{d}} = (R^n)^{\frac{k}{d}} = 1,
\]
and the exponent $n/d$ is a positive integer less than~$n$.

%% -----File: 016.png---Folio 10-------

But if $k$ and~$n$ are relatively prime, i.e., have no common divisor $>1$,
$R^k$ is a primitive $n$th root of unity. To prove this, we must show that
$(R^k)^{l} \ne 1$ if $l$~is a positive integer~$<n$. By De Moivre's theorem,
\index{Relatively prime}%
\[
R^{kl} = \cos\frac{2kl\pi}{n}
     + i \sin\frac{2kl\pi}{n}.
\]
If this were equal to unity, $2kl\pi/n$ would be a multiple of~$2\pi$, and hence
$kl$ a multiple of~$n$. Since $k$ is relatively prime to~$n$, the second factor~$l$
would be a multiple of~$n$, whereas $0 < l < n$.
\end{Proof}


\begin{Exercises}{Page10}

\begin{Problems}

\item[1.] Show that the primitive cube roots of unity are $\omega$ and~$\omega^{2}$.

\item[2.]  For $R$ given by~\Eq{7}, prove that the primitive $n$th
roots of unity are (i)~for $n=6$,
$R$, $R^5$; (ii)~for $n=8$, $R$, $R^3$, $R^5$, $R^7$; (iii)~for $n=12$, $R$, $R^5$, $R^7$, $R^{11}$.

\item[3.]  When $n$ is a prime, prove that any $n$th root of unity, other than~$1$, is primitive.

\item[4.]  Let $R$ be a primitive $n$th root \Eq{7} of unity, where $n$ is a product of two different
primes $p$ and~$q$. Show that $R, \dotsc, R^n$ are primitive with the exception of $R^p$, $R^{2p}, \dotsc,
R^{qp}$, whose $q$th powers are unity, and $R^q$, $R^{2q}, \dotsc, R^{pq}$, whose $p$th powers are unity.
These two sets of exceptions have only $R^{pq}$ in common. Hence there are exactly
$pq - p - q + 1$ primitive $n$th roots of unity.

\item[5.]  Find the number of primitive $n$th roots of unity if $n$ is a square of a prime~$p$.

\item[6.]  Extend Ex.~4 to the case in which $n$ is a product of three distinct primes.

\item[7.]  If $R$ is a primitive $15$th root \Eq{7} of unity, verify that $R^3$, $R^6$, $R^9$, $R^{12}$ are the primitive
fifth roots of unity, and $R^5$ and~$R^{10}$ are the primitive cube roots of unity. Show
that their eight products by pairs give all the primitive $15$th roots of unity.

\item[8.]  If $\rho$ is any primitive $n$th root of unity, prove that $\rho$, $\rho^2, \dots, \rho^n$ are distinct and
give all the $n$th roots of unity. Of these show that $\rho^k$ is a primitive $n$th root of unity
if and only if $k$ is relatively prime to~$n$.

\item[9.]  Show that the six primitive $18$th roots of unity are the negatives of the primitive
ninth roots of unity.
\end{Problems}
\end{Exercises}

%% -----File: 017.png---Folio 11-------


\Chapter[Theorems on Roots of Equations]
{II}{Elementary Theorems on the Roots of an Equation}

\Section{12.}{Quadratic Equation} If $a$, $b$, $c$ are given numbers, $a \ne 0$,
\index{Quadratic equation}%
\[
ax^2 + bx + c = 0  \quad (a \ne 0)
\Tag{1}
\]
is called a \emph{quadratic equation} or equation of the second degree. The
reader is familiar with the following method of solution by ``completing
the square.''  Multiply the terms of the equation by~$4a$, and transpose
the constant term; then
\[
4a^{2}x^2 + 4abx = -4ac.
\]
Adding $b^2$ to complete the square, we get
\[
(2ax + b)^2 = \Delta,\qquad \Delta = b^2 - 4ac,
\]
\[
x_{1} = \frac{-b + \sqrt{\Delta}}{2a}\qquad
x_{2} = \frac{-b - \sqrt{\Delta}}{2a}
\Tag{2}
\]

By addition and multiplication, we find that
\[
x_{1} + x_{2} = \frac{-b}{a},\qquad
x_{1}   x_{2} = \frac{ c}{a}.
\Tag{3}
\]
Hence for all values of the variable~$x$,
\[
a(x - x_1)(x - x_2)
  \equiv ax^2 - a(x_1 + x_2)x + ax_1 x_2
  \equiv ax^2 + bx + c,
\Tag{4}
\]
the sign $\equiv$ being used instead of~$=$ since these functions of~$x$ are \emph{identically
equal}, i.e., the coefficients of like powers of~$x$ are the same. We speak
of $a(x - x_1)(x - x_2)$ as the \emph{factored form} of the quadratic function $ax^2 + bx + c$,
and of $x - x_1$ and $x - x_2$ as its \emph{linear factors}.
\index{Factored form}%
\index{Identity}%
\index{Linear factors}%
\index{Symbol!a@{$\equiv$\IndAdd{identically equal to}}}% [** PP: Manually alphabetized]

In \Eq{4} we assign to~$x$ the values $x_1$ and~$x_2$ in turn, and see that
\[
0 = ax_1^2 + bx_1 + c,\qquad
0 = ax_2^2 + bx_2 + c.
\]

Hence the values \Eq{2} are actually the roots of equation~\Eq{1}.

We call $\Delta = b^2 - 4ac$ the \emph{discriminant} of the function $ax^2 + bx + c$ or
\index{Discriminant!of quadratic}%
of the corresponding equation~\Eq{1}. If $\Delta = 0$, the roots~\Eq{2} are evidently
equal, so that, by~\Eq{4}, $ax^2 + bx + c$ is the square of $\sqrt{a}(x - x_1)$, and conversely.
%% -----File: 018.png---Folio 12-------
We thus obtain the useful result that $ax^2 + bx + c$ \emph{is a perfect
square (of a linear function of $x$) if and only if $b^2 = 4ac$ (\emph{i.e.}, if its discriminant
is zero)}.
\index{Quadratic function a square}%

Consider a \emph{real} quadratic equation, i.e., one whose coefficients $a$, $b$, $c$
are all real numbers. Then if $\Delta$ is positive, the two roots~\Eq{2} are real.
But if $\Delta$ is negative, the roots are conjugate imaginaries~(§2).

When the coefficients of a quadratic equation \Eq{1} are any complex
numbers, $\Delta$ has two complex square roots~(§9), so that the roots \Eq{2} of
\Eq{1} are complex numbers, which need not be conjugate.

\begin{Remark}
For example, the discriminant of $x^2 - 2x + c$ is $\Delta = 4(1 - c)$. If $c = 1$, then $\Delta = 0$ and
$x^2 - 2x + 1 \equiv (x - 1)^2$ is a perfect square, and the roots $1$, $1$ of $x^2 - 2x + 1 = 0$ are equal.
If $c = 0$, $\Delta = 4$ is positive and the roots $0$ and~$2$ of $x^2 - 2x \equiv x(x - 2) = 0$ are real. If $c = 2$,
$\Delta = -4$ is negative and the roots $1 ± \sqrt{-1}$ of $x^2 - 2x + 2 = 0$ are conjugate complex
numbers. The roots of $x^2 - x + 1 + i = 0$ are $i$ and~$1 - i$, and are not conjugate.
\index{Discriminant!of quadratic}%
\end{Remark}


% [** PP: No ToC entry in original]
\Section[Polynomial]
{13.}{Integral Rational Function, Polynomial.} If $n$ is a positive integer
and $c_0$, $c_1, \dotsc, c_n$ are constants (real or imaginary),
\index{Integral!rational function}%
\index{Polynomial}%
\[
f(x) \equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_{n-1} x + c_n
\]
is called a \emph{polynomial} in~$x$ of \emph{degree}~$n$, or also an \emph{integral rational function}
of~$x$ of degree~$n$. It is given the abbreviated notation~$f(x)$, just as the
logarithm of $x + 2$ is written $\log(x + 2)$.
\index{Symbol!b@{$f(x)$\IndAdd{polynomial}}}% [** PP: Manually alphabetized]

If $c_0 \ne 0$, $f(x) = 0$ is an equation of degree~$n$. If $n = 3$, it is often called
a \emph{cubic equation}; and, if $n = 4$, a \emph{quartic equation}. For brevity, we often
speak of an equation all of whose coefficients are real as a \emph{real equation}.
\index{Real equation}%


\Section[Remainder Theorem]
{14.}{The Remainder Theorem.}
\index{Remainder theorem}%
\begin{Thm}
If a polynomial $f(x)$ be divided by
$x - c$ until a remainder independent of~$x$ is obtained, this remainder is equal
to~$f(c)$, which is the value of~$f(x)$ when $x = c$.
\end{Thm}

Denote the remainder by~$r$ and the quotient by~$q(x)$. Since the
dividend is~$f(x)$ and the divisor is~$x - c$, we have
\[
f(x) \equiv (x-c)q(x) + r,
\]
identically in~$x$. Taking $x = c$, we obtain $f(c) = r$.

If $r = 0$, the division is exact. Hence we have proved also the following
useful theorem.

\begin{Theorem}[The Factor Theorem.]
If $f(c)$ is zero, the polynomial $f(x)$ has the
factor $x - c$. In other words, if $c$ is a root of $f(x) = 0$, $x - c$ is a factor of~$f(x)$.
\end{Theorem}
\index{Factor theorem}%

%% -----File: 019.png---Folio 13-------

For example, $2$~is a root of $x^3 - 8 = 0$, so that $x - 2$ is a factor of $x^3 - 8$. Another
illustration is furnished by formula (4). % [** PP: Added period]


\begin{Exercises}{Page13}

Without actual division find the remainder when
\begin{Problems}
\item[1.]  $x^4 - 3x^2 - x - 6$ is divided by $x + 3$.

\item[2.]  $x^3 - 3x^2 + 6x - 5$ is divided by $x - 3$.
\end{Problems}

Without actual division show that
\begin{Problems}
\item[3.]  $18x^{10} + 19x^5 + 1$ is divisible by $x + 1$.

\item[4.]  $2x^4 - x^3 - 6x^2 + 4x - 8$ is divisible by $x - 2$ and $x + 2$.

\item[5.]  $x^4 - 3x^3 + 3x^2 - 3x + 2$ is divisible by $x - 1$ and $x - 2$.

\item[6.]  $r^3 - 1$, $r^4 - 1$, $r^5 - 1$ are divisible by $r - 1$.

\item[7.]  By performing the indicated multiplication, verify that
\[
r^n - 1 \equiv (r - 1)(r^{n-1} + r^{n-2} + \dotsb + r + 1).
\]

\item[8.]  In the last identity replace $r$ by~$x/y$, multiply by~$y^n$, and derive
\[
x^n - y^n \equiv (x-y)(x^{n-1} + x^{n-2}y + \dotsb + xy^{n-2} + y^{n-1}).
\]

\item[9.]  In the identity of Exercise~8 replace $y$ by $-y$, and derive
\begin{align*}
x^n + y^n
  &\equiv (x+y)(x^{n-1} - x^{n-2} y + \dotsb - xy^{n-2} + y^{n-1}),
   \quad \text{$n$~odd}; \\
x^n - y^n
  &\equiv (x+y)(x^{n-1} - x^{n-2} y + \dotsb + xy^{n-2} - y^{n-1}),
   \quad\text{$n$~even}.
\end{align*}
\end{Problems}

Verify by the Factor Theorem that $x + y$ is a factor.
\begin{Problems}
\item[10.]  If $a$, $ar$, $ar^2, \dotsc, ar^{n-1}$ are $n$ numbers in \emph{geometrical progression} (the ratio of any
term to the preceding being a constant $r \ne 1$), prove by Exercise~7 that their sum is
equal to
\index{Geometrical!progression}%
\[
\frac{a(r^n - 1)}{r - 1}.
\]

\item[11.]  At the end of each of $n$ years a man deposits in a savings bank $a$~dollars. With
annual compound interest at~4\%, show that his account at the end of $n$~years will be
\index{Compound interest}%
\[
\frac{a}{.04} \bigl\{(1.04)^n - 1\bigr\}
\]
dollars. Hint: The final deposit draws no interest; the prior deposit will amount to
$a(1.04)$ dollars; the deposit preceding that will amount to $a(1.04)^2$ dollars, etc. Hence
apply Exercise~10 for $r = 1.04$.
\end{Problems}
\end{Exercises}


\Section{15.}{Synthetic Division} The labor of computing the value of a polynomial
in~$x$ for an assigned value of~$x$ may be shortened by a simple device.
To find the value of
\index{Synthetic division}%
\[
x^4 + 3x^3 - 2x - 5
\]
%% -----File: 020.png---Folio 14-------
for $x = 2$, note that $x^4 = x·x^3 = 2x^3$, so that the sum of the first two terms
of the polynomial is~$5x^3$. To $5x^3 = 5·2^2x$ we add the next term~$-2x$ and
obtain~$18x$ or~$36$. Combining~$36$ with the final term~$-5$, we obtain the
desired value~$31$.

This computation may be arranged systematically as follows. After
supplying zero coefficients of missing powers of~$x$, we write the coefficients
in a line, ignoring the powers of~$x$.
\[
\begin{array}{rRRRRPc}
1 & 3 &  0 & -2 & -5 && \Lcol{2} \\
\cline{7-7}
  & 2 & 10 & 20 & 36 && \\
\cline{1-5}
1 & 5 & 10 & 18 & 31 && \Strut
\end{array}
\]
First we bring down the first coefficient~$1$. Then we multiply it by the
given value~$2$ and enter the product~$2$ directly under the second coefficient~$3$,
add and write the sum~$5$ below. Similarly, we enter the product of
$5$ by~$2$ under the third coefficient~$0$, add and write the sum~$10$ below; etc.
The final number~$31$ in the third line is the value of the polynomial when
$x = 2$. The remaining numbers in this third line are the coefficients, in
their proper order, of the quotient
\[
x^3 + 5x^2 + 10x + 18,
\]
which would be obtained by the ordinary long division of the given polynomial
by $x - 2$.

We shall now prove that this process, called \emph{synthetic division}, enables
us to find the quotient and remainder when any polynomial $f(x)$ is divided
by $x - c$. Write
\index{Quotient by synthetic division}%
\begin{align*}
f(x) &\equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_n, \\
\intertext{and let the constant remainder be $r$ and the quotient be}
q(x) &\equiv b_0 x^{n-1} + b_1 x^{n-2} + \dotsb + b_{n-1}.
\end{align*}

By comparing the coefficients of $f(x)$ with those in
\begin{multline*}
(x - c)q(x) + r
  \equiv b_0 x^n
  + (b_1 - cb_0) x^{n-1} \\
  + (b_2 - cb_1) x^{n-2} + \dotsb
  + (b_{n-1} - cb_{n-2}) x
  + r - cb_{n-1},
\end{multline*}
we obtain relations which become, after transposition of terms,
\[
b_0 = a_0,\
b_1 = a_1 + cb_0,\
b_2 = a_2 + cb_1, \dotsc,\
b_{n-1} = a_{n-1} + cb_{n-2},\
r = a_n + cb_{n-1}.
\]
%% -----File: 021.png---Folio 15-------
The steps in the work of computing the~$b$'s may be tabulated as follows:
\[
\begin{array}{rRRRWWPc}
a_0 & a_1  & a_2  & \cdots & a_{n-1}  & a_n      && \Lcol{c} \\
\cline{8-8}
    & cb_0 & cb_1 & \cdots & cb_{n-2} & cb_{n-1} &&          \\
\cline{1-6}
b_0 & b_1  & b_2  & \cdots & b_{n-1}, & r        && \Strut
\end{array}
\]
In the second space below $a_0$ we write $b_0$ (which is equal to $a_0$). We
multiply $b_0$ by $c$ and enter the product directly under $a_1$, add and write
the sum $b_1$ below it. Next we multiply $b_1$ by $c$ and enter the product
directly under $a_2$, add and write the sum $b_2$ below it; etc.


\begin{Exercises}{Page15}

Work each of the following exercises by synthetic division.
\begin{Problems}
\item[1.] Divide $x^3 + 3x^2 - 2x - 5$ by $x-2$.

\item[2.] Divide $2x^5 - x^3 + 2x - 1$ by $x+2$.

\item[3.] Divide $x^3 + 6x^2 + 10x - 1$ by $x - 0.09$.

\item[4.] Find the quotient of $x^3 - 5x^2 - 2x + 24$ by $x-4$, and then divide the quotient by
$x-3$. What are the roots of $x^3 - 5x^2 - 2x + 24 = 0$?

\item[5.] Given that $x^4 - 2x^3 - 7x^2 + 8x + 12 = 0$ has the roots $-1$ and~$2$, find the quadratic
equation whose roots are the remaining two roots of the given equation, and find these
roots.

\item[6.] If $x^4 - 2x^3 - 12x^2 + 10x + 3 = 0$ has the roots $1$ and~$-3$, find the remaining two roots.

\item[7.] Find the quotient of $2x^4 - x^3 - 6x^2 + 4x - 8$ by $x^2 - 4$.

\item[8.] Find the quotient of $x^4 - 3x^3 + 3x^2 - 3x + 2$ by $x^2 - 3x + 2$.

\item[9.] Solve Exercises 1, 2, 3, 6, 7 of~§14 by synthetic division.
\end{Problems}
\end{Exercises}


\Section{16.}{Factored Form of a Polynomial} Consider a polynomial %[** PP: Typo polynominal]
\index{Factored form}%
\[
f(x) \equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_n  \quad (c_0 \ne 0),
\]
whose leading coefficient $c_0$ is not zero. If $f(x) = 0$ has the root~$\alpha_1$, which
may be any complex number, the Factor Theorem shows that $f(x)$ has the
factor $x - \alpha_1$, so that
\[
f(x) \equiv (x - \alpha_1)Q(x),\quad
Q(x) \equiv c_0 x^{n-1} + c_1' x^{n-2} + \dotsb + c_{\alpha-1}'.
\]
If $Q(x) = 0$ has the root~$\alpha_2$, then
\[
Q(x) \equiv (x - \alpha_2)Q_1(x),\quad
f(x) \equiv (x - \alpha_1)(x - \alpha_2)Q_1(x).
\]
If $Q_1(x) = 0$ has the root~$\alpha_3$, etc., we finally get
\[
f(x) \equiv c_0(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n).
\Tag{5}
\]

We shall deduce several important conclusions from the preceding
discussion. First, suppose that the equation $f(x)=0$ of degree~$n$ is known
%% -----File: 022.png---Folio 16-------
to have $n$~distinct roots $\alpha_1, \dotsc, \alpha_n$. In $f(x) \equiv (x-\alpha_1)Q(x)$ take $x=\alpha_2$;
then $0=(\alpha_2-\alpha_1)Q(\alpha_2)$, whence $Q(\alpha_2)=0$ and $Q(x)=0$ has the root $\alpha_2$.
Similarly, $Q_1(x)=0$ has the root $\alpha_3$, etc. Thus all of the assumptions
(each introduced by an ``if'') made in the above discussion have been
justified and we have the conclusion~\Eq{5}. Hence \emph{if an equation $f(x)=0$
of degree~$n$ has $n$~distinct roots $\alpha_1, \dotsc, \alpha_n$, $f(x)$ can be expressed in the
factored form~\Eq{5}}.
\index{Number!of roots}%

It follows readily that the equation can not have a root~$\alpha$ different
from $\alpha_1, \dotsc, \alpha_n$. For, if it did, the left member of~\Eq{5} is zero when
$x=\alpha$ and hence one of the factors of the right member must then be zero,
say $\alpha-\alpha_j = 0$, whence the root~$\alpha$ is equal to~$\alpha_j$. We have now proved
the following important result.

\begin{Theorem}
An equation of degree $n$ cannot have more than $n$ distinct roots.
\end{Theorem}


\Section[Multiple Roots]
{17.}{Multiple Roots.\protect\footnotemark}\addtocounter{footnote}{1}%
  \footnotetext{Multiple roots are treated by calculus in~§58.}%
\addtocounter{footnote}{-1}%
\index{Multiple roots}%
Equalities may occur among the $\alpha$'s in~\Eq{5}.
Suppose that exactly $m_1$ of the $\alpha$'s (including~$\alpha_1$) are equal to~$\alpha_1$; that
$\alpha_2 \ne \alpha_1$, while exactly~$m_2$ of the~$\alpha$'s are equal to~$\alpha_2$; etc. Then \Eq{5} becomes
\[
f(x) \equiv
  c_0(x-\alpha_1)^{m_1}
     (x-\alpha_2)^{m_2}  \dotsm
     (x-\alpha_k)^{m_k}, \quad m_1 + m_2 + \dotsb + m_k = n,
\Tag{6}
\]
where $\alpha_1, \dotsc, \alpha_k$ are distinct. We then call~$\alpha_1$ a \emph{root of multiplicity}~$m_1$
of $f(x)= 0$, $\alpha_2$ a root of multiplicity~$m_2$, etc. In other words, $\alpha_1$ is a root
of multiplicity~$m_1$ of $f(x)=0$ if $f(x)$ is exactly divisible by $(x-\alpha_1)^{m_1}$, but is
not divisible by $(x-\alpha_1)^{m_1+1}$. We call $\alpha_1$ also an $m_1$-\emph{fold root}. In the
particular cases $m_1=1$, $2$, and~$3$, we also speak of $\alpha_1$ as a \emph{simple root}, \emph{double
root}, and \emph{triple root}, respectively. For example, $4$~is a simple root, $3$~a
\index{Double root|see{Discriminant}}%
\index{Multiplicity of root}%
\index{Simple root}%
\index{Triple root}%
double root, $-2$~a triple root, and $6$ a root of multiplicity~$4$ (or a $4$-fold
root) of the equation
\[
7(x-4)(x-3)^2(x+2)^3(x-6)^4 = 0
\]
of degree~$10$ which has no further root. This example illustrates the
next theorem, which follows from~\Eq{6} exactly as the theorem in~§16
followed from~\Eq{5}.

\begin{Theorem}
An equation of degree~$n$ cannot have more than $n$~roots,
a root of multiplicity~$m$ being counted as $m$~roots.
\end{Theorem}


\Section{18.}{Identical Polynomials}
\index{Identical polynomials}%
\begin{Thm}
If two polynomials in~$x$,
\[
a_0 x^n + a_1 x^{n-1} + \dotsb + a_n,\qquad
b_0 x^n + b_1 x^{n-1} + \dotsb + b_n,
\]
%% -----File: 023.png---Folio 17-------
each of degree~$n$, are equal in value for more than $n$~distinct values of~$x$, they
are term by term identical, i.e., $a_0 = b_0$, $a_1 = b_1, \dotsc, a_n = b_n$.
\end{Thm}

For, taking their difference and writing $c_0 = a_0 - b_0, \dotsc, c_n = a_n - b_n$,
we have
\index{Number!of roots}%
\[
c_0 x^n + c_1 x^{n-1} + \dotsb + c_n = 0
\]
for more than $n$~distinct values of~$x$. If $c_0 \ne 0$, we would have a contradiction
with the theorem in~§16. Hence $c_0 = 0$. If $c_1 \ne 0$, we would have
a contradiction with the same theorem with $n$~replaced by~$n-1$. Hence
$c_1 = 0$, etc.  Thus $a_0 = b_0$, $a_1 = b_1$, etc.


\begin{Exercises}{Page17}

\begin{Problems}
\item[1.]  Find a cubic equation having the roots $0$, $1$, $2$.

\item[2.]  Find a quartic equation having the roots $±1$, $±2$.

\item[3.]  Find a quartic equation having the two double roots $3$ and~$-3$.

\item[4.]  Find a quartic equation having the root~$2$ and the triple root~$1$.

\item[5.]  What is the condition that $ax^2+bx+c=0$ shall have a double root?

\item[6.]  If $a_0 x^n + \dotsb + a_n = 0$ has more than $n$~distinct roots, each coefficient is zero.

\item[7.]  Why is there a single answer to each of Exercises 1--4, if the coefficient of the
highest power of the unknown be taken equal to unity? State and answer the corresponding
general question.
\end{Problems}
\end{Exercises}


\Section[Fundamental Theorem of Algebra]
{19.}{The Fundamental Theorem of Algebra.}
\index{Fundamental theorem of algebra}%
\begin{Thm}
Every algebraic equation
with complex coefficients has a complex \(real or imaginary\) root.
\end{Thm}

This theorem, which is proved in the Appendix, implies that
\begin{Thm}%
every
equation of degree~$n$ has exactly $n$~roots if a root of multiplicity~$m$ be counted
as $m$~roots.
\end{Thm}%
In other words,
\begin{Thm}%
\index{Integral!rational function}%
\index{Linear factors}%
every integral rational function of degree~$n$
is a product of $n$~linear factors.
\end{Thm}%
For, in~§16, equations $f(x)=0$, $Q(x)=0$,
$Q_1(x)=0, \dotsc$ each has a root, so that \Eq{5} and~\Eq{6} hold.


% [** PP: ToC entry matches unit title; using running head]
\Section[Relations between Roots and Coefficients]
{20.}{Relations between the Roots and the Coefficients.} In~§12 we
found the sum and the product of the two roots of any quadratic equation
and then deduced the factored form of the equation. We now apply
the reverse process to any equation
\index{Relations between roots and coefficients}%
\begin{align*}
f(x) &\equiv c_0 x^n + c_1 x^{n-1} + \dotsb + c_n = 0\qquad (c_0 \ne 0),
\Tag{7} \\
\intertext{whose factored form is}
f(x) &\equiv c_0(x-\alpha_1)(x-\alpha_2) \dotsm (x-\alpha_n).
\Tag{8}
\end{align*}
Our next step is to find the expanded form of this product. The following
special products may be found by actual multiplication:
%% -----File: 024.png---Folio 18-------
\begin{align*}% [** PP: Aligning, breaking second line]
(x - \alpha_1)(x - \alpha_2)
  &\equiv x^2 - (\alpha_1 + \alpha_2)x + \alpha_1\alpha_2, \\
%[** PP: Typo (x_1 - \alpha_1)]
(x - \alpha_1)(x - \alpha_2)(x - \alpha_3)
  &\equiv x^3 - (\alpha_1 + \alpha_2 + \alpha_3)x^2 \\
  &\qquad     + (\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3)x
              - \alpha_1\alpha_2\alpha_3.
\end{align*}
These identities are the cases $n = 2$ and $n = 3$ of the following general
formula:
\begin{multline*}
%[** PP: Typo (x_1 - \alpha_1)]
(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n)
  \equiv x^n
  - (\alpha_1 + \dotsb + \alpha_n)x^{n-1} \\
  + (\alpha_1\alpha_2 + \alpha_1\alpha_3 + \alpha_2\alpha_3 + \dotsb
   + \alpha_{n-1}\alpha_n)x^{n-2} \\
  - (\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \dotsb
   + \alpha_{n-2}\alpha_{n-1}\alpha_n)x^{n-3} \\
  + \dotsb + (-1)^n \alpha_1\alpha_2 \dotsm \alpha_n,
\Tag{9}
\end{multline*}
the quantities in parentheses being described in the theorem below. If
we multiply each member of~\Eq{9} by $x - \alpha_{n+1}$, it is not much trouble to verify
that the resulting identity can be derived from~\Eq{9} by changing $n$ into
$n+1$, so that \Eq{9} is proved true by mathematical induction. Hence the
quotient of \Eq{7} by $c_0$ is term by term identical with~\Eq{9}, so that
\[
\begin{aligned}
\alpha_1 + \alpha_2 + \dotsb + \alpha_n &= -c_1 / c_0, \\
%
\alpha_1\alpha_2 + \alpha_1\alpha_3
                 + \alpha_2\alpha_3 + \dotsb
                 + \alpha_{n-1}\alpha_n &= c_2 / c_0, \\
%
\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_2\alpha_4 + \dotsb
                         + \alpha_{n-2}\alpha_{n-1}\alpha_n &= -c_3 / c_0, \\
&\vdots\\ %<tb>
\alpha_1\alpha_2 \dotsm \alpha_{n-1}\alpha_n &= (-1)^n c_n / c_0.
\end{aligned}
\Tag{10}
\]
These results may be expressed in the following words:

\begin{Theorem}
If $\alpha_1, \dotsc, \alpha_n$ are the roots of equation~\Eq{7}, the sum of the
roots is equal to~$-c_1 / c_0$, the sum of the products of the roots taken two at
a time is equal to~$c_2 / c_0$, the sum of the products of the roots taken three at a
time is equal to~$-c_3 / c_0$, etc.; finally, the product of all the roots is equal to~$(-1)^n c_n / c_0$.
\index{Product of roots}%
\index{Sum of!products of roots}%
\index{Sum of!roots}%
\end{Theorem}

Since we may divide the terms of our equation~\Eq{7} by~$c_0$, the essential
part of our theorem is contained in the following simpler statement:

\begin{Corollary}
In an equation in~$x$ of degree~$n$, in which the coefficient
of~$x^n$ is unity, the sum of the $n$~roots is equal to the negative of the coefficient
of~$x^{n-1}$, the sum of the products of the roots two at a time is equal to the coefficient
of~$x^{n-2}$, etc.; finally the product of all the roots is equal to the constant
term or its negative, according as $n$ is even or odd.
\end{Corollary}

\begin{Remark}
For example, in a cubic equation having the roots $2$, $2$, $5$, and having unity as the
coefficient of~$x^3$, the coefficient of~$x$ is $2·2 + 2·5 + 2·5 = 24$.
\end{Remark}

%% -----File: 025.png---Folio 19-------


\begin{Exercises}{Page19}

\begin{Problems}
\item[1.]  Find a cubic equation having the roots $1$, $2$, $3$.

\item[2.]  Find a quartic equation having the double roots $2$ and~$-2$.

\item[3.]  Solve $x^4 - 6x^3 + 13x^2 - 12x + 4 = 0$, which has two double roots.

\item[4.]  Prove that one root of $x^3 + px^2 + qx + r = 0$ is the negative of another root if and
only if $r = pq$.

\item[5.]  Solve $4x^3 - 16x^2 - 9x + 36 = 0$, given that one root is the negative of another.

\item[6.]  Solve $x^3 - 9x^2 + 23x - 15 = 0$, given that one root is the triple of another.

\item[7.]  Solve $x^4 - 6x^3 + 12x^2 - 10x + 3 = 0$, which has a triple root.

\item[8.]  Solve $x^3 - 14x^2 - 84x + 216 = 0$, whose roots are in geometrical progression, i.e.,
with a common ratio $r$ [say $m/r$, $m$, $mr$].
\index{Geometrical!progression}%

\item[9.]  Solve $x^3 - 3x^2 - 13x + 15 = 0$, whose roots are in arithmetical progression, i.e.,
with a common difference $d$ [say $m-d$, $m$, $m+d$].
\index{Arithmetical progression}%

\item[10.]  Solve $x^4 - 2x^3 - 21x^2 + 22x + 40 = 0$, whose roots are in arithmetical progression.
[Denote them by $c-3b$, $c-b$, $c+b$, $c+3b$, with the common difference $2b$]. % [** PP: Added period]

\item[11.]  Find a quadratic equation whose roots are the squares of the roots of
$x^2-px+q = 0$.

\item[12.]  Find a quadratic equation whose roots are the cubes of the roots of $x^2 - px + q = 0$.
Hint: $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$.

\item[13.]  If $\alpha$ and~$\beta$ are the roots of $x^2 - px + q = 0$, find an equation whose roots are (i)~$\alpha^2 / \beta$;
and~$\beta^2 / \alpha$;  (ii)~$\alpha^3\beta$ and~$\alpha\beta^3$;  (iii)~$\alpha+1 / \beta$ and~$\beta + 1 / \alpha$.

\item[14.]  Find a necessary and sufficient condition that the roots, taken in some order,
of $x^3 + px^2 + qx + r = 0$ shall be in geometrical progression.

\item[15.]  Solve $x^3 - 28x + 48 = 0$, given that two roots differ by~$2$.
\end{Problems}
\end{Exercises}


\Section{21.}{Imaginary Roots occur in Pairs} The two roots of a real quadratic
equation whose discriminant is negative are conjugate imaginaries~(§12).
This fact illustrates the following useful result.
\index{Imaginary!roots}%
\index{Surd roots in pairs}% [** PP: Original entry points to page 20]

\begin{Theorem}
If an algebraic equation with real coefficients has the root $a+bi$,
where $a$ and~$b$ are real and $b \ne 0$, it has also the root~$a-bi$.
\end{Theorem}

Let the equation be $f(x)= 0$ and divide $f(x)$ by
\[
(x-a)^2 + b^2 \equiv (x - a-bi)(x - a+bi)
\Tag{11}
\]
until we reach a remainder $rx + s$ whose degree in~$x$ is less than the degree
of the divisor. Since the coefficients of the dividend and divisor are all
real, those of the quotient~$Q(x)$ and remainder are real.  We have
\[
f(x) \equiv Q(x)\bigl\{(x-a)^2 + b^2\bigr\} + rx + s,
\]
identically in~$x$. This identity is true in particular when $x = a+bi$, so
that
\[
0 = r(a+bi)+s = ra+s+rbi.
\]
%% -----File: 026.png---Folio 20-------
Since all of the letters, other than~$i$, denote real numbers, we have~(§2)
$ra + s = 0$, $rb = 0$. But $b \ne 0$. Hence $r = 0$, and then $s = 0$. Hence $f(x)$
is exactly divisible by the function~\Eq{11}, so that $f(x) = 0$ has the root $a - bi$.

The theorem may be applied to the real quotient~$Q(x)$. We obtain
the

\begin{Corollary}
If a real algebraic equation has an imaginary root of
multiplicity~$m$, the conjugate imaginary of this root is a root of multiplicity~$m$.
\index{Multiplicity of root}%
\end{Corollary}

Counting a root of multiplicity~$m$ as $m$~roots, we see that a real equation
cannot have an odd number of imaginary roots. Hence by~§19, \emph{a real
equation of odd degree has at least one real root}.
\index{Real equation}%

Of the $n$~linear factors of a real integral rational function of degree~$n$
(§19), those having imaginary coefficients may be paired as in~\Eq{11}.
Hence
\begin{Thm}%
\index{Integral!rational function}%
every integral rational function with real coefficients can be expressed
as a product of real linear and real quadratic factors.
\end{Thm}


\begin{Exercises}{Page20}

\begin{Problems}
\item[1.] Solve $x^3 - 3x^2 - 6x - 20 = 0$, one root being $-1 + \sqrt{-3}$.

\item[2.] Solve $x^4 - 4x^3 + 5x^2 - 2x - 2 = 0$, one root being $1-i$.

\item[3.] Find a cubic equation with real coefficients two of whose roots are $1$ and $3+2i$.

\item[4.] If a real cubic equation $x^3 - 6x^2 + \dotsb = 0$ has the root $1 +\sqrt{-5}$, what are the
remaining roots? Find the complete equation.

\item[5.] If an equation with \emph{rational} coefficients has a root $a + \sqrt{b}$, where $a$ and $b$ are
rational, but $\sqrt{b}$ is irrational, prove that it has the root $a - \sqrt{b}$. [Use the method of~§21.]

\item[6.] Solve $x^4 - 4x^3 + 4x - 1 = 0$, one root being $2 + \sqrt{3}$.

\item[7.] Solve $x^3 - (4 + \sqrt{3})x^2 + (5 + 4\sqrt{3})x - 5\sqrt{3} = 0$, having the root $\sqrt{3}$.

\item[8.] Solve the equation in Ex.~7, given that it has the root $2+i$.

\item[9.] Find a cubic equation with rational coefficients having the roots $\frac{1}{2}, \frac{1}{2} + \sqrt{2}$.

\item[10.] Given that $x^4 - 2x^3 - 5x^2 - 6x + 2 = 0$ has the root $2 - \sqrt{3}$, find another root and
by means of the sum and the product of the four roots deduce, without division, the
quadratic equation satisfied by the remaining two roots.

\item[11.] Granted that a certain cubic equation has the root~$2$ and no real root different
from~$2$, does it have two imaginary roots?

\item[12.] Granted that a certain quartic equation has the roots $2 ± 3i$, and no imaginary
roots different from them, does it have two real roots?

\item[13.] By means of the proof of Ex.~5, may we conclude as at the end of~§21 that
every integral rational function with rational coefficients can be expressed as a product
of linear and quadratic factors with rational coefficients?
\end{Problems}
\end{Exercises}

%% -----File: 027.png---Folio 21-------


\Section{22.}{Upper Limit to the Real Roots} Any number which exceeds
\index{Upper limit to roots|(}%
all real roots of a real equation is called an \emph{upper limit to the real roots}.
We shall prove two theorems which enable us to find readily upper limits
to the real roots. For some equations Theorem~I gives a better (smaller)
upper limit than Theorem~II; for other equations, the reverse is true.
Evidently any positive number is an upper limit to the real roots of an
equation having no negative coefficients.


\begin{Theorem}[Theorem~I.]
If, in a real equation
\[
f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_n = 0  \qquad (a_0 > 0),
\]
the first negative coefficient is preceded by $k$ coefficients which are positive or
zero, and if $G$~denotes the greatest of the numerical values of the negative
coefficients, then each real root is less than $1 + \sqrt[k]{G / a_0}$.
\end{Theorem}

\begin{Remark}
For example, in $x^5 + 4x^4 - 7x^2 - 40x + 1 = 0$, $G = 40$ and $k = 3$ since we must supply
the coefficient zero to the missing power~$x^3$. Thus the theorem asserts that each root
is less than $1 + \sqrt[3]{40}$ and therefore less than~$4.42$. Hence $4.42$ is an upper limit to the
roots.
\end{Remark}

\begin{Proof}
For positive values of~$x$, $f(x)$ will be reduced in value or remain
unchanged if we omit the terms $a_1 x^{n-1}, \dotsc, a_{k-1}x^{n-k+1}$ (which are
positive or zero), and if we change each later coefficient $a_k, \dotsc, a_n$ to~$-G$.
Hence
\[
f(x) \geqq a_0 x^n - G(x^{n-k} + x^{n-k-1} + \dotsb + x + 1).
\]
But, by Ex.~7 of~§14,
\[
x^{n-k} + \dotsb + x + 1 \equiv \frac{x^{n-k+1} - 1}{x - 1},
\]
if $x \ne 1$. Furthermore,
\[
a_0 x^n - G\left(\frac{x^{n-k+1} - 1}{x - 1}\right)
  \equiv \frac{x^{n-k+1}\{a_0 x^{k-1}(x-1) - G\} + G}{x - 1}.
\]
Hence, if $x>1$,
\begin{align*}
f(x) &> \frac{x^{n-k+1}\bigl\{a_0 x^{k-1}(x-1) - G\bigr\}}{x - 1}, \\
f(x) &> \frac{x^{n-k+1}\bigl\{a_0 (x-1)^k - G\bigr\}}{x - 1}.
\end{align*}
Thus, for $x>1$, $f(x)>0$ and $x$ is not a root if $a_0 (x-1)^k - G \geqq 0$, which
is true if $x \geqq 1 + \sqrt[k]{G / a_0}$.
\end{Proof}

%% -----File: 028.png---Folio 22-------


% [** PP: No ToC entry]
\Section{23.}{Another Upper Limit to the Roots}\hfill\break
% [** PP: Format Theorem heading manually to avoid \BeforeSkip]
{\indent\normalfont\normalsize\scshape Theorem~II.}\quad
\begin{Thm}%
If, in a real algebraic equation in which the coefficient
of the highest power of the unknown is positive, the numerical value of each
negative coefficient be divided by the sum of all the positive coefficients which
precede it, the greatest quotient so obtained increased by unity is an upper
limit to the roots.
\end{Thm}

\begin{Remark}
For the example in~§22, the quotients are $7/(1+4)$ and~$40/5$, so that Theorem~II
asserts that $1+8$ or~$9$ is an upper limit to the roots. Theorem~I gave the better upper
limit~$4.42$. But for $x^3 + 8x^2 - 9x + c^2 = 0$, Theorem~I gives the upper limit~$4$, while
Theorem~II gives the better upper limit~$2$.

We first give the proof for the case of the equation
\[
f(x) \equiv p_4 x^4 - p_3 x^3 + p_2 x^2 - p_1 x + p_0 = 0
\]
in which each $p_i$ is positive.  In view of the identities
\[
x^4 \equiv (x-1) (x^3 + x^2 + x + 1) + 1,\qquad
x^2 \equiv (x-1) (x+1) + 1,
\]
$f(x)$ is equal to the sum of the terms
\begin{alignat*}{3}
p_4 (x-1) x^3 &+{}& p_4 (x-1)x^2 + p_4 (x-1)x &+{}& p_4 (x-1) &+ p_4, \\
 {} - p_3 x^3 &   &              + p_2 (x-1)x &+{}& p_2 (x-1) &+ p_2, \\
              &   &                {} - p_1 x &   &           &+ p_0.
\end{alignat*}
If $x>1$, negative terms occur only in the first and third columns, while the sum of the
terms in each of these two columns will be $\geqq 0$ if
\[
p_4 (x-1) - p_3 \geqq 0,\quad (p_4 + p_2 )(x-1) - p_1 \geqq 0.
\]
Hence $f(x) > 0$ and $x$ is not a root if
\[
x \geqq 1 + \frac{p_3}{p_4},\quad
x \geqq 1 + \frac{p_1}{p_4 + p_2}.
\]
This proves the theorem for the present equation.

Next, let $f(x)$ be modified by changing its constant term to~$-p_0$. We modify the
above proof by employing the sum $(p_4 + p_2)x - p_0$ of all the terms in the corresponding
last two columns. This sum will be $>0$ if $x > p_0 / (p_4 + p_2)$, which is true if
\[
x \geqq 1 + \frac{p_0}{p_4 + p_2}.
\]
\end{Remark}

To extend this method of proof to the general case
\[
f(x) \equiv a_n x^n + \dotsb + a_0\qquad (a_n > 0),
\]
we have only to employ suitable general notations.  Let the negative
coefficients be $a_{k_1}, \dotsc, a_{k_t} $, where $k_1 > k_2 > \dots > k_t$.   For  each   positive
%% -----File: 029.png---Folio 23-------
integer~$m$ which is $\leqq n$ and distinct from $k_1, \dotsc, k_t$, we replace~$x^m$
by the equal value
\[
d(x^{m-1} + x^{m-2} + \dotsb + x + 1) + 1
\]
where $d \equiv x-1$. Let $F(x)$ denote the polynomial in~$x$, with coefficients
involving~$d$, which is obtained from~$f(x)$ by these replacements. Let
$x>1$, so that $d$ is positive. Thus the terms $a_{k_i} x^{k_i}$ are the only negative
quantities occurring in~$F(x)$. If $k_i > 0$, the terms of~$F(x)$ which involve
explicitly the power $x^{k_i}$ are $a_{k_i} x^{k_i}$ and the $a_m d x^{k_i}$ for the various positive
coefficients~$a_m$ which precede~$a_{k_i}$. The sum of these terms will be $\geqq 0$
if $a_{k_i} + d \sum a_m \geqq 0$, i.e., if
\[
x \geqq 1 + \frac{-a_{k_i}}{\sum a_m}.
\]

There is an additional case if $k_t = 0$, i.e., if $a_0$ is negative. Then the
terms of~$F(x)$ not involving~$x$ explicitly are $a_0$ and the $a_m (d+1)$ for the
various positive coefficients~$a_m$. Their sum, $a_0 + x \sum a_m$, will be $> 0$ if
\[
x > \frac{-a_0}{\sum a_m},
\]
which is true if
\[
x \geqq 1 + \frac{-a_0}{\sum a_m}.
\]
\index{Upper limit to roots|)}%


\begin{Exercises}{Page23}

Apply the methods of both §22 and~§23 to find an upper limit to the roots~of
\begin{Problems}
\item[1.]  $4x^5 - 8x^4 + 22x^3 + 98x^2 - 73x + 5 = 0$.

\item[2.]  $x^4 - 5x^3 + 7x^2 - 8x + 1 = 0$.

\item[3.]  $x^7 + 3x^6 - 4x^5 + 5x^4 - 6x^3 - 7x^2 - 8 = 0$.

\item[4.]  $x^7 + 2x^5 + 4x^4 - 8x^2 - 32 = 0$.

\item[5.]  A lower limit to the negative roots of $f(x) = 0$ may be found by applying our
theorems to $f(-x) = 0$, i.e., to the equation derived from $f(x) = 0$ by replacing~$x$ by~$-x$.
Find a lower limit to the negative roots in Exs.\ 2, 3,~4.
\index{Lower limit to roots}%

\item[6.]  Prove that every real root of a real equation $f(x) = 0$ is less than $1 + g / a_0$ if $a_0 > 0$,
where $g$ denotes the greatest of the numerical values of $a_1, \dotsc, a_n$. Hint: if $x>0$,
\[
a_0 x^n + a_1 x^{n-1} + \dotsb \geqq a_0 x^n - g(x^{n-1} + \dotsb + x + 1).
\]
Proceed as in~§22 with $k = 1$.

\item[7.]  Prove that $1 + g \div |a_0|$ is an upper limit for the moduli of all complex roots of any
equation $f(x)=0$ with complex coefficients, where $g$ is the greatest of the values $|a_1|,
\dotsc, |a_n|$, and $|a|$ denotes the modulus of~$a$. Hint: use Ex.~5 of~§8.
\index{Symbol!c@{$\lvert a\rvert$\IndAdd{modulus}}}% [** PP: Manually alphabetized]
\end{Problems}
\end{Exercises}

%% -----File: 030.png---Folio 24-------


\Section{24.}{Integral Roots}
\index{Integral!roots|(}%
\begin{Thm}
For an equation all of whose coefficients are integers,
any integral root is an exact divisor of the constant term.
\end{Thm}

For, if~$x$ is an integer such that
\[
a_0 x^n + \dotsb + a_{n-1}x + a_n = 0,
\Tag{12}
\]
where the $a$'s are all integers, then, by transposing terms, we obtain
\[
x(-a_0x^{n-1} - \dotsb - a_{n-1}) = a_n.
\]
Thus $x$ is an exact divisor of~$a_n$ since the quotient is the integer given by
the quantity in parenthesis.

\begin{Example}[1.]
Find all the integral roots of
\[
x^3 + x^2 - 3x + 9 = 0.
\]
\end{Example}

\begin{Solution}
The exact divisors of the constant term~$9$ are $±1$, $±3$,~$±9$. By trial,
no one of $±1$, $3$ is a root. Next, we find that~$-3$ is a root by synthetic division~(§15):
\[
\begin{array}{rRRRPc}
1 &  1  & -3 &  9 && \Lcol{-3} \\
\cline{6-6}
  & -3  &  6 & -9 && \\
\cline{1-4}
1 & -2  &  3 &  0 && \Strut
\end{array}
\]

Hence the quotient is $x^2 - 2x + 3$, which is zero for $x = 1 ±\sqrt{-2}$. Thus~$-3$ is the
only integral root.
\end{Solution}

When the constant term has numerous exact divisors, some device
may simplify the application of the theorem.

\begin{Example}[2.\protect\footnotemark] Find all the integral roots of
\[
y^3 + 12y^2 - 32y - 256 = 0.
\]
  \footnotetext{This problem is needed for the solution~(§48) of a certain quartic equation.}
\end{Example}

\begin{Solution}
Since all the terms except~$y^3$ are divisible by~$2$, an integral root~$y$ must
be divisible by~$2$. Since all the terms except~$y^3$ are now divisible by~$2^4$, we have $y = 4z$,
where $z$ is an integer. Removing the factor~$2^6$ from the equation in~$z$, we obtain
\[
z^3 + 3z^2 - 2z - 4 = 0.
\]
An integral root must divide the constant term~$4$. Hence, if there are any integral
roots, they occur among the numbers $±1$, $±2$, $±4$. By trial, $-1$ is found to be a
root:
\[
\begin{array}{rRRRPc}
1 &  3  & -2 & -4 && \Lcol{-1} \\
\cline{6-6}
  & -1  & -2 &  4 && \\
\cline{1-4}
1 &  2  & -4 &  0 && \Strut
\end{array}
\]
%% -----File: 031.png---Folio 25-------
Hence the quotient is $z^2 + 2z - 4$, which is zero for $z = -1 ± \sqrt{5}$. Thus $y = 4z = -4$
is the only integral root of the proposed equation.
\end{Solution}


\begin{Exercises}{Page25}

Find all the integral roots of
\begin{Problems}[2]
\item[1.] $x^3 + 8x^2 + 13x + 6 = 0$.

\item[2.] $x^3 - 5x^2 - 2x + 24 = 0$.

\ResetCols{2}

\item[3.] $x^3 - 10x^2 + 27x - 18 = 0$.

\item[4.] $x^4 + 4x^3 + 8x + 32 = 0$.

\ResetCols{2}

\item[5.] The equation in Ex.~4 of~§23.

\end{Problems}
\end{Exercises}


% [** PP: No ToC entry]
\Section{25.}{Newton's Method for Integral Roots}  In~§24 we proved that
an integral root~$x$ of equation~\Eq{12} having integral coefficients must be
an exact divisor of~$a_n$. Similarly, if we transpose all but the last two
terms of~\Eq{12}, we see that $a_{n-1}x + a_n$ must be divisible by~$x^2$, and hence
$a_{n-1} + a_n/x$ divisible by~$x$. By transposing all but the last three terms
of~\Eq{12}, we see that their sum must be divisible by~$x^3$, and hence $a_{n-2} +
(a_{n-1} + a_n/x)/x$ divisible by~$x$. We thus obtain a series of conditions
of divisibility which an integral root must satisfy. The final sum
$a_0 + a_{1}/x + \dotsb$ must not merely be divisible by~$x$, but be actually zero,
since it is the quotient of the function~\Eq{12} by~$x^n$.

In practice, we must test in turn the various divisors~$x$ of~$a_n$. If a
chosen~$x$ is not a root, that fact will be disclosed by one of the conditions
mentioned. Newton's method is quicker than synthetic division since
it usually detects early and throws out wrong guesses as to a root, whereas
in synthetic division the decision comes only at the final step.

\begin{Remark}
For example, the divisor~$-3$ of the constant term of
\[
f(x) \equiv x^4 - 9x^3 + 24x^2 - 23x + 15 = 0
\Tag{13}
\]
is not a root since $-23 + 15/(-3) = -28$ is not divisible by~$-3$. To show that none
of the tests fails for~$3$, so that $3$ is a root, we may arrange the work systematically as
follows:
\[
\begin{array}{rRRRRPl}
 1 & -9 & 24 & -23 & 15 && \Lcol{\quad3} \\
-1 &  6 & -6 &   5 &    && \Lcol{\text{(divisor)}} \\
\cline{1-4}\cline{7-7}
 0 & -3 & 18 & -18 &    && \Strut
\end{array}
\Tag{14}
\]
First we divide the final coefficient~$15$ by~$3$, place the quotient~$5$ directly under the coefficient~$-23$,
and add. Next, we divide this sum~$-18$ by~$3$, place the quotient~$-6$
directly under the coefficient~$24$, and add. After two more such steps we obtain the
sum zero, so that $3$ is a root.

It is instructive to obtain the preceding process by suitably modifying synthetic
division. First, we replace~$x$ by~$1/y$ in~\Eq{13}, multiply each term by~$y^4$, and obtain
\[
15y^4 - 23y^3 + 24y^2 - 9y + 1 = 0.
\]
%% -----File: 032.png---Folio 26-------
We may test this for the root $y = \frac{1}{3}$, which corresponds to the root $x = 3$ of~\Eq{13}, by
ordinary synthetic division:
\[
\begin{array}{rRRRRPl}
15 & -23 & 24 & -9 &  1 && \Lcol{\quad\frac{1}{3}} \\
   &   5 & -6 &  6 & -1 && \Lcol{\text{(multiplier)}} \\
\cline{1-5}\cline{7-7}
15 & -18 & 18 & -3 &  0 && \Strut
\end{array}
\]
The coefficients in the last two lines (after omitting~$15$) are the same as those of the last
two lines in~\Eq{14} read in reverse order. This should be the case since we have here
multiplied the same numbers by~$\frac{1}{3}$ that we divided by~$3$ in~\Eq{14}. The numbers in
the present third line are the coefficients of the quotient~(§15). Since we equate the
quotient to zero for the applications, we may replace these coefficients by the numbers
in the second line which are the products of the former numbers by~$\frac{1}{3}$. The numbers
in the second line of~\Eq{14} are the negatives of the coefficients of the quotient of~$f(x)$
by $x-3$.
\end{Remark}

\begin{Example}
Find all the integral roots of equation~\Eq{13}.
\end{Example}

\begin{Solution}
For a negative value of~$x$, each term is positive. Hence all the real
roots are positive. By~§23, $10$ is an upper limit to the roots. By~§24, any integral
root is an exact divisor of the constant term~$15$. Hence the integral roots, if any, occur
among the numbers $1$, $3$,~$5$. Since $f(1) = 8$, $1$~is not a root. By~\Eq{14}, $3$~is a root. Proceeding
similarly with the quotient by~$x-3$, whose coefficients are the negatives of the
numbers in the second line of~\Eq{14}, we find that $5$ is a root.
\end{Solution}


\begin{Exercises}{}% [** PP: No answers]

\begin{Problems}
\item[1.] Solve Exs.~1--4 of~§24 by Newton's method.

\item[2.] Prove that, in extending the process~\Eq{14} to the general equation~\Eq{12}, we may
employ the final equations in~§15 with $r = 0$ and write
\[
\begin{array}{rRRRWWNPl}
 a_0 &  a_1 &  a_2 & \ldots &  a_{n-2} &  a_{n-1} & a_n && \Lcol{\quad c} \\
-b_0 & -b_1 & -b_2 & \ldots & -b_{n-2} & -b_{n-1} &     && \Lcol{\text{(divisor)}} \\
\cline{1-6}\cline{9-9}
   0 & -c{b_0} & -cb_1 & \ldots & -cb_{n-3} & -cb_{n-2} &   && \Strut
\end{array}
\]
Here the quotient, $-b_{n-1}$, of~$a_n$ by~$c$ is placed directly under $a_{n-1}$, and added to it to
yield the sum $-cb_{n-2}$, etc.
\end{Problems}
\end{Exercises}


% [** PP: No ToC entry]
\Section{26.}{Another Method for Integral Roots} An integral divisor~$d$ of the constant term is not a root if $d-m$ is not a divisor of~$f(m)$, where $m$ is
any chosen integer. For, if $d$ is a root of $f(x)=0$, then
\[
f(x) \equiv (x-d)Q(x),
\]
where $Q(x)$ is a polynomial having integral coefficients~(§15). Hence
$f(m) = (m-d) q$, where $q$ is the integer~$Q(m)$.

%% -----File: 033.png---Folio 27-------

\begin{Remark}
In the example of~§25, take $d = 15$, $m=1$.  Since $f(1)=8$ is not divisible by $15-1=14$, $15$ is not an integral root.

Consider the more difficult example
\[
f(x) \equiv x^3 - 20x^2 + 164x - 400 = 0,
\]
whose constant term has many divisors. There is evidently no negative root, while
$21$ is an upper limit to the roots. The positive divisors less than~$21$ of $400 = 2^4 5^2$ are
$d = 1$, $2$, $4$, $8$, $16$, $5$, $10$,~$20$. First, take $m=1$ and note that $f(1) = -255 = -3·5·17$. The
corresponding values of $d-1$ are $0$, $1$, $3$, $7$, $15$, $4$, $9$,~$19$; of these, $7$, $4$, $9$,~$19$ are not divisors
of~$f(1)$, so that $d = 8$, $5$, $10$ and~$20$ are not roots. Next, take $m=2$ and note that $f(2)= -144$
is not divisible by $16-2 = 14$. Hence $16$ is not a root. Incidentally, $d=1$ and
$d=2$ were excluded since $f(d)\neq 0$. There remains only $d=4$, which is a root.
\end{Remark}

In case there are numerous divisors within the limits to the roots, it
is usually a waste of time to list all these divisors. For, if a divisor is
found to be a root, it is preferable to employ henceforth the quotient,
as was done in the example in~§25.


\begin{Exercises}{Page27}

Find all the integral roots of
\begin{Problems}
\item[1.] $x^4 - 2x^3 - 21x^2 + 22x + 40 = 0$.

\item[2.] $y^3 - 9y^2 - 24y + 216 = 0$.

\item[3.] $x^4 - 23x^3 + 187x^2 - 653x + 936 = 0$.

\item[4.] $x^5 + 47x^4 + 423x^3 + 140x^2 + 1213x - 420 = 0$.

\item[5.] $x^5 - 34x^3 + 29x^2 + 212x - 300 = 0$.
\end{Problems}
\end{Exercises}
\index{Integral!roots|)}%


\Section{27.}{Rational Roots}
\index{Rational roots}%
\begin{Thm}
If an equation with integral coefficients
\[
c_0 x^n + c_1 x^{n-1} + \dotsb + c_{n-1} x + c_n = 0
\Tag{15}
\]
has the rational root~$a/b$, where $a$ and~$b$ are integers without a common divisor
$>1$, then $a$ is an exact divisor of~$c_n$, and~$b$ is an exact divisor of~$c_0$.
\end{Thm}

Insert the value $a/b$ of~$x$ and multiply all terms of the equation by~$b^n$.
We obtain
\[
c_0 a^n + c_1 a^{n-1} b + \dotsb + c_{n-1} ab^{n-1} + c_n b^n = 0.
\]
Since $a$ divides all the terms preceding the last term, it divides that term.
But $a$ has no divisor in common with~$b^n$; hence $a$ divides~$c_n$. Similarly,
$b$~divides all the terms after the first term and hence divides~$c_0$.

\begin{Example}
Find all the rational roots of
\[
2x^3 - 7x^2 + 10x - 6 = 0.
\]
\end{Example}

%% -----File: 034.png---Folio 28-------

\begin{Solution}
By the theorem, the denominator of any rational root~$x$ is a divisor of~$2$.
Hence $y = 2x$ is an integer. Multiplying the terms of our equation by~$4$, we obtain
\index{Transformed equation}%
\[
y^3 - 7y^2 + 20y - 24 = 0.
\]
There is evidently no negative root. By either of the tests in §§22,~23, an upper limit
to the positive roots of our equation in~$x$ is $1+7/2$, so that $y<9$. Hence the only
possible values of an integral root~$y$ are $1$, $2$, $3$, $4$, $6$,~$8$. Since~$1$ and~$2$ are not roots, we try~$3$:
\[
\begin{array}{rRRRPc}
 1 & -7 & 20 & -24 && \Lcol{3} \\
\cline{6-6}
-1 &  4 & -8 &     && \\
\cline{1-3}
 0 & -3 & 12 &     && \Strut
\end{array}
\]
Hence $3$ is a root and the remaining roots satisfy the equation $y^2 - 4y + 8 = 0$ and are
$2±2i$. Thus the only rational root of the proposed equation is $x=3/2$.
\end{Solution}

If $c_0=1$, then $b=±1$ and~$a/b$ is an integer. Hence we have the
\begin{Corollary}
Any rational root of an equation with integral coefficients,
that of the highest power of the unknown being unity, is an integer.
\end{Corollary}

Given any equation with integral coefficients
\[
a_0 y^n + a_1 y^{n-1} + \dotsb + a_n = 0,
\]
we multiply each term by~${a_0}^{n-1}$, write $a_0 y = x$, and obtain an equation~\Eq{15}
with integral coefficients, in which the coefficient~$c_0$ of~$x^n$ is now unity.
By the Corollary, each rational root~$x$ is an integer. Hence we need only
find all the integral roots~$x$ and divide them by~$a_0$ to obtain all the rational
roots~$y$ of the proposed equation.

Frequently it is sufficient (and of course simpler) to set $ky = x$, where
$k$ is a suitably chosen integer less than~$a_0$.


\begin{Exercises}{Page28}

Find all of the rational roots of
\begin{Problems}[2]

\item[1.] $y^4 -\frac{40}{3}y^3 + \frac{130}{3}y^2 - 40y + 9 = 0$.

\item[2.] $6y^3 - 11y^2 + 6y - 1 = 0$.

\ResetCols{1}

\item[3.] $108y^3 - 270y^2 - 42y + 1 = 0$. [Use $k = 6$.]

\item[4.] $32y^3 - 6y - 1 = 0$.  [Use the least~$k$.]

\ResetCols{2}

\item[5.] $96y^3 - 16y^2 - 6y + 1 = 0$.

\item[6.] $24y^3 - 2y^2 - 5y + 1 = 0$.

\ResetCols{2}

\item[7.] $y^3 - \frac{1}{2}y^2 - 2y + 1 = 0$.

\item[8.] $y^3 - \frac{2}{3}y^2 + 3y - 2 = 0$.

\ResetCols{1}

\item[9.] Solve Exs.~2--6 by replacing $y$ by~$1/x$.
\end{Problems}

Find the equations whose roots are the products of 6 by the roots of
\begin{Problems}[2]

\item[10.] $y^2 - 2y - \frac{1}{3} = 0$.

\item[11.] $y^3 - \frac{1}{2}y^2 - \frac{1}{3}y + \frac{1}{4} = 0$.
\end{Problems}
\end{Exercises}

%% -----File: 035.png---Folio 29-------


\Chapter{III}{Constructions with Ruler and Compasses}
\index{Geometrical!construction|(}% [** PP: Using index subitem]

% [** PP: No ToC entry]
\Section{28.}{Impossible Constructions} We shall prove that it is not possible,
by the methods of Euclidean geometry, to trisect all angles, or to construct
a regular polygon of~$7$ or $9$~sides. The proof, which is beyond the
scope of elementary geometry, is based on principles of the theory of
equations. Moreover, the discussion will show that a regular polygon
of $17$~sides can be constructed with ruler and compasses, a fact not suspected
during the twenty centuries from Euclid to Gauss.

\Section{29.}{Graphical Solution of a Quadratic Equation} If $a$
and~$b$ are constructible, and
\index{Quadratic equation!graphical solution}%
\begin{flalign*}% [** PP: Hack to center over the narrowed text block]
\makebox[\linewidth-3.25in][c]{$x^2 - ax + b = 0$}&& & &&
\Tag{1}
\end{flalign*}
%Illustration: \textsc{Fig}.~6
\begin{wrapfigure}{r}{2.75in}
\hfil\Input{035a}
\end{wrapfigure}
has real coefficients and real roots, the roots
can be constructed with ruler and compasses
as follows. Draw a circle having as a diameter
the line~$BQ$ joining the points $B = (0, 1)$
and $Q = (a, b)$ in Fig.~6. Then \emph{the
abscissas $ON$ and~$OM$ of the points of intersection
of this circle with the $x$-axis are the
roots of~\Eq{1}}.

For, the center of the circle is $\bigl(a/2, (b+1)/2\bigr)$; the square of~$BQ$ is
$a^2 + (b-1)^2$; hence the equation of the circle is
\[
\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b+1}{2}\right)^2
  = \frac{a^2 + (b-1)^2}{4}.
\]

This is found to reduce to~\Eq{1} when $y = 0$, which proves the theorem.

When the circle is tangent to the $x$-axis, so that $M$ and~$N$ coincide,
the two roots are equal. When the circle does not cut the $x$-axis, or
when $Q$ coincides with~$B$, the roots are imaginary.

Another construction follows from~§30.

%% -----File: 036.png---Folio 30-------

\begin{Exercises}{Page30}

Solve graphically:
\begin{Problems}[3]

\item[1.] $x^2 - 5x + 4 = 0$.

\item[2.] $x^2 + 5x + 4 = 0$.

\item[3.] $x^2 + 5x - 4 = 0$.

\ResetCols{3}

\item[4.] $x^2 - 5x - 4 = 0$.

\item[5.] $x^2 - 4x + 4 = 0$.

\item[6.] $x^2 - 3x + 4 = 0$.

\end{Problems}
\end{Exercises}


\Section{30.}{Analytic Criterion for Constructibility} The first step in our
consideration of a problem proposed for construction consists in formulating
the problem analytically. In some instances elementary algebra
suffices for this formulation. For example, in the ancient problem of
the duplication of a cube, we take as a unit of length a side of the given
cube, and seek the length~$x$ of a side of another cube whose volume is
double that of the given cube; hence
\[
x^3 = 2.
\Tag{2}
\]

But usually it is convenient to employ analytic geometry as in~§29;
a point is determined by its coordinates $x$ and~$y$ with reference to fixed
rectangular axes; a straight line is determined by an equation of the
first degree, a circle by one of the second degree, in the coordinates of the
general point on it.  Hence we are concerned with certain numbers,
some being the coordinates of points, others being the coefficients of equations,
and still others expressing lengths, areas or volumes.  These numbers
may be said to define analytically the various geometric elements
involved.

\begin{Criterion}
A proposed construction is possible by ruler and compasses
if and only if the numbers which define analytically the desired geometric
elements can be derived from those defining the given elements by a
finite number of rational operations and extractions of real square
roots.
\end{Criterion}
\index{Square roots}%

\begin{Remark}
In §29 we were given the numbers $a$ and~$b$, and constructed lines of lengths
\[
\tfrac{1}{2}(a ± \sqrt{a^2 -4b)}.
\]
\end{Remark}

\begin{Proof}
First, we grant the condition stated in the criterion and prove
that the construction is possible with ruler and compasses. For, a rational
function of given quantities is obtained from them by additions, subtractions,
multiplications, and divisions. The construction of the sum
or difference of two segments is obvious. The construction, by means
of parallel lines, of a segment whose length~$p$ is equal to the product $a·b$
of the lengths of two given segments is shown in Fig.~7; that for the quotient
%% -----File: 037.png---Folio 31-------
$q = a/b$ in Fig.~8. Finally, a segment of length $s = \sqrt{n}$ may be constructed,
as in Fig.~9, by drawing a semicircle on a diameter composed
of two segments of lengths $1$ and~$n$, and then drawing a perpendicular
to the diameter at the point which separates the two segments. Or we
may construct a root of $x^2 - n = 0$ by~§29.
%[Illustration: Fig. 7]
%[Illustration: Fig. 8]
%[Illustration: Fig. 9]
\begin{figure*}[hbt]
\begin{center}
\Input{037a}\hfil
\Input{037b}\hfil\hfil
\Input{037c}
\end{center}
\end{figure*}

Second, suppose that the proposed construction is possible with ruler
and compasses. The straight lines and circles drawn in making the construction
are located by means of points either initially given or obtained
as the intersections of two straight lines, a straight line and a circle, or
two circles. Since the axes of coordinates are at our choice, we may
assume that the $y$-axis is not parallel to any of the straight lines employed
in the construction. Then the equation of any one of our lines is
\[
y = mx + b.
\Tag{3}
\]

Let $y = m'x + b'$ be the equation of another of our lines which intersects~\Eq{3}.
The coordinates of their point of intersection are
\[
x = \frac{b' - b}{m - m'},\qquad
y = \frac{mb' - m'b}{m - m'},
\]
which are rational functions of the coefficients of the equations of the
two lines.

Suppose that a line~\Eq{3} intersects the circle
\[
(x - c)^2 + (y - d)^2 = r^2,
\]
with the center $(c, d)$ and radius~$r$. To find the coordinates of the points
of intersection, we eliminate~$y$ between the equations and obtain a quadratic
equation for~$x$. Thus $x$ (and hence also $mx + b$ or~$y$) involves no
%% -----File: 038.png---Folio 32-------
irrationality other than a real square root, besides real irrationalities
present in $m$, $b$, $c$, $d$,~$r$.
\index{Square roots}%

Finally, the intersections of two circles are given by the intersections
of one of them with their common chord, so that this case reduces to the
preceding.
\end{Proof}

\begin{Remark}
For example, a side of a regular pentagon inscribed in a circle of radius unity is
(Ex.~2 of~§37)
\[
s = \tfrac{1}{2}\sqrt{10 - 2\sqrt{5}},
\Tag{4}
\]
which is a number of the type mentioned in the criterion. Hence a regular pentagon
can be constructed by ruler and compasses (see the example above
quoted).
\end{Remark}


\Section{31.}{Cubic Equations with a Constructible Root} We saw that the
\index{Cubic equation}%
problem of the duplication of a cube led to a cubic equation~\Eq{2}. We
shall later show that each of the problems, to trisect an angle, and to construct
regular polygons of $7$ and~$9$ sides with ruler and compasses, leads
to a cubic equation. We shall be in a position to treat all of these problems
as soon as we have proved the following general result.

\begin{Theorem}
It is not possible to construct with ruler and compasses a
line whose length is a root or the negative of a root of a cubic equation with
rational coefficients having no rational root.
\end{Theorem}

Suppose that $x_1$ is a root of
\[
x^3 + \alpha x^2 + \beta x + \gamma = 0 \qquad
\text{($\alpha$, $\beta$, $\gamma$ rational)}
\Tag{5}
\]
such that a line of length $x_1$ or~$-x_1$ can be constructed with ruler and compasses;
we shall prove that one of the roots of~\Eq{5} is rational. We have
only to discuss the case in which $x_1$ is irrational.

By the criterion in~§30, since the given numbers in this problem are
$\alpha$, $\beta$, $\gamma$, all rational, $x_1$ can be obtained by a finite number of rational
operations and extractions of real square roots, performed upon rational
numbers or numbers derived from them by such operations. Thus $x_1$
involves one or more real square roots, but no further irrationalities.

As in the case of~\Eq{4}, there may be superimposed radicals. Such a
two-story radical which is not expressible as a rational function, with
rational coefficients, of a finite number of square roots of positive rational
numbers is said to be a radical of \emph{order}~2. In general, an $n$-story radical
is said to be of order~$n$ if it is not expressible as a rational function, with
\index{Order of radical}%
rational coefficients, of radicals each with fewer than $n$~superimposed
radicals, the innermost ones affecting positive rational numbers.

%% -----File: 039.png---Folio 33-------

We agree to simplify $x_1$ by making all possible replacements of certain
types that are sufficiently illustrated by the following numerical examples.

If $x_1$ involves $\sqrt{3}$, $\sqrt{5}$, and~$\sqrt{15}$, we agree to replace $\sqrt{15}$ by~$\sqrt{3}·\sqrt{5}$.
If $x_1 = s - 7t$, where $s$ is given by~\Eq{4} and
\[
t = \tfrac{1}{2} \sqrt{10 + 2\sqrt{5}},
\]
so that $st = \sqrt{5}$, we agree to write $x_1$ in the form $s - 7\sqrt{5}/s$, which involves
a single radical of order~$2$ and no new radical of lower order. Finally,
we agree to replace $\sqrt{4 - 2\sqrt{3}}$ by its simpler form $\sqrt{3}-1$.

After all possible simplifications of these types have been made, the
resulting expressions have the following properties (to be cited as our
agreements): no one of the radicals of highest order~$n$ in~$x_1$ is equal to
a rational function, with rational coefficients, of the remaining radicals
of order~$n$ and the radicals of lower orders, while no one of the radicals
of order $n-1$ is equal to a rational function of the remaining radicals of
order $n-1$ and the radicals of lower orders, etc.

Let $\sqrt{k}$ be a radical of highest order~$n$ in~$x_1$. Then
\[
x_1 = \frac{a + b \sqrt{k}}{c + d \sqrt{k}},
\]
where $a$, $b$, $c$, $d$ do not involve $\sqrt{k}$, but may involve other radicals. If
$d = 0$, then $c \neq 0$ and we write $e$ for $a/c$, $f$ for $b/c$, and get
\begin{flalign*}
&& x_1 &= e + f\sqrt{k}, && \Rightmark{(f \neq 0)}
\Tag{6}
\end{flalign*}
where neither $e$ nor~$f$ involves~$\sqrt{k}$. If $d \neq 0$, we derive~\Eq{6} by multiplying
the numerator and denominator of the fraction for~$x_1$ by $c - d\sqrt{k}$, which
is not zero since $\sqrt{k} = c/d$ would contradict our above agreements.

By hypothesis, \Eq{6} is a root of equation~\Eq{5}. After expanding the
powers and replacing the square of $\sqrt{k}$ by~$k$, we see that
\[
(e + f \sqrt{k})^3
  + \alpha(e + f \sqrt{k})^2
  +  \beta(e + f \sqrt{k})
  + \gamma = A + B\sqrt{k},
\Tag{7}
\]
where $A$ and~$B$ are certain polynomials in $e$, $f$,~$k$ and the rational numbers
$\alpha$, $\beta$,~$\gamma$. Thus $A + B \sqrt{k} = 0$. If $B \neq 0$, $\sqrt{k} = -A/B$ is a rational function,
with rational coefficients, of the radicals, other than~$\sqrt{k}$, in~$x_1$, contrary
to our agreements. Hence $B = 0$ and therefore $A = 0$.

When $e - f \sqrt{k}$ is substituted for~$x$ in the cubic function~\Eq{5}, the result
%% -----File: 040.png---Folio 34-------
is the left member of~\Eq{7} with $\sqrt{k}$ replaced by~$-\sqrt{k}$, and hence the result
is $A - B\sqrt{k}$. But $A = B = 0$. This shows that
\[
x_2 = e - f \sqrt{k}
\Tag{8}
\]
is a new root of our cubic equation. Since the sum of the three roots
is equal to~$-\alpha$ by~§20, the third root is
\[
x_3 = -\alpha - x_1 - x_2 = -\alpha - 2e.
\Tag{9}
\]

Now $\alpha$ is rational. If also $e$ is rational, $x_3$ is a rational root and we have
reached our goal. We next make the assumption that $e$ is irrational
and show that it leads to a contradiction. Since $e$ is a component part
of the constructible root~\Eq{6}, its only irrationalities are square roots.
Let $\sqrt{s}$ be one of the radicals of highest order in~$e$. By the argument
which led to~\Eq{6}, we may write $e = e' + f'\sqrt{s}$, whence, by~\Eq{9},
\begin{flalign*}
&& x_3 &= g + h \sqrt{s},&& \Rightmark{(h \neq 0)}
\Tag{9'}
\end{flalign*}
where neither $g$ nor~$h$ involves~$\sqrt{s}$. Then by the argument which led
to~\Eq{8}, $g - h \sqrt{s}$ is a root, different from~$x_3$, of our cubic equation, and hence
is equal to $x_1$ or~$x_2$ since there are only three roots~(§16). Thus
\[
g - h \sqrt{s} = e ± f \sqrt{k}.
\]

By definition, $\sqrt{s}$ is one of the radicals occurring in $e$. Also, by~\Eq{9'},
every radical occurring in $g$ or~$h$ occurs in $x_3$ and hence in $e = \frac{1}{2}(-\alpha - x_3)$,
by~(9), $\alpha$ being rational. Hence $\sqrt{k}$ is expressible rationally in terms
of the remaining radicals occurring in $e$ and~$f$, and hence in~$x_1$, whose value
is given by~\Eq{6}. But this contradicts one of our agreements.


\Section{32.}{Trisection of an Angle} For a given angle~$A$, we can construct
\index{Trisection of angle}%
with ruler and compasses a line of length $\cos A$ or $-\cos A$, namely the
adjacent leg of a right triangle, with hypotenuse unity, formed by dropping
a perpendicular from a point in one side of $A$ to the other, produced if
necessary. If it were possible to trisect angle~$A$, i.e., construct the angle
$A/3$ with ruler and compasses, we could as before construct a line whose
length is $±\cos(A/3)$. Hence if we show that this last cannot be done
when the only given geometric elements are the angle~$A$ and a line of
unit length, we shall have proved that the angle~$A$ cannot be trisected.
We shall give the proof for $A = 120°$.

We employ the trigonometric identity
\[
\cos A = 4 \cos^3 \frac{A}{3} - 3 \cos \frac{A}{3}.
\]
%% -----File: 041.png---Folio 35-------
Multiply each term by~$2$ and write $x$ for $2\cos(A/3)$. Thus
\[
x^3 - 3x = 2\cos A.
\Tag{10}
\]

For $A = 120°$, $\cos A = -\frac{1}{2}$ and~\Eq{10} becomes
\[
x^3 - 3x + 1 = 0.
\Tag{11}
\]

Any rational root is an integer~(§27) which is an exact divisor of the
constant term~(§24). By trial, neither $+1$ nor~$-1$ is a root. Hence
\Eq{11} has no rational root. Hence~(§31)
\begin{Thm}%
it is not possible to trisect all
angles with ruler and compasses.
\end{Thm}

\begin{Remark}
Certain angles, like $90°$, $180°$, can be trisected. When $A=180°$, the equation
\Eq{10} becomes $x^3 - 3x = -2$ and has the rational root $x = 1$. It is the rationality of a
root which accounts for the possibility of trisecting this special angle~$180°$.
\end{Remark}


\Section[Duplication of a Cube]
{33.}{Regular Polygon of $9$~Sides, Duplication of a Cube.} Since angle
\index{Duplication of cube}%
$120°$ cannot be trisected with ruler and compasses~(§32), angle~$40°$ cannot
be so constructed in terms of angle~$120°$ and the line of unit length
as the given geometric elements. Since the former of these elements
and its cosine are constructible when the latter is given, we may take
the line of unit length as the only given element. In a regular polygon
of $9$~sides, the angle subtended at the center by one side is $\frac{1}{9}·360° = 40°$.
Hence \emph{a regular polygon of $9$~sides cannot be constructed with ruler and compasses}.
\index{Regular polygon!09@{$9$~sides}}%
Here, as in similar subsequent statements where the given
elements are not specified, the only such element is the line of unit length.

A rational root of $x^3 = 2$ is an integer~(§27) which is an exact divisor
of~$2$. The cubes of $±1$ and~$±2$ are distinct from~$2$. Hence there is no
rational root. Hence (§§30,~31)
\begin{Thm}%
it is not possible to duplicate a cube with
ruler and compasses.
\end{Thm}


% [** PP: ToC entry reads Regular Polygons of 7, 9, 17, and n Sides]
\Section{34.}{Regular Polygon of $7$~Sides} If we could construct with ruler
and compasses an angle~$B$ containing $360/7$ degrees, we could so construct
a line of length $x = 2 \cos B$. Since $7B = 360°$, $\cos 3B = \cos 4B$.
But
\index{Regular polygon!07@{$7$~sides}|(}%
\begin{align*}
2 \cos 3B &= 2(4 \cos^3 B - 3 \cos B) = x^3 - 3x, \\
2 \cos 4B &= 2(2 \cos^2 2B - 1) = 4(2 \cos^2 B - 1)^2 - 2 = (x^2 - 2)^2 - 2.
\end{align*}
Hence
\[
0 = x^4 - 4x^2 + 2 - (x^3 - 3x) = (x - 2)(x^3 + x^2 - 2x - 1).
\]
But $x = 2$ would give $\cos B = 1$, whereas $B$ is acute. Hence
\[
x^3 + x^2 - 2x - 1 = 0.
\Tag{12}
\]

%% -----File: 042.png---Folio 36-------

Since this has no rational root, \emph{it is impossible to construct a regular
polygon of $7$~sides with ruler and compasses}.


% [** PP: No ToC entry]
\Section{35.}{Regular Polygon of $7$~Sides and Roots of Unity} If
\index{Roots of unity}%
\[
R = \cos\frac{2\pi}{7} + i \sin\frac{2\pi}{7},
\]
we saw in~§10 that $R$, $R^2$, $R^3$, $R^4$, $R^5$, $R^6$, $R^7 = 1$ give all the roots of $y^7 = 1$
and are complex numbers represented by the vertices of a regular polygon
of $7$~sides inscribed in a circle of radius unity and center at the origin of
coordinates. By~§6,
\[
\frac{1}{R} = \cos\frac{2\pi}{7} -i \sin\frac{2\pi}{7},\qquad
R + \frac{1}{R} = 2\cos\frac{2\pi}{7}.
\]

We saw in~§34 that $2 \cos(2\pi/7)$ is one of the roots of the cubic
equation~\Eq{12}. This equation can be derived in a new manner by utilizing
the preceding remarks on $7$th roots of unity. Our purpose is not primarily
to derive~\Eq{12} again, but to illustrate some principles necessary in the
general theory of the construction of regular polygons.

Removing from $y^7 - 1$ the factor $y-1$, we get
\[
y^6 + y^5 + y^4 + y^3 + y^2 + y + 1 = 0,
\Tag{13}
\]
whose roots are $R$, $R^2, \dotsc, R^6$. Since we know that $R+1/R$ is one of
the roots of the cubic equation~\Eq{12}, it is a natural step to make the substitution
\[
y+ \frac{1}{y} = x
\Tag{14}
\]
in~\Eq{13}. After dividing its terms by~$y^3$, we have
\[
\left(y^3 + \frac{1}{y^3}\right) +
\left(y^2 + \frac{1}{y^2}\right) +
\left(y   + \frac{1}{y  }\right) + 1=0.
\Tag{13'}
\]
By squaring and cubing the members of~\Eq{14}, we see that
\[
y^2 + \frac{1}{y^2} = x^2 -  2,\qquad
y^3 + \frac{1}{y^3} = x^3 - 3x.
\Tag{15}
\]
Substituting these values in~\Eq{13'}, we obtain
\[
x^3 + x^2 - 2x - 1 = 0.
\tag{12}% [** PP: [sic], equation repeated]
\]
That is, the substitution~\Eq{14} converts equation~\Eq{13} into~\Eq{12}.
\index{Regular polygon!07@{$7$~sides}|)}%

%% -----File: 043.png---Folio 37-------

If in~\Eq{14} we assign to~$y$ the six values $R, \dotsc, R^6$, we obtain only
three distinct values of~$x$:
\[
x_1 = R   + \frac{1}{R}   = R   + R^6, \quad
x_2 = R^2 + \frac{1}{R^2} = R^2 + R^5, \quad
x_3 = R^3 + \frac{1}{R^3} = R^3 + R^4.
\Tag{16}
\]

In order to illustrate a general method of the theory of regular polygons,
we start with the preceding sums of the six roots in pairs and find
the cubic equation having these sums as its roots. For this purpose we
need to calculate
\[
x_1 + x_2 + x_3, \qquad
x_1 x_2 + x_1 x_3 + x_2 x_3, \qquad
x_1 x_2 x_3.
\]
First, by~\Eq{16},
\[
x_1 + x_2 + x_3 = R + R^2 + \dotsb + R^6 = -1,
\]
since $R, \dotsc, R^6$ are the roots of~\Eq{13}. Similarly,
\begin{gather*}
x_1 x_2 + x_1 x_3 + x_2 x_3 = 2(R + R^2 + \dotsb +R^6) = -2, \\
x_1 x_2 x_3 = 2 + R + R^2 + \dotsb + R^6 = 1.
\end{gather*}

Consequently~(§20), the cubic having $x_1$, $x_2$, $x_3$ as roots is~\Eq{12}.


\Section{36.}{Reciprocal Equations} Any algebraic equation such that the
reciprocal of each root is itself a root of the same multiplicity is called a
\emph{reciprocal equation}.
\index{Reciprocal equation}%

\begin{Remark}
The equation $y^7 - 1 = 0$ is a reciprocal equation, since if $r$ is any root, $1/r$ is evidently
also a root. Since~\Eq{13} has the same roots as this equation, with the exception of unity
which is its own reciprocal, \Eq{13} is also a reciprocal equation.
\end{Remark}

If $r$ is any root $\neq 0$ of any equation
\[
f(y) \equiv y^n + \dotsb + c = 0,
\]
$1/r$ is a root of $f(1/y)=0$ and hence of
\[
y^n f \left(\frac{1}{y}\right) \equiv 1 + \dotsb + cy^n = 0.
\]
If the former is a reciprocal equation, it has also the root~$1/r$, so that every
root of the former is a root of the latter equation. Hence, by~§18, the
left member of the latter is identical % [** PP: Typo indentical]
with $cf(y)$. Equating the constant
terms, we have $c^2 = 1$, $c= ±1$. Hence
\[
y^n f\left(\frac{1}{y}\right) \equiv ±f(y).
\Tag{17}
\]
%% -----File: 044.png---Folio 38-------
Thus if $p_i y^{n-i}$ is a term of~$f(y)$, also $±p_i y^i$ is a term. Hence
\[
f(y) \equiv y^n ± 1 + p_1(y^{n-1} ± y) + p_2 (y^{n-2} ± y^2) + \dotsb.
\Tag{18'}
\]

If $n$ is \emph{odd}, $n = 2t+1$, the final term is $p_t(y^{t+1} ± y^t)$, and $y ± 1$ is a factor
of $f(y)$. In view of~\Eq{17}, the quotient
\[
Q(y) \equiv \frac{f(y)}{y±1}
\]
has the property that
\[
y^{n-1} Q \left(\frac{1}{y}\right) \equiv Q(y).
\]
Comparing this with~\Eq{17}, which implied~\Eq{18'}, we see that $Q(y)=0$ is
a reciprocal equation of the type
\[
y^{2t} + 1
  + c_1 (y^{2t-1} + y)
  + c_2 (y^{2t-2} + y^2) + \dotsb
  + c_{t-1} (y^{t+1} + y^{t-1})
  + c_t y^t = 0.
\Tag{18}
\]

If $n$ is \emph{even}, $n = 2t$, and if the upper sign holds in~\Eq{17}, then~\Eq{18'} is
of the form~\Eq{18}. Next, let the lower sign hold in~\Eq{17}. Since a term
$p_t y^t$ would imply a term~$-p_t y^t$, we have $p_t = 0$. The final term in~\Eq{18'}
is therefore $p_{t-1} (y^{t+1} - y^{t-1})$. Hence $f(y)$ has the factor $y^2-1$. The
quotient $q(y) \equiv f(y)/(y^2-1)$ has the property that
\[
y^{n-2} q \left(\frac{1}{y}\right) \equiv q(y).
\]
Comparing this with~\Eq{17} as before, we see that $q(y)=0$ is of the form~\Eq{18}
where now $2t = n -2$. Hence, at least after removing one or both
of the factors $y±1$, \emph{any reciprocal equation may be given the form~\Eq{18}}.

The method by which~\Eq{13} was reduced to a cubic equation may be
used to reduce any equation~\Eq{18} to an equation in~$x$ of half the degree.
First, we divide the terms of~\Eq{18} by~$y^t$ and obtain
\index{Cubic equation}% [** PP: Original entry points to Folio 40]
\[
\left(y^t + \frac{1}{y^t}\right)
  + c_1 \left(y^{t-1} + \frac{1}{y^{t-1}}\right) + \dotsb
  + c_{t-1} \left(y + \frac{1}{y}\right) + c_t = 0.
\]
Next, we perform the substitution~\Eq{14} by either of the following methods:
We may make use of the relation
\[
y^k + \frac{1}{y^k}
  = x \left(y^{k-1} + \frac{1}{y^{k-1}}\right)
    - \left(y^{k-2} + \frac{1}{y^{k-2}}\right)
\]
to compute the values of $y^k + 1/y^k$ in terms of~$x$, starting with the special
%% -----File: 045.png---Folio 39-------
cases \Eq{14} and~\Eq{15}. For example,
\[
\begin{split}
y^4 + \frac{1}{y^4}
  &= x \left(y^3 + \frac{1}{y^3}\right)
     - \left(y^2 + \frac{1}{y^2}\right) \\
  &= x (x^3-3x) - (x^2-2) = x^4 - 4x^2 + 2. % [** PP: Added .]
\end{split}
\]
Or we may employ the explicit formula~\Eq{19} of~§107 for the sum $y^k + 1/y^k$
of the $k$th~powers of the roots~$y$ and~$1/y$ of $y^2 - xy + 1 = 0$.


% [** PP: No separate ToC entry]
\Section[Regular Polygon of $9$~Sides]
{37.}{Regular Polygon of $9$~Sides and Roots of Unity.} If
\index{Regular polygon!09@{$9$~sides}}%
\index{Roots of unity}%
\[
R = \cos\frac{2\pi}{9} + i \sin\frac{ 2\pi}{9},
\]
the powers $R$, $R^2$, $R^4$, $R^5$, $R^7$, $R^8$, are the primitive ninth roots of unity~(§11).
They are therefore the roots of
\[
\frac{y^9 - 1}{y^3 - 1} = y^6 + y^3 + 1 = 0.
\Tag{19}
\]
Dividing the terms of this reciprocal equation by~$y^3$ and applying the second
relation~\Eq{15}, we obtain our former cubic equation~\Eq{11}.


\begin{Exercises}{Page40}

\begin{Problems}
\item[1.] Show by~\Eq{16} that the roots of~\Eq{12} are $2\cos 2\pi/7$, $2\cos 4\pi/7$, $2\cos 6\pi/7$.

\begin{minipage}[b]{\linewidth-2.25in}
\item[2.] The imaginary fifth roots of unity satisfy
$y^4 + y^3 + y^2 + y + 1 = 0$, which by the substitution~\Eq{14} becomes
$x^2 + x - 1 = 0$.   It has the root
\[
R + \frac{1}{R} = 2 \cos\frac{2\pi}{5} = \frac{1}{2}(\sqrt{5}-1).
\]
In a circle of radius unity and center~$O$ draw two perpendicular
diameters $AOA'$, $BOB'$. With the middle
point~$M$ of~$OA'$ as center and radius~$MB$ draw a circle
cutting~$OA$ at~$C$ (Fig.~10). Show that $OC$ and~$BC$
are the sides~$s_{10}$ and~$s_5$ of the inscribed regular decagon
and pentagon respectively. Hints:
\end{minipage}%
% Illustration: \textsc{Fig}. 10
\hfill\Input{045a}
\index{Regular!decagon}%
\index{Regular!pentagon}%
\begin{align*}
MB &= \tfrac{1}{2}\sqrt{5},\qquad
OC  = \tfrac{1}{2}(\sqrt{5} - 1),\qquad
BC  = \sqrt{1+OC^2} = \tfrac{1}{2}\sqrt{10 - 2\sqrt{5}}, \\
s_{10}  &=  2 \sin 18° = 2 \cos\frac{2\pi}{5} = OC, \\
{s_5}^2 &= (2 \sin 36°)^2 = 2\left(1 - \cos\frac{2\pi}{5}\right)
  = \frac{1}{4}(10 - 2\sqrt{5}), \qquad s_5 = BC.
\end{align*}

%% -----File: 046.png---Folio 40-------

\item[3.] If $R$ is a root of~\Eq{19} verify as at the end of~§35 that $R+R^8$, $R^2+R^7$, and $R^4+R^5$
are the roots of~\Eq{11}.

\item[4.] Hence show that the roots of~\Eq{11} are $2\cos 2\pi/9$, $2\cos 4\pi/9$, $2\cos 8\pi/9$.

\item[5.] Reduce $y^{11} = 1$ to an equation of degree~$5$ in~$x$.

\item[6.] Solve $y^5 - 7y^4 + y^3 - y^2 + 7y - 1 = 0$ by radicals. [One root is~$1$.]

\item[7.] After finding so easily in \ChapRef{I} the trigonometric forms of the complex roots
of unity, why do we now go to so much additional trouble to find them algebraically?

\item[8.] Prove that every real root of $x^4 + ax^2 + b = 0$ can be constructed with ruler and
compasses, given lines of lengths $a$ and~$b$.

\item[9.] Show that the real roots of $x^3 - px - q = 0$ are the abscissas of the intersections
of the parabola $y = x^2$ and the circle through the origin with the center
$(\frac{1}{2}q, \frac{1}{2} + \frac{1}{2}p)$.
\end{Problems}

Prove that it is impossible, with ruler and compasses:
\begin{Problems}
\item[10.] To construct a straight line representing the distance from the circular base
of a hemisphere to the parallel plane which bisects the hemisphere.

\item[11.] To construct lines representing the lengths of the edges of an existing rectangular
parallelopiped having a diagonal of length~$5$, surface area~$24$, and volume~$1$, $2$, $3$, or~$5$.

\item[12.] To trisect an angle whose cosine is $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, $\frac{1}{8}$ or~$p/q$, where $p$ and~$q$ ($q>1$) are
integers without a common factor, and $q$ is not divisible by a cube.
\index{Trisection of angle}%
\end{Problems}

Prove algebraically that it is possible, with ruler and compasses:
\begin{Problems}
\item[13.] To trisect an angle whose cosine is $(4a^3 - 3ab^2)/b^3$, where the integer~$a$ is numerically
less than the integer~$b$; for example, $\cos^{-1} 11/16$ if $a = -1$, $b = 4$.

\item[14.] To construct the legs of a right triangle, given its area and hypotenuse.

\item[15.] To construct the third side of a triangle, given two sides and its area.

\item[16.] To locate the point~$P$ on the side $BC=1$ of a given square $ABCD$ such that
the straight line $AP$ cuts $DC$ produced at a point~$Q$ for which the
length of $PQ$ is a given
number~$g$. Show that $y=BP$ is a root of a reciprocal quartic equation, and solve it
when $g = 10$.
\end{Problems}
\end{Exercises}


% [** PP: No ToC entry]
\Section{38.}{The Periods of Roots of Unity} Before taking up the regular
\index{Roots of unity!periods of}%
polygon of $17$~sides, we first explain another method of finding the pairs
of imaginary seventh roots of unity $R$ and~$R^6$, $R^2$ and~$R^5$, $R^3$ and~$R^4$,
employed in~\Eq{16}. To this end we seek a positive integer~$g$ such that
the six roots can be arranged in the order
\[
R,\quad  R^g,\quad  R^{g^2},\quad  R^{g^3},\quad  R^{g^4},\quad  R^{g^5},
\Tag{20}
\]
where each term is the $g$th power of its predecessor. Trying $g = 2$, we find
that the fourth term would then be $R^8 = R$. Hence $g\neq 2$. Trying $g = 3$,
we obtain
\[
R,\quad  R^3,\quad  R^2,\quad  R^6,\quad  R^4,\quad  R^5,
\Tag{21}
\]
where each term is the cube of its predecessor.

%% -----File: 047.png---Folio 41-------

To define three \emph{periods}, each of two terms,
\[
R   + R^6,\qquad
R^2 + R^5,\qquad
R^3 + R^4,
\Tag{16'}
\]
we select the first term~$R$ of~\Eq{21} and the third term $R^6$ after it and add
them, then the second term $R^3$ and the third term $R^4$ after it, and finally
$R^2$ and the third term $R^5$ after it.

We may also define two periods, each of three terms,
\[
z_1 = R   + R^2 + R^4,\qquad
z_2 = R^3 + R^6 + R^5,
\]
by taking alternate terms in~\Eq{21}.

\begin{Remark}
Since $z_1 + z_2 = -1$, $z_1 z_2 = 3 + R + \dotsb + R^6 = 2$, $z_1$ and $z_2$ are the roots of $z^2 + z + 2 = 0$.
Then $R$, $R^2$, $R^4$ are the roots of $w^3 - z_1w^2 + z_2w - 1 = 0$.
\end{Remark}


% [** PP: No separate ToC entry]
\Section{39.}{Regular Polygon of $17$~Sides} Let $R$ be a root $\neq 1$ of $x^{17} = 1$.
Then
\index{Regular polygon!17@{$17$~sides}|(}%
\[
\frac{R^{17} - 1}{R - 1} = R^{16} + R^{15} + \dotsb + R + 1 = 0.
\]
As in §38, we may take $g=3$ and arrange the roots $R, \dotsc, R^{16}$ so that
each is the cube of its predecessor:
\[
R,\      R^3,\    R^9,\    R^{10},\
R^{13},\ R^5,\    R^{15},\ R^{11},\
R^{16},\ R^{14},\ R^8,\    R^7,\
R^4,\    R^{12},\ R^2,\    R^6.
\]

Taking alternate terms, we get the two periods, each of eight terms,
\begin{align*}
y_1 &= R   + R^9    + R^{13} + R^{15} + R^{16} + R^8 + R^4    + R^2, \\
y_2 &= R^3 + R^{10} + R^5    + R^{11} + R^{14} + R^7 + R^{12} + R^6.
\end{align*}
Hence $y_1 + y_2 = -1$. We find that $y_1 y_2 = 4(R + \dotsb + R^{16}) = -4$. Thus
\[
y_1,\ y_2 \quad\text{satisfy}\quad y^2 + y - 4 = 0.
\Tag{22}
\]

Taking alternate terms in~$y_1$, we obtain the two periods
\[
z_1 = R   + R^{13} + R^{16} + R^4, \qquad
z_2 = R^9 + R^{15} + R^8    + R^2.
\]
Taking alternate terms in~$y_2$, we get the two periods
\[
w_1 = R^3    + R^5    + R^{14} + R^{12}, \qquad
w_2 = R^{10} + R^{11} + R^7    + R^6.
\]
Thus $z_1 + z_2 = y_1$, $w_1 + w_2 = y_2$. We find that $z_1 z_2 = w_1 w_2 = -1$. Hence
\begin{align*}
z_1,\ z_2 &\quad\text{satisfy}\quad z^2 - y_1 z - 1 = 0,
\Tag{23} \\
w_1,\ w_2 &\quad\text{satisfy}\quad w^2 - y_2 w - 1 = 0.
\Tag{24}
\end{align*}

%% -----File: 048.png---Folio 42-------

Taking alternate terms in~$z_1$, we obtain the periods
\[
v_1 = R + R^{16}, \qquad v_2 = R^{13} + R^4.
\]
Now, $v_1 + v_2 = z_1$, $v_1v_2 = w_1$. Hence
\begin{align*}
v_1,\ v_2 &\quad\text{satisfy}\quad  v^2 - z_1v + w_1 = 0,
\Tag{25} \\
R,\ R^{16} &\quad\text{satisfy}\quad \rho^2 - v_1\rho + 1 = 0.
\Tag{26}
\end{align*}

Hence we can find $R$ by solving a series of quadratic equations. Which
of the sixteen values of~$R$ we shall thus obtain depends upon which root
of~\Eq{22} is called $y_1$ and which~$y_2$, and similarly in \Eq{23}--\Eq{26}. We shall now
show what choice is to be made in each such case in order that we shall
finally get the value of the particular root
\[
R = \cos\frac{2\pi}{17} + i \sin\frac{2\pi}{17}.
\]
Then
\begin{alignat*}{4}
\frac{1}{R}
    &= \cos \frac{2\pi}{17} - i \sin \frac{2\pi}{17}, &\qquad
v_1 &= R &+{}& \frac{1}{R} &&= 2 \cos \frac{2\pi}{17}, \\
%
R^4 &= \cos \frac{8\pi}{17} + i \sin \frac{8\pi}{17}, &
v_2 &= R^4 &+{}& \frac{1}{R^4} &&= 2 \cos \frac{8\pi}{17}.
\end{alignat*}
Hence $v_1 > v_2 > 0$, and therefore $z_1 = v_1 + v_2 > 0$. Similarly,
\begin{align*}
w_1 &= R^3 + \frac{1}{R^3} + R^5 + \frac{1}{R^5}
     = 2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17}
     = 2 \cos \frac{6\pi}{17} - 2 \cos \frac{7\pi}{17} > 0, \\
%
y_2 &= 2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17}
     + 2 \cos \frac{12\pi}{17} + 2 \cos \frac{14\pi}{17} < 0,
\end{align*}
since only the first cosine in $y_2$ is positive and it is numerically less than
the third. But $y_1 y_2 = -4$. Hence $y_1>0$. Thus \Eq{22}--\Eq{24} give
\begin{align*}
y_1 &= \tfrac{1}{2}( \sqrt{17}-1), &
y_2 &= \tfrac{1}{2}(-\sqrt{17}-1), \\
%
z_1 &= \tfrac{1}{2}y_1 + \sqrt{1 + \tfrac{1}{4}y_1^2}, &
w_1 &= \tfrac{1}{2}y_2 + \sqrt{1 + \tfrac{1}{4}y_2^2}.
\end{align*}

We may readily construct segments of these lengths. Evidently
$\sqrt{17}$ is the length of the hypotenuse of a right triangle whose legs are of
lengths $1$ and~$4$, while for the radical in~$z_1$ we employ legs of lengths $1$
and~$\frac{1}{2}y_1$. We thus obtain segments representing the coefficients of the
%% -----File: 049.png---Folio 43-------
quadratic equation~\Eq{25}. Its roots may be constructed as in~§29. The
larger root is
\[
v_1 = 2 \cos\frac{2\pi}{17}.
\]
Hence we can construct angle $2\pi/17$ with ruler and compasses, and therefore
a regular polygon of $17$~sides.


% [** PP: No ToC entry]
\Section{40.}{Construction of a Regular Polygon of $17$~Sides} In a circle of
radius unity, construct two perpendicular
diameters $AB$, $CD$,
and draw tangents at $A$, $D$,
which intersect at~$S$ (Fig.~11).
Find the point~$E$ in~$AS$ for which
$AE = \frac{1}{4} AS$, by means of two bisections.
Then
\[
AE = \tfrac{1}{4}, \qquad
OE = \tfrac{1}{4} \sqrt{17}.
\]
%[Illustration: \textsc{Fig.} 11]
\begin{figure*}[b]
\begin{center}
\Input{049a}
\end{center}
\end{figure*}

\noindent Let the circle with center~$E$
and radius~$OE$ cut~$AS$ at~$F$ and~$F'$.
Then
\begin{align*}
AF  &= EF  - EA = OE - \tfrac{1}{4} =  \tfrac{1}{2} y_1, \\
AF' &= EF' + EA = OE + \tfrac{1}{4} = -\tfrac{1}{2} y_2, \\
OF  &= \sqrt{OA^2 + AF^2} = \sqrt{1 + \tfrac{1}{4} y_1^2}, \qquad
OF' = \sqrt{1 + \tfrac{1}{4} y_2^2}.
\end{align*}
Let the circle with center~$F$ and radius~$FO$ cut~$AS$ at~$H$, outside of~$F'F$;
that with center~$F'$ and radius~$F'O$ cut~$AS$ at~$H'$ between $F'$ and~$F$. Then
\begin{align*}
AH &= AF + FH = AF + OF
    = \tfrac{1}{2} y_1 + \sqrt{1 + \tfrac{1}{4} y_1^2} = z_1, \\
AH' &= F'H' - F'A = OF'- AF'= w_1.
\end{align*}

It remains to construct the roots of equation~\Eq{25}. This will be done
as in~§29. Draw $HTQ$ parallel to $AO$ and intersecting $OC$ produced at~$T$.
Make $TQ = AH'$. Draw a circle having as diameter the line $BQ$
joining $B = (0,1)$ with $Q = (z_1, w_1)$. The abscissas $ON$ and $OM$ of the intersections
of this circle with the $x$-axis $OT$ are the roots of~\Eq{25}. Hence
the larger root~$v_1$ is $OM = 2 \cos(2\pi/17)$.

%% -----File: 050.png---Folio 44-------

Let the perpendicular bisector $LP$ of $OM$ cut the initial circle of unit
radius at~$P$. Then
\[
\cos LOP = OL = \cos\frac{2\pi}{17},\qquad LOP=\frac{2\pi}{17}.
\]

Hence the chord~$CP$ is a side of the inscribed regular polygon of $17$~sides,
constructed with ruler and compasses.
\index{Regular polygon!17@{$17$~sides}|)}%


% [** PP: No separate ToC entry]
\Section{41.}{Regular Polygon of $n$ Sides} If $n$ be a prime such that $n-1$ is
\index{Regular polygon!n@{$n$~sides}}%
\index{Roots of unity}%
a power $2^h$ of~$2$ (as is the case when $n = 3$, $5$, $17$), the $n-1$ imaginary $n$th
roots of unity can be separated into 2~sets each of $2^{h-1}$ roots, each of these
sets subdivided into 2~sets each of $2^{h-2}$ roots, etc., until we reach the pairs
$R$, $1/R$ and $R^2$, $1/R^2$, etc., and in fact\footnote
  {See the author's article ``Constructions with ruler and compasses; regular polygons,''
  in \textit{Monographs on Topics of Modern Mathematics}, Longmans, Green and Co.,
1911, p.~374.}
in such a manner that we have a
series of quadratic equations, the coefficients of any one of which depend
only upon the roots of quadratic equations preceding it in the series.
Note that this was the case for $n = 17$ and for $n = 5$. It is in this manner
that it can be proved that the roots of $x^n = 1$ can be found in terms of
square roots, so that a regular polygon of $n$~sides can be inscribed by ruler
and compasses, provided $n$ be a prime of the form $2^h + 1$.

If $n$ be a product of distinct primes of this form, or $2^k$ times such a
product (for example, $n = 15$, $30$ or~$6$), or if $n = 2^m$ ($m > 1$), it follows readily
(see Ex.~1 below) that we can inscribe with ruler and compasses a regular
polygon of $n$~sides. But this is impossible for all other values of~$n$.


\begin{Exercises}{Page44}

\begin{Problems}
\item[1.] If $a$ and~$b$ are relatively prime numbers, so that their greatest common divisor
is unity, we can find integers $c$ and~$d$ such that $ac + bd = 1$. Show that, if regular polygons
of $a$ and~$b$ sides can be constructed and hence angles $2\pi/a$ and $2\pi/b$, a regular
polygon of $a·b$ sides can be derived.

\item[2.] If $p = 2^h + 1$ is a prime, $h$ is a power of~$2$. For $h = 2^0$, $2^1$, $2^2$, $2^3$, the values of~$p$
are $3$, $5$, $17$, $257$ and are primes. [Show that $h$ cannot have an odd factor other than
unity.]

\item[3.] For $13$th roots of unity find the least~$g$~(§38), write out the three periods each
of four terms, and find the cubic equation having them as roots.
\index{Cubic equation}%

\item[4.] For the primitive ninth roots of unity find the least~$g$ and write out the three
periods each of two terms.
\end{Problems}

Solve the following reciprocal equations:
\index{Reciprocal equation}%
\begin{Problems}[2]
\item[5.] $y^4 + 4y^3 - 3y^2 + 4y + 1 = 0$.

\item[6.] $y^5 - 4y^4 + y^3 + y^2 - 4y + 1 = 0$.

\ResetCols{2}

\item[7.] $2y^6 - 5y^5 + 4y^4 - 4y^2 + 5y - 2 = 0$.

\item[\qquad8.] $y^5 + 1 = 31(y + 1)^5$.
\end{Problems}
\end{Exercises}
\index{Geometrical!construction|)}% [** PP: Using index subitem]

%% -----File: 051.png---Folio 45-------


\Chapter[Cubic and Quartic Equations]
{IV}{Solution of Cubic and Quartic Equations; Their Discriminants}
\index{Cubic equation}%

% [** PP: No ToC entry]
\Section{42.}{Reduced Cubic Equation} If, in the general cubic equation
\index{Cubic equation!reduced}%
\[
x^3 + bx^2 + cx + d = 0,
\Tag{1}
\]
we set $x = y-b/3$, we obtain the \emph{reduced cubic equation}
\[
y^3 + py + q = 0,
\Tag{2}
\]
lacking the square of the unknown~$y$, where
\[
p = c - \frac{b^2}{3}, \qquad
q = d - \frac{bc}{3} + \frac{2b^3}{27}.
\Tag{3}
\]

After finding the roots $y_1$, $y_2$, $y_3$ of~\Eq{2}, we shall know the roots of~\Eq{1}:
\[
x_1 = y_1 - \frac{b}{3}, \qquad
x_2 = y_2 - \frac{b}{3}, \qquad
x_3 = y_3 - \frac{b}{3}.
\Tag{4}
\]


\Section[Algebraic Solution of a Cubic]
{43.}{Algebraic Solution of the Reduced Cubic Equation.}   We shall
employ the method which is essentially the same as that given by Vieta
in~1591. We make the substitution
\[
y = z - \frac{p}{3z}
\Tag{5}
\]
in~\Eq{2} and obtain
\[
z^3 - \frac{p^3}{27z^3} + q = 0,
\]
since the terms in $z$ cancel, and likewise the terms in~$1/z$. Thus
\[
z^6 + qz^3 - \frac{p^3}{27} = 0.
\Tag{6}
\]
Solving this as a quadratic equation for~$z^3$, we obtain
\[
z^3 = -\frac{q}{2} ±\sqrt{R},\qquad
R = \left(\frac{p}{3}\right)^3 + \left(\frac{q}{2}\right)^2.
\Tag{7}
\]

%% -----File: 052.png---Folio 46-------

By §8, any number has three cube roots, two of which are the products
of the remaining one by the imaginary cube roots of unity:
\[
\omega   = -\tfrac{1}{2} + \tfrac{1}{2} \sqrt{3}i,\qquad
\omega^2 = -\tfrac{1}{2} - \tfrac{1}{2} \sqrt{3}i.
\Tag{8}
\]
We can choose particular cube roots
\[
A = \sqrt[3]{-\frac{q}{2} + \sqrt{R}},\qquad
B = \sqrt[3]{-\frac{q}{2} - \sqrt{R}},
\Tag{9}
\]
such that $AB = -p/3$, since the product of the numbers under the cube
root radicals is equal to~$(-p/3)^3$. Hence the six values of~$z$ are
\[
A,\quad \omega A,\quad \omega^2 A,\quad
B,\quad \omega B,\quad \omega^2 B.
\]
These can be paired so that the product of the two in each pair is~$-p/3$: %[** PP: Punctuation indistinct]
\[
AB = -\frac{p}{3},\qquad
\omega A·\omega^2 B = -\frac{p}{3},\qquad
\omega^2 A·\omega B = -\frac{p}{3}.
\]
Hence with any root~$z$ is paired a root equal to~$-p/(3z)$. By~\Eq{5}, the sum
of the two is a value of~$y$. Hence the \emph{three} values of~$y$ are
\[
y_1 = A + B,\qquad
y_2 = \omega A + \omega^2 B,\qquad
y_3 = \omega^2 A + \omega B.
\Tag{10}
\]

It is easy to verify that these numbers are actually roots of~\Eq{2}. For
example, since $\omega^3 = 1$, the cube of $y_2$ is
\[
A^3 + B^3 + 3\omega A^2 B + 3\omega^2 AB^2
  = -q - p(\omega A + \omega^2 B) = -q - py_2,
\]
by~\Eq{9} and $AB = -p/3$.

The numbers~\Eq{10} are known as \emph{Cardan's formulas} for the roots of a
\index{Cardan's formulas}%
reduced cubic equation~\Eq{2}. The expression $A + B$ for a root was first
published by Cardan in his \textit{Ars Magna} of~1545, although he had obtained
it from Tartaglia under promise of secrecy.

\begin{Example}
Solve $y^3 - 15y - 126 = 0$.
\end{Example}

\begin{Solution}
The substitution~\Eq{5} is here $y = z + 5/z$.  We get
\[
z^6 - 126z^3 + 125 = 0,\qquad
z^3 = 1 \text{ or } 125.
\]
The pairs of values of~$z$ whose product is $5$ are $1$ and~$5$, $\omega$ and $5\omega^2$, $\omega^2$ and $5\omega$. Their
sums $6$, $\omega + 5\omega^2$, and $\omega^2 + 5\omega$ give the three roots.
\end{Solution}

\begin{Exercises}{Page46}

Solve the equations:

\begin{Problems}[2]

\item[1.] $y^3 - 18y + 35 = 0$.

\item[2.] $x^3 + 6x^2 + 3x + 18 = 0$.

\ResetCols{2}

\item[3.] $y^3 - 2y + 4 = 0$.

\item[4.] $28x^3 + 9x^2 - 1 = 0$.
\end{Problems}
\end{Exercises}

%% -----File: 053.png---Folio 47-------

\Section{44.}{Discriminant} The product of the squares of the differences of
the roots of any equation in which the coefficient of the highest power of
the unknown is unity shall be called the \emph{discriminant} of the equation.
For the reduced cubic~\Eq{2}, the discriminant is
\index{Discriminant!of cubic}%
\[
(y_1 - y_2)^2 (y_1 - y_3)^2 (y_2 - y_3)^2 = -4p^3 - 27q^2,
\Tag{11}
\]
a result which should be memorized in view of its important applications.
It is proved by means of~\Eq{10} and $\omega^3 = 1$, $\omega^2 + \omega + 1 = 0$, as follows:
\begin{gather*}
y_1 - y_2 = (1-\omega)(A-\omega^2 B), \qquad
y_1 - y_3 = (1-\omega^2)(A-\omega B), \\
y_2 - y_3 = (\omega -\omega^2) (A-B), \\
(1-\omega)(1-\omega^2) = 3, \quad
\omega - \omega^2 = \sqrt{3}i.
\end{gather*}
Since $1$, $\omega$, $\omega^2$ are the cube roots of unity,
\[
(x-1)(x-\omega)(x-\omega^2) \equiv x^3 - 1,
\]
identically in~$x$. Taking $x = A/B$, we see that
\[
(A-B)(A-\omega B)(A-\omega^2 B) = A^3 - B^3 = 2 \sqrt{R},
\]
by~\Eq{9}. Hence
\[
(y_1-y_2)(y_1-y_3)(y_2-y_3) = 6\sqrt{3}\sqrt{R}i.
\]
Squaring, we get~\Eq{11}, since  $-108R = -4p^3 - 27q^2$ by~\Eq{7}. For later
use, we note that the discriminant of the reduced cubic is equal to $-108 R$.

\emph{The discriminant $\Delta$ of the general cubic~\Eq{1} is equal to the discriminant
of the corresponding reduced cubic~\Eq{2}.} For, by~\Eq{4},
\[
x_1 - x_2 = y_1 - y_2, \qquad
x_1 - x_3 = y_1 - y_3, \qquad
x_2 - x_3 = y_2 - y_3.
\]

Inserting in~\Eq{11} the values of $p$ and~$q$ given by~\Eq{3}, we get
\[
\Delta = 18bcd - 4b^3 d + b^2 c^2 - 4c^3 - 27d^2.
\Tag{12}
\]

\begin{Remark}
It is sometimes convenient to employ a cubic equation
\[
ax^3 + bx^2 + cx +d = 0 \quad   (a \neq 0),
\Tag{13}
\]
in which the coefficient of $x^3$ has not been made unity by division. The product~$P$
of the squares of the differences of its roots is evidently derived from~\Eq{12} by replacing
$b$, $c$, $d$ by $b/a$, $c/a$, $d/a$. Hence
\[
a^4 P = 18 abcd - 4b^3 d + b^2 c^2 - 4ac^3 - 27a^2 d^2.
\Tag{14}
\]
This expression (and not $P$ itself) is called the discriminant of~\Eq{13}.
\end{Remark}

%% -----File: 054.png---Folio 48-------


\Section[Number of Real Roots of a Cubic]
{45.}{Number of Real Roots of a Cubic Equation.}
\index{Cubic equation!number of real roots}% [** PP: Added ``of'']
\index{Number!of roots}%
\begin{Thm}
A cubic equation
with real coefficients has three distinct real roots if its discriminant~$\Delta$ is positive,
a single real root and two conjugate imaginary roots if $\Delta$ is negative, and at
least two equal real roots if $\Delta$ is zero.
\end{Thm}

If the roots $x_1$, $x_2$, $x_3$ are all real and distinct, the square of the difference
of any two is positive and hence $\Delta$ is positive.

If $x_1$ and $x_2$ are conjugate imaginaries and hence $x_3$ is real~(§21),
$(x_1 -x_2)^2$ is negative. Since $x_1 - x_3$ and $x_2 - x_3$ are conjugate imaginaries,
their product is positive. Hence $\Delta$ is negative.

If $x_1 = x_2$, $\Delta$ is zero. If $x_2$ were imaginary, its conjugate would be
equal to~$x_3$ by~§21, and $x_2$, $x_3$ would be the roots of a real quadratic
equation. The remaining factor $x - x_1$ of the cubic would have real
coefficients, whereas $x_1 = x_2$ is imaginary. Hence the equal roots must
be real.

Our theorem now follows from these three results by formal logic.
For example, if $\Delta$ is positive, the roots are all real and distinct, since
otherwise either two would be imaginary and $\Delta$ would be negative, or two
would be equal and $\Delta$ would be zero.


\begin{Exercises}{Page48}

Compute the discriminant and find the number of real roots of
\begin{Problems}[2]
\item[1.] $y^3 - 2y - 4 = 0$.

\item[2.] $y^3 - 15y + 4 = 0$.

\ResetCols{2}

\item[3.] $y^3 - 27y + 54 = 0$.

\item[4.] $x^3 + 4x^2 - 11x + 6 = 0$.

\ResetCols{1}

\item[5.] Show by means of~§21 that a double root of a real cubic is real.
\end{Problems}
\end{Exercises}


% [** PP: No ToC entry]
\Section{46.}{Irreducible Case} When the roots of a real cubic equation are
all real and distinct, the discriminant $\Delta$ is positive and $R = -\Delta/108$ is
negative, so that Cardan's formulas present the values of the roots in a
\index{Cardan's formulas}%
\index{Cube root}%
\index{Irreducible case}%
form involving cube roots of imaginaries. This is called the irreducible
case since it may be shown that a cube root of a general complex number
cannot be expressed in the form $a + bi$, where $a$ and $b$ involve only real
radicals.\footnote
  {Author's \textit{Elementary Theory of Equations,} pp.~35, 36.} % [** PP: Added . after pp]
While we cannot always find these cube roots algebraically,
we have learned how to find them trigonometrically~(§8).

\begin{Example}
Solve the cubic equation~\Eq{2} when $p = -12,\, q = -8\sqrt{2}$.
\end{Example}

\begin{Solution}
By~\Eq{7}, $R = -32$ Hence formulas~\Eq{9} become
\[
A = \sqrt[3]{4\sqrt{2} + 4\sqrt{2}i},\qquad
B = \sqrt[3]{4\sqrt{2} - 4\sqrt{2}i}.
\]
%% -----File: 055.png---Folio 49-------
The values of~$A$ were found in~§8. The values of~$B$ are evidently the conjugate imaginaries
of the values of~$A$. Hence the roots are
\[
4\cos{15°},\quad
4\cos{135°},\quad
4\cos{255°}.
\]
\end{Solution}


\begin{Exercises}{Page49}
\begin{Problems}[2]

\item[1.] Solve $y^3 -15y+4=0$.

\item[2.] Solve $y^3 -2y-1=0$.

\ResetCols{2}

\item[3.] Solve $y^3 -7y+7=0$.

\item[4.] Solve $x^3+ 3x^2 -2x-5=0$.

\ResetCols{2}

\item[5.] Solve $x^3 +x^2 -2x-1=0$.

\item[6.] Solve $x^3 +4x^2 -7=0$.

\end{Problems}
\end{Exercises}


\Section[Trigonometric Solution of a Cubic]
{47.}{Trigonometric Solution of a Cubic Equation with $\Delta>0$.} When
\index{Cubic equation!trigonometric solution}%
the roots of a real cubic equation are all real, i.e., if $R$ is negative, they
can be computed simultaneously by means of a table of cosines with much
less labor than required by Cardan's formulas. To this end we write
the trigonometric identity
\[
\cos 3A = 4\cos^3 A - 3\cos A
\]
in the form
\[
z^3 - \tfrac{3}{4}z - \tfrac{1}{4}\cos 3A = 0\qquad  (z = \cos A).
\]
In the given cubic $y^3 + py + q = 0$ take $y=nz$; then
\[
z^3 + \frac{p}{n^2}z + \frac{q}{n^3} = 0,
\]
which will be identical with the former equation in~$z$ if
\[
n = \sqrt{-\tfrac{4}{3}p},\quad
\cos{3A} = -\tfrac{1}{2}q ÷ \sqrt{-p^{3}/27}.
\]
Since $R = p^3/27 + q^2/4$ is negative, $p$ must be negative, so that $n$ is real
and the value of $\cos{3A}$ is real and numerically less than unity. Hence
we can find $3A$ from a table of cosines. The three values of $z$ are then
\[
\cos A,\qquad
\cos(A + 120°),\qquad
\cos(A+240°).
\]
Multiplying these by~$n$, we obtain the three roots~$y$ correct to a number
of decimal places which depends on the tables used.


\begin{Exercises}{}

\begin{Problems}
\item[1.] For $y^3 - 2y - 1 =0$, show that $n^2 =8/3$, $\cos{3A} = \sqrt{27/32}$, $3A=23° 17' 0''$,
$\cos A = 0.99084$, $\cos (A+120°) = -0.61237$, $\cos (A+240°) = -0.37847$, and that the
roots~$y$ are $1.61804$, $-1$, $-0.61804$.

\item[2.] Solve Exs.\ 1, 3, 4, 5, 6 of~§46 by trigonometry.
\end{Problems}
\end{Exercises}

%% -----File: 056.png---Folio 50-------


% [** PP: ToC reads ``Ferrari's and Descartes' Solution of a Quartic'']
\Section{48.}{Ferrari's Solution of the Quartic Equation} The general quartic
equation
\index{Quartic equation|(}%
\[
x^4 +bx^3 +cx^2 +dx+e=0,
\Tag{15}
\]
or equation of degree four, becomes after transposition of terms
\[
x^4 + bx^3 = -cx^2 - dx - e.
\]
The left member contains two of the terms of the square of $x^2 +\tfrac{1}{2}bx$.
Hence by completing the square, we get
\[
(x^2 + \tfrac{1}{2}bx)^2 = (\tfrac{1}{4}b^2 -c)x^2 - dx - e.
\]

Adding $(x^2+ \frac{1}{2}bx)y+ \frac{1}{4}y^2$ to each member, we obtain
\[
(x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y)^2
  = (\tfrac{1}{4}b^2 - c + y)x^2
  + (\tfrac{1}{2}by - d)x
  + \tfrac{1}{4}y^2 - e.
\Tag{16}
\]
The second member is a perfect square of a linear function of $x$ if and
only if its discriminant is zero~(§12):
\[
(\tfrac{1}{2}by - d)^2 - 4(\tfrac{1}{4}b^2 - c + y)(\tfrac{1}{4}y^2 - e) = 0,
\]
which may be written in the form
\[
y^3 - cy^2 + (bd - 4e)y - b^{2}e + 4ce - d^2 = 0.
\Tag{17}
\]

Choose any root~$y$ of this \emph{resolvent cubic equation~\Eq{17}}. Then the
right member of~\Eq{16} is the square of a linear function, say $mx+n$. Thus
\index{Resolvent cubic}%
\[
x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y =  mx+n \quad\text{or}\quad
x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y = -mx-n.
\Tag{18}
\]
The roots of these quadratic equations are the four roots of~\Eq{16} and
hence of the equivalent equation~\Eq{15}. This method of solution is due
to Ferrari (1522--1565).

\begin{Example}
Solve $x^4 +2x^3 -12x^2 -10x+3 = 0$.
\end{Example}

\begin{Solution}
Here $b=2$, $c=-12$, $d=-10$, $e = 3$. Hence~\Eq{17} becomes
\[
y^3 + 12y^2 - 32y - 256 = 0,
\]
which by Ex.~2 of~§24 has the root $y=-4$. Our quartic may be written in the form
\[
(x^2 + x)^2 = 13x^2 + 10x-3.
\]
Adding $(x^2 +x)(-4)+4$ to each member, we get
\begin{gather*}
(x^2 + x - 2)^2 = 9x^2 + 6x + 1 = (3x + 1)^2, \\
x^2 +  x - 2 = ±(3x + 1),\qquad
x^2 - 2x - 3 = 0 \text{ or } x^2 + 4x - 1 = 0,
\end{gather*}
whose roots are $3$, $-1$, $-2± \sqrt{5}$. As a check, note that the sum of the roots is~$-2$.
\end{Solution}

%% -----File: 057.png---Folio 51-------


\begin{Exercises}{Page51}

\begin{Problems}
\item[1.] Solve $x^4 - 8x^3 + 9x^2 + 8x - 10 = 0$. Note that~\Eq{17} is $(y - 9) (y^2 - 24) = 0$.

\item[2.] Solve $x^4 - 2x^3 - 7x^2 + 8x + 12 = 0$. Since the right member of~\Eq{16} is
$(8 + y) (x^2 - x) + \frac{1}{4} y^2 - 12$, use $y = -8$.

\item[3.] Solve $x^4 - 3x^2 + 6x - 2 = 0$.

\item[4.] Solve $x^4 - 2x^2 - 8x - 3 = 0$.

\item[5.] Solve $x^4 - 10x^2 - 20x - 16 = 0$.

\end{Problems}
\end{Exercises}


\Section[Resolvent Cubic]
{49.}{Roots of the Resolvent Cubic Equation.} Let $y_1$ be the root~$y$
which was employed in~§48. Let $x_1$ and~$x_2$ be the roots of the first
quadratic equation~\Eq{18}, and $x_3$ and~$x_4$ the roots of the second. Then
\index{Resolvent cubic}%
\[
x_1 x_2 = \tfrac{1}{2} y_1 - n,\qquad
x_3 x_4 = \tfrac{1}{2} y_1 + n,\qquad
x_1 x_2 + x_3 x_4 = y_1.
\]
If, instead of $y_1$, another root $y_2$ or~$y_3$ of the resolvent cubic~\Eq{17} had been
employed in~§48, quadratic equations different from~\Eq{18} would have
been obtained, such, however, that their four roots are $x_1$, $x_2$, $x_3$, $x_4$, paired
in a new manner. The root which is paired with $x_1$ is $x_2$ or $x_3$ or~$x_4$. It
is now plausible that the values of the three $y$'s are
\[
y_1 = x_1 x_2 + x_3 x_4,\qquad
y_2 = x_1 x_3 + x_2 x_4,\qquad
y_3 = x_1 x_4 + x_2 x_3.
\Tag{19}
\]

To give a more formal proof that the $y$'s given by~\Eq{19} are the roots
of~\Eq{17}, we employ~(§20)
\begin{gather*}
x_1 + x_2 + x_3 + x_4 = -b,\qquad
x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 = -d, \\
x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 = c,\qquad
 x_1 x_2 x_3 x_4 = e.
\end{gather*}

From these four relations we conclude that
\begin{gather*}
y_1 + y_2 + y_3 = c,\\
\begin{split}
y_1 y_2 + y_1 y_3 + y_2 y_3
  &= (x_1 + x_2 + x_3 + x_4) (x_1 x_2 x_3 + \dotsb + x_2 x_3 x_4)
   - 4x_1 x_2 x_3 x_4 \\
&= bd - 4e,
\end{split} \\
\begin{split}
y_1 y_2 y_3
  &= (x_1 x_2 x_3 + \dotsb )^2
   + x_1 x_2 x_3 x_4 \bigl\{(x_1 + \dotsb )^2 - 4( x_1 x_2 + \dotsb) \bigr\} \\
  &= d^2 + e ( b^2 - 4c ).
\end{split}
\end{gather*}
Hence (§20) $y_1$, $y_2$, $y_3$ are the roots of the cubic equation~\Eq{17}.


% [** PP: No ToC entry]
\Section{50.}{Discriminant} The discriminant $\Delta$ of the quartic equation~\Eq{15}
\index{Discriminant!of quartic}%
is defined to be the product of the squares of the differences of its roots:
\[
\Delta = ( x_1 - x_2 )^2 ( x_1 - x_3 )^2 ( x_1 - x_4 )^2
         ( x_2 - x_3 )^2 ( x_2 - x_4 )^2 ( x_3 - x_4 )^2.
\]

%% -----File: 058.png---Folio 52-------

The fact that $\Delta$ is equal to the discriminant of the resolvent cubic
equation~\Eq{17} follows at once from~\Eq{19}, by which
\begin{align*}
y_1 - y_2 &= (x_1-x_4)(x_2-x_3),\qquad
  y_1-y_3 = (x_1-x_3)(x_2-x_4), \\
%
y_2 - y_3 &= (x_1-x_2)(x_3-x_4),\qquad
  (y_1-y_2)^2 (y_1-y_3)^2 (y_2- y_3)^2 = \Delta.
\end{align*}
Hence (§44) $\Delta$ is equal to the discriminant $-4p^3 - 27q^2$ of the reduced
cubic $Y^3 + pY + q = 0$, obtained from~\Eq{17} by setting $y = Y + c/3$. Thus
\[
p = bd - 4e - \tfrac{1}{3} c^2,\qquad
q = -b^2 e + \tfrac{1}{3} bcd + \tfrac{8}{3} ce - d^2 - \tfrac{2}{27} c^3.
\Tag{20}
\]

\begin{Theorem}
The discriminant of any quartic equation~\Eq{15} is equal to
the discriminant of its resolvent cubic equation and therefore is equal to the
discriminant $-4p^3 - 27q^2$ of the corresponding reduced cubic $Y^3 + pY + q = 0$,
whose coefficients have the values~\Eq{20}.
\end{Theorem}


\begin{Exercises}{}
\index{Number!of roots}%

\begin{Problems}
\item[1.] Find the discriminant of $x^4 - 3x^3 + x^2 + 3x - 2 = 0$ and show that the equation
has a multiple root.

\item[2.] Show by its discriminant that $x^4 - 8x^3 + 22x^2 - 24x + 9 = 0$ has a multiple root.

\item[3.] If a real quartic equation has two pairs of conjugate imaginary roots, show that
its discriminant~$\Delta$ is positive. Hence prove that, if $\Delta<0$, there are exactly two real
roots.

\item[4.] Hence show that $x^4 - 3x^3 + 3x^2 - 3x + 2 = 0$ has two real and two imaginary roots.
\end{Problems}
\end{Exercises}


% [** PP: No separate ToC entry]
\Section{51.}{Descartes' Solution of the Quartic Equation} Replacing $x$ by
$z - b/4$ in the general quartic~\Eq{15}, we obtain the \emph{reduced} quartic equation
\[
z^4 + qz^2 + rz + s = 0,
\Tag{21}
\]
lacking the term with~$z^3$. We shall prove that we can express the left
member of~\Eq{21} as the product of two quadratic factors
\[
(z^2 + 2kz + l)(z^2 - 2kz + m)
  = z^4 + (l + m - 4k^2)z^2 + 2k(m - l)z + lm.
\]
The conditions are
\[
l + m - 4k^2 = q,\qquad
2k(m-l)=r,\qquad
lm = s.
\]
If $k\neq0$, the first two give
\[
2l = q + 4k^2 - \frac{r}{2k},\qquad
2m = q + 4k^2 + \frac{r}{2k}.
\]
Inserting these values in $2l·2m = 4s$, we obtain
\[
64k^6 + 32qk^4 + 4(q^2 - 4s)k^2 - r^2 = 0.
\Tag{22}
\]
%% -----File: 059.png---Folio 53-------
The latter may be solved as a cubic equation for~$k^2$. Any root $k^2 \neq 0$
gives a pair of quadratic factors of~\Eq{21}:
\[
z^2 ± 2kz + \tfrac{1}{2}q + 2k^2 \mp \frac{r}{4k}.
\Tag{23}
\]
The four roots of these two quadratic functions are the four roots of~\Eq{21}.
This method of Descartes (1596--1650) therefore succeeds unless every
root of~\Eq{22} is zero, whence $q = s = r = 0$, so that \Eq{12} is the trivial equation
$z^4 = 0$.

\begin{Remark}
For example, consider $z^4 - 3z^2 + 6z - 2 = 0$. Then \Eq{22} becomes
\[
64k^6 - 3·32k^4 + 4·17k^2 - 36 = 0.
\]
The value $k^2 = 1$ gives the factors $z^2 + 2z - 1$, $z^2 - 2z + 2$. Equating these to zero, we
find the four roots $-1 ± \sqrt{2}$, $1± \sqrt{-1}$.
\end{Remark}


% [** PP: No ToC entry]
\Section{52.}{Symmetrical Form of Descartes' Solution} To obtain this symmetrical
form, we use all three roots $k_1^2$, $k_2^2$, $k_3^2$ of~\Eq{22}. Then
\[
k_1^2 + k_2^2 + k_3^2 = -\tfrac{1}{2}q,\qquad
k_1^2 k_2^2 k_3^2 = \frac{r^2}{64}.
\]
It is at our choice as to which square root of~$k_1^2$ is denoted by~$+k_1$ and
which by~$-k_1$, and likewise as to $±k_2$, $±k_3$. For our purposes any
choice of these signs is suitable provided the choice give
\[
k_1 k_2 k_3 = -\frac{r}{8}.
\Tag{24}
\]

Let $k_1 \neq 0$. The quadratic function~\Eq{23} is zero for $k = k_1$ if
\[
(z ± k_1)^2 = -\frac{q}{2} - k_1^2 ± \frac{r}{4k_1}
  = k_2^2 + k_3^2 \mp \frac{8k_1k_2k_3}{4k_1}
  = (k_2 \mp k_3)^2.
\]
Hence the four roots of the quartic equation~\Eq{21} are
\[
 k_1 + k_2 + k_3,\qquad
 k_1 - k_2 - k_3,\qquad
-k_1 + k_2 - k_3,\qquad
-k_1 - k_2 + k_3.
\Tag{25}
\]


\begin{Exercises}{}

\begin{Problems}
\item[1.]  Solve Exs.~4, 5 of~§48 by the method of Descartes.

\item[2.]  By writing $y_1$, $y_2$, $y_3$ for the roots $k_1^2$, $k_2^2$, $k_3^2$ of
\[
64y^3 + 32qy^2 + 4(q^2 - 4s)y - r^2 = 0,
\Tag{26}
\]
show that the four roots of~\Eq{21} are the values of
\[
z = \sqrt{y_1} + \sqrt{y_2} + \sqrt{y_3}
\Tag{27}
\]
%% -----File: 060.png---Folio 54-------
for all combinations of the square roots for which
\[
\sqrt{y_1}·\sqrt{y_2}·\sqrt{y_3} = -\frac{r}{8}.
\Tag{28}
\]

\item[3.] Euler (1707--1783) solved~\Eq{21} by assuming that it has a root of the form~\Eq{27}.
Square~\Eq{27}, transpose the terms free of radicals, square again, replace the last factor
of $8\sqrt{y_1 y_2 y_3}\,(\sqrt{y_1} + \sqrt{y_2} + \sqrt{y_3})$ by~$z$, and identify the resulting quartic in~$z$ with~\Eq{21}.
Show that $y_1$, $y_2$, $y_3$ are the roots of~\Eq{26} and that relation~\Eq{28} holds.

\item[4.] Find the six differences of the roots~\Eq{25} and verify that the discriminant $\Delta$ of~\Eq{21}
is equal to the quotient of the discriminant of~\Eq{26} by~$4^6$.

\item[5.] In the theory of the inflexion points of a plane cubic curve there occurs the
equation
\[
z^4 - Sz^2 - \tfrac{4}{3}Tz - \tfrac{1}{12}S^2 = 0.
\]
Show that~\Eq{26} now becomes
\[
\left(y - \frac{S}{6}\right)^3 = C,\qquad
C \equiv \left(\frac{T}{6}\right)^2 - \left(\frac{S}{6}\right)^3,
\]
and that the roots of the quartic equation are
\[
±\sqrt{\tfrac{1}{6}S + \sqrt[3]{C}}
±\sqrt{\tfrac{1}{6}S + \omega\sqrt[3]{C}}
±\sqrt{\tfrac{1}{6}S + \omega^{2}\sqrt[3]C},
\]
where $\omega$ is an imaginary cube root of unity and the signs are to be chosen so that the product of the three summands is equal to~$+\tfrac{1}{6}T$.
\end{Problems}
\end{Exercises}


\begin{Exercises}[MISCELLANEOUS~]{Page54}

\begin{Problems}
\item[1.] Find the coordinates of the single real point of intersection of the parabola
$y = x^2$ and the hyperbola $xy - 4x + y + 6 = 0$.

\item[2.] Show that the abscissas of the points of intersection of $y=x^2$ and
$ax^2 - xy + y^2 - x - (a+5)y - 6 = 0$
are the roots of $x^4 - x^3 - 5x^2 - x - 6 = 0$. Compute the discriminant
of the latter and show that only two of the four points of intersection are real.

\item[3.] Find the coordinates of the two real points in Ex.~2.

\item[4.] A right prism of height~$h$ has a square base whose side is~$b$ and whose diagonal
is therefore $b\sqrt{2}$. If $v$ denotes the volume and $d$ a diagonal of the prism, $v = hb^2$ and
$d^2 = h^2 + (b\sqrt{2})^2$. Multiply the last equation by~$h$ and replace $hb^2$ by~$v$. Hence
$h^3 - d^2h + 2v = 0$.
Its discriminant is zero if $d = 3\sqrt{3}$, $v = 27$; find~$h$.

\item[5.] Find the admissible values of~$h$ in Ex.~4 when $d = 12$, $v = 332.5$.

\item[6.] Find a necessary and sufficient condition that quartic equation~\Eq{15} shall have
one root the negative of another root.

Hint: $(x_1 + x_2)(x_3 + x_4) = q - y_1$. Hence substitute $q$ for~$y$ in~\Eq{17}.
\index{Quartic equation|)}%

\item[7.] In the study of parabolic orbits occurs the equation
\[% [** PP: Displayed for better line breaking.]
\tan\tfrac{1}{2}v + \tfrac{1}{3}\tan^3 \tfrac{1}{2}v = t.
\]
Prove that there is a single real root and that it has the same sign as~$t$.

\item[8.] In the problem of three astronomical bodies occurs the equation $x^3 + ax + 2 = 0$.
Prove that it has three real roots if and only if $a\leqq{-3}$.
\end{Problems}
\end{Exercises}

%% -----File: 061.png---Folio 55-------


% [** PP: Not matching running head]
\Chapter{V}{The Graph of an Equation}

\index{Graphs|(}%
\Section[Use of Graphs]
{53.}{Use of Graphs in the Theory of Equations.} To find geometrically
the real roots of a real equation $f(x)=0$, we construct a graph of $y=f(x)$
and measure the distances from the origin~$O$ to the intersections of the
graph and the $x$-axis, whose equation is $y=0$.

%%{Illustration} \textsc{Fig. 12}
\begin{wrapfigure}[16]{r}{2.25in}
\quad\Input{061a}
\end{wrapfigure}
For example to find geometrically the real
roots of
\[
x^2 - 6x - 3 = 0,
\Tag{1}
\]
we equate the left member to~$y$ and make a
graph of
\[
y = x^2 - 6x - 3.
\Tag{1'}
\]
We obtain the parabola in Fig.~12. Of the
points shown, $P$ has the \emph{abscissa}
\index{Abscissa}%
$x = OQ = 4$
and the \emph{ordinate}
\index{Ordinate}%
$y = -QP = -11$. From the
points of intersection of $y = 0$ (the $x$-axis $OX$)
with the parabola, we obtain the approximate
values $6.46$ and $-0.46$ of the roots of~\Eq{1}.


\begin{Exercises}{}

\begin{Problems}
\item[1.] Find graphically the real roots of $x^2 - 6x + 7 = 0$.
\index{Quadratic equation!graphical solution}%
\end{Problems}

Hint: For each $x$, $y = x^2 - 6x + 7$ exceeds the $y$ in~\Eq{1'} by~$10$, so that the new graph
is obtained by shifting the parabola in Fig.~12 upward $10$~units, leaving the axes~$OX$
and~$OY$ unchanged. What amounts to the same thing, but is simpler to do, we leave
the parabola and~$OY$ unchanged, and move the axis~$OX$ downward $10$~units.

\begin{Problems}
\item[2.] Discuss graphically the reality of the roots of $x^2 - 6x + 12 = 0$.

\item[3.] Find graphically the roots of $x^2 - 6x + 9 = 0$.
\end{Problems}
\end{Exercises}

%%Illustration \textsc{Fig}.~13
\begin{wrapfigure}[21]{l}{1.25in}
\Input{062a}
\end{wrapfigure}
\Section{54.}{Caution in Plotting} If the example set were
\index{Plotting}%
\[
y = 8x^4 - 14x^3 - 9x^2 + 11x - 2,
\Tag{2}
\]
one might use successive integral values of~$x$, obtain the points $(-2, 180)$,
%% -----File: 062.png---Folio 56-------
$(-1, 0)$, $(0, -2)$, $(1, -6)$, $(2, 0)$, $(3, 220)$, all but the
first and last of which are shown (by crosses) in Fig.~13,
and be tempted to conclude that the graph is a
\Shape{U}-shaped curve approximately like that in Fig.~12
and that there are just two real roots, $-1$ and~$2$, of
\[
8x^4 - 14x^3 - 9x^2 + 11x - 2 = 0.
\Tag{2'}
\]
But both of these conclusions would be false. In fact,
the graph is a \Shape{W}-shaped curve (Fig.~13) and the
additional real roots are $\frac{1}{4}$ and~$\frac{1}{2}$.

This example shows that it is often necessary to
employ also values of~$x$ which are not integers. The
purpose of the example was, however, not to point
out this obvious fact, but rather to emphasize the
chance of serious error in sketching a curve through a
number of points, however numerous. The true curve
between two points below the $x$-axis may not cross the
$x$-axis, or may have a peak and actually cross the $x$-axis
twice, or may be an \Shape{M}-shaped curve crossing it four
times, etc.

%%Illustration \textsc{Fig}.~14
\begin{wrapfigure}{r}{1.5in}
\Input{062b}
\end{wrapfigure}
For example, the graph (Fig.~14) of
\[
y = x^3 + 4x^2 - 11
\Tag{3}
\]
crosses the $x$-axis only once; but this fact cannot
be established by a graph located by a number
of points, however numerous, whose abscissas
are chosen at random.

We shall find that correct conclusions regarding
the number of real roots may be deduced
from a graph whose bend points~(§55) have been
located.


\Section{55.}{Bend Points} A point (like $M$ or~$M'$ in
Fig.~14) is called a \emph{bend point} of the graph of
\index{Bend point}%
$y=f(x)$ if the tangent to the graph at that point
is horizontal and if all of the adjacent points of
the graph lie below the tangent or all above the
tangent.   The first, but not the second, condition
%% -----File: 063.png---Folio 57-------
is satisfied by the point~$O$ of the graph of $y = x^3$ given in Fig.~15 (see~§57).
In the language of the calculus, $f(x)$ has a (relative) maximum or
minimum value at the abscissa of a bend point on the graph of $y=f(x)$.
\index{Maximum}%
\index{Minimum}%
%[Illustration: Fig. 15]
%[Illustration: Fig. 16]
\begin{figure*}[hbt]
\begin{center}
\Input{063a}\hfil
\Input{063b}
\end{center}
\end{figure*}

Let $P = (x, y)$ and $Q = (x+h, Y)$ be two points on the graph, sketched
in Fig.~16, of $y=f(x)$. By the \emph{slope} of a straight line is meant the tangent
of the angle between the line and the $x$-axis, measured counter-clockwise
from the latter. In Fig.~16, the slope of the straight line~$PQ$ is
\index{Slope}%
\[
\frac{Y - y}{h} = \frac{f(x+h) - f(x)}{h}.
\Tag{4}
\]

For equation~\Eq{3}, $f(x) = x^3 + 4x^2 - 11$. Hence
\begin{align*}
f(x+h) &= (x+h)^3 + 4(x+h)^2 - 11 \\
       &= x^3 + 4x^2 - 11 + (3x^2 +8x)h + (3x+4)h^2 + h^3.
\end{align*}
The slope~\Eq{4} of the secant~$PQ$ is therefore here
\index{Derivative|(}%
\[
3x^2 + 8x + (3x+4)h + h^2.
\]
Now let the point~$Q$ move along the graph toward~$P$. Then $h$ approaches
the value zero and the secant~$PQ$ approaches the tangent at~$P$. The
slope of the tangent at~$P$ is therefore the corresponding limit $3x^2 + 8x$
of the preceding expression. We call $3x^2 + 8x$ the \emph{derivative} of $x^3 + 4x^2 - 11$.

%% -----File: 064.png---Folio 58-------

In particular, if $P$ is a bend point, the slope of the (horizontal) tangent
at~$P$ is zero, whence $3x^2 + 8x = 0$, $x = 0$ or $x = -\tfrac{8}{3}$. Equation~\Eq{3} gives
the corresponding values of~$y$.   The resulting points
\[
M = (0, -11),\qquad  M' = (-\tfrac{8}{3}, -\tfrac{41}{27})
\]
are easily shown to be bend points. Indeed, for $x>0$ and for $x$ between
$-4$ and~$0$, $x^2(x+4)$ is positive, and hence $f(x) > -11$ for such values of~$x$,
so that the function~\Eq{3} has a relative minimum at $x = 0$. Similarly,
there is a relative maximum at $x = -\tfrac{8}{3}$. We may also employ the general
method of~§59 to show that $M$ and~$M'$ are bend points. Since these bend
points are both below the $x$-axis we are now certain that the graph
crosses the $x$-axis only once.

The use of the bend points insures greater accuracy to the graph than
the use of dozens of points whose abscissas are taken at random.

\Section{56.}{Derivatives} We shall now find the slope of the tangent to the
graph of $y=f(x)$, where $f(x)$ is any polynomial
\[
\Tag{5}
f(x) = a_0 x^n + a_1 x^{n-1} + \dotsb + a_{n-1} x + a_n.
\]
We need the expansion of $f(x+h)$ in powers of~$x$. By the binomial
theorem,
\begin{align*}
a_0 (x+h)^n
  &= a_0 x^n + na_0 x^{n-1} h + \frac{n(n-1)}{2} a_0 x^{n-2} h^2 + \dotsb, \\
a_1 (x+h)^{n-1}
  &= a_1 x^{n-1} + (n-1)a_1 x^{n-2} h
   + \frac{(n-1)(n-2)}{2} a_1 x^{n-3} h^2 + \dotsb, \\
\multispan{2}{\dotfill} \\
a_{n-2} (x+h)^2
  &= a_{n-2} x^2 +2a_{n-2} xh + a_{n-2} h^2, \\
a_{n-1} (x+h)
  &= a_{n-1} x + a_{n-1} h, \\
a_n &= a_n.
\end{align*}
The sum of the left members is evidently $f(x+h)$. On the right, the
sum of the first terms (i.e., those free of~$h$) is~$f(x)$. The sum of the coefficients
of~$h$ is denoted by~$f'(x)$, the sum of the coefficients of $\tfrac{1}{2} h^2$ is denoted
by $f''(x), \dotsc$, the sum of the coefficients of
\[
\frac{h^k}{1·2\dotsm k}
\]
%% -----File: 065.png---Folio 59-------
is denoted by~$f^{(k)}(x)$. Thus
\begin{align*}
f'(x)  &= na_0 x^{n-1} + (n-1)a_1 x^{n-2} + \dotsb + 2a_{n-2} x + a_{n-1},
\Tag{6} \\
f''(x) &= n(n-1)a_0 x^{n-2} + (n-1)(n-2)a_1 x^{n-3} + \dotsb + 2a_{n-2},
\Tag{7}
\intertext{etc. Hence we have}
f(x+h) &= f(x) + f'(x)h + f''(x) \frac{h^2}{1·2} + f'''(x) \frac{h^3}{1·2·3}
\Tag{8} \\
  & \qquad + \dotsb + f^{(r)}(x) \frac{h^r}{r!}
           + \dotsb + f^{(n)}(x) \frac{h^n}{n!},
\end{align*}
where $r!$ is the symbol, read $r$~\emph{factorial}, for the product $1·2·3\dotsm(r-1)r$.
Here $r$ is a positive integer, but we include the case $r = 0$ by the definition,
$0! = 1$.
\index{Symbol!d@{$r$"!\IndAdd{factorial}}}% [** PP: Manually alphabetized]

This formula~\Eq{8} is known as \emph{Taylor's theorem} for the present case of
\index{Taylor's theorem}%
a polynomial~$f(x)$ of degree~$n$. We call $f'(x)$ the (\emph{first}) \emph{derivative of~$f(x)$},
and $f''(x)$ the \emph{second derivative} of~$f(x)$, etc.
Concerning the fact that
$f''(x)$ is equal to the first derivative of $f'(x)$ and that, in general, the $k$th
derivative $f^{(k)}(x)$ of $f(x)$ is equal to the first derivative of $f^{(k-1)}(x)$, see
Exs.~6--9 of the next set.
\index{Symbol!e@{$f^{(k)}(x)$\IndAdd{$k$th derivative}}}% [** PP: Manually alphabetized]

In view of~\Eq{8}, the limit of~\Eq{4} as $h$~approaches zero is $f'(x)$. Hence
\begin{Thm}%
$f'(x)$ is the slope of the tangent to the graph of $y=f(x)$ at the point~$(x, y)$.
\end{Thm}
\index{Slope}%

In \Eq{5} and~\Eq{6}, let every $a$ be zero except~$a_0$. Thus the derivative of
$a_0 x^n$ is $na_0 x^{n-1}$, and hence is obtained by multiplying the given term by
its exponent~$n$ and then diminishing its exponent by unity. For example,
the derivative of~$2x^3$ is~$6x^2$.

Moreover, the derivative of $f(x)$ is equal to the sum of the derivatives
of its separate terms. Thus the derivative of $x^3 + 4x^2 - 11$ is $3x^2 + 8x$,
as found also in~§55.


\begin{Exercises}{Page59}

\begin{Problems}
\item[1.] Show that the slope of the tangent to $y = 8x^3 - 22x^2 + 13x - 2$ at $(x, y)$ is
$24x^2 - 44x + 13$, and that the bend points are $(0.37, 0.203)$, $(1.46, -5.03)$, approximately.
Draw the graph.

\item[2.] Prove that the bend points of $y = x^3 - 2x - 5$ are $(.82, -6.09)$, $(-.82$, $-3.91)$, % [** PP: Allow line break between coordinates]
approximately. Draw the graph and locate the real roots.

\item[3.] Find the bend points of $y = x^3 + 6x^2 + 8x + 8$. Locate the real roots.

\item[4.] Locate the real roots of $f(x) = x^4 + x^3 - x - 2 = 0$.

Hints: The abscissas of the bend points are the roots of $f'(x) = 4x^3 + 3x^2 - 1 = 0$.
The bend points of $y = f'(x)$ are $(0, -1)$ and $(-\frac{1}{2}, -\frac{3}{4})$, so that $f'(x)= 0$ has a single real
root (it is just less than $\frac{1}{2}$). The single bend point of $y=f(x)$ is $(\frac{1}{2}, -\frac{37}{16})$, approximately.

%% -----File: 066.png---Folio 60-------

\item[5.] Locate the real roots of $x^6 - 7x^4 - 3x^2 + 7 = 0$.

\item[6.] Prove that $f''(x)$, given by~\Eq{7}, is equal to the first derivative of~$f'(x)$.

\item[7.] If $f(x) = f_1(x) + f_2(x)$, prove that the $k$th derivative of~$f$ is equal to the sum of
the $k$th derivatives of $f_1$ and~$f_2$. Use~\Eq{8}.

\item[8.] Prove that $f^{(k)}(x)$ is equal to the first derivative of $f^{(k-1)}(x)$. Hint: prove this
for $f = ax^m$; then prove that it is true for $f=f_1 + f_2$ if true for $f_1$ and~$f_2$.

\item[9.] Find the third derivative of $x^6 + 5x^4$ by forming successive first derivatives;
also that of $2x^5 - 7x^3 + x$.

\item[10.] Prove that if $g$ and~$k$ are polynomials in~$x$, the derivative of $gk$ is $g'k + gk'$. Hint:
multiply the members of $g(x+h) = g(x) + g'(x)h + \dotsb$ and $k(x+h) = k(x) + k'(x)h + \dotsb$
and use~\Eq{8} for $f = gk$.

\end{Problems}
\end{Exercises}


\Section{57.}{Horizontal Tangents} If $(x, y)$ is a bend point of the graph of
\index{Tangents}%
$y=f(x)$, then, by definition, the slope of the tangent at $(x, y)$ is zero.
Hence~(§56), the abscissa~$x$ is a root of $f'(x)=0$. In Exs.~1--5 of the
preceding set, it was true that, conversely, any real root of $f'(x)=0$ is
the abscissa of a bend point. However, this is not always the case.
We shall now consider in detail an example illustrating this fact. The
example is the one merely mentioned in~§55 to indicate the need of the
second requirement made in our definition of a bend point.

The graph (Fig.~15) of $y = x^3$ has no bend point since $x^3$ increases when
$x$~increases. Nevertheless, the derivative $3x^2$ of~$x^3$ is zero for the real
value $x = 0$. The tangent to the curve at $(0, 0)$ is the horizontal line
$y=0$. It may be thought of as the limiting position of a secant through~$O$
which meets the curve in two further points, seen to be equidistant
from~$O$. When one, and hence also the other, of the latter points approaches~$O$,
the secant approaches the position of tangency. In this sense the
tangent at~$O$ is said to meet the curve in three coincident points, their
abscissas being the three coinciding roots of $x^3 = 0$. In the language of~§17,
$x^3 = 0$ has the triple root $x = 0$. The subject of bend points, to which
we recur in~§59, has thus led us to a digression on the important subject
of multiple roots.
\index{Derivative|)}%


\Section{58.}{Multiple Roots} In~\Eq{8} replace $x$ by~$\alpha$, and $h$ by $x-\alpha$. Then
\index{Multiple roots}%
\begin{align*}
f(x) &= f(\alpha) + f'(\alpha) (x-\alpha)
      + f''(\alpha)  \frac{(x-\alpha)^2}{1·2}
      + f'''(\alpha) \frac{(x-\alpha)^3}{1·2·3} + \dotsb
\Tag{9} \\
&\phantom{{} = f(\alpha)}
   {} + f^{(m-1)}(\alpha) \frac{(x-\alpha)^{m-1}}{(m-1)!}
      + f^{(m)}(\alpha)   \frac{(x-\alpha)^m}{m!} + \dotsb.
\end{align*}
By definition~(§17) $\alpha$ is a root of $f(x)=0$ of multiplicity~$m$ if $f(x)$ is exactly
%% -----File: 067.png---Folio 61-------
divisible by $(x-\alpha)^m$, but not by $(x-\alpha)^{m+1}$. Hence \emph{$\alpha$~is a root of multiplicity~$m$
of $f(x) = 0$ if and only if}
\index{Multiplicity of root}%
\[
f(\alpha) = 0,\quad
f'(\alpha) = 0,\quad
f''(\alpha) = 0, \dotsc,\quad
f^{(m-1)}(\alpha) = 0,\quad
f^{(m)}(\alpha) \ne 0.
\Tag{10}
\]

\begin{Remark}
For example, $x^4 + 2x^3 =0$ has the triple root $x = 0$ since $0$ is a root, and since the
first and second derivatives $4x^3 +6x^2$ and $12x^2 +12x$ are zero for $x = 0$, while the third
derivative $24x + 12$ is not zero for $x = 0$.
\end{Remark}

If in~\Eq{9} we replace $f$ by~$f'$ and hence $f^{(k)}$ by~$f^{(k+1)}$, or if we differentiate
every term with respect to~$x$, we see by either method that
\begin{multline*}
f'(x) = f'(\alpha) + f''(\alpha)(x-\alpha) + \dotsb
  + f^{(m-1)} (\alpha)\frac{(x-\alpha)^{m-2}}{(m-2)!} \\
  + f^{(m)}(\alpha)\frac{(x-\alpha)^{m-1}}{(m-1)!} + \dotsb.
\Tag{11}
\end{multline*}

Let $f(x)$ and $f'(x)$ have the common factor $(x-\alpha)^{m-1}$, but not the common
factor $(x-\alpha)^m$, where $m>1$.  Since~\Eq{11} has the factor $(x-\alpha)^{m-1}$, we
have $f'(\alpha) = 0, \dotsc, f^{(m-1)}(\alpha) = 0$. Since also $f(x)$ has the factor $x-\alpha$,
evidently $f(\alpha)=0$. Then, by~\Eq{9}, $f(x)$ has the factor $(x-\alpha)^m$, which,
by hypothesis, is not also a factor of $f'(x)$.  Hence, in~\Eq{11}, $f^{(m)}(\alpha)\ne 0$.
Thus, by~\Eq{10}, $\alpha$ is a root of $f(x)=0$ of multiplicity~$m$.

Conversely, let $\alpha$ be a root of $f(x)=0$ of multiplicity~$m$. Then relations~\Eq{10}
hold, and hence, by~\Eq{11}, $f'(x)$ is divisible by $(x-\alpha)^{m-1}$, but
not by $(x-\alpha)^m$. Thus $f(x)$ and $f'(x)$ have the common factor $(x-\alpha)^{m-1}$,
but not the common factor $(x-\alpha)^m$.

We have now proved the following useful result.

\begin{Theorem}
If $f(x)$ and $f'(x)$ have a greatest common divisor $g(x)$
\index{Greatest common divisor}%
involving~$x$, a root of $g(x)=0$ of multiplicity $m-1$ is a root of $f(x)=0$ of
multiplicity~$m$, and conversely any root of $f(x) =0$ of multiplicity~$m$ is a root
of $g(x)=0$ of multiplicity $m-1$.
\end{Theorem}

In view of this theorem, the problem of finding all the multiple roots
of $f(x)=0$ and the multiplicity of each multiple root is reduced to the
problem of finding the roots of $g(x)=0$ and the multiplicity of each.

\begin{Remark}
For example, let $f(x) = x^3 - 2x^2 - 4x + 8$. Then
\[
f'(x) = 3x^2 - 4x - 4,\qquad
9f(x) = f'(x)(3x-2) - 32(x-2).
\]
Since $x - 2$ is a factor of $f'(x)$, it may be taken to be the greatest common divisor of~$f(x)$
and~$f'(x)$, the choice of the constant factor~$c$ in $c(x-2)$ being here immaterial. Hence~$2$
is a double root of $f(x)=0$, while the remaining root~$-2$ is a simple root.
\end{Remark}

%% -----File: 068.png---Folio 62-------

\begin{Exercises}{Page62}

\begin{Problems}
\item[1.] Prove that $x^3 - 7x^2 + 15x - 9 = 0$ has a double root.

\item[2.] Show that $x^4 - 8x^2 + 16 = 0$ has two double roots.

\item[3.] Prove that $x^4 - 6x^2 - 8x - 3 = 0$ has a triple root.

\item[4.] Test $x^4 - 8x^3 + 22x^2 - 24x + 9 = 0$ for multiple roots.

\item[5.] Test $x^3 - 6x^2 + 11x - 6 = 0$ for multiple roots.

\item[6.] Test $x^4 - 9x^3 + 9x^2 + 81x - 162 = 0$ for multiple roots.
\end{Problems}
\end{Exercises}


\Section{59.}{Ordinary and Inflexion Tangents} The equation of the straight
\index{Inflexion|(}%
\index{Tangents}%
line through the point $(\alpha, \beta)$ with the slope~$s$ is $y-\beta = s(x -\alpha)$. The slope
of the tangent to the graph of $y=f(x)$ at the point $(\alpha, \beta)$ on it is $s=f'(\alpha)$
by~§56. Also, $\beta=f(\alpha)$. Hence the equation of the tangent is
\[
y = f(\alpha) + f'(\alpha)(x-\alpha).
\Tag{12}
\]

By subtracting the members of this equation from the corresponding
members of equation~\Eq{9}, we see that the abscissas~$x$ of the points of intersection
of the graph of $y=f(x)$ with its tangent satisfy the equation
\begin{multline*}
f''(\alpha)\frac{(x-\alpha)^2}{2!}
  + f'''(\alpha)\frac{(x-\alpha)^3}{3!} + \dotsb
  + f^{(m-1)}(\alpha)\frac{(x-\alpha)^{m-1}}{(m-1)!} \\
  + f^{(m)}(\alpha)\frac{(x-\alpha)^m}{m!} + \dotsb = 0.
\end{multline*}
Here the term containing $f^{(m-1)}(\alpha)$ must evidently be suppressed if $m = 2$,
since the term containing $f^{(m)}(\alpha)$ then coincides with the first term.

If $\alpha$ is a root of multiplicity~$m$ of this equation, i.e., if the left member
is divisible by $(x -\alpha)^m$, but not by $(x-\alpha)^{m+1}$, the point $(\alpha, \beta)$ is counted
as $m$~coincident points of intersection of the curve with its tangent (just
as in the case of $y=x^3$ and its tangent $y = 0$ in~§57). This will be the case
if and only if
\[
f''(\alpha)=0,\qquad
f'''(\alpha) =0, \dotsc,\qquad
f^{(m-1)}(\alpha)=0,\qquad
f^{(m)}(\alpha)\ne 0,
\Tag{13}
\]
in which $m>1$ and, as explained above, only the final relation $f''(\alpha)\ne 0$
is retained if $m= 2$. If $m=3$, the conditions are $f''(\alpha)= 0$, $f^{(3)}(\alpha) \ne 0$.

\begin{Remark}
For example, if $f(x) = x^4$ and $\alpha = 0$, then $f''(0) =f'''(0) =0$, $f^{(4)} (0) = 24\ne 0$, so that
$m=4$. The graph of $y = x^4$ is a \Shape{U}-shaped curve, whose intersection with the tangent
(the $x$-axis) at $(0, 0)$ is counted as four coincident points of intersection.
\end{Remark}

Given $f(x)$ and $\alpha$, we can find, as in the preceding example, the value
of~$m$ for which relations~\Eq{13} hold. We then apply the
%% -----File: 069.png---Folio 63-------
\begin{Theorem}
If $m$ is even \($m>0$\), the points of the curve in the vicinity
of the point of tangency $(\alpha, \beta)$ are all on the same side of the tangent, which
is then called an \emph{\textbf{ordinary tangent}}. But if $m$ is odd \($m>1$\), the curve crosses
the tangent at the point of tangency $(\alpha, \beta)$, and this point is called an \emph{\textbf{inflexion
point}}, while the tangent is called an \emph{\textbf{inflexion tangent}}.
\end{Theorem}

\begin{Remark}
For example, in Fig.~15, $OX$ is an inflexion tangent, while the tangent at any point
except~$O$ is an ordinary tangent. In Figs.~18, 19, 20, the tangents at the points marked
by crosses are ordinary tangents, but the tangent at the point midway between them
and on the $y$-axis is an inflexion tangent.
\end{Remark}

To simplify the proof, we first take as new axes lines parallel to the
old axes and intersecting at $(\alpha, \beta)$. In other words, we set $x-\alpha = X$,
$y-\beta=Y$, where $X$, $Y$ are the coordinates of $(x, y)$ referred to the new
axes. Since $\beta = f(\alpha)$, the tangent~\Eq{12} becomes $Y = f'(\alpha)X$, while, by~\Eq{9},
$y = f(x) = \beta + f'(\alpha)(x-\alpha) + \dotsb$ becomes
\[
Y = f'(\alpha)X + f''(\alpha)\frac{X^2}{2} + \dotsb
  = f'(\alpha)X + f^{(m)}(\alpha)\frac{X^m}{m!} + \dotsb,
\]
after omitting terms which are zero by~\Eq{13}.

%Illustration: Fig. 17
\begin{wrapfigure}[15]{r}{2.375in}
\hfil\Input{069a}
\end{wrapfigure}
To simplify further the algebraic work,
we pass to oblique axes,\footnote
  {Since the earlier $x$, $y$ do not occur in~\Eq{14} and the new equation of the tangent,
  we shall designate the final coordinates by $x$, $y$ without confusion.}
the new $y$-axis
coinciding with the $Y$-axis, while the new
$x$-axis is the tangent, the angle between
which and the $X$-axis is designated by $\theta$.
Then
\[
\tan\theta=f'(\alpha).
\]
By Fig.~17,
\[
X   = x\cos\theta,\qquad
Y-y = f'(\alpha)X.
\]

Hence when expressed in terms of the
new coordinates $x$, $y$, the tangent is $y = 0$, while the equation~\Eq{14} of the
curve becomes
\[
y = cx^m + dx^{m+1} + \dotsb,\qquad
c = \frac{f^{(m)}(\alpha)\cos^m \theta}{m!} \ne 0.
\]

For $x$ sufficiently small numerically, whether positive or negative,
the sum of the terms after $cx^m$ is insignificant in comparison with $cx^{m}$,
%% -----File: 070.png---Folio 64-------
so that $y$ has the same sign as $cx^m$~(§64). Hence, if $m$ is even, the points
of the curve in the vicinity of the origin and on both sides of it are all
on the same side of the $x$-axis, i.e., the tangent. But, if $m$ is odd, the points
with small positive abscissas~$x$ lie on one side of the $x$-axis and those with
numerically small negative abscissas lie on the opposite side.

Our transformations of coordinates changed the equations of the
curve and of its tangent, but did not change the curve itself and its tangent.
Hence our theorem is proved.

By our theorem, $\alpha$ is the abscissa of an inflexion point of the graph
of $y=f(x)$ if and only if conditions~\Eq{13} hold with $m$~odd ($m>1$). These
conditions include neither $f(\alpha) = 0$ nor $f'(\alpha)=0$, in contrast with~\Eq{10}.
In the theory of equations we are primarily interested in the abscissas
$\alpha$ of only those points of inflexion whose inflexion tangents are horizontal,
and are interested in them, because we must exclude such roots $\alpha$ of
$f'(x)=0$ when seeking the abscissas of bend points, which are the important
points for our purposes. A point on the graph at which the tangent is
both horizontal and an ordinary tangent is a bend point by the definition
in~§55. Hence if we apply our theorem to the special case $f'(\alpha)=0$,
we obtain the following

\begin{Criterion}
Any root $\alpha$ of $f'(x) = 0$ is the abscissa of a bend point
of the graph of $y = f(x)$ or of a point with a horizontal inflexion tangent according
as the value of~$m$ for which relations~\Eq{13} hold is even or odd.
\index{Bend point}%
\end{Criterion}

\begin{Remark}
For example, if $f(x) = x^4$, then $\alpha = 0$ and $m = 4$, so that $(0, 0)$ is a bend point of the
\Shape{U}-shaped graph of $y=x^4$. If $f(x)=x^3$, then $\alpha = 0$ and $m = 3$, so that $(0, 0)$ is a point
with a horizontal inflexion tangent $(OX$ in Fig.~15) of the graph of $y = x^3$.
\end{Remark}


\begin{Exercises}{Page64}

\begin{Problems}

\item[1.] If $f(x) = 3x^5 + 5x^3 + 4$, the only real root of $f'(x)=0$ is $x = 0$. Show that $(0, 4)$
is an inflexion point, and thus that there is no bend point and hence that $f(x)=0$ has a
single real root.

\item[2.] Prove that $x^3 - 3x^2 + 3x + c = 0$ has an inflexion point, but no bend point.

\item[3.] Show that $x^5 - 10x^3 - 20x^2 - 15x + c = 0$ has two bend points and no horizontal
inflexion tangents.

\item[4.] Prove that $3x^5 - 40x^3 + 240x + c = 0$ has no bend point, but has two horizontal
inflexion tangents.

\item[5.] Prove that any function $x^3 - 3\alpha x^2 + \dotsb$ of the third degree can be written in
\index{Cubic equation!reduced}% [** PP: Not using page range]
the form $f(x) = (x-\alpha)^3 + ax + b$. The straight line having the equation $y = ax+b$ meets
the graph of $y=f(x)$ in three coincident points with the abscissa $\alpha$ and hence is an
inflexion tangent. If we take new axes of coordinates parallel to the old and intersecting
at the new origin $(\alpha, 0)$, i.e., if we make the transformation $x = X+\alpha$, $y = Y$,
%% -----File: 071.png---Folio 65-------
of coordinates, we see that the equation $f(x)=0$ becomes a reduced cubic equation
$X^3 + pX + q = 0$~(§42).

\item[6.] Find the inflexion  tangent to $y = x^3 + 6x^2 - 3x + 1$ and transform
$x^3 + 6x^2 - 3x + 1 = 0$ into a reduced cubic equation.
\end{Problems}
\end{Exercises}
\index{Inflexion|)}% [** PP: Original page range is 62--64]


\Section[Real Roots of a Cubic Equation]
{60.}{Real Roots of a Real Cubic Equation.} It suffices to consider
\begin{flalign*}
&& f(x) &= x^3 - 3lx + q && \Rightmark{(l \neq 0),}
\end{flalign*}
in view of Ex.~5 above. Then $f' = 3 (x^2 - l)$, $f'' = 6x$. If $l<0$, there is
no bend point and the cubic equation $f(x)=0$ has a single real root.
If $l>0$, there are two bend points
\[
( \sqrt{l}, q - 2l\sqrt{l}),\qquad
(-\sqrt{l}, q + 2l\sqrt{l}),
\]
which are shown by crosses in Figs.~18--20 for the graph of $y=f(x)$ in the
\index{Cubic equation!graph of}%
\index{Cubic equation!number of real roots}% [** PP: Added ``of'']
three possible cases specified by the inequalities shown below the figures.
For a large positive~$x$, the term~$x^3$ in~$f(x)$ predominates, so that the graph
contains a point high up in the first quadrant,
thence extends downward to the
right-hand bend point, then ascends to
the left-hand bend point, and finally descends.
As a check, the graph contains
a point far down in the third quadrant,
since for $x$ negative, but sufficiently large
numerically, the term $x^3$ predominates and the sign of~$y$ is negative.
% [** Illustrations]
% \caption{$q \geq  2l\sqrt{l}$} FIG. 18
% \caption{$q \leq -2l\sqrt{l}$} FIG. 19
\begin{figure*}[htb]
\begin{center}
\Input{071a}\hfill
\Input{071b}
\end{center}
\end{figure*}

% [** Illustration]
% \caption{$-2l\sqrt{l} < q < 2l\sqrt{l}$}  Fig. 20
\begin{wrapfigure}[9]{r}{2.6in}
\hfill\raisebox{\baselineskip}{\Input{071c}}
\end{wrapfigure}
If the equality sign holds in Fig.~18 or Fig.~19, a necessary and sufficient
condition for which is $q^2 = 4l^3$, one of the bend points is on the $x$-axis, and
the cubic equation has a double root. The inequalities in Fig.~20 hold
if and only if $q^2 < 4l^3$, which implies that $l>0$. Hence \emph{$x^3 - 3lx + q = 0$
has three distinct real roots if and only if $q^2 < 4l^3$, a single real root if and
only if $q^2 > 4l^3$, a double root \(necessarily real\) if and only if $q^2 = 4l^3$ and $l\neq 0$,
and a triple root if $q^2 = 4l^3 = 0$}.
\index{Discriminant!of cubic}%

%% -----File: 072.png---Folio 66-------


\begin{Exercises}{Page66}

Find the bend points, sketch the graph, and find the number of real roots of
\begin{Problems}[2]
\item[1.] $x^3 + 2x - 4 = 0$.

\item[2.] $x^3 - 7x + 7 = 0$.

\item[3.] $x^3 - 2x - 1 = 0$.

\item[4.] $x^3 + 6x^2 - 3x + 1 = 0$.
\end{Problems}
\begin{Problems}
\item[5.] Prove that the inflexion point of $y = x^3 - 3lx + q$ is $(0, q)$.

\item[6.] Show that the theorem in the text is equivalent to that in~§45.

\item[7.] Prove that, if $m$ and~$n$ are positive odd integers and $m>n$, $x^m + px^n + q = 0$ has
no bend point and hence has a single real root if $p>0$; but, if $p<0$, it has just two
bend points which are on the same side or opposite sides of the $x$-axis according as
\[
\left(\frac{np}{m}\right)^m + \left(\frac{nq}{m-n}\right)^{m-n}
\]
is positive or negative, so that the number of real roots is $1$ or~$3$ in the respective cases.

\item[8.] Draw the graph of $y = x^4 - x^2$. By finding its intersections with the line $y = mx + b$, solve $x^4 - x^2 - mx - b= 0$.

\item[9.] Prove that, if $p$ and~$q$ are positive, $x^{2m} - px^{2n} + q = 0$ has four distinct real roots,
two pairs of equal roots, or no real root, according as
\[
\left(\frac{np}{m}\right)^m - \left(\frac{nq}{m-n}\right)^{m-n} > 0,
\quad\text{${} = 0$,\quad or\quad ${} < 0$}.
\]

\item[10.] Prove that no straight line crosses the graph of $y = f(x)$ in more than $n$~points if
the degree~$n$ of the real polynomial $f(x)$ exceeds unity. [Apply~§16.] This fact serves as a check on the accuracy of a graph.
\end{Problems}
\end{Exercises}


\Section[Continuity]
{61.}{Definition of Continuity of a Polynomial.} Hitherto we have
located certain points of the graph of $y=f(x)$, where $f(x)$ is a polynomial
in~$x$ with real coefficients, and taken the liberty to join them by a continuous
curve.
\index{Continuity}%

A polynomial $f(x)$ with real coefficients shall be called \emph{continuous at}
$x = a$, where $a$ is a real constant, if the difference
\[
D = f(a+h) - f(a)
\]
is numerically less than any assigned positive number~$p$ for all real values
of~$h$ sufficiently small numerically.


% [** PP: No ToC entry]
\Section[Continuity of Polynomials]
{62.}{Any Polynomial $f(x)$ with real Coefficients is continuous at $x = a$,
where $a$ is any real Constant.} Taylor's formula~\Eq{8} gives
\index{Polynomial}%
\[
D = f'(a) h + \frac{f''(a)}{1·2} h^2 + \dotsb
  + \frac{f^{(n)}(a)}{1·2\dotsm n} h^n.
\]
This polynomial is a special case of
\[
F = a_1 h + a_2 h^2 + \dotsb + a_n h^n.
\]
%% -----File: 073.png---Folio 67-------
We shall prove that, \emph{if $a_1, \dotsc, a_n$ are all real, $F$ is numerically less than
any assigned positive number~$p$ for all real values of~$h$ sufficiently small
numerically}. Denote by~$g$ the greatest numerical value of $a_1, \dotsc, a_n$.
If $h$ is numerically less than~$k$, where $k<1$, we see that $F$ is numerically less than
\[
g(k + k^2 + \dotsb + k^n) < g\frac{k}{1-k} < p,\qquad
\text{if } k < \frac{p}{p + g}.
\]
Hence a real polynomial $f(x)$ is continuous at every real value of~$x$. But
the function $\tan x$ is not continuous at $x=90°$~(§63).


\Section[Condition for a Root Between $a$ and~$b$]
{63.}{Root between $a$ and~$b$ if $f(a)$ and~$f(b)$ have opposite Signs.}
\begin{Thm}
If
the coefficients of a polynomial $f(x)$ are real and if $a$ and~$b$ are real numbers
such that $f(a)$ and~$f(b)$ have opposite signs, the equation $f(x) = 0$ has at least
one real root between $a$ and~$b$; in fact, an odd number of such roots, if an
$m$-fold root is counted as $m$~roots.
\end{Thm}

%[Illustration: \textsc{Fig}. 21] [** PP: Move to top of paragraph]
\begin{wrapfigure}[15]{r}{1.3in}
\hfill\Input{073a}
\end{wrapfigure}
The only argument\footnote
  {An arithmetical proof based upon a refined theory of irrational numbers is given
  in Weber's \textit{Lehrbuch der Algebra}, ed.~2, vol.~1, p.~123.}
given here (other than that in Ex.~5 below) is
one based upon geometrical intuition. We are stating that, if the points
\[
\bigl(a, f(a)\bigr),\qquad
\bigl(b, f(b)\bigr)
\]
lie on opposite sides of the $x$-axis, the graph of $y=f(x)$ crosses the $x$-axis
once, or an odd number of times, between the vertical lines through
these two points. Indeed, the part of the graph between
these verticals is a continuous curve having one
and only one point on each intermediate vertical line,
since the function has a single value for each value
of~$x$.

This would not follow for the graph of $y^2 = x$, which
is a parabola with the $x$-axis as its axis. It may not
cross the $x$-axis between the two initial vertical lines,
but cross at a point to the left of each.

A like theorem does not hold for $f(x) = \tan x$, when
$x$ is measured in radians and $0 < a < \pi/2 < b < \pi$, since
$\tan x$ is not continuous at $x=\pi/2$. When $t$ increases
from $a$ to $\pi/2$, $\tan x$ increases without limit. When
$x$ decreases from $b$ to~$\pi/2$, $\tan x$ decreases without
limit. There is no root between $a$ and~$b$ of $\tan x = 0$.

%% -----File: 074.png---Folio 68-------


\begin{Exercises}{}

\begin{Problems}
\item[1.] Prove that $8x^3 - 4x^2 - 18x + 9 = 0$ has a root between $0$ and~$1$, one between $1$ and~$2$,
and one between $-2$ and~$-1$.

\item[2.] Prove that $16x^4 - 24x^2 + 16x - 3 = 0$ has a triple root between $0$ and~$1$, and a
simple root between $-2$ and~$-1$.

\item[3.] Prove that if $a < b < c \dotsb < l$, and $\alpha$, $\beta, \dotsc, \lambda$ are positive, these quantities
being all real,
\[
\frac{\alpha}{x-a} +
\frac{\beta}{x-b} +
\frac{\gamma}{x-c} + \dotsb +
\frac{\lambda}{x-l} + t = 0
\]
has a real root between $a$ and~$b$, one between $b$ and~$c, \dotsc$ one between $k$ and~$l$, and
if $t$ is negative one greater than~$l$, but if $t$ is positive one less than~$a$.

\item[4.] Verify that the equation in Ex.~3 has no imaginary root by substituting $r+si$
and $r-si$ in turn for~$x$, and subtracting the results.

\item[5.] Admitting that an equation $f(x) \equiv x^n + \dotsb = 0$ with real coefficients has $n$~roots,
show algebraically that there is a real root between $a$ and~$b$ if $f(a)$ and~$f(b)$ have opposite
signs. Note that a pair of conjugate imaginary roots $c ± di$ are the roots of
\[
(x-c)^2 + d^2 = 0
\]
and that this quadratic function is positive if $x$ is real. Hence if $x_1, \dotsc, x_r$ are the
real roots and
\[
\phi(x) \equiv (x-x_1) \dotsm (x-x_r),
\]
then $\phi(a)$ and~$\phi(b)$ have opposite signs. Thus $a-x_i$ and $b-x_i$ have opposite signs for
at least one real root~$x_i$. (Lagrange.)
\end{Problems}
\end{Exercises}


\Section[Sign of a Polynomial at Infinity]
{64.}{Sign of a Polynomial.} Given a polynomial
\index{Polynomial!sign of}%
\index{Sign of polynomial}%
\begin{flalign*}
&& f(x) &= a_0 x^n + a_1 x^{n-1} + \dotsb + a_n && \Rightmark{(a_0 \neq 0)}
\end{flalign*}
with real coefficients, we can find a positive number~$P$ such that $f(x)$ has
the same sign as $a_0 x^n$ when $x>P$. In fact,
\[
f(x) = x^n (a_0 + \phi),\qquad
\phi = \frac{a_1}{x} + \frac{a_2}{x^2} + \dotsb + \frac{a_n}{x^n}.
\]
By the result in~§62, the numerical value of $\phi$ is less than that of~$a_0$
when $1/x$ is positive and less than a sufficiently small positive number,
say $1/P$, and hence when $x>P$. Then $a_0 + \phi$ has the same sign as~$a_0$,
and hence $f(x)$ the same sign as~$a_0 x^n$.

The last result holds also when $x$ is a negative number sufficiently large
numerically. For, if we set $x=-X$, the former case shows that $f(-X)$
has the same sign as $(-1)^n a_0 X^n$ when $X$ is a sufficiently large positive
number.

%% -----File: 075.png---Folio 69-------

We shall therefore say briefly that, for $x = +\infty$, $f(x)$ has the same
sign as~$a_0$; while, for $x = -\infty$, $f(x)$ has the same sign as $a_0$ if $n$ is even,
but the sign opposite to $a_0$ if $n$ is odd.


\begin{Exercises}{}
\index{Number!of roots}%

\begin{Problems}
\item[1.] Prove that $x^3 + ax^2 + bx - 4 = 0$ has a positive real root [use $x=0$ and $x=+\infty$].

\item[2.] Prove that $x^3 + ax^2 + bx + 4 = 0$ has a negative real root [use $x=0$ and $x=-\infty$].

\item[3.] Prove that if $a_0 > 0$ and $n$ is odd, $a_0x^n + \dotsb + a_n = 0$ has a real root of sign opposite
to the sign of $a_n$ [use $x = -\infty$, $0$, $+\infty$].

\item[4.] Prove that $x^4 + ax^3 + bx^2 + cx - 4 = 0$ has a positive and a negative root.

\item[5.] Show that any equation of even degree $n$ in which the coefficient of $x^n$ and the
constant term are of opposite signs has a positive and a negative root.
\end{Problems}
\end{Exercises}


\Section{65.}{Rolle's Theorem}
\index{Rolle's theorem}%
\index{Root@{Root between $a$ and~$b$}}%
\begin{Thm}
Between two consecutive real roots $a$ and~$b$ of $f(x)=0$,
there is an odd number of real roots of $f'(x) = 0$, a root of multiplicity~$m$
being counted as $m$~roots.
\end{Thm}

Let
\begin{flalign*}
&& f(x) &\equiv (x-a)^r(x-b)^s Q(x), && \Rightmark{a<b,}
\end{flalign*}
where $Q(x)$ is a polynomial divisible by neither $x-a$ nor $x-b$. Then
by the rule for the derivative of a product (§56, Ex.~10),
\index{Derivative}%
\[
\frac{(x-a)(x-b)f'(x)}{f(x)}
  \equiv r(x-b) + s(x-a) + (x-a)(x-b) \frac{Q'(x)}{Q(x)}.
\]
The second member has the value $r(a-b) < 0$ for $x = a$ and the value
$s(b-a) > 0$ for $x=b$, and hence vanishes an odd number of times between
$a$ and $b$~(§63). But, in the left member, $(x-a)(x-b)$ and~$f(x)$ remain
of constant sign between $a$ and~$b$, since $f(x) = 0$ has no root between $a$
and~$b$. Hence $f'(x)$ vanishes an odd number of times.

\begin{Corollary}
Between two consecutive %[** PP: Typo consective]
real roots $\alpha$ and~$\beta$ of $f'(x) = 0$
there occurs at most one real root of $f(x) = 0$.
\end{Corollary}

For, if there were two such real roots $a$ and~$b$ of $f(x) = 0$, the theorem
shows that $f'(x) = 0$ would have a real root between $a$ and~$b$ and hence between
$\alpha$ and~$\beta$, contrary to hypothesis.

Applying also~§63 we obtain the
\begin{Criterion}
If $\alpha$ and~$\beta$ are consecutive real roots of $f'(x) = 0$, then $f(x) = 0$
has a single real root between $\alpha$ and~$\beta$ if $f(\alpha)$ and~$f(\beta)$ have opposite signs,
but no root if they have like signs. At most one real root of $f(x) = 0$ is greater
than the greatest real root of $f'(x) = 0$, and at most one real root of $f(x) = 0$ is
less than the least real root of $f'(x) = 0$.
\end{Criterion}

%% -----File: 076.png---Folio 70-------

If $f(\alpha) = 0$ for our root $\alpha$ of $f'(x) = 0, \alpha$ is a multiple root of $f(x) = 0$ and it would be removed before the criterion is applied.

\begin{Example}
For $f(x) = 3x^5 - 25x^3 + 60x - 20$,
\[
\tfrac{1}{15}f'(x) = x^4 - 5x^2 + 4 = (x^2 - 1)(x^2 - 4).
\]
Hence the roots of $f'(x)=0$ are $± 1,\, ± 2$. Now
{\footnotesize
\[
f(-\infty) = -\infty,\
f(-2) = -36,\
f(-1) = -58,\
f(1) = 18,\
f(2) = -4,\
f(+\infty) = +\infty.
\]}%
Hence there is a single real root in each of the intervals
\[
(-1, 1),\quad (1, 2),\quad (2, +\infty),
\]
and two imaginary roots. The three real roots are positive.
\end{Example}


\begin{Exercises}{}

\begin{Problems}
\item[1.] Prove that $x^5 - 5x + 2 = 0$ has 1~negative, 2~positive and 2~imaginary roots.

\item[2.] Prove that $x^6 + x - 1 = 0$ has 1~negative, 1~positive and 4~imaginary roots.

\item[3.] Show that $x^5 - 3x^3 + 2x^2 - 5 = 0$ has two imaginary roots, and a real root in each
of the intervals $(-2, -1.5)$, $(-1.5, -1)$, $(1, 2)$.

\item[4.] Prove that $4x^5 - 3x^4 - 2x^2 + 4x - 10 = 0$ has a single real root.

\item[5.] Show that, if $f^{(k)}(x) = 0$ has imaginary roots, $f(x) = 0$ has imaginary roots.

\item[6.] Derive Rolle's theorem from the fact that there is an odd number of bend points
between $a$ and~$b$, the abscissa of each being a root of $f'(x) = 0$ of odd multiplicity, while
the abscissa of an inflexion point with a horizontal tangent is a root of $f'(x) = 0$ of even
multiplicity.
\end{Problems}
\end{Exercises}
\index{Graphs|)}%

%% -----File: 077.png---Folio 71-------


% [** PP: ToC entry reads Isolation of the Real Roots]
\Chapter[Isolation of Real Roots]
{VI}{Isolation of the Real Roots of a Real Equation}

% [** PP: No ToC entry]
\Section{66.}{Purpose and Methods of Isolating the Real Roots} In the next
chapter we shall explain processes of computing the real roots of a given
real equation to any assigned number of decimal places. Each such
method requires some preliminary information concerning the root to
be computed. For example, it would be sufficient to know that the root
is between $4$ and~$5$, provided there be no other root between the same
limits. But in the contrary case, narrower limits are necessary, such
as $4$ and~$4.3$, with the further fact that only one root is between these new
limits. Then that root is said to be \emph{isolated}.
\index{Isolation of roots}% [** PP: Original has no page range]

\begin{Remark}
If an equation has a single positive root and a single negative root, the real roots
are isolated, since there is a single root between $-\infty$ and~$0$, and a single one between
$0$ and~$+\infty$. However, for the practical purpose of their computation, we shall need
narrower limits, sufficient to fix the first significant figure of each root, for example
$-40$ and~$-30$, or $20$ and~$30$.
\end{Remark}

We may isolate the real roots of $f(x)=0$ by means of the graph of
$y=f(x)$. But to obtain a reliable graph, we saw in \ChapRef{V} that we
must employ the bend points, whose abscissas occur among the roots of % [** PP: Added `of']
$f'(x)=0$. Since the latter equation is of degree $n-1$ when $f(x)=0$ is of
degree~$n$, this method is usually impracticable when $n$ exceeds~$3$. The
method based on Rolle's theorem~(§65) is open to the same objection.

The most effective method is that due to Sturm~(§68). We shall,
however, begin with Descartes' rule of signs since it is so easily applied.
Unfortunately it rarely tells us the exact number of real roots.


\Section{67.}{Descartes' Rule of Signs} Two consecutive terms of a real polynomial
\index{Descartes' rule of signs}%
or equation are said to present a \emph{variation of sign} if their coefficients
have unlike signs. By the variations of sign of a real polynomial or equation
we mean all the variations presented by consecutive terms.
\index{Variation of sign}%

\begin{Remark}
Thus, in $x^5 - 2x^3 - 4x^2 + 3 = 0$, the first two terms present a variation of sign, and
likewise the last two terms. The number of variations of sign of the equation is two.
\end{Remark}

%% -----File: 078.png---Folio 72-------

\begin{Theorem}[Descartes' Rule]
\index{Number!of roots|(}%
The number of positive real roots of an equation
with real coefficients is either equal to the number of its variations of sign
or is less than that number by a positive even integer. A root of multiplicity~$m$
is here counted as $m$~roots.
\end{Theorem}

\begin{Remark}
For example, $x^6 - 3x^2 + x + 1 = 0$ has either two or no positive roots, the exact number
not being found. But $3x^3 - x - 1 = 0$ has exactly one positive root, which is a simple
root.
\end{Remark}

Descartes' rule will be derived in §73 as a corollary to Budan's theorem.
The following elementary proof\footnote
  {The proofs given in college algebras are mere verifications of special cases.}
was communicated to the author by
Professor D.~R. Curtiss.

Consider any real polynomial
\[
f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_l x^{n-l}
\qquad (a_0 \ne 0,\ a_l \ne 0).
\]

Let $r$ be a positive real number. By actual multiplication,
\[
F(x) \equiv (x-r)f(x)
     \equiv A_0 x^{n+1} + A_1 x^n + \dotsb + A_{l+1}x^{n-l},
\]
where
\[
A_0 = a_0,\quad
A_1 = a_1 - ra_0,\quad
A_2 = a_2 - ra_1, \dotsc,
A_l = a_l - ra_{l-1},\quad
A_{l+1} = -r a_l.
\]
In $f(x)$ let $a_{k_1}$ be the first non-vanishing coefficient of different sign from~$a_0$,
let $a_{k_2}$ be the first non-vanishing coefficient following $a_{k_1}$ and of the
same sign as~$a_0$, etc., the last such term, $a_{k_v}$, being either $a_l$ or of the same
sign as~$a_l$. Evidently $v$ is the number of variations of sign of~$f(x)$.

\begin{Remark}
For example, if $f(x) \equiv 2x^6 + 3x^5 - 4x^4 - 6x^3 + 7x$, we have $v=2$, $a_{k_1} = a_2 = -4$, $a_{k_2} = a_5 = 7$.
Note that $a_4 = 0$ since $x^2$ is absent.
\end{Remark}

The numbers $A_0, A_{k_1}, \dotsc, A_{k_v}, A_{l+1}$ are all different from zero and
have the same signs as $a_0, a_{k_1}, \dotsc, a_{k_v}, -a_l$, respectively. This is
obviously true for $A_0 = a_0$ and $A_{l+1}= -ra_l$. Next, $A_{k_i}$ is the sum of the
non-vanishing number $a_{k_i}$ and the number $-ra_{k_i - 1}$, which is either zero
or else of the same sign as $a_{k_i}$ since $a_{k_i - 1}$ is either zero or of opposite sign
to~$a_{k_i}$. Hence the sum~$A_{k_i}$ is not zero and has the same sign as~$a_{k_i}$.

By hypothesis, each of the numbers $a_0, a_{k_1}, \dotsc, a_{k_v}$ after the first
is of opposite sign to its predecessor, while $-a_l$ is of opposite sign to~$a_{k_v}$.
Hence each term after the first in the sequence $A_0, A_{k_1}, \dotsc, A_{k_v}, A_{l+1}$
is of opposite sign to its predecessor. Thus these terms present $v+1$
variations of sign. We conclude that $F(x)$ has at least one more variation
of sign than $f(x)$. But we may go further and prove the following

%% -----File: 079.png---Folio 73-------

\begin{Lemma}
The number of variations of sign of~$F(x)$ is equal to that of
$f(x)$ increased by some positive odd integer.
\end{Lemma}

For, the sequence $A_0, A_1, \dotsc, A_{k_1}$  has an odd number of variations
of sign since its first and last terms are of opposite sign; and similarly
for the $v$~sequences
\[
\begin{array}{c@{\quad}c@{\,}c@{\,}c}
A_{k_1}, & A_{k_1 + 1}, & \dotsc, & A_{k_2}; \\
\Dots{4} \\
A_{k_v}, & A_{k_v + 1}, & \dotsc, & A_{l+1}.
\end{array}
\]
The total number of variations of sign of the entire sequence $A_0, A_1,\dotsc,
A_{l+1}$ is evidently the sum of the numbers of variations of sign for the
$v+1$ partial sequences indicated above, and is thus the sum of $v+1$ positive
odd integers. Since each such odd integer may be expressed as $1$
plus~$0$ or a positive even integer, the sum mentioned is equal to $v+1$ plus~$0$
or a positive even integer, i.e., to $v$ plus a positive odd integer.

To prove Descartes' rule of signs, consider first the case in which $f(x)= 0$
has no positive real roots, i.e., no real root between $0$ and~$+\infty$. Then
$f(0)$ and $f(\infty$) are of the same sign~(§63), and hence the first and last
coefficients of $f(x)$ are of the same sign.\footnote
  {In case $f(x)$ has a factor $x^{n-l}$, we use the polynomial $f(x)/x^{n-l}$ instead of $f(x)$ in
  this argument.}
Thus $f(x)$ has either no variations
of sign or an even number of them, as Descartes' rule requires.

Next, let $f(x)= 0$ have the positive real roots $r_1,\dotsc, r_k$ and no others.
A root of multiplicity~$m$ occurs here $m$~times, so that the $r$'s need not be
distinct. Then
\[
f(x) \equiv (x - r_1)\dotsm (x - r_k)\phi(x),
\]
where $\phi(x)$ is a polynomial with real coefficients such that $\phi(x)=0$ has
no positive real roots. We saw in the preceding paragraph that $\phi(x)$
has either no variations of sign or an even number of them. By the
Lemma, the product $(x - r_k)\phi(x)$ has as the number of its variations of
sign the number for $\phi(x)$ increased by a positive odd integer. Similarly
when we introduce each new factor $x - r_i$. Hence the number of variations
of sign of the final product $f(x)$ is equal to that of $\phi(x)$ increased
by $k$~positive odd integers, i.e., by $k$ plus $0$ or a positive even integer.
Since $\phi(x)$ has either no variations of sign or an even number of them,
the number of variations of sign of $f(x)$ is $k$ plus $0$ or a positive even integer,
a result equivalent to our statement of Descartes' rule.

%% -----File: 080.png---Folio 74-------

If $-p$ is a negative root of $f(x)= 0$, then $p$ is a positive root of $f(-x)= 0$.
Hence we obtain the

\begin{Corollary}
\index{Number!of negative roots}%
The number of \emph{negative} roots of $f(x)=0$ is either equal
to the number of variations of sign of $f(-x)$ or is less than that number
by a positive even integer.
\end{Corollary}

\begin{Remark}
For example, $x^4 + 3x^3 + x - 1 = 0$ has a single negative root, which is a simple root,
since $x^4 - 3x^3 - x - 1 = 0$ has a single positive root.
\end{Remark}

As indicated in Exs.~10, 11 below, Descartes' rule may be used to isolate
the roots.


\begin{Exercises}{Page74}

Prove by Descartes' rule the statements in Exs.~1--8, 12,~15.
\begin{Problems}

\item[1.] An equation all of whose coefficients are of like sign has no positive root. Why
is this self-evident?

\item[2.] There is no negative root of an equation, like $x^5 - 2x^4 - 3x^2 + 7x - 5 = 0$, in which
the coefficients of the odd powers of~$x$ are of like sign, and the coefficients of the even
powers (including the constant term) are of the opposite sign. Verify by taking $x= -p$,
where $p$ is positive.

\item[3.] $x^3 + a^2 x + b^2 = 0$ has two imaginary roots if $b\ne 0$.

\item[4.] For $n$~even, $x^n - 1 = 0$ has only two real roots.

\item[5.] For $n$~odd, $x^n - 1 = 0$ has only one real root.

\item[6.] For $n$~even, $x^n + 1 = 0$ has no real root; for $n$~odd, only one.

\item[7.] $x^4 + 12x^2 + 5x - 9 = 0$ has just two imaginary roots.

\item[8.] $x^4 + a^2 x^2 + b^2 x - c^2 = 0$ ($c\ne 0$) has just two imaginary roots.

\item[9.] Descartes' rule enables us to find the exact number of positive roots only when
all the coefficients are of like sign or when
\[
f(x) = x^n + p_1 x^{n-1} + \dotsb + p_{n-s} x^s
     - p_{n-s+1} x^{s-1} - \dotsb - p_n = 0,
\]
each $p_i$ being $\geqq 0$. Without using that rule, show that the latter equation has one
and only one positive root~$r$. Hints: There is a positive root~$r$ by~§63 ($a=0$, $b=\infty$).
Denote by~$P(x)$ the quotient of the sum of the positive terms by~$x^s$, and by $-N(x)$
that of the negative terms. Then $N(x)$ is a sum of powers of~$1/x$ with positive coefficients.
\begin{align*}
\text{If}\quad x>r,\qquad P(x)>P(r),\qquad N(x)<N(r),\qquad f(x)>0; \\
\text{If}\quad x<r,\qquad P(x)<P(r),\qquad N(x)>N(r),\qquad f(x)<0.
\end{align*}

\item[10.] Prove that we obtain an upper limit to the number of real roots of $f(x)=0$
between $a$ and~$b$, if we set
\[
x = \frac{a+by}{1+y}\qquad
\left(\therefore y=\frac{x-a}{b-x}\right),
\]
multiply by $(1+y)^n$, and apply Descartes' rule to the resulting equation in~$y$.

%% -----File: 081.png---Folio 75-------

\item[11.] Show by the method of Ex.~10 that there is a single root between $2$ and~$4$ of
$x^3 + x^2 - 17x + 15 = 0$. Here we have $27y^3 + 3y^2 - 23y - 7 = 0$.

\item[12.] In the astronomical problem of three bodies occurs the equation
\[
r^5 + (3 - \mu)r^4 + (3 - 2\mu )r^3 - \mu r^2 - 2\mu r - \mu = 0,
\]
where $0 < \mu < 1$.   Why is there a single positive real root?

\item[13.] Prove that $x^5 + x^3 - x^2 + 2x - 3 = 0$ has four imaginary roots by applying Descartes'
rule to the equation in~$y$ whose roots are the squares of the roots of the former.
Transpose the odd powers, square each new member, and replace $x^2$ by~$y$.

\item[14.] As in Ex.~13 prove that $x^3 + x^2 + 8x + 6 = 0$ has imaginary roots.

\item[15.] If a real equation $f(x)=0$ of degree~$n$ has $n$~real roots, the number of positive
roots is exactly equal to the number~$V$ of variations of sign. Hint: consider also
$f(-x)$.

\item[16.] Show that $x^3 - x^2 + 2x + 1 = 0$ has no positive root. Hint: multiply by $x + 1$.
\end{Problems}
\end{Exercises}


\Section{68.}{Sturm's Method} Let $f(x) = 0$ be an equation with real coefficients,
\index{Sturm's functions|(}%
and $f'(x)$ the first derivative of~$f(x)$. The first step of the usual process
of finding the greatest common divisor of $f(x)$ and $f'(x)$, if it exists, consists
\index{Greatest common divisor}%
in dividing~$f$ by~$f'$ until we obtain a remainder $r(x)$, whose degree
is less than that of~$f'$. Then, if $q_1$ is the quotient, we have $f = q_1 f' + r$.
Instead of dividing~$f'$ by~$r$, as in the greatest common divisor process, and
proceeding further in that manner, we write $f_2 = -r$, divide~$f'$ by~$f_2$, and
denote by~$f_3$ the remainder with its sign changed. Thus
\[
f   = q_1 f'  - f_2,\qquad
f'  = q_2 f_2 - f_3,\qquad
f_2 = q_3 f_3 - f_4,\dotsc.
\]

The latter equations, in which each remainder is exhibited as the negative
of a polynomial~$f_i$, yield a modified process, just as effective as the
usual process, of finding the greatest common divisor~$G$ of $f(x)$ and~$f'(x)$ if
it exists.

Suppose that $-f_4$ is the first constant remainder. If $f_4 = 0$, then $f_3 = G$,
since $f_3$ divides~$f_2$ and hence also $f'$ and~$f$ (as shown by using our above
equations in reverse order); while, conversely, any common divisor of
$f$ and~$f'$ divides $f_2$ and hence also~$f_3$.

But if $f_4$ is a constant $\ne 0$, $f$ and~$f'$ have no common divisor involving~$x$.
This case arises if and only if $f(x) = 0$ has no multiple root~(§58),
and is the only case considered in~§§69--71.

Before stating Sturm's theorem in general, we shall state it for a
numerical case and illustrate its use.

%% -----File: 082.png---Folio 76-------


\begin{Example} $f(x) =x^3 +4x^2 -7$. Then $f'=3x^2 +8x$,
\begin{alignat*}{3}
f   &= (\tfrac{1}{3}x + \tfrac{4}{9})f'- f_2, &&\qquad
f_2 &\equiv{}& \tfrac{32}{9}x + 7, \\
f'  &= (\tfrac{27}{32}x + \tfrac{603}{1024})f_2 - f_3, &&\qquad
f_3 &={}& \tfrac{4221}{1024}.
\end{alignat*}

For\footnote
  {Before going further, check that the preceding relations hold when $x= 1$ by inserting
the computed values of $f$, $f'$, $f_2$ for $x =1$. Experience shows that most students make some
error in finding $f_2, f_3, \dotsc$, so that checking is essential.}
$x = 1$, the signs of $f$, $f'$, $f_2$, $f_3$, are  ${}-{}+{}+{}+{}$, showing a single variation of
consecutive signs. For $x = 2$, the signs are ${}+{}+{}+{}+{}$, showing no variation of sign.
Sturm's theorem states that there is a \emph{single} real root between $1$ and~$2$. For $x= -\infty$,
the signs are ${}-{}+{}-{}+{}$, showing 3~variations of sign. The theorem states that there
are $3-1=2$ real roots between $-\infty$ and~$1$. Similarly,
\[
\begin{array}{c@{\quad}|c|c}
 x & \text{Signs} & \text{Variations} \\
\hline
-1 &    {}-{}-{}+{}+{} & 1 \\
-2 &    {}+{}-{}-{}+{} & 2 \\
-3 &    {}+{}+{}-{}+{} & 2 \\
-4 &    {}-{}+{}-{}+{} & 3
\end{array}
\]
Hence there is a single real root between $-2$ and~$-1$, and a single one between $-4$
and~$-3$. Each real root has now been \emph{isolated} since we have found two numbers
such that a single real root lies between these two numbers or is equal to one of them.

Some of the preceding computation was unnecessary. After isolating a root between
$-2$ and~$-1$, we know that the remaining root is isolated between $-\infty$ and~$-2$. But
before we can compute it by Horner's method, we need closer limits for it. For that
purpose it is unnecessary to find the signs of all four functions, but merely the sign
of~$f$~(§63).
\end{Example}


\Section{69.}{Sturm's Theorem}
\begin{Thm}
Let $f(x) =0$ be an equation with real coefficients
and without multiple roots. Modify the usual process of seeking the greatest
common divisor of $f(x)$ and its first derivative\footnote
  {The notation $f_1$ instead of the usual~$f'$, and similarly $f_0$ instead of~$f$, is used to regularize
  the notation of all the~$f$'s, and enables us to write any one of the equations~\Eq{1}
  in the single notation~\Eq{3}.}
$f_1(x)$ by exhibiting each
remainder as the negative of a polynomial~$f_i$:
\[
f   = q_1 f_1 - f_2,\
f_1 = q_2 f_2 - f_3,\
f_2 = q_3 f_3 - f_4, \dotsc,\
f_{n-2} = q_{n-1}f_{n-1} - f_n,
\Tag{1}
\]
where\footnote
  {If the division process did not yield ultimately a constant remainder $\ne 0$, $f$ and~$f_1$
  would have a common factor involving~$x$, and hence $f(x) =0$ a multiple root.}
$f_n$ is a constant $\ne 0$. If $a$ and~$b$ are real numbers, $a<b$, neither
%% -----File: 083.png---Folio 77-------
a root of $f(x) = 0$, the number of real roots of $f(x) = 0$ between $a$ and~$b$ is equal
to the excess of the number of variations of sign of
\[
f(x),\quad f_1(x),\quad f_2(x), \dotsc, f_{n-1}(x),\quad f_n
\Tag{2}
\]
for $x = a$ over the number of variations of sign for $x = b$. Terms which vanish
are to be dropped out before counting the variations of sign.
\end{Thm}

For brevity, let $V_x$ denote the number of variations of sign of the
numbers~\Eq{2} when $x$ is a particular real number not a root of $f(x)= 0$.

First, if $x_1$ and~$x_2$ are real numbers such that no one of the continuous
functions~\Eq{2} vanishes for a value of~$x$ between $x_1$ and~$x_2$ or for $x = x_1$ or
$x = x_2$, the values of any one of these functions for $x = x_1$ and $x = x_2$ are
both positive or both negative~(§63), and therefore $V_{x_1} = V_{x_2}$.

Second, let $\rho$ be a root of $f_i(x) = 0$, where $1 \leqq i < n$. Then
\[
f_{i-1}(x) = q_i f_i(x) - f_{i+1}(x)
\Tag{3}
\]
and the equations~\Eq{1} following this one show that $f_{i-1}(x)$ and~$f_i(x)$ have
no common divisor involving~$x$ (since it would divide the constant~$f_n$).
By hypothesis, $f_i(x)$ has the factor $x-\rho$. Hence $f_{i-1}(x)$ does not have
this factor $x-\rho$. Thus, by~\Eq{3},
\[
f_{i-1}(\rho) = -f_{i+1}(\rho) \ne 0.
\]
Hence, if $p$ is a sufficiently small positive number, the values of
\[
f_{i-1}(x),\quad f_i(x),\quad f_{i+1}(x)
\]
for $x = \rho - p$ show just one variation of sign, since the first and third
values are of opposite sign, and for $x = \rho + p$ show just one variation of
sign, and therefore show no change in the number of variations of sign
for the two values of~$x$.

It follows from the first and second cases that $V_\alpha = V_\beta$ if $\alpha$ and~$\beta$ are
real numbers for neither of which any one of the functions~\Eq{2} vanishes
and such that no root of $f(x) = 0$ lies between $\alpha$ and~$\beta$.

Third, let $r$ be a root of $f(x) = 0$. By Taylor's theorem~\Eq{8} of~§56,
\begin{align*}
f(r - p) &=     -pf'(r) + \tfrac{1}{2} p^2 f''(r) - \dotsb, \\
f(r + p) &= \Neg pf'(r) + \tfrac{1}{2} p^2 f''(r) + \dotsb.
\end{align*}
If $p$ is a sufficiently small positive number, each of these polynomials in~$p$
has the same sign as its first term. For, after removing the factor~$p$,
%% -----File: 084.png---Folio 78-------
we obtain a quotient of the form $a_0 + s$, where $s = a_1 p + a_2 p^2 + \dotsb$ is
numerically less than~$a_0$ for all values of~$p$ sufficiently small~(§62). Hence
if $f'(r)$ is positive, $f(r-p)$ is negative and $f(r+p)$ is positive, so that the
terms $f(x)$, $f_1(x)\equiv f'(x)$ have the signs ${}-{}+{}$ for $x = r-p$ and the signs
${}+{}+{}$ for $x = r+p$. If $f'(r)$ is negative, these signs are ${}+{}-{}$ and~${}-{}-{}$
respectively. In each case, $f(x)$, $f_1(x)$ show one more variation of sign
for $x = r-p$ than for $x = r+p$. Evidently $p$ may be chosen so small that
no one of the functions $f_1(x), \dotsc, f_n$ vanishes for either $x=r-p$ or $x=r+p$,
and such that $f_1(x)$ does not vanish for a value of~$x$ between $r-p$ and~$r+p$,
so that $f(x) = 0$ has the single real root~$r$ between these limits~(§65).
Hence by the first and second cases, $f_1, \dotsc, f_n$ show the same number
of variations of sign for $x = r-p$ as for $x = r+p$. Thus, for the entire
series of functions~\Eq{2}, we have
\[
V_{r-p} - V_{r+p} = 1.
\Tag{4}
\]

The real roots of $f(x)= 0$ within the main interval from $a$ to~$b$ (i.e., the % [** PP: Not italicizing i.e.]
aggregate of numbers between $a$ and~$b$) separate it into intervals. By
the earlier result, $V_x$ has the same value for all numbers in the same
interval. By the present result~\Eq{4}, the value~$V_x$ in any interval exceeds
the value for the next interval by unity. Hence $V_a$ exceeds~$V_b$ by the
number of real roots between $a$ and~$b$.
\index{Interval}%

\begin{Corollary}
If $a<b$, then $V_a\geqq V_{b}$.
\end{Corollary}

A violation of this Corollary usually indicates an error in the computation
of Sturm's functions~\Eq{2}.


\begin{Exercises}{Page78}

Isolate by Sturm's theorem the real roots of
\begin{Problems}[2]
\item[1.] $x^3 +2x +20 = 0$.

\item[2.] $x^3 +x-3 = 0$.
\end{Problems}
\end{Exercises}


% [** PP: No ToC entry]
\Section{70.}{Simplifications of Sturm's Functions} In order to avoid fractions,
we may first multiply $f(x)$ by a \emph{positive} constant before dividing it by
by $f_1(x)$, and similarly multiply~$f_1$ by a positive constant before dividing it
by~$f_2$, etc. Moreover, we may remove from any~$f_i$ any factor~$k_i$ which is
either a positive constant or a polynomial in~$x$ positive for\footnote
  {Usually we would require that $k_i$ be positive for all values of~$x$, since we usually
wish to employ the limits $-\infty$ and~$+\infty$.}
$a\leqq x \leqq b$,
and use the remaining factor~$F_i$ as the next divisor.

To prove that Sturm's theorem remains true when these modified
%% -----File: 085.png---Folio 79-------
functions $f$, $F_1, \dotsc, F_m$ are employed in place of functions~\Eq{2}, consider
the equations replacing~\Eq{1}:
\begin{align*}
f_1 = k_1 F_1,\qquad
c_2 f   &= q_1 F_1 - k_2 F_2,\qquad c_3 F_1 = q_2 F_2 - k_3 F_3, \\
c_4 F_2 &= q_3 F_3 - k_4 F_4, \dotsc,
c_m F_{m-2} = q_{m-1} F_{m-1} - k_m F_m,
\end{align*}
in which $c_2, c_3, \dotsc$ are positive constants and $F_m$ is a constant $\ne 0$. A
common divisor (involving~$x$) of~$F_{i-1}$ and~$F_i$ would divide $F_{i-2},\dotsc,
F_2, F_1$, $f$,~$f_1$, whereas $f(x)=0$ has no multiple roots. Hence if $\rho$ is a root
of $F_i(x)=0$, then $F_{i-1}(\rho)\ne 0$ and
\[
c_{i+1} F_{i-1}(\rho) = -k_{i+1}(\rho) F_{i+1}(\rho),\qquad
c_{i+1}>0,\qquad k_{i+1}(\rho)>0.
\]
Thus $F_{i-1}$ and~$F_{i+1}$ have opposite signs for $x = \rho$. We proceed as in~§69.


\begin{Example}[1.]
If $f(x) = x^3 + 6x - 10$, $f_1 = 3(x^2 + 2)$ is always positive. Hence we may
employ $f$ and~$F_1 = 1$. For $x =-\infty$, there is one variation of sign; for $x =+\infty$, no
variation. Hence there is a single real root; it lies between $1$ and~$2$.
\end{Example}

\begin{Example}[2.]
\index{Cubic equation!number of real roots}% [** PP: Added ``of'']
If $f(x) = 2x^4 - 13x^2 - 10x - 19$, we may take
\[
f_1 = 4x^3 - 13x - 5.
\]
Then
\[
2f  = xf_1 - f_2,\qquad
f_2 = 13x^2 + 15x + 38 = 13(x + \tfrac{15}{26})^2 + \tfrac{1751}{52}.
\]
Since $f_2$ is always positive, we need go no further (we may take $F_2 = 1$). For $x =-\infty$,
the signs are ${}+{}-{}+{}$; for $x = +\infty$, ${}+{}+{}+{}$. Hence there are two real roots. The
signs for $x = 0$ are ${}-{}-{}+{}$. Hence one real root is positive and the other negative.
\end{Example}


\begin{Exercises}{Page79}

Isolate by Sturm's theorem the real roots of
\begin{Problems}[2]

\item[1.] $x^3 + 3x^2 - 2x - 5 = 0$.

\item[2.] $x^4 + 12x^2 + 5x - 9 = 0$.

\ResetCols{2}

\item[3.] $x^3 - 7x - 7 = 0$.

\item[4.] $3x^4 - 6x^2 + 8x - 3 = 0$.

\ResetCols{1}

\item[5.] $x^6 + 6x^5 - 30x^2 - 12x - 9 = 0$ [stop with~$f_2$].

\item[6.] $x^4 - 8x^3 + 25x^2 - 36x + 8 = 0$.

\item[7.] For $f = x^3 + px + q$ ($p\ne 0$), show that $f_1 = 3x^2 + p$, $f_2 = -2px - 3q$,
\[
4p^2 f_1 = (-6px + 9q)f_2 - f_3,\quad
f_3 = -4p^3 - 27q^2,
\]
so that $f_3$ is the discriminant~$\Delta$~(§44).  Let $[p]$ denote the sign of~$p$. Then the signs
of $f$, $f_1$, $f_2$, $f_3$ are
\begin{align*}
&{}-{}+{}+ [p]\ [\Delta]\quad \text{for $x = -\infty$}, \\
&{}+{}+{}- [p]\ [\Delta]\quad \text{for $x = +\infty$}.
\end{align*}
For $\Delta$ negative there is a single real root. For $\Delta$ positive and therefore $p$~negative,
there are three distinct real roots. For $\Delta = 0$, $f_2$~is a divisor of~$f_1$ and~$f$, so that
$x = -3q/(2p)$ is a double root.

%% -----File: 086.png---Folio 80-------

\item[8.] Prove that if one of Sturm's functions has $p$~imaginary roots, the initial equation
has at least $p$~imaginary roots.

\item[9.] State Sturm's theorem so as to include the possibility of~$a$, or~$b$, or both $a$ and~$b$ being roots of $f(x)=0$.
\end{Problems}
\end{Exercises}


% [** PP: ToC reads Sturm's Functions for a General Quartic Equation]
\Section{71.}{Sturm's Functions for a Quartic Equation} For the reduced quartic
equation $f(z) =0$,
\index{Quartic equation|(}%
\[
\left\{
\begin{aligned}
f   &= z^4 + qz^2 + rz + s, \\
f_1 &= 4z^3 + 2qz + r, \\
f_2 &= -2qz^2 - 3rz - 4s.
\end{aligned}
\right.
\Tag{5}
\]
Let $q\ne 0$ and divide $q^2 f_1$ by~$f_2$. The negative of the remainder is
\[
f_3 = Lz - 12rs - rq^2,\qquad
L = 8qs - 2q^3 - 9r^2.
\Tag{6}
\]
Let $L\ne 0$. Then $f_4$ is a constant which is zero if and only if $f=0$ has
multiple roots, i.e., if its discriminant~$\Delta$ is zero. We therefore desire~$f_4$ % [** PP: Not italicizing i.e.]
expressed as a multiple of~$\Delta$. By~§50,
\[
\Delta = -4P^3 - 27Q^2,\qquad
P = -4s - \frac{q^2}{3},\qquad
Q = \tfrac{8}{3}qs - r^2 - \tfrac{2}{27}q^3.
\Tag{7}
\]
We may employ $P$ and~$Q$ to eliminate
\[
4s  = -P - \frac{q^2}{3},\qquad
r^2 = -Q - \tfrac{2}{3}qP - \tfrac{8}{27}q^3.
\Tag{8}
\]
We divide $L^2 f_2$ by
\[
f_3 = Lz + 3rP,\qquad
L = 9Q + 4qP.
\Tag{9}
\]
The negative of the remainder\footnote
  {Found directly by the Remainder Theorem~(§14) by inserting the root $z = -3rP/L$
  of $f_3=0$ into $L^2 f_2$.}
is
\[
18r^2 qP^2 - 9r^2 LP + 4sL^2 = q^2 \Delta.
\Tag{10}
\]
The left member is easily reduced to~$q^2\Delta$. Inserting the values~\Eq{8} and
replacing $L^2$ by $L(9Q + 4qP)$, we get
\[
-18qQP^2 - 12q^2 P^3 - \tfrac{16}{3}q^4 P^2 + 2qP^2 L
  + \tfrac{4}{3}q^3 PL - 3q^2 QL.
\]
Replacing $L$ by its value~\Eq{9}, we get~$q^2\Delta$. Hence we may take
\[
f_4 = \Delta.
\Tag{11}
\]
Hence if $qL\Delta\ne 0$, we may take \Eq{5}, \Eq{9}, \Eq{11} as Sturm's functions.

%% -----File: 087.png---Folio 81-------

Denote the sign of~$q$ by~$[q]$. The signs of Sturm's functions are
\[
\begin{array}{rrrrr@{\quad}l}
+ & - & -[q] & -[L] & [\Delta] & \text{for $x = -\infty$}, \\
+ & + & -[q] &  [L] & [\Delta] & \text{for $x = +\infty$}.
\end{array}
\]

First, let $\Delta > 0$. If $q$ is negative and $L$ is positive, the signs are
${}+{}-{}+{}-{}+{}$ and~${}+{}+{}+{}+{}+{}$, so that there are four real roots. In each
of the remaining three cases for $q$ and~$L$, there are two variations of sign
in either of the two series and hence there is no real root.

Next, let $\Delta < 0$. In each of the three cases in which $q$ and~$L$ are not
both positive, there are three variations of sign in the first series and one
variation in the second, and hence just two real roots. If $q$ and~$L$ are
both positive, the number of variations is~$1$ in the first series and~$3$ in the
second, so that this case is excluded by the Corollary to Sturm's theorem.
To give a direct proof, note that, by the value of~$L$ in~\Eq{6}, $L>0$, $q>0$
imply $4s > q^2$, i.e., $s>0$, and hence, by~\Eq{7}, $P$~is negative, so that each term
of~\Eq{10} is $\geqq 0$, whence $\Delta > 0$. % [** PP: Added space]

Hence, if $qL\Delta \ne 0$, there are four distinct real roots if and only if $\Delta$
and~$L$ are positive, and $q$~negative; two distinct real and two imaginary
roots if and only if $\Delta$ is negative.

Combining this result with that in Ex.~4 below, we obtain the

\begin{Theorem}
If the discriminant~$\Delta$ of $z^4 + qz^2 + rz + s = 0$ is negative, there
are two distinct real roots and two imaginary roots; if $\Delta > 0$, $q<0$, $L>0$,
four distinct real roots; if $\Delta > 0$ and either $q\geqq 0$ or $L\leqq 0$, no real roots.
Here $L = 8qs - 2q^3 - 9r^2$.
\end{Theorem}
\index{Discriminant!of quartic}%
\index{Quartic equation|)}%

Our discussion furnished also the series of Sturm functions, which
may be used in isolating the roots.


\begin{Exercises}{}

\begin{Problems}

\item[1.] If $q\Delta\ne 0$, $L = 0$, then $f_3 = 3rP$ is not zero (there being no multiple root) and its
sign is immaterial in determining the number of real roots. Prove that there are just
two real roots if $q<0$, and none if $q>0$. By~\Eq{10}, $q$ has the same sign as~$\Delta$.

\item[2.] If $r\Delta\ne 0$, $q = 0$, obtain~$-f_3$ by substituting $z = -4s/(3r)$ in~$f_1$. Show that we may
take $f_3 = r\Delta$ and that there are just two real roots if $\Delta < 0$, and no real
roots if $\Delta >0$.

\item[3.] If $\Delta \ne 0$, $q = r = 0$, prove that there are just two real roots if $\Delta<0$, and no real
roots if $\Delta > 0$. Since $\Delta = 256s^3$, check by solving $z^4 + s = 0$.

\item[4.] If $\Delta \ne 0$, $qL = 0$, there are just two real roots if $\Delta < 0$, and no real roots if $\Delta > 0$.
[Combine the results in Exs.~1--3.]

\item[5.] Apply the theorem to Exs.~2, 4, 6 of~§70.

\item[6.] Isolate the real roots of Exs.~3, 4, 5 of~§48.
\end{Problems}
\end{Exercises}

%% -----File: 088.png---Folio 82-------


% [** PP: No ToC entry]
\Section[Sturm's Theorem for Multiple Roots]
{72.}{Sturm's Theorem for the Case of Multiple Roots.} We might
\index{Multiple roots}%
remove the multiple roots by dividing $f(x)$ by\footnote
  {The degree of~$f(x)$ is not~$n$,
  nor was it necessarily~$n$ in~§69.}
$f_n(x)$, the greatest common
divisor of $f(x)$ and $f_1 = f'(x)$; but this would involve considerable
work, besides wasting the valuable information in hand. As before, we
suppose $f(a)$ and $f(b)$ different from zero. We have equations~\Eq{1} in
which $f_n$ is now not a constant.

\begin{Thm}%
The difference $V_a - V_b$ is the number of real roots between $a$ and~$b$, each
multiple root being counted only once.
\end{Thm}

If $\rho$ is a root of $f_i(x) = 0$, but not a multiple root of $f(x)= 0$, then $f_{i-1}(\rho)$
is not zero. For, if it were zero, $x-\rho$ would by~\Eq{1} be a common factor
of $f$ and~$f_1$. We may now proceed as in the second case in~§69.

The third case requires a modified proof only when $r$ is a multiple root.
Let $r$ be a root of multiplicity~$m$, $m\geqq 2$. Then $f(r)$, $f'(r), \dotsc, f^{(m-1)}(r)$
are zero and, by Taylor's theorem,
\begin{align*}
f(r+p)  &= \frac{p^m}{1·2\dotsm m} f^{(m)}(r) + \dotsb, \\
f'(r+p) &= \frac{p^{m-1}}{1·2\dotsm (m-1)} f^{(m)}(r) + \dotsb.
\end{align*}
These have like signs if $p$ is a positive number so small that the signs of
the polynomials are those of their first terms. Similarly, $f(r-p)$ and
$f'(r-p)$ have opposite signs. Hence $f$ and~$f_1$ show one more variation
of sign for $x = r-p$ than for $x = r+p$. Now $(x-r)^{m-1}$ is a factor of~$f$
and~$f_1$ and hence, by~\Eq{1}, of $f_2, \dotsc, f_n$. Let their quotients by this
factor be $\phi, \phi_1, \dotsc, \phi_n$. Then equations~\Eq{1} hold after the $f'$s are replaced
by the~$\phi$'s. Taking $p$ so small that $\phi_1(x) = 0$ has no root between $r-p$
and $r+p$, we see by the first and second cases in~§69 that $\phi_1, \dotsc, \phi_n$
show the same number of variations of sign for $x = r-p$ as for $x = r+p$.
The same is true for $f_1, \dotsc, f_n$ since the products of $\phi_1, \dotsc, \phi_n$ by
$(x-r)^{m-1}$ have for a given~$x$ the same signs as $\phi_1, \dotsc, \phi_n $ or the same
signs as $-\phi_1, \dotsc, -\phi_n$. But the latter series evidently shows the
same number of variations of sign as $\phi_1, \dotsc, \phi_n$. Hence~\Eq{4} is proved
and consequently the present theorem.
\index{Sturm's functions|)}%

%% -----File: 089.png---Folio 83-------


\begin{Exercises}{Page83}

\begin{Problems}

\item[1.] For $f = x^4 - 8x^2 + 16$, prove that $F_1 = x^3 - 4x$, $F_2 = x^2 - 4$, $F_1 = xF_2$. Hence $n = 2$.
Verify that $V_{-\infty} = 2$, $V_{\infty} = 0$, and that there are just two real roots, each a double
root.
\end{Problems}

Discuss similarly the following equations.
\begin{Problems}[3]
\item[2.] $x^4 - 5x^3 + 9x^2 - 7x + 2 = 0$.

\item[3.] $x^4 + 2x^3 - 3x^2 - 4x + 4 = 0$.

\item[4.] $x^4 - x^2 - 2x + 2 = 0$.
\end{Problems}
\end{Exercises}


\Section{73.}{Budan's Theorem}
\index{Budan's theorem}%
\index{Derivative}%
\begin{Thm}%
Let $a$ and~$b$ be real numbers, $a<b$, neither\footnote
  {In case $a$ or~$b$
  is a root of $f(x)=0$, the theorem holds if we count the number of
  roots $>a$ and $\leqq b$. This inclusive theorem has been proved, by means of Rolle's
  theorem, by A.~Hurwitz, \textit{Mathematische Annalen}, Vol.~71, 1912, p.~584, who extended
  Budan's theorem from the case of a polynomial to a function~$f(x)$ which is real and
  regular for $a \leqq x < b$.}
a root of $f(x) = 0$, an equation of degree~$n$ with real coefficients. Let $V_a$
denote the number of variations of sign of
\[
f(x),\qquad f'(x),\qquad f''(x),\qquad \dotsc,\qquad f^{(n)}(x)
\Tag{12}
\]
for $x = a$, after vanishing terms have been deleted. Then $V_a - V_b$ is either
the number of real roots of $f(x) = 0$ between $a$ and~$b$ or exceeds the number
of those roots by a positive even integer. A root of multiplicity~$m$ is here
counted as $m$~roots.
\end{Thm}

\begin{Remark}
For example, if $f(x) = x^3 - 7x - 7$, then $f' = 3x^2 - 7$, $f'' = 6x$, $f''' = 6$. Their values
for $x = 3$, $4$, $-2$, $-1$ are tabulated below.
\[
\begin{array}{r|rrrr|c}
 x &  f & f' & f'' & f''' & \text{Variations} \\
\hline
 3 & -1 & 20 &  18 &    6 & 1 \\
 4 & 29 & 41 &  24 &    6 & 0 \\
-2 & -1 &  5 & -12 &    6 & 3 \\
-1 & -1 & -4 &  -6 &    6 & 1
\end{array}
\]
Hence the theorem shows that there is a single real root between $3$ and~$4$, and two
or no real roots between $-2$ and~$-1$. The theorem does not tell us the exact number
of roots between the latter limits. To decide this ambiguity, note that $f(-3/2) = +1/8$,
so that there is a single real root between $-2$ and~$-1.5$, and a single one between
$-1.5$ and~$-1$.
\end{Remark}

The proof is quite simple if no term of the series~\Eq{12} vanishes for
$x = a$ or for $x = b$ and if no two consecutive terms vanish for the same
value of~$x$ between $a$ and~$b$. Indeed, if no one of the terms vanishes for
$x_1 \leqq x\leqq x_2$, then $V_{x_1} = V_{x_2}$, since any term has the  same sign for $x = x_1$
as for $x = x_2$. Next, let $r$ be a root of $f^{(i)}(x)=0$, $a<r<b$. By hypothesis,
%% -----File: 090.png---Folio 84-------
the first derivative $f^{(i+1)}(x)$ of $f^{(i)}(x)$ is not zero for $x=r$. As in the third
step (now actually the case $i = 0$) in~§69, $f^{(i)}(x)$ and $f^{(i+1)}(x)$ show one
more variation of sign for $x = r-p$ than for $x = r+p$, where $p$ is a sufficiently
small positive number. If $i>0$, $f^{(i)}$ is preceded by a term $f^{(i-1)}$ in~\Eq{12}.
By hypothesis, $f^{(i-1)}(x) \ne 0$ for $x=r$ and hence has the same sign for
$x = r - p$ and $x = r + p$ when $p$ is sufficiently small. For these values of~$x$,
$f^{(i)}(x)$ has opposite signs. Hence $f^{(i-1)}$ and $f^{(i)}$ show one more or one less
variation of sign for $x = r-p$ than for $x = r+p$, so that $f^{(i-1)}$, $f^{(i)}$, $f^{(i+1)}$ show
two more variations or the same number of variations of sign.

Next, let no term of the series~\Eq{12} vanish for $x = a$ or for $x = b$, but
let several successive % [** PP: Typo succssive]
terms
\[
f^{(i)}(x),\qquad f^{(i+1)}(x), \dotsc, f^{(i+j-1)}(x)
\Tag{13}
\]
all vanish for a value~$r$ of~$x$ between $a$ and~$b$, while $f^{(i+j)}(r)$ is not zero,
but is say positive.\footnote
  {If negative, all signs in the table below are to be changed; but the conclusion holds.}
Let $I_1$ be the interval between $r-p$ and~$r$, and $I_2$
the interval between $r$ and~$r+p$. Let the positive number~$p$ be so small
that no one of the functions~\Eq{13} or $f^{(i+j)}(x)$ is zero in these intervals, so
that the last function remains positive. Hence $f^{(i+j-1)}(x)$ increases with~$x$
(since its derivative is positive) and is therefore negative in~$I_1$
and positive in~$I_2$. Thus $f^{(i+j-2)}(x)$ decreases in~$I_1$ and increases in~$I_2$
and hence is positive in each interval. In this manner we may verify the
signs in the following table:
\[
\begin{array}{c|cccccccc}
\multicolumn{2}{r}{f^{(i)}} & f^{(i+1)} & f^{(i+2)} & \ldots &
  f^{(i+j-3)} & f^{(i+j-2)} & f^{(i+j-1)} & f^{(i+j)} \\
%
I_1 & (-)^{j} & (-)^{j-1} & (-)^{j-2} & \ldots &
  - & + & - & +  \\
I_2 & + & + & + & \ldots & + & + & + & +
\end{array}
\]
Hence these functions show $j$~variations of sign in~$I_1$ and none in~$I_2$.

If $i>0$, the first term of~\Eq{13} is preceded by a function $f^{(i-1)}(x)$ which
is not zero for $x = r$, and hence not zero in $I_1$ or $I_2$ if $p$ is sufficiently small.
If $j$ is even, the signs of $f^{(i-1)}$ and $f^{(i)}$ are ${}+{}+{}$ or ${}-{}+{}$ in both $I_1$ and~$I_2$,
showing no loss in the number of variations of sign. If $j$ is odd, their
signs are
\[
\begin{array}{c|ccc}
I_1 & {}+{}-{} & \raisebox{-12pt}{\smash{\text{or}}} & {}-{}-{} \\
I_2 & {}+{}+{} &                                     & {}-{}+{}
\end{array}
\]
so that there is a loss or gain of a single variation of sign. Hence
\[
f^{(i-1)}, \qquad
f^{(i)},   \qquad
f^{(i+1)}  \quad \dotsc,\qquad
f^{(i+j)}
\]
%% -----File: 091.png---Folio 85-------
show a loss of $j$~variations of sign if $j$ is even, and a loss of $j±1$ if $j$ is odd,
and hence always a loss of an even number $\geqq 0$ of variations of sign.

If $i=0$, $f^{(i)}\equiv f$ has $r$ as a $j$-fold root and the functions in the table show
$j$~more variations of sign for $x = r-p$ than for $x = r+p$.

Thus, when no one of the functions~\Eq{12} vanishes for $x=a$ or for $x=b$,
the theorem follows as at the end of~§69 (with unity replaced by the
multiplicity of a root).

Finally, let one of the functions~\Eq{12}, other than $f(x)$ itself, vanish for
$x= a$ or for $x = b$. If $\delta$ is a sufficiently small positive number, all of the
$N$~roots of $f(x)=0$ between $a$ and~$b$ lie between $a+\delta$ and $b-\delta$, and for
the latter values no one of the functions~\Eq{12} is zero. By the above
proof,
\begin{gather*}
V_{a+\delta} - V_{b-\delta} = N + 2t, \\
V_a - V_{a + \delta} = 2j,\qquad
V_{b - \delta} - V_b = 2s,
\end{gather*}
where $t$, $j$, $s$ are integers $\geqq 0$. Hence $V_a - V_b = N+2(t+j+s)$.

Descartes' rule of signs~(§67) is a corollary to Budan's theorem. Consider
any equation with real coefficients
\index{Descartes' rule of signs}%
\[
f(x) \equiv a_0 x^n + a_1 x^{n-1} + \dotsb + a_{n-1}x + a_n = 0,
\]
having $a_n \ne 0$. For $x = 0$ the functions~\Eq{12} have the same signs as
\[
a_n,\qquad a_{n-1},\qquad \dotsc, \quad a_1,\qquad a_0.
\]
Hence $V_0$ is equal to the number~$V$ of variations of sign of~$f(x)$.

For $x= +\infty$, the functions all have the same sign, which is that of~$a_0$.
Thus $V_0 - V_{\infty} = V$ is either the number of positive roots or exceeds that
number by a positive even integer. Finally, Descartes' rule holds if
$a_n = 0$, as shown by removing the factors~$x$.
\index{Number!of roots|)}%


\begin{Exercises}{Page85}

Isolate by Budan's theorem the real roots of
\begin{Problems}[2]

\item[1.] $x^3 -x^2 -2x+1=0$.

\item[2.] $x^3 +3x^2 -2x-5=0$.

\ResetCols{1}

\item[3.] Prove that if $f{(a)}\ne 0$, $V_a$ equals the number of real roots $>a$ or exceeds that number by an even integer.

\item[4.] Prove that there is no root greater than a number making each of the functions~\Eq{12}
positive, if the leading coefficient of $f(x)$ is positive. (Newton.)

\item[5.] Hence verify that $x^4 -4x^3 - 3x + 23 = 0$ has no root~$>4$.

\item[6.] Show that $x^4 - 4x^3 + x^2 + 6x + 2 = 0$ has no root~$>3$.
\end{Problems}
\end{Exercises}

%% -----File: 092.png---Folio 86-------


\Chapter{VII}{Solution of Numerical Equations}
\index{Solution of numerical equations|(}%


\Section[Horner's Method]
{74.}{Horner's Method.\protect\footnotemark}\addtocounter{footnote}{1}%
  \footnotetext{W.~G. Horner, London Philosophical Transactions, 1819. Earlier (1804) by
  P.~Ruffini. See Bulletin American Math.\ Society, May, 1911.}
\addtocounter{footnote}{-1}%
\index{Horner's method}%
\index{Synthetic division|(}%
After we have isolated a real root of a real
equation by one of the methods in \ChapRef{VI}, we can compute the root
to any desired number of decimal places either by Horner's method,
which is available only for polynomial equations, or by Newton's method~(§75),
which is applicable also to logarithmic, trigonometric, and other
equations.

To find the root between $2$ and~$3$ of
\[
x^3 - 2x - 5 = 0,
\Tag{1}
\]
set $x = 2+p$. Direct substitution gives the \emph{transformed equation} for~$p$:
\index{Transformed equation}%
\[
p^3 + 6p^2 + 10p - 1 = 0.
\Tag{2}
\]
The method just used is laborious especially for equations of high degree.
We next explain a simpler method. Since $p = x -2$,
\[
x^3 - 2x - 5 \equiv (x-2)^3 + 6(x-2)^2 + 10(x-2) - 1,
\]
identically in~$x$. Hence $-1$ is the remainder obtained when the given
polynomial $x^3 - 2x - 5$ is divided by $x-2$. By inspection, the quotient~$Q$
is equal to
\[
(x-2)^2 + 6(x-2) + 10.
\]
Hence $10$ is the remainder obtained when $Q$ is divided by $x-2$. The
new quotient is equal to $(x-2) + 6$, and another division gives the
remainder~$6$. Hence to find the coefficients $6$, $10$, $-1$ of the terms following~$p^3$
in the transformed equation~\Eq{2}, we have only to divide the given
polynomial $x^3 - 2x - 5$ by $x-2$, the quotient~$Q$ by $x-2$, etc., and take
the remainders in reverse order. However, when this work is performed
by synthetic division~(§15) as tabulated below, no reversal of order is
%% -----File: 093.png---Folio 87-------
necessary, since the coefficients then appear on the page in their desired
order.
\[
\begin{array}{rRRRPc}
1 & 0 & -2 & -5 && \Lcol{2} \\
\cline{6-6}
  & 2 &  4 &  4 &&          \\
\cline{1-4}
1 & 2 &  \Rcol{2} & \mathbf{-1} && \\
  & 2 &  \Rcol{8} &             && \\
\cline{1-3}
1 & \Rcol{4} & \mathbf{10} &    && \\
  & \Rcol{2} &             &    && \\
\cline{1-2}
1 & \mathbf{6} &           &    &&
\end{array}
\]

Thus $1$, $6$, $10$, $-1$ are the coefficients of the desired equation~\Eq{2}.

To obtain an approximation to the decimal~$p$, we ignore for the moment
the terms involving $p^3$ and~$p^2$; then by $10p - 1 = 0$, $p=0.1$. But this
value is too large since the terms ignored are all positive. For $p=0.09$,
the polynomial in~\Eq{2} is found to be negative, while for $p=0.1$ it was just
seen to be positive. Hence $p = 0.09+h$, where $h$ is of the denomination
thousandths. The coefficients $1$, $6.27, \dotsc$ of the transformed equation
for~$h$ appear in heavy type just under the first zigzag line in the following
scheme:
\[
\begin{array}{r<{\qquad}l Pr@{}l Pr@{}l r@{}l}
1 & 6    && 10 &        && -1 &         &&   \Lcol{0.09}  \\
\cline{10-10}
  & 0.09 &&  0 & .5481  &&  0 & .949329 &&                \\
\cline{1-8}
1 & 6.09 && 10 & .5481 &\Lcol{}& \mathbf{-0} & \mathbf{.050671} && \\
  & 0.09 &&  0 & .5562 &\Lcol{}&             &                  && \\
\cline{1-5}
1 & 6.18 &\Lcol{}& \mathbf{11} & \mathbf{.1043} & &  &\Rcol{}& &
\smash[b]{\raisebox{-1.5ex}{$\dfrac{0.05}{11.1}$}} \\
  & 0.09 &\Lcol{}&             &                & &  &\Rcol{}& & \\
\cline{1-2}
1 & \textbf{6.27} &&           &                & &  &\Rcol{}& $=\,$&$0.004$\\
\cline{9-10}
  & 0.004 && 0 & .025096 && 0 & .044517584     &  &     \\
\cline{1-8}
1 & 6.274 &&11 & .129396 &\Lcol{}& \mathbf{-0} & \mathbf{.006153416} && \\
  & 0.004 && 0 & .025112 &\Lcol{}&             &                     && \\
\cline{1-5}
1 & 6.278 &\Lcol{}& \mathbf{11} & \mathbf{.154508} && & && \\
  & 0.004 &\Lcol{}&             &                  && & && \\
\cline{1-2}
1 & \textbf{6.282} &&           &                  && & &&
\end{array}
\]
Hence $x=2.094+t$, where $t$ is a root of
\[
t^3 + 6.282t^2 + 11.154508t - 0.006153416 = 0.
\]
By the last two terms, $t$ is between $0.0005$ and~$0.0006$. Then the value
%% -----File: 094.png---Folio 88-------
of $C\equiv t^3 +6.282t^2$ is found to lie between $0.00000157$ and~$0.00000227$.
Hence we may ignore~$C$ provided the constant term be reduced by an
amount between these limits. Whichever of the two limits we use, we
obtain the same dividend below correct to 6~decimal places.
\[
\begin{array}{r>{\quad}r<{\quad}l@{}l}
\multicolumn{1}{r|}{1\xatop{1}.\xatop{1}\xatop{5}4508}
& 0.006151 & \multicolumn{1}{|l@{}}{0.000551} & =t \\
\cline{1-1} \cline{3-3}
& 5577 & \\[-8pt]
& \settowidth{\TmpLen}{5577}\rule{\TmpLen}{0.5pt} & \\
&  574 & \\
&  558 & \\[-8pt]
& \settowidth{\TmpLen}{558}\rule{\TmpLen}{0.5pt} & \\
&   16 & \\
&   11 & \\[-8pt]
& \settowidth{\TmpLen}{11}\rule{\TmpLen}{0.5pt} & \\
&    5 &
\end{array}
\]
Since the quotient is $0.0005+$, only two decimal places of the divisor are
used, except to see by inspection how much is to be carried when making
the first multiplication. Hence we mark a cross above the figure~$5$ in
the hundredths place of the divisor and use only~$11.15$. Before making
the multiplication by the second significant figure~$5$ of the quotient~$t$,
we mark a cross over the figure~$1$ in the tenths place of the divisor and
hence use only~$11.1$. Thus $x = 2.0945514+$, with doubt only as to whether
the last figure should be $4$ or~$5$.

If we require a greater number of decimal places, it is not necessary
to go back and construct a new transformed equation from the equation
in~$t$. We have only to revise our preceding dividend on the basis of our
present better value of~$t$. We now know that $t$ is between $0.000551$ and
$0.000552$. To compute the new value of the correction~$C$, in which we
may evidently ignore~$t^3$, we use logarithms.
\[
% [** PP: Alignment: log integer mantissa = integer mantissa, x 2]
\begin{array}{rr@{}l@{}>{{}}c<{{}}@{}r@{}l@{}>{\qquad}rr@{}l@{}>{{}}c<{{}}@{}r@{}l@{}}
\log & 5&.51   & = &  &.74115 & \log & 5&.52   & = && .74194 \\
\therefore
\log & 5&.51^2 & = & 1&.48230 &\therefore
                                \log & 5&.52^2 & = &1&.48388 \\
\log & 6&.282  & = &  &.79810 & \log & 6&.282  & = & &.79810 \\
\cline{5-6} \cline{11-12}
\log & 190&.72 & = & 2&.28040 & \log &191&.42  & = &2&.28198 \Strut
\end{array}
\]
Hence $C$ is between $0.000001907$ and $0.000001915$. Whichever of the two
limits we use, we obtain the same new dividend below correct to 8~decimal
places.
%% -----File: 095.png---Folio 89-------
\[
\begin{array}{r>{\quad}r<{\quad}l@{}l}
\multicolumn{1}{r|}{1\xatop{1}.\xatop{1}\xatop{5}\xatop{4}\xatop{5}08}
& 0.00615150 & \multicolumn{1}{|l@{}}{0.00055148} \\
\cline{1-1} \cline{3-3}
& 557725 & \\[-8pt]
& \settowidth{\TmpLen}{557725}\rule{\TmpLen}{0.5pt} & \\
&  57425 & \\
&  55773 & \\[-8pt]
& \settowidth{\TmpLen}{55773}\rule{\TmpLen}{0.5pt} & \\
&   1652 & \\
&   1115 & \\[-8pt]
& \settowidth{\TmpLen}{1115}\rule{\TmpLen}{0.5pt} & \\
&    537 & \\
&    446 & \\[-8pt]
& \settowidth{\TmpLen}{446}\rule{\TmpLen}{0.5pt} & \\
&     91 & \\
&     89 & \\[-8pt]
& \settowidth{\TmpLen}{89}\rule{\TmpLen}{0.5pt} & \\
&      2 &
\end{array}
\]
Hence, finally, $x=2.094551482$, with doubt only as to the last figure.


\begin{Exercises}{Page89}

(The number of transformations made by synthetic division should be about half
the number of significant figures desired for a root.)

By one of the methods in \ChapRef{VI}, isolate each real root of the following equations,
and compute each real root to 5~decimal places.
\begin{Problems}[2]

\item[1.] $x^3 +2x+20=0$.

\item[2.] $x^3 +3x^2 -2x-5=0$.

\ResetCols{2}

\item[3.] $x^3 +x^2 -2x-1=0$.

\item[4.] $x^4 +4x^3 -17.5x^2 -18x+58.5=0$.

\ResetCols{2}

\item[5.] $x^4 -11,727x+40,385=0$.

\item[6.] $x^3 =10$.
\end{Problems}

Find to 7~decimal places all the real roots of
\begin{Problems}[2]
\item[7.] $x^3 +4x^2 -7=0$.

\item[8.] $x^3 -7x-7=0$.
\end{Problems}

Find to 8~decimal places
\begin{Problems}
\item[9.] The root between $2$ and~$3$ of $x^3 -x-9=0$ (make only 3~transformations).

\item[10.] The real cube root of~$7.976$.

\item[11.] The abscissa of the real point of intersection of the conics $y=x^2$, $xy+x+3y-6=0$.

\item[12.] Find to 3~decimal places the abscissas of the points of intersection of $x^2+y^2=9$,
$y=x^2-x$.

\item[13.] A sphere two feet in diameter is formed of a kind of wood a cubic foot of which
weighs two-thirds as much as a cubic foot of water (i.e., the specific gravity of the wood
is~$2/3$). Find to four significant figures the depth~$h$ to which the floating sphere
will sink in water.
\index{Specific gravity}%

Hints: The volume of a sphere of radius~$r$ is~$\tfrac{4}{3}\pi r^3$. Hence our sphere whose radius
%% -----File: 096.png---Folio 90-------
is $1$~foot weighs as much as $\tfrac{4}{3}\pi·\tfrac{2}{3}$ cubic feet of water. The volume of the submerged
portion of the sphere is $\pi h^2 (r-\tfrac{1}{3}h)$ cubic feet. Since this is also the volume of the displaced
water, its value for $r=1$ must equal $\tfrac{4}{3}\pi·\tfrac{2}{3}$. Hence
$h^3 - 3h^2 + \tfrac{8}{3} = 0$.

\item[14.] If the specific gravity of cork is~$1/4$, find to four significant figures how far a
cork sphere two feet in diameter will sink in water.

\item[15.] Compute $\cos 20°$ to four decimal places by use of
\[
\cos 3A = 4\cos^3 A - 3\cos A,\qquad
\cos 60° = \tfrac{1}{2}.
\]

\item[16.] Three intersecting edges of a rectangular parallelopiped are of lengths $6$,~$8$,
and $10$~feet. If the volume is increased by $300$~cubic feet by equal elongations of the
edges, find the elongation to three decimal places.

\item[17.] Given that the volume of a right circular cylinder is $\alpha\pi$  and the total area of
its surface is~$2\beta\pi$, prove that the radius~$r$ of its base is a root of $r^3 - \beta r + \alpha = 0$. If $\alpha = 56$,
$\beta = 28$, find to four decimal places the two positive roots~$r$. The corresponding altitude
is~$\alpha/r^2$.

\item[18.] What rate of interest is implied in an offer to sell a house for \$2700 cash, or
in annual installments each of \$1000 payable 1,~2, and 3~years from date?
\index{Compound interest|(}%

Hint: The amount of \$2700 with interest for 3~years should be equal to the sum
of the first payment with interest for 2~years, the amount of the second payment with
interest for 1~year, and the third payment. Hence if $r$ is the rate of interest and we
write~$x$ for~$1+r$, we have
\[
2700x^3 = 1000x^2 + 1000x + 1000.
\]

\item[19.] Find the rate of interest implied in an offer to sell a house for \$3500 cash, or in
annual installments each of \$1000 payable 1,~2, 3, and 4~years from date.

\item[20.] Find the rate of interest implied in an offer to sell a house for \$3500 cash, or
\$4000 payable in annual installments each of \$1000, the first payable now.
\index{Compound interest|)}%
\end{Problems}
\end{Exercises}


% [** PP: Splitting combined ToC entries]
\Section{75.}{Newton's Method} Prior to 1676, Newton\footnote
  {Isaac Newton, \textit{Opuscula},~I, 1794, p.~10, p.~37.}
\index{Newton's!method of solution|(}%
had already found
the root between $2$ and~$3$ of equation~\Eq{1}. He replaced~$x$ by~$2+p$ and
obtained~\Eq{2}. Since $p$ is a decimal, he neglected the terms in $p^3$ and~$p^2$,
and hence obtained $p=0.1$, approximately. Replacing $p$ by $0.1 + q$ in~\Eq{2},
he obtained
\[
q^3 + 6.3q^2 + 11.23q + 0.061 = 0.
\]
Dividing $-0.061$ by~$11.23$, he obtained $-0.0054$ as the approximate
value of~$q$. Neglecting $q^3$ and replacing $q$ by $-0.0054 + r$, he obtained
\[
6.3r^2 + 11.16196r + 0.000541708 = 0.
\]
Dropping $6.3r^2$, he found $r$ and hence
\[
x = 2 + 0.1 - 0.0054 - 0.00004853 = 2.09455147,
\]
%% -----File: 097.png---Folio 91-------
of which all figures but the last are correct~(§74). But the method will
not often lead so quickly to so accurate a value of the root.

Newton used the close approximation $0.1$ to~$p$, in spite of the fact
that this value exceeds the root~$p$ and hence led to a negative correction
at the next step. This is in contrast with Horner's method in which each
correction is positive, so that each approximation must be chosen less
than the root, as $0.09$ for~$p$.

Newton's method may be presented in the following general form,
which is applicable to any equation $f(x) = 0$, whether $f(x)$ is a polynomial
or not. Given an approximate value~$a$ of a real root, we can usually
find a closer approximation $a+h$ to the root by neglecting the powers
$h^2$, $h^3, \dotsc$ of the small number~$h$ in Taylor's formula~(§56)
\[
f(a+h) = f(a) + f'(a)h + f''(a) \frac{h^2}{2} + \dotsb
\]
and hence by taking
\[
f(a) + f'(a)h = 0,\qquad h = \frac{-f(a)}{f'(a)}.
\]
We then repeat the process with $a_1 = a+h$ in place of the former~$a$.

Thus in Newton's example, $f(x) = x^3 - 2x - 5$, we have, for $a=2$,
\begin{align*}
h   &= \frac{-f(2)}{f'(2)}
     = \frac{1}{10},\qquad
 a_1 = a+h = 2.1, \\
%
h_1 &= \frac{-f(2.1)}{f'(2.1)}
     = \frac{-0.061}{11.23}
     = -0.0054\dotsc.
\end{align*}


% [** PP: Split ToC entry, part 2]
\Section[Algebraic and Graphical Discussion]
{76.}{Graphical Discussion of Newton's Method.} Using rectangular
coördinates, consider the graph of $y = f(x)$ and the point~$P$ on it with the
abscissa $OQ = a$ (Fig.~22). Let the tangent at~$P$ meet the $x$-axis at~$T$
%[** Illustrations FIG. 22    FIG. 23]
\begin{figure*}[hbt]
\begin{center}
\Input{097a}\hfill
\Input{097b}
\end{center}
\end{figure*}
%% -----File: 098.png---Folio 92-------
and let the graph meet the $x$-axis at~$S$. Take $h = QT$, the subtangent.
Then
\begin{align*}
QP=f(a),\qquad f'(a)
  &= \tan XTP = \frac{-f(a)}{h}, \\
h &= \frac{-f(a)}{f'(a)}.
\end{align*}
In the graph in Fig.~22, $OT=a+h$ is a better approximation to the
root $OS$ than $OQ=a$. The next step (indicated by dotted lines) gives a
still better approximation~$OT_1$.

If, however, we had begun with the abscissa~$a$ of a point~$P_1$ in Fig.~22
near a bend point, the subtangent would be very large and the method
would probably fail to give a better approximation. Failure is certain
if we use a point~$P_2$ such that a single bend point lies between it and~$S$.

We are concerned with the approximation to a root previously isolated
as the only real root between two given numbers $\alpha$ and~$\beta$. These should
be chosen so nearly equal that $f'(x)=0$ has no real root between $\alpha$ and~$\beta$,
and hence $f(x)=y$ has no bend point between $\alpha$ and~$\beta$. Further, if $f''(x)=0$
has a root between our limits, our graph will have an inflexion point with
an abscissa between $\alpha$ and~$\beta$, and the method will likely fail (Fig.~23).

Let, therefore, neither $f'(x)$ nor $f''(x)$ vanish between $\alpha$ and~$\beta$. Since
$f''$ preserves its sign in the interval from $\alpha$ to~$\beta$, while $f$ changes in sign,
$f''$ and~$f$ will have the same sign for one end point. According as the
abscissa of this point is $\alpha$ or~$\beta$, we take $a=\alpha$ or $a=\beta$ for the first step of
Newton's process. In fact, the tangent at one of the end points meets
the $x$-axis at a point~$T$ with an abscissa within the interval from $\alpha$ to~$\beta$.
If $f'(x)$ is positive in the interval, so that the tangent makes an acute
angle with the $x$-axis, we have Fig.~24 or Fig.~25; if $f'$ is negative, Fig.~26 or Fig.~22.
%[** Illustration \textsc{Fig.~24}  \textsc{Fig.~25}  \textsc{Fig.~26}]
\begin{figure*}[hbt]
\begin{center}
\Input{098a}\hfill
\Input{098b}\hfill
\Input{098c}
\end{center}
\end{figure*}

%% -----File: 099.png---Folio 93-------

\begin{Remark}
In Newton's example, the graph between the points with the abscissas $\alpha = 2$ and
$\beta = 3$ is of the type in Fig.~24, but more nearly like a vertical straight line. In view
of this feature of the graph, we may safely take $a=\alpha$, as did Newton, although our
general procedure would be to take $a = \beta$. The next step, however, accords with our
present process; we have $\alpha=2$, $\beta=2.1$ in Fig.~24 and hence we now take $a=\beta$, getting
\[
\frac{0.061}{11.23} = 0.0054
\]
as the subtangent, and hence $2.1 - 0.0054$ as the approximate root.
\end{Remark}

If we have secured (as in Fig.~24 or Fig.~26) a better upper limit to the
root than~$\beta$, we may take the abscissa~$c$ of the intersection of the chord
$AB$ with the $x$-axis as a better lower limit than~$\alpha$. By similar triangles,
\[
-f(\alpha) : c - \alpha = f(\beta) : \beta - c,
\]
whence
\[
c = \frac{\alpha f(\beta) - \beta f(\alpha)}{f(\beta) - f(\alpha)}.
\Tag{3}
\]
This method of finding the value of~$c$ intermediate to $\alpha$ and~$\beta$ is called the
method of interpolation (\emph{regula falsi}).
\index{Interpolation}%
\index{Regula falsi}%

\begin{Remark}
In Newton's example, $\alpha= 2$, $\beta=2.1$,
\[
f(\alpha) = -1,\qquad
f(\beta)  = 0.061,\qquad
c = 2.0942.
\]
\end{Remark}

The advantage of having $c$ at each step is that we know a close limit
of the error made in the approximation to the root.

We may combine the various possible cases discussed into one:

\begin{Thm}%
If $f(x)=0$ has a single real root between $\alpha$ and~$\beta$, and $f'(x) = 0$, $f''(x) = 0$
have no real root between $\alpha$ and~$\beta$, and if we designate by~$\beta$ that one of the
numbers $\alpha$ and~$\beta$ for which $f(\beta)$ and $f''(\beta)$ have the same sign, then the root
lies in the narrower interval from~$c$ to $\beta - f(\beta)/f'(\beta)$, where~$c$ is given by~\Eq{3}.
\end{Thm}

It is possible to prove\footnote
  {Weber's \textit{Algebra}, 2d~ed.,~I, pp.~380--382; \textit{Kleines Lehrbuch der Algebra}, 1912, p.~163.}
this theorem algebraically and to show that by
repeated applications of it we can obtain two limits $\alpha'$, $\beta'$ between which
the root lies, such that $\alpha' - \beta'$ is numerically less than any assigned positive
number. Hence the root can be found in this manner to any desired accuracy.


\begin{Example}
$f(x) = x^3 - 2x^2 - 2$,\quad $\alpha = 2\tfrac{1}{4}$,\quad $\beta = 2\tfrac{1}{2}$. Then
\[
f(\alpha) = -\tfrac{47}{64},\qquad f(\beta) = \tfrac{9}{8}.
\]
%% -----File: 100.png---Folio 94-------
Neither of the roots $0$, $4/3$ of $f'(x)=0$ lies between $\alpha$ and~$\beta$, so that $f(x)=0$ has a single
real root between these limits~(§65). Nor is the root~$\tfrac{2}{3}$ of $f''(x)=0$ within these limits.
The conditions of the theorem are therefore satisfied. For $\alpha<x<\beta$, the graph is of
the type in Fig.~24. We find that approximately
\begin{gather*}
c = \tfrac{559}{238} = 2.3487,\qquad
\beta_1 = \beta - \frac{f(\beta)}{f'(\beta)} = 2.3714, \\
\beta_1 - \frac{f(\beta_{1})}{f'(\beta_{1})} = 2.3597.
\end{gather*}
For $x = 2.3593$, $f(x) = -0.00003$. We therefore have the root to four decimal places.
For $a = 2.3593$,
\[
f'(a) = 7.2620,\qquad
a - \frac{f(a)}{f'(a)} = 2.3593041,
\]
which is the value of the root correct to 7~decimal places. We at once verify that the
result is greater than the root in view of our work and Fig.~24, while if we change the
final digit from $1$ to~$0$, $f(x)$ is negative.
\end{Example}


\begin{Exercises}{Page94}

\begin{Problems}
\item[1.] For $f(x) = x^4 + x^3 - 3x^2 - x - 4$, show by Descartes' rule of signs that $f'(x)=0$
and $f''(x)=0$ each have a single positive root and that neither has a root between $1$
and~$2$.  Which of the values $1$ and~$2$ should be taken as~$\beta$?

\item[2.] When seeking a root between $2$ and~$3$ of $x^3 - x - 9 = 0$, which value should be
taken as~$\beta$?
\end{Problems}
\end{Exercises}


% [** PP: Split ToC entry, part 3]
\Section[Systematic Computation]
{77.}{Systematic Computation of Roots by Newton's Method.} By way
of illustration we shall compute to 7~decimal places a positive root of
\[
f(x) = x^4 + x^3 - 3x^2 - x - 4 = 0.
\]

Since $f(1) = -6$, $f(2) = 6$, there is a real root between $1$ and~$2$. Since
\[
f'(x) = 4x^3 + 3x^2 - 6x - 1,\qquad
f'(1) = 0,\qquad
f'(2) = 31,
\]
the graph of $y=f(x)$ is approximately horizontal near $(1, -6)$ and approximately
vertical near $(2, 6)$. Hence the root is much nearer to $2$ than to~$1$.
Thus in applying Newton's method we employ $a = 2$ as the first approximation
to the root. The correction~$h$ is then
\[
h = \frac{-f(2)}{f'(2)} = \frac{-6}{31} = -0.2 \ldots .
\]
The work of performing the substitutions $x = 2+d$, $d = -0.2 + e, \dotsc$,
to find the transformed equations satisfied by $d$, $e, \dotsc$, is done by synthetic
%% -----File: 101.png---Folio 95-------
division, exactly as in Horner's method, except that some of the
multipliers are now negative: see \hyperref[table:1]{Table~1},
page~\pageref*{table:1}.
% The structure below is so large it's almost certain to cause a
% horribly-underfull page. Instead, float it, setting up bidirectional
% anchors, and slightly change the wording of the previous sentence.
\phantomsection\label{tableanchor:1}%

\begin{table*}[hp]
% ** PP: Dear future maintainer of this table code:
% The P column specifier has fixed width 1/8in and serves as
% padding. The idea is to set columns of numbers in stanzas rlP,
% with no inter-column space, and with the decimal points in the
% left-aligned column. \Lcol and \Rcol are wrappers for center-aligned
% columns with a vertical bar at the corresponding side. The floating
% sub-tables at the right side are rather terribly hacked; to get the
% ``underlines'' of correct width, entries are wrapped in left-aligned
% makebox commands. In short, this structure is a nest of hacks.
%
{\small% Coax width into text block
\begin{gather*}% Commit further abuse of semantics....
\begin{array}{r<{\qquad} r@{}l@{}P<{\qquad} *{3}{>{\quad}r@{}l@{}P} r@{}lP}
 1 & 1  &&& -3  &&&       -1 &&&          -4 &&  &\Lcol{2} & \\
\cline{14-14}
   & 2  &&&  6  &&&        6 &&&          10 &&  &&\Strut \\
\cline{1-11}
 1 & 3  &&&  3  &&& 5 &\Rcol{}&&  \mathbf{6} &&  &&\Strut \\
   & 2  &&& 10  &&& 26&\Rcol{}&&             &&  && \\
\cline{1-9}
 1 & 5  &&& 13 &\Rcol{}&& \mathbf{31} &&&    &&  &&\Strut \\
   & 2  &&& 14 &\Rcol{}&&             &&&    &&  && \\
\cline{1-6}
 1 & 7 &\Rcol{}&& \mathbf{27} &&&     &&&    &&  &&\Strut \\
   & 2 &\Rcol{}&&             &&&     &&&    &&  && \\
\cline{1-3}
\mathbf{1} & \mathbf{9}\Strut
              &&&             &&&     &&&    &&  &\Lcol{\makebox[1em][l]{$-0.2$}} \\
\cline{14-15}
   &      -0&.2&&        -1&.76&&
                                -5&.048&&  -5&.1904&  &&\Strut \\
\cline{2-12}
   &       8&.8&&        25&.24&&
                         25&.952&\Lcol{}&
                          \mathbf{0}&\mathbf{.8096}&  &&\Strut \\
%
   &     -0&.2 &&       -1&.72 &&
                        -4&.704 &\Lcol{}&    &&  && \\
\cline{2-9}
%
   &      8&.6 && 23&.52 &\Rcol{}&
             \mathbf{21}&\mathbf{.248} &&    &&  && \Strut \\
%
   &     -0&.2 && -1&.68 &\Rcol{}&
                        &              &&    &&  && \\
\cline{2-7}
   & 8&.4 &\Rcol{}&
         \mathbf{21}&\mathbf{.84}&    &&&    &&  &&
  \Lcol{\makebox[0pt][l]{\smash[b]{\raisebox{-1ex}{$\dfrac{-0.8096}{21.248}$}}}} \\
%
   &-0&.2 &\Rcol{}&
                    &            &    &&&    &&  && \Lcol{}  \\
\cline{2-4}
 \mathbf{1} & \mathbf{8}&\mathbf{.2}&&
                    &            &    &&&    &&  && \Lcol{\makebox[0pt][l]{$= -0.04$}}&&\ \\
\cline{14-16}
   & -0&.04 &&    -0&.3264       &
                           & -0&.860544 &  &-0&.81549824 &&  & \\
\cline{2-12}
   &  8&.16 &&    21&.5136       &
                           & 20&.387456 &\Rcol{}&
                   \mathbf{-0}&\mathbf{.00589824}\Strut &&  & \\
%
   & -0&.04 &&    -0&.3248       &
                           & -0&.847552 &\Rcol{}&
                              &                          &&  & \\
\cline{2-10}
   &  8&.12 &&    21&.1888
                   &\Rcol{}& \mathbf{19}&\mathbf{.539904} &&&&&&\Strut \\
%
   & -0&.04 &&    -0&.3232
                   &\Rcol{}&            &                 &&&&&& \\
\cline{2-7}
   &  8&.08 &\Rcol{}& \mathbf{20}&\mathbf{.8656}
                   &       &            &                 &&&&&&\Strut \\
   & -0&.04 &\Rcol{}&& &&& &&& && \\
\cline{2-4}
 \mathbf{1} & \mathbf{8}&\mathbf{.04} &&
                   &       &            &                 &&&\Strut&&
\smash{\makebox[0pt][r]{$g=\dfrac{0.005898}{19.54} = .000302$}} &&
\end{array} \\[24pt]
\text{\hyperref[tableanchor:1]{\scshape Table~1.}}% Caption...!
\end{gather*}}% End of \small
\phantomsection\label{table:1}%
\end{table*}

The root is $2 - 0.2 - 0.04 + 0.000302 = 1.760302$, in which the last
figure is in slight doubt. Indeed, it can be proved that \emph{if the final fraction~$g$,
when expressed as a decimal, has $k$~zeros between the decimal point and the
first significant figure, the division may be safely carried to $2k$~decimal places}.
In our example $k=3$, so that we retained 6~decimal places in~$g$.
\index{Synthetic division|)}%

To proceed independently of this rule, we note that $g$ is obviously
between $0.00030$ and $0.00031$. Then the value of $g^4 + 8.04g^3 + 20.8656g^2$
%% -----File: 102.png---Folio 96-------
is found to lie between $0.000001878$ and $0.000002006$. Whichever of these
limits we use as a correction by which to reduce the constant term, we
obtain the same dividend below correct to 6~decimal places.
\[
\begin{array}{r>{\quad}r<{\quad}l@{}l}
\multicolumn{1}{r|}{\xatop{1}\xatop{9}.\xatop{5}\xatop{3}9904}
& 0.005896 & \multicolumn{1}{|l@{}}{0.0003017} \\
\cline{1-1} \cline{3-3}
  & 005862 & \\[-8pt]
& \settowidth{\TmpLen}{005862}\rule{\TmpLen}{0.5pt} & \\
  &     34 & \\
  &     20 & \\[-8pt]
& \settowidth{\TmpLen}{20}\rule{\TmpLen}{0.5pt} & \\
  &     14 & \\
  &     14 & \\[-8pt]
& \settowidth{\TmpLen}{14}\rule{\TmpLen}{0.5pt} & \\
\end{array}
\]
Hence the root is $1.7603017$ to 7~decimal places.


\begin{Exercises}{Page96}

\begin{Problems}
\item[1.] Find to 8~decimal places the root between $2$ and~$3$ of $x^3 - x - 9 = 0$.

\item[2.] Find to 7~decimal places the root between $2$ and~$3$ of $x^3 - 2x^2 - 2 = 0$.

\item[3.] Find the real cube root of $7.976$ to 6~decimal places.

\item[4.] Explain by Taylor's expansion of $f(2+d)$ why the values of
\[
f(2),\qquad  f'(2),\qquad
\tfrac{1}{2}f''(2),\qquad
\frac{1}{2·3} f'''(2),\qquad
\frac{1}{2·3·4} f''''(2)
\]
are in reverse order the coefficients of the transformed equation
\[
d^4 + 9d^3 + 27d^2 + 31d + 6 = 0,
\]
obtained in the Example in the text, and printed in heavy type.

\item[5.] The method commonly used to find the positive square root of~$n$ by a computing
machine consists in dividing~$n$ by an assumed approximate value~$a$ of the square root
and taking half the sum of~$a$ and the quotient as a better approximation. Show that
the latter agrees with the value of $a+h$ given by applying Newton's method to
$f(x) = x^2-n$.
\index{Square roots}%
\end{Problems}
\end{Exercises}


% [** PP: Split ToC entry, part 4]
\Section[For Functions not Polynomials]
{78.}{Newton's Method for Functions not Polynomials.}\hfill\break
% [** PP: Format Example manually to avoid \BeforeSkip]
{\indent\normalfont\small\scshape Example~1.\quad\upshape
Find the angle~$x$ at the center of a circle subtended by a chord which
cuts off a segment whose area is one-eighth of that of the circle.}

\begin{Solution}
If $x$ is measured in radians and if $r$ is the radius, the area of the segment
is equal to the left member of
\[
\tfrac{1}{2} r^2(x - \sin x) = \tfrac{1}{8} \pi r^2,
\]
whence
\[
x - \sin x = \tfrac{1}{4} \pi.
\]
%% -----File: 103.png---Folio 97-------
By means of a graph of $y = \sin x$ and the straight line represented by $y = x-\tfrac{1}{4}\pi$, we
see that the abscissa of their point of intersection is approximately $1.78$~radians or~$102°$.
Thus $a = 102°$ is a first approximation to the root of
\[
f(x) \equiv x - \sin x - \tfrac{1}{4} \pi = 0.
\]
By Newton's method a better approximation is $a+h$, where\footnote
  {The derivative of $\sin x$ is~$\cos x$. We need the limit of
\index{Derivative}%
  \[
  \frac{\sin(x+2k) - \sin x}{2k}
    \equiv \frac{2\cos \tfrac{1}{2}(2x + 2k) \sin \tfrac{1}{2}(2k)}{2k}
    \equiv \frac{\cos(x+k) \sin k}{k}
  \]
  as $2k$ approaches zero. Since the ratio of $\sin k$ to~$k$ approaches~$1$, the limit is $\cos x$.}
\[
h = \frac{-f(a)}{f'(a)}
  = \frac{-a + \sin a + \tfrac{1}{4} \pi}{1 - \cos a}.
\]
\begin{gather*}
\begin{array}{@{}r@{}c@{}l@{}l@{}}
         \sin 102°   = &      & 0.9781 & \\
\tfrac{1}{4}(3.1416) = &      & 0.7854 & \\
\cline{3-3}
                       &      & 1.7635 & \\
& \\
                  102° &{}={} & 1.7802 & \text{ radians} \\
\cline{2-3}
                       &{}-{} & 0.0167 &
\end{array}
\qquad
\begin{aligned}
    \cos 102° &= -0.2079 \\
1 - \cos 102° &= \Neg 1.2079 \\
& \\
h   &= \frac{-0.0167}{1.2079} = -0.0138 \\
a_1 &= a+h = 1.7664
\end{aligned} \\ % Top-level end-of-line
h_1 = \frac{-f(a_1)}{f'(a_1)}
    = \frac{-1.7664 + 0.9809 + 0.7854}{1.1944}
    = -0.0001.
\end{gather*}
Hence $x = a_1 + h_1 = 1.7663$~radians, or $101° 12'$. % [** PP: Added period]
\end{Solution}


\begin{Example}[2.\protect\footnotemark]
\index{Interpolation}%
  \footnotetext{This Ex.~2, which should be contrasted with Ex.~3, is solved by interpolation
  since that method is simpler than Newton's method in this special case.}
Solve $x - \log x = 7$, the logarithm being to base~$10$.
\end{Example}

\begin{Solution}
Evidently $x$ exceeds~$7$ by a positive decimal which is the value of~$\log x$.
Hence in a table of common logarithms, we seek a number~$x$ between $7$ and~$8$ whose
logarithm coincides approximately with the decimal part of~$x$. We read off the values
in the second column.
\[
\begin{array}{c|c|c}
x      &   \log x   & x - \log x \\
\hline
7.897  &   0.89746  &  6.99954   \\
7.898  &   0.89752  &  7.00048
\end{array}
\]
By the final column the ratio of interpolation is~$46/94$. Hence $x = 7.8975$ to four
decimal places.
\end{Solution}

%% -----File: 104.png---Folio 98-------


\begin{Example}[3.]
Solve $2x-\log x = 7$, the logarithm being to base~10.
\end{Example}

\begin{Solution}
Evidently $x$ is a little less than~$4$. A table of common logarithms shows
at once that a fair approximation to $x$ is~$a=3.8$. Write
\[
f(x) \equiv 2x - \log x - 7,\qquad
\log x = M \log_e x,\qquad
M = 0.4343.
\]
By calculus, the derivative of $\log_e x$ is~$1/x$. Hence
\begin{gather*}
f'(x) = 2 - \frac{M}{x},\qquad f'(a) = 2 - 0.1143 = 1.8857, \\
f(a) = 0.6 - \log 3.8 = 0.6 - 0.57978 = 0.02022, \\
-h = \frac{f(a)}{f'(a)} = 0.0107,\qquad a_1 = a + h = 3.7893, \\
f(a_1) = 0.000041,\qquad f(3.7892) = -0.000148. \\
\frac{148}{189} × 0.0001 = 0.000078,\qquad x = 3.789278.
\end{gather*}
All figures of~$x$ are correct as shown by Vega's table of logarithms to 10~places.
\end{Solution}


\begin{Exercises}{Page98}

Find the angle~$x$ at the center of a circle subtended by a chord which cuts off a segment
whose ratio to the circle is
\begin{Problems}[2]

\item[1.] $\frac{1}{4}$.

\item[2.] $\frac{3}{8}$.
\end{Problems}

When the logarithms are to base~10,
\begin{Problems}[2]
\item[3.] Solve $2x - \log x = 9$.

\item[4.] Solve $3x - \log x = 9$.

\ResetCols{1}

\item[5.] Find the angle just $>15°$ for which
$\frac{1}{2} \sin x + \sin 2x = 0.64$.

\item[6.] Find the angle just $>72°$ for which
$x - \frac{1}{2} \sin x = \frac{1}{4} \pi$.

\item[7.] Find all solutions of Ex.~5 by replacing $\sin 2x$ by $2\sin x\cos x$, squaring, and
solving the quartic equation for~$\cos x$.

\item[8.] Solve similarly $\sin x + \sin 2x = 1.2$.

\item[9.] Find $x$ to 6~decimal places in $\sin x = x - 2$.

\item[10.] Find $x$ to 5~decimal places in $x = 3\log_e x$.
\end{Problems}
\end{Exercises}
\index{Newton's!method of solution|)}%


\Section{79.}{Imaginary Roots} To find the imaginary roots $x+yi$ of an equation
$f(z)=0$ with real coefficients, expand $f(x+yi)$ by Taylor's theorem;
we get
\index{Imaginary!roots}%
\[
f(x) + f'(x)yi
     - f''(x)  \frac{y^2}{1·2}
     - f'''(x) \frac{y^3 i}{1·2·3} + \dotsb = 0.
\]

%% -----File: 105.png---Folio 99-------

Since $x$ and~$y$ are to be real, and $y \ne 0$,
\[
\left\{
\begin{aligned}
f(x) - f''(x) \frac{y^2}{1·2} + f''''(x) \frac{y^4}{1·2·3·4} - \dotsb = 0, \\
f'(x) - f'''(x) \frac{y^2}{1·2·3} + f^{(5)}(x)\frac{y^4}{5!} - \dotsb = 0.
\end{aligned}\right.
\Tag{4}
\]

In the Example and Exercises below, $f(z)$ is of degree~$4$ or less. Then
the second equation~\Eq{4} is linear in~$y^2$. Substituting the resulting value
of~$y^2$ in the first equation~\Eq{4}, we obtain an equation $E(x) = 0$, whose real
roots may be found by one of the preceding methods. If the degree of
$f(z)$ exceeds~$4$, we may find $E(x) = 0$ by eliminating $y^2$ between the two
equations~\Eq{4} by one of the methods to be explained in \ChapRef{X}.


\begin{Example}
For $f(z) = z^4 - z + 1$, equations~\Eq{4} are
\[
x^4 - x + 1 - 6x^2 y^2 + y^4 = 0,\qquad
4x^3 - 1 - 4xy^2 = 0.
\]
Thus
\[
y^2 = x^2 - \frac{1}{4x},\qquad
-4x^6 + x^2 + \frac{1}{16} = 0.
\]
The cubic equation in $x^2$ has the single real root
\[
x^2 = 0.528727,\qquad  x = ±0.72714.
\]
Then $y^2 = 0.184912$ or $0.87254$, and
\[
z = x+yi
  = 0.72714 ± 0.43001i,\qquad
   -0.72714 ± 0.93409i.
\]
\end{Example}


\begin{Exercises}{Page99}

Find the imaginary roots of
\begin{Problems}[2]
\item[1.] $z^3 - 2z - 5 = 0$.

\item[2.] $28z^3 + 9z^2 - 1 = 0$.

\ResetCols{2}

\item[3.] $z^4 - 3z^2 - 6z = 2$.

\item[4.] $z^4 - 4z^3 + 11z^2 - 14z + 10 = 0$.

\ResetCols{1}

\item[5.] $z^4 - 4z^3 + 9z^2 - 16z + 20 = 0$. Hint:
\[
E(x) \equiv x(x - 2)(16x^4 - 64x^3 + 136x^2 - 144x + 65) = 0,
\]
and the last factor becomes $(w^2 + 1)(w^2 + 9)$ for $2x = w + 2$.
\end{Problems}
\end{Exercises}

\begin{Note}
If we know a real root~$r$ of a cubic equation $f(z)=0$, we may remove the
factor $z-r$ and solve the resulting quadratic equation. When, as usual, $r$ involves
several decimal places, this method is laborious and unsatisfactory. But we may utilize
a device, explained in the author's \textit{Elementary Theory of Equations}, pp.~119--121, §§6,~7.
As there explained, a similar device may be used when we know two real roots of a
quartic equation.
\end{Note}

%% -----File: 106.png---Folio 100-------


\begin{Exercises}[MISCELLANEOUS~]{Page100}

(Give answers to 6~decimal places, unless the contrary is stated.)
\begin{Problems}

\item[1.] What arc of a circle is double its chord?

\item[2.] What arc of a circle is double the distance from the center of the circle to the
chord of the arc?

\item[3.] If $A$ and~$B$ are the points of contact of two tangents to a circle of radius unity
from a point~$P$ without it, and if arc $AB$ is equal to~$PA$, find the length of the arc.

\item[4.] Find the angle at the center of a circle of a sector which is bisected by its chord.

\item[5.] Find the radius of the smallest hollow iron sphere, with air exhausted, which will
float in water if its shell is $1$~inch thick and the specific gravity of iron is~$7.5$.

\item[6.] From one end of a diameter of a circle draw a chord which bisects the semicircle.

\item[7.] The equation $x \tan x = c$ occurs in the theory of vibrating strings. Its approximate
solutions may be found from the graphs $y = \cot x$, $y = x/c$.   Find $x$ when $c = 1$.

\item[8.] The equation $\tan x = x$ occurs in the study of the vibrations of air in a spherical
cavity. From an approximate solution $x_1 = 1.5\pi$, we obtain successively better approximations
$x_2 = \tan^{-1} x_1 = 1.4334 \pi$, $x_3 = \tan^{-1} x_2, \dotsc$. Find the first three solutions to
4~decimal places.

\item[9.] Find to 3~decimal places the first five solutions of
\[
\tan x = \frac{2x}{2-x^2},
\]
which occurs in the theory of vibrations in a conical pipe.

\item[10.] $4 \tau x^3 - (3x - 1)^2 = 0$ arises in the study of the isothermals of a gas. Find its
roots when (i)~$\tau = 0.002$ and (ii)~$\tau = 0.99$.

\item[11.] Solve $x^x = 100$.

\item[12.] Solve $x = 10\log x$.

\item[13.] Solve $x + \log x = x \log x$.

\item[14.] Solve Kepler's equation $M = x - e \sin x$ when $M = 332° 28' 54.8''$,
 $e = 14° 3' 20''$.

\item[15.] In what time would a sum of money at 6\% interest compounded annually
amount to as much as the same sum at simple interest at~8\%?
\index{Compound interest}%

\item[16.] In a semicircle of diameter~$x$ is inscribed a quadrilateral with sides $a$, $b$, $c$,~$x$;
then $x^3 - (a^2 + b^2 + c^2) x - 2abc = 0$ (I.~Newton).  Given $a = 2$, $b = 3$, $c = 4$, find~$x$.

\item[17.] What rate of interest is implied in an offer to sell a house for \$9000~cash, or
\$1000~down and \$3000 at the end of each year for three years?
\end{Problems}
\end{Exercises}
\index{Solution of numerical equations|)}%

%% -----File: 107.png---Folio 101-------


\Chapter{VIII}{Determinants; Systems of Linear Equations}
\index{Determinants|(}%
\index{Linear equations!system|(}%

% [** PP: ToC entry reads Solution of 2 or 3 Linear Equations by Determinants]
\Section[Solution of $2$~Linear Equations by Determinants]
{80.}{Solution of Two Linear Equations by Determinants of Order~$2$.}
Assume that there is a pair of numbers $x$ and~$y$ for which
\[
\left\{
\begin{aligned}
a_1 x + b_1 y &= k_1, \\
a_2 x + b_2 y &= k_2.
\end{aligned}
\right.
\Tag{1}
\]
Multiply the members of the first equation by~$b_2$ and those of the second
equation by~$-b_1$, and add the resulting equations. We get
\[
(a_1 b_2 - a_2 b_1)x = k_1 b_2 - k_2 b_1.
\]
Employing the respective multipliers $-a_2$ and~$a_1$, we get
\[
(a_1 b_2 - a_2 b_1)y = a_1 k_2 - a_2 k_1.
\]
The common multiplier of $x$ and~$y$ is
\[
a_1 b_2 - a_2 b_1,
\Tag{2}
\]
and is denoted by the symbol
\[
\begin{vmatrix}
  a_1 & b_1 \\
  a_2 & b_2
\end{vmatrix},
\Tag{2'}
\]
which is called a \emph{determinant of the second order}, and also called the determinant
of the coefficients of $x$ and~$y$ in equations~\Eq{1}. The results above
may now be written in the form
\[
\begin{vmatrix}
 a_1 & b_1 \\
 a_2 & b_2
\end{vmatrix} x
=
\begin{vmatrix}
 k_1 & b_1 \\
 k_2 & b_2
\end{vmatrix},\qquad
\begin{vmatrix}
 a_1 & b_1 \\
 a_2 & b_2
\end{vmatrix} y
=
\begin{vmatrix}
 a_1 & k_1 \\
 a_2 & k_2
\end{vmatrix}.
\Tag{3}
\]
We shall call $k_1$ and~$k_2$ the known terms of our equations~\Eq{1}. Hence,
\emph{if $D$ is the determinant of the coefficients of the unknowns, the product of~$D$ by
any one of the unknowns is equal to the determinant obtained from~$D$ by
substituting the known terms in place of the coefficients of that unknown}.

%% -----File: 108.png---Folio 102-------

If $D \ne 0$, relations~\Eq{3} uniquely determine values of $x$ and~$y$:
\[
x = \frac{k_1 b_2 - k_2 b_1}{D},\qquad
y = \frac{a_1 k_2 - a_2 k_1}{D},
\]
and these values satisfy equations~\Eq{1}; for example,
\[
a_1 x + b_1 y = \frac{(a_1 b_2 - a_2 b_1)k_1}{D} = k_1.
\]
Hence our equations~\Eq{1} have been solved by determinants when $D \ne 0$.
We shall treat in~§96 the more troublesome case in which $D = 0$.


\begin{Example}
For $2x - 3y = -4$, $6x - 2y = 2$, we have
\begin{align*}
&&
\begin{vmatrix}
 2 & -3 \\
 6 & -2
\end{vmatrix} x
  &=
\begin{vmatrix}
-4 & -3 \\
\Neg 2 & -2
\end{vmatrix},  & 14x &= 14,\qquad x = 1, && \\
&&
14y &=
\begin{vmatrix}
\Neg 2 & -4 \\
\Neg 6 & \Neg 2
\end{vmatrix} = 28, & y &= 2.
\end{align*}
\end{Example}


\begin{Exercises}{Page102}

Solve by determinants the following systems of equations:
\begin{Problems}[3]
\item[1.] $\begin{System}[\,]{2}
 8x &-{}&  y &= 34, \\
  x &+{}& 8y &= 53.
\end{System}$

\item[2.] $\begin{System}[\,]{2}
 3x &+{}& 4y &= 10, \\
 4x &+{}&  y &=  9.
\end{System}$

\item[3.] $\begin{System}[\,]{2}
 ax &+{}& by &= a^2, \\
 bx &-{}& ay &= ab.
\end{System}$
\end{Problems}
\end{Exercises}


% [** PP: No separate ToC entry]
\Section[Solution of $3$~Linear Equations by Determinants]
{81.}{Solution of Three Linear Equations by Determinants of Order~$3$.}

Consider a system of three linear equations
\[
\begin{aligned}
a_1 x + b_1 y + c_1 z &= k_1, \\
a_2 x + b_2 y + c_2 z &= k_2, \\
a_3 x + b_3 y + c_3 z &= k_3.
\end{aligned}
\Tag{4}
\]
Multiply the members of the first, second and third equations by
\[
b_2 c_3 - b_3 c_2,\qquad
b_3 c_1 - b_1 c_3,\qquad
b_1 c_2 - b_2 c_1,
\Tag{5}
\]
respectively, and add the resulting equations. We obtain an equation
in which the coefficients of $y$ and~$z$ are found to be zero, while the coefficient
of~$x$ is
\[
a_1 b_2 c_3 -
a_1 b_3 c_2 +
%
a_2 b_3 c_1 -
a_2 b_1 c_3 +
%
a_3 b_1 c_2 -
a_3 b_2 c_1.
\Tag{6}
\]
%% -----File: 109.png---Folio 103-------
Such an expression is called a \emph{determinant of the third order} and denoted
by the symbol
\[
\begin{vmatrix}
  a_1 & b_1 & c_1 \\
  a_2 & b_2 & c_2 \\
  a_3 & b_3 & c_3
\end{vmatrix}.
\Tag{6'}
\]

The nine numbers $a_1, \dotsc, c_3$ are called the \emph{elements} of the determinant.
In the symbol these elements lie in three (horizontal) \emph{rows}, and
also in three (vertical) \emph{columns}. Thus $a_2$, $b_2$, $c_2$ are the elements of the
second row, while the three $c$'s are the elements of the third column.
\index{Determinants!columns}%
\index{Determinants!elements}%
\index{Determinants!rows}%

The equation (free of $y$ and~$z$), obtained above, may now be written
as
\[
\begin{vmatrix}
  a_1 & b_1 & c_1 \\
  a_2 & b_2 & c_2 \\
  a_3 & b_3 & c_3
\end{vmatrix} x
=
\begin{vmatrix}
  k_1 & b_1 & c_1 \\
  k_2 & b_2 & c_2 \\
  k_3 & b_3 & c_3
\end{vmatrix},
\]
since the right member was the sum of the products of the expressions~\Eq{5}
by $k_1$, $k_2$, $k_3$, and hence may be derived from~\Eq{6} by replacing the
$a$'s by the~$k$'s. Thus the theorem of~§80 holds here as regards the
unknown~$x$. We shall later prove, without the laborious computations
just employed, that the theorem holds for all three unknowns.
\index{Linear equations!system|)}%


% [** PP: No ToC entry, matching running head]
\Section[Signs of the Terms of a Determinant]
{82.}{The Signs of the Terms of a Determinant of Order~$3$.}  In the
\index{Determinants!signs of terms|(}%
six terms of our determinant~\Eq{6}, the letters $a$, $b$, $c$ were always written
in this sequence, while the subscripts are the six possible arrangements
of the numbers $1$, $2$, $3$. The first term $a_1 b_2 c_3$ shall be called the \emph{diagonal
term}, since it is the product of the elements in the main diagonal running
\index{Determinants!diagonal term}%
from the upper left-hand corner to the lower right-hand corner of the
symbol~\Eq{6'} for the determinant. The subscripts in the term $-a_1 b_3 c_2$
are derived from those of the diagonal term by interchanging $2$ and~$3$,
and the minus sign is to be associated with the fact that an odd number
(here one) of interchanges of subscripts were used. To obtain the arrangement
$2$, $3$, $1$ of the subscripts in the term $+a_2 b_3 c_1$ from the natural order
$1$, $2$, $3$ (in the diagonal term), we may first interchange $1$ and~$2$, obtaining
the arrangement $2$, $1$, $3$, and then interchange $1$ and~$3$; an even number
(two) of interchanges of subscripts were used and the sign of the term
is plus.

While the arrangement $1$, $3$, $2$ was obtained from $1$, $2$, $3$ by one interchange
$(2, 3)$, we may obtain it by applying in succession the three interchanges
%% -----File: 110.png---Folio 104-------
$(1, 2)$, $(1, 3)$, $(1, 2)$, and in many new ways. To show that the
number of interchanges which will produce the final arrangement $1$, $3$, $2$
is odd in every case, note that each of the three possible interchanges,
viz., $(1, 2)$, $(1, 3)$, and~$(2, 3)$, changes the sign of the product
\[
P = (x_1 - x_2) (x_1 - x_3) (x_2 - x_3),
\]
where the $x$'s are arbitrary variables. Thus a succession of $k$~interchanges
yields $P$ or~$-P$ according as $k$ is even or odd. Starting with the arrangement
$1$, $2$, $3$ and applying $k$~successive interchanges, suppose that we
obtain the final arrangement $1$, $3$, $2$. But if in $P$ we replace the subscripts
$1$, $2$, $3$ by $1$, $3$, $2$, respectively, i.e., if we interchange $2$ and~$3$, we obtain~$-P$.
Hence $k$ is odd. We have therefore proved the following rule
of signs:

\begin{Thm}%
Although the arrangement $r$, $s$, $t$ of the subscripts in any term $±a_r b_s c_t$ of
the determinant may be obtained from the arrangement $1$, $2$, $3$ by various
successions of interchanges, the number of these interchanges is either always
an even number and then the sign of the term is plus or always an odd number
and then the sign of the term is minus.
\end{Thm}


\begin{Exercises}{}

Apply the rule of signs to all terms of
\begin{Problems}[2]

\item[1.] Determinant~\Eq{6}.

\item[2.] Determinant $a_1 b_2 - a_2 b_1$.
\end{Problems}
\end{Exercises}


\Section[Even and Odd Arrangements]
{83.}{Number of Interchanges always Even or always Odd.}  We now
extend the result in~§82 to the case of $n$~variables $x_1, \dotsc, x_n$. The
product of all of their differences $x_i - x_j$ ($i<j$) is
\begin{align*}
P = (x_1 - x_2)(x_1 - x_3)  \dotsm &(x_1 - x_n) \\
          {} · (x_2 - x_3)  \dotsm &(x_2 - x_n) \\
                            \vdots & \\
                              {} · &(x_{n-1} - x_n).
\end{align*}
Interchange any two subscripts $i$ and~$j$. The factors which involve neither
$i$ nor~$j$ are unaltered. The factor $(x_i - x_j)$ involving both is changed in
sign. The remaining factors may be paired to form the products
\[
±(x_i - x_k)(x_j - x_k)\qquad (k = 1, \dotsc, n;\quad k \ne i,\ k \ne j).
\]
Such a product is unaltered. Hence $P$ is changed in sign.

Suppose that an arrangement $i_1, i_2, \dotsc, i_n$ can be obtained from
%% -----File: 111.png---Folio 105-------
$1, 2, \dotsc$, $n$ by using $m$ successive interchanges and also by $t$~successive
interchanges. Make these interchanges on the subscripts in~$P$; the
resulting functions are equal to~$(-1)^m P$ and~$(-1)^t P$, respectively. But
the resulting functions are identical since either can be obtained at one
step from~$P$ by replacing the subscript~$1$ by~$i_1$, $2$ by~$i_2$, \ldots, $n$ by~$i_n$. Hence
\[
(-1)^m P \equiv (-1)^t P,
\]
so that $m$ and~$t$ are both even or both odd.

Thus \emph{if the same arrangement is derived from $1$, $2, \dotsc, n$ by $m$~successive
interchanges as by $t$~successive interchanges, then $m$ and~$t$ are both even or
both odd.}


\Section{84.}{Definition of a Determinant of Order~$n$}  We define a determinant
of order~$4$ to be
\[
\begin{vmatrix}
  a_1 & b_1 & c_1 & d_1 \\
  a_2 & b_2 & c_2 & d_2 \\
  a_3 & b_3 & c_3 & d_3 \\
  a_4 & b_4 & c_4 & d_4
\end{vmatrix}
  = \sum_{(24)} ± a_q b_r c_s d_t,
\Tag{7}
\]
where $q, r, s, t$ is any one of the $24$~arrangements of $1, 2, 3, 4$, and the
sign of the corresponding term is $+$ or~$-$ according as an even or odd
number of interchanges are needed to derive this arrangement $q, r, s, t$
from $1, 2, 3, 4$. Although different numbers of interchanges will produce
the same arrangement $q, r, s, t$ from $1, 2, 3, 4$, these numbers are all even
or all odd, as just proved, so that the sign is fully determined.

We have seen that the analogous definitions of determinants of orders
$2$ and~$3$ lead to our earlier expressions~\Eq{2} and~\Eq{6}.

We will have no difficulty in extending the definition to a determinant
of general order~$n$ as soon as we decide upon a proper notation for the $n^2$
elements. The subscripts $1, 2, \dotsc, n$ may be used as before to specify
the rows. But the alphabet does not contain $n$~letters with which to
specify the columns. The use of $e', e'', \dotsc, e^{(n)}$ for this purpose would
conflict with the notation for derivatives and besides be very awkward
when exponents are used. It is customary in mathematical journals
and scientific books (a custom not always followed in introductory text
books, to the distinct disadvantage of the reader) to denote the $n$~letters
used to distinguish the $n$~columns by $e_1, e_2, \dotsc, e_n$ (or some other letter
with the same subscripts) and to prefix (but see~§85) such a subscript by
%% -----File: 112.png---Folio 106-------
the new subscript indicating the row. The symbol for the determinant
is therefore
\[
D = \begin{vmatrix}
e_{11} & e_{12} & \cdots & e_{1n} \\
e_{21} & e_{22} & \cdots & e_{2n} \\
\Dots{4} \\
e_{n1} & e_{n2} & \cdots & e_{nn}
\end{vmatrix}.
\Tag{8}
\]
By definition this shall mean the sum of the $n(n-1) \dotsm 2·1$ terms
\[
(-1)^i e_{{i_1}1}  e_{{i_2}2}  \dotsm  e_{{i_n}n}
\Tag{9}
\]
in which $i_1, i_2, \dotsc, i_n$ is an arrangement of $1, 2, \dotsc, n$, derived from
$1, 2, \dotsc, n$ by $i$~interchanges. Any term~\Eq{9} of the determinant~\Eq{8} is,
apart from sign, the product of $n$~factors, one and only one from each column,
and one and only one from each row.
\index{Determinants!interchanges}%

For example, if we take $n=4$ and write $a_j, b_j, c_j, d_j$ for $e_{j1}, e_{j2}, e_{j3}, e_{j4}$,
the symbol~\Eq{8} becomes~\Eq{7} and the general term~\Eq{9} becomes the general
term $(-1)^i a_{i_1} b_{i_2} c_{i_3} d_{i_4}$ of the second member of~\Eq{7}.


\begin{Exercises}{Page106}

\begin{Problems}

\item[1.] Find the six terms involving $a_2$ in the determinant~\Eq{7}.

\item[2.] What are the signs of $a_3b_5c_2d_1e_4$, $a_5b_4c_3d_2e_1$ in a determinant of order five?
\index{Determinants!signs of terms|)}%

\item[3.] Show that the arrangement $4, 1, 3, 2$ may be obtained from $1, 2, 3, 4$ by use of
the two successive interchanges $(1, 4)$, $(1, 2)$, and also by use of the four successive
interchanges $(1, 4)$, $(1, 3)$, $(1, 2)$, $(2, 3)$.

\item[4.] Write out the six terms of~\Eq{8} for $n = 3$, rearrange the factors of each term so that
the new first subscripts shall be in the order $1, 2, 3$, and verify that the resulting six
terms are those of the determinant~$D'$ in~§85 for $n = 3$.
\end{Problems}
\end{Exercises}


\Section{85.}{Interchange of Rows and Columns}
\begin{Thm}
Any determinant is not
altered in value if in its symbol we replace the elements of the first, second,
$\dotsc, n$th rows by the elements which formerly appeared in the same order
in the first, second, $\dotsc, n$th columns, or briefly if we interchange the corresponding
rows and columns.
\end{Thm}% No \par
For example,
\[
\begin{vmatrix}
a & b \\
c & d
\end{vmatrix}
= ad - bc =
\begin{vmatrix}
a & c\\
b & d
\end{vmatrix}.
\]

We are to prove that the determinant~$D$ given by~\Eq{8} is equal to
\[
D' =
\begin{vmatrix}
e_{11} & e_{21} & \cdots & e_{n1} \\
e_{12} & e_{22} & \cdots & e_{n2} \\
\Dots{4} \\
e_{1n} & e_{2n} & \cdots & e_{nn}
\end{vmatrix}.
\]
%% -----File: 113.png---Folio 107-------
If we give to $D'$ a more familiar aspect by writing $e_{ik} = a_{ki}$ for each element
so that, as in~\Eq{8}, the row subscript precedes instead of follows the column
subscript, the definition of the determinant in terms of the $a$'s gives $D'$
in terms of the $e$'s as the sum of all expressions
\[
(-1)^i e_{1{k_1}} e_{2{k_2}} \dotsm e_{n{k_n}},
\]
% [** PP: Added comma after k_2]
in which $k_1, k_2, \dotsc, k_n$ is an arrangement of $1, 2, \dotsc, n$, derived from
the latter sequence by $i$~interchanges.

As for the terms of~$D$, without altering~\Eq{9}, we may rearrange its factors
so that the first subscripts shall appear in the order $1, 2, \dotsc, n$, and
obtain
\[
(-1)^i e_{1{k_1}} e_{2{k_2}} \dotsm e_{n{k_n}}.
\]
This can be done by performing in reverse order the $i$~successive interchanges
of the letters~$e$ corresponding to the $i$~successive interchanges
which were used to derive the arrangement $i_1, i_2, \dotsc, i_n$ of the first
subscripts from the arrangement $1, 2, \dotsc, n$. Thus the new second
subscripts $k_1, \dotsc, k_n$ are derived from the old second subscripts $1, \dotsc, n$
by $i$~interchanges. The resulting signed product is therefore a term
of~$D'$. Hence $D = D'$.


% [** PP: Split combined ToC entry]
\Section{86.}{Interchange of Two Columns}
\begin{Thm}
A determinant is merely changed
in sign by the interchange of any two of its columns.
\end{Thm}% No \par
For example,
\[
D = \begin{vmatrix}
a & b \\
c & d
\end{vmatrix}
= ad-bc,\qquad
\Delta = \begin{vmatrix}
b & a \\
d & c
\end{vmatrix}
= bc - ad = -D.
\]

Let $\Delta$ be the determinant derived from~\Eq{8} by the interchange of the
$r$th and $s$th columns. The terms of~$\Delta$ are therefore obtained from the
terms~\Eq{9} of~$D$ by interchanging $r$ and~$s$ in the series of second subscripts.
Interchange the $r$th and $s$th letters~$e$ to restore the second subscripts
to their natural order. Since the first subscripts have undergone an
interchange, the negative of any term of~$\Delta$ is a term of~$D$, and $\Delta = -D$.


% [** PP: No separate ToC entry]
\Section{87.}{Interchange of Two Rows}
\index{Determinants!interchanges}%
\begin{Thm}
A determinant~$D$ is merely changed
in sign by the interchange of any two rows.
\end{Thm}

Let $\Delta$ be the determinant obtained from~$D$ by interchanging the $r$th
and $s$th rows. By interchanging the rows and columns in~$D$ and in~$\Delta$,
we get two determinants $D'$ and~$\Delta'$, either of which may be derived from
the other by the interchange of the $r$th and $s$th columns. Hence, by
§§85,~86,
\[
\Delta = \Delta' = -D' = -D.
\]

%% -----File: 114.png---Folio 108-------


% [** No ToC entry]
\Section{88.}{Two Rows or Two Columns Alike}
\begin{Thm}
A determinant is zero if any
two of its rows or any two of its columns are alike.
\end{Thm}

For, by the interchange of the two like rows or two like columns, the
determinant is evidently unaltered, and yet must change in sign by §§86,~87.
Hence $D = -D$, $D = 0$.


\begin{Exercises}{}

\begin{Problems}

\item[1.] Prove that the equation of the straight line determined by the two distinct
points $(x_1, y_1)$ and $(x_2, y_2)$ is
\[
\begin{vmatrix}
x   & y   & 1 \\
x_1 & y_1 & 1 \\
x_2 & y_2 & 1
\end{vmatrix} = 0.
\]

\item[2.] Show that
\[
\begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix}
=
\begin{vmatrix}
a_2 & c_2 & b_2 \\
a_1 & c_1 & b_1 \\
a_3 & c_3 & b_3
\end{vmatrix}
=
\begin{vmatrix}
a_3 & a_1 & a_2 \\
b_3 & b_1 & b_2 \\
c_3 & c_1 & c_2
\end{vmatrix}.
\]
\end{Problems}

By use of the Factor Theorem~(§14) and the diagonal term, prove that
\begin{Problems}

\item[3.]
\[
\begin{vmatrix}
1   & 1   & 1 \\
a   & b   & c \\
a^2 & b^2 & c^2
\end{vmatrix} = (b-a)(c-a)(c-b).
\]

\item[4.]
\[
\begin{vmatrix}
1     & 1     & \cdots & 1 \\
x_1   & x_2   & \cdots & x_n \\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
\Dots{4} \\
x_1^{n-1} & x_2^{n-1} & \cdots &  x_n^{n-1}
\end{vmatrix}
  = \prod^n_{\substack{i, j=1 \\ i>j}}(x_i-x_j).
\]

This is known as the determinant of Vandermonde, who discussed it in~1770. The
symbol on the right means the product of all factors of the type indicated.
\index{Determinant!of Vandermonde}%

\item[5.] Prove that a skew-symmetric determinant of odd order is zero:
\index{Determinants!skew symmetric}%
\[
\begin{vmatrix}
\Neg 0 & \Neg a & \Neg b \\
    -a & \Neg 0 & \Neg c \\
    -b &     -c & \Neg 0
\end{vmatrix} = 0,
\qquad
\begin{vmatrix}
\Neg 0 & \Neg a & \Neg b & \Neg c & \Neg d \\
    -a & \Neg 0 & \Neg e & \Neg f & \Neg g \\
    -b &     -e & \Neg 0 & \Neg h & \Neg j \\
    -c &     -f &     -h & \Neg 0 & \Neg k \\
    -d &     -g &     -j &     -k & \Neg 0
\end{vmatrix} = 0.
\]
\end{Problems}
\end{Exercises}

%% -----File: 115.png---Folio 109-------


\Section{89.}{Minors} The determinant of order $n-1$ obtained by erasing
(or covering up) the row and column crossing at a given element of a
determinant of order~$n$ is called the \emph{minor} of that element.
\index{Determinants!minors}%

\begin{Remark}
For example, in the determinant~\Eq{6'} of order~$3$, the minors of $b_1$, $b_2$, $b_3$ are respectively
\[
B_1 = \begin{vmatrix}
a_2 & c_2 \\
a_3 & c_3
\end{vmatrix},\quad
%
B_2 = \begin{vmatrix}
a_1 & c_1 \\
a_3 & c_3
\end{vmatrix},\quad
%
B_3 = \begin{vmatrix}
a_1 & c_1 \\
a_2 & c_2
\end{vmatrix}.
\]
Again, \Eq{6'} is the minor of~$d_4$ in the determinant of order~$4$ given by~\Eq{7}.
\end{Remark}


\Section[Expansion]
{90.}{Expansion According to the Elements of a Row or Column.} In
\index{Determinants!expansion}%
\[
D = \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix},
\Tag{6'}
\]
denote the minor of any element by the corresponding capital letter,
so that $b_1$ has the minor~$B_1$, $b_3$ has the minor~$B_3$, etc., as in~§89. We
shall prove that
\begin{align*}
&& D &= \Neg a_1A_1 - b_1B_1 + c_1C_1, &
   D &= \Neg a_1A_1 - a_2A_2 + a_3A_3, && \\
%
&& D &= -a_2A_2 + b_2B_2 - c_2C_2, &
   D &= -b_1B_1 + b_2B_2 - b_3B_3, && \\
%
&& D &= \Neg a_3A_3 - b_3B_3 + c_3C_3, &
   D &= \Neg c_1C_1 - c_2C_2 + c_3C_3. &&
\end{align*}
The three relations at the left (or right) are expressed in words by saying
that a \emph{determinant~$D$ of the third order may be expanded according to the
elements of the first, second or third row} (\emph{or column}). To obtain the expansion,
we multiply each element of the row (or column) by the minor of
the element, prefix the proper sign to the product, and add the signed
products. The signs are alternately $+$ and~$-$, as in the diagram
\[
\begin{matrix}
+ & - & + \\
- & + & - \\
+ & - & +
\end{matrix}
\]

\begin{Remark}
For example, by expansion according to the second column,
\[
\begin{vmatrix}
1 & 4 & 5 \\
2 & 0 & 3 \\
3 & 0 & 9
\end{vmatrix}
= -4 \begin{vmatrix}
2 & 3 \\
3 & 9
\end{vmatrix}
= -4 × 9 = -36.
\]

%% -----File: 116.png---Folio 110-------

Similarly the value of the determinant~\Eq{7} of order~$4$ may be found by expansion
according to the elements of the fourth column:
\[
-d_1 \begin{vmatrix}
a_2 & b_2  & c_2 \\
a_3 & b_3  & c_3 \\
a_4 & b_4  & c_4
\end{vmatrix}
%
+ d_2 \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_3 & b_3 & c_3 \\
a_4 & b_4 & c_4
\end{vmatrix}
%
- d_3 \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_4 & b_4 & c_4
\end{vmatrix}
%
+ d_4 \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix}.
\]
\end{Remark}

We shall now prove that \emph{any determinant~$D$ of order~$n$ may be expanded
according to the elements of any row or any column}.

Let $E_{ij}$ denote the minor of~$e_{ij}$ in~$D$, given by~\Eq{8}, so that $E_{ij}$ is
obtained by erasing the $i$th~row and $j$th~column of~$D$.

(\emph{i}) We first prove that
\[
D = e_{11}E_{11} - e_{21}E_{21} + e_{31}E_{31} - \dotsb
      + (-1)^{n-1} e_{n1}E_{n1},
\Tag{10}
\]
so that $D$ may be expanded according to the elements of its first column.
By~\Eq{9} the terms of~$D$ having the factor~$e_{11}$ are of the form
\[
(-1)^i e_{11} e_{{i_2}2} \dotsm e_{{i_n}n},
\]
where $1, i_2, \dotsc, i_n$ is an arrangement of $1, 2, \dotsc, n$, obtained from the
latter by $i$~interchanges, so that $i_2, \dotsc, i_n$ is an arrangement of $2, \dotsc, n$,
derived from the latter by $i$~interchanges. After removing from each
term the common factor~$e_{11}$ and adding the quotients, we obtain a sum
which, by definition, is the value of the determinant~$E_{11}$ of order~$n-1$.
Hence the terms of~$D$ having the factor~$e_{11}$ may all be combined into
$e_{11}E_{11}$, which is the first part of~\Eq{10}.

We next prove that the terms of~$D$ having the factor~$e_{21}$ may be combined
into~$-e_{21}E_{21}$, which is the second part of~\Eq{10}. For, if $\Delta$ be the
determinant obtained from~$D$ by interchanging its first and second rows,
the result just proved shows that the terms of~$\Delta$ having the factor~$e_{21}$
may be combined into the product of~$e_{21}$ by the minor
\[
\begin{vmatrix}
e_{12} & e_{13} & \cdots & e_{1n} \\
e_{32} & e_{33} & \cdots & e_{3n} \\
\Dots{4} \\
e_{n2} & e_{n3} & \cdots & e_{nn}
\end{vmatrix}
\]
of~$e_{21}$ in~$\Delta$. Now this minor is identical with the minor~$E_{21}$ of~$e_{21}$ in~$D$.
But $\Delta = -D$~(§87). Hence the terms of~$D$ having the factor~$e_{21}$ may be
%% -----File: 117.png---Folio 111-------
combined into $-e_{21} E_{21}$. Similarly, the terms of~$D$ having the factor~$e_{31}$
may be combined into $e_{31} E_{31}$, etc., as in~\Eq{10}.

(\emph{ii}) We next prove that~$D$ may be expanded according to the elements
of its $k$th~column ($k > 1$):
\[
D = \sum_{j=1}^n (-1)^{j+k} e_{jk} E_{jk}.
\Tag{11}
\]
Consider the determinant~$\delta$ derived from~$D$ by moving the $k$th~column
over the earlier columns until it becomes the new first column. Since
this may be done by $k-1$ interchanges of adjacent columns, $\delta = (-1)^{k-1} D$.
The minors of the elements $e_{1k}, \dotsc, e_{nk}$ in the first column of~$\delta$ are evidently
the minors $E_{1k}, \dotsc, E_{nk}$ of $e_{1k}, \dotsc, e_{nk}$ in~$D$. Hence, by~\Eq{10},
\[
\delta = e_{1k} E_{1k} - e_{2k} E_{2k} + \dotsb
            + (-1)^{n-1} e_{nk} E_{nk}
       = \sum^n_{j=1} (-1)^{j+1} e_{jk} E_{jk}.
\]
Thus $D=(-1)^{k-1} \delta$ has the desired value~\Eq{11}.

(\emph{iii}) Finally, $D$ may be expanded according to the elements of its
$k$th~row:
\[
D = \sum^n_{j=1} (-1)^{j+k} e_{kj} E_{kj}.
\]
In fact, by Case~(\emph{ii}), the latter is the expansion of the equal determinant~$D'$
in~§85 according to the elements of its $k$th~column.


\Section{91.}{Removal of Factors}
\index{Determinants!removal of factor}%
\begin{Thm}
A common factor of all of the elements of the
same row or same column of a determinant may be divided out of the elements
and placed as a factor before the new determinant.
\end{Thm}

In other words, if all of the elements of a row or column are divided
by~$n$, the value of the determinant is divided by~$n$.  For example,
\[
\begin{vmatrix}
na_1 & nb_1 \\
 a_2 &  b_2
\end{vmatrix}
= n \begin{vmatrix}
a_1 & b_1 \\
a_2 & b_2
\end{vmatrix},\qquad
\begin{vmatrix}
a_1 & nb_1 & c_1 \\
a_2 & nb_2 & c_2 \\
a_3 & nb_3 & c_3
\end{vmatrix}
= n \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix}.
\]

Proof is made by expanding the determinants according to the elements
of the row or column in question and noting that the minors are the same
for the two determinants.  Thus the second equation is equivalent to
\[
-(nb_1)B_1 + (nb_2)B_2 - (nb_3)B_3 = n(-b_1B_1 + b_2B_2 - b_3B_3),
\]
where $B_i$ denotes the minor of~$b_i$ in the final determinant.

%% -----File: 118.png---Folio 112-------

% [** PP: Heading stubbornly refuses to start on a new page]
\clearpage
\begin{Exercises}{Page112}

\begin{Problems}[2]

\item[1.] $\ds
\begin{vmatrix}
3a & 3b & 3c \\
5a & 5b & 5c \\
 d &  e &  f
\end{vmatrix} = 0$.

\item[2.] $\ds
\begin{vmatrix}
2r & l & 3r \\
2s & m & 3s \\
2t & n & 3t
\end{vmatrix} = 0$.
\end{Problems}

Expand by the shortest method and evaluate
\begin{Problems}[2]

\item[3.] $\ds
\begin{vmatrix}
2 & 7 & 3 \\
5 & 9 & 8 \\
0 & 3 & 0
\end{vmatrix}$.

\item[4.] $\ds
\begin{vmatrix}
5 & 7 & 0 \\
6 & 8 & 0 \\
3 & 9 & 4
\end{vmatrix}$.

\ResetCols{1}

\item[5.] $\ds
\begin{vmatrix}
a   & b   & c   & d \\
a^2 & b^2 & c^2 & d^2 \\
a^3 & b^3 & c^3 & d^3 \\
a^4 & b^4 & c^4 & d^4
\end{vmatrix}
= abcd(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$.
\end{Problems}
\end{Exercises}


\Section{92.}{Sum of Determinants}
\index{Determinants!sum of}%
\begin{Thm}
A determinant having $a_1+q_1$, $a_2+q_2, \dotsc$ as
the elements of a column is equal to the sum of the determinant having $a_1$,
$a_2, \dotsc$ as the elements of the corresponding column and the determinant
having $q_1$, $q_2, \dotsc$ as the elements of that column, while the elements of the
remaining columns of each determinant are the same as in the given determinant.
\end{Thm}

For example,
\[
\begin{vmatrix}
a_1 + q_1 & b_1 & c_1 \\
a_2 + q_2 & b_2 & c_2 \\
a_3 + q_3 & b_3 & c_3
\end{vmatrix}
= \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix}
+ \begin{vmatrix}
q_1 & b_1 & c_1 \\
q_2 & b_2 & c_2 \\
q_3 & b_3 & c_3
\end{vmatrix}.
\]

To prove the theorem we have only to expand the three determinants
according to the elements of the column in question (the first column in
the example) and note that the minors are the same for all three determinants.
Hence $a_1 + q_1$ is multiplied by the same minor that $a_1$ and~$q_1$
are multiplied by separately, and similarly for $a_2 + q_2$, etc.

The similar theorem concerning the splitting of the elements of any
row into two parts is proved by expanding the three determinants according
to the elements of the row in question. For example,
\[
\begin{vmatrix}
a+r & b+s \\
c   & d
\end{vmatrix}
= \begin{vmatrix}
a & b \\
c & d
\end{vmatrix}
+ \begin{vmatrix}
r & s \\
c & d
\end{vmatrix}.
\]

%% -----File: 119.png---Folio 113-------


\Section{93.}{Addition of Columns or Rows}
\index{Determinants!addition of columns}%
\begin{Thm}
A determinant is not changed
in value if we add to the elements of any column the products of the corresponding
elements of another column by the same arbitrary number.
\end{Thm}

Let $a_1$, $a_2, \dotsc$ be the elements to which we add the products of the
elements $b_1$, $b_2, \dotsc$ by~$n$. We apply~§92 with $q_1=nb_1$, $q_2=nb_2, \dotsc$.
Thus the modified determinant is equal to the sum of the initial determinant
and a determinant having $b_1$, $b_2, \dotsc$ in one column and $nb_1$, $nb_2,
\dotsc$ in another column. But~(§91) the latter determinant is equal to
the product of~$n$ by a determinant with two columns alike and hence
is zero~(§88). For example,
\[
\begin{vmatrix}
a_1 + nb_1 & b_1 & c_1 \\
a_2 + nb_2 & b_2 & c_2 \\
a_3 + nb_3 & b_3 & c_3
\end{vmatrix}
= \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix}
+ n\begin{vmatrix}
b_1 & b_1 & c_1 \\
b_2 & b_2 & c_2 \\
b_3 & b_3 & c_3
\end{vmatrix},
\]
and the last determinant is zero.

Similarly, \emph{a determinant is not changed in value if we add to the elements
of any row the products of the corresponding elements of another row by the
same arbitrary number}.

\begin{Remark}
For example,
\[
\begin{vmatrix}
a + nc & b + nd \\
c      & d
\end{vmatrix}
= \begin{vmatrix}
a & b \\
c & d
\end{vmatrix}
+ n \begin{vmatrix}
c & d \\
c & d
\end{vmatrix}
= \begin{vmatrix}
a & b \\
c & d
\end{vmatrix}.
\]
\end{Remark}


\begin{Example}
Evaluate the first determinant below.
\[
\begin{vmatrix}
1 &     -2 & 1 \\
1 & \Neg 2 & 3 \\
6 & \Neg 4 & 3
\end{vmatrix}
= \begin{vmatrix}
1 &  0 & 1 \\
1 &  8 & 3 \\
6 & 10 & 3
\end{vmatrix}
= \begin{vmatrix}
\Neg 0 &  0 & 1 \\
    -2 &  8 & 3 \\
\Neg 3 & 10 & 3
\end{vmatrix}
= \begin{vmatrix}
    -2 &  8 \\
\Neg 3 & 10
\end{vmatrix} = -44.
\]
\end{Example}

\begin{Solution}
First we add to the elements of the second column the products of the
elements of the last column by~$2$. In the resulting second determinant, we add to the
elements of the first column the products of the elements of the third column by~$-1$.
Finally, we expand the resulting third determinant according to the elements of its
first row.
\end{Solution}


\begin{Exercises}{}

\begin{Problems}
\item[1.] Prove that
\[
\begin{vmatrix}
b   + c   & c   + a   & a   + b   \\
b_1 + c_1 & c_1 + a_1 & a_1 + b_1 \\
b_2 + c_2 & c_2 + a_2 & a_2 + b_2
\end{vmatrix}
= 2 \begin{vmatrix}
a   & b   & c   \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix}
\]
\end{Problems}

%% -----File: 120.png---Folio 114-------

By reducing to a determinant of order~$3$, etc., prove that
\begin{Problems}
\item[2.] $\ds
\begin{vmatrix}
1   & 1   & 1   & 1   \\
a   & b   & c   & d   \\
a^2 & b^2 & c^2 & d^2 \\
a^3 & b^3 & c^3 & d^3
\end{vmatrix}
= (a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$.
\bigskip % [** PP: Explicit vertical space]

\ResetCols{2}

\item[3.] $\ds
\begin{vmatrix}
2 &      -1 & \Neg 3 &      -2 \\
1 &  \Neg 7 & \Neg 1 &      -1 \\
3 &  \Neg 5 &     -5 &  \Neg 3 \\
4 &      -3 & \Neg 2 &      -1
\end{vmatrix} = -42$.

\item[4.] $\ds
\begin{vmatrix}
1 & 1 & \phantom{1}1 & \phantom{1}1 \\
1 & 2 & \phantom{1}3 & \phantom{1}4 \\
1 & 3 & \phantom{1}6 & 10 \\
1 & 4 &           10 & 20
\end{vmatrix} = 1$.

\end{Problems}
\end{Exercises}


\Section[System of $n$~Linear Equations in $n$~Unknowns]
{94.}{System of $n$~Linear Equations in $n$~Unknowns with $D \neq 0$.} In
\index{Linear equations!system|(}%
\[
\begin{matrix}
a_{11} x_1 + a_{12} x_2 + \dotsb + a_{1n} x_n = k_1, \\
\Dots{1} \\
a_{n1} x_1 + a_{n2} x_2 + \dotsb + a_{nn} x_n = k_n,
\end{matrix}
\Tag{12}
\]
let $D$ denote the determinant of the coefficients of the $n$~unknowns:
\[
D = \begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
\Dots{4} \\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}.
\]
Then
{\small
\[
Dx_1 = \begin{vmatrix}
a_{11}x_1 & a_{12} & \cdots & a_{1n} \\
\Dots{4} \\
a_{n1}x_1 & a_{n2} & \cdots & a_{nn} \\
\end{vmatrix}
= \begin{vmatrix} % [** PP: Removed commas after first column]
a_{11}x_1 + a_{12}x_2 + \dotsb + a_{1n}x_n & a_{12} & \cdots & a_{1n} \\
\Dots{4} \\
a_{n1}x_1 + a_{n2}x_2 + \dotsb + a_{nn}x_n & a_{n2} & \cdots & a_{nn}
\end{vmatrix},
\]}%
where the second determinant was derived from the first by adding to
the elements of the first column the products of the corresponding elements
of the second column by~$x_2$, etc., and finally the products of the elements
of the last column by~$x_n$. The elements of the new first column are equal
to $k_1, \dotsc, k_n$ by~\Eq{12}. In this manner, we find that
\[
Dx_1 = K_1,\qquad
Dx_2 = K_2,\qquad \dotsc,\qquad
Dx_n = K_n,
\Tag{13}
\]
in which $K_i$ is derived from~$D$ by substituting $k_1, \dotsc, k_n$ for the elements
$a_{1i}, \dotsc, a_{ni}$ of the $i$th~column of~$D$, whence
\[
K_1 = \begin{vmatrix}
k_1 & a_{12} & \cdots & a_{1n} \\
\Dots{4} \\
k_n & a_{n2} & \cdots & a_{nn}
\end{vmatrix}, \dotsc\qquad
K_n = \begin{vmatrix}
a_{11} & \cdots & a_{1n-1} & k_1 \\
\Dots{4} \\
a_{n1} & \cdots & a_{nn-1} & k_n
\end{vmatrix}.
\]

%% -----File: 121.png---Folio 115-------

If $D \neq 0$, the unique values of $x_1, \dotsc, x_n$ determined by division
from~\Eq{13} actually satisfy equations~\Eq{12}. For instance, the first equation
is satisfied since
\[
k_1D - a_{11} K_1 - a_{12} K_2 - \dotsb - a_{1n} K_n
= \begin{vmatrix}
k_1 & a_{11} & a_{12} & \cdots & a_{1n} \\
k_1 & a_{11} & a_{12} & \cdots & a_{1n} \\
k_2 & a_{21} & a_{22} & \cdots & a_{2n} \\
\Dots{5} \\
k_n & a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix},
\]
as shown by expansion according to the elements of the first row; and
the determinant is zero, having two rows alike.

\begin{Theorem}
If $D$ denotes the determinant of the coefficients of the
$n$~unknowns in a system of $n$~linear equations, the product of $D$ by any one
of the unknowns is equal to the determinant obtained from $D$ by substituting
the known terms in place of the coefficients of that unknown. If $D \neq 0$, we
obtain the unique values of the unknowns by division by~$D$.
\end{Theorem}

We have therefore given a complete proof of the results stated and
illustrated in §80,~§81. % [** PP: Retaining instead of §§80,~81.]
Another proof is suggested in Ex.~7 below.
The theorem was discovered by induction in 1750 by G.~Cramer.


\begin{Exercises}{Page115}

Solve by determinants the following systems of equations (reducing each determinant
to one having zero as the value of every element but one in a row or column,
as in the example in~§93).
\begin{Problems}[2]

\item[1.] $\begin{System}{3}
  x &+{}&  y &+{}&  z &= 11, \\
 2x &-{}& 6y &-{}&  z &=  0, \\
 3x &+{}& 4y &+{}& 2z &=  0.
\end{System}$

\item[2.] $\begin{System}{3}
 x &+{}&  y &+{}&  z &=  0, \\
 x &+{}& 2y &+{}& 3z &= -1, \\
 x &+{}& 3y &+{}& 6z &=  0.
\end{System}$

\ResetCols{2}

\item[3.] $\begin{System}{3}
 x &-{}& 2y &+{}&  z &= 12, \\
 x &+{}& 2y &+{}& 3z &= 48, \\
6x &+{}& 4y &+{}& 3z &= 84.
\end{System}$

% [** PP: Matching original alignment; not aligning like variables]
\item[4.] $\begin{System}{2}
3x &-{}& 2y &=  7, \\
3y &-{}& 2z &=  6, \\
3z &-{}& 2x &= -1.
\end{System}$

\ResetCols{2}

\item[5.] $\begin{System}{4}
x &+{}&  y &+{}&   z &+{}&   w &=  1, \\
x &+{}& 2y &+{}&  3z &+{}&  4w &= 11, \\
x &+{}& 3y &+{}&  6z &+{}& 10w &= 26, \\
x &+{}& 4y &+{}& 10z &+{}& 20w &= 47.
\end{System}$

\item[6.] $\begin{System}{4}
2x &-{}&  y &+{}& 3z &-{}& 2w &= 4, \\
 x &+{}& 7y &+{}&  z &-{}&  w &= 2, \\
3x &+{}& 5y &-{}& 5z &+{}& 3w &= 0, \\
4x &-{}& 3y &+{}& 2z &-{}&  w &= 5.
\end{System}$

\ResetCols{1}

\item[7.] Prove the first relation~\Eq{13} by multiplying the members of the first equation~\Eq{12}
by~$A_{11}$, those of the second equation by $-A_{21}, \dotsc$, those of the $n$th equation by
$(-1)^{n-1}A_{n1}$, and adding, where $A_{ij}$ by denotes the minor of~$a_{ij}$ in~$D$. Hint: The resulting
coefficient of~$x_2$ is the expansion, according to the elements of its first column, of a determinant
derived from $D$ by replacing $a_{11}$ by~$a_{12}$, $\dotsc$, $a_{n1}$ by~$a_{n2}$.
\end{Problems}
\end{Exercises}

%% -----File: 122.png---Folio 116-------


\Section[Rank]
{95.}{Rank of a Determinant.} If we erase from a determinant~$D$ of
order~$n$ all but $r$~rows and all but $r$~columns, we obtain a determinant
of order~$r$ called an \emph{$r$-rowed minor of~$D$}. In particular, any element is
regarded as a one-rowed minor, and $D$ itself is regarded as an $n$-rowed
minor.
\index{Determinants!minors}%

If a determinant~$D$ of order~$n$ is not zero, it is said to be of \emph{rank~$n$}.
If, for $0 < r < n$, some $r$-rowed minor of~$D$ is not zero, while every $(r+1)$-rowed
minor is zero, $D$ is said to be of \emph{rank~$r$}. It is said to be of rank
zero if every element is zero.
\index{Determinants!rank}%

\begin{Remark}
For example, a determinant~$D$ of order~$3$ is of rank~$3$ if $D \neq 0$; of rank~$2$ if $D = 0$,
but some two-rowed minor is not zero; of rank~$1$ if every two-rowed minor is zero,
but some element is not zero. Again, every three-rowed minor of
\[
\begin{vmatrix}
a & b & c & d \\
e & f & g & h \\
a & b & c & d \\
e & f & g & h
\end{vmatrix}
\]
is zero since two pairs of its rows are alike. Hence it is of rank~$2$ if some two-rowed
minor is not zero. But it is of rank~$1$ if $a, b, c, d$ are not all zero and are proportional
to $e, f, g, h$,  since all two-rowed minors are then zero.
\end{Remark}


\Section[System of $n$~Linear Equations in $n$~Unknowns]
{96.}{System of $n$~Linear Equations in $n$~Unknowns with $D=0$.} We
shall now discuss the equations~\Eq{12} for the troublesome case (previously
ignored) in which the determinant~$D$ of the coefficients of the unknowns
is zero. In view of~\Eq{13}, the given equations are evidently inconsistent
if any one of the determinants $K_1, \dotsc, K_n$ is not zero. But if $D$ and
these $K$'s are all zero, our former results~\Eq{13} give us no information
concerning the unknowns~$x_i$, and we resort to the following

\begin{Theorem}
Let the determinant~$D$ of the coefficients of the unknowns
in equations~\Eq{12} be of rank~$r$, $r<n$. If the determinants~$K$ obtained from
the $(r+1)$-rowed minors of~$D$ by replacing the elements of any column by
the corresponding known terms~$k_i$ are not all zero, the equations are inconsistent.
But if these determinants $K$ are all zero, the $r$~equations involving
the elements of a non-vanishing $r$-rowed minor of $D$ determine uniquely $r$
of the unknowns as linear functions of the remaining $n-r$ unknowns, which
are independent variables, and the expressions for these $r$~unknowns satisfy
also the remaining $n-r$ equations.
\end{Theorem}

%% -----File: 123.png---Folio 117-------

\begin{Remark}
Consider for example the three equations~\Eq{4} in the unknowns $x, y, z$. Five cases
arise:

% [** PP: Not italicizing i.e.]
\begin{itemize}
\item[($\alpha$)] $D$ of rank~$3$, i.e., $D \neq 0$.

\item[($\beta$)] $D$ of rank $2$ (i.e., $D=0$, but some two-rowed minor $\neq 0$), and
\[
K_1 = \begin{vmatrix}
k_1 & b_1 & c_1 \\
k_2 & b_2 & c_2 \\
k_3 & b_3 & c_3
\end{vmatrix},\quad
%
K_2 = \begin{vmatrix}
a_1 & k_1 & c_1 \\
a_2 & k_2 & c_2 \\
a_3 & k_3 & c_3
\end{vmatrix},\quad
%
K_3 = \begin{vmatrix}
a_1 & b_1 & k_1 \\
a_2 & b_2 & k_2 \\
a_3 & b_3 & k_3
\end{vmatrix}
\]
not all zero.

\item[($\gamma$)] $D$ of rank~$2$ and $K_1$, $K_2$, $K_3$ all zero.

\item[($\delta$)] $D$ of rank~$1$ (i.e., every two-rowed minor $= 0$, but some element $\neq 0$), and
\[
\begin{vmatrix}
a_i & k_i \\
a_j & k_j
\end{vmatrix},\qquad
%
\begin{vmatrix}
b_i & k_i \\
b_j & k_j
\end{vmatrix},\qquad
\begin{vmatrix}
c_i & k_i \\
c_j & k_j
\end{vmatrix}\qquad \text{($i$, $j$ chosen from $1$, $2$, $3$)}
\]
not all zero; there are nine such determinants~$K$.

\item[($\epsilon$)] $D$ of rank~$1$, and all nine of the two-rowed determinants~$K$ zero.
\end{itemize}

In case~($\alpha$) the equations have a single set of solutions~(§94). In cases ($\beta$) and~($\delta$)
there is no set of solutions. For ($\beta$) the proof follows from~\Eq{13}. In case~($\gamma$) one
of the equations is a linear combination of the other two; for example, if $a_1b_2 - a_2b_1 \neq 0$,
the first two equations determine $x$ and~$y$ as linear functions of~$z$ (as shown by transposing
the terms in~$z$ and solving the resulting equations for $x$ and~$y$), and the resulting
values of $x$ and~$y$ satisfy the third equation identically as to~$z$. Finally, in case~($\epsilon$),
two of the equations are obtained by multiplying the remaining one by constants.

The reader acquainted with the elements of solid analytic geometry will see that
the planes represented by the three equations have the following relations:
\begin{itemize}

\item[($\alpha$)] The three planes intersect in a single point.

\item[($\beta$)] Two of the planes intersect in a line parallel to the third plane.

\item[($\gamma$)] The three planes intersect in a common line.

\item[($\delta$)] The three planes are parallel and not all coincident.

\item[($\epsilon$)] The three planes coincide.

\end{itemize}
\end{Remark}

The remarks preceding our theorem furnish an illustration (the case
$r=n-1$) of the following

\begin{Lemma}[1.] If every $(r+1)$-rowed minor~$M$ formed from certain $r+1$
rows of~$D$ is zero, the corresponding $r+1$ equations~\Eq{12} are inconsistent
provided there is a non-vanishing determinant~$K$ formed from any~$M$
by replacing the elements of any column by the corresponding known
terms~$k_i$.
\end{Lemma}

For concreteness,\footnote
  {All other cases may be reduced to this one by rearranging the $n$~equations and
relabelling the unknowns (replacing $x_3$ by the new~$x_1$, for example).}
let the rows in question be the first $r+1$ and let
%% -----File: 124.png---Folio 118-------
\[
K = \begin{vmatrix}
a_{11}   & \cdots & a_{1r}   & k_1 \\
\Dots{4} \\
a_{r+11} & \cdots & a_{r+1r} & k_{r+1}
\end{vmatrix} \neq 0.
\]
Let $d_1, \dotsc, d_{r+1}$ be the minors of $k_1, \dotsc, k_{r+1}$ in~$K$. Multiply the
first $r+1$ equations~\Eq{12} by $d_1$, $-d_2, \dotsc, (-1)^rd_{r+1}$, respectively, and
add. The right member of the resulting equation is the expansion of~$±K$.
The coefficient of~$x_s$ is the expansion of
\[
±\begin{vmatrix}
a_{11}   & \cdots & a_{1r}   & a_{1s} \\
\Dots{4} \\
a_{r+11} & \cdots & a_{r+1r} & a_{r+1s}
\end{vmatrix}
\]
and is zero, being an~$M$ if $s>r$, and having two columns identical if $s \leqq r$.
Hence $0 = ±K$. Thus if $K \neq 0$, the equations are inconsistent.

\begin{Lemma}[2.] If all of the determinants $M$ and~$K$ in Lemma~1 are zero,
but an $r$-rowed minor of an~$M$ is not zero, one of the corresponding $r+1$
equations is a linear combination of the remaining $r$ equations.
\end{Lemma}

As before let the $r+1$ rows in question be the first $r+1$. Let the
non-vanishing $r$-rowed minor be
\[
d_{r+1} = \begin{vmatrix}
a_{11} & \cdots & a_{1r} \\
\Dots{3} \\
a_{r1} & \cdots & a_{rr}
\end{vmatrix} \neq 0.
\Tag{14}
\]
Let the functions obtained by transposing the terms~$k_i$ in~\Eq{12} be
\[
L_i \equiv a_{i1} x_1 + a_{i2} x_2 + \dotsb + a_{in} x_n - k_i.
\]
By the multiplication made in the proof of Lemma~1,
\[
d_1L_1 - d_2L_2 + \dotsb + (-1)^rd_{r+1}L_{r+1} = \mp K = 0.
\]
Hence $L_{r+1}$ is a linear combination of $L_1, \dotsc, L_r$.

The first part of the theorem is true by Lemma~1. The second part
is readily proved by means of Lemma~2. Let~\Eq{14} be the non-vanishing
$r$-rowed minor of~$D$. For $s>r$, the $s$th equation is a linear combination
of the first $r$~equations, and hence is satisfied by any set of solutions of
the latter. In the latter transpose the terms involving $x_{r+1}, \dotsc, x_n$.
Since the determinant of the coefficients of $x_1, \dotsc, x_r$ is not zero,~§94
shows that $x_1, \dotsc, x_r$ are uniquely determined linear functions of
$x_{r+1}, \dotsc, x_n$ (which enter from the new right members).

%% -----File: 125.png---Folio 119-------


\begin{Exercises}{Page119}

Apply the theorem to the following four systems of equations and check the conclusions:
\begin{Problems}[2]

\item[1.] $\begin{System}{3}
2x&+{}&  y&+{}& 3z &=  1, \\
4x&+{}& 2y&-{}&  z &= -3, \\
2x&+{}&  y&-{}& 4z &= -4.
\end{System}$

\item[2.] $\begin{System}{3}
2x&+{}&  y&+{}& 3z &= 1, \\
4x&+{}& 2y&-{}&  z &= 3, \\
2x&+{}&  y&-{}& 4z &= 4.
\end{System}$

\ResetCols{2}

\item[3.] $\begin{System}{3}
 x&-{}&  3y&+{}&  4z &= 1, \\
4x&-{}& 12y&+{}& 16z &= 3, \\
3x&-{}&  9y&+{}& 12z &= 3.
\end{System}$

\item[4.] $\begin{System}{3}
 x&-{}&  3y&+{}&  4z &= 1, \\
4x&-{}& 12y&+{}& 16z &= 4, \\
3x&-{}&  9y&+{}& 12z &= 3.
\end{System}$

\ResetCols{1}

\item[5.] Discuss the system
\[
\begin{System}{3}
ax&+{}&  y&+{}&  z &= a-3, \\
 x&+{}& ay&+{}&  z &=  -2, \\
 x&+{}&  y&+{}& az &=  -2,
\end{System}
\]
when (\emph{i})~$a = 1$; (\emph{ii})~$a = -2$; (\emph{iii})~$a \neq 1$, $-2$, obtaining the simplest forms of the
unknowns.

\item[6.] Discuss the system
\[
\begin{System}{3}
 x  &+{}&    y&+{}&    z &= 1,  \\
ax  &+{}&   by&+{}&   cz &= k,  \\
a^2x&+{}& b^2y&+{}& c^2z &= k^2,
\end{System}
\]
when (\emph{i})~$a$, $b$, $c$ are distinct; (\emph{ii}) $a = b \neq c$; (\emph{iii}) $a = b = c$.
\end{Problems}
\end{Exercises}


\Section[Homogeneous Equations]% [** PP: Allow line to break in ... list]
{97.}{Homogeneous Linear Equations.} When the known terms $k_1$, $\dotsc,
k_n$ in~\Eq{12} are all zero, the equations are called \emph{homogeneous}. The determinants~$K$
are now all zero, so that the $n$~homogeneous equations are never
inconsistent. This is also evident from the fact that they have the set
of solutions $x_1 = 0, \dotsc, x_n = 0$. By~\Eq{13}, there is no further set of solutions
if $D \neq 0$. If $D = 0$, there are further sets of solutions. This is
shown by the theorem of~§96 which now takes the following simpler form.

\begin{Thm}%
If the determinant~$D$ of the coefficients of $n$~linear homogeneous equations
in $n$~unknowns is of rank~$r$, $r<n$, the $r$~equations involving the elements of a
non-vanishing $r$-rowed minor of~$D$ determine uniquely $r$ of the unknowns as
linear functions of the remaining $n-r$ unknowns, which are independent variables,
and the expressions for these $r$~unknowns satisfy also the remaining
$n-r$ equations.
\end{Thm}

The particular case mentioned is the much used theorem:

\begin{Thm}%
\index{Linear equations!homogeneous}%
A necessary and sufficient condition that $n$~linear homogeneous equations
in $n$~unknowns shall have a set of solutions, other than the trivial one in which
each unknown is zero, is that the determinant of the coefficients be zero.
\end{Thm}

%% -----File: 126.png---Folio 120-------


\begin{Exercises}{Page120}

Discuss the following systems of equations:
\begin{Problems}[2]
\item[1.] $\begin{System}{3}
x &+{}&  y &+{}& 3z &= 0,\\
x &+{}& 2y &+{}& 2z &= 0,\\
x &+{}& 5y &-{}&  z &= 0.
\end{System}$

\item[2.] $\begin{System}{3}
 2x &-{}&   y &+{}&  4z &= 0,\\
  x &+{}&  3y &-{}&  2z &= 0,\\
  x &-{}& 11y &+{}& 14z &= 0.
\end{System}$

\ResetCols{2}

\item[3.] $\begin{System}{3}
 x &-{}&  3y &+{}&  4z &= 0,\\
4x &-{}& 12y &+{}& 16z &= 0,\\
3x &-{}&  9y &+{}& 12z &= 0.
\end{System}$

\item[4.] $\begin{System}{4}
6x &+{}& 4y &+{}& 3z &-{}& 84w &= 0,\\
 x &+{}& 2y &+{}& 3z &-{}& 48w &= 0,\\
 x &-{}& 2y &+{}&  z &-{}& 12w &= 0,\\
4x &+{}& 4y &-{}&  z &-{}& 24w &= 0.
\end{System}$

\ResetCols{1}

\item[5.] $\begin{System}{4}
2x &+{}&  3y &-{}&  4z &+{}&  5w &= 0,\\
3x &+{}&  5y &-{}&   z &+{}&  2w &= 0,\\
7x &+{}& 11y &-{}&  9z &+{}& 12w &= 0,\\
3x &+{}&  4y &-{}& 11z &+{}& 13w &= 0.
\end{System}$
\end{Problems}
\end{Exercises}


% [** PP: ToC entry reads System of $m$~Linear Equations in $n$~Unknowns,
% Matrix and Augmented Matrix]
\Section{98.}{System of $m$~Linear Equations in $n$~Unknowns} The case $m<n$
may be treated by means of the lemmas in~§96. If $m>n$, we select
any $n$ of the equations and apply to them the theorems of §§94,~96. If
they are found to be inconsistent, the entire system is evidently inconsistent.
But if the $n$~equations are consistent, and if $r$ is the rank of the
determinant of their coefficients, we obtain $r$ of the unknowns expressed
as linear functions of the remaining $n-r$ unknowns. Substituting these
values of these $r$~unknowns in the remaining equations, we obtain a
system of $m-n$ linear equations in $n-r$ unknowns. Treating this system
in the same manner, we ultimately either find that the proposed
$m$~equations are consistent and obtain the general set of solutions,
or find that they are inconsistent. To decide in advance whether the
former or latter of these cases will arise, we have only to find the maximum
order~$r$ of a non-vanishing $r$-rowed determinant formed from the
coefficients of the unknowns, taken in the regular order in which they
occur in the equations, and ascertain whether or not the corresponding
($r+1$)-rowed determinants~$K$, formed as in~§96, are all zero.

The last result may be expressed simply by employing the terminology
of matrices. The system of coefficients of the unknowns in any set of
linear equations
\[
\begin{matrix}
a_{11} x_1 + \dotsb + a_{1n} x_n = k_1, \\
\Dots{1} \\
a_{m1} x_1 + \dotsb + a_{mn} x_n = k_m,
\end{matrix}
\Tag{15}
\]
arranged as they occur in the equations, is called the \emph{matrix} of the coefficients,
and is denoted by
\index{Matrix}%
\[
A = \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
\Dots{4} \\
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{pmatrix}.
\]
%% -----File: 127.png---Folio 121-------
By annexing the column composed of the known terms $k_i$ we obtain
the so-called \emph{augmented matrix}
\index{Matrix!augmented}%
\[
B = \begin{pmatrix}
a_{11} & a_{12} & \cdots & a_{1n} & k_1\\
\Dots{5}\\
a_{m1} & a_{m2} & \cdots & a_{mn} & k_m
\end{pmatrix}.
\]

The definitions of an $r$-rowed minor (determinant) of a matrix and of
the rank of a matrix are entirely analogous to the definitions in~§95.

In view of Lemma~1 in~§96, our equations~\Eq{15} are inconsistent if
$B$ is of rank $r+1$ and $A$ is of rank $\leqq r$. By Lemma~2, if $A$ and~$B$ are both
of rank~$r$, all of our equations are linear combinations of~$r$ of them. Noting
also that the rank~$r$ of~$A$ cannot exceed the rank of~$B$, since every minor
of~$A$ is a minor of~$B$, and hence a non-vanishing $r$-rowed minor of~$A$ is a
minor of~$B$, so that the rank of~$B$ is not less than~$r$, we have the following

\begin{Theorem}
A system of $m$~linear equations in $n$~unknowns is consistent
if and only if the rank of the matrix of the coefficients of the unknowns is
equal to the rank of the augmented matrix. If the rank of both matrices is~$r$,
certain $r$ of the equations determine uniquely $r$ of the unknowns as linear
functions of the remaining $n - r$ unknowns, which are independent variables,
and the expressions for these $r$~unknowns satisfy also the remaining
$m - r$ equations.
\index{Determinants!rank}%
\end{Theorem}

When $m = n+1$, $B$ has an $m$-rowed minor called the \emph{determinant of
the square matrix~$B$}. If this determinant is not zero, $B$ is of rank~$m$.
Since $A$ has no $m$-rowed minor, its rank is less than~$m$. Hence we obtain
the

\begin{Corollary}
Any system of $n+1$ linear equations in $n$~unknowns is
inconsistent if the determinant of the augmented matrix is not zero.
\end{Corollary}


\begin{Exercises}{Page121}

Discuss the following systems of equations:
\begin{Problems}[2]
\item[1.] $\begin{System}{3}
 2x &+{}&  y &+{}& 3z &= 1,\\
 4x &+{}& 2y &-{}&  z &= -3,\\
 2x &+{}&  y &-{}& 4z &= -4,\\
10x &+{}& 5y &-{}& 6z &= -10.
\end{System}$

\item[2.] $\begin{System}{3}
2x &-{}&  y &+{}& 3z &= 2,\\
 x &+{}& 7y &+{}& z  &= 1,\\
3x &+{}& 5y &-{}& 5z &= a,\\
4x &-{}& 3y &+{}& 2z &= 1.
\end{System}$

\ResetCols{2}

\item[3.] $\begin{System}{3}
4x &-{}&  y &+{}&  z &= 5,\\
2x &-{}& 3y &+{}& 5z &= 1,\\
 x &+{}&  y &-{}& 2z &= 2,\\
5x &   &    &-{}&  z &= 2.
\end{System}$

\item[4.] $\begin{System}{2}
 4x &-{}& 5y &=  2,\\
 2x &+{}& 3y &= 12,\\
10x &-{}& 7y &= 16.
\end{System}$

\ResetCols{1}

\item[5.] Prove the Corollary by multiplying the known terms by $x_{n+1}=1$ and applying~§97
with $n$ replaced by $n+1$.

\item[6.] Prove that if the matrix of the coefficients of any system of linear homogeneous
\index{Linear equations!homogeneous}%
equations in $n$~unknowns is of rank~$r$, the values of certain $n-r$ of the unknowns may be
%% -----File: 128.png---Folio 122-------
assigned at pleasure and the others will then be uniquely determined and satisfy all of the equations.
\end{Problems}
\end{Exercises}
\index{Linear equations!system|)}% [** PP: Original page range is 114--121]


\Section{99.}{Complementary Minors} The determinant
\index{Determinants!complementary minors}%
\[
D = \begin{vmatrix}
a_1 & b_1 & c_1 & d_1 \\
a_2 & b_2 & c_2 & d_2 \\
a_3 & b_3 & c_3 & d_3 \\
a_4 & b_4 & c_4 & d_4
\end{vmatrix}
\Tag{16}
\]
is said to have the \emph{two-rowed complementary minors}
\[
M = \begin{vmatrix}
a_1 & b_1 \\
a_3 & b_3
\end{vmatrix},\qquad
M' = \begin{vmatrix}
c_2 & d_2 \\
c_4 & d_4
\end{vmatrix},
\]
since either is obtained by erasing from~$D$ all the rows and columns having
an element which occurs in the other.

In general, if we erase from a determinant~$D$ of order~$n$ all but $r$~rows
and all but $r$~columns, we obtain a determinant~$M$ of order~$r$ called an
$r$-rowed minor of~$D$. But if we had erased from~$D$ the $r$~rows and $r$~columns
previously kept, we would have obtained an ($n-r$)-rowed minor of~$D$
called the \emph{minor complementary to~$M$}. In particular, any element is
regarded as a one-rowed minor and is complementary to its minor (of
order $n-1$).


% [** PP: Next two paragraphs have single ToC entry, Laplace's Development]
\Section{100.}{Laplace's Development by Columns}
\index{Determinants!Laplace's development|(}%
\begin{Thm}
Any determinant~$D$ is
equal to the sum of all the signed products $±MM'$, where $M$ is an $r$-rowed
minor having its elements in the first $r$~columns of~$D$, and $M'$ is the minor
complementary to~$M$, while the sign is $+$ or~$-$ according as an even or odd
number of interchanges of rows of~$D$ will bring $M$ into the position occupied
by the minor~$M_1$ whose elements lie in the first $r$~rows and first $r$~columns
of~$D$.
\end{Thm}

\begin{Remark}
For $r = 1$, this development becomes the known expansion of $D$ according to the
elements of the first column~(§90); here $M_1 = e_{11}$.

If $r=2$ and $D$ is the determinant~\Eq{16},
\begin{align*}
D &= \begin{vmatrix}
a_1 & b_1 \\
a_2 & b_2
\end{vmatrix} · \begin{vmatrix}
c_3 & d_3 \\
c_4 & d_4
\end{vmatrix} - \begin{vmatrix}
a_1 & b_1 \\
a_3 & b_3
\end{vmatrix} · \begin{vmatrix}
c_2 & d_2 \\
c_4 & d_4
\end{vmatrix} + \begin{vmatrix}
a_1 & b_1 \\
a_4 & b_4
\end{vmatrix} · \begin{vmatrix}
c_2 & d_2 \\
c_3 & d_3
\end{vmatrix} \\[1ex]
 &{}+ \begin{vmatrix}
a_2 & b_2 \\
a_3 & b_3
\end{vmatrix} · \begin{vmatrix}
c_1 & d_1 \\
c_4 & d_4
\end{vmatrix} - \begin{vmatrix}
a_2 & b_2 \\
a_4 & b_4
\end{vmatrix} · \begin{vmatrix}
c_1 & d_1 \\
c_3 & d_3
\end{vmatrix} + \begin{vmatrix}
a_3 & b_3 \\
a_4 & b_4
\end{vmatrix} · \begin{vmatrix}
c_1 & d_1 \\
c_2 & d_2
\end{vmatrix}.
\end{align*}
%% -----File: 129.png---Folio 123-------
The first product in the development is $M_1M_1'$; the second product is $-MM'$ (in the
notations of~§99), and the sign is minus since the interchange of the second and third
rows of $D$ brings this $M$ into the position of~$M_1$. The sign of the third product in
the development is plus since two interchanges of rows of~$D$ bring the first factor
into the position of~$M_1$.
\end{Remark}

If $D$ is the determinant~\Eq{8}, then
\[
M_1 = \begin{vmatrix}
e_{11} & \cdots & e_{1r} \\
\Dots{3} \\
e_{r1} & \cdots & e_{rr}
\end{vmatrix},\qquad
M_1' = \begin{vmatrix}
e_{r+1 r+1} & \cdots & e_{r+1 n}\\
\Dots{3} \\
e_{n r+1} & \cdots & e_{nn}
\end{vmatrix}.
\]

Any term of the product $M_1M_1'$ is of the type
\[
  (-1)^i e_{{i_1}1} e_{{i_2}2} \dotsm e_{{i_r}r}
· (-1)^j e_{i_{r+1}{r+1}} \dotsm e_{{i_n}n},
\Tag{17}
\]
where $i_1, \dotsc, i_r$ is an arrangement of $1, \dotsc, r$ derived from $1, \dotsc, r$
by $i$~interchanges, while $i_{r+1}, \dotsc, i_n$ is an arrangement of $r+1, \dotsc, n$
derived by $j$~interchanges. Hence $i_1, \dotsc, i_n$ is an arrangement of
$1, \dotsc, n$ derived by $i+j$ interchanges, so that the product~\Eq{17} is a term
of $D$ with the proper sign.

It now follows from~§87 that any term of any of the products~$±MM'$
mentioned in the theorem is a term of~$D$. Clearly we do not obtain twice
in this manner the same term of~$D$.

Conversely, any term~$t$ of~$D$ occurs in one of the products~$±MM'$.
Indeed, $t$ contains as factors $r$~elements from the first $r$~columns of~$D$,
no two being in the same row, and the product of these is, except perhaps
as to sign, a term of some minor~$M$. Thus $t$ is a term of $MM'$ or
of~$-MM'$. In view of the earlier discussion, the sign of $t$ is that of the
corresponding term in $±MM'$, where the latter sign is given by the
theorem.


% [** PP: No separate ToC entry]
\Section{101.}{Laplace's Development by Rows} There is a Laplace development
of $D$ in which the $r$-rowed minors~$M$ have their elements in the first
$r$~rows of~$D$, instead of in the first $r$~columns as in~§100. To prove this,
we have only to apply~§100 to the equal determinant obtained by interchanging
the rows and columns of~$D$.

There are more general (but less used) Laplace developments in which
the $r$-rowed minors~$M$ have their elements in any chosen $r$~columns (or
rows) of~$D$. It is simpler to apply the earlier developments to the determinant
$±D$ having the elements of the chosen $r$~columns (or rows) in the
new first $r$~columns (or rows).
\index{Determinants!Laplace's development|)}%

%% -----File: 130.png---Folio 124-------


\begin{Exercises}{}
\begin{Problems}
\item[1.] Prove that
\[
\begin{vmatrix}
a & b & c & d \\
e & f & g & h \\
0 & 0 & j & k \\
0 & 0 & l & m
\end{vmatrix} = \begin{vmatrix}
a & b \\
e & f
\end{vmatrix} · \begin{vmatrix}
j & k \\
l & m
\end{vmatrix}.
\]

\item[2.] By employing $2$-rowed minors from the first two rows, show that
\[
\frac{1}{2}\begin{vmatrix}
a & b & c & d \\
e & f & g & h \\
a & b & c & d \\
e & f & g & h
\end{vmatrix} = \begin{vmatrix}
a & b \\
e & f
\end{vmatrix} · \begin{vmatrix}
c & d \\
g & h
\end{vmatrix} - \begin{vmatrix}
a & c \\
e & g
\end{vmatrix} · \begin{vmatrix}
b & d \\
f & h
\end{vmatrix} + \begin{vmatrix}
a & d \\
e & h
\end{vmatrix} · \begin{vmatrix}
b & c \\
f & g
\end{vmatrix} = 0.
\]

\item[3.] By employing $2$-rowed minors from the first two columns of the $4$-rowed determinant
in Ex.~2, show that the products in Laplace's development cancel.
\end{Problems}
\end{Exercises}


\Section{102.}{Product of Determinants}
\index{Determinants!product of}%
\begin{Thm}
The product of two determinants of
the same order is equal to a determinant of like order in which the element
of the $r$th row and $c$th column is the sum of the products of the elements of
the $r$th row of the first determinant by the corresponding elements of the $c$th
column of the second determinant.
\end{Thm}

\begin{Remark}
For example,
\[
\begin{vmatrix}
a & b \\
c & d
\end{vmatrix} · \begin{vmatrix}
e & f \\
g & h
\end{vmatrix} = \begin{vmatrix}
ae + bg & af + bh \\
ce + dg & cf + dh
\end{vmatrix}.
\Tag{18}
\]
\end{Remark}

While for brevity we shall give the proof for determinants of order~$3$,
the method is seen to apply to determinants of any order. By Laplace's
development with $r = 3$~(§101), we have
\[
\begin{vmatrix}
\Neg a_1 & \Neg b_1 & \Neg c_1 & 0   & 0   & 0   \\
\Neg a_2 & \Neg b_2 & \Neg c_2 & 0   & 0   & 0   \\
\Neg a_3 & \Neg b_3 & \Neg c_3 & 0   & 0   & 0   \\
      -1 & \Neg   0 & \Neg   0 & e_1 & f_1 & g_1 \\
\Neg   0 &       -1 & \Neg   0 & e_2 & f_2 & g_2 \\
\Neg   0 & \Neg   0 &       -1 & e_3 & f_3 & g_3
\end{vmatrix} = \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix} · \begin{vmatrix}
e_1 & f_1 & g_1 \\
e_2 & f_2 & g_2 \\
e_3 & f_3 & g_3
\end{vmatrix}.
\Tag{19}
\]

%% -----File: 131.png---Folio 125-------

In the determinant of order~$6$, add to the elements of the fourth, fifth,
and sixth columns the products of the elements of the first column by
$e_1$, $f_1$, $g_1$, respectively (and hence introduce zeros in place of the former
elements $e_1$, $f_1$, $g_1$). Next, add to the elements of the fourth, fifth, and
sixth columns the products of the elements of the second column by
$e_2$, $f_2$, $g_2$, respectively. Finally, add to the elements of the fourth, fifth,
and sixth columns the products of the elements of the third column by
$e_3$, $f_3$, $g_3$, respectively.    The new determinant is
\[
\begin{vmatrix}
\Neg a_1 & \Neg b_1 & \Neg c_1 &
    a_1e_1+b_1e_2+c_1e_3 & a_1f_1+b_1f_2+c_1f_3 & a_1g_1+b_1g_2+c_1g_3 \\
%
\Neg a_2 & \Neg b_2 & \Neg c_2 &
    a_2e_1+b_2e_2+c_2e_3 & a_2f_1+b_2f_2+c_2f_3 & a_2g_1+b_2g_2+c_2g_3 \\
%
\Neg a_3 & \Neg b_3 & \Neg c_3 &
    a_3e_1+b_3e_2+c_3e_3 & a_3f_1+b_3f_2+c_3f_3 & a_3g_1+b_3g_2+c_3g_3 \\
%
    -1 & \Neg 0 & \Neg 0 & 0 & 0 & 0 \\
\Neg 0 &     -1 & \Neg 0 & 0 & 0 & 0 \\
\Neg 0 & \Neg 0 &     -1 & 0 & 0 & 0
\end{vmatrix}.
\]
By Laplace's development (or by expansion according to the elements of
the last row, etc.), this is equal to the $3$-rowed minor whose elements
are the long sums. Hence this minor is equal to the product in the right
member of~\Eq{19}.


\begin{Exercises}{}

\begin{Problems}

\item[1.]  Prove~\Eq{18} by means of~§92.

\item[2.]  Prove that, if $s_i = \alpha^i + \beta^i + \gamma^i$,
\[
\begin{vmatrix}
   1     &     1      &     1   \\
\alpha   &   \beta    &  \gamma \\
\alpha^2 &   \beta^2  &  \gamma^2
\end{vmatrix} · \begin{vmatrix}
1 & \alpha &  \alpha^2 \\
1 & \beta  &  \beta^2  \\
1 & \gamma &  \gamma^2
\end{vmatrix} = \begin{vmatrix}
3   & s_1 & s_2 \\
s_1 & s_2 & s_3 \\
s_2 & s_3 & s_4
\end{vmatrix}.
\]

\item[3.] If $A_i$, $B_i$, $C_i$ are the minors of $a_i$, $b_i$, $c_i$ in the determinant~$D$ defined by the second
factor below, prove that
\[
\begin{vmatrix}
\Neg A_1 &     -A_2 & \Neg A_3 \\
    -B_1 & \Neg B_2 &     -B_3 \\
\Neg C_1 &     -C_2 & \Neg C_3
\end{vmatrix} · \begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{vmatrix} = \begin{vmatrix}
D & 0 & 0 \\
0 & D & 0 \\
0 & 0 & D
\end{vmatrix}.
\]
Hence the first factor is equal to $D^2$ if $D \neq 0$.

%% -----File: 132.png---Folio 126-------

\item[4.] Express $(a^2 + b^2 + c^2 + d^2)(e^2 + f^2 + g^2 + h^2)$ as a sum of four squares by writing
\index{Sum of!four squares}%
\[
\begin{vmatrix}
\Neg a + bi & c + di \\
    -c + di & a - bi
\end{vmatrix} · \begin{vmatrix}
\Neg e + fi & g + hi \\
    -g + hi & e - fi
\end{vmatrix}
\]
as a determinant of order~$2$ similar to each factor. Hint: If $k'$ denotes the conjugate
of the complex number~$k$, each of the three determinants is of the form
\[
\begin{vmatrix}
\Neg k  &   l \\
    -l' &   k'
\end{vmatrix}.
\]
\end{Problems}
\end{Exercises}


\begin{Exercises}[MISCELLANEOUS~]{Page126}

\begin{Problems}

\item[1.] Solve
\[
\begin{System}{3}
  ax &+{}&   by &+{}&   cz &= k,\\
a^2x &+{}& b^2y &+{}& c^2z &= k^2,\\
a^4x &+{}& b^4y &+{}& c^4z &= k^4
\end{System}
\]
by determinants for~$x$, treating all cases.

\item[2.] In three linear homogeneous equations in four unknowns, prove that the values
of the unknowns are proportional to four determinants of order~$3$ formed from the
coefficients.
\end{Problems}

Factor the following determinants:
\begin{Problems}[2]

\item[3.] $\ds
\begin{vmatrix}
1 & a & bc \\
1 & b & ca \\
1 & c & ab
\end{vmatrix}$.

\item[4.] $\ds
\begin{vmatrix}
x & x^2 & yz \\
y & y^2 & xz \\
z & z^2 & xy
\end{vmatrix} = \begin{vmatrix}
x^2 & x^3 & 1 \\
y^2 & y^3 & 1 \\
z^2 & z^3 & 1
\end{vmatrix}$.

\ResetCols{1}

\item[5.]
\[
\begin{vmatrix}
a & b & c \\
c & a & b \\
b & c & a
\end{vmatrix} = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega),
\]
where $\omega$ is an imaginary cube root of unity.

\ResetCols{2}

\item[6.] $\ds
\begin{vmatrix}
a & b & c & d \\
b & a & d & c \\
c & d & a & b \\
d & c & b & a
\end{vmatrix}$.

\item[7.] $\ds
\begin{vmatrix}
a & b & c & d \\
d & a & b & c \\
c & d & a & b \\
b & c & d & a
\end{vmatrix}$.

\ResetCols{1}

% [** PP: Using array for better surrounding space in wide matrix]
\item[8.] If the points $(x_1, y_1), \dotsc, (x_4, y_4)$ lie on a circle, prove that
\[
\left|\begin{array}{cccc}
x_1^2 + y_1^2  & x_1 & y_1 & 1 \\
\Dots{4} \\
x_4^2 + y_4^2  & x_4 & y_4 & 1
\end{array}\right| = 0.
\]

%% -----File: 133.png---Folio 127-------

\item[9.] Prove that
\begin{gather*}
\begin{vmatrix}
aa' + bb' + cc' & ea' + fb' + gc' \\
ae' + bf' + cg' & ee' + ff' + gg'
\end{vmatrix} \\
%
{} = \begin{vmatrix}
a & b \\
e & f
\end{vmatrix} · \begin{vmatrix}
a' & b' \\
e' & f'
\end{vmatrix} + \begin{vmatrix}
a & c \\
e & g
\end{vmatrix} · \begin{vmatrix}
a' & c' \\
e' & g'
\end{vmatrix} + \begin{vmatrix}
b & c \\
f & g
\end{vmatrix} · \begin{vmatrix}
b' & c' \\
f' & g'
\end{vmatrix}.
\end{gather*}

\item[10.] Prove that the cubic equation
\index{Cubic equation}%
\[
D(x) \equiv \begin{vmatrix}
a-x &  b   &  c \\
b   & f-x  &  g \\
c   &  g   & h-x
\end{vmatrix} = 0
\]
has only real roots. Hints:
\begin{gather*}
D(x) · D(-x) = \left|\begin{array}{lll}
a^2+b^2+c^2-x^2 &   ab+bf+cg         &  ac+bg+ch \\
ab+bf+cg        &   b^2+f^2+g^2-x^2  &  bc+fg+gh \\
ac+bg+ch        &   bc+fg+gh         &  c^2+g^2+h^2-x^2
\end{array}\right| \\
%
{} = -x^6+x^4(a^2+f^2+h^2+2b^2+2c^2+2g^2) - x^2(D_1+D_2+D_3)+ D^2(0),
\end{gather*}
where $D_3$ denotes the first determinant in Ex.~9 with all accents removed and with
$e = b$, while $D_1$ and~$D_2$ are analogous minors of elements in the main diagonal of the
present determinant of order~$3$ with $x = 0$. Hence the coefficient of~$-x^2$ is a sum of
squares. Since the function of degree~$6$ is not zero for a negative value of~$x^2$, $D(x)=0$
has no purely imaginary root. If it had an imaginary root $r+si$, then $D(x+r)=0$
would have a purely imaginary root~$si$. But $D(x+r)$ is of the form $D(x)$ with $a$, $f$, $h$
replaced by $a-r$, $f-r$, $h-r$. Hence $D(x)=0$ has only real roots. The method is
applicable to such determinants of order~$n$.

\item[11.] If $a_1, \dotsc, a_n$ are distinct, solve the system of equations
\[
\frac{x_1}{k_i-a_1} + \frac{x_2}{k_i-a_2} + \dotsb
  + \frac{x_n}{k_i - a_n} = 1\qquad (i=1, \dotsc, n).
\]

Hint: Regard $k_1, \dotsc, k_n$ as the roots of an equation of degree~$n$ in $k$ formed from
the typical one above by substituting~$k$ for~$k_i$ and clearing of fractions; write $k = a_j-t$,
and consider the product of the roots of $t^n + \dotsb = 0$. Hence find~$x_j$.

\item[12.] Solve the equation
\[
\begin{vmatrix}
a+x &   x &   x \\
  x & b+x &   x \\
  x &   x & c+x
\end{vmatrix} = 0.
\]
\end{Problems}
\end{Exercises}
\index{Determinants|)}%

%% -----File: 134.png---Folio 128-------


\Chapter{IX}{Symmetric Functions}

\index{Sigma function|(}%
\index{Symmetric functions|(}%

% [** PP: Unit has two separate ToC entries]
\Section{103.}{Sigma Functions, Elementary Symmetric Functions} A rational
function of the independent variables $x_1, x_2, \dotsc, x_n$ is said to be \emph{symmetric}
in them if it is unaltered by the interchange of any two of the variables.
For example,
\[
x_1^2 + x_2^2 + x_3^2 +4x_1 + 4x_2 +4x_3
\]
is a symmetric polynomial in $x_1$, $x_2$, $x_3$; the sum of the first three terms
is denoted by $\Sigma x_1^2$ and the sum of the last three by $4\Sigma x_1$. In general,
if $t$ is a rational function of $x_1, \dotsc, x_n, \Sigma t$ denotes the sum of $t$ and all
of the distinct functions obtained from $t$ by permutations of the variables;
such a $\Sigma$-function (read \emph{sigma function}) is symmetric in $x_1, \dotsc, x_n$.
\index{Symbol!f@{$\Sigma$\IndAdd{symmetric function}}}% [** PP: Manually alphabetized]

For example, if there are three independent variables $\alpha$, $\beta$, $\gamma$,
\begin{gather*}
\begin{aligned}
\Sigma \alpha\beta
  &= \alpha\beta + \alpha\gamma + \beta\gamma,\qquad &
\Sigma \alpha^2\beta
  &= \alpha^2\beta  + \alpha\beta^2
  + \alpha^2\gamma + \alpha\gamma^2
  + \beta^2 \gamma + \beta \gamma^2, \\
%
\Sigma \frac{1}{\alpha}
  &= \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}, &
\Sigma \frac{\beta}{\alpha}
  &= \frac{\beta}{\alpha}  + \frac{\alpha}{\beta} +
     \frac{\beta}{\gamma}  + \frac{\gamma}{\beta} +
     \frac{\alpha}{\gamma} + \frac{\gamma}{\alpha},
\end{aligned} \\ % Top-level break
%
\Sigma \frac{\alpha^2 + \beta^2}{\alpha\beta}
  = \frac{\alpha^2 + \beta^2}{\alpha\beta}
  + \frac{\alpha^2 + \gamma^2}{\alpha\gamma}
  + \frac{\beta^2  + \gamma^2}{\beta\gamma}.
\end{gather*}

In particular, $\Sigma \alpha = \alpha + \beta + \gamma$, $\Sigma \alpha\beta$, and $\alpha\beta\gamma$ are called the three \emph{elementary
symmetric functions} of $\alpha$, $\beta$, $\gamma$. In general,
\index{Elementary symmetric function}%
\[
\Sigma \alpha_1,\quad
\Sigma \alpha_1\alpha_2,\quad
\Sigma \alpha_1\alpha_2\alpha_3, \dotsc,\quad
\Sigma \alpha_1\alpha_2 \dotsm \alpha_{n-1},\quad
\alpha_1 \alpha_2 \dotsc \alpha_n
\]
are the elementary symmetric functions of $\alpha_1$, $\alpha_2, \dotsc, \alpha_n$. In~§20 they
were written out more fully and proved to be equal to $-c_1$, $c_2$, $-c_3, \dotsc,
(-1)^n c_n$ if $\alpha_1, \dotsc, \alpha_n$ are the roots of the equation
\[
x^n + c_1x^{n-1} + c_2x^{n-2} + \dotsb + c_n = 0
\Tag{1}
\]
whose leading coefficient is unity.

%% -----File: 135.png---Folio 129-------


\begin{Exercises}{}

If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + r = 0$,
so that $\Sigma \alpha = -p$, $\Sigma \alpha\beta = q$, and $\alpha\beta\gamma = -r$,
prove that
\begin{Problems}

\item[1.] $(\Sigma \alpha)^2 = \Sigma \alpha^2 + 2\Sigma \alpha\beta$, whence $\Sigma \alpha^2 = p^2 - 2q$.

\item[2.] $\Sigma \alpha · \Sigma \alpha\beta = \Sigma \alpha^2\beta + 3\alpha\beta\gamma$, whence $\Sigma \alpha^2\beta = 3r-pq$.

\item[3.] $\Sigma \alpha^2\beta\gamma = pr$.

\item[4.] $\Sigma \alpha^2\beta^2 = (\Sigma \alpha\beta)^2 - 2\alpha\beta\gamma\Sigma \alpha = q^2 - 2pr$.
\end{Problems}

If $\alpha$, $\beta$, $\gamma$, $\delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, prove that
\begin{Problems}
\item[5.]\qquad\qquad $\Sigma \dfrac{1}{\alpha} = \dfrac{-r}{s}$,\qquad
  $\Sigma \dfrac{1}{\alpha\beta} = \dfrac{q}{s}$,\qquad
  $\Sigma \dfrac{1}{\alpha^2} = \dfrac{r^2 - 2qs}{s^2}$. \\
Hint: Compute the sum, sum of the products two at a time, and sum of the squares
of the roots of the equation
\[
1 + py + qy^2 + ry^3 + sy^4 = 0,
\]
obtained by replacing~$x$ by~$1/y$ in the given quartic equation.

\item[6.] $\Sigma \dfrac{\beta}{\alpha} = \Sigma \alpha · \Sigma \dfrac{1}{\alpha} - 4 = \dfrac{pr}{s} - 4$.

\item[7.] $\Sigma \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \Sigma \dfrac{\beta}{\alpha}$.

\item[8.] $\Sigma \dfrac{\beta\gamma}{\alpha^2} = \Sigma \alpha\beta · \Sigma \dfrac{1}{\alpha^2} - \Sigma \dfrac{\beta}{\alpha} = \dfrac{1}{s^2} (qr^2 - 2q^2s - prs + 4s^2)$.

\item[9.] $\Sigma \dfrac{\gamma}{\alpha\beta} = \dfrac{3r - pq}{s}$.
\end{Problems}
\end{Exercises}


\Section[Fundamental Theorem]
{104.}{Fundamental Theorem on Symmetric Functions.}
\begin{Thm}
Any polynomial
symmetric in $x_1, \dotsc, x_n$ is equal to an integral rational function,
with integral coefficients, of the elementary symmetric functions
\[
E_1 = \Sigma x_1,\qquad
E_2 = \Sigma x_1x_2,\qquad
E_3 = \Sigma x_1x_2x_3,\dotsc,\qquad
E_n = x_1x_2 \dotsm x_n
\Tag{2}
\]
and the coefficients of the given polynomial. In particular, any symmetric
polynomial with integral coefficients is equal to a polynomial in the elementary
symmetric functions with integral coefficients.
\end{Thm}

\begin{Remark}
For example, if $n = 2$,
\[
rx_1^2 + rx_2^2 + sx_1 + sx_2 \equiv r(E_1^2 - 2E_2) + sE_1.
\]
In case $r$ and~$s$ are integers, the resulting polynomial in $E_1$ and~$E_2$ has integral coefficients.
\end{Remark}

The theorem is most frequently used in the equivalent form:

%% -----File: 136.png---Folio 130-------

\emph{Any polynomial symmetric in the roots of an equation,
\[
x^n - E_1 x^{n-1} + E_2 x^{n-2} - \dotsb + (-1)^n E_n = 0,
\]
is equal to an integral rational function, with integral coefficients, of the coefficients
of the equation and the coefficients of the polynomial.}

It is this precise theorem that is required in all parts of modern algebra
and the theory of numbers, where attention to the nature of the coefficients
is vital, rather than the inadequate, oft-quoted, theorem that any symmetric
function of the roots is expressible (rationally) in terms of the
coefficients.

It suffices to prove the theorem for any homogeneous symmetric polynomial~$S$,
i.e., one expressible as a sum of terms % [** PP: Not italicizing i.e.]
\[
h = ax_1^{k_1} x_2^{k_2} \dotsm x_n^{k_n}
\]
of constant total degree $k = k_1 + k_2 + \dotsb + k_n$ in the~$x$'s. Evidently
we may assume that no two terms of~$S$ have the same set of exponents
$k_1, \dotsc, k_n$ (since such terms may be combined into a single one). We
shall say that $h$ is \emph{higher} than the term $bx_1^{l_1} x_2^{l_2} \dotsm x_n^{l_n}$ if $k_1>l_1$, or if
$k_1 = l_1$, $k_2>l_2$, or if $k_1 = l_1$, $k_2 = l_2$, $k_3>l_3, \dotsc$, so that the first one of
the differences $k_1-l_1$, $k_2-l_2$, $k_3-l_3, \dotsc$ which is not zero is positive.

We first prove that, if the above term $h$ is the highest term of~$S$, then
\[
k_1 \geqq k_2 \geqq k_3 \dotsb \geqq k_n.
\]
For, if $k_1<k_2$, the symmetric polynomial~$S$ would contain the term
\[
ax_1^{k_2} x_2^{k_1} x_3^{k_3} \dotsm x_n^{k_n},
\]
which is higher than~$h$. If $k_2<k_3$, $S$ would contain the term
\[
ax_1^{k_1} x_2^{k_3} x_3^{k_2} \dotsm x_n^{k_n},
\]
which is higher than~$h$, etc.

If the highest term in another homogeneous symmetric polynomial~$S'$
is
\[
h' = a'x_1^{k'_1} x_2^{k'_2} \dotsm x_n^{k'_n},
\]
and that of $S$ is $h$, then the highest term in their product~$SS'$ is
\[
hh' = aa'x_1^{k_1 + k'_1} \dotsm x_n^{k_n + k'_n}.
\]
%% -----File: 137.png---Folio 131-------
Indeed, suppose that $SS'$ has a term, higher than~$hh'$,
\[
cx_1^{l_1 + l'_1} \dotsm x_n^{l_n + l'_n},
\Tag{3}
\]
which is either a product of terms
\[
t  = b x_1^{l_1}  \dotsm x_n^{l_n},\qquad
t' = b'x_1^{l'_1} \dotsm x_n^{l'_n}
\]
of $S$ and~$S'$ respectively, or is a sum of such products. Since~\Eq{3} is higher
than~$hh'$, the first one of the differences
\[
l_1 + l'_1 - k_1 -k'_1, \dotsc, l_n + l'_n - k_n -k'_n
\]
which is not zero is positive. But, either all of the differences $l_1 - k_1,
\dotsc, l_n - k_n$ are zero or the first one which is not zero is negative, since
$h$ is either identical with $t$ or is higher than~$t$. Likewise for the differences
$l'_1 - k'_1, \dotsc, l'_n - k'_n$.   We therefore have a contradiction.

It follows at once that the highest term in a product of any number
of homogeneous symmetric polynomials is the product of their highest
terms. Now the highest terms in $E_1$, $E_2$, $E_3, \dotsc, E_n$, given by~\Eq{2}, are
\[
x_1,  \qquad  x_1x_2,\qquad  x_1x_2x_3,\quad\dotsc,\qquad
x_1x_2 \dotsm x_n,
\]
respectively. Hence the highest term in $E_1^{a_1} E_2^{a_2} \dotsm E_n^{a_n}$ is
\[
x_1^{a_1 + a_2 + \dotsb + a_n}
x_2^{a_2 + \dotsb + a_n} \dotsm
x_n^{a_n}.
\]

Thus the highest term in
\[
\sigma = aE_1^{k_1 - k_2} E_2^{k_2 - k_3} \dotsm E_{n-1}^{k_{n-1} - k_n}E_n^{k_n}
\]
is~$h$. Hence $S_1 = S - \sigma$ is a homogeneous symmetric polynomial of the
same total degree~$k$ as~$S$ and having a highest term~$h_1$ not as high as~$h$.
As before, we form a product~$\sigma_1$ of the $E$'s whose highest term is this~$h_1$.
Then $S_2 = S_1 - \sigma_1$ is a homogeneous symmetric polynomial of total degree~$k$
and with a highest term $h_2$ not as high as~$h_1$. We must finally reach
a difference $S_t - \sigma_t$ which is identically zero. Indeed, there is only a
finite number of products of powers of $x_1, \dotsc, x_n$ of total degree~$k$.
Among these are the parts $h'$, $h'_1$, $h'_2, \dotsc$ of~$h$, $h_1$, $h_2, \dotsc$ with the coefficients
suppressed. Since each $h_i$ is not as high as~$h_{i-1}$, the $h'$, $h'_1$, $h'_2, \dotsc$ are
all distinct. Hence there is only a finite number of~$h_i$. Since $S_t - \sigma_t \equiv 0$,
\[
S = \sigma + S_1
  = \sigma + \sigma_1 + S_2 = \dotsb
  = \sigma + \sigma_1 + \sigma_2 + \dotsb + \sigma_t.
\]
Hence $S$ is a polynomial in $E_1$, $E_2, \dotsc, E_n$ and $a$, $b, \dotsc$, with integral
coefficients.

%% -----File: 138.png---Folio 132-------

\begin{Example}[1.]
If  $S = \Sigma x_1^2 x_2^2 x_3$ and $n>4$,  we have
\begin{align*}
\sigma &= E_2E_3
  = S + 3 \Sigma x_1^2 x_2 x_3 x_4 + 10 \Sigma x_1x_2x_3x_4x_5, \\
%
S_1 &= S - \sigma
  = -3 \Sigma x_1^2 x_2 x_3 x_4 - 10 \Sigma x_1x_2x_3x_4x_5, \\
%
\sigma_1 &= -3 E_1E_4
  = -3 (\Sigma x_1^2 x_2 x_3 x_4 + 5 \Sigma x_1x_2x_3x_4x_5), \\
%
S_2 &= S_1 - \sigma_1
  = 5 \Sigma x_1x_2x_3x_4x_5
  = 5E_5, \\
%
S &= \sigma + S_1
  = \sigma + \sigma_1 + S_2
  = E_2 E_3 - 3 E_1 E_4 + 5E_5.
\end{align*}
\end{Example}


\begin{Example}[2.]
If  $S = \Sigma x_1^3 x_2 x_3$ and $n> 4$,
\begin{align*}
\sigma = E_1^2 E_3
  &= E_1(\Sigma x_1^2 x_2 x_3 + 4 \Sigma x_1 x_2 x_3 x_4) \\
%
  &= \Sigma x_1^3 x_2 x_3 + 2 \Sigma x_1^2 x_2^2 x_3
 + 3 \Sigma x_1^2 x_2 x_3 x_4 \\
%
  & \phantom{{}=\Sigma x_1^3x_2x_3 }
 + 4 (\Sigma x_1^2 x_2 x_3 x_4 + 5 \Sigma x_1x_2x_3x_4x_5),
\end{align*}
\[
S_1 = S - \sigma
  = -2  \Sigma x_1^2 x_2^2 x_3 - 7 \Sigma x_1^2 x_2 x_3 x_4
    -20 \Sigma x_1x_2x_3x_4x_5.
\]
Take $\sigma_1 = -2 E_2E_3$ and proceed as in Ex.~1.
\end{Example}

\begin{Example}[3.]
By examples 1 and~2, if $n>4$,
\[
a \Sigma x_1^2 x_2^2 x_3 +
b \Sigma x_1^3 x_2 x_3
  = bE_1^2 E_3 - (3a + b)E_1E_4 + (a - 2b)E_2E_3 + 5(a + b)E_5.
\]
\end{Example}


% [** PP: ToC entry matches unit title, using running head]
\Section[Functions Symmetric in all but One Root]
{105.}{Rational Functions Symmetric in all but One of the Roots.}
\index{Symmetric functions!in all but one root|(}%
\begin{Thm}
If
$P$ is a rational function of the roots of an equation $f(x)= 0$ of degree~$n$ and
if $P$ is symmetric in $n-1$ of the roots, then $P$ is equal to a rational function,
with integral coefficients, of the remaining root and the coefficients of $f(x)$ and~$P$.
\end{Thm}

\begin{Remark}
For example, $P = r \alpha_1 + \alpha_2^2 + \alpha_3^2 + \dotsb + \alpha_n^2$ is symmetric in $\alpha_2, \dotsc, \alpha_n$, and
\[
P = r \alpha_1 + \Sigma \alpha_1^2 - \alpha_1^2
  = c_1^2 - 2c_2 + r \alpha_1 - \alpha_1^2,
\]
if $\alpha_1, \dotsc, \alpha_n$ are the roots of equation~\Eq{1}.
\end{Remark}

Since\footnote
  {If $N/D$ is symmetric in $\alpha_1$, $\alpha_2$, and the polynomials $N$ and~$D$ have no common
  factor, while $N$ becomes~$N'$ and $D$ becomes~$D'$ when $\alpha_1$, $\alpha_2$ are interchanged, then
  $ND'\equiv DN'$. Thus $N$ divides~$N'$ and both are of the same degree. Hence $N'=cN,
  D' = cD$, where $c$ is a constant. By again interchanging $\alpha_1$, $\alpha_2$, we obtain $N$ from~$N'$,
  whence $N = cN' = c^2N$, $c^2 = 1$. If $c = -1$, we take $\alpha_1 = \alpha_2$ and see that $N = N'= -N$,
  $N = 0$, whence $N$ has the factor $\alpha_1 - \alpha_2$. Similarly, $D$ has the same factor, contrary to
  hypothesis. Hence $c = +1$ and $N$ and~$D$ are each symmetric in $\alpha_1$, $\alpha_2$.}
any symmetric rational function is the quotient of two symmetric
polynomials, the above theorem will follow if proved for the case in which
the words rational function are in both places replaced by polynomial.

If $\alpha_1$ is the remaining root, the polynomial~$P$ is symmetric in the roots
$\alpha_2, \dotsc, \alpha_n$ of $f(x)/(x - \alpha_1) = 0$, an equation of degree $n-1$ whose coefficients
are polynomials in $\alpha_1$, $c_1, \dotsc, c_n$ with integral coefficients. Hence~(§104),
$P$ is equal to a polynomial, with integral coefficients, in $\alpha_1$, $c_1, \dotsc, c_n$
and the coefficients of~$P$.

%% -----File: 139.png---Folio 133-------

\begin{Example}
If $\alpha$, $\beta$, $\gamma$ are the roots of $f(x) \equiv x^3 + px^2 + qx + r = 0$, find
\[
\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta}
  = \frac{\alpha^2 +  \beta^2}{\alpha + \beta}
  + \frac{\alpha^2 + \gamma^2}{\alpha + \gamma}
  + \frac{\beta^2  + \gamma^2}{\beta  + \gamma}.
\]
\end{Example}

\begin{Solution}
Since $\beta^2 + \gamma^2 = p^2 - 2q - \alpha^2$,\quad $\beta + \gamma = -p - \alpha$,
\[
\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta}
  = \Sigma \frac{p^2 - 2q - \alpha^2}{-p - \alpha}
  = \Sigma \left(\alpha - p + \frac{2q}{\alpha + p}\right)
  = - p - 3p + 2q \Sigma \frac{1}{\alpha + p}.
\]
But $\alpha + p$, $\beta + p$, $\gamma + p$ are the roots $y_1$, $y_2$, $y_3$ of the cubic equation obtained from
\index{Cubic equation|(}% [** PP: Using page range]
$f(x)=0$ by the substitution $x + p = y$, i.e., $x = y-p$. The resulting equation is % [** PP: Not italicizing i.e.]
\[
y^3 - 2py^2 + (p^2 + q)y + r - pq = 0.
\]
Since we desire the sum of the reciprocals of $y_1$, $y_2$, $y_3$, we set $y = 1/z$ and find the sum
of the roots $z_1$, $z_2$, $z_3$ of
\[
1 - 2pz + (p^2 + q)z^2 + (r - pq)z^3 = 0.
\]
Hence
\[
\Sigma \frac{1}{\alpha + p}
  = \Sigma \frac{1}{y_1}
  = \Sigma z_1 = \frac{p^2 + q}{pq - r},\qquad
\Sigma \frac{\alpha^2 + \beta^2}{\alpha + \beta}
  = \frac{2q^2 - 2p^2q + 4pr}{pq - r}.
\]
\end{Solution}


\begin{Exercises}{Page133}

[In Exs.~1--12, $\alpha$, $\beta$, $\gamma$ are the roots of $f(x) = x^3 + px^2 + qx + r = 0$.]

Using $\beta\gamma + \alpha(\beta + \gamma) = q$, find
\begin{Problems}[2]
\item[1.] $\Sigma \dfrac{\beta\gamma + \alpha^2}{\beta + \gamma}$, % [** PP: Added ,]

\item[2.] $\Sigma \dfrac{3\beta\gamma - 2\alpha^2}{\beta + \gamma - \alpha}$.

\ResetCols{1}

\item[3.]  Why would the use of $\beta\gamma = -r/\alpha$ complicate Exs.\ 1,~2?  Verify that
\[
\beta\gamma
  = \frac{-r}{\alpha}
  = \frac{f(\alpha) - r}{\alpha}
  = \alpha^2 + p \alpha + q.
\]

\item[4.] Why would you use $\beta\gamma = -r/\alpha$ in finding
  $\Sigma \dfrac{\beta^2 + \gamma^2}{\beta\gamma + c}$?

\ResetCols{3}

\item[5.] Find $\Sigma (\beta + \gamma)^2$.

\item[6.] Find $\Sigma (\alpha + \beta - \gamma)^3$.

\item[7.] Find $\smash{\Sigma \left(\dfrac{\beta - \gamma}{\beta + \gamma}\right)^2}$.

\ResetCols{1}

\item[8.] Find a necessary and sufficient condition on the coefficients that the roots, in
some order, shall be in harmonic progression.
Hint:  If $\dfrac{1}{\alpha} + \dfrac{1}{\gamma} = \dfrac{2}{\beta}$, then $\dfrac{-3r}{q} - \beta = 0$,
and conversely. Hence the condition is
\[
\left(\frac{-3r}{q} - \alpha\right)
\left(\frac{-3r}{q} -  \beta\right)
\left(\frac{-3r}{q} - \gamma\right)
  = f\left(\frac{-3r}{q}\right)
  = 0.
\]

%% -----File: 140.png---Folio 134-------

\item[9.] Find the cubic equation with the roots
$\beta\gamma  - \dfrac{1}{\alpha}$,
$\alpha\gamma - \dfrac{1}{\beta}$,
$\alpha\beta  - \dfrac{1}{\gamma}$.
Hint:  since these are $(-r - 1)/\alpha$, etc., make the substitution $(-r - 1)/x = y$.
\end{Problems}

Find the substitution which replaces the given cubic equation by one with the roots
\begin{Problems}

\item[10.]
$\alpha\beta  + \alpha\gamma$,
$\alpha\beta  + \beta\gamma$,
$\alpha\gamma + \beta\gamma$.

\ResetCols{2}

\item[11.] $\dfrac{2\alpha - 1}{\beta + \gamma - \alpha}$, etc.

\item[12.] $\dfrac{\beta\gamma + 3\alpha^2}{\beta + \gamma - 2\alpha}$, etc.
\end{Problems}

If $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, find
\begin{Problems}[2]

\item[13.] $\Sigma\dfrac{\beta^2 + \gamma^2 + \delta^2}{\beta + \gamma + \delta}$.

\item[14.] $\Sigma\dfrac{\beta\gamma + \beta\delta + \gamma\delta}{\beta + \gamma + \delta - 3}$.

\ResetCols{1}

\item[15.] Prove that if $y_1$, $y_2$, $y_3$ are the roots of $y^3 + py + q = 0$, the equation with the roots
$z_1 = (y_2 - y_3)^2$, $z_2 = (y_1 - y_3)^2$, $z_3 = (y_1 - y_2)^2$ is
\index{Equation for differences of roots!squares of differences}%
\[
z^3 + 6pz^2 + 9p^2 z + 4p^3 + 27q^2 = 0.
\]
Hints: since $z_1 = \Sigma y_1^2 - 2y_2y_3 - y_1^2 = -2p + 2q/y_1 - y_1^2$, etc., we set $z = -2p + 2q/y - y^2$.
By the given equation, $y^2 + p + q/y = 0$. Thus the desired substitution is $z = -p + 3q/y$,
$y = 3q/(z + p)$.

\item[16.] Hence find the discriminant of the reduced cubic equation.
\index{Cubic equation|)}%
\index{Discriminant!of cubic}%

\item[17.] If $x_1, \dotsc, x_n$ are the roots of $f(x)=0$, show that
\[
\Sigma \frac{1}{x_1 - c} = \frac{-f'(c)}{f(c)}.
\]
Hint: $x_1 - c = y_1, \dotsc, x_n - c = y_n$ are the roots of
\[
f(c+y) = f(c) + yf'(c) + y^2(\quad)+ \dotsb = 0,
\]
as shown by Taylor's theorem. Or we may employ~\Eq{5} below % [** PP: Added `below']
for $x = c$.
\end{Problems}
\end{Exercises}
\index{Symmetric functions!in all but one root|)}%


% [** PP: ToC entry appended with ``Newton's Identities'']
\Section{106.}{Sums of Like Powers of the Roots} If $\alpha_1, \dotsc, \alpha_n$ are the roots of
\index{Sum of!like powers of roots|(}%
\[
f(x) \equiv x^n + c_1 x^{n-1} + c_2x^{n-2} + \dotsb + c_n = 0,
\Tag{1}
\]
we write $s_1$ for~$\Sigma \alpha_1$, $s_2$ for~$\Sigma \alpha_1^2$, and, in general,
\index{Symbol!g@{$s_k$\IndAdd{sum of $k$th powers}}}% [** PP: Manually alphabetized]
\[
s_k = \Sigma \alpha_1^k
    = \alpha_1^k + \alpha_2^k + \dotsb + \alpha_n^k.
\]

The factored form of~\Eq{1} is
\[
f(x) \equiv (x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_n).
\Tag{4}
\]
%% -----File: 141.png---Folio 135-------
The derivative $f'(x)$ of this product is found by multiplying the derivative
(unity) of each factor by the product of the remaining factors and adding
the results. Hence
\index{Derivative}%
\begin{gather*}
f'(x) = (x - \alpha_2) \dotsm (x - \alpha_n)
      + (x - \alpha_1)(x - \alpha_3) \dotsm (x - \alpha_n) + \dotsb, \\
f'(x) \equiv \frac{f(x)}{x - \alpha_1} + \frac{f(x)}{x - \alpha_2} + \dotsb
           + \frac{f(x)}{x - \alpha_n}.
\Tag{5}
\end{gather*}

If $\alpha$ is any root of~\Eq{1}, $f(\alpha)= 0$ and
\begin{align*}
\frac{f(x)}{x - \alpha}
  &= \frac{f(x) - f(\alpha)}{x - \alpha}
   = \frac{x^n - \alpha^n}{x - \alpha}
   + c_1\frac{x^{n-1} - \alpha^{n-1}}{x - \alpha} + \dotsb
   + c_{n-1}\frac{x - \alpha}{x - \alpha} \\
%
  &= x^{n-1} + \alpha x^{n-2} + \alpha^2 x^{n-3} + \dotsb
   + c_1(x^{n-2} + \alpha x^{n-3} + \dotsb) \\
  &\hphantom{{}=x^{n-1}} + c_2(x^{n-3} + \dotsb) + \dotsb,
\end{align*}
\[
\begin{split}
\frac{f(x)}{x - \alpha}
  &= x^{n-1} + (\alpha + c_1)x^{n-2}
             + (\alpha^2 + c_1 \alpha + c_2)x^{n-3} + \dotsb \\
  &+ (\alpha^k + c_1 \alpha^{k-1} + c_2 \alpha^{k-2} + \dotsb
             + c_{k-1} \alpha + c_k)x^{n-k-1} + \dotsb.
\end{split}
\Tag{6}
\]
Taking $\alpha$ to be $\alpha_1, \dotsc, \alpha_n$ in turn, adding the results, and applying~\Eq{5}, we
obtain
\begin{align*}
f'(x) = nx^{n-1}
  &+ (s_1 + nc_1)x^{n-2} + (s_2 + c_1 s_1 + nc_2)x^{n-3} + \dotsb \\
  &+ (s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb
   + c_{k-1} s_1 + nc_k)x^{n-k-1} + \dotsb.
\end{align*}

The derivative of~\Eq{1} is found at once by the rules of calculus (or by~§56)
to be
\[
f'(x) = nx^{n-1} + (n-1)c_1 x^{n-2} + (n-2)c_2 x^{n-3} + \dotsb
      + (n-k)c_k x^{n-k-1} + \dotsb.
\]
Since this expression is identical term by term with the preceding, we have
\[
\begin{aligned}
&s_1 + c_1 = 0,\qquad s_2 + c_1 s_1 + 2c_2 = 0, \dotsc, \\
&s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_{k-1} s_1 + kc_k = 0\quad
  (k \leqq n - 1).
\end{aligned}
\Tag{7}
\]

We may therefore find in turn $s_1$, $s_2, \dotsc, s_{n-1}$:
\[
s_1 = -c_1,\qquad
s_2 = c_1^2 - 2c_2,\qquad
s_3 = -c_1^3 + 3c_1 c_2 - 3c_3, \dotsc.
\Tag{8}
\]

To find $s_n$, replace $x$ in~\Eq{1} by $\alpha_1, \dotsc, \alpha_n$ in turn and add the resulting
equations. We get
\[
s_n + c_1 s_{n-1} + c_2 s_{n-2} + \dotsb + c_{n-1} s_1 + nc_n = 0.
\Tag{9}
\]

%% -----File: 142.png---Folio 136-------

We may combine \Eq{7} and~\Eq{9} into
\[
s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_{k-1} s_1 + kc_k = 0\qquad
  (k=1, 2, \dotsc, n).
\Tag{10}
\]

This set of formulas~\Eq{10} will be referred to as \emph{Newton's identities}.
\index{Newton's!identities}%
The student should be able to write them down from memory and, when
writing them, should always check the final one~\Eq{9} by deriving it as above.

To derive a formula which shall enable us to compute the $s_k$ for
$k>n$, we multiply~\Eq{1} by~$x^{k-n}$, take $x = \alpha_1, \dotsc, x = \alpha_n$ in turn, and add
the resulting equations. We get
\[
s_k + c_1 s_{k-1} + c_2 s_{k-2} + \dotsb + c_n s_{k-n} = 0\qquad (k>n).
\Tag{11}
\]

Instead of memorizing this formula, it is preferable to deduce it for the
particular equation for which it is needed, thus avoiding errors of substitution
as well as confusion with~\Eq{10}.


\begin{Example}
Find $s_k$ for $x^n - 1 = 0$.
\end{Example}
\index{Roots of unity}%

\begin{Solution}
Comparing our equation with~\Eq{1}, we have $c_1 = 0, \dotsc, c_{n-1} = 0$, $c_n = -1$.
Hence in~\Eq{10} for $k<n$, each $c$ is zero and $s_k = 0$. But, for $k = n$, \Eq{10} becomes $s_n - n = 0$.
We may check the latter by substituting each root $\alpha_1, \dotsc, \alpha_n$ in our given equation
and adding. Finally, to find $s_l$ when $l>n$, multiply our equation by~$x^{l-n}$. In the
resulting equation $x^l - x^{l-n} = 0$ we substitute each root, add, and obtain $s_l - s_{l-n} = 0$.
Hence from $s_l$ we obtain an equal $s$ by subtracting $n$ from~$l$. After repeated subtractions,
we reach a value~$k$ for which $1 \leqq k \leqq n$. Since $s_k = 0$ or $n$ according as $k<n$ or
$k = n$, it follows that $s_l = 0$ or $n$ according as $l$ is not or is divisible by~$n$.
\end{Solution}


\begin{Exercises}{Page136}

\begin{Problems}
\item[1.] For a cubic equation, $s_4 = c_1^4 - 4c_1^2 c_2 + 4c_1 c_3 + 2c_2^2$.

\item[2.] For an equation of degree $\geqq 4$, $s_4 = c_1^4 - 4c_1^2 c_2 + 4c_1 c_3 + 2c_2^2- 4c_4$.

\item[3.] Find $s_2$, $s_3$, $s_4$, $s_5$ for $x^2 - px + q = 0$.

\item[4.] Find $s_k$ for $x^5 - 3 = 0$.

\item[5.] Find $s_2$, $s_3$, $s_6$, $s_7$ for $x^5 - px + q = 0$.
\end{Problems}
\end{Exercises}


\Section[Waring's Formula]
{107.}{Waring's Formula for $s_k$ in Terms of the Coefficients.} While
\index{Waring's formula}%
we have learned how to find $s_1$, $s_2$, $s_3, \dotsc$ in turn by Newton's identities,
it is occasionally useful to know an explicit expression for~$s_k$, where $k$
has an arbitrary value. The formula in question is applied ordinarily
only to a quadratic equation
\[
x^2 + px + q = 0.
\]
Accordingly we shall treat this case in detail. If its roots are $\alpha$ and~$\beta$,
then
\[
x^2 + px + q \equiv (x - \alpha)(x - \beta).
\]
%% -----File: 143.png---Folio 137-------
Replace $x$ by~$1/y$ and multiply by~$y^2$. We get
\[
1 + py + qy^2 \equiv (1 - \alpha y)(1 - \beta y).
\Tag{12}
\]
Taking derivatives, we have
\[
p + 2qy \equiv - \alpha (1 - \beta y) - \beta (1 - \alpha y).
\]
Change of signs and division by the members of~\Eq{12} gives
\[
\frac{-p - 2qy}{1 + py + qy^2}
  \equiv \frac{\alpha}{1 - \alpha y} + \frac{\beta}{1 - \beta y}.
\Tag{13}
\]

The identity in Ex.~7, §14, with $n$ changed to~$k$, may be written in
the form
\[
\frac{1}{1 - r} \equiv 1 + r + r^2 + \dotsb + r^{k-1} + \frac{r^k}{1-r}.
\Tag{14}
\]
Take $r = \alpha y$ and multiply the resulting terms by~$\alpha$; thus
\begin{align*}
\frac{\alpha}{1 - \alpha y}
  &= \alpha + \alpha^2 y + \dotsb + \alpha^k y^{k-1}
            + \frac{\alpha^{k+1} y^k}{1 - \alpha y}. \\
\intertext{Similarly,}
\frac{\beta}{1 - \beta y}
  &=  \beta + \beta^2 y + \dotsb + \beta^k y^{k-1}
            + \frac{\beta^{k+1} y^k}{1 - \beta y}.
\end{align*}
To show that on adding, and writing $s_k$ for $\alpha^k + \beta^k$, we obtain~\Eq{15}, we need
the sum of the final fractions, which by~\Eq{12} is
\[
\frac{\phi y^k}{(1 - \alpha y)(1 - \beta y)}
  = \frac{\phi y^k}{1 + py + qy^2},\qquad
\phi \equiv \alpha^{k+1}(1 - \beta y) + \beta^{k+1}(1 - \alpha y).
\]
Hence
\[
\frac{\alpha}{1 - \alpha y} + \frac{\beta}{1 - \beta y}
  = s_1 + s_2 y + \dotsb + s_k y^{k-1} + \frac{\phi y^k}{1 + py + qy^2},
\Tag{15}
\]
where the exact expression for $\phi$ is immaterial.

Next, we seek an expansion of the fraction in the left member of~\Eq{13}.
Its denominator will be identical with that in~\Eq{14} if we choose
$r = -py - qy^2$.    Evidently~\Eq{14} may be written in the compact form
\[
\frac{1}{1 - r} \equiv \sum_{t=0}^{k-1} r^t + \frac{r^k}{1 - r}.
\]
Hence it becomes
\[
\frac{1}{1 + py + qy^2}
  = \sum_{t=0}^{k-1} (-1)^t(py + qy^2)^t + \frac{\psi y^k}{1 + py + qy^2},
\]
%% -----File: 144.png---Folio 138-------
where $\psi = (-p - qy)^k$, although no use will be made of the particular
form of the polynomial~$\psi$. By the binomial theorem,
\[
(py + qy^2)^t = \sum \frac{(g+h)!}{g!h!} (py)^g(qy^2)^h,
\]
where the summation extends over all sets of integers $g$ and~$h$, each $\geqq 0$,
for which $g + h = t$, while $g!$~denotes the product of $1$, $2, \dotsc, g$ if $g \geqq 1$,
but denotes unity if $g = 0$. Hence
\begin{gather*}
\frac{-p - 2qy}{1 + py + qy^2}
  = (p + 2qy) \sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^hy^{g+2h} + E,
\Tag{16} \\
E \equiv \frac{(-p - 2qy) \psi y^k}{1 + py + qy^2},
\end{gather*}
where the summation extends over all sets of integers $g$ and~$h$, each $\geqq 0$,
for which $g+h \leqq k-1$.

Since the left members of \Eq{15} and~\Eq{16} are identically equal by~\Eq{13},
their right members must be identical, so that the coefficients of $y^{k-1}$
in them must be equal.\footnote
  {In fact, the $(k-1)$th derivatives of the two right members are identical, and we
  obtain the indicated result by substituting $y = 0$ in these two derivatives and equating
  the results. Note that the final terms in both \Eq{15} and~\Eq{16} have~$y$ as a factor of their
  $(k-1)$th derivatives.} % end footnote
Hence the coefficient~$s_k$ of~$y^{k-1}$ in~\Eq{15} is equal
to the coefficient of~$y^{k-1}$ in~\Eq{16}, which is made up of two parts, corresponding
to the two terms of the factor $p + 2qy$. When we use the constant
term~$p$, we must employ from~$\sum$ in~\Eq{16} the terms in which the exponent
of~$y$ is equal to~$k-1$. But when we use the other term~$2qy$, we must
employ from~$\sum$ the terms in which the exponent of~$y$ is equal to~$k-2$, in
order to obtain the combined exponent~$k-1$ of~$y$. Hence $s_k$ is equal to
the sum of the following two parts:
\begin{align*}
 p\sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^h && (g+2h &= k-1),\\
2q\sum (-1)^{g+h+1} \frac{(g+h)!}{g!h!} p^gq^h && (g+2h &= k-2).
\end{align*}
In the upper sum, write $i$ for~$g+1$, and $j$ for~$h$. In the lower sum, write
$i$ for~$g$, and $j$ for~$h+1$. Hence
\[
s_k =  \sum (-1)^{i+j} \frac{(i+j-1)!}{(i-1)!j!} p^iq^j
    + 2\sum (-1)^{i+j} \frac{(i+j-1)!}{i!(j-1)!} p^iq^j,
\]
%% -----File: 145.png---Folio 139-------
where now each summation extends over all sets of integers $i$ and~$j$, each
$\geqq 0$, for which
\[
i + 2j = k.
\Tag{17}
\]

Finally, we may combine our two sums. Multiply the numerator
and denominator of the first fraction by~$i$, and those of the second fraction
by~$j$. Thus
\[
s_k = k\sum (-1)^{i+j} \frac{(i+j-1)!}{i!j!} p^iq^j,
\Tag{18}
\]
since the present fraction occurred first multiplied by~$i$ and second multiplied
by~$2j$, and, by~\Eq{17}, the sum of these multipliers is equal to~$k$. Our
final result is~\Eq{18}, where the summation extends over all sets of integers
$i$ and~$j$, each $\geqq 0$, satisfying~\Eq{17}.

\begin{Remark}
If we replace $i$ by its value $k-2j$, and change the sign of~$p$, we obtain from~\Eq{18}
the result that \emph{the sum of the $k$th powers of the roots of $x^2 - px + q = 0$ is equal to}
\index{Quadratic equation!sum of powers of roots}%
\begin{align*}
s_k &= k\sum_{j=0}^K (-1)^j \frac{(k-j-1)!}{(k-2j)!j!} p^{k-2j} q^j
\Tag{19} \\
    &= p^k - kp^{k-2} q + \frac{k(k-3)}{1·2} p^{k-4} q^2
           - \frac{k(k-4)(k-5)}{1·2·3} p^{k-6} q^3 + \dotsb, % [** PP: . -> ,]
\end{align*}
where $K$ is the largest integer not exceeding~$k/2$.

The product of the roots is equal to~$q$. Hence if $x$ denotes one root, the second
root is~$q/x$. Thus $s_k = x^k + (q/x)^k$. Again, the sum of the roots is $x + q/x = p$. Regard
$q$ as given and $p$ as unknown. Hence, if $c$ is an arbitrary constant, the equation
\[
p^k - kqp^{k-2} + \frac{k(k-3)}{1·2} q^2p^{k-4} - \dotsb = c
\Tag{20}
\]
is transformed by the substitution $p = x + q/x$ into
\[
x^k + \left(\frac{q}{x}\right)^k = c.
\]
Hence equation~\Eq{20} may be solved for $p$ by radicals by the method employed in~§43
for a cubic equation.
\end{Remark}

%% -----File: 146.png---Folio 140-------

The above proof applies\footnote
  {See the author's \textit{Elementary Theory of Equations}, pp.~72--74, where there is given
  also a shorter proof by means of infinite series.}
without essential change to any equation
$x^n + c_1x^{n-1} + \dotsb + c_n = 0$ and leads to the following formula for the sum
of the $k$th~powers of its roots:
\[
s_k = k\sum (-1)^{r_1 + \dotsb + r_n}
      \frac{(r_1 + \dotsb + r_n-1)!}{r_1! \dotsm r_n!}
      c_1^{r_1} \dotsm c_n^{r_n},
\Tag{21}
\]
where the summation extends over all sets of integers $r_1, \dotsc, r_n$, each
$\geqq 0$, for which $r_1 + 2r_2 + 3r_3 + \dotsb + nr_n = k$. This result~\Eq{21} is known
as \emph{Waring's formula} and was published by him in 1762.

\begin{Example}
Let $n=3$, $k=4$. Then $r_1 + 2r_2 + 3r_3 = 4$ and
\begin{align*}
(r_1, r_2, r_3)
  &= (4, 0, 0),\qquad
     (2, 1, 0),\qquad
     (1, 0, 1),\qquad
     (0, 2, 0), \\
s_4 &= 4(\frac{3!}{4!} c_1^4 - \frac{2!}{2!1!} c_1^2c_2
       + \frac{1!}{1!1!} c_1c_3 + \frac{1!}{2!} c_2^2) \\
    &= c_1^4 - 4c_1^2c_2 + 4c_1c_3 + 2c_2^2.
\end{align*}
\end{Example}


\begin{Exercises}{Page140}

\begin{Problems}
\item[1.] For the quadratic $x^2 - px + q = 0$ write out the expressions for $s_2$, $s_3$, $s_4$, $s_5$ given by~\Eq{19},
and compare with those obtained from Newton's identities (Ex.~3, §106).

\item[2.] Find $s_4$ for a quartic equation by Waring's formula.

\item[3.] For $k=5$, \Eq{20} becomes De Moivre's quintic $p^5 - 5qp^3 + 5q^2p = c$. Solve it by
radicals for~$p$.
\index{De Moivre's!quintic}% [** PP: Using subitem]

\item[4.] Solve \Eq{20} by radicals when $k=7$.
\end{Problems}
\end{Exercises}


% [** PP: No ToC entry, using running head]
\Section[Computation of Sigma Functions]
{108.}{$\Sigma$-functions Expressed in Terms of the Functions~$s_k$.} Since
we have learned two methods of expressing the $s_k$ in terms of the coefficients,
it is desirable to learn how to express any $\Sigma$-polynomial (and
hence any symmetric function) in terms of the~$s_k$.

By performing the indicated multiplication, we find that
\[
s_a s_b \equiv \Sigma \alpha_1^a · \Sigma \alpha_1^b
  = \Sigma \alpha_1^{a+b} + m\Sigma \alpha_1^a \alpha_2^b,
\]
where $m = 1$ if $a\neq b$, $m=2$ if $a=b$. Transposing the first term, which
is equal to $s_{a+b}$, and dividing by~$m$, we obtain
\[
\Sigma \alpha_1^a \alpha_2^b = \frac{1}{m} (s_a s_b - s_{a+b}).
\Tag{22}
\]

%% -----File: 147.png---Folio 141-------

In order to compute $\Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2$ in terms of the~$s_k$, we form the product
\[
\Sigma \alpha_1^4 · \Sigma \alpha_1^3 \alpha_2^2
  = \Sigma \alpha_1^7 \alpha_2^2 + \Sigma \alpha_1^6 \alpha_2^3
  + \Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2.
\]
Making three applications of~\Eq{22}, we get
\[
s_4(s_3s_2 - s_5)
  = (s_7s_2 - s_9) + (s_6s_3 - s_9) + \Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2.
\]
Hence
\[
\Sigma \alpha_1^4 \alpha_2^3 \alpha_3^2
  = s_2s_3s_4 - s_2s_7 - s_3s_6 - s_4s_5 + 2s_9.
\]


\begin{Exercises}{Page141}

For a quartic equation, express in terms of the $s_k$ and ultimately in terms of the
coefficients $c_1, \dotsc, c_4$:

\begin{Problems}[4]

\item[1.] $\Sigma \alpha_1^2 \alpha_2^2$.

\item[2.] $\Sigma \alpha_1^3 \alpha_2$.

\item[3.] $\Sigma \alpha_1^2 \alpha_2 \alpha_3$.

\item[4.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3^2$.

\ResetCols{1}

\item[5.] If $a\geqq b > c > 0$, prove that
\[
\Sigma \alpha_1^a \alpha_2^b \alpha_3^c
  = \frac{1}{m} (s_a s_b s_c - s_a s_{b+c}
               - s_b s_{a+c} - s_c s_{a+b} + 2s_{a+b+c}),
\]
where $m = 1$ if $a > b$, $m = 2$ if $a = b$.

\item[6.] $\Sigma \alpha_1^a \alpha_2^b \alpha_3^b
  = \frac{1}{2}(s_a s_b^2 - s_as_{2b} - 2s_b s_{a+b} + 2s_{a+2b})$,\qquad $a > b > 0$.

\item[7.] $\Sigma \alpha_1^a \alpha_2^a \alpha_3^a
  = \frac{1}{6}(s_a^3 - 3s_a s_{2a} + 2s_{3a})$,\qquad $a > 0$.
\end{Problems}
\end{Exercises}


\Section{109.}{Computation of Symmetric Functions} The method last explained
is practicable when a term of the $\Sigma$-function involves only a few distinct
roots, the largeness of the exponents not introducing a difficulty in the
initial work of expressing the $\Sigma$-function in terms of the~$s_k$.

But when a term of the $\Sigma$-function involves a large number of roots
with small exponents, we resort to a method suggested by~§104, which
tells us which auxiliary simpler symmetric functions %[** PP: Typo fuctions]
should be multiplied
together to produce our $\Sigma$-function along with simpler ones.

\begin{Remark}
For example, to find $\Sigma x_1^2x_2x_3x_4$, when $n > 4$, we employ
\begin{gather*}
E_1E_4 \equiv \Sigma x_1 · \Sigma x_1x_2x_3x_4
  = \Sigma x_1^2x_2x_3x_4 + 5 \Sigma x_1x_2x_3x_4x_5, \\
\Sigma x_1^2x_2x_3x_4 = E_1E_4 - 5E_5.
\end{gather*}

To find $\Sigma x_1^2x_2^2x_3^2x_4$, employ $E_3E_4 = \Sigma x_1x_2x_3 · \Sigma x_1x_2x_3x_4$.

When many such products of $\Sigma$-functions are to be computed, it will save time
in the long run to learn and apply the ``method of leaders'' explained in the author's
\textit{Elementary Theory of Equations}, pp.~64--65.
\end{Remark}

%% -----File: 148.png---Folio 142-------


\begin{Exercises}[MISCELLANEOUS~]{Page142}

Express in terms of the coefficients $c_1, \dotsc, c_n$:
\begin{Problems}[4]

\item[1.] $\Sigma \alpha_1^2 \alpha_2 \alpha_3$.

\item[2.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3$.

% [** PP: Force into four columns]
\item[3.] \makebox[0pt][l]{$\Sigma \alpha_1^2 \alpha_2^2 \alpha_3 \alpha_4$.}

\item[\quad4.] $\Sigma \alpha_1^2 \alpha_2^2 \alpha_3^2$.
\end{Problems}

If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3 + px^2 + qx + r = 0$,
find a cubic equation with the roots
\begin{Problems}[3]
\item[5.] $\alpha^2$, $\beta^2$, $\gamma^2$.

\item[6.] $\alpha\beta$, $\alpha\gamma$, $\beta\gamma$.

\item[7.] $\dfrac{2}{\alpha}$, $\dfrac{2}{\beta}$, $\dfrac{2}{\gamma}$.

\ResetCols{2}

\item[8.] $\alpha^2 + \beta^2$,  $\alpha^2 + \gamma^2$, $\beta^2 + \gamma^2$.

\item[9.] $\alpha^2 + \alpha\beta + \beta^2$, etc.
\end{Problems}

If $\alpha$, $\beta$, $\gamma$, $\delta$ are the roots of $x^4 + px^3 + qx^2 + rx + s = 0$, find
\begin{Problems}
\item[10.] $\Sigma \dfrac{\beta}{\alpha}
  = \Sigma \dfrac{\beta + \gamma + \delta}{\alpha}
  = \Sigma \dfrac{-p - \alpha}{\alpha}
  = -4 - p \Sigma \frac{1}{\alpha}$.

\item[11.] $\Sigma \dfrac{\beta}{\alpha^2}$. Use
  $\Sigma \dfrac{1}{\alpha}·\Sigma \dfrac{\beta}{\alpha}
  = \Sigma \dfrac{\beta}{\alpha^2}
 + 3\Sigma \dfrac{1}{\alpha}
 + 2\Sigma \dfrac{\gamma}{\alpha\beta}$.

\item[12.] Express $\Sigma \alpha_1^a \alpha_2^b \alpha_3^c \alpha_4^d$ in terms of the $s_k$ when (\emph{i})~$a>b>c>d>0$, and (\emph{ii})~when
$a=b=c=d$.

\item[13.] By solving the first $k$ of Newton's identities~\Eq{10} as a system of linear equations,
find an expression in the form of a determinant (\emph{i})~for $s_k$ in terms of
$c_1, \dotsc, c_k$, and
(\emph{ii})~for $c_k$ in terms of $s_1, \dotsc, s_k$.

\item[14.]  One set of $n$~numbers is a mere rearrangement of another set if $s_1, \dotsc, s_n$
have the same values for each set.
\end{Problems}
\end{Exercises}
\index{Symmetric functions|)}%
\index{Sigma function|)}%
\index{Sum of!like powers of roots|)}%

%% -----File: 149.png---Folio 143-------


% [** PP: Not matching running head]
\Chapter{X}{Elimination, Resultants And Discriminants}
\index{Elimination|(}%
\index{Resultant|(}%

% [** PP: Next four units have combined ToC entry]
\Section{110.}{Elimination} If the two equations
\begin{flalign*}
&& ax+b &= 0,\qquad cx+d = 0 && \Rightmark{(a \ne 0,\ c \ne 0)}
\end{flalign*}
are simultaneous, i.e., if $x$ has the same value in each, then % [** PP: Not italicizing i.e.]
\[
x = -\frac{b}{a} = -\frac{d}{c}, \qquad
R \equiv ad - bc = 0,
\]
and conversely. Hence a necessary and sufficient condition that the
equations have a common root is $R = 0$. We call $R$ the \emph{resultant} (or
\emph{eliminant}) of the two equations.

The result of eliminating~$x$ between the two equations might equally
well have been written in the form $bc - ad = 0$. But the arbitrary selection
of~$R$ as the resultant, rather than the product of~$R$ by some constant,
as~$-1$, is a matter of more importance than is apparent at first sight. For,
we seek a \emph{definite} function of the coefficients $a$, $b$, $c$, $d$ of the \emph{functions}
$ax+b$, $cx+d$, and not merely a property $R = 0$ or $R\ne 0$ of the corresponding
\emph{equations}. Accordingly, we shall lay down the definition in~§111,
which, as the reader may verify, leads to~$R$ in our present example.

Methods of elimination which seem plausible often yield not $R$ itself,
but the product of~$R$ by an extraneous function of the coefficients. This
point (illustrated in~§114) indicates that the subject demands a more
careful treatment than is often given.


% [** No separate ToC entry]
\Section[Resultant of Two Polynomials]
{111.}{Resultant of Two Polynomials in~$x$.}   Let
\begin{flalign*}
&&
\left\{
\begin{aligned}
f(x) &= a_0x^m + a_1x^{m-1} + \dotsb + a_m \\
g(x) &= \;b_0x^n + \;b_1x^{n-1} + \dotsb + \, b_n
\end{aligned}
\right.
&&
\begin{aligned}
&\Rightmark{(a_0 \ne 0),} \\
&\Rightmark{(b_0 \ne 0)}
\end{aligned}
\Tag{1}
\end{flalign*}
be two polynomials of degrees $m$ and~$n$. Let $\alpha_1, \dotsc, \alpha_m$ be the roots
of $f(x) = 0$. Since $\alpha_1$ is a root of $g(x)=0$ only when $g(\alpha_1)=0$, the two
equations have a root in common if and only if the product
\[
g(\alpha_1)g(\alpha_2) \dotsm g(\alpha_m)
\]
%% -----File: 150.png---Folio 144-------
is zero. This symmetric function of the roots of $f(x)= 0$ is of degree~$n$
in any one root and hence is expressible as a polynomial of degree~$n$ in the
elementary symmetric functions~(§104), which are equal to  $-a_1/a_0$,
$a_2/a_0, \dotsc$. To be rid of the denominators~$a_0$, it therefore suffices to
multiply our polynomial by~$a_0^n$. We therefore define
\[
R(f, g) = a_0^n g(\alpha_1)g(\alpha_2) \dotsm g(\alpha_m)
\Tag{2}
\]
to be the \emph{resultant} of $f$ and~$g$. It equals an integral rational function of
$a_0, \dotsc, a_m$, $b_0, \dotsc, b_n$ with integral coefficients.
\index{Symbol!h@{$R(f, g)$\IndAdd{resultant}}}% [** PP: Manually alphabetized]


\begin{Exercises}{}

\begin{Problems}
\item[1.] If $m = 1$, $n = 2$,\quad
  $R(f, g) = b_0 a_1^2 - b_1 a_0 a_1 + b_2 a_0^2$.

\item[2.] If $m = 2$, $n = 1$,\quad
  $R(f,g) = a_0(b_0\alpha_1 + b_1)(b_0\alpha_2 + b_1)
          = a_0 b_1^2 - a_1 b_0 b_1 + a_2 b_0^2$, since
\[
 a_0(\alpha_1 + \alpha_2) = -a_1,\qquad a_0 \alpha_1 \alpha_2 = a_2.
\]

\item[3.] If $\beta_1, \dotsc, \beta_n$ are the roots of $g(x)= 0$, so that
\[
g(\alpha_i)
  = b_0(\alpha_i - \beta_1)(\alpha_i - \beta_2) \dotsm (\alpha_i - \beta_n),
\]
then
\[
\begin{array}{r@{}l}
R(f,g) = a_0^n b_0^m  %[** PP: Added ')' after \beta_2 below]
  & (\alpha_1 - \beta_1)(\alpha_1 - \beta_2) \dotsm (\alpha_1 - \beta_n) \\
 ·& (\alpha_2 - \beta_1)(\alpha_2 - \beta_2) \dotsm (\alpha_2 - \beta_n) \\
  & \Dots{1} \\
 ·& (\alpha_m - \beta_1)(\alpha_m - \beta_2) \dotsm (\alpha_m - \beta_n).
\end{array}
\]
Multiplying together the differences in each column, we see that
\[
R(f, g) = (-1)^{mn} b_0^m f(\beta_1)f(\beta_2) \dotsm f(\beta_n)
  = (-1)^{mn} R(g, f).
\]

\item[4.] If $m=2$, $n=1$,\quad
  $R(g,f) = b_0^2 f( -b_1/b_0) = a_0 b_1^2 - a_1 b_0 b_1 + a_2 b_0^2$,
which is equal to $R(f,g)$
by Ex.~2. This illustrates the final result in Ex.~3.

\item[5.] If $m=n=2$, %[*Equation split to new line]
\begin{align*}
R(f,g) &= a_0^2 b_0^2 \alpha_1^2 \alpha_2^2
        + a_0^2 b_0 b_1 \alpha_1 \alpha_2 (\alpha_1 + \alpha_2) \\
%
       &\quad + a_0^2 b_0 b_2(\alpha_1^2 + \alpha_2^2)
              + a_0^2 b_1^2 \alpha_1 \alpha_2
              + a_0^2 b_1 b_2(\alpha_1 + \alpha_2) + a_0^2 b_2^2 \\
%
       &= b_0^2 a_2^2 - b_0 b_1 a_1 a_2 + b_0 b_2(a_1^2 - 2a_0 a_2)
        + b_1^2 a_0 a_2 - b_1 b_2 a_0 a_1 + a_0^2 b_2^2.
\end{align*}
This equals $R(g,f)$, since it is unaltered when the $a$'s and~$b$'s are interchanged.

\item[6.] Prove by~\Eq{2} that $R$ is homogeneous and of total degree~$m$ in $b_0, \dotsc, b_n$; and
by Ex.~3, that $R$ is homogeneous and of total degree~$n$ in $a_0, \dotsc, a_m$. Show that $R$
has the terms $a_0^n b_n^m$ and $(-1)^{mn} b_0^m a_m^n$.

\item[7.] $R(f, g_1 g_2) = R(f, g_1) · R(f, g_2)$.

\item[8.] $R(f, x^n) = (-1)^{mn} R(x^n, f) = (-1)^{mn} a_m^n$.
\end{Problems}
\end{Exercises}

%% -----File: 151.png---Folio 145-------


% [** PP: No separate ToC entry]
\Section[Sylvester's Method of Elimination]
{112.}{Sylvester's Dialytic Method of Elimination.\protect\footnotemark}%
\addtocounter{footnote}{1}%
  \footnotetext{Given without proof by Sylvester, \textit{Philosophical Magazine}, 1840, p.~132.}
\addtocounter{footnote}{-1}%
Let the equations
\index{Sylvester's eliminant}%
\[
f(x) \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad
g(x) \equiv b_0x^2 + b_1x + b_2 = 0
\]
have a common root~$x$. Multiply the first equation by $x$ and the second
by $x^2$ and $x$ in turn. We now have five equations
\begin{alignat*}{2}
a_0x^4 + a_1x^3 &+ a_2x^2 + a_3x       &&= 0, \\
         a_0x^3 &+ a_1x^2 + a_2x + a_3 &&= 0, \\
b_0x^4 + b_1x^3 &+ b_2x^2              &&= 0, \\
         b_0x^3 &+ b_1x^2 + b_2x       &&= 0, \\
                &\phantom{{}+{}} b_0x^2 + b_1x + b_2 &&= 0,
\end{alignat*}
which are linear and homogeneous in $x^4$, $x^3$, $x^2$, $x$,~$1$. Hence~(§97)
\[
F = \begin{vmatrix}
  a_0 & a_1 & a_2 & a_3 &   0 \\
    0 & a_0 & a_1 & a_2 & a_3 \\
  b_0 & b_1 & b_2 &   0 &   0 \\
    0 & b_0 & b_1 & b_2 &   0 \\
    0 &   0 & b_0 & b_1 & b_2
    \end{vmatrix}
\Tag{3}
\]
must be zero. Next, if $F=0$, there exist~(§97) values which, when
substituted for $x^4$, $x^3$, $x^2$, $x$ and~$1$, satisfy the five equations. But why is
the value for $x^4$ the fourth power of the value for~$x$, that for $x^3$ the cube of
the value for~$x$, etc.? Since the direct verification of these facts would
be very laborious, we resort to a device to show that, conversely, if $F=0$
the two given equations have a root in common.

In~\Eq{3} replace $a_3$ by $a_3 - z$ and consider the equation
\[
\begin{vmatrix}
  a_0 & a_1 & a_2 & a_3-z & 0     \\
    0 & a_0 & a_1 & a_2   & a_3-z \\
  b_0 & b_1 & b_2 &   0   & 0     \\
    0 & b_0 & b_1 & b_2   & 0     \\
    0 &   0 & b_0 & b_1   & b_2
\end{vmatrix}
= 0.
\Tag{4}
\]
To prove that it has the roots $f(\beta_1)$ and $f(\beta_2)$, where $\beta_1$ and~$\beta_2$ are the roots
of $g(x) = 0$, we take $z = f(\beta_i)$ and prove that the determinant is then equal
to zero. For, if we add to the last column the products of the elements
%% -----File: 152.png---Folio 146-------
of the first four columns by $\beta_i^4$, $\beta_i^3$, $\beta_i^2$, $\beta_i$, respectively, we find that all
of the elements of the new last column are zero.

Since \Eq{4} reduces to~\Eq{3} for $z=0$, it is of the form
\[
b_0^3z^2 + kz + F = 0,
\]
in which the value of $k$ is immaterial. By considering the product of
the roots of this quadratic equation, we see that
\[
F = b_0^3 f(\beta_1) f(\beta_2).
\]
Hence the Sylvester determinant~$F$ is the resultant $R(g, f)$ and hence
is the resultant $R(f, g)$, since $mn$ is here even (Ex.~3,~§111).
\index{Determinants|(}%

In general, if the equations are
\[
f(x) \equiv a_0x^m + \dotsb + a_m = 0,\qquad
g(x) \equiv b_0x^n + \dotsb + b_n = 0,
\]
we multiply the first equation by $x^{n-1}$, $x^{n-2}, \dotsc, x$, $1$, in turn, and the
second by $x^{m-1}$, $x^{m-2}, \dotsc, x$, $1$, in turn. We obtain $n+m$ equations
which are linear and homogeneous in the $m+n$ quantities $x^{m+n-1}, \dotsc,
x$, $1$. Hence the determinant
\[
F=
\left|
\begin{array}{ccccccccccc}
  a_0 & a_1 & a_2 & \Dots{2} & a_m & 0 & \Dots{3} & 0 \\
  0 & a_0 & a_1 & a_2 & \Dots{2} & a_m & 0 & \Dots{2} & 0 \\
  0 &   0 & a_0 & a_1 & a_2 & \Dots{2} & a_m  & 0 & \dots & 0 \\
  \Dots{11} \\
  0 & \Dots{2} & 0 & a_0 & a_1 & a_2 & \Dots{3} & a_m \\
  b_0 & b_1 & \Dots{3} & b_n & 0 & \Dots{3} & 0 \\
  0 & b_0 & b_1 & \Dots{4} & b_n & \Dots{2} & 0 \\
  \Dots{11} \\
  0 & \dots & 0 & b_0 & b_1 & \Dots{5} & b_n
\end{array}
\right|
% Now set braces, using empty arrays to get correct vertical alignment
{\setlength{\arraycolsep}{0pt}
\begin{array}{c}
\left.
\begin{array}{c}
\\
\\
\\
\\
\\
\end{array}
\right\} \text{$n$~rows} \\
\left.
\begin{array}{c}
\\
\\
\\
\\
\end{array}
\right\} \text{$m$~rows}
\end{array}}
\Tag{5}
\]
is zero. It may be shown to be equal to the resultant $R(f, g)$, whether
$mn$ is even or odd, by the method employed in the above case $m = 3$, $n = 2$.

We may also prove as follows that if $F = 0$ the equations $f = 0$ and
$g = 0$ have a common root. Since $F$ was obtained as the determinant
of the coefficients of
\[
x^{n-1}f, \dotsc, xf, f,\qquad
x^{m-1}g, \dotsc, xg, g,
\]
$F = 0$ implies, by~§96, Lemma~2, the existence of a linear relation
\[
B_0x^{n-1}f + \dotsb + B_{n-2}xf + B_{n-1}f +
A_0x^{m-1}g + \dotsb + A_{m-2}xg + A_{m-1}g \equiv 0,
\]
%% -----File: 153.png---Folio 147-------
identically in~$x$, with constant coefficients $B_0, \dotsc, A_{m-1}$ not all zero.
In other words, $\beta f + \alpha g \equiv 0$, where
\[
\alpha\equiv A_0x^{m-1} + \dotsb + A_{m-2}x + A_{m-1},\quad
\beta \equiv B_0x^{n-1} + \dotsb + B_{n-2}x + B_{n-1}.
\Tag{6}
\]

Neither $\alpha$ nor~$\beta$ is identically zero. For, if $\alpha\equiv 0$, for example, then
$\beta f\equiv 0$ and $\beta\equiv 0$, whereas the $A_i$ and~$B_i$ are not all zero.

Consider the factored forms of $f$, $g$, $\alpha$, $\beta$. Suppose that $f$ and~$g$ have
no common linear factor. The highest power of each linear factor occurring
in $f$ divides $\alpha g \equiv -\beta f$ and hence divides~$\alpha$. Thus $f$ divides~$\alpha$, whereas
$f$ is of higher degree than~$\alpha$. Hence our assumption that $f = 0$ and $g = 0$
have no common root has led to a contradiction.

\begin{Remark}
A similar idea is involved in the method of elimination due to Euler (1707--1783).
If $f=0$ and $g=0$ have a common root~$c$, then $f\equiv (x-c)\alpha$, $-g\equiv (x-c)\beta$, identically in~$x$,
where $\alpha$ and~$\beta$ are polynomials in~$x$ of degrees $m-1$ and~$n-1$, respectively. Give
them the notations~\Eq{6}. In the identity $\beta f + \alpha g \equiv 0$, the coefficient of each power of~$x$
is zero. Hence
\index{Euler's eliminant}%
\[
\begin{array}{l@{}r@{}l@{}r@{}l}
a_0B_0        &           &{}+b_0A_0        &                             &=0\\
a_1B_0+a_0B_1 &           &{}+b_1A_0+b_0A_1 &                             &=0\\
\Dots{5}\\
              & a_mB_{n-2}+a_{m-1}B_{n-1} & & {}+b_nA_{m-2}+b_{n-1}A_{m-1}&=0\\
              &                a_mB_{n-1} & &                {}+b_nA_{m-1}&=0.
\end{array}
\]
Since these $m+n$ linear homogeneous equations in the unknowns $B_0, \dotsc, B_{n-1}$, $A_0, \dotsc,
A_{m-1}$ have a set of solutions not all zero, the determinant of the coefficients is zero. %[** PP: , -> .]
By interchanging the rows and columns, we obtain the determinant~\Eq{5}.
% [** PP: Added.]
\end{Remark}


\begin{Exercises}{}

\begin{Problems}
\item[1.] For $m=n=2$, show that the resultant is
\[
R=
\begin{vmatrix}
  a_0  & a_1 & a_2 & 0   \\
    0  & a_0 & a_1 & a_2 \\
  b_0  & b_1 & b_2 & 0   \\
    0  & b_0 & b_1 & b_2
\end{vmatrix}
\]
Interchange the second and third rows, apply Laplace's development, and prove that
\index{Determinants!Laplace's development|(}%
\[
R= (a_0b_2)^2 - (a_0b_1)(a_1b_2),
\]
where $(a_0b_2)$ denotes $a_0b_2 - a_2b_0$, etc.

%% -----File: 154.png---Folio 148-------

\item[2.] For $m = n = 3$, write down the resultant~$R$ and, by interchanges of rows, derive
the second determinant in
\[
R=
\begin{vmatrix}
a_0 & a_1 & a_2 & a_3 &   0 &   0 \\
  0 & a_0 & a_1 & a_2 & a_3 &   0 \\
  0 &   0 & a_0 & a_1 & a_2 & a_3 \\
b_0 & b_1 & b_2 & b_3 &   0 &   0 \\
  0 & b_0 & b_1 & b_2 & b_3 &   0 \\
  0 &   0 & b_0 & b_1 & b_2 & b_3
\end{vmatrix}
= -
\begin{vmatrix}
a_0 & a_1 & a_2 & a_3 &   0 &   0 \\
b_0 & b_1 & b_2 & b_3 &   0 &   0 \\
  0 & a_0 & a_1 & a_2 & a_3 &   0 \\
  0 & b_0 & b_1 & b_2 & b_3 &   0 \\
  0 &   0 & a_0 & a_1 & a_2 & a_3 \\
  0 &   0 & b_0 & b_1 & b_2 & b_3
\end{vmatrix}
\]
To the second determinant apply Laplace's development, selecting minors from the
first two rows, and to the complementary minors apply a similar development. This
may be done by inspection and the following value of~$-R$ will be obtained:
\[
\begin{split}
 & (a_0b_1) \bigl\{(a_1b_2)(a_2b_3) - (a_1b_3)^2+(a_2b_3)(a_0b_3)\bigr\} \\
-& (a_0b_2) \bigl\{(a_0b_2)(a_2b_3) - (a_0b_3)(a_1b_3)\bigr\} \\
+& (a_0b_3) \bigl\{(a_0b_1)(a_2b_3) - (a_0b_3)^2\bigr\}.
\end{split}
\]
The third term of the first line and the first term of the last line are alike. Hence,
changing the signs,
\index{Determinants!Laplace's development|)}% [** PP: Original page range is 147--149]
\[
\begin{split}
R = (a_0b_3)^3
  &- 2(a_0b_1)(a_0b_3)(a_2b_3) - (a_0b_2)(a_0b_3)(a_1b_3) \\
  &+ (a_0b_2)^2 (a_2b_3) + (a_0b_1)(a_1b_3)^2 - (a_0b_1)(a_1b_2)(a_2b_3).
\end{split}
\]
\end{Problems}

Other methods of simplifying Sylvester's determinant~\Eq{5} are given in~§113.
\end{Exercises}


% [** PP: No separate ToC entry]
\Section{113.}{Bézout's Method of Elimination} When the two equations are
of the same degree, the method published by Bézout in~1764 will be clear
from the example
\index{Bézout's eliminant}%
\[
f \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad
g \equiv b_0x^3 + b_1x^2 + b_2x + b_3 = 0.
\]
Then
\[
\begin{gathered}
a_0g - b_0f, \\
(a_0x + a_1)g - (b_0x + b_1)f, \\
(a_0x^2 + a_1x + a_2)g - (b_0x^2 + b_1x + b_2)f
\end{gathered}
\Tag{7}
\]
are equal respectively to
\begin{alignat*}{4}
(a_0b_1)x^2 &   &                 &+{}& (a_0b_2)\;x      &+{}& (a_0b_3) &= 0,\\
(a_0b_2)x^2 &{}+{}& \bigl\{(a_0b_3) &+{}& (a_1b_2)\bigr\}x &+{}& (a_1b_3) &= 0,
\Tag{8} \\
(a_0b_3)x^2 &   &                 &+{}& (a_1b_3)\;x      &+{}& (a_2b_3) &= 0,
\end{alignat*}
%% -----File: 155.png---Folio 149-------
where $(a_0b_1) = a_0b_1 - a_1b_0$, etc. The determinant of the coefficients is the
negative of the resultant $R(f, g)$. Indeed, the negative of the determinant
is easily verified to have the expansion given at the end of Ex.~2
just above.

\begin{Remark}
To give a more instructive proof of the last fact, note that, by~\Eq{7}, equations~\Eq{8}
are linear combinations of
\[
x^2f = 0,\qquad xf = 0,\qquad f = 0,\qquad
x^2g = 0,\qquad xg = 0,\qquad g = 0,
\]
the latter being the equations used in Sylvester's method of elimination. The determinant
\index{Sylvester's eliminant}%
of the coefficients in these six equations is the first determinant~$R$ in Ex.~2 just
above. The operations carried out to obtain equations~\Eq{8} are seen to correspond
step for step to the following operations on determinants. To the products of the elements
of the fourth row by $a_0$ add the products of the elements of the 1st, 2nd, 3rd,
5th, 6th rows by $-b_0$, $-b_1$, $-b_2$, $a_1$, $a_2$ respectively [corresponding to the formation
of the third function~\Eq{7}]. To the products of the elements of the fifth row by $a_0$ add
the products of the elements of the 2nd, 3rd, 6th rows by $-b_0$, $-b_1$, $a_1$ respectively [corresponding
to the second function~\Eq{7}]. Finally, to the products of the elements of the
sixth row by $a_0$ add the products of the elements of the third row by $-b_0$ [corresponding
to $a_0g - b_0f$]. Hence
\[
a_0^3R =
\left|
\begin{array}{ccc;{2pt/2pt}ccc;{2pt/2pt}}
a_0 & a_1 & \Bare{a_2} &    a_3   &       0           & \Bare{0}   \\
  0 & a_0 & \Bare{a_1} &    a_2   &      a_3          & \Bare{0}   \\
  0 &   0 & \Bare{a_0} &    a_1   &      a_2          & \Bare{a_3} \\
  0 &   0 &      0     & (a_0b_3) &    (a_1b_3)       & (a_2b_3)   \\
  0 &   0 &      0     & (a_0b_2) & (a_0b_3)+(a_1b_2) & (a_1b_3)   \\
  0 &   0 &      0     & (a_0b_1) &    (a_0b_2)       & (a_0b_3)
\end{array}\; % Explicit space before right \vert
\right|,
\]
so that $R$ is equal to the $3$-rowed minor enclosed by the dots. The method of Bézout
therefore suggests a definite process for the reduction of Sylvester's determinant of
order~$2n$ (when $m = n$) to one of order~$n$.

Next, for equations of different degrees, consider the example
\[
f \equiv a_0x^4 + a_1x^3 + a_2x^2 + a_3x + a_4,\qquad
g \equiv b_0x^2 + b_1x + b_2.
\]
Then
\[
a_0x^2g - b_0f,\qquad  (a_0x + a_1)x^2g - (b_0x + b_1)f
\]
are equal respectively to
\[
\begin{aligned}
&(a_0b_1)x^3 + (a_0b_2)x^2 - a_3b_0x - a_4b_0,\\
&(a_0b_2)x^3 + \bigl\{(a_1b_2) - a_3b_0\bigr\}x^2
             - \bigl\{a_3b_1 + a_4b_0\bigr\}x - a_4b_1.
\end{aligned}
\]
The determinant of the coefficients of $x^3$, $x^2$, $x$, $1$ in these two functions and $xg$, $g$, after
the first and second rows are interchanged, is the determinant of order~$4$ enclosed by
dots in the second determinant below. Hence it is the resultant $R(f, g)$.

%% -----File: 156.png---Folio 150-------

As in the former example, we shall indicate the corresponding operations on Sylvester's
determinant
\index{Sylvester's eliminant}%
\[
R=
\begin{vmatrix}
 a_0  &   a_1  &   a_2  &   a_3  &   a_4   &   0   \\
 0    &   a_0  &   a_1  &   a_2  &   a_3   &   a_4 \\
 b_0  &   b_1  &   b_2  &   0    &   0     &   0   \\
 0    &   b_0  &   b_1  &   b_2  &   0     &   0   \\
 0    &   0    &   b_0  &   b_1  &   b_2   &   0   \\
 0    &   0    &   0    &   b_0  &   b_1   &   b_2
\end{vmatrix}
\]
Multiply the elements of the third and fourth rows by~$a_0$. In the resulting determinant
$a_0^2R$, add to the elements of the third row the products of the elements of the first,
second and fourth rows by $-b_0$, $-b_1$, $a_1/a_0$ respectively. Add to the elements of the
fourth row the products of those of the second by $-b_0$. We get
\[
a_0^2R =
\left|
\begin{array}{cc;{2pt/2pt}cccc;{2pt/2pt}}
 a_0 & \Bare{a_1} &   a_2    &      a_3        &  a_4           & \Bare{0}  \\
 0   & \Bare{a_0} &   a_1    &      a_2        &  a_3           & \Bare{a_4} \\
 0   &      0     & (a_0b_2) & (a_1b_2)-a_3b_0 & -a_3b_1-a_4b_0 & -a_4b_1 \\
 0   &      0     & (a_0b_1) &    (a_0b_2)     & -a_3b_0        & -a_4b_0 \\
 0   &      0     &   b_0    &      b_1        &  b_2           &    0    \\
 0   &      0     &    0     &      b_0        &  b_1           &   b_2   \\
 \end{array}\;
\right|
\]
Hence $R$ is equal to the minor enclosed by dots.
\end{Remark}


\begin{Exercises}{}

\begin{Problems}

\item[1.] For $m=3$, $n=2$, apply to Sylvester's determinant~$R$ exactly the same operations
as used in the last case in~§113 and obtain
\[
R=
\begin{vmatrix}
(a_0b_2) & (a_1b_2)-a_3b_0 & -a_3b_1 \\
(a_0b_1) & (a_0b_2)        & -a_3b_0 \\
 b_0     &   b_1           &   b_2
\end{vmatrix}.
\]

\item[2.] For $m=n=4$, reduce Sylvester's~$R$ (as in the first case in~§113) to
\[
\begin{vmatrix}
(a_0b_1) & (a_0b_2)            & (a_0b_3)          & (a_0b_4) \\
(a_0b_2) & (a_0b_3) + (a_1b_2) & (a_0b_4)+(a_1b_3) & (a_1b_4) \\
(a_0b_3) & (a_0b_4)+(a_1b_3)   & (a_1b_4)+(a_2b_3) & (a_2b_4) \\
(a_0b_4) & (a_1b_4)            & (a_2b_4)          & (a_3b_4)
\end{vmatrix}.
\]
\end{Problems}
\end{Exercises}

%% -----File: 157.png---Folio 151-------


% [** PP: No ToC entry]
\Section{114.}{General Theorem on Elimination}
\begin{Thm}
If any method of eliminating~$x$
between two equations in~$x$ leads to a relation $F = 0$, where $F$ is a polynomial
in the coefficients, then $F$ has as a factor the true resultant of the equations.
\end{Thm}

Some of the preceding proofs become simpler if this theorem is applied.
For example, determinant~\Eq{3} is divisible by the resultant~$R$. Since the
diagonal term of~\Eq{3} is a term $a_0^2b_2^3$ of~$R$ (Ex.~6,~§111), $F$ is identical
with~$R$.

The preceding general theorem is proved in the author's \textit{Elementary
Theory of Equations}, pp.~152--4. We shall here merely verify the theorem
in an instructive special case. Let
\[
f \equiv a_0x^3 + a_1x^2 + a_2x + a_3 = 0,\qquad
g \equiv b_0x^3 + b_1x^2 + b_2x + b_3 = 0
\]
have a common root $x\ne 0$. Then
\begin{align*}
-b_0f + a_0g    &= (a_0b_1)x^2 + (a_0b_2)x + (a_0b_3), \\
(b_3f - a_3g)/x &= (a_0b_3)x^2 + (a_1b_3)x + (a_2b_3).
\end{align*}
By Ex.~1 of~§112, the resultant of these two quadratic functions is
\[
F=
\begin{vmatrix}
(a_0b_3) & (a_0b_1) \\
(a_2b_3) & (a_0b_3)
\end{vmatrix}^2
-
\begin{vmatrix}
(a_0b_3) & (a_0b_1) \\
(a_1b_3) & (a_0b_2)
\end{vmatrix}
·
\begin{vmatrix}
(a_1b_3) & (a_0b_2) \\
(a_2b_3) & (a_0b_3)
\end{vmatrix}.
\]
This is, however, not the resultant~$R$ of the cubic functions $f$,~$g$. To show
that $(a_0b_3)$ is an extraneous factor, note that the terms of $F$ not having
this factor explicitly are
\index{Elimination!extraneous factor}%
\[
(a_0b_1) (a_2b_3) \bigl\{(a_0b_1)(a_2b_3) - (a_0b_2)(a_1b_3)\bigr\}.
\]
The quantity in brackets is equal to $-(a_0b_3)(a_1b_2)$, since, as in Ex.~2
of~§101,
\[
0 = \tfrac{1}{2}
\begin{vmatrix}
 a_0 & a_1 & a_2 & a_3 \\
 b_0 & b_1 & b_2 & b_3 \\
 a_0 & a_1 & a_2 & a_3 \\
 b_0 & b_1 & b_2 & b_3
\end{vmatrix}
= (a_0b_1)(a_2b_3) - (a_0b_2)(a_1b_3) + (a_0b_3)(a_1b_2).
\]
We now see that $F=(a_0b_3)R$, where $R$ is given in Ex.~2 of~§112. This
method of elimination therefore introduces an extraneous factor $(a_0b_3)$.
The student should employ only methods of elimination (such as those
due to Sylvester, Euler, and Bézout) which have been proved to lead
to the true resultant.

%% -----File: 158.png---Folio 152-------


\begin{Exercises}{Page152}

Find the result of eliminating~$x$ and hence find all sets of common solutions of
\begin{Problems}
\item[1.] $x^2-y^2=9$,  $xy = 5y$.

\item[2.] $x^2 + y^2 = 25$, $x^2 + 3(c-1)x + c(y^2 - 25) = 0$.

\item[3.] When $x^2 + ax + b = 0$ has a double root, what $3$-rowed determinant is zero?

\item[4.]  Find the roots of $x^6 + 3x^4 + 32x^3 + 67x^2 + 32x + 65 = 0$ by~§79.
\end{Problems}
\end{Exercises}


\Section{115.}{Discriminants} Let $\alpha_1, \dotsc, \alpha_m$ be the roots of
\index{Discriminant|(}%
\begin{flalign*}
&& f(x) &\equiv a_0x^m + a_1x^{m-1} + \dotsb + a_m = 0
    && \Rightmark{(a_0\ne 0),}
\Tag{9} \\
\intertext{so that}
&& f(x) &\equiv a_0(x - \alpha_1)(x - \alpha_2) \dotsm (x - \alpha_m).
\Tag{10}
\end{flalign*}
As in~§44, we define the discriminant of~\Eq{9} to be
\[
D = a_0^{2m-2}(\alpha_1 - \alpha_2)^2(\alpha_1 - \alpha_3)^2 \dotsm
              (\alpha_1 - \alpha_m)^2(\alpha_2 - \alpha_3)^2 \dotsm
              (\alpha_{m-1} - \alpha_m)^2.
\]
Evidently $D$ is unaltered by the interchange of any two roots. Since the
degree in any root is $2(m-1)$, the symmetric function~$D$ is equal to a
polynomial in $a_0, \dotsc, a_m$. Indeed, $a_0^{2m-2}$ is the lowest power of $a_0$
sufficient to cancel the denominators introduced by replacing $\Sigma \alpha_1$ by
$-a_1/a_0, \dotsc, \alpha_1\alpha_2\dotsm \alpha_m$ by~$±a_m/a_0$. By differentiating~\Eq{10}, we see
that
\begin{align*}
f'(\alpha_1)
  &= a_0(\alpha_1 - \alpha_2)(\alpha_1 - \alpha_3) \dotsm (\alpha_1 - \alpha_m),\\
f'(\alpha_2)
  &= a_0(\alpha_2 - \alpha_1)(\alpha_2 - \alpha_3) \dotsm (\alpha_2 - \alpha_m),\\
f'(\alpha_3)
  &= a_0(\alpha_3 - \alpha_1)(\alpha_3 - \alpha_2)(\alpha_3 - \alpha_4) \dotsm
        (\alpha_3 - \alpha_m),
\end{align*}
etc.  Hence
\begin{align*}
 a_0^{m-1}f'(\alpha_1) \dotsm f'(\alpha_m)
  &= a_0^{2m-1} (-1)^{1 + 2 + \dotsb + m-1}
     (\alpha_1 - \alpha_2)^2\dotsm (\alpha_{m-1} - \alpha_m)^2 \\
  &= (-1)^{\frac{m(m-1)}{2}} a_0 D.
\end{align*}
By~\Eq{2}, the left member is the resultant of $f(x)$, $f'(x)$. Hence
\[
D = (-1)^{\frac{m(m-1)}{2}} \frac{1}{a_0} R(f, f').
\Tag{11}
\]

%% -----File: 159.png---Folio 153-------


\begin{Exercises}{}
\index{Cubic equation|(}%

\begin{Problems}

\item[1.] Show that the discriminant of $f \equiv y^3 + py + q = 0$ is $-4p^3 - 27q^2$ by evaluating
the determinant of order five for $R(f, f')$.

\item[2.] Prove that the discriminant of the product of two functions is equal to the product
of their discriminants multiplied by the square of their resultant. Hint: use
the expressions in terms of the differences of the roots.

\item[3.] For $a_0 = 1$, show that the discriminant is equal to
\[
\begin{vmatrix}
1 & \alpha_1  & \alpha_1^2 & \cdots & \alpha_1^{m-1} \\
1 & \alpha_2  & \alpha_2^2 & \cdots & \alpha_2^{m-1} \\
\Dots{5}                      \\
1 & \alpha_m  & \alpha_m^2 & \cdots & \alpha_m^{m-1}
\end{vmatrix}^2
=
\begin{vmatrix}
 s_0 &      s_1 & s_2     & \cdots & s_{m-1}\\
 s_1 &      s_2 & s_3     & \cdots & s_m \\
\Dots{5}\\
 s_{m-1} &  s_m & s_{m+1} & \cdots & s_{2m-2}
\end{vmatrix}
\]
where $s_i = \alpha_1^i + \dotsb + \alpha_m^i$. See Ex.~4,~§88; Ex.~2,~§102.

\item[4.]  Hence verify that the discriminant of $x^3 + px + q = 0$ is equal to
\[
\begin{vmatrix}
  3 &   0 & -2p   \\
  0 & -2p & -3q   \\
-2p & -3q &  2p^2
\end{vmatrix}
= -4p^3 - 27q^2.
\]

\item[5.]  By means of Ex.~1,~§113, show that the discriminant of $a_0x^3 + a_1x^2 + a_2x + a_3 = 0$ is
\[% [** PP: Not breaking]
-\begin{vmatrix}
  2a_0a_2 & a_1a_2 + 3a_0a_3 & 2a_1a_3 \\
  a_1     & 2a_2             & 3a_3    \\
  3a_0    & 2a_1             & a_2
\end{vmatrix} %\\
= 18a_0a_1a_2a_3 - 4a_0a_2^3 - 4a_1^3a_3 + a_1^2a_2^2 - 27a_0^2a_3^2.
\]
\end{Problems}
\end{Exercises}
\index{Discriminant|)}%


\begin{Exercises}[MISCELLANEOUS~]{Page153}

\begin{Problems}
\item[1.]  Find the equation whose roots are the abscissas of the points of intersection
of two general conics.

\item[2.]  Find a necessary and sufficient condition that
\[
 f(x) \equiv x^4 + px^3 + qx^2 + rx + s = 0
\]
shall have one root the negative of another root. When this condition is satisfied,
what are the quadratic factors of~$f(x)$? Apply to Ex.~4,~§74. Hint: add and subtract
$f(x)$ and~$f(-x)$.

\item[3.]  Solve $f(x) \equiv x^4 - 6x^3 + 13x^2 - 14x + 6 = 0$, given that two roots $\alpha$ and~$\beta$ are such
that $2\alpha + \beta = 5$. Hint: $f(x)$ and $f(5-2x)$ have a common factor.

\item[4.]  Solve $x^3 + px + q = 0$ by eliminating $x$ between it and $x^2 + vx + w = y$ by the greatest
common divisor process, and choosing $v$ and~$w$ so that in the resulting cubic equation
for $y$ the coefficients of $y$ and~$y^2$ are zero.  The next to the last step of the elimination
%% -----File: 160.png---Folio 154-------
gives $x$ as a rational function of~$y$. (Tschirnhausen, \textit{Acta Erudit.}, Lipsiae,~II, 1683,
p.~204.)

\item[5.] Find the preceding $y$-cubic as follows. Multiply $x^2 + vx + w = y$ by~$x$ and replace
$x^3$ by~$-px-q$; then multiply the resulting quadratic equation in~$x$ by~$x$ and replace
$x^3$ by its value. The determinant of the coefficients of $x^2$, $x$, $1$ must vanish.

\item[6.] Eliminate $y$ between $y^3 = v$, $x = ry + sy^2$, and get
\[
x^3 - 3rsvx - (r^3v + s^3v^2) = 0.
\]
Take $s=1$ and choose %[** PP: Typo chose]
$r$ and~$v$ so that this equation shall be identical with $x^3 + px + q = 0$,
and hence solve the latter. (Euler,~1764.)

\item[7.] Eliminate $y$ between $y^3 = v$, $x = f + ey + y^2$ and get
\[
\begin{vmatrix}
 1  &  e  & f-x \\
 e  & f-x &  v  \\
f-x &  v  & ev
\end{vmatrix}
=0.
\]
This cubic equation in $x$ may be identified with the general cubic equation by choice
of $e$, $f$, $v$. % [** PP: , -> .]
Hence solve the latter.

\item[8.] Determine $r$, $s$ and~$v$ so that the resultant of
\[
y^3 = v,\qquad y = \frac{x+r}{y+s}
\]
shall be identical with $x^3 + px + q = 0$. (Bézout,~1762.)

\item[9.] Show that the reduction of a cubic equation in~$x$ to the form $y^3 = v$ by the substitution
\[
x = \frac{r + sy}{1 + y}
\]
is not essentially different from the method of Ex.~7. [Multiply the numerator and
denominator of~$x$ by $1 - y + y^2$.]

\item[10.] Prove that the equation whose roots are the $n(n-1)$ differences $x_j-x_k$ of the
roots of $f(x)=0$ may be obtained by eliminating $x$ between the latter and $f(x+y)=0$
and deleting from the eliminant the factor~$y^n$ (arising from $y = x_j - x_j = 0$). The
equation free of this factor may be obtained by eliminating~$x$ between $f(x)=0$ and
\index{Equation for differences of roots}%
\[
\bigl\{f(x+y) - f(x)\bigr\}/y
  = f'(x) + f''(x)\frac{y}{1·2} + \dotsb
  + f^{(n)}(x)\frac{y^{n-1}}{1·2\dotsm n} = 0.
\]
This eliminant involves only even powers of~$y$, so that if we set $y^2 = z$ we obtain an
equation in~$z$ having as its roots the squares of the differences of the roots of $f(x)=0$.
\index{Equation for differences of roots!squares of differences}%
(Lagrange \textit{Résolution des équations}, 1798,~§8.)

\item[11.] Compute by Ex.~10 the $z$-equation when $f(x) = x^3 + px + q$.
\end{Problems}
\end{Exercises}
\index{Cubic equation|)}%
\index{Determinants|)}% [** PP: Original entry range is 146--153]
\index{Elimination|)}%
\index{Resultant|)}%

%% -----File: 161.png---Folio 155-------


\Appendix
\index{Fundamental theorem of algebra|(}%

\begin{Theorem}
An equation of degree~$n$ with any complex coefficients
\[
f(z) \equiv z^n + a_1 z^{n-1} + \dotsb + a_n = 0
\]
has a complex \(real or imaginary\) root.
\end{Theorem}

Write $z = x+iy$ where $x$ and~$y$ are real, and similarly $a_1 = c_1 + id_1$, etc.
By means of the binomial theorem, we may express any power of~$z$ in the
form $X+iY$. Hence
\[
f(z) = \phi(x,y) + i\psi(x,y),
\Tag{1}
\]
where $\phi$ and~$\psi$ are polynomials with real coefficients.

The first proof of the fundamental theorem was given by Gauss in
1799 and simplified by him in~1849. This simplified proof consists in
showing that the two curves represented by $\phi(x, y) = 0$ and $\psi(x, y) = 0$
have at least one point $(x_1, y_1)$ in common, so that $z_1 = x_1 + iy_1$ is a root
of $f(z)= 0$. This proof is given in \ChapRef{V} of the author's \textit{Elementary
Theory of Equations}.

We here give a shorter proof, the initial idea of which was suggested,
but not fully developed, by Cauchy.\footnote
  {For a history of the fundamental theorem, see \textit{Encyclopédie des sciences mathématiques},
  tome~I, vol.~II, pp.~189--205.}

\begin{Lemma}[1.]
$a_1 h + a_2 h^2 + \dotsb + a_n h^n$ is less in absolute value than any
assigned positive number~$p$ for all complex values of $h$ sufficiently small in
absolute value.
\end{Lemma}

The proof differs from that of the auxiliary theorem in~§62 only in
reading ``in absolute value'' for ``numerically.''

We shall employ the notation $|z|$ for the absolute value $+\sqrt{x^2 + y^2}$ of
$z = x + iy$.
\index{Symbol!c@{$\lvert a\rvert$\IndAdd{modulus}}}% [** PP: Manually alphabetized]

%% -----File: 162.png---Folio 156-------

\begin{Lemma}[2.]
Given any positive number~$P$, we can find a positive number~$R$
such that $|f(z)| > P$ if $|z| \geqq R$.
\end{Lemma}

The proof is analogous to that in~§64. We have
\[
f(z) = z^n(1+D),\qquad
D \equiv a_1\left(\frac{1}{z}\right) + \dotsb
       + a_n\left(\frac{1}{z}\right)^n.
\]
Since (Ex.~5,~§8) the absolute value of a sum of two complex numbers
is equal to or greater than the difference of their absolute values, we have
\[
|f(z)| \geqq |z|^n \bigl[1 - |D|\bigr].
\]

Let $p$ be any assigned positive number~$<1$. Applying Lemma~1 with
$h$ replaced by~$1/z$, we see that $|D| < p$ if $|1/z|$ is sufficiently small, i.e., % [** PP: Not italicizing i.e.]
if $\rho\equiv |z|$ is sufficiently large. Then
\[
|f(z)| > \rho^n(1-p) \geqq P
\]
if $\rho^n \geqq P/(1-p)$, which is true if
\[
\rho \geqq \sqrt[n]{\frac{P}{1-p}} \equiv R.
\]
This proves Lemma~2.

\begin{Lemma}[3.]
Given a complex number a such that $f(a) \ne 0$, we can find
a complex number~$z$ for which $|f(z)| < |f(a)|$.
\end{Lemma}

Write $z = a+h$. By Taylor's theorem~\Eq{8} of~§56,
\[
f(a+h) = f(a) + f'(a)h + \dotsb + f^{(r)}(a)·\frac{h^r}{r!} + \dotsb
       + f^{(n)}(a)·\frac{h^n}{n!}.
\]
Not all of the values $f'(a)$, $f''(a), \dotsc$ are zero since $f^{(n)}(a) = n!$. Let
$f^{(r)}(a)$ be the first one of these values which is not zero. Then
\[
\frac{f(a+h)}{f(a)}
  = 1 + \frac{f^{(r)}(a)}{f(a)}·\frac{h^r}{r!} + \dotsb
      + \frac{f^{(n)}(a)}{f(a)}·\frac{h^n}{n!}.
\]
Writing the second member in the simpler notation
\[
g(h) \equiv 1 + bh^r + ch^{r+1} + \dotsb + lh^n,\qquad  b\ne 0,
\]
we shall prove that a complex value of~$h$ may be found such that $|g(h)| < 1$.
Then the absolute value of $f(z)/f(a)$ will be~$<1$ and Lemma~3 proved.
To find such a value of~$h$, write $h$ and~$b$ in their trigonometric forms~(§4)
\[
h = \rho(\cos \theta + i \sin \theta),\qquad
b =  |b|(\cos \beta  + i \sin \beta).
\]
%% -----File: 163.png---Folio 157-------
Then by~§5,~§7,% [** PP: Retaining instead of §§5,~7]
\[
bh^r = |b| \rho^r \bigl\{\cos(\beta+r\theta) + i\sin (\beta+r\theta)\bigr\}.
\]
Since $h$ is at our choice, $\rho$ and angle~$\theta$ are at our choice. We choose~$\theta$
so that $b + r\theta = 180°$. Then the quantity in brackets reduces to~$-1$,
whence
\[
g(h) = (1 - |b|\rho^r) + h^r(ch + \dotsb + lh^{n-r}).
\]
By Lemma~1, we may choose $\rho$ so small that
\[
|ch + \dotsb + lh^{n-r}| < |b|.
\]
By taking $\rho$ still smaller if necessary, we may assume at the same time
that $|b| \rho^r < 1$. Then
\[
|g(h)| < (1 - |b|\rho^r) + \rho^r|b|,\qquad  |g(h)|<1.
\]

\paragraph{\indent Minimum Value of a Continuous Function.} Let $F(x)$ be any polynomial
\index{Continuity}%
\index{Minimum}%
with real coefficients. Among the real values of~$x$ for which
$2\leqq x\leqq 3$, there is at least one value~$x_1$ for which $F(x)$ takes its minimum
value~$F(x_1)$, i.e., % [** PP: Not italicizing i.e.]
for which $F(x_1)\leqq F(x)$ for all real values of~$x$ such that
$2\leqq x\leqq 3$. This becomes intuitive geometrically. The portion of the
graph of $y = F(x)$ which extends from its point with the abscissa~$2$ to its
point with the abscissa~$3$ either has a lowest point or else has several
equally low points, each lower than all the remaining points. The arithmetic
proof depends upon the fact that $F(x)$ is continuous for each~$x$
between $2$ and~$3$ inclusive~(§62). The proof is rather delicate and is
omitted since the theorem for functions of one variable~$x$ is mentioned
here only by way of introduction to our case of functions of two variables.

We are interested in the analogous question for
\[
G(x,y) = \phi^2(x, y) + \psi^2(x, y),
\]
which, by~\Eq{1}, is the square of~$|f(z)|$. As in the elements of solid analytic
geometry, consider the surface represented by $Z = G(x,y)$ and the right
circular cylinder $x^2 + y^2 = R^2$. Of the points on the first surface and on
or within their curve of intersection there is a lowest point or there are
several equally low lowest points, possibly an infinite number of them.
Expressed arithmetically, among all the pairs of real numbers $x$,~$y$ for
%% -----File: 164.png---Folio 158-------
which $x^2 + y^2\leqq R^2$, there is\footnote
  {Harkness and Morley, \textit{Introduction to the Theory of Analytic Functions}, p.~79,
  prove that a real function of two variables which is continuous throughout % [** PP: Typo thoroughout]
  a closed
  region has a minimum value at some point of the region.}
at least one pair $x_1$,~$y_1$ for which the
polynomial $G(x,y)$ takes a minimum value $G(x_1, y_1)$, i.e., for which % [** PP: Not italicizing i.e.]
$G(x_1, y_1) \leqq G(x, y)$ for all pairs of real numbers $x$,~$y$ for which $x^2 + y^2 \leqq R^2$.

\paragraph{\indent Proof of the Fundamental Theorem.} Let $z'$ denote any complex
number for which $f(z')\ne 0$. Let $P$ denote any positive number exceeding
$|f(z')|$. Determine $R$ as in Lemma~2. In it the condition $|z|\geqq R$ may
be interpreted geometrically to imply that the point $(x,y)$ representing
$z = x + iy$ is outside or on the circle~$C$ having the equation $x^2 + y^2 = R^2$.
Lemma~2 thus states that, if $z$ is represented by any point outside or on
the circle~$C$, then $|f(z)|>P$. In other words, if $|f(z)|\leqq P$, the point
representing~$z$ is inside circle~$C$. In particular, the point representing~$z'$
is inside circle~$C$.

In view of the preceding section on minimum value, we have
\[
G(x_1,y_1) \leqq G(x,y)
\]
for all pairs of real numbers $x$,~$y$ for which $x^2 + y^2\leqq R^2$, where $x_1$,~$y_1$ is one
such pair. Write $z_1$ for $x_1 + iy_1$. Since $|f(z)|^2 = G(x,y)$, we have
\[
|f(z_1)|\leqq |f(z)|
\]
for all $z$'s represented by points on or within circle~$C$. Since $z'$ is represented
by such a point,
\[
|f(z_1)|\leqq |f(z')| < P.
\Tag{2}
\]

This number $z_1$ is a root of $f(z)=0$. For, if $f(z_1)\neq 0$, Lemma~3 shows
that there would exist a complex number $z$ for which
\[
|f(z)| < |f(z_1)|.
\Tag{3}
\]
Then $|f(z)| < P$ by~\Eq{2}, so that the point representing~$z$ is inside circle~$C$,
as shown above. By the statement preceding~\Eq{2},
\[
|f(z_1)|\leqq |f(z)|.
\]
But this contradicts~\Eq{3}. Hence the fundamental theorem is proved.
\index{Fundamental theorem of algebra|)}%

%% -----File: 165.png---Folio 159-------

\PrintAnswers

\begin{Answers}[5]{Page2}%, 3}

\item[1.] $3i$.

\item[2.] $2$.

\item[3.] $-20 + 20i$.

\item[4.] $-\frac{2}{3}$.

\item[5.] $(8 + 2\sqrt{3})$.

\ResetCols{2}

\item[6.] $\frac{1}{5}(6 + \sqrt{5}) + \frac{1}{5}(2\sqrt{5} - 3)i$.

\item[7.] $\dfrac{-9}{13} + \dfrac{19}{13} i$.

\ResetCols{2}

\item[8.] $\dfrac{a^2 - b^2}{a^2 + b^2} + \dfrac{2ab}{a^2 + b^2}i$.

\item[10.] Yes.

\ResetCols{2}

\item[13.] $3$, $4$ and $-3$, $-4$.

\item[14.] $±(5 + 6i)$.

\ResetCols{2}

\item[15.] $±(3 - 2i)$.

\item[16.] $±\bigl[c + d + (c - d)i\bigr]$.
\end{Answers}


\begin{Answers}[1]{Page6}%, 7}

\item[2.] $-3$, $-3\omega$, $-3\omega^2$;\quad $i$, $\omega i$;\quad $\omega^2 i$;\quad
   $R = \cos 40° + i\sin 40°$, $\omega R$, $\omega^2 R$.

\item[3.] $±(1 + i)/\sqrt{2}$;\quad  $±(1 - i)/\sqrt{2}$;\quad $±\omega^2$.
\end{Answers}


\begin{Answers}[1]{Page9}

\item[4.] $-1$, $\cos A + i \sin A$ ($A=36°$, $108°$, $252°$, $324°$).

\item[6.] $R^3$, $R^6$, $R^9$.
\end{Answers}


\begin{Answers}{Page10}

\item[5.] $p(p-1)$.

\item[6.] $(p-1)(q-1)(r-1)$ if $n=pqr$.
\end{Answers}


\begin{Answers}{Page13}

\item[1.] $51$.

\item[2.] $13$.
\end{Answers}


\begin{Answers}{Page15}

\item[1.] Rem.~$11$, quot.~$x^2 + 5x + 8$.

\item[2.] $-61$, $2x^4 - 4x^3 + 7x^2 - 14x + 30$.

\ResetCols{1}

\item[3.] $-0.050671$, $x^2 + 6.09x + 10.5481$.

\item[4.] $x^2 - x - 6$, $x+2$;\quad $4$, $3$, $-2$.

\item[5.] $x^2 - x - 6 = 0$, $3$, $-2$.

\ResetCols{3}

\item[6.] $2±\sqrt{5}$.

\item[7.] $2x^2 - x + 2$.

\item[8.] $x^2 + 1$.
\end{Answers}


\begin{Answers}{Page17}

\item[1.] $x^3 - 3x^2 + 2x = 0$.

\item[2.] $x^4 - 5x^2 + 4 = 0$.

\ResetCols{2}

\item[3.] $x^4 - 18x^2 + 81 = 0$.

\item[4.] $x^4 - 5x^3 + 9x^2 - 7x + 2 = 0$.

\ResetCols{2}

\item[5.] $b^2 = 4ac$.

\item[7.] By theorem in~§18.
\end{Answers}

%% -----File: 166.png---Folio 160-------


\begin{Answers}{Page19}

\item[1.] $x^3 - 6x^2 + 11x - 6 = 0$.

\item[2.] $x^4 - 8x^2 + 16 = 0$.

\ResetCols{4}

\item[3.] $1$, $2$.

\item[5.] $4$, $\tfrac{3}{2}$, $-\tfrac{3}{2}$.

\item[6.] $1$, $3$, $5$.

\item[7.] $1$, $1$, $1$, $3$.

\ResetCols{3}

\item[8.] $2$, $-6$, $18$.

\item[9.] $-3$, $1$, $5$.

\item[10.] $5$, $2$, $-1$, $-4$.

\ResetCols{2}

\item[11.] $y^2 - (p^2 - 2q)y + q^2 = 0$.

\item[12.] $y^2 - (p^3 - 3pq)y + q^3 = 0$.

\ResetCols{1}

\item[13.] (i) $y^2 - y(p^3 - 3pq)/q + q = 0$. \\
  (ii) $y^2 - q(p^2 - 2q)y + q^4 = 0$. \\
  (iii) $y^2 - (p + p/q)y + 2 + q + 1/q = 0$.

\ResetCols{2}

\item[14.] $p^3r = q^3$.

\item[15.] $2$, $4$, $-6$.
\end{Answers}


\begin{Answers}{Page20}

\item[1.] $5$, $-1±\sqrt{-3}$.

\item[2.] $1±i$, $1±\sqrt{2}$.

\ResetCols{2}

\item[3.] $x^3 - 7x^2 + 19x - 13 = 0$.

\item[4.] $4$, $1-\sqrt{-5}$, $x^3 - 6x^2 + 14x - 24 = 0$.

\ResetCols{2}

\item[6.] $± 1$, $2±\sqrt{3}$.

\item[7.] $\sqrt{3}$, $2±i$.

\ResetCols{2}

\item[9.] $x^3 - \tfrac{3}{2}x^2 - \tfrac{5}{4}x + \tfrac{7}{8} = 0$.

\item[10.] $2 + \sqrt{3}$, $x^2 + 2x + 2 = 0$.

\ResetCols{2}

\item[11,] \textbf{12.}~Not necessarily.

\item[13.] No.
\end{Answers}


\begin{Answers}[5]{Page23}

\item[1.] $19\tfrac{1}{4}$, $3$.

\item[2.] $6$.

\item[3.] $2$.

\item[4.] $3$.

\item[5.] $0, -7, -\tfrac{7}{3}$.
\end{Answers}


\begin{Answers}{Page25}

\item[1.] $-1$, $-1$, $-6$.

\item[2.] $-2$, $3$, $4$.

\ResetCols{3}

\item[3.] $1$, $3$, $6$.

\item[4.] $-2$, $-4$.

\item[5.] None.
\end{Answers}


\begin{Answers}{Page27}

\item[1.] $2$, $-1$, $-4$, $5$.

\item[2.] $9$.

\ResetCols{3}

\item[3.] $8$, $9$.

\item[4.] $-12$, $-35$.

\item[5.] $2$, $2$, $-3$.

\end{Answers}


\begin{Answers}[4]{Page28}

\item[1.] $1$, $3$, $9$, $\frac{1}{3}$.

\item[2.] $1$, $\tfrac{1}{2}$, $\tfrac{1}{3}$.

\item[3.] $-\tfrac{1}{6}$.

\item[4.] $\tfrac{1}{2}$, $-\tfrac{1}{4}$, $-\tfrac{1}{4}$.

\ResetCols{4}

\item[5.] $\tfrac{1}{4}$, $-\tfrac{1}{4}$, $\tfrac{1}{6}$.

\item[6.] $-\tfrac{1}{2}$, $\tfrac{1}{3}$, $\tfrac{1}{4}$.

\item[7.] $\tfrac{1}{2}$.

\item[8.] $\tfrac{2}{3}$.

\ResetCols{2}

\item[10.] $x^2 - 12x - 12 = 0$.

\item[11.] $x^3 - 3x^2 - 12x + 54 = 0$.
\end{Answers}


\begin{Answers}[3]{Page30}

\item[1.] $1$, $4$.

\item[2.] $-1$, $-4$.

\item[3.] $0.7$, $-5.7$.

\ResetCols{3}

\item[4.] $-0.7$, $5.7$.

\item[5.] $2$, $2$.

\item[6.] Imaginary.
\end{Answers}


\begin{Answers}[1]{Page40}

\item[5.] $x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$.

\item[6.] $-\tfrac{1}{2}(1±\sqrt{-3})$, $\tfrac{1}{2}(7±\sqrt{45})$.

\item[10.] See~\Eq{11},~§32.

\item[11.] Edges roots of $x^3 - 7x^2 + 12x - v = 0$, all real~(§45) and irrational.

\item[14.] $\Delta = \text{area}$, $c = \text{hypotenuse}$, squares of legs $\tfrac{1}{2}(c^2 ± \sqrt{c^4 - 16\Delta^2})$.

\item[15.] $\Delta$ area, $a$, $b$ given sides, square third side is $a^2 + b^2 ± 2\sqrt{a^2b^2 - 4\Delta^2}$.

\item[16.] $y^4 - 2y^3 + (2 - g^2)y^2 - 2y + 1 = 0$, pos.\ roots $0.09125$, $10.95862$.
\end{Answers}

%% -----File: 167.png---Folio 161-------


\begin{Answers}[1]{Page44}

\item[3.] $g=2$, $R + R^8 + R^{12} + R^5$, etc., $z^3 + z^2 - 4z + 1 = 0$.

\item[4.] $g=2$, $R+R^8$, $R^2+R^7$, $R^4+R^5$.

\ResetCols{2}

\item[5.] $\tfrac{1}{2}(1±\sqrt{-3})$, $\tfrac{1}{2}(-5±\sqrt{21})$.

\item[6.] $-1$, $2±\sqrt{3}$, $\tfrac{1}{2} ± \tfrac{1}{2}\sqrt{-3}$.

\ResetCols{2}

\item[7.] $1$, $1$, $1$, $-1$, $\tfrac{1}{4}(1±\sqrt{-15})$.

\item[8.] $-1$, $-2$, $-\tfrac{1}{2}$, $\tfrac{1}{6}(-5±\sqrt{-11})$.

\end{Answers}


\begin{Answers}{Page46}

\item[1.] $-5$, $\tfrac{1}{2}(5±\sqrt{-3})$.

\item[2.] $-6$, $±\sqrt{-3}$.

\ResetCols{2}

\item[3.] $-2$, $1± i$.

\item[4.] $\tfrac{1}{4}$, $\tfrac{1}{7}(-2±\sqrt{-3})$.

\end{Answers}


\begin{Answers}{Page48}

\item[1.] $\Delta = -400$, one.

\item[2.] $\Delta = 4 · 27 · 121$, three.

\ResetCols{2}

\item[3.] $\Delta = 0$, two.

\item[4.] $\Delta = 0$, two.

\end{Answers}


\begin{Answers}{Page49}

\item[1.] $-4$, $2±\sqrt{3}$.

\item[2.] See Ex.~1,~§47.

\ResetCols{2}

\item[3.] $1.3569$, $1.6920$, $-3.0489$. %[** PP: Horizontal layout]

\item[4.] $-1.201639$, $1.330058$, $-3.128419$.

\ResetCols{2}

\item[5.] $1.24698$, $-1.80194$, $-0.44504$.

\item[6.] $1.1642$, $-1.7729$, $-3.3914$.

\end{Answers}


\begin{Answers}[3]{Page51}

\item[1.] $1$, $-1$, $4±\sqrt{6}$.

\item[2.] $-1$, $-2$, $2$, $3$.

\item[3.] $1± i$, $-1±\sqrt{2}$.

\ResetCols{2}

\item[4.] $1±\sqrt{2}$, $-1±\sqrt{-2}$.

\item[5.] $4$, $-2$, $-1± i$.

\end{Answers}


\begin{Answers}{Page54}% (bottom)}

\item[1.] $(-3, 9)$.

\item[2.] $\Delta=-250000$, $x=3$, $-2$, $±i$.

\ResetCols{3}

\item[3.] $(3,9)$, $(-2,4)$.

\item[4.] $h=3$.

\item[5.] $6.856$, $7$.

\end{Answers}


\begin{Answers}[1]{Page59}%, 60}

\item[2.] $2.1$.

\item[3.] $(-0.845, 4.921)$, $(-3.155, 11.079)$;\quad between $-4$ and~$-5$.

\item[4.] $1.1$, $-1.3$.  %[** PP: Added .]

\item[5.] Between $0$ and~$1$, $0$ and~$-1$, $2.5$ and~$3$, $-2.5$ and~$-3$.

\item[9.] $120(x^3 + x)$, $120x^2 - 42$.

\end{Answers}


\begin{Answers}[3]{Page62}

\item[1.] $3$.

\item[2.] $2$, $-2$.

\item[3.] $-1$.

\ResetCols{3}

\item[4.] Double roots, $1$, $3$.

\item[5.] None.

\item[6.] $3$, $3$, $-3$, $6$.

\end{Answers}


\begin{Answers}{Page64}%, 65}

\item[3.] Use Ex.~3, p.~62, abscissas $-1$, $3$.

\item[\qquad4.] Use Ex.~2, p.~62.

\ResetCols{1}

\item[6.] $y = -15x - 7$, $X^3 - 15X + 23 = 0$.

\end{Answers}

%% -----File: 168.png---Folio 162-------


\begin{Answers}{Page66}

\item[1.] One real.

\item[2.] $(±\sqrt{\frac{7}{3}}, 7\mp\frac{14}{3}\sqrt{\frac{7}{3}})$, three real.

\ResetCols{2}

\item[3.] $(±\sqrt\frac{2}{3}, -1\mp\frac{4}{3}\sqrt{\frac{2}{3}})$, three.

\item[4.] $(-2±\sqrt{5}, 23\mp10\sqrt{5})$, one.
\end{Answers}


\begin{Answers}[1]{Page74}%, 75}

\item[13.] $y^5 + 2y^4 + 5y^3 + 3y^2 - 2y - 9 = 0$.

\item[14.] $y^3 + 15y^2 + 52y - 36 = 0$.

\end{Answers}


\begin{Answers}{Page78}

\item[1.] One, between $-2$ and~$-3$.

\item[2.] One, between $1$ and~$2$.

\end{Answers}


\begin{Answers}{Page79}%, 80}

\item[1.] $(-4, -3)$, $(-2, -1)$, $(1, 2)$.

\item[2.] $(-2, -1)$, $(0, 1)$.

\ResetCols{2}

\item[3.] $(-2, -1.5)$, $(-1.5, -1)$, $(3, 4)$.

\item[4.] $(-2, -1)$, $(0, 1)$.

\ResetCols{2}

\item[5.] $(-7, -6)$, $(1, 2)$.

\item[6.] $(0, 1)$, $(3, 4)$.

\end{Answers}


\begin{Answers}[3]{Page83}

\item[2.] $1$, $1$, $1$, $2$.

\item[3.] $1$, $1$, $-2$, $-2$.

\item[4.] $1$, $1$, two imaginary.

\end{Answers}


\begin{Answers}{Page85}

\item[1.] $(-2, -1)$, $(0, 1)$, $(1, 2)$.

\item[2.] $(-4, -3)$, $(-2, -1)$, $(1, 2)$.

\end{Answers}


\begin{Answers}{Page89}%, 90}

\item[1.] Single, $-2.46955$.

\item[2.]  $-1.20164$, $1.33006$, $-3.12842$.

\ResetCols{2}

\item[3.] $1.24698$, $-1.80194$, $-0.44504$.

\item[4.] $± 2.1213203$, $\Neg2.1231056$, $-6.1231056$.

\ResetCols{2}

\item[5.] $3.45592$, $21.43067$.

\item[6.] $2.15443$.

\ResetCols{1}

\item[7.] $-1.7728656$, $\Neg1.1642479$, $-3.3913823$.

\item[8.] $\Neg3.0489173$, $-1.3568958$, $-1.6920215$.

\ResetCols{2}

\item[9.] $2.24004099$.

\item[10.] $1.997997997$.

\ResetCols{2}

\item[11.] $1.094551482$.

\item[12.] $2.059$, $-1.228$.

\ResetCols{2}

\item[13.] $1.2261$.

\item[14.] $0.6527 = \text{reciprocal of } 2 \cos 40°$.

\ResetCols{3}

\item[15.] $0.9397$.

\item[16.] $1.3500$.

\item[17.] $2.7138$, $3.3840$.

\ResetCols{3}

\item[18.] $5.46\%$.

\item[19.] $5.57\%$.

\item[20.] $9.70\%$.

\end{Answers}


\begin{Answers}{Page94}

\item[1.] $2$.

\item[2.] $3$.

\end{Answers}
%% -----File: 169.png---Folio 163-------


\begin{Answers}[3]{Page96}

\item[1.] $2.24004099$.

\item[2.] $2.3593041$.

\item[3.] $1.997998$.

\end{Answers}


\begin{Answers}[4]{Page98}

\item[1.] $132° 20.7'$.

\item[2.] $157° 12'$.

\item[3.] $4.8425364$.

\item[4.] $3.1668771$.

\ResetCols{1}

\item[5,] \textbf{7.}~$15° 16\tfrac{1}{2}'$,
  $85° 56\tfrac{1}{2}'$,
  $212° 49'$,
  $225° 57'$.

\ResetCols{4}

\item[6.] $72° 17'$.

\item[8.] $5° 56\tfrac{1}{2}'$, $25° 18'$.

\item[9.] $2.5541949$.

\item[10.] $1.85718$.

\end{Answers}


\begin{Answers}{Page99}

\item[1.] $-1.04727± 1.13594 i$.

\item[2.] $-\frac{2}{7} ± \frac{1}{7}\sqrt{3}i$.

\ResetCols{3}

\item[3.] $-1±i$.

\item[4.] $1±i$, $1±2i$.

\item[5.] $2±i$, $±2i$.

\end{Answers}


\begin{Answers}{Page100}

\item[1.] $217° 12' 27.4'' = 3.790988$ radians.

\item[2.] $42° 20' 47\tfrac{1}{4}''$ doubled.

\ResetCols{3}

\item[3.] $133° 33.8'$.

\item[4.] $108° 36' 14''$.

\item[5.] $21.468212$.

\ResetCols{2}

\item[6.] Angle at center $47° 39' 13''$.

\item[7.] $49° 17' 36.5''$.

\ResetCols{1}

\item[8.] $1.4303\pi$, $2.4590\pi$, $3.4709\pi$;\quad $257° 27' 12.225''$ more exact than first.

\item[9.]  $x/\pi = 0.6625, 1.891, 2.930, 3.948, 4.959$.

\item[10.] (i) $0.327739$, $0.339224$, $1124.333037$. \\
  (ii) $0.250279$, $0.894609$, $1.127839$. \hfill\break
  % [** PP: If \\ above, LaTeX thinks next token is an optional argument]
  [Set $x = 1 + y$, $y = 1/z$ and solve by trigonometry.]

\ResetCols{2}

\item[11.] $3.597285$.

\item[12.] $10$, $1.371288$.

\ResetCols{2}

\item[13.] $0.326878$, $12.267305$.

\item[14.] $324° 16' 29.55''$.

\ResetCols{3}

\item[15.] $10$~yr.\ $4$~mo.\ $0$~days.

\item[\qquad16.] $6.074674$.

\item[17.] $6.13$\%.

\end{Answers}


\begin{Answers}[3]{Page102}

\item[1.] $x = 5$, $y = 6$.

\item[2.] $x = 2$, $y = 1$.

\item[3.] $x = a$, $y = 0$.

\end{Answers}


\begin{Answers}[1]{Page106}

\item[1.] $-a_2b_1c_3d_4 + a_2b_1c_4d_3 + a_2b_3c_1d_4
          - a_2b_3c_4d_1 - a_2b_4c_1d_3 + a_2b_4c_3d_1$.

\item[2.] $+$, $+$.
\end{Answers}


\begin{Answers}{Page112}

\item[3.]  $-3$.

\item[4.]  $-8$.

\end{Answers}


\begin{Answers}{Page115}

\item[1.] $x = -8$, $y = -7$, $z = 26$.

\item[2.] $x = 3$, $y = -5$, $z = 2$.

\ResetCols{2}

\item[3.] $x = 6$, $y = 3$, $z = 12$.

\item[4.] $x = 5$, $y = 4$, $z = 3$.

\ResetCols{2}

\item[5.] $x = -5$, $y = 3$, $z = 2$, $w = 1$.

\item[6.] $x = 1$, $y = z = 0$, $w = -1$.

\end{Answers}

%% -----File: 170.png---Folio 164-------


\begin{Answers}[1]{Page119}

\item[1.] Consistent: $y = -8/7 - 2x$, $z = 5/7$ (common line).

\ResetCols{2}

\item[2.] Inconsistent, case $(\beta)$.

\item[3.] Inconsistent (two parallel planes).

\ResetCols{1}

\item[4.] Consistent (single plane).

\item[5.] % [** PP: Changing ;s to .s]
    (i) $z = -x-y-2$.  \hfill
    (ii) inconsistent. \hfill
    (iii) $x = \dfrac{a - 1}{a + 2}$, $y = z =\dfrac{-3}{a + 2}$.

\item[6.]
    (i)  $x = \dfrac{(k-b)(c-k)}{(a-b)(c-a)}$.
    (ii) $y = \dfrac{k-c}{a-c}-x$,
         $z = \dfrac{a-k}{a-c}$ if $k=a$ or $k=c$, but
         inconsistent if $k$ is different from $a$ and~$c$.
    (iii) $z = 1 - x - y$ if $k=a$, inconsistent if $k\ne a$.

\end{Answers}


\begin{Answers}{Page120}

\item[1.] $r = 2$, $x:y:z = -4:1:1$.

\item[2.] $r = 2$, $x:y:z = -10:8:7$.

\ResetCols{2}

\item[3.] $r = 1$, two unknowns arbitrary.

\item[4.] $r = 3$, $x:y:z:w = 6:3:12:1$.

\ResetCols{1}

\item[5.] $r = 2$, $z = -\frac{11}{3} x - \frac{19}{3} y$,
                   $w = -\frac{10}{3} x - \frac{17}{3} y$.

\end{Answers}


\begin{Answers}[1]{Page121}

\item[1.]  Ranks of $A$ and~$B$ are~$2$;\quad $y = -8/7 - 2x, z = 5/7$.

\item[2.]  Consistent only when $a = -225/61$ and then $x = -\dfrac{5}{61}$, $y = \dfrac{3}{61}$, $z = \dfrac{45}{61}$.

\item[3.]  Rank of~$A$ is~$2$, rank of~$B$ is~$3$, inconsistent.

\item[4.]  $A$ and~$B$ of rank~$2$, $x = 3$, $y = 2$.

\end{Answers}


\begin{Answers}[1]{Page126}%, 127}

\item[1.] $x = \dfrac{k(b-k)(c-k)(k+b+c)}{a(b-a)(c-a)(a+b+c)}$, if $a$, $b$, $c$ are distinct and not zero and their
sum $\neq 0$.   If $a = b \neq c$, $ac \ne 0$, equations are inconsistent unless $k = 0$, $a$, $c$, or $-a-c$,
and then $y = \dfrac{k(c-k)}{a(c-a)} - x$, $z = \dfrac{k(k-a)}{c(c-a)}$, $x$~arbitrary.

\ResetCols{2}

\item[3.] $(a-b)(b-c)(c-a)$.

\item[4.] $(x-y)(y-z)(z-x)(xy + yz + zx)$.

\ResetCols{1}

\item[6.] $(a+b+c+d)(a+b-c-d)(a-b-c+d)(a-b+c-d)$.

\item[7.] $(a+b+c+d)(a-b+c-d)(a+bi-c-di)(a-bi-c+di)$.

\item[11.] $\ds x_j = (k_1-a_j)\dotsm(k_n-a_j)
  \div \prod\limits^n_{\substack{s=1 \\ s\neq j}} (a_s-a_j)$.

\item[12.] $x(ab + ac + bc) = -abc$.

\end{Answers}

%% -----File: 171.png---Folio 165-------


\begin{Answers}{Page133}%,134}

\item[1.] $\dfrac{p^4 - 3p^2q + 5pr + q^2}{r - pq}$.

\item[\qquad2.] $\dfrac{(5p^2-12q)(p^2-4q)}{4(p^3 - 4pq + 8r)} - \dfrac{13}{4}p$.

\ResetCols{2}

\item[5.] $2p^2-2q$.

\item[\qquad6.] $24r-p^3$.

\ResetCols{2}

\item[7.] $\dfrac{3p^2q^2 - 4p^3r - 4q^3 - 2pqr - 9r^2}{(r - pq)^2}$.

\item[\qquad8.] $27r^2 - 9pqr + 2q^3 = 0$.

\ResetCols{2}

\item[10.] $y = q+r/x$.

\item[11.] $x = \dfrac{1-py}{2+2y}$.

\ResetCols{2}

\item[12.] $y = \dfrac{4x^2 + px + q}{-3x-p}$, see~§112.

\item[13.] $\dfrac{2q(p^3 + 2pq - r)}{p^2q - pr + s} - 5p$, see Ex.~17.

\end{Answers}


\begin{Answers}{Page136}

\item[3.] $s_2 = p^2 - 2q$, \\
  $s_3 = p^3 - 3pq$, \\
  $s_4 = p^4 - 4p^2q + 2q^2$, \\
  $s_5 = p^5 - 5p^3q + 5pq^2$.

\item[4.] $s_{5n} = 5·3^n$, \\
  $s_k = 0$ if $k$ is not divisible by~$5$.

\item[5.] All zero.

\end{Answers}


\begin{Answers}[1]{Page140}

\item[2.] See Ex.~2, p.~136.

\item[3.] $\epsilon^j     \sqrt[5]{\frac{1}{2}c + \sqrt{Q}}
         + \epsilon^{5-j} \sqrt[5]{\frac{1}{2}c - \sqrt{Q}}$,\quad
   $Q = \frac{1}{4}c^2 - q^5$\hfill ($j=0$, $1$, $2$, $3$, $4$).

\item[4.] $\epsilon^j     \sqrt[7]{\frac{1}{2}c + \sqrt{Q}}
         + \epsilon^{7-j} \sqrt[7]{\frac{1}{2}c - \sqrt{Q}}$,\quad
   $Q = \frac{1}{4}c^2 - q^7$\hfill ($j=0$, $1,\dotsc, 6$).

\end{Answers}


\begin{Answers}{Page141}

\item[1.] $c_2^2 - 2c_1c_3 + 2c_4$.

\item[2.] $c_1^2c_2 - 2c_2^2 - c_1c_3 + 4c_4$.

\ResetCols{2}

\item[3.] $c_1c_3 - 4c_4$.

\item[4.] $c_3^2 - 2c_2c_4$.

\end{Answers}


\begin{Answers}{Page142}

\item[1.] $c_1c_3 - 4c_4$ if $n>3$, $c_1c_3$ if $n=3$.

\item[\qquad2.] $3c_1c_4 - c_2c_3 - 5c_5$.

\ResetCols{2}

\item[3.] $c_2c_4 - 4c_1c_5 + 9c_6$.

\item[\qquad4.] $c_3^2 - 2c_2c_4 + 2c_1c_5 - 2c_6$.

\ResetCols{1}

\item[5.] $y^3 - (p^2 - 2q)y^2 + (q^2 - 2pr)y - r^2 = 0$.

\ResetCols{2}

\item[6.] $y^3 - qy^2 + pry - r^2 = 0$.

\item[7.] $ry^3 + 2qy^2 + 4py + 8 = 0$.

\ResetCols{2}

\item[8.] Eliminate $x$ by $y = s_2 - x^2$.

\item[9.] Use $p^2 - q + px = y$.

\ResetCols{2}

\item[10.] $-4 + pr/s$.

\item[11.] $(rs - pr^2 + 2pqs)/s^2$.

\ResetCols{1}

\item[12.] (i) $s_a s_b s_c s_d - \Sigma s_a s_b s_{c+d}
   + 2\Sigma s_a s_{b+c+d} + \Sigma s_{a+b} s_{c+d} - 6s_{a+b+c+d}$. \\
  (ii) $\tfrac{1}{24}(s_a^4 - 6s_a^2s_{2a}
   + 8s_as_{3a} + 3s_{2a}^2 - 6s_{4a})$.

%% -----File: 172.png---Folio 166-------

\item[13.] (i)
    \[
    s_k = - \left|
    \begin{array}{cccccc}
       1      & 0      & 0      & \ldots & 0      & c_1 \\
       c_1    & 1      & 0      & \ldots & 0      & 2c_2   \\
       c_2    & c_1    & 1      & \ldots & 0      & 3c_3   \\
       c_3    & c_2    & c_1    & \ldots & 0      & 4c_4   \\
       \Dots{6} \\
      c_{k-1} &c_{k-2} &c_{k-3} & \ldots & c_1    & kc_k
     \end{array}\;\right|,\quad
   s_3 = -
    \begin{vmatrix}
     1   & 0   & c_1\\
     c_1 & 1   & 2c_2\\
     c_2 & c_1 & 3c_3
     \end{vmatrix},
     \]
    where all but the last term in the main diagonal is~$1$, and all terms above the
    diagonal are zero except those in the last column. If $k>n$, we must take
    $c_j =0 \quad (j>n)$.

    (ii)
    \[
     k!\,c_k = - \left|
     \begin{array}{cccccc}
       1       & 0       & 0       & \ldots & 0      & s_1    \\
       s_1     & 2       & 0       & \ldots & 0      & s_2    \\
       s_2     & s_1     & 3       & \ldots & 0      & s_3    \\
       \Dots{6} \\
     \ s_{k-1} & s_{k-2} & s_{k-3} & \ldots & s_1    & s_k
     \end{array}\;\right|,\quad
     3!\,c_3 = -
     \begin{vmatrix}
      1    & 0   & s_1\\
      s_1  & 2   & s_2\\
      s_2  & s_1 & s_3
     \end{vmatrix}.
    \]

\end{Answers}


\begin{Answers}[1]{Page152}

\item[1.] $y^2(16 - y^2)$;\quad $y=0$, $x=±3$;\quad $y=±4$, $x=+5$.

\item[2.] $(c-1)^2(y^2 - 25)(y^2 - 16)$. If $c\neq 1$, $y=±5$, $x=0$;\quad $y=±4$, $x=+3$.

\ResetCols{2}

\item[3.]
$\begin{vmatrix}
 1 & a & b \\
 2 & a & 0 \\
 0 & 2 & a
\end{vmatrix} = 4b-a^2$.

\item[4.] $2±3i$, $-2±i$, $±i$.
  $\vphantom{\begin{vmatrix}1\\ 1\\ 1\end{vmatrix}}$

\end{Answers}


\begin{Answers}[1]{Page153}%, 154}

\item[2.] $pqr - p^2s - r^2 = 0$, $x^2 + r/p$, $x^2 + px + ps/r$.

\ResetCols{2}

\item[3.] $1$, $3$, $1± i$.

\item[11.] See Ex.~15, p.~134.

\end{Answers}

%% -----File: 173.png---Folio 167-------

\PrintIndex

\iffalse

\title % INDEX

Numbers refer to pages.

\begin(theindex)

\item Abscissa, 55

\item Absolute value, 3

\item Amplitude, 3

\item Argument, 3

\item Arithmetical progression, 19

\indexspace

\item Bend point, 56, 64

\item Bézout's eliminant, 148

\item Budan's theorem, 83

\indexspace

\item Cardan's formulas, 46, 48

\item Complex number, 1
  \subitem geometrical representation, 3, 6, 7
  \subitem trigonometric form, 3

\item Compound interest, 13, 90, 100

\item Conjugate, 1

\item Continuity, 66, 157

\item Cube root, 5, 48
  \subitem of unity, 3, 4

\item Cubic equation, 32, 40, 45, 127, 134, 153--4
  \subitem graph of, 65
  \subitem number real roots, 48, 65, 79
  \subitem reduced, 45, 64--5
  \subitem trigonometric solution, 49

\indexspace

\item De Moivre's quintic, 140
  \subitem theorem, 5

\item Derivative, 57--60, 69, 83, 97, 135

\item Descartes' rule of signs, 71, 85

\item Determinants, 101--27, 146--53
  \subitem addition of columns, 113
  \subitem columns, 103
  \subitem complementary minors, 122
  \subitem diagonal term, 103
  \subitem elements, 103
  \subitem expansion, 109
  \subitem interchanges, 106, 107
  \subitem Laplace's development, 122--3, 147--9
  \subitem minors, 109, 116
  \subitem of Vandermonde, 108
  \subitem product of, 124
  \subitem rank, 116, 121

\item Determinants, removal of factor, 111
  \subitem rows, 103
  \subitem signs of terms, 103--6
  \subitem skew symmetric, 108
  \subitem sum of, 112

\item Discriminant, 152--3
  \subitem of cubic, 47, 65, 134
  \subitem of quadratic, 11, 12
  \subitem of quartic, 51, 81

\item Double root, 16 (see Discriminant)

\item Duplication of cube, 35

\indexspace

\item Elementary symmetric function, 128

\item Elimination, 143--154
  \subitem extraneous factor, 151

\item Equation for differences of roots, 154
  \subitem squares of differences, 134, 154

\item Euler's eliminant, 147

\indexspace

\item Factor theorem, 12

\item Factored form, 11, 15

\item Fundamental theorem of algebra, 17, 155--8

\indexspace

\item Geometrical construction, 29--44
  \subitem progression, 13, 19

\item Graphs, 55--70

\item Greatest common divisor, 61, 75

\indexspace

\item Horner's method, 86

\indexspace

\item Identical polynomials, 16

\item Identity, 11

\item Imaginary, 2
  \subitem roots, 19, 98

\item Inflexion, 62--64

\item Integral rational function, 12, 17, 20
  \subitem roots, 24--27

\item Interpolation, 93, 97

\item Interval, 78

\item Irreducible case, 48

\item Isolation of roots, 71

%% -----File: 174.png---Folio 168-------

\item Linear equations, system, 101--3, 114--21
  \subitem homogeneous, 119, 121

\item Linear factors, 11, 17

\item Lower limit to roots, 23

\indexspace

\item Matrix, 120
  \subitem augmented, 121

\item Maximum, 57

\item Minimum, 57, 157

\item Modulus, 3

\item Multiple roots, 16, 60, 82

\item Multiplicity of root, 16, 20, 61

\indexspace

\item Newton's identities, 136
  \subitem method of solution, 90--98

\item Number of roots, 16, 17, 48, 52, 69, 72--85
  \subitem of negative roots, 74

\indexspace

\item Order of radical, 32

\item Ordinate, 55

\indexspace

\item Plotting, 55

\item Polynomial, 12, 66
  \subitem sign of, 68

\item Primitive root of unity, 9

\item Product of roots, 18

\item Pure imaginary, 2

\indexspace

\item Quadratic equation, 11
  \subitem graphical solution, 29, 55
  \subitem sum of powers of roots, 139

\item Quadratic function a square, 12

\item Quartic equation, 50--54, 80--81

\item Quotient by synthetic division, 14

\indexspace

\item Rational roots, 27

\item Real equation, 12, 20

\item Reciprocal equation, 37, 44

\item Regula falsi, 93

\item Regular polygon, 8
  \subitem 7 sides, 35--36
  \subitem 9 sides, 35, 39
  \subitem 17 sides, 41--44
  \subitem \textit(n) sides, 44

\item Regular decagon, 39
  \subitem pentagon, 39

\item Relations between roots and coefficients, 17

\item Relatively prime, 9, 10

\item Remainder theorem, 12

\item Resolvent cubic, 50, 51

\item Resultant, 143--154

\item Rolle's theorem, 69

\item Root between \textit(a) and \textit(b), 67

\item Roots of unity, 8, 36, 39, 44, 136
  \subitem periods of, 40

\item Roots, \textit(n)th, 7

\indexspace

\item Sigma function, 128--142

\item Sign of polynomial, 68

\item Simple root, 16

\item Slope, 57, 59

\item Solution of numerical equations, 86--100

\item Specific gravity, 89

\item Square roots, 1, 30, 31, 96

\item Sturm's functions, 75--82

\item Sum of four squares, 126
  \subitem like powers of roots, 134--142
  \subitem products of roots, 18
  \subitem roots, 18

\item Surd roots in pairs, 20

\item Sylvester's eliminant, 145, 149, 150

\item Symbol \equiv, 11; \textit(f)(\textit(x)), 12; $$\textit(a)$$,23, 155; \textit(r)!, 59;
\textit(f^(k)(x)), 59; \Sigma, 128; \textit(s_k), 134; \textit(R)(\textit(j), \textit(g)),
144

\item Symmetric functions, 128--142
  in all but one root, 132--4

\item Synthetic division, 13, 86--95

\indexspace

\item Tangents, 60, 62

\item Taylor's theorem, 59

\item Transformed equation, 28, 86

\item Triple root, 16

\item Trisection of angle, 34, 40

\indexspace

\item Upper limit to roots, 21--23

\indexspace

\item Variation of sign, 71

\indexspace

\item Waring's formula, 136

\end(theindex)
\fi


% LICENSE

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