% ----------------------------------------------------------------------

\chapter{Classical algebra and geometry}
\label{alggeo}
\index{classical algebra}
\index{algebra!classical}
\index{classical geometry}
\index{geometry!classical}

\index{arithmetic}
\index{angle!congruent}
\index{parallel lines}
\index{line!parallel}
Probably every book must suppose something of its reader.  This book
supposes, or affects to suppose, little other than that its reader reads
English and has a strong aptitude for mathematics, but it does assume
that the reader has learned the simplest elements of classical
arithmetic, algebra and geometry from his youth:  that $1+1=2$; why
$(2)(3)=6$; what it means when a letter like~$x$ stands in the place of
an unspecified number; the technique to solve that $3x - 2 = 7$; how to
read the functional notation $f(x)$; which quantity a square root~$\sqrt
x$ is; what to make of the several congruent angles that attend a line
when the line intersects some parallels; and so on.  Even so, some basic
points of algebra and geometry seem worth touching briefly here.  The
book starts fast with these.

% ----------------------------------------------------------------------

\section{Basic arithmetic relationships}
\label{alggeo:222}
\index{arithmetic}

This section states some arithmetical rules.

\subsection[Commutivity, associativity, distributivity]{Commutivity,
  associativity, distributivity, identity and inversion}
\label{alggeo:222.1}
\index{commutivity}
\index{associativity}
\index{distributivity}
\index{identity!arithmetic}
\index{inversion!arithmetic}
\index{zero}
\index{$0$ (zero)}
\index{one}
\index{$1$ (one)}
\index{unity}

\index{rectangle}
\index{box}
\index{area}
\index{volume}
\index{identifying property}
\index{inversion!multiplicative}
\index{commutivity!multiplicative}
\index{associativity!multiplicative}
\index{identity!multiplicative}
\index{inversion!additive}
\index{commutivity!additive}
\index{associativity!additive}
\index{identity!additive}
\index{distributivity}
Table~\ref{alggeo:222:table} lists several arithmetical rules,\footnote{%
  \cite[\S~1.2]{Shilov}\cite[chapter~1]{Spiegel}
}
each of
which applies not only to real numbers but equally to the complex
numbers of \S~\ref{alggeo:225}.
\begin{table}
  \caption{Basic properties of arithmetic.}
  \label{alggeo:222:table}
  \bc
    \[
      \br{rcll}
        a+b&=&b+a&\mbox{Additive commutivity}\\
        a+(b+c)&=&(a+b)+c&\mbox{Additive associativity}\\
        a+0=0+a&=&a&\mbox{Additive identity}\\
        a+(-a)&=&0&\mbox{Additive inversion}\\
        ab&=&ba&\mbox{Multiplicative commutivity}\\
        (a)(bc)&=&(ab)(c)&\mbox{Multiplicative associativity}\\
        (a)(1)=(1)(a)&=&a&\mbox{Multiplicative identity}\\
        (a)(1/a)&=&1&\mbox{Multiplicative inversion}\\
        (a)(b+c)&=&ab+ac&\mbox{Distributivity}
      \er
    \]
  \ec
\end{table}
Most of the rules are appreciated at once if the meaning of the symbols
is understood.  In the case of multiplicative commutivity, one imagines
a rectangle with sides of lengths~$a$ and~$b$, then the same rectangle
turned on its side, as in Fig.~\ref{alggeo:222:20}: since the area of
the rectangle is the same in either case, and since the area is the
length times the width in either case (the area is more or less a matter
of counting the little squares), evidently multiplicative commutivity
holds.
\begin{figure}
  \caption{Multiplicative commutivity.}
  \label{alggeo:222:20}
  \bc
    {
      \nc\xax{-5}
      \nc\xbx{-2}
      \nc\xcx{ 5}
      \nc\xdx{ 2}
      \begin{pspicture}(\xax,\xbx)(\xcx,\xdx)
        \psset{dimen=middle}
        \small
        %\psframe[linewidth=0.5pt](\xax,\xbx)(\xcx,\xdx)
        \nc\xxg{2.0}
        \nc\xxu{1.0}
        \nc\xxv{0.6}
        \nc\xxs{0.4}
        \nc\xxp{6}
        \nc\xxq{4}
        \nc\xxb{ \psline[linewidth=0.5pt](-\xxu,0)(\xxu,0) }
        \nc\xxc{ \psline[linewidth=0.5pt](0,-\xxv)(0,\xxv) }
        \nc\xxa{
          \multips(0,-\xxv)(0,\xxs){\xxq}{\xxb}
          \multips(-\xxu,0)(\xxs,0){\xxp}{\xxc}
          \psframe[linewidth=2.0pt](-\xxu,-\xxv)(\xxu,\xxv)
        }
        \rput(-\xxg,0){
          \rput{ 0}(0,0){\xxa}
          \uput[d](0,-\xxv){$a$}
          \uput[l](-\xxu,0){$b$}
        }
        \rput( \xxg,0){
          \rput{90}(0,0){\xxa}
          \uput[d](0,-\xxu){$b$}
          \uput[l](-\xxv,0){$a$}
        }
      \end{pspicture}
    }
  \ec
\end{figure}
A similar argument validates multiplicative associativity, except that
here one computes the \emph{volume} of a three-dimensional rectangular
box, which box one turns various ways.

\index{inversion!multiplicative}
\index{multiplicative inversion}
\index{reciprocal}
Multiplicative inversion lacks an obvious interpretation when
$a=0$.  Loosely,
\[
  \frac{1}{0}=\infty.
\]
But since $3/0 = \infty$ also, surely either the zero or the infinity,
or both, somehow differ in the latter case.%
\footnote{%
  Weierstrass, Kronecker, Dedekind and Frege, among others, spent much
  of the nineteenth century intensely debating the implications of this
  very question.  The applied book you are reading however will treat
  the matter in a more relaxed manner than did these mathematical titans
  of yesteryear.
}

\index{C and C++}
Looking ahead in the book, we note that the multiplicative properties do
not always hold for more general linear transformations.  For example,
matrix multiplication is not commutative and vector cross-multiplication
is not associative.  Where associativity does not hold and parentheses
do not otherwise group, right-to-left association is notationally
implicit:%
\footnote{%
  The fine~C and~C++ programming languages unfortunately are stuck with
  the reverse order of association, along with division inharmoniously
  on the same level of syntactic precedence as multiplication.  Standard
  mathematical notation is more elegant:
  \[
    abc/uvw = \frac{(a)(bc)}{(u)(vw)}.
  \]
}$\mbox{}^,$%
\footnote{%
  The nonassociative \emph{cross product} $\ve B \times \ve C$ is
  introduced in \S~\ref{vector:220.30}.
}
\[
  \ve A \times \ve B \times \ve C = \ve A \times ( \ve B \times \ve C ).
\]
The sense of it is that the thing on the left ($\ve A\times\mbox{}$) \emph{operates}
on the thing on the right ($\ve B \times \ve C$).  (In the rare case
in which the question arises, you may wish to use parentheses
anyway.)

\subsection{Negative numbers}
\label{alggeo:222.20}
\label{negative number}

Consider that
\bqb
  (+a)(+b) &=& +ab, \\
  (+a)(-b) &=& -ab, \\
  (-a)(+b) &=& -ab, \\
  (-a)(-b) &=& +ab.
\eqb
The first three of the four equations probably are unsurprising, but the
last is interesting.  Why would a negative count~$-a$ of a negative
quantity~$-b$ come to a positive product~$+ab$?  To see why, consider
the progression
\settowidth\tla{$\ds+$}
\bqb
  &\vdots& \\
  (+3)(-b) &=& -3b, \\
  (+2)(-b) &=& -2b, \\
  (+1)(-b) &=& -1b, \\
  ( 0)(-b) &=& \makebox[\tla]{}0b, \\
  (-1)(-b) &=& +1b, \\
  (-2)(-b) &=& +2b, \\
  (-3)(-b) &=& +3b, \\
  &\vdots&
\eqb
The logic of arithmetic demands that the product of two negative numbers
be positive for this reason.

\subsection{Inequality}
\label{alggeo:222.6}
\index{inequality}

If%
\footnote{%
  Few readers attempting this book will need to be reminded that~$<$
  means ``is less than,'' that~$>$ means ``is greater than,'' or
  that~$\le$ and~$\ge$ respectively mean ``is less than or equal to''
  and ``is greater than or equal to.''
}
\[
  a < b,
\]
then necessarily
\[
a + x < b + x.
\]
However, the relationship between~$ua$ and~$ub$ depends on the sign of~$u$:
\bqb
  ua < ub && \mbox{if $u > 0$;} \\
  ua > ub && \mbox{if $u < 0$.}
\eqb
Also,
\settowidth\tla{and}
\bqb
  \frac{1}{a} > \frac{1}{b} &&
    \mbox{if $a > 0$ \makebox[\tla][c]{or} $b < 0$;} \\
  \frac{1}{a} < \frac{1}{b} && \mbox{if $a < 0$ and $b > 0$.}
\eqb
See further \S~\ref{alggeo:224power}.

\subsection{The change of variable}
\label{alggeo:222.40}
\index{variable!change of}
\index{change of variable}

\index{variable!assignment}
\index{assignment}
\index{$\leftarrow$}
\index{C and C++}
The mathematician often finds it convenient \emph{to change variables,}
introducing new symbols to stand in place of old.  For this we have the
\emph{change of variable} or \emph{assignment} notation%
\footnote{%
  There appears to exist no broadly established standard applied mathematical
  notation for the change of variable, other than the~$=$ equal sign,
  which regrettably does not always fill the role well.  One can indeed
  use the equal sign, but then what does the change of variable $k=k+1$
  mean?  It looks like an impossible assertion that~$k$ and $k+1$ were
  the same.  The notation $k\la k+1$ by contrast is unambiguous,
  incrementing~$k$ by one.  Nevertheless, admittedly, the latter
  notation has seen only scattered use in the literature.

  The~C and~C++ programming languages use~\texttt{==} for equality
  and~\texttt{=} for assignment (change of variable), as the reader may be
  aware.
}
\[
  Q \la P,
\]
which means, ``in place of~$P$, put~$Q$''; or, ``let~$Q$ now equal~$P$.''
For example, if $a^2 + b^2 = c^2$, then the change of variable $2\mu \la a$
yields the new form $(2\mu)^2 + b^2 = c^2$.

\index{variable!definition notation for}
\index{definition notation}
\index{$\equiv$}
Similar to the change of variable notation is the \emph{definition} notation
\[
  Q \equiv P.
\]
This means, ``let the new symbol~$Q$ represent~$P$.''%
\footnote{%
  One would never write, $k\equiv k+1$.  Even $k\la k+1$ can confuse
  readers inasmuch as it appears to imply two different values for the same
  symbol~$k$, but the latter notation is sometimes used anyway when
  new symbols are unwanted or because more precise alternatives (like
  $k_n = k_{n-1} + 1$) seem overwrought.
  Still, usually it is better to introduce a new symbol, as in $j\la k+1$.

  In some books,~$\equiv$ is printed
  %as~$\stackrel{\triangle}{=}$.
  as~$\triangleq$.
}

The two notations logically mean about the same thing.  Subjectively, $Q
\equiv P$ identifies a quantity~$P$ sufficiently interesting to be given
a permanent name~$Q$; whereas $Q\la P$ implies nothing especially
interesting about~$P$ or~$Q$, but just introduces a (perhaps temporary)
new symbol~$Q$ to ease the algebra.  These concepts grow clearer as
examples of the usage arise in the book.

% ----------------------------------------------------------------------

\section{Quadratics}
\label{alggeo:240}
\index{quadratics}

\index{factorization}
\index{root}
\index{constant expression}
\index{linear expression}
\index{quadratic expression}
\index{cubic expression}
\index{quartic expression}
\index{quintic expression}
\index{order}
Differences and sums of squares are conveniently factored as
\bq{alggeo:224:30}
  \index{square!sum or difference of}
  \begin{split}
    a^2-b^2 &= (a+b)(a-b), \\
    a^2+b^2 &= (a+ib)(a-ib), \\
    a^2-2ab+b^2 &= (a-b)^2, \\
    a^2+2ab+b^2 &= (a+b)^2
  \end{split}
\eq
(where~$i$ is the \emph{imaginary unit,} a number defined such that $i^2
= -1$, introduced in \S~\ref{alggeo:225} below).  Useful as these four
forms are, however, none of them can directly factor the more general
quadratic%
\footnote{%
  The adjective \emph{quadratic} refers to the algebra of expressions in
  which no term has greater than second order.  Examples of quadratic
  expressions include~$x^2$, $2x^2-7x+3$ and $x^2+2xy+y^2$.  By
  contrast, the expressions $x^3-1$ and~$5x^2y$ are \emph{cubic} not
  quadratic because they contain third-order terms.  First-order
  expressions like $x+1$ are \emph{linear;} zeroth-order expressions
  like~$3$ are \emph{constant.}  Expressions of fourth and fifth order
  are \emph{quartic} and \emph{quintic,} respectively.
  (If not already clear from the context, \emph{order} basically
  refers to the number of variables multiplied together in a term.  The
  term $5x^2y=5[x][x][y]$ is of third order, for instance.)
}
expression%
\footnote{%
  The~$\beta$ and~$\gamma$ are Greek letters, the full roster of which
  you can find in appendix~\ref{greek}\@.
}
\[
  z^2 - 2\beta z + \gamma^2.
\]
To factor this, we \emph{complete the square,} writing,
\bqb
  \index{square, completing the}
  \index{completing the square}
  z^2 - 2\beta z + \gamma^2
  &=& z^2 - 2\beta z + \gamma^2
    + (\beta^2 - \gamma^2) - (\beta^2 - \gamma^2) \xn\\
  &=& z^2 - 2\beta z + \beta^2 - (\beta^2 - \gamma^2) \xn\\
  &=& (z-\beta)^2 - (\beta^2 - \gamma^2).
\eqb
The expression evidently has \emph{roots}---that is, it has values
of~$z$ that null the expression---where
\[
  (z-\beta)^2 = (\beta^2 - \gamma^2),
\]
or in other words where%
\footnote{%
  The symbol~$\pm$ means~``$+$ or~$-$.''  In conjunction with this
  symbol, the alternate symbol~$\mp$ occasionally also appears,
  meaning~``$-$ or~$+$''---which is the same thing except that, where
  the two symbols appear together, $(\pm z) + (\mp z) = 0$.
}
\bq{alggeo:240:quad}
  \index{quadratic formula}
  \index{root extraction!from a quadratic polynomial}
  z = \beta \pm \sqrt{\beta^2 - \gamma^2}.
\eq
This suggests the factoring that
\bq{alggeo:240:50}
  z^2 - 2\beta z + \gamma^2
  = (z-z_1)(z-z_2),
\eq
where~$z_1$ and~$z_2$ are the two values of~$z$ given
by~(\ref{alggeo:240:quad}).%
\footnote{%
  It suggests it because the expressions on the left and right sides
  of~(\ref{alggeo:240:50}) are each quadratic and because the two
  expressions appear to share the same roots.
}
Substituting into the equation the values of~$z_1$ and~$z_2$ and
simplifying proves the suggestion correct.

It follows that the two solutions of the quadratic equation
\bq{alggeo:240:quadeq}
  z^2 = 2\beta z - \gamma^2
\eq
are those given by~(\ref{alggeo:240:quad}), which is called \emph{the
quadratic formula.}%
\footnote{%
  The form of the quadratic formula which usually appears in print
  is
  \[
    x = \frac{-b \pm \sqrt{b^2-4ac}}{2a},
  \]
  which solves the quadratic $ax^2 + bx + c = 0$.  However, this writer
  finds the form~(\ref{alggeo:240:quad}) easier to remember.  For
  example, by~(\ref{alggeo:240:quad}) in light
  of~(\ref{alggeo:240:quadeq}), the quadratic
  \[
    z^2 = 3z - 2
  \]
  has the solutions
  \[
    z = \frac{3}{2} \pm \sqrt{
      \left( \frac{3}{2} \right)^2 - 2
    } = 1\ \mbox{or}\ 2.
  \]
}
(\emph{Cubic} and \emph{quartic formulas} also exist to extract the
roots of polynomials respectively of third and fourth order, but they
are much harder.  See chapter~\ref{cubic} and its
Tables~\ref{cubic:cubic-table} and~\ref{cubic:quartic-table}.)

% ----------------------------------------------------------------------

\section{Integer and series notation}
\label{alggeo:227}
\index{series}
\index{series!notation for}
\index{integer}

\index{series!sum of}
\index{series!product of}
\index{sum}
\index{summation}
\index{product}
\index{multiplication}
\index{dummy variable}
\index{index}
\index{index!of summation}
\index{summation!index of}
\index{index!of multiplication}
\index{multiplication!index of}
\index{loop counter}
Sums and products of series arise so often in mathematical work that one
finds it convenient to define terse notations to express them.  The
summation notation
\[
  \sum_{k=a}^{b} f(k)
\]
means to let~$k$ equal each of the integers $a,a+1,a+2,\ldots,b$ in
turn, evaluating the function $f(k)$ at each~$k$ and then adding up the
several $f(k)$.  For example,%
\footnote{%
  What's that~$\mbox{0x56}$?  Answer:  it is a \emph{hexadecimal
  numeral} that represents the same number the familiar, decimal numeral~$86$
  represents.  It is an eighty-six.  The book's preface explains why the
  book gives such numbers in hexadecimal.  Appendix~\ref{hex} tells how
  to read the numerals, if you do not already know.
}
\[
  \sum_{k=3}^{6} k^2 = 3^2 + 4^2 + 5^2 + 6^2 = \mr{0x56}.
\]
The similar multiplication notation
\[
  \prod_{j=a}^{b} f(j)
\]
means \emph{to multiply} the several $f(j)$ rather than to add them.
The symbols~$\sum$ and~$\prod$ are respectively the Greek letters for~S
and~P, writ large, and may be regarded as standing for ``Sum'' and
``Product.''  The~$j$ or~$k$ is a \emph{dummy variable, index of
summation} or \emph{loop counter}---a variable with no independent
existence, used only to facilitate the addition or multiplication of the
series.%
\footnote{%
  Section~\ref{integ:240} speaks further of the dummy variable.
}
(Nothing prevents one from writing~$\prod_k$ rather than~$\prod_j$,
incidentally.  For a dummy variable, one can use any letter one likes.
However, the general habit of writing~$\sum_k$ and~$\prod_j$ proves
convenient at least in \S~\ref{drvtv:250.30} and chapter~\ref{taylor}, so we
start now.)

\index{\char 33}
\index{factorial}
The product shorthand
\bqb
  n!    &\equiv& \prod_{j=1}^{n} j,\\
  n!/m! &\equiv& \prod_{j=m+1}^{n} j,
\eqb
is very frequently used.  The notation~$n!$ is pronounced ``$n$
factorial.'' Regarding the notation $n!/m!$, this can of course be
regarded correctly as~$n!$ divided by~$m!$, but it usually proves more
amenable to regard the notation as a single unit.%
\footnote{%
  One reason among others for this is that factorials rapidly multiply
  to extremely large sizes, overflowing computer registers during
  numerical computation.  If you can avoid unnecessary multiplication by
  regarding $n!/m!$ as a single unit, it helps.
}

\index{series!multiplication order of}
Because multiplication in its more general sense as linear
transformation (\S~\ref{matrix:120.10})
is not always commutative, we specify that
\[
  \prod_{j=a}^{b} f(j) = [f(b)] [f(b-1)] [f(b-2)]
  \cdots [f(a+2)] [f(a+1)] [f(a)]
\]
rather than the reverse order of multiplication.%
\footnote{%
  The extant mathematical literature seems to lack an established
  standard on the order of multiplication implied by the ``$\prod$''
  symbol, but this is the order we will use in this book.
}
Multiplication proceeds from right to left.  In the event that the
reverse order of multiplication is needed, we will use the notation
\[
  \coprod_{j=a}^{b} f(j) = [f(a)] [f(a+1)] [f(a+2)]
  \cdots [f(b-2)] [f(b-1)] [f(b)].
\]

Note that for the sake of definitional consistency,
\[
  \begin{split}
    \sum _{k=N+1}^{N} f(k) &= 0 + \sum _{k=N+1}^{N} f(k) = 0, \\
    \prod_{j=N+1}^{N} f(j) &= (1) \prod_{j=N+1}^{N} f(j) = 1.
  \end{split}
\]
This means among other things that
\bq{alggeo:227:40}
  0!=1.
\eq

\index{belonging}
\index{membership}
\index{set}
\index{set notation}
\index{$\in$}
\index{$\in \mathbb Z$}
\index{Fortran}
Context tends to make the notation
\[
  N,j,k \in \mathbb Z
\]
unnecessary, but if used (as here and in \S~\ref{alggeo:224}) it states
explicitly that~$N$, $j$ and~$k$ are integers.  (The symbol~$\mathbb Z$
represents%
\footnote{%
  The letter~$\mathbb Z$ recalls the transitive and intransitive German
  verb \emph{z\"ahlen,} ``to count.''
}
the set of all integers: $\mathbb Z \equiv
\{\ldots,-5,-4,-3,-2,-1,0,1,2,3,4,5,\ldots\}$.  The symbol~$\in$ means
``belongs to'' or ``is a member of.''  Integers conventionally get
the letters%
\footnote{%
  Though Fortran is perhaps less widely used a computer programming
  language than it once was, it dominated applied-mathematical computer
  programming for decades, during which the standard way to declare an
  integral variable to the Fortran compiler was simply to let its name
  begin with~\texttt{I}, \texttt{J}, \texttt{K}, \texttt{L}, \texttt{M}
  or~\texttt{N}; so, this alphabetical convention is fairly
  well cemented in practice.
}%
~$i$, $j$, $k$, $m$, $n$, $M$ and~$N$ when available---though~$i$
sometimes is avoided because the same letter represents the
imaginary unit of \S~\ref{alggeo:225}.  Where additional letters are
needed~$\ell$, $p$ and~$q$, plus the capitals of these and the earlier
listed letters, can be pressed into service, occasionally joined
even by~$r$ and~$s$.  Greek letters are avoided, as---ironically in light
of the symbol~$\mathbb Z$---are the Roman letters~$x$, $y$ and~$z$.  Refer
to appendix~\ref{greek}\@.)

On first encounter, the~$\sum$ and~$\prod$ notation seems a bit
overwrought, whether or not the $\in\mathbb Z$ notation also is used.
Admittedly it is easier for the beginner to read
``$f(1)+f(2)+\cdots+f(N)$'' than ``$\sum_{k=1}^{N} f(k)$.'' However,
experience shows the latter notation to be extremely useful in
expressing more sophisticated mathematical ideas.  We will use such
notation extensively in this book.

% ----------------------------------------------------------------------

\section{The arithmetic series}
\label{alggeo:229}
\index{arithmetic series}
\index{series!arithmetic}

A simple yet useful application of the series sum of \S~\ref{alggeo:227}
is the \emph{arithmetic%
\footnote{%
  As an adjective, the word is pronounced ``arithMETic.''
}
series}
\[
  \sum_{k=a}^{b} k = a + (a+1) + (a+2) + \cdots + b.
\]
Pairing~$a$ with~$b$, then $a+1$ with $b-1$, then $a+2$ with $b-2$,
etc., the average of each pair is $[a+b]/2$; thus the average of the
entire series is $[a+b]/2$.  (The pairing may or may not leave an
unpaired element at the series midpoint $k=[a+b]/2$, but this changes
nothing.)  The series has $b-a+1$ terms.  Hence,
\bq{alggeo:229:10}
  \sum_{k=a}^{b} k = (b-a+1)\frac{a+b}{2}.
\eq

Success with this arithmetic series leads one to wonder about the
\emph{geometric series} $\sum_{k=0}^\infty z^k$.
Section~\ref{alggeo:228.30} addresses that point.

% ----------------------------------------------------------------------

\section{Powers and roots}
\label{alggeo:224}
\index{power}

This necessarily tedious section discusses powers and roots.
It offers no surprises.  Table~\ref{alggeo:224:t1} summarizes its
definitions and results.  Readers seeking more rewarding reading may
prefer just to glance at the table and then to skip directly to the start of
the next section.
\begin{table}
  \caption{Power properties and definitions.}
  \label{alggeo:224:t1}
  \index{power!properties of}
  \bqb
    \ds z^n &\equiv& \prod_{j=1}^{n}z, \ \ n \ge 0 \\
    z&=&(z^{1/n})^n=(z^n)^{1/n} \\
    \sqrt z &\equiv& z^{1/2} \\
    (uv)^a &=& u^a v^a \\
    z^{p/q} &=& (z^{1/q})^p=(z^p)^{1/q} \\
    z^{ab} &=& (z^a)^b = (z^b)^a \\
    z^{a+b} &=& z^az^b \\
    z^{a-b} &=& \frac{z^a}{z^b} \\
    z^{-b} &=& \frac{1}{z^b} \\
    j,n,p,q &\in& \mathbb Z
  \eqb
\end{table}

\index{integer}
In this section, the exponents
\[
  j,k,m,n,p,q,r,s \in \mathbb Z
\]
are integers,
but the exponents~$a$ and~$b$ are arbitrary real numbers.
(What is a \emph{real number?}  Section~\ref{alggeo:225} will explain;
but, meanwhile, you can think of a real number as just a number,
like~4, $-5/2$ or~$\sqrt{3}$.  There are also \emph{complex numbers}
like $7+i4$, for which this section's results---and, indeed, most of the
chapter's and book's results---turn out to be equally valid; but except
in eqn.~\ref{alggeo:224:30} we have not met this~$i$ yet, so you need not
worry about it for now.)

\subsection{Notation and integral powers}
\label{alggeo:224int}

\index{power!notation for}
\index{power!integral}
\index{exponent}
The power notation
\[
  z^n
\]
indicates the number~$z$, multiplied by itself~$n$ times.  More
formally, when the \emph{exponent}~$n$ is a nonnegative integer,%
\footnote{%
  % Strictly, the comma here should go inside the quotes, but in this
  % context that might confuse the reader, so we bend the style rule
  % here.
  The symbol~``$\equiv$'' means~``$=$'', but further usually
  indicates that the expression on its right serves to define the
  expression on its left.  Refer to \S~\ref{alggeo:222.40}.
}
\bq{alggeo:224:10}
  z^n \equiv \prod_{j=1}^{n}z.
\eq
For example,%
\footnote{\label{alggeo:224int:fn1}%
  The case~$0^0$ is interesting because it lacks an obvious
  interpretation.  The specific interpretation depends on the nature and
  meaning of the two zeros.  For interest, if $E \equiv 1/\ep$, then
  \[
    \lim_{\ep\rightarrow 0^+} \ep^\ep
    = \lim_{E\rightarrow \infty} \left(\frac{1}{E}\right)^{1/E}
    = \lim_{E\rightarrow \infty} E^{-1/E}
    = \lim_{E\rightarrow \infty} e^{-(\ln E)/E}
    = e^0 = 1.
  \]
}
\bqb
  z^3 &=& (z)(z)(z), \\
  z^2 &=& (z)(z), \\
  z^1 &=& z, \\
  z^0 &=& 1.
\eqb
Notice that in general,
\[
  z^{n-1} = \frac{z^n}{z}.
\]
This leads us to extend the definition to negative integral powers with
\bq{alggeo:224:0}
  z^{-n} = \frac{1}{z^n}.
\eq
From the foregoing it is plain that
\bq{alggeo:224:1}
  \begin{split}
    z^{m+n} &= z^mz^n, \\
    z^{m-n} &= \frac{z^m}{z^n},
  \end{split}
\eq
for any integral~$m$ and~$n$.
For similar reasons,
\bq{alggeo:224:2}
  z^{mn}=(z^m)^n=(z^n)^m.
\eq
On the other hand, from multiplicative associativity and commutivity,
\bq{alggeo:224:2a}
  (uv)^n = u^n v^n.
\eq

\subsection{Roots}
\label{alggeo:224roots}
\index{root}
\index{power!fractional}

Fractional powers are not something we have defined yet, so for
consistency with~(\ref{alggeo:224:2}) we let
\[
  (u^{1/n})^n = u.
\]
This has $u^{1/n}$ as the number which, raised to the $n$th power,
yields~$u$.  Setting
\[
  v = u^{1/n},
\]
it follows by successive steps that
\bqb
  v^n &=& u, \\
  (v^n)^{1/n} &=& u^{1/n}, \\
  (v^n)^{1/n} &=& v.
\eqb
Taking the~$u$ and~$v$ formulas together, then,
\bq{alggeo:224:11}
  (z^{1/n})^n = z = (z^n)^{1/n}
\eq
for any~$z$ and integral~$n$.

\index{square root}
The number $z^{1/n}$ is called the \emph{$n$th root} of~$z$---or in the
very common case that $n=2$, the \emph{square root} of~$z$, often
written as
\[
  \sqrt{z}.
\]
When~$z$ is real and nonnegative, the last notation usually implicitly
is taken to mean the real, nonnegative square root.  In any case, the
power and root operations mutually invert one another.

\index{power!real}
\index{real number!approximation of as a ratio of integers}
\index{ratio}
What about powers expressible neither as~$n$ nor as $1/n$, such as the~$3/2$
power?  If~$z$ and~$w$ are numbers related by
\[
  w^q=z,
\]
then
\[
  w^{pq}=z^p.
\]
Taking the $q$th root,
\[
  w^p=(z^p)^{1/q}.
\]
But $w=z^{1/q}$, so this has that
\[
  (z^{1/q})^p=(z^p)^{1/q},
\]
which is to say that it does not matter whether one applies the power or
the root first:  the result is the same.  Extending~(\ref{alggeo:224:2})
therefore, we define $z^{p/q}$ such that
\bq{alggeo:224:12}
  (z^{1/q})^p=z^{p/q}=(z^p)^{1/q}.
\eq
Since one can arbitrarily closely approximate any real number by a ratio
of integers,~(\ref{alggeo:224:12}) implies a power definition for all
real exponents.

Equation~(\ref{alggeo:224:12}) is this subsection's main result.
However, \S~\ref{alggeo:224pppp} will find it useful if we can also
show here that
\bq{alggeo:224:5}
  (z^{1/q})^{1/s}=z^{1/qs}=(z^{1/s})^{1/q}.
\eq
The proof is straightforward.  If
\[
  w \equiv z^{1/qs},
\]
then raising to the~$qs$ power yields that
\[
  (w^s)^q = z.
\]
Successively taking the $q$th and $s$th roots gives that
\[
  w = (z^{1/q})^{1/s}.
\]
By identical reasoning,
\[
  w = (z^{1/s})^{1/q}.
\]
But since $w \equiv z^{1/qs}$, the last two equations
imply~(\ref{alggeo:224:5}), as we have sought.

\subsection{Powers of products and powers of powers}
\label{alggeo:224pppp}
\index{power!of a product}
\index{power!of a power}

Per~(\ref{alggeo:224:2a}),
\[
  (uv)^p = u^p v^p.
\]
Raising this equation
to the $1/q$ power, we have that
\bqb
  (uv)^{p/q} &=& \left[u^p v^p\right]^{1/q} \\
  &=& \left[(u^p)^{q/q} (v^p)^{q/q}\right]^{1/q} \\
  &=& \left[(u^{p/q})^q (v^{p/q})^q\right]^{1/q} \\
  &=& \left[(u^{p/q}) (v^{p/q})\right]^{q/q} \\
  &=& u^{p/q} v^{p/q}.
\eqb
But as argued already in \S~\ref{alggeo:224roots}, some ratio $p/q$ of
integers exists to approach any real number~$a$ with arbitrary
precision, so the last means that
\bq{alggeo:224:4}
  (uv)^a = u^a v^a
\eq
for any real~$a$.

On the other hand, per~(\ref{alggeo:224:2}),
\[
  z^{pr} = (z^p)^r.
\]
Raising this equation to the $1/qs$ power and
applying~(\ref{alggeo:224:2}), (\ref{alggeo:224:12}) and
% bad break
(\ref{alggeo:224:5}) to reorder the powers, we have that
\[
  z^{(p/q)(r/s)} = (z^{p/q})^{r/s}.
\]
By identical reasoning,
\[
  z^{(p/q)(r/s)} = (z^{r/s})^{p/q}.
\]
Again as before, $p/q$ and $r/s$ approximate real numbers, so
\bq{alggeo:224:4a}
  (z^a)^b=z^{ab}=(z^b)^a
\eq
for any real~$a$ and~$b$.

\subsection{Sums of exponents}
\label{alggeo:224sdp}
\index{exponent!sum of}

With~(\ref{alggeo:224:1}), (\ref{alggeo:224:4})
and~(\ref{alggeo:224:4a}), one can reason that
\[
  z^{(p/q)+(r/s)} = (z^{ps+rq})^{1/qs} =
  (z^{ps}z^{rq})^{1/qs} = z^{p/q}z^{r/s},
\]
or in other words that
\bq{alggeo:224:6}
  z^{a+b} = z^az^b.
\eq
In the case that $a=-b$,
\[
  1 = z^{-b+b} = z^{-b}z^b,
\]
which implies that
\bq{alggeo:224:7}
  z^{-b} = \frac{1}{z^b}.
\eq
But then replacing $-b \la b$ in~(\ref{alggeo:224:6}) leads to
\[
  z^{a-b} = z^az^{-b},
\]
which according to~(\ref{alggeo:224:7}) is that
\bq{alggeo:224:8}
  z^{a-b} = \frac{z^a}{z^b}.
\eq

\subsection{Summary and remarks}
\label{alggeo:224summary}

Table~\ref{alggeo:224:t1} on page~\pageref{alggeo:224:t1} summarizes
the section's definitions and results.

Looking ahead to \S~\ref{alggeo:225}, \S~\ref{trig:280} and
chapter~\ref{cexp}, we observe that nothing in the foregoing analysis
requires the base variables~$z$, $w$, $u$ and~$v$ to be real numbers; if
complex (\S~\ref{alggeo:225}), the formulas remain valid.  Still, the
analysis does imply that the various exponents~$m$, $n$, $p/q$, $a$, $b$
and so on are real numbers.  We shall remove this restriction later,
purposely defining the action of a complex exponent to comport with the
results found here.  With such a definition the results apply not only
for all bases but also for all exponents, real or complex.

\subsection{Power-related inequality}
\label{alggeo:224power}
\index{inequality!power-related}
\index{power-related inequality}

If
\[
  0 < x < y
\]
are real numbers (for this subsection alone of the section does not
apply to the complex numbers of \S~\ref{alggeo:225}), then
inductively---since $0 < (x)(x) < (y)(y)$, $0 < (x)(x^2) < (y)(y^2)$,
and so on---we have that $0 < x^p < y^p$ for positive, real,
integral~$p$.  Moreover, the implication is reversible, so $0 < x^{1/q}
< y^{1/q}$, too.  Combining these with $a = p/q$ and recalling
\S~\ref{alggeo:222.6},
\bqb
  0 < x^a < y^a && \mbox{if $a > 0$,} \\
  0 < x^a = y^a && \mbox{if $a = 0$,} \\
  0 < y^a < x^a && \mbox{if $a < 0$.}
\eqb
Similar reasoning has further that%
\settowidth\tlb{$0 < x < 1$}%
\bqb
  1 < x < x^a && \mbox{if \makebox[\tlb][l]{$x > 1$} and $a > 1$,} \\
  1 < x^a < x && \mbox{if \makebox[\tlb][l]{$x > 1$} and $0 < a < 1$,} \\
  0 < x^a < x < 1 && \mbox{if \makebox[\tlb][l]{$0 < x < 1$} and $a > 1$,} \\
  0 < x < x^a < 1 && \mbox{if \makebox[\tlb][l]{$0 < x < 1$} and $0 < a < 1$,}
\eqb
among others.

% ----------------------------------------------------------------------

\section{Multiplying and dividing power series}
\label{alggeo:228}
\index{power series}
\index{polynomial}
\index{Laurent series}
\index{Laurent, Pierre Alphonse (1813--1854)}

\index{power series!with negative powers}
A \emph{power series}%
\footnote{\label{alggeo:228fn1}%
  Another name for the \emph{power series} is \emph{polynomial.}
  The word ``polynomial'' usually connotes a power series with a finite
  number of terms, but the two names in fact refer to essentially the
  same thing.

  Professional mathematicians use the terms more precisely.
  Equation~(\ref{alggeo:228:05}), they call---or at any rate some of
  them call---a ``power series'' only if
  $a_k=0$ for all $k<0$---in other words, technically, not if it
  includes negative powers of~$z$.  They call it a ``polynomial'' only
  if it is a ``power series'' with a finite number of terms.  They
  call~(\ref{alggeo:228:05}) in general a \emph{Laurent series.}

  The name ``Laurent series'' is a name we shall meet again in
  \S~\ref{taylor:380}.  In the meantime however we admit that the
  professionals have vaguely daunted us by adding to the name some
  pretty sophisticated connotations, to the point that we applied
  mathematicians (at least in the author's country) seem to feel somehow
  unlicensed actually to use the name.  We tend to
  call~(\ref{alggeo:228:05}) a ``power series with negative powers,'' or
  just ``a power series.''

  This book follows the last usage.  You however can
  call~(\ref{alggeo:228:05}) a \emph{Laurent series} if you prefer (and
  if you pronounce it right: ``lor-ON'').  That is after all exactly
  what it is.  Nevertheless, if you do use the name ``Laurent series,''
  be prepared for some people subjectively---for no particular
  reason---to expect you to establish complex radii of convergence, to
  sketch some annulus in the Argand plane, and/or to engage in other
  maybe unnecessary formalities.  If that is not what you seek, then you
  may find it better just to call the thing by the less lofty name of
  ``power series''---or better, if it has a finite number of terms, by
  the even humbler name of ``polynomial.''

  Semantics.
  % bad, bad break  (To remove the following \pagebreak indirectly causes a
  %   page with almost nothing on it to appear.  This however is subject
  %   to the content of the entire chapter up to and about this point.
  %   It is a temperamental bad break.)
  \pagebreak
  All these names mean about the same thing, but one is expected
  most
  carefully
  always to give the right name in the right
  place.  What a bother!  (Someone once told the writer that the
  Japanese language can give different names to the same object,
  depending on whether the \emph{speaker} is male or female.  The
  power-series terminology seems to share a spirit of that kin.)  If you
  seek just one word for the thing, the writer recommends that you call
  it a ``power series'' and then not worry too much about it until
  someone objects.  When someone does object, you can snow him with the
  big word ``Laurent series,'' instead.

  The experienced scientist or engineer may notice that the above
  vocabulary omits the name ``Taylor series.''  The vocabulary omits the
  name because that name fortunately remains unconfused in usage---it
  means quite specifically
  a power series without negative powers and tends to connote a
  representation of some particular function of interest---as we shall
  see in chapter~\ref{taylor}.%
}
is a weighted sum of integral powers:
\bq{alggeo:228:05}
  A(z) = \sum_{k=-\infty}^{\infty} a_kz^k,
\eq
in which the several weights~$a_k$ are arbitrary constants.  This
section discusses the multiplication and division of power series.

\subsection{Multiplying power series}
\label{alggeo:228.10}
\index{power series!multiplying}

Given two power series
\bq{alggeo:228:07}
  \begin{split}
    A(z) &= \sum_{k=-\infty}^{\infty} a_kz^k, \\
    B(z) &= \sum_{k=-\infty}^{\infty} b_kz^k,
  \end{split}
\eq
the product of the two series is evidently
\settoheight\tla{\scriptsize$k=-\infty\ j=-\infty$}
\bq{alggeo:228:prod}
  P(z) \equiv A(z)B(z) =
    \sum_{k=-\infty\rule{0em}{\tla}}^{\infty}
    \left[
    \left(\sum_{j=-\infty\rule{0em}{\tla}}^{\infty} a_jb_{k-j}\right)
    z^k \right].
\eq

\subsection{Dividing power series}
\label{alggeo:228.20}
\index{power series!division of}
\index{long division}

\index{remainder}
\index{quotient}
\index{dividend}
\index{divisor}
\index{numerator}
\index{denominator}
The quotient $Q(z) = B(z)/A(z)$ of two power series is a little
harder to calculate, and there are at least two ways to do it.
Section~\ref{alggeo:228.50} below will do it by matching coefficients,
but this subsection does it by long division.  For example,
\bqb
  \frac{2z^2 - 3z + 3}{z-2}
  &=& \frac{2z^2 - 4z}{z-2} + \frac{z+3}{z-2} = 2z + \frac{z+3}{z-2} \\
  &=& 2z + \frac{z-2}{z-2} + \frac{5}{z-2} = 2z + 1 + \frac{5}{z-2}.
\eqb
The strategy is to take the dividend%
\footnote{%
  If $Q(z)$ is a \emph{quotient} and $R(z)$ a \emph{remainder,} then
  $B(z)$ is a \emph{dividend} (or \emph{numerator}) and $A(z)$ a
  \emph{divisor} (or \emph{denominator}).  Such are the
  Latin-derived names of the parts of a long division.
}
$B(z)$ piece by piece, purposely choosing pieces easily divided by
$A(z)$.

If you feel that you understand the example, then that is really all
there is to it, and you can skip over several pages of thick notation
straight to \S~\ref{alggeo:228.30} if you like.  Indeed, to skip is
recommended to many or most readers---though, if you do skip, you might
nonetheless glance along the way at Tables~\ref{alggeo:228:tbl-down}
and~\ref{alggeo:228:tbl-up}, which summarize and formalize the procedure
the example has used and which also include the clever, alternate
procedure of \S~\ref{alggeo:228.50}.

Formally, we prepare the long division $B(z)/A(z)$ by writing,
\bq{alggeo:228:20}
  B(z) = A(z)Q_n(z) + R_n(z),
\eq
where $R_n(z)$ is a \emph{remainder} (being the part of $B[z]$
\emph{remaining} to be divided); and
\bq{alggeo:228:08}
  \begin{split}
    A(z) &= \sum_{k=-\infty}^{K} a_kz^k,\ \ a_K \neq 0,\\
    B(z) &= \sum_{k=-\infty}^{N} b_kz^k, \\
    R_N(z) &= B(z), \\
    Q_N(z) &= 0, \\
    R_n(z) &= \sum_{k=-\infty}^{n} r_{nk}z^k, \\
    Q_n(z) &= \sum_{k=n-K+1}^{N-K} q_{k}z^k,
  \end{split}
\eq
where~$K$ and~$N$ identify the greatest orders~$k$ of~$z^k$ present in
$A(z)$ and $B(z)$, respectively.

Well, that is a lot of symbology.  What does it mean?  The
key to understanding it lies in understanding~(\ref{alggeo:228:20}),
which is not one but several equations---one equation for each value
of~$n$, where $n=N,N-1,N-2,\ldots$\,.  The dividend $B(z)$ and the divisor
$A(z)$ stay the same from one~$n$ to the next, but the quotient $Q_n(z)$
and the remainder $R_n(z)$ change.  At start, $Q_N(z)=0$ while
$R_N(z)=B(z)$, but the thrust of the long division process is to build
$Q_n(z)$ up by wearing~$R_n(z)$ down.  The goal is to grind $R_n(z)$
away to nothing, to make it disappear as $n\ra-\infty$.

As in the example, we pursue the goal by choosing from $R_n(z)$ an
easily divisible piece containing the whole high-order term of~$R_n(z)$.
The piece we choose is $(r_{nn}/a_K)z^{n-K}A(z)$, which we add and
subtract from~(\ref{alggeo:228:20}) to obtain the form
\[
  B(z) = A(z)\left[Q_n(z)+\frac{r_{nn}}{a_K}z^{n-K}\right] +
  \left[ R_n(z) - \frac{r_{nn}}{a_K}z^{n-K}A(z) \right].
\]
Matching this equation against the desired iterate
\[
  B(z) = A(z)Q_{n-1}(z) + R_{n-1}(z)
\]
and observing from the definition of $Q_n(z)$ that
$Q_{n-1}(z) = Q_n(z) +
\linebreak % bad break
q_{n-K}z^{n-K}$, we find that
\bq{alggeo:228:25}
  \begin{split}
  q_{n-K} &= \frac{r_{nn}}{a_K}, \\
  R_{n-1}(z) &= R_n(z) - q_{n-K} z^{n-K}A(z),
  \end{split}
\eq
where no term remains in $R_{n-1}(z)$ higher than a~$z^{n-1}$
term.

To begin the actual long division, we initialize
\[
  R_N(z) = B(z),
\]
for which~(\ref{alggeo:228:20}) is trivially true if
$Q_N(z)=0$.  Then we iterate per~(\ref{alggeo:228:25}) as many times as
desired.  If an infinite number of times, then so long as $R_n(z)$ tends
to vanish as $n \ra -\infty$, it follows from~(\ref{alggeo:228:20}) that
\bq{alggeo:228:30}
  \frac{B(z)}{A(z)} = Q_{-\infty}(z).
\eq
Iterating only a finite number of times leaves a remainder,
\bq{alggeo:228:35}
  \frac{B(z)}{A(z)} = Q_{n}(z) + \frac{R_{n}(z)}{A(z)},
\eq
except that it may happen that $R_{n}(z) = 0$ for sufficiently small~$n$.

Table~\ref{alggeo:228:tbl-down} summarizes the long-division
procedure.%
\footnote{\cite[\S~3.2]{SRW}}
\begin{table}
  \caption{Dividing power series through successively smaller powers.}
  \label{alggeo:228:tbl-down}
  \index{long division!procedure for}
  \bqb
    B(z) &=& A(z)Q_n(z) + R_n(z) \\
    A(z) &=& \sum_{k=-\infty}^{K} a_kz^k,\ \ a_K \neq 0 \\
    B(z) &=& \sum_{k=-\infty}^{N} b_kz^k \\
    R_N(z) &=& B(z) \\
    Q_N(z) &=& 0 \\
    R_n(z) &=& \sum_{k=-\infty}^{n} r_{nk}z^k \\
    Q_n(z) &=& \sum_{k=n-K+1}^{N-K} q_{k}z^k \\
    q_{n-K} &=& \frac{r_{nn}}{a_K}
      = \frac{1}{a_K} \left(
        b_n - \sum_{k=n-K+1\rule{0em}{\tla}}^{N-K} a_{n-k} q_k
      \right) \\
    R_{n-1}(z) &=& R_n(z) - q_{n-K} z^{n-K}A(z) \\
    \frac{B(z)}{A(z)} &=& Q_{-\infty}(z)
  \eqb
\end{table}
In its $q_{n-K}$ equation, the table includes also the result of
\S~\ref{alggeo:228.50} below.

\index{residual order}
\index{order!residual}
The foregoing algebra is probably punishing enough; but if not, then
one can further observe in light of Table~\ref{alggeo:228:tbl-down} that
if%
\footnote{%
  The notations~$K_o$, $a_k$ and~$z^k$ are usually pronounced,
  respectively, as ``$K$ naught,'' ``$a$ sub $k$'' and ``$z$ to
  the~$k$'' (or, more fully, ``$z$ to the $k$th power'')---at least in
  the author's country.
}
\bqb
  A(z) &=& \sum_{k=K_o}^{K} a_kz^k, \\
  B(z) &=& \sum_{k=N_o}^{N} b_kz^k,
\eqb
then
\bq{alggeo:228:37}
  R_n(z) = \sum_{k=n-(K-K_o)+1}^n r_{nk} z^k
    \ \ \mbox{for all $n < N_o+(K-K_o)$.}
\eq
That is, the remainder has residual order one less than the divisor has.
The reason for this, of course, is that we have strategically planned
the long-division iteration precisely to cause the divisor's leading
term to cancel the remainder's leading term at each step.%
\footnote{%
  If a more formal demonstration of~(\ref{alggeo:228:37}) is wanted,
  then consider per~(\ref{alggeo:228:25}) that
  \[
    R_{m-1}(z) = R_m(z) - \frac{r_{mm}}{a_K} z^{m-K}A(z).
  \]
  If the least-order term of $R_m(z)$ is a~$z^{N_o}$ term (as clearly is
  the case at least for the initial remainder $R_N[z] = B[z]$), then
  according to the equation so also must the least-order term of
  $R_{m-1}(z)$ be a~$z^{N_o}$ term, unless an even lower-order term is
  contributed by the product $z^{m-K}A(z)$.  But that very product's
  term of least order is a $z^{m-(K-K_o)}$ term.  Under these
  conditions, evidently the least-order term of $R_{m-1}(z)$ is a
  $z^{m-(K-K_o)}$ term when $m-(K-K_o) \le N_o$; otherwise
  a~$z^{N_o}$ term.  This is better stated after the change of variable
  $n+1\la m$: the least-order term of $R_n(z)$ is a $z^{n-(K-K_o)+1}$
  term when $n < N_o + (K-K_o)$; otherwise a~$z^{N_o}$ term.

  The greatest-order term of $R_n(z)$ is by definition a~$z^n$ term.
  So, in summary, when~$n < N_o + (K-K_o)$, the terms of $R_n(z)$ run
  from $z^{n-(K-K_o)+1}$ through~$z^n$, which is exactly the
  claim~(\ref{alggeo:228:37}) makes.
}
(If not clear from the context, a polynomial's \emph{residual order} is
the difference between the least and greatest orders of its several
terms.  For example, the residual order of $9x^5-7x^4+6x^3$ is two
because $5-3=2$---or, if you prefer, because
$9x^5-7x^4+6x^3=[x^3][9x^2-7x+6]$, where $9x^2-7x+6$ is of second
order.\footnote{%
  But what of $0x^5-7x^4+6x^3$ with its leading null coefficient?  Is
  \emph{this} polynomial's residual order also two?

  Answer:  that depends on what you mean.  The strictly semantic
  question of what a mere phrase ought to signify is not always very
  interesting.  After all, an infinite number of practically irrelevant
  semantic distinctions \emph{could} be drawn.  The applied
  mathematician lacks the time.

  % This is a somewhat interesting paragraph.  Whether it belongs in the
  % book at all remains to be seen, but it does not belong here.
  %Remember: in the late nineteenth and early twentieth centuries, the
  %mathematics profession made an heroic attempt to establish a complete
  %body of such definitions.  Though honorable, that attempt failed
  %(\S~\ref{intro:284}).  This does not mean that you and I should now
  %turn our backs on all mathematical rigor, but it does suggest that we
  %might be wise to approach some of the more abstract demands of
  %\emph{professional} mathematical rigor with a certain measure of
  %applied practicality.

  Anyway, whatever semantics might eventually be settled upon, at
  least~(\ref{alggeo:228:37}) and Table~\ref{alggeo:228:tbl-down} remain
  unambiguous.%
})

The long-division procedure of Table~\ref{alggeo:228:tbl-down} extends
the quotient $Q_n(z)$ through successively smaller powers of~$z$.  Often,
however, one prefers to extend the quotient through successively
\emph{larger} powers of~$z$, where a~$z^K$ term is $A(z)$'s term of
least rather than greatest order.  In this case, the long division goes
by the complementary rules of
Table~\ref{alggeo:228:tbl-up}.
\begin{table}
  \caption{Dividing power series through successively larger powers.}
  \label{alggeo:228:tbl-up}
  \index{long division!procedure for}
  \bqb
    B(z) &=& A(z)Q_n(z) + R_n(z) \\
    A(z) &=& \sum_{k=K}^{\infty} a_kz^k,\ \ a_K \neq 0 \\
    B(z) &=& \sum_{k=N}^{\infty} b_kz^k \\
    R_N(z) &=& B(z) \\
    Q_N(z) &=& 0 \\
    R_n(z) &=& \sum_{k=n}^{\infty} r_{nk}z^k \\
    Q_n(z) &=& \sum_{k=N-K}^{n-K-1} q_{k}z^k \\
    q_{n-K} &=& \frac{r_{nn}}{a_K}
      = \frac{1}{a_K} \left(
        b_n - \sum_{k=N-K\rule{0em}{\tla}}^{n-K-1} a_{n-k} q_k
      \right) \\
    R_{n+1}(z) &=& R_n(z) - q_{n-K} z^{n-K}A(z) \\
    \frac{B(z)}{A(z)} &=& Q_{\infty}(z)
  \eqb
\end{table}

\subsection{Dividing power series by matching coefficients}
\label{alggeo:228.50}
\index{power series!division of by matching coefficients}
\index{division!by matching coefficients}
\index{coefficient!matching of}
\index{matching coefficients}

\index{quotient}
There is another, sometimes quicker way to divide power series than by
the long division of \S~\ref{alggeo:228.20}.  One can divide them by
matching coefficients.%
\footnote{\cite{Kohler-lecture}\cite[\S~2.5]{Fisher}}
If
\bq{alggeo:228:60}
  Q_\infty(z) = \frac{B(z)}{A(z)},
\eq
where
\bqb
  A(z) &=& \sum_{k=K}^\infty a_kz^k, \ \ a_K \neq 0, \\
  B(z) &=& \sum_{k=N}^\infty b_kz^k
\eqb
are known and
\[
  Q_\infty(z) = \sum_{k=N-K}^\infty q_kz^k
\]
is to be calculated, then one can rearrange~(\ref{alggeo:228:60}) as
that
\[
  A(z)Q_\infty(z) = B(z).
\]
Expanding the rearranged equation's left side according
to~(\ref{alggeo:228:prod}) and changing indices suitably on both sides,
\settoheight\tla{\scriptsize $k$}
\[
  \sum_{n=N\rule{0em}{\tla}}^{\infty} \left[
    \left( \sum_{k=N-K\rule{0em}{\tla}}^{n-K} a_{n-k} q_k \right)
    z^n
  \right]
  = \sum_{n=N}^\infty b_nz^n.
\]
But for this to hold for all~$z$, the coefficients must match for
each~$n$:
\[
  \sum_{k=N-K\rule{0em}{\tla}}^{n-K}
  a_{n-k} q_k
  = b_n,
  \ \ n \ge N.
\]
Transferring all terms but $a_Kq_{n-K}$ to the equation's right side and
dividing by~$a_K$, we have that
\bq{alggeo:228:65}
  q_{n-K} = \frac{1}{a_K} \left(
    b_n - \sum_{k=N-K\rule{0em}{\tla}}^{n-K-1} a_{n-k} q_k
  \right),
  \ \ n \ge N.
\eq
Equation~(\ref{alggeo:228:65}) computes the coefficients of $Q(z)$, each
coefficient depending not on any remainder but directly on the
coefficients earlier computed.

The coefficient-matching technique of this subsection is easily adapted
to the division of series in decreasing, rather than increasing, powers
of~$z$.  Tables~\ref{alggeo:228:tbl-down} and~\ref{alggeo:228:tbl-up}
incorporate the technique both ways.

Admittedly, the fact that~(\ref{alggeo:228:65}) yields a sequence of
coefficients does not necessarily mean that the resulting power series
$Q_\infty(z)$ converges to some definite value over a given domain.
Consider for instance~(\ref{alggeo:228:45}), which diverges when%
\footnote{%
  See footnote~\ref{alggeo:228:fn1}.
}
$\left|z\right| > 1$, even though all its coefficients are known.
At least~(\ref{alggeo:228:65}) is correct when $Q_\infty(z)$
does converge.  Even when $Q_\infty(z)$ as such does not converge,
however, often what interest us are only the series' first several terms
\[
  Q_n(z) = \sum_{k=N-K}^{n-K-1} q_kz^k.
\]
In this case, in light of~(\ref{alggeo:228:20}),
\bq{alggeo:228:68}
  Q_\infty(z) = \frac{B(z)}{A(z)} = Q_n(z) + \frac{R_n(z)}{A(z)}
\eq
and convergence is not an issue.  Solving~(\ref{alggeo:228:68})
or~(\ref{alggeo:228:20}) for $R_n(z)$,
\bq{alggeo:228:70}
  R_n(z) = B(z) - A(z)Q_n(z).
\eq

\subsection
[Common quotients and the geometric series]
{Common power-series quotients and the geometric series}
\label{alggeo:228.30}
\index{power series!common quotients of}

Frequently encountered power-series quotients, calculated by the long
division of \S~\ref{alggeo:228.20}, computed by the coefficient
matching of \S~\ref{alggeo:228.50}, and/or verified by multiplying,
include%
\footnote{\label{alggeo:228:fn1}%
  The notation~$\left|z\right|$ represents the magnitude of~$z$.  For
  example, $\left|5\right| = 5$ and $\left|8\right| = 8$, but also
  $\left|-5\right| = 5$ and $\left|-8\right| = 8$.
}
\bq{alggeo:228:40}
  \frac{1}{1\pm z} =
    \begin{cases}
      \ds\sum_{k=0}^\infty (\mp z)^k, &\left|z\right| < 1; \\
      \ds-\!\!\!\sum_{k=-\infty}^{-1}\!\!\!(\mp z)^k, &\left|z\right| > 1.
    \end{cases}
\eq

\index{geometric series}
\index{series!geometric}
Equation~(\ref{alggeo:228:40}) almost incidentally answers a question
which has arisen in \S~\ref{alggeo:229} and which often arises in
practice: to what total does the infinite \emph{geometric series}
$\sum_{k=0}^{\infty} z^k$, $\left|z\right|<1$, sum?  Answer: it sums exactly to
$1/(1-z)$.  However, there is a simpler, more aesthetic, more
instructive way to
demonstrate the same thing, as follows.  Let
\[
  S_0 \equiv \sum_{k=0}^\infty z^{k}, \ \ \left|z\right| < 1.
\]
Multiplying by~$z$ yields that
\[
  zS_0 = \sum_{k=1}^\infty z^{k}.
\]
Subtracting the latter equation from the former leaves that
\[
  (1-z)S_0 = 1,
\]
which, after dividing by $1-z$, implies that
\bq{alggeo:228:45}
  S_0 \equiv \sum_{k=0}^\infty z^{k} = \frac{1}{1-z}, \ \ \left|z\right| < 1,
\eq
as was to be demonstrated.

\subsection{Variations on the geometric series}
\label{alggeo:228.40}
\index{geometric series!variations on}
\index{series!geometric, variations on}
\index{power series!extending the technique of}

\index{Planck, Max (1858--1947)}
\index{blackbody radiation}
Besides being more aesthetic than the long division
of \S~\ref{alggeo:228.20}, the difference technique of
\S~\ref{alggeo:228.30} permits one to extend the basic geometric series
in several ways.  For instance, one can compute the sum
\[
  S_1 \equiv \sum_{k=0}^\infty k z^{k}, \ \ \left|z\right| < 1
\]
(which arises in, among others, Planck's quantum blackbody radiation
calculation%
\footnote{\cite{McMahon}}%
) as follows.  Multiply the unknown~$S_1$ by~$z$,
producing
\[
  zS_1 = \sum_{k=0}^\infty k z^{k+1} = \sum_{k=1}^\infty (k-1) z^k.
\]
Subtract~$zS_1$ from~$S_1$, leaving
\[
  (1-z)S_1 = \sum_{k=0}^\infty k z^{k} - \sum_{k=1}^\infty (k-1)z^k
  = \sum_{k=1}^\infty z^k = z \sum_{k=0}^\infty z^k = \frac{z}{1-z},
\]
where we have used~(\ref{alggeo:228:45}) to collapse the last sum.
Dividing by $1-z$,
\bq{alggeo:228:50}
  S_1 \equiv \sum_{k=0}^\infty k z^{k} = \frac{z}{(1-z)^2},
  \ \ \left|z\right| < 1,
\eq
which was to be found.

Further series of the kind, such as $\sum_k k^2 z^k$, can be calculated
in like manner as the need for them arises.
Introducing the derivative, though, chapter~\ref{drvtv} does it better:%
\footnote{%
  It does not really matter, but you can regard~$k^n$ to be unity---that
  is, $k^n=1$---when $n=0$ and $k=0$, though $n=0$ technically lies
  outside the domain of~(\ref{alggeo:228:80}) as expressed.  See also
  footnote~\ref{alggeo:224int:fn1}.
}
\bq{alggeo:228:80}
  S_n \equiv \sum_{k=0}^\infty k^n z^{k} = z\frac{dS_{n-1}}{dz},
    \ \ n \in \mathbb Z,\ n > 0;
\eq
except that you must first read chapter~\ref{drvtv} or otherwise know about
derivatives to understand this.%
\footnote{\label{alggeo:228:fn2}%
  This of course is a forward reference.
  Logically,~(\ref{alggeo:228:80}) belongs in or after chapter~\ref{drvtv},
  but none of the earlier chapters use it, so it is kept here with the
  rest of the geometric-series math.  See chapter~\ref{drvtv}'s
  footnote~\ref{drvtv:250:fn1}.%
}
See also \S~\ref{taylor:314}.

% ----------------------------------------------------------------------

% bad break
\section[Constants and variables]{Indeterminate constants, independent
vari- \linebreak ables and dependent variables}
\label{alggeo:231}
\index{constant}
\index{constant, indeterminate}
\index{variable}
\index{variable!independent}
\index{variable!dependent}

\index{sound}
Mathematical models use \emph{indeterminate constants,}
\emph{independent variables} and \emph{dependent variables.}  The three
are best illustrated by example as follows.  Consider the time~$t$ a
sound needs to travel from its source to a distant listener:
\[
  t=\frac{\Delta r}{v_\mr{sound}},
\]
where~$\Delta r$ is the distance from source to listener
and~$v_\mr{sound}$ is the speed of sound.  Here,~$v_\mr{sound}$ is an
indeterminate constant (given particular atmospheric conditions, it does
not vary),~$\Delta r$ is an independent variable, and~$t$ is a dependent
variable.  The model gives~$t$ as a function of~$\Delta r$; so, if you
tell the model how far the listener sits from the sound source, then the
model returns the time the sound needs to propagate from one to the
other.  Regarding the third quantity, the indeterminate
constant~$v_\mr{sound}$, one conceives of this as having a definite,
fixed value; yet, oddly, notwithstanding that the value is (or is
thought of as) fixed, the model's abstract validity may not depend on
whether one actually knows what the value is (if I tell you that sound
goes at~$\mbox{350\:\mr{m/s}}$, but later you find out that the real
figure is~$\mbox{331\:\mr{m/s}}$, this probably does not ruin the
theoretical part of your analysis; you may only have to recalculate
numerically).  Knowing the value is not the point.  The point is that
conceptually there pre\"exists some correct figure for the indeterminate
constant; that sound goes at some constant speed---whatever it is---and
that one can calculate the delay in terms of this.%
\footnote{%
  Besides the quantities themselves, there is also the manner in which,
  or pattern by which, the quantities relate to one another.  The
  philosophy that attends this distinction lies mostly beyond the book's
  scope, but it still seems worth a footnote to quote Bertrand Russell
  (1872--1970) on the subject:
  \begin{quote}
    Given any propositional concept, or any unity \ldots, which may in
    the limit be simple, its constituents are in general of two sorts:
    (1)~those which may be replaced by anything else whatever without
    destroying the unity of the whole; (2)~those which have not this
    property.  Thus in ``the death of Caesar,'' anything else may be
    substituted for Caesar, but a proper name must not be substituted
    for \emph{death,} and hardly anything can be substituted for
    \emph{of.}  Of the unity in question, the former class of
    constituents will be called \emph{terms,} the latter
    \emph{concepts\mdots} [Emphases in the original.]\cite[appendix~A,
    \S~482]{Russell:1903}
  \end{quote}
  Sections~\ref{alggeo:250} and~\ref{alggeo:225}, and a few later ones,
  glance upon the matter.
}

\index{concert hall}
Though the three kinds of quantity remain in some sense distinct, still,
which particular quantity one regards as an indeterminate constant, as
an independent variable, or as a dependent variable may depend less upon
any property of the quantity itself---or of the thing the quantity
quantifies---than upon the mathematician's point of view.  Moreover, the
mathematician's point of view can waver.  The same model in the example
would remain valid if atmospheric conditions were
changing~($v_\mr{sound}$ would then be an independent variable) or if
the model were used in designing a musical concert hall%
\footnote{%
  As a child, were you ever let to read one of those trendy, second-rate
  arithmetic textbooks that had you calculate such as the quantity of
  air in an astronaut's round helmet?  One could have calculated the
  quantity of water in a kitchen's mixing bowl just as well, but
  astronauts' helmets are so much more interesting than bowls, you see.
  (Whether you will have endured the condescending frivolity
  specifically of the ersatz astronaut's textbook depends largely on
  when and where you were raised.  Trends will come and go, but maybe
  you will have met another year's version of the same kind of thing.)

  So, what of the concert hall?
  The chance that the typical reader will ever specify the dimensions of
  a real musical concert hall is of course vanishingly small.  However,
  it is the idea of the example that matters here, because the chance
  that the typical reader will ever specify \emph{something}
  technical is quite large.  Although sophisticated models with many
  factors and terms do indeed play a large role in engineering, the
  great majority of practical engineering calculations---for quick,
  day-to-day decisions where small sums of money and negligible risk to
  life are at stake, or for pro\"emial or exploratory analysis---are
  done with models hardly more sophisticated than the one shown here.
  So, maybe the concert-hall example is not so unreasonable, after all.
}
to suffer a maximum acceptable sound time lag from the stage to the
hall's back row ($t$ would then be an independent variable;~$\Delta r$,
dependent).  Occasionally one goes so far as deliberately to shift one's
point of view in mid-analysis---now regarding as an independent
variable, for instance, that which one a moment ago had regarded as an
indeterminate constant (a typical case of such a shift arising in the
solution of differential equations by the method of unknown
coefficients, \S~\ref{inttx:240}).

It matters which symbol represents which of the three kinds of quantity
in part because, in calculus, one analyzes how change in independent
variables affects dependent variables as indeterminate constants remain
fixed.

(Section~\ref{alggeo:227} has introduced the dummy variable, which the
present section's threefold taxonomy seems to exclude.  However, in fact,
most dummy variables are just independent variables---a few are
dependent variables---whose scope is restricted to a particular
expression.  Such a dummy variable does not seem very ``independent,''
of course; but its dependence is on the operator controlling the
expression, not on some other variable within the expression.  Within
the expression, the dummy variable fills the role of an independent
variable; without, it fills no role because logically it does not exist
there.  Refer to \S\S~\ref{alggeo:227} and~\ref{integ:240}.)

% ----------------------------------------------------------------------

\section{Exponentials and logarithms}
\label{alggeo:230}
\index{exponential}

\index{exponent}
In \S~\ref{alggeo:224} we have considered the power operation~$z^a,$ where
(in \S~\ref{alggeo:231}'s language) the independent variable~$z$ is the base
and the indeterminate constant~$a$ is the exponent.  There is another way to
view the power operation, however.  One can view it as the \emph{exponential}
operation
\[
  a^z,
\]
where the variable~$z$ is the exponent and the constant~$a$ is the
base.

\subsection{The logarithm}
\label{alggeo:230.10}
\index{logarithm}

The exponential operation follows the same laws the power operation
follows; but, because the variable of interest is now the exponent
rather than the base, the inverse operation is not the root but rather
the \emph{logarithm:}
\bq{alggeo:230:logdef}
  \log_a (a^z) = z.
\eq
The logarithm $\log_a w$ answers the question, ``To what power must I
raise~$a$ to get~$w$?''

Raising~$a$ to the power of the last equation, we have that
\[
  a^{\log_a (a^z)} = a^z.
\]
With the change of variable $w \la a^z$, this is that
\bq{alggeo:230:logdef2}
  a^{\log_a w} = w.
\eq
Thus, the exponential and logarithmic operations mutually invert one
another.

\subsection{Properties of the logarithm}
\label{alggeo:230.20}
\index{logarithm!properties of}

The basic properties of the logarithm include that
\bqa
  \log_a uv &=& \log_a u + \log_a v,
    \label{alggeo:230:310}\\
  \log_a \frac{u}{v} &=& \log_a u - \log_a v,
    \label{alggeo:230:311}\\
  \log_a (w^z) &=& z \log_a w,
    \label{alggeo:230:315}\\
  w^z &=& a^{z\log_a w},
    \label{alggeo:230:316}\\
  \log_b w &=& \frac{ \log_a w }{ \log_a b }.
    \label{alggeo:230:318}
\eqa
Of these,~(\ref{alggeo:230:310}) follows from the steps
\bqb
  (uv) &=& (u)(v), \\
  (a^{\log_a uv}) &=& (a^{\log_a u})(a^{\log_a v}), \\
  a^{\log_a uv} &=& a^{\log_a u + \log_a v};
\eqb
and~(\ref{alggeo:230:311}) follows by similar reasoning.
Equations~(\ref{alggeo:230:315}) and~(\ref{alggeo:230:316}) follow from
the steps
\bqb
  \settowidth\tla{$\ds a^{\log_a (w^z)}$}
  \settowidth\tla{$\ds a^{\log_a (w^z)}$}
  w^z = \makebox[\tla][r]{$\ds(w^z)$} &=& (w)^z, \\
  w^z = a^{\log_a (w^z)} &=& (a^{\log_a w})^z, \\
  w^z = a^{\log_a (w^z)} &=& a^{z \log_a w}.
\eqb
Equation~(\ref{alggeo:230:318}) follows from the steps
\bqb
  w &=& b^{\log_b w}, \\
  \log_a w &=& \log_a( b^{\log_b w}), \\
  \log_a w &=& \log_b w \log_a b.
\eqb

Among other purposes,~(\ref{alggeo:230:310})
through~(\ref{alggeo:230:318}) serve respectively to transform products
to sums, quotients to differences, powers to products, exponentials to
differently based exponentials, and logarithms to differently based logarithms.
Table~\ref{alggeo:230:tbl} repeats the equations along
with~(\ref{alggeo:230:logdef}) and~(\ref{alggeo:230:logdef2}) (which also
emerge as restricted forms of eqns.~\ref{alggeo:230:315}
and~\ref{alggeo:230:316}), thus summarizing the general properties of the
logarithm.
\begin{table}
  \caption{General properties of the logarithm.}
  \label{alggeo:230:tbl}
  \bqb
    \log_a uv &=& \log_a u + \log_a v \\
    \log_a \frac{u}{v} &=& \log_a u - \log_a v \\
    \log_a (w^z) &=& z \log_a w \\
    w^z &=& a^{z\log_a w} \\
    \log_b w &=& \frac{ \log_a w }{ \log_a b } \\
    \log_a (a^z) &=& z \\
    w &=& a^{\log_a w}
  \eqb
\end{table}

% ----------------------------------------------------------------------

\section{The triangle}
\label{alggeo:323}
\index{geometry}
\index{triangle}

This section develops several facts about the triangle.\footnote{%
  Fashion seems to ask a writer to burden the plain word ``triangle''
  with various accurate but not-very-helpful adjectives like ``planar''
  and ``Euclidean.''  We like planes and Euclid (\S~\ref{alggeo:2235})
  but would resist the fashion.  Readers already know what a triangle
  is.%
}

\subsection{Area}
\label{alggeo:323.10}
\index{area}
\index{triangle!area of}

\index{right triangle}
\index{triangle!right}
\index{rectangle!splitting of down the diagonal}
The area of a right triangle%
\footnote{%
  As the reader likely knows, a \emph{right triangle} is a triangle, one
  of whose three angles is perfectly square.
}
is half the area of the corresponding rectangle.  This is seen by
splitting a rectangle down its diagonal into a pair of right triangles
of equal size.  The fact that \emph{any} triangle's area is half its
base length times its height is seen by dropping a perpendicular from
one point of the triangle to the opposite side (see
Fig.~\ref{intro:284:fig} on page~\pageref{intro:284:fig}),
thereby dividing the
triangle into two right triangles, for each of which the fact is true.
In algebraic symbols,
\bq{alggeo:323:10}
  A=\frac{bh}{2},
\eq
where~$A$ stands for area,~$b$ for base length, and~$h$ for
perpendicular height.

\subsection{The triangle inequalities}
\label{alggeo:323.20}
\index{triangle inequalities}
\index{proof!by sketch}
\index{sketch, proof by}

Any two sides of a triangle together are longer than the third alone,
which itself is longer than the difference between the two.  In symbols,
\bq{alggeo:323:20}
  \left|a-b\right| < c < a+b,
\eq
where~$a$, $b$ and~$c$ are the lengths of a triangle's three sides.
These are the \emph{triangle inequalities.}  The truth of the sum
inequality, that $c < a+b$, is seen by sketching some triangle on a sheet of
paper and asking: if~$c$ is the direct route between two points and
$a+b$ is an indirect route, then how can $a+b$ not be longer?  Of course
the sum inequality is equally good on any of the triangle's three sides,
so one can write that $a<c+b$ and $b<c+a$ just as well as that $c<a+b$.
Rearranging the~$a$ and~$b$ inequalities, we have that $a-b<c$ and
$b-a<c$, which together say that $\left|a-b\right|<c$.  The last is the
difference inequality, completing~(\ref{alggeo:323:20})'s proof.%
\footnote{%
  Section~\ref{mtxinv:447} proves the triangle inequalities more
  generally, though regrettably without recourse to this subsection's
  properly picturesque geometrical argument.
}

\subsection{The sum of interior angles}
\label{alggeo:323.30}
\index{angle}
\index{angle!sum of}
\index{angle!interior}
\index{triangle}
\index{polygon}

\index{angle!of a triangle}
\index{$2\pi$}
\index{belonging}
\index{membership}
\index{set}
\index{set notation}
\index{$\in$}
A triangle's three interior angles%
\footnote{%
  Most readers will already know the
  notation~$2\pi$ and its meaning as the angle of full revolution.  The
  notation is properly introduced in \S\S~\ref{trig:226}, \ref{trig:260}
  and~\ref{taylor:355} at any rate.  Briefly nevertheless, the
  symbol~$2\pi$ represents a complete turn, a full circle, a spin to
  face the same direction as before.  Hence (for instance)~$2\pi/4$
  represents a square turn or right angle.

  You may be used to the notation~$360^\circ$ in place of~$2\pi$; but,
  for the reasons explained in appendix~\ref{hex} and in
  footnote~\ref{trig:260:fn1} of chapter~\ref{trig}, this book tends to
  avoid the notation~$360^\circ$.
}
sum to $2\pi/2$.  One way to see the
truth of this fact is to imagine a small car rolling along one of the
triangle's sides.  Reaching the corner, the car turns to travel along
the next side, and so on round all three corners to complete a circuit,
returning to the start.  Since the car again faces the original
direction, we reason that it has turned a total of~$2\pi$, a full
revolution.  But the angle~$\phi$ the car turns at a corner and the
triangle's inner angle~$\psi$ there together form the straight angle
$2\pi/2$ (the sharper the inner angle, the more the car turns: see
Fig.~\ref{alggeo:323:f70}).
\begin{figure}
  \caption{The sum of a triangle's inner angles: turning at the corner.}
  \label{alggeo:323:f70}
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\end{figure}
In mathematical notation,
\bqb
  \phi_1+\phi_2+\phi_3 &=& 2\pi;\\
  \phi_k+\psi_k &=& \frac{2\pi}{2},\ \ k \in \{1, 2, 3\};
\eqb
where~$\psi_k$ and~$\phi_k$ are respectively the triangle's inner angles
and the angles through which the car turns; and where the \emph{set
notation}%
\footnote{%
  Applied mathematicians tend to less enthusiasm than professional
  mathematicians do over set notation like the membership symbol~$\in$,
  but such notation still finds use in applications now and again as,
  for example, it does in this instance.%
  % The rest of this footnote is pretty good but is omitted from the
  % book for now.  It probably at least does not belong here.
  %Though set notation indeed proves useful in applications now and
  %again, you will find no introduction to set theory as such in this
  %book, which has for instance not a single Venn
  %diagram~\cite[\S~2-1]{Papoulis}.  The reason is that the explicit,
  %formal, systematic development of set theory largely serves a pure
  %mathematical program which just is not the aim of this non-Kantian
  %book.
  %
  %Respectable mathematicians like the great Gottlob Frege
  %(1848--1925)---admittedly no Kantian he!---and his successors might
  %still have preferred even this book to treat set notation more
  %formally than it does; but for better or for worse this is Black's
  %book, not Frege's, and Frege was no engineer, whereas Black happens to
  %be, so there is that difference in perspective.  At any rate, Black
  %judges this book's approach to be preferable under the circumstance.
}
$k \in \{1, 2, 3\}$, met in \S~\ref{alggeo:227}, means that the
last equation holds whether $k=1$, 2 or~3.  Solving the latter equation
for~$\phi_k$ and substituting into the former yields that
\bq{alggeo:323:50}
  \psi_1+\psi_2+\psi_3 = \frac{2\pi}{2},
\eq
which was to be demonstrated.

\index{angle!of a polygon}
Extending the same technique to the case of an $n$-sided polygon,
\bqb
  \sum_{k=1}^n \phi_k &=& 2\pi,\\
  \phi_k+\psi_k &=& \frac{2\pi}{2}.
\eqb
Solving the latter equation for~$\phi_k$ and substituting into the
former, we have that
\[
  \sum_{k=1}^n \left( \frac{2\pi}{2} - \psi_k \right) = 2\pi,
\]
or in other words that
\bq{alggeo:323:60}
  \sum_{k=1}^n \psi_k = (n-2)\frac{2\pi}{2}.
\eq
Equation~(\ref{alggeo:323:50}) is then seen to be a special case
of~(\ref{alggeo:323:60}) with $n=3$.

\subsection{The Pythagorean theorem}
\label{alggeo:223}
\index{Pythagorean theorem}
\index{Pythagoras (c.~580--c.~500~B.C.)}

Along with Euler's formula~(\ref{cexp:euler}), the fundamental
theorem of calculus
% bad break
(\ref{integ:antider}), Cauchy's integral
formula~(\ref{taylor:cauchy}), Fourier's equation~(\ref{fouri:eqn}) and
maybe a few others, the Pythagorean theorem is one of the most famous
results in all of mathematics.

\index{diagonal}
\index{perpendicular}
\index{similar triangles}
\index{triangle!similar}
\index{right triangle}
\index{triangle!right}
\index{angle}
\index{scaling}
\index{rotation}
\index{reflection}
\index{proportion}
The introduction to chapter~\ref{intro} has proved the theorem.\footnote{%
  \index{Euclid (325--265~B.C.)}%
  The elegant proof of chapter~\ref{intro}
  is simpler than the one famously given by the
  ancient geometer Euclid, yet more appealing than alternate proofs
  often found in print.  Whether Euclid was acquainted with either of
  the two Pythagorean proofs the book you are reading gives, or indeed
  was acquainted with both, this writer does not know; but it is
  possible \cite[``Pythagorean theorem,'' 02:32, 31 March 2006]{wikip}
  that Euclid chose his proof because it comported better with the
  restricted set of geometrical elements with which he permitted himself
  to work.
  Be that as it may, the present writer encountered the proof of
  chapter~\ref{intro}
  somewhere years ago and has never seen it in print
  since (but has not looked for it, either), so can claim no credit for
  originating it.  Unfortunately the citation is now long lost.  A
  current, electronic source for the proof of
  chapter~\ref{intro}
  is~\cite{wikip} as cited earlier in this footnote.
}
Alternately and less concisely, Fig.~\ref{alggeo:223:37} too can prove
it as follows.
\begin{figure}
  \caption{A right triangle subdivided.}
  \label{alggeo:223:37}
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              \rput( 0.02,0){$c$}%
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              \psline(\squarb,0)(\squarb,\squarb)(-\squarb,\squarb)(-\squarb,-\squarb)(0,-\squarb)
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          \psline(0,0)(0,-\squareanglelength)
          \rput( 2.40,-0.30){$t$}
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\end{figure}%
In the figure are three right triangles: a small one~$sha$ on the left;
a medium one~$htb$ on the right; and, if the small and medium are joined
together, a large one~$abc$ overall.
The small triangle shares an angle~$sa$ with the large which, in light
of~(\ref{alggeo:323:50}) and of that both are right
triangles, means that \emph{all three angles} of the small triangle
equal the corresponding three angles of the large, which in turn means
that the small and large triangles are \emph{similar} to one another;
that is, the small triangle is merely a scaled, rotated, possibly
reflected version of the large.  The medium triangle shares an
angle~$tb$ with the large which likewise means that the medium and
large triangles are similar.

Insofar as triangles are similar (\S~\ref{alggeo:2235}), corresponding
ratios of their sides must be equal:
\[
  \frac{a}{c}=\frac{s}{a};\quad
  \frac{b}{c}=\frac{t}{b}.
\]
Rearranging factors,
\[
  \frac{a^2}{c}=s;\quad
  \frac{b^2}{c}=t.
\]
The sum of the last line's two equations is that
\[
  \frac{a^2+b^2}{c}=s+t;
\]
or rather, because $s+t=c$, that
\[
  \frac{a^2+b^2}{c}=c,
\]
which, when multiplied by~$c$, yields the Pythagorean
theorem~(\ref{alggeo:pythag}),
\[
  a^2+b^2=c^2.
\]
This paragraph's proof has found no new conclusion but one is glad to
discover that the Pythagorean theorem seems to remain true no matter how
one reaches it.\footnote{%
  The source of this proof is an otherwise forgotten book that stood on
  a shelf in the writer's high-school library in~1983.  The writer still
  remembers the proof but has long since lost the citation.%
}

\index{Pythagorean theorem!in three dimensions}
\index{diagonal!three-dimensional}
\index{altitude}
\index{diagonal}
\index{hypotenuse}
\index{leg}
The Pythagorean theorem is readily extended into three dimensions as
\bq{alggeo:223:40}
  a^2 + b^2 + h^2 = r^2,
\eq
where~$h$ is an altitude perpendicular to both~$a$ and~$b$
and thus also to~$c$;
and where~$r$ is the corresponding three-dimensional diagonal---the
diagonal of the right triangle whose legs are~$c$ and~$h$.
Inasmuch as~(\ref{alggeo:pythag}) applies to any right triangle, it
follows that $c^2+h^2=r^2$, which equation expands directly to
yield~(\ref{alggeo:223:40}).

\subsection{Congruence and Euclid's similar proportionality}
\label{alggeo:2235}
\index{Euclid (325--265~B.C.)}
\index{proportionality}
\index{similarity}
\index{congruence}

\index{assertion}
\index{unproved assertion}
\index{angle}
\index{scaling}
\index{rotation}
\index{reflection}
The careful reader might have noticed the presence of an unproved
assertion in the last subsection.  The assertion was this: that, if each
of a triangle's angles equals the corresponding angle of another
triangle, then the one triangle is a scaled, rotated, possibly reflected
version of the other; or, stated another way, that, if the respective
angles are equal, then the respective sides though maybe not equal are
at least proportionate.

Does such an assertion not require proof?

\index{intuition}
\index{geometrical intuition}
\index{proof!necessity of}
\index{attitude}
Whether such an assertion requires proof depends on the vividness of
one's geometrical intuition and/or on one's attitude toward the kind of
assertion the assertion is.  If one's attitude is an applied attitude
and if intuition finds the assertion obvious enough, then the assertion
may not require proof.

Even given an applied attitude, though, it's a borderline case, isn't
it?  Fortunately, the ancient geometer Euclid has supplied us a proof in
book~VI of his \emph{Elements}~\cite{Euclid}\@.  Though we will not
fully detail Euclid's proof we can digest it as follows.

\index{parallelogram}
\index{parallel sides}
\index{side!parallel}
\index{length!equal}
Fig.~\ref{alggeo:2235:fig} arranges two triangles~$pqr$
and~$stu$ whose respective angles are equal.
\begin{figure}
  \caption{Proportionality of similar triangles (or the AA criterion).}
  \label{alggeo:2235:fig}
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        \rput(2.90,0.80){$u$}
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        \rput(-1.01,2.72){$u_1$}
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\end{figure}
Extending the indicated sides and exploiting the equal angles, the
figure extrudes a parallelogram from the two triangles.  Being opposite
sides of a parallelogram,
\bq{alggeo:2235.20}
  q_1 = q;\quad
  u_1 = u.
\eq
Because\footnote{The pure mathematician who loves Euclid can abhor such
shortcuts!  ``Because''?!  Where is the proof that the lines are
parallel, the purist wants to know?  Where is the proof that opposite
sides of a parallelogram are equal in length?  Notwithstanding, busily
rushing ahead, we applicationists fail to notice the purist's
consternation.}~$q$ runs parallel to $q_1+t$,
\[
  \frac{r}{u_1} = \frac{p}{s}.
\]
Applying~(\ref{alggeo:2235.20}),
\[
  \frac{r}{u} = \frac{p}{s}.
\]
Rearranging factors,
\bq{alggeo:2235.30}
  \frac{r}{p} = \frac{u}{s}.
\eq
By the same kind of argument,
\[
  \frac{t}{s} = \frac{q}{p},
\]
or, reversed and inverted,
\bq{alggeo:2235.32}
  \frac{p}{q} = \frac{s}{t}.
\eq
Multiplying~(\ref{alggeo:2235.30}) by~(\ref{alggeo:2235.32}),
\bq{alggeo:2235.34}
  \frac{r}{q} = \frac{u}{t}.
\eq
Together,~(\ref{alggeo:2235.30}), (\ref{alggeo:2235.32})
and~(\ref{alggeo:2235.34}) complete Euclid's proof.  Proceeding
nevertheless if you wish, one may further rearrange the three equations
respectively to read,
\[
  \frac{r}{u} = \frac{p}{s},\quad
  \frac{p}{s} = \frac{q}{t},\quad
  \frac{q}{t} = \frac{r}{u};
\]
or, in other words,
\bq{2235:40}
  \frac{p}{s} = \frac{q}{t} = \frac{r}{u},
\eq
which, using this book's modern applied notation (which Euclid lacked)
condenses Euclid's elegant finding into a single line.

\index{effort}
\index{mental energy}
\index{logical exercise}
The finding is indeed elegant.  However, as earlier asked, to an
applicationist, is the finding's proof \emph{necessary?} Could one not
instead just look at Fig.~\ref{alggeo:2235:fig} and \emph{see} that the
triangles are proportionate?\footnote{%
  The modern professional mathematician might answer, ``Sure, if you are
  an engineer or the like, then go ahead.  However, if you wish to be a
  proper analyst, then no.  We have a theory of \emph{metric spaces.}
  We have theories of other things, too.  You needn't just skip all
  that.''

  This book, a book of applied mathematics, will not go the route of
  metric spaces.  However,~\cite{Shilov} (which though not open source
  has long been available in an inexpensive paperback edition) offers
  the interested reader a comparatively accessible account of such.%
}
(By \emph{see,} we do not precisely mean, ``see with the eyes.'' The
eyes do help but what we here mean is \emph{to perceive mentally} as in
chapter~\ref{intro}.)

After all, if these triangles were not proportionate, then how else
should they be?

Until Euclid's proof had been given, would any mathematician (whether
applied or professional) who had sketched a triangle on a sheet of
paper really have conjectured a different result?  Was the
investigation's outcome ever in doubt?  To put it another way: suppose
that the investigation had found that the triangles were
disproportionate; would its finding have been believed?  Or is the
proof merely a salutary logical exercise, a noble sport whose object is
to hunt down and annihilate every premise one can possibly do without?

Effort and mental energy are not unlimited resources.  Time spent in
speculative foundational investigations may be regretted, if that time
is as a consequence later unavailable to devote to the acquisition of
more advanced, more practical mathematical techniques.  Mathematics is
after all far too vast for a single mathematician to master the whole of
it during a single lifetime.  One must prioritize.

The question is left for the reader to contemplate.

\index{mirror}
Meanwhile, a pair of definitions: as \S~\ref{alggeo:223} has already
suggested, two triangles are \emph{similar} if their respective angles
are equal, as, for example, in Fig.~\ref{alggeo:2235:fig}; and two
triangles are \emph{congruent} (this definition is new) if they are
similar and are equal in size.  Note that every triangle is congruent
to its own reflection (as in a mirror).  Regarding similarity, the
three triangles of Fig.~\ref{alggeo:223:37} are mutually
similar but not congruent.

\index{side-side-side (SSS)}
\index{SSS (side-side-side)}
\index{side-angle-side (SAS)}
\index{SAS (side-angle-side)}
\index{angle-side-angle (ASA)}
\index{ASA (angle-side-angle)}
\index{angle-angle-side (AAS)}
\index{AAS (angle-angle-side)}
Two triangles are congruent to one another if they meet any of the
following criteria:
\begin{itemize}
  \item SSS (side-side-side);
  \item SAS (side-angle-side);
  \item ASA (angle-side-angle);
  \item AAS (angle-angle-side).
\end{itemize}
The~SSS criterion requires that each of the triangles have three sides
as long as the other triangle's (for example, a triangle whose
sides are~5, 6 and~8 cm long and a triangle whose sides are~6, 5 and~8
cm long together satisfy the criterion).  The~SAS criterion requires
that each of the triangles have two sides as long as the other
triangle's and, where a triangle's two sides meet, one equal angle (for
example, two triangles, each of whose sides of~6 and~8 cm meet to form a
two-hour or~$30^\circ$ angle, together satisfy the criterion).  The~ASA
criterion requires that each of the triangles have two equal angles
joined by one equal side.  The~AAS criterion requires that each of the
triangles have two equal angles plus one equal side, this side however
being another than the side that joins the two angles.

\index{angle-angle (AA)}
\index{AA (angle-angle)}
\index{angle-angle-angle (AAA)}
\index{AAA (angle-angle-angle)}
As Euclid has proved earlier in the subsection, two triangles are
similar (though not necessarily congruent) if they meet this criterion:
\begin{itemize}
  \item AA (angle-angle).
\end{itemize}
Why is the criterion not~AAA?  Answer: because AA suffices, for,
according to~(\ref{alggeo:323:50}), AA implies~AAA.

\index{side-side-angle (SSA)}
\index{SSA (side-side-angle)}
The criterion
\begin{itemize}
  \item SSA (side-side-angle),
\end{itemize}
depicted in Fig.~\ref{alggeo:2235:fig}, is tricky: it does not quite
establish congruence in and of itself, but if \emph{three} triangles
mutually meet the~SSA criterion then at least two of them are congruent.
\begin{figure}
  \caption{The SSA and AAS criteria.}
  \label{alggeo:2235:figssa}
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  \ec
\end{figure}

Besides~SSA, Fig.~\ref{alggeo:2235:fig} also illustrates the~AAS
criterion.  Illustration of criteria besides~SSA and~AAS is left as an
exercise.

See also \S~\ref{trig:270}.

% ----------------------------------------------------------------------

\section{Functions}
\label{alggeo:250}
\index{function}

\index{mapping}
Briefly, a \emph{function} is a mapping from one number (or vector of
several numbers, \S\S~\ref{matrix:120} and~\ref{taylor:370}) to another.
This book is not the place for a gentle introduction to the concept of
the function; but as an example, $f(x) \equiv x^2 - 1$ is a function
which maps~1 to~0 and~$-3$ to~$8$, among others.

\index{domain}
\index{range}
When discussing functions, one often speaks of \emph{domains} and
\emph{ranges.}  The \emph{domain} of a function is the set of numbers
one can put into it.  The \emph{range} of a function is the
corresponding set of numbers one can get out of it.  In the example, if
the domain is restricted to real~$x$ such that $\left|x\right|\le
3$---that is, such that $-3 \le x \le 3$---then the corresponding range
is $-1 \le f(x) \le 8$.

\index{inverse!of a function}
\index{function!inverse of}
\index{invertibility}
\index{noninvertibility}
\index{$\approx$}
If $y=f(x)$, then the notation $f^{-1}(\cdot)$ indicates the
\emph{inverse} of the function $f(\cdot)$ such that
\bq{alggeo:250:30}
  \begin{split}
    f^{-1}[f(x)] &= x, \\
    f[f^{-1}(y)] &= y,
  \end{split}
\eq
thus swapping the function's range with its domain.  In the example,
\linebreak % bad break
$f(-3)=8$,
so $f^{-1}(8)=-3$.  Unfortunately, $f^{-1}(8)=3$ as well,
making the example's function $f(x)$ strictly \emph{noninvertible} over
its given domain; but, fortunately, various attendant considerations
(whether these be theoretical considerations, as in \S~\ref{taylor:320},
or merely practical considerations) tend in concrete circumstances to
distinguish between candidates like the example's~$-3$ and~$3$,
rendering many such functions as $f(x)$ effectively invertible in
context.
Therefore, the applied mathematician may not always have to
worry too much about strict invertibility.

Inconsistently, inversion's notation $f^{-1}(\cdot)$ clashes with the
% bad break
similar-look\-ing
but different-meaning notation $f^2(\cdot) \equiv
[f(\cdot)]^2$, whereas $f^{-1}(\cdot) \neq [f(\cdot)]^{-1}$.  Both
notations are conventional and both are used in this book.

\index{root}
\index{zero}
\index{pole}
\index{pole!multiple}
\index{multiple pole}
\index{singularity}
\index{divergence to infinity}
Other terms that arise when discussing functions are \emph{root} (or
\emph{zero}), \emph{singularity} and \emph{pole}.  A \emph{root} (or
\emph{zero}) of a function is a domain point at which the function
evaluates to zero (the example has roots at $x=\pm 1$).  A
\emph{singularity} of a function is a domain point at which the
function's output \emph{diverges;} that is, where the function's output
goes infinite.%
\footnote{%
  Here is one example of the book's deliberate lack of formal
  mathematical rigor.  A more precise formalism to say that ``the
  function's output goes infinite'' might be that
  \[
    \lim_{x\ra x_o}\left|f(x)\right| = \infty,
  \]
  and yet preciser formalisms than this are conceivable, and
  occasionally even are useful.  Be that as it may, the applied
  mathematician tends to avoid such formalisms \emph{where there seems no
  immediate need for them.}
}
A \emph{pole} is a singularity that behaves locally like $1/x$ (rather
than, for example, like $1/\sqrt x$).  A singularity that behaves as
$1/x^N$ is a \emph{multiple pole}, which (\S~\ref{inttx:260.20}) can be
thought of as~$N$ poles.  The example's function $f(x)$ has no
singularities for finite~$x$; however, the function $h(x)\equiv 1/(x^2-1)$ has
poles at $x=\pm 1$.

\index{branch point}
\index{essential singularity}
\index{singularity!essential}
(Besides the root, the singularity and the pole, there is also the
troublesome \emph{branch point,} an infamous example of which is $z=0$
in the function $g[z]\equiv\sqrt{z}$.
Branch points are important, but the book must lay a more extensive
foundation before introducing them properly in \S~\ref{taylor:325}.%
\footnote{%
  There is further the \emph{essential singularity,} an example of which
  is $z=0$ in $p(z)\equiv\exp(1/z)$, but typically the best way to handle so
  unreasonable a singularity is to change a variable, as
  $w \la 1/z$, or otherwise to frame the problem such that one need not
  approach the singularity.
  % diagn:  Check the vague reference that follows.
  Except tangentially much later when it treats asymptotic series,
  this book will have little to say of the essential singularity.
})

% ----------------------------------------------------------------------

\section{Complex numbers (introduction)}
\label{alggeo:225}
\index{number}
\index{real number}
\index{imaginary number}
\index{complex number}
\index{number!real}
\index{number!imaginary}
\index{number!complex}

\index{square root}
\index{unit}
\index{unit!real}
\index{unit!imaginary}
\index{imaginary unit}
\index{conjugate}
\index{elf}
\index{$i$}
Section~\ref{alggeo:224roots} has introduced square roots.  What it has
not done is to tell us how to regard a quantity like $\sqrt{-1}$.  Since
there exists no real number~$i$ such that
\bq{alggeo:225:i}
  i^2 = -1
\eq
and since the quantity~$i$ thus defined is found to be critically
important across broad domains of higher mathematics, we
accept~(\ref{alggeo:225:i}) as the definition of a fundamentally new
kind of quantity, \emph{the imaginary number.}%
\footnote{%
  The English word \emph{imaginary} is evocative, but perhaps not
  of quite the right concept in this usage.  Imaginary numbers
  are not to mathematics as, say, imaginary elfs are to the physical world.
  In the physical world, imaginary elfs are (presumably) not substantial
  objects.  However, in the mathematical realm, imaginary numbers
  \emph{are} substantial.  The word \emph{imaginary} in the mathematical
  sense is thus more of a technical term than a descriptive adjective.

  The number~$i$ is just a concept, of course, but then so is the
  number~$1$ (though you and I have often met one \emph{of
  something}---one apple, one chair, one summer afternoon,
  etc.---neither of us has ever met just~1).  In
  Platonic~\cite[chapter~2]{Feser} or Fregean~\cite{Frege} terms,~$i$ is
  literally no
  less valid a \emph{form} than~$1$ is.  The reason imaginary numbers
  have been called ``imaginary'' probably has to do with the fact that
  they emerge from mathematical operations only, never directly from
  counting things.  Notice, however, that the number 1/2 never emerges
  directly from counting things, either.  If for some reason the $i$year
  were offered as a unit of time, then the period separating your
  fourteenth and twenty-first birthdays could have been measured as
  $-i7$~$i$years.  Madness?  No, let us not call it that; let us call it
  a useful formalism, rather.

  The unpersuaded reader is asked to suspend judgment a while.  He will
  soon see the use.
}

Imaginary numbers are given their own number line, plotted at right
angles to the familiar real number line as in
Fig.~\ref{alggeo:225:fig}.
\begin{figure}
  \caption[The complex (or Argand) plane.]{The complex (or Argand)
  plane, and a complex number~$z=2+i1$ therein.}
  \label{alggeo:225:fig}
  \index{complex plane}
  \index{Argand plane}
  \index{Argand, Jean-Robert (1768--1822)}
  \bc
    {
      \nc\xax{-5}
      \nc\xbx{-3}
      \nc\xcx{ 5}
      \nc\xdx{ 3}
      \begin{pspicture}(\xax,\xbx)(\xcx,\xdx)
        \psset{dimen=middle}
        \small
        %\psframe[linewidth=0.5pt](\xax,\xbx)(\xcx,\xdx)
        \nc\xx{2.5}
        \nc\xxb{1.0}
        \nc\xxc{2.0}
        \nc\xxd{0.1}
        \nc\xxe{0.25}
        \nc\xxf{0.25}
        \nc\xxp{1.3}
        \nc\xxq{26.565}
        \nc\xxqq{13.283}
        \nc\xxr{2.2361}
        \nc\xxrr{1.1180}
        \nc\xxrrr{2.3}
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        \nc\xxlsep{0.20}
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          \psline[linewidth=1.0pt]{<->}(-\xx,0)(\xx,0)
          \psline[linewidth=0.5pt]( \xxb,-\xxd)( \xxb,\xxd)
          \psline[linewidth=0.5pt](-\xxb,-\xxd)(-\xxb,\xxd)
          \psline[linewidth=0.5pt]( \xxc,-\xxd)( \xxc,\xxd)
          \psline[linewidth=0.5pt](-\xxc,-\xxd)(-\xxc,\xxd)
        }
        \xxa
        \rput{90}(0,0){\xxa}
        \rput[r](-\xxe,-2){$-i2$}
        \rput[r](-\xxe,-1){$-i $}
        \rput[r](-\xxe, 1){$ i $}
        \rput[r](-\xxe, 2){$ i2$}
        \rput[t](-2,-\xxf){$-2$}
        \rput[t](-1,-\xxf){$-1$}
        \rput[t]( 1,-\xxf){$ 1$}
        \rput[t]( 2,-\xxf){$ 2$}
        \psline[linewidth=2.0pt]{cc-*}(0,0)(2,1)
        \psarc[linewidth=0.5pt]{->}(0,0){\xxp}{0}{\xxq}
        \rput{\xxqq}(0,0){
          \uput[r](\xxp,0){
            \rput{*0}(0,0){$\phi$}
          }
        }
        \rput{\xxq}(0,0){
          \uput{\xxlsep}[u](\xxrr,0){
            \rput{*0}(0,0){$\rho$}
          }
          \uput{\xxlsep}[r](\xxrrr,0){
            \rput{*0}(0,0){$z$}
          }
        }
        \rput[l](\xxs,0){$\Re(z)$}
        \rput[b](0,\xxs){$i\Im(z)$}
      \end{pspicture}
    }
  \ec
\end{figure}
The sum of a real number~$x$ and an imaginary number~$iy$ is the
\emph{complex number}
\[
  z = x + iy.
\]
The \emph{conjugate}~$z^{*}$ of this complex number is defined to be%
\footnote{%
  For some inscrutable reason, in the author's country at least,
  professional mathematicians seem universally to write~$\overline z$
  instead of~$z^{*}$, whereas rising engineers take the mathematicians'
  courses at school and then, having passed those courses,
  promptly start writing~$z^{*}$
  for the rest of their lives.  The writer has his preference between
  the two notations and this book reflects it, but the curiously
  absolute character of the notational split is interesting as a social
  phenomenon.
}
\[
  z^{*} = x - iy.
\]
The \emph{magnitude} (or \emph{modulus,} or \emph{absolute value})
$\left|z\right|$
of the complex number is defined to be the length~$\rho$ in
Fig.~\ref{alggeo:225:fig}, which per the Pythagorean theorem
(\S~\ref{alggeo:223})
is such that
\bq{225:10}
  \index{magnitude}
  \index{modulus}
  \index{absolute value}
  \index{complex number!magnitude of}
  \left|z\right|^2 = x^2 + y^2.
\eq
The \emph{phase} $\arg z$ of the complex number is defined to be the
angle~$\phi$ in the figure, which in terms of the trigonometric
functions of \S~\ref{trig:226}%
\footnote{\label{alggeo:225:fn1}%
  This is a forward reference, returned by footnote~\ref{trig:226:fn1}
  of chapter~\ref{trig}.  If the equation does not make sense to you yet for
  this reason, skip it for now.  The important point is that $\arg z$ is
  the angle~$\phi$ in the figure.
}
is such that
\bq{225:11}
  \index{phase}
  \index{complex number!phase of}
  \index{arg}
  \tan (\arg z) = \frac{y}{x}.
\eq
Specifically to extract the real and imaginary parts of a complex
number, the notations
\begin{equation}
  \label{225:12}
  \index{real part}
  \index{imaginary part}
  \index{complex number!real part of}
  \index{complex number!imaginary part of}
  \begin{split}
    \Re(z) &= x, \\
    \Im(z) &= y,
  \end{split}
\end{equation}
are conventionally recognized (although often the symbols $\Re[\cdot]$
and $\Im[\cdot]$ are written~$\mr{Re}[\cdot]$ and~$\mr{Im}[\cdot]$,
particularly when printed by hand).  For example, if $z=2+i1$, then
$\Re(z)=2$, $\Im(z)=1$, $|z|=\sqrt{5}$, and $\arg z = \arctan(1/2)$.

\subsection[Rectangular complex multiplication]{Multiplication and
division of complex numbers in rectangular form}
\label{alggeo:225.1}
\index{multiplication}
\index{division}
\index{complex number!multiplication and division}

Several elementary properties of complex numbers are readily seen if the
fact that $i^2 = -1$ is kept in mind, including that
\bqa
  z_1z_2 &=& (x_1x_2 - y_1y_2) + i(y_1x_2 + x_1y_2),
    \label{alggeo:225:10}\\
  \frac{z_1}{z_2} &=& \frac{x_1+iy_1}{x_2+iy_2}
    = \left(\frac{x_2-iy_2}{x_2-iy_2}\right)\frac{x_1+iy_1}{x_2+iy_2}
    \xn\\
  &=& \frac{(x_1x_2+y_1y_2)+i(y_1x_2-x_1y_2)}{x_2^2+y_2^2}.
    \label{alggeo:225:20}
\eqa
It is a curious fact that
\bq{alggeo:225:23}
  \frac{1}{i} = -i.
\eq
It is a useful fact that
\bq{alggeo:225:24}
  z^{*}z = x^2 + y^2 = \left| z \right|^2
\eq
(the curious fact, eqn.~\ref{alggeo:225:23}, is useful, too).
Sometimes convenient are the forms
\bq{alggeo:225:25}
  \begin{split}
    \Re(z) &= \frac{z + z^{*}}{2}, \\
    \Im(z) &= \frac{z - z^{*}}{i2},
  \end{split}
\eq
trivially proved.
Surprising and significant is that, though~$\sqrt{-1} = i$ has revealed
a previously unknown class of quantity in~$i$, nevertheless,
$\sqrt{i}=(1+i)/\sqrt{2}$ reveals no further new class of quantity, but
only a quantity expressible%
\footnote{\cite[\S~I:22-5]{Feynman}}
in terms of~1 and~$i$, as~(\ref{alggeo:225:10}) verifies
(and upon which \S~\ref{trig:280} will shed a clearer light).

\subsection{Complex conjugation}
\label{alggeo:225.2}
\index{conjugate}
\index{conjugation}
\index{complex conjugation}
\index{complex number!conjugating}

\index{induction}
\index{proof!by induction}
An important property of complex numbers descends subtly from the
fact that
\[
  i^2 = -1 = (-i)^2.
\]
If one defined some number $j\equiv-i$, asserting that~$j$ not~$i$ were
the true imaginary unit,%
\footnote{\cite{WADavis-conversation}\cite[\S~I:22-5]{Feynman}}
then one would find that
\[
  (-j)^2 = -1 = j^2,
\]
and thus that all the basic properties of complex numbers in the~$j$
system held just as well as they did in the~$i$ system.  The units~$i$
and~$j$ would differ indeed, but would perfectly mirror one another in
every respect.

That is the basic idea.  To establish it symbolically needs a page or
so of slightly abstract algebra as follows, the goal of which will be
to show that $[f(z)]^{*} = f(z^{*})$ for some unspecified function
$f(z)$ with specified properties.  To begin with, if
\[
  z = x+iy,
\]
then
\[
  z^{*} = x-iy
\]
by definition.  Proposing that $(z^{k-1})^{*}=(z^{*})^{k-1}$ (which may
or may not be true but for the moment we assume it), we can write,
\bqb
  z^{k-1}       &=& s_{k-1}+it_{k-1}, \\
  (z^{*})^{k-1} &=& s_{k-1}-it_{k-1},
\eqb
where $s_{k-1}$ and $t_{k-1}$ are symbols introduced respectively to
represent the real and imaginary parts of $z^{k-1}$.  Multiplying the
former equation by $z=x+iy$ and the latter by $z^{*}=x-iy$, we have that
\bqb
  z^k       &=& (xs_{k-1}-yt_{k-1})+i(ys_{k-1}+xt_{k-1}), \\
  (z^{*})^k &=& (xs_{k-1}-yt_{k-1})-i(ys_{k-1}+xt_{k-1}).
\eqb
With the definitions that $s_k \equiv xs_{k-1}-yt_{k-1}$ and $t_k \equiv
ys_{k-1}+xt_{k-1}$, this is written more succinctly,
\bqb
  z^k       &=& s_k+it_k, \\
  (z^{*})^k &=& s_k-it_k.
\eqb
In other words, if $(z^{k-1})^{*}=(z^{*})^{k-1}$, then it necessarily
follows that $(z^k)^{*}=(z^{*})^k$.
Solving the definitions of $s_k$ and $t_k$ for $s_{k-1}$ and $t_{k-1}$
yields the reverse definitions that
$s_{k-1} = (xs_k+yt_k)/(x^2+y^2)$ and
$t_{k-1} = (-ys_k+xt_k)/(x^2+y^2)$.
Therefore, except when $z=x+iy$ happens to be null or infinite, the
implication is reversible by reverse reasoning, so by mathematical
induction%
\footnote{%
  \emph{Mathematical induction} is an elegant old technique for the
  construction of mathematical proofs.  Section~\ref{taylor:314}
  elaborates on the technique and offers a more extensive example.
  Beyond the present book, a very good introduction to mathematical
  induction is found in~\cite{Hamming}.
}
we have that
\bq{alggeo:225:21}
  (z^k)^{*}=(z^{*})^k
\eq
for all integral~$k$.  We have also from~$(\ref{alggeo:225:10})$
that
\bq{alggeo:225:22}
  (z_1z_2)^{*} = z_1^{*}z_2^{*}
\eq
for any complex~$z_1$ and~$z_2$.

Consequences of~(\ref{alggeo:225:21}) and~(\ref{alggeo:225:22}) include
that if
\bqa
  f(z) &\equiv& \sum_{k=-\infty}^{\infty} (a_k+ib_k) (z-z_o)^k, \label{alggeo:225:laurent} \\
  f^{*}(z) &\equiv& \sum_{k=-\infty}^{\infty} (a_k-ib_k) (z-z_o^{*})^k, \label{alggeo:225:laurent2}
\eqa
where~$a_k$ and~$b_k$ are real and imaginary parts of the coefficients
peculiar to the function $f(\cdot)$, then
\bq{alggeo:225:conj}
  \left[f(z)\right]^{*} = f^{*}(z^{*}).
\eq
In the common case in which $b_k=0$ for all~$k$ and in
which $z_o=x_o$ is a real number,~$f(\cdot)$ and~$f^{*}(\cdot)$ are the
same function, so~(\ref{alggeo:225:conj}) reduces to the desired form
\bq{alggeo:225:conj2}
  \left[f(z)\right]^{*} = f(z^{*}),
\eq
which says that the effect of conjugating the function's input is merely
to conjugate its output.

Equation~(\ref{alggeo:225:conj2}) expresses a significant, general rule
of complex numbers and complex variables which is better explained in
words than in mathematical symbols.  The rule is this: for most
equations and systems of equations used to model physical systems,
\emph{one can produce an equally valid alternate model simply by
simultaneously conjugating all the complex quantities
present.}%
\footnote{\cite{Hamming}\cite{Spiegel}}

\subsection{Power series and analytic functions (preview)}
\label{alggeo:225.3}
\index{power series}
\index{function!analytic}
\index{analytic function}

Equation~(\ref{alggeo:225:laurent}) expresses a general power series%
\footnote{\cite[\S~10.8]{Hildebrand}}
in $z-z_o$.  Such power series have broad application.%
\footnote{%
  That is a pretty impressive-sounding statement: ``Such power series
  have broad application.''  However, molecules, air and words also have
  ``broad application''; merely stating the fact does not tell us much.
  In fact the general power series is a sort of one-size-fits-all
  mathematical latex glove, which can be stretched to fit around almost
  any function of interest.
  Admittedly nevertheless, what grips one's attention here is not so
  much in the general form~(\ref{alggeo:225:laurent}) of the series as
  it is in the specific choice of~$a_k$ and~$b_k$, which this section
  does not discuss.

  Observe that the Taylor series (which this section also does not
  discuss:  see \S~\ref{taylor:310}) is a power series with
  $a_k=b_k=0$ for $k<0$.%
}
It happens in practice that most functions of interest in modeling
physical phenomena can conveniently be constructed (at least over a
local domain) as power series with suitable choices of~$a_k$, $b_k$
and~$z_o$.\footnote{%
  But see also the Fourier series of chapter~\ref{fours} which, by a
  different approach, can construct many functions over \emph{nonlocal}
  domains.%
}

The property~(\ref{alggeo:225:conj}) applies to all such functions,
with~(\ref{alggeo:225:conj2}) also applying to those for which $b_k=0$ and
$z_o=x_o$.  The property the two equations represent is called the
\emph{conjugation property.}  Basically, it says that if one replaces
all the~$i$ in some mathematical model with~$-i$, then the resulting
conjugate model is equally as valid as the original.%
\footnote{%
  To illustrate, from the fact that $(1+i2)(2+i3) + (1-i) = -3+i6$, the
  conjugation property infers immediately that $(1-i2)(2-i3) + (1+i) = -3-i6$.
  Observe however that no such property holds for the real
  parts: $(-1+i2)(-2+i3) + (-1-i) \neq 3+i6$.
}

Such functions, whether $b_k=0$ and $z_o=x_o$ or not, are \emph{analytic
functions} (\S~\ref{taylor:320}).  In the formal mathematical
definition, a function is analytic which is infinitely differentiable
(chapter~\ref{drvtv}) in the immediate domain neighborhood of interest.
However, for applications a fair working definition of the analytic
function might be ``a function expressible as a power series.''
Chapter~\ref{taylor} elaborates.  All power series are infinitely
differentiable except at their poles.

\index{function!nonanalytic}
\index{nonanalytic function}
There nevertheless exist one common group of functions which cannot be
constructed as power series.  These all have to do with the parts of
complex numbers and have been introduced in this very section: the
magnitude $\left|\cdot\right|$; the phase $\arg(\cdot)$; the conjugate
$(\cdot)^{*}$; and the real and imaginary parts $\Re(\cdot)$ and
$\Im(\cdot)$.  These functions are not analytic and do not in general
obey the conjugation property.  Also not analytic are the Heaviside unit
step $u(t)$ and the Dirac delta $\delta(t)$ (\S~\ref{integ:670}),
used to model discontinuities explicitly.

We shall have more to say about analytic functions in chapter~\ref{taylor}.
We shall have more to say about complex numbers in
\S\S~\ref{trig:280}, \ref{drvtv:230.35},
and~\ref{drvtv:240},
and much more yet in chapter~\ref{cexp}.

