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

\chapter{The Taylor series}
\label{taylor}
\index{Taylor series}
\index{series!Taylor}
\index{Taylor, Brook (1685--1731)}

\index{function!fitting of}
The Taylor series is a power series that fits a function in a
limited domain neighborhood.  Fitting a function in such a way
brings at least two advantages:
\bi
  \item it lets us take derivatives and integrals in the same
    straightforward way~(\ref{drvtv:240:polyderivz})
    one can take them given any power series; and
  \item it implies a simple procedure to calculate values of the
    function numerically.
\ei
This chapter introduces the Taylor series and some of its incidents.  It
also derives Cauchy's integral formula.  The chapter's early sections
prepare the ground for the treatment of the Taylor series proper in
\S~\ref{taylor:310}.%
\footnote{
  Because even at the applied level the proper derivation of the Taylor
  series involves mathematical induction, analytic continuation and the
  matter of convergence domains, no balance of rigor the chapter might
  strike seems wholly satisfactory.  The chapter errs maybe toward too
  much rigor; for, with a little less, most of \S\S~\ref{taylor:314},
  \ref{taylor:317}, \ref{taylor:320} and~\ref{taylor:330} would cease to
  be necessary.  For the impatient, to read only the following sections
  might not be an unreasonable way to shorten the chapter: \S\S~%
  \ref{taylor:310},
  \ref{taylor:325},
  \ref{taylor:350},
  \ref{taylor:315} and~%
  \ref{taylor:355},
  plus the introduction of \S~\ref{taylor:314}.

  From another point of view, the chapter errs maybe toward too little
  rigor.  Some pretty constructs of pure mathematics serve the Taylor
  series and Cauchy's integral formula.  However, such constructs drive
  the applied mathematician on too long a detour (a detour
  appendix~\ref{purec} briefly overviews).  The chapter as written
  represents the most nearly satisfactory compromise the writer has been
  able to attain.%
}

(The chapter's early sections, \S\S~\ref{taylor:314}
and~\ref{taylor:317}, are thick with tiny algebraic details.  The reader
who does not wish, for now, to pick through tiny algebraic details can
safely just skim the two sections and then turn ahead to start reading
in \S~\ref{taylor:310}.  Notwithstanding, the reader who would not skip
details, howsoever tiny the details might be, can sharpen his pencil and
continue to \S~\ref{taylor:314}, next.)

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

\section{The power-series expansion of $1/(1-z)^{n+1}$}
\label{taylor:314}
\index{expansion of $1/(1-z)^{n+1}$}

Before approaching the Taylor series proper in \S~\ref{taylor:310}, we
shall find it both interesting and useful to demonstrate that
\bq{taylor:314:70}
  \frac{1}{(1-z)^{n+1}} = \sum_{k=0}^{\infty} \cmb{n+k}{n} z^k,
  \ \ n\ge 0.
\eq
The demonstration comes in three stages.  Of the
three, it is the second stage (\S~\ref{taylor:314.20}) which actually
proves~(\ref{taylor:314:70}).  The first stage
(\S~\ref{taylor:314.10}) comes up with the formula for the second stage
to prove.  The third stage (\S~\ref{taylor:314.25}) establishes the
sum's convergence.  In all the section,
\[
  i,j,k,m,n,K \in \mathbb Z.
\]

\subsection{The formula}
\label{taylor:314.10}

In \S~\ref{alggeo:228.30} we found that
\[
  \frac{1}{1-z} = \sum_{k=0}^{\infty} z^k = 1 + z + z^2 + z^3 + \cdots
\]
for $\left|z\right| < 1$.  What about $1/(1-z)^2$, $1/(1-z)^3$, $1/(1-z)^4$, and so
on?  By the long-division procedure of Table~\ref{alggeo:228:tbl-up},
one can calculate the first few terms of $1/(1-z)^2$ to be
\[
\frac{1}{(1-z)^2} = \frac{1}{1-2z+z^2}
  = 1 + 2z + 3z^2 + 4z^3 + \cdots
\]
whose coefficients $1,2,3,4,\ldots$ happen to be the numbers down the
first diagonal of Pascal's triangle (Fig.~\ref{drvtv:pasc} on
page~\pageref{drvtv:pasc}; see also Fig.~\ref{drvtv:pasc0}).  Dividing
$1/(1-z)^3$ seems to produce the coefficients $1,3,6,\mbox{0xA},\ldots$
down the second diagonal; dividing $1/(1-z)^4$, the coefficients down
the third.  A curious pattern seems to emerge, worth investigating more
closely.  The pattern recommends the conjecture~(\ref{taylor:314:70}).

To motivate the conjecture a bit more formally (though without actually
proving it yet), suppose
that $1/(1-z)^{n+1}$, $n \ge 0$, is expandable in the power series
\bq{taylor:314:10}
  \frac{1}{(1-z)^{n+1}} = \sum_{k=0}^{\infty} a_{nk} z^k,
\eq
where the~$a_{nk}$ are coefficients to be determined.
Multiplying by $1-z$, we have that
\[
  \frac{1}{(1-z)^n} = \sum_{k=0}^{\infty}
  [a_{nk}-a_{n(k-1)}] z^k.
\]
This is to say that
\[
  a_{(n-1)k} = a_{nk}-a_{n(k-1)},
\]
or in other words that
\bq{taylor:314:30}
  a_{n(k-1)} + a_{(n-1)k} = a_{nk}.
\eq
Thinking of Pascal's triangle,~(\ref{taylor:314:30}) reminds
one of a formula of Table~\ref{drvtv:220:tbl}, transcribed here in the
symbols
\bq{taylor:314:50}
  \cmb{m-1}{j-1} + \cmb{m-1}{j} = \cmb{m}{j},
\eq
except that~(\ref{taylor:314:30}) is not $a_{(m-1)(j-1)} + a_{(m-1)j} =
a_{mj}$.
%except that~(\ref{taylor:314:30}) is not $a_{(j-1)(m-j)} + a_{j(m-j-1)} =
%a_{j(m-j)}$.

\index{false try}
Various changes of variable are possible to
make~(\ref{taylor:314:50}) better match~(\ref{taylor:314:30}).  We might
try at first a few false ones, but eventually the change
\[
  \begin{split}
    n + k &\la m, \\
    k &\la j,
  \end{split}
\]
recommends itself.  Thus changing in~(\ref{taylor:314:50}) gives
\[
  \cmb{n+k-1}{k-1} + \cmb{n+k-1}{k} = \cmb{n+k}{k}.
\]
Transforming according to a rule of
Table~\ref{drvtv:220:tbl}, this is
\bq{taylor:314:51}
  \cmb{n+[k-1]}{n} + \cmb{[n-1]+k}{n-1} = \cmb{n+k}{n},
\eq
which fits~(\ref{taylor:314:30}) perfectly if
\bq{taylor:314:60}
  a_{nk} = \cmb{n+k}{n}.
\eq
Hence we conjecture that~(\ref{taylor:314:60}), applied
to~(\ref{taylor:314:10}), would make~(\ref{taylor:314:70}) true.

Equation~(\ref{taylor:314:70}) is thus suggestive.  It works at least
for the important case of $n=0$; this much is easy to test.  In light
of~(\ref{taylor:314:30}), it seems to imply a relationship between the
$1/(1-z)^{n+1}$ series and the $1/(1-z)^n$ series for any~$n$.  But
\emph{to seem} is not \emph{to be.}  At this point, all we can say is
that~(\ref{taylor:314:70}) seems right.  We will establish that it is
right in the next subsection.

\subsection{The proof by induction}
\label{taylor:314.20}
\index{induction}
\index{proof!by induction}

Equation~(\ref{taylor:314:70}) is proved by induction as follows.
Consider the sum
\bq{taylor:315:10}
  S_n \equiv \sum_{k=0}^{\infty} \cmb{n+k}{n} z^k.
\eq
Multiplying by $1-z$ yields that
\[
  (1-z)S_n = \sum_{k=0}^{\infty}
    \left[ \cmb{n+k}{n} - \cmb{n+[k-1]}{n} \right]
    z^k.
\]
Per~(\ref{taylor:314:51}), this is
\bq{taylor:315:11}
  (1-z)S_n = \sum_{k=0}^{\infty}
    \cmb{[n-1]+k}{n-1} z^k.
\eq

Now suppose that~(\ref{taylor:314:70}) is true for $n=i-1$ (where~$i$
denotes an integer rather than the imaginary unit):
\bq{taylor:315:15}
  \frac{1}{(1-z)^i} = \sum_{k=0}^{\infty}
    \cmb{[i-1]+k}{i-1} z^k.
\eq
In light of~(\ref{taylor:315:11}), this means that
\[
  \frac{1}{(1-z)^i} = (1-z)S_i.
\]
Dividing by $1-z$,
\[
  \frac{1}{(1-z)^{i+1}} = S_i.
\]
Applying~(\ref{taylor:315:10}),
\bq{taylor:315:20}
  \frac{1}{(1-z)^{i+1}} =
    \sum_{k=0}^{\infty} \cmb{i+k}{i} z^k.
\eq
Evidently~(\ref{taylor:315:15}) implies~(\ref{taylor:315:20}).  In other
words, if~(\ref{taylor:314:70}) is true for $n=i-1$, then it is also
true for $n=i$.  Thus \emph{by induction,} if it is true for any one~$n$,
then it is also true for all greater~$n$.

The ``if'' in the last sentence is important.  Like all inductions, this
one needs at least one \emph{start case} to be valid (many inductions
actually need a consecutive pair of start cases).  The $n=0$ supplies
the start case
\[
  \frac{1}{(1-z)^{0+1}} =
    \sum_{k=0}^{\infty}
      \cmb{k}{0} z^k
    = \sum_{k=0}^{\infty} z^k,
\]
which per~(\ref{alggeo:228:45}) we know to be true.

\subsection{Convergence}
\label{taylor:314.25}
\index{convergence}

\index{iff}
\index{Weierstrass, Karl Wilhelm Theodor (1815--1897)}
The question remains as to the domain over which the
sum~(\ref{taylor:314:70}) converges.%
\footnote{
  The meaning of the verb \emph{to converge} may seem clear enough from
  the context and from earlier references, but if explanation here
  helps: a series converges if and only if it approaches a specific,
  finite value after many terms.  A more rigorous way of saying the
  same thing is as follows: the series
  \[
    S = \sum_{k=0}^{\infty} \tau_k
  \]
  converges iff (if and only if), for all possible positive
  constants~$\ep$, there exists a finite $K \ge -1$ such that
  \[
    \left|\sum_{k=K+1}^{n} \tau_k \right| < \ep,
  \]
  for all $n \ge K$ (of course it is also required that the~$\tau_k$
  be finite, but you knew that already).

  The professional mathematical literature calls such convergence
  ``uniform convergence,'' distinguishing it through a test devised by
  Weierstrass from the weaker ``pointwise convergence''
  \cite[\S~1.5]{Andrews}\@.  The applied mathematician can profit
  by learning the professional view in the matter but
  the effect of trying to teach the professional view in a book like
  this would not be pleasing.  Here, we avoid error by keeping a
  clear view of the physical phenomena the mathematics is meant to
  model.

  It is interesting nevertheless to consider an example of an integral
  for which convergence is not so simple, such as Frullani's integral of
  \S~\ref{inttx:460}.
}
To answer the question, consider that per Table~\ref{drvtv:220:tbl},
\[
  \cmb{m}{j} = \frac{m}{m-j}\cmb{m-1}{j}, \ \ m > 0.
\]
With the substitution $n + k \la m$, $n \la j$, this means that
\[
  \cmb{n+k}{n} =
    \frac{n+k}{k}
  \cmb{n+[k-1]}{n},
\]
or more tersely,
\[
  a_{nk} = \frac{n+k}{k}
  a_{n(k-1)},
\]
where
\[
  a_{nk} \equiv \cmb{n+k}{n}
\]
are the coefficients of the power series~(\ref{taylor:314:70}).
Rearranging factors,
\bq{taylor:314:80}
  \frac{a_{nk}}{a_{n(k-1)}} = \frac{n+k}{k} = 1 + \frac{n}{k}.
\eq

\index{majorization}
Multiplying~(\ref{taylor:314:80}) by $z^k/z^{k-1}$ gives the ratio
\[
  \frac{a_{nk}z^k}{a_{n(k-1)}z^{k-1}} = \left(1 + \frac{n}{k}\right)z,
\]
which is to say that the $k$th term of~(\ref{taylor:314:70}) is $(1 +
n/k)z$ times the $(k-1)$th term.  So long as the criterion%
\footnote{
  Although one need not ask the question to understand the proof, the
  reader may nevertheless wonder why the simpler $\left|(1 + n/k)z\right| < 1$ is
  not given as a criterion.  The surprising answer is that not all series
  $\sum \tau_k$ with $\left|\tau_k/\tau_{k-1}\right| < 1$ converge!  For example,
  the extremely simple $\sum 1/k$ does not converge.  As we see however,
  all series $\sum \tau_k$ with $\left|\tau_k/\tau_{k-1}\right| < 1-\delta$ do
  converge.  The distinction is subtle but rather important.

  The really curious reader may now ask why $\sum 1/k$ does not
  converge.  Answer: it \emph{majorizes} $\int_1^{x} (1/\tau)\,d\tau =
  \ln x$.  See~(\ref{cexp:225:dln}) and \S~\ref{taylor:316}.
}
\[
  \left|\left(1 + \frac{n}{k}\right)z\right| \le 1-\delta
\]
is satisfied for all sufficiently large $k>K$---where $0 < \delta \ll 1$
is a small positive constant---then the series evidently converges (see
\S~\ref{alggeo:228.30} and eqn.~\ref{trig:278:triangle}).  But we can
bind $1 + n/k$ as close to unity as desired by making~$K$ sufficiently
large, so to meet the criterion it suffices that
\bq{taylor:314:83}
  \left|z\right|<1.
\eq
The bound~(\ref{taylor:314:83}) thus establishes a sure convergence
domain for~(\ref{taylor:314:70}).

\subsection{General remarks on mathematical induction}
\label{taylor:314.30}

We have proven~(\ref{taylor:314:70}) by means of a mathematical
induction.  The virtue of induction as practiced in
\S~\ref{taylor:314.20} is that it makes a logically clean, airtight case
for a formula.  Its vice is that it conceals the subjective process
which has led the mathematician to consider the formula in the first
place.  Once you obtain a formula somehow, maybe you can prove it by
induction; but the induction probably does not help you to obtain the
formula!  A good inductive proof usually begins by motivating the
formula proven, as in \S~\ref{taylor:314.10}.

\index{Hamming, Richard~W.\ (1915--1998)}
\index{rigor}
\index{applied mathematics}
\index{mathematics!applied}
Richard~W.\ Hamming once said of mathematical induction,
\begin{quote}
  The theoretical difficulty the student has with mathematical induction
  arises from the reluctance to ask seriously, ``How could I prove a
  formula for an infinite number of cases when I know that testing a
  finite number of cases is not enough?''  Once you really face this
  question, you will understand the ideas behind mathematical induction.
  It is only when you grasp the problem clearly that the method becomes
  clear.\ \cite[\S~2.3]{Hamming}
\end{quote}
Hamming also wrote,
\begin{quote}
  The function of rigor is mainly critical and is seldom constructive.
  Rigor is the hygiene of mathematics, which is needed to protect us
  against careless thinking.\ \cite[\S~1.6]{Hamming}
\end{quote}
The applied mathematician may tend to avoid rigor for which he finds no
immediate use, but he does not disdain mathematical rigor on principle.
The style lies in exercising rigor at the right level for the problem at
hand.  Hamming, a professional mathematician who sympathized
with the applied mathematician's needs, wrote further,
\begin{quote}
  Ideally, when teaching a topic the degree of rigor should follow the
  student's perceived need for it\mdots It is necessary to require a
  gradually rising level of rigor so that when faced with a real need
  for it you are not left helpless.  As a result, [one cannot teach] a
  uniform level of rigor, but rather a gradually rising level.
  Logically, this is indefensible, but psychologically there is little
  else that can be done.\ \cite[\S~1.6]{Hamming}
\end{quote}
Applied mathematics holds that the practice \emph{is} defensible, on the
ground that the math serves the model; but Hamming nevertheless makes a
pertinent point.

Mathematical induction is a broadly applicable technique for
constructing mathematical proofs.  We will not always write inductions
out as explicitly in this book as we have done in the present
section---often we will leave the induction as an implicit exercise for
the interested reader---but this section's example at least lays out the
general pattern of the technique.

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

\section{Shifting a power series' expansion point}
\label{taylor:317}
\index{Taylor series!converting a power series to}
\index{expansion point!shifting of}
\index{shifting an expansion point}
\index{power series!shifting the expansion point of}

One more question we should treat before approaching the Taylor
series proper in \S~\ref{taylor:310} concerns the shifting of a power
series' expansion point.  How can the expansion point of the
power series
\bqa
  f(z) &=& \sum_{k=K}^{\infty} (a_k)(z-z_o)^k,
  \label{taylor:317:30}
  \\
  \ds (k,K) &\in& \mathbb Z, \ \ K\le 0,
  \xn
\eqa
which may have terms of negative order, be shifted from $z=z_o$ to $z=z_1$?

The first step in answering the question is straightforward: one
rewrites
% bad break
(\ref{taylor:317:30}) in the form
\[
  f(z) = \sum_{k=K}^{\infty}
    (a_k)([z-z_1]-[z_o-z_1])^k,
\]
then changes the variables
\bq{taylor:317:50}
  \begin{split}
    w &\equiv \frac{z-z_1}{z_o-z_1}, \\
    c_k &\equiv [-(z_o-z_1)]^ka_k,
  \end{split}
\eq
to obtain
\bq{taylor:317:51}
  f(z) = \sum_{k=K}^{\infty}
    (c_k)(1-w)^k.
\eq
Splitting the $k<0$ terms from the $k \ge 0$
terms in~(\ref{taylor:317:51}), we have that
\bqa
  f(z) &=& f_-(z) + f_+(z),  \label{taylor:317:52} \\
  f_-(z) &\equiv& \sum_{k=0}^{-(K+1)} \frac{c_{[-(k+1)]}}{(1-w)^{k+1}}, \xn\\
  f_+(z) &\equiv& \sum_{k=0}^{\infty} (c_k)(1-w)^k. \xn
\eqa
Of the two subseries, the $f_-(z)$ is expanded term by term
using~(\ref{taylor:314:70}), after which combining like powers of~$w$
yields the form
\bq{taylor:317:52q}
  \begin{split}
    f_-(z) &= \sum_{j=0}^{\infty} q_j w^j, \\
    q_j &\equiv \sum_{n=0}^{-(K+1)} \left(c_{[-(n+1)]}\right)\cmb{n+j}{n}.
  \end{split}
\eq
The $f_+(z)$ is even simpler to expand: one need only
multiply the series out term by term per~(\ref{drvtv:230:binthe}),
combining like powers of~$w$ to reach the form
\bq{taylor:317:52p}
  \begin{split}
    f_+(z) &= \sum_{j=0}^{\infty} p_j w^j, \\
    p_j &\equiv \sum_{n=j}^{\infty}(-)^j(c_n)\cmb{n}{j}.
  \end{split}
\eq
% A test of the above has been programmed and tried.  It seems to check
% out, at least for the test parameters.

Equations~(\ref{taylor:317:30}) through~(\ref{taylor:317:52p}) serve
to shift a power series' expansion point, calculating the
coefficients of a power series for $f(z)$ about $z=z_1$, given those of
a power series about $z=z_o$.  Notice that---unlike the original,
$z=z_o$ power series---the new, $z=z_1$ power series has terms
$(z-z_1)^k$ only for $k\ge 0$; it has no terms of negative order.
At the price per~(\ref{taylor:314:83}) of restricting the convergence
domain to $\left|w\right|<1$, our shift of the expansion point away from the
pole at $z=z_o$ has resolved the $k<0$ terms.  Moreover, though one
recognizes the price, we actually pay the price only to the extent to
which there are $k<0$ terms---which often there aren't.

\index{forbidden point}
The method fails if $z=z_1$ happens to be a pole or other nonanalytic
point of $f(z)$.  The convergence domain vanishes as~$z_1$ approaches
such a forbidden point.  (Examples of such forbidden points include
$z=0$ in $h[z] = 1/z$ and in $g[z]=\sqrt z$.  See \S\S~\ref{taylor:320}
through~\ref{taylor:350}.)  Furthermore, even if~$z_1$ does represent a
fully analytic point of $f(z)$, it also must lie within the convergence
domain of the original, $z=z_o$ series for the shift to be trustworthy
as derived.

The attentive reader might observe that we have formally
established the convergence neither of $f_-(z)$
in~(\ref{taylor:317:52q}) nor of $f_+(z)$ in~(\ref{taylor:317:52p}).
Regarding the former convergence,
that of $f_-(z)$, we have strategically framed the problem so that one
needn't worry about it, running the sum in~(\ref{taylor:317:30}) from
the finite $k=K \le 0$ rather than from the infinite $k=-\infty$; and
since according to~(\ref{taylor:314:83}) each term of the original
$f_-(z)$ of~(\ref{taylor:317:52}) converges for $\left|w\right|<1$, the
reconstituted $f_-(z)$ of~(\ref{taylor:317:52q}) safely converges in the
same domain.  The latter convergence, that of $f_+(z)$, is harder to
establish in the abstract because that subseries has an infinite number
of terms.  As we will see by pursuing a different line of argument in
\S~\ref{taylor:310}, however, the $f_+(z)$ of~(\ref{taylor:317:52p}) can
be nothing other than the Taylor series about $z=z_1$ of the function
$f_+(z)$ in any event, enjoying the same convergence domain any such
Taylor series enjoys.%
\footnote{
  A rigorous argument can be constructed without appeal to
  \S~\ref{taylor:310} if desired, from the ratio $n/(n-k)$
  of Table~\ref{drvtv:220:tbl}, which ratio approaches unity
  with increasing~$n$.  A more elegant rigorous argument can be
  made indirectly by way of a complex contour integral.  In applied
  mathematics, however, one does not normally try to shift the expansion
  point of an \emph{unspecified} function $f(z)$, anyway.  Rather, one
  shifts the expansion point of some concrete function like $\sin z$ or
  $\ln(1-z)$.  The imagined difficulty (if any) vanishes in the concrete
  case.  Appealing to \S~\ref{taylor:310}, the important point is the
  one made in the narrative: $f_+(z)$ can be nothing other than the
  Taylor series in any event.
}

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

\section{Expanding functions in Taylor series}
\label{taylor:310}
\index{Taylor series}
\index{series!Taylor}
\index{Taylor, Brook (1685--1731)}

Having prepared the ground, we now stand in position to treat the
Taylor series proper.  The treatment begins with a question: if you had
to express some function $f(z)$ by a power series
\[
  f(z) = \sum_{k=0}^{\infty} (a_k)(z-z_o)^k,
\]
with terms of nonnegative order $k \ge 0$ only, how would you do it?
The procedure of \S~\ref{taylor:314} worked well enough in the case of
$f(z)=1/(1-z)^{n+1}$, but it is not immediately obvious that the same
procedure would work more generally.  What if $f(z)=\sin z$, for example?%
\footnote{
  The actual Taylor series for $\sin z$ is given in \S~\ref{taylor:315}.
}

Fortunately a different way to attack the power-series expansion problem
is known.  It works by asking the question: what power series, having
terms of nonnegative order only, most resembles $f(z)$ in the immediate
neighborhood of $z=z_o$?  To resemble $f(z)$, the desired power series
should have $a_0 = f(z_o)$; otherwise it would not have the right value
at $z=z_o$.  Then it should have $a_1=f'(z_o)$ for the right slope.
Then, $a_2=f''(z_o)/2$ for the right second derivative, and so on.  With
this procedure,
\bq{taylor:310:20}
  f(z) = \sum_{k=0}^{\infty}
    \left(\left.\frac {d^k\!f}
    {dz^k}\right|_{z=z_o}\right)\frac{(z-z_o)^k}{k!}.
\eq
Equation~(\ref{taylor:310:20}) is the \emph{Taylor series.}  Where it
converges, it has all the same derivatives $f(z)$ has, so if $f(z)$
is infinitely differentiable then the Taylor series is an exact
representation of the function.%
\footnote{\label{taylor:310:fn1}%
  The professional mathematician reading such words is likely to blanch.
  To him, such words want rigor.  To him, such words want pages and
  whole chapters
  \cite{Arnold:1997}\cite{Fisher}\cite{Spiegel}\cite{Shilov}\cite{Hildebrand}\cite{Shilov}
  of rigor.

  Before submitting unreservedly to the professional's scruples in
  the matter, however, let us not forget (\S~\ref{intro:284.2}) that the
  professional's approach is founded in postulation, whereas that ours
  is founded in physical metaphor.  Our means differ from his for this
  reason alone.

  Still, an alternate way to think about the Taylor series' sufficiency
  might interest some readers.  It begins with an infinitely
  differentiable function $F(z)$ and its Taylor series $f(z)$
  about~$z_o$, letting $\Delta F(z) \equiv F(z) - f(z)$ be the part of
  $F(z)$ not representable as a Taylor series about~$z_o$.

  If $\Delta F(z)$ is the part of $F(z)$ not representable as a Taylor
  series, then $\Delta F(z_o)$ and all its derivatives at~$z_o$ must
  be identically zero (otherwise by the Taylor series formula of
  eqn.~\ref{taylor:310:20}, one could construct a nonzero Taylor series
  for $\Delta F[z_o]$ from the nonzero derivatives).  However, if $F(z)$
  is infinitely differentiable and if \emph{all} the derivatives of
  $\Delta F(z)$ are zero at $z=z_o$ then---by the alternate, unbalanced
  definition~(\ref{drvtv:defunbal}) of the derivative---%
  %from \S~\ref{drvtv:240},
  all the derivatives must
  also be zero at $z=z_o\pm\ep$, whence also at $z=z_o\pm2\ep$, and so
  on.  This means that $\Delta F(z) = 0$.  In other words, there is no
  part of $F(z)$ not representable as a Taylor series.

  A more formal way to make the same argument would be to suppose that
  $\left. d^n\,\Delta F/dz^n \right|_{z=z_o+\ep} = h$ for some integer
  $n \ge 0$; whereas that this would mean that $\left. d^{n+1}\,\Delta
  F/dz^{n+1} \right|_{z=z_o} = h/\ep$; but that, inasmuch as the latter
  is one of the derivatives of $\Delta F(z)$ at $z=z_o$, it follows that
  $h=0$.

  The interested reader can fill the details in, but basically that is
  how the alternate argument begins.
  After all, if the first, second, third derivatives and so forth,
  evaluated at some expansion point, indicate anything at all, then must
  they not indicate how the function and its several derivatives will
  evolve from that point?  And if they do indicate that, then what could
  null derivatives indicate but null evolution?  Yet such arguments even
  if accepted are satisfactory only from a certain point of view, and
  yet slightly less so once one considers the asymptotic series
  % diagn: check and possibly add more to this reference
  of \S~\ref{prob:760} later in the book.

  A more elegant rigorous argument, preferred by the professional
  mathematicians~\cite{Kohler-lecture}\cite{Arnold:1997}\cite{Fisher} but needing
  significant theoretical preparation, involves integrating over a
  complex contour about the expansion point.  Appendix~\ref{purec}
  sketches that proof.

  The discussion of rigor is confined here to a footnote
  not to deprecate rigor as such, but to deprecate insistence on rigor
  which serves little known purpose in applications.  Applied
  mathematicians normally regard mathematical functions to be imprecise
  analogs of, or metaphors for,
  physical quantities of interest.  Since the functions are imprecise
  analogs in any case, the applied mathematician is logically free
  implicitly \emph{to define} the functions he uses to be Taylor series in
  the first place; that is, to restrict the set of infinitely
  differentiable functions used in the model to the subset of such
  functions representable as Taylor series.  With such an implicit
  definition, whether there actually exist any infinitely differentiable
  functions not representable as Taylor series is more or less beside
  the point---at least until a concrete need for such a hypothetical
  function should present itself.

  In applied mathematics, the definitions serve the model, not the other
  way around.  Other than to divert the interested reader to
  appendix~\ref{purec}, this footnote will leave the matter in that
  form.

  (It is entertaining incidentally to consider \cite[``Extremum'']{EWW}
  the Taylor series of the function $\sin[1/x]$---though in practice
  this particular function is readily expanded after the obvious change
  of variable $u \la 1/x$.)
}

\index{Maclaurin series}
\index{Maclaurin, Colin (1698--1746)}
The Taylor series is not guaranteed to converge outside some
neighborhood near $z=z_o$, but where it does converge it is precise.

When $z_o=0$, the series is also called the \emph{Maclaurin
series.}  By either name, the series is a construct of great importance
and tremendous practical value, as we shall soon see.

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

\section{Analytic continuation}
\label{taylor:320}
\index{analytic continuation}

\index{analytic function}
\index{function!analytic}
\index{infinite differentiability}
\index{Taylor series!transposition of to a different expansion point}
As earlier mentioned in \S~\ref{alggeo:225.3}, an \emph{analytic
function} is a function which is infinitely differentiable in the domain
neighborhood of interest---or, maybe more appropriately for our applied
purpose, a function expressible as a Taylor series in that
neighborhood.  As we have seen, only one Taylor series about~$z_o$ is
possible for a given function $f(z)$:
\[
  f(z) = \sum_{k=0}^{\infty} (a_k) (z-z_o)^k.
\]
However, nothing prevents one from transposing the series to a different
expansion point $z=z_1$ by the method of \S~\ref{taylor:317},
except that the transposed series may there enjoy a different
convergence domain.  As it happens, this section's purpose finds it
convenient to swap symbols $z_o \lra z_1$, transposing rather from
expansion about $z=z_1$ to expansion about $z=z_o$.  In the swapped
notation, so long as the expansion point $z=z_o$ lies fully within
(neither outside nor right on the edge of) the $z=z_1$ series'
convergence domain, the two series evidently describe the selfsame
underlying analytic function.

\index{neighborhood}
\index{domain neighborhood}
Since an analytic function $f(z)$ is infinitely differentiable and
enjoys a unique Taylor expansion $f_o(z-z_o)=f(z)$ about each
point~$z_o$ in its domain, it follows that if two Taylor series
$f_1(z-z_1)$ and $f_2(z-z_2)$ find even a small neighborhood
$\left|z-z_o\right| < \ep$ which lies in the domain of both, then the
two can both be transposed to the common $z=z_o$ expansion point.
If the two are found to have the same Taylor series there, then~$f_1$
and~$f_2$ both represent the same function.  Moreover, if a series~$f_3$
is found whose domain overlaps that of~$f_2$, then a series~$f_4$ whose
domain overlaps that of~$f_3$, and so on, and if each pair in the chain
matches at least in a small neighborhood in its region of overlap, then
the whole chain of overlapping series necessarily represents the same
underlying analytic function $f(z)$.  The series~$f_1$ and the
series~$f_n$ represent the same analytic function even if their domains
do not directly overlap at all.

\index{pole}
\index{nonanalytic point}
\index{forbidden point}
\index{single-valued function}
\index{multiple-valued function}
\index{function!single- and multiple-valued}
\index{Argand domain and range planes}
This is a manifestation of the principle of \emph{analytic
continuation.}  The principle holds that if two analytic functions are
the same within some domain neighborhood $\left|z-z_o\right| < \ep$, then they are
the same everywhere.%
\footnote{
  The writer hesitates to mention that he is given to
  understand~\cite{Spiegel} that the domain neighborhood can technically
  be reduced to a domain contour of nonzero length but zero width.
  Having never met a significant application of this extension of the
  principle, the writer has neither researched the extension's proof nor
  asserted its truth.  He does not especially recommend that the reader
  worry over the point.  The domain neighborhood $\left|z-z_o\right| < \ep$
  suffices.
}
Observe however that the principle fails at poles and other nonanalytic
points, because the function is not differentiable there.

The result of \S~\ref{taylor:317}, which shows general power series to
be expressible as Taylor series except at their poles and other
nonanalytic points, extends the analytic continuation principle to cover
power series in general, including power series with terms of negative
order.

Now, observe: though all convergent power series are indeed analytic,
one need not actually expand every analytic function in a power series.
Sums, products and ratios of analytic functions are hardly less
differentiable than the functions themselves---as also, by the derivative
chain rule, is an analytic function of analytic functions.  For example,
where $g(z)$ and $h(z)$ are analytic, there also is $f(z) \equiv
g(z)/h(z)$ analytic (except perhaps at isolated points where $h[z] =
0$).  Besides, given Taylor series for $g(z)$ and $h(z)$ one can make a
power series for $f(z)$ by long division if desired, so that is all
right.  Section~\ref{taylor:385} speaks further on the point.

The subject of analyticity is rightly a matter of deep concern to the
professional mathematician.  It is also a long wedge which drives pure
and applied mathematics apart.  When the professional mathematician
speaks generally of a ``function,'' he means \emph{any function at all.}
One can construct some pretty unreasonable functions if one wants
to, such as
\bqb
  f( [2k+1]{2^m} ) &\equiv& (-)^m, \ \ (k,m) \in \mathbb Z; \\
  f(z) &\equiv& 0\ \mbox{otherwise.}
\eqb
However, neither functions like this~$f(z)$
nor more subtly unreasonable functions normally arise in the modeling of
physical phenomena.  When such functions do arise, one transforms,
approximates, reduces, replaces and/or avoids them.  The full theory
which classifies and encompasses---or explicitly excludes---such
functions is thus of limited interest to the applied mathematician, and
this book does not cover it.%
\footnote{\label{taylor:320:fn10}%
  Many books do cover it in varying degrees,
  including~\cite{Fisher}\cite{Spiegel}\cite{Hildebrand} and numerous
  others.
  The foundations of the pure theory of a complex variable, though
  abstract, are beautiful, and though they do not comfortably fit a book
  like this even an applied mathematician can profit substantially by
  studying them.  The few pages of appendix~\ref{purec} trace only the
  pure theory's main thread.  However that may be, the
  pure theory is probably best appreciated after one already understands
  its chief conclusions.  Though not for the explicit purpose of serving
  the pure theory, the present chapter does develop just such an
  understanding.
}

%% Hm.  I have mixed my metaphors here.  How to unmix them?
%Though any mathematician, pure or applied, can profit substantially by
%studying the pure theory of analytic functions of a complex variable,
%the applied mathematician is advised not to let the pure theory snare
%him---or, at least, the reader of this book is advised not to let it
%snare him.  The pure theory is a deep hole to dive into.  One finds
%oneself---splash!---underwater down there, swimming complex contours of
%various dimly seen shapes in pursuit of the relevant theorems, which
%theorems in turn buoy other theorems, unrooted in application.  The
%fascination of the pure theory, the daunting aspect of the hole,
%conspire to delay and often effectively to prevent the applied
%mathematician from exploring the day-lit mathematical landscape past the
%hole's far rim.  That's not good.  Avoiding the hole itself, this book
%leads the reader about the rim along a safe path at a prudent pace, then
%onward.
%
%To the reader who still wants to dive straightway into the hole: the
%writer salutes him.  Will he hear one more word of counsel?  Applied
%mathematics usually concerns itself with the specific functions used to
%model some physical phenomenon, not with functions generally.%
%\footnote{
%  There is the question of a solution's uniqueness, which asks whether
%  any function other than a particular $f(z)$ solves a given set of
%  equations.  However, the question is typically answered by
%  contradiction, assuming falsely that there existed an $\tilde f(z)
%  \neq f(z)$, then forming the difference $\Delta f(z) \equiv \tilde
%  f(z) - f(z)$ and showing that $\Delta f(z)$ cannot but everywhere be
%  null.
%}
%Before the reader dives, let him consider whether there is some specific
%function he knows, useful in modeling a physical phenomenon, which
%function the applied theory inadequately treats, for which function one
%cannot or should not rearrange the problem to evade the difficulty.
%Section~\ref{taylor:385} contrives some ideas along these lines, but at
%the moment the writer writes these words, he can think of no such
%function.  The reader who, too, can think of none%
%\footnote{
%  The reader who \emph{can} think of one is most cordially asked to
%  e-mail it to the author, at the address on the title page's back.
%}
%might consider delaying the dive.

This does not mean that the scientist or engineer never encounters
nonanalytic functions.  On the contrary, he encounters several, but they
are not subtle: $\left|z\right|$; $\arg z$; $z^{*}$; $\Re(z)$; $\Im(z)$; $u(t)$;
$\delta(t)$.  Refer to \S\S~\ref{alggeo:225} and~\ref{integ:670}.  Such
functions are nonanalytic either because they lack proper derivatives in
the Argand plane according to~(\ref{drvtv:defz}) or because one has
defined them only over a real domain.

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

\section{Branch points}
\label{taylor:325}
\index{branch point}

The function $g(z)\equiv\sqrt z$ is an interesting, troublesome function.
Its derivative is $dg/dz = 1/2\sqrt z$, so even though the function is
finite at $z=0$, its derivative is not finite there.  Evidently $g(z)$
has a nonanalytic point at $z=0$, yet the point is not a pole.  What is
it?

\index{contour}
\index{domain contour}
\index{range contour}
We call it a \emph{branch point.}  The defining characteristic of the
branch point is that, given a function $f(z)$ with such a point at
$z=z_o$, if one encircles%
\footnote{
  For readers whose native language is not English, ``to encircle''
  means ``to surround'' or ``to enclose.''  The verb does not require
  the boundary to have the shape of an actual, geometrical circle;
  any closed shape suffices.  However, the circle is a typical shape,
  probably the most fitting shape to imagine when thinking of the
  concept abstractly.
}
the point once alone (that is, without also encircling some other
branch point) by a closed contour in the Argand domain plane, while
simultaneously tracking $f(z)$ in the Argand range plane---and if one
demands that~$z$ and $f(z)$ move smoothly, that neither of these
suddenly skip
from one spot to another---then one finds that $f(z)$ ends in a
different place than it began, even though~$z$ itself has returned
precisely to its own starting point.  The range contour remains open
even though the domain contour is closed.
\begin{quote}
  In complex analysis, a branch point may be thought of informally as a
  point~$z_o$ at which a ``multiple-valued function'' changes values
  when one winds once around~$z_o$.\footnote{\cite[``Branch point,'' 18:10, 16
  May 2006]{wikip}}
  % The above citation style differs slightly from the book's usual.
  % Wikipedia citation just draws too much attention, with its timestamp
  % and all.
\end{quote}

\index{single-valued function}
\index{multiple-valued function}
\index{function!single- and multiple-valued}
\index{visual dimension or axis}
\index{dimension!visual}
\index{axis!visual}
An analytic function like $g(z)\equiv\sqrt z$ having a branch point evidently
is not single-valued.  It is multiple-valued.  For a single~$z$ more
than one distinct $g(z)$ is possible, as Fig.~\ref{taylor:325fig} suggests.
\begin{figure}
  \caption{A co\"ordinated evolution of~$z$ and $g(z)\equiv \sqrt{z}$.}
  \label{taylor:325fig}
  \bc
    \nc\fxa{-5.25} \nc\fxb{6.9}
    \nc\fya{-1.95} \nc\fyb{2.45}
    \nc\ttax{2.6}
    \nc\ttb{1.7}
    \nc\ttc{1.3}
    \nc\ttd{0.8}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      %\localscalebox{1.0}{1.0}
      {%
        \small%
        \psset{dimen=middle}%
        \psset{linewidth=2.0pt}%
        \nc\ttdrawaxes{{%
          \psset{linewidth=0.5pt}%
          \psline(-\ttb,0)(\ttb,0)%
          \psline(0,-\ttb)(0,\ttb)%
        }}%
        \rput(-\ttax,0){%
          \ttdrawaxes%
          \uput[r](\ttb,0){$\Re\{z\}$}%
          \uput[u](0,\ttb){$\Im\{z\}$}%
          \rput(\ttc,0)\psdot%
          \psarc{->}(0,0){\ttc}{0}{351}%
          \rput{*0}(0,0){$\pscrossbwithborder$}%
        }%
        \rput(\ttax,0){%
          \ttdrawaxes%
          \uput[r](\ttb,0){$\ds\Re\left\{g(z)\equiv{\sqrt z}\right\}$}%
          \uput[u](0,\ttb){$\ds\Im\left\{g(z)\equiv{\sqrt z}\right\}$}%
          \rput(\ttd,0)\psdot%
          \psarc{->}(0,0){\ttd}{0}{175.5}%
        }%
      }%
    \end{pspicture}
  \ec
\end{figure}
(Looking at the figure, incidentally, the perceptive reader might ask
why the figure does not merely plot $g[z]$ against~$z$ on a single pair
of axes.  Of course, the reader knows why, but the question is worth
asking, anyway.  The answer is that the figure would indeed like to
plot $g[z]$ against~$z$ on a single pair of axes but cannot because four
dimensions would be required for the visual!  Three dimensions are
sometimes let to overburden two-dimensional paper---as in
Fig.~\ref{integ:fig-sphere} for example---but four dimensions are too
many; so, instead, Fig.~\ref{taylor:325fig} co\"ordinates a pair of
plots at two visual dimensions per plot.  There is irony in this, for
the real and imaginary parts of a scalar together constitute only a
single actual dimension, but no one seems to know how to display an
imaginary part \emph{visually} without using an extra axis.)

\index{pole}
An analytic function like $h(z)\equiv 1/z$, by contrast to $g(z)$,
is single-valued even
though it has a pole.  This function does not suffer the syndrome
described.  When a domain contour encircles a pole of $h(z)$ or of any
other function that has a pole,
the corresponding
range contour is properly closed.  Poles do not cause their functions to
be multiple-valued and, thus, \emph{poles are not branch points.}

Evidently $f(z)\equiv(z-z_o)^a$ has a branch point at $z=z_o$ if and
only if~$a$ is not an integer.  If $f(z)$ does have a
branch point---if~$a$ is not an integer---then the mathematician must
draw a distinction between $z_1=z_o+\rho e^{i\phi}$ and $z_2=z_o+\rho
e^{i(\phi+2\pi)}$, \emph{even though the two are exactly the same
number.}  Indeed $z_1=z_2$, but paradoxically $f(z_1)\neq f(z_2)$.

This is difficult.  It is confusing, too, until one realizes that the
fact of a branch point says nothing whatsoever about the argument~$z$.
As far as~$z$ is concerned, there really is no distinction between
$z_1=z_o+\rho e^{i\phi}$ and $z_2=z_o+\rho e^{i(\phi+2\pi)}$---none at all.
What draws the distinction is the multiple-valued function $f(z)$ which
uses the argument.

\index{Pfufnik, Gorbag~J.}
It is as though I had a mad colleague who called me Thaddeus~H.\ Black,
until one day I happened to walk past behind his desk (rather than in
front as I usually did), whereupon for some reason he began calling me
Gorbag~J.\ Pfufnik.  I had not changed at all, but now the colleague calls
me by a different name.  The change isn't really in me, is it?  It's in
my colleague, who seems to suffer a branch point.  If it is important to
me to be sure that my colleague really is addressing me when he cries,
``Pfufnik!'' then I had better keep a running count of how many times I
have turned about his desk, hadn't I, even though the number of turns is
personally of no import to me.

\index{branch point!strategy to avoid}
The usual analysis strategy when one encounters a branch point is simply
to avoid the point.  Where an integral follows a closed contour as in
\S~\ref{taylor:350}, the strategy is to compose the contour to exclude
the branch point, to shut it out.  Such a strategy of avoidance usually
prospers.%
\footnote{
  Traditionally associated with branch points in complex variable theory
  are the notions of \emph{branch cuts} and \emph{Riemann sheets.} These
  ideas are interesting, but are not central to the analysis as
  developed in this book and are not covered here.  The interested
  reader might consult a book on complex variables or advanced
  calculus like~\cite{Hildebrand}, among many others.
}

Curiously, the function $p(z) \equiv \ln z$ has a branch point at $z =
0$ despite that the function's order is zero (\S~\ref{cexp:228}).
By contrast, the order of $g(z)\equiv \sqrt z$ was $1/2$, a noninteger,
so a branch point was to be expected there; whereas a branch point in a
zeroth-order function like $\ln(\cdot)$ comes perhaps as a
surprise---see Fig.~\ref{taylor:325figln}.
\begin{figure}
  \caption{A co\"ordinated evolution of~$z$ and $p(z)\equiv \ln{z}$.}
  \label{taylor:325figln}
  \bc
    \nc\fxa{-5.25} \nc\fxb{6.9}
    \nc\fya{-1.95} \nc\fyb{2.45}
    % scale = 0.8*0.8/1.3; |z| = (1.3/0.8)^2
    \nc\ttax{2.6}
    \nc\ttb{1.7}
    \nc\ttc{1.3}
    \nc\ttd{0.478038464769675}
    \nc\ttf{1.6} % clipped
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      %\localscalebox{1.0}{1.0}
      {%
        \small%
        \psset{dimen=middle}%
        \psset{linewidth=2.0pt}%
        \nc\ttdrawaxes{{%
          \psset{linewidth=0.5pt}%
          \psline(-\ttb,0)(\ttb,0)%
          \psline(0,-\ttb)(0,\ttb)%
        }}%
        \rput(-\ttax,0){%
          \ttdrawaxes%
          \uput[r](\ttb,0){$\Re\{z\}$}%
          \uput[u](0,\ttb){$\Im\{z\}$}%
          \rput(\ttc,0)\psdot%
          \psarc{->}(0,0){\ttc}{0}{351}%
          \rput{*0}(0,0){$\pscrossbwithborder$}%
        }%
        \rput(\ttax,0){%
          \ttdrawaxes%
          \uput[r](\ttb,0){$\ds\Re\left\{p(z)\equiv{\ln z}\right\}$}%
          \uput[u](0,\ttb){$\ds\Im\left\{p(z)\equiv{\ln z}\right\}$}%
          \rput(\ttd,0)\psdot%
          \psline{->}(\ttd,0)(\ttd,\ttf)
        }%
      }%
    \end{pspicture}
  \ec
\end{figure}
Fortunately, as \S~\ref{taylor:350} will soon show, the branch point
of~$\ln(\cdot)$ is not a point one needs to avoid.  On the contrary, one
often explicitly seeks out such a point.  Before addressing that
interesting matter, though, let us turn attention briefly to the
definitions of entire and meromorphic functions, next, and after that to
extrema over a complex domain.

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

\section{Entire and meromorphic functions}
\label{taylor:330}
\index{function!entire}
\index{function!meromorphic}
\index{entire function}
\index{meromorphic function}

Though an applied mathematician is unwise to let abstract definitions
enthrall his thinking, pure mathematics nevertheless brings some
technical definitions the applied mathematician can use.  Two such
are the definitions of \emph{entire} and \emph{meromorphic} functions.%
\footnote{\cite{EWW}}

A function $f(z)$ which is analytic for all finite~$z$ is an
\emph{entire function.}\footnote{\cite[chapter~6]{Spiegel}}
Examples include $f(z) = z^2$ and $f(z) = \exp
z$, but not $f(z) = 1/z$ which has a pole at
$z=0$.

\index{essential singularity}
\index{singularity!essential}
A function $f(z)$ which is analytic for all finite~$z$ except at
isolated poles (which can be $n$-fold poles if~$n$ is a
finite, positive integer), which has no branch points,
of which no circle of finite radius in the Argand domain plane
encompasses an infinity of poles, is a \emph{meromorphic
function.}\footnote{The definition follows~\cite[\S~1.1]{Lebedev}.
At least one competent author~\cite[chapter~6]{Spiegel} however seems
(inadvertently?) to exclude functions with an infinite number of poles
like the $P(z)\equiv\sum_{k=0}^\infty [-]^k/[k!(z+k)]$
of~\cite{Lebedev}.  Nevertheless, \emph{according to the book you are
now reading,} a function like~$P(z)$ remains meromorphic because, though
it has an infinity of poles, it does not crowd this infinity into any
finite domain.}
Examples include $f(z) = 1/z$, $f(z) = 1/(z+2) + 1/(z-1)^3 +
2z^2$ and $f(z) = \tan z$---the last of which has an infinite number of
poles, but of which the poles nowhere cluster in infinite numbers.  The
function $f(z) = \tan (1/z)$ is not meromorphic since it has an infinite
number of poles within (for instance) the Argand unit circle.
Even the function $f(z)
= \exp (1/z)$ is not meromorphic: it has only the one, isolated
nonanalytic point at $z=0$, and that point is no branch point; but the
point is an \emph{essential singularity,} having the character of an
infinitifold ($\infty$-fold) pole.%
\footnote{\cite{Kohler-lecture}}

Incidentally, if it seems unclear that the singularities of $\tan z$ are
actual poles, then consider that
\[
  \tan z = \frac{\sin z}{\cos z} = -\frac{\cos w}{\sin w},
\]
wherein we have changed the variable
\[
  w \la z - (2n+1)\frac{2\pi}{4}, \ \ n \in \mathbb Z.
\]
Section~\ref{taylor:315} and its Table~\ref{taylor:315:tbl}, below,
give Taylor series for $\cos z$ and $\sin z$, with which
\[
  \tan z =
  \frac{-1+w^2/2-w^4/\mbox{0x18}-\cdots}{w-w^3/6+w^5/\mbox{0x78}-\cdots}.
\]
By long division,
\[
  \tan z = -\frac{1}{w} +
  \frac{w/3-w^3/\mbox{0x1E}+\cdots}{1-w^2/6+w^4/\mbox{0x78}-\cdots}.
\]
(On the other hand, if it is unclear that $z = [2n+1][2\pi/4]$ are
the only singularities $\tan z$ has---that it has no singularities of
which $\Im[z] \neq 0$---then consider that the singularities of $\tan z$
occur where $\cos z=0$, which by Euler's formula,
eqn.~\ref{cexp:250:cos}, occurs where $\exp[+iz] = \exp[-iz]$.  This in
turn is possible only if $\left|\exp[+iz]\right| =
\left|\exp[-iz]\right|$, which happens only for real~$z$.)

Sections~\ref{taylor:380}, \ref{taylor:385} and~\ref{inttx:260} speak
further of the matter.

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

\section{Extrema over a complex domain}
\label{taylor:335}
\index{extremum}

If a function $f(z)$ is expanded by~(\ref{taylor:310:20}) or by other
means about an analytic expansion point $z=z_o$ such that
\[
  f(z) = f(z_o) + \sum_{k=1}^\infty (a_k) (z-z_o)^k;
\]
and if
\bqb
  a_k &=& 0 \ \ \ \mbox{for $k < K$, but} \\
  a_K &\neq& 0, \\
  (k,K) &\in& \mathbb Z, \ \ 0 < K < \infty,
\eqb
such that~$a_K$ is the series' first nonzero
coefficient; then, in the immediate neighborhood of the expansion
point,\footnote{%
  The pure logicist might prefer to express this in the
  $\delta$-$\epsilon$ style of \S~\ref{drvtv:240.38}, especially
  since~$z$ and~$z_o$ might have physical units of measure attached, in
  which case the inequality that $\left|z-z_o\right| \ll 1$ though
  suggestive would be strictly meaningless.  However, the practical
  applicationist is probably not so fussy.

  What the notation intends to specify, and what the applied
  mathematician understands it to mean, is that~$z$ lie infinitesimally
  close to~$z_o$, or at any rate that~$z$ lie sufficiently close
  to~$z_o$ to emphasize the behavior described.  Just \emph{how}
  close~$z$ should lie is not the point.  If in doubt, go closer!

  In case the reader is still unsatisfied, here it is in
  $\delta$-$\epsilon$ style: for any positive quantity~$\ep$ whose
  physical units of measure (if any) are compatible as follows, there
  exists a positive quantity~$\delta$ such that $|[f(z_o) + (a_K)
  (z-z_o)^K - f(z)]/[z-z_o]^K| < \ep$ for all $|z-z_o| < \delta$.
  %or, if you prefer,
  %$
  %  \forall\ep\in\mathbb R>0,
  %  \exists\delta\in\mathbb R>0:
  %  \forall z\in\mathbb C,
  %  z_o\in\mathbb C,
  %  |[f(z_o) + (a_K) (z-z_o)^K - f(z)]/[z-z_o]^K| < \ep
  %  \Rightarrow
  %  |z-z_o| < \delta
  %$,
  %where~$\mathbb R$ is the set of all real numbers,~%
  %$\mathbb C$ is the set of all complex numbers,~%
  %$\forall$ means ``for all,''~%
  %$\exists$ means ``there exists,'' and~%
  %$\Rightarrow$ means ``implies that''---%
  %though the writer like most engineers seldom uses
  %$\forall\exists$-type logical notation~\cite[appendix~A]{Shilov} and
  %cannot promise that he has here used it well.
  Nevertheless, in applications, the narrative's briefer style, $|z-z_o|
  \ll 1$, probably suffices.%
}
\[
  f(z) \approx f(z_o) + (a_K) (z-z_o)^K,
  \ \ \left|z-z_o\right| \ll 1.
\]
Changing $\rho' e^{i\phi'} \la z-z_o$, this is
\bq{taylor:335:10}
  f(z) \approx f(z_o) + a_K\left(\rho'\right)^K e^{iK\phi'},
  \ \ 0 \le \rho' \ll 1.
\eq
Evidently one can shift the output of an analytic function $f(z)$
slightly in any desired Argand direction by shifting slightly the
function's input~$z$.  Specifically according to~(\ref{taylor:335:10}),
to shift $f(z)$ by $\Delta f \approx \ep e^{i\psi}$, one can shift~$z$
by $\Delta z \approx (\ep/a_K)^{1/K}e^{i(\psi+n2\pi)/K}$, $n \in \mathbb Z$.  Except at a
nonanalytic point of $f(z)$ or in the trivial case that~$f(z)$ were
everywhere constant, this always works---even where $[df/dz]_{z=z_o}=0$.

That by varying an analytic function's input one can smoothly shift the
function's output in any desired Argand
direction has the significant consequence that neither the
real nor the imaginary part of the function---nor for that matter any
linear combination $\Re[e^{-i\omega} f(z)]$ of the real and imaginary
parts---can have an extremum within the interior of a domain over which
the function is fully analytic.  That is, \emph{a function's extrema
over a bounded analytic domain never lie within the domain's interior
but always on its boundary}%
\footnote{
  Professional mathematicians tend to define the domain and its boundary
  more carefully.
}$\mbox{}^,$%
\footnote{
  \cite{deSturler-lecture}%
  \cite{Kohler-lecture}
}
(except, as earlier mentioned, in the trivial case of an~$f[z]$ that is
everywhere constant).

The last consequence is perhaps unexpected.  It would not have been so
for a real domain bounded by a pair of end points, but for an analytic
domain bounded by a complex contour that is the way it is.
When looking for a complex function's extrema, one need not search an
analytic domain's interior, for if the domain contains no poles nor any
other nonanalytic points then, apparently, the domain's boundary is the
only place an extremum can exist.

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

\section{Cauchy's integral formula}
\label{taylor:350}
\index{Cauchy's integral formula}
\index{Cauchy, Augustin Louis (1789--1857)}
\index{integral!closed complex contour}
\index{contour integration!closed complex}

In \S~\ref{integ:260} we considered the problem of vector contour
integration, in which the sum value of an integration depends not only
on the integration's endpoints but also on the path, or \emph{contour,}
over which the integration is done, as in Fig.~\ref{integ:260:fig}.
Because real scalars are confined to a single line, no alternate choice
of path is possible where the variable of integration is a real scalar,
so the contour problem does not arise in that case.  It does however
arise where the variable of integration is a \emph{complex} scalar,
because there again different paths are possible.  Refer to the Argand
plane of Fig.~\ref{alggeo:225:fig}.

Consider the integral
\bq{taylor:350:10}
  S_n = \int_{z_1}^{z_2} z^{n-1} \,dz, \ \ n \in \mathbb Z.
\eq
If~$z$ were always a real number, then by the antiderivative
(\S~\ref{integ:230}) this integral would evaluate to $(z_2^n-z_1^n)/n$;
or, in the case of $n=0$, to $\ln(z_2/z_1)$ [though complications could
still arise if $n<0$ and~$z_2$ differed in sign from~$z_1$].  Inasmuch
as~$z$ is complex, however, the correct evaluation is less obvious.  To
evaluate the integral sensibly in the latter case, one must consider
some specific path of integration in the Argand plane.  One must also
consider the meaning of the symbol~$dz$.

\subsection{The meaning of the symbol~$dz$}
\label{taylor:350.10}
\index{$dz$}

\index{infinitesimal!dropping of when negligible}
The symbol~$dz$ represents an infinitesimal step in some direction in
the Argand plane:
\bqb
  dz &=& \left[z+dz\right] - \left[z\right] \\
  &=& \left[(\rho+d\rho)e^{i(\phi+d\phi)}\right]
    -\left[\rho e^{i\phi}\right] \\
  &=& \left[(\rho+d\rho)e^{i\,d\phi}e^{i\phi}\right]
    -\left[\rho e^{i\phi}\right] \\
  &=& \left[(\rho+d\rho)(1+i\,d\phi)e^{i\phi}\right]
    -\left[\rho e^{i\phi}\right].
\eqb
Since the product of two infinitesimals is negligible even
on an infinitesimal scale,
we can drop the $d\rho\,d\phi$ term.%
\footnote{
  The dropping of second-order infinitesimals like $d\rho\,d\phi$,
  added to first order infinitesimals like~$d\rho$, is a standard
  calculus technique.  One cannot \emph{always} drop them, however.
  Occasionally one encounters a sum in which not only do the finite
  terms cancel, but also the first-order infinitesimals.  In such a
  case, the second-order infinitesimals dominate and cannot be dropped.
  An example of the type is
  \[
    \lim_{\ep\ra 0}\frac{(1-\ep)^3+3(1+\ep)-4}{\ep^2} =
    \lim_{\ep\ra 0}\frac{(1-3\ep+3\ep^2)+(3+3\ep)-4}{\ep^2} = 3.
  \]

  One typically notices that such a case has arisen when the dropping
  of second-order infinitesimals has left an ambiguous~$0/0$.
  To fix the problem, you simply go back to the step during which you
  dropped the infinitesimal and you restore it, then you proceed
  from there.  Otherwise there isn't much point in carrying
  second-order infinitesimals around.  In the relatively uncommon event
  that you need them, you'll know it.  The math itself will tell you.
}
After canceling finite terms, we are left with the peculiar but
fine formula
\bq{taylor:350:dz}
  \index{$dz$}
  dz = (d\rho+i\rho\,d\phi)e^{i\phi}.
\eq

\subsection{Integrating along the contour}
\label{taylor:350.20}
\index{integral!complex contour}
\index{contour integration!complex}
\index{contour!complex}

Now consider the integration~(\ref{taylor:350:10}) along the contour of
Fig.~\ref{taylor:350:fig1}.
\begin{figure}
  \caption[A complex contour of integration in two segments.]{A
    contour of integration in the Argand plane, in two segments:
    constant-$\rho$ ($z_a$ to~$z_b$); and constant-$\phi$ ($z_b$ to~$z_c$).}
  \label{taylor:350:fig1}
  \bc
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  \ec
\end{figure}
In\-te\-grat\-ing along the constant-$\phi$ segment,
\bqb
  \int_{z_b}^{z_c} z^{n-1} \,dz
  &=&
    \int_{\rho_b}^{\rho_c} (\rho e^{i\phi})^{n-1} (d\rho+i\rho\,d\phi)e^{i\phi}
  \\&=&
    \int_{\rho_b}^{\rho_c} (\rho e^{i\phi})^{n-1} (d\rho)e^{i\phi}
  \\&=&
    e^{in\phi}\int_{\rho_b}^{\rho_c} \rho^{n-1} d\rho
  \\&=&
    \frac{e^{in\phi}}{n}\left(\rho_c^n-\rho_b^n\right)
  \\&=&
    \frac{z_c^n-z_b^n}{n}.
\eqb
In\-te\-grat\-ing along the constant-$\rho$ arc,
\bqb
  \int_{z_a}^{z_b} z^{n-1} \,dz
  &=&
    \int_{\phi_a}^{\phi_b} (\rho e^{i\phi})^{n-1} (d\rho+i\rho\,d\phi)e^{i\phi}
  \\&=&
    \int_{\phi_a}^{\phi_b} (\rho e^{i\phi})^{n-1} (i\rho\,d\phi)e^{i\phi}
  \\&=&
    i\rho^{n}\int_{\phi_a}^{\phi_b} e^{in\phi} \,d\phi
  \\&=&
    \frac{i\rho^n}{in}\left(e^{in\phi_b}-e^{in\phi_a}\right)
  \\&=&
    \frac{z_b^n-z_a^n}{n}.
\eqb
Adding the two, we have that
\[
  \int_{z_a}^{z_c} z^{n-1} \,dz
  = \frac{z_c^n-z_a^n}{n},
\]
surprisingly the same as for real~$z$.  Moreover, contrary to the
diagram but nevertheless fairly obviously, nothing prevents one from
specifying an alternate path from~$z_a$ to~$z_c$ that reverses the
sequence, traversing a constant-$\phi$ segment first and then a
constant-$\rho$ arc afterward; as long as $n\neq 0$ and, if $n <
0$, the path observes to avoid the point $z=0$, such a change apparently
would not alter the last result.  Either way, since any path of
integration between any two complex numbers~$z_1$ and~$z_2$ is
approximated arbitrarily closely per~(\ref{taylor:350:dz}) by a
succession of short constant-$\rho$ and constant-$\phi$ elements, it
follows generally that
\bq{taylor:350:20}
  \int_{z_1}^{z_2} z^{n-1} \,dz
  = \frac{z_2^n-z_1^n}{n},\ \ n \in \mathbb Z, \ n\neq 0.
\eq

The applied mathematician might reasonably ask,
``Was~(\ref{taylor:350:20}) really worth the trouble?  We knew
\emph{that} already.  It's the same as for real numbers.''

Well, we really didn't know it before deriving it, but the point is well
taken nevertheless.  However, notice the exemption of $n=0$.
Equation~(\ref{taylor:350:20}) does not hold in that case.  Consider the
$n=0$ integral
\[
  S_0 = \int_{z_1}^{z_2} \frac{dz}{z}.
\]
Following the same steps as before and using~(\ref{cexp:225:dln})
and~(\ref{alggeo:230:311}), we find that
\bq{taylor:350:31}
  \int_{\rho_1}^{\rho_2} \frac{dz}{z}
  = \int_{\rho_1}^{\rho_2}
    \frac{(d\rho+i\rho\,d\phi)e^{i\phi}}{\rho e^{i\phi}}
  = \int_{\rho_1}^{\rho_2} \frac{d\rho}{\rho}
  = \ln\frac{\rho_2}{\rho_1}.
\eq
This is always real-valued, but otherwise it brings no surprise.
However,
\bq{taylor:350:32}
  \int_{\phi_1}^{\phi_2} \frac{dz}{z}
  = \int_{\phi_1}^{\phi_2}
    \frac{(d\rho+i\rho\,d\phi)e^{i\phi}}{\rho e^{i\phi}}
  = i\int_{\phi_1}^{\phi_2} d\phi
  = i (\phi_2-\phi_1).
\eq
The odd thing about this is in what happens when the contour closes a
complete loop in the Argand plane about the $z=0$ pole.  In this case,
$\phi_2=\phi_1+2\pi$, so
\[
  S_0 = i2\pi
\]
\emph{even though the integration ends where it begins.}

Generalizing, we have that
\bq{taylor:cauchy0}
  \renewcommand{\arraystretch}{2.0}
  \br{rcl}
    \ds\oint (z-z_o)^{n-1} \,dz &=& 0,\ \ n \in \mathbb Z, \ n \neq 0; \\
    \ds\oint \frac{dz}{z-z_o} &=& i2\pi;
  \er
\eq
where as in \S~\ref{integ:260} the symbol~$\oint$ represents integration
about a closed contour that ends where it begins, and where it is
implied that the contour loops positively (counterclockwise, in the
direction of increasing~$\phi$) exactly once about the $z=z_o$ pole.

Notice that the formula's~$i2\pi$ does not depend on the precise path of
integration, but only on the fact that the path loops once positively
about the pole.  Notice also that nothing in the derivation
of~(\ref{taylor:350:20}) actually requires that~$n$ be an integer, so
one can write,
\bq{taylor:350:21}
  \int_{z_1}^{z_2} z^{a-1} \,dz
  = \frac{z_2^a-z_1^a}{a},\ \ a\neq 0.
\eq
However,~(\ref{taylor:cauchy0}) does not hold in the latter case;
its integral comes to zero for nonintegral~$a$ only if the contour
does not enclose the branch point at $z=z_o$.

\index{pole}
\index{nonanalytic point}
For a closed contour \emph{which encloses no
pole or other nonanalytic point,}
% bad break
(\ref{taylor:350:21}) has that $\oint z^{a-1} \,dz = 0,$ or with
the change of variable $z-z_o \la z$,
\[
  \oint (z-z_o)^{a-1} \,dz = 0.
\]
But because any analytic function can be expanded in the form $f(z) =
\sum_k (c_k)(z-z_o)^{a_k-1}$ (which is just a Taylor series if the~$a_k$
happen to be positive integers), this means that
\bq{taylor:cauchyf}
  \oint f(z) \,dz = 0
\eq
if $f(z)$ is everywhere analytic within the contour.%
\footnote{
  The careful reader will observe that~(\ref{taylor:cauchyf})'s
  derivation does not explicitly handle an $f(z)$ represented by a
  Taylor series with an infinite number of terms and a finite
  convergence domain (for example, $f[z] = \ln[1-z]$).  However, by
  \S~\ref{taylor:317} one can transpose such a series from~$z_o$ to an
  overlapping convergence domain about~$z_1$.  Let the contour's interior
  be divided into several cells, each of which is small enough
  to enjoy a single convergence domain.  Integrate about each cell.
  Because the cells share boundaries within the contour's interior, each
  interior boundary is integrated twice, once in each direction,
  canceling.  The original contour---each piece of which is an exterior
  boundary of some cell---is integrated once piecewise.  This is the
  basis on which a more rigorous proof is constructed.
}

\subsection{The formula}
\label{taylor:350.30}

The combination of~(\ref{taylor:cauchy0}) and~(\ref{taylor:cauchyf}) is
powerful.  Consider the closed contour integral
\[
  \oint \frac{f(z)}{z-z_o} \,dz,
\]
where the contour encloses no nonanalytic point of $f(z)$ itself but
does enclose the pole of $f(z)/(z-z_o)$ at $z=z_o$.  If the contour were
a tiny circle of infinitesimal radius about the pole, then the
integrand would reduce to $f(z_o)/(z-z_o)$; and then
per~(\ref{taylor:cauchy0}),
\bq{taylor:cauchy}
  \oint \frac{f(z)}{z-z_o} \,dz = i2\pi f(z_o).
\eq
But if the contour were not an infinitesimal circle but rather the
larger contour of Fig.~\ref{taylor:350:fig2}?
\begin{figure}
  \caption{A Cauchy contour integral.}
  \label{taylor:350:fig2}
  \index{contour!complex}
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\end{figure}
In this case, if the dashed detour which excludes the pole is taken,
then according to~(\ref{taylor:cauchyf}) the resulting integral totals
zero; but the two straight integral segments evidently cancel; and
similarly as we have just reasoned, the \emph{reverse-directed} integral
about the tiny detour circle is $-i2\pi f(z_o)$; so to bring the
total integral to zero the integral about the main contour must be
$i2\pi f(z_o)$.  Thus,~(\ref{taylor:cauchy}) holds for any
positively-directed contour which once encloses a pole and no other
nonanalytic point, whether the contour be small or large.
Equation~(\ref{taylor:cauchy}) is \emph{Cauchy's integral formula.}

\index{linear superposition}
\index{superposition}
\index{residue}
\index{regular part}
If the contour encloses multiple poles (\S\S~\ref{alggeo:250}
and~\ref{inttx:260.20}), then by the principle of linear superposition
(\S~\ref{integ:240.05}),
\bq{taylor:cauchyn}
  \oint \left[ f_o(z) + \sum_k \frac{f_k(z)}{z-z_k} \right] \,dz
    = i2\pi \sum_k f_k(z_k),
\eq
where the $f_o(z)$ is a \emph{regular part};%
\footnote{\cite[\S~1.1]{Lebedev}}
and again, where neither $f_o(z)$ nor any of the several $f_k(z)$ has a pole
or other nonanalytic point within (or on) the contour.  The values
$f_k(z_k)$, which represent the strengths of the poles, are called
\emph{residues}.  In words,~(\ref{taylor:cauchyn}) says that an integral
about a closed contour in the Argand plane comes to~$i2\pi$ times the
sum of the residues of the poles (if any) thus enclosed.  (Note however
that eqn.~\ref{taylor:cauchyn} does not handle branch points.  If there
is a branch point, the contour must exclude it or the formula will not
work.)

As we shall see in \S~\ref{inttx:250}, whether in the form
of~(\ref{taylor:cauchy}) or of~(\ref{taylor:cauchyn}) Cauchy's integral
formula is an extremely useful result.%
\footnote{%
  \cite[\S~10.6]{Hildebrand}%
  \cite{Spiegel}%
  \cite[``Cauchy's integral formula,'' 14:13, 20 April 2006]{wikip}
}

\subsection{Enclosing a multiple pole}
\label{taylor:350.40}
\index{pole!multiple, enclosing a}
\index{multiple pole!enclosing}

\index{complex contour!about a multiple pole}
\index{closed contour!about a multiple pole}
\index{contour!complex, about a multiple pole}
When a complex contour of integration encloses a double, triple or other
$n$-fold pole, the integration can be written,
\[
  S = \oint \frac{f(z)}{(z-z_o)^{m+1}} \,dz, \ \ m \in \mathbb Z, \ m \ge 0,
\]
where $m+1=n$.  Expanding $f(z)$ in a Taylor
series~(\ref{taylor:310:20}) about $z=z_o$,
\[
  S = \oint \sum_{k=0}^{\infty}
    \left(\left.\frac {d^k\!f}
    {dz^k}\right|_{z=z_o}\right)
    \frac{dz}{(k!)(z-z_o)^{m-k+1}}.
\]
But according to~(\ref{taylor:cauchy0}), only the $k=m$ term
contributes, so
\bqb
  S &=& \oint \left(\left.\frac {d^m\!f}
    {dz^m}\right|_{z=z_o}\right)
    \frac{dz}{(m!)(z-z_o)} \\
  \\&=& \frac{1}{m!}\left(\left.\frac {d^m\!f}
    {dz^m}\right|_{z=z_o}\right)
    \oint\frac{dz}{(z-z_o)} \\
  \\&=& \frac{i2\pi}{m!}\left(\left.\frac {d^m\!f}
    {dz^m}\right|_{z=z_o}\right),
\eqb
where the integral is evaluated in the last step according
to~(\ref{taylor:cauchy}).  Altogether,
\bq{taylor:350:30}
  \oint \frac{f(z)}{(z-z_o)^{m+1}} \,dz
  = \frac{i2\pi}{m!}\left(\left.\frac {d^m\!f}
    {dz^m}\right|_{z=z_o}\right),\ \ m \in \mathbb Z, \ m \ge 0.
\eq
Equation~(\ref{taylor:350:30}) evaluates a contour integral about an
$n$-fold pole as~(\ref{taylor:cauchy}) does about a single pole.  (When
$m=0$, the two equations are the
same.)%
\footnote{\cite{Kohler-lecture}\cite{Spiegel}}

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

\section{Taylor series for specific functions}
\label{taylor:315}
\index{Taylor series!for specific functions}

With the general Taylor series formula~(\ref{taylor:310:20}), the
derivatives of Tables~\ref{cexp:drv} and~\ref{cexp:drvi}, and the
observation from~(\ref{drvtv:240.30:10}) that
\[
  \frac{d(z^a)}{dz} = az^{a-1},
\]
one can calculate Taylor series for many functions.  For instance,
expanding about $z=1$,
\settowidth\tla{$\ds\left. \frac{-(-)^k(k-1)!}{z^k} \right|_{z=1}$}
\settowidth\tlb{$-1$}
\bqb
  \left. \ln z \right|_{z=1} =
    \makebox[\tla][r]
    {$\ds\left. \ln z \right|_{z=1}$}
    &=& \makebox[\tlb][r]{$0$}, \\
  \left. \frac{d}{dz}\ln z \right|_{z=1} =
    \makebox[\tla][r]
    {$\ds\left. \frac{1}{z} \right|_{z=1}$}
    &=& \makebox[\tlb][r]{$1$}, \\
  \left. \frac{d^2}{dz^2}\ln z \right|_{z=1} =
    \makebox[\tla][r]
    {$\ds\left. \frac{-1}{z^2} \right|_{z=1}$}
    &=& \makebox[\tlb][r]{$-1$}, \\
  \left. \frac{d^3}{dz^3}\ln z \right|_{z=1} =
    \makebox[\tla][r]
    {$\ds\left. \frac{2}{z^3} \right|_{z=1}$}
    &=& \makebox[\tlb][r]{$2$}, \\
  &\vdots& \\
  \left. \frac{d^k}{dz^k}\ln z \right|_{z=1} =
    \makebox[\tla][r]
    {$\ds\left. \frac{-(-)^k(k-1)!}{z^k} \right|_{z=1}$}
    &=& -(-)^k(k-1)!,\ \ k > 0.
\eqb
With these derivatives, the Taylor series about $z=1$ is
\[
  \ln z = \sum_{k=1}^{\infty}
    \left[-(-)^k(k-1)!
    \right]\frac{(z-1)^k}{k!}
  = -\sum_{k=1}^{\infty}
    \frac{(1-z)^k}{k},
\]
evidently convergent for $\left|1-z\right| < 1$.  (And if~$z$ lies outside the
convergence domain?  Several strategies are then possible.  One can
expand the Taylor series about a different point; but cleverer and
easier is to take advantage of some convenient relationship like $\ln w
= -\ln[1/w]$.  Section~\ref{taylor:316.80} elaborates.)  Using such Taylor
series, one can relatively efficiently calculate actual numerical values
for $\ln z$ and many other functions.

Table~\ref{taylor:315:tbl} lists Taylor series for a few functions of
interest.  All the series converge for $\left|z\right|<1$.  The $\exp z$, $\sin z$
and $\cos z$ series converge for all complex~$z$.
\begin{table}
  \caption{Taylor series.}
  \label{taylor:315:tbl}
  \bqb
    f(z) &=& \sum_{k=0}^{\infty}
      \left(\left.\frac {d^k\!f}
      {dz^k}\right|_{z=z_o}\right)\prod_{j=1}^k\frac{z-z_o}{j} \\
    (1+z)^{a-1} &=& \sum_{k=0}^\infty \prod_{j=1}^k
      \left(\frac{a}{j}-1\right)z \\
    \exp z &=& \sum_{k=0}^{\infty} \prod_{j=1}^k \frac{z}{j}
      = \sum_{k=0}^{\infty} \frac{z^k}{k!} \\
    \sin z &=& \sum_{k=0}^{\infty} \left[ z \prod_{j=1}^k \frac{-z^2}{(2j)(2j+1)} \right] \\
    \cos z &=& \sum_{k=0}^{\infty} \prod_{j=1}^k \frac{-z^2}{(2j-1)(2j)} \\
    \sinh z &=& \sum_{k=0}^{\infty} \left[ z \prod_{j=1}^k \frac{z^2}{(2j)(2j+1)} \right] \\
    \cosh z &=& \sum_{k=0}^{\infty} \prod_{j=1}^k \frac{z^2}{(2j-1)(2j)} \\
    -\ln (1-z) &=& \sum_{k=1}^{\infty} \frac{1}{k} \prod_{j=1}^k z
      = \sum_{k=1}^{\infty} \frac{z^k}{k} \\
    \arctan z &=& \sum_{k=0}^{\infty} \frac 1{2k+1} \left[ z \prod_{j=1}^k (-z^2) \right]
      = \sum_{k=0}^{\infty} \frac{(-)^k z^{2k+1}}{2k+1}
  \eqb
\end{table}
Among the several series, the series for $\arctan z$ is computed
indirectly%
\footnote{\cite[\S~11-7]{Shenk}}
by way of Table~\ref{cexp:drvi} and~(\ref{alggeo:228:40}):
\bqb
  \arctan z &=& \int_0^z \frac{1}{1+w^2} \,dw \\
    &=& \int_0^z \sum_{k=0}^\infty (-)^k w^{2k} \,dw \\
    &=& \sum_{k=0}^\infty \frac{(-)^k z^{2k+1}}{2k+1}.
\eqb

\index{Taylor expansion, first-order}
\index{approximation to first order}
\index{first-order approximation}
\index{sine!approximation of to first order}
\index{exponential!approximation of to first order}
It is interesting to observe from Table~\ref{taylor:315:tbl} the useful
first-order approximations that
\bq{taylor:315:60}
  \settowidth\tla{$\exp z$}
  \begin{split}
    \lim_{z \ra 0} \makebox[\tla][r]{$\exp z$} &= 1 + z, \\
    \lim_{z \ra 0} \makebox[\tla][r]{$\sin z$} &= z,
  \end{split}
\eq
among others.

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

\section{Error bounds}
\label{taylor:316}
\index{bound on a power series}
\index{power series!bounds on}
\index{error bound}

\index{term!finite number of}
One naturally cannot actually sum a Taylor series to an infinite
number of terms.  One must add some finite number of terms, then
quit---which raises the question: how many terms are enough?  How can
one know that one has added adequately many terms; that the remaining
terms, which constitute the tail of the series, are sufficiently
insignificant?  How can one set error bounds on the truncated sum?

\subsection{Examples}
\label{taylor:316.20}

\index{alternating signs}
\index{sign!alternating}
\index{partial sum}
\index{sum!partial}
\index{truncation}
\index{series!truncation of}
Some series alternate sign.  For these it is easy if the numbers
involved happen to be real.  For example, from
Table~\ref{taylor:315:tbl},
\[
  \ln \frac 3 2 = \ln \left( 1 + \frac 1 2 \right)
  = \frac 1{(1)(2^1)} - \frac 1{(2)(2^2)}
  + \frac 1{(3)(2^3)} - \frac 1{(4)(2^4)}
  + \cdots
\]
Each term is smaller in magnitude than the last, so the true value of
$\ln(3/2)$ necessarily lies between the sum of the series to~$n$ terms
and the sum to $n+1$ terms.  The last and next partial sums bound the
result.  Up to but not including the fourth-order term, for instance,
\[
  \renewcommand\arraystretch{2.0}
  \br{c}
    \ds S_4 - \frac 1{(4)(2^4)} < \ln \frac 3 2 < S_4, \\
    \ds S_4 = \frac 1{(1)(2^1)} - \frac 1{(2)(2^2)} + \frac 1{(3)(2^3)}.
  \er
\]

\index{majorization}
Other series however do not alternate sign.  For example,
\bqb
  \ln 2 &=& -\ln \frac 1 2  = -\ln \left( 1 - \frac 1 2 \right) = S_5 + R_5, \\
  S_5 &=& \frac 1{(1)(2^1)} + \frac 1{(2)(2^2)}
  + \frac 1{(3)(2^3)} + \frac 1{(4)(2^4)}, \\
  R_5 &=& \frac 1{(5)(2^5)} + \frac 1{(6)(2^6)} + \cdots
\eqb
The basic technique in such a case is to find a replacement series (or
integral)~$R_n'$ which one can collapse analytically, each of whose terms
equals or exceeds in magnitude the corresponding term of~$R_n$.  For the
example, one might choose
\[
  R_5' = \frac 1 5 \sum_{k=5}^\infty \frac 1{2^k} = \frac{2}{(5)(2^5)},
\]
wherein~(\ref{alggeo:228:45}) had been used to collapse the summation.  Then,
\[
  S_5 < \ln 2 < S_5 + R_5'.
\]
For real $0 \le x < 1$ generally,
\bqb
  S_n &<& -\ln(1-x) \makebox[\arraycolsep]{} < \makebox[\arraycolsep]{} S_n + R_n', \\
  S_n &\equiv& \sum_{k=1}^{n-1} \frac{x^k}{k}, \\
  R_n' &\equiv& \sum_{k=n}^{\infty} \frac{x^k}{n}
    = \frac{x^{n}}{(n)(1-x)}.
\eqb

Many variations and refinements are possible, some of which we will
meet in the rest of the section, but that is the basic technique: to add
several terms of the series to establish a lower bound, then to
overestimate the remainder of the series to establish an upper bound.
The overestimate~$R_n'$ \emph{majorizes} the series' true remainder~$R_n$.
Notice that the~$R_n'$ in the example is a fairly small number, and that
it would have been a lot smaller yet had we included a few more terms
in~$S_n$ (for instance, $n = \mr{0x40}$ would have bound $\ln 2$ tighter
than the limit of a computer's typical \texttt{double}-type
floating-point accuracy).  The technique usually works well in practice
for this reason.

\subsection{Majorization}
\label{taylor:316.25}
\index{majorization}

\emph{To majorize} in mathematics is to be, or to replace by virtue of
being, everywhere at least as great as.
This is best explained by example.  Consider the summation
\[
  S = \sum_{k=1}^\infty \frac 1{k^2}
    = 1 + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} + \cdots
\]
The exact value this summation totals to is unknown to us, but the
summation does rather resemble the integral (refer to
Table~\ref{integ:basic-antider})
\[
  I = \int_{1}^\infty \frac {dx}{x^2}
    = \left.-\frac{1}{x}\right|_1^\infty = 1.
\]
Figure~\ref{taylor:316:fig-maj} plots~$S$ and~$I$ together as areas---or
more precisely, plots $S-1$ and~$I$ together as areas (the summation's
first term is omitted).
% This figure, here commented out, was a reasonably good figure but it
% did not show the majorization quite correctly.
%\begin{figure}
%  \caption[Majorization.]
%    {The area~$S$ under the stairstep curve $f_S(x)$ majorizes the area~$I$ under the smooth curve $f_I(x)$.}
%  \label{taylor:316:fig-maj}
%  \bc
%    \nc\fxa{-1.0} \nc\fxb{9.0}
%    \nc\fya{-0.7} \nc\fyb{3.2}
%    \nc\xxa{-0.5} \nc\xxb{8.1}
%    \nc\xya{-0.3} \nc\xyb{2.5}
%    \nc\xxyt{0.2cm}
%    \nc\xxt[1]{\psline[linewidth=0.5pt](0,0)(0,-\xxyt)\uput[d](0,-\xxyt){#1}}
%    \nc\xyt[1]{\psline[linewidth=0.5pt](0,0)(-\xxyt,0){\psset{labelsep=0.12cm}\uput[l](-\xxyt,0){#1}}}
%    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
%      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
%      {
%        \small
%        \psset{dimen=middle}
%        {
%          \psset{linewidth=0.5pt}
%          \psline(\xxa,0)(\xxb,0) \uput[r](\xxb,0){$x$}
%          \psline(0,\xya)(0,\xyb) \uput[u](0,\xyb){$y$}
%          \rput(2.58,0.23){$y\!=\!1/x^2$}
%          \psline(3.22,0.20)(3.50,0.20)(3.78,0.32)
%        }
%        {
%          \psset{linewidth=2.0pt,unit=1.7cm}
%          \psplot[linestyle=dashed,plotpoints=200]{1.0}{4.6}
%            { 1.0 x dup mul div }
%          \psline{-C}( 1,0.00000)(1,1.00000)
%          \psline{C-C}(1,1.00000)(2,1.00000) \psline{C-c}(2,1.00000)(2,0.25000)
%          \psline{c-C}(2,0.25000)(3,0.25000) \psline{C-c}(3,0.25000)(3,0.11111)
%          \psline{c-C}(3,0.11111)(4,0.11111) \psline{C-c}(4,0.11111)(4,0.06250)
%          \psline{c-}( 4,0.06250)(4.6,0.06250)
%          {
%            \psset{labelsep=0.10cm}
%            \uput[u](1.5,1.00000){$y=1$}
%            \uput[u](2.5,0.25000){$y=1/2^2$}
%            \uput[u](3.5,0.11111){$y=1/3^2$}
%            \uput[u](4.5,0.06250){$y=1/4^2$}
%          }
%          \rput(0,1){\xyt{$1$}}
%          \rput(1,0){\xxt{$1$}}
%          \rput(2,0){\xxt{$2$}}
%          \rput(3,0){\xxt{$3$}}
%          \rput(4,0){\xxt{$4$}}
%        }
%      }
%    \end{pspicture}
%  \ec
%\end{figure}
\begin{figure}
  \caption[Majorization.]
    {Majorization.  The area~$I$ between the dashed curve and the~$x$
    axis majorizes the area $S-1$ between the stairstep curve and
    the~$x$ axis, because the height of the dashed curve is everywhere
    at least as great as that of the stairstep curve.}
  \label{taylor:316:fig-maj}
  \bc
    \nc\fxa{-1.0} \nc\fxb{11.10}
    \nc\fya{-0.9} \nc\fyb{ 5.00}
    \nc\xxa{-0.2} \nc\xxb{ 3.15}
    \nc\xya{-0.1} \nc\xyb{ 1.30}
    \nc\xxyt{0.20cm}
    \nc\xxt[1]{\psline[linewidth=0.5pt](0,0)(0,-\xxyt)\uput[d](0,-\xxyt){#1}}
    \nc\xyt[1]{\psline[linewidth=0.5pt](0,0)(-\xxyt,0){\psset{labelsep=0.12cm}\uput[l](-\xxyt,0){#1}}}
    \nc\xxc{3.12}
    \nc\xxd{0.86}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
      {
        \small
        \psset{dimen=middle,unit=3.3cm}
        {
          \psset{linewidth=0.5pt}
          \psline(\xxa,0)(\xxb,0) \uput[r](\xxb,0){$x$}
          \psline(0,\xya)(0,\xyb) \uput[u](0,\xyb){$y$}
          \rput(2.55,0.28){$y=1/x^2$}
          \psline(2.32,0.28)(2.20,0.28)(2.16,0.22)
        }
        {
          \psset{linewidth=2.0pt}
          \psplot[linestyle=dashed,plotpoints=200]{1.0}{\xxc}
            { 1.0 x dup mul div }
          \psplot[linewidth=0.5pt,linestyle=dotted,plotpoints=50]{\xxd}{1.0}
            { 1.0 x dup mul div }
          \psline[linestyle=dashed]{-c}(1,0.25)(1,1)
          \psline{-C}( 1,0.00000)(1,0.25000)
          \psline{C-c}(1,0.25000)(2,0.25000) \psline{c-C}(2,0.25000)(2,0.11111)
          \psline{C-c}(2,0.11111)(3,0.11111) \psline{c-C}(3,0.11111)(3,0.06250)
          \psline{C-}( 3,0.06250)(\xxc,0.06250)
          {
            \psset{linewidth=0.5pt,labelsep=0.12cm}
            \psline(0,0.25000)(-0.04,0.25000)(-0.09,0.28750)(-0.13,0.28750)\rput[r](-0.15,0.28750){$1/2^2$}
            \psline(0,0.11111)(-0.04,0.11111)(-0.09,0.17500)(-0.13,0.17500)\rput[r](-0.15,0.17500){$1/3^2$}
            \psline(0,0.06250)(-0.13,0.06250)\rput[r](-0.15,0.06250){$1/4^2$}
          }
          {
            \psset{linewidth=0.5pt,linestyle=dotted}
            \psline(0,1.00000)(1,1.00000)
            \psline(0,0.25000)(1,0.25000)
            \psline(0,0.11111)(1,0.11111)
            \psline(1,0.11111)(2,0.11111)
            \psline(0,0.06250)(1,0.06250)
            \psline(1,0.06250)(2,0.06250)
            \psline(1.996,0.06250)(3,0.06250)
          }
          \rput(0,1){\xyt{$1$}}
          \rput(1,0){\xxt{$1$}}
          \rput(2,0){\xxt{$2$}}
          \rput(3,0){\xxt{$3$}}
        }
      }
    \end{pspicture}
  \ec
\end{figure}
As the plot shows, the unknown area $S-1$ cannot possibly be as
great as the known area~$I$.  In symbols, $S-1 < I = 1$; or,
\[
  S < 2.
\]
The integral~$I$ majorizes the summation $S-1$, thus guaranteeing the
absolute upper limit on~$S$.  (Of course $S < 2$ is a very loose limit,
but that isn't the point of the example.  In practical
calculation, one would let a computer add many terms of the series
first numerically, and only then majorize the remainder.  Even so,
cleverer ways to majorize the remainder of this particular series
will occur to the reader, such as in representing the terms
graphically---not as flat-topped rectangles---but as slant-topped
trapezoids, shifted in the figure a half unit rightward.)

\index{integral!and series summation}
\index{summation!compared to integration}
\index{minorization}
Majorization serves surely to bound an unknown quantity by a larger,
known quantity.  Reflecting, \emph{minorization}%
\footnote{
  The author does not remember ever encountering the word
  \emph{minorization} heretofore in print, but as a reflection of
  \emph{majorization} the word seems logical.  This book at least will
  use the word where needed.  You can use it too if you like.
}
serves surely to bound an unknown quantity by a smaller, known quantity.
The quantities in question are often integrals and/or series summations,
the two of which are akin as Fig.~\ref{taylor:316:fig-maj} illustrates.
The choice of whether to majorize a particular unknown quantity by an
integral or by a series summation depends on the convenience of the
problem at hand.

\index{harmonic series}
\index{series!harmonic}
The series~$S$ of this subsection is interesting, incidentally.  It is a
\emph{harmonic series} rather than a power series, because though its
terms do decrease in magnitude it has no~$z^k$ factor (or seen from
another point of view, it does have a~$z^k$ factor, but $z=1$), and
the ratio of adjacent terms' magnitudes approaches unity as~$k$ grows.
Harmonic series can be hard to sum accurately,
but clever majorization can help (and if majorization does not help
enough, then series transformations---%
% diagn
which unfortunately lie beyond this edition's treatment---%
can help even more).

\subsection{Geometric majorization}
\label{taylor:316.30}
\index{majorization!geometric}
\index{geometric majorization}
\index{geometric series!majorization by}
\index{series!geometric}

\index{residual}
Harmonic series can be hard to sum as \S~\ref{taylor:316.25} has
observed, but more common than harmonic series are true power series,
easier to sum in that they include a~$z^k$ factor in each term.  There is
no one, ideal bound that works equally well for all power series.
However, the point of establishing a bound is not to sum a power series
exactly but rather to fence the sum within some sufficiently (rather
than optimally) small neighborhood.  A simple, general bound which works
quite adequately for most power series encountered in practice,
including among many others all the Taylor series of
Table~\ref{taylor:315:tbl}, is the \emph{geometric majorization}
\bq{taylor:316:30}
  \left|\ep_n\right| < \frac{\left|\tau_n\right|}{1-\left|\rho_n\right|}.
\eq
Here,~$\tau_n$ represents the power series' $n$th-order
term (in Table~\ref{taylor:315:tbl}'s series for $\exp z$, for
example, $\tau_n = z^n/[n!]$).  The~$\left|\rho_n\right|$ is a positive real number
chosen, preferably as small as possible, such that
{
  \settowidth\tla{for at least one}
  \bqa
    \left| \frac{\tau_{k+1}}{\tau_k} \right| &\le& \left|\rho_n\right|
      \ \ \mbox{\makebox[\tla][r]{for all} $k \ge n$,} \label{taylor:316:33} \\
    \left| \frac{\tau_{k+1}}{\tau_k} \right| &<& \left|\rho_n\right|
      \ \ \mbox{for at least one $k \ge n$,} \xn \\
    0 &<& \left|\rho_n\right| \makebox[\arraycolsep]{} < \makebox[\arraycolsep]{} 1;
  \eqa
}%
which is to say, more or less, such that each term in the series' tail
is smaller than the last by at least a factor of~$\left|\rho_n\right|$.  Given these
definitions, if%
\footnote{
  Some scientists and engineers---as, for example, the authors
  of~\cite{Peterson/Mittra:1986} and even this writer in earlier
  years---prefer to define $\ep_n \equiv S_n - S_\infty$, oppositely as
  we define it here.  This choice is a matter of taste.  Professional
  mathematicians---as, for example, the author of~\cite{vdVorst}---seem
  to tend toward the $\ep_n \equiv S_\infty - S_n$
  of~(\ref{taylor:316:34}).
}
\bq{taylor:316:34}
  \begin{split}
    S_n &\equiv \sum_{k=0}^{n-1} \tau_k, \\
    \ep_n &\equiv S_\infty - S_n,
  \end{split}
\eq
where~$S_\infty$ represents the true, exact (but uncalculatable,
unknown) infinite series sum, then~(\ref{alggeo:228:45})
and~(\ref{trig:278:triangle}) imply the geometric
majorization~(\ref{taylor:316:30}).

\index{logarithm, natural!error of}
\index{natural logarithm!error of}
If the last paragraph seems abstract, a pair of concrete examples should
serve to clarify.  First, if the Taylor series
\[
  -\ln (1-z) = \sum_{k=1}^{\infty} \frac{z^k}{k}
\]
of Table~\ref{taylor:315:tbl} is truncated before the $n$th-order term,
then
\bqb
  -\ln (1-z) &\approx& \sum_{k=1}^{n-1} \frac{z^k}{k}, \\
  \left|\ep_n\right| &<& \frac{\left|z^n\right|/n}{1-\left|z\right|},
\eqb
where~$\ep_n$ is the error in the truncated sum.%
\footnote{
  This particular error bound fails for $n=0$, but that is no flaw.
  There is no reason to use the error bound for $n=0$ when, merely by
  taking one or two more terms into the truncated sum, one can quite
  conveniently let $n=1$ or $n=2$.
}
Here, $\left|\tau_{k+1}/\tau_k\right| = [k/(k+1)]\left|z\right| <
\left|z\right|$ for all $k \ge n > 0$,
so we have chosen $\left|\rho_n\right| = \left|z\right|$.

\index{exponential!natural, error of}
\index{natural exponential!error of}
Second, if the Taylor series
\[
  \exp z = \sum_{k=0}^{\infty} \prod_{j=1}^k \frac{z}{j}
    = \sum_{k=0}^{\infty} \frac{z^k}{k!}
\]
also of Table~\ref{taylor:315:tbl} is truncated before the $n$th-order
term, and if we choose to stipulate that
\[
  n+1 > \left|z\right|,
\]
then
\bqb
  \exp z &\approx& \sum_{k=0}^{n-1} \prod_{j=1}^k \frac{z}{j}
    = \sum_{k=0}^{n-1} \frac{z^k}{k!}, \\
  \left|\ep_n\right| &<& \frac{\left|z^n\right|/n!}{1-\left|z\right|/(n+1)}.
\eqb
Here, $\left|\tau_{k+1}/\tau_k\right| = \left|z\right|/(k+1)$, whose maximum value for all
$k\ge n$ occurs when $k=n$, so we have chosen $\left|\rho_n\right| = \left|z\right|/(n+1)$.

\subsection{Calculation outside the fast convergence domain}
\label{taylor:316.80}
\index{slow convergence}
\index{convergence!slow}
\index{domain}
\index{convergence!domain of}

\index{function!entire}
\index{entire function}
Used directly, the Taylor series of Table~\ref{taylor:315:tbl} tend to
converge slowly for some values of~$z$ and not at all for others.  The
series for $-\ln(1-z)$ and $(1+z)^{a-1}$ for instance each converge for
$\left| z \right| < 1$ (though slowly for $\left| z \right| \approx 1$);
whereas each series diverges when asked to compute a quantity like $-\ln
3$ or $3^{a-1}$ directly.  To shift the series' expansion points per
\S~\ref{taylor:317} is one way to seek convergence, but for nonentire
functions~(\S~\ref{taylor:330}) like these a more probably profitable
strategy is to find and exploit some property of the functions to
transform their arguments, such as
\bqb
  -\ln \gamma &=& \ln\frac{1}{\gamma}, \\
  \gamma^{a-1} &=& \frac{1}{(1/\gamma)^{a-1}},
\eqb
which leave the respective Taylor series to compute quantities like
$-\ln(1/3)$ and $(1/3)^{a-1}$ they can handle.

Let $f(1+\zeta)$ be a function whose Taylor series about $\zeta=0$
converges for $\left|\zeta\right| < 1$ and which obeys properties of the
forms%
\footnote{\label{taylor:316.80fn}%
  This paragraph's notation is necessarily abstract.  To make it seem
  more concrete, consider that the function $f(1+\zeta)=-\ln(1-z)$ has
  $\zeta = -z$, $f(\gamma) = g[f(1/\gamma)] = -f(1/\gamma)$ and
  $f(\alpha\gamma) = h[f(\alpha),f(\gamma)] = f(\alpha)+f(\gamma)$; and
  that the function $f(1+\zeta)=(1+z)^{a-1}$ has $\zeta = z$, $f(\gamma)
  = g[f(1/\gamma)] = 1/f(1/\gamma)$ and $f(\alpha\gamma) =
  h[f(\alpha),f(\gamma)]=f(\alpha)f(\gamma)$.
}
\bq{taylor:316:83}
  \begin{split}
    f(\gamma) &= g\left[f\left(\frac{1}{\gamma}\right)\right], \\
    f(\alpha\gamma) &= h\left[f(\alpha),f(\gamma)\right],
  \end{split}
\eq
where $g[\cdot]$ and $h[\cdot,\cdot]$ are functions we know how
to compute like $g[\cdot]=-[\cdot]$ or $g[\cdot]=1/[\cdot]$; and like
$h[\cdot,\cdot]=[\cdot]+[\cdot]$ or $h[\cdot,\cdot]=[\cdot][\cdot]$.
Identifying
\bq{taylor:316:84}
  \begin{split}
    \frac{1}{\gamma} &= 1 + \zeta, \\
    \gamma &= \frac{1}{1 + \zeta}, \\
    \frac{1-\gamma}{\gamma} &= \zeta,
  \end{split}
\eq
we have that
\bq{taylor:316:85}
  f(\gamma) = g\left[f\left(1+\frac{1-\gamma}{\gamma}\right)\right],
\eq
whose convergence domain $\left|\zeta\right| < 1$
is $\left|1-\gamma\right|/\left|\gamma\right| < 1$, which is
$\left|\gamma-1\right| < \left|\gamma\right|$ or in other words
\[
  \Re(\gamma) > \frac 1 2.
\]
Although the transformation from~$\zeta$ to~$\gamma$ has not lifted the
convergence limit altogether, we see that it has apparently opened the
limit to a broader domain.

For example, if $f(\gamma) = \ln\gamma$ and $\Re(\gamma) > 1/2$,
then\footnote{See footnote~\ref{taylor:316.80fn}.} $g[\cdot]=-[\cdot]$
and thus, according to~(\ref{taylor:316:85}),
\[
  \begin{split}
    \ln\gamma &= -\ln\left(1+\frac{1-\gamma}{\gamma}\right)\\
      &= -\ln(1-z), \ \ z\equiv\frac{\gamma-1}{\gamma},
  \end{split}
\]
a formulation that lets one apply Table~\ref{taylor:315:tbl} to
calculate, say, $\ln 3$.
% The writer has checked ln 3 by this method on the computer.

\index{domain!sidestepping a}
Though this writer knows no way to lift the convergence limit
altogether that does not cause more problems than it solves, one can
take advantage of the $h[\cdot,\cdot]$ property of~(\ref{taylor:316:83})
to sidestep the limit, computing $f(\omega)$ indirectly for any~$\omega
\neq 0$ by any of several tactics.  One nonoptimal but entirely
effective tactic is represented by the equations
\bq{taylor:316:90}
  \begin{split}
    \omega &\equiv i^n2^m\gamma, \\
    \left|\Im(\gamma)\right| &\le \Re(\gamma), \\
    1 \le \Re(\gamma) &< 2, \\
    m,n &\in \mathbb Z,
  \end{split}
\eq
whereupon the formula
\bq{taylor:316:92}
  f(\omega) = h[f(i^n2^m), f(\gamma)]
\eq
calculates $f(\omega)$ fast for any $\omega \neq 0$---provided only
that we have other means to compute $f(i^n2^m)$, which not infrequently
we do.%
\footnote{
  Equation~(\ref{taylor:316:92}) admittedly leaves open the question of
  how to compute~$f(i^n2^m)$, but at least for the functions this
  subsection has used as examples this is
  not hard.  For the logarithm, $-\ln(i^n2^m)=m\ln(1/2)-in(2\pi/4)$.
  For the power, $(i^n2^m)^{a-1}=\cis[(n2\pi/4)(a-1)]/[(1/2)^{a-1}]^m$.
  The sine and cosine in the cis function are each calculated directly
  by Taylor series
  (possibly with the help of Table~\ref{trig:228:table}),
  as are the numbers $\ln(1/2)$ and $(1/2)^{a-1}$.  The number~$2\pi$,
  we have not calculated yet, but will in \S~\ref{taylor:355}.
}

Notice how~(\ref{taylor:316:90}) fences~$\gamma$ within a comfortable
zone, keeping~$\gamma$ moderately small in magnitude but never too near
the $\Re(\gamma) = 1/2$ frontier in the Argand plane.  In theory all
finite~$\gamma$ rightward of the frontier let the Taylor series
converge, but extreme~$\gamma$ of any kind let the series converge only
slowly (and due to compound floating-point rounding error
perhaps inaccurately)
inasmuch as they imply that $\left|\zeta\right| \approx 1$.  Besides
allowing all~$\omega\neq 0$, the tactic~(\ref{taylor:316:90}) also thus
significantly speeds series convergence.

The method and tactic of~(\ref{taylor:316:83})
through~(\ref{taylor:316:92}) are useful in themselves and also
illustrative generally.  Of course, most nonentire functions lack
properties of the specific kinds that~(\ref{taylor:316:83}) demands, but
such functions may have other properties one can analogously exploit.%
\footnote{
  To draw another example from Table~\ref{taylor:315:tbl}, consider that
  \[
    \begin{split}
      \arctan \omega &= \alpha + \arctan \zeta, \\
      \zeta &\equiv \frac{
        \omega\cos\alpha - \sin\alpha
      }{
        \omega\sin\alpha + \cos\alpha
      },
    \end{split}
  \]
  where $\arctan \omega$ is interpreted as the geometrical angle the
  vector $\vu x + \vu y \omega$ makes with~$\vu x$.  Axes are rotated
  per~(\ref{trig:240:rot}) through some angle~$\alpha$ to reduce the
  tangent from~$\omega$ to~$\zeta$, where $\arctan \zeta$ is
  interpreted as the geometrical angle the vector $\vu x + \vu y \omega
  = \vu x' (\omega\sin\alpha+\cos\alpha) +\vu y'
  (\omega\cos\alpha-\sin\alpha)$ makes with~$\vu x'$, thus causing the
  Taylor series to converge faster or indeed to converge at all.
  % The writer has checked this example with omega=3.0, alpha=pi/3, on
  % the computer.

  Any number of further examples and tactics of the kind will occur to
  the creative reader, shrinking a function's argument by some
  convenient means before feeding the argument to the function's Taylor
  series.
}

\subsection{Nonconvergent series}
\label{taylor:316.70}

Variants of this section's techniques can be used to prove that a series
does not converge at all.  For example,
\[
  \sum_{k=1}^{\infty} \frac{1}{k}
\]
does not converge because
\[
  \frac 1 k > \int_{k}^{k+1} \frac{d\tau}{\tau};
\]
hence,
\[
  \sum_{k=1}^{\infty} \frac{1}{k}
  > \sum_{k=1}^{\infty} \int_{k}^{k+1} \frac{d\tau}{\tau}
  = \int_{1}^\infty \frac{d\tau}{\tau}
  = \ln\infty.
\]

\subsection{Remarks}
\label{taylor:316.95}

The study of error bounds is not a matter of rules and formulas so much
as of ideas, suggestions and tactics.  As far as the writer knows,
there is no such thing
as an optimal error bound---with sufficient cleverness, some tighter
bound can usually be discovered---but often easier and more effective
than cleverness is simply to add a few extra terms into the series
before truncating it (that is, to increase~$n$ a little).  To eliminate
the error entirely usually demands adding an infinite number of terms,
which is impossible; but since eliminating the error entirely also
requires recording the sum to infinite precision, which is impossible
anyway, eliminating the error entirely is not normally a goal one seeks.
To eliminate the error to the $\mbox{0x34}$-bit (sixteen-decimal place)
precision of a computer's \texttt{double}-type floating-point
representation typically requires something like~$\mbox{0x34}$
terms---if the series be wisely composed and if care be taken to
keep~$z$ moderately small and reasonably distant from the edge of the
series' convergence domain.  Besides, few engineering applications
really use much more than~$\mbox{0x10}$ bits (five decimal places) in
any case.  Perfect precision is impossible, but adequate precision is
usually not hard to achieve.

Occasionally nonetheless a series arises for which even adequate
precision is quite hard to achieve.  An infamous example is
\[
  S = -\sum_{k=1}^\infty \frac{(-)^k}{\sqrt k}
    = 1 - \frac 1{\sqrt 2} + \frac 1{\sqrt 3} - \frac 1{\sqrt 4} + \cdots,
\]
which obviously converges, but sum it if you can!  It is not easy to do.
% diagn: if and when the summing technique for this sqrt-series is added
% to the book, add a reference to it here

Before closing the section, we ought to arrest one potential agent of
terminological confusion.  The ``error'' in a series summation's error
bounds is unrelated to the error of probability theory (chapter~\ref{prob}).  The English
word ``error'' is thus overloaded here.  A series sum converges to a
definite value, and to the same value every time the series is summed;
no chance is involved.  It is just that we do not necessarily know
exactly what that value is.  What we can do, by this section's
techniques or perhaps by other methods, is to establish a definite
neighborhood in which the unknown value is sure to lie; and we can make
that neighborhood as tight as we want, merely by including a sufficient
number of terms in the sum.

The topic of series error bounds is what
G.~S.\ Brown refers to as ``trick-based.''%
\footnote{\cite{Brown-conversation}}
There is no general answer to the error-bound problem, but there are
several techniques which help, some of which this section has
introduced.  Other techniques, we shall meet later in the book as the
need for them arises.

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

\section{Calculating~$2\pi$}
\label{taylor:355}
\index{$2\pi$!calculating}

The Taylor series for $\arctan z$ in Table~\ref{taylor:315:tbl} implies
a neat way of calculating the constant~$2\pi$.  We already know that
$\tan(2\pi/8) = 1$, or in other words that
\[
  \arctan 1 = \frac{2\pi}{8}.
\]
Applying the Taylor series, we have that
\bq{taylor:355:10}
  2\pi = 8\sum_{k=0}^{\infty} \frac{(-)^k}{2k+1}.
\eq
The series~(\ref{taylor:355:10}) is simple but converges extremely
slowly.  Much faster convergence is given by angles smaller than
$2\pi/8$.  For example, from Table~\ref{trig:hourtable},
\[
  \arctan \frac{\sqrt 3 - 1}{\sqrt 3 + 1} = \frac{2\pi}{\mbox{0x18}}.
\]
Applying the Taylor series at this angle, we have that%
\footnote{
  The writer is given to understand that clever mathematicians have
  invented subtle, still much faster-converging iterative schemes
  toward~$2\pi$.  However, there is fast and there is fast.  The
  relatively straightforward series this section gives converges to
  the best accuracy of your computer's floating-point register within a
  paltry~\mbox{0x40} iterations or so---and, after all, you only need to
  compute the numerical value of~$2\pi$ once.

  Admittedly, the writer supposes that useful lessons lurk in the
  clever mathematics underlying the subtle schemes, but such schemes are
  not covered here.
}
\bq{taylor:355:15}
  2\pi = \mbox{0x18}\sum_{k=0}^{\infty} \frac{(-)^k}{2k+1}
    \left(\frac{\sqrt 3 - 1}{\sqrt 3 + 1}\right)^{2k+1}
    \approx \mbox{0x6.487F}.
\eq

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

\section{Odd and even functions}
\label{taylor:365}
\index{function!odd or even}
\index{odd function}
\index{even function}

An \emph{odd function} is one for which $f(-z)=-f(z)$.  Any function
whose Taylor series about $z_o=0$ includes only odd-order terms is
an odd function.  Examples of odd functions include~$z^3$
and $\sin z$.

An \emph{even function} is one for which $f(-z)=f(z)$.  Any function
whose Taylor series about $z_o=0$ includes only even-order terms is
an even function.  Examples of even functions include~$z^2$
and $\cos z$.

\index{symmetry}
Odd and even functions are interesting because of the symmetry they
bring---the plot of a real-valued odd function being symmetric about a
point, the plot of a real-valued even function being symmetric about a
line.  Many functions are neither odd nor even, of course, but one can
always split an analytic function into two components---one odd, the
other even---by the simple expedient of sorting the odd-order terms from
the even-order in the function's Taylor series.  For example,
$\exp z = \sinh z + \cosh z$.
Alternately,
\bq{taylor:365:10}
  \begin{split}
    f(z) &= f_{\mr{odd}}(z) +  f_{\mr{even}}(z), \\
    f_{\mr{odd}}(z) &= \frac{f(z)-f(-z)}{2}, \\
    f_{\mr{even}}(z) &= \frac{f(z)+f(-z)}{2},
  \end{split}
\eq
the latter two lines of which are verified by substituting $-z\la z$ and
observing the definitions at the section's head of odd and even, and then
the first line of which is verified by adding the latter two.

Section~\ref{fouri:110.65} will have more to say about odd and even
functions.

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

\section{Trigonometric poles}
\label{taylor:382}
\index{trigonometric function!poles of}
\index{pole!of a trigonometric function}

The singularities of the trigonometric functions are single poles of
residue~$\pm 1$ or~$\pm i$.  For the circular trigonometrics, all the
poles lie along the real number line; for the hyperbolic trigonometrics,
along the imaginary.  Specifically, of the eight trigonometric functions
\[
  \renewcommand\arraystretch{2.0}
  \br{c}
    \ds\frac{1}{\sin z},
    \ds\frac{1}{\cos z},
    \ds\frac{1}{\tan z},
    \ds\tan z,
  \\
    \ds\frac{1}{\sinh z},
    \ds\frac{1}{\cosh z},
    \ds\frac{1}{\tanh z},
    \ds\tanh z,
  \er
\]
the poles and their respective residues are
\settowidth\tla{\scriptsize$z-i(k-1/2)\pi$}
\bq{taylor:382:10}
  \begin{split}
    \left.\frac{z-k\pi}{\sin z}\right|_{\makebox[\tla][l]{\scriptsize$z=k\pi$}}
      &= (-)^k, \\
    \left.\frac{z-(k-1/2)\pi}{\cos z}\right|_{\makebox[\tla][l]{\scriptsize$z=(k-1/2)\pi$}}
      &= (-)^k, \\
    \left.\frac{z-k\pi}{\tan z}\right|_{\makebox[\tla][l]{\scriptsize$z=k\pi$}}
      &= 1, \\
    \left.[z-(k-1/2)\pi]\tan z\right|_{\makebox[\tla][l]{\scriptsize$z=(k-1/2)\pi$}}
      &= -1, \\
    \left.\frac{z-ik\pi}{\sinh z}\right|_{\makebox[\tla][l]{\scriptsize$z=ik\pi$}}
      &= (-)^k, \\
    \left.\frac{z-i(k-1/2)\pi}{\cosh z}\right|_{\makebox[\tla][l]{\scriptsize$z=i(k-1/2)\pi$}}
      &= (-)^ki, \\
    \left.\frac{z-ik\pi}{\tanh z}\right|_{\makebox[\tla][l]{\scriptsize$z=ik\pi$}}
      &= 1, \\
    \left.[z-i(k-1/2)\pi]\tanh z\right|_{\makebox[\tla][l]{\scriptsize$z=i(k-1/2)\pi$}}
      &= 1, \\
    k &\in \mathbb Z.
  \end{split}
\eq

\index{l'H\^opital's rule}
To support~(\ref{taylor:382:10})'s claims, we shall marshal the
identities of Tables~\ref{cexp:tbl-prop} and~\ref{cexp:drv} plus
l'H\^opital's rule~(\ref{drvtv:260:lhopital}).  Before calculating
residues and such, however, we should like to verify that
the poles~(\ref{taylor:382:10}) lists are in fact the only poles that
there are; that we have forgotten no poles.  Consider for instance the
function $1/\sin z = i2/(e^{iz}-e^{-iz})$.  This function evidently
goes infinite only when $e^{iz}=e^{-iz}$, which is possible only for
real~$z$; but for real~$z$, the sine function's very definition
establishes the poles $z=k\pi$ (refer to Fig.~\ref{trig:226:f1}).  With
the observations from Table~\ref{cexp:tbl-prop} that $i\sinh z=\sin iz$
and $\cosh z=\cos iz$, similar reasoning for each of the eight
trigonometrics forbids poles other than those~(\ref{taylor:382:10})
lists.  Satisfied that we have forgotten no poles, therefore, we finally
apply l'H\^opital's rule to each of the ratios
\[
  %\renewcommand\arraystretch{2.0}
  \br{c}
    \ds\frac{z-k\pi}{\sin z},
    \ds\frac{z-(k-1/2)\pi}{\cos z},
    \ds\frac{z-k\pi}{\tan z},
    \ds\frac{z-(k-1/2)\pi}{1/\tan z},
  \\
    \ds\frac{z-ik\pi}{\sinh z},
    \ds\frac{z-i(k-1/2)\pi}{\cosh z},
    \ds\frac{z-ik\pi}{\tanh z},
    \ds\frac{z-i(k-1/2)\pi}{1/\tanh z}
  \er
\]
to reach~(\ref{taylor:382:10}).

\index{function!meromorphic}
\index{meromorphic function}
Trigonometric poles evidently are special only in that a trigonometric
function has an infinite number of them.  The poles are ordinary, single
poles, with residues, subject to Cauchy's integral formula and so on.
The trigonometrics are meromorphic functions (\S~\ref{taylor:330}) for
this reason.%
\footnote{\cite{Kohler-lecture}}

\index{function!entire}
\index{entire function}
The six simpler trigonometrics, $\sin z$, $\cos z$, $\sinh z$, $\cosh z$,
$\cis z$ and
\linebreak % bad break
$\exp z$---conspicuously excluded from this section's gang
of eight---have no poles for finite~$z$ because~$e^{\pm iz}$
and~$e^{\pm z}$ are finite.  These
simpler trigonometrics are not only meromorphic but also entire.
Observe however that the \emph{inverse} trigonometrics are
multiple-valued and have branch points, and thus are not meromorphic at
all.

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

\section{The Laurent series}
\label{taylor:380}
\index{Laurent series}
\index{Laurent, Pierre Alphonse (1813--1854)}

\index{pole}
\index{long division}
\index{regular part}
Any analytic function can be expanded in a Taylor series, but never
about a pole or branch point of the function.  Sometimes one
nevertheless wants to expand at least about a pole.  Consider for
example expanding
\bq{taylor:380:06}
  f(z) = \frac{e^{-z}}{1-\cos z}
\eq
about the function's pole at $z=0$.  Expanding dividend and divisor
separately,
\bqb
  f(z) &=& \frac{
    1 - z + z^2/2 - z^3/6 + \cdots
  }{
    z^2/2 - z^4/\mbox{0x18} + \cdots
  }
  \\&=& \frac{
    \sum_{j=0}^\infty \left[
      (-)^j z^{j}/j!
    \right]
  }{
    -\sum_{k=1}^\infty (-)^k z^{2k}/(2k)!
  }
  \\&=& \frac{
    \sum_{k=0}^\infty \left[
      -z^{2k}/(2k)! + z^{2k+1}/(2k+1)!
    \right]
  }{
    \sum_{k=1}^\infty (-)^k z^{2k}/(2k)!
  }.
\eqb
By long division,
\bqb
  f(z) &=& \frac{2}{z^2} - \frac{2}{z} + \left\{
      \left[ -\frac{2}{z^2} + \frac{2}{z} \right]
      \sum_{k=1}^\infty \frac{(-)^k z^{2k}}{(2k)!}
    \right.\\&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\mbox{}
      + \sum_{k=0}^\infty \left[
        -\frac{z^{2k}}{(2k)!} + \frac{z^{2k+1}}{(2k+1)!}
      \right]
    \right\}
    \left/ \sum_{k=1}^\infty \frac{(-)^k z^{2k}}{(2k)!} \right.
  \\&=& \frac{2}{z^2} - \frac{2}{z} + \left\{
      \sum_{k=1}^\infty
      \left[
        - \frac{(-)^k 2 z^{2k-2}}{(2k)!}
        + \frac{(-)^k 2 z^{2k-1}}{(2k)!}
      \right]
    \right.\\&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\mbox{}
      + \sum_{k=0}^\infty \left[
        -\frac{z^{2k}}{(2k)!} + \frac{z^{2k+1}}{(2k+1)!}
      \right]
    \right\}
    \left/ \sum_{k=1}^\infty \frac{(-)^k z^{2k}}{(2k)!} \right.
  \\&=& \frac{2}{z^2} - \frac{2}{z} + \left\{
      \sum_{k=0}^\infty
      \left[
        \frac{(-)^k 2 z^{2k}}{(2k+2)!}
        - \frac{(-)^k 2 z^{2k+1}}{(2k+2)!}
      \right]
    \right.\\&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\mbox{}
      + \sum_{k=0}^\infty \left[
        -\frac{z^{2k}}{(2k)!} + \frac{z^{2k+1}}{(2k+1)!}
      \right]
    \right\}
    \left/ \sum_{k=1}^\infty \frac{(-)^k z^{2k}}{(2k)!} \right.
  \\&=& \frac{2}{z^2} - \frac{2}{z} + \sum_{k=0}^\infty \left[
      \frac{ -(2k+1)(2k+2) + (-)^k 2 }{(2k+2)!}z^{2k}
    \right.\\&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\mbox{}
      + \frac{ (2k+2) - (-)^k 2 }{(2k+2)!}z^{2k+1}
    \right]
    \left/ \sum_{k=1}^\infty \frac{(-)^k z^{2k}}{(2k)!}. \right.
\eqb
The remainder's $k=0$ terms now disappear as intended; so, factoring
$z^2/z^2$ from the division leaves
\bqa
  f(z) &=& \frac{2}{z^2} - \frac{2}{z} + \sum_{k=0}^\infty \left[
      \frac{ (2k+3)(2k+4) + (-)^k 2 }{(2k+4)!}z^{2k}
    \right.\xn\\&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\mbox{}
      - \frac{ (2k+4) + (-)^k 2 }{(2k+4)!}z^{2k+1}
    \right]
    \left/ \sum_{k=0}^\infty \frac{(-)^k z^{2k}}{(2k+2)!}. \right.
  \xn\\\label{taylor:380:30}
\eqa
One can continue dividing to extract further terms if desired, and if
all the terms
\[
  f(z) = \frac{2}{z^2} - \frac{2}{z} + \frac{7}{6} - \frac{z}{2} + \cdots
\]
are extracted the result is the \emph{Laurent series} proper,
\bq{taylor:380:laurent}
  f(z) = \sum_{k=K}^\infty (a_k) (z-z_o)^k, \ \ (k,K) \in \mathbb Z, \ K \le 0.
\eq
However for many purposes (as in eqn.~\ref{taylor:380:30}) the partial
Laurent series
\bqa
  f(z) &=& \sum_{k=K}^{-1} (a_k) (z-z_o)^k + \frac{
    \sum_{k=0}^\infty (b_k) (z-z_o)^k
  }{
    \sum_{k=0}^\infty (c_k) (z-z_o)^k
  },
  \label{taylor:380:partial}
  \\
  (k,K) &\in& \mathbb Z, \ \ K \le 0, \ c_0 \neq 0,
  \xn
\eqa
suffices and may even be preferable.  In either form,
\bq{taylor:380:40}
  f(z) = \sum_{k=K}^{-1} (a_k) (z-z_o)^k + f_o(z-z_o), \ \ (k,K) \in
  \mathbb Z, \ K \le 0,
\eq
where, unlike $f(z)$, $f_o(z-z_o)$ is analytic at $z=z_o$.  The
$f_o(z-z_o)$ of~(\ref{taylor:380:40}) is $f(z)$'s \emph{regular part} at
$z=z_o$.

The ordinary Taylor series diverges at a function's pole.  Handling the
pole separately, the Laurent series remedies this defect.%
\footnote{
  The professional mathematician's treatment of the Laurent series
  usually seems to begin by defining an annular convergence domain (a
  convergence domain bounded without by a large circle and within by a
  small) in the Argand plane.  From an applied point of view
  however what interests us is the basic technique to remove the poles
  from an otherwise analytic function.%
}$\mbox{}^,$\footnote{%
  \cite[\S~10.8]{Hildebrand}%
  \cite[\S~2.5]{Fisher}
}

Sections~\ref{inttx:250} and~\ref{inttx:260} tell more about poles
generally, including multiple poles like the one in the example
here.

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

\section{Taylor series in $1/z$}
\label{taylor:385}
\index{Taylor series!in $1/z$}

A little imagination helps the Taylor series a lot.  The Laurent series
of \S~\ref{taylor:380} represents one way to extend the Taylor series.
Several other ways are possible.  The typical trouble one has with the
Taylor series is that a function's poles and branch points limit the
series' convergence domain.  Thinking flexibly, however, one can often
evade the trouble.

\index{essential singularity}
\index{singularity!essential}
Consider the function
\[
  f(z) = \frac{\sin(1/z)}{\cos z}.
\]
This function has a nonanalytic point of a most peculiar nature at
$z=0$.  The point is an essential singularity, and one cannot expand the
function directly about it.  One could expand the function directly
about some other point like $z=1$, but calculating the Taylor
coefficients would take a lot of effort and, even then, the resulting
series would suffer a straitly limited convergence domain.  All that
however tries too hard.  Depending on the application, it may suffice to
write,
\[
  f(z) = \frac{\sin w}{\cos z}, \ \ w \equiv \frac 1 z.
\]
This is
\[
  f(z) = \frac
    {z^{-1} - z^{-3}/3! + z^{-5}/5! - \cdots}
    {1 - z^2/2! + z^4/4! - \cdots},
\]
which is all one needs to calculate $f(z)$ numerically---and may be all
one needs for analysis, too.

As an example of a different kind, consider
\[
  g(z) = \frac{1}{(z-2)^2}.
\]
Most often, one needs no Taylor series to handle such a function
(one simply does the indicated arithmetic).  Suppose however that
a Taylor series specifically about $z=0$ were indeed needed for some
reason.  Then by~(\ref{taylor:314:70}) and~(\ref{drvtv:220:30}),
\[
  g(z) = \frac{1/4}{(1-z/2)^2}
    = \frac 1 4 \sum_{k=0}^\infty \cmb{1+k}{1} \bigg[\frac z 2\bigg]^k
    = \sum_{k=0}^\infty \frac{k+1}{2^{k+2}} z^k,
\]
That expansion is good only for $\left|z\right| < 2$, but for $\left|z\right| > 2$ we also
have that
\[
  g(z) = \frac{1/z^2}{(1-2/z)^2}
    = \frac 1 {z^2} \sum_{k=0}^\infty \cmb{1+k}{1} \left[\frac 2 z\right]^k
    = \sum_{k=2}^\infty \frac{2^{k-2}(k-1)}{z^k},
\]
which expands in negative rather than positive powers of~$z$.  Note that
we have computed the two series for $g(z)$ without ever actually taking
a derivative.

Neither of the section's examples is especially interesting in itself,
but their point is that it often pays to think flexibly in extending and
applying the Taylor series.  One is not required immediately to take the
Taylor series of a function as it presents itself; one can first
change variables or otherwise rewrite the function in some convenient
way, and then take the Taylor series either of the whole function at once
or of pieces of it separately.  One can expand in negative powers of~$z$
equally validly as in positive powers.  And, though taking derivatives
per~(\ref{taylor:310:20}) may be the canonical way to determine Taylor
coefficients, any effective means to find the coefficients suffices.

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

\section{The multidimensional Taylor series}
\label{taylor:370}
\index{Taylor series!multidimensional}

\index{cross-term}
\index{term!cross-}
\index{cross-derivative}
\index{derivative!cross-}
Equation~(\ref{taylor:310:20}) has given the Taylor series for functions
of a single variable.  The idea of the Taylor series does not differ
where there are two or more independent variables, only the details are
a little more complicated.  For example, consider the function
$f(z_1,z_2) = z_1^2 + z_1z_2 + 2z_2$,
which has terms~$z_1^2$ and~$2z_2$---these we understand---but
also has the cross-term~$z_1z_2$ for which the relevant derivative
is the cross-derivative $\partial^2\!f/\partial z_1\,\partial z_2$.
Where two or more independent variables are involved, one must account
for the cross-derivatives, too.

With this idea in mind, the multidimensional Taylor series is
\bq{taylor:370:10}
  f(\ve z) = \sum_{\ve k}
    \left(\left.\frac {\partial^{\ve k}\!f}
    {\partial\ve z^{\ve k}}\right|_{\ve z=\ve z_o}\right)
    \frac{(\ve z-\ve z_o)^{\ve k}}{\ve k!}.
\eq

Well, that's neat.  What does it mean?
\bi
  \item \index{vector}\index{vector!generalized}
    The~$\ve z$ is a \emph{vector}%
    \footnote{
      In this generalized sense of the word, a \emph{vector} is an
      ordered set of~$N$ elements.  The geometrical vector $\ve v = \vu
      x x + \vu y y + \vu z z$ of \S~\ref{trig:230}, then, is
      a vector with $N=3$, $v_1=x$, $v_2=y$ and $v_3=z$.  (Generalized
      vectors of arbitrary~$N$ will figure prominently in the book from
      chapter~\ref{matrix} onward.)
    }
    incorporating the several independent variables
    $z_1,z_2,\ldots,z_N$.
  \item \index{vector!integer}\index{vector!nonnegative integer}
    The~$\ve k$ is a nonnegative integer vector of~$N$
    counters---$k_1,k_2,\ldots,k_N$---one for each of the independent
    variables.  Each of the~$k_n$ runs independently from~$0$
    to~$\infty$, and every permutation is possible.  For example, if
    $N=2$ then
    \bqb
      \ve k &=& (k_1,k_2) \\
      &=& (0,0),(0,1),(0,2),(0,3),\ldots; \\
      &&  (1,0),(1,1),(1,2),(1,3),\ldots; \\
      &&  (2,0),(2,1),(2,2),(2,3),\ldots; \\
      &&  (3,0),(3,1),(3,2),(3,3),\ldots; \\
      &&  \ldots
    \eqb
  \item The $\partial^{\ve k}\!f/\partial\ve z^{\ve k}$ represents the $\ve k$th
    cross-derivative of $f(\ve z)$, meaning that
    \[
      \frac{\partial^{\ve k}\!f}{\partial\ve z^{\ve k}} \equiv
      \left(\prod_{n=1}^{N}\frac{\partial^{k_n}}{(\partial z_n)^{k_n}}\right)f.
    \]
  \item The $(\ve z-\ve z_o)^{\ve k}$ represents
    \[
      (\ve z-\ve z_o)^{\ve k} \equiv \prod_{n=1}^N (z_n-z_{on})^{k_n}.
    \]
  \item The~$\ve k!$ represents
    \[
      \ve k! \equiv \prod_{n=1}^N k_n!.
    \]
\ei
With these definitions, the multidimensional Taylor
series~(\ref{taylor:370:10}) yields all the right derivatives and
cross-derivatives at the expansion point $\ve z = \ve z_o$.  Thus within
some convergence domain about $\ve z = \ve z_o$, the multidimensional
Taylor series~(\ref{taylor:370:10}) represents a function $f(\ve z)$ as
accurately as the simple Taylor series~(\ref{taylor:310:20}) represents
a function $f(z)$, and for the same reason.

