\chapter{The Fourier series}
\label{fours}
\index{Fourier series}
\index{series!Fourier}
\index{Fourier, Jean Baptiste Joseph\\(1768--1830)} % bad break

% This chapter-leading paragraph is a bit arch, but I've not yet
% been able to think of anything better to write here.
\index{Kelvin, Lord (1824--1907)}
It might be said that, among advanced mathematical techniques,
none is so useful, and few so appealing, as the one Lord Kelvin has
acclaimed ``a great mathematical poem.''%
\footnote{\cite[chapter~17]{JJH}}
It is the Fourier transform, which this chapter and the next will develop.
This first of the two chapters brings the Fourier transform in its
primitive guise as the \emph{Fourier series.}

\index{square wave}
\index{wave!square}
\index{superposition!of sinusoids}
\index{sinusoid!superposition of}
\index{waveform!repeating}
\index{repeating waveform}
\index{period}
The Fourier series is an analog of the Taylor series of chapter~\ref{taylor}
but meant for \emph{repeating waveforms,} functions $f(t)$ of which
\bq{fours:000:11}
  f(t) = f(t+nT_1), \ \Im(T_1) = 0, \ T_1 \neq 0,
  \ \mbox{for all $n \in \mathbb Z$},
\eq
where~$T_1$ is the waveform's characteristic \emph{period.} Examples
include the \emph{square wave} of
Fig.~\ref{fours:000:fig10}.
\begin{figure}
  \caption{A square wave.}
  \label{fours:000:fig10}
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\end{figure}
A Fourier series expands such a repeating waveform as a superposition
of complex exponentials (or, equivalently, if the waveform is real, as a
superposition of sinusoids).

\index{waveform!approximation of}
\index{primary frequency}
\index{frequency!primary}
Suppose that you wanted to approximate the square wave of
Fig.~\ref{fours:000:fig10} by a single sinusoid.  You might try the
sinusoid at the top of Fig.~\ref{fours:000:fig20}---which is not very
convincing, maybe, but if you added to the sinusoid another, suitably
scaled sinusoid of thrice the frequency then you would obtain the
somewhat better fitting curve in the figure's middle.  The curve at the
figure's bottom would yet result after you had added in four more
sinusoids respectively of five, seven, nine and eleven times the primary
frequency.  Algebraically,%
\bqa
  f(t) &=& \frac{8A}{2\pi}\bigg[
    \cos\frac{(2\pi) t}{T_1}
    - \frac 1 3 \cos\frac{3(2\pi) t}{T_1}
  \xn\\&&\ \ \ \ \ \ \ \ \ \ \mbox{}
    + \frac 1 5 \cos\frac{5(2\pi) t}{T_1}
    - \frac 1 7 \cos\frac{7(2\pi) t}{T_1}
    + \cdots
  \bigg].
  \label{fours:000:20}
\eqa%
\begin{figure}
  \caption[Superpositions of sinusoids.]{Superpositions of one, two and
    six sinusoids to approximate the square wave of
    Fig.~\ref{fours:000:fig10}.}
  \label{fours:000:fig20}
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\end{figure}%
How faithfully~(\ref{fours:000:20}) really represents the repeating
waveform and why its coefficients happen to be $1, -\frac 1 3, \frac 1
5, -\frac 1 7, \ldots$ are among the questions this chapter will try to
answer; but, visually at least, it looks as though superimposing
sinusoids worked.

The chapter begins in preliminaries, starting with a discussion of
Parseval's principle.

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

\section{Parseval's principle}
\label{fours:080}
\index{Parseval's principle}
\index{Parseval, Marc-Antoine (1755--1836)}

\index{step in every direction}
\emph{Parseval's principle} is that \emph{a step in every direction is
no step at all.}  In the Argand plane (Fig.~\ref{alggeo:225:fig}),
stipulated that
\bq{fours:080:08}
  \begin{split}
    \Delta\omega\, T_1 &= 2\pi, \\
    \Im(\Delta\omega) &= 0, \\
    \Im(t_o) &= 0, \\
    \Im(T_1) &= 0, \\
    T_1 &\neq 0,
  \end{split}
\eq
and also that%
\footnote{
  That $2 \le N$ is a redundant requirement, since
  (\ref{fours:080:09})'s other lines imply it, but it doesn't hurt to
  state it anyway.
}
\bq{fours:080:09}
  \begin{split}
    j,n,N &\in \mathbb Z, \\
    n &\neq 0, \\
    \left| n \right| &< N, \\
    2 &\le N,
  \end{split}
\eq
the principle is expressed algebraically as that%
\footnote{
  An expression like $t_o \pm T_1/2$ means $t_o \pm (T_1/2)$, here and
  elsewhere in the book.
}
\bq{fours:080:10}
  \int_{t_o-T_1/2}^{t_o+T_1/2} e^{in \,\Delta\omega\, \tau} \,d\tau = 0
\eq
or alternately in discrete form as that
\bq{fours:080:15}
  \sum_{j=0}^{N-1} e^{i 2\pi nj/N} = 0.
\eq

\index{parameter}
\index{dimension}
\index{physical unit}
\index{unit!physical}
Because the product $\Delta\omega\, T_1 = 2\pi$ relates~$\Delta\omega$
to~$T_1$, the symbols~$\Delta\omega$ and~$T_1$ together represent
in~(\ref{fours:080:08}) and~(\ref{fours:080:10}) not two but only
one parameter; you cannot set them independently, for the one merely
inverts the other.
If~$T_1$ bears physical units then these
will typically be units of time (seconds, for instance),
whereupon~$\Delta\omega$ will bear the corresponding units of angular
frequency (such as radians per second).  The frame offset~$t_o$ and the
dummy variable~$\tau$ must have the same dimensions%
\footnote{
  The term \emph{dimension} in this context refers to the kind of
  physical unit.  A quantity like~$T_1$ for example, measurable in seconds
  or years (but not, say, in kilograms or dollars), has dimensions of
  time.  An automobile's speed having dimensions of length divided by
  time can be expressed in miles per hour as well as in meters per
  second but not directly, say, in volts per centimeter; and so on.
}%
~$T_1$ has.  This matter is discussed further in \S~\ref{fours:085}.

\index{physical insight}
\index{insight}
\index{symmetry!appeal to}
To prove~(\ref{fours:080:10}) symbolically is easy: one merely carries
out the indicated integration.  To prove~(\ref{fours:080:15})
symbolically is not much harder: one replaces the complex exponential
$e^{i 2\pi nj/N}$ by $\lim_{\ep \ra 0^{+}} e^{(i - \ep)2\pi nj/N}$ and
then uses~(\ref{alggeo:228:45}) to evaluate the summation.
Notwithstanding, we can do better, for an alternate, more edifying,
physically more insightful explanation of the two equations is possible
as follows.  Because~$n$ is a nonzero integer,~(\ref{fours:080:10})
and~(\ref{fours:080:15}) represent sums of steps in every
direction---that is, steps in every phase---in the Argand plane (more
precisely, eqn.~\ref{fours:080:15} represents a sum over a discrete but
balanced, uniformly spaced selection of phases).  An appeal to symmetry
forbids such sums from favoring any one phase $n \,\Delta\omega\, \tau$
or $2\pi nj/N$ over any other.  This being the case, how could the sums
of~(\ref{fours:080:10}) and~(\ref{fours:080:15}) come to any totals
other than zero?  The plain answer is that they can come to no other
totals.  A step in every direction is indeed no step at all.  This is
why~(\ref{fours:080:10}) and~(\ref{fours:080:15}) are so.%
\footnote{
  The writer unfortunately knows of no conventionally established
  name for Parseval's principle.  The name \emph{Parseval's principle}
  seems as apt as any and this is the name this book will use.

  A pedagogical knot seems to tangle Marc-Antoine Parseval's various
  namesakes.  Because Parseval's principle can be extracted as a special
  case from Parseval's theorem (eqn.~\ref{fouri:110:parseval}
  in the next chapter), the literature sometimes indiscriminately
  applies the name ``Parseval's theorem'' to both.  This is fine as far
  as it goes, but the knot arrives when one needs Parseval's principle
  to derive the Fourier series, which one needs to derive the Fourier
  transform, which one needs in turn to derive Parseval's theorem, at
  least as this book develops them.  The way to untie the knot is to
  give Parseval's principle its own name and to let it stand as an
  independent result.
}

We have actually already met Parseval's principle, informally, in
\S~\ref{inttx:260.20}.
There is also \emph{Parseval's theorem} to come in
\S~\ref{fouri:110.55}.

One can translate Parseval's principle from the Argand realm to the
analogous realm of geometrical vectors, if needed, in the obvious way.

%Parseval's principle is a specialization of a more general mathematical
%result named ``Parseval's theorem,'' the latter of which refers to
%a formalized abstraction of the mathematics%
%\footnote{\cite[eqns.~2-40 and~2-41]{Couch}}
%that descends from the engineering observation that, given lengthy
%enough a chance and hefty enough a temporary storage element (like a
%counterweight, flywheel, buffer, holding tank, pressure vessel, resonant
%element, tuned stub or electrical capacitor), one can insert and/or
%extract mechanical or electrical power to and/or from a wave, and that
%one can do so a single, discrete frequency at a time; whereas one cannot
%separate the power that accords to sources operating at the same
%frequency unless the sources are twain and happen to operate in
%quadrature.  The trouble in developing the full Parseval's theorem from
%the first is that, if one prefers not to develop it in a conceptual
%vacuum before concepts like ``power'' or ``quadrature'' have been
%properly introduced, one will still want the theorem's main idea,
%Parseval's principle, before treating power and quadrature.  So, which
%comes first?  Power and quadrature, or Parseval's theorem?  Our answer
%is to introduce Parseval's principle first and then to let the full
%theorem, which is not central and will be almost obvious later on in any
%case, take care of itself.

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

\section{Time, space and frequency}
\label{fours:085}
\index{time}
\index{space}
\index{frequency}

\index{frequency!cyclic}
\index{frequency!angular}
\index{cyclic frequency}
\index{angular frequency}
\index{dimension}
\index{engine}
\index{internal-combustion engine}
\index{RPM}
\index{hertz}
A \emph{frequency} is the inverse of an associated period of time,
expressing the useful concept of the rate at which a cycle repeats.  For
example, an internal-combustion engine whose crankshaft revolves once
every~20 milliseconds---which is to say, once every $1/3000$ of a
minute---runs thereby at a frequency of~3000 revolutions per
minute (RPM) or, in other words,~3000 cycles per minute.
Frequency however comes in two styles: cyclic frequency (as in
the engine's example), conventionally represented by letters like~$\nu$
and~$f$; and angular frequency, by letters like~$\omega$ and~$k$.
If~$T$, $\nu$ and~$\omega$ are letters taken to stand respectively for a
period of time, for the associated cyclic frequency, and for the
associated angular frequency, then by definition
\bq{fours:085:10}
  \begin{split}
    \nu T &= 1, \\
    \omega T &= 2\pi, \\
    \omega &= 2\pi \nu.
  \end{split}
\eq
The period~$T$ will bear units of time like seconds or minutes.  The cyclic
frequency~$\nu$ will bear units of inverse time like cycles
per second (hertz)
or cycles per minute.%
\footnote{%
  Notice incidentally, contrary to the improper verbal usage one
  sometimes hears, that there is no such thing as a ``hert.''  Rather,
  ``Hertz'' is somebody's name.  The uncapitalized form ``hertz'' thus
  is singular as well as plural.%
}
The angular frequency~$\omega$ will bear units of
inverse time like radians per second or radians per minute.

\index{cycle}
\index{radian}
\index{second}
\index{minute}
\index{unit!of measure}
\index{measure, unit of}
The last brings us to a point that has confused many students of
science and engineering: if $T=20\ \mbox{ms}$ and $\nu=3000\
\mbox{cycles}/\mbox{min}$, then why not $\nu
T = (3000\:\mbox{cycles}/\mbox{min})(20\:\mbox{ms}) =
6.0\times 10^4\:
\mbox{cycle}\cdot\mbox{ms}/\mbox{min}=1\:\mbox{cycle}\neq 1?$
The answer is that a cycle is not conventionally held
to be a unit of measure and, thus, does not conventionally enter
into the arithmetic.\footnote{%
  % diagn: review this new footnote
  The writer believes the convention to be wise.  The reason behind the
  convention is not easy to articulate (though the narrative will try
  to articulate it, anyway), but experience does seem to support the
  convention nevertheless.  Little is gained, and convenience is lost,
  when one---contrary to convention---treats a countable entity like a
  cycle as one would treat an arbitrary quantity of physical reference
  like a second.  The cycle and the second are not things of the same
  kind.  As such, they tend not to approve treatment of the same kind,
  even if such treatment is possible.%
}
Minimally correct usage is rather that
\[
  \nu T =
  (3000\:\mbox{min}^{-1})(20\:\mbox{ms}) =
  6.0\times 10^4\frac{\mbox{ms}}{\mbox{min}}=1.
\]
As for the word ``cycle,'' one can include the word in the above line if
one wishes to include it, but with the understanding that the word lacks
arithmetic significance.  Arithmetically, one can drop the word at any
stage.

\index{countability}
\index{baseball}
It follows, perhaps unexpectedly,
that the cycle per minute and the radian per minute do
not arithmetically differ from one another.  Arithmetically,
counterintuitively,
\[
  \frac{1}{60}\,\frac{\mbox{cycle}}{\mbox{second}} =
  1\frac{\mbox{cycle}}{\mbox{minute}} =
  1\frac{\mbox{radian}}{\mbox{minute}} =
  \frac{1}{60}\,\frac{\mbox{radian}}{\mbox{second}}.
\]
This looks obviously wrong, of course, but don't worry: it is a mere
tautology which, though perhaps literally accurate (we will explain why),
expresses no very profound idea.  Its sole point is that both the
cycle per minute and the radian per minute, interpreted as units of
measure, are units of $[\mbox{minute}]^{-1}$; whereas---in the context
of phrases like ``cycle per minute'' and ``radian per minute''---the
words ``cycle'' and ``radian'' are verbal cues that, in and of
themselves, play no actual part in the mathematics.  \emph{This is not
because the cycle and the radian were ephemeral but rather because the
minute is unfundamental.}

The minute, a unit of measure representing a definite but arbitrary
quantity of time, requires arithmetical representation.  The
cycle and the radian, by contrast, are nonarbitrary,
discrete, inherently countable things; and, where things are counted, it
is ultimately up to the mathematician to interpret the count (consider
for instance that nine baseball caps may imply nine baseball players and
one baseball team, but that there is nothing in the number nine itself
to tell us so).  To distinguish angular frequencies from cyclic
frequencies, it remains to the mathematician to lend factors of~$2\pi$
where needed.

If, nevertheless, you think the last displayed equation just too weird,
then
don't write it that way; but think of, say, gear ratios.  A gear ratio
might be 3:1 or 5:1 or whatever, but the ratio is unitless.  You can
\emph{say}~``3.0 turns of the small gear per turn of the large gear,''
but this manner of speaking
does not make the ``turn of the small gear per turn of the large
gear'' a proper unit of measure.  The ``cycle per radian'' is in the
same sense likewise not a proper unit of measure.  (Now, if you
\emph{still} think the last displayed equation just too
weird---well, it \emph{is} weird.  You can ignore the equation, instead
interpreting the expression ``cycle per radian'' as a way of naming the
number~$2\pi$.  This sort of works, but beware that such an interpretation
does not extend very well to other nonunit units like ``decibel'' and is
not the interpretation the writer recommends.  Also beware: an
expression like $\sin\{[2\pi/4]\:\mbox{radians}\} =
\sin[2\pi/4] = 1$ means something sensible whereas an expression like
$\sin\{[2\pi/4]\:\mbox{minutes}\} = \mbox{??}$ probably does
not.\footnote{The word ``minute'' here refers to the
period of time.  There is, separately, an angle called by the same
English word, ``minute,'' but that is not what we are talking about
here.}  Anyway, if ``radian'' is taken to be~1---as it must be taken if
$\sin\{[2\pi/4]\:\mbox{radians}\}$ is to come out right---then ``cycle''
must be taken to be~$2\pi$, which does not quite square with
eqn.~\ref{fours:085:10}, does it?  No, the problem is that the radian
and the cycle are no units of measure.)\footnote{%
  Some recent undergraduate engineering textbooks have taken to
  the style of
  \[
    E = \frac{Q}{Cd}\ [\mbox{volts/meter}].
  \]
  The intent seems to be to encourage undergraduates to include units of
  measure with their engineering quantities, as
  \[
    E = \frac{Q}{Cd} = 5.29\:\mbox{volts}/\mbox{meter}.
  \]
  Unfortunately, my own, occasional experience at teaching
  undergraduates suggests that undergraduates tend to read the textbook
  as though it had read
  \[
    E = \left[\frac{Q}{Cd}\right]\left[1.0\:\frac{\mbox{volt}}{\mbox{meter}}\right],
  \]
  which is wrong and whose resultant confusion compounds, wasting
  hours of the undergraduates' time.  It seems to me preferable to
  insist that undergraduates learn from the first the correct meaning of
  an unadorned equation like%
  \[
    E = \frac{Q}{Cd},
  \]
  where, say,
  $Q=13.3\ \mbox{volt}\cdot\mbox{sec}/\mbox{ohm}$,
  $C=0.470\ \mbox{sec}/\mbox{ohm}$, and
  $d=5.35\ \mbox{cm}$; and that they grasp the need not to write
  algebraically perplexing falsities such as that ``$d\ \mbox{cm} =
  5.35\ \mbox{cm}$''---perplexing falsities which, unfortunately, the
  textbook style in question inadvertently encourages them to write.

  When during an engineering lecture it becomes pedagogically necessary
  to associate units of measure to a symbolic equation, my own practice
  at the blackboard has been to write
  \[
    E = \frac{Q}{Cd},\quad E\ :\ [\mbox{volts/meter}].
  \]
  Done sparingly, this seems to achieve the desired effect, though in
  other instances the unadorned style is preferred.
}

The word ``frequency'' without a qualifying adjective is usually taken
in English to mean cyclic frequency unless the surrounding context
implies otherwise.  Notwithstanding, interestingly, experience seems to
find angular frequency to be oftener the more natural or convenient to
use.

\index{frequency!spatial}
\index{spatial frequency}
\index{wave number}
Frequencies exist in space as well as in time:
\bq{fours:085:15}
  k\lambda = 2\pi.
\eq
Here,~$\lambda$ is a \emph{wavelength} measured in meters or other units
of length.  The \emph{wave number}%
\footnote{%
  One could wish for a better name for the thing than \emph{wave
  number.}  By whatever name, the wave number~$k$ is no integer,
  notwithstanding that the letter~$k$ tends to represent integers in
  other contexts.%
}%
~$k$ is an angular spatial frequency measured in units like radians per
meter; that is, $[\mbox{meter}]^{-1}$.  (Oddly, no conventional symbol
for \emph{cyclic} spatial frequency seems to be current.  The literature
mostly just uses $k/2\pi$ which, in light of the potential for confusion
between~$\nu$ and~$\omega$ in the temporal domain, is probably for the
best.)

\index{propagation speed}
\index{speed!of propagation}
Where a wave propagates the propagation speed
\bq{fours:085:20}
  v = \frac{\lambda}{T} = \frac{\omega}{k}
\eq
relates periods and frequencies in space and time.

\index{dimensionlessness}
Now, we must admit that we fibbed when we said (or implied) that~$T$ had to have
dimensions of time.  Physically, that is the usual interpretation, but
mathematically~$T$ (and~$T_1$, $t$, $t_o$, $\tau$, etc.)\ can bear any
units and indeed is not required to bear units at all, a fact to which
\S~\ref{fours:080} has alluded.  The only mathematical requirement is
that the product $\omega T = 2\pi$ (or $\Delta\omega\, T_1 = 2\pi$ or
the like, as appropriate) be dimensionless.  However, when~$T$ has
dimensions of length rather than of time it is conventional---indeed, it
is practically mandatory if one wishes to be understood---to change
$\lambda \la T$ and $k \la \omega$ as this section has done, though the
essential Fourier mathematics is the same regardless of $T$'s dimensions
(if any) or of whether alternate symbols like~$\lambda$ and~$k$ are
used.

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

\section{Several useful pulses}
\label{fours:095}
\index{pulse}
\index{pulse!useful}
\index{pulse!square}
\index{square pulse}
\index{pulse!triangular}
\index{triangular pulse}
\index{pulse!raised-cosine}
\index{raised-cosine pulse}
\index{pulse!Gaussian}
\index{Gaussian pulse}

\index{Dirac delta function!implementation of}
\index{delta function, Dirac!implementation of}
\index{$\Pi$ as the rectangular pulse}
\index{$\Lambda$ as the triangular pulse}
\index{$\Psi$ as the raised-cosine pulse}
\index{$\Omega$ as the Gaussian pulse}
The Dirac delta of \S~\ref{integ:670} and Fig.~\ref{integ:670:fig-d}
is among other reasons useful for the unit area it covers, but for some
purposes its curve is too sharp.  One occasionally finds it expedient to
substitute the \emph{square,} \emph{triangular} or \emph{raised-cosine
pulse} of Fig.~\ref{fours:095:fig1},
\bq{fours:095:10}
  \settowidth\tlb{$\left[1 + \cos (\pi t) \right]/2$}
  \begin{split}
    \Pi   (t) &\equiv \begin{cases}
      \makebox[\tlb][l]{$1$} &\mbox{if $\left|t\right| < 1/2$,} \\
      \makebox[\tlb][l]{$1/2$} &\mbox{if $\left|t\right| = 1/2$,} \\
      0 &\mbox{otherwise;}
    \end{cases} \\
    \Lambda(t) &\equiv \begin{cases}
      \makebox[\tlb][l]{$1 - \left| t \right|$} &\mbox{if $\left|t\right| \le 1$,} \\
      0 &\mbox{otherwise;}
    \end{cases} \\
    \Psi(t) &\equiv \begin{cases}
      \makebox[\tlb][l]{$\left[1 + \cos (\pi t) \right]/2$} &\mbox{if $\left|t\right| \le 1$,} \\
      0 &\mbox{otherwise;}
    \end{cases}
  \end{split}
\eq
for the Dirac delta,
\begin{figure}
  \caption{The square, triangular, raised-cosine and Gaussian pulses.}
  \label{fours:095:fig1}
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\end{figure}
each of which pulses evidently shares Dirac's property that
\bq{fours:095:20}
  \settowidth\tla{$\Psi$}
  \begin{split}
    \int_{-\infty}^\infty \frac{1}{T}
      \makebox[\tla][c]{$\delta $}\left(\frac{\tau - t_o}{T}\right) \,d\tau &= 1, \\
    \int_{-\infty}^\infty \frac{1}{T}
      \makebox[\tla][c]{$\Pi    $}\left(\frac{\tau - t_o}{T}\right) \,d\tau &= 1, \\
    \int_{-\infty}^\infty \frac{1}{T}
      \makebox[\tla][c]{$\Lambda$}\left(\frac{\tau - t_o}{T}\right) \,d\tau &= 1, \\
    \int_{-\infty}^\infty \frac{1}{T}
      \makebox[\tla][c]{$\Psi   $}\left(\frac{\tau - t_o}{T}\right) \,d\tau &= 1,
  \end{split}
\eq
for any real $T>0$ and real~$t_o$.  In the limit,
\bq{fours:095:30}
  \settowidth\tla{$\Psi$}
  \begin{split}
    \lim_{T \ra 0^{+}} \frac{1}{T}
      \makebox[\tla][c]{$\Pi    $}\left(\frac{t - t_o}{T}\right)
      &= \delta(t - t_o), \\
    \lim_{T \ra 0^{+}} \frac{1}{T}
      \makebox[\tla][c]{$\Lambda$}\left(\frac{t - t_o}{T}\right)
      &= \delta(t - t_o), \\
    \lim_{T \ra 0^{+}} \frac{1}{T}
      \makebox[\tla][c]{$\Psi   $}\left(\frac{t - t_o}{T}\right)
      &= \delta(t - t_o),
  \end{split}
\eq
constituting at least three possible implementations of the Dirac delta
in case such an implementation were needed.  Looking ahead, if we may
further abuse the Greek capitals to let them represent pulses whose
shapes they accidentally resemble, then a fourth, subtler
implementation---more complicated to handle but analytic
(\S~\ref{taylor:320}) and therefore preferable for some purposes---is
the \emph{Gaussian pulse}
\bqa
  \lim_{T \ra 0^{+}} \frac{1}{T}
    \Omega\left(\frac{t - t_o}{T}\right)
    &=& \delta(t - t_o),
  \label{fours:095:33} \\
  \Omega(t) &\equiv& \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{t^2}{2}\right),\xn
\eqa
the mathematics of which \S~\ref{fouri:130} and
% diagn: check the following reference.
%chapters~\ref{specf} and~\ref{prob}
chapter~\ref{prob}
will begin to unfold.
%but see Fig.~\ref{prob:normdist-fig} on page~\pageref{prob:normdist-fig}
%for a plot.

\index{signal!discretely sampled}
The three nonanalytic pulses share a convenient property
for all~$t$:
\bq{fours:095:40}
  \settowidth\tla{$\Psi$}
  \begin{split}
    \sum_{m=-\infty}^\infty \makebox[\tla][c]{$\Pi    $}(t-m) &= 1; \\
    \sum_{m=-\infty}^\infty \makebox[\tla][c]{$\Lambda$}(t-m) &= 1; \\
    \sum_{m=-\infty}^\infty \makebox[\tla][c]{$\Psi   $}(t-m) &= 1.
  \end{split}
\eq
This property makes the three pulses especially useful in the rendering
of discretely sampled signals.
%The fourth pulse, $\Omega(t)$, lacks the property~(\ref{fours:095:40})
%but is analytic (\S~\ref{taylor:320}), which the others are not.
%\index{Archimedes (287--212~B.C.)}%
%\index{Cantor, Georg (1845--1918)}%

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

\section[Expanding waveforms in Fourier series]%
  {Expanding repeating waveforms in Fourier series}
\label{fours:100}
\index{Fourier series}
\index{series!Fourier}

\index{sinusoid}
\index{complex exponential}
\index{exponential, complex}
\index{Fourier coefficient}
\index{coefficient!Fourier}
The Fourier series represents a repeating waveform~(\ref{fours:000:11})
%like the waveform of Fig.~\ref{fours:000:fig10}
as a superposition of sinusoids.  More precisely, inasmuch as Euler's
formula~(\ref{cexp:250:cos}) makes a sinusoid a sum of two complex
exponentials, the Fourier series supposes that a repeating waveform were
a superposition
\bq{fours:100:10}
  f(t) = \sum_{j=-\infty}^{\infty} a_j e^{ij \,\Delta\omega\, t}
\eq
of many complex exponentials, in which~(\ref{fours:080:08}) is obeyed yet in
which neither the several Fourier coefficients~$a_j$ nor the
waveform $f(t)$ itself need be real.  Whether one can properly represent
every repeating waveform as a superposition~(\ref{fours:100:10}) of
complex exponentials is a question \S\S~\ref{fours:100.80}
and~\ref{fours:170} will address later; but, at least to the extent to
which one can properly represent such a waveform, we will now assert
that one can recover any or all of the waveform's Fourier
coefficients~$a_j$ by choosing an arbitrary frame offset~$t_o$ ($t_o =
0$ is a typical choice) and then integrating
\bq{fours:100:15}
  a_j = \frac{1}{T_1} \int_{t_o-T_1/2}^{t_o+T_1/2}
    e^{-ij\,\Delta\omega\,\tau} f(\tau) \,d\tau.
\eq

\subsection{Derivation of the Fourier-coefficient formula}
\label{fours:100.10}

\index{integration!over a complete cycle}
\index{cycle!integration over a complete}
\index{frequency!shifting of}
\index{Fourier coefficient!recovery of}
But why should~(\ref{fours:100:15}) work?  How is it to recover a Fourier
coefficient~$a_j$?  The answer is that it recovers a Fourier
coefficient~$a_j$ by isolating it, and that it isolates it by shifting
frequencies and integrating.

Equation~(\ref{fours:100:10}) has proposed to express a repeating
waveform as a series of complex exponentials, each exponential of the
form $a_j e^{ij \,\Delta\omega\, t}$ in which~$a_j$ is a weight to be
determined.  Unfortunately,~(\ref{fours:100:10}) can hardly be very
useful until the several~$a_j$ are determined, whereas how to
determine~$a_j$ from~(\ref{fours:100:10}) for a given value of~$j$ is
not immediately obvious.

The trouble with using~(\ref{fours:100:10}) to determine the
several coefficients~$a_j$ is that it includes all the terms
of the series and, hence, all the coefficients~$a_j$ at once.  To
determine~$a_j$ for a given value of~$j$, one should like to suppress
the entire series except the single element $a_j e^{ij \,\Delta\omega\,
t}$, isolating this one element for analysis.  Fortunately, Parseval's
principle~(\ref{fours:080:10}) gives us a way to do this, as we shall
soon see.

Now, to prove~(\ref{fours:100:15}) we mean to use~(\ref{fours:100:15}),
a seemingly questionable act.  Nothing prevents us however from taking
only the right side of~(\ref{fours:100:15})---not as an equation but as
a mere expression---and doing some algebra with it to see where the
algebra leads, for if the algebra should lead to the \emph{left} side
of~(\ref{fours:100:15}) then we should have proven the equation.
Accordingly, changing dummy variables $\tau \la t$ and $\ell \la j$
in~(\ref{fours:100:10}) and then substituting
into~(\ref{fours:100:15})'s right side the resulting expression for
$f(\tau)$, we have by successive steps that
\bqb
  \lefteqn{\frac{1}{T_1} \int_{t_o-T_1/2}^{t_o+T_1/2}
    e^{-ij\,\Delta\omega\,\tau} f(\tau) \,d\tau} &&
  \\&=&
    \frac{1}{T_1} \int_{t_o-T_1/2}^{t_o+T_1/2}
    e^{-ij\,\Delta\omega\,\tau}
    \sum_{\ell = -\infty}^{\infty} a_\ell e^{i\ell \,\Delta\omega\, \tau } \,d\tau
  \\&=&
    \frac{1}{T_1}
      \sum_{\ell = -\infty}^{\infty} a_\ell \int_{t_o-T_1/2}^{t_o+T_1/2}
      e^{i(\ell-j)\,\Delta\omega\,\tau}
      \,d\tau
  \\&=&
    \frac{a_j}{T_1}
      \int_{t_o-T_1/2}^{t_o+T_1/2}
      e^{i(j-j)\,\Delta\omega\,\tau}
      \,d\tau
  \\&=&
    \frac{a_j}{T_1}
      \int_{t_o-T_1/2}^{t_o+T_1/2} \,d\tau
  = a_j,
\eqb
in which Parseval's principle~(\ref{fours:080:10}) has killed all
but the $\ell=j$ term in the summation.  Thus is~(\ref{fours:100:15})
proved.

Except maybe to the extent to which one would like to examine
convergence (see the next paragraph), the idea behind the
proof remains more interesting than the proof itself, for one
would like to know not only the fact that~(\ref{fours:100:15}) is true
but also the thought which leads one to propose the equation in the
first place.  The thought is as follows.  Assuming
that~(\ref{fours:100:10}) can indeed represent the waveform $f(t)$
properly, one observes that the transforming factor
$e^{-ij\,\Delta\omega\,\tau}$ of~(\ref{fours:100:15}) serves to shift
the waveform's $j$th component $a_j e^{ij \,\Delta\omega\, t}$---whose
angular frequency is evidently $\omega = j \,\Delta\omega$---down to a
frequency of zero, incidentally shifting the waveform's several other
components to various nonzero frequencies as well.  Significantly, the
transforming factor leaves each shifted frequency to be a whole multiple
of the waveform's fundamental frequency~$\Delta\omega$.  By Parseval's
principle, (\ref{fours:100:15})'s integral then kills all the thus
frequency-shifted components except the zero-shifted one \emph{by
integrating the components over complete cycles,} passing only the
zero-shifted component which, once shifted, has no cycle.  Such is the
thought which has given rise to the equation.

Before approving the proof's interchange of summation and integration, a
pure mathematician would probably have preferred to establish
conditions under which the summation and integration should each
converge.  To the applied mathematician however, the establishment of
\emph{general} conditions turns out to be an unrewarding
exercise,\footnote{%
  \index{Dirichlet, J.~Peter Gustav Lejeune (1805-1859)}%
  The conditions conventionally observed among professional
  mathematicians seem to be known as the \emph{Dirichlet conditions.}
  As far as this writer can tell, the Dirichlet conditions lie pretty
  distant from applications---not that there aren't concrete
  applications that transgress them (for example in stochastics), but
  rather that the failure of~(\ref{fours:100:15}) to converge in a given
  concrete application is more readily apparent by less abstract means
  than Dirichlet's.

  This book could merely \emph{list} the Dirichlet conditions without
  proof; but, since the book is a book of derivations, it will decline
  to do that.
  The conditions look plausible.  We'll leave it at that.

  The writer suspects that few readers will ever encounter a concrete
  application that really wants the Dirichlet conditions, but one never
  knows.  The interested reader can pursue Dirichlet elsewhere.  (Where?
  No recommendation.  No book on the writer's shelf seems strong enough
  on Dirichlet to recommend.)%
}
so we will let the matter pass with this remark: nothing prevents one
from treating~(\ref{fours:100:10}) as
\[
  f(t) = \lim_{J\ra\infty}
    \sum_{j=-(J-1)}^{J-1} a_j e^{ij \,\Delta\omega\, t},
\]
which manages the convergence problem (to the extent to which it even is
a problem) in most cases of practical interest.  Further work on the
convergence problem is left to the charge of the concerned reader, but
see also \S\S~\ref{integ:240.10}, \ref{integ:240.20}
and~\ref{conclu:convergence}.

\subsection{The square wave}
\label{fours:100.20}
\index{square wave}
\index{wave!square}

According to~(\ref{fours:100:15}), the Fourier coefficients of
Fig.~\ref{fours:000:fig10}'s square wave are, if $t_o=T_1/4$ is chosen
and by successive steps,
\bqb
  a_j &=&
    \frac{1}{T_1} \int_{-T_1/4}^{3T_1/4}
    e^{-ij\,\Delta\omega\,\tau} f(\tau) \,d\tau
  \\&=&
    \frac{A}{T_1} \left[
      \mbox{$\ds
        \int_{-T_1/4}^{T_1/4}
        -\int_{T_1/4}^{3T_1/4}
      $}
    \right]
    e^{-ij\,\Delta\omega\,\tau} \,d\tau
  \\&=&
    \frac{iA}{2\pi j}  e^{-ij\,\Delta\omega\,\tau} \left[
      \bigg|_{-T_1/4}^{T_1/4} - \bigg|_{T_1/4}^{3T_1/4}
    \right].
\eqb
But%
{%
  \settowidth\tla{\scriptsize $-T_1/4$}%
  \bqb
    \left. e^{-ij\,\Delta\omega\,\tau} \right|_{\tau =                                -T_1/4  }  =
    \left. e^{-ij\,\Delta\omega\,\tau} \right|_{\tau = \makebox[\tla][l]{\scriptsize $3T_1/4$}} &=& i^j, \\
    \left. e^{-ij\,\Delta\omega\,\tau} \right|_{\tau = \makebox[\tla][l]{\scriptsize $ T_1/4$}} &=& (-i)^j,
  \eqb%
}%
so
\bqb
  \lefteqn{
    e^{-ij\,\Delta\omega\,\tau} \left[
      \bigg|_{-T_1/4}^{T_1/4} - \bigg|_{T_1/4}^{3T_1/4}
    \right]
  } &&
  \\&=& [(-i)^j - i^j] - [i^j - (-i)^j]
     =  2[(-i)^j - i^j]
  \\&=& \mbox{$\ldots,-i4,0,i4,0,-i4,0,i4,\ldots$ for $j=\ldots,-3,-2,-1,0,1,2,3,\ldots$}
\eqb
Therefore,
\bq{fours:100:20}
  \begin{split}
    a_j &= \left[(-i)^j - i^j\right] \frac{i2A}{2\pi j} \\
    &= \begin{cases}
      (-)^{(j-1)/2}4A/2\pi j &\mbox{for odd~$j$,} \\
      0 &\mbox{for even~$j$}
    \end{cases}
  \end{split}
\eq
are the square wave's Fourier coefficients which, when the coefficients
are applied to~(\ref{fours:100:10}) and when~(\ref{cexp:250:cos}) is invoked,
indeed yield the specific series of sinusoids~(\ref{fours:000:20}) and
Fig.~\ref{fours:000:fig20} have proposed.

\subsection{The rectangular pulse train}
\label{fours:100.30}
\index{pulse train}
\index{pulse train!rectangular}
\index{rectangular pulse train}

The square wave of \S~\ref{fours:100.20} is an important, canonical case
and~(\ref{fours:000:20}) is arguably worth memorizing.  After the square
wave, however, an endless variety of repeating waveforms present
themselves.  Section~\ref{fours:100.20} has exampled how to compute
their Fourier series.

\index{duty cycle}
One variant on the square wave is nonetheless interesting enough to
attract special attention.  This variant is the \emph{pulse train} of
Fig.~\ref{fours:100:fig3},
\bq{fours:100:27}
  f(t) = A \sum_{j=-\infty}^{\infty} \Pi\left(\frac{t-jT_1}{\eta T_1}\right);
\eq
where $\Pi(\cdot)$ is the square pulse of~(\ref{fours:095:10}); the
symbol~$A$ represents the pulse's full height rather than the
half-height of Fig.~\ref{fours:000:fig10}; and the dimensionless factor
$0 \le \eta \le 1$ is the train's \emph{duty cycle,} the fraction of
each cycle its pulse is as it were on duty.
\begin{figure}
  \caption{A rectangular pulse train.}
  \label{fours:100:fig3}
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                  \rput[l](\xxmb,0){$\eta T_1$}%
                }%
              }%
            }%
          }
        }
        {
          \psset{linewidth=2.0pt}
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             (-1.10\tla,0)(-1.10\tla,\tlb)(-0.90\tla,\tlb)(-0.90\tla,0)%
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             ( 0.90\tla,0)( 0.90\tla,\tlb)( 1.10\tla,\tlb)( 1.10\tla,0)%
             ( 1.35\tla,0)
        }
      }
    \end{pspicture}
  \ec
\end{figure}
By the routine of \S~\ref{fours:100.20},
\bqb
  a_j &=&
    \frac{1}{T_1} \int_{-T_1/2}^{T_1/2}
    e^{-ij\,\Delta\omega\,\tau} f(\tau) \,d\tau
  \\&=&
    \frac{A}{T_1} \int_{-\eta T_1/2}^{\eta T_1/2}
    e^{-ij\,\Delta\omega\,\tau} \,d\tau
  \\&=&
    \frac{iA}{2\pi j}  e^{-ij\,\Delta\omega\,\tau}
      \bigg|_{-\eta T_1/2}^{\eta T_1/2}
  =
    \frac{2A}{2\pi j} \sin \frac{2\pi\eta j}{2}
\eqb
for $j \neq 0$.  On the other hand,
\[
  a_0 =
    \frac{1}{T_1} \int_{-T_1/2}^{T_1/2}
      f(\tau) \,d\tau
  =
    \frac{A}{T_1} \int_{-\eta T_1/2}^{\eta T_1/2} \,d\tau
  = \eta A
\]
is the waveform's mean value.  Altogether for the pulse train,
\bq{fours:100:30}
  a_j = \begin{cases}
    \ds \frac{2A}{2\pi j} \sin \frac{2\pi\eta j}{2} & \mbox{if $j \neq 0$,} \\
    \eta A & \mbox{if $j = 0$}
  \end{cases}
\eq
(though eqn.~\ref{fours:160:05} will improve the notation later).

\index{area}
An especially interesting special case occurs when the duty cycle grows
very short.  Since $\lim_{\eta \ra 0^{+}} \sin (2\pi\eta j/2) = 2\pi\eta
j/2$ according to~(\ref{taylor:315:60}), it follows
from~(\ref{fours:100:30}) that
\bq{fours:100:35}
  \lim_{\eta \ra 0^{+}} a_j = \eta A,
\eq
the same for every index~$j$.  As the duty cycle~$\eta$ tends to vanish the
pulse tends to disappear and the Fourier coefficients along with it; but
we can compensate for vanishing duty if we wish by increasing the
pulse's amplitude~$A$ proportionally, maintaining the product
\bq{fours:100:36}
  \eta T_1 A = 1
\eq
of the pulse's width~$\eta T_1$ and its height~$A$, thus preserving unit
area%
\footnote{
  In light of the discussion of time, space and frequency in
  \S~\ref{fours:085}, we should clarify that we do not here mean a
  physical area measurable in square meters or the like.  We merely mean
  the dimensionless product of the width (probably measured in units of
  time like seconds) and the height (correspondingly probably measured
  in units of frequency like inverse seconds) of the rectangle a single
  pulse encloses in Fig.~\ref{fours:100:fig3}.  Though it is not a
  physical area the rectangle one sketches on paper to represent
  it, as in the figure, of course does have an area.  The word
  \emph{area} here is meant in the latter sense.
}
under the pulse.  In the limit $\eta \ra 0^{+}$, the pulse then by
definition becomes the Dirac delta of Fig.~\ref{integ:670:fig-d}, and the
pulse train by construction becomes the \emph{Dirac delta pulse train} of
Fig.~\ref{fours:100:fig4}.
\begin{figure}
  \caption{A Dirac delta pulse train.}
  \label{fours:100:fig4}
  \index{Dirac delta pulse train}
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\end{figure}
Enforcing~(\ref{fours:100:36}) on~(\ref{fours:100:35}) yields the
Dirac delta pulse train's Fourier coefficients
\bq{fours:100:37}
  a_j = \frac{1}{T_1}.
\eq

\subsection{Linearity and sufficiency}
\label{fours:100.80}
\index{Fourier series!sufficiency of}
\index{Fourier series!linearity of}
\index{linearity!of the Fourier series}

The Fourier series is evidently linear according to the rules of
\S~\ref{integ:240.05}.  That is, if the Fourier coefficients of $f_1(t)$
are~$a_{j1}$ and the Fourier coefficients of $f_2(t)$ are~$a_{j2}$, and
if the two waveforms $f_1(t)$ and $f_2(t)$ share the same fundamental
period~$T_1$, then the Fourier coefficients of $f(t) = f_1(t) + f_2(t)$
are $a_j = a_{j1} + a_{j2}$.  Likewise, the Fourier coefficients of
$\alpha f(t)$ are~$\alpha a_j$ and the Fourier coefficients of the null
waveform $f_{\mr{null}}(t) \equiv 0$ are themselves null, thus
satisfying the conditions of linearity.

All this however supposes that the Fourier series actually works.%
\footnote{
  The remainder of this dense subsection can be regarded as optional reading.
}
% diagn: The following bad break depends not only on the wording of the
%   sentence but probably also on the index of the foregoing footnote.
%   It is thus a chancy bad break, needing monitoring.
% bad break
\linebreak
Though Fig.~\ref{fours:000:fig20} is suggestive, the figure alone hardly
serves to demonstrate that every repeating waveform were representable
as a Fourier series.  To try to consider every repeating waveform at
once would be too much to try at first in any case, so let us start from a
more limited question: does there exist any continuous, repeating
waveform%
\footnote{
  As elsewhere in the book, the notation $f(t) \neq 0$ here forbids only
  the all-zero waveform.  It does not forbid waveforms like $f(t) = A
  \sin \omega t$ that happen to take a zero value at certain values
  of~$t$.
}
$f(t) \neq 0$ of period~$T_1$ whose Fourier coefficients $a_j =
0$ are identically zero?

If the waveform $f(t)$ in question is continuous then nothing prevents
us from discretizing~(\ref{fours:100:15}) as
\bqb
  a_j &=& \lim_{M \ra \infty}
    \frac{1}{T_1} \sum_{\ell=-M}^M
    e^{(-ij\,\Delta\omega)(t_o + \ell\,\Delta\tau_M)}
    f(t_o + \ell\,\Delta\tau_M) \,\Delta\tau_M,
  \\
  \Delta\tau_M &\equiv& \frac{T_1}{2M+1},
\eqb
and further discretizing the waveform itself as
% The $\,\Pi$ here is odd, but the equation does not look right without it.
\[
  f(t) = \lim_{M \ra \infty} \sum_{p=-\infty}^\infty
    f(t_o + p\,\Delta\tau_M)
    \,\Pi\left[\frac{t - (t_o + p\,\Delta\tau_M)}{\Delta\tau_M}\right],
\]
in which $\Pi[\cdot]$ is the square pulse of~(\ref{fours:095:10}).
Substituting the discretized waveform into the discretized formula
for~$a_j$, we have that
\settoheight\tlj{{\scriptsize $M$}}
\bqb
  a_j &=& \lim_{M \ra \infty}
    \frac{\Delta\tau_M}{T_1} \sum_{\ell=-M}^M
    \sum_{\rule{0em}{\tlj}p=-\infty}^\infty
    e^{(-ij\,\Delta\omega)(t_o + \ell\,\Delta\tau_M)}
    f(t_o + p\,\Delta\tau_M)
    \Pi(\ell-p)
  \\&=& \lim_{M \ra \infty}
    \frac{\Delta\tau_M}{T_1} \sum_{\ell=-M}^M
    e^{(-ij\,\Delta\omega)(t_o + \ell\,\Delta\tau_M)}
    f(t_o + \ell\,\Delta\tau_M).
\eqb
If we define the $(2M+1)$-element vectors and $(2M+1)\times(2M+1)$
matrix
\[
  \begin{split}
    [\ve f_M]_\ell &\equiv f(t_o + \ell\,\Delta\tau_M), \\
    [\ve a_M]_j &\equiv a_j, \\
    [C_M]_{j\ell} &\equiv
    \frac{\Delta\tau_M}{T_1}e^{(-ij\,\Delta\omega)(t_o + \ell\,\Delta\tau_M)}, \\
    -M \le (j,\ell) &\le M,
  \end{split}
\]
then matrix notation renders the last equation as
\[
  \lim_{M \ra \infty} \ve a_M = \lim_{M \ra \infty} C_M \ve f_M,
\]
whereby
\[
  \lim_{M \ra \infty} \ve f_M = \lim_{M \ra \infty} C_M^{-1} \ve a_M,
\]
assuming that~$C_M$ is invertible.

But is~$C_M$ invertible?  This seems a hard question to answer until we
realize that the rows of~$C_M$ consist of sampled complex exponentials
which repeat over the interval~$T_1$ and thus stand subject to
Parseval's principle~(\ref{fours:080:15}).  Realizing this, we can do
better than merely to state that~$C_M$ is invertible: we can write down
its actual inverse,
\[
  [C_M^{-1}]_{\ell j} =
    \frac{T_1}{(2M+1)\,\Delta\tau_M}e^{(+ij\,\Delta\omega)(t_o + \ell\,\Delta\tau_M)},
\]
such that%
\footnote{
  Equation~(\ref{matrix:180:IMC}) has defined the notation~$I_{-M}^M$,
  representing a $(2M+1)$-dimensional identity matrix whose string of
  ones extends along its main diagonal from $j=\ell=-M$ through
  $j=\ell=M$.
}
$C_MC_M^{-1} = I_{-M}^M$ and thus per~(\ref{mtxinv:220:20}) also that
$C_M^{-1}C_M = I_{-M}^M$.  So, the answer to our question is that,
yes,~$C_M$ is invertible.

Because~$C_M$ is invertible, \S~\ref{eigen:370} has it that neither~$\ve
f_M$ nor~$\ve a_M$ can be null unless both are.  In the limit $M \ra
\infty$, this implies that no continuous, repeating waveform $f(t)
\neq 0$ exists whose Fourier coefficients $a_j = 0$ are identically
zero.

Now consider a continuous, repeating waveform $F(t)$ and its Fourier
series $f(t)$.  Let $\Delta F(t) \equiv F(t) - f(t)$ be the part of
$F(t)$ unrepresentable as a Fourier series, continuous because both
$F(t)$ and $f(t)$ are continuous.  Being continuous and unrepresentable
as a Fourier series, $\Delta F(t)$ has null Fourier coefficients; but as
the last paragraph has concluded this can only be so if $\Delta F(t) =
0$.  Hence, $\Delta F(t) = 0$ indeed, which implies%
\footnote{
  Chapter~\ref{taylor}'s footnote~\ref{taylor:310:fn1} has argued in a
  similar style, earlier in the book.
}
that $f(t) = F(t)$.  In other words, \emph{every continuous, repeating
waveform is representable as a Fourier series.}

And what of discontinuous waveforms?  Well, the square wave of
Figs.
% bad break
\ref{fours:000:fig10} and~\ref{fours:000:fig20} this chapter has
posed as its principal example is a repeating waveform but, of course,
not a continuous one.  A truly discontinuous waveform would
admittedly invalidate the discretization above of $f(t)$, but see:
nothing prevents us from approximating the square wave's
discontinuity by an arbitrarily steep slope, whereupon this subsection's
conclusion again applies.%
\footnote{%
  Where this subsection's conclusion cannot be made to apply is where
  unreasonable waveforms like $A\sin[B/\sin\omega t]$ come into play.
  We will leave to the professional mathematician the classification of
  such unreasonable waveforms, the investigation of the waveforms'
  Fourier series and the provision of greater rigor generally.

  One can object to the subsection's reliance on discretization, yet
  discretization is a useful technique, and to the extent to which pure
  mathematics has not yet recognized and formalized it, \emph{maybe} that
  suggests---in the spirit of~\cite{Robinson}---that some interested
  professional mathematician has more work to do, whenever he gets
  around to it.  Or maybe it doesn't.  Meanwhile, a lengthy, alternate,
  more rigorous proof that does not appeal to discretization is found
  in~\cite[chapter~3]{HFDavis}.%
}

The better, more subtle, more complete answer to the question though is
that a discontinuity incurs Gibbs' phenomenon, which \S~\ref{fours:170}
will derive.

\subsection{The trigonometric form}
\label{fours:100.90}
\index{Fourier series!in trigonometric form}

It is usually best, or at least neatest and cleanest, and moreover more
evocative, to calculate Fourier coefficients and express Fourier series
in terms of complex exponentials as~(\ref{fours:100:10})
and~(\ref{fours:100:15}) do.  Occasionally, though, when the repeating
waveform $f(t)$ is real, one prefers to work in sines and cosines rather
than in complex exponentials.  One writes~(\ref{fours:100:10}) by
Euler's formula~(\ref{cexp:euler}) as
\[
  f(t) =
    a_0 + \sum_{j=1}^{\infty} \left[
      (a_j+a_{-j})\cos j \,\Delta\omega\, t
      + i(a_j-a_{-j})\sin j \,\Delta\omega\, t
    \right].
\]
Then, superimposing coefficients in~(\ref{fours:100:15}),
\bq{fours:100:91}
  \settowidth\tla{$i(a_j - a_{-j})$}
  \begin{split}
    a_0 &= \frac{1}{T_1} \int_{t_o-T_1/2}^{t_o+T_1/2}
      f(\tau) \,d\tau,
    \\
    b_j \equiv \makebox[\tla][r]{$(a_j + a_{-j})$} &=
      \frac{2}{T_1} \int_{t_o-T_1/2}^{t_o+T_1/2}
      \cos(j \,\Delta\omega\, \tau) f(\tau) \,d\tau,
    \\
    c_j \equiv \makebox[\tla][r]{$i(a_j - a_{-j})$} &=
      \frac{2}{T_1} \int_{t_o-T_1/2}^{t_o+T_1/2}
      \sin(j \,\Delta\omega\, \tau) f(\tau) \,d\tau,
  \end{split}
\eq
which give the Fourier series the trigonometric form
\bq{fours:100:90}
  \begin{split}
    f(t) &=
        a_0 + \sum_{j=1}^{\infty} \left(
          b_j\cos j \,\Delta\omega\, t
          + c_j\sin j \,\Delta\omega\, t
        \right).
  \end{split}
\eq

\index{waveform!real}
The complex conjugate of~(\ref{fours:100:15}) is
\[
  a_j^{*} = \frac{1}{T_1} \int_{t_o-T_1/2}^{t_o+T_1/2}
    e^{+ij\,\Delta\omega\,\tau} f^{*}(\tau) \,d\tau.
\]
If the waveform happens to be real then $f^{*}(t) = f(t)$, which in
light of the last equation and~(\ref{fours:100:15}) implies that
\bq{fours:100:93}
  a_{-j} = a_j^{*} \ \mbox{if} \ \Im[f(t)] = 0.
\eq
Combining~(\ref{fours:100:91}) and~(\ref{fours:100:93}), we have that
\bq{fours:100:94}
  \left.
    \setlength\arraycolsep{0.30\arraycolsep}
    \renewcommand\arraystretch{1.3}
    \br{rcr}
      b_j &=&  2\Re(a_j) \\
      c_j &=& -2\Im(a_j)
    \er
  \right\}
  \mbox{if} \ \Im[f(t)] = 0.
\eq

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

\section{The sine-argument function}
\label{fours:160}
\index{sine-argument function}

\index{sinc function}
\index{sine-argument function!Taylor series for}
\index{Taylor series!for the sine-argument function}
Equation~(\ref{fours:100:30}) gives the pulse train of
Fig.~\ref{fours:100:fig3} its Fourier coefficients, but a better
notation for~(\ref{fours:100:30}) is
\bq{fours:160:05}
  a_j = \eta A \sinarg \frac{2\pi\eta j}{2},
\eq
where
\bq{fours:160:10}
  \sinarg z \equiv \frac{\sin z}{z}
\eq
is the \emph{sine-argument function,}%
\footnote{
  Many (including the author himself in other contexts) call it the
  \emph{sinc function,} denoting it $\sinc(\cdot)$ and pronouncing it as
  ``sink.''  Unfortunately, some
  \cite[\S~4.3]{Phillips/Parr}%
  \cite[\S~2.2]{Couch}%
  \cite{Octave}
  use the $\sinc(\cdot)$ notation for another function,
  \[
    \sinc_\mr{alternate} z \equiv \sinarg\frac{2\pi z}{2}
      = \frac{\sin(2\pi z/2)}{2\pi z/2}.
  \]
  The unambiguous $\sinarg(\cdot)$ suits this particular book
  better, anyway, so this is the notation we will use.
}
plotted in Fig.~\ref{fours:160:fig}.
\begin{figure}
  \caption{The sine-argument function.}
  \label{fours:160:fig}
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\end{figure}
The function's Taylor series is
\bq{fours:160:20}
  \sinarg z = \sum_{j=0}^{\infty} \left[ \prod_{m=1}^j \frac{-z^2}{(2m)(2m+1)} \right],
\eq
the Taylor series of $\sin z$ from Table~\ref{taylor:315:tbl}, divided
by~$z$.

This section introduces the sine-argument function and some of its
properties, plus also the related sine integral.%
\footnote{
  Readers interested in Gibbs' phenomenon, \S~\ref{fours:170}, will
  read the present section because Gibbs depends on its results.
  Among other readers however some, less interested in special functions
  than in basic Fourier theory, may find this section unprofitably
  tedious.  They can skip ahead to the start of the next chapter without
  great loss.
}

\subsection{Derivative and integral}
\label{fours:160.20}
\index{sine-argument function!derivative of}
\index{sine-argument function!integral of}
\index{derivative!of the sine-argument function}
\index{integral!of the sine-argument function}
\index{sine integral}

\index{sine integral!Taylor series for}
\index{Taylor series!for the sine integral}
The sine-argument function's derivative is computed from the
definition
% bad break
(\ref{fours:160:10}) and the derivative product
rule~(\ref{drvtv:proddiv}) to be
\bq{fours:160:30}
  \frac{d}{dz} \sinarg z = \frac{\cos z - \sinarg z}{z}.
\eq
The function's integral is expressed as a Taylor series after
integrating the function's own Taylor series~(\ref{fours:160:20}) term
by term to obtain the form
\bq{fours:160:35}
  \sinint z \equiv \int_0^z \sinarg \tau \,d\tau =
    \sum_{j=0}^{\infty} \left[ \frac{z}{2j+1}
      \prod_{m=1}^j \frac{-z^2}{(2m)(2m+1)} \right],
\eq
plotted in Fig.~\ref{fours:160:figi}.  Convention gives this integrated
function its own name and notation: it calls it the \emph{sine
integral}%
\footnote{\cite[\S~3.3]{Lebedev}}%
$\mbox{}^{,}$%
\footnote{
  The name ``sine-argument'' incidentally seems to have been
  back-constructed from the name ``sine integral.''
}
and denotes it by $\sinint(\cdot)$.
\begin{figure}
  \caption{The sine integral.}
  \label{fours:160:figi}
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  \ec
\end{figure}

\subsection{Properties of the sine-argument function}
\label{fours:160.30}
\index{sine-argument function!properties of}

Sine-argument properties include the following.
\bi
  \item
    The sine-argument function is real over the real domain.  That is,
    if $\Im(t) = 0$ then $\Im(\sinarg t) = 0$.
  \item
    The zeros of $\sinarg z$ occur at $z = n\pi$, $n \neq 0$, $n \in
    \mathbb Z$.
  \item
    \index{extremum!of the sine-argument function}
    \index{extremum!global}
    It is that $\left|\sinarg t\right| < 1$ over the real domain $\Im(t) = 0$
    except at the global maximum $t = 0$, where
    \bq{fours:160:38}
      \sinarg 0 = 1.
    \eq
  \item
    \index{alternating signs}
    \index{sign!alternating}
    Over the real domain $\Im(t) = 0$, the function $\sinarg t$ alternates
    between distinct, positive and negative lobes.  Specifically,
    $(-)^n\sinarg(\pm t) > 0$ over $n\pi < t < (n+1)\pi$ for each
    $n \ge 0$, $n \in \mathbb Z$.
  \item
    Each of the sine-argument's lobes has but a single peak.  That is,
    over the real domain $\Im(t) = 0$, the derivative $(d/dt)\sinarg t =
    0$ is zero at only a single value of~$t$ on each lobe.
  \item
    The sine-argument function and its derivative converge toward
    \bq{fours:160:40}
      \begin{split}
        \lim_{t \ra \pm \infty} \sinarg t &= 0, \\
        \lim_{t \ra \pm \infty} \frac{d}{dt} \sinarg t &= 0.
      \end{split}
    \eq
\ei
Some of these properties are obvious in light of the sine-argument function's
definition~(\ref{fours:160:10}).  Among the less obvious properties,
that $\left|\sinarg t\right| < 1$ says merely that $\left|\sin t\right| <
\left|t\right|$ for nonzero~$t$; which must be true since~$t$,
interpreted as an angle---which is to say, as a curved distance about a
unit circle---can
hardly be shorter than $\sin t$, interpreted as the corresponding
direct shortcut to the axis (see Fig.~\ref{trig:226:f1}).  For $t=0$,
(\ref{taylor:315:60}) obtains---or, if you prefer,~(\ref{fours:160:20}).

\index{sketch, proof by}
\index{proof!by sketch}
That each of the sine-argument function's lobes should have but a single peak
seems right in view of Fig.~\ref{fours:160:fig} but is nontrivial to
prove.  To assert that each lobe has but a single peak is to assert that
$(d/dt)\sinarg t = 0$ exactly once in each lobe; or, equivalently---after
setting (\ref{fours:160:30})'s left side to zero, multiplying
by $z^2/\cos z$ and changing $t \la z$---it is to assert that
\[
  \tan t = t
\]
exactly once in each interval
\[
  n\pi \le t < (n+1)\pi, \ n \ge 0,
\]
for $t \ge 0$; and similarly for $t \le 0$.  But according to
Table~\ref{cexp:drv}
\[
  \frac{d}{dt} \tan t = \frac{1}{\cos^2 t} \ge 1,
\]
whereas $dt/dt = 1$, implying that $\tan t$ is everywhere at least as
steep as~$t$ is---and, well, the most concise way to finish the argument
is to draw a picture of it, as in Fig.~\ref{fours:160:fig3}, where the
curves evidently cannot but intersect exactly once in each interval.
\begin{figure}
  \caption{The points at which~$t$ intersects $\tan t$.}
  \label{fours:160:fig3}
  \index{tangent!compared against its argument}
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  \ec
\end{figure}

\subsection{Properties of the sine integral}
\label{fours:160.33}
\index{sine integral!properties of}

Properties of the sine integral $\sinint t$ of~(\ref{fours:160:35})
include the following.
\bi
  \item
    Over the real domain $\Im(t) = 0$, the sine integral $\sinint t$ is
    positive for positive~$t$, negative for negative~$t$ and, of course,
    zero for $t=0$.
  \item
    \index{extremum!local}
    The local extrema of $\sinint t$ over the real domain $\Im(t) = 0$
    occur at the zeros of $\sinarg t$.
  \item
    \index{extremum!of the sine integral}
    The global maximum and minimum of $\sinint t$ over the real domain
    $\Im(t) = 0$ occur respectively at the first positive and negative
    zeros of $\sinarg t$, which are $t = \pm\pi$.
  \item
    The sine integral converges toward
    \bq{fours:160:41}
      \lim_{t \ra \pm \infty} \sinint t = \pm\frac{2\pi}{4}.
    \eq
\ei
That the sine integral should reach its local extrema at the
sine-argument's zeros ought to be obvious to the extent to which the
concept of integration is understood.  To explain the other properties
it helps first to have expressed the sine integral in the form
\bqb
  \sinint t
    &=& S_n + \int_{n\pi}^t \sinarg\tau \,d\tau, \\
  S_n &\equiv& \sum_{j=0}^{n-1} U_j, \\
  U_j &\equiv& \int_{j\pi}^{(j+1)\pi} \sinarg\tau \,d\tau, \\
  n\pi &\le& t \makebox[\arraycolsep]{}
    < \makebox[\arraycolsep]{} (n+1)\pi, \\
  0 &\le& n, \ \ (j,n) \in \mathbb Z,
\eqb
where each partial integral~$U_j$ integrates over a single lobe of the
sine-argument.  The several~$U_j$ alternate in sign but, because each
lobe majorizes the next (\S~\ref{taylor:316.25})---that is, because,%
\footnote{
  More rigorously, to give the reason perfectly unambiguously, one could
  fuss here for a third of a page or so over signs, edges and the like.
  To do so is left as an exercise to those who aspire to the pure
  mathematics profession.
}
%\footnote{
%  To be extra precise: it is so because
%  $\left| \sinarg \tau \right| > \left| \sinarg \tau+\pi \right|$
%  for all $\tau \ge 0$ except $\tau = n\pi$, $n > 0$, for which
%  $\left| \sinarg \tau \right| = \left| \sinarg \tau+\pi \right| = 0$.
%  To be extra, extra precise, one could take special note of the case
%  that $n=0$, but this is all rather too much pedantry for a book on
%  applied mathematics; and the narrative, as written, adequately
%  explains the concept without the help of footnotes like this one.
%}
in the integrand,
$\left| \sinarg \tau \right| \ge \left| \sinarg \tau+\pi \right|$
for all $\tau \ge 0$---the magnitude of the area under each lobe exceeds
that under the next, such that%
{%
  \setlength\arraycolsep{0.40\arraycolsep}%
  \bqb
    0 &\le& (-)^j\int_{j\pi}^t \sinarg \tau \,d\tau
      < (-)^jU_j < (-)^{j-1}U_{j-1}, \\
    j\pi &\le& t \makebox[\arraycolsep]{}
      < \makebox[\arraycolsep]{} (j+1)\pi, \\
    0 &\le& j, \ j \in \mathbb Z
  \eqb%
}%
(except that the $U_{j-1}$ term of the inequality does not apply when
$j=0$, since there is no~$U_{-1}$) and thus that
\bqb
  &
    0 = S_0 < S_{2m} < S_{2m+2} < S_{\infty} < S_{2m+3} < S_{2m+1} < S_1
  &\\&
    \mbox{for all $m>0$, $m \in \mathbb Z$;}
  &
\eqb
or in other words that
\[
  0 < S_0 < S_2 < S_4 < S_6 < S_8 < \cdots < S_\infty < \cdots < S_9 < S_7 < S_5 < S_3 < S_1.
\]
The foregoing applies only when $t \ge 0$ but naturally one can reason
similarly for $t \le 0$, concluding that the integral's global maximum
and minimum over the real domain occur respectively at the sine-argument
function's first positive and negative zeros, $t = \pm \pi$; and
further concluding that the integral is positive for all positive~$t$
and negative for all negative~$t$.

Equation~(\ref{fours:160:41}) wants some cleverness to calculate and
will be the subject of the next subsection.

\subsection{The sine integral's limit by complex contour}
\label{fours:160.35}
\index{sine integral!evaluation of by complex contour}

% This paragraph seems worth saving here.
%
%\index{Wilbraham, Henry (1825--1883)}
%The first two lines of~(\ref{fours:160:40}) are plain enough but its
%last line is nonobvious.  The sine-argument function is arguably the book's first instance
%of a \emph{special function,} though because~(\ref{fours:160:10})
%expresses it in closed form it has not traditionally been classified as
%such.  We will have much to say about special functions in
%[chapters not yet written],
%but for the moment we will only observe that one often wants unusually
%clever techniques to discover constants like the~%
%$2\pi/4$ of~(\ref{fours:160:40}).  Fortunately, some unusually clever
%mathematicians have toiled, in the words of G.~S.\ Brown, ``a very long
%time, in very dark rooms,''%
%\footnote{\cite{Brown-lecture}}
%over the past few centuries, precisely to solve problems like this one.
%Other such problems have found their solutions quite by accident when a
%mathematician (often an applied one), working on something else
%entirely, had the considerable wit to realize that the answer to some
%apparently unrelated but nonetheless interesting question had surfaced
%unexpectedly in his algebra.  The specific problem
%of~(\ref{fours:160:40}) found its solution in~1848 and~1899 in the
%latter manner at the hands respectively of Henry Wilbraham and
%J.~Willard Gibbs.  Section~\ref{fours:170} will explain.

Equation~(\ref{fours:160:41}) has proposed that the sine integral
converges toward a value of $2\pi/4$, but why?  The integral's Taylor
series~(\ref{fours:160:35}) is impractical to compute for large~$t$ and is
useless for $t \ra \infty$, so it cannot answer the question.  To
evaluate the integral in the infinite limit, we shall have to think of
something cleverer.

Noticing per~(\ref{cexp:250:sin}) that
\[
  \sinarg z = \frac{e^{+iz} - e^{-iz}}{i2z},
\]
rather than trying to integrate the sine-argument function all at once
let us first try to integrate just one of its two complex terms, leaving
the other term aside to handle later, for the moment computing only
\[
  I_1 \equiv \int_0^\infty \frac{e^{iz} \,dz}{i2z}.
\]
To compute the integral~$I_1$, we will apply the closed-contour
technique of \S~\ref{inttx:250}, choosing a contour in the Argand plane
that incorporates~$I_1$ but shuts out the integrand's pole at $z=0$.

\index{false try}
Many contours are possible and one is unlikely to find an amenable contour
on the first attempt, but perhaps after several false tries we discover
and choose the contour of Fig.~\ref{fours:160:fig-contour}.
\begin{figure}
  \caption{A complex contour about which to integrate $e^{iz}/i2z$.}
  \label{fours:160:fig-contour}
  \index{contour!complex}
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\end{figure}
The integral about the inner semicircle of this contour is
\[
  I_6 = \int_{C_6} \frac{e^{iz}\,dz}{i2z}
  = \lim_{\rho \ra 0^{+}} \int_{2\pi/2}^0
    \frac{e^{iz}(i\rho e^{i\phi} \, d\phi)}{i2(\rho e^{i\phi})}
  = \int_{2\pi/2}^0 \frac{e^{i0} \, d\phi}{2}
  = -\frac{2\pi}{4}.
\]
The integral across the contour's top segment is
\[
  I_3 = \int_{C_3} \frac{e^{iz} \,dz}{i2z}
  = \lim_{a \ra \infty} \int_{a}^{-a}
    \frac{e^{i(x+ia)} \,dx}{i2z}
  = \lim_{a \ra \infty} \int_{-a}^{a}
    \frac{-e^{ix}e^{-a} \,dx}{i2z},
\]
from which, according to the continuous triangle
inequality~(\ref{inttx:250:triangle}),
\[
  \left| I_3 \right|
  \le \lim_{a \ra \infty} \int_{-a}^{a}
    \left| \frac{-e^{ix}e^{-a} \,dx}{i2z} \right|
  = \lim_{a \ra \infty} \int_{-a}^{a}
    \frac{e^{-a} \,dx}{2 \left| z \right|};
\]
which, since $0 < a \le \left| z \right|$ across the segment, we can
weaken to read
\[
  \left| I_3 \right|
  \le \lim_{a \ra \infty} \int_{-a}^{a}
    \frac{e^{-a} \,dx}{2a}
  = \lim_{a \ra \infty} e^{-a} = 0,
\]
only possible if
\[
  I_3 = 0.
\]
The integral up the contour's right segment is
\[
  I_2 = \int_{C_2} \frac{e^{iz} \,dz}{i2z}
  = \lim_{a \ra \infty} \int_{0}^{a}
    \frac{e^{i(a+iy)} \,dy}{2z}
  = \lim_{a \ra \infty} \int_{0}^{a}
    \frac{e^{ia}e^{-y} \,dy}{2z},
\]
from which, according to the continuous triangle inequality,
\[
  \left| I_2 \right| \le
  \lim_{a \ra \infty} \int_{0}^{a}
    \left| \frac{e^{ia}e^{-y} \,dy}{2z} \right|
  = \lim_{a \ra \infty} \int_{0}^{a}
    \frac{e^{-y} \,dy}{2\left| z \right|};
\]
which, since $0 < a \le \left| z \right|$ across the segment, we can
weaken to read
\[
  \left| I_2 \right|
  \le \lim_{a \ra \infty} \int_{0}^{a}
    \frac{e^{-y} \,dy}{2a}
  = \lim_{a \ra \infty} \frac{1}{2a} = 0,
\]
only possible if
\[
  I_2 = 0.
\]
The integral down the contour's left segment is
\[
  I_4 = 0
\]
for like reason.
Because the contour encloses no pole,
\[
  \oint \frac{e^{iz} \,dz}{i2z}
  = I_1 + I_2 + I_3 + I_4 + I_5 + I_6
  = 0,
\]
which in light of the foregoing calculations implies that
\[
  I_1 + I_5 = \frac{2\pi}{4}.
\]
Now,
\[
  I_1
  = \int_{C_1} \frac{e^{iz} \,dz}{i2z}
  = \int_0^\infty \frac{e^{ix} \,dx}{i2x}
\]
is the integral we wanted to compute in the first place, but what is
that~$I_5$?  Answer:
\[
  I_5
  = \int_{C_5} \frac{e^{iz \,dz}}{i2z}
  = \int_{-\infty}^0 \frac{e^{ix} \,dx}{i2x};
\]
or, changing $-x \la x$,
\[
  I_5 = \int_0^{\infty} \frac{-e^{-ix} \,dx}{i2x},
\]
which fortuitously happens to integrate the heretofore neglected term
of the sine-argument function we started with.  Thus,
\[
  \lim_{t \ra \infty} \sinint t
    = \int_0^\infty \sinarg x \,dx
    = \int_0^\infty \frac{e^{+ix} - e^{-ix}}{i2x} \,dx
    = I_1 + I_5 = \frac{2\pi}{4},
\]
which was to be computed.%
\footnote{
  Integration by closed contour is a subtle technique, is it not?  What
  a finesse this subsection's calculation has been!  The author rather
  strongly sympathizes with the reader who still somehow cannot quite
  believe that contour integration actually works, but in the case of
  the sine integral another, quite independent method to evaluate the
  integral is known and it finds the same number $2\pi/4$.  The
  interested reader can extract this other method from Gibbs' calculation
  in \S~\ref{fours:170}, which refers a sine integral to the known
  amplitude of a square wave.

  We said that it was fortuitous that~$I_5$, which we did not know how
  to eliminate, turned out to be something we needed anyway; but is it
  really merely fortuitous, once one has grasped the technique?  An
  integration of $-e^{-iz}/i2z$ is precisely the sort of thing an
  experienced applied mathematician would expect to fall out as a
  byproduct of the contour integration of $e^{iz}/i2z$.  The trick is
  to discover the contour from which it actually does fall out, the
  discovery being a process of informed trial and error.
}

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

\section{Gibbs' phenomenon}
\label{fours:170}
\index{Wilbraham, Henry (1825--1883)}
\index{Gibbs, Josiah Willard (1839--1903)}
\index{Gibbs phenomenon}

\index{discontinuous waveform}
\index{waveform!discontinuous}
Section~\ref{fours:100.80} has shown how the Fourier series suffices to
represent a continuous, repeating waveform.  Paradoxically, the
chapter's examples have been of discontinuous waveforms like the square
wave.  At least in Fig.~\ref{fours:000:fig20} the Fourier series seems
to work for such discontinuous waveforms, though we have never exactly
demonstrated that it should work for them, or how.  So, what does all
this imply?

\index{overshot}
\index{oscillation}
In one sense, it does not imply much of anything.  One can
represent a discontinuity by a relatively sharp continuity---as for
instance one can represent the Dirac delta of Fig.~\ref{integ:670:fig-d}
by the triangular pulse of Fig.~\ref{fours:095:fig1}, with its sloped
edges, if~$T$ in~(\ref{fours:095:30}) is sufficiently small---and,
considered in this light, the Fourier series works.
Mathematically however one is more likely to approximate a Fourier
series by truncating it after some finite number~$N$ of terms; and,
indeed, so-called%
\footnote{
  So called because they pass low frequencies while suppressing high
  ones, though systems encountered in practice admittedly typically
  suffer a middle frequency domain through which frequencies are only
  partly suppressed.
}
``low-pass'' physical systems that naturally suppress
high frequencies%
\footnote{\cite[\S~15.2]{JJH}}
are common, in which case to truncate the series is
more or less the right thing to do.  Yet, a significant thing happens
when one truncates the Fourier series.  \emph{At a discontinuity, the
Fourier series oscillates and overshoots.}%
\footnote{\cite{Khamsi}}

\index{ringing}
Henry Wilbraham investigated this phenomenon as early as~1848\@.
\linebreak % bad break
J.~Willard Gibbs
explored its engineering implications in~1899.%
\footnote{%
  \cite{Wilbraham:1848}%
  \cite{Gibbs:1899}%
}
Let us along with them refer to the square wave
of Fig.~\ref{fours:000:fig20} on page~\pageref{fours:000:fig20}.  As
further Fourier components are added the Fourier waveform better
approximates the square wave, but, as we said, it oscillates about and
overshoots---it ``rings about'' in the electrical engineer's
vernacular---the square wave's discontinuities (the verb ``to ring''
here recalls the ringing of a bell or steel beam).  This oscillation
and overshot turn out to be irreducible, and moreover they can have
significant physical effects.

\index{integration!as summation}
\index{summation!as integration}
Changing $t - T_1/4 \la t$ in~(\ref{fours:000:20}) to delay the square
wave by a quarter cycle yields that
\[
  f(t) = \frac{8A}{2\pi}
    \sum_{j=0}^{\infty} \frac{1}{2j+1}
    \sin \left[\frac{(2j+1)(2\pi)t}{T_1}\right],
\]
which we can, if we like, write as
\[
  f(t) = \lim_{N \ra \infty} \frac{8A}{2\pi}
    \sum_{j=0}^{N-1} \frac{1}{2j+1}
    \sin \left[\frac{(2j+1)(2\pi)t}{T_1}\right].
\]
Again changing
\[
  \Delta v \la \frac{2(2\pi)t}{T_1}
\]
makes this
\[
  f\left[\frac{T_1}{2(2\pi)} \,\Delta v\right]
    = \lim_{N\ra\infty} \frac{4A}{2\pi}
    \sum_{j=0}^{N-1}
    \sinarg \left[\left(j+\frac{1}{2}\right)\,\Delta v\right]
    \,\Delta v.
\]
Stipulating that~$\Delta v$ be infinitesimal,
\[
  0 < \Delta v \ll 1,
\]
(which in light of the definition of~$\Delta v$ is to stipulate that $0
< t \ll T_1$) such that $dv \equiv \Delta v$ and, therefore, that the
summation become an integration; and further defining
\[
  u \equiv N \,\Delta v;
\]
we have that
\bq{fours:170:20}
   \lim_{N\ra\infty} f\left[\frac{T_1}{2(2\pi)N} u\right]
    = \frac{4A}{2\pi} \int_0^u \sinarg v \,dv = \frac{4A}{2\pi} \sinint u.
\eq
Equation~(\ref{fours:160:41}) gives us that $\lim_{u \ra \infty} \sinint u =
2\pi/4$, so~(\ref{fours:170:20}) as it should has it that $f(t) \approx
A$ when%
\footnote{
  Here is an exotic symbol:~$\gnapprox$.  It means what it appears to
  mean, that $t > 0$ and $t \not\approx 0$.
}
$t \gnapprox 0$.  When $t \approx 0$ however it gives the waveform
locally the sine integral's shape of Fig.~\ref{fours:160:figi}.

\index{engineer}
\index{mechanical engineer}
\index{electrical engineer}
Though unexpected the effect can and does actually arise in physical
systems.  When it does, the maximum value of $f(t)$ is of interest to
mechanical and electrical engineers among others because, if an element
in an engineered system will overshoot its designed position, the
engineer wants to allow safely for the overshot.  According to
\S~\ref{fours:160.33}, the sine integral $\sinint u$ reaches its maximum
at
\[
  u = \frac{2\pi}{2},
\]
where according to~(\ref{fours:160:35})
% An Octave script to calculate f_max:
% f=0;
% term=4/(2*pi);
% jmax=12;
% for j = 0:jmax
%   if (j>0)
%     term *= -(2*pi/2)^2/((2*j)*(2*j+1));
%   end
%   f += ((2*pi/2)/(2*j+1)) * term
% end
\[
  f_\mr{max} = \frac{4A}{2\pi} \sinint \frac{2\pi}{2} =
    \frac{4A}{2\pi}
    \sum_{j=0}^{\infty} \left[ \frac{2\pi/2}{2j+1}
      \prod_{m=1}^j \frac{-(2\pi/2)^2}{(2m)(2m+1)} \right]
  \approx (\mbox{0x1.2DD2})A.
\]
This overshot, peaking momentarily at $(\mbox{0x1.2DD2})A$, and the
associated sine-integral ringing constitute \emph{Gibbs' phenomenon,}
as Fig.~\ref{fours:170:fig10} depicts.
\begin{figure}
  \caption{Gibbs' phenomenon.}
  \label{fours:170:fig10}
  \bc
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    \nc\xxyab{1.2}
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    \nc\xxm{0.25}
    \nc\xxo{-0.10}
    \nc\xxgibbs{1.17898}
    \nc\fxa{-5.0} \nc\fxb{5.0}
    \nc\fya{-1.6} \nc\fyb{2.1}
    \begin{pspicture}(\fxa,\fya)(\fxb,\fyb)
      %\psframe[linewidth=0.5pt,dimen=outer](\fxa,\fya)(\fxb,\fyb)
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      {
        \small
        \psset{dimen=middle}
        {
          \psset{linewidth=0.5pt}
          \psline(-\xxxab,0)(\xxxab,0)
          \psline(0,-\xxyab)(0,\xxyab)
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          \uput[u](0,\xxyab){$f(t)$}
          \rput(\xxo\tla,\tlb){\psline(-\xxl,0)(\xxl,0)\psline{->}(0,\xxm)(0,0)}
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        {
          \psset{linewidth=2.0pt}
          \psline(-1.35\tla,-\tlb)(-1.00\tla,-\tlb)
          \psline(-1.00\tla, \tlb)(-0.50\tla, \tlb)
          \psline(-0.50\tla,-\tlb)( 0.00\tla,-\tlb)
          \psline( 0.00\tla, \tlb)( 0.50\tla, \tlb)
          \psline( 0.50\tla,-\tlb)( 1.00\tla,-\tlb)
          \psline( 1.00\tla, \tlb)( 1.35\tla, \tlb)
          \nc\xxgibbsline[1]{\psline{c-c}(#1\tla,-\xxgibbs\tlb)(#1\tla,\xxgibbs\tlb)}
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          \xxgibbsline{-0.50}
          \xxgibbsline{ 0.00}
          \xxgibbsline{ 0.50}
          \xxgibbsline{ 1.00}
        }
      }
    \end{pspicture}
  \ec
\end{figure}

We have said that Gibbs' phenomenon is irreducible, and indeed strictly
this is so: a true discontinuity, if it is to obey Fourier, must
overshoot according to Gibbs.  Admittedly as earlier alluded, one can
sometimes substantially evade Gibbs by softening a discontinuity's edge,
giving it a steep but not vertical slope and maybe rounding its corners
a little;%
\footnote{
  If the applied mathematician is especially exacting he might represent
  a discontinuity by the cumulative normal distribution
  function~(\ref{prob:100:40})
  or maybe (if slightly less exacting) by an arctangent,
  and indeed there are times at which he might do so.  However,
  such extra-fine mathematical craftsmanship is unnecessary to this
  section's purpose.
}
or, alternately, by rolling the Fourier series off gradually
rather than truncating it exactly at~$N$ terms.  Engineers may do one or
the other, or both, explicitly or implicitly, which is why the full
Gibbs is not always observed in engineered systems.  Nature
may do likewise.  Neither however is the point.  The point is that sharp
discontinuities do not behave in the manner one might na\"ively have
expected, yet that one can still analyze them profitably, adapting this
section's subtle technique as the circumstance might demand.  A good
engineer or other applied mathematician will make himself aware of
Gibbs' phenomenon and of the mathematics behind it for this reason.

% ----------------------------------------------------------------------
% This section didn't quite work.  Maybe some day.
%
%\section{Waveforms of limited domain}
%\label{fours:180}
%
%A waveform of limited domain can be modeled as a repeating waveform.  If
%it can be modeled as a repeating waveform, then its Fourier series can
%normally be expressed.
%
%\begin{figure}
%  \caption{Here.}
%  \label{fours:180fig1}
%  \bc
%    \nc\fxa{-5.0} \nc\fxb{5.0}
%    \nc\fya{-3.0} \nc\fyb{3.0}
%    \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,linewidth=2.0pt}%
%        \nc\ttb{4.65}
%        \nc\ttc{1.5}
%        \nc\ttd{1.63636363636364}
%        {%
%          \psset{linewidth=0.5pt}%
%          \psline(-\ttb,0)(\ttb,0)%
%          \uput[r](\ttb,0){$t$}%
%        }%
%        {%
%          \psline(-\ttd,-\ttc)(-\ttd,\ttc)%
%          \psline( \ttd,-\ttc)( \ttd,\ttc)%
%        }%
%        \nc\tta{4.5}
%        \psplot[plotpoints=500]{-\ttd}{\ttd}{%
%          /twopi    6.28318530717959 def
%          /domega 110.0  def
%          /ampl     0.40 def
%          %x domega mul cos 1.0 mul ampl mul -0.87833 mul
%          x domega mul sin 1.0 mul ampl mul -0.42645 mul
%          %x domega mul 2.0 mul cos 1.0 mul ampl mul  1.36897 mul 2.0 div
%          x domega mul 2.0 mul sin 1.0 mul ampl mul -0.00542 mul 2.0 div
%          %x domega mul 3.0 mul cos 1.0 mul ampl mul -1.00534 mul 3.0 div
%          x domega mul 3.0 mul sin 1.0 mul ampl mul -0.58394 mul 3.0 div
%          %x domega mul 4.0 mul cos 1.0 mul ampl mul -0.29542 mul 4.0 div
%          x domega mul 4.0 mul sin 1.0 mul ampl mul  0.55244 mul 4.0 div
%          %x domega mul 5.0 mul cos 1.0 mul ampl mul -0.92348 mul 5.0 div
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%          add add add add %add add add add add
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%    \end{pspicture}
%  \ec
%\end{figure}

