%%%%		-*- Mode: LaTeX -*-
%%%% Author:    J. Robert Buchanan
%%%% Address:   Department of Mathematics
%%%%		Millersville University
%%%%            P.O. Box 1002
%%%%            Millersville, PA 17551-0302
%%%% Phone:     717-871-7305
%%%% FAX:	717-871-7948
%%%% Email:     Robert.Buchanan@millersville.edu
%%%%
%%%% Date:      06/30/2022
%%%%
%%%% Purpose:   Convergence of Fourier Series
%%%%

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\title{Convergence of Fourier Series}
\subtitle{MATH 467 \textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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\institute[Millersville University of Pennsylvania]
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{
  Department of Mathematics
}
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\subject{Talks}
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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will explore the questions:
\begin{itemize}
\item What are the conditions that guarantee the Fourier series of a given
function $f(x)$ converges?
%
\item If the Fourier series of a given function $f(x)$ converges, does it
converge to the value of $f(x)$ at a given $x$?
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Short Answer to First Question}

A Fourier series will converge for a large class of functions, though
we will prove convergence only for the class of \textbf{piecewise
  smooth} functions.

\vspace*{0.10in}

\uncover<2->{%
\begin{definition}
A function $f(x)$ is
\textbf{piecewise continuous}
on $[a,b]$ (or $(a,b)$)
if there are finitely many points
$a=x_{0} < x_{1} < x_{2} < x_{3} < \cdots < x_{n} = b$,
such that
\begin{itemize}
\item $f(x)$ is continuous on $(x_{i-1},x_{i})$
for all $i = 1, 2, \ldots, n$,
%
\item the one-sided  limits $\displaystyle \lim_{x \to x_{i}^{-}}
  f(x)$ and $\displaystyle \lim_{x \to x_{i}^{+}} f(x)$
exist for all $i = 1, 2, \ldots, n-1$, and
%
\item the one-sided limits $\displaystyle \lim_{x \to a^{+}} f(x)$ and
  $\displaystyle \lim_{x \to b^{-}} f(x)$ both exist.
\end{itemize}
\end{definition}
}

\end{frame}
%%

\begin{frame}\frametitle{Remarks}

\begin{itemize}
\item The limits mentioned in the definition of piecewise continuous
  must be finite, real numbers.
\item Function $f(x)$ is piecewise continuous on $(-\infty,\infty)$
if it is piecewise continuous on every finite interval
$[a,b]$.
\item Function $f(x)$ is \textbf{piecewise smooth}
on $[a,b]$ if $f'(x)$ is piecewise continuous on $[a,b]$.
\item Since any piecewise continuous function is integrable, the
  Fourier coefficients of any piecewise continuous function are
  well-defined (but this is not enough to show the Fourier series converges).
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Notation}

For any $c \in [a,b]$, define
\[
\lim_{x \to c^{+}} f(x) = f(c+)
\quad\mbox{and}\quad
\lim_{x \to c^{-}} f(x) = f(c-)	.
\]
Function $f(x)$ is continuous at $c$ if and only if $f(c+) = f(c-) = f(c)$.

\end{frame}
%%

\begin{frame}\frametitle{Examples}

Determine which of the following functions is piecewise smooth,
piecewise continuous, or neither on $[-L,L]$ for some $L > 0$.
\begin{align*}
f_{1}(x) &= \left\{ \begin{array}{rl}
L	& \mbox{if $-L \le x < 0$}	\\
x	& \mbox{if $0 \le x \le L$}
\end{array} \right.	\\
f_{2}(x) &= \left\{ \begin{array}{rl}
x^{2}	& \mbox{if $-L \le x < 0$}	\\
x-1	& \mbox{if $0 \le x \le L$}
\end{array} \right.	\\
f_{3}(x) &= \cos \frac{1}{x}	\\
f_{4}(x) &= x \cos \frac{1}{x}	\\
f_{5}(x) &= x^{3/5}	\\
f_{6}(x) &= x^{6/5}	\\
f_{7}(x) &= \frac{1}{x}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Discussion (1 of 7)}

\[
f_{1}(x) = \left\{ \begin{array}{rl}
L	& \mbox{if $-L \le x < 0$}	\\
x	& \mbox{if $0 \le x \le L$}
\end{array} \right.
\]
Function $f_{1}(x)$ is piecewise continuous and piecewise smooth on
$[-L,L]$ with
\begin{align*}
f_{1}(-L+) &= f_{1}(0-) = f_{1}(L-) = L	\\
f_{1}(0+) &= 0	\\
f_{1}'(x) &= \left\{ \begin{array}{rl}
0	& \mbox{if $-L < x < 0$}	\\
1	& \mbox{if $0 < x < L$}
\end{array} \right.	\\
f_{1}'(-L+) &= f_{1}'(0-) = 0	\\
f_{1}'(0+) &= f_{1}'(L-) = 1	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Discussion (2 of 7)}

\[
f_{2}(x) = \left\{ \begin{array}{rl}
x^{2}	& \mbox{if $-L \le x < 0$}	\\
x-1	& \mbox{if $0 \le x \le L$}
\end{array} \right.
\]
Function $f_{2}(x)$ is piecewise continuous and piecewise smooth on
$[-L,L]$ with
\begin{align*}
f_{2}(-L+) &= L^{2}	\\
f_{2}(0-) &= 0	\\
f_{2}(0+) &= -1	\\
f_{2}(L-) &= L-1	\\
f_{2}'(x) &= \left\{ \begin{array}{rl}
2 x	& \mbox{if $-L < x < 0$}	\\
1	& \mbox{if $0 < x < L$}
\end{array} \right.	\\
f_{2}'(-L+) &= -2 L	\\
f_{2}'(0-) &= 0	\\
f_{2}'(0+) &= f_{2}'(L-) = 1	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Discussion (3 of 7)}

\[
f_{3}(x) = \cos \frac{1}{x}
\]
Function $f_{3}(x)$ is neither piecewise continuous nor piecewise smooth on
$[-L,L]$ since $f_{3}(0-)$ does not exist.

\vspace*{0.25in}

\[
f_{3}'(x) = \frac{1}{x^{2}} \sin \frac{1}{x}
\]
Note that $f_{3}'(0-)$ does not exist.

\end{frame}
%%

\begin{frame}\frametitle{Discussion (4 of 7)}

\[
f_{4}(x) = x \, \cos \frac{1}{x}
\]
Function $f_{4}(x)$ is piecewise continuous but not piecewise smooth on
$[-L,L]$.
\begin{align*}
f_{4}(-L+) &= -L \, \cos \frac{1}{L}	\\
f_{4}(0-) &= f_{4}(0+) = 0	\\
f_{4}(L-) &= L \, \cos \frac{1}{L}
\end{align*}

\vspace*{0.25in}

\[
f_{4}'(x) = \cos \frac{1}{x} - \frac{1}{x} \sin \frac{1}{x}
\]
Note that $f_{4}'(0-)$ does not exist.

\end{frame}
%%

\begin{frame}\frametitle{Discussion (5 of 7)}

\[
f_{5}(x) = x^{3/5}
\]
Function $f_{5}(x)$ is continuous but not piecewise smooth on
$[-L,L]$.

\vspace*{0.25in}

\[
f_{5}'(x) = \frac{3}{5 x^{2/5}}
\]
Note that $f_{5}'(0-)$ does not exist.

\end{frame}
%%

\begin{frame}\frametitle{Discussion (6 of 7)}

\[
f_{6}(x) = x^{6/5}
\]
Function $f_{6}(x)$ is continuous and piecewise smooth on
$[-L,L]$.

\vspace*{0.25in}

\[
f_{6}'(x) = \frac{6}{5} x^{1/5}
\]

\end{frame}
%%

\begin{frame}\frametitle{Discussion (7 of 7)}

\[
f_{7}(x) = \frac{1}{x}
\]
Function $f_{7}(x)$ is neither piecewise continuous nor piecewise smooth on
$[-L,L]$.
Note that $f_{7}(0-)$ does not exist.

\vspace*{0.25in}

\[
f_{7}'(x) = -\frac{1}{x^{2}}
\]
Note that $f_{7}'(0-)$ does not exist.

\end{frame}
%%

\begin{frame}\frametitle{Dirichlet Convergence Theorem}

\begin{theorem}
Assume that $f(x)$ is a
piecewise smooth function on the interval $[-L,L]$ extended to
$(-\infty,\infty)$ periodically with period $2L$.
Then the  Fourier series of $f(x)$ converges for all $x$ to the value
\[
\frac{1}{2} \left( f(x+) + f(x-) \right)	.
\]
\end{theorem}

\vspace*{0.10in}

\uncover<2->{%
\textbf{Remarks}:
\begin{itemize}
\item If $f$ is continuous at $x=x_{0}$ then the Fourier series
  converges to $f(x_{0})$ when $x=x_{0}$.
\item If $f$ has a jump or removable discontinuity at $x=x_{0}$, the
  Fourier series converges to the average of the limits of $f$ from
  the left and right at $x=x_{0}$.
\end{itemize}
}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the piecewise-defined function
\[
f(x) = \left\{ \begin{array}{ll}
x & \mbox{if $-1 \le x < 0$,}	\\
0 & \mbox{if $0 \le x < 1$.}
\end{array} \right.
\]
Its Fourier series representation is
\[
f(x) \sim -\frac{1}{4} + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \pi} \sin (n
              \pi x)
	+ \sum_{n=1}^{\infty} \frac{2}{(2n-1)^{2} \pi^{2}} \cos ((2n-1) \pi x)	.
\]
Sketch the graph of the Fourier series.

\end{frame}
%%

\begin{frame}\frametitle{Graph}

\begin{center}
\resizebox{\textwidth}{!}{\includegraphics{example01fs}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Application: Finding the Sum of a Series}

If an infinite series is made up of Fourier coefficients for some
function, the function can be used to sum the series.

\vspace*{0.10in}

\uncover<2->{%
\begin{example}
\begin{enumerate}
\item Find the Fourier series for $f(x) = |x|$ on $[-\pi,\pi]$.
\item Use the Fourier series and $f(x)$ to find
\[
\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}}	.
\]
\end{enumerate}
\end{example}
}

\end{frame}
%%

\begin{frame}\frametitle{Solution}

Function $f(x) = |x|$ is an even function.
\begin{align*}
a_{0} &= \frac{2}{\pi} \int_{0}^{\pi} x \, dx =	\pi	\\
a_{n} &= \frac{2}{\pi} \int_{0}^{\pi} x \cos (n x) \, dx = \left\{
\begin{array}{rl}
-4/(n^{2} \pi) & \mbox{if $n$ is odd,}	\\
0 & \mbox{if $n$ is even.}
\end{array} \right.
\end{align*}
Fourier series:
\begin{align*}
|x|
	&=
\frac{\pi}{2} - \frac{4}{\pi} \sum_{n=1}^{\infty}
\frac{1}{(2n-1)^{2}} \cos ((2n-1) x)	\\
0	&=
\frac{\pi}{2} - \frac{4}{\pi} \sum_{n=1}^{\infty}
\frac{1}{(2n-1)^{2}} \cos ((2n-1) (0))	\\
\frac{\pi^{2}}{8} &= \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Application: Finding the Sum of a Series}

Consider the function $\displaystyle f(x) = \left\{ \begin{array}{rl}
-1 & \mbox{if $-\pi \le x < 0$}	\\
0 & \mbox{if $x = 0$}	\\
1 & \mbox{if $0 < x \le \pi$.}
\end{array} \right.$

\begin{enumerate}
\item Find the Fourier series for $f(x)$ on $[-\pi,\pi]$.
\item Sketch the graph of the Fourier series for $f(x)$.
\item Use the Fourier series and $f(x)$ to find
\[
\sum_{k=1}^{\infty} \frac{(-1)^{k}}{2k-1}	.
\]
\end{enumerate}

\end{frame}
%%

\begin{frame}\frametitle{Solution}

Function $f(x)$ is an odd function.
\begin{align*}
b_{n} &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin (n x) \, dx
  \\
	&= \frac{-1}{\pi} \int_{-\pi}^{0} \sin (n x) \, dx +
\frac{1}{\pi} \int_{0}^{\pi} \sin (n x) \, dx	\\
	&= \frac{2}{\pi} \int_{0}^{\pi} \sin (n x) \, dx	\\
	&= \frac{2}{n \pi} (1 - \cos (n \pi))	\\
	&= \left\{ \begin{array}{rl}
4/(n \pi) & \mbox{if $n$ is odd,}	\\
0 & \mbox{if $n$ is even.}
\end{array} \right.
\end{align*}
Fourier series:
\[
f(x) = \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{1}{2n-1} \sin ((2n-1) x)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Graph}

\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{example03fs}}
\end{center}

\textbf{Question}: why does the Fourier series converge to $f(x)$ on
$(-\pi,\pi)$?

\end{frame}
%%

\begin{frame}\frametitle{Summing the Series}

Let $x = \pi/2$.
\begin{align*}
f(x) &= \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{1}{2n-1} \sin
           ((2n-1) x)	\\
\uncover<2->{%
1	&= \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{1}{2n-1} \sin
           \left( (2n-1) \frac{\pi}{2} \right)	\\
\uncover<3->{%
\frac{\pi}{4}	&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2n-1}	\\
-\frac{\pi}{4}	&= \sum_{n=1}^{\infty} \frac{(-1)^{n}}{2n-1}
}
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Partial Sums}

\begin{center}
\resizebox{\textwidth}{!}{\includegraphics{gibbs}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Detailed View}

\begin{center}
\resizebox{\textwidth}{!}{\includegraphics{gibbsdetail}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Gibbs Phenomenon}

\begin{itemize}
\item Oscillation of the partial sums of a Fourier series near a jump
  discontinuity is called the \textbf{Gibbs phenomenon}.
\item It was first mathematically explained by Josiah Willard Gibbs,
  though others had considered it (including Albert A.~Michelson of
  the Michelson-Morley experiment).
\item We will outline an explanation of the Gibbs phenomenon.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Partial Sum}

Denote the $N$th partial sum of the Fourier series as
\[
s_{N}(x)
	=
\frac{4}{\pi} \sum_{n=1}^{N} \frac{1}{2n-1} \sin ((2n-1) x)	.
\]
A calculus argument can locate the maxima in the graph of $s_{N}(x)$
near $x = 0$.
\uncover<2->{%
\begin{align*}
s_{N}'(x)
	&=
\frac{4}{\pi} \sum_{n=1}^{N} \cos ((2n-1) x)	\\
\uncover<3->{%
(\sin x) s_{N}'(x)
	&=
\frac{4}{\pi} \sum_{n=1}^{N} \sin x \cos ((2n-1) x)
}
\end{align*}
}
\uncover<4->{%
Use a product-to-sum formula on the right-hand side.
}

\end{frame}
%%

\begin{frame}\frametitle{Critical Numbers}

\begin{align*}
(\sin x) s_{N}'(x)
	&=
\frac{4}{\pi} \sum_{n=1}^{N} \cos ((2n-1) x) \sin x	\\
	&=
\frac{2}{\pi} \sum_{n=1}^{N} \left( \sin (2n x) - \sin (2(n-1) x)
              \right)	\\
\uncover<2->{%
	&=
\frac{2}{\pi} \sin (2N x)
}
\end{align*}
\uncover<3->{%
If $0 < x < \pi$ then $s_{N}'(x) = 0$ if and only if $\sin (2N x) =
0$.
Thus the critical numbers in $(0, \pi)$ are
\[
x = \frac{m \pi}{2 N}
\quad\mbox{for $m = 1, 2, \ldots, 2N-1$.}
\]
}

\end{frame}
%%

\begin{frame}\frametitle{Maxima or Minima?}

Apply the Second Derivative Test to determine whether the critical
numbers are maxima or minima.
\begin{align*}
(\sin x) s_{N}'(x)
	&=
\frac{2}{\pi} \sin (2N x)	\\
(\cos x) s_{N}'(x) + (\sin x) s_{N}''(x)
	&=
\frac{4N}{\pi} \cos (2N x)
\end{align*}
Let $x = \pi/(2N)$, the critical number closest to $x=0$.
\begin{align*}
\left( \cos \frac{\pi}{2N} \right) s_{N}' \left( \frac{\pi}{2N}
  \right) + \left( \sin \frac{\pi}{2N} \right) s_{N}'' \left(
  \frac{\pi}{2N} \right)
	&=
\frac{4N}{\pi} \cos \pi	\\
\left( \sin \frac{\pi}{2N} \right) s_{N}'' \left( \frac{\pi}{2N} \right)
	&=
-\frac{4N}{\pi}
\end{align*}
This implies $s_{N}(\pi/2N)$ is a local maximum.

\end{frame}
%%

\begin{frame}\frametitle{Value of Local Maximum}

\begin{align*}
s_{N}\left( \frac{\pi}{2N} \right)
	&=
\frac{4}{\pi} \sum_{n=1}^{N} \frac{1}{2n-1} \sin \left( \frac{(2n-1) \pi}{2N} \right)	\\
	&=
\frac{2}{\pi} \sum_{n=1}^{N} \frac{2N \pi}{(2n-1)N \pi} \sin \left(
              \frac{(2n-1) \pi}{2N} \right)	\\
\uncover<2->{%
	&=
\frac{2}{\pi} \sum_{n=1}^{N} \frac{2N}{(2n-1) \pi} \sin \left(
              \frac{(2n-1) \pi}{2N} \right) \, \frac{\pi}{N}
}
\end{align*}
\uncover<3->{%
\textbf{Remark}: the term inside the summation is a Riemann sum with
$\Delta x = \pi/N$, $w_{n} = (2n-1) \pi/(2N)$, and $\displaystyle f(x)
= \frac{1}{x} \sin x$.
}

\end{frame}
%%

\begin{frame}\frametitle{Limit of Riemann Sum}

\begin{align*}
\lim_{N \to \infty} s_{N}\left( \frac{\pi}{2N} \right)
	&= \lim_{N \to \infty} \frac{2}{\pi} \sum_{n=1}^{N} \frac{2N}{(2n-1) \pi} \sin \left(
              \frac{(2n-1) \pi}{2N} \right) \, \frac{\pi}{N}	\\
	&= \frac{2}{\pi} \int_{0}^{\pi} \frac{\sin x}{x} \, dx	\\
	&\approx 1.179
\end{align*}
\uncover<2->{%
\textbf{Remark}: thus no matter how many terms are included in the
Fourier series there is an $x \to 0^{+}$ for which $f(x) \approx 1.179
> 1$.
}

\end{frame}
%%

\begin{frame}\frametitle{Pointwise \textit{vs}. Uniform Convergence}

\begin{definition}
Suppose $\{ f_{n} \}_{n=1}^{\infty}$ is a sequence of functions
defined on domain $D$ and the sequence of values $\{ f_{n}(x)
\}_{n=1}^{\infty}$ converges for each $x \in S \subset D$.
Then $\{ f_{n} \}_{n=1}^{\infty}$ is said to \textbf{converge
  pointwise on $S$ to $f$} defined by
\[
f(x) = \lim_{n \to \infty} f_{n}(x)
\quad\mbox{for $x \in S$.}
\]
\end{definition}

While the Fourier series converges \textbf{pointwise} to $f(x)$ it
does not converge \textbf{uniformly} to $f(x)$.

\end{frame}
%%

\begin{frame}\frametitle{Uniform Convergence}

\begin{definition}
A sequence of functions $\{ f_{n} \}_{n=1}^{\infty}$ defined on domain
$D$ \textbf{converges uniformly to $f$ on $D$} provided there exists a
sequence of positive real numbers $\{ \epsilon_{n} \}_{n=1}^{\infty}$
for which $\displaystyle \lim_{n \to \infty} \epsilon_{n} = 0$ and
\[
\left| f_{n}(x) - f(x) \right| < \epsilon_{n}
\]
for all $n \in \mathbb{N}$ and for all $x \in D$.
\end{definition}

\vspace*{0.10in}

\textbf{Remark}: an infinite series converges uniformly if its
sequence of partial sums converges uniformly.

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

Sequence $\{ x^{n} e^{-n x} \}_{n=1}^{\infty}$ converges uniformly to
$f(x) = 0$ on $D = [0,\infty)$.
\begin{center}
\resizebox{0.90\textwidth}{!}{\includegraphics{uniform}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Justification}

\begin{itemize}
\item Let $f_{n}(x) = x^{n} e^{-n x}$ for $n \in \mathbb{N}$ and $x
  \ge 0$.
\item By l'H\^{o}pital's Rule $\displaystyle \lim_{n \to \infty}
  f_{n}(x) = \lim_{n \to \infty} \frac{x^{n}}{e^{n x}} = 0$ for all $x
  \ge 0$, thus $f_{n}(x) \to 0$ pointwise for all $x \ge 0$.
Define $f(x) = 0$.
\item The First Derivative Test shows that $0 \le f_{n}(x) \le f_{n}(1) =
  e^{-n}$ for all $x \ge 0$, so let $\epsilon_{n} = e^{-n}$ and note
  that $\displaystyle \lim_{n \to \infty} \epsilon_{n} = 0$.
\item The sequence of functions $f_{n}(x) \to f(x)$ uniformly for $x
  \ge 0$ since $|f_{n}(x) - f(0)| = f_{n}(x) \le \epsilon_{n}$ for all
  $n \in \mathbb{N}$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Examples (1 of 3)}

Sequence $\{ x^{n} \}_{n=1}^{\infty}$ converges pointwise on $[0,1]$ to
  $\displaystyle f(x) =
\left\{ \begin{array}{rl}
0 & \mbox{if $0 \le x < 1$}	\\
1 & \mbox{if $x = 1$}
\end{array} \right.$
but does not converge uniformly.

\end{frame}
%%

\begin{frame}\frametitle{Justification}

\begin{itemize}
\item Let $x \in [0, 1)$ then $\displaystyle \lim_{n \to \infty} x^{n}
  = 0$.
Also, $\lim_{n \to \infty} 1^{n} = 1$, so $x^{n} \to f(x)$ pointwise
for $0 \le x \le 1$.
\item Let $\{ \epsilon_{n} \}_{n=1}^{\infty}$ be any sequence of
  positive real numbers such that $\displaystyle \lim_{n \to \infty}
  \epsilon_{n} = 0$.
Without loss of generality assume $0 < \epsilon_{n} < 1$ for all $n
\in \mathbb{N}$.
\item Note that $0 < 1 - \epsilon_{n} < 1$ for all $n \in \mathbb{N}$
  and thus $0 < (1 - \epsilon_{n})^{1/n} < 1$ as well.
Therefore for all $n \in \mathbb{N}$ there exists $x_{n}$ such that $(1
- \epsilon_{n})^{1/n} < x_{n} < 1$.
\item Consider $x_{n}^{n} > 1 - \epsilon_{n} > 0$ and thus $x_{n}^{n}
  \not\to 0$ as $n \to \infty$ and the convergence is not uniform.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Examples (2 of 3)}

The sequence $\{ x^{n} \}_{n=1}^{\infty}$ converges uniformly to $f(x) = 0$
    on $[0,b]$ for any $0 < b < 1$.

\end{frame}
%%

\begin{frame}\frametitle{Justification}

\begin{itemize}
\item As in the previous example $x^{n} \to 0$ pointwise for all $0
  \le x \le b < 1$.
\item Define $\epsilon_{n} = b^{n}$ for all $n \in \mathbb{N}$, then
$\displaystyle \lim_{n \to \infty} \epsilon_{n} = 0$.
\item Note that $x^{n} \le b^{n}$ for all $0 \le x \le b$, or
  equivalently, $| x^{n} - 0 | \le \epsilon_{n}$ for all $0 \le x \le
  b$.
Hence the convergence is uniform. 
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Examples (3 of 3)}

The infinite series $\displaystyle \sum_{n=1}^{\infty} x^{n}$
    converges pointwise on $(-1,1)$ to $\displaystyle f(x) =
    \frac{x}{1-x}$ but not uniformly.

\end{frame}
%%

\begin{frame}\frametitle{Justification}

\begin{itemize}
\item The $N$th partial sum of the series is
\[
f_{N}(x) = x + x^{2} + \cdots + x^{N} = \frac{1 - x^{N+1}}{1-x} - 1	.
\]
\item For any $-1 < x < 1$,
\[
\lim_{N \to \infty} f_{N}(x) = \lim_{N \to \infty} \left[ \frac{1 -
    x^{N+1}}{1-x} - 1 \right] = \frac{1}{1-x} - 1 = \frac{x}{1-x} = f(x)	,
\]
(this proves pointwise convergence).
\item<2-> Consider
\begin{align*}
\lim_{x \to 1^{-}} | f_{N}(x) - f(x) |
	&=
\lim_{x \to 1^{-}} \left| \frac{1 - x^{N+1}}{1-x} - 1 - \frac{x}{1-x} \right|	\\
	&=
\lim_{x \to 1^{-}} \frac{x^{N+1}}{1-x} = \infty	,
\end{align*}
so the convergence is not uniform.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Properties Preserved by Uniform Convergence}

Suppose the series $\displaystyle \sum_{n=1}^{\infty} u_{n}(x)$
converges uniformly to its sum $u(x)$ on $[a,b]$.
\begin{itemize}
\item If for each $n$, $u_{n}(x)$ is continuous on $[a,b]$,
then the sum $u(x)$ is continuous on $[a,b]$.
\item If for each $n$, $u_{n}(x)$ is integrable on $[a,b]$, then the sum
$u(x)$ is integrable on $[a,b]$, and
\[
\int_{a}^{b} u(x) \, dx
	=
\int_{a}^{b} \left( \sum_{n=1}^{\infty} u_{n}(x) \right) \, dx
	=
\sum_{n=1}^{\infty} \int_{a}^{b} u_{n}(x) \, dx	.
\]
\item If for each $n$, $u_{n}'(x)$ exists and $\sum_{n=1}^{\infty}
  u_{n}'(x)$ converges uniformly on $[a,b]$,
then the sum $u(x)$ is differentiable on
$[a,b]$ and the derivative can be obtained by differentiating
the series term by term,
\[
u'(x)
	=
\left( \sum_{n=1}^{\infty} u_{n}(x) \right)'
	=
\sum_{n=1}^{\infty} u_{n}'(x)
\quad\mbox{for all $x \in [a,b]$.}
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Weierstrass $M$-Test}

The following theorem provides a convenient means of determining
whether an infinite series converges uniformly.
\begin{theorem}
Let $\sum_{n=1}^{\infty} u_{n}(x)$ be a
series of functions defined on an interval $[a,b]$ and suppose
that for each $n$ there is a non-negative number $M_{n}$ such that
$| u_{n}(x) | \le M_{n}$ for all $x \in [a,b]$ and
$\displaystyle \sum_{n=1}^{\infty} M_{n}$
converges, then $\displaystyle \sum_{n=1}^{\infty} u_{n}(x)$ converges
uniformly on $[a,b]$.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Show that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}} \cos (n
x)$ converges uniformly on $(-\infty,\infty)$.

\vspace*{0.10in}

\uncover<2->{%
Since $\frac{1}{n^{2}} | \cos ( n x ) | \le \frac{1}{n^{2}}$ for all
$x \in \mathbb{R}$ and
\[
\sum_{n=1}^{\infty} \frac{1}{n^{2}}
\]
converges, then the series converges uniformly for all $x \in \mathbb{R}$.
}

\end{frame}
%%

\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Sections $3.6$--$3.7$
\item Exercises: 11--19
\end{itemize}

\end{frame}
%%

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\end{document}

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