\begin{frame}\frametitle{Partial Sums}

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

\end{frame}
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\begin{frame}\frametitle{Detailed View}

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

\end{frame}
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\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}
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\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}
%%

