% ch02slidesa.tex

\begin{slide}
\begin{center}\large\bf
Fourier Series
\end{center}

\begin{itemize}
\item {Fourier Sine Series}
\item {Fourier Cosine Series}
\item {Fourier Series}
  \begin{itemize}
  \item {Convergence of Fourier Series for $2T$-Periodic Functions}
  \item {Convergence of Half-Range Expansions: Cosine Series}
  \item {Convergence of Half-Range Expansions: Sine Series}
\end{itemize}
\item {Sawtooth Wave}
\item {Triangular Wave}
\item {Parseval's Identity and Bessel's Inequality}
\item {Complex Fourier Series}
\item {Dirichlet Kernel and Convergence}
\end{itemize}

\end{slide}

\begin{slide}
\header{Fourier Sine Series}

\bigskip

\BF{Definition}.  Consider the orthogonal system $\{ \sin \left(\frac{n\pi x}{T}\right)\}_{n=1}^\infty$ on $[-T,T]$. A Fourier sine series with coefficients $\{b_n\}_{n=1}^\infty$  is the expression
$$ F(x) = \sum_{n=1}^\infty b_n \sin \left(\frac{n\pi x}{T}\right)$$

\BF{Theorem}. A Fourier sine series $F(x)$ is an odd $2T$-periodic function.

\bigskip

\def\AA{\sin \left(\frac{n\pi x}{T}\right)}

\BF{Theorem}. The coefficients $\{ b_n\}_{n=1}^\infty$ in a Fourier sine series $F(x)$
are determined by the formulas (inner product on $[-T,T]$)
$$ b_n = \frac{\innerprod{F,\AA}}{\innerprod{\AA,\AA}} =
 \frac{2}{T}\int_0^T F(x)\AA dx.
$$

\end{slide}

\begin{slide}
\header{Fourier Cosine Series}

\bigskip

\def\BB{\cos \left(\frac{m\pi x}{T}\right)}

\BF{Definition}.  Consider the orthogonal system $\{ \BB \}_{m=0}^\infty$ on $[-T,T]$. A Fourier cosine series with coefficients $\{a_m\}_{m=0}^\infty$  is the expression
$$ F(x) = \sum_{m=0}^\infty a_m \BB $$

\BF{Theorem}. A Fourier cosine series $F(x)$ is an even $2T$-periodic function.

\bigskip


\BF{Theorem}. The coefficients $\{ a_m\}_{m=0}^\infty$ in a Fourier cosine series $F(x)$
are determined by the formulas (inner product on $[-T,T]$)
$$ a_m = \frac{\innerprod{F,\BB}}{\innerprod{\BB,\BB}} = \left\{ \begin{array}{lr}
 \frac{2}{T}\int_0^T F(x)\BB dx & m>0, \\[2ex]
 \frac{1}{T}\int_0^T F(x)\ dx & m=0.
 \end{array}\right.
$$

\end{slide}

\begin{slide}
\header{Fourier Series}

\bigskip

\def\AA{\sin \left(\frac{n\pi x}{T}\right)}
\def\BB{\cos \left(\frac{m\pi x}{T}\right)}

\BF{Definition}.  Consider the orthogonal system $\{ \BB \}_{m=0}^\infty$, $\{ \AA \}_{n=1}^\infty$ , on $[-T,T]$. A Fourier series with coefficients $\{a_m\}_{m=0}^\infty$, $\{b_n\}_{n=1}^\infty$  is the expression
$$ F(x) = \sum_{m=0}^\infty a_m \BB + \sum_{n=1}^\infty b_n \AA $$

\BF{Theorem}. A Fourier series $F(x)$ is a $2T$-periodic function.

\bigskip

\BF{Theorem}. The coefficients $\{a_m\}_{m=0}^\infty$, $\{b_n\}_{n=1}^\infty$ in a Fourier series $F(x)$
are determined by the formulas (inner product on $[-T,T]$)
$$ a_m = \frac{\innerprod{F,\BB}}{\innerprod{\BB,\BB}} = \left\{ \begin{array}{lr}
 \frac{1}{T}\int_{-T}^T F(x)\BB dx & m>0, \\[2ex]
 \frac{1}{2T}\int_{-T}^T F(x)\ dx & m=0.
 \end{array}\right.
$$

$$ b_n = \frac{\innerprod{F,\AA}}{\innerprod{\AA,\AA}} =
 \frac{1}{T}\int_{-T}^T F(x)\AA dx.
$$
\end{slide}

\begin{slide}
\header{Convergence of Fourier Series for $2T$-Periodic Functions}

\bigskip

The Fourier series of a $2T$-periodic piecewise smooth function $f(x)$ is
$$ a_0 + \sum_{n=1}^\infty \left( a_n \cos\left(\frac{n\pi x}{T}\right) + b_n \sin\left(\frac{n\pi x}{T}\right) \right)
$$
where
$$ a_0 = \frac{1}{2T}\int_{-T}^T f(x)dx,$$
$$ a_n = \frac{1}{T}\int_{-T}^T f(x)\cos\left(\frac{n\pi x}{T}\right)dx,$$
$$ b_n = \frac{1}{T}\int_{-T}^T f(x)\sin\left(\frac{n\pi x}{T}\right)dx.$$
The series converges to $f(x)$ at points of continuity of $f$ and to $\frac{f(x+)+f(x-)}{2}$ otherwise.
\end{slide}

\begin{slide}
\header{Convergence of Half-Range Expansions: Cosine Series}

\bigskip

The Fourier cosine series of a piecewise smooth function $f(x)$ on $[0,T]$ is the even $2T$-periodic function
$$ a_0 + \sum_{n=1}^\infty a_n \cos\left(\frac{n\pi x}{T}\right)
$$
where
$$ a_0 = \frac{1}{T}\int_{0}^T f(x)dx,$$
$$ a_n = \frac{2}{T}\int_{0}^T f(x)\cos\left(\frac{n\pi x}{T}\right)dx.$$
The series converges on $0<x<T$ to $f(x)$ at points of continuity of $f$ and to $\frac{f(x+)+f(x-)}{2}$ otherwise.


\end{slide}

\begin{slide}
\header{Convergence of Half-Range Expansions: Sine Series}

\bigskip

The Fourier sine series of a piecewise smooth function $f(x)$ on $[0,T]$ is the odd $2T$-periodic function
$$  \sum_{n=1}^\infty  b_n \sin\left(\frac{n\pi x}{T}\right)
$$
where
$$ b_n = \frac{2}{T}\int_{0}^T f(x)\sin\left(\frac{n\pi x}{T}\right)dx.$$
The series converges on $0<x<T$ to $f(x)$ at points of continuity of $f$ and to $\frac{f(x+)+f(x-)}{2}$ otherwise.


\end{slide}

\begin{slide}\header{Sawtooth Wave}

\medskip

\BF{Definition}. The \DF{sawtooth wave} is the odd $2\pi$-periodic function defined on $-\pi \le x \le \pi$ by the formula
$$\sawtooth(x) = \left\{ \begin{array}{l@{\quad}r} \frac12(\pi-x) & 0 < x \le \pi, \\[2ex] \frac12(-\pi-x) & -\pi \le x < 0, \\[2ex] 0 & x=0. \end{array}\right.$$






\BF{Theorem}. The sawtooth wave has Fourier sine series
$$\sawtooth(x) = \sum_{n=1}^\infty \frac{1}{n}\, \sin nx.$$
\end{slide}

\begin{slide}
\header{Triangular Wave}

\medskip

\BF{Definition}. The \DF{triangular wave} is the even $2\pi$-periodic function defined on $-\pi \le x \le \pi$ by the formula
$$\twave(x) = \left\{ \begin{array}{l@{\quad}r} \pi-x & 0 < x \le \pi, \\[2ex] \pi+x & -\pi \le x \le 0. \end{array}\right.$$




\BF{Theorem}. The triangular wave has Fourier cosine series
$$\twave(x) = \frac{\pi}{2}+\frac{4}{\pi}\sum_{k=0}^\infty \frac{1}{(2k+1)^2}\, \cos (2k+1)x.$$
\end{slide}

\begin{slide}
\header{ParsevaL's Identity and Bessel's Inequality}

\bigskip

\BF{Theorem}. (Bessel's Inequality)
$$ a_0^2 + \frac12\sum_{n=1}^\infty \left( a_n^2 + b_n^2\right) \le \frac{1}{2T}\int_{-T}^T |f(x)|^2dx$$

\BF{Theorem}. (Parseval's Identity)
$$\frac{1}{2T}\int_{-T}^T |f(x)|^2dx = a_0^2 + \frac12\sum_{n=1}^\infty \left( a_n^2 + b_n^2\right) $$

\BF{Theorem}. Parseval's identity for the sawtooth function implies
$$\frac{\pi^2}{12} = \frac12\sum_{n=1}^\infty \frac{1}{n^2}$$

\end{slide}

\begin{slide}
\header{Complex Fourier Series}
\bigskip

\BF{Definition}. Let $f(x)$ be $2T$-periodic and piecewise smooth. The complex Fourier series of $f$ is
$$ \sum_{n=-\infty}^\infty c_n e^{\frac{in\pi x}{T}},\quad c_n = \frac{1}{2T}\int_{-T}^T f(x)e^{\frac{-in\pi x}{T}}dx$$

\BF{Theorem}. The complex series converges to $f(x)$ at points of continuity of $f$  and to $\frac{f(x+)+f(x-)}{2}$ otherwise.

\bigskip

\BF{Theorem}. (Complex Parseval Identity)
$$\frac{1}{2T}\int_{-T}^T |f(x)|^2dx = \sum_{n=-\infty}^\infty |c_n|^2$$
\end{slide}

\begin{slide}
\header{Dirichlet Kernel and Convergence}

\bigskip

\BF{Theorem}. (Dirichlet Kernel Identity)
$$\dd \frac12+ \cos u + \cos 2u + \cdots + \cos nu = \frac{\sin\left( \left(n+\frac12\right)u \right)}{2\sin\left(\frac12 u\right)}$$

\BF{Theorem}. (Riemann-Lebesgue) \newline
For piecewise continuous $g(x)$,
$\dd \lim_{N\to\infty} \int_{-\pi}^{\pi} g(x)\sin (Nx) dx =0$.

\medskip

{\small\sf \BF{Proof}: Integration theory implies it suffices to establish the result for smooth $g$.
Integrate by parts to obtain $\frac{1}{n}(g(-\pi)-g(\pi))(-1)^n+\frac{1}{n}\int_{-\pi}^\pi g(x)\cos(nx)dx$. Letting $n\to\infty$ implies the result.}

\bigskip

\BF{Theorem}. Let $f(x)$ be $2\pi$-periodic and smooth on the whole real line. Then
the Fourier series of $f(x)$ converges uniformly to $f(x)$.
\end{slide}

\begin{slide}
\header{Convergence Proof}

\bigskip

\BF{STEP 1}. Let $s_N(x)$ denote the Fourier series partial sum. Using Dirichlet's kernel formula, we verify the identity
$$ f(x)-s_N(x) = \frac{1}{\pi}\int_{x-\pi}^{x+\pi}  (f(x)-f(x+w))\left(
 \frac{\sin((N+1/2)w)}{2\sin(w/2)}
\right)dw
$$
\bigskip

\BF{STEP 2}. The integrand $I$ is re-written as
$$ I = \frac{f(x)-f(x+w)}{w}\frac{w}{2\sin(w/2)}\sin((N+1/2)w).$$
\bigskip

\BF{STEP 3}. The function $\dd g(w)=\frac{f(x)-f(x+w)}{w}\frac{w}{\sin(w/2)}$ is piecewise continuous. Apply the Riemann-Lebesgue Theorem to complete the proof of the theorem.
\end{slide}

\begin{slide}
\header{Gibbs' Phenomena}

\medskip

\BF{Engineering Interpretation}: The graph of $f(x)$ and the graph of \newline
$a_0+ \sum_{n=1}^N (a_n \cos nx + b_n\sin nx)$ \newline
are identical to pixel resolution, provided $N$ is sufficiently large. Computers can therefore graph $f(x)$ using a truncated Fourier series.

\medskip

 If $f(x)$ is only piecewise smooth, then pointwise convergence is still true, at points of continuity of $f$, but uniformity of the convergence fails near discontinuities of $f$ and $f'$. Gibbs discovered the fixed-jump artifact, which appears at discontinuities of $f$.


\end{slide} 