MATH 414 Lecture 10

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Pointwise Convergence of Fourier Series

Theorem. Let $f(v)$ be a $2\pi$-periodic piecewise function with Fourier series $a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx)+b_n\sin (nx)$. Let $S_n(x)=a_0+\sum _{n=1}^N{a}_n\cos (nx)+b_n\sin (nx)$ be the partial sum. For fixed $x$,

$\underset{N\to \mathrm{\infty }}{\lim }{S}_N(x)=\{\begin{array}{ll}f(x)& f\text{is continuous at}x\\ \frac{f(x^{+})+f(x^{-})}{2}& f\text{has a jump at}x\end{array}$

1. $F(x)=\underset{N\to \mathrm{\infty }}{\lim }{S}_N(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx)+b_n\sin (nx)$.
2. $f(x)$ is the original function.

Sketch of Proof.

1. Riemann-Lebesgue Lemma: The following holds if $f$ is piecewise-continuous: $\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f(x){\mathrm{e}}^{inx}\mathrm{d}x& =0\\ \underset{n\to \mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f(x)\sin (nx)\mathrm{d}x& =0\end{array}$
2. Get partial sums of an integral: $\begin{array}{rl}{S}_N(x)& =\underset{-\pi }{\overset{\pi }{\int }}f(u)D_N(x-u)\mathrm{d}u\\ D_N(t)& =\frac{1}{2\pi }\frac{\sin \left((N+\frac{1}{2})t\right)}{\sin \left(\frac{x}{2}\right)}\end{array}$
3. Get error
4. Use R-L lemma to finish up

quod erat demonstrandum

Interesting example: $\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{2k-1}=\frac{\pi }{4}$

Example

Let $f(x)=x^2$. The fourier series $F(x)\frac{{\pi }^2}{3}+\sum _{n=1}^{\mathrm{\infty }}\frac{4(-1{)}^n}{n^2}\cos (nx)$ for $x\in [-\pi ,\pi ]$ represents the $2\pi$-periodic extension of $f$.

Now $f(0)=0$, so $F(0)=\frac{{\pi }^2}{3}+\sum _{n=1}^{\mathrm{\infty }}\frac{4(-1{)}^n}{n^2}\cancel{\cos (nx)}=0$.

Rewritten, $\frac{{\pi }^2}{12}=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n^2}$

Interesting trivia: rectifiers in a radio or audio signal system give an envelope of the amplitude of the current (e.g. a vu-meter on a recording set)

Another Example

Consider periodic extension of $f(x)=\pi -x$ for $x\in [0,2\pi ]$.

$F(x)=\sum _{k=1}^{\mathrm{\infty }}\frac{2\sin (nx)}{n}$

At $x=\frac{\pi }{2}$, we get $\frac{\pi }{4}=\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k+1}}{2k-1}$