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 $a_{0}+\sum _{n=1}^{\infty }a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}$. Let $S_{n}(x)=a_{0}+\sum _{n=1}^{N}a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}$ be the partial sum. For fixed $x$,

$\lim _{N\to \infty }S_{N}(x)={\begin{cases}f(x)&f\ {\mbox{is continuous at}}\ x\\{\frac {f(x^{+})+f(x^{-})}{2}}&f\ {\mbox{has a jump at}}\ x\end{cases}}$

1. $F(x)=\lim _{N\to \infty }S_{N}(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}$.
2. $f(x)$ is the original function.

Sketch of Proof.

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

quod erat demonstrandum

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

Example

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

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

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

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 $x\in \left[0,2\pi \right]$.

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

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