MATH 414 Lecture 12

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Three Types of Periodic Extensions

Standard $\pi$-periodic (a regular Fourier Series)

Even $2\pi$-periodic extension (Cosine Series)

Odd $2\pi$-periodic extension (Sine Series)

Uniform Convergence

Let $f$ be a $2\pi$-periodic function with Fourier Series $a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx)+b_n\sin (nx)$. If $S_N(x)=a_0+\sum _{n=1}^N{a}_n\cos (nx)+b_n\sin (nx)$, then $S_N$ converges uniformly to $f$ on $x\in \mathbb{ℝ}$ if and only if for every $ϵ>0$, there is a $N_0$ that does not depend on $x$ such that $|f(x)-S_N(x)|<ϵ$ for all $x$, provided $N>N_0$.

Theorem. If $f$ is piecewise smooth and continuous (no jumps, but it can have corners, then $S_N(x)$ converges uniformly to $f(x)$.

Proof. [to be discussed later]

quod erat demonstrandum

#18 Convolution

$(f*g)(x):=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)g(x-t)\mathrm{d}t$

Suppose this has the fourier series $\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\sigma }_n{\mathrm{e}}^{inx}$, where ${\sigma }_n=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}(f*g)(x){\mathrm{e}}^{-inx}\mathrm{d}x$

BIG HINT: replace $(f*g)(x)$ with its definition

Fubini's theorem will be useful when changing integrals.

Parseval's Theorem

Let $f\in L^2[-\pi ,\pi ]$. This implies that $f$ has finite energy over a finite interval.

An $L^2$ equality:

• Let $f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}a-n\cos (nx)+b_n\sin (nx)$
• Alternatively, $f(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{\mathrm{e}}^{inx}$

$\underset{-\pi }{\overset{\pi }{\int }}{|f(t)|}^2\mathrm{d}t=2\pi \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|c_n\right|}^2=2\pi {\left|a_0\right|}^2+\pi \sum _{n=1}^{\mathrm{\infty }}{\left|a_n\right|}^2+{\left|b_n\right|}^2$

Example

$f(x)=\pi -x$. Consider the even extension of $f$ from $-\pi$ to $\pi$.

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

1. $\underset{-\pi }{\overset{\pi }{\int }}{|f(x)|}^2\mathrm{d}x=2\underset{0}{\overset{\pi }{\int }}{(\pi -x)}^2\mathrm{d}x=\frac{2}{3}{\pi }^3$
2. $a_0=\frac{\pi }{2}$, $b_n=0$, $a_{oddn}=0$

$\underset{-\pi }{\overset{\pi }{\int }}{|f(x)|}^2\mathrm{d}x=\frac{2}{3}{\pi }^3=2\pi {\left(\frac{\pi }{2}\right)}^2+\left(\sum _{k=1}^{\mathrm{\infty }}\frac{16}{(2k-1)4}\right)\pi$

Consider the odd extension of $f$ from $-\pi$ to $\pi$.

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

1. $a_n=0$, $b_n=\frac{2}{n}$

$\underset{-\pi }{\overset{\pi }{\int }}{|f(x)|}^2\mathrm{d}x=2\underset{0}{\overset{\pi }{\int }}{(\pi -x)}^2\mathrm{d}x=\frac{2}{3}{\pi }^3=\pi \left(\sum _{n=1}^{\mathrm{\infty }}\frac{4}{n^2}\right)$

We get $\frac{{\pi }^2}{6}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$

Punchline

If $f(t)=\sum c_n{\mathrm{e}}^{int}$, then the total energy in the wave is equal to the sum of energies of all its modes:

$\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}{|f(t)|}^2\mathrm{d}t=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\left|c_n\right|}^2$