MATH 414 Lecture 7

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Standard Interval

$-\pi \le x\le \pi$

$\begin{array}{rl}f(x)& =a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx)+b_n\sin (nx)\\ a_0& =\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\mathrm{d}x\\ a_n& =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos (nx)\mathrm{d}x\\ b_n& =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\sin (nx)\mathrm{d}x\end{array}$

Scaled Interval

$-c\le x\le c$

If we let $t=\frac{\pi }{c}x$, then $x=\frac{c}{\pi }$, and evaluating our expressions with respect to $t$ yields.

$\begin{array}{rl}f(x)& =a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos \left(\frac{n\pi }{c}x\right)+b_n\sin \left(\frac{n\pi }{c}x\right)\\ a_0& =\frac{1}{2a}\underset{-a}{\overset{a}{\int }}f(x)\mathrm{d}x\\ a_n& =\frac{1}{a}\underset{-a}{\overset{a}{\int }}f(x)\cos \left(\frac{n\pi }{a}x\right)\mathrm{d}x\\ b_n& =\frac{1}{a}\underset{-a}{\overset{a}{\int }}f(x)\sin \left(\frac{n\pi }{a}x\right)\mathrm{d}x\end{array}$

Example

$f(x)=\{\begin{array}{ll}1& 0\le x\le 1\\ 0& \text{otherwise}\end{array}x\in [-2,2]$

In this case, $a=2$, so

$a_0=\frac{1}{2\cdot 2}\underset{-2}{\overset{2}{\int }}f(x)\mathrm{d}x=\frac{1}{4}$

$a_n=\frac{1}{2}\underset{-2}{\overset{2}{\int }}f(x)\cos \left(\frac{n\pi x}{2}\right)\mathrm{d}x=\frac{1}{2}\underset{0}{\overset{1}{\int }}\cos \left(\frac{n\pi x}{2}\right)\mathrm{d}x=\frac{1}{n\pi }\sin \left(\frac{n\pi }{2}\right)$

$b_n=\frac{1}{2}\underset{-2}{\overset{2}{\int }}f(x)\sin \left(\frac{n\pi x}{2}\right)\mathrm{d}x=\frac{1}{2}\underset{0}{\overset{1}{\int }}\sin \left(\frac{n\pi x}{2}\right)\mathrm{d}x=\frac{1}{n\pi }(1-\cos \left(\frac{n\pi }{2}\right))$

Hence

$f(x)=\frac{1}{4}+\sum _{n=1}^{\mathrm{\infty }}\frac{\sin \left(\frac{n\pi }{2}\right)}{npi}\cos \left(\frac{n\pi x}{2}\right)+\frac{1-\cos \left(\frac{n\pi }{2}\right)}{n\pi }\sin \left(\frac{n\pi x}{2}\right)$

Shifted Interval

Lemma. Let $F(x)$ be a $2\pi$-periodic function. Then

$\underset{-\pi +c}{\overset{\pi +c}{\int }}F(x)\mathrm{d}x=\underset{-\pi }{\overset{\pi }{\int }}F(x)\mathrm{d}x=\underset{0}{\overset{\pi }{\int }}F(x)\mathrm{d}x$

In other words, $\underset{-\pi +c}{\overset{\pi +c}{\int }}F(x)\mathrm{d}x$ is independent of $c$.

Proof. Let $G(c)=\underset{-\pi +c}{\overset{\pi +c}{\int }}F(x)\mathrm{d}x$.

$\begin{array}{rl}\frac{\mathrm{d}G}{\mathrm{d}c}& =\frac{\mathrm{d}}{\mathrm{d}c}(\underset{0}{\overset{\pi +c}{\int }}F(x)\mathrm{d}x)-\frac{\mathrm{d}}{\mathrm{d}c}(\underset{0}{\overset{c-\pi }{\int }}F(x)\mathrm{d}x)\\ & =F(\pi +c)-F(c-\pi )\\ & =F(\pi +c)-f(c+\pi )\\ & =0\end{array}$

quod erat demonstrandum

Periodic Extension

Suppose $f(x)$ is defined on $0\le x\le 2\pi$ and undefined elsewhere.

A periodic extension basically copies the defined part of the function into the undefined part

If the period of a function is $T$, then

$f(t)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos \left(\frac{2\pi nt}{T}\right)+b_n\sin \left(\frac{2\pi nt}{T}\right)$
$a_n=\frac{1}{\left(\frac{T}{2}\right)}\underset{0}{\overset{T}{\int }}f(t)\cos \left(\frac{2\pi nt}{n}\right)$
$b_n=\frac{1}{\left(\frac{T}{2}\right)}\underset{0}{\overset{T}{\int }}f(t)\sin \left(\frac{2\pi nt}{n}\right)$

Example

For example, take $f(x)=x$

Then $f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos nx+b_n\sin nx$

$b_n=\frac{1}{\pi }\underset{0}{\overset{2\pi }{\int }}x\sin nx\mathrm{d}x$

etc...

In the end, we have $f(x)=\pi -\sum _{n=1}^{\mathrm{\infty }}\frac{2\sin nx}{n}$