MATH 414 Lecture 7

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Standard Interval

$-\pi \leq x\leq \pi$

{\begin{aligned}f(x)&=a_{0}+\sum _{n=1}^{\infty }a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}\\a_{0}&={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)\,\mathrm {d} x\\a_{n}&={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\,\cos {(n\,x)}\,\mathrm {d} x\\b_{n}&={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\,\sin {(n\,x)}\,\mathrm {d} x\end{aligned}}

Scaled Interval

$-c\leq x\leq c$

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

{\begin{aligned}f(x)&=a_{0}+\sum _{n=1}^{\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}}\int _{-a}^{a}f(x)\,\mathrm {d} x\\a_{n}&={\frac {1}{a}}\int _{-a}^{a}f(x)\,\cos {\left({\frac {n\,\pi }{a}}\,x\right)}\,\mathrm {d} x\\b_{n}&={\frac {1}{a}}\int _{-a}^{a}f(x)\,\sin {\left({\frac {n\,\pi }{a}}\,x\right)}\,\mathrm {d} x\end{aligned}}

Example

$f(x)={\begin{cases}1&0\leq x\leq 1\\0&{\mbox{otherwise}}\end{cases}}\quad \quad x\in \left[-2,2\right]$

In this case, $a=2$ , so

$a_{0}={\frac {1}{2\cdot 2}}\int _{-2}^{2}f(x)\,\mathrm {d} x={\frac {1}{4}}$

$a_{n}={\frac {1}{2}}\int _{-2}^{2}f(x)\,\cos {\left({\frac {n\,\pi \,x}{2}}\right)}\,\mathrm {d} x={\frac {1}{2}}\int _{0}^{1}\,\cos {\left({\frac {n\,\pi \,x}{2}}\right)}\,\mathrm {d} x={\frac {1}{n\,\pi }}\,\sin {\left({\frac {n\,\pi }{2}}\right)}$

$b_{n}={\frac {1}{2}}\int _{-2}^{2}f(x)\,\sin {\left({\frac {n\,\pi \,x}{2}}\right)}\,\mathrm {d} x={\frac {1}{2}}\int _{0}^{1}\,\sin {\left({\frac {n\,\pi \,x}{2}}\right)}\,\mathrm {d} x={\frac {1}{n\,\pi }}\,\left(1-\cos {\left({\frac {n\,\pi }{2}}\right)}\right)$

Hence

f(x)={\frac {1}{4}}+\sum _{n=1}^{\infty }{\frac {\sin {\left({\frac {n\,\pi }{2}}\right)}}{n\,pi}}\,\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

\int _{-\pi +c}^{\pi +c}F(x)\,\mathrm {d} x=\int _{-\pi }^{\pi }F(x)\,\mathrm {d} x=\int _{0}^{\pi }F(x)\,\mathrm {d} x\,\!

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

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

{\begin{aligned}{\frac {\mathrm {d} G}{\mathrm {d} c}}&={\frac {\mathrm {d} }{\mathrm {d} c}}\left(\int _{0}^{\pi +c}F(x)\,\mathrm {d} x\right)-{\frac {\mathrm {d} }{\mathrm {d} c}}\left(\int _{0}^{c-\pi }F(x)\,\mathrm {d} x\right)\\&=F(\pi +c)-F(c-\pi )\\&=F(\pi +c)-f(c+\pi )\\&=0\end{aligned}}

quod erat demonstrandum

Periodic Extension

Suppose $f(x)$ is defined on $0\leq x\leq 2\pi$ 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

$f(t)=a_{0}+\sum _{n=1}^{\infty }a_{n}\,\cos {\left({\frac {2\pi \,n\,t}{T}}\right)}+b_{n}\,\sin {\left({\frac {2\pi \,n\,t}{T}}\right)}$
$a_{n}={\frac {1}{\left({\frac {T}{2}}\right)}}\int _{0}^{T}f(t)\,\cos {\left({\frac {2\pi \,n\,t}{n}}\right)}$
$b_{n}={\frac {1}{\left({\frac {T}{2}}\right)}}\int _{0}^{T}f(t)\,\sin {\left({\frac {2\pi \,n\,t}{n}}\right)}$