MATH 414 Lecture 9

« previous | Monday, February 3, 2014 | next »

Complex Fourier Series

Example

$f(x)=x$ on interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \in ( 0, \2 \pi )}

Find complex series of form

$f(x)=\sum _{n=-\infty }^{\infty }c_{n}\,\mathrm {e} ^{i\,n\,x}$

$c_{n}={\frac {1}{2\pi }}\int _{0}^{2\pi }f(x)\,\mathrm {e} ^{-i\,n\,x}\,\mathrm {d} x={\frac {1}{2\pi }}x\,\mathrm {e} ^{-i\,n\,x}\,\mathrm {d} x$

Integrating by parts gives

${\begin{aligned}c_{n}&={\frac {1}{2\pi }}\left(\left.{\frac {x\,\mathrm {e} ^{-i\,n\,x}}{-i\,n}}\right|_{0}^{2\pi }-{\cancel {\int _{0}^{2\pi }{\frac {\mathrm {e} ^{i\,n\,x}}{-i\,n}}\,\mathrm {d} x}}\right)\\&={\frac {1}{-i\,n}}\,\mathrm {e} ^{-2\pi \,i\,n}\end{aligned}}$

Observe that $\mathrm {e} ^{i\,\theta }$ is $2\pi$-periodic, and when $\theta =2\pi \,n$, we get $\mathrm {e} ^{2\pi \,n\,i}=1$. Hence we can cancel that integrated term and simplify $c_{n}$ to just

$c_{n}={\frac {i}{n}}$

$c_{0}={\frac {1}{2\pi }}\int _{0}^{2\pi }x\,\mathrm {e} ^{-i\,0\,x}\,\mathrm {d} x=\pi$

Therefore $f(x)=\pi +\sum _{n=-\infty ;n\neq 0}^{\infty }{\frac {i\,\mathrm {e} ^{i\,n\,x}}{n}}$

Piecewise Continuous Functions

Consider interval $x\in \left[a,b\right]$. $f$ is piecewise continuous on $[a,b]$ if and only if $f$ is continuous except for a finite number of "jump" type discontinuities.

Piecewise Smooth Functions

$f$ is piecewise smooth (pws) on $[a,b]$ if and only if $f$ is piecewise continuous on $[a,b]$ and at the jumps, both left and right derivatives exist.

The left and right derivatives of a function $f$ are defined as follows:

${\begin{aligned}f'_{L}(x_{0})=f'(x_{0}^{-})&=\lim _{x\to x_{0}^{-}}\left({\frac {f(x)-f(x_{0}^{-})}{x-x_{0}}}\right)\\f'_{R}(x_{0})=f'(x_{0}^{+})&=\lim _{x\to x_{0}^{+}}\left({\frac {f(x)-f(x_{0}^{+})}{x-x_{0}}}\right)\end{aligned}}$

Example: Square Wave

$f(x)={\begin{cases}1&{\mbox{even multiples}}\ [0,\pi ]0&{\mbox{odd multiples}}\ [\pi ,2\pi ]\end{cases}}$

Left and right derivatives at points of discontinuities are both 0, but this does not mean the function is differentiable at $x_{0}$!

Theorem. Let $f(x)$ be a $2\pi$-periodic piecewise smooth function, and let $a_{0}+\sum _{n=1}^{\infty }a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}$ be its Fourier series.

Then $\lim _{N\to \infty }\left(a_{0}+\sum _{n=1}^{N}a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}\right)={\begin{cases}f(x)&f(x)\ {\mbox{is point of continuity}}{\frac {f(x^{+})+f(x^{-})}{2}}&f(x)\ {\mbox{is a jump discontinuity}}\end{cases}}$

Proof.

quod erat demonstrandum

Gibbs Phenomenon: Fourier series seems to diverge from function slightly just before function discontinuity