MATH 414 Lecture 9

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Complex Fourier Series

Example

f(x)=x on interval Failed to parse (syntax error): {\displaystyle x \in ( 0, \2 \pi )}

Find complex series of form

f(x)=n=cneinx

cn=12π02πf(x)einxdx=12πxeinxdx

Integrating by parts gives

cn=12π(xeinxin|02π02πeinxindx)=1ine2πin

Observe that eiθ is 2π-periodic, and when θ=2πn, we get e2πni=1. Hence we can cancel that integrated term and simplify cn to just

cn=in


c0=12π02πxei0xdx=π


Therefore f(x)=π+n=;n0ieinxn

Piecewise Continuous Functions

Consider interval x[a,b]. 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:

f'L(x0)=f(x0)=limxx0(f(x)f(x0)xx0)f'R(x0)=f(x0+)=limxx0+(f(x)f(x0+)xx0)

Example: Square Wave

f(x)={1even multiples [0,π]0odd multiples [π,2π]

Left and right derivatives at points of discontinuities are both 0, but this does not mean the function is differentiable at x0!


Theorem. Let f(x) be a 2π-periodic piecewise smooth function, and let a0+n=1ancos(nx)+bnsin(nx) be its Fourier series.

Then limN(a0+n=1Nancos(nx)+bnsin(nx))={f(x)f(x) is point of continuityf(x+)+f(x)2f(x) is a jump discontinuity

Proof.

quod erat demonstrandum


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