Fourier Example: Difference between revisions

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<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>
<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math>
[[Image:Fourier.gif]]


==References:==
==References:==
[[Fourier Series: Basic Results]]
[[Fourier Series: Basic Results]]

Revision as of 01:36, 19 January 2010

Find the Fourier Series of the function:


f(x)={0,−π≤x<0π,0≤x≤π


Solution


Here we have



ao=12π(∫−π00dx+∫0ππdx)=π2


an=∫0ππcos(nx)dx=0,n≥1,


and


bn=∫0ππsin⁡(nx)dx=1n(1−cos(xπ))=1n(1−(−1)n)


We obtain b2n = 0 and


b2n+1=22n+1


Therefore, the Fourier series of f(x) is

f(x)=π2+2(sin(x)+sin(3x)3+sin(5x)5+...)

References:

Fourier Series: Basic Results