Fourier Example: Difference between revisions
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Find the Fourier Series of the function: | Find the Fourier Series of the function: | ||
:<math>f(x) = \begin{cases}0,& -\pi<x<0\\ | :<math>f(x) = \begin{cases}0,& -\pi<x<0\\ | ||
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\end{cases}</math> | \end{cases}</math> | ||
:<math>b_n = \frac{1}{2 | :<math>b_n = \frac{1}{2\pi}\int_{0}^\pi \pi\sin(nx)\, dx, = \frac{1}{n}(1-cos(x\pi))=\frac{1}{n}(1-(-1)^n)</math> | ||
We obtain b_2n = 0 and | |||
<math>b_{2n+1}=\frac{2}{2n+1}</math> | |||
Therefore, the Fourier series of f(x) is | |||
<math>f(x)=\frac{\pi}{2}+2(sin(x)+\frac{sin(3x)}{3}+\frac{sin(5x)}{5}+...)</math> |
Revision as of 00:41, 19 January 2010
Find the Fourier Series of the function:
We obtain b_2n = 0 and
Therefore, the Fourier series of f(x) is