Fourier Example: Difference between revisions
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\end{cases}</math> |
\end{cases}</math> |
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''' |
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== Solution == |
== Solution == |
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''' |
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Here we have |
Here we have |
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⚫ | |||
<math>a_o=\frac{1}{2\pi}(\int_{-\pi}^00\ dx+\int_{0}^\pi\pi\ dx)=\frac{\pi}{2}</math> |
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<math>a_n=\int_{0}^\pi\pi cos(nx)\ dx=0, n\ge1,</math> |
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and |
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<math>b_{2n+1}=\frac{2}{2n+1}</math> |
<math>b_{2n+1}=\frac{2}{2n+1}</math> |
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Therefore, the Fourier series of f(x) is |
Therefore, the Fourier series of f(x) is |
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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> |
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==Solution Graph== |
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[[Image:Fourier.gif]] |
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==***BONUS*** VIDEO! (For people completely lost on Fourier Series)== |
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http://www.youtube.com/watch?v=nXEqrOt-nB8 |
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==References:== |
==References:== |
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[[Fourier Series: Basic Results]] |
[[Fourier Series: Basic Results]] |
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==Readers== |
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[[Lau, Chris|Christopher Garrison Lau I]] |
Latest revision as of 14:26, 25 January 2010
Find the Fourier Series of the function:
Solution
Here we have
and
We obtain = 0 and
Therefore, the Fourier series of f(x) is
Solution Graph
***BONUS*** VIDEO! (For people completely lost on Fourier Series)
http://www.youtube.com/watch?v=nXEqrOt-nB8