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- Thread starter misterme09
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nicksauce

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if n = 1, then it's the integral of sin^2x, which is certainly not zero over -pi to pi.

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That's true. Okay, but for bn where n isn't 1, bn is zero, correct?

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nicksauce

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That's true. Okay, but for bn where n isn't 1, bn is zero, correct?

Correct

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A fourier series is a series that has a period equal to part of a qualified f(x) from -L to L or 0 to 2l.

Yes, the fourier series of Sin[x] with L = 2n Pi where n is an element of Z is simply Sin[x].

But the fourier series of Sin[x] with L =/= 2n Pi is NOT Sin[x].

Look at the graph of Sin[x] from 0 to Pi/2. If you wanted a function that repeated that period, Sin[x] would no-where near work.

The first few terms of that fourier series would be: (4 Cos[1/2] Sin[x])/(-1 + \[Pi]^2) + (

8 Cos[1] Sin[2 x])/(-4 + \[Pi]^2) + (

12 Cos[3/2] Sin[3 x])/(-9 + \[Pi]^2)

According to mathematica. (It's slightly incorrect, since the outcome of the periods are off. Working on it)

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