Evaluating sum using Fourier Series

  • Thread starter tedwillis
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  • #1
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First, I've had to find the Fourier series of [tex]F(t) = |sin(t)|,[/tex] which I've calculated as

[tex]f(t) = \frac{2}{\pi} + \sum_{n=1}^{\infty}\frac{4cos(2nt)}{\pi-4\pi n^2}[/tex]

I'm pretty sure that's right, but now I need to evaluate the sum using the above Fourier series:
[tex]\sum_{n=1}^{\infty}\frac{(-1)^n}{4n^2-1}[/tex]

I don't really have any clue about where to start.
 

Answers and Replies

  • #2
HallsofIvy
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Well, if
[tex]f(t)= \frac{2}{\pi}+ \sum_{n= 1}^\infty \frac{4cos(2nt)}{\pi- 4\pi n^2}[/tex]
it follows that
[tex]-\frac{4}{\pi}\sum_{n=1}^\infty \frac{cos(2nt)}{4n^2- 1}= f(t)- \frac{\pi}{2}[/tex]
so that
[tex]\sum_{n=1}^\infty \frac{cos(2nt)}{4n^2- 1}= \frac{\pi}{4}(\frac{\pi}{2}- f(t))[/tex]

Now, what should you choose t to be?
 

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