How Can Fourier Series Demonstrate the Summation Identity Ʃ1/(2m+1)^2 = pi^2/8?

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zheng89120
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Homework Statement



Consider the function:

f(x) = {0 if 0<x<L/2
x-L/2 if L/2<x<L}

Define a periodic extension, obtain the complex Fourier series, and show that Ʃ1/(2m+1)^2 = pi^2/8...

Homework Equations



complex Fourier series

The Attempt at a Solution



I defined it as an even function by reflecting the function over the y-axis.

I did some calculations which yielded a complex Fourier series coefficient of:

cn = L[ exp(-i*pi*n)/(-2i*pi*n) + exp(-i*pi*n/2)/(pi2*n2) ]

not sure if this is correct, and how to get the fact that Ʃ1/(2m+1)2 = pi2/8

P.S. Sorry I forgot to add that they wanted: Define a periodic extension over period 2L
 
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zheng89120 said:

Homework Statement



Consider the function:

f(x) = {0 if 0<x<L/2
x-L/2 if L/2<x<L}

Define a periodic extension, obtain the complex Fourier series, and show that Ʃ1/(2m+1)^2 = pi^2/8...

Homework Equations



complex Fourier series

The Attempt at a Solution



I defined it as an even function by reflecting the function over the y-axis.
I don't think this is what they had in mind.

By extending the function, I believe they wanted you to repeat the same pattern that's in the interval [0, L].
zheng89120 said:
I did some calculations which yielded a complex Fourier series coefficient of:

cn = L[ exp(-i*pi*n)/(-2i*pi*n) + exp(-i*pi*n/2)/(pi2*n2) ]

not sure if this is correct, and how to get the fact that Ʃ1/(2m+1)2 = pi2/8