# How to prove $(1-2x)^2=1/3+4/\pi^2\sum_1^\infty \cos(2n x \pi)/n^2$ for $x \in [0,1)$?

I tried to prove that $$(1-2x)^2=1/3+4/\pi^2\sum_1^\infty \cos(2n x \pi)/n^2$$ for $x \in [0,1)$ with Fourier analysis, but I just found a Fourier series which defines the function. I also found the fourier series of $\cos(2n x \pi)$.

I don't think these results are helpful.

Any suggestions on how to prove this equation?

-
I don't know how to do it... –  tomerg Jan 7 '12 at 8:26
Do you know about cosine Fourier series? –  Hans Lundmark Jan 7 '12 at 8:30
Briefly (......) –  tomerg Jan 7 '12 at 8:36

-