Consider a 2pi-periodic function f(x) = |x| for -pi ≤ x ≤ pi
a) Compute the Fourier series of the function f.
b) Prove that (from n=1 to n=infinity)∑ 1/(2k-1)^2 = pi^2/8.
**note all "sums" from here on out will be defined from n = 1 to n=infinity
The Attempt at a Solution
For part a we start with the definition of the Fourier series
f(x) = 1/2*a_0 + ∑(a_n*cos(n*x)+b_n*sin(n*x))
Since f is an even function, we know that b_n = 0.
a_n = (2/pi)*integral from 0 to pi of (x*cos(nx))dx
a_n = 2((-1)^n-1)/(pi*n^2)
a_0 = 1/pi * integral from -pi to pi of |x| dx = pi
So we have the following fourier series for f(x) and the answer for part a:
pi/2 + ∑2((-1)^n-1)/(pi*n^2) * cos(nx)
For part b, we set x = 0, and find
|x| = pi/2 + ∑2((-1)^n-1)/(pi*n^2) * cos(nx)
=> -pi/2 = ∑2((-1)^n-1)/(pi*n^2)
=> -pi^2/4 = ∑((-1)^n-1)/n^2
Divide both sides by -2
=> pi^2/8 = ∑((-1)^n-1)/(-2*n^2)
But I'm not quite sure how to get the right side (the sum) similar to ∑ 1/(2k-1)^2