# Fourier Analysis Question

Gold Member

## Homework Statement

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

Last edited:

## Answers and Replies

BvU
Science Advisor
Homework Helper
You wonder how ## {1\over 1} + 0 + {1\over 9} + 0 + {1\over 25} ... ## (n = 1, 2, 3, 4, 5) can be equal to
## {1\over 1} + {1\over 9} + {1\over 25} ... ## (k = 1, 2, 3) ?

• RJLiberator
Gold Member
Ah, I see.

So, in the end, I was doing things right here.

It was my inability to manipulate the problem to come to the conclusion.

Tough, tough course.

BvU
Science Advisor
Homework Helper
Good thing there is PF • RJLiberator