# 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:

Related Calculus and Beyond Homework Help News on Phys.org
BvU
Homework Helper
2019 Award
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) ?

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