# Fourier series problem

1. Dec 2, 2013

### Nikitin

1. The problem statement, all variables and given/known data
https://wiki.math.ntnu.no/_media/tma4120/2013h/tma4120_h11.pdf

Check out the solution to problem 4b)

My question is: Why do they set $b_n = \frac{2}{\pi} \int_{0}^{\pi}(...)dx$ instead of $b_n = \frac{1}{\pi} \int_{0}^{\pi} (...)dx$?

Ie, why did they multiply the integral with 2? Did they find the fourier integral to the odd expansion of $sin(x)^2$? Because that's the only way for sin(x)^2 to be defined by solely a sine fourier expansion?

EDIT: Ooops, posted in wrong section. Please move to calculus and beyond homework forum! sorry!

Last edited: Dec 2, 2013
2. Dec 2, 2013

### LCKurtz

Yes, that's exactly it. Since they only have sine functions for the eigenfunction expansion, the FS must represent an odd function so they take the odd extension of $\sin^2(x)$ and use the half range formulas.