How to expand f(x)=x^2, -pi<x<pi as fourier series

[blueyellow]

Homework Statement

how to expand f(x)=x^2, -pi<x<pi as Fourier series
answer at back of the book says:
f(x)=((pi^2)/3)+4sigma(from 1 to infinity) [((-1)^n)/(n^2)]cos nx
i tried the a(n) b(n) stuff but i don't see where they get the n^2 in the denominator from

when i integrated to try to find a(n) i get n on the denominator

 

[HallsofIvy]

It seems pretty straight forward to me. In order to get rid of that "$x^2$" you will need to integrate by parts- twice. And each time you will get a 1/n integrating the trig function. If you want more detail on where you went wrong, we will have to see what you did. HOW did you integrate
$$\int_{-\pi}^\pi x^2 cos(x) dx$$?

 

[blueyellow]

integral of x^2 cos ((n*pi*x)/L) dx

using partial integration
integral=first term*integral of second term -[integral of (first term differentiated*integral of second term)]
first term=x^2
second term=cos...
=integral (from -pi to pi) x^2 cos((n*pi*x)/2pi) dx
=integral (from -pi to pi) x^2 cos((n*x)/2) dx
=[(2x^2)/n sin ((nx)/2)- integral of 2x sin (nx/2)] dx
=[(2x^2)/n sin ((nx)/2)- [-4x/n cos (nx/2) + 4/n cos (nx)]dx

putting in the limits
integral=(4pi^2)/n

a(n)=1/L *integral
=2pi/n

that term at the beginning of the answer is a0/2, right?
because if n=0 is put into 2pi/n, it'll be infinity!