Fourier Series Q: periodic function f(x) = x^2 for -6 <= x <= 6

[Mujo]

Hi,

There is a problem that's giving me trouble.
For a periodic function f that has a period of 12, where f(x) = x^2 for -6 <= x <= 6,
then the Fourier series of f is the following: a0 + sum[n=1 to inf] an*cos(n*pi*x/6) + bn*sin(n*pi*x/6).

Then it is asked for a0, an and bn.


The work done so far:
I've figured that bn is 0 but I can't get a0 and an.
I've been trying to use the following formulas to no avail.

a0 = 1/L integral(-L to L) f(x) dx
an = 1/L integral(-L to L) f(x)cos(n*pi*x/L)dx

Thanks a lot!

 

[Deleted member 4993]

Hi,

There is a problem that's giving me trouble.
For a periodic function fthat has a period of 12, where f(x) = x^2 for -6 <= x <= 6,
then the Fourier series of f is the following: a0 + sum[n=1 to inf] an*cos(n*pi*x/6) + bn*sin(n*pi*x/6).

Then it is asked for a0, an and bn.


The work done so far:
I've figured that bn is 0 but I can't get a0 and an.
I've been trying to use the following formulas to no avail.

a0 = 1/L integral(-L to L) f(x) dx
an = 1/L integral(-L to L) f(x)cos(n*pi*x/L)dx

Thanks a lot!

Are you saying you cannot evaluate:

\(\displaystyle \displaystyle{\dfrac{1}{L}\int_{-L}^L x^2 dx}\) correctly?