## Varification needed for small trigonometrical Fourier series, PRESSING

Hi,
I have x(t) = 1/2 + cos(t) + cos(2t)

so I can see that a0 = 1/2
and that it is an even function so there is no bn
Also that T = 2pi so
an = 2/2pi ∫02pi x(t).cos(nω0t) dt

but when I integrate this I get an = 0 yet I've been told that the answer is

x(t) = 1/2 + Ʃn = 12 cos(nω0t)

which would mean that an = 1, but I'm not sure how.

Anyone?

Thanks heaps!

EDIT: I just found another related problem I'm struggling with if anyone's interested:

 PhysOrg.com engineering news on PhysOrg.com >> Researchers use light projector and single-pixel detectors to create 3-D images>> GPS solution provides 3-minute tsunami alerts>> Single-pixel power: Scientists make 3-D images without a camera
 Yeah, after this was inactive I posted it in the homework help section: http://www.physicsforums.com/showthr...=1#post4173472

Mentor

## Varification needed for small trigonometrical Fourier series, PRESSING

 Quote by toneboy1 Yeah, after this was inactive I posted it in the homework help section: http://www.physicsforums.com/showthr...=1#post4173472
Please do not multiple post like that. If you wanted this thread moved, click the Report button on your post and ask the Mentors to move it.

 Tags fourier, series, trigonometric, varify