Varification needed for small trigonometrical Fourier series, PRESSING

toneboy1

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 b_n
Also that T = 2pi so
a_n = 2/2pi ∫₀^2pi x(t).cos(nω₀t) dt

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

x(t) = 1/2 + Ʃ_{n = 1}² cos(nω₀t)

which would mean that a_n = 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:
http://www.physicsforums.com/showthread.php?t=654607

rude man

This has been adequately addressed already. Your x(t) is already a Fourier series. If you computed the coefficients to be different you made a math error.

toneboy1

Yeah, after this was inactive I posted it in the homework help section: http://www.physicsforums.com/showthr...=1#post4173472

berkeman

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.