Fourier Series, just a quick question

• Sep 24th 2010, 11:27 PM
ikurwae89
Fourier Series, just a quick question
Given f(t) = cos(t) for 0 < t < pi and f(t)= 0 for -pi<t<0

Find it's fourier series?

In the solution it has 1/2 cos(t) + \$\displaystyle Sigma\$2/pi (n/n^2-1) for no =2,4,6 ...

My only concern is how did they get 1/2 cos(t)???

Is the 1/2 cos(t) just a "fix up" term so that it covers aras such as t=0, t=pi etc ??

Thanks
• Sep 25th 2010, 03:25 PM
ikurwae89
I'll make my self more clear.

Suppose you were to find, the FS of f(t)= cos(t) for 0<t<pi and f(t)=0 -pi<t<0

We would get FS(t) = 1/2 Cos(t) + Sigma 2/pi (n/n^2-1) for n = even

Now how on earth do I get 1/2 cos(t) from my Integeral coefficients?

My A0 =0 An=0 but Bn= Sigma .....

?????

Unless I say, 1/2 cos(t) must be there due to the fact that at t=0 I must get f(t) = 1/2 ???? which follows from Dirichlets Theorem ?
• Sep 25th 2010, 03:38 PM
A Beautiful Mind
I don't come bearing the answer to your question, but I do think it would be easier for people who are knowledgeable enough to provide one would like your question formatted in LaTeX. It makes things much more easier to understand.

Latex Math Symbols
• Sep 28th 2010, 03:44 AM
ikurwae89
Darn, Latex is hard.

My only concern is why 1/2 cos(t) when Ao=0???