Thread: 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) + 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

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 ?

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

Darn, Latex is hard.

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