# Math Help - Fourier Series

1. ## Fourier Series

I want to find the fourier series for the following function:
f(t)= cost (t) t>=0 and = 0 cos(t)<0
Now since the function is 2pi periodic and even the sin coeff will be 0 and I need only to calculate the fourier cosine.
But what should I use for limits?Becuase the coefficient,defined by the integral between the function and cos(kt), will be equal to cos(kt)sin(t),which is 0 if the integral is defined from 0 to pi os 2pi=>the fourier cosine to be equal to 0.Any help?

2. Originally Posted by StefanM
I want to find the fourier series for the following function:
f(t)= cost (t) t>=0 and = 0 cos(t)<0

This is pretty unclear: did you mean f(t) = cos(t) when t >= 0 , and f(t) = 0 when t < 0, or you did mean f(t) = cos(t) whenever cos(t) >= 0 ,and

f (t) = 0 otherwise? I think it should be the second one since then the function is clearly periodic, but you better make it clearer.

Tonio

Now since the function is 2pi periodic and even the sin coeff will be 0 and I need only to calculate the fourier cosine.
But what should I use for limits?Becuase the coefficient,defined by the integral between the function and cos(kt), will be equal to cos(kt)sin(t),which is 0 if the integral is defined from 0 to pi os 2pi=>the fourier cosine to be equal to 0.Any help?
.

3. the second part.....f(t) = cos(t) whenever cos(t) >= 0 ,and f (t) = 0 otherwise

4. Originally Posted by StefanM
the second part.....f(t) = cos(t) whenever cos(t) >= 0 ,and f (t) = 0 otherwise

Ok, but then why do you think the limits of the integrals defining the coefficients are between 0 and 2*pi? If you sketch the function f(t) you can see it is

periodic 2*pi, but different from zero ONLY in (-pi/2 , pi/2) , so these must be the integrals' limits times 1/pi each.

Unfortunately we have no latex anymore so it's impossilbe for me to write down my mind, but if you do it you'll get the integral of cos(t)cos(nt) , which equals

sin(n-1)x/2(n-1) + sin(n+1)x/2(n+1) , and then you have to distinguish between n = 2 (mod 4) and n = 0 (mod 4) , which gives you

2/(n^2-1) and -2/(n^2-1), resp. (and all the time divided by pi)...

Tonio