Fourier Cosine series of cos(x)

[sarahisme]

Hello peoples,

I think this is a trick question... well sort of :P

http://img133.imageshack.us/img133/472/picture8ox1.png

for part (a) i get that the cosine Fourier Series for f(x) = cos(x) to be:

http://img138.imageshack.us/img138/6114/picture9sq2.png

i hope that is ok, but its part (b) that is troubling me...

is all that happens as http://img138.imageshack.us/img138/7436/picture10gf7.png is that the cosine Fourier series of cos(x) goes to 0?

i am guessing i am missing some trick to this question?

Cheers! :D

Sarah

 

[jpr0]

When you let \alpha\to\pi then the interval you're computing the Fourier series on becomes [-\pi,\pi]. The original function you're given, \cos(x) is exactly periodic on this interval, and so you should find that your Fourier series has only coefficient, for n=1, which corresponds to cos(x).

 

[sarahisme]

When you let \alpha\to\pi then the interval you're computing the Fourier series on becomes [-\pi,\pi]. The original function you're given, \cos(x) is exactly periodic on this interval, and so you should find that your Fourier series has only coefficient, for n=1, which corresponds to cos(x).

ah yep i see now. i need to be more careful with the n = 1 term when alpha = pi

thanks for the help! :D

Sarah