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Fourier Cosine series of cos(x)

  • Thread starter sarahisme
  • Start date
Hello peoples,

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

http://img133.imageshack.us/img133/472/picture8ox1.png [Broken]

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 [Broken]

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 [Broken] 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
 
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When you let [itex]\alpha\to\pi[/itex] then the interval you're computing the fourier series on becomes [itex][-\pi,\pi][/itex]. The original function you're given, [itex]\cos(x)[/itex] is exactly periodic on this interval, and so you should find that your fourier series has only coefficient, for [itex]n=1[/itex], which corresponds to [itex]cos(x)[/itex].
 
jpr0 said:
When you let [itex]\alpha\to\pi[/itex] then the interval you're computing the fourier series on becomes [itex][-\pi,\pi][/itex]. The original function you're given, [itex]\cos(x)[/itex] is exactly periodic on this interval, and so you should find that your fourier series has only coefficient, for [itex]n=1[/itex], which corresponds to [itex]cos(x)[/itex].
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
 

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