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Fourier Cosine series of cos(x)

  1. Oct 14, 2006 #1
    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

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 14, 2006 #2
    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].
  4. Oct 14, 2006 #3
    ah yep i see now. i need to be more careful with the n = 1 term when alpha = pi

    thanks for the help! :D

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