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Problem for Fourier series

  1. Dec 6, 2005 #1
    [​IMG] [​IMG]
    It's solution is [​IMG]
    What I know is that insert 2n-1 instead of n
    plz... explain to me
    I don't know this reason

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    Last edited: Dec 6, 2005
  2. jcsd
  3. Dec 6, 2005 #2


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    Please make an effort to provide a more complete statement of your problem.
  4. Dec 6, 2005 #3
    First I'm sorry that I can't write english well.
    I have a problem's solution.
    But I don't understand solution
    For example Another problem's solution is n instead of 2n-1
    But this problem's solution uses 2n-1 instead of n
    I understood that 2n-1 is odd
    but why do 2n-1 instead of n ?
  5. Dec 6, 2005 #4


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    When x= 1 or -1, 1-|x|= 0. Suppose you had just [itex]\frac{n\pi}{2}x[/itex]. Then for even values of n, x= 1, you would have [itex]cos(m\pi)[/itex] (m= n/2) which is either -1 or 1. By restricting the numerator to be odd, you are making sure that you always have, for x= 1, [itex]cos(\frac{\pi}{2})[/itex], [itex]cos(\frac{3\pi}{2})[/itex], etc.

    By the way, how does one do Fourier Series in precalculus?
    Last edited: Dec 7, 2005
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