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Math Help - Question about Fourier Series

  1. #1
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    Question about Fourier Series

    Hi

    Can someone explain these two cases

    i have a final Fourier series that is:
    f(t) \sim \frac{\pi}{2} + \frac{4}{\pi} \sum_{i=1}^\infty \frac{cos(2n+1)t}{(2n+1)^2} which was simplified from \frac{1}{\pi}[\frac{2}{n^2} - \frac{2(-1)^{n}}{n^2}].

    when n -> odd \frac{4}{\pi n^2}
    when n -> even 0

    and other one f(t) \sim \frac{8}{\pi} \sum_{i=1}^\infty \frac{sin(2n-1)t}{(2n-1)} which was simplified from \frac{2}{\pi n}(2-2(-1)^{n})

    when n -> even 0
    when n -> odd \frac{8}{\pi n}

    when do i use (2n+1) to replace n? and when do i use (2n-1) to replace n?

    P.S
    Last edited by Paymemoney; March 4th 2011 at 05:02 PM.
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  2. #2
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    For a start, if you are summing a function of n from 1 to infinity then the notation should say \sum_{n=1}^\infty\,, not \sum_{i=1}^\infty.

    If you want to sum over all the odd numbers, then you can either call them 2n1 and start at n=1, or you can use 2n+1 but then you must start at n=0.

    So in those Fourier series, the cosine sum should have the summation \sum_{n=0}^\infty\,, and the sine sum should have the summation \sum_{n=1}^\infty.
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  3. #3
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    ``````````````
    Last edited by Paymemoney; March 5th 2011 at 08:01 PM.
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  4. #4
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    Quote Originally Posted by Opalg View Post
    So in those Fourier series, the cosine sum should have the summation \sum_{n=0}^\infty\,, and the sine sum should have the summation \sum_{n=1}^\infty.
    Is this always the case or are there times were this does not apply?

    Ok, theres another question i have done and i don't understand why the answer i got for b_n when n is odd is
     \frac{-1}{\pi} \sum_{n=1}^\infty \frac{1}{2n} sin(2n\pi t)

    whereas my answer is:
    \frac{-1}{\pi} \sum_{n=1}^\infty \frac{1}{n} sin(n\pi t)


    The question is

     y   =   \left\{{-1 ---> {\text -1 \leq t < 0}\atop<br />
          {t -------->{\text-1 \leq t < 0}\atop<br />
   {f(t+2) ------> {\text {for all t}}}\right

    P.S
    Last edited by Paymemoney; March 6th 2011 at 12:25 AM.
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