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Math Help - Fourier series

  1. #1
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    Fourier series

    How do they simplify this l don't understand.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nyasha View Post
    How do they simplify this l don't understand.
    It doesn't. Take n=1.
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  3. #3
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    I would agree with Drexel28 that the expression on the LHS does not simplify to the RHS. However, n = 1 is an identity, so far as I can see:

    \dfrac{8}{1\pi}\,\sin(\pi/2)=8/\pi, for the LHS.

    \dfrac{8}{1\pi}\,\dfrac{(-1)^{2}}{2-1}=8/\pi, for the RHS.

    A better number is n=2, because then you get a zero on the LHS, and there's no way to get a zero on the RHS.

    You can simplify the LHS, but not the way they've done. As it turns out, the LHS is zero whenever n is even. So substitute 2k-1=n to obtain

    \dfrac{8}{(2k-1)\pi}\sin((2k-1)\pi/2)=\dfrac{8(-1)^{k-1}}{(2k-1)\pi}.

    The last step there you can do because the sin function oscillates between plus and minus one at odd-integer multiples of \pi/2.

    Make sense?
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  4. #4
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    error
    Last edited by Archie Meade; November 30th 2010 at 05:09 AM.
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  5. #5
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    Quote Originally Posted by nyasha View Post
    How do they simplify this l don't understand.
    2n-1 is always odd for n=1, 2, 3.....

    \displaystyle\frac{8}{n{\pi}}sin\left[\frac{n{\pi}}{2}\right]=\frac{8}{n{\pi}}(-1)^k

    for n odd, as it is zero when n is even.

    k=even for multiples of {\pi} being \frac{4n-3}{2}

    k=odd for multiples of {\pi} being \frac{4n-1}{2}


    sin\frac{\pi}2}=sin\frac{5{\pi}}{2}=sin\frac{9{\pi  }}{2}=....=sin\frac{(4n-3){\pi}}{2}=1,\;\;\;n=1,\;2,\;3,....

    sin\frac{3{\pi}}{2}=sin\frac{7{\pi}}{2}=sin\frac{1  1{\pi}}{2}=.....=sin\frac{(4n-1){\pi}}{2}=-1,\;\;\;n=1,\;2,\;3,.....


    As the evaluation of Fourier co-efficients is involved, should you have something similar to

    \displaystyle\frac{8}{n{\pi}}\frac{sin\frac{(2n-1){\pi}}{2}}{2n-1}=\frac{8}{n{\pi}}\frac{(-1)^{n+1}}{2n-1} ?

    or

    \displaystyle\frac{8}{n{\pi}}\frac{sin(n{\pi}-\frac{\pi}{2})}{2n-1}=\frac{8}{n{\pi}}\frac{(-1)^{n+1}}{2n-1}
    Last edited by Archie Meade; December 1st 2010 at 02:27 AM.
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