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Math Help - parseval law prove

  1. #1
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    parseval law prove

    http://i50.tinypic.com/a9pkpe.gif

    1.cant see how the input in *
    gives the result
    ??

    there is no mathmatical description
    Last edited by transgalactic; December 24th 2009 at 08:37 AM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    http://i50.tinypic.com/a9pkpe.gif

    1.cant see how the input in *
    gives the result
    ??

    there is no mathmatical description

    1) Define everything that appears in that file: what's f, e_k, etc.

    2) Write clearer

    3) Write way, waaaay larger: evern with the + is hard to distinguish what's going on there..

    4) If possible, much preferable to use LaTeX to write mathematics.

    Tonio
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  3. #3
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    i cant get this expression
    http://i50.tinypic.com/2aq6pu.jpg

    why when we input this basis we get this expression

    i cant see how its done mathematickly
    Last edited by transgalactic; December 24th 2009 at 12:09 PM.
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    i cant get this expression
    http://i50.tinypic.com/2aq6pu.jpg
    The functions \frac{1}{\sqrt(2\pi)}, \frac{cos(nx)}{\sqrt{2\pi}}, and \frac{sin(nx)}{\sqrt{2\pi}} form an "orthonormal basis". That means specifically, that that the "inner product", defined here as the integral of the product of two functions, is 0 for any two distinct functions of this set and 1 for the product of a function with itself. If f(x) is written as a combination of those functions, multiplying it by itself and then integrating gives the coefficient, squared, for each such function it self- that's the reason for the sum of " a_n^2+ b_n^2.

    I don't understand your reference to "two sums". \sum_{n=0}^\infty{a_n cos(nx)}+ \sum_{n=0}^\infty b_n cos(nx) can as easily be written as the single sum [tex]\sum_{n=0}^\infty (a_n sin(nx)+ b_n cos(nx)). Similarly \sum_{n=0}^\infty (a_n^2+ b_n^2) can be written as the two sums \sum_{n=0}^\infty a_n^2+ \sum_{n=0}^\infty b_n^2.
    Last edited by Plato; December 24th 2009 at 12:37 PM. Reason: LaTeX
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  5. #5
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    ok why then we have the norm divided by pi
    ?
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  6. #6
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    We don't divide the norm by pi, we divide each sine or cosine by \sqrt{\pi}. That way, when we multiply one by itself and integrate, to find the norm, we have the entire product multiplied by \pi. Of course the integral of sin^2(nx) is ]int_0^{2\pi} sin^2(nx)dx= \frac{1}{2}\int_0^{2\pi} (1- cos(2nx))dx which gives \frac{1}{2}(x- \frac{1}{2}sin(2nx)) evaluated between 0 and 2\pi. Since sin(2nx) is 0 at 0 and at 2\pi, that integral is \pi. A similar calculation gives the same result for cos(nx). Dividing each "basis" function by \sqrt{\pi} divides the integral of the square by \pi and so gives an integral (norm) of 1.

    If we start with the "constant" function, 1, the integral is \int_0^{2\pi} 1 dx= 2\pi so we divide that by \sqrt{2\pi}.

    That way our "basis" is an orthonormal basis.
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  7. #7
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    thanks
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