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Math Help - Bessel function J1

  1. #1
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    Bessel function J1

    Bessel function J1-bessel.jpg
    Last edited by mr fantastic; August 16th 2010 at 10:10 PM. Reason: Re-titled.
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  2. #2
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    You have the Bessel function of the first kind

    \displaystyle J_1(x) = \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n!(n+1)!2^{2n+1}}

    The first derivative is

    \displaystyle J_1'(x) = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!(n+1)!2^{2n+1}}\frac{d}{dx}x^{2n+1} = \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)}{n!(n+1)!2^{2n+1}}x^{2n}

    Analogously, the second derivative is

    \displaystyle J_1''(x) = \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)}{n!(n+1)!2^{2n+1}}\frac{d}{dx}x^{2n} = \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)(2n)}{n!(n+1)!2^{2n+1}}x^{2n-1}

    Plugging those in in the equation you will get


    \displaystyle  \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)(2n)}{n!(n+1)!2^{2n+1}}x^{2n+1} + \sum_{n=0}^{\infty}\frac{(-1)^n(2n+1)}{n!(n+1)!2^{2n+1}}x^{2n+1} +(x^2-1)\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n!(n+1)!2^{2n+1}}

    All you need to do now is multiply out the x^2-1 on the third series, equate the summing ranges for all three series (you will have to take out the first 2 terms of the first two series) and then you should get some cancellation and eventually 0.
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