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Math Help - Deriving Bessel generating function

  1. #1
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    Exclamation Deriving Bessel generating function

    So I am trying to prove that

    exp\left[\frac{x}{2}(t-\frac{1}{t}\right]=\sum^{\infty}_{n=-\infty} J_{n}(x)t^{n}

    What I did so far is to set t=e^{i \theta} and I came down to something like that using fourier series
    exp\left[\frac{x}{2}(t-\frac{1}{t}\right]=\frac{2i\sin(\pi\sin \theta)}{\pi}\left[\frac{1}{2i\sin \theta} + \sum^{\infty}_{n=1} \frac{(-1)^{n}}{n^{2}-\sin^{2} \theta} (i\sin\theta\cos nx - n\sin nx)\right]

    I am not even sure it is correct or if it is the right way, I am just confused as to where to go from here. Any suggestions??? I really need help this thing is due soon. Any help is welcome!!!
    Last edited by ynn6871; April 29th 2009 at 10:38 AM.
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  2. #2
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    One hint

    We have:
    \exp{\frac{zt}{2}} = \sum_{m=0}^\infty \frac{z^m t^m}{2^m m!}

    and

    \exp\left(-\frac{z}{2t}\right) = \sum_{k=0}^\infty (-1)^k\frac{z^k t^{-k}}{2^k k!}

    Because the above series are uniformly convergent we have:
    \exp\frac{z}{2}\left(t - \frac{1}{t}\right) = \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{(-1)^k}{k!m!} \left( \frac{z}{2} \right)^{k+m} t^{m-k}

    From this point, you can easily see what to do
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  3. #3
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    Thanks, I can't believe I didn't think of that!!!
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  4. #4
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    Now using that generating function formula I am supposed to prove the following,

    J_{n}(x+y)=\sum^{\infty}_{k=-\infty} J_{k}(x)J_{n-k}(x)

    I came down to something like

    exp\left[\frac{(x+y)}{2}(t-\frac{1}{t})\right]= \sum^{\infty}_{n=-\infty} J_{n}(x)t^{n}\sum^{\infty}_{m=-\infty} J_{m}(y)t^{m}

    Again, I have no idea if I am going in the right direction or where to go from here in order to prove this thing. Any suggestion???
    Last edited by ynn6871; April 29th 2009 at 04:14 PM.
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  5. #5
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    It is a good start. You should get to the point where you have two power series: \sum \alpha_n t^n and \sum \beta_n t^n which are equal for t in some open set... Then we can say, that their coefficients are equal.

    The coefficients of this series shoud be (of course) J_{n}(x+y) and \sum^{\infty}_{k=-\infty} J_{k}(x)J_{n-k}(x)
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