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Math Help - Showing the bessel function is entire

  1. #1
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    Showing the bessel function is entire

    Please refer to attached image.

    I guess just by looking at it, you can tell it's always defined and thus always differentiable.

    However, I'm not precisely sure how to show this? I was thinking perhaps applying the Louiville's theorem.

    Do I have to evaluate the sum, and show that it converges? I really could use some direction, as I am unsure which theorems to apply,
    or further yet, how to apply them to a sum.

    As for the second question, I tried applying the quotient rule, but things got really messy and I couldn't get the answer. Was this the incorrect method, or just me making a silly mistake somewhere in my working? Surely there is a way to simplify the expression.

    Any help is very much appreciated!

    Thanks
    Attached Thumbnails Attached Thumbnails Showing the bessel function is entire-exercise-q.jpg  
    Last edited by 99.95; September 23rd 2013 at 01:41 AM.
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  2. #2
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    Re: Showing the bessel function is entire

    Hey 99.95.

    You could show that the series converges by say a ratio test. If you show convergence for all z then you have shown its entire.

    As for showing that it satisfies the DE, do a term by term differentiation twice and then show that all terms of the infinite series cancel out when you substitute the derivatives in with the DE equation. (Remember d/dz z^n = nz^(n-1) even when z is complex).
    Thanks from 99.95 and topsquark
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  3. #3
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    Re: Showing the bessel function is entire

    Moment of clarification,
    in the given sum/function, are both j and x treated as variables, i.e will I need to do partial differentiation, or, are j and n parameters and x is the only variable?
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  4. #4
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    Re: Showing the bessel function is entire

    Only z is a true variable: you must fix the order value n to be some constant.
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