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Power series identity

  1. Dec 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Show
    [tex]e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=-\infty}J_n(x)t^n[/tex]


    2. Relevant equations
    [tex]J_k(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{(n+k)!n!}(\frac{x}{2})^{2n+k}[/tex]


    3. The attempt at a solution
    Power series product
    [tex](\sum^{\infty}_{n=0}a_n)\cdot (\sum^{\infty}_{n=0} b_n)=\sum^{\infty}_{n=0}c_n[/tex]
    where
    [tex]\sum^n_{i=0}a_ib_{n-i}[/tex]
    [tex](\sum^{\infty}_{n=0}\frac{1}{n!}(\frac{x}{2})^nt^n)\cdot (\sum^{\infty}_{n=0} \frac{(-1)^n}{n!}(\frac{x}{2})^nt^{2i-n})=\sum^{\infty}_{n=0}c_n[/tex]
    where
    [tex]c_n=\sum^{n}_{i=0}\frac{(-1)^{n-i}}{i!(n-i)!}(\frac{x}{2})^nt^{2i-n}[/tex]
    So
    [tex]e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=0}\sum^{n}_{i=0} \frac {(-1)^{n-i}}{i!(n-i)!}(\frac{x}{2})^nt^{2i-n}[/tex]
    I don't see how to get from here
    [tex]e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=-\infty}J_n(x)t^n[/tex]
     
  2. jcsd
  3. Dec 24, 2012 #2

    haruspex

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    You'll need to extract t from the inner sum. To do that, you need to change the summation variables so that only one of them (the outer sum variable) shows up in t's exponent. It's a bit messy because it can take either sign, so it might be best to handle the ranges 2i>=n, 2i<n separately.
     
  4. Dec 24, 2012 #3
    Is there some other way to do it. Easier? Here I go from Taylor to Laurent series.
     
  5. Dec 24, 2012 #4

    haruspex

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    It's probably not that hard. Substitute for n with n=2i-k. Your double sum is over n>=i, so that becomes i>=k. Sum over i first, and see if that produces Jk(x). I doubt there's an easier way.
     
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