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Writing this series as a hypergeometric series

  1. Apr 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Write [tex]\displaystyle \sum_{k=0}^{\infty} \frac{1}{9^k (\frac{2}{3})_k} \frac{w^{3k}}{k!}[/tex] in terms of the Gauss hypergeometric series of the form [itex]_2 F_1(a,b;c;z)[/itex].

    2. Relevant equations

    The Gauss hypergeometric series is http://img200.imageshack.us/img200/5992/gauss.png [Broken]

    3. The attempt at a solution

    It's nearly a series of that form if I put [itex]z=w^3[/itex] and [itex]k=n[/itex] but how do I get the [itex]9^{-k} = 3^{-k}3^{-k}[/itex] factors in terms of shifted factorials (that is if I need to)?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 5, 2012 #2
    That term does not go into the factorials, it goes into z^n.
  4. Apr 5, 2012 #3
    Ah of course. So if I put [itex]z=\frac{w^3}{9}[/itex] then the series can be written as [itex]_2 F_1 (a,b ; \frac{2}{3} ; \frac{w^3}{9})[/itex] for some [itex]a[/itex] and [itex]b[/itex] with [itex](a)_n(b)_n = 1[/itex] for all [itex]n=0,1,2,...[/itex] Can I just pick [itex]a=b=0[/itex]?
  5. Apr 5, 2012 #4
    I would do some extra checking to be sure that that's right. Can you plug the solution into the hypergeometric differential equation with a=b=0 and see if it solves it?
  6. Apr 5, 2012 #5
    Actually [tex](0)_n (0)_n \neq 1[/tex] for n=0,1,2,... so how do I get 2 shifted factorials to equal 1?
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