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Hypergeometric function problem

  1. Jan 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate
    [tex]_2F_1(\frac{1}{2},\frac{1}{2},\frac{3}{2};x)[/tex]


    2. Relevant equations
    [tex]_2F_1(a,b,c;x)=\sum^{\infty}_{n=0}\frac{(a)_n(b)_n}{n!(c)_n}x^n[/tex]
    [tex](a)_n=a(a+1)...(a+n-1)[/tex]


    3. The attempt at a solution
    [tex](\frac{1}{2})_n=\frac{1}{2}\frac{3}{2}\frac{5}{2}...\frac{2n-1}{2}[/tex]
    [tex](\frac{3}{2})_n=\frac{3}{2}\frac{5}{2}\frac{7}{2}...\frac{2n+1}{2}[/tex]
    From this relations
    [tex]\frac{(\frac{1}{2})_n}{(\frac{3}{2})_n}=\frac{1}{2n+1}[/tex]
    But I don't see how to calculate this to the end...
     
  2. jcsd
  3. Jan 1, 2013 #2

    haruspex

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    Can you write [tex](\frac{1}{2})_n[/tex] in terms of factorials and powers of 2?
     
  4. Jan 1, 2013 #3
    Yes I could
    [tex](\frac{1}{2})_n=\frac{(2n-1)!!}{2^n}[/tex]
    [tex](\frac{3}{2})_n=\frac{(2n+1)!!}{2^{n-1}}[/tex]
    So I get
    [tex]_2F_1(\frac{1}{2},\frac{1}{2},\frac{3}{2};x)=\sum^{\infty}_{n=0}\frac{(2n-1)!!}{(2n+1)2^{n+1}n!}x^n=\sum^{\infty}_{n=0}\frac{(2n-1)!!}{2(2n+1)(2n)!!}x^n[/tex]
    And I don't know what to do know.
     
  5. Jan 1, 2013 #4

    lurflurf

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    Hint:

    [tex]\arcsin (x) \sim \sum_{k=0}^\infty \dfrac{(2k)!x^{2k+1}}{4^k (k!)^2 (2k+1)} \sim \sum_{k=0}^\infty \dfrac{(2k-1)!!x^{2k+1}}{(2k)!!(2k+1)}[/tex]

    I am not sure how to motivate this though. It can be difficult to recognize familiar functions by their power series.
     
  6. Jan 2, 2013 #5

    haruspex

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    Try writing out the general term of the expansion of (1-x)-1/2
     
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