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Please help in Gamma function to series conversion.

  1. Nov 22, 2009 #1
    Can you show me how to get the series representation of [tex]\Gamma[/tex](n-3/2+1)?

    For example [tex]\Gamma[/tex](n+3/2+1)=[tex]\frac{(2n+3)(2n+1)!}{2^{2n+2}.n!}[/tex].

    I cannot figure out how to write a series with:

    n=0 => [tex]\Gamma[/tex](0-3/2+1)= -2[tex]\sqrt{\pi}[/tex]

    n=1 => [tex]\Gamma[/tex](1-3/2+1)= [tex]\sqrt{\pi}[/tex]

    n=2 => [tex]\Gamma[/tex](2-3/2+1)= 1/2[tex]\sqrt{\pi}[/tex]

    n=3 => [tex]\Gamma[/tex](3-3/2+1)= 4/3[tex]\sqrt{\pi}[/tex]

    This is not homework. I have spent 2 days on this and can't figure it out!!!


  2. jcsd
  3. Nov 22, 2009 #2


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    If I understand you correctly, the trick is to replace n by n-3:
    [tex]\Gamma(n - \frac 3 2 +1) = \Gamma((n-3) + \frac 3 2 + 1) =\frac{(2(n-3)+3)(2(n-3)+1)!}{2^{2(n-3)+2}(n-3)!}=\frac{(2n-3)(2n-5)!}{2^{2n-4}(n-3)!}[/tex]

    At least I think this works for [itex]n \ge 3[/itex].
  4. Nov 22, 2009 #3
    Thanks for trying. But that was where I got really stuck. I can easily for find the series representation for n=1,2,3,4......... It is the n=0 that is the problem. The series solution has to cover n=0,1,2.......

    Actually I am working on the problem is to show

    J-3/2(x)=[tex]\sqrt{\frac{2}{\pi x}}[/tex][[tex]\frac{-cos(x)}{x}[/tex]-sin(x)]

    Where this is solution of Bessel function.
  5. Nov 23, 2009 #4
    Please, even if you have any suggestion, I would like to listen to it. I am very despirate!!!! I absolutely ran out of ideas!!!!
  6. Nov 23, 2009 #5


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  7. Nov 23, 2009 #6

    I am studying on my own, this question only on the first section, before a lot of the information your link shows. I have to spend some time looking at this. Yes I did look at this before.

    The question in the book ask both J(3/2) and J(-3/2). The possitive is relative easy because the gamma function always stay possitive.
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