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Gamma Function

  1. Nov 26, 2009 #1
    k*Γ((n-1)/2 + 1)=Γ(n/2 + 1)

    I need to solve for k, and I'm having some difficulty manipulating the gamma function to obtain my desired result. Any properties, hints or help would be greatly appreciated.
     
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  3. Nov 26, 2009 #2
    The Gamma function relates to the factoria function for a positive integer by [tex]\Gamma
    (n+1) =n![/tex] If things like n/2 are a problem, we have [tex]\Gamma((n+1)/2+1)=(n+1)/2*\Gamma\((n+1)/2)[/tex]
     
    Last edited: Nov 26, 2009
  4. Nov 26, 2009 #3

    HallsofIvy

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    I must be completely misunderstanding the question.
    [tex]k= \frac{\Gamma(\frac{n}{2}+ 1)}{\Gamma(\frac{n-1}{2}+ 1)}[/tex]
    What more do you want? To reduce the right side to a single gamma function?
     
  5. Nov 26, 2009 #4
    Yes, sorry. Obviously that could be a solution and I'll live with that solution if it's the best I can get; however, I'm pretty sure k can be reduced to just a simple expression of n without the Gamma function hanging around.

    And Robert Ihnot, that's pretty much all I've been using and a bit of the Beta Function, but was unable to get very far last night. I kept getting a Γ(-1/2) and I can't work with that.
     
  6. Nov 26, 2009 #5
    You can work with that using Euler's Reflective formula: [tex]\Gamma(1-z)\Gamma(z) [/tex]
    [tex]=\pi divided by sin(\pi(z))[/tex]
     
    Last edited: Nov 26, 2009
  7. Nov 26, 2009 #6
    As it turns out I had made an error early on in the problem and it turns out I need to find k for k*Γ((n+1)/2 + 1)=Γ(n/2 + 1) instead of k*Γ((n-1)/2 + 1)=Γ(n/2 + 1). Which now seems much more promising and I should be able to find a solution with the Beta Function. Now I just need to find the solution to
    B((n/2) + 1, 1/2)
     
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