Gamma Function

  • #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.
 

Answers and Replies

  • #2
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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:
  • #3
HallsofIvy
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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.
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?
 
  • #4
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?

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.
 
  • #5
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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]
 
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  • #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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