Gamma Function

**Matthollyw00d** · Nov26-09, 02:44 AM

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.

**robert Ihnot** · Nov26-09, 03:32 AM

The gamma function relates to the factoria function for a positive integer by $LaTeX Code: \\Gamma <BR>(n+1) =n!$ If things like n/2 are a problem, we have $LaTeX Code: \\Gamma((n+1)/2+1)=(n+1)/2*\\Gamma\\((n+1)/2)$

**HallsofIvy** · Nov26-09, 05:19 AM

Originally Posted by Matthollyw00d

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.
$LaTeX Code: k= \\frac{\\Gamma(\\frac{n}{2}+ 1)}{\\Gamma(\\frac{n-1}{2}+ 1)}$
What more do you want? To reduce the right side to a single gamma function?

**Matthollyw00d** · Nov26-09, 01:05 PM

Originally Posted by HallsofIvy

I must be completely misunderstanding the question.
$LaTeX Code: k= \\frac{\\Gamma(\\frac{n}{2}+ 1)}{\\Gamma(\\frac{n-1}{2}+ 1)}$
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.

**robert Ihnot** · Nov26-09, 05:24 PM

You can work with that using Euler's Reflective formula: $LaTeX Code: \\Gamma(1-z)\\Gamma(z)$
$LaTeX Code: =\\pi divided by sin(\\pi(z))$

**Matthollyw00d** · Nov26-09, 10:11 PM

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)