Solving for k With Gamma Function

[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.

 

[robert Ihnot]

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

 

[HallsofIvy]

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

 

[Matthollyw00d]

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

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

 

[Matthollyw00d]

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)