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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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- Thread starter Matthollyw00d
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- #1

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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.

- #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]

(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]

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- #3

HallsofIvy

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I must be completely misunderstanding the question.

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.

[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

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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?

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]

[tex]=\pi divided by sin(\pi(z))[/tex]

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- #6

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B((n/2) + 1, 1/2)

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