I need to simplify Γ(2α) in Γ(α) terms where α is a real number greater than 1. (I need to cancel out Γ( . ) terms in a simplification). Shall accept a product term with Γ(α), like (2α - 1)*...αΓ(α). Thanks in advance.
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What's the actual question? – Angina Seng Feb 27 '19 at 06:01
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So you want help proving that $\Gamma(2\alpha)=(2\alpha-1)\cdots\alpha\Gamma(\alpha)$? – Arthur Feb 27 '19 at 06:04
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@LordSharktheUnknown I need to cancel out Γ( . ) terms in a simplification – Dovini Jayasinghe Feb 27 '19 at 06:06
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@Arthur Of course yes – Dovini Jayasinghe Feb 27 '19 at 06:06
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Take the product $$ (2\alpha-1)\cdots\alpha\Gamma(\alpha) $$ and work your way from the right. The two rightmost terms are $\alpha\Gamma(\alpha)$. By the defining property of the $\Gamma$ function, this is equal to $\Gamma(\alpha+1)$.
Write down the total product after this simplification. What are the two rightmost terms? Can that be simplified? What is the resulting total product? Can that be simplified? How far can you keep going? What's the end result?
PS. This does, of course, assume that $\alpha$ is a positive integer.
Arthur
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Thanks for the answer. But in my situation, α is a real number which is greater than 1. It's known that Γ(α+1) = αΓ(α) works for any α, either it's positive integer valued or positive real valued. The following link provides the formula. (page 2 / 20 ) https://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf – Dovini Jayasinghe Feb 27 '19 at 06:19
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(2α−1)(2α−2)(2α−3)⋯(α+2)(α+1)αΓ(α) makes no sense how it works for a real valued α – Dovini Jayasinghe Feb 27 '19 at 06:21
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@DoviniJayasinghe Yes, that relation still holds for any valid input. But if $\alpha$ is not a natural number, then note that the identity we're after is actually of the form $\Gamma(2\alpha)=(2\alpha-1)\cdots(2\alpha-n)\Gamma(2\alpha-n)$. The last term here won't be the $\alpha$ you're looking for. – Arthur Feb 27 '19 at 06:22
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