How Do You Solve Gamma Function Identities?

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SUMMARY

The discussion focuses on solving identities related to the Gamma function, specifically the identities involving negative integers and half-integers. The third identity, Γ(-k+1/2) = (2√π)/(2^{-2k}) (Γ(-2k)/Γ(-k)), and the first identity, Γ(1+(-k-1/2)) = (-k-1/2)Γ(-k-1/2), are explored but initially do not yield satisfactory results. The participant suggests that combining these identities with an induction argument may provide a clearer path to a solution.

PREREQUISITES
  • Understanding of Gamma function properties and identities
  • Familiarity with mathematical induction techniques
  • Basic knowledge of complex analysis concepts
  • Experience with mathematical problem-solving strategies
NEXT STEPS
  • Study the derivation and applications of Gamma function identities
  • Learn about mathematical induction and its use in proofs
  • Explore advanced topics in complex analysis related to the Gamma function
  • Investigate other special functions and their relationships with the Gamma function
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Students and researchers in mathematics, particularly those focusing on complex analysis, special functions, and mathematical proofs involving the Gamma function.

binbagsss
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Homework Statement



To show:

gammques.png

Homework Equations



gamide.png

The Attempt at a Solution


To be honest, I'm pretty stuck.

I could try to use the third identity:
##\Gamma(-k+\frac{1}{2})=\frac{2\sqrt{\pi}}{2^{-2k}}\frac{\Gamma(-2k)}{\Gamma(-k)} ##

but this doesn't really seem to get me anywhere.

I could also try to use the first identity, by adding and subtracting a 1/2:

##\Gamma(1+(-k-\frac{1}{2}))=(-k-\frac{1}{2})\Gamma(-k-\frac{1}{2})##

Which again doesn't seem to help..

Thanks in advance.
 
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binbagsss said:

Homework Statement



To show:

View attachment 211112

Homework Equations



View attachment 211111

The Attempt at a Solution


To be honest, I'm pretty stuck.

I could try to use the third identity:
##\Gamma(-k+\frac{1}{2})=\frac{2\sqrt{\pi}}{2^{-2k}}\frac{\Gamma(-2k)}{\Gamma(-k)} ##

but this doesn't really seem to get me anywhere.

I could also try to use the first identity, by adding and subtracting a 1/2:

##\Gamma(1+(-k-\frac{1}{2}))=(-k-\frac{1}{2})\Gamma(-k-\frac{1}{2})##

Which again doesn't seem to help..
It (the last version) helps a lot if you combine it with an induction argument.
 

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