Gamma Function < proving

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  • #1
homad2000
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Hello, I need help proving this:

[URL]http://mathworld.wolfram.com/images/equations/GammaFunction/Inline177.gif[/URL] = [URL]http://mathworld.wolfram.com/images/equations/GammaFunction/Inline179.gif[/URL]

Homework Statement





Homework Equations





The Attempt at a Solution

 
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Answers and Replies

  • #3
Dick
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You should be able to find gamma(1/2) easily enough. A change of variable makes the integral representation a gaussian. Now use gamma(z+1)=z*gamma(z) to find gamma(-1/2). Use that to find gamma(-3/2) etc. Use induction for the general case.
 
  • #4
dextercioby
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I'm so dumb, it can be proved by induction, of course.

[tex] \Gamma\left(\frac{1}{2}-n\right) = \frac{(-1)^{n} 2^{n}}{(2n-1)!!} \sqrt{\pi} [/tex]

Induction says that, if, for a specific n, P(n) is true and you manage to show that P(n+1) is true as well, then P(n) is true for all n from N.

So

[tex] \Gamma\left(\frac{1}{2}-(n+1)\right) = \Gamma\left(\left(\frac{1}{2}-n\right) -1\right) = \frac{1}{\left(\frac{1}{2}-n\right)-1}}\Gamma\left(\frac{1}{2}-n\right) = \frac{(-1)\cdot 2}{(2n+1)} \Gamma\left(\frac{1}{2}-n\right) [/tex]

Now plug what it's in the hypothesis, and you'll get your answer.
 
  • #5
dextercioby
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Now use what I wrote to get a proof of your own for the "sister" equality

[tex] \Gamma\left(\frac{1}{2}+n\right) = \frac{(2n-1)!!}{2^n}\, \sqrt{\pi} [/tex]
 
  • #6
homad2000
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brilliant!! i haven't thought of induction!! but i solved it using another identity:

%20formula%20}\operatorname{\Gamma}(\alpha)\operatorname{\Gamma}(1-\alpha)=\pi\,\csc\,\pi\alpha..gif
 

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  • %20formula%20}\operatorname{\Gamma}(\alpha)\operatorname{\Gamma}(1-\alpha)=\pi\,\csc\,\pi\alpha..gif
    %20formula%20}\operatorname{\Gamma}(\alpha)\operatorname{\Gamma}(1-\alpha)=\pi\,\csc\,\pi\alpha..gif
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  • #7
dextercioby
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You wrote something that came out red as an error message. You can write it with our LaTeX code, or you may put a screenshot of your work.
 
  • #8
homad2000
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using the identity:[tex]\Gamma[/tex](x) [tex]\Gamma[/tex](1-x) = [tex]\pi[/tex] / sin(pi *x)
 
  • #9
dextercioby
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Yes, it's obviously easier that way.
 

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