Please help in Gamma function to series conversion.

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SUMMARY

The discussion focuses on deriving the series representation of the Gamma function, specifically \Gamma(n - \frac{3}{2} + 1). The user, Alan, seeks assistance in formulating this series for various values of n, particularly for n=0, which yields -2\sqrt{\pi}. A proposed solution involves substituting n with n-3, leading to the expression \Gamma(n - \frac{3}{2} + 1) = \frac{(2n-3)(2n-5)!}{2^{2n-4}(n-3)!}, applicable for n ≥ 3. The conversation also touches on related Bessel functions, indicating a broader context of mathematical exploration.

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yungman
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Can you show me how to get the series representation of \Gamma(n-3/2+1)?


For example \Gamma(n+3/2+1)=\frac{(2n+3)(2n+1)!}{2^{2n+2}.n!}.


I cannot figure out how to write a series with:

n=0 => \Gamma(0-3/2+1)= -2\sqrt{\pi}

n=1 => \Gamma(1-3/2+1)= \sqrt{\pi}

n=2 => \Gamma(2-3/2+1)= 1/2\sqrt{\pi}

n=3 => \Gamma(3-3/2+1)= 4/3\sqrt{\pi}

This is not homework. I have spent 2 days on this and can't figure it out!

Thanks

Alan
 
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If I understand you correctly, the trick is to replace n by n-3:
\Gamma(n - \frac 3 2 +1) = \Gamma((n-3) + \frac 3 2 + 1) =\frac{(2(n-3)+3)(2(n-3)+1)!}{2^{2(n-3)+2}(n-3)!}=\frac{(2n-3)(2n-5)!}{2^{2n-4}(n-3)!}

At least I think this works for n \ge 3.
 
LCKurtz said:
If I understand you correctly, the trick is to replace n by n-3:
\Gamma(n - \frac 3 2 +1) = \Gamma((n-3) + \frac 3 2 + 1) =\frac{(2(n-3)+3)(2(n-3)+1)!}{2^{2(n-3)+2}(n-3)!}=\frac{(2n-3)(2n-5)!}{2^{2n-4}(n-3)!}

At least I think this works for n \ge 3.

Thanks for trying. But that was where I got really stuck. I can easily for find the series representation for n=1,2,3,4... It is the n=0 that is the problem. The series solution has to cover n=0,1,2...

Actually I am working on the problem is to show

J-3/2(x)=\sqrt{\frac{2}{\pi x}}[\frac{-cos(x)}{x}-sin(x)]

Where this is solution of Bessel function.
 
Please, even if you have any suggestion, I would like to listen to it. I am very despirate! I absolutely ran out of ideas!
 
LCKurtz said:
It has been many years since I looked at Bessel functions, so I can't help you more. I assume you have already looked at resources on the web such as this wikipedia article:

http://en.wikipedia.org/wiki/Bessel_function#Bessel_functions_of_the_first_kind_:_J.CE.B1

If you haven't already looked there you might find something useful. Good luck.

Thanks

I am studying on my own, this question only on the first section, before a lot of the information your link shows. I have to spend some time looking at this. Yes I did look at this before.

The question in the book ask both J(3/2) and J(-3/2). The possitive is relative easy because the gamma function always stay possitive.
 

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