Please help in Gamma function to series conversion.

[yungman]

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

 

[LCKurtz]

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 $n \ge 3$.

 

[yungman]

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.

 

[yungman]

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]

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.

 

[yungman]

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.