1. Nov 22, 2009

### 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

2. Nov 22, 2009

### 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 $n \ge 3$.

3. Nov 22, 2009

### yungman

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.

4. Nov 23, 2009

### 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!!!!

5. Nov 23, 2009

### LCKurtz

6. Nov 23, 2009

### yungman

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.