1. Nov 23, 2009

### yungman

I have been working on this for a few days and cannot prove this:

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

Main reason is $$\Gamma$$(n-3/2+1) give a negative value for n=0 and possitive value for n=1,2,3.... I cannot find a series representation of this gamma function.

Please advice me how to solve this problem. This is not a school homework.

thanks a million

Alan

Last edited: Nov 23, 2009
2. Nov 23, 2009

### g_edgar

Maybe instead of the series representation for the Bessel function: show that your right-hand side satisfies Bessel's differential equation, and has the proper initial values, so that it therefore equals the Bessel function required.

3. Nov 24, 2009

### matematikawan

For n > 0 we apply this formula
$$\Gamma (n+1) = n \Gamma (n)$$

but if n < 0 we apply this formula
$$\Gamma (n)=\frac{\Gamma (n+1)}{n}$$

4. Nov 24, 2009

### yungman

I know this formula!!!!!!! This is embarassing!!!! How can I over looked this and spent 3 days on this.....Even joined two more math forums!!!! I even plug in the numbers and hope this is not that simple!!!! I use

$$\Gamma (-3/2)=\frac{\Gamma (-1/2)}{-3/2}$$ all the time!!! Just never try with n in it!!!!

Thanks a million.......Even though you make me look really really bad!!!!!

Cheers.
Alan