# Gamma function

• April 17th 2010, 09:48 PM
yungman
Gamma function
I want to verify the procedure of finding . $\Gamma(n+p+1)$. with p =-ve. This is usually found in Bessel's equation. It is well talked about if p=+ve. But books I have don't even talking about in general how to find the series representation when p=-ve. I worked this out and I want to verify with you guys whether I am correct for some simple numbers of p= -1/2, -3/2,-5/2.
I use $\Gamma(n+p+1) = \frac{\Gamma(n+p+2)}{ (n+p+1) }$ as the bases to expand the equations below.

I start using this identity of p=1/2

$\Gamma(n+\frac{1}{2}+1) = ((n-1)+\frac{1}{2}+1)\Gamma[(n-1)+\frac{1}{2}+1]$

$\Gamma(n+\frac{1}{2}+1) =((n-1)+\frac{1}{2}+1)((n-2)+\frac{1}{2}+1)....((n-n)+\frac{1}{2}+1)\Gamma(0+\frac{1}{2}+1) = \frac{1}{2} \sqrt{\pi}$

$\Gamma(n+\frac{1}{2}+1) = (\frac{0+1+2}{2})(\frac{2+1+2}{2})(\frac{4+1+2}{2} )...(\frac{(2n-2)+1+2}{2})\Gamma(\frac{0+1+2}{2})$

$\Rightarrow \Gamma(n+\frac{1}{2}+1) = (\frac{1}{2})(\frac{3}{2})(\frac{5}{2}).......(\fr ac{2n+1}{2})\Gamma(\frac{1}{2}) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1} n!}$

For p=-1/2, I use the result from above and add on to it:
$\Gamma(n-\frac{1}{2}+1) = \frac{\Gamma(n-\frac{1}{2}+2)}{ (n-\frac{1}{2}+1) } = \frac{\Gamma(n+\frac{1}{2}+1)}{ (n-\frac{1}{2}+1) } = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (n-\frac{1}{2}+1) } = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (\frac{2n-1+2}{2}) }$

$\Rightarrow \Gamma(n-\frac{1}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n}n! (2n+1) } = \frac{(2n)!\sqrt{\pi}}{2^{2n}n! }$

For p=-3/2
$\Gamma(n-\frac{3}{2}+1) = \frac{\Gamma(n-\frac{3}{2}+2)}{ (n-\frac{3}{2}+1) } = \frac{\Gamma(n-\frac{1}{2}+1)}{ (n-\frac{3}{2}+1) } = \frac{\Gamma(n+\frac{1}{2}+1)}{ (n-\frac{3}{2}+1) (n-\frac{1}{2}+1) }$

$\Rightarrow \Gamma(n-\frac{3}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (n-\frac{3}{2}+1)(n-\frac{1}{2}+1) } = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (\frac{2n-3+2}{2}) (\frac{2n-1+2}{2})}$

$\Rightarrow \Gamma(n-\frac{3}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n-1}n! (2n-1)(2n+1) } = \frac{(2n)!\sqrt{\pi}}{2^{2n-1}n! (2n-1)}$

For p=-5/2
$\Gamma(n-\frac{5}{2}+1) = \frac{\Gamma(n-\frac{3}{2}+1)}{ (n-\frac{5}{2}+1) } = \frac{\Gamma(n+\frac{1}{2}+1)}{ (n-\frac{5}{2}+1) (n-\frac{3}{2}+1) (n-\frac{1}{2}+1) }$

$\Gamma(n-\frac{5}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n+1}n! (\frac{2n-3}{2}) (\frac{2n-1}{2})(\frac{2n+1}{2})}$

$\Rightarrow \Gamma(n-\frac{5}{2}+1) = \frac{(2n+1)!\sqrt{\pi}}{2^{2n-2}n! (2n-3)(2n-1) (2n+1)} = \frac{(2n)!\sqrt{\pi}}{2^{2n-2}n! (2n-3)(2n-1)}$

Please tell me whether I did this correct or not.
• April 19th 2010, 09:14 AM
yungman