- #1
yungman
- 5,741
- 291
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] = ((n-n)+\frac{1}{2}+1)((n-n+1)+\frac{1}{2}+1)...[(n-1)+\frac{1}{2}+1]\Gamma(0+\frac{1}{2}+1)
\Rightarrow \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})
\Gamma(\frac{3}{2})=\frac{1}{2}\Gamma (\frac{1}{2})
\Rightarrow \Gamma(n+\frac{1}{2}+1) = (\frac{1}{2})(\frac{3}{2})(\frac{5}{2})...(\frac{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 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. There is not example or homework of this in a few of my books but we do use it in Bessel equation expansion. I just work out this formulas.
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] = ((n-n)+\frac{1}{2}+1)((n-n+1)+\frac{1}{2}+1)...[(n-1)+\frac{1}{2}+1]\Gamma(0+\frac{1}{2}+1)
\Rightarrow \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})
\Gamma(\frac{3}{2})=\frac{1}{2}\Gamma (\frac{1}{2})
\Rightarrow \Gamma(n+\frac{1}{2}+1) = (\frac{1}{2})(\frac{3}{2})(\frac{5}{2})...(\frac{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 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. There is not example or homework of this in a few of my books but we do use it in Bessel equation expansion. I just work out this formulas.
Last edited: