- #1
yungman
- 5,718
- 240
Can you show me how to get the series representation of [tex]\Gamma[/tex](n-3/2+1)?
For example [tex]\Gamma[/tex](n+3/2+1)=[tex]\frac{(2n+3)(2n+1)!}{2^{2n+2}.n!}[/tex].
I cannot figure out how to write a series with:
n=0 => [tex]\Gamma[/tex](0-3/2+1)= -2[tex]\sqrt{\pi}[/tex]
n=1 => [tex]\Gamma[/tex](1-3/2+1)= [tex]\sqrt{\pi}[/tex]
n=2 => [tex]\Gamma[/tex](2-3/2+1)= 1/2[tex]\sqrt{\pi}[/tex]
n=3 => [tex]\Gamma[/tex](3-3/2+1)= 4/3[tex]\sqrt{\pi}[/tex]
This is not homework. I have spent 2 days on this and can't figure it out!
Thanks
Alan
For example [tex]\Gamma[/tex](n+3/2+1)=[tex]\frac{(2n+3)(2n+1)!}{2^{2n+2}.n!}[/tex].
I cannot figure out how to write a series with:
n=0 => [tex]\Gamma[/tex](0-3/2+1)= -2[tex]\sqrt{\pi}[/tex]
n=1 => [tex]\Gamma[/tex](1-3/2+1)= [tex]\sqrt{\pi}[/tex]
n=2 => [tex]\Gamma[/tex](2-3/2+1)= 1/2[tex]\sqrt{\pi}[/tex]
n=3 => [tex]\Gamma[/tex](3-3/2+1)= 4/3[tex]\sqrt{\pi}[/tex]
This is not homework. I have spent 2 days on this and can't figure it out!
Thanks
Alan