Math Help Forum - View Single Post - gamma and e^(-x^2)

galactus · #10 December 31st, 2008, 09:55 PM

Well, Kriz, let' see. I have been working on it while I was waiting on your reponnse and gong to the bathroom.

It is getting late, Kriz. I am going to skip some major steps and join you tomorrow. Gotta go. But I have this so far. Need to explain some steps further if need be.

From $B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)}$

Then,

${\Gamma}(p+\frac{1}{2})=\frac{{\Gamma}(p)\cdot {\Gamma}(\frac{1}{2})}{B(p,1/2)}$

$=\frac{(p-1)!\sqrt{\pi}}{\frac{4^{p}(p!)^{2}}{p(2p)!}}$

$\boxed{=\frac{(2p)!\sqrt{\pi}}{4^{p}\cdot p!}}$

We need to show that $B(p,1/2)=\frac{4^{n}(n!)^{2}}{n(2n)!}$ which I think can be gotten from $\frac{4^{p}}{2}\cdot B(p,p)$

because $B(p,p)=\frac{{\Gamma}(p){\Gamma}(p)}{{\Gamma}(2p)}=\frac{(p-1)!(p-1)!}{(2p-1)!}$

Therefore, we get $=\frac{(p!)^{2}\cdot 2p}{p^{2}\cdot (2p)!}$

$=\frac{2(p!)^{2}}{p(2p)!}$

$B(p,1/2)=\frac{4^{p}}{2}\cdot \frac{2(p!)^{2}}{p(2p)!}$

Plugging this in the above gives:

$\frac{(n-1)!\sqrt{\pi}}{\frac{(p!)^{2}4^{p}}{p(2p)!}}=\frac{(2p)!\sqrt{\pi}}{4^{p}p!}$

Hope you can follow that. I will touch it up tomorrow.

It's getting late and my brain is beginning to see p's and q's everywhere.

Cheers, Kriz