# Gamma Function

• Mar 24th 2011, 06:21 AM
Cairo
Gamma Function
How is $\Gamma(50\frac{1}{2})$ evaluated iteratively?
• Mar 24th 2011, 06:27 AM
chisigma
Quote:

Originally Posted by Cairo
How is $\Gamma(50\frac{1}{2})$ evaluated iteratively?

Please explain the notation $50\frac{1}{2}$... is it a [real] number?...

Kind regards

$\chi$ $\sigma$
• Mar 24th 2011, 08:23 AM
Cairo
Sorry.

Yes it is real - as in $\Gamma(n+\frac{1}{2})$.
• Mar 24th 2011, 01:05 PM
Drexel28
Quote:

Originally Posted by Cairo
How is $\Gamma(50\frac{1}{2})$ evaluated iteratively?

I'm not quite sure how 'iteratively' you want it. But, you can use the formula $\displaystyle \Gamma\left(z+\frac{1}{2}\right)=\frac{\Gamma(2z)} {\Gamma(z)}2^{1-2z}\sqrt{\pi}$. If you'd like to know a proof of this, I can type one up.
• Mar 24th 2011, 01:14 PM
chisigma
Quote:

Originally Posted by Cairo
Sorry.

Yes it is real - as in $\Gamma(n+\frac{1}{2})$.

From the property ...

$\displaystyle \Gamma(1+x) = x\ \Gamma(x)$ (1)

... You derive that...

$\displaystyle \Gamma(2+x) = x\ (1+x)\ \Gamma(x)$

$\displaystyle \Gamma(3+x) = x\ (1+x)\ (2+x)\ \Gamma(x)$

...

$\displaystyle \Gamma(n+x) = x\ (1+x)\ ... \ (n-1+x)\ \Gamma(x)$ (2)

Now if You set in (2) $x=\frac{1}{2}$ and take into account that is $\Gamma(\frac{1}{2})= \sqrt{\pi}$ You obtain...

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

Kind regards

$\chi$ $\sigma$
• Mar 24th 2011, 01:14 PM
Cairo
Quote:

Originally Posted by Drexel28
I'm not quite sure how 'iteratively' you want it. But, you can use the formula $\displaystyle \Gamma\left(z+\frac{1}{2}\right)=\frac{\Gamma(2z)} {\Gamma(z)}2^{1-2z}\sqrt{\pi}$. If you'd like to know a proof of this, I can type one up.

Thanks for this Drexel.

I managed to find something similar on Wikipaedia, but it is the proof I am after.

Can you supply one?
• Mar 24th 2011, 01:15 PM
Cairo
Thanks chisigma.

But is that the same expression that Drexel gets?
• Mar 24th 2011, 09:40 PM
Drexel28
Quote:

Originally Posted by Cairo
Thanks for this Drexel.

I managed to find something similar on Wikipaedia, but it is the proof I am after.

Can you supply one?

Certainly. But, my original proof didn't work out. So I found an alternate...but it's fairly technical. You might find $\chi\sigma$'s suggestion more useful. Also, I did find an alternate proof on page 22 of this book (a very good book by the wayP may be found for a more general result. Regardless...

For $z>0$ define $\displaystyle f(z)=\frac{\Gamma(2z)}{\Gamma\left(z+\frac{1}{2}\r ight)}2^{1-2z}\sqrt{\pi}$. Note firstly then that $\displaystyle f(1)=\frac{\Gamma(2)}{\Gamma(\frac{1}{2})}2^{-1}\sqrt{\pi}=1$. Moreover,

\displaystyle \begin{aligned}f(z+1) &=\frac{\Gamma(2z+2)}{\Gamma\left(\left(z+\frac{1} {2}\right)+1\right)}2^{1-2z-2}\sqrt{\pi}\\ &= \frac{1}{4}\frac{\left(2z+1\right)\left(2z\right)} {z+\frac{1}{2}}\frac{\Gamma(2z)}{\Gamma\left(z+\fr ac{1}{2}\right)}2^{1-2z}\sqrt{\pi}\\ &=z\frac{\Gamma(2z)}{\Gamma\left(z+\frac{1}{2}\rig ht)}2^{1-2z}\sqrt{\pi}\\ &=zf(z)\end{aligned}

Finally, note that $\log(f(z))=\log\left(\Gamma(2z)\right)-\log\left(\Gamma\left(z+\frac{1}{2}\right)\right)+ \left(1-2z\right)\log(2)+\frac{1}{2}\log(\pi)$. So that $\frac{d^2}{dz^2}\log(f(z))=4\psi'(2z)-\psi\left(z+\frac{1}{2}\right)$ where $\psi$ is the digamma function. But, a quick check (using digamma's series expression) shows then that $4\psi'(2z)-\psi\left(z+\frac{1}{2}\right)>0$ and thus $\log(f(z))$ is convex, and so $f(z)$ is log convex. The result then follows from the Bohr-Mollerup theorem.