Simplifying the expression involving instances of Gamma function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:36:15Z http://mathoverflow.net/feeds/question/82745 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82745/simplifying-the-expression-involving-instances-of-gamma-function Simplifying the expression involving instances of Gamma function Dragisa Zunic 2011-12-05T22:47:18Z 2012-01-03T08:45:57Z <p>Is it possible to simplify the following expression involving instances of Gamma function:</p> <p>$$E(p)=\frac{\frac{\Gamma(\frac{p+1}{2})}{\Gamma(\frac{p+2}{2})}} {(\frac{\Gamma(\frac{p+1}{p})^2}{\Gamma(\frac{p+2}{p})})^{\frac{p+2}{2}}}$$</p> <p>where $p$ is rational (or even real) and $p\geq2$. The bottom part of expression $E$ comes from the formula for the area of a superellipse, i.e., supercircle:</p> <p>$$\mid x\mid ^p + \mid y \mid ^p =r^p,\ p\geq 2$$</p> <p>and the rest is related to that also. Thanx in advance.</p> http://mathoverflow.net/questions/82745/simplifying-the-expression-involving-instances-of-gamma-function/82765#82765 Answer by Robert Israel for Simplifying the expression involving instances of Gamma function Robert Israel 2011-12-06T03:23:13Z 2011-12-06T03:23:13Z <p>The only parts of this that can be simplified at all are $\Gamma \left( {\frac {p+1}{p}} \right) ={\frac {\Gamma \left( \frac{1}{p} \right) }{p}}$, and similarly for $\Gamma\left(\frac{p+2}{p}\right)$ and $\Gamma\left(\frac{p+2}{2}\right)$</p> http://mathoverflow.net/questions/82745/simplifying-the-expression-involving-instances-of-gamma-function/84791#84791 Answer by Eric Naslund for Simplifying the expression involving instances of Gamma function Eric Naslund 2012-01-03T08:45:57Z 2012-01-03T08:45:57Z <p>I guess it depends on what you mean by simplify. We could rewrite things in terms of (generalized) central binomial coefficients:</p> <p>First the denominator: Notice that $$\frac{\Gamma\left(1+\frac{1}{p}\right)^{2}}{\Gamma\left(1+\frac{2}{p}\right)}=\binom{\frac{2}{p}}{\frac{1}{p}}^{-1}=\frac{1}{2p}\frac{\Gamma\left(\frac{1}{p}\right)^{2}}{\Gamma\left(\frac{2}{p}\right)}.$$ For the numerator $$\frac{\Gamma\left(\frac{p+1}{2}\right)}{\Gamma\left(\frac{p+2}{2}\right)}=\frac{\Gamma\left(\frac{p+1}{2}\right)^{2}}{\Gamma\left(\frac{p+1}{2}+\frac{1}{2}\right)\Gamma\left(\frac{p+1}{2}\right)}=\frac{\Gamma\left(\frac{p+1}{2}\right)^{2}}{\sqrt{\pi}2^{-p}\Gamma\left(p+1\right)}=\frac{2^{p}}{p\sqrt{\pi}}\binom{p-1}{\frac{p-1}{2}}^{-1}$$ so the fraction becomes $$\frac{2^{p}}{p\sqrt{\pi}}\binom{\frac{2}{p}}{\frac{1}{p}}^{\frac{p+2}{2}}\biggr/\binom{p-1}{\frac{p-1}{2}}.$$ You could also write it using the beta function, then it is $$\frac{2^{\frac{3p+2}{2}}p^{\frac{p+2}{2}}}{\sqrt{\pi}}\frac{\text{B}\left(\frac{1}{p},\frac{1}{p}\right)^{\frac{p+2}{2}}}{\text{B}\left(\frac{p+1}{2},\frac{p+1}{2}\right)}.$$ To clean it up, it feels like you need a nicer way to write $\Gamma\left(\frac{1}{p}\right)^{p}$. It seems to look like a multinomial coefficient. </p> <p>Now, there is a way to rewrite everything as a multidimensional integral over a simplex, and I find this to be the cleanest way to rewrite it. This is related to a generalization of the Beta Function. Tell me if this interests you, and I can include it.</p>