3
$\begingroup$

Is it possible to simplify the following expression involving instances of Gamma function:

$$E(p)=\frac{\frac{\Gamma(\frac{p+1}{2})}{\Gamma(\frac{p+2}{2})}} {\left(\frac{\Gamma(\frac{p+1}{p})^2}{\Gamma(\frac{p+2}{p})}\right)^{\frac{p+2}{2}}}$$

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:

$$\mid x\mid ^p + \mid y \mid ^p =r^p,\ p\geq 2$$

and the rest is related to that also. Thanx in advance.

$\endgroup$
3
  • $\begingroup$ Either I'm missing the easy things, or David is missing the different denominators. Can you say more David? Gerhard "Ask Me About System Design" Paseman, 2011.12.05 $\endgroup$ Dec 6, 2011 at 1:36
  • $\begingroup$ Duh, stupid me. Of course... $\endgroup$
    – David Roberts
    Dec 6, 2011 at 6:07
  • $\begingroup$ So I imagine it can't be made much simpler. I also tried using MatLab simplify command, but it didn't give any revolutionary results.. Thank you. $\endgroup$ Dec 9, 2011 at 8:24

2 Answers 2

4
$\begingroup$

I guess it depends on what you mean by simplify. We could rewrite things in terms of (generalized) central binomial coefficients:

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.

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.

$\endgroup$
1
  • $\begingroup$ Thank you Eric, I think this might be enough for me for now. $\endgroup$ Jan 23, 2012 at 19:10
2
$\begingroup$

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 $\Gamma\left(\frac{p+2}{2}\right)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.