2
$\begingroup$

I wish to prove the following inequality: $$A \leq B$$ where $$A=\Gamma(1+\frac{2}{\beta})\cdot \frac{q^{\frac{1}{\beta}}}{1+\beta}\cdot {}_2F_1(1,1+\frac{2}{\beta};2+\frac{1}{\beta};1-q) $$ $$B=(\Gamma(1+\frac{1}{\beta}))^2\cdot \frac{(1-q^{\frac{1}{\beta}})}{1-q}$$ the parameters $q$ and $\beta$ satisfy following conditions:

$\beta \geq 1$,

$0 \leq q \leq 1$,

$\Gamma(x)$ the gamma function

and ${}_2F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}$ the hypergeometric function

The maximum of the quotient $\frac{A}{B}$ is obtained when $\beta=1$, i then tried unsuccessfully to express it by the product of 2 functions bounded by 1. Also it seems difficult to have the derivative as $\beta$ is in the F. Perhaps the hypergeom can simplify with the gamma function?

Any help would be appreciated.

$\endgroup$
  • $\begingroup$ Is $\gamma(x)$ the usual gamma function $\Gamma(x)$? I have only seen the small gamma notation for the lower incomplete gamma function en.wikipedia.org/wiki/Incomplete_gamma_function $\endgroup$ – Yuriy S Feb 13 '18 at 21:30
  • $\begingroup$ yes it's the gamma function, thanks for your reminder $\endgroup$ – Xingheng Liu Feb 23 '18 at 8:30
  • $\begingroup$ Have you looked at the Wikipedia page for the Hypergeometric function? It has a lot of identities, some of them contain Gamma function as well. $\endgroup$ – Yuriy S Feb 23 '18 at 8:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.