How to prove the quotient of confluent hypergeometric functions of adjacent orders is convex?

Asked Apr 26 '22 at 00:53

Active Apr 26 '22 at 03:41

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Denote $F(;n;x)$ as the confluent hypergeometric function $_0F_1$, i.e. $F(;n;x)=\sum\limits_{k=0}^{\infty}\frac{x^k (n-1)!}{(n+k-1)!k!}$. How to prove $\frac{F(;n+1;x)}{F(;n;x)}$ is a convex function of $x$ when $x>0$ and $n$ is a positive integer?

The sign of the second derivative of $\frac{F(;n+1;x)}{F(;n;x)}$ depends on the following factor $[2\tilde{F}^3(;n+1;x)-3\tilde{F}(;n;x)\tilde{F}(;n+1;x)\tilde{F}(;n+2;x)+\tilde{F}^2(;n;x)\tilde{F}(;n+3;x)]$, where regularized confluent hypergeometric function $\tilde{F}(;n;x)=\frac{F(;n;x)}{(n-1)!}$.

edited Apr 26 '22 at 03:41

asked Apr 26 '22 at 00:53

Ricardo

At least, this is true for $x=0$ – Claude Leibovici Apr 26 '22 at 01:31
Yes, at $x=0$, it's simple considering https://functions.wolfram.com/HypergeometricFunctions/Hypergeometric0F1Regularized/07/01/01/0004/ – Ricardo Apr 26 '22 at 03:11
Yes, I know. The problem looks to be difficult. – Claude Leibovici Apr 26 '22 at 03:27
But intuitively and visually, this conclusion is not surprising. I'm kinda disappointed it's not simple to prove it. I was even thinking about using Mathematical Induction... – Ricardo Apr 26 '22 at 03:38
Or it can be asked as, when $n$ is a positive integer, is $\left[\sum\limits_{p=0}^q\frac{2(p+n+1)n(4+3p-2q)+p(4+3p-2q)-q+2!}{(q-p)!(n-p+q)!p!(n+p+1)!(n+p+2)!(2n+p+1)!}\right]$ nonnegative for any given $q$? – Ricardo Apr 28 '22 at 00:32

How to prove the quotient of confluent hypergeometric functions of adjacent orders is convex?

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