3
$\begingroup$

Suppose $F$ is the Gaussian hypergeometric function 2F1, $x\in \mathbb{N}$, $t>x \in \mathbb{N}$, $\mu \in (0,1)$. Is the function: $$f(\mu)=\mu^{x+1} F(x,-t+x,2+x,\mu),$$ convex in $\mu$? I've done some numerical examples that seem to verify this with the following manipulate Mathematica code (apologies for the mess):

Manipulate[
    Plot[
        {
            Subscript[\[Mu], 1]^(x + 1) b*
            Hypergeometric2F1[x, -t + x, 2 + x, Subscript[\[Mu], 1]]
            + (1 - b) Subscript[\[Mu], 2]^(x + 1)*
            Hypergeometric2F1[x, -t + x, 2 + x, Subscript[\[Mu], 2]],

            (b*Subscript[\[Mu], 1] + (1 - b) Subscript[\[Mu], 2])^(x + 1)
            Hypergeometric2F1[
                x, -t + x, 2 + x, (b*Subscript[\[Mu], 1] + (1 - b) Subscript[\[Mu], 2])
            ]
        },
        {b, 0, 1}
    ],
    {t, 1, 80, 1},
    {x, 1, 80, 1},
    {Subscript[\[Mu], 1], 0, 1},
    {Subscript[\[Mu], 2], 0, 1}
]

EDIT: Forgot to add $x+1$ instead of $x$.

Improve this question
$\endgroup$
3
  • $\begingroup$ I am not sure that in the reference below there is an exact answer to the question. $\endgroup$
    – Sergei
    Commented Dec 17, 2014 at 6:59
  • 1
    $\begingroup$ I have read though it as well, and haven't been able to see how it answers my question either. Am currently reading "Representations and inequalities for generalized hypergeometric functions" by Dmitrii Karp. I think there may be some monotonic results that might help. $\endgroup$
    – Austin Bren
    Commented Dec 17, 2014 at 22:56
  • $\begingroup$ Glad to hear it. D.Karp is my former student and co-author. You may find many useful facts from his papers. May be start with log-convexity? $\endgroup$
    – Sergei
    Commented Dec 18, 2014 at 13:59

1 Answer 1

Reset to default
1
$\begingroup$

See

Miller, Sanford S.(1-SUNY3); Mocanu, Petru T.(R-CLUJ) Univalence of Gaussian and confluent hypergeometric functions. Proc. Amer. Math. Soc. 110 (1990), no. 2, 333–342.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .