(I am not sure if I use correct tags for this problem)

I need to find the maximum of $$F(l,m) = (-1)^m2^{-2l-m} {}_2F_1(-l, -l-m, 1, -3)$$ when $l$, $m \in \mathbb N_+$ are positive integers. Here ${}_2F_1$ stands for the hypergeometric function and below $C_l^j=\binom{l}j$.

It can equivalently be expressed by finite summation ($l$, $m$ are integers) $$F(l,m) = \frac{(-1)^m}{2^{2l+m}} \sum_{j=1}^l C_l^jC_{l+m}^j(-3)^j, $$ or by Jacobi polynomials $$F(l,m) = \frac{P_l^{(0,m)}(-1/2)}{(-2)^mP_l^{(0,m)}(1)}, $$ where $P_l^{(0,m)}(x) = \frac{(-1)^l}{2^ll!}(1+t)^{-m}\frac{d^l}{dt^l}[(1-t)^l(1+t)^{m+l}]$ is the Rodrigues's formula for Jacobi polynomial of order $(0, m)$.

Numerical results show that the maximum may be $F(1,1) = 5/8$; for large $m$ and $l$ the absolute value $|F(l,m)|$ decreases quickly; the following estimate for Jacobi polynomial seems can only prove that $F(l,m)$ is uniformly bounded: $$\max_{-1 \leq x \leq 1}\pi\sqrt{1-x^2}(1-x)^m[p_n^{(\alpha,\beta)}(x)]^2 \leq 2e(2+\sqrt{\alpha^2+\beta^2}), $$ where $p_n^{(\alpha, \beta)}$ is the normalization of $P_n^{(\alpha, \beta)}$. However the upper bound from this estimate is simply too large.

  • $\begingroup$ It would also be very helpful if we can prove that the upper bound of $F(l, m)$ is strictly below $1$. I think it would be much more easier. $\endgroup$ – gregarki khayal Dec 1 '16 at 14:31
  • $\begingroup$ Is $C_k^n=\binom{n}k$? $\endgroup$ – T. Amdeberhan Dec 1 '16 at 14:43
  • $\begingroup$ @T.Amdeberhan yes, already edited. $\endgroup$ – gregarki khayal Dec 1 '16 at 15:45

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