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In this entry I asked for the real-rootedness of a polynomial, and two very interesting answers were given: one using Malo's theorem and the other a clever rewriting of the expression using Jacobi Polynomials which are known to be real-rooted.

Now, in the middle of a reasoning I had to somehow prove that the following polynomials are real-rooted:

$$Q_{m,n}(t) = \sum_{j=0}^m \binom{m}{j}\binom{n}{j} \binom{t-j}{m+n+1}$$

defined for $m\leq n$, integers.

Observe that where in the last question one had $t^j$, here one has a $\binom{t-j}{m+n+1}$.

Of course, it has been verified numerically for the first small cases and it is true and, in fact, it seems to be true that all the roots are $\leq m+n$. It's possible to prove that many of the roots are integers, which suggests one may "factor out" something like $\binom{t-m}{n+1}$ out of the expression. The problem is that the remaining factor does not seem to be friendly at all.

This leads me to the following questions:

1) Any ideas to prove the real-rootedness of this thing?

2) (Much wider and maybe not useful for this concrete problem but interesting on its own) There's a theorem of Schur (see here Theorem 1) that gives a sufficient condition for a polynomial to be real rooted. I want to know if there's any kind of generalization in the case one changes the basis $\{1,x,x^2,\ldots\}$ for example with $\{\binom{x}{0},\binom{x}{1},\binom{x}{2},\ldots\}$.

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First, we write the polynomial in hypergeometric form $$ Q_{m,n}(t)=\frac{(-1)^{m+n+1} \Gamma (m+n-t+1) }{\Gamma (-t) \Gamma (m+n+2)}\, _3F_2\left({-m,-n,m+n-t+1\atop 1,-t};1\right). $$ Applying Thomae's transformation we get (see eq. 1.3 in W. N. Bailey, Contiguous Hypergeometric Functions of the Type 3F2(1)): $$ Q_{m,n}(t)=\frac{(-1)^{n+1} m! \Gamma (m+n-t+1) }{\Gamma (m+n+2) \Gamma (m-t)}\, _3F_2\left({-m,n+1,-m-n+t\atop 1,-m};1\right). $$ Here the polynomial is understood as $\displaystyle{\lim_{\epsilon\to 0}{}_3F_2\left({-m,n+1,-m-n+t\atop 1,-m+\epsilon};1\right) }$, so $-m$ in the numerator and denominator of the hypergeometric function can not be cancelled.

The hypergeometric polynomial in the last expression can be expressed in terms of Hahn polynomials, which are discrete orthogonal polynomials (see M. Ismail, Classical and Quantum Orthogonal Polynomials in One Variable): enter image description here Thus $$ Q_{m,n}(t)=(-1)^{n+1}\frac{ (m-t)_{n+1} }{(m+1)_{n+1}}\,Q_m(m+n-t;0,n-m,m). $$ It is known that zeroes of Hahn polynomials are real and simple. In particular, one can see that some roots are indeed integers due to the factor $(m-t)_{n+1} $, in agreement with OP's observation.

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  • $\begingroup$ Thank you very much for your answer. I was not aware of the existence of Hahn polynomials. Doing some manipulations, I think I can give a very nice simplification to this proof if I could demonstrate that for $a,b$ nonnegative integers: $$\binom{a+b-1}{a-1} {}_3F_2\left( {{1-a, b, -t} \atop {1,b+1}} ; 1\right) = {}_3F_2\left( {{1-a, b, t+1} \atop { 1, 1-a}};1\right).$$ However I can not see how to apply Thomae's transform in any suitable way to go from one to the other side. $\endgroup$ – Luis Ferroni Feb 12 at 17:07
  • $\begingroup$ Also, since I'll be using this result as a lemma in a paper I would like to acknowledge you, if that's not a problem to you, of course. :) $\endgroup$ – Luis Ferroni Feb 12 at 17:11
  • $\begingroup$ See eq. 1.3 in W. N. Bailey, Contiguous Hypergeometric Functions of the Type 3F2(1). Try different combinations of parameters, one of them bound to work. Acknowledging is ok. $\endgroup$ – Nemo Feb 12 at 21:10

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