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Let $f(z)\in\mathbb{C}[z]$ have all its zeros on the line $\Re(z)=\alpha$ for some $\alpha\in\mathbb{R}$. It is an elementary fact (equivalent to Lemma 9.13 here) that if $u\in\mathbb{C}$ and $|u|=1$, then all the zeros of $g(z):=f(z+1)-uf(z-1)$ also lie on the line $\Re(z)=\alpha$. Moreover, $\deg g(z)=\deg f(z)$ unless $u=1$.

Let $u_1,u_2,\dots$ be any sequence of complex numbers on the circle $|z|=1$ such that no $u_i=1$. Let $n\geq 1$, and set $f_{1,n}(z)= z^n$. For $j\geq 1$ define inductively $f_{j+1,n}(z)=f_{j,n}(z+1)-u_jf_{j,n}(z-1)$. What can we say about the limiting behavior of $f_{m,n}(z)$ as $m,n\to\infty$? Will it always approach some scaled version of $\cosh(z)$? For one instance of this behavior, see Theorem 11.7 of the above link.

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  • $\begingroup$ The answer seems to depend strongly on the choice of the sequence $u_j$ and on the relation between $m$ and $n$ as they tend to $\infty$. You should add some assumptions on these for existence of a limit after scaling. $\endgroup$
    – Alexandre Eremenko
    Commented Apr 3, 2021 at 23:15
  • $\begingroup$ I assume somehow you can relate this to the Appell Hermite polynomials, and that you've explored that already. // Btw, thanks for the answers to some of my questions you've provided and the many resources on the Net. Having been in the thick of it, would you have any more to add to (or subtract from) the discussions in the MO-Q mathoverflow.net/questions/111970/…? $\endgroup$
    – Tom Copeland
    Commented Apr 5, 2021 at 0:18
  • $\begingroup$ @AlexandreEremenko, I looked at the uniform distribution on $|z|=1$ and also the eigenvalues of a random $n\times n$ unitary matrix (Haar measure), always with $m=n$, and seemed in both cases to be getting $\cosh(z)$. Admittedly pretty flimsy evidence. I was hoping that for some distribution on $|z|=1$ I would get the same statistical behavior as for the conjectured behavior of the zeros of the Riemann zeta function in the critical strip, or even (really wishful thinking) a simple modification of $\zeta(z)$ itself. $\endgroup$
    – Richard Stanley
    Commented Apr 6, 2021 at 2:02
  • $\begingroup$ @TomCopeland, of course I was familiar with Rota's statement, but I never understood, nor did I ask him, what it meant. $\endgroup$
    – Richard Stanley
    Commented Apr 6, 2021 at 2:06
  • $\begingroup$ As to your question, perhaps something can be gleaned from "On Riemann's zeta function" by Bump and Ng (eudml.org/doc/173728) and "Binomial Polynomials Mimicking Riemann’s Zeta Function" by Coffey and Lettington (arxiv.org/pdf/1703.09251.pdf). $\endgroup$
    – Tom Copeland
    Commented May 17, 2021 at 19:48

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