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Denote the unsigned Stirling numbers of the first kind by $s(n,j)$.

Question. Is it true that the polynomials $$P_n(x)=\sum_{j\geq0}s(n,j)\binom{x}j$$ have only real roots?

Note. Obviously, the roots of $F_n(x)=\sum_{j\geq0}s(n,j)\,x^j=\prod_{k=0}^{n-1}(x+k)$ are all real (integers).

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    $\begingroup$ for what it worth, $P_n(x)$ is a constant term of the Laurent series $(1+z)^x\prod_{j=0}^{n-1}(z^{-1}+j)$. $\endgroup$
    – Fedor Petrov
    Jun 6, 2018 at 7:20
  • $\begingroup$ Consider $P_k(x)=(1+\alpha)(1+2\alpha)\cdots(1+(k-1)\alpha)\cdot{x\choose n}$, where $\alpha$ is the operator sending ${x\choose k}$ to ${x\choose k-1}$. That is, $\alpha$ sends $f(x)$ to $f(x+1)-f(x)$. Maybe we can induct on $k$ with some results on linear combinations of real-rooted polynomials with interlacing roots. $\endgroup$
    – MTyson
    Jun 7, 2018 at 17:35
  • $\begingroup$ The Bell polynomials, with the s(n,j) in the top formula replaced with Stirling numbers of the second kind, are known to have real zeros. Replacing $x^k \to \binom{x}{k}$ does not preserve real rootedness, starting with n=6. $\endgroup$
    – skbmoore
    Jun 7, 2018 at 18:52
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    $\begingroup$ The associated Laguerre polynomials might also obey the property where a polynomial has its $x \to \binom{x}{k} $ and retain real rootedness. I checked those with the coefficients of $L_n^a(x)$ for n=1 to 80, a=-2 to +2 by 1/10. For $a<-1/4(12+\sqrt{14})$ you are guaranteed complex roots. This is mentioned because a class of polynomials might have an easier proof than a lone example. $\endgroup$
    – skbmoore
    Jun 7, 2018 at 20:43
  • $\begingroup$ Do this family of polynomials have a name? $\endgroup$
    – Hvjurthuk
    Feb 20 at 19:02

1 Answer 1

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Fix $n$ and consider $P_k(x)=(1+\alpha)(1+2\alpha)\cdots(1+(k-1)\alpha)\cdot{x\choose n}$, where $\alpha$ is the operator sending ${x\choose k}$ to ${x\choose k-1}$. That is, $\alpha$ sends $f(x)$ to $f(x+1)-f(x)$. We'll show by induction that each $P_k(x)$ has all real roots and that the distance between consecutive roots is at least $1$. The claim is trivial for $k=1$.

Assume the claim holds for $P_k(x)$. Then $P_{k+1}(x)=(1+k\alpha)P_k(x)$ is the linear combination $kP_k(x+1)-(k-1)P_k(x)$. Let the roots of $P_k(x)$ be $a_1<\cdots<a_n$. The roots of $P_k(x+1)$ are $a_1-1<\cdots<a_n-1$, which interlace with those of $P_k(x)$ by the induction assumption. In the notation of Fisk's "Polynomials, roots, and interlacing", $P_k(x+1)\underline\ll P_k(x)$. Let the roots of $P_{k+1}(x)$ be $b_1\le\cdots\le b_n$. It follows from (the proof of) Prop 1.35 that all the roots are real and that $b_1\le a_1-1$ and $a_{i-1}\le b_i\le a_i-1$ for all $i>1$. Therefore $b_i-b_{i-1}\ge a_{i-1}-(a_{i-1}-1)=1$, which completes the induction.

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