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The following question on polynomials arose as a potentially helpful intermediate step on a proof of a Theorem that I want to demonstrate. Its statement is quite elementary, and I can think of a number of applications it might have to the study of Hilbert polynomials in commutative algebra and certain combinatorial sequences arising in discrete geometry.

Conjecture 1: Let $(c_0,\ldots,c_n)$ be non-negative integers and consider the polynomial $P(x) = \sum_{i=0}^n c_i \binom{x+n-i}{n}$. Assume that $P(x)$ has at least one negative real-root and denote by $-r$ the largest (in absolute value) negative real-root of $P(x)$. Then either $\lfloor r \rfloor = 0$ or $-\lfloor r\rfloor$ is a root of $P(x)$ too.

A weaker statement that could also be useful:

Conjecture 1': Let $(c_0,\ldots,c_n)$ be non-negative integers and consider the polynomial $P(x) = \sum_{i=0}^n c_i \binom{x+n-i}{n}$. If $P(-1)\neq 0$, then the negative real roots of $P(x)$ (if any) lie in the interval $(-1,0)$.

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    $\begingroup$ No. If $P(x) = x+2$, then the sequence $(c_0,c_1)$ is $(2,-1)$ and the non-negativity assumption does not hold. $\endgroup$
    – Luis Ferroni
    Commented Jun 14, 2023 at 13:30
  • $\begingroup$ Ah, sorry of course, I see, I'll delete that comment. Roots of Ehrhart polynomials are studied in arxiv.org/abs/math/0402148 (which you probably are aware of already...) $\endgroup$
    – Sam Hopkins
    Commented Jun 14, 2023 at 13:33

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Conjecture 1' looked a bit easier to test. Unless I've misinterpreted the question (not impossible), I believe the conjecture is false. If we look at the case $n=2$, the polynomial is $$\begin{align} P(x)&=c_0\binom{x+2}{2}+c_1\binom{x+1}{2}+c_2\binom{x}{2} \\ &=\frac12\left(c_0(x+2)(x+1)+c_1 (x+1)x + c_2 x(x-1)\right) \\ &=\frac12\left(\left(c_0+c_1+c_2 \right)x^2+\left(3c_0+c_1-c_2 \right)x+2c_0 \right) \end{align}$$

Taking $\left(c_0,c_1,c_2\right)=\left(18,1,1\right)$ gives $$P(x)=10x^2+27x+18=(2x+3)(5x+6)$$

with two negative roots, each less than $-1$.

Is there a reformulation that would be worth checking?

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  • $\begingroup$ I think you're absolutely right. I was too confident that my statement would be true for all polynomials that were positive in the basis I described, but seemingly one should impose additional assumptions (such as being the Hilbert series of a standard graded domain, or an Ehrhart polynomial). I think in those cases I can prove that both of my conjectures hold, but I need those extra conditions in an essential way. A modification that might be worth trying, though, is that of imposing $c_0=1$. $\endgroup$
    – Luis Ferroni
    Commented Jun 15, 2023 at 15:58
  • $\begingroup$ No, I'm pretty sure that just puts the burden on $c_1$. So make $c_0=1$, $c_1$ gigantic, and all the rest tiny. This will work for $n\geqslant 3$. $\endgroup$
    – Dave Benson
    Commented Jun 15, 2023 at 19:03
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It seems to me that anything of the form $(c_0,\dots,c_n)=(\text{gigantic,tiny,...,tiny,non-zero tiny})$ will give you roots close to $-n,\dots,-1$, and since $c_n\ne 0$, we have $P(-1)\ne 0$. As an example, $(100,0,1)$ gives roots $(-299\pm\sqrt{8601})/202$, or roughly $-1.94$ and $-1.02$.

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