Real rootedness of a polynomial with binomial coefficients

Question

It is possible to show using diverse techniques that the following polynomial:

$$P_n(x)=1 + \binom{n}{2} x + \binom{n}{4} x^2 + \binom{n}{6} x^3 + \binom{n}{8} x^4 +\ldots + \binom{n}{2\lfloor\tfrac{n}{2}\rfloor} x^{\lfloor \frac{n}{2}\rfloor},$$

is real-rooted. For instance, it is an $s$-Eulerian polynomial, for which Savage and and Visontai in [1] have proved real-rootedness.

Here I ask for a very similar polynomial, which I think might be solved with some other technique that perhaps I am overlooking.

$$Q_n(x)=1 + \left(\binom{n}{2}-n\right) x + \binom{n}{4} x^2 + \binom{n}{6} x^3 + \binom{n}{8} x^4 +\ldots + \binom{n}{2\lfloor\tfrac{n}{2}\rfloor} x^{\lfloor \frac{n}{2}\rfloor},$$

This polynomial $Q_n(x) = P_n(x) - nx$ is the Ehrhart $h^*$-polynomial of the hypersimplex $\Delta_{2,n}$. There are several conjectures regarding unimodality/log-concavity/real-rootedness for the $h^*$-polynomial of an arbitrary hypersimplex $\Delta_{k,n}$, but for $k > 2$ the formulas are much more complicated.

Answer 1

Let's consider $n\geqslant 5$ case only ($n=1,2,3,4$ are straightforward). Then $Q_n$ has non-negative coefficients and we care on the number of negative roots of $Q_n$.

We have $2P_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n$. So, we should prove that $$2Q_n(-x)=(1+i\sqrt{x})^n+(1-i\sqrt{x})^n+2nx$$ has $\lfloor n/2\rfloor$ positive roots. Denote $x=\tan^2 t$, $0<t<\pi/2$ and $$h(t):=Q_n(\tan^2 t)=\frac{\cos nt}{\cos^n t}+n\tan^2 t=\frac{\cos nt+n\sin^2 t\cos^{n-2}t}{\cos^n t}.$$

Denote $a=n/2-1$ and note that the maximal value $(1-x)x^a$ over $x\in [0,1]$ is obtained for $x=a/(a+1)$ and equals $\frac{1}{(a+1)(1+1/a)^a}<\frac{1}{2(a+1)}=\frac1n$, applying this for $x=\cos^2 t$ we see that $0\leqslant n\sin^2 t\cos^{n-2}t<1$, thus the signs of $h(t)$ at the points $\pi k/n$, $k=0, 1,\ldots, \lfloor n/2\rfloor$ interchange and $h$ has at least $\lfloor n/2\rfloor$ distinct roots on $(0,\pi/2)$, as desired.