Positivity of the coefficients of the Ehrhart polynomial of a cross-polytope

Question

Question 35996 asks about the Ehrhart polynomial $i_d(n)$ of the standard regular cross-polytope. It can be defined equivalently by $$ \sum_{n\geq 0}i_d(n)x^n = \frac{(1+x)^d}{(1-x)^{d+1}}. $$ It can be shown that the coefficients of $i_d(n)$ are positive, using Theorem 3.2 of http://math.mit.edu/~rstan/papers/cycles.pdf to show that all zeros of $i_d(n)$ have real part $-1/2$. Is there some "positive" formula for $i_d(n)$ that makes it transparent that the coefficients are positive? Or at least, is there another proof that doesn't involve the zeros of $i_d(n)$?

Answer 1

Here is a very simple way to show the positivity. Define $$f(d,x) = (1+x)^d/(1-x)^{d+1}.$$ Then, by induction $$ \frac{\partial^t f(d,x)}{\partial\, d^t} = f(d,x) \, \ln\biggl(\frac{1+x}{1-x}\biggr)^t.$$ Putting in $d=0$ we have that the Taylor series of $f(d,x)$ with respect to $d$ is $$ f(d,x) = \sum_{t=0}^\infty \frac{1}{t!} (1-x)^{-1} \ln\biggl(\frac{1+x}{1-x}\biggr)^t\,d^t. $$ Both $(1-x)^{-1}$ and $\ln\Bigl(\frac{1+x}{1-x}\Bigr)$ have non-negative Taylor coefficients, which completes the proof.

In summary, the coefficient of $x^nd^t$ is $2^t/t!$ times the coefficient of $x^n$ in $$\biggl(\sum_{k\ge 0} x^k\biggr) \biggl(\sum_{k\ge 0} \frac{x^{2k+1}}{2k+1}\biggr)^t. $$

Answer 2

From the binomial theorem, $(1+x)^d=\sum_{j\geq0}\binom{d}jx^j$ while $$\frac1{(1-x)^{d+1}}=\sum_{k\geq0}\binom{d+k}kx^k.$$ Therefore, by Cauchy Product formula and as a polynomial in $n$, $i_d(n)$ takes the form $$i_d(n)=\sum_{k\geq0}\binom{d}{n-k}\binom{d+k}k \qquad \text{or} \qquad i_d(n)=\sum_{k\geq0}2^k\binom{d}k\binom{n}k.$$ If we employ Zeilberger's algorithm, we find the recurrence $$ (d+2)\cdot i_{d+2}(n)=(2n+1)\cdot i_{d+1}(n)+(d+1)\cdot i_d(n). \tag1$$ One may now proceed by induction on $d$, with $$i_1(n)=2n+1 \qquad \text{and} \qquad i_2(n)=2n^2+2n+1.$$ An aside: Equation (1) reveals $n=-\frac12$ is a root of $i_d(n)$ for all $d$ odd.