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

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 $$i_d(n)$$?

• Possibly related?: mathoverflow.net/questions/308178/… – Sam Hopkins Feb 8 at 20:25
• @SamHopkins: It is related but not so useful in answering my question, since positivity is proved by the same Theorem 3.2. – Richard Stanley Feb 8 at 20:32
• I don't understand the difficulty: 1/(1-x) has Maclaurin series with positive coefficients, so that so do all its powers (which are known anyway), and multiplying by $(1+x)^d$ preserves this. Since the radius of convergence of the thing on the right is $1$, uniqueness yields the result. Moreover, it also follows that the coefficients form a log concave sequence ... (since those of $1/(1-x)$ and of $1+x$ do). – David Handelman Feb 8 at 22:19
• Now I see: $i_d (n)$ is supposed to be a polynomial itself in $x$. So ignore my remarks ... – David Handelman Feb 9 at 0:37
• It would be nice to include hyperlink to mentioned MO 35996 question mathoverflow.net/questions/35996/ehrhart-polynomial – Alexander Chervov Feb 9 at 19:26

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 $$\biggl(\sum_{k\ge 0} x^k\biggr) \biggl(\sum_{k\ge 0} \frac{x^{2k+1}}{2k+1}\biggr)^t.$$