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 $i\_d(n)$?



Positivity of the coefficients of the Ehrhart polynomial of a cross-polytope
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
$$\biggl(\sum\_{k\ge 0} x^k\biggr)
 \biggl(\sum\_{k\ge 0} \frac{x^{2k+1}}{2k+1}\biggr)^t.
$$


