Q:

Pade approximant to exponential function

Suppose:
a) $p(z)$ is an even degree polynomial (of degree $k = 2j$) with real coefficients;
b) $p(0) = 1$;
c) $p(z)$ and $p(-z)$ have no roots in common anywhere in the complex plane;
d) $f(z) = p(z)/p(-z)$ is a Pade approximant to $\exp(z) = e^z$, such that
    the Taylor expansion of $f(z)$ agrees with that of $\exp(z)$ up to $(2k)$th order.
Are there simple proofs of the following conjectures:
i)  the coefficients of $p(z)$ are all positive
ii) $f(x) \le \exp(x)$ for all nonnegative $x$
iii) $p(x) p(-x) \exp(ax)$ has all positive coefficients in its Taylor expansion for any $a \ge 1$
iv) $p(x) p(-x)$ has no real roots.
Comments:
By Descartes rule of signs, (i) implies $p(z)$ has no positive roots.
  By a theorem of Laguerre (Ouvres, Tome 1), (iii) would imply (iv)

A:

Well, there are explicit expressions for the numerator $N_{pq}(z)$ and denominator $D_{pq}(z)$ of the $(p,q)$ Padé approximant for $\exp(z)$:
$\displaystyle N_{pq}(z)=\sum_{j=0}^p \frac{(p+q-j)!p!}{(p+q)!j!(p-j)!}z^j$
$\displaystyle D_{pq}(z)=\sum_{j=0}^q \frac{(p+q-j)!q!}{(p+q)!j!(q-j)!}(-z)^j$
from which the Padé approximant is $R_{pq}(z)=\frac{N_{pq}(z)}{D_{pq}(z)}$.
You should be able to prove (i)-(iv) from these expressions.

