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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)

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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.

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(i) is immediate, the others (if true) would seem to require some work. – Kevin O'Bryant Dec 31 '15 at 5:01

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