Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.