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I've found (as have others), that for some analytic functions, a Padé approximant of it has an infinite convergence radius, whereas its associated Taylor series has a finite convergence radius. $f(x)=\sqrt{1+x^2}$ appears to be one such function. My questions are:

1) Is there any function where the Taylor series has the largest convergence radius of all associated Padé approximants? If so, is the Taylor series radius strictly larger, or only equal to the convergence radius of other Padé approximants (i.e. excluding the Taylor series itself)?

2) If not, is there any function that is analytic everywhere, and yet for which there is no (limit of) Padé approximant(s) that has an infinite convergence radius?

It would be both very cool and very useful if there is always a (limit of) Padé approximant(s) that has an infinite convergence radius for any function that is analytic everywhere, though I haven't the slightest how one checks/analyzes convergence of Padé approximants if the degrees of numerator and denominator both approach infinity. :)

One extra question, if there is always such a Padé approximant:

3) Is there always a numerically stable method of computing this approximant up to a finite order?

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  • $\begingroup$ I don't have a good answer to your questions (I'm still trying to read the book by Baker and Graves-Morris), but I wish to direct your attention to the theorem by Montessus de Ballore, which guarantees the uniform convergence of a sequence of Padé approximants of increasing numerator degree and fixed denominator degree, assuming that the function being approximated is meromorphic within a disk, and all the poles of the function within said disk are simple. $\endgroup$
    – J. M. isn't a mathematician
    Commented Aug 30, 2010 at 13:10
  • $\begingroup$ Look at the function more closely - it has a branch cut (in the complex plane). So, over the complex, you won't have good convergence properties. Over the real line, sure, it will behave very well. But that seems more like an accident, no? Just rotate your function by a quarter turn, and now things are ugly again. Pade works best when you have simple poles, not branch cuts. $\endgroup$
    – Jacques Carette
    Commented Sep 9, 2010 at 0:44

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Don't be lured into thinking Pade approximates are 'nice'. Here is why:

Notation:

Let $[L/M]_f(z)$ be the Pade approximation $\frac{p_n(z)}{q_m(z)}$ to $f$ where $\text{deg}(p_n)\leq n$ and $\text{deg}(q_m)\leq m$, $q_m(0)=1$.

Parital Answer:

1) The partial theta function $h_q(z)$ with $q=e^{iz}$ and $z/2\pi$ real and irrational is a counterexample by Lubinsky and Saff. It is analytic in $|z|<1$ but there is no subsequence of pade approximants $[L/M]_f(z)$ with $M$ fixed that converges for all $|z|<1$. See Pade Approximants by Baker and Braves Morris, p279.

2) A theorem by Wallin shows that there is an entire function such that the approximant sequence $[L/L]_f(z)$, $L\in\mathbb{N}$ is unbounded in $z\neq0$. (I cannot remember a reference for this.) However, I don't have an answer if you consider pade sequences $[L/M]_f(z)$ where $L$ and $M$ are two independent sequences.

3) For special sequences of Pade approximations (i.e. $[L/L]_f(z)$ -the diagonal sequence) there are numerically stable ways to calculate the Pade approximant. Not sure in general.

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  • $\begingroup$ Wynn ε is the standard method if you're interested only in evaluation of the diagonal approximants; getting the coefficients of the rational function require the solution of an appropriate Toeplitz system built from the series coefficients. For following arbitrary paths on the Padé table, Rutishauser's quotient-difference algorithm is used, but it is unstable in inexact arithmetic. $\endgroup$
    – J. M. isn't a mathematician
    Commented Sep 8, 2010 at 20:40

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