Re: Padé approximants?

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

On Sun, Apr 07, 2019 at 02:46:23PM +0200, Dirk Laurie wrote:
> I can't find a built-in function that computes Padé approximants of a
> power series.
> 
> e.g. pade(1 + x + 1/2*x^2 + 1/6*x^3 + 1/24*x^4 + O(x^5)) → (x^2 + 6*x
> + 12)/(x^2 - 6*x + 12)
> 
> It is of course quite easy to program for oneself in the nondegenerate
> case (epsilon algorithm) but I wonder if it may be disguised as a
> special case of a more general function.

It is called bestapprPade:

? bestapprPade(1+x+1/2*x^2+1/6*x^3+1/24*x^4+O(x^5))
%19 = (x^2+6*x+12)/(x^2-6*x+12)

Full disclosure:
Henri Padé was a teacher at the French Lycée where I had been a student
a century later.

Cheers,
Bill.