Re: Padé approximants?

[Dirk Laurie <dirk.laurie@gmail.com>]

Thanks. I see it is sorted under 4: NUMBER THEORETICAL functions
whereas I was looking for it in 10: POLYNOMIALS and power series.

-- Dirk

Op So. 7 Apr. 2019 om 15:32 het John Scott <jscott@posteo.net> geskryf:
>
> On Sunday, April 7, 2019 8:46:23 AM EDT Dirk Laurie wrote:
> > 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)
>
> Pari can compute Padé approximants. The function is called bestapprPade(), and
> happens to work with the same syntax you described.
>
> ? bestapprPade(1+x+1/2*x^2 + 1/6*x^3 + 1/24*x^4 + O(x^5))
> %1 = (x^2 + 6*x + 12)/(x^2 - 6*x + 12)