Karim Belabas on Sat, 15 Jul 2017 09:12:42 +0200

 Re: Laurent polynomials instead of fractions

```* Emmanuel ROYER [2017-07-14 05:46]:
> Dear pari-users,
>
> How can I ask pari-gp to write the following expression
>
> ((xi^2 - 2*xi + 1)/xi) + ((-2*xi^4 + 8*xi^3 - 12*xi^2 + 8*xi -
> 2)/xi^2)*q + O(q^2)
>
> with the q-coefficients as Laurent polynomials in xi instead of fractions?

? f = ((xi^2 - 2*xi + 1)/xi) + ((-2*xi^4 + 8*xi^3 - 12*xi^2 + 8*xi - 2)/xi^2)*q + O(q^2);

For display purposes, bivariate objects (esp. power series) quickly
become unreadable, so transforming them to vectors (only for display)

? Vec(f)
%2 = [(xi^2 - 2*xi + 1)/xi, (-2*xi^4 + 8*xi^3 - 12*xi^2 + 8*xi - 2)/xi^2]

Now to display rational function as Laurent series, you can either
convert them to power series :

? Vec(f) * (1+O(xi^10))
%3 = [xi^-1 - 2 + xi + O(xi^9), -2*xi^-2 + 8*xi^-1 - 12 + 8*xi - 2*xi^2 + O(xi^8)]

or remove denominators (usually better) :

? [v = valuation(f,xi), Vec(f) / xi^v]
%4 = [-2, [xi^3 - 2*xi^2 + xi, -2*xi^4 + 8*xi^3 - 12*xi^2 + 8*xi - 2]]

(I'm displaying both the valuation and the renormalized coeffs here). With
this technique, there is no real need to convert to a vector, you can stick
to the power series

? f / xi^v
%5 = (xi^3 - 2*xi^2 + xi) + (-2*xi^4 + 8*xi^3 - 12*xi^2 + 8*xi - 2)*q + O(q^2)

Cheers,

K.B.
--
Karim Belabas, IMB (UMR 5251)  Tel: (+33) (0)5 40 00 26 17
Universite de Bordeaux         Fax: (+33) (0)5 40 00 21 23
351, cours de la Liberation    http://www.math.u-bordeaux.fr/~kbelabas/
F-33405 Talence (France)       http://pari.math.u-bordeaux.fr/  [PARI/GP]
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