Re: Laurent polynomials instead of fractions

[Karim Belabas <Karim.Belabas@math.u-bordeaux.fr>]

* Dirk Laurie [2017-07-15 13:55]:
> 2017-07-15 9:12 GMT+02:00 Karim Belabas <Karim.Belabas@math.u-bordeaux.fr>:
> 
> > ? [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)
> 
> I didn't know 'valuation', was it been in Pari-GP in about 2005 when I
> learnt it?

Sure.

> The help is no more informative than the name:
> 
> ?valuation
> valuation(x,p): valuation of x with respect to p.

The *short* help (which, by design, is as terse as possible).

The *long* help is probably what you're looking for:

(14:25) gp > ??valuation
valuation(x,p):

   Computes the highest exponent of p dividing x.   If p is of type integer,  x
must be an integer,  an intmod whose modulus is divisible by p,  a fraction,  a
q-adic  number  with  q = p,  or a polynomial or power series in which case the
valuation is the minimum of the valuation of the coefficients.

   If  p  is  of  type  polynomial,   x  must be of type polynomial or rational
function, and also a power series if x is a monomial. Finally, the valuation of
a vector, complex or quadratic number is the minimum of the component
valuations.

   If  x  =  0,   the result is +oo if x is an exact object.   If x is a p-adic
numbers or power series, the result is the exponent of the zero. Any other type
combinations gives an error.



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]
`