Hi all
thanks for the answers. Here is simple example. It's in fact
intersection of two
projective curves:
a cuspidal cubic curve y^2*z-x^3=0 and its polar w.r.t. (2:1:1) :
2*y*z + y^2 - 2*3*x^2.
when substituting z=1 i used:
solve([y^2-x^3==0, 2*y +y^2 -2*3*x^2==0], x, y).
The general case will be also intersection of two homogenous
polynomials of degrees n,n-1 with
three variables.
Thanks
Michael
On Jun 6, 1:34 pm, John Cremona <john.crem...@gmail.com> wrote:
> Can the original poster provide a (simple) example of the kind of set
> of equations he wants to solve? For example, are they polynomials in
> several variables, or more exotic? In the case of polynomial
> equations it is more likely that (perhaps via Singular) the
> multiplicities can be obtained.
>
> John Cremona
>
> On Jun 6, 8:22 am, simon.k...@uni-jena.de wrote:
>
> > Hi!
>
> > On 6 Jun., 05:45, Robert Dodier <robert.dod...@gmail.com> wrote:
>
> > > CVS log claims this bug was fixed recently (between 5.17 & 5.18).
> > > Here's what I get with Maxima from CVS (5.18+).
>
> > > ...
>
> > Very good! So, ticket #6228 can be closed when the new maxima version
> > is in Sage.
>
> > But I think we should now come back to the original poster's question:
> > - Can Sage provide the multiplicities for the solutions of a *set* of
> > nonlinear equations?
>
> > Can it?
> > At least, "multiplicities=True" seems to have no effect in "solve":
> > sage: solve((x^2-1)^3==0, x, multiplicities=True)
> > ([x == -1, x == 1], [3, 3])
> > sage: solve(((x^2-1)^3==0,(x^2-1)^3==0), x, multiplicities=True)
> > [[x == 1], [x == -1]]
>
> > This time, it is all Sage's fault, because maxima gives the right
> > answer:
> > sage: maxima.eval('solve(((z^2-1)^3,(z^2-1)^3),z)')
> > '[z=-1,z=1]'
> > sage: maxima.eval('multiplicities')
> > '[3,3]'
>
> > Hence, it is not the same as ticket #6228.
> > Therefore I opened a new one,
> > namelyhttp://trac.sagemath.org/sage_trac/ticket/6231
>
> > Cheers,
> > Simon
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