That's pretty disturbing because those complex roots should have
multiplicity 3.
Is this a known bug?
-Marshall Hampton
On Jun 5, 1:05 pm, simon.k...@uni-jena.de wrote:
> Hi Michael,
>
> On 5 Jun., 18:26, Michael Friedman <ois3...@gmail.com> wrote:
>
> > I'm pretty new to Sage, so I'm sorry in advance for the trivial
> > question.
> > I have a set of (non-linear) equations, and I need to find the
> > multiplicity of each solution. How do I do it?
>
> First of all, solving a nonlinear eqution is not a trivial question
> IMHO.
>
> There are various useful ways of getting help from Sage. One is
> "search_def". When I did
> sage: search_def('multiplicities')
> I got three replies, two of them in the module
> sage.symbolic.expression: There is a multiplicity option for the
> methods "roots" and "solve".
>
> So, you could define some symbolic expression and apply the solve
> method, e.g.:
> sage: z = var('z')
> sage: E=(z^3-1)^3
> sage: E.solve(z, multiplicities=True)
> ([z == (sqrt(3)*I - 1)/2, z == (-sqrt(3)*I - 1)/2, z == 1], [1, 1,
> 3])
> sage: E.roots(z, multiplicities=True)
> [((sqrt(3)*I - 1)/2, 1), ((-sqrt(3)*I - 1)/2, 1), (1, 3)]
>
> Apparently the "multiplicities" parameter is also available in the
> "solve" function:
> sage: solve(E==0,multiplicities=True)
> ([z == (sqrt(3)*I - 1)/2, z == (-sqrt(3)*I - 1)/2, z == 1], [1, 1,
> 3])
>
> I hope the multiplicities 1, 1 and 3 are correct (didn't think about
> it, but it seems a bit odd to me).
>
> Unfortunately that option seems to be not documented in the "solve"
> function. But, if you want to see the documentation, do
> sage: solve?
>
> Best regards,
> Simon
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