Hi Christophe,
On Fri, Jun 26, 2009 at 11:23:41AM +0200, Christophe wrote:
>
> Hello,
> the following code only gives [1] but I would also have the
> multilplicity of the root 1 (which is 3) :
> ===
> x = sympy.Symbol('x')
> f = '(x-1)**3'
> sympy.solve(f, x)
> ===
>
> I
Hello,
the following code only gives [1] but I would also have the
multilplicity of the root 1 (which is 3) :
===
x = sympy.Symbol('x')
f = '(x-1)**3'
sympy.solve(f, x)
===
Is there a way to know that ?
Christophe.
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Y
