I'm not sure if this helps your situation or not, but if you are interested in the roots of "f(x)=0", then using roots has a much more predictable behaviour. So for example: sage: expr=(x^3+10*x^2+11*x+8) sage: expr.roots() <snip> sage: expr.roots(ring=RR) [(-8.86042628425072, 1)] sage: expr.roots(ring=CC) [(-8.86042628425072, 1), (-0.569786857874640 - 0.760417012386730*I, 1), (-0.569786857874640 + 0.760417012386730*I, 1)]
This way, you're giving just enough hint to SAGE to give you exactly
the answer you are looking for.
Soroosh
On Mon, 23 Aug 2010 20:42:39 +0100
robin hankin <hankin.ro...@gmail.com> wrote:
> Hi
>
> thanks for your earlier answers.
>
> I quite often do this:
>
> sage: solve(x^3 + 10*x^2+11*x+8==0,x)
> [snip]
>
> Then I realize that the analytic solution is rather complicated.
> So I want a numerical approximation.
>
> I tried this:
>
> roots = solve(x^3+10*x^2+11*x+8==0,x)
> sage: roots
> [x == -1/2*(1/3*sqrt(926) - 613/27)^(1/3)*(I*sqrt(3) + 1) -
> 1/18*(-67*I*sqrt(3) + 67)/(1/3*sqrt(926) - 613/27)^(1/3) - 10/3, x ==
> -1/2*(1/3*sqrt(926) - 613/27)^(1/3)*(-I*sqrt(3) + 1) -
> 1/18*(67*I*sqrt(3) + 67)/(1/3*sqrt(926) - 613/27)^(1/3) - 10/3, x ==
> (1/3*sqrt(926) - 613/27)^(1/3) + 67/9/(1/3*sqrt(926) - 613/27)^(1/3) -
> 10/3]
> sage: N(roots)
>
>
> but this returns an error ("too many values to unpack").
>
>
> The best I can do is
>
> N(roots[1].rhs())
>
> but this is just one at a time. How do I make N() operate on all of
> roots? Or is there a much neater way of accomplishing the same thing?
>
> cheers
>
> rksh
>
>
>
>
>
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