On Mon, 20 Feb 2012 10:39:11 -0500
Michael Orlitzky <[email protected]> wrote:
> On 02/19/12 19:58, Mike wrote:
> > When I run:
> >
> > x,y=var('x,y', domain=RR)
> > solve(2.0*x+3.0*y==4.0, y)
> >
> > I get
> >
> > [y == -2/3*x + 4/3]
> >
> > but I would like to get
> >
> > [y == -0.666666666666666*x + 1.3333333333333]
> >
> > How can I do this?
> >
>
> Wild guess: the float coefficients are coerced to QQ, because
> otherwise numerical inaccuracy would prevent us from finding a
> solution. For example,
>
> (0.333333... + 0.666666...)*x
>
> might not equal x.
>
> For a workaround, someone recently showed me this. You would call
> `symbolic_approx` on your result.
>
> ---
>
> class NumericEvaluator(Converter):
>
> def arithmetic(self, ex, operator):
> return reduce(operator, map(self, ex.operands()))
>
> def pyobject(self, ex, obj):
> return ex.n()
>
> def symbol(self, ex):
> return SR(ex)
>
> def symbolic_approx(expr):
> ne = NumericEvaluator()
> return ne(expr)
>
Or you can do this:
sage: t = -2/3*x + 4/3
sage: t._convert(RR)
-0.666666666666667*x + 1.33333333333333
Cheers,
Burcin
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