On 11 January 2012 14:45, Aaron Meurer <[email protected]> wrote:

> I think it's reasonable to keep it as it is now.  The reason is that
> we can then say that expr.subs(x, sol) will give 0.  This wouldn't
> hold for solutions based on continuous extension.
>
> I was going to say that you could use cancel() to get rid of these,
> but that only holds for rational functions. For example, sin(x)/x - 1
> has a "zero" at x = 0, but the only way to get this from solve is to
> give check=False. So I think we should keep it like this, but document
> that if you want the others, you should pass check=False to solve().
>
I've added a documentation issue
http://code.google.com/p/sympy/issues/detail?id=2981

>
> In [31]: solve(sin(x)/x, x)
> Out[31]: []
>
> In [32]: solve(sin(x)/x, x, check=False)
> Out[32]: [0]
>
> This is kind of analogous the force option to many of the simplify()
> options.  By default, we are rigorous, and only do the transformation
> if it is valid.  But we allow force=True to give the algebraic
> solution. Rigorously, x/(x**2 + x) is not equal to 1/(x + 1), but many
> fields consider them to be equivalent for the sake of simplification
> of the theory.
>
> Aaron Meurer
>
> On Wed, Jan 11, 2012 at 6:28 AM, Chris Smith <[email protected]> wrote:
> >> No, no, it is continuous  because the limit when x-->0 exists (equals
> >> 0), and the same as a value of function at this point, 0**2/0 (which by
> >> definition is equal 0).
> >>
> > In earlier discussions we decided not to allow solutions of equations
> > that set any denominator to zero even though the limit at that point
> > might exist. I wonder if that should be changed so that if the limit
> > exists and is the same from both directions the solution is returned;
> >
> >>>> e=(x**2*(1/x - z**2/x))
> >>>> e.expand()
> > -x*z**2 + x
> >>>> solve(_,x)
> > [0]
> >>>> solve(e,x)
> > []
> >>>> e.subs(x,0)
> > 0
> >
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