I observe following behavior:
In [113]: solve([Le((-1 + x),(-2 + x+x**3)), Assume(x,Q.real)], x,
relational=False)
Out[113]: [[1, ∞)]
In [114]: solve([Le((-1 + x)/(-2 + x+x**3),1), Assume(x,Q.real)], x,
relational=False)
NotImplementedError (...)
But there is a workaround: one must put the denominators to the second
side, observing their sign. This requires to write some useful (but
probably missing) routines, so it will take me a while. I will post an
example soon.
This will be a step forward from polynomials. Many more expressions in
real life can be simplified to a single fraction using together().
Btw: maybe the function should behave as if relational was set to
False by default; what do you think?
On Nov 18, 9:15 pm, Mateusz Paprocki <matt...@gmail.com> wrote:
> Hi,
>
>
>
>
>
>
>
>
>
> On Thu, Nov 18, 2010 at 11:51:59AM -0800, Filip Dominec wrote:
>
> > On Nov 18, 7:02 pm, Mateusz Paprocki <matt...@gmail.com> wrote:
> > > it can be easily extended
> > > to support rational functions and absolute values (somewhere I have
> > > preliminary code for this).
>
> > Cool. Right now I am using sympy to solve some calculations for
> > geometrical optics. These problems boiled down to a system of rational
> > equations with several variables. However, there are also several
> > constraints which would be handled the most efficient way if I could
> > simply calculate intersection of all the intervals for which the
> > inequalities hold.
>
> > This is my motivation to use the inequality solver. I would be happy
> > if I could get the fresh version from git, test it on a real problem
> > and report how it works.
>
> > > I think it shouldn't be very hard to write a function for
> > > converting relational to interval form (where it makes sense).
>
> > I expect it would not be hard, but I have not oriented in the code yet
> > to try it myself.
>
> Actually, it is possible to get intervals, you just need to set
> 'relational' flag to False, e.g.:
>
> In [1]: ieqs = [(x-1)*(x-2)*(x-3) >= 0, (x+1)*(x-2) >= 0]
>
> In [2]: solve(ieqs + [Assume(x, Q.real)], x, relational=False)
> Out[2]: [{2}, [3, ∞)]
>
> As you can see solve() can handle systems of inequalities (allowed
> operators are ==, !=, >, >=, <, <=). The inequality solver is run
> when at least one relational operator is encountered. If you set
> relational=False together with a complex variable, then you will
> get the result in relational form anyway.
>
> > Filip
>
> > --
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> --
> Mateusz
>
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