But you can't create a constant Poly without giving the domain. Try
something like the following (and, again, trace the code to see if there is
a more direct way of learning the domain):
>>> Poly(1.2+var('x'))
Poly(1.0*x + 1.2, x, domain='RR')
>>> Poly(1+var('x'))
Poly(x + 1, x, domain='ZZ')
>>> Poly(1/S(3)+var('x'))
Poly(x + 1/3, x, domain='QQ')
>>> _.domain
QQ
On Friday, June 27, 2014 10:44:06 AM UTC-5, Aaron Meurer wrote:
>
> If you create a constant Poly, it picks the domain. You should look at
> the source to see how it picks that.
>
> Aaron Meurer
>
> On Sat, Jun 21, 2014 at 10:48 AM, Christophe Bal <proj...@gmail.com
> <javascript:>> wrote:
> > Hello.
> >
> > I do not know if it is the case but I think that sympy should have a
> domain
> > method for expressions. This will avoid error like for example float
> > calculations raising to 1 for example that wi-ould not be a natural but
> a
> > float.
> >
> > C.
> >
> >
> > 2014-06-21 17:08 GMT+02:00 Saurabh Jha <saurab...@gmail.com
> <javascript:>>:
> >
> >> I think that makes sense. I think this functionality is already
> >> implemented in Polys module in some form. I am not able to pin point
> it.
> >>
> >>
> >>>
> >>> First, it' unnecessarily slow.
> >>> Just check whether f == round(f) for floats.
> >>>
> >>> Second, the question whether a float is integral or not borders on
> >>> madness - you never know whether the answer "it's a natural number"
> >>> comes from the round-off lottery or is genuine.
> >>>
> >>> Third, regular expressions aren't going to give any meaningful results
> >>> for symbolic expressions, and the "Sym" in SymPy defines its mission
> >>> statement: symbolic math.
> >>>
> >>> If you pursue a way to find out whether a *symbolic expression* is in
> >>> ZZ, QQ, Complex, or whatever, then that would be worth thinking about.
> >>
> >>
> >>
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