expr1 = parse_expr('x**3 - 3*x')
expr2 = parse_expr('-x-1')
print(expr1.subs({x: p_infinity}).evalf())
print(expr1.evalf(subs={x: p_infinity}))
print(expr2.subs({x: p_infinity}).evalf())
print(expr2.evalf(subs={x: p_infinity}))
output:
nan
+inf
-inf
+inf
Question was why method subs and arg subs in evalf() work different for
+-oo.
Anyway, thanks for the tip with limits. Works great.
среда, 4 декабря 2013 г., 22:11:31 UTC+4 пользователь Aaron Meurer написал:
>
> I just checked evalf and it returns the right answers, inf and -inf.
> So apparently it's smarter than I had thought.
>
> Aaron Meurer
>
> On Wed, Dec 4, 2013 at 5:04 AM, Sergey Kirpichev
> <[email protected]<javascript:>>
> wrote:
> >
> > On Wednesday, December 4, 2013 7:13:09 AM UTC+4, Aaron Meurer wrote:
> >>
> >> evalf won't help
> >> either: it's use is for numerical evaluation, but numerical inf
> >> behaves the same way.
> >
> >
> > Are you sure that evalf isn't broken here?
> >
> >
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