While we are on this topic, I must mention that directed infinities
are missing from Sympy. Are there any plans of incorporating this in
the near future?

On Apr 19, 1:10 am, "Aaron S. Meurer" <[email protected]> wrote:
> I think this is the correct way of thinking about it. Basically, the 
> expression is not well-defined to any one point, so we return nan.  A rule of 
> thumb is that any of the l'Hopital's rule "indeterminate forms" should return 
> nan.
>
> The one exception to this rule is that we follow Python's convention that 
> 0**0 == 1 (seehttp://code.google.com/p/sympy/issues/detail?id=2260).
>
> Aaron Meurer
>
> On Apr 18, 2011, at 12:07 PM, Tom Bachmann wrote:
>
>
>
>
>
>
>
> > Imho there are (at least) two ways of thinking about this. One
> > consistent way is the following:
>
> > Let c, d be "extended numbers" (reals, complexes, oo, zoo, etc).
> > Let f(x,y) be any function of two "ordinary numbers".
>
> > We define f(c,d) = e if for all expressions e1, e2 such that limit(e1)
> > = c, limit(e2) = d we have limit(f(e1, e2)) = e.
> > We define f(c,d) = nan if there exist expressions e1, e2, e3, e4 with
> > limit(e1) = c, limit(e2) = c, limit(e3) = d, limit(e4) = d and
> > limit(f(e1, e3)) != limit(f(e2, e4)).
> > [It is understood here that all limits are of the same kind, e.g. all
> > as x->0 in C or all es x->oo in R, or whatever.]
>
> > In particular then for f(x,y) = x - y, it follows that zoo-zoo = nan,
> > since e1=z, e2=z, e3=z+1, e4=z have the desired properties.
>
> > A slightly less ad-hoc way of thinking about this is using filters,
> > c.f. the wiki [1].
>
> > Actually I think that no matter what the "high-powered" definition is
> > going to be, the first property I described should be whenever the
> > result is not nan.
>
> > Obviously not more involved in sympy than you are, I'm just reporting
> > what I have caught up, hence take my words with a (big) grain of
> > salt :-).
>
> > [1]https://github.com/sympy/sympy/wiki/Infinities-and-Singularities
>
> > On 18 Apr., 17:33, Saptarshi Mandal <[email protected]> wrote:
> >> I am just having difficulty wrapping my head around zoo - zoo. Complex
> >> infinity should be unsigned. And even if you did write something like
> >> this should this be nan or zoo?
>
> >> Otoh, since zoo - zoo doesnt make sense, S.NaN should actually be the
> >> correct answer. Can someone more knowledgeable about this enlighten
> >> me?
>
> >> On Apr 18, 4:36 pm, "Alexey U. Gudchenko" <[email protected]> wrote:
>
> >>> 18.04.2011 08:20, smichr пишет:
>
> >>>> Does zoo absorb everything in an addition or multiplication (except
> >>>> maybe oo or another zoo)?
>
> >>> It is related with this old 
> >>> issue:http://code.google.com/p/sympy/issues/detail?id=360
>
> >>> And with this open pull 
> >>> request:https://github.com/sympy/sympy/pull/173/files
>
> >>> E.g. Those assertions have quit right behavior:
>
> >>> assert zoo == zoo
> >>> assert zoo != oo
> >>> assert 1*zoo == zoo
> >>> assert 2*zoo == zoo
> >>> assert 1 != zoo
> >>> assert zoo != 1
> >>> assert zoo != Symbol("x")**3
> >>> assert zoo + 1 == zoo + 1
> >>> assert zoo + 1 == zoo
> >>> assert 2 + zoo == zoo
> >>> assert 3*zoo + 2 == zoo
> >>> assert 1/zoo == 0
>
> >>> x = Symbol('x')
>
> >>> assert zoo + zoo == nan
> >>> assert zoo - zoo == nan
> >>> assert zoo + oo == nan
> >>> assert zoo - oo == nan
>
> >>> assert zoo + I == zoo
> >>> assert zoo * I == zoo
> >>> assert zoo + x + 2 + 3*I == zoo + x
> >>> assert zoo * x * 2 * 3*I * -oo == zoo * x
>
> >>> assert zoo * (1+I) == zoo
> >>> assert zoo * (4+I+x) != zoo
>
> >>> assert zoo * 0 == nan
> >>> assert zoo * nan == nan
> >>> assert 0 * zoo == nan
> >>> assert nan * zoo == nan
> >>> assert zoo + nan == nan
>
> >>> assert zoo**zoo == nan
> >>> assert zoo**0   == nan
> >>> assert zoo**2   == zoo
> >>> assert zoo**(-1) == 0
>
> >>> assert abs(zoo) == oo
>
> >>> If anyone disagree with those assertions please let us know.
>
> >>> --
> >>> Alexey U.
>
> > --
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