On Thu, Jun 25, 2009 at 4:45 AM, Ronan Lamy<[email protected]> wrote: > >> Wolfram says that log(0) = -infinity. >> >> > Otherwise, limit(1/exp(log(-x), x, >> > 0, "+") = +infinity ! >> >> Check this link: >> http://www53.wolframalpha.com/input/?i=+limit+(1%2Fexp(log(-x))+as+x-%3E0%2B > > I don't think Mathematica computes limits by substituting x with 0 in > the expression, so it's not that relevant. My problem with using > log(0) = -infinity is the following: > > In [19]: (1/f(log(-x))).subs(x,0).subs(f, exp) > Out[19]: ∞ > > In [20]: (1/f(log(-x))).subs(f, exp).subs(x,0) > Out[20]: -∞ > > Different results for the same expression depending on the order of the > substitutions.
The problem here is that 1/0 returns positive infinity as opposed to complex infinity. In Mathematica both evaluations yield complex infinity. Fredrik --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sympy?hl=en -~----------~----~----~----~------~----~------~--~---
