Your explanation makes eminent sense! I did not think of this at all: surely a term 1^oo does not drop out of the sky, it has a ‚history‘. Thanks!
On Sun 7. Nov 2021 at 02:44 Oscar Benjamin <[email protected]> wrote: > On Sat, 6 Nov 2021 at 16:55, Peter Stahlecker > <[email protected]> wrote: > > > > I am just a hobby mathematician, but it seems to me like this: > > > > 1^oo := lim(1^n) = lim(1) = 1. > > > > The other 'limits' seem to me to be an inadmissible 'exchange' of limits: > > 1 != (1 + 1/n) for any finite n > > When implementing the internals of a computer algebra system you have > to be careful because the "input" expressions might arrive indirectly. > You have to think: what possible previous operations could have > resulted in the situation I have now which is to evaluate 1**oo? There > are many possible answers to that and they correspond to different > final answers. In general using oo in expressions is not well defined > if we don't specify how the limit should be taken but in certain cases > the result is the same in any case. Otherwise as the zen of Python > says: > """ > In the face of ambiguity, refuse the temptation to guess. > """ > > -- > Oscar > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To view this discussion on the web visit > https://groups.google.com/d/msgid/sympy/CAHVvXxTWVCrdQvHzpPuUh7Uz1JiP6j2HC%2BNsSpSm5YiY3jtK%2BA%40mail.gmail.com > . > -- Best regards, Peter Stahlecker -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/CABKqA0aKhzjM5fNgsegbfPNy1ehbRM8y66E9xHgqNMq962bSpA%40mail.gmail.com.
