> On 11/3/2013 4:53 AM, Robin Chapman wrote: > > On 02/11/2013 21:38, Virgil wrote: > >> In article <firstname.lastname@example.org>, > >> PotatoSauce <email@example.com> wrote: > >> > >>> In practice, 0^0 = 1 works just fine. > >> > >> Unless one wants 0^X to be continuous at X = 0. > > > > Not every function is continuous .... > > > > Should summation forms > > > Sum k -oo to oo ( a_k * x^k ) > > a_k = 0 when |k| > M for some natural number M > > > over the reals be continuous?
I'm missing the relevance of your question.
Are you suggesting that 0^0 = 1 doesn't work in the reals because 1/x is not defined (and therefore not continuous) over all reals?
(That's a_k = [k = -1] for all k.)
Or are you really asking if 1/x over the reals somehow _should_ be continuous over the reals even though it isn't?
Either way, I'm still missing the relevance.
A third interpretation (that you are just contributing to the general nervous cluelessness) would give relevance but I don't expect that from you.