On 11/3/2013 8:32 AM, Jussi Piitulainen wrote: > fom writes: > >> 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. >
It had merely been a poor choice.
That is a polynomial form.
If 0^0 =/= 1, then a_0 * x^0 does not have a uniform interpretation for all possible values of x.
I had understood the exchanges in terms of deciding priority with respect to a continuous 0^x.
I chose a general form because so much mathematics is written that way. I simply put a bound on non-zero coefficients in keeping with practice for finitary mathematics. But you are correct that for x=0 there is an essential discontinuity for every term with a negative index and non-zero coefficient.