
Re: The ambiguity of 0^0 on N
Posted:
Sep 19, 2013 4:27 PM


On Thu, 19 Sep 2013, Dan Christensen wrote:
> Yeah, it could get ugly, but until we can actually prove that 0^0=1 from > first principles, this would be the prudent thing to do.
I think, the easiest "first principle" is what another raeder proposed for natural numbers:
n^0 = 1 n^(k+1) = n·n^k
(And 0^k for k>0 is *not* an axiom but can be proven.)
No special case, no exception. And it reflects as a formula what mathematical common sense says:
n^0 is 1 because it is an empty product. 0^k is 0 because one factor is zero *if* there is at least one.
One could, of course, have started with the induction on n instead of k:
0^0 = x (to be determined) 0^k = 0 for k>0 (n+1)^k = Sum(i=0,...,k) C(k,i)·n^i·1^(ki) C(,) is binominal coeff. and an additional rule because here the definition of 1^1 is dependent on itself.
To evaluate this, one needs
1^k = Sum(i=0,...,k) C(k,i)·0^i·1^(ni) = C(k,0)·x·1^k
x = 1/C(0,0) = 1
 Helmut Richter

