On 11/02/2013 08:55 PM, Virgil wrote: > In article <email@example.com>, > "Michael F. Stemper" <firstname.lastname@example.org> wrote: > >> On 11/02/2013 04:38 PM, Virgil wrote: >>> In article <email@example.com>, >>> PotatoSauce <firstname.lastname@example.org> wrote: >>> >>>> In practice, 0^0 = 1 works just fine. >>> >>> Unless one wants 0^X to be continuous at X = 0. >> >> Which is normally not a requirement when working in the >> naturals -- the context of this discussion. I don't think >> that anybody's said that 0^0 can or should be defined for >> the reals. > > But why should one have to have different definitions for integers and > reals?
Because when you define the naturals as the finite cardinals, the definition of exponentiation gives 0^0 = 1 without any special handling.
Because the binomial theorem works without special cases if you define 0^0 = 1 in the naturals.
Because it's an empty product.
Because the power rule for differentiation doesn't need any special cases if you define 0^0 = 1 in the naturals.
Because Knuth said so.
-- Michael F. Stemper Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read.