> it is (in some sense) arbitrary that X^Y means Y -> X or its size (cardinal exponentiation)
Are you referring to the notation? I don’t think notation is what’s being contested.
> in analysis may mean exp(y log x)
Doesn’t work with a base of 0.
> We accept that for natural X,Y except (0,0)
Why “except”?
> that does not imply that they should agree at (0,0), because the definitions are simply not the same.
The definition of exponentiation of reals typically starts with exponentiation of naturals as a given (see Baby Rudin).
> The usual way to transport a definition from a discrete domain to a continuous one is a technique called analytic continuation. I am curious if there is an analytic continuation of the discrete X^Y which contains 0^0=1
Analytic continuation refers to something else, but I get what you’re trying to say.
The answer is simple: Real exponentiation is an extension of natural exponentiation. Hence, it should have the same value as the latter wherever the latter is defined.
Yes, I’ve read that SE question before. It’s good that you brought it up. I recommend reading the comments under the accepted answer and the other answers as well.
Are you referring to the notation? I don’t think notation is what’s being contested.
> in analysis may mean exp(y log x)
Doesn’t work with a base of 0.
> We accept that for natural X,Y except (0,0)
Why “except”?
> that does not imply that they should agree at (0,0), because the definitions are simply not the same.
The definition of exponentiation of reals typically starts with exponentiation of naturals as a given (see Baby Rudin).
> The usual way to transport a definition from a discrete domain to a continuous one is a technique called analytic continuation. I am curious if there is an analytic continuation of the discrete X^Y which contains 0^0=1
Analytic continuation refers to something else, but I get what you’re trying to say.
The answer is simple: Real exponentiation is an extension of natural exponentiation. Hence, it should have the same value as the latter wherever the latter is defined.
Yes, I’ve read that SE question before. It’s good that you brought it up. I recommend reading the comments under the accepted answer and the other answers as well.