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Re:0^0 by thomasahle on Thursday April 03, 2014 @04:35AM (#7521602)
What do you think the definition of ^ is then?

The ^ operator is defined as:

  0^0 = 1
  x^y = exp(ylog(x))  if  x != 0
  0^y = 0             if  y != 0
Or similarly set theoretically.
Re:0^0 by Chattered on Thursday April 03, 2014 @08:09AM (#7522244)
I'm happier starting with the natural numbers, and just defining m^n as m multiplied by itself n times. Everyone gets that this is the point of exponentiation of natural numbers, and it has the obvious recursive definition:

    m^0 = 1
    m^n = m^(n-1).
And now 0^0 = 1 follows immediately, just as it does for 0+0 = 0 and the recursive definition of addition in terms of successor.

After that, I consider the question of what x^y means for non naturals as motivated largely by considerations of algebraic and analytic extension. We get negative exponents simply by extending the operation with the requirement that

    m^0 = 1
    m^(n+p) = m^n*m^p.
We get rational exponents by extending the operation with the requirement that

    (m^n)^p = m^(n*p).
We get real exponents by extending the operation to the limits where they exist. We get complex exponents by extending the operation to the largest analytic extension.

The tension then is that the real and complex extensions don't "confirm" the original reasonable setting of 0^0=1, but I personally find it more objectionable poking a hole in the function, when it was so clear in the case of natural numbers that 0^0 should be 1 (and see the set theoretic and category theoretic accounts for more reasons why this is clear).

Re:0^0 by tripzilch on Sunday April 06, 2014 @06:06AM (#7540467)
Most real numbers only exist in fever dreams and acid trips, anyway.
Re:0^0 by pavelrub on Thursday April 03, 2014 @05:02AM (#7521689)
Whatever it is, 2^1=2 is not part of it - it follows from it.
Re:0^0 by thomasahle on Thursday April 03, 2014 @06:40AM (#7521959)
Uhm, anything that is part of a definition follows from that definition.

`A and B implies A' and so on.

In the end a function, in the mathematical sense, is just a long (possibly infinite) table of `from' and `to' values. We could argue that one definition of ^ is more or less complex than another, for some definition of complexity, but I'm not sure which would win, and if it would be useful.

Re:0^0 by pavelrub on Thursday April 03, 2014 @10:26AM (#7523009)
"Follows" in the sense that it is not an axiom - it has a proof of more than one line in the sense of the Hilbert system. Seeing functions as collections of ordered pairs tells us nothing here.

Your definition of ^ doesn't directly tell me that 2^1=2. It does tell me however that 0^0=1.

Re:0^0 by Aardwolf on Thursday April 03, 2014 @05:07AM (#7521706)
Isn't that a circular definition, since, exp is e^x?
Re:0^0 by thomasahle on Thursday April 03, 2014 @06:38AM (#7521956)
We can just define exp as:

  exp(x) = 1 + x/1! + x*x/2! + x*x*x/3! + ...
Re:0^0 by taejo on Thursday April 03, 2014 @05:51AM (#7521828)
If one defines x^y this way, one usually defines exp by a power series, and then defines e as the number such that e^x = exp(x).
Re:0^0 by taejo on Friday April 04, 2014 @07:32AM (#7530575)
Or of course just e = exp(1)