4 ms·

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 s

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.

pavelrub 13y ago
Whatever it is, 2^1=2 is not part of it - it follows from it.
- thomasahle 13y ago
  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.
  - pavelrub 13y ago
    "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.
Aardwolf 13y ago
Isn't that a circular definition, since, exp is e^x?
- taejo 13y ago
  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).
  - taejo 13y ago
    Or of course just e = exp(1)
- thomasahle 13y ago
  We can just define exp as: exp(x) = 1 + x/1! + x*x/2! + x*x*x/3! + ...
Chattered 13y ago
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).
- tripzilch 13y ago
  Most real numbers only exist in fever dreams and acid trips, anyway.

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.

pavelrub 13y ago

Whatever it is, 2^1=2 is not part of it - it follows from it.

thomasahle 13y ago
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.
- pavelrub 13y ago
  "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.

Aardwolf 13y ago

Isn't that a circular definition, since, exp is e^x?

taejo 13y ago
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).
- taejo 13y ago
  Or of course just e = exp(1)
thomasahle 13y ago
We can just define exp as: exp(x) = 1 + x/1! + x*x/2! + x*x*x/3! + ...

Chattered 13y ago

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).

tripzilch 13y ago
Most real numbers only exist in fever dreams and acid trips, anyway.