1. Dec 15, 2008

### Tac-Tics

Earlier today, I came up with an explanation of why 0^0 is undefined in terms of properties of exponentiation. In it, I was treating exponentiation as a function from R^2 to R. Then, it occurred to me that the gradient of a function f(x,y) = x^y would be a horrible nightmare. Perhaps something very deserving of an extra credit question on some finals or something.

Anyway, I took a stab at the problem, and I wanted to double check I got the right answer.

Let grad f = (df/dx, df/dy). (The d's should be dels, but I'm lazy right now).

(grad f)(0, y) is undefined for all y <= 0.
(df/dy)(x, 0) = 0, for all x /= 0.
(df/dy)(x, y) = ln(y) y^x, for all other x and y
(df/dx)(x, 0) = 0, for all x /= 0
(df/dx)(x, y) = y x^(y-1), for y > 0, x /= 0.

It's really a mess! Exponentiation is a deadly mine field of holes and infinities! I even messed up on paper, saying when y = -1, df/dx = ln(x), which is true, but the gradient is complex at that point.

Can anyone find any mistakes I made? Does anyone have any other simple functions with crazy behavior like this? I think it's crazy at least =-)

2. Dec 15, 2008

### NoMoreExams

Maybe this is a minute point but your f is not R^2 -> R, for one thing that would imply that (0,0) is in your domain which you are claiming is impossible. Also you are implying that (-1, 1/2) is in your domain but that would map to C not R.

3. Dec 15, 2008

### Tac-Tics

This is exactly why I posed to this forum... because I knew I'd miss something.

I realized that what I really meant was f was a partial function on R^2. I totally missed the negative numbers to non-integral powers bit, though. Maybe it would make more sense to extend it to C^2->C, where the gradient would be a little more relaxed.