- #1
Tac-Tics
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- 7
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 =-)
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 =-)