09/24/2007, 05:42 PM
The reason why I ask is that you have to be very very careful when discussing all fixed points of exponentials, because there are so many branches in every direction.
For example lets define a generalized version of the infinitely iterated exponential like this:
where W_k is the k-th branch of the Lambert W-function. Using this definition of the infinitely iterated exponential function, we can describe all of the solutions of \( y = x^y \) as \( y = {}^{\infty}x_{(k0)} \) indexed by k, but all of the solutions of \( y^{1/y} = x \) as \( y = {}^{\infty}x_{(0j)} \) indexed by j. So by fixed-points I assume you mean the former, and not the latter. I could be wrong though, so please clarify.
Andrew Robbins
For example lets define a generalized version of the infinitely iterated exponential like this:
\(
{}^{\infty}x_{(kj)} = e^{-W_k(-\log(x)-2\pi i j)}
\)
{}^{\infty}x_{(kj)} = e^{-W_k(-\log(x)-2\pi i j)}
\)
where W_k is the k-th branch of the Lambert W-function. Using this definition of the infinitely iterated exponential function, we can describe all of the solutions of \( y = x^y \) as \( y = {}^{\infty}x_{(k0)} \) indexed by k, but all of the solutions of \( y^{1/y} = x \) as \( y = {}^{\infty}x_{(0j)} \) indexed by j. So by fixed-points I assume you mean the former, and not the latter. I could be wrong though, so please clarify.
Andrew Robbins
