Ooooooooooooo
This looks really promising. It sucks the paper is structured in the exact opposite direction, lol. I know how to go from repelling to attracting fast, that's not too hard, as iterating the exponential will always converge, but for the repelling case this isn't so clear. I'll have to experiment with this. This analysis is a little adjacent to what I was looking for, but I think there might be something I can unearth from here.
MO, suggested that I look at \(f(y)\) about \(y = e\) because there should be a branching point centered around that point. By this you can define a function holomorphic on \(|\log(y)| \neq 1\) with a fixed/branch point at \(e\) (that may be holomorphic there), such that:
\[
f(y) : \{|\log(y)| < 1\} \to \{|\log(y)| > 1\}\\
\]
Up to here everything is possible. Someone on MO suggests I try to develop a series about \(e\) from here. Which honestly sounds like a closed form using W-Lambert and one of its branches is really possible from here. (See how the inverse of \(w^{1/w}\) is expressible in W-lambert.) I'd just have to choose a different branch of the inverse. This sounds troubling, because I hate W-Lambert, lol. I'll see if I can go on a deepdive using W-lambert.
let:
\[
F(y^{1/y}) = y\,\,\text{for}\,|\log(y)| > 1\\
\]
Where \(F\) is some W-lambert atrocity. This would actually solve my problem, and I wouldn't need \(f\). If I have an attracting fixed point, take the self root and run an inverse W-lambert trick. YA, this should work.
Now to read up how to run W-Lambert. Skipped all those lessons, lmao.
EDIT:
I answered my own question on MO at https://mathoverflow.net/questions/42615...attracting
This entirely solves the problem. But I don't know what branch of the Lambert function to use as of now.
Does anyone have a good formula for how the various branches of Lambert relate to the Lambert formula for the inverse of:
\[
g(y) = y^{1/y}\\
\]
This looks really promising. It sucks the paper is structured in the exact opposite direction, lol. I know how to go from repelling to attracting fast, that's not too hard, as iterating the exponential will always converge, but for the repelling case this isn't so clear. I'll have to experiment with this. This analysis is a little adjacent to what I was looking for, but I think there might be something I can unearth from here.
MO, suggested that I look at \(f(y)\) about \(y = e\) because there should be a branching point centered around that point. By this you can define a function holomorphic on \(|\log(y)| \neq 1\) with a fixed/branch point at \(e\) (that may be holomorphic there), such that:
\[
f(y) : \{|\log(y)| < 1\} \to \{|\log(y)| > 1\}\\
\]
Up to here everything is possible. Someone on MO suggests I try to develop a series about \(e\) from here. Which honestly sounds like a closed form using W-Lambert and one of its branches is really possible from here. (See how the inverse of \(w^{1/w}\) is expressible in W-lambert.) I'd just have to choose a different branch of the inverse. This sounds troubling, because I hate W-Lambert, lol. I'll see if I can go on a deepdive using W-lambert.
let:
\[
F(y^{1/y}) = y\,\,\text{for}\,|\log(y)| > 1\\
\]
Where \(F\) is some W-lambert atrocity. This would actually solve my problem, and I wouldn't need \(f\). If I have an attracting fixed point, take the self root and run an inverse W-lambert trick. YA, this should work.
Now to read up how to run W-Lambert. Skipped all those lessons, lmao.
EDIT:
I answered my own question on MO at https://mathoverflow.net/questions/42615...attracting
This entirely solves the problem. But I don't know what branch of the Lambert function to use as of now.
Does anyone have a good formula for how the various branches of Lambert relate to the Lambert formula for the inverse of:
\[
g(y) = y^{1/y}\\
\]
