So, this may seem like an odd question, but can anyone describe for me the function that takes the attracting fixed point to the repelling fixed point? I currently have a subpar manner of calculating this, and I'm wondering if there is a quicker way.
Now, to explain, let \(|\log(y)| < 1\), so for example \(y=2\). I want a function \(f\) such that \(f(2) = 4\), and similarly for the other fixed point pairs.
That is:
\[
f(y)^{1/f(y)} = y^{1/y}\\
\]
or more specifically: \(\log(f(y))/f(y) = \log(y)/y\) for the principal branch of the log.
While \(|\log(y) | < 1\) and \(|\log f(y) | > 1\)?
I feel there must be some literature on this function. So far, I've been running the protocol:
\[
f(y) = \lim_{n\to\infty} \log^{\circ n}_{y^{1/y}}(e)\\
\]
Which works pretty well, but I'm not even sure works in the general sense. Is \(e\) always in the attracting orbit of \(\log_{y^{1/y}}(z)\) of the repelling fixed point \(f(y)\)? I feel like there may be counter examples, but I'm too lazy, and my code seems to be working
This problems been annoying me. And it's either something gnawingly stupid that I missed, or it's a rather deep problem.
To reiterate the question. I'm looking for a function:
\[
\begin{align}
f : \{y \in \mathbb{C}\,|\,|\log(y)| < 1\} &\to \{ y \in \mathbb{C}\,|\, |\log(y)| > 1\}\\
f(y)^{1/f(y)} &= y^{1/y}\\
\end{align}
\]
I might ask this on MO, but god knows they're all a bunch of c**** there.
Regards, James
EDIT:
here's the MO question link https://mathoverflow.net/questions/42615...attracting
I'm expecting a bunch of c***** to discredit my question by tomorrow. I hate MO.
Now, to explain, let \(|\log(y)| < 1\), so for example \(y=2\). I want a function \(f\) such that \(f(2) = 4\), and similarly for the other fixed point pairs.
That is:
\[
f(y)^{1/f(y)} = y^{1/y}\\
\]
or more specifically: \(\log(f(y))/f(y) = \log(y)/y\) for the principal branch of the log.
While \(|\log(y) | < 1\) and \(|\log f(y) | > 1\)?
I feel there must be some literature on this function. So far, I've been running the protocol:
\[
f(y) = \lim_{n\to\infty} \log^{\circ n}_{y^{1/y}}(e)\\
\]
Which works pretty well, but I'm not even sure works in the general sense. Is \(e\) always in the attracting orbit of \(\log_{y^{1/y}}(z)\) of the repelling fixed point \(f(y)\)? I feel like there may be counter examples, but I'm too lazy, and my code seems to be working
This problems been annoying me. And it's either something gnawingly stupid that I missed, or it's a rather deep problem.
To reiterate the question. I'm looking for a function:
\[
\begin{align}
f : \{y \in \mathbb{C}\,|\,|\log(y)| < 1\} &\to \{ y \in \mathbb{C}\,|\, |\log(y)| > 1\}\\
f(y)^{1/f(y)} &= y^{1/y}\\
\end{align}
\]
I might ask this on MO, but god knows they're all a bunch of c**** there.
Regards, James
EDIT:
here's the MO question link https://mathoverflow.net/questions/42615...attracting
I'm expecting a bunch of c***** to discredit my question by tomorrow. I hate MO.
