Let's take a beta function:
\[
\beta(s+1) = e^{\beta(s)}/(1+e^{-s})\\
\]
Let's call a curve \(\gamma\) such that \(\gamma = \{|\beta'(s)| = 1\}\).
Let's secondarily call a curve \(\rho\), such that for sheldon/Kneser's function \(\rho = \{|\text{tet}_K'(s)| = 1\}\). The value \(\frac{d}{dz}\log(f(z+1)) = \frac{\frac{d}{dz}e^{f(z)}}{e^{f(z)}} = f'(z)\). We know from the Kneser equation that:
\[
|\text{tet}_K'(s)| = 1
\]
Is a continuous and analytic arc. It encircles zero. If we write the function beta which looks like these things we should be fine. But we are testing where and when:
\[
|\beta'(z)| \approx |f'(z)|\\
\]
Where we can uniquely define \(f\) and \(f'\) as the functions such that:
\[
\begin{align}
f(z+1) &= e^{f(z)}\\
|f'(z)| &= 1 = |\frac{d}{dz}\log(f(z+1))| = |\frac{f'(z+1)}{f(z+1)}|\\
\end{align}
\]
This forces a single variable solution. Which turns out to be Kneser's solution near \(z \approx 0\). We're integrating around zero here.
The central idea is that if \(|\beta'(z)| = 1\) creates a path; then this maps to a circle on Kneser's tetration around \(\text{tet}_K(-1) = 0\).
This is all I'm willing to share before I have my code right. But I see how to turn any \(\beta\) function into any tetration
You can also see Kneser as the unique function \(f\) such that:
\[
|f'(z+1)| =|e^{f(z)}|\\
\]
if we assume that \(|f'(z)| =1\).........
This is explained through basic level set calculus....
\[
\beta(s+1) = e^{\beta(s)}/(1+e^{-s})\\
\]
Let's call a curve \(\gamma\) such that \(\gamma = \{|\beta'(s)| = 1\}\).
Let's secondarily call a curve \(\rho\), such that for sheldon/Kneser's function \(\rho = \{|\text{tet}_K'(s)| = 1\}\). The value \(\frac{d}{dz}\log(f(z+1)) = \frac{\frac{d}{dz}e^{f(z)}}{e^{f(z)}} = f'(z)\). We know from the Kneser equation that:
\[
|\text{tet}_K'(s)| = 1
\]
Is a continuous and analytic arc. It encircles zero. If we write the function beta which looks like these things we should be fine. But we are testing where and when:
\[
|\beta'(z)| \approx |f'(z)|\\
\]
Where we can uniquely define \(f\) and \(f'\) as the functions such that:
\[
\begin{align}
f(z+1) &= e^{f(z)}\\
|f'(z)| &= 1 = |\frac{d}{dz}\log(f(z+1))| = |\frac{f'(z+1)}{f(z+1)}|\\
\end{align}
\]
This forces a single variable solution. Which turns out to be Kneser's solution near \(z \approx 0\). We're integrating around zero here.
The central idea is that if \(|\beta'(z)| = 1\) creates a path; then this maps to a circle on Kneser's tetration around \(\text{tet}_K(-1) = 0\).
This is all I'm willing to share before I have my code right. But I see how to turn any \(\beta\) function into any tetration
You can also see Kneser as the unique function \(f\) such that:
\[
|f'(z+1)| =|e^{f(z)}|\\
\]
if we assume that \(|f'(z)| =1\).........
This is explained through basic level set calculus....
