ALRIGHT! I GOT IT!
The following function \(\psi(z)\) takes Tetration to the beta method with multiplier \(1\). I'm not going to go into too much detail how I found this, but I've pulled 2 all nighters testing my hypothesis and it appears to be working.
We begin by restricting \(\Im(s) > 0\); a similar analysis works for the lower half plane, just exchange \(L\) (and \(P\)) for its conjugate. We start by writing a very very strange beast:
\[
\psi(s) = \Omega_{j=1}^\infty \frac{e^z}{1+e^{-\log^{\circ j}(s)}}\,\bullet z \Big{|}_{z=P}\\
\]
Now, this object only converges when we have the value \(z=P\). \(P\) is the unique value such that:
\[
\frac{e^P}{1+1/L} = P\\
\]
And is nearest to \(L\) to satisfy this. Numerical calculation has:
\[
P=0.1568938457.... + i0.8419125707...
\]
This is just your run of the mill infinite composition. So to check convergence we check that:
\[
\sum_{j=1}^\infty \left| \frac{e^P}{1+e^{-\log^{\circ j}(s)}} - P\right| < \infty\\
\]
Now we know that for large \(j > J\) that \(\log^{\circ j}(s) = L + \frac{(\log^{\circ j-1}(s)-L)}{L} + O(s-L)^2\): Which ends up looking like \(L + O(1/L^j)\). This is pretty standard in dynamics; the upper half plane \(\Im(s) > 0\) is in the attracting basin of the fixed point \(L\). Then from here, we know that:
\[
\frac{e^P}{1+e^{-\log^{\circ j}(s)}} = P + O(\frac{1}{L^j})\\
\]
Which is a convergent series. It's a little technical to show this specific function converges; but I can show it; it's just ugly as hell. The rigorous statement is that, if we assume \(z_j \to P\), and \(\sum_j |z_j - P| < \infty\) then:
\[
\sum_{j=1}^\infty \left| \frac{e^{z_j}}{1+e^{-\log^{\circ j}(s)}} - P\right| < \infty\\
\]
This is precisely what our infinite composition looks like; so we're golden and:
\[
\psi(s)\,\,\text{is holomorphic for}\,\,\Im(s) > 0\\
\]
We can equally find this function for \(\Im(s) < 0\), just write \(\overline{\psi(\overline{s})} = \psi(s)\).
Now this function satisfies:
\[
\psi(e^s) = \frac{e^{\psi(s)}}{1+e^{-s}}
\]
So If I write:
\[
\psi(\text{tet}_K(z)) = h(z)\\
\]
Then:
\[
h(z+1) = \psi(e^{\text{tet}_K(z)}) = \frac{e^{h(z)}}{1+e^{-\text{tet}_K(z)}}\\
\]
This is the unique equation satisfied by \(\beta\) though; and therefore:
\[
\psi(\text{tet}_K(z)) = \beta(s)\\
\]
Once we've made the appropriate change of variables. This becomes increasingly difficult, but should certainly be doable. I may have to edit somethings; but we're really fucking close. Using Kneser's slog will probably make an appearance at some point to get this perfect... I hope not, but I still remain optimistic nonetheless.
For large \(\Re(z)\) we have that \(\psi(z) \to z\), as expected; and we map \(\psi(L) = P\) and \(\psi(\overline{L}) = \overline{P}\)--which is exactly as expected. I was screwing up earlier because I was trying to map the level sets exactly--through a conjugation. This is much cleaner; and maps the level set to whatever Kneser is doing here... But the fact the level set is so good is a great marker. It means that:
\[
\left|\psi'(\text{tet}_K(z))\big{|}_{z \in \gamma}\right| = \left|\frac{1}{\text{tet}_K'(z)}\big{|}_{z \in \gamma}\right|
\]
This describes an unknown level set with a known level set. It might make sense here to change the level set from \(|\beta'(z)| = 1\) into something more relaxed. Nonetheless the function \(\psi\) is constructed. And it's fucking beautiful! I'm working on coding it at the moment, and graphing it. Currently my code is super super super fucking slow; but it's kinda getting the job done.
The math should be solid for this; but I haven't proven everything yet. This is what I like to refer to as a "fringe infinite composition"--so you have to be very careful with how you evaluate it.
Finding \(\psi\) seems super easy now, and by which; we have the inverse:
\[
\text{tet}_K(z) = \psi^{-1}(\beta(s))\\
\]
The variable change in \(s\) to \(z\) is the hard part for the moment, I'll figure it out
. One step at a time!
And I think that these should be relatively untaxing in a computational sense. We can develop the taylor series of \(\psi\) near infinity rather efficiently because \(\psi(z) \approx z\), and the drop off should be like \(\psi(z) = z + O(e^{-z})\)....
This is "converting beta to Kneser" at it's finest. We don't even need Kneser to do it!
Regards, James
HERE'S A BEAUTIFUL GRAPH FROM \(|\Re(z)| < 3\) and \(|\Im(z)| < 3\) of the function \(\psi\) such that:
\[
\psi(e^z) = \frac{e^{\psi(z)}}{1+e^{-z}}\\
\]
Here's an even bigger graph for \(|\Re(z)| < 8\) and \(|\Im(z)| < 8\)
The hypothesis for now is that:
\[
\psi(z) = s\\
\]
where under the \(\beta\) mapping we have that \(z \mapsto e^z\), we map to \(\beta(s+1) = \frac{e^{\beta(s)}}{1+e^{-s}}\). I need some time to think on this more though, but this is still crazy that \(\psi\) is so easily found!!!!!!!!! Even if this isn't the \(\psi\) we're exactly looking for; it's really close!
The exact formula we are seeing right now is that:
\[
\psi(\text{tet}_K(z+1)) = \beta(\text{tet}_K(z) + 1)\\
\]
Assuming that \(\text{tet}_K(z) \approx L\) this formula is super accurate. I'm having trouble going from here to the next step. I'm going to focus on coding \(\psi^{-1}\) first; which does the opposite transformation...
I'm still thinking on this, but I think I'm nearly there.
The following function \(\psi(z)\) takes Tetration to the beta method with multiplier \(1\). I'm not going to go into too much detail how I found this, but I've pulled 2 all nighters testing my hypothesis and it appears to be working.
We begin by restricting \(\Im(s) > 0\); a similar analysis works for the lower half plane, just exchange \(L\) (and \(P\)) for its conjugate. We start by writing a very very strange beast:
\[
\psi(s) = \Omega_{j=1}^\infty \frac{e^z}{1+e^{-\log^{\circ j}(s)}}\,\bullet z \Big{|}_{z=P}\\
\]
Now, this object only converges when we have the value \(z=P\). \(P\) is the unique value such that:
\[
\frac{e^P}{1+1/L} = P\\
\]
And is nearest to \(L\) to satisfy this. Numerical calculation has:
\[
P=0.1568938457.... + i0.8419125707...
\]
This is just your run of the mill infinite composition. So to check convergence we check that:
\[
\sum_{j=1}^\infty \left| \frac{e^P}{1+e^{-\log^{\circ j}(s)}} - P\right| < \infty\\
\]
Now we know that for large \(j > J\) that \(\log^{\circ j}(s) = L + \frac{(\log^{\circ j-1}(s)-L)}{L} + O(s-L)^2\): Which ends up looking like \(L + O(1/L^j)\). This is pretty standard in dynamics; the upper half plane \(\Im(s) > 0\) is in the attracting basin of the fixed point \(L\). Then from here, we know that:
\[
\frac{e^P}{1+e^{-\log^{\circ j}(s)}} = P + O(\frac{1}{L^j})\\
\]
Which is a convergent series. It's a little technical to show this specific function converges; but I can show it; it's just ugly as hell. The rigorous statement is that, if we assume \(z_j \to P\), and \(\sum_j |z_j - P| < \infty\) then:
\[
\sum_{j=1}^\infty \left| \frac{e^{z_j}}{1+e^{-\log^{\circ j}(s)}} - P\right| < \infty\\
\]
This is precisely what our infinite composition looks like; so we're golden and:
\[
\psi(s)\,\,\text{is holomorphic for}\,\,\Im(s) > 0\\
\]
We can equally find this function for \(\Im(s) < 0\), just write \(\overline{\psi(\overline{s})} = \psi(s)\).
Now this function satisfies:
\[
\psi(e^s) = \frac{e^{\psi(s)}}{1+e^{-s}}
\]
So If I write:
\[
\psi(\text{tet}_K(z)) = h(z)\\
\]
Then:
\[
h(z+1) = \psi(e^{\text{tet}_K(z)}) = \frac{e^{h(z)}}{1+e^{-\text{tet}_K(z)}}\\
\]
This is the unique equation satisfied by \(\beta\) though; and therefore:
\[
\psi(\text{tet}_K(z)) = \beta(s)\\
\]
Once we've made the appropriate change of variables. This becomes increasingly difficult, but should certainly be doable. I may have to edit somethings; but we're really fucking close. Using Kneser's slog will probably make an appearance at some point to get this perfect... I hope not, but I still remain optimistic nonetheless.
For large \(\Re(z)\) we have that \(\psi(z) \to z\), as expected; and we map \(\psi(L) = P\) and \(\psi(\overline{L}) = \overline{P}\)--which is exactly as expected. I was screwing up earlier because I was trying to map the level sets exactly--through a conjugation. This is much cleaner; and maps the level set to whatever Kneser is doing here... But the fact the level set is so good is a great marker. It means that:
\[
\left|\psi'(\text{tet}_K(z))\big{|}_{z \in \gamma}\right| = \left|\frac{1}{\text{tet}_K'(z)}\big{|}_{z \in \gamma}\right|
\]
This describes an unknown level set with a known level set. It might make sense here to change the level set from \(|\beta'(z)| = 1\) into something more relaxed. Nonetheless the function \(\psi\) is constructed. And it's fucking beautiful! I'm working on coding it at the moment, and graphing it. Currently my code is super super super fucking slow; but it's kinda getting the job done.
The math should be solid for this; but I haven't proven everything yet. This is what I like to refer to as a "fringe infinite composition"--so you have to be very careful with how you evaluate it.
Finding \(\psi\) seems super easy now, and by which; we have the inverse:
\[
\text{tet}_K(z) = \psi^{-1}(\beta(s))\\
\]
The variable change in \(s\) to \(z\) is the hard part for the moment, I'll figure it out
. One step at a time!And I think that these should be relatively untaxing in a computational sense. We can develop the taylor series of \(\psi\) near infinity rather efficiently because \(\psi(z) \approx z\), and the drop off should be like \(\psi(z) = z + O(e^{-z})\)....
This is "converting beta to Kneser" at it's finest. We don't even need Kneser to do it!
Regards, James
HERE'S A BEAUTIFUL GRAPH FROM \(|\Re(z)| < 3\) and \(|\Im(z)| < 3\) of the function \(\psi\) such that:
\[
\psi(e^z) = \frac{e^{\psi(z)}}{1+e^{-z}}\\
\]
Here's an even bigger graph for \(|\Re(z)| < 8\) and \(|\Im(z)| < 8\)
The hypothesis for now is that:
\[
\psi(z) = s\\
\]
where under the \(\beta\) mapping we have that \(z \mapsto e^z\), we map to \(\beta(s+1) = \frac{e^{\beta(s)}}{1+e^{-s}}\). I need some time to think on this more though, but this is still crazy that \(\psi\) is so easily found!!!!!!!!! Even if this isn't the \(\psi\) we're exactly looking for; it's really close!
The exact formula we are seeing right now is that:
\[
\psi(\text{tet}_K(z+1)) = \beta(\text{tet}_K(z) + 1)\\
\]
Assuming that \(\text{tet}_K(z) \approx L\) this formula is super accurate. I'm having trouble going from here to the next step. I'm going to focus on coding \(\psi^{-1}\) first; which does the opposite transformation...
I'm still thinking on this, but I think I'm nearly there.
