So, I screwed up my graphs of \(\psi\) above. I forgot that \(\psi\) is \(2 \pi i\) periodic. I was wondering why my code wasn't as accurate as it should be, and it's because:
\[
\psi(e^{z+2\pi i}) = \frac{e^{\psi(z+2 \pi i)}}{1+e^{-z-2\pi i}} = \frac{e^{\psi(z)}}{1+e^{-z}}\\
\]
Whereby, we can assume that \(\psi(z+2\pi i) = \psi(z)\). This creates a MUCH MORE accurate code base. Where as earlier, my code would get inaccurate near \(\Im(z) \approx \pi\). Sorry, I'm still learning as I go.
But here is a large graph of \(\psi(z)\) for \(|\Re(z)| < 10\) and \(|\Im(z)| < 10\). This code base always satisfies:
\[
\psi(e^z) = \frac{e^{\psi(z)}}{1+e^{-z}}\\
\]
No matter how large the imaginary part gets! The \(\psi\) function kind of behaves like Kneser's super logarithm. And it's pretty counter intuitive that \(\text{slog}_K(z+2\pi i) = \text{slog}_K(z)\). We get a similar kind of mapping here. But we're mapping:
\[
\psi(e^z) = e^{\psi(z)} + O(e^{-z})\\
\]
But it still suffers the periodic condition.
And does so for 100 digit accuracy. My earlier graphs would start to crap out past \(\Im(z) = \pi\)--but still look relatively good. This solves that problem entirely. It shows much more clearly how bizarre \(\psi(z)\) actually looks! We have a branching problem at each line \(|\Im(z)| = \pi k\) for all \(k \in \mathbb{Z}\). Which hilights precisely where \(e^{z} = -1\) and where we switch from \(L\) to \(\overline{L}\); which is as expected.
This means, the critical strip for \(\psi\) is for \(0 < \Im(z) < \pi\), and if you calculate this, you calculate everywhere in the upper/lower half planes.
Again, I'm still going along as I see it. But this appears to be working. I'll try to write the hard math later. For the moment I'm just checking the numbers support my hypothesis...
This function has plenty of branching problems, you can notice them off hand pretty fast. This really isn't that big of a deal, especially considering this is tetration. This function satisfies:
\[
\begin{align}
\psi(z+2\pi i) &= \psi(z)\\
\psi(e^z) &= \frac{e^{\psi(z)}}{1+e^{-z}}\\
\end{align}
\]
And it defines an analytic function for \(\Im(z) \neq k\pi \) for all \(k \in \mathbb{Z}\). As we only care about \(\psi\) in this domain, we are fine, because it works for what I hope to set up as the \(\psi\) conjugation of \(\beta\)... There are better ways to define the function \(\psi\), but for our purposes this is fine--and it highlights the perfection of our infinite composition formula much clearer.
The key point, I forgot when defining the infinite composition, is that \(\log( \{0 < \Im(z) < \pi\}) = \{0 < \Im(z) < \pi\}\) and this is a much cleaner way of thinking of it when we want to extend to the complex plane. This lines up perfectly with the poles of \(\frac{1}{1+e^{-z}}\) along the boundary.
So anyway, here is \(\psi\) to the best of my ability! It sure is an ugly fucker!
We can from here, draw a more stable \(\psi\) which only has it's flip at the real axis. This is done by noticing we assumed that:
\[
\psi(z+2\pi i) = \psi(z)\\
\]
When all the formula is really telling us is that:
\[
e^{\psi(z+2\pi i)} = e^{\psi(z)}\\
\]
This means that:
\[
\psi(z+2 \pi i) = \psi(z) + 2 \pi i k\\
\]
If we pay attention to the imaginary parts we gather, we can get a cleaner \(\psi\), which is witnessed by the above graphs from the post above, before this. We shouldn't worry too too much about this. But we definitely need to notice it. Without the assumption of periodicity, we should get that \(\psi(z)\) is holomorphic for \(\Im(z) > 0\). The thing is.... this doesn't happen. The natural result has this problem exactly at \(\Im(z) = \pi\); so the math is telling us that \(k = 0\). Which makes sense if we think of these objects as curves in \(\mathbb{C}\).
I'm still a little loss for words. So, for the moment, I'm confident I can construct:
\[
\psi(z)\,\,\text{is holomorphic for}\,\,0 <|\Im(z)| < \pi
\]
But I think there's a much cleaner numerical procedure; which should find the imaginary values we pick up as we go further in the imaginary axis.
\[
\psi(e^{z+2\pi i}) = \frac{e^{\psi(z+2 \pi i)}}{1+e^{-z-2\pi i}} = \frac{e^{\psi(z)}}{1+e^{-z}}\\
\]
Whereby, we can assume that \(\psi(z+2\pi i) = \psi(z)\). This creates a MUCH MORE accurate code base. Where as earlier, my code would get inaccurate near \(\Im(z) \approx \pi\). Sorry, I'm still learning as I go.
But here is a large graph of \(\psi(z)\) for \(|\Re(z)| < 10\) and \(|\Im(z)| < 10\). This code base always satisfies:
\[
\psi(e^z) = \frac{e^{\psi(z)}}{1+e^{-z}}\\
\]
No matter how large the imaginary part gets! The \(\psi\) function kind of behaves like Kneser's super logarithm. And it's pretty counter intuitive that \(\text{slog}_K(z+2\pi i) = \text{slog}_K(z)\). We get a similar kind of mapping here. But we're mapping:
\[
\psi(e^z) = e^{\psi(z)} + O(e^{-z})\\
\]
But it still suffers the periodic condition.
And does so for 100 digit accuracy. My earlier graphs would start to crap out past \(\Im(z) = \pi\)--but still look relatively good. This solves that problem entirely. It shows much more clearly how bizarre \(\psi(z)\) actually looks! We have a branching problem at each line \(|\Im(z)| = \pi k\) for all \(k \in \mathbb{Z}\). Which hilights precisely where \(e^{z} = -1\) and where we switch from \(L\) to \(\overline{L}\); which is as expected.
This means, the critical strip for \(\psi\) is for \(0 < \Im(z) < \pi\), and if you calculate this, you calculate everywhere in the upper/lower half planes.
Again, I'm still going along as I see it. But this appears to be working. I'll try to write the hard math later. For the moment I'm just checking the numbers support my hypothesis...
This function has plenty of branching problems, you can notice them off hand pretty fast. This really isn't that big of a deal, especially considering this is tetration. This function satisfies:
\[
\begin{align}
\psi(z+2\pi i) &= \psi(z)\\
\psi(e^z) &= \frac{e^{\psi(z)}}{1+e^{-z}}\\
\end{align}
\]
And it defines an analytic function for \(\Im(z) \neq k\pi \) for all \(k \in \mathbb{Z}\). As we only care about \(\psi\) in this domain, we are fine, because it works for what I hope to set up as the \(\psi\) conjugation of \(\beta\)... There are better ways to define the function \(\psi\), but for our purposes this is fine--and it highlights the perfection of our infinite composition formula much clearer.
The key point, I forgot when defining the infinite composition, is that \(\log( \{0 < \Im(z) < \pi\}) = \{0 < \Im(z) < \pi\}\) and this is a much cleaner way of thinking of it when we want to extend to the complex plane. This lines up perfectly with the poles of \(\frac{1}{1+e^{-z}}\) along the boundary.
So anyway, here is \(\psi\) to the best of my ability! It sure is an ugly fucker!
We can from here, draw a more stable \(\psi\) which only has it's flip at the real axis. This is done by noticing we assumed that:
\[
\psi(z+2\pi i) = \psi(z)\\
\]
When all the formula is really telling us is that:
\[
e^{\psi(z+2\pi i)} = e^{\psi(z)}\\
\]
This means that:
\[
\psi(z+2 \pi i) = \psi(z) + 2 \pi i k\\
\]
If we pay attention to the imaginary parts we gather, we can get a cleaner \(\psi\), which is witnessed by the above graphs from the post above, before this. We shouldn't worry too too much about this. But we definitely need to notice it. Without the assumption of periodicity, we should get that \(\psi(z)\) is holomorphic for \(\Im(z) > 0\). The thing is.... this doesn't happen. The natural result has this problem exactly at \(\Im(z) = \pi\); so the math is telling us that \(k = 0\). Which makes sense if we think of these objects as curves in \(\mathbb{C}\).
I'm still a little loss for words. So, for the moment, I'm confident I can construct:
\[
\psi(z)\,\,\text{is holomorphic for}\,\,0 <|\Im(z)| < \pi
\]
But I think there's a much cleaner numerical procedure; which should find the imaginary values we pick up as we go further in the imaginary axis.
