04/13/2021, 06:21 AM
Hey, Tommy!
So I've worked through this a lot. I love what you're asking! I'd like to clarify a couple of things. The first being,
\(
\beta_\lambda(s) \to \infty\,\,\text{as}\,\,\Re(s) \to \infty\\
\)
(Which is a non-trivial result about \( e^z \) in disguise.)
And as such the sequence of convergents, starting with \( \tau^0 = 0 \),
\(
\tau_\lambda^{n+1}(s) = -\log(1+e^{-\lambda s}) + \log(1+ \frac{\tau_\lambda^n(s+1)}{\beta_\lambda(s+1)})\\
\)
Converge to zero in the neighborhood of \( \Re(s) = +\infty \). Everything equals zero at positive infinity here. When we make the change of variables \( s= -\log(w)/\lambda \), this means certain pathways towards zero converge towards tetration (specifically these path ways \( w \mapsto e^{-\lambda n}w \)). Pulling back with logs is perfectly possible here. Everything gets very large--superexponentially-- so logarithms reign supreme. And we've added a geometric sequence \( w \to e^{-\lambda}w \); which converges like Banach's theorem.
What this does is make a tetration , somewhere way off in the right half plane.
Now when we are taking our logs (what you call optimistic) is mostly just a proof of *existence* of such logarithms. I have no f******n clue how to compute this. This is an \( \exists \) statement. That is all. Some sequence of logs works. And I'll die to the death to prove it. There's going to be branch cuts/singularities dependent on \( \lambda \).
Skipping a bunch of steps. When we get the tetration we want; the principal used is if \( C \) is some curve in \( \mathbb{C} \) and \( \log( C) \) is the real line--then \( C \) must be the real-line; if \( \log(z) \) has a branch cut along the negative real axis. This is not an optimistic result; it's necessary. There are some things I'm not 100% on in my paper--this is not one of them.
We then argue, if \( \text{tet}_\beta(s_0) =1 \), then \( \log(\text{tet}_\beta(s_0)) = 2\pi i k \). Which, from there being no real valued curve \( C \); the result must follow. This requires nothing new. I can write it all out if you want, but it's just a statement about logarithms,
I'm also 99% sure my preliminary solution is analytic. It was built solely from these identities. The real trouble--and what I'm worried might fail--is when I start pasting solutions together using \( \lambda(s) = \frac{1}{\sqrt{1+s}} \).
This means if I write,
\(
\beta_{\log(2)}(x) = \Omega_{j=1}^\infty \frac{e^z}{2^{j-x} + 1}\,\bullet z\\
\)
Where,
\(
\beta_{\log(2)}(x+1) = \frac{e^{\beta_{\log(2)}(x)}}{2^{-x} +1}\\
\)
Then,
\(
F_{\log(2)}(x) = \lim_{n\to\infty} \log^{\circ n}(\beta_{\log(2)}(x+n))\\
\)
Is analytic--and does not equal \( F_{\log(3)}(x) \). Where,
\(
F_{\log(3)}(x) = \lim_{n\to\infty} \log^{\circ n}(\beta_{\log(3)}(x+n))\\
\)
And,
\(
\beta_{\log(3)}(x+1) = \frac{e^{\beta_{\log(3)}(x)}}{3^{-x} +1}\\
\)
It's its own unique solution to the Tetration equation. This I'm pretty much certain of.
The idea, and the parts I'm less sure of, but still pretty sure, are pasting these solutions together to get the actual tetration we want.
To accent why this matters is simple.
\(
F_{\lambda}(z)\,\,\text{has singularites when}\,\,z= j + (2k+1)\frac{\pi i}{\lambda}\,\,\text{for}\,\,j,k \in \mathbb{Z}\,j\ge 1\\
\)
So we want to take a varying \( \lambda \) while we iterate the logarithm. We have to avoid a whole swath of singularities somehow. I get what you mean by optimistic, but it's optimistic for a very different reason than you expressed. Once you have it way off in the right half plane--the pull-back is perfectly manageable. I need to double check different things. Once my mapping works at \( \Re(s) = +\infty \) it works everywhere. The trouble is getting it to work at \( \Re(s) = +\infty \).
I'm pretty sure I have it though. I can't produce numbers which disagree. Everything just overflows though when I try to get the tetration. But, again, that's just how bad a coder I am. I'm still trying to figure out how to get some workable code. I'll get it.
Honestly though, as to what you're writing. After constructing this tetration I'm convinced there are multiple paths of construction. And your Incomplete Gamma function method may be possible. As an identity I'd keep in mind. Make sure your function \( H(s) \) satisfies,
\(
e^{H(s)}/H(s+1) \to 1\\
\)
Or even better,
\(
\log H(s+1) - H(s) \to |s|^{-\epsilon} \to 0\\
\)
I've been fiddling with these a lot--and I think this is the way to do it.
Regards, James
PS: I could go on for hours about this solution, Tommy.
So I've worked through this a lot. I love what you're asking! I'd like to clarify a couple of things. The first being,
\(
\beta_\lambda(s) \to \infty\,\,\text{as}\,\,\Re(s) \to \infty\\
\)
(Which is a non-trivial result about \( e^z \) in disguise.)
And as such the sequence of convergents, starting with \( \tau^0 = 0 \),
\(
\tau_\lambda^{n+1}(s) = -\log(1+e^{-\lambda s}) + \log(1+ \frac{\tau_\lambda^n(s+1)}{\beta_\lambda(s+1)})\\
\)
Converge to zero in the neighborhood of \( \Re(s) = +\infty \). Everything equals zero at positive infinity here. When we make the change of variables \( s= -\log(w)/\lambda \), this means certain pathways towards zero converge towards tetration (specifically these path ways \( w \mapsto e^{-\lambda n}w \)). Pulling back with logs is perfectly possible here. Everything gets very large--superexponentially-- so logarithms reign supreme. And we've added a geometric sequence \( w \to e^{-\lambda}w \); which converges like Banach's theorem.
What this does is make a tetration , somewhere way off in the right half plane.
Now when we are taking our logs (what you call optimistic) is mostly just a proof of *existence* of such logarithms. I have no f******n clue how to compute this. This is an \( \exists \) statement. That is all. Some sequence of logs works. And I'll die to the death to prove it. There's going to be branch cuts/singularities dependent on \( \lambda \).
Skipping a bunch of steps. When we get the tetration we want; the principal used is if \( C \) is some curve in \( \mathbb{C} \) and \( \log( C) \) is the real line--then \( C \) must be the real-line; if \( \log(z) \) has a branch cut along the negative real axis. This is not an optimistic result; it's necessary. There are some things I'm not 100% on in my paper--this is not one of them.
We then argue, if \( \text{tet}_\beta(s_0) =1 \), then \( \log(\text{tet}_\beta(s_0)) = 2\pi i k \). Which, from there being no real valued curve \( C \); the result must follow. This requires nothing new. I can write it all out if you want, but it's just a statement about logarithms,
I'm also 99% sure my preliminary solution is analytic. It was built solely from these identities. The real trouble--and what I'm worried might fail--is when I start pasting solutions together using \( \lambda(s) = \frac{1}{\sqrt{1+s}} \).
This means if I write,
\(
\beta_{\log(2)}(x) = \Omega_{j=1}^\infty \frac{e^z}{2^{j-x} + 1}\,\bullet z\\
\)
Where,
\(
\beta_{\log(2)}(x+1) = \frac{e^{\beta_{\log(2)}(x)}}{2^{-x} +1}\\
\)
Then,
\(
F_{\log(2)}(x) = \lim_{n\to\infty} \log^{\circ n}(\beta_{\log(2)}(x+n))\\
\)
Is analytic--and does not equal \( F_{\log(3)}(x) \). Where,
\(
F_{\log(3)}(x) = \lim_{n\to\infty} \log^{\circ n}(\beta_{\log(3)}(x+n))\\
\)
And,
\(
\beta_{\log(3)}(x+1) = \frac{e^{\beta_{\log(3)}(x)}}{3^{-x} +1}\\
\)
It's its own unique solution to the Tetration equation. This I'm pretty much certain of.
The idea, and the parts I'm less sure of, but still pretty sure, are pasting these solutions together to get the actual tetration we want.
To accent why this matters is simple.
\(
F_{\lambda}(z)\,\,\text{has singularites when}\,\,z= j + (2k+1)\frac{\pi i}{\lambda}\,\,\text{for}\,\,j,k \in \mathbb{Z}\,j\ge 1\\
\)
So we want to take a varying \( \lambda \) while we iterate the logarithm. We have to avoid a whole swath of singularities somehow. I get what you mean by optimistic, but it's optimistic for a very different reason than you expressed. Once you have it way off in the right half plane--the pull-back is perfectly manageable. I need to double check different things. Once my mapping works at \( \Re(s) = +\infty \) it works everywhere. The trouble is getting it to work at \( \Re(s) = +\infty \).
I'm pretty sure I have it though. I can't produce numbers which disagree. Everything just overflows though when I try to get the tetration. But, again, that's just how bad a coder I am. I'm still trying to figure out how to get some workable code. I'll get it.
Honestly though, as to what you're writing. After constructing this tetration I'm convinced there are multiple paths of construction. And your Incomplete Gamma function method may be possible. As an identity I'd keep in mind. Make sure your function \( H(s) \) satisfies,
\(
e^{H(s)}/H(s+1) \to 1\\
\)
Or even better,
\(
\log H(s+1) - H(s) \to |s|^{-\epsilon} \to 0\\
\)
I've been fiddling with these a lot--and I think this is the way to do it.
Regards, James
PS: I could go on for hours about this solution, Tommy.