(10/26/2021, 07:03 PM)Ember Edison Wrote: 1. So which is more "analytic", the beta method or Kneser's method? I can't relate Sheldon's discovery to the complex plane
2. And do you have any idea to get beta method Super-logarithm and super-root?
3. I was very surprised by the robustness of the beta method, and I'm still narrowing down the base to see if the beta method crashes completely first, or if Pari-GP crashes first. Does the beta method have a limit when base -> 0? Can remove the singularity 0?
Well Kneser is definitely more analytic for \(b=1\); unless \(\lambda \to 0\) we get that \(\text{tet}_{\lambda,1}(s)\) converges well enough. I'm not sure at the moment if it will be as analytic; and my heuristics seem to imply IF it converges and is analytic, it'll probably actually be Kneser! But I'm not sure at the moment. When you start making \(\lambda = 0.0001\) or smaller; either we overflow; or branch out (though the branching is strict to the real-line and the essential singularities). The branching seems to start to isolate itself to the real line; and everywhere else we look fairly holomorphic and in a neighborhood of \(L\) or \(L^*\) our familiar fixed points. To such an extent, you can say \(\lim_{|s|\to\infty} \lim_{\lambda \to 0} \text{tet}_{\lambda ,1}(s) \to L\) for \(\pi/2 < \arg(s) < \pi\)--which would imply its Kneser per Paulsen & Cowgill.
But for the moment, if you take,
\[
F(s) = \text{tet}_{1,1}(s)\,\,\text{the}\, 2 \pi i\, \text{ periodic tetration base}\,e
\]
Then Sheldon has demonstrated rather clearly that it's \(\mathcal{C}^\infty\) on \(\mathbb{R}\) with a collection of branching points everywhere it starts to blow up. So it leaves me with the original statement I had in my paper; which is that,
\[
F(s)\,\,\text{is holomorphic on}\,\mathbb{C}/\mathcal{B}\\
\]
Where,
\[
\int_{\mathcal{B}} \,dA = 0\\
\]
Or that its points of discontinuities/branches/singularities are measure zero in \(\mathbb{R}^2\) using the standard Lebesgue area measure. I'm confident this is the best you can get; and I'm still confident in this result; I just assumed it'd be analytic on \(\mathbb{R}\). You can show this result because the logarithmic singularities/branches cannot be dense in \(\mathbb{C}\)--because \(F(s) = \beta(s) + \tau(s)\) and you can remove the singularity at \(\Re(s) = \infty\) so that \(\tau(\infty) = 0\). And from there it must converge in a neighborhood of \(\Re(s) = \infty\).
I had thought that the pullback would be better than it is--because small iterations of graphs looks very very good. For larger iterations, I thought my code was failing, and the pretty pictures were correct; but the truth is, when you make a very accurate graph you can see the branching problems quite clearly--as per the above graph.
Nonetheless; that is defunct towards the final theorem, which, at best is worded less than desirable and "almost right". I'm going to rework my paper a good amount; probably start from nearly scratch.
2.
The best I can give you is to invert the infinite composition and apply the inverse limit. So, let's do this real quick:
\[
\begin{align}
\beta(s) &= \Omega_{j=1}^\infty \frac{e^{bz}}{e^{\lambda(j-s)}+1}\,\bullet z\\
&= {\displaystyle \frac{e^{\displaystyle \frac{be^{\displaystyle \frac{be^{\displaystyle ...}}{e^{\lambda(3-s)}+1}}}{e^{\lambda(2-s)}+1}}}{e^{\lambda(1-s)}+1}}
\end{align}
\]
Make the change of variables \(w = e^{\lambda s}\) to get \(g(w) = g(e^{\lambda s}) = \beta(s)\), which is holomorphic on the disk \(|w| < e^{\lambda}\) and fixes zero. Invert this to get \(g^{-1}(w)\) so that: \(g^{-1}(w) = e^{\lambda \beta^{-1}(s)}\).
Now all these operations aren't too computationally exhausting. From here we run the limit:
\[
\text{slog}(s+s_0) = \lim_{n\to\infty} \beta^{-1}(\exp^{n}(s)) - n\\
\]
This should converge to the appropriate limit on \(\mathbb{R}\), not sure about in the complex plane. There may be some trouble depending on how "holomorphic" the beta method is for each \(b\) and \(\lambda\).
3.
So to get the gist of the limit it's helpful to remember what the beta method is.
\[
\beta_{\lambda ,b}(s) = \Omega_{j=1}^\infty \dfrac{e^{bz}}{e^{\lambda(j-s)}+1}\,\bullet z\\
\]
And this thing converges everywhere that the sum,
\[
\sum_{j=1}^\infty \dfrac{e^{bz}}{e^{\lambda(j-s)}+1}\\
\]
Converges compactly normally. Now, this just means it converges on compact subsets with the supremum norm of each element of the sum. This instantly tells us \(\Re \lambda > 0\), as this is the dominant form of the series and gives it its geometric convergence. The value \(s\) just can't hit a singularity; which removes a countable amount of points. And in this sum, b really doesn't do anything.... So with this in mind, what happens when \(b=0\). Well the expression looks like this:
\[
\Omega_{j=1}^\infty \dfrac{1}{e^{\lambda(j-s)}+1}\,\bullet z
\]
And since there is no z term, we get a "null" composition. There's no composition being made. So... this means that the only term involved is the first term and...:
\[
\beta_{\lambda,0}(s) = \frac{1}{e^{\lambda(1-s)}+1}\\
\]
And, additionally, it will be holomorphic in a neighborhood of \(b=0\), because the above sum converges compactly normally.
The only REAL trouble point is:
\[
\lambda = 0\\
\]
So now, if you want to talk about the trouble of \(^z 0\), remember that you are actually limiting \(b \to -\infty\); which is definitely an anomalous point. And in this case,
\[
\lim_{\Re(b) \to - \infty} \beta_{\lambda,b}(s) = 0\,\,\text{...sort of, this happens at least for}\,\lambda,s\in\mathbb{R}^+\\
\]
The key again, is where does:
\[
\lim_{\Re(b)\to-\infty} \sum_{j=1}^\infty \dfrac{e^{bz}}{e^{\lambda(j-s)}+1} = 0\\
\]
And filtering out what happens when \(z \approx 0\), obviously the final result depends on where z is negative or positive. And it can only get more complicated...
So at this point, it's safe to say this will be really whacky as you limit the appropriate variables.... But I don't really see any step of the method crashing unless we hit a branching point or we have s hitting a singularity or we are limiting \(\lambda \to 0\) or \(|b|\to\infty\).