Hey, Tommy
It's a little arbitrary that I use \( \sqrt{1+s} \); but I'll explain it again.
The function \( \beta_\lambda(s) \) has singularities at the points \( s = j + (2k+1)\pi i / \lambda \); so when we write,
\(
F_\lambda(s) = \lim_{n\to\infty} \log^{\circ n} \beta_\lambda(s+n)\\
\)
This will inherently have branch-cuts/singularities along the points \( j + (2k+1)\pi i / \lambda \) for \( j,k \in \mathbb{Z} \). But other than that, this construction works very well. So what we want to do, is move \( \lambda \) while we're taking this limit.
The way I proved this works was essentially with \( \lambda = 1/\sqrt{1+s} \); but it need not be this function. If \( \lambda : \mathbb{R}^+ \to \mathbb{R}^+ \) and,
\(
\beta_{\lambda(s)}(s) : \{s \in \mathbb{C}\,|\,|\arg(s)| < \theta < \pi/2\} \to \mathbb{C}\\
\)
Where \( \lim_{s\to\infty} \lambda(s) = \mathcal{O}(s^{-\epsilon}) \) for \( 0 < \epsilon < 1 \); then the construction works. And additionally; they will produce the same functions (because I use Banach's Fixed Point Theorem to construct these things).
The way I think about it is as a Riemann mapping. We are going to move \( \lambda \) while we're iterating, to avoid the singularities. So for example, the solutions to the equation,
\(
s = j + (2k+1) \pi i \sqrt{1+s}\\
\)
Get further and further out in the complex plane. And we can find a sector \( S_\theta = \{s \in \mathbb{C}\,|\,|\arg(s)| < \theta < \pi/2 \} \), in which,
\(
\beta_{1/\sqrt{1+s}}(s) : S_\theta \to \mathbb{C}\\
\)
And is holomorphic here. This function still acts as an asymptotic solution to tetration, but it doesn't have a convenient functional equation. Now when we do our construction with the iterated logarithms; we can get;
\(
F(s) : S_\theta \to \mathbb{C}\\
\)
But what's so great about this, is that for all \( s \in \mathbb{C} \) there exists an \( n \) such that \( s+n \in S_\theta \). So we can pull this back with logarithms; and extend \( F \) to \( s \in\mathbb{C}/(-\infty,-2] \) for an appropriate normalization constant \( x_0 \approx 2 \).
Now the reason I'm presuming this isn't Kneser's tetration is because; as we limit the imaginary argument to infinity; we are approaching the boundary of \( F_\lambda(s) \) (The almost cylinder where it's holomorphic), which as I mentioned earlier, is a wall of singularities/branch cuts. Which implies \( \text{tet}_\beta(s) \not \to L \) as \( \Im(s) \to \infty \). Instead, it should display no normality condition. I'm having trouble making this heuristic a rigorous proof; but I'm getting there.
So all in all, you don't really need \( \lambda = 1/\sqrt{1+s} \); you could also choose \( \lambda = 1/(1+s)^{1/3} \) just as well; and this will produce the same tetration (albeit, you should have a different normalization constant).
Another way to think about this, which is a bit of an abuse of notation; but it works,
\(
\text{tet}_\beta(s) = F_\lambda(s+x_0)|_{\lambda = 0}\\
\)
Where we write this limit as,
\(
\text{tet}_\beta(s) = \lim_{n\to\infty}\lim_{\lambda \to 0} \log^{\circ n} \beta_\lambda(s+x_0+n)\,\,\text{where}\,\,\lambda = \mathcal{O}(n^{-\epsilon})\,\,\text{for}\,\,0 < \epsilon < 1\\
\)
I hope that clears it up. I mostly just chose \( \lambda = 1 / \sqrt{1+s} \) because it's simple and effective. This function effectively moves all the singularities from \( s = j+ (2k+1)\pi i/\lambda \) to \( \Im(s) = \pm \infty \); which is what I mean by Riemann Mapping.
Regards, James
Additionally Tommy, we can think about this as a Riemann mapping on a whole bunch of tetration solutions.
If I take \( \lambda = 1+i \) and construct \( F_{1+i}(s) \) I get something like this,
Which has singularities on its boundary; will have a period of \( \ell = 2 \pi i / 1+i \); and we want to use this to create the right tetration. We want to move the boundary of the cylinder to \( \Im(s) = \pm \infty \). But how do we do this?
We do it by moving \( \lambda \) while we take \( n\to\infty \), it's that simple. The function \( \lambda = 1/\sqrt{1+s} \) is just one of many functions which will work.
It's a little arbitrary that I use \( \sqrt{1+s} \); but I'll explain it again.
The function \( \beta_\lambda(s) \) has singularities at the points \( s = j + (2k+1)\pi i / \lambda \); so when we write,
\(
F_\lambda(s) = \lim_{n\to\infty} \log^{\circ n} \beta_\lambda(s+n)\\
\)
This will inherently have branch-cuts/singularities along the points \( j + (2k+1)\pi i / \lambda \) for \( j,k \in \mathbb{Z} \). But other than that, this construction works very well. So what we want to do, is move \( \lambda \) while we're taking this limit.
The way I proved this works was essentially with \( \lambda = 1/\sqrt{1+s} \); but it need not be this function. If \( \lambda : \mathbb{R}^+ \to \mathbb{R}^+ \) and,
\(
\beta_{\lambda(s)}(s) : \{s \in \mathbb{C}\,|\,|\arg(s)| < \theta < \pi/2\} \to \mathbb{C}\\
\)
Where \( \lim_{s\to\infty} \lambda(s) = \mathcal{O}(s^{-\epsilon}) \) for \( 0 < \epsilon < 1 \); then the construction works. And additionally; they will produce the same functions (because I use Banach's Fixed Point Theorem to construct these things).
The way I think about it is as a Riemann mapping. We are going to move \( \lambda \) while we're iterating, to avoid the singularities. So for example, the solutions to the equation,
\(
s = j + (2k+1) \pi i \sqrt{1+s}\\
\)
Get further and further out in the complex plane. And we can find a sector \( S_\theta = \{s \in \mathbb{C}\,|\,|\arg(s)| < \theta < \pi/2 \} \), in which,
\(
\beta_{1/\sqrt{1+s}}(s) : S_\theta \to \mathbb{C}\\
\)
And is holomorphic here. This function still acts as an asymptotic solution to tetration, but it doesn't have a convenient functional equation. Now when we do our construction with the iterated logarithms; we can get;
\(
F(s) : S_\theta \to \mathbb{C}\\
\)
But what's so great about this, is that for all \( s \in \mathbb{C} \) there exists an \( n \) such that \( s+n \in S_\theta \). So we can pull this back with logarithms; and extend \( F \) to \( s \in\mathbb{C}/(-\infty,-2] \) for an appropriate normalization constant \( x_0 \approx 2 \).
Now the reason I'm presuming this isn't Kneser's tetration is because; as we limit the imaginary argument to infinity; we are approaching the boundary of \( F_\lambda(s) \) (The almost cylinder where it's holomorphic), which as I mentioned earlier, is a wall of singularities/branch cuts. Which implies \( \text{tet}_\beta(s) \not \to L \) as \( \Im(s) \to \infty \). Instead, it should display no normality condition. I'm having trouble making this heuristic a rigorous proof; but I'm getting there.
So all in all, you don't really need \( \lambda = 1/\sqrt{1+s} \); you could also choose \( \lambda = 1/(1+s)^{1/3} \) just as well; and this will produce the same tetration (albeit, you should have a different normalization constant).
Another way to think about this, which is a bit of an abuse of notation; but it works,
\(
\text{tet}_\beta(s) = F_\lambda(s+x_0)|_{\lambda = 0}\\
\)
Where we write this limit as,
\(
\text{tet}_\beta(s) = \lim_{n\to\infty}\lim_{\lambda \to 0} \log^{\circ n} \beta_\lambda(s+x_0+n)\,\,\text{where}\,\,\lambda = \mathcal{O}(n^{-\epsilon})\,\,\text{for}\,\,0 < \epsilon < 1\\
\)
I hope that clears it up. I mostly just chose \( \lambda = 1 / \sqrt{1+s} \) because it's simple and effective. This function effectively moves all the singularities from \( s = j+ (2k+1)\pi i/\lambda \) to \( \Im(s) = \pm \infty \); which is what I mean by Riemann Mapping.
Regards, James
Additionally Tommy, we can think about this as a Riemann mapping on a whole bunch of tetration solutions.
If I take \( \lambda = 1+i \) and construct \( F_{1+i}(s) \) I get something like this,
Which has singularities on its boundary; will have a period of \( \ell = 2 \pi i / 1+i \); and we want to use this to create the right tetration. We want to move the boundary of the cylinder to \( \Im(s) = \pm \infty \). But how do we do this?
We do it by moving \( \lambda \) while we take \( n\to\infty \), it's that simple. The function \( \lambda = 1/\sqrt{1+s} \) is just one of many functions which will work.
