# Tetration Forum

You're currently viewing a stripped down version of our content. View the full version with proper formatting.
So I've been running through my head a question. But in order to resolve that question, I need to better understand Kneser's construction. The premise of this post is to talk about constructing pentation from Kneser's construction.

Now, I know that,

$
\text{tet}_{\text{Kneser}}(s) : \mathcal{H} \to \mathbb{C}\\
$

Where $\mathcal{H} = \{\Im(s) > 0\}$. And I know it tends to a fixed point $L$ as $\Re(s) \to -\infty$. (Or is it multiple fixed points?) Either way, this convergence must be geometric. Giving us, what I'll call,

$
\Phi_{\text{Kneser}}(s) = \Omega_{j=1}^\infty \text{tet}_{\text{Kneser}}(s-j+z) \bullet z : \mathcal{H} \to \mathbb{C}\\
$

Satisfying the functional equation,

$
\Phi(s+1) = \text{tet}(s+\Phi(s))\\
$

And it tends to $L$ as $s \to -\infty$ exactly like $\text{tet}(s-1 + L)$.

Proving this converges will be a little troublesome if we don't have good control over Kneser's solution; but for the moment assume it's good enough. It'll definitely work for large enough $T$ with $\Re(s) < -T$. The idea then being,

$
\text{slog}^{\circ n} \Phi(s+n) \to \text{pent}(s)\\
$

Now, I know what you're thinking. Won't the same thing happen that happened with tetration happen with this method? And I'd say probably. Now instead of looking for $\log(0)$'s though, we'd be looking for $\text{slog}(L)$'s. So where ever $\Phi(s_0) = L$ we get ourselves into a whole lot of trouble.  The only real benefit I can think in this circumstance is that $\Phi$ is not periodic. And that $\Phi(s) \to L$ but it shouldn't equal $L$, or at least, it should be well controlled where it does. This depends on Kneser though, does it ever attain the value $L$ other than at $-\infty$?

I really wish there was more supplemental literature on Kneser's construction other than what's available on this forum... -_-

Nonetheless, I think this might have a better chance at converging than my tetration function constructed with $\phi$. It's just a hunch though.
(02/14/2021, 04:28 AM)JmsNxn Wrote: [ -> ]So I've been running through my head a question. But in order to resolve that question, I need to better understand Kneser's construction. The premise of this post is to talk about constructing pentation from Kneser's construction.

Now, I know that,

$\text{tet}_{\text{Kneser}}(s):\mathcal{H}\to\mathbb{C}$

Where $\mathcal{H} = \{\Im(s) > 0\}$. And I know it tends to a fixed point $L$ as $\Re(s) \to -\infty$. (Or is it multiple fixed points?)
...This depends on Kneser though, does it ever attain the value $L$ other than at $-\infty$?
...
I really wish there was more supplemental literature on Kneser's construction other than what's available on this forum... -_-
Hey James,
The two fixed points are L, L* in the upper/lower halves of the complex plane.  Kneser tends to L as $\Im(z)\to\infty;\;\;\Re(z)\to-\infty$
Wherever $\text{tet}(z)=L+2n\pi i;\;\;\text{tet}(z+1)=L$ and this happens an infinite number of times in the complex plane
See this mathstack post for a good readable overview of Kneser.
Interesting, so we'd have branches of $\text{slog}$ that are holomorphic in a neighborhood of $L$ then. Unlike with the logarithm where no branch is holomorphic in a neighborhood of $0$. I think the trouble with this construction will be getting $\Phi$ to be well behaved as we grow $\Re(s)$. And not to mention, getting general growth lemmas on the various branches of $\text{slog}$. The only work around I see would be to take the inverse iteration,

$
\text{pent} = \text{tet}_{\text{Kneser}}^{\circ n} \Phi(s-n)\\
$

Where of course, we'd have to modify $\Phi$ to converge in this circumstance. And here we'd probably lose any chance of it being real-valued. I'm going to keep this on a backburner and come back to it later. I'm going to stay focused on $\mathcal{C}^\infty$ proofs for the moment.