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 Questions about Kneser... JmsNxn Long Time Fellow Posts: 419 Threads: 81 Joined: Dec 2010 02/14/2021, 04:28 AM (This post was last modified: 02/14/2021, 04:33 AM by JmsNxn.) 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. sheldonison Long Time Fellow Posts: 664 Threads: 23 Joined: Oct 2008 02/14/2021, 05:00 PM (This post was last modified: 02/14/2021, 05:22 PM by sheldonison.) (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. - Sheldon JmsNxn Long Time Fellow Posts: 419 Threads: 81 Joined: Dec 2010 02/16/2021, 12:46 AM (This post was last modified: 02/16/2021, 06:52 AM by JmsNxn.) 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. « Next Oldest | Next Newest »

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