• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 tommy's singularity theorem and connection to kneser and gaussian method tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 09/18/2021, 12:06 PM (This post was last modified: 09/18/2021, 12:11 PM by tommy1729.) Let K(s) = exp(K(s-1)) be the Kneser solution. Let G(s) = exp(G(s-1)) be the gaussian method tetration. Conjecture : K(s + a(s)) = G(s) , G(s + b(s)) = K(s) , where a(s) and b(s) are one-periodic analytic functions and one of a(s),b(s) is entire. Lemma 1 : both K(s) and G(s) are analytic solutions. Lemma 2 : from lemma 1 , there exists analytic periodic functions c(s),d(s) such that K(s + c(s)) = G(s) , G(s + d(s)) = K(s). ( ofcourse s + c(s) and s + d(s) are functional inverses ) Lemma 3 : G(s) is analytic where erf(s) is close to 1. ( triangle or sector ) Lemma 4 : tommy's singularity theorem : Let tet(s) be analytic tetration such that when tet(s) is defined , so is tet(s + r) for real r >= 0. Let tet(s) have a singularity at s = z and tet(s+1) has no singularity at s = z. by the functional equation this implies that tet(z) is a logaritmic singularity. It follows by induction : if tet(s) has a singularity at s = z that is not a logaritmic ( ln or ln ln or ln ln ln or ... ) then tet(s+n) is also a singarity for all integer n > 0... or any integer n actually. therefore , for any analytic tetration all non log-type singularities are 1 periodic ! Lemma 5 : IF tet(s) has non-log-type singularities then tet(s) = K(s + sing(s)) where sing is a 1 periodic function with singularities. It seems to follow that  lemma 6 : IF  tet(s) has no non-log-type singularities then tet(s) = K(s + theta(s)) where theta is a 1 periodic function without singularities. the harder thing is to exclude log-type singul from theta(s). Assuming those log-type are excluded from theta(s) in lemma 6 ; lemma 7 : ...  Since G(s) has no periodic singularities ( the triangle where erf converges fast to 1 forbids it ) , it follows that  G(s) = K(s + theta(s)) for entire theta(s). ( again : the harder thing is to exclude log-type singul from theta(s) , the starting conjecture is a bit weaker such that inv( s + theta(s) ) can be entire ... and i assume that makes s + theta(s) have the log-type sing then. if s + theta(s) really is entire then i assume inv( s + theta(s) ) has log-type singularities and branches due to s + theta(s) being flat ( derivat = 0 ) ... it seems to follow from the above that those " flat branches " must be log-type sing as well ... THE REAL HARD PART is to exclude that both s + theta(s) and inv* have both log-type sing ! ) - more or less - QED regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,491 Threads: 355 Joined: Feb 2009 09/18/2021, 12:13 PM ofcourse this also applies to generalizations of the gaussian method, and probably many others ( like beta ). regards tommy1729 JmsNxn Long Time Fellow Posts: 568 Threads: 95 Joined: Dec 2010 09/20/2021, 04:29 AM (This post was last modified: 09/20/2021, 05:33 AM by JmsNxn.) I'm a little wary of this tommy, for the reason that you avoid Sheldon's expression. But for the most part I agree with you, excepting lemma 6 & 7--you lost me there. This doesn't feel natural to me, as it avoids a lot of the talk about recurrent values. And additionally, the point of the beta method is that it's a construction void of theta mappings--but we can still use theta mappings as a lens. I'm going to give a little walk through of what I think will happen when we look at the theta mapping for the gaussian/beta method. Let $\text{slog}_K$ be kneser's slog, which is holomorphic on $\mathbb{C} / \mathcal{L}$ where, $ \mathcal{L} = \{ q \in \mathbb{C}\,|\,\text{the set}\,\{\exp^{\circ n}(q)\,|\,n\in\mathbb{N}\}\,\text{is finite}\}\\$ This would be the periodic/cyclic points and their preimages. Now at each of these points $q$ we must have, $ \text{slog}_K(q) = \infty\,\,\text{because}\\ \text{slog}_K(\exp^{\circ n}(q)) = \text{slog}_K(q) + n\\$ But there are only finite amounts of values in the set $\{\exp^{\circ n}(q)\,|\,n\in\mathbb{N}\}$; and so this must be a singularity--thus we have a singularity. So now, when you are talking about your mapping $a(s)$; let's assume it's entire, and 1-periodic. Then, $ a(s) = \text{slog}_K(G(s)) - s\\$ At least, locally (for an appropriate $\text{slog}_K$). Now, what happens when $G(s_0) = q$? We get either a singularity, or a zero. This can be proved by cases. Assume that $\lim_{s\to s_0} \text{slog}_K(G(s)) - s \neq 0$ then this limit is just $\infty- s_0$--thus a is not entire. Assuming this limit tends to zero is the only work around. And if we assume this, and that $a(s)$ is entire, then, $ G(s_0) = q \,\,\text{implies}\,\,\text{tet}_K(s_0) = q\\$ Or that, we are mapping the same recurrent points to the same recurrent points.  If we call $\mathcal{C} = \{s \in \mathbb{C}\,|\, \text{tet}_K(s) \in \mathcal{L}\}$. Then, in a crisper notation, it means $G|_{\mathcal{C}} = \text{tet}_K|_{\mathcal{C}}$. But then $\lim_{\Im(s) \to \infty} G(s) = L$, by just following a sequence along these recurrent points. Now granted, this is still possible--but it smells like it's just going to be a proof that it's Kneser, or the assumption a is entire is wrong. I'm going to follow Sheldon's argument to show that if a is entire, then it's kneser, otherwise the assumption that a is entire is wrong. But first, just a brief comment. The trouble is that--the moment that we have $\text{tet}_K(s_0) \in \mathcal{L}$ and that $G(s_0) \not\in \mathcal{L}$ we lose a nice way of comparing these things functionally. And because of that, we really do need the singularities to make the tetration different. This isn't as much of a problem as it seems though. Second to this argument, I suggest rereading Sheldon's comment about how if you have a 1-periodic entire  function $a$ such that, $ \text{tet}_K(z+a(z)) : \mathbb{C}/(-\infty,-2] \to \mathbb{C}\\$ Then the only one which is holomorphic in the upperhalf plane, and lower half plane, and real valued, is the constant value $a = 0$--actually you can pretty much just follow Wiliam Paulsen and Samuel Cowgill. The difference is that Sheldon's version is a tad stronger in that, if what you are saying is true, then $z+a(z)$ is an entire transcendental function, and omits only one value. This means it hits the values $z+a(z) = -2,-3,-4,$ infinitely often (subtracting perhaps one value). This would mean necessarily our tetrations would have singularities in the upper half plane; they would be of this log type, but singularities nonetheless. I'm still of the contention this shouldn't happen, but let's continue figuring it out. The key is, we need it to have singularities for any of this to work flawlessly in the upper-half plane. In fact, we need it to have branch cuts just as well. But remember, tetration is very much a lala land where anything can happen at infinity, lol. And the idea that $a(s) \to \infty$ (as some kind of singularity) as $s \to s_0$, it's possible that, $ \lim_{s\to s_0} \text{tet}_K(s + a(s)) \to q = G(s_0)\\$ These points pop up infinitely often near infinity; whose to say that for $s_n \to s_0$ that $\text{tet}_K(s_n+a(s_n))$ doesn't follow an orbit which closes in around $q$. I can construct one right now. Let $q$ be a cyclic point of order 3, and assume that $G(s_0) = q$; find a point $u_0$ such that $\text{tet}_K(u_0+s_0) = q$ and assume that $a(s_n) \to \infty$. If i were to take $a(s_n) = u_0 + 3n + 1/n$ then definitely, $ \lim_{n\to\infty} \text{tet}_K(s_n+a(s_n)) = q\\$ I mean to say, it's not impossible for this to happen. It should probably branch, and have a branch cut at the point $s_0$. There's definitely a lot of chaos there, but nothing about this is impossible so far. And now, here is where you have to hear me out. The function $z+a(z)$ (for it to follow the story as we've told it so far, before sheldon brought up theta mappings) is that it is 1-periodic and has singularities + branch cuts about the singularities (the recurrent points)... Now before you go how that doesn't make sense, and would make errors and everything. Remember, log(z) has branchcuts and a singularity, but $\exp(\log(z))$ is an entire function... The theta mappings, if you so will allow, play the role of the logarithm; and we can't go talking about the logarithm until we understand the exponential. We're in a very similar situation here. Theta mappings are not the answer, because our theta mappings (tommy's gaussian/beta) have singularities/ branch cuts. This is initially how I even approached this problem. We're going to dodge theta mappings entirely. We're not going to use them to construct this, because it's a hopeless affair. Additionally, if we could classify/construct these tetrations with theta mappings, someone would've done it already. Instead, theta mappings are perfect for Kneser and its off shoots--but it's an incomplete picture of tetration. We've found previously unmapped territory--and to talk about it using theta mappings (as tho it hasn't been mostly discovered) will be a waste of time. The goal is to throw a wrench in the gears. As a lens, it is infinitely valuable though. Imagine theta mappings, as one of the ways to look at the picture; and it can be enlightening, but the other views will remove the distortion. I think talking about theta mappings is central to how tetration works; but it needs to be viewed as a lens. It's a way to look at beta/gaussian; but isn't the whole picture. And if we're staying consistent with everything we've talked about, we must have that, $ z+a(z)\,\,\text{has branch cuts and singularities...}\\ \text{and that doesn't invalidate any of our work so far.}\\$ And I want to really express this. The theta mapping arguments do not destroy our work so far. But they do add another way of justifying/sussing out singularities. But we're looking for a crazy 1-periodic function $\theta$, which has singularities and branching problems and, $ \text{tet}_K(z+\theta(z)) = \text{tet}_\beta(z)\\$ And, no one has shown that's impossible yet. In fact, I believe I've shown it is possible--as sketchy as the proof is at the moment. But from what you've concluded, I feel you don't grasp it; as you are assuming $a$ is entire. That's just not gonna work. Thats no different from the old theta mapping theorem of this forum; it's just gonna be kneser if its holomorphic in the upper half plane. My main argument is that the theta mapping picture is not the whole picture of tetration. We need singularities/branch cuts in theta. That's the whole point of the method. Nonetheless, this was an interesting post, Tommy. It definitely got my brain rattling. Thanks for keeping this discussion alive. Otherwise it would just be me going in cycles like $\exp$ on $\mathcal{L}$, lol. Sincere regards, James. « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Tommy's Gaussian method. tommy1729 24 3,939 11/11/2021, 12:58 AM Last Post: JmsNxn The Generalized Gaussian Method (GGM) tommy1729 2 380 10/28/2021, 12:07 PM Last Post: tommy1729 Arguments for the beta method not being Kneser's method JmsNxn 54 6,572 10/23/2021, 03:13 AM Last Post: sheldonison Some "Theorem" on the generalized superfunction Leo.W 44 9,639 09/24/2021, 04:37 PM Last Post: Leo.W " tommy quaternion " tommy1729 14 5,310 09/16/2021, 11:34 PM Last Post: tommy1729 Why the beta-method is non-zero in the upper half plane JmsNxn 0 334 09/01/2021, 01:57 AM Last Post: JmsNxn Improved infinite composition method tommy1729 5 1,275 07/10/2021, 04:07 AM Last Post: JmsNxn Generalized Kneser superfunction trick (the iterated limit definition) MphLee 25 8,228 05/26/2021, 11:55 PM Last Post: MphLee Alternative manners of expressing Kneser JmsNxn 1 896 03/19/2021, 01:02 AM Last Post: JmsNxn A different approach to the base-change method JmsNxn 0 726 03/17/2021, 11:15 PM Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)