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 I thought I'd compile some of the things I know... JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 01/19/2021, 05:10 AM (This post was last modified: 01/19/2021, 10:59 PM by JmsNxn.) I'm going to explain some things about my tetration and the auxiliary function $\phi$ I use to construct it. These are mostly things I either can prove but aren't so hard to prove, so I didn't include them; or things I can't quite prove yet, but I'm probably not far off the mark. First of all, I'd like to clarify the definition of $\phi$. If we call $h_j(s,z) = e^{s-j+z}$, then: $ \phi(s,z) = \lim_{n\to\infty} h_1(s,h_2(s,...h_n(s,z)))$ Now, $\frac{d}{dz} \phi(s,z) = 0$, so this limit converges uniformly on compact subsets, in $s \in \mathbb{C}$ and for $z \in \mathbb{C}$; but it converges to a constant in $z$. Tommy raised a good point in his thread, where this wasn't apparently obvious to him, where as it is to me. If we have an alternative solution to, $ f(x+1) = e^{x+f(x)}$ And $|f(x-n)| \le M \in \mathbb{R}^+$ for all $n\in\mathbb{N}$; then, $ f(x) = \phi(x)$ Which can be seen because, $ f(x) = h_1(x,h_2(x,...h_n(x,f(x-n))))$ And since $f(x-n)$ is a normal sequence in $n$ this converges flatly. *** As to the branch cuts; my solution probably (99% sure) has branch cuts other than the one at $(-\infty,-2]$. This is a bit harder to identify, but I'll explain. Suppose there are no branch cuts and $e\uparrow\uparrow s$ is holomorphic for $\Re(s) > 0$. Then, firstly, $ e\uparrow \uparrow s = \phi(s) + \tau(s)\\$ and, $ \lim_{|s|\to\infty} \frac{\tau(s)}{s} = 1\,\,\text{while}\,\,|\arg(s)| < \pi/2\\$ Therefore for very large $s$; I'm pretty sure I can show that, $ |\frac{1}{e \uparrow \uparrow s}| \le C|s|\\$ This implies that, $ \vartheta(x) = \sum_{n=0}^\infty \frac{1}{e\uparrow\uparrow n} \frac{(-x)^n}{n!}\\$ Is mellin transformable, which means, $ \Gamma(s)/ e\uparrow\uparrow (-s) = \int_0^\infty \vartheta(x)x^{s-1}\,dx\,\,\text{for}\,\,0 < \Re(s) < 1\\$ And now we run into a whole lot of trouble. First of all, it's been discussed on this forum that this probably doesn't converge. I can actually prove it now--this cannot converge. If two holomorphic functions on $\mathbb{C}_{\Re(s) > 0}$, say $f,h$, which satisfy these types of bounds, then: $ f(n) = h(n) \Leftrightarrow f(s) = h(s)\\$ If these function agree on the naturals they agree everywhere. Here's where we encounter problems. Huge problems. Suppose that, $ e \uparrow \uparrow s_0 = e \uparrow \uparrow s_1\\$ Then surely $f = 1/e\uparrow \uparrow s + s_0$ and $h =1/ e\uparrow \uparrow s + s_1$ equal on the natural numbers, therefore, $f = h$ But! $e \uparrow \uparrow s$ IS NOT PERIODIC. Therefore this function must be injective. But it can't be injective. As per (I forget the name of the authors, but it can be found in Milnor's Complex Dynamics) the orbits of the exponential are dense.  This means the sequence $e \uparrow \uparrow \mathcal{N} + n$ for any neighborhood $\mathcal{N}\subset \mathbb{C}$ is dense. This can't happen if $e\uparrow \uparrow s$ is injective. CONTRADICTION. Therefore $e \uparrow \uparrow s$ CANNOT be holomorphic on $\Re(s) > 0$. It has branch-cuts. Or rather, the nicer way of saying this is, $e \uparrow \uparrow s$ has zeroes for $\Re(s) > 0$. Now these branch cuts work as you play your logarithms, which is; each one looks like $\log(0)$; for convenience we'll make the convention that the $\log(z)$ is holomorphic for $|\arg(z)| < \pi$ in a neighborhood of the singularity. This way the branch cuts are a straight, horizontal line from the point of branching off to $-\infty$. From this we partition $e\uparrow\uparrow s$ in the left-half plane into vertical sections. And each section limits to a fixed point of $\exp$ as $\Re(s) \to -\infty$. I hope the reader can see this is not much different than what happens in Kneser's case. This is largely why I think this solution is Kneser's solution. *** The next thing I'd like to say, or what I can think of; is a formula for the derivative of $e \uparrow \uparrow s$. Suppose we've isolated ourselves to one of these horizontal strips $\mathcal{S}$ in which $e\uparrow \uparrow s$ is holomorphic. Then,$e\uparrow\uparrow s \to L$ as $\Re(s) \to -\infty$ where $e^L = L$; and it tends to this as $L + e^{Ls}$. This is shown because functions (in this case $\log$) when iterated tend to fixed points geometrically like the multiplier. There exists the formula in this strip, $ \frac{d}{ds} e \uparrow \uparrow s = Ae^{Ls} \prod_{j=0}^\infty \frac{e \uparrow \uparrow (s-j)}{L}\\$ For some $A \in \mathbb{C}$. This implies our derivative is non-zero in this horizontal strip. Proving this theorem is just out of my grasp, because I've yet to understand the branch-cuts yet. But it's derived from the relationship, $ \frac{d}{ds} e \uparrow \uparrow s +1 = (e \uparrow \uparrow s +1) \frac{d}{ds}e \uparrow \uparrow s\\$ And the geometric convergence of the above product, which when attached with $e^{Ls}$, satisfies this functional equation... This is one of those things I'm pretty sure I can prove but I'm not 100%. *** Anyone have any questions, I'm happy to answer them here. There's a lot of things I did not publish. I cut a lot of fat. I've always prescribed to the iceberg theory of writing; perhaps to my detriment... ...........I'll add more as I think of things and scrubble through my notes. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 01/22/2021, 08:50 AM (This post was last modified: 01/22/2021, 08:54 AM by sheldonison.) (01/19/2021, 05:10 AM)JmsNxn Wrote: As to the branch cuts; my solution probably (99% sure) has branch cuts other than the one at $(-\infty,-2]$. This is a bit harder to identify, but I'll explain... But! $e \uparrow \uparrow s$ IS NOT PERIODIC. Therefore this function must be injective. But it can't be injective. As per (I forget the name of the authors, but it can be found in Milnor's Complex Dynamics) the orbits of the exponential are dense.  This means the sequence $e \uparrow \uparrow \mathcal{N} + n$ for any neighborhood $\mathcal{N}\subset \mathbb{C}$ is dense. This can't happen if $e\uparrow \uparrow s$ is injective. CONTRADICTION. Therefore $e \uparrow \uparrow s$ CANNOT be holomorphic on $\Re(s) > 0$. It has branch-cuts. Or rather, the nicer way of saying this is, $e \uparrow \uparrow s$ has zeroes for $\Re(s) > 0$. ... I hope the reader can see this is not much different than what happens in Kneser's case. This is largely why I think this solution is Kneser's solution.I'm going to call James' solution $\text{Tet}_\phi(z)$, which is generated from James' entire function $\phi(z)$.  And Kneser's solution $\text{Tet}_k$, and conjecture that therefore they are not equal, although both solutions have $\text{Tet}(z+1)=\exp(\text{Tet}(z)$ at the real axis. Kneser doesn't have any singularities in the upper half of the complex plane.  Kneser's $\text{Tet}_k$ is analytic except for singularities at integers $\forall \mathbb{W}\leq-2$.   My understanding is that James' Tetration sould be defined in terms of $\phi$ as follows: $\text{Tet}_\phi(z+k)=\lim_{n\to\infty}\ln^{[\circ n]}(\phi(z+n))$, where k is chosen so Tet(0)=1 We can write an equation for when there are singularities for $\ln(\ln(\phi(z)))$, by starting with  $\phi(z+1)=\exp(\phi(z))\cdot\exp(z+1)=\exp(\phi(z)+z+1)$ So then taking the logarithm once $\ln(\phi(z+1))=\phi(z)+z+1$ And taking the logarithm twice there are singularities where $\phi(z)+z+1=0$ There is a trivial version at the real axis, approximately at z=-1.3018, phi(z)=0.3018.  But how do we find the values for these singularities in the complex plane, other than the trivial 2pi i periodic examples?  Finding these values is complicated...  My conjecture is that there are an infinite number of complex values leading to singularities in $\ln(\ln(\phi)))$ as $\Re(z)$ increases where phi(z)+z+1~=0 ...  More later ... - Sheldon JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 01/22/2021, 10:21 PM (This post was last modified: 01/22/2021, 10:23 PM by JmsNxn.) (01/22/2021, 08:50 AM)sheldonison Wrote: My understanding is that James' Tetration sould be defined in terms of $\phi$ as follows: $\text{Tet}_\phi(z+k)=\lim_{n\to\infty}\ln^{[\circ n]}(\phi(z+n))$, where k is chosen so Tet(0)=1 We can write an equation for when there are singularities for $\ln(\ln(\phi(z)))$, by starting with  $\phi(z+1)=\exp(\phi(z))\cdot\exp(z+1)=\exp(\phi(z)+z+1)$ So then taking the logarithm once $\ln(\phi(z+1))=\phi(z)+z+1$ And taking the logarithm twice there are singularities where $\phi(z)+z+1=0$ There is a trivial version at the real axis, approximately at z=-1.3018, phi(z)=0.3018.  But how do we find the values for these singularities in the complex plane, other than the trivial 2pi i periodic examples?  Finding these values is complicated...  My conjecture is that there are an infinite number of complex values leading to singularities in $\ln(\ln(\phi)))$ as $\Re(z)$ increases where phi(z)+z+1~=0 ...  More later ... Yes, that's the gist of the solution; except you've shifted $\phi$ by one, it should be $\phi(s+1) = e^{s+\phi(s)}$. I concur whole heartedly there are singularities in the complex plane. I have no idea how to find them though; but the idea that, $ \phi(s) + s= 0\\$ has solutions with arbitrarily large $\Re(s)$ seems like a very smart way of going about it. I seem to have mistaken my tetrations.  I thought Kneser had singularities other than at the negative integers less than one... I must be thinking of a different tetration which had branch cuts all over the place; i.e: the tetration had zeroes... sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 01/23/2021, 03:21 AM (This post was last modified: 01/23/2021, 03:35 AM by sheldonison.) (01/22/2021, 10:21 PM)JmsNxn Wrote: Yes, that's the gist of the solution; except you've shifted $\phi$ by one, it should be $\phi(s+1) = e^{s+\phi(s)}$. I concur whole heartedly there are singularities in the complex plane....ah, yes, I have my phi shifted by 1.  With your definition the limit as s gets arbitrarily negative is phi(s)=exp(s-1), whereas the other way the limit is phi(s)=exp(s), so I can see why I didn't catch my error, since that limit also leads to a very schroeder like equation with a formal series $F(x)=x+a_{2}x^2+a_{3}x^3+...;\;\phi(s)=F(\exp(s))$ as opposed to $F(x)=\frac{x}{e}+a_{2}x^2+a_{3}x^3+...;\;\phi(s)=F(\exp(s))$ which is a little less like a formal schroeder series ... anyway the resulting phi is the same entire function either way, just shifted by 1. - Sheldon sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 01/24/2021, 01:28 AM (This post was last modified: 01/24/2021, 01:39 AM by sheldonison.) so, $\phi(s+1)=e^{\phi(s)+s$, and we're looking for singularities where $\ln(\ln(\phi(s+1)))$ has a singularity where  $\ln(\phi(s)+1)=0;\;\;\;\phi+s=0;$ cause then $\ln(\phi(s)+s)$ has a singularity. The conjecture is that somewhere nearby the neighborhood of the point where $\phi(s-1)+s-1=2(n-1)\pi i$, then for a reasonably small value of x, especially as the integer n gets arbitrarily large, we will have $\phi(s+x)+s+x=0$, and we will have our singularity.  The idea is we are looking for singularities where Re(s) is positive so Re(phi(s)) must be negative to counter it and generate the desired zero value.  So that's why the singularities are near odd multiples of pi i for phi(s-1)+s-1.  Meanwhile, at the real axis phi is getting quite large, so this is a way to find the places where |phi| is actually relatively small.  The conjecture is further that there is a 1:1 correspondence with with these values of s, and the singularities in ln(ln(phi(s+1)) closest to the real axis. Then here are the first 10 singularities ... notice that as conjectured, the singularities are pretty close to where phi(s-1)+s-1=(2n-1)pi i, just a bit closer to the real axis, and slightly smaller in real magnitude as well.  So then these singularities act a lot like the singularities in the base change tetration, or in Peter Walker's slog, that we have discussed before on this forum.  These functions are $C_\infty$ and conjectured to be nowhere analytic since the imaginary part of the value of s where there is a singularity gets arbitrarily small as n gets larger, and since the singularities quickly get arbitrarily close together as n increases as well, so as n increases, $\text{Tet}_\phi(s+k)=\lim_{n\to\infty}\ln^{[\circ n]}(\phi(s+n))$ we encounter a wall of singularities. Code:s1,s2 = 3.08188+1.191682*I, 2.69716+1.019425*I;  phi((s1-1)+s1-1= 1pi*i; phi(s2)+s2=0 |s1-s2|=0.421523 s1,s2 = 3.12347+0.563422*I, 3.09274+0.531645*I;  phi((s1-1)+s1-1= 3pi*i; phi(s2)+s2=0 |s1-s2|=0.044203 s1,s2 = 3.21853+0.435711*I, 3.20564+0.419751*I;  phi((s1-1)+s1-1= 5pi*i; phi(s2)+s2=0 |s1-s2|=0.020517 s1,s2 = 3.27739+0.374627*I, 3.26990+0.364341*I;  phi((s1-1)+s1-1= 7pi*i; phi(s2)+s2=0 |s1-s2|=0.012728 s1,s2 = 3.31871+0.337247*I, 3.31366+0.329806*I;  phi((s1-1)+s1-1= 9pi*i; phi(s2)+s2=0 |s1-s2|=0.008994 s1,s2 = 3.35001+0.311396*I, 3.34630+0.305640*I;  phi((s1-1)+s1-1=11pi*i; phi(s2)+s2=0 |s1-s2|=0.006847 s1,s2 = 3.37493+0.292151*I, 3.37206+0.287496*I;  phi((s1-1)+s1-1=13pi*i; phi(s2)+s2=0 |s1-s2|=0.005470 s1,s2 = 3.39548+0.277098*I, 3.39317+0.273215*I;  phi((s1-1)+s1-1=15pi*i; phi(s2)+s2=0 |s1-s2|=0.004521 s1,s2 = 3.41286+0.264900*I, 3.41094+0.261585*I;  phi((s1-1)+s1-1=17pi*i; phi(s2)+s2=0 |s1-s2|=0.003831 s1,s2 = 3.42786+0.254750*I, 3.42623+0.251868*I;  phi((s1-1)+s1-1=19pi*i; phi(s2)+s2=0 |s1-s2|=0.003309 - Sheldon JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 01/24/2021, 06:00 AM (This post was last modified: 01/24/2021, 07:10 AM by JmsNxn.) Hey, Sheldon. So if I'm reading you correctly, we are trying to look at solutions of; $ \phi(s) + s = (2k+1)\pi i\\$ Then, for $|x| < \delta$, $ \phi(s+x) + s+x = 0\\$ Therefore these points cluster when doing the iteration, $ \tau \mapsto s + \log(1+\frac{\tau(s+1)}{\phi(s+1)})\\$ Because we get too close to $\log(0)$ at the real line! Fascinating! Fascinating! I have made a grave mistake; and I think I can prove it! Quick question though; do you mind looking at the iteration: $ \tau(x+i\pi/2) = x+i\pi/2 +\log(1+ \frac{\tau(x+1+i\pi/2)}{\phi(x+1+i\pi/2)})\\$ For $\tau_0(x + i\pi/2) = 0$ and $\tau_1(x+i\pi/2) = x+i\pi/2$. If this converges there may be something else I can say. And I may not be entirely wrong. What if it's just $C^{\infty}$ on the real line (the singularities cluster there); but in the complex plane we don't necessarily always suffer from this problem. What if $(-\infty,-2]$ is not the branch cut but $\mathbb{R}$ is, and so is $\mathbb{R} + 2\pi i k$? Naturally it's nowhere analytic on the branch-cut and it can't converge uniformly there (what you see as clustering of singularities) (but in the complex plane the singularities cluster towards the real line; away from from a compact set excluding $2\pi i$ multiples of $\mathbb{R}$!). I think it may still be holomorphic on $\mathbb{C}$ minus branch cuts; just $\mathbb{R}$ is a branch cut... (LAST EDIT I SWEAR!) Think: the log's can't force the imaginary part to zero because we keep on adding in an imaginary part... not sure how to explain it. Long story short: I was so focused on $0 < \Im(s) < 2\pi$ I forgot about the end points. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 01/24/2021, 05:05 PM (This post was last modified: 01/24/2021, 05:16 PM by sheldonison.) (01/24/2021, 06:00 AM)JmsNxn Wrote: Hey, Sheldon. So if I'm reading you correctly, we are trying to look at solutions of; $\phi(s)+s=(2k+1)\pi i\\$ Then, for $|x|<\delta$, $\phi(s+x+1)+s+1+x=0\\$I updated it to $\phi(s+x+1)$, but yeah.  The conjecture is that given a suitable definition of these values of s, and what "near" means, then the closest singularities to the real line of $\ln(\ln(\phi))$ are near (s+1), and there is a 1:1 correspondence between those singularities and integers k>=0, and that as k increasers these singularities cluster arbitrarily close together, and arbitrarily close to the real axis. For my calculations, I used Newton's method twice, once to find the values of s for each value of k by starting with the value of s for k-1.  And then I used Newton's method a second time to find the nearby value where $\phi(s+x+1)+s+1+x=0$ - Sheldon JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 01/25/2021, 01:01 AM (This post was last modified: 01/25/2021, 01:07 AM by JmsNxn.) Hey, Sheldon. So I've thought of a different way of phrasing your conjecture, which may be easier to prove. It's based off where I see the flaw in my proof. If we write, for $0 \le y \le 2\pi$ $ f(y,t) = e^{-iy}\phi(t+iy)\\$ Then, $ \lim_{t \to -\infty} \frac{f(y,t)}{e^{t-1}} = 1\\$ and, $ \frac{d}{dy}|f(t,y)| = 0\,\,\text{at least when}\,\, y = 0,\pi,2\pi\\$ To see this, $ \frac{d}{dy} |f(t+1,y)| = \frac{d}{dy}|e^{t+e^{iy}f(t,y)}| = 0\\ \frac{d}{dy} \Re(e^{iy}f(t,y)) = \frac{d}{dy} \big(\cos(y)\Re f(t,y) - \sin(y) \Im f(t,y)\big) = 0\\ \sin(y)\Re f(t,y) + \cos(y)\frac{d}{dy}\Re f(t,y) - \cos(y)\Im f(t,y) - \sin (y) \frac{d}{dy}\Im f(t,y) = 0\\ $ Now when $y = 0, \pi , 2\pi$, we know $\sin(y) = 0$ and $\Im f(t,y) = 0$ ($f$ is real valued here.) So we only care about, $ \frac{d}{dy}\Re f(t,y)= \frac{d}{dy}|f(t,y)|\cos(\arg f(t,y)) = 0$ Assuming $\frac{d}{dy}|f(t,y)| = 0$ implies $\frac{d}{dy}|f(t+1,y)| = 0$ when $y = 0, \pi , 2\pi$. Since $\frac{d}{dy}|f(t,y)| = 0$ as $\Re(t) \to -\infty$. We're done. Now, I INCORRECTLY assumed these are the only places where we have zeroes in $y$. What actually happens, and what you are noticing, is that, $ \frac{d}{dy}|_{y=y_k} |f(t,y)| = 0\\$ has solutions, but as $t$ grows, these solutions $y_k \to 0$. Which means, we get a whole bunch of minima clustering near the real axis. And where there are minima we get solutions to $\phi(s) + s = 0$. From your conjecture we can derive this (as it hits $\phi(s) + s = 0$ then $\phi(s+1) =1$, which implies there's a minima somewhere about there), and from this we can derive your conjecture; if we hit minima we should approach $\phi(s) + s =0$. Now, the conjecture I implore, which equates to this; is that for all these $y_k$ they cluster towards the real line. This would imply, $ |f(t,y)| \ge |f(t,\pi)|\,\,\text{while}\,\, \delta < y < 2\pi - \delta\\$ For $\delta \to 0$ as $t \to \infty$. I.e: there are no more minima/maxima, in this strip for large enough $t$ other than the minima at $\pi$. This would allow us to show, $ e \uparrow \uparrow s : \{ 0 < \Im(s) < 2\pi\} \to \mathbb{C}\\$ is analytic; but at the real line, it cannot be. Nor at multiples of $2 \pi ik + \mathbb{R}$. This actually meshes very well with the things I can say; it makes a lot of sense frankly. JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 01/25/2021, 10:08 PM (This post was last modified: 01/25/2021, 10:13 PM by JmsNxn.) So I believe I've managed to prove that the minima/maxima cluster towards the real line; giving arbitrarily close singularities to the real line as the real argument of $\phi$ grows. And, additionally, that solutions of $\phi(s) + s = 0$ necessarily cluster towards the real line; and cannot happen in the strip $\delta < \Im(s) < 2\pi - \delta$ for large enough $\Re(s) > T$ (here $\delta$ depends on $T$, and as $T\to\infty$ we get $\delta \to 0$). Luckily I so worded the main results of this paper that this was never not a possibility. But I had implicitly assumed it would be analytic on $\mathbb{R}$. I'm in the process of cleaning this up; and trying to make the argument a bit more solid than I have it right now. But I believe that this tetration should look like this, $e \uparrow \uparrow s : \mathbb{C}/(\mathbb{R} + 2 \pi i k) \to \mathbb{C}$ is analytic $e \uparrow \uparrow t + 2\pi i k$ is at least continuously differentiable--definitely probably nowhere analytic (but I can't prove it's nowhere analytic--can't really say much). This of course still meshes with the statement $e \uparrow \uparrow s$ is holomorphic on $\mathbb{C}/\mathcal{L}$ for $\mathcal{L}$ a nowhere dense set. It just looks a lot different than I was expecting. sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 01/25/2021, 11:54 PM (01/25/2021, 10:08 PM)JmsNxn Wrote: So I believe I've managed to prove that the minima/maxima cluster towards the real line; giving arbitrarily close singularities to the real line as the real argument of $\phi$ grows. And, additionally, that solutions of $\phi(s) + s = 0$ necessarily cluster towards the real line... I'm in the process of cleaning this up; and trying to make the argument a bit more solid than I have it right now....I look forward to seeing your updated paper. - Sheldon « Next Oldest | Next Newest »

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