Math overflow question on fractional exponential iterations sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 03/28/2018, 07:57 PM (This post was last modified: 03/28/2018, 09:33 PM by sheldonison.) Dmytro Taranovsky asks in his post, https://mathoverflow.net/questions/28381...rent-bases ... " ... Let a and b be real numbers above e^{1/e}, and c exp(1/e), then my conjecture is that the limit holds.  Unfortunately, Peter Walker's solution is also conjectured to be nowhere analytic; see: https://math.stackexchange.com/questions...lytic-slog (2) That for Kneser's solution, there are counter examples and the limit does not hold, even with the restriction that c\;\exp_b^d(x)$ (3) Given any analytic solution for base(a), then if we desire a tetration solution for base(b) which has the desired property then I conjecture that the tetration base b is nowhere analytic!  The conjecture is also that the slog for b would be given by a modification of Peter Walker's h function. - Sheldon sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 03/29/2018, 10:13 PM (This post was last modified: 03/29/2018, 10:18 PM by sheldonison.) (03/28/2018, 07:57 PM)sheldonison Wrote: ... Let a and b be real numbers above e^{1/e}, and c\;\exp_b(x)\;\;\;$ but only if x is big enough when a1, and helps one understand why Walker's solution works. - Sheldon JmsNxn Ultimate Fellow Posts: 1,178 Threads: 123 Joined: Dec 2010 03/30/2018, 07:37 PM (This post was last modified: 03/30/2018, 08:47 PM by JmsNxn.) It seems upsetting that the only tetrations that could satisfy this are non-analytic. I'm not prone to believe this, only because it doesn't look nice... I'm wondering if we can look at it the following way $\exp_b^{d}(x) = \exp_b^{c}(\exp_b^{\delta}(x))$ Then the question boils into whether, for all $\delta, \delta' >0$ then $\exp^c_{b+\delta}(x) = o(\exp_b^{c+\delta'}(x))$ I think we can show this is true when $c \in \mathbb{N}$. By induction, first, for $c = 0$ the induction step is obvious. Namely $x = o (\exp_b^{\delta'}(x))$. Suppose the result holds for $c = n$ then (taking $<$ to mean asymptotically less than): $\exp_{b+\delta}^{n+1}(x) =\exp^{n}_{b+\delta}(\exp_{b+\delta}(x))<\exp_{b}^{n+\delta'/2}(\exp_{b+\delta}(x)) < \exp_{b}^{n+\delta'/2}(\exp_b^{1+\delta'/2}(x)) = \exp_b^{n+\delta'+1}(x)$ Sadly, I can't think of anyway to generalize this to non-integral $c$.... I'm thinking, a nice way to look at it from here is to look at root functions of the $\exp$ function. But then we'd need an implication $f^{\circ n}(x) = o(g^{\circ n}(x)) \Rightarrow f(x) = o(g(x))$, which looks like it could be true for monotonically growing unbounded functions. But that's probably too easy, we'd probably need a nice condition on the root functions for that to be true. EDIT: It appears I made a fruitful mistake in the above proof. The base induction step would have to be (1) $\exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x)$, not the obvious one $x < \exp^{\delta'}(x)$. This is the base step I should have used. The proof then says if this base step (1) holds the result holds for all natural $c$, namely $\exp_{b+\delta}^n(x) < \exp_b^{n+\delta'}(x)$. And I think with some finesse we can show that this implies it's true for root functions of $\exp_b$, which should leave for a proof where $c \in \mathbb{Q}$. Then perhaps a density argument may work on non rational $c$. I'll work on this more later, but I think maybe we can reduce this entire problem into the condition that if for all $\delta,\delta'>0$ we have $\exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x)$ then it follows that $\exp_{b+\delta}^c(x) < \exp_b^{c+\delta'}(x)$. ...We'll probably have to assume that $\exp_b^c(x)$ is monotone non-decreasing in $x$ and unbounded, or at least, eventually monotone non-decreasing. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 03/31/2018, 04:19 AM (This post was last modified: 03/31/2018, 04:29 PM by sheldonison.) (03/30/2018, 07:37 PM)JmsNxn Wrote: It seems upsetting that the only tetrations that could satisfy this are non-analytic. I'm not prone to believe this, only because it doesn't look nice... There's an old thread http://math.eretrandre.org/tetrationforu...hp?tid=236 I started the thread in 2009, before I had written generic programs for analytic tetration for any base, and I was using an excel spreadsheet to approximate analytic tetration.  I estimated that as x gets arbitrarily large $\text{slog}_2(x)-\text{slog}_e(x)\approx1.1282$ The 2009 thread continues on to discuss what I called "the wobble"... Credit needs to go to William Paulsen and Samuel Cowgill in their upcoming paper which discusses these issues more rigorously than I can.  But it was quickly clear in the 2009 thread that there is an inherent wobble when comparing tetration bases; this was apparent for bases a little bit bigger than eta=exp(1/e) using straightforward techniques on an excel spreadsheet.  The limit as x get arbitrarily large does not converge to a simple number like the 1.1282 estimate, but instead converges to a 1-cyclic function near that value.  For base(2) and for base(e), if you use Kneser's construction; then the 1-cyclic limit is graphed below. $\text{slog}_e(\text{tet}_2(x))-x$     On this forum, other ideas like "the base change function" were discussed, where you use Peter Walker's idea to define tetration base(a) from tetration base(b).  For example, you could define tetration base(2) from Kneser's tetration base(e).  The relevant equations might look something like this.  But the "h" function below is conjectured to be nowhere analytic, even though Walker proved it is $C^{\infty}$ for the case in his paper.  Walker defined the base(e) slog from the Abel function for iterating $x\mapsto\exp(x)-1$.  This is mathematically conjugate (or exactly equivalent) to iterating base eta.  $y\mapsto\eta^y\;\;\eta=\exp(1/e)\;\;x=\frac{y}{e}-1$ $h_n (x)=\ln_b^{[n]}(\exp_a^{[n]}(x))$ $h(x)=\lim_{n\to\infty}h_n (x)$ $\text{slog}_b(x)=\text{slog}_a(h(x))-\text{slog}_a(h(1)));$ /* constant to guarantee slog_b(1)=0 */ - Sheldon JmsNxn Ultimate Fellow Posts: 1,178 Threads: 123 Joined: Dec 2010 04/01/2018, 03:09 AM (This post was last modified: 04/01/2018, 06:22 AM by JmsNxn.) Okay, so the question and your intuition relates to the old base change formula and that it failed to be analytic. That makes sense, but is disappointing to think we're going to lose this property if we choose an analytic solution. So what this slog limit is saying is that for ''good'' analytic tetrations: $f(x)=\exp_{b+\delta}^{c}(x) - \exp^{c+\delta'}_b(x)$ changes sign infinitely often (given $\delta,\delta'<\epsilon$)? This reminds me of something. I've dealt with those limits before and felt discouraged at an ability to prove uniform convergence. Given holomorphic $f,g : \mathbb{D} \to \mathbb{D}$ where $f(0) = g(0) = 0$, when trying to find a function $\Psi:\mathbb{D}\to\mathbb{D}$ such that $\Psi(f(z)) = g(\Psi(z))$, the natural choice is $\Psi(z) =\lim_{n\to\infty} g^{-n}(f^{n}(z))$ (which never seems to work). But it sure does look nice. The only way this works, I found, is to assume $f'(0) = g'(0) = \lambda$ and take the Schroder function of both functions $h_0, h_1$ where $h_0(f(z)) = \lambda h_0(z)$ and $h_1(g(z)) = \lambda h_1(z)$ and then $\Psi(z) = h_1^{-1}(h_0(z))$ which works locally. Then the above limit for $\Psi$ is convergent. But we had to sacrifice a lot to get there.  Of course if we're working on a non simply connected set $H$ instead of $\mathbb{D}$ and we assumed that $f,g$ had no fixed points on this set, this could work. But tetration takes $\mathbb{C}/\{z \in (-\infty,-2)\}\to \mathbb{C}$, so it probably has fixed points (maybe this is provable). Which should guarantee a base change function $h$ is non extendable to $\mathbb{C}/\{z \in (-\infty,-2)\,\}$. This is kinda' helping me understand why these functions fail to be analytic. No conjugation can change the multiplier value and clearly $^ze$ will have a different multiplier at its fixed point as $^z 2$ will have at its fixed point. I'll have to read Walker's paper. The only work around I had to this was working with Schroder functions and when dealing with the real line where there are no fixed points I can't imagine a manner of getting a nice uniform convergence. I'm still wondering if I can prove that if $\exp_{b+\delta}(x) < \exp_b^{1+\delta'}(x)$  then $\exp_{b+\delta}^c(x) < \exp_b^{c + \delta'}(x)$ which could then be a condition for tetration to be non-analytic.  Still seems like a lot of this is up in the air though. I apologize if this has me a bit scatter brained. « Next Oldest | Next Newest »

 Possibly Related Threads… Thread Author Replies Views Last Post Bridging fractional iteration and fractional calculus Daniel 5 28 Today, 03:25 AM Last Post: JmsNxn Fractional Integration Caleb 11 366 02/10/2023, 03:49 AM Last Post: JmsNxn digit extracting iterations tommy1729 0 57 02/05/2023, 11:08 PM Last Post: tommy1729 Matrix question for Gottfried Daniel 6 528 12/10/2022, 09:33 PM Last Post: MphLee Discussing fractional iterates of $$f(z) = e^z-1$$ JmsNxn 2 360 11/22/2022, 03:52 AM Last Post: JmsNxn Qs on extension of continuous iterations from analytic functs to non-analytic Leo.W 18 2,725 09/18/2022, 09:37 PM Last Post: tommy1729 Fibonacci as iteration of fractional linear function bo198214 48 5,892 09/14/2022, 08:05 AM Last Post: Gottfried The iterational paradise of fractional linear functions bo198214 7 913 08/07/2022, 04:41 PM Last Post: bo198214 Describing the beta method using fractional linear transformations JmsNxn 5 816 08/07/2022, 12:15 PM Last Post: JmsNxn Apropos "fix"point: are the fractional iterations from there "fix" as well? Gottfried 12 1,699 07/19/2022, 03:18 AM Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)