Does tetration take the right half plane to itself? JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 05/10/2017, 07:46 PM (This post was last modified: 05/10/2017, 07:49 PM by JmsNxn.) This question is rather straightforward. Let $1 < \alpha < \eta$. We know that there exists a unique bounded tetration function $F(z) = \alpha \uparrow^2 z$ that is holomorphic for $\Re(z) > 0$. But is $\Re (F(z)) > 0$? Does the bounded tetration function take the right half plane to itself? If this is true, this manages to prove a lot of things. First of all, it follows that tetration has only one fixed point $\omega$ such that $F(\omega) = \omega$, that is geometrically attracting $0 0}$.  All of the orbits of $F$ tend to this fixed point. This implies we can construct a complex iteration $F^{\circ z}(\zeta):\mathbb{C}_{\Re(z) > 0} \times \mathbb{C}_{\Re(\zeta)>0} \to \mathbb{C}_{\Re(\zeta)>0}$. Therefore giving us pentation $F^{\circ z}(1) = \alpha \uparrow^{3} z$ that ALSO takes the right half plane into itself. Therefore it ALSO has a unique fixed point, this fixed point is ALSO geometrically attracting, all of the orbits of pentation tend to this fixed point, and now we can rinse and repeat to construct hexation, so on and so forth.  The great part about this is how it qualifies the sequence of bounded analytic hyper-operators. It gives a lot of great properties.  We get the following things for free. $\alpha \uparrow^n z$ for $n\ge 2$ is holomorphic in $z$ and analytic for $1 < \alpha < \eta$  (I still haven't really managed to show the hyper-operators are analytic in the base argument, only continuous; but with this lemma, it follows trivially).  $\alpha \uparrow^n z : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$ (before I simply wrote they send to the complex plane and focused on their behaviour in $\mathbb{R}^+$). $\alpha \uparrow^n z$ has a purely imaginary period (something that gives a lot of cool things, like a Fourier series representation for example). if $\omega_{n-1}$ is the fixed point of $\uparrow^{n-1}$ or the limit at infinity of $\uparrow^n$, then $|\alpha \uparrow^n z - \omega_{n-1}|< C q^x$ for $x = \Re(z)$ for some $0 < q < 1$. (Exponential decay is always nice.) ...(I could go on)... And most importantly, these functions satisfy the holy grail of functional equations $\alpha \uparrow^n (\alpha \uparrow^{n+1} z) = \alpha \uparrow^{n+1} (z+1)$ The functional equation is something I could never truly get perfect, because I only managed to show it on the real positive line, without sending the right half plane to itself, the composition in the complex plane is technically ill defined. So all in all, I've boiled a whollllllllleeeee swash of questions into one question. Does $\alpha \uparrow^2 z : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$? If anyone's curious, I can explain how I've approached the question. It's a little convoluted, so I'll leave it out till someone asks me. JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 05/10/2017, 08:22 PM (This post was last modified: 05/10/2017, 09:22 PM by JmsNxn.) Of course, as soon as I post this I've essentially proved it using the Fourier expansion. Now I'm just hoping Sheldon comes on and can enlighten me a bit more as to why this works. First and foremost $\alpha \uparrow^2 z = L - \sum_{n=1}^\infty a_n \lambda^{nz}$ where $\alpha^L = L$ and $\lambda = \log(L)$ If $h(\lambda z) = \alpha^{h(z)}$ is the inverse schroder function, normalized so that $h'(0) = 1$, then $a_n = -h^{(n)}(0) h^{-1}(1)^n$ where  $h^{-1}(1)^n(-1)^n > 0$ and, as Sheldon mentions, but I haven't seen a fully rigorous proof, $h^{(n)}(0)(-1)^{n+1} > 0$ for $n \ge 1$ therefore $a_n>0$. Now it follows that $|\alpha \uparrow^2 z - L| \le \sum_{n=1}^\infty a_n \lambda^{nx} \le \sum_{n=1}^\infty a_n = L -1$ and tetration is contained in a disk about $L$ that lies in the right half plane. NOW, all I need is $h^{(n)}(0)(-1)^{n+1} > 0$ for $n \ge 1$. Sheldon has claimed this, but in order to put it in the paper, I'll probably either need to prove it myself, or have a good reference. In a short enough, well thought out manner, that is rigorous enough to put in the finalized version of my paper (credit will be given, of course). sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/15/2017, 08:16 PM (This post was last modified: 05/15/2017, 09:35 PM by sheldonison.) Hello James, https://math.stackexchange.com/questions...29#1996129 Presumably, what you're interested in is the details of the proof sketch in my two answers. Read the answer beginning with, "This answer is an attempt to explain lemma1. Why is ..."  I will send you an email with a repeat of my private message sent Friday/today and then you can contact me and we can talk off line.  Basically, h2 is the first approximation of the Schroder function.  And h2 is fully monotonic.  Then we show the h_m sequence is also fully monotonoic.  And then we make the convergence rate of the h_m sequence more rigorous (lemma5) to show the h_m sequence converges to h(z), the Schroder function.  There are some more steps.  But I think its good.  As I said, feel free to contact me offline. - Sheldon sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/15/2017, 09:00 PM (This post was last modified: 05/15/2017, 11:25 PM by sheldonison.) (05/10/2017, 07:46 PM)JmsNxn Wrote: This question is rather straightforward. Let $1 < \alpha < \eta$. We know that there exists a unique bounded tetration function $F(z) = \alpha \uparrow^2 z$ that is holomorphic for $\Re(z) > 0$. But is $\Re (F(z)) > 0$? Does the bounded tetration function take the right half plane to itself?So are you asking if it is a 1-1 mapping from the right half of the plane to itself?  Certainly there's a singularity at z=-2.   Lets think about base sqrt(2), with two fixed points, 2 and 4. As $\Re(z)$ gets arbitrarily large $\text{sexp}_{\sqrt{2}}(z) \approx 2$  At the imaginary axis, sexp(0)=1.  At half the period, sexp(period/2) the value would have grown to something like 2.468.  In between, imag(z) gets to about +/- 0.7208. So at the imaginary axis, it will roughly "circle" or loop around the fixed point of 2 an infinite number of times, as follows:  This very roughly circular path is what the right half of the plane is mapped to!   Of course, it would be more complicated if I repeated this experiment at the singularity at z=-2 and again mapped one imaginary period, but I don't see the plane mapped to itself at z=-2 either.  The imaginary period is $=\frac{-2\pi i}{\ln(\ln(2))} \approx$ 17.143     - Sheldon JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 05/16/2017, 04:09 AM (This post was last modified: 05/16/2017, 04:11 AM by JmsNxn.) Your second response confuses me. That's VERY close to what I am asking, but it is slightly off. It is DEFINITELY NOT 1-1, first of all. Tetration is periodic, so there's no hope in hell of that, I wouldn't dare say that. Tetration IS injective modulo a period (which is easy to prove). I do know that tetration does live in the right half plane, using experimental arguments. And I can prove it if the Schroder function of $\alpha^z$ is fully monotone about zero, then tetration lives in a small disk about the fixed point of the exponential (that small disk residing in the right half plane).  Yes, I reread a bunch of your posts regarding it. I read them very slow, and I now know why the inverse Schroder function is fully monotone. It is much simpler than I thought! MUCH simpler than what I was thinking. We just have to use that the composition of two fully monotone functions is fully monotone, and that the uniform limit of fully monotone functions is fully monotone. Where "fully monotone functions" are defined as having "completely monotone first derivative." It easily follows from this that $\log^{\circ n}(\lambda^n z + L)$ is fully monotone, it uniformly converges to the inverse Schroder function, and therefore the inverse Schroder function is a fully monotone function. Therefore, by the above arguments $^z\alpha : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$, it actually sends the right half plane to a small disk in the right half plane (but that's extraneous). Therefore $\alpha \uparrow^n z : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$.  If perhaps you are confused. I'm betting when you read the paper it'll be a lot clearer. I'll be sure to make a head nod and an accreditation to you. I'll PM you to ask how to reference you. I'm rebuilding a paper I wrote two years ago, and adding grout to everything, so there are no leaks and no holes. JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 05/16/2017, 04:46 AM (This post was last modified: 05/16/2017, 04:59 AM by JmsNxn.) Rereading your second post, this much can be proved VERY easily! The fact that tetration cycles around the fixed point can exactly be related to the exponential map. Let $0 < \lambda < 1$, if $\lambda^{z}: \mathbb{C}_{\Re(z)>0} \to \mathbb{D}$ (where $\mathbb{D}$ is the unit disk), It is periodic (first of all) and rotates around zero as we increase the imaginary argument, it tends to zero as $\Re(z) \to \infty$, and is INJECTIVE modulo its period. We get the exact same thing with tetration $^z\alpha$, substituting $0 = L$ and $\lambda = \log(L)$ and $\mathbb{D} = \{z \in \mathbb{C} : |z-L| \le |L-1|$. This is biholomorphic to the previous scenario via the Schroder function. It's no surprise that it behaves how you described. I always wondered what kind of crazy fractals it performed when it cycled around, but that much I couldn't plot. I just knew that many fractional iterations created a weird $\lambda^z$ on some weird simply connected domain. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 05/16/2017, 03:34 PM (05/16/2017, 04:09 AM)JmsNxn Wrote: ...  It easily follows from this that $\log^{\circ n}(\lambda^n z + L)$ is fully monotone, it uniformly converges to the inverse Schroder function, and therefore the inverse Schroder function is a fully monotone function...Good luck with your paper.  We need more rigorous iterated exponentiation papers. I just wondered what your approach to show the sequence "uniformly converges".  Is there an easy theorem, or did you want to go with a more complicated approach something like my lemma5?  Just curious.  Also, you probably meant  $\log_b^{\circ n}(\lambda^n z + L)$ where b is the tetration base. - Sheldon JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 05/16/2017, 08:46 PM (This post was last modified: 05/16/2017, 08:48 PM by JmsNxn.) (05/16/2017, 03:34 PM)sheldonison Wrote: (05/16/2017, 04:09 AM)JmsNxn Wrote: ...  It easily follows from this that $\log^{\circ n}(\lambda^n z + L)$ is fully monotone, it uniformly converges to the inverse Schroder function, and therefore the inverse Schroder function is a fully monotone function...Good luck with your paper.  We need more rigorous iterated exponentiation papers. I just wondered what your approach to show the sequence "uniformly converges".  Is there an easy theorem, or did you want to go with a more complicated approach something like my lemma5?  Just curious.  Also, you probably meant  $\log_b^{\circ n}(\lambda^n z + L)$ where b is the tetration base. Yeah, I meantt $\log_b$. This is actually a well known result on how to represent the inverse Schroder function. Bo used it before, and that's how I first learnt about it.  Essentially if $f(z_0) = z_0$ and $0 < |f'(z_0)| < 1$, then not only does $\lim_{n\to\infty} \frac{f^{\circ n}(z) - z_0}{\lambda^n} \to \Psi(z)$ for $|z- z_0|<\delta$, where $\Psi$ is the Schroder function.  We also get that, if $g = f^{-1}$ $\lim_{n\to\infty} g^{\circ n}(\lambda^n z + z_0) \to \Psi^{-1}(z)$ for $|z-z_0| < \delta$, for $\delta$ sufficiently small. This is all you really need. « Next Oldest | Next Newest »

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