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 eta as branchpoint of tetrational bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/04/2011, 08:22 AM (06/03/2011, 10:57 PM)mike3 Wrote: (06/02/2011, 02:04 PM)bo198214 Wrote: The kneser tetration $b\mapsto b[4]p$ is real on the real axis $b>\eta$, which implies that $\overline{b} [4] p = \overline{b[4]p}$ (conjugation). So approaching from above or below is just conjugate to each other. So what one need to show imho is that the imaginary part will not tend to zero when approaching the real axis at $b<\eta$. _Or_ show that it is not conjugate-symmetric there. Maybe I was not insistent enough on that: If we have a function $f$ that is real-analytic on any interval (a,b) and you continue it through any path $\gamma(t)$ in the upper halfplane. Then for the conintuation in the lower halfplane $\overline{\gamma}$ we have: $f(\overline{\gamma(t)}) = \overline{f(\gamma(t))}$, i.e. in simple words: f is conjugate-symmetric *everywhere* mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/04/2011, 09:08 AM (06/04/2011, 08:22 AM)bo198214 Wrote: (06/03/2011, 10:57 PM)mike3 Wrote: (06/02/2011, 02:04 PM)bo198214 Wrote: The kneser tetration $b\mapsto b[4]p$ is real on the real axis $b>\eta$, which implies that $\overline{b} [4] p = \overline{b[4]p}$ (conjugation). So approaching from above or below is just conjugate to each other. So what one need to show imho is that the imaginary part will not tend to zero when approaching the real axis at $b<\eta$. _Or_ show that it is not conjugate-symmetric there. Maybe I was not insistent enough on that: If we have a function $f$ that is real-analytic on any interval (a,b) and you continue it through any path $\gamma(t)$ in the upper halfplane. Then for the conintuation in the lower halfplane $\overline{\gamma}$ we have: $f(\overline{\gamma(t)}) = \overline{f(\gamma(t))}$, i.e. in simple words: f is conjugate-symmetric *everywhere* Correct. So showing that it is not conjugate-symmetric at all would seem to work, no? Thus if it behaves like the regular iteration at fixed point 2 in the upper halfplane, and like that at fixed point 4 in the lower halfplane, then it would seem it would not be conjugate-symmetric (note the difference in behaviors implied: the function would be bounded in the upper-right quadrant, while not so in the lower-right), thus not real-valued for real heights greater than -2. However, I wonder if the deviation from real-valuedness may be relatively small, which is why the pure regular "seems to work so well" for $1 < b < \eta$. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/04/2011, 09:19 AM (06/04/2011, 09:08 AM)mike3 Wrote: So showing that it is not conjugate-symmetric at all would seem to work, no? Then the Kneser-tetration would not be (real-)analytic in the base for base > eta. Do you think that???? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/04/2011, 09:50 AM (This post was last modified: 06/04/2011, 09:52 AM by mike3.) (06/04/2011, 09:19 AM)bo198214 Wrote: (06/04/2011, 09:08 AM)mike3 Wrote: So showing that it is not conjugate-symmetric at all would seem to work, no? Then the Kneser-tetration would not be (real-)analytic in the base for base > eta. Do you think that???? Oops, sorry, I see now, you're talking about in the base. I was referring to in the height Never mind, just missed that. But anyway, wouldn't showing that it was not conjugate-symmetric in the height when $b < \eta$ work? sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 06/04/2011, 01:56 PM (06/02/2011, 01:19 PM)bo198214 Wrote: (06/02/2011, 12:51 PM)bo198214 Wrote: Lets see what happens, with the fixpoints $M^\pm(b)$ when moving on the circle $b=\beta(t)=\eta+ (\eta-\sqrt{2})e^{\pi i t}$ for $-1 in the next post. Ok, this happens when moving the base in the lower halfplane $t\in (-1,0)$. The blue curve is the movment of the upper fixpoint $M^+(\beta(t))$ and the red one is the lower fixpoint $M^-(\beta(t))$. At t=0 we have the two complex fixpoints with real part around 2.5.Henryk, I calculated what happens to the fixed points, in going from circle base $=2\eta+\sqrt{2}$ to base$=\sqrt{2}$. Moving along the lower circular path, "underneath" eta, I can seamless morph the upper repelling fixed point 2.478+0.8518i seamlessly to a repelling fixed point of 4.0. But the lower fixed point, which starts out at 2.478-0.8518i, and eventually becomes a fixed point of 2.0, somewhere early along the path, around 7 degees/180, the repelling fixed point of 2.478-0.8518i becomes an attracting fixed point of 2.435 - 0.8239i, before continuing on as an attracting fixed point, moving towards 2.0 Mike has spent more time thinking about and graphing "repelling"/"attracting" fixed point combinations in the complex plane, as well as "repelling"/"repelling" fixed point combinations. I'm relying here entirely on my intuition. When we combine "repelling"/"repelling" fixed points, we get a single horizontal line at real axis (which is no longer real valued), which includes the singularities corresponding to log(0), log(log(0)), log(log(log(0))), just as in the Kneser solution for real sexp(z). The real axis is is where the two functions "mesh" into each other. The upper half of the plane corresponds to the upper fixed point, and the lower half of the plane corresponds to the lower fixed point. My guess is that when you have an upper repelling fixed point, and a lower attracting fixed point, you still get that horizontal line at the real axis, with its singularities, but you get other horizontal lines all in the lower half of the complex plane, where the singularities at log(0), log(log(0)), log(log(log(0))) show up periodically because of the periodic nature of the attracting fixed point. But this would only be in the lower half of the plane. My intuition fails me, in knowing what happens to the upper repelling fixed point approaching 4.0, and the lower attracting fixed point approaching 2.0, at the very last step, when we get to b=sqrt(2). It doesn't seem to match the [tex]\text{newsexp}_{\sqrt{2}}(z), that I posted earlier. Perhaps its yet another solution, that is not real valued at the real axis .... By the way, I believe these complex base sexp functions are exactly what Mike is posting about. In principle, I think I know how to calculate a Kneser mapping pair, with a different theta(z) for each of the functions, where the two theta(z) functions merge to a new combined function. It is likely next on my tetration "todo" list. - Sheldon bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/04/2011, 02:10 PM (06/04/2011, 01:56 PM)sheldonison Wrote: I calculated what happens to the fixed points, in going from circle base $=2\eta+\sqrt{2}$ to base$=\sqrt{2}$. Moving along the lower circular path, "underneath" eta, I can seamless morph the upper repelling fixed point 2.478+0.8518i seamlessly to a repelling fixed point of 4.0. But the lower fixed point, which starts out at 2.478-0.8518i, and eventually becomes a fixed point of 2.0, somewhere early along the path, around 7 degees/180, the repelling fixed point of 2.478-0.8518i becomes an attracting fixed point of 2.435 - 0.8239i, before continuing on as an attracting fixed point, moving towards 2.0 The switch repelling/attracting imho takes place on the Shell-Tron-boundary. sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 06/04/2011, 03:05 PM (This post was last modified: 06/04/2011, 03:17 PM by sheldonison.) (06/04/2011, 02:10 PM)bo198214 Wrote: (06/04/2011, 01:56 PM)sheldonison Wrote: I calculated what happens to the fixed points, in going from circle base $=2\eta+\sqrt{2}$ to base$=\sqrt{2}$. Moving along the lower circular path, "underneath" eta, I can seamless morph the upper repelling fixed point 2.478+0.8518i seamlessly to a repelling fixed point of 4.0. But the lower fixed point, which starts out at 2.478-0.8518i, and eventually becomes a fixed point of 2.0, somewhere early along the path, around 7 degees/180, the repelling fixed point of 2.478-0.8518i becomes an attracting fixed point of 2.435 - 0.8239i, before continuing on as an attracting fixed point, moving towards 2.0 The switch repelling/attracting imho takes place on the Shell-Tron-boundary.Yeah, I think that's a big change in the function there. Do you have a reference on the "shell tron region"? I seem to remember reading about it many times, "Kidney bean shaped" region or something, but I don't know what the theory tells us happens. Also, Kneser mappings probably won't work from attracting fixed points, unless there are repeating regions of superexponential growth as real(x) grows. For example, the lower sexp superfunction of eta definitely lacks such superexponential growth, as real(x) grows. Also, the lower superfunction for sexp sqrt(2). Instead, both of those functions converge to the fixed point as real(x) grows, for all imag(z). - Sheldon bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/04/2011, 06:59 PM (This post was last modified: 06/04/2011, 07:00 PM by bo198214.) (06/04/2011, 03:05 PM)sheldonison Wrote: Yeah, I think that's a big change in the function there. Do you have a reference on the "shell tron region"? I seem to remember reading about it many times, "Kidney bean shaped" region or something, but I don't know what the theory tells us happens. For you I just created a first shot of the Shell-Thron region entry on the Hyperoperations Wiki . (anyone may) feel free to improve and extend Quote:Also, Kneser mappings probably won't work from attracting fixed points, unless there are repeating regions of superexponential growth as real(x) grows. For example, the lower sexp superfunction of eta definitely lacks such superexponential growth, as real(x) grows. Also, the lower superfunction for sexp sqrt(2). Instead, both of those functions converge to the fixed point as real(x) grows, for all imag(z). Oh this might indicate that the Kneser mapping is only possible for (and continuable to) bases outside the Shell-Thron region! bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/04/2011, 07:25 PM (06/04/2011, 09:50 AM)mike3 Wrote: I see now, you're talking about in the base. I was referring to in the height Never mind, just missed that. But anyway, wouldn't showing that it was not conjugate-symmetric in the height when $b < \eta$ work? And it seems we both confused to a certain extend base and height. Ok, I see your (two) points now: Given that it converges in the upper halfplane to the upper fixpoint and correspondingly in the lower halfplane, then 1. Its not conjugate symmetric when approaching the real axis below eta, and hence can not be real valued. 2. it approaches a different function from above and below Yes, this seems not to contradict the conjugacy when approaching from above and below. So we somehow need to verify your assumption, that it converges to the upper fixpoint for positive imaginary infinity. Perhaps I will write a Wiki article about perturbed Fatou coordinates, which may help to decide this question. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/04/2011, 10:16 PM (This post was last modified: 06/04/2011, 10:35 PM by bo198214.) (06/04/2011, 07:25 PM)bo198214 Wrote: So we somehow need to verify your assumption, that it converges to the upper fixpoint for positive imaginary infinity. Perhaps I will write a Wiki article about perturbed Fatou coordinates, which may help to decide this question. So, I just finished writing up, whats in the article of Shishikura. So far only quotes, its quite difficult, but I hope you see that he talks about the same thing we do. You find it here on the Hyperoperations Wiki. His $f_0$ corresponds to our $\exp_\eta$. His $f$ (which corresponds to our $\exp_{\eta+\eps}$) has not two conjugate fixpoints, but one at 0 and one near zero. His function $\varphi_f$ corresponds to our superfunction. And his function $\tilde{\mathcal{R}}$ corresponds to our 1-periodic $\theta$. You can use the "Discussion" page whenever there something needs clarification. « Next Oldest | Next Newest »

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