Timeline for Is $\lim_{z\to0} \exp_{\sqrt{2}}^{\circ z}(\xi)$ continuous?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 8, 2021 at 8:50 | history | edited | YCor |
edited tags
|
|
| Jul 22, 2017 at 8:05 | comment | added | user78249 | ...Quite frankly, there could only be a single point $\xi$ on $[2,2+4\pi i/\log2]$ that does not belong to the basin of $f$. | |
| Jul 22, 2017 at 7:20 | comment | added | user78249 | @GeraldEdgar Do you have a proof that $[2,2 + 4\pi i /\log2] \subset A$? Or are you basing this off of numerical calculation? I've been trying to prove this fruitlessly, and so I am still doubting that $f^{\circ 1/2}$ does not exist. | |
| Jul 15, 2017 at 1:44 | comment | added | user78249 | So you are saying $f^{\circ z}(\xi)$ exists solely if $\Re(z) > 1$? I should note, I am not referencing the construction for $\Re(z) > 0$, this is just based on my own proofs (which may be too hand wavey, as you seem to be pointing out). So you may be on to something. Perhaps I should focus on $\Re(z) > 1$. This seems a little disappointing, it implies there is no function $g$ such that $g(g(\xi)) = \sqrt{2}^\xi$. I'll have to go over my work more clearly and reanalyze all the pieces. | |
| Jul 12, 2017 at 18:07 | comment | added | Gerald Edgar | In fact, I claim that $f^{\circ(1/2)}(\xi)$ cannot be continuous on that segment. $f^{\circ 3}(\xi)$ is inside a nice neighborhood of $2$, we see what $f^{\circ(3.5)}(\xi)$ looks like from the Schröder function. Then take inverses to see $f^{\circ(2.5)}(\xi)$ and $f^{\circ(1.5)}(\xi)$ which are nice closed curves, and $f^{\circ(0.5)}(\xi)$ which cannot be continuous and agree at the two endpoints. | |
| Jul 12, 2017 at 17:50 | comment | added | Gerald Edgar | "Now it is no obvious fact, but..." I guess that existence is what I am doubting. $A$ is an open set. The line segment from $2$ to $2+4\pi i/\log 2$ lies in the basin of attraction of $2$, so by definition of $A$ that line segment is in $A$. Maybe it is easier to believe in the holomorphic function in two variables only where $\mathrm{Re}\; z > 1$. It goes down to $z=0$ for $\xi$ in a neighborhood of $2$, but I question it holding for all $\xi \in A$. | |
| Jul 12, 2017 at 7:27 | comment | added | user78249 | Thinking about it, maybe if this result is open, one should actually prove $G(\xi) = \xi$ to prove if $\xi \in A$ then $\xi + 4k\pi i/\log(2) \not \in A$. Because I have a quasiproof which may actually give this result; I just assumed it wrong because I never considered this consequence a possibility (too good to be true). | |
| Jul 12, 2017 at 0:42 | comment | added | user78249 | @GeraldEdgar, that sounds right. I think one can show using a theorem in Milnor's Complex Dynamics that it's dense--I'll pick up the book and double check. The trouble is $A$ is somewhat more mysterious. I've been mostly dealing with dynamics on the real line, but for complex things get trickier to visualize. I have been drawing mental comparisons between $\sin^{\circ t}$ on the real line and the such (both periodic, might have something in common). Also, IFF $G\Big{|}_{A} = \text{Id}\Big{|}_{A}$, then $A$ has only one set of periods in it, which describes it a bit more | |
| Jul 11, 2017 at 13:42 | comment | added | Gerald Edgar | I did some numerical computations. As far as I could tell from the pictures, $f^{\circ k}(\xi) \to 2$ for all complex numbers $\xi$. Well, I know that is wrong since $f(4) = 4$. But I guess the pictures suggest the set with $f^{\circ k}(\xi) \to 2$ is dense in $\mathbb C$. And that doesn't tell us anything about $A$. | |
| Jul 10, 2017 at 22:19 | comment | added | user78249 | Sadly, I've never seen a nice picture of the Fatou set of $f$ (is there some app to produce your own?) so it's hard to even get intuition. I've never found a proof detailing this fact. But you may be on to something here. Maybe if $\xi \in A$ then necessarily $\xi + 4k\pi i/\log(2) \not\in A$ for $k \neq 0$ and in fact $G(\xi) = \xi$ for $\xi \in A$. Any anomalies get wiped out because $f$ would be univalent on $A$. I kind of just thought this too good to be true, if I'm being frank. | |
| Jul 10, 2017 at 21:00 | comment | added | Gerald Edgar | Do we know: if $\xi \in A$, then also $\xi+4\pi i/\log(2) \in A$ | |
| Jul 10, 2017 at 2:39 | history | edited | user78249 | CC BY-SA 3.0 |
added 124 characters in body
|
| Jul 10, 2017 at 2:30 | history | asked | user78249 | CC BY-SA 3.0 |