Timeline for Iterated Circumcircle
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2010 at 11:59 | vote | accept | Joseph O'Rourke | ||
| Jun 15, 2010 at 4:59 | comment | added | Victor Protsak | Ian: Thank you for confirming it. I will have to look up Atkinson's theorem, for as I wrote to Joseph, I am far from an expert. I guess it was the right decision to tack on "ergodic theory" tag:) $$ $$ Re "rewriting this comment": There are a few tricks you should know: (a) Type your comment in the answer window at the bottom and get an instantaneous preview, then copy-paste it into the comment window. (b) You can access the exact text of your comment, including latex markup, through your user page (recent activity -> comments). (c) Before posting the comment, copy it into the clipboard. | |
| Jun 15, 2010 at 4:47 | comment | added | Victor Protsak | @Joseph: You are most welcome! It's a nice little problem. You are correct, when I wrote it I didn't see a way to prove or disprove it, although I $\textit{suspected}$ that iterations a.s. do not converge (because the size needn't go to $0$). I would defer to Ian, who is much more competent in this area than I, for the final pronouncement. | |
| Jun 14, 2010 at 12:07 | comment | added | Ian Morris | (I am getting fed up with rewriting this comment due to typos...! Third time lucky....) We can write $\sum_{k=1}^n \ln |2\cos \psi_k|$ as an ergodic sum with respect to a transformation of $\{1,2\}^{\mathbb{N}}\times [0,\pi]$, (where the first co-ordinate corresponds to the stochastic choice at each stage) so I think that it follows from G. Atkinson's theorem on cocycle recurrence that $\prod_{k=1}^n |2\cos \phi_k|$ generically diverges, accumulating at both 0 and $\infty$. | |
| Jun 14, 2010 at 12:03 | comment | added | Joseph O'Rourke | @Victor: Thanks for your perspicacious analysis! "and it remains unclear whether the sizes of the triangles go to zero"---I gather this means it is unclear whether e.g. nearly all triangles converge? | |
| Jun 14, 2010 at 10:22 | comment | added | Victor Protsak | It's not a mistake. For $\phi_n$, that follows from its definition (half the angle in a triangle). For $\psi_n$, since the absolute value of the cosine function is $\pi$-periodic, so we only need to keep track of $\psi_n (\mod\pi).$ | |
| Jun 14, 2010 at 9:21 | comment | added | Ian Morris | I'm a bit confused as to why $\phi_n$ is distributed over $[0,\pi/2]$ and $\psi_n$ is distributed over $[0,\pi]$. Is this a mistake, or have I missed something? | |
| Jun 14, 2010 at 7:36 | history | edited | Victor Protsak | CC BY-SA 2.5 |
edited for style and expanded
|
| Jun 14, 2010 at 3:07 | history | answered | Victor Protsak | CC BY-SA 2.5 |