Timeline for Convergence of an iterated sequence
Current License: CC BY-SA 3.0
15 events
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| Mar 9, 2018 at 21:44 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 7, 2018 at 21:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 8, 2018 at 20:58 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 9, 2017 at 20:33 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 9, 2017 at 20:17 | answer | added | Anthony Quas | timeline score: 2 | |
| Nov 9, 2017 at 11:01 | comment | added | user111097 | @AnthonyQuas Could you please explain a bit more how the above arguments could be applied to the case $p\neq 1/2$? By the way, why do you assume the slope $|\epsilon|<1$? Thank you very much! | |
| Nov 3, 2017 at 10:35 | history | edited | YCor |
edited tags
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| Nov 3, 2017 at 9:22 | answer | added | user111097 | timeline score: 0 | |
| Nov 3, 2017 at 6:07 | history | edited | Martin Sleziak |
removed deprecated (geometry) tag - see the tag info: http://mathoverflow.net/tags/geometry/info; if there are some other geometry-related tags which are suitable, please use some of them instead
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| Nov 3, 2017 at 1:39 | comment | added | Anthony Quas | It occurs to me that even for $p\ne1/2$, the subdivision is still determined by the slope, so this remains a one-dimensional problem (and the map is piecewise rational). It should be straightforward enough to answer your question then. | |
| Nov 2, 2017 at 23:49 | comment | added | Anthony Quas | Any half-plane whose intersection with the square has measure 1/2 goes through the centre of the square (proof: rotate by 180 degrees around the centre of the square). This means that after step 1, the two pieces are completely defined by the slope of the line joining them. I will rewrite the square as $[-1/2,1/2]^2$. If the slope is $|\epsilon|<1$, I claim the barycentres of the pieces are at $\pm(-\epsilon/6,1/4-\epsilon^2/12)$. If my calcs are right, this shows the new slope is $2\epsilon/(3-\epsilon^2)$. So the slopes $\epsilon=\pm1$ are unstable fixpts; and $\epsilon=0$ is attracting. | |
| Nov 2, 2017 at 23:35 | comment | added | user111097 | @AnthonyQuas Actually, this problem is linked to the Voronoi diagram that is used in solving an (semi-discrete) optimal transport problem, where the marginal distributions are given by $\mu=p\delta_{(x_1,y_1)}+(1-p)\delta_{(x_2,y_2)}$ and $\nu=\mathcal{Les}$ | |
| Nov 2, 2017 at 23:33 | comment | added | user111097 | @AnthonyQuas Thanks a lot for the quick reply. Could you please specify a bit more? | |
| Nov 2, 2017 at 23:01 | comment | added | Anthony Quas | In the case where p=1/2, this is a one-dimensional problem: the line separating the barycentres always goes through (1/2,1/2). Have you studied convergence or otherwise in this setting? | |
| Nov 2, 2017 at 21:32 | history | asked | user111097 | CC BY-SA 3.0 |