Timeline for Ordering periodic orbits
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 9, 2015 at 5:33 | comment | added | Anthony Quas | Nothing is happening here. He/she is just taking the set $\{x_0,x_1,\ldots,x_{2n}\}$ and writing it in increasing order as $\{z_1,z_2,\ldots,z_{2n+1}\}$ with $z_1<z_2<\ldots$. The remainder of the proof will show either $z_1=x_{2n}$, $z_2=x_{2n-2}$ etc. or $z_1=x_{2n-1}$, $z_2=x_{2n-3}$ etc. | |
| Sep 9, 2015 at 1:36 | comment | added | user80064 | I think that this is possible, because the orbit is an invariant set, then $f(\{z_{i}, i=1,2,\dots,2n+1\}) \subset \{x_{k}, i=0,2,\dots,2n\}$. However i don't know if this is answer | |
| Sep 9, 2015 at 1:24 | review | First posts | |||
| Sep 9, 2015 at 5:03 | |||||
| Sep 9, 2015 at 1:24 | history | asked | user80064 | CC BY-SA 3.0 |