Timeline for Confusion on the assumption when discussing the kneading invariants for unimodal maps
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 26, 2020 at 16:05 | comment | added | Anthony Quas | The point is $f$ is a decreasing map from $I$ to $I$, so that $f^2$ is an increasing map from $I$ to $I$. It's pretty easy to see that if you have an increasing map, everything converges to a fixed point. Probably math.stackexchange.com would be a better place if you want to ask about that. | |
| Jun 26, 2020 at 14:00 | comment | added | JacobsonRadical | @AnthonyQuas could you please elaborate a little bit why every orbit in $I$ converges either to a fixed point or a period two orbit? | |
| Jun 26, 2020 at 13:51 | comment | added | JacobsonRadical | @AnthonyQuas brilliant! Post an answer so that I can accept and vote? | |
| Jun 26, 2020 at 2:34 | comment | added | Anthony Quas | In this case, the interval $I=[c,f(c)]$ is mapped into itself and the restriction of $f$ to $I$ is decreasing. This means that every orbit in $I$ converges either to a fixed point or a period two orbit. For points in $[0,c]$, the orbit is either monotonic (and so converges to a fixed point), or it eventually leaves $[0,c]$, whereupon it enters $I$. For points in $[f(c),1]$, the image is in $[0,f^2(c)]\subset [0,c]\cup I$, so that we have shown that every orbit approaches a fixed point or period 2 orbit. Pretty dull... | |
| Jun 25, 2020 at 20:06 | review | First posts | |||
| Jun 25, 2020 at 20:30 | |||||
| Jun 25, 2020 at 19:56 | history | asked | JacobsonRadical | CC BY-SA 4.0 |