Timeline for Proven chaos in logistic maps
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5, 2022 at 8:59 | comment | added | coudy | If the parameter $r=3.8$ is chaotic, then there is 100% chance that an initial value chosen at random will have iterates that do not accumulate on a periodic point. I don't know if the parameter 3.8 is chaotic. Now, there is a parameter as close as you want to 3.8 (e.g. $< 10^{-100000000}$ from r) and an initial value as close as you want from the one you have given, such that the iterates of this initial value converge to a periodic orbit. Since a computer can't distinguish between very close values, one can't discard such a situation on the basis of a simulation. | |
| Feb 4, 2022 at 23:15 | comment | added | bobuhito | I wasn't saying there was any contradiction. My point was that my 50% still seems theoretically possible. I am just a math amateur...I can't tell whether you are agreeing or instead knocking it down to 0%. | |
| Feb 4, 2022 at 22:48 | comment | added | coudy | Indeed there may be an attracting periodic point that attracts an open set of points. This does not prevent the existence in the complement of that open set of a compact uncountable invariant set $K$ containing a point whose limit set is $K$. | |
| Feb 4, 2022 at 22:37 | history | edited | coudy | CC BY-SA 4.0 |
add reference as requested
|
| Feb 4, 2022 at 11:24 | comment | added | bobuhito | "There is a countable number of periodic orbits so the set of all these orbits is of zero Lebesgue measure" - But the set of points which converge to these periodic orbits can be uncountable (with non-zero measure), right? And, as I said in my first comment, please trust me to use enough bits of precision (I know the Lyapunov exponent is 0.4, so used 10,000 bits for 10,000 iterations before and currently using 1,000,000 bits for 1,000,000 iterations). | |
| Feb 4, 2022 at 7:39 | comment | added | coudy | From the theoretical viewpoint, there is a countable number of periodic orbits so the set of all these orbits is of zero Lebesgue measure. Practically, the computer works at a given precision and so iterates the map on a finite set of numbers. For such a system, all points are preperiodic. You can see an example of this dichotomy with the Arnold's cat map, which is chaotic but which appears to be periodic when simulated on a computer; see en.wikipedia.org/wiki/Arnold%27s_cat_map#The_discrete_cat_map and jstor.org/stable/2324989 | |
| Feb 3, 2022 at 18:55 | comment | added | bobuhito | I expect to catch and fix any rounding errors (program uses arbitrary precision and I therefore simply rerun all simulations with fewer bits to see/check my precision margin). It sounds like my 3.8 does leave some moderate probability of hitting a periodic cycle, let me say 50%. If I run a million iterations with no repeats, would that make a significant dent in this 50%? or are the cycles with most measure much longer than a million? | |
| Feb 3, 2022 at 12:53 | history | answered | coudy | CC BY-SA 4.0 |