Timeline for Radial behavior of dynamical map $x_{n+1}=2x_ny_n$, $y_{n+1}=1-2x_n^2$
Current License: CC BY-SA 4.0
16 events
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| Jan 5 at 18:25 | comment | added | Saúl Pilatowsky-Cameo | Ah yes, of course, thanks! | |
| Jan 5 at 18:19 | comment | added | fedja | @SaúlPilatowsky-Cameo The Jacobian must agree with the ratio of the densities at the point and at its image, not with the density at the initial point itself, to make the measure invariant. So, if the invariant measure is $w(z)dA(z)$, we must have $\frac{w(Tz)}{w(z)}=\frac 1{J(z)}$, not $w(z)=\frac 1{J(z)}$ as you seem to suggest. Does that make it clearer? | |
| Jan 5 at 18:07 | comment | added | Saúl Pilatowsky-Cameo | @fedja Silly question: Could you elaborate on why an invariant measure goes like $1/(1-r^2)$? I would naively think that it has to go as 1/Jacobian | |
| Jan 5 at 6:43 | comment | added | fedja | @KarlFabian I added the solution to the topological version of the problem in a separate answer. That is not what the OP wanted, of course, but you may still find it interesting. It also demonstrates a few ideas I'm trying to play with and the points I'm stuck at. | |
| Jan 3 at 10:05 | comment | added | Karl Fabian | You are right for the logarithm, and I can't strictly refute it. Anyway, I added an update to my answer to address the step size distribution which is quite strange, so I'm still not convinced that the $\sqrt{n}$ argument is true. | |
| Jan 3 at 0:06 | comment | added | fedja | @KarlFabian Yes. but I talked about the logarithm of the product, didn't I? I mean the random walk is adding, not multiplying. So yeah, I believe that $1-r_n^2$ is of order $e^{-\sqrt n}$ most of the time. Can you refute that? | |
| Jan 2 at 22:51 | comment | added | Karl Fabian | But as we are looking at a product, would your heuristic then give rather $e^{\sqrt{n}}$ ? | |
| Jan 2 at 22:25 | comment | added | fedja | @KarlFabian Well, yes and no: the doubling dynamics doesn't yield independent r.v., of course, but if you look at the Fourier side, you can easily show that the growth is typically at most $\sqrt n$ and, if you believe that certain estimates for lacunary series can be extended to almost lacunary series, then it cannot be much slower either. Technically you are absolutely correct: it is just as flawed as the idea that primes behave like random numbers, but I explicitly stated that it is a heuristics (=educated guess), not a proof of anything. | |
| Jan 2 at 19:20 | comment | added | Karl Fabian | Very interesting, but your $\sqrt{n}$ heuristics is flawed because the random walk steps are not independent. I looked at the same quantity with Mathematica: SSin[x_]:=Abs[If[x<1*^-15, x,Sin[x]]] phi0=phi=RandomReal[{0,2 Pi}]; prod=0; nl=NestList[ (phi=Mod[phi,2 Pi];prod+=Log[4]+2Log[ SSin[phi*=2]])&,1,100000]; ListPlot[nl,PlotTheme->"Detailed",AspectRatio->1/3,PlotLabel->Style["Subscript[[Phi], 0] = "<>ToString[phi0],24],FrameLabel->{ Style["n",20],Style["Subscript[log, 10 ]Subscript[S, n]",20]}] | |
| Jan 2 at 18:40 | comment | added | Saúl Pilatowsky-Cameo | About the numerics, I use Julia's BigFloat, and setprecision :) | |
| Jan 2 at 18:37 | comment | added | Saúl Pilatowsky-Cameo | I agree with you about the random walk. There is an interesting thing: if you are at the boundary, the angular map is ergodic (angle doubling map plus an extra rotation). Heuristically, close to the boundary, the map in polar coordinates looks like a skew-product system, where $\log(1-r_n^2)$ is like a deterministic random walk driven by the angle doubling. I was hoping to use known results on those (e.g. academic.oup.com/jlms/article-abstract/s2-13/3/486/804419 ). But of course what I said is not exactly true, as even close to the boundary the angular equation depends slightly on r. | |
| Jan 2 at 18:27 | comment | added | fedja | @SaúlPilatowsky-Cameo BTW, if not a secret, what program are you using for arbitrary precision computing? | |
| Jan 2 at 18:25 | comment | added | fedja | @SaúlPilatowsky-Cameo Excellent! You certainly approached it in a smarter way than I. :-). In general, I believe that the behavior of $\log(1-r_n^2)$ should be pretty much the same as of the random walk given by $X_{k+1}=X_k+S_k-e^{X_k}$ where $S_k$ are i.i.d. with some reasonable mean zero distribution. That random walk returns to a bounded domain infinitely often with probability $1$. No proofs yet, just some heuristics. | |
| Jan 2 at 18:01 | comment | added | Saúl Pilatowsky-Cameo | Re plots and numerical precision: the plots in the question are run using 10000 decimal precision, (which is not very optimal, but ensures that there is no rounding errors). | |
| Jan 2 at 16:11 | history | edited | fedja | CC BY-SA 4.0 |
added 4 characters in body
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| Jan 2 at 15:26 | history | answered | fedja | CC BY-SA 4.0 |