13
$\begingroup$

Let $a_1=0$ and let $ - \ln(2) < a_2 < \ln(2) $

Define

$$ a_n = a_{n-1}^3 - a_{n-2} $$

Then

$$ \sup_{n>2} a_n = a_2 $$

And

$$ \inf_{n>2} a_n = - a_2 $$

How to prove that ?

Improve this question
$\endgroup$
10
  • 8
    $\begingroup$ Could you explain where this problem comes from and why you made such a conjecture ? $\endgroup$
    – M. Dus
    Commented Dec 24, 2019 at 17:33
  • 6
    $\begingroup$ Why people are closing this question? It seems an interesting example to me $\endgroup$
    – Pietro Majer
    Commented Dec 25, 2019 at 13:12
  • 1
    $\begingroup$ The associated continuous dynamical system is a Liénard system; general theorems on these imply that it has a unique limit cycle. Also see the van der Pol oscillator for behaviour of the same kind. $\endgroup$
    – user41593
    Commented Mar 23, 2021 at 10:42
  • 1
    $\begingroup$ The associated differential equation is just $(\dot x, \dot y)=(y,y^3-x)$ (cf. Robert's answer), and your recurrence is a discretization of such system. I think Smale's horseshoe theory fully explains all the observations about almost-periodic and periodic points; cf. the van der Pol equation in deterministic chaos for an essentially similar case study. But I have no idea about your claimed inequality. The general theory on Liénard systems (Massera's theorem and others) shows that there is a unique limit cycle, the picture should be more or less the same as for vdP. $\endgroup$
    – user41593
    Commented Mar 23, 2021 at 15:58
  • 1
    $\begingroup$ I should add that limit cycles of these systems are known not to be algebraic, so the answer to your question on finding the mystery constant is most likely no. But they are amenable to Runge-Kutta methods, and asymptotics as the perturbation tends to 0 are known. I now see I maybe should have gathered all my comments in the form of an answer, but to me Henri's answer already says the most important thing there is to know. $\endgroup$
    – user41593
    Commented Mar 25, 2021 at 9:38

3 Answers 3

Reset to default
11
$\begingroup$

The question is really about the iteration behaviour of the maps $(x,y) \mapsto (y, y^3-x)$ with various starting points. We have a fixed point $(0,0)$ and a $6$-cycle $$(1,0),(0, -1), (-1, -1), (-1, 0), (0, 1), (1, 1)$$ It appears that for something like $-0.797 < x < 0.797$, $(0,x)$ is on an invariant curve. For example, here are $10000$ iterates starting at $(0,0.796)$:

enter image description here

For starting values with slightly larger $x$, the process seems to become unstable.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you Robert. You anticipated my 2 next questions. Me and my mentor ( tommy1729 who gave me this idea ) . The value ln(2) can indeed be stretched to a higher value. A first remark is that tommy noted that the computation or estimate of this number strongly depends on the numerical precision. With 8 digits or so it is hard to get above ln(2) ; you get the illusion of divergence. A closed form for this number is ofcourse desired ; the 1st followup question. Secondly is there convergence or divergence at that number ? Or stated differently : do all valid x lie on an open or a closed interval? $\endgroup$
    – mick
    Commented Dec 25, 2019 at 10:04
  • $\begingroup$ Merry Xmas from me and tommy. I wonder how it is possible to get a score of +6 but also 3 close votes :/ Does not make sense to me. $\endgroup$
    – mick
    Commented Dec 25, 2019 at 10:07
8
$\begingroup$

For $b:=a_2$, the GCD of the polynomials $a_{10}=a_{10}(b)$ and $a_9(b)+b$ is $$b \left(b^2+1\right) \left(b^{78}-b^{76}+b^{74}-b^{72}-8 b^{70}+8 b^{68}-8 b^{66}+8 b^{64}+28 b^{62}-28 b^{60}+28 b^{58}-28 b^{56}-59 b^{54}+59 b^{52}-59 b^{50}+59 b^{48}+85 b^{46}-85 b^{44}+85 b^{42}-85 b^{40}-86 b^{38}+86 b^{36}-86 b^{34}+86 b^{32}+61 b^{30}-61 b^{28}+61 b^{26}-60 b^{24}-30 b^{22}+30 b^{20}-30 b^{18}+27 b^{16}+9 b^{14}-9 b^{12}+9 b^{10}-6 b^8-3 b^6+3 b^4-3 b^2+1\right). $$ The only root of this GCD in $(0,1)$ is $b_*\approx0.637295\in(0,\ln2)$.

So, for $a_2=b_*$ we have $a_{10}=0=a_1$ and $-a_9=a_{11}=a_2$, so that the sequence $(a_n)$ is of period $9$; we also have $\max_n a_n=a_2$ and $\min_n a_n=-a_2$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you for your answer. There are probably infinitely many cycles. Marry Xmas. $\endgroup$
    – mick
    Commented Dec 25, 2019 at 10:13
6
$\begingroup$

I encountered a similar problem with the iteration $u_n=\sqrt{u_{n-1}^2+1}-u_{n-2}$, where there is fundamentally a $9$-cycle in the shape of a maple leaf (replace the +1 by 0 to see this). I asked an expert 20 years ago who told me that using the KAM (Kolmogorov-Arnold-Moser) theorem, one could prove that the "curve" drawn above by R. Israel is not a curve at all, but a very thin strip of width 0.07 or something. Don't ask me for the proof, I have no idea.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Very interesting. Im not an expert with KAM, I only heard of it. The type of recursions that give these kind of behaviours like in the OP or the one you give are conjectured to have a pattern by my mentor. However more investigation is needed even for the precise statement of the conjectures. There might be a way without KAM. I will let you know if I can find such a way ( I and some friends have ideas that might work ). Thank you for your contribution !! +1 $\endgroup$
    – mick
    Commented Mar 22, 2021 at 23:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .