4
$\begingroup$

Let's think about some finite field $\mathbb{F}$. Is it possible to construct a map

$x[n+1] = \mathcal{P}(x[n], x[n-1],...,x[n-k]), \ \ \ \forall x\in\mathbb{F} $

where $\mathcal{P}$ - polynomial with coeffs $\in\mathbb{F}$, so it's behavior be non-periodic, but chaotic,
so x[n] be "jumping" randomly over $\mathbb{F}$?
(As $x[n+1] =1-2\cdot x[n]^2$ on [-1,1])

$\endgroup$
2
  • $\begingroup$ Polynomials can only have a fixed number of variables, so perhaps you mean $\mathcal P(x[n],x[n-1],\dots,x[n-k])$? (Any reason you are not using subscsripts? $x[n]$ is programming notation, not a math notation.) $\endgroup$ – Thomas Andrews May 23 '13 at 15:45
  • $\begingroup$ @Thomas Andrews Sorry... My mind immersed in Mathematica $\endgroup$ – lesobrod May 23 '13 at 15:48
5
$\begingroup$

If $x[n+1]$ depends on $d$ of the previous values in the sequence (i.e. $x[n+1]=P(x[n],x[n-1],\ldots,x[n-d+1])$, the answer is negative. There are only $|\mathbb{F}|^d$ possible $d$-tuples of the values, so once any of them repeats (Dirichlet's principle), the subsequent values will start repeating too, making the function eventually-periodic.

$\endgroup$
4
  • $\begingroup$ Yes, now I see clearly. $\endgroup$ – lesobrod May 23 '13 at 15:54
  • $\begingroup$ Hmm Are there other ways to make simple chaotic finite state system? $\endgroup$ – lesobrod May 23 '13 at 15:55
  • 1
    $\begingroup$ @lesobrod no, because of the exact same reasonning... $\endgroup$ – Glougloubarbaki May 23 '13 at 16:01
  • 1
    $\begingroup$ So what about nonconservative system: $x[n+1]=P(n,x[n-1],...)$ ? Any chance for finite chaos... $\endgroup$ – lesobrod May 23 '13 at 16:04
2
$\begingroup$

Let $f:F^k \to F$ be a function on any finite set $F$, then the sequence defined recursively by $x_{n}=f(x_{n-1},x_{n-2},...,x_{n-k})$ can only take a finite number of values. So it must be periodic, maybe only after a certain rank.

But your question is not trivial, though. Assuming the field $F$ to be infinite but of finite characteristic, for example the p-adic numbers, can you define chaotic in such a topology?

$\endgroup$
1
  • $\begingroup$ Thank you, I'll try to play with $p$-adic monsters... $\endgroup$ – lesobrod May 23 '13 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.