# Chaotic iterative example needed

I'm using a very simple numerical method to find solutions to an equation. Start with an equation $\operatorname{f}(x)=0$ that you need to solve. Rearrange to give $x=\operatorname{g}(x)$ and then use the recurrence relation $x_{n+1} = \operatorname{g}(x_n)$ to hopefully tend towards a solution.

I'm trying to find an interesting example where starting with $x_0 < \ell$, but very close to $\ell$, gives a sequences tending towards one limit, while starting with $x_0 > \ell$, but very close to $\ell$, gives another. (Where $\operatorname{f}(\ell) \neq 0)$ I can find unintesting examples where negative $x_0$ gives a negative limit and positive $x_0$ gives a positive limit.

Can anyone suggest an interesting example. For example, for some $x_0 > 2$, very close to $2$. gives a sequence tending towards a negative limit and for some $x_0 < 2$, very close to $2$ gives a sequence tending towards a positive limit.

EDIT: I am looking for a single, smooth function $\operatorname{g}$. I want a good old-fashioned elementary function. If possible, a polynomial or rational function would be great.

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Perhaps this is an interesting example: math.stackexchange.com/questions/281082/… – Gottfried Helms Jun 23 '13 at 18:29
@GottfriedHelms I'm working over the reals. – Fly by Night Jun 23 '13 at 18:30
Can you check your usage of positive and negative? I believe that strict adherence to your condition is not possible. See my answer. – Calvin Lin Jun 23 '13 at 18:33
@CalvinLin I edited my post to change this before you wrote your comment. See edit history. – Fly by Night Jun 23 '13 at 18:41
Please do not assume that things get done immediately as it occurs, which would be your assumption of strictly following the history. I had to think about it, AND type up an answer before I made that comment. If you look at when my answer was typed up, it occurred before you made your edit. – Calvin Lin Jun 23 '13 at 18:44

Let try to take it stepwise.

As your title says, I guess you are seeking for some type of (deterministic) chaotic behaviour on a one-dimensional iterative process that sends from $\Bbb R$ to $\Bbb R$. Also I guess this because you are seeking for initial condition sensitivity as is in deterministic chaos a prrequisite.

For such system behaviour, necessary is:

(1) that $g$ is non-linear with respect to $x_n$ and/or some $x_{n-i}$ (associative memory) where $i<n$ and a positive integer

(2) that there are at least $2$ discrete elementary modes (better $3$); an intuition would be to have the second mode either via memory constructed into the non-linear memory parts $x_{n-i}$ or if not possible then as a separate variable or variables, then $\overrightarrow x_n$ a vector (this possibly fits to your case).

An alternative would be indeed to try a fractal version where a process sends from $\Bbb C$ to $\Bbb C$, complex domain.

Are these conditions something you could accept in your approach?

A posterior

what the answer distinguishes is not an issue of the Real domain, rather key is the fact that your proposition seems to pursue only one elementary mode, chaos is provoked by a certain type of self organisation at higher order that demands at least 2-3 elementary modes that cooperate in this self-organisation process. A one-modal equation will not provide this precondition.

Either you would need more dimmensions via a vector over Real e.g.

$$\overrightarrow x_{n+1}(=[x_{n+1,1},x_{n+1,2},x_{n+1,3},\dots]^T)=\overrightarrow g(x_{n,1},x_{n,2},x_{n,3},\dots)$$

or such memory over the Real e.g.

$$x_{n+1}(=[x_{n},x_{n-1},x_{n-2},\dots]^T)=g(x_{n},x_{n-1},x_{n-2},\dots)$$

or both and in both case $g$ a proper non-linear function. Your case probably would possibly fit in (2).

In case you regard (2) suggest you have a look on the simple example of the logistic map.

Hope that the above Latex Art works is correct.

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Would it be fair to say that the essence of your answer is that continuous functions on the reals are just too well-behaved to produce the sort of behavior the asker desires? – dfeuer Jun 23 '13 at 19:02
see above extension – al-Hwarizmi Jun 23 '13 at 19:31

I may be missing something, but if you start Newton's method for $f(x)=x^2-a$ with $x_0<0$ it will converge to $-\sqrt a$ and with $x_0>0$ it will converge to $\sqrt a$. That's the simplest example I know.

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But I'm not using Newton's method. – Fly by Night Jun 23 '13 at 19:05
Note also that OP wanted the positive and negative to flip. You have the 'negative' converging to a negative root, and the 'positive' converging to a positive root. – Calvin Lin Jun 23 '13 at 19:31
@FlybyNight, Newton's method is a fixed-point method given by a rational function, as you wished. – lhf Jun 23 '13 at 19:49
@lhf I explicitly gave the numerical method that I'm using in my post. No calculus is involved. You start with an equation, say $x\!\operatorname{e}^x=1$. You rearrange to give $x=\operatorname{e}^{-x}$. You then apply the recurrence relation $x_{n+1} = \operatorname{e}^{-x_n}$ for some initial guess $x_0$. In this case, choosing $x_0=1$ gives $x_n \to 0.567143290$ (9dp) for sufficiently large $n$. – Fly by Night Jun 24 '13 at 17:13
@FlybyNight, I wasn't clear what you meant by rearrange. – lhf Jun 24 '13 at 18:49

Call $u\approx4.8742$ the smallest positive solution of the equation $u\cos u=\pi/4$ and consider sequences $(x_n)$ defined by $x_{n+1}=g(x_n)$ where the function $g$ is $$g(x)=2x\cos(ux/2).$$ Then:

• If $x_0=2$, then $x_1=\pi/u$ and $x_n=0$ for every $n\geqslant2$.
• If $x_0\lt2$ with $x_0$ close to $2$, then $x_2\gt0$ with $x_2$ close to $0$ and $x_n\to z$ where $z=2\pi/(3u)$ is the smallest positive root of $x=g(x)$, that is, $z\approx0.4297$.
• If $x_0\gt2$ with $x_0$ close to $2$, then $x_2\lt0$ with $x_2$ close to $0$ and $x_n\to-z$ where $z$ is the largest negative root of $x=g(x)$.
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It's not possible for a sequence such that every positive (or $x>2$) initial value converge to a negative fixed point (say $a$) and every negative (or $x<2$) initial value converges to positive fixed point (say $b)$. Just consider the starting value which is the fixed point $a$ (which is negative), and we see that it converges to the negative fixed point (and vice versa if the condition was $x>-2$).

Otherwise, take $g(x) = \begin{cases} -5 & x>2, x \neq 5 \\ 5 & x = 5 \\ 5 & x < 2, x \neq -5 \\ -5 x = - 5 \end{cases}$.

Does what you want.

Another example

$g(x) = \begin{cases} -5 & 2 < x < 3\\ 5 & 1 < x < 2 \\ g_1(x) & x>3 \\ g_2 (x) & x< 1 \end{cases}$

where $g_1$ has converges to $5$ and $g_2$ converges to $-5$.

If you want it to be smooth, you can replace the region around $2$ with a straight line, and just chose initial values that are outside of this region (since you said for specifically chosen initial values).

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Thanks for the reply. I didn't want it for all $x_0 < \ell$ and all $x_0 > \ell$. I should have said. I want an example, for example $x_0 = 1.999$ and I get a positive limit and $x_0 = 2.001$ and I get a negative limit. – Fly by Night Jun 23 '13 at 18:33
@FlybyNight Sure, then this is such an example. You can make it as interesting as you want it to be. – Calvin Lin Jun 23 '13 at 18:34
I'm working on numerical solutions. I had smooth, elementary functions in mind. I would have no need to use numerical methods to solve the original problem that gave this $\operatorname{g}$. – Fly by Night Jun 23 '13 at 18:38
@FlybyNight I've given you a more general example. You can replace it with whatever smooth elementary functions you want to smooth it out. – Calvin Lin Jun 23 '13 at 18:39
Thank you, but it is not what I was looking for. – Fly by Night Jun 23 '13 at 18:40