In response to this question, I was told that a certain sequence $a_n$ is "chaotic" and "wandering." The particular sequence in question is defined by $a_0 = z, z \in \mathbb{C}$ and $a_{n+1} = {a_0}^{a_n}$, that is, $z, z^z, z^{z^{z}} ...$ I was told it is "chaotic" because "you don't know what they do (or what they will do next). They may decide to converge after the 1000000000-th term or go to ∞, for example). Precisely because you don't know (unless you calculate the next terms), you cannot characterize them as either [convergent or divergent]."

To me, this description really makes absolutely no sense. By that logic, it seems to me that every sequence of the form $a_n = f^n(x)$ or even $a_n = f(n)$ is "chaotic" and "wandering." To me this entire notion of a "chaotic" or "wandering" sequence is complete and utter nonsense. How else am I (or anyone else) supposed to know the next values of a sequence unless I explicitly calculate them. I suppose one could approach this by saying that the values do exist, but we are "not allowed" to calculate them. However, this doesn't make much sense: if we are not allowed to calculate the values of a sequence than why should we bother defining the concept at all?

In summary, I am asking about $2$ things: first, what exactly is a "chaotic" sequence? Second, and more important to me, is the sequence $z^{z^{z^{...}}}$ really chaotic?

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    $\begingroup$ I can only answer the first part right now, but Devaney's book (a standard reference on the topic) has a more detailed definition than that given above, although the captured meaning is roughly the same. A chaotic map $f:V\to V$ is a map satisfying 3 properties: 1) Sensitive dependence on initial conditions, 2) $f$ is topologically transitive, 3) The periodic points form a dense set in $V$. Point 3 shouldn't need further explanation, point 1 means that there is a positive $\delta$ s.t. for any $x\in V$ and for any neighborhood $N$ of $x$, there exist $y\in N$ and $n\geq 0$ such that... $\endgroup$
    – MonadBoy
    Apr 25, 2016 at 4:48
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    $\begingroup$ ...$|f^n(x)-f^n(y)|>\delta$. Point 2 means that for any pair of open sets $A,B\subset V$ there is a positive $k$ such that $f^k(A)\cap B \neq \emptyset.$ I don't have the time to check if $f(z)=z^z$ fulfills these conditions, but if it does, it is chaotic by Devaney's (title: "An Introduction to Chaotic Dynamical Systems", 2nd edition) definitions. $\endgroup$
    – MonadBoy
    Apr 25, 2016 at 4:50
  • $\begingroup$ If the answer below was helpful to you, then you should upvote and accept it. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. For more information read this. $\endgroup$
    – user127032
    Apr 27, 2016 at 20:07
  • $\begingroup$ @MonadBoy, I don't think the "chaos game" mapping with scale factor = 1/2, which generates the Sierpinski gasket, has any periodic points. Does that mean it isn't chaotic? It's unpredictable in sequence, yet entirely predictable what the complete set will approach as number of iterations approaches infinity. $\endgroup$
    – Phil Goetz
    Jan 11 at 20:57

1 Answer 1


In summary, I am asking about 2 things:

first, what exactly is a "chaotic" sequence?

The commenters above gave you some definitions you can work with to find the answer to that.

Second, and more important to me, is the sequence $z^{z^{...}}$ really chaotic?

It can be, for certain values $c=z_0$. But no it's not, (generally) for all $c\in\mathbb{C}$.

And of course, now is a good time to check with Daniel Geisler's tetration map (*). This map shows you exactly for which $z=c$ the iterated exponential can be "chaotic": Precisely for those $c$ which are black on the By Period figure.

(*) Geisler's map of tetration is of course equivalent to Corless' map. The two maps satisfy: $\phi(Corless)=Geisler$, with $\phi=\exp(z/\exp(z))$, the potential map of the iterated exponential.


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