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Let $z \in \mathbb{C}.$ Let $t = W(-\ln z)$ where $W$ is the Lambert W Function. Define the sequence $a_n$ by $a_0 = z$ and $a_{n+1} = z^{a_n}$ for $n \geq 1$, that is to say $a_n$ is the sequence $z, z^z, z^{z^z} ...$ This is sometimes called the iterated exponential with base $z$.

Here I am trying to prove (or disprove) $2$ things:

  1. For every $x \in \mathbb{R}$ there is $y^\star \in [0,\infty)$ such that for any $y \in \mathbb{R}$ with $|y| > y^\star$ the iterated exponential with base $z = x+ yi$ converges to a set of $3$ periodic accumulation points. As $|y| \to \infty$ the $3$ points approach the orbit $\lbrace 0, 1, z \rbrace$

  2. Let $E = (e^{1/e} , \infty) \cup \lbrace s \in \mathbb{C}:|t| = |W(-\ln s)| = 1 $ and $t^n \ne 1$ for all $n \in \mathbb{N} \rbrace$ If $z \in \mathbb{C} \setminus E$, the iterated exponential is bounded and there exists $k \in \mathbb{N}$ such that the iterated exponential converges to a set of $k$ periodic accumulation points.

I have been studying Daniel Geisler's tetration map; my first question is essentially a formalization of some observations I have made. My second question has proven to be more problematic. I have been experimenting numerically for several years, and I have observed that an iterated exponential sometimes gets "slingshot" to a neighborhood of $\infty$. When this happens it can be very difficult to tell what the next terms will be.

The main thing I have tried in these situations is a kind of asymptotic analysis. The next term $a_{n+1}$ is considered as the product $z^{\Re(a_n)}(z^i)^{\Im(a_n)}$. There are $36$ possibilities determined by the signs of $\Re(a_n), \Im(a_n)$ and whether $|z|, |z^i|$ are greater than, less than, or equal to $1$. Typically, though each factor is either $0, \infty,$ or else it just keeps going around the unit circle and doesn't approach any fixed value. So there are really only $3$ possibilities: $0, \infty, 0\cdot\infty$.

If I'm "lucky" enough to get $a_{n+1} \sim 0$, I can easily see that the "slingshot to $\infty$" is followed by (approximately) $\lbrace 0, 1, z, z^z, \ldots \rbrace$ Otherwise I get stuck in limbo: it may be that the entire sequence is diverging to $\infty$. However, I have never been able to find specific examples of this, other than $(e^{1/e} , \infty)$. If I get $a_{n+1}\sim \infty$ this basically puts me back at square one; I can ask the same question about $a_{n+2}$, namely, is it $0, \infty$, or something else. So, in most (if not all) cases, I feel I have to explicitly calculate the next terms; otherwise I have no way of knowing if the whole sequence diverges to $\infty$, or if some subsequent term is very close to $0$.

Another major problem is that the next term often is so big that it causes an overflow error on my calculator. I have tried calculating the natural logs of the terms, using the iteration $b_0 = \ln z, b_{n+1} = e^{b_n}\ln z$, but sometimes it doesn't help, because I still get overflow errors. An example of this is $z = -2.5.$ After just 6 terms $a_n$ is on the order of $10^{26649}$ and $b_n$ has an almost identical value after 7 terms.

Note: all of my work so far assumes the use of the principal branches of the natural log and the Lambert W function.

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    $\begingroup$ Just one question: "in case 3b (...) has period k for some $k \in \mathbb N$" - do you really mean "periodic" and thus a set of k periodic accumulation points to which the iteration converges? Please pardon, if I'm messing some things up? $\endgroup$ Commented Jun 11, 2016 at 3:47
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    $\begingroup$ @GottfriedHelms Yes, that is what I mean. I'm the one who should be apologizing, because I wasn't exactly sure how to say that, so instead I said something which apparently caused some confusion. $\endgroup$
    – cpiegore
    Commented Jun 11, 2016 at 3:52
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    $\begingroup$ @GottfriedHelms I would also like to point out that I am purposefully omitting case 3b from consideration, because in my opinion it is the most "troublesome" of the 4 cases. $\endgroup$
    – cpiegore
    Commented Jun 11, 2016 at 3:54
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    $\begingroup$ you didn't talk so much of the different branches of $\log(z)$ and $W(z)$, and how changing of branch of $\log$ during the iteration $a_{n+1} = z^{a_n} = e^{a_n \log(z)}$ affects the result. and how do you prove $3a$ ? $\endgroup$
    – reuns
    Commented Jun 11, 2016 at 5:46
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    $\begingroup$ @user1952009 First, I am assuming that any work done on this is made using the principal branch of the natural log. This is because, as far as I can tell, the orbits produced by the principal log are the most studied and most well understood. Also, the other branches of $W(z)$ appear to have no relationship to the iterated exponential. When defining $a_n$ using different branches of the log, I have found there are 2 different ways to do this, which lead to completely different results. $\endgroup$
    – cpiegore
    Commented Jun 11, 2016 at 14:54

1 Answer 1

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This is an attempt at proving statement 1. Showing that the conclusion holds for any $y \in \mathbb{R}$ with $|y| > y^\star$ basically boils down to proving $\lim\limits_{|y| \to \infty}(x+yi)^{x+yi} = 0$ for fixed $x$. Without loss of generality we may assume $y >0$ since $\bar z^{\bar z} = \overline{z^z}$. We may also ignore the argument of $(x+yi)^{x+yi}$ and show that $|(x+yi)^{x+yi}| \to 0$

$|(x+yi)^{x+yi}| = (x^2+y^2)^{x/2}e^{y\arctan(x/y)-y\pi/2}$. Since $x$ is fixed $(x^2+y^2)^{x/2} \sim y^{x}$ and since $y\arctan(x/y) \to x$, as $y \to \infty$ we have $e^{y\arctan(x/y)-y\pi/2} \sim e^{-y} \implies (x+yi)^{x+yi} \sim y^{x}e^{-y} \to 0$ since $e^{-y} \to 0$ way faster than $y^{x} \to \infty$.

I am still not sure how to show there is a smallest value $y^\star$ for which the conclusion is true. Nor do I know of a formula, or even an algorithm, to find it, other than a lot of numerical experimentation.

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    $\begingroup$ Shouldn't it be $(x^2+y^2)^{x/2} \sim y^{\color{red}{x}}$? $\endgroup$
    – celtschk
    Commented Jun 19, 2016 at 4:24
  • $\begingroup$ No because $x$ is fixed, ie, $x$ is a constant with respect to $y$, ie, I am only taking the limit with respect to $y$ $\endgroup$
    – cpiegore
    Commented Jun 19, 2016 at 4:51
  • $\begingroup$ Okay, I admit I am not perfect. I may have misunderstood the use of the tilde symbol. $\endgroup$
    – cpiegore
    Commented Jun 19, 2016 at 4:55
  • $\begingroup$ Actually it should be $(x^2+y^2)^{x/2} \sim y^{x/2}$, but the limit is still $0$ since $e^y$ grows faster than any power of $y$ $\endgroup$
    – cpiegore
    Commented Jun 19, 2016 at 5:01
  • $\begingroup$ Why $x/2$? As far as I can tell, $(y^2)^{x/2} = y^x$. Am I missing something? But of course the limit is still $0$ that way. $\endgroup$
    – celtschk
    Commented Jun 19, 2016 at 5:17

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