Let $z_0 \in \mathbb{C}$ be a complex number, and define from it the infinite sequence $z_n = e^{z_{n-1}}$.

Question: in general, what can we say about the properties of the sequence $\{z_n\}$?

I know that if $z_0 \in \mathbb{R}$, then $\{z_n\}$ goes to $+\infty$ very rapidly.

Also, there are infinitely many $z_0$ that are fixed points of $e^z$- according to Wolfram Alpha, all values of the form $-W_n(-1)$ with $n\in \mathbb{Z}$ work.

I would guess that for all $m>1$, there are also infinitely many $z_0$ for which the sequence has a period of $m$ (though I don't know if that's true).

But all results so far were given for special values of $z_0$ (a set of measure 0). What can we say about the sequence for a general $z_0$? Does it usually diverge to $\infty$, or converge to a fixed point, or does it have weirder behavior?

I've tried to check it on Python (with $z_0 = i$ for example), and the process seems very numerically unstable, so it's hard to say what the analytic behavior is from the simulation.


2 Answers 2


It's a deep issue. For an overall graph see:


For the fixed points, see this question by Gottfried Helms.

The 1-period fixed points are given by $z_k=-W_k(-1)$, where $W$ is the Lambert map. Gottfried, through reverse iteration has calculated p-periodic points for $p>1$, for many $p$, in this post.

This hints towards the idea that there are infinitely many $p$-periodic points for any $p$, although this is still an open question (The periodic fixed points are shown in yellow on the first graph).

The graph for the iterates is a Cantor Bouquet type of fractal and since bouquets repeat by splitting down (bouquet "fingers") to an infinite level (depending on resolution), this suggests that there is a continuum of $p$-periodics for any $p>1$.

1-periodics will congregate around the main bouquet. 2-periodics congregate around the main sub-bouquet "fingers", 3-periodics around the sub-sub-bouquet fingers, and so on (yellow points on the graph).

To find 1-periodics, you need to solve the equation $z=\exp(z)$, whose solution is given as above, by $W$. To find 2-periodics, you need to solve the equation $z=\exp(\exp(z))$, which can only be done using numerical methods - like Gottfried's. For 3-periodics, you need to solve $z=\exp(\exp(\exp(z)))$, wtc.

If you decide to use numerics, you have to be aware that you won't be getting all of the $p$-periodic points at once, unless you iterate all branches of the inverse (complex branch $\ln_k$ in this case) - like Gottfried does.

Then, for a given period you index them, according to the $\ln$ branch, as $z_k$, $k\in\mathbb{Z}$.

To get a sense of $p$-periodics in the iterates fractal, you need to distinguish between different $p$ periods, so if you scan the plane through iterating $\exp$, you can store the iterates for given $z_0$ in an array and postprocess it to assign it a period $p$ if the iterates repeat every $p$ steps in the array. Then color them accordingly (The fractal shown in the first link does not distinguish between fixed points of different periods. It just colors yellow anything that eventually repeats).

Edit: For more info on this type of fractal - Julia Sets as iterates of $\lambda\exp(z)$ in general, see Devaney's website, for example, where he shows that it contains indecomposable continua.

Indecomposable continua are large areas which can cover the entire complex plane. Orbits of points inside those continua can be chaotic. For example, this fractal has $\lambda=1>1/e$ and as such the Julia Set has exploded through a Knaster explosion and all the regions after the main tip of the Cantor Bouquet, form an indecomposable continuum. Points from this main contiuum (exploded tip of the main feature) seem to converge spiral-like towards the two main features left and right of the main Bouquet: The two yellow spirals, left and right of the Bouquet. Other tip explosions, lead to successively deeper yellow fixed points of higher periods, inside the bouquet fingers.

  • $\begingroup$ Hi Yiannis! In answers to my questions I understand L.Rempe-Gillen, that the question, whether for a) for any $n$ there are n-periodic points, and that b) for any $n$ they can be "indexed" by a $\mathbb Z^n$ index, has been examined and has been answered to the positive (...cntd...) $\endgroup$ Sep 14, 2020 at 20:37
  • $\begingroup$ (...cntd...) (however, the examination according to the texts used another formalism from mine but the results are equivalent). The only "fuzzy" observation is, that for the index $0$ the notation must be expanded to $0^+$ and $0^-$ convention. I think it was in the last of the Rempe-Gillen answers/comments here in MSE or MO, didn't check yet... $\endgroup$ Sep 14, 2020 at 20:39
  • $\begingroup$ Greetings, Gottfried. Do you have any reference for these examinations and how they were done? It'd be very interesting if anyone proved that there are infinitely many for any $p>1$. Can you expand on that $0^+$, $0^-$ thingy or on any of the related issues? $\endgroup$
    – user127032
    Sep 15, 2020 at 0:04
  • $\begingroup$ Yiannis, please pardon my delay, my head is only peripheral in this math currently due to intense work on improving rooms in my flat.Anyway - I had always difficulties with the mapping of my concept (iterated branched logarithm) to that studies of the iterated exponential and their properties or vice versa. First: the property that for each $n$-periodicity there exist infinitely many exemplars of periods/cycles is referred by Bergweiler ("Iteration of meromorphic functions") back to Baker and Misuriewicz ... $\endgroup$ Sep 16, 2020 at 16:41
  • 1
    $\begingroup$ @GottfriedHelms Thanks Gottfried. I have Bergweiler's article, but don't remember anything related to $p$-periodic fixed points. As for Devaney, then yes, his string $s$ is just an itenerary sequence for the "fingers" in Cantor Bouquets. So, following the itenerary, basically one can track the orbit of an initial point's iteration. In your scheme, you are iterating the inverse relation, so I'd expect that Devaney's string should be read in reverse to match your results. In other words, with your method, I expect that you get the inverse string of Devaney's. Thanks again. $\endgroup$
    – user127032
    Sep 16, 2020 at 16:52

Obligatory "not an answer but some shed light". Also, I apologize for the visually lengthy post.

Clearly, likewise with most types of recursion on the complex plain, fractals are inevitable (as Yiannis Galidakis points out).

To investigate the matter, I quickly wrote up a program to track which points on the complex plane were "hit" by an iteration (represented by luminosity), starting at an initially random value. For each of my tests, I simulated $10^7$ different starting values for $z_0$. and iterated them up to $10^3$ times. Here are the results.

scale = 40 (right most-pixel is has real-value of 40)


Not much to see here, other than $4$ attractors (seemingly on or near the real line?) being visible.

scale = 4


Here, we start to see some structure emerge. Interestingly, the right-most concentration (which I assume is an 'attractor'), can be seen to have a ring-like structure. Just eye-balling, it seems these points are at real value $\approx 0, 1, 2.5$.

scale = 2


The final image shows that there seems to be at least one or more 'ring' of luminosity around each attractor. I'm assuming that the 'incompleteness' (for lack of a better term), of each 'ring' is because I've only chosen initial points from the rendered region. I imagine that taking a larger subset for initial points would close the gap.

  • $\begingroup$ How confident are you that your calculations are numericlaly stable? $10^3$ iterations of an exponential is a lot. $\endgroup$ Sep 14, 2020 at 17:04
  • $\begingroup$ To second QiaochuYuan's comment: my standard in Pari/GP is 200 dec digits precision, but with 10^3 iterations I think I'd re-compute examples with 1000 ro 2000 digits precision. Moreover, and much more problematic is, that many many random values in this segment of complex numbers run quickly into iterates of the form $10^a+b i$ with a in the 10000 (or million or...) and yet drive back to the "human daily access" numbers... but cannot add to the shown luninosity since your never won't get their next iterates... $\endgroup$ Sep 14, 2020 at 21:17
  • $\begingroup$ @QiaochuYuan I completely agree that $10^3$ is overkill (and potentially extortionate) l for my level of precision. I probably should have put more thought into that. I had the patience to sit through about $10^{10}$ CPU operations and with $10^7$ of that going to the number of initial values, that left me with $10^3$ iterations for each. $\endgroup$
    – Graviton
    Sep 15, 2020 at 8:52
  • 1
    $\begingroup$ @GottfriedHelms Absolutely. This was less so of rigorous-research and more so of a "I was bored and curious if a fractal would show up". $\endgroup$
    – Graviton
    Sep 15, 2020 at 8:54

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