Let $z \in \mathbb{C}$ and let $W$ be the Lambert W function. I am trying to prove that:

If $t = W(-\log z)$, $|t| = 1$, and $t^n \ne 1$ for all $n \in \mathbb{N}$ (ie, $t$ is not a root of unity) then $z^{z^{z^{...}}}$ does not converge.

I have investigated this numerically in previous posts and I am almost certain my conclusion is correct, but I am not sure how to actually prove it.

My first thought was to use the definition of a non-convergent sequence: there exists $\epsilon > 0$ for every $N \in \mathbb{N}$ such that if $n > N$ than $|a_n - L| \ge \epsilon$. But I am having a hard time getting started on this, because I am not sure what sort of $\epsilon$ would work.

It may also help to use this: $z^c = c\implies z = c^{1/c} \implies z^{-1} = c^{-1/c} \implies 1/z = (1/c)^{1/c} \\ \implies -\log z = \frac{\log(1/c)}{c} \implies -\log z = e^{\log(1/c)}\log(1/c) \\ \implies \log(1/c) = W(-\log z) \implies 1/c = e^{W(-\log z)} \\ \implies \frac{1}{c} = \frac{-\log z}{W(-\log z)} \\ \implies c = \frac{W(-\log z)}{-\log z}$

If I understand correctly, this shows that if the sequence $z^{z^{z^{...}}}$ converges, than it must converge to $\frac{W(-\log z)}{-\log z}$.

  • 1
    $\begingroup$ Presumably, uou mean $t=W(-\log z)$ and $|t|=1$? Otherwise, the condition on $t$ doesn't affect the behavior of the sequence at all. $\endgroup$ – Thomas Andrews Aug 17 '16 at 18:44
  • $\begingroup$ No, I mean $t = W(-\log z)$ and $|t| = 1$ and $t^n \ne 1$. I learned the hard way that the case $|t| = 1$ actually splits into 2 sub-cases which behave quite differently from each other. $\endgroup$ – cpiegore Aug 17 '16 at 18:49
  • $\begingroup$ "I have investigated this numerically in previous posts" -- it would be good to link those posts. $\endgroup$ – 6005 Aug 17 '16 at 18:52
  • $\begingroup$ If $|t| = 1$ and $t^n = 1$ than it is convergent. On the other hand if $t^n \ne 1$ it appears the sequence does not converge. I already have a proof for the first case. Here I am interested specifically in the second case. $\endgroup$ – cpiegore Aug 17 '16 at 18:52
  • $\begingroup$ @cpiegore Yes, we understand that. I think Thomas Andrews was referring to that you wrote before $|t| = |W(-\log z)|$ instead of $t = W(-\log z)$. Anyway, consider linking to the old posts in your question. $\endgroup$ – 6005 Aug 17 '16 at 18:54

As already indicated in an comment or answer to a related earlier question[1] [2] I'd like to hint to the feature of continued fractions and the partial convergents.

Empirically it seems, that one aspect of the continued fraction-ansatz might be a promising base for a proof of the OP's conjecture. The key is here, that -beginning at some $z_0$ - iterations of a certain height $h$ ( $h$ taken from the convergents of the continued fraction), are returning to the neighbourhood of $z_0$ . And moreover, for all convergents of higher indexes that neighbourhood is even nearer to $z_0$, so that the orbit returns to all previous coordinates nearer and nearer - and thus we already cannot talk of "convergence to the fixpoint" - just by that property alone.

Even more: if started at the coordinate $z_{-1}=0$ then this means also, that some iterations run also into "neighbourhoods" of the infinity: namely that which were the approximating the pre-image $z_{-2}$ of $z_{-1}=0$ . This can be taken as argument to talk about a really diverging orbit ("slingshots to infinity").
In answers to earlier question I've already provided pictures for that orbits; I'll focus here only on the data at the convergents of the continued fraction.

(Redescribeing the OP's variables): define $c$ as a rational or as irrational number suitable for use to define $t$ on the boundary of the complex unit disc : $\quad t \overset{\text{def}}= \exp(2 \pi î / c) \quad $ which also ensures, that $\quad |t|=1. \quad$
From this $\quad z \overset{\text{def}}=\exp(t \exp(-t)) \; $ and a fixpoint is $\quad u \overset{\text{def}}=\omega = \exp(t) \; $ such that $\quad z^u = u. \quad$
The OP discusses the orbit $$z_0=1,z_1=z, \{z_{k+1}=z^{z_k } \}_{k=1}^\infty$$ for bases $z$ generated by irrational values of $c$ - which define $t$ as complex unit-roots of irrational order.
The question here is, how to prove, that the orbit does not "converge", which means, the visible (interpolated) curve painted from the orbit-coordinates in the complex plot does not contract towards the fixpoint.

Observation: All empirical observations show that the orbit is roughly circular or spiralling around the fixpoint. THis means, with some iteration-heights $h$ (depending on $c$) $$ | z_{k+h} - z_k| \lt | z_{k+i} - z_k| \text{ for all } 0 \lt i \lt h$$ and the orbits returns to the neighbourhood of an earlier iterate.

But while this is not yet an indication, that this would not form an approximate spiral which is but contracting to the fixpoint, the key is here, that taking $h$ from the $j'th$ partial convergent $w_j$ from the continued fraction $\mathcal{Q}_c$ of $c$, then we have also the property that $$ | z_{k+{w_{j+2}}} - z_k| \lt | z_{k+{w_j}} - z_k| \text{ for all } j $$ This means mainly that it cannot happen that the entire orbit contracts to the fixpoints - instead it must -arbitrarily near! - return to the earliest iterates, which means with some high numbers $h$ near to $z_0=1$ as well as to $z_{-1}=0$ but also to $z_{-2} \to \infty$ which means, there is a (sub-) sequence of $z_h$ with increasing $h$ which diverges (to negative infinity).

To give more intuition I show the continued fractions, the convergents and the distances to $z_0=1$ for some irrational $c$.

 \\  0.898979485566

 Q = contfrac(c)
 \\ = [0, 1, 8, 1, 8, 1, 8, 1, 8, 1,...

\\ the used partial convergents $w_j$ are the numbers in the first row of
\\ the following table:

\\ [0 1 8  9 80 89 792 881 7840 8721 77608 86329 ...]
\\ [1 1 9 10 89 99 881 980 8721 9701 86329 96030 ...]

\\ Using an iteration-height-difference $h$ and $h + w_j$ we find, 
\\ that the iterates are very near

\\ Table:
  j   Q_j  w_j  d_j= | z_h - z_0|  !  j   Q_j     w_j    d_j=  | z_h - z_0|
  0  0      0                  0   !  1     1       1          0.595311020000
  2  8      8     0.915341230229   !  3     1       9         0.0614965157660
  4  8     80     0.159012534234   !  5     1      89        0.00684710148528
  6  8    792    0.0161906381488   !  7     1     881       0.000750667481285
  8  8   7840   0.00172273459390   !  9     1    8721      0.0000819516167718
 10  8  77608  0.000187006854713   ! 11     1   86329     0.00000895053455733

  $Q_j$  are the entries of the continued fraction of $c$
  $w_j$ are the convergents; if we take them as iteration-height differences,
        we get improving nearer returns to $z_0$ as we increase $j$ 
  $d_j = | z_h - z_0|$ are the distances of the interesting iterates. $h$ is here
     the value, which is given by $w_j$ 

The same table occurs for all tested $c$ .

I think, if it can be proven, that the observed property of the improving approximation of high-iterates $z_h$ to $z_0$ by increasing of the index of the partial convergents of the continued fraction of $c$ is really a systematic feature, then the OP's conjecture should be proven by this.

Here is the table for approximations $d_h$ to infinity, where the $h$ indicate the preimage of the iterates of the previous table (which approximate to $z_0 = 0$).

\\ Table:
  j   Q_j  w_j-1  d_j= | z_h - z_0|  !  j  Q_j  w_j-1   d_j=  | z_h - z_0|
   2  0      7     6.6803521816      !  3   1      8     4.7090153033
   4  8     79     9.4432625097      !  5   1     88     9.8251150006
   6  8    791    13.4101809798      !  7   1    880    15.0068356503
   8  8   7839    17.9648387628      !  9   1   8720    20.1868266387
  10  8  77607    22.7784296865      ! 11   1  86328    25.3613790368
  $Q_j$  are the entries of the continued fraction of $c$
  $w_j-1$ are the convergents minus one; if we take them as iteration-height differences,
        we get improving nearer returns to $z_{-1}\to \infty$ as we increase $j$ 
  $d_j = | z_h - z_0|$ are the distances of the interesting iterates. $h$ is here
     the value, which is given by $w_j -1$, and iterations begin at $z_0=0$        

Conclusion 2: The above table shows, that we have sub-sequences which even diverge to infinity, which -conditional on the validity of the continued-fraction feature for this problem of infinite orbit beginning at $0$ or at $1$ - proves the conjecture of the OP.


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