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Let $s$ be any complex number, $t = e^s$ and $z = t^{1/t}$. Define the sequence $(a_n)_{n\in\mathbb{N}}$ by $a_0 = z $ and $a_{n+1} = z^{a_n} $ for $n \geq 0$, that is to say $a_n$ is the sequence $z$, $z^z$, $z^{z^z}$, $z^{z^{z^{z}}}$ and so on.

I want to show that the sequence $(a_n)_{n\in\mathbb{N}}$ converges to $t$ if and only if $s$ lies in the unit disk. I know that when the sequence converges the limit is $\frac{W(-\ln(z))}{-\ln(z)}$ where $W$ is the Lambert W function.

I have verified the above statements numerically for several thousand values of $z$ but I have no idea how to actually prove it.

I graphed the natural logs of the limits on my computer. I got what appeared to be the unit disk.

If we take the log of the limit we get $ln(\frac{W(−ln(z)}{−ln(z)})=-ln(−ln(z))−W(−ln(z))+ln(−ln(z))=−W(−ln(z))$

So what I want to prove is equivalent to showing the sequence $a_n$ is convergent if and only if $|W(−ln(z))| \leq 1$

I was reading the Wikipedia article on the Lambert W Function and I found this proof that the limit $c$, when it exists, is $c= \frac{W(-ln(z))}{-ln(z)}$

$z^c = c\implies z = c^{1/c} \implies z^{-1} = c^{-1/c} \implies 1/z = (1/c)^{1/c} \implies -ln(z) = \frac{ln(1/c)}{c} \implies -ln(z) = e^{ln(1/c)}ln(1/c) \implies ln(1/c) = W(-ln(z)) \implies 1/c = e^{W(-ln(z))} \implies \frac{1}{c} = \frac{-ln(z)}{W(-ln(z)} \implies c = \frac{W(-ln(z)}{-ln(z)}$

I can only assume that at least 1 step is not justified when $|W(-ln(z)| > 1$ though I am not sure which one. I think that part of the problem is the equation $z^c=c$ has a solution for every non-zero complex number $c$ while the sequence $a_n$ only converges for certain special values of z. In other words the convergence of the $a_n$ is a sufficient but not necessary condition for the existence of a solution to the equation.

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  • $\begingroup$ @Did Here $w$ is the Lambert W function; it is not an undefined variable. $\endgroup$ – cpiegore Feb 13 '16 at 4:45
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    $\begingroup$ What is $u^z$ when $u$ and $z$ are complex numbers? $\endgroup$ – Did Feb 13 '16 at 4:46
  • $\begingroup$ @Did In polar coordinates $(re^{iu})^(se^{iv}) = (re^{iu})^s*(re^{iu}))^(e^{iv}) = ?$ $\endgroup$ – cpiegore Feb 13 '16 at 4:56
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    $\begingroup$ Except that, unless $z$ is an integer, this leads to possibly infinitely many values for $u^z$... For example, $u=1$ and $z=\sqrt2$ would yield that $u^z$ is $\exp(2i\pi n\sqrt2)$ for every integer $n$, which is a dense set of points on the unit circle. This example indicates the question is in serious need of explanations. $\endgroup$ – Did Feb 13 '16 at 6:48
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    $\begingroup$ Maybe start with complex number $a$ (to be thought of as $\log z$), then iterate using function $t \mapsto \exp(at)$ instead of function $x \mapsto z^x$. $\endgroup$ – GEdgar Feb 15 '16 at 16:23
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$$a_0 = z\qquad \qquad \qquad a_{n+1} = z^{a_n}$$

let $b_n = \ln a_n$ so $a_n = e^{b_n}$ and $$b_{n+1} = \ln \left(z^{e^{b_n}}\right) = e^{b_n} \ln z$$

if $b_n$ converges to $b$ then $b = e^b \ln z = - e^{b} (-\ln z)$ so

$$b = e^b \ln z = W(-\ln z)$$

where $W$ is (one of the branches of ?) the Lambert function.

let $c_n = b_n - b$ so $$c_{n+1} = e^{(b + c_n)} \ln z - b= (e^b \ln z) e^{c_n} - b = b e^{c_n} - b = b (e^{c_n} - 1)$$

suppose $|b| > 1$ and $\ln z \ne 0$. then suppose $b_n \to b$ thus $c_n \to 0$ so $e^{c_n}-1 \sim c_n$ so $c_{n+1} \sim b c_n $ which clearly cannot converge except if $c_0 = 0$ which would imply $c_n = 0$ for every $n$ which is not the case because $b_0 = \ln z$ so $b_1 = e^{\ln z} \ln z \ne b_0$ so $c_1 \ne c_0$ $\implies$ a contradiction.

hence if $\ln z \ne 0$ $$|b| = \left|W(-\ln z)\right| \le 1$$ is a necessary condition for the convergence of $(b_n)$ and $(a_n)$

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  • $\begingroup$ Can you show that $|b| \leq 1$ is also a sufficient condition for the convergence of the sequences? $\endgroup$ – cpiegore Feb 18 '16 at 23:07
  • $\begingroup$ Also, a few some minor points: first $b = ln(a)$ is actually equal to $-W(-ln(z))$. Second I believe it is not necessary to assume $ln(z) \ne 0$ or $z \ne 1$; $W(0)$ is perfectly well defined since $0*e^0 =0$ and $0 \leq 1$. Looks good otherwise. $\endgroup$ – cpiegore Feb 18 '16 at 23:11
  • $\begingroup$ Can you show that $|b| \leq 1$ is also a sufficient condition for the convergence of the sequences? $\endgroup$ – cpiegore Feb 20 '16 at 14:35
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    $\begingroup$ @cpiegore : as I shown it reduces to solving the convergence of $c_{n+1} = b(e^{c_n}-1)$ which is easy in $\mathbb{R}$ but much less in $\mathbb{C}$. it is probably already solved somewhere so make a research. and you'll probably have to show by induction that $Re(c_n)$ and $|c_n|$ let $c_n$ converge, maybe by studying $arg(c_n)$ ? you could also simulate the sequence on a computer to see what is happening when it converges or diverges (and I don't think it converges for most $|b| = 1$). $\endgroup$ – reuns Feb 20 '16 at 22:58
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I was reading the Wikipedia article on the Lambert W Function and I found this proof that the limit $c$, when it exists, is $c= \frac{w(-ln(z))}{-ln(z)}$

$z^c = c\implies z = c^{1/c} \implies z^{-1} = c^{-1/c} \implies 1/z = (1/c)^{1/c} \implies -ln(z) = \frac{ln(1/c)}{c} \implies -ln(z) = e^{ln(1/c)}ln(1/c) \implies ln(1/c) = W(-ln(z)) \implies 1/c = e^{W(-ln(z))} \implies \frac{1}{c} = \frac{-ln(z)}{W(-ln(z)} \implies c = \frac{W(-ln(z)}{-ln(z)}$

I can only assume that at least 1 step is not justified when $|W(-ln(z)| > 1$ though I am not sure which one.

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