Does logging infinitely converge?

Trying to evaluate $$\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))$$For some fixed $x$ produces a complex answer that appears to converge, at least sometimes.

So I want a proof that this converges for either some $x$, no $x$, or all $x$.

If it converges for all $x$ or some $x$, what does it converge to?

If it diverges, is there a way we can evaluate it like we evaluate diverging sums?

And after all of that, does it appear to converge to the same value, no matter what $x$ value we start with?

I know $\ln(z)=\ln(|z|)+i\arg(z)$, but I can't repeat this process without a given $z$. (where $z$ is complex).

A similar post of mine found here does not answer my question and focuses more on the limits, calculus, and infinites.

This question asks for consideration from a complex-analysis point of view, considering convergence of value in the complex plane.

• why are you repeating your own question? math.stackexchange.com/questions/1587322/… Dec 24, 2015 at 1:42
• @FedePoncio I am not repeating my own question. This one is more focused on the complex solutions and behavior of the function. Dec 24, 2015 at 1:43
• @FedePoncio The last one was more focused on limits and infinite. Dec 24, 2015 at 1:43
• @FedePoncio If you can draw answers to these question (the questions specified in this post) from my previous post, I will consider deleting this question. Dec 24, 2015 at 2:02
• Where this converges, it must (by continuity) converge to a fixed point, that is, a $z$ satisfying $\ln(z)=z$. Dec 24, 2015 at 4:59

I can't answer your question for a complex number, but if $x\in\Bbb R^+$ you can only iterate $\log$ finitely many times before it becomes complex.

Let $b_0=e$ and $b_{n+1}=e^{b_n}$. Let $x\in[1,\infty)$, $a_0=x$, $a_{n+1}=\log a_n$, and let $N$ be the largest integer such that $b_N\le x<b_{N+1}$.

So, $$b_N\le a_0<b_{N+1}$$.

Iterating $\log$ $N$ times on this inequality gives: $$b_0\le a_N<b_1$$. In other words, \begin{align} e\le a_N<e^e & \implies 1\le a_{N+1}<e\\ & \implies 0\le a_{N+2}<1. \end{align}

So, $a_{N+3}$ will be negative, and the next iteration will be complex. If you're taking $\log$ as a real-valued function, you can only iterate it finitely many times before it becomes undefined.

• I've already noted this question to concern the complex answers (in particular). Thank you trying though. Dec 24, 2015 at 2:14
• Ah I see. A suggestion if I may: it's much less confusing to readers if you just edit your original question, rather than creating an entirely new post. Dec 24, 2015 at 2:19
• I believe, however, that the two questions are so different in what they are asking that they should be in two different posts. As you may have noticed, the previous answer is simple, using calculus definitions on how we define things. This one removes all restrictions and asks for a complex solution. Dec 24, 2015 at 2:21

This answer is formed from numerical research only! Nevertheless I will try to prove my result.

Given sequence $a_n$, defined recursively by

$$\begin{cases}a_1=x\\a_n=\ln(a_{n-1})\end{cases}$$

is convergent for all $x$ different than

$$0,\;1,\;e,\;e^e,\;e^{e^e},\;...$$

And its limit is given by

$$\lim_{n\to\infty} a_n \approx 0.318132 + 1.33724i\quad\text{for }\Re(x)\geqslant0$$ $$\lim_{n\to\infty} a_n \approx 0.318132 - 1.33724i\quad\text{for }\Re(x)<0$$

These two constants are roots of the equation $$g=\ln(g)$$ which are equal to, respectively, $$-W_{-1}(-1)\quad\text{and}\quad-W_0(-1)$$

where $W_k$ is the $k$-th branch of the Lambert W-function.

• What about other branches? Dec 27, 2015 at 17:13
• @SimpleArt all values $-W_{k}(-1)$ where $k\ne0,-1$ does not satisfy $g=\ln(g)$ Dec 27, 2015 at 17:28
• Oh, well, that sucks. Dec 27, 2015 at 17:29
• @SimpleArt this behavior is pretty predictable. Complex logaritm is multivalued and you use the principal branch of it, when using other branches there will be solutions in other branches of Lambert W-function. For example when taking $\ln(x)=\ln|x|+i\arg(x)+2\pi$ we will have solutions in $-W_{-2}(-1)$ and $-W_{1}(-1)$. Dec 27, 2015 at 17:32
• Oh, yes, that is true. Dec 27, 2015 at 17:33