Does logging infinitely converge?

Solution 1:

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.

Solution 2:

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.