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I don't know the best way to describe it in technical terms, but what is the result of a continuing logarithm of $z$, for example:

$$\log(\log(\dots\log(z)))$$

Where it is taking the logarithm of the logarithm and so on, for an infinite amount of times?

How would this type of thing behave? Does it converge, go off to infinity or an infinitesimal? Does the resulting behavior depend on whether $z$ is imaginary or real? Positive or negative?

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    $\begingroup$ The term you are looking for is "composition." We call $f(g(z))$ the composition of $f$ and $g$ and I would describe this as "the limit of repetitive self-compositions" $\endgroup$
    – Stella Biderman
    Feb 1, 2017 at 22:48
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    $\begingroup$ @StellaBiderman Better: Iteration. An iteration is repeated compositions of the same function. $\endgroup$
    – Simply Beautiful Art
    Feb 1, 2017 at 22:48
  • $\begingroup$ @SimplyBeautifulArt haha. I almost wrote "iterated composition." Good to know that's a Thing. $\endgroup$
    – Stella Biderman
    Feb 1, 2017 at 23:00
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    $\begingroup$ Proposal to merge this with my old question I found: Does logging infinitely converge? $\endgroup$
    – Simply Beautiful Art
    Feb 1, 2017 at 23:06

2 Answers 2

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If $\Im(z)$ (the imaginary part of $z$) is greater than or equal to zero, then this converges to roughly $z_\infty = 0.318152 + 1.33724 i$.

If $\Im(z)<0$ it converges to $z_\infty = 0.318152 - 1.33724 i$.

This of course relies on the usual branch cut for $\log z$.

The exceptional cases are any cases where a finite number of itetations lands on $1$ (or starting with $z=0$. These include $1, e, e^e$, and so forth.

However, there are isolated points for which the iterated log neither goes to infinity nor converges. For example, for any $z$ such that $$e^z = \log z \neq z$$ the iterated log oscillates between $z$ and $\log z$. I think there are such points; for example, there is an unstable 2-cycle fixed point at roughly $$ z= 0.883998 + 6.922346 i $$

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  • $\begingroup$ Lol, I too find proving convergence a hard task. $\endgroup$
    – Simply Beautiful Art
    Feb 1, 2017 at 22:48
  • $\begingroup$ The branch choice should rule out such points. For example your example cannot have this property since $\Im \log(z) \in (-\pi,\pi]$ so $\log(\log(z)) \not= z$ since $\Im z = 6.92 > \pi$. $\endgroup$
    – Winther
    Feb 1, 2017 at 23:35
  • $\begingroup$ @Winther You are right, my example point does not work. I'll have to see where I went wrong. $\endgroup$
    – Mark Fischler
    Feb 2, 2017 at 0:26
  • $\begingroup$ @Winther WHenever I zero in an apparent fixed point, the second iteration comes out off by $2\pi i$, which I take as a sign that branch cuts are preventing it from happening. For example, $$\log(\log(0.883998+6.92275i))\approx 0.883998+6.92275i-2\pi i$$ $\endgroup$
    – Mark Fischler
    Feb 2, 2017 at 1:29
  • $\begingroup$ Exactly. That happens since we are working with a fixed branch of the logarithm which constrains the imaginary part to be in a certain band. The equation $e^z = \log(z)$ do have infinitely many solutions, however only two of them have an imaginary part corresponding to any particular branch choice $\text{Log}(z)$ so if $\text{Log}(\text{Log}(z)) = z$ then $\text{Log}(z) = z$ which equals one of the two fixpoints it can converge to. I think the only way we can have cycles is if we could make a different branch choice for the logarithm at each step (for which we can make any sequence diverge). $\endgroup$
    – Winther
    Feb 2, 2017 at 18:12
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Iff it converges, then

$$a=\log(\log(\dots\log(z)\dots))=\log(a)$$

Thus,

$$a=\log(a)$$

$$e^a=a\implies-1=-ae^{-a}\implies a=-W_k(-1)\stackrel{k=0}\approx0.318-1.337i$$

is the Lambert W function. Note that this is constant, and dependent only on whether or not the choice of $z$ converge and where $z$ is. The position of $z$ will determine which branch of the Lambert W function it will converge to. A couple notes:

If we $z$ is a perfect super power of $e$, then it diverges due to $\log(0)$. That is, $z\ne e^{e^{e^{\dots}}}$ and $z\ne0$.

I am pretty sure it converges everywhere else, with the exception of $z=-W(-1)$.

Not finished with the rest: $\color{white}{Let \$z=re^{i\theta}\$. Then we have \$\$\log(z)=\log(r)+i\theta=r_1e^{i\theta_1}\$\$ We can see that \$r_1=\sqrt{\log^2(r)+\theta^2}\$ and \$\theta_1=\arg(\log(r)+i\theta)\$. From this, we can discern that \$r\$ must be bounded, and likewise, we are left to look at $\theta$. As \$r\to r'\$ be the limit of \$r\$, \$\theta\$ also approaches a limit}$

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  • $\begingroup$ "I am pretty sure it diverges everywhere else though" Why is that? $\endgroup$
    – Winther
    Feb 1, 2017 at 22:04
  • $\begingroup$ @Winther Memory, because I think I asked a similar question before. $\endgroup$
    – Simply Beautiful Art
    Feb 1, 2017 at 22:04
  • $\begingroup$ That's not a very convincing argument. $\endgroup$
    – Winther
    Feb 1, 2017 at 22:04
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    $\begingroup$ A big big issue is the domain of definition... for example, $x$ has already to be above 15 or so for $ln(ln(ln(ln(x))))$ to be defined... (I prefer to work with normalized notation "ln"). $\endgroup$
    – Jean Marie
    Feb 1, 2017 at 22:08
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    $\begingroup$ A small numerical experiment shows that it converges for all input not on the form $1,e,e^e,\ldots$. This is using the principal branch. $\endgroup$
    – Winther
    Feb 1, 2017 at 22:17

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