Without using a computer, find
$$\lim_{n\to\infty} \underbrace{\ln(\ln(\cdots\ln x))}_{n\text{ times}}$$
for $x\in\mathbb{R}^+\setminus\{1\}$

Question

Without using a computer, find
$$\lim_{n\to\infty} \underbrace{\ln(\ln(\cdots\ln x))}_{n\text{ times}}$$
for $x\in\mathbb{R}^+\setminus\{1\}$

theopenstudyowl · Answer

impossible

anonymous · Answer

Oh and if anyone's wondering, $\mathbb{R}^+\setminus\{1\}$ is the set of all positive real numbers excluding $1$.

ikram002p · Answer

ok i'll start by first conjecture, is it infinity ?

anonymous · Answer

It's not, it actually converges to a finite number in the complex plane.

ikram002p · Answer

oh wait as n goes to infinity i thought as x goes to :P

ganeshie8 · Answer

$$u = e^u$$

anonymous · Answer

To be fair, I've been using Mathematica to find the limit, but I was wondering if there's an analytical method to finding a closed form (if it exists).

ganeshie8 · Answer

wolfram gives answer in terms of product log function

anonymous · Answer

I suppose that's as closed as it's going to get :)

ganeshie8 · Answer

$$\lim_{n\to\infty} \underbrace{\ln(\ln(\cdots\ln x))}_{n\text{ times}} = -W_k(-1)$$

kainui · Answer

I was thinking there might be a closed form, let me try and show you how far I can get:

$$y=\ln(\ln(\cdots))$$
log both sides, the infinite side doesn't change, so that's just y still.
$$\ln(y)=\ln(\ln(\cdots))=y$$$$y=e^y$$Do some algebra on it:$$-ye^{-y}=-1$$ Here's the fun part that might be adjustable! We rewrite that right part:

$$-1 = i*i = i*e^{i \pi}$$

So now we have ALMOST something nice and invertible but not quite.

$$-ye^{-y}=ie^{i \pi}$$ Maybe there's something to do to play around with this to get a nice closed form I have some ideas give me a minute.

anonymous · Answer

If it helps, the approximate value of the limit is $0.318132+1.33724 i$.

kainui · Answer

Whoops I also made a mistake and wrote $e^{i \pi}$ when it should be $e^{i \pi/2}$

kainui · Answer

We can see from this though: (using the slightly different $-1=(-i)(-i)$
$$-ye^{-y}=-ie^{-i \pi/2}\approx-i\frac{\pi}{2}e^{-i \pi/2}$$$$y \approx i\frac{\pi}{2} \approx 1.57079i\ \approx 0.318132+1.33724 i$$

So it's fairly close but not quite there. It seems to be multivalued though, I'm still playing with this it's interesting.

kainui · Answer

Yeah I guess as much as I wanna play around with this, it probably doesn't have a closed form since this also doesn't have a closed form: http://en.wikipedia.org/wiki/Omega_constant