A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Without using a computer, find
\[\lim_{n\to\infty} \underbrace{\ln(\ln(\cdots\ln x))}_{n\text{ times}}\]
for \(x\in\mathbb{R}^+\setminus\{1\}\)
anonymous
 one year ago
Without using a computer, find \[\lim_{n\to\infty} \underbrace{\ln(\ln(\cdots\ln x))}_{n\text{ times}}\] for \(x\in\mathbb{R}^+\setminus\{1\}\)

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh and if anyone's wondering, \(\mathbb{R}^+\setminus\{1\}\) is the set of all positive real numbers excluding \(1\).

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0ok i'll start by first conjecture, is it infinity ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's not, it actually converges to a finite number in the complex plane.

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0oh wait as n goes to infinity i thought as x goes to :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0To 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
 one year ago
Best ResponseYou've already chosen the best response.2wolfram gives answer in terms of product log function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I suppose that's as closed as it's going to get :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{n\to\infty} \underbrace{\ln(\ln(\cdots\ln x))}_{n\text{ times}} = W_k(1)\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0If it helps, the approximate value of the limit is \(0.318132+1.33724 i\).

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Whoops I also made a mistake and wrote \(e^{i \pi}\) when it should be \(e^{i \pi/2}\)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1We can see from this though: (using the slightly different \(1=(i)(i)\) \[ye^{y}=ie^{i \pi/2}\approxi\frac{\pi}{2}e^{i \pi/2}\]\[y \approx i\frac{\pi}{2} \approx 1.57079i\ \approx 0.318132+1.33724 i\] 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
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.