anonymous
  • anonymous
Without using a computer, find \[\lim_{n\to\infty} \underbrace{\ln(\ln(\cdots\ln x))}_{n\text{ times}}\] for \(x\in\mathbb{R}^+\setminus\{1\}\)
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
theopenstudyowl
  • theopenstudyowl
impossible
1 Attachment
anonymous
  • anonymous
Oh and if anyone's wondering, \(\mathbb{R}^+\setminus\{1\}\) is the set of all positive real numbers excluding \(1\).
ikram002p
  • ikram002p
ok i'll start by first conjecture, is it infinity ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
It's not, it actually converges to a finite number in the complex plane.
ikram002p
  • ikram002p
oh wait as n goes to infinity i thought as x goes to :P
ganeshie8
  • ganeshie8
\[u = e^u\]
anonymous
  • anonymous
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
  • ganeshie8
wolfram gives answer in terms of product log function
anonymous
  • anonymous
I suppose that's as closed as it's going to get :)
ganeshie8
  • ganeshie8
\[\lim_{n\to\infty} \underbrace{\ln(\ln(\cdots\ln x))}_{n\text{ times}} = -W_k(-1)\]
Kainui
  • Kainui
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
  • anonymous
If it helps, the approximate value of the limit is \(0.318132+1.33724 i\).
Kainui
  • Kainui
Whoops I also made a mistake and wrote \(e^{i \pi}\) when it should be \(e^{i \pi/2}\)
Kainui
  • Kainui
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
  • Kainui
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.