# infinite power tower

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• July 23rd 2009, 02:38 AM
Random Variable
Quote:

Originally Posted by pomp
Sorry, but the function is not defined on this range, so I don't see how it can be continuous. By defining it in the way that you have it is equivalent to the expression involving the Lambert W function that I gave in the PDF, the point of putting it in this form was to show that the function is only defined for certain values of x.

You would need to define the function differently for it to make sense over the range $x \in (e^{1/e},\infty)$

Yep. I'm an idiot.
• July 23rd 2009, 08:32 AM
CaptainBlack
Quote:

Originally Posted by Random Variable
Let $f(x) = x^{x^{x^{x^{...}}}}$

What is the maximum value of x such that f(x) is finite?

You have been reading the current edition of Mathematical Gazette and I claim my £64,000.

CB
• July 23rd 2009, 08:37 AM
pomp
Quote:

Originally Posted by chisigma
a) $x>1$

In this case for all $n>0$ is $f_{n} (x) >1$ so that is...

$f(x)= \lim_{n \rightarrow \infty} f_{n} (x) = +\infty$
...
Anywhere also in this case the answer to the question proposed by R.V. is $x=1$ ...

Kind regards

$\chi$ $\sigma$

Although this seems the intuitive answer, it is actually incorrect. You're in good company though, Eisenstein himself actually quite naively stated the same when working on the function. If you have the time I suggest having a quick skim through the PDF near the top of this thread.

EDIT: As a quick test to reassure yourself, iterate $f_n(x)$ for $x=\sqrt{2}$. Clearly x>1 but you should note some limiting behaviour..

Spoiler:
$\lim_{n \to \infty} f_n(\sqrt{2}) = 2$
• July 23rd 2009, 09:31 PM
Deco
Logic?
Without even looking at any numbers, I can logically see that $x=1$ must be true. Assuming any situation, $a^b$, for any $b<1$ results in decay. Unless of course $a<1$ aswell, but that would work contrary to obtaining a maximum. For $b>1$, $a<1$, $a^b \rightarrow 0$. For both variables being positive also results in $a^b \rightarrow \infty$.

Therefore the only answer that would allow for a base being raised an infinite amount of times to the same value would be when $x=1$

I know this isn't rigorous, but god-damnit it's right.
• July 23rd 2009, 09:33 PM
pomp
Quote:

Originally Posted by Deco
I know this isn't rigorous, but god-damnit it's right.

No it isn't, refer to other posts in this thread.
• July 23rd 2009, 09:59 PM
Deco
Quote:

Clearly x>1 but you should note some limiting behaviour..
Are there no bounds to this? This would imply $10^{10^{10^{10^{...}}}}$ doesn't approach $\infty$? Am I the only only having a hard time to grasp this. And since $x>1$, doesn't that mean there is no maximum value of $x$?
• July 23rd 2009, 10:12 PM
pomp
Quote:

Originally Posted by Deco
Are there no bounds to this? This would imply $10^{10^{10^{10^{...}}}}$ doesn't approach $\infty$? Am I the only only having a hard time to grasp this. And since $x>1$, doesn't that mean there is no maximum value of $x$?

There is an upper bound as well as a lower bound. It was stated in another post and in the PDF near the begining of this thread that the domain of convergence is $[e^{-e},e^{1/e}]$
• July 23rd 2009, 10:40 PM
Random Variable
I still can't believe I stated that the function is continous where it's not defined. Thanks for bumping this thread and reminding me. (Doh)
• July 24th 2009, 01:01 AM
pomp
Quote:

Originally Posted by Random Variable
I still can't believe I stated that the function is continous where it's not defined. Thanks for bumping this thread and reminding me. (Doh)

haha sorry! On the bright side, at least you managed to get your head around the counter intuitive part! Seems to be throwing some people off.
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12