1. ## infinite power tower

Let $\displaystyle f(x) = x^{x^{x^{x^{...}}}}$

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

2. I wrote a small paper on this for a 2nd year project. The aim was to give a 30 minute easy to follow seminar to our peers.

[Hope it answers your question] Note that this was only the handout for the seminar so it isn't conclusive. Also, it isn't the final draught (cant find it )so there may be a few mistakes

EDIT: Didn't realise which forum this was posted in and so thought you were asking for help

3. Originally Posted by pomp
I wrote a small paper on this for a 2nd year project. The aim was to give a 30 minute easy to follow seminar to our peers.

[Hope it answers your question] Note that this was only the handout for the seminar so it isn't conclusive. Also, it isn't the final draught (cant find it )so there may be a few mistakes

EDIT: Didn't realise which forum this was posted in and so thought you were asking for help
The explanation I've seen is different but it involves a bit of handwaving. Your explanantion is much more thorough.

4. Originally Posted by Random Variable
The explanation I've seen is different but it involves a bit of handwaving. Your explanantion is much more thorough.
Ah man I'm really interested as to how this goes.

Does anyone else find it really really hard to approach a problem when they've already worked on it in a different way, from a different angle?

5. Originally Posted by pomp
Ah man I'm really interested as to how this goes.

Does anyone else find it really really hard to approach a problem when they've already worked on it in a different way, from a different angle?
Spoiler:

$\displaystyle f(x) = x^{x^{x^{x^{...}}}}$ can be written as $\displaystyle f(x)=x^{f(x)}$

Now find $\displaystyle f'(x)$ and see for what value of x it blows up to infinity.

6. for $\displaystyle x\rightarrow 0 , f'(x) \rightarrow \infty$ ?

7. Originally Posted by pickslides
for $\displaystyle x\rightarrow 0 , f'(x) \rightarrow \infty$ ?
I'm not sure if f'(x) is approaching anything as x goes to zero.

EDIT: You can't approach infinity because infinity is not a number. Poor choice of words on my part.

But I don't think f'(x) is going to infinity or approaching any number as x goes to zero.

8. I can see why it works in this case, but simply proving that $\displaystyle f'(x) \rightarrow \infty$ for some $\displaystyle x$ doesn't prove that $\displaystyle f$ is not defined for any values of $\displaystyle x$ greater than this. $\displaystyle \tan(x)$ comes to mind

9. Originally Posted by pomp
I can see why it works in this case, but simply proving that $\displaystyle f'(x) \rightarrow \infty$ for some $\displaystyle x$ doesn't prove that $\displaystyle f$ is not defined for any values of $\displaystyle x$ greater than this. $\displaystyle \tan(x)$ comes to mind
Is it valid in this case because f(x) is continous for all positive real numbers?

Did you find the the value of $\displaystyle f(e^{1/e})$? I got $\displaystyle f(e^{1/e})=e$

10. Originally Posted by Random Variable
Is it valid in this case because f(x) is continous for all positive real numbers?

Did you find the the value of $\displaystyle f(e^{1/e})$? I got $\displaystyle f(e^{1/e})=e$
$\displaystyle f(x) = \frac{1}{x}$ satisfies that criteria.

Yeh that's what I got, $\displaystyle f(x^{1/x})=x$ for $\displaystyle \forall$ $\displaystyle x \in [e^{-e}, e^{1/e}]$

11. Originally Posted by pomp
$\displaystyle f(x) = \frac{1}{x}$ satisfies that criteria.

Yeh that's what I got, $\displaystyle f(x^{1/x})=x$ for $\displaystyle \forall$ $\displaystyle x \in [e^{-e}, e^{1/e}]$
How about because it's a continuous increasing function for all values greater than or equal to $\displaystyle e^{1/e}$ ?

EDIT: Or just that it's continuous on the interval $\displaystyle [e^{1/e}, \infty)$

12. Originally Posted by Random Variable
Or just that it's continuous on the interval $\displaystyle [e^{1/e}, \infty)$
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 $\displaystyle x \in (e^{1/e},\infty)$

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

What is the maximum value of x such that f(x) is finite?
The problem is quite elementary if the approach is correct. By definition for $\displaystyle x>0$ is...

$\displaystyle x^{x}= e^{x \ln x}$

... so that the f(x) you have proposed is...

$\displaystyle f(x)= \lim_{n \rightarrow \infty} e^{x^{n} \ln x}=$

$\displaystyle = 1$ if $\displaystyle 0<x\le 1$

$\displaystyle = +\infty$ if $\displaystyle x>1$

... so that the answer to your question is $\displaystyle x=1$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

14. Originally Posted by chisigma
The problem is quite elementary if the approach is correct. By definition for $\displaystyle x>0$ is...

$\displaystyle x^{x}= e^{x \ln x}$

... so that the f(x) you have proposed is...

$\displaystyle f(x)= \lim_{n \rightarrow \infty} e^{x^{n} \ln x}=$

$\displaystyle = 1$ if $\displaystyle 0<x\le 1$

$\displaystyle = +\infty$ if $\displaystyle x>1$

... so that the answer to your question is $\displaystyle x=1$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
You seem to have misinterpreted the definition of f.

$\displaystyle \lim_{n \to \infty} e^{x^n \ln{x}} = \lim_{n \to \infty} x^{x^n} \neq x^{x^{x^{x^. . . }}}$

Power towers are read from top to bottom e.g

$\displaystyle 2^{2^{2^2}} = 2^ {2^{4}} = 2^{16} \neq 4^{2^2} = 16^2$

Where on the left hand side of the not equal sign I have calculated correctly, from top to bottom, and on the right hand side I have calculated incorrectly, from bottom to top.

15. All right!... if power towers are to be read from top to bottom, then we can construct a sequence of functions of $\displaystyle x$ recursively defined as...

$\displaystyle f_{n+1} (x) = x^{f_{n} (x)}$ with $\displaystyle f_{0} (x)=1$

... and we define f(x) as...

$\displaystyle f(x)= \lim_{n \rightarrow \infty} f_{n} (x)$ (1)

Now we have three possibilities...

a) $\displaystyle x>1$

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

$\displaystyle f(x)= \lim_{n \rightarrow \infty} f_{n} (x) = +\infty$

b) $\displaystyle x=1$

In this case for all $\displaystyle n$ is $\displaystyle f_{n} (1) = 1$ so that is $\displaystyle f(1)=1$

c) $\displaystyle 0 < x < 1$

In this case for all $\displaystyle n>0$ is $\displaystyle 0<f_{n} (x)<1$ but the limit (1) doesn't exist, so that $\displaystyle f(x)$ is undefined. For demonstrate this you can easily demonstrate that ...

$\displaystyle \lim_{x \rightarrow 0+} f_{n} (x) = 1$ for n even, $\displaystyle \lim_{x \rightarrow 0+} f_{n} (x)=0$ for n odd

Anywhere also in this case the answer to the question proposed by R.V. is $\displaystyle x=1$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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