# Infinite Exponent towers

1. Jan 28, 2005

### The Divine Zephyr

Infinite Exponent "towers"

Solve for x: $$x^{x^{x^{x^{...}}}}=2$$

Last edited: Jan 28, 2005
2. Jan 28, 2005

### vincentchan

if x=1, x^x^x^x.... =1
if x >1, x^x^x^x.... = infinity
so x is undefine

3. Jan 28, 2005

### dextercioby

You can say that the equation does not admit a real solution...

Daniel.

P.S.The problem would be interesting to consider and solve in $\mathbb{C}$...

4. Jan 28, 2005

### dextercioby

$$x^{x^{x^{x^{...}}}}=2$$

Daniel.

5. Jan 28, 2005

### saltydog

Teteration

Let the iterated exponent (a teteration) be called LHS (left hand side)

Then LHS=y=2
but the iterated exponent is also equal to LHS so that:
LHS^y=2

but y=2
so that:

x^2=2
or:

x=Sqrt[2]

Yea, I know it's hard to grasp. I need to work on it too.

SD

6. Jan 28, 2005

### The Divine Zephyr

So far, I got that

if x=1, my LHS=1
if 0<x<1, LHS converges to 1
if x>1, LHS diverges

I plugged it in on a calculator and it divirged into infinity...

Go for it, I'll think about that, too.

Last edited: Jan 28, 2005
7. Jan 28, 2005

### dextercioby

Try to see whether it verifies the equation...

Daniel.

Last edited: Jan 28, 2005
8. Jan 28, 2005

### The Divine Zephyr

It doesn't, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}$$ does infact come very close to 2 (I think exactly, dont have my 89 with me...), however. But as soon as more terms are piled on, it spirals into infinity.

9. Jan 28, 2005

### hypermorphism

This is obvious for the expression $$((x^x)^x)^x...$$
but how did you show convergence for
$$x^{x^{x^{x^\cdots}}}$$
?
For the first, you get the sequence 2^{-1}, 2^{-1/2}, 2^{-1/4},...,2^{-1/2^i} but for the second the form gets ugly:
2^{-1}, 2^{-1/2}, {sqrt{2}/2}^sqrt{2}, {1/2}^{{sqrt{2}/2}^sqrt{2}}, ...
My forays with Windows calculator are giving me oscillations, so I can't be sure it doesn't converge to something less than 1.

Last edited: Jan 28, 2005
10. Jan 28, 2005

### The Divine Zephyr

I'm sorry, the correct expression for me would have been 0<x<1 will converge to 1

11. Jan 28, 2005

### hypermorphism

2^{-1} is less than 1. :) But I'm not sure the power tower sequence it generates converges to 1, as I can't find a general ith term for the sequence.

12. Jan 28, 2005

### dextercioby

SQRT [2] IS THE CORRECT ANSWER...

Generally

$$x^{x^{x^{x^{...}}}}=a$$

Has the solution

$$x=a^{\frac{1}{a}}$$

Iteration & logarithmation to show it...

Daniel.

13. Jan 28, 2005

### The Divine Zephyr

I see where you are coming from, but I cant see it working the the equation...

SQRT [2]^SQRT [2]^SQRT [2]=2, but as soon as more SQRT [2]s are stacked, it flies off the mark...

14. Jan 28, 2005

### dextercioby

No,it doesn't,trust me...Do you approximate results (intermediary) ??If so,then that's why it may jump over 2...

Daniel.

15. Jan 28, 2005

### hypermorphism

Hmm. Mathworld has an expression for the general solution of the infinite power tower at http://mathworld.wolfram.com/PowerTower.html , but its not as simple as a^{1/a}. However, the equation seems to verify that the power tower of sqrt(2) converges to 2.
Regarding dex, it's true. sqrt(2) is an irrational number so it can't be rationed about like a finite decimal on a calculator. :)

Last edited: Jan 28, 2005
16. Jan 28, 2005

### The Divine Zephyr

I idnt approximate. I got 2 for a stack of 3 sqrt twos, but as I did $$x^{x^{x^{x^{...}}}}$$, it went to infinity.

$$\log_{x}2=x^{x^{x^{x^{...}}}}$$

Can you show me how your solution was attained?

I see the link, I'll go check it out.

17. Jan 28, 2005

### vincentchan

dextercioby:
are you having a bad day?? ......I'll show you if x=2^1/2, then x^x^x^x >2,

let x=2^1/2

x^x^x^x=x^(x*x*x) = x^(2*x) = 2^(1/2*2)^x = 2^x >2

18. Jan 28, 2005

### Hurkyl

Staff Emeritus
vincent: a^a^a^a means

$$a^{a^{a^{a}}}$$

not

$$(((a^a)^a)^a)^a$$

Or, written flatly, it's x^x^x^x := x^(x^(x^x)))

Last edited: Jan 28, 2005
19. Jan 28, 2005

### hypermorphism

This step is wrong. We're finding the result of x^(x^(x^...)), not ((x^x)^x)^...
Ie., (sqrt(2)^sqrt(2))^sqrt(2) = sqrt(2)^(sqrt(2)*sqrt(2)) = sqrt(2)^2 = 2,
but there is no similar way to simplify sqrt(2)^(sqrt(2)^sqrt(2)).

20. Jan 28, 2005

### vincentchan

so the expression is:
x^(x^(x^(x^(x.....)??
i was keep doing
x^x^x^x^x......

idoit me
for your $$a^{a^{a^{a}}}$$
it really depend on how ppls read.....