Exploring Infinite Exponent Towers: Solving for x in x^x^x^x^...=2

[The Divine Zephyr]

Infinite Exponent "towers"

Please help me solve this problem; I don't even know how to start...

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

 

[vincentchan]

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

 

[dextercioby]

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

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 $\mathbb{C}$...

 

[dextercioby]

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

Daniel.

 

[saltydog]

Teteration

Please help me solve this problem; I don't even know how to start...

Solve for x: $$((((((x^x)^x)^x)^x)^x)^ ...)=2$$

LaTeX can't seem to stack exponents, like they should be
(x^x^x^x^x^x^x^x^x^x^x^x^...)

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

 

[The Divine Zephyr]

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

So far, I got that

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

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]

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

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

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

 

[dextercioby]

x=Sqrt[2]

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

Try to see whether it verifies the equation...

Daniel.

 

[The Divine Zephyr]

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

 

[hypermorphism]

if x<1, LHS converges to 1

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.

 

[The Divine Zephyr]

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.

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

 

[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.

 

[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.

 

[The Divine Zephyr]

I see where you are coming from, but I can't 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...

 

[dextercioby]

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

Daniel.

 

[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. :)

 

[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.

 

[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

 

[Hurkyl]

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)))

 

[hypermorphism]

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)

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)).

 

[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...

 

[dextercioby]

Anyways,the problem asked for the solution of the equation.Not for divagation regarding "power tower"...

Vincentchan,are YOU having a bad day??

Daniel.

 

[StatusX]

$$\sqrt{2}$$ is obviously incorrect. It diverges, just like all x>1. I'm guessing you arrived at that answer by something like this:

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

The problem is that you are assuming $$\infty=\infty$$ (since the power tower diverges), and this isn't always true. By the same logic, I could say:

$$s = 1 + 2 + 4 + 8 + ...$$
$$2s = 2 + 4 + 8 + ... = s - 1$$
$$s = -1$$

 

[dextercioby]

$$\sqrt{2}$$ is obviously incorrect. It diverges, just like all x>1. I'm guessing you arrived at that answer by something like this:

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

The problem is that you are assuming $$\infty=\infty$$, since the power tower diverges, and this isn't always true.By the same logic, I could say:

$$s = 1 + 2 + 4 + 8 + ...$$
$$2s = 2 + 4 + 8 + ... = s - 1$$
$$s = -1$$

YOU ARE TERRIBLY WRONG!

Please,do not GUESS WHAT I AM THINKING...

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

THIS EQUATION HAS THE SOLUTION i specified in the post with lots of red...

Daniel.

 

[Hurkyl]

It's fairly easy to prove the sequence $\sqrt{2} \uparrow \uparrow n$ converges as $n \rightarrow \infty$: it's a bounded, monotone sequence.

 

[learningphysics]

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.

Careful how you plug this into your calculator...

Are you calculating:
$$x^{x^{x^{x^{...}}}}$$

Or are you calculating

$$(((x^x)^x)^x)^x...$$

The first one goes to 2 if x=sqrt(2). The second one does not.

The way to plug it into the calculator is like this:

$$a_0=\sqrt{2}^\sqrt{2}$$
$$a_1=\sqrt{2}^{(a_0)}$$
$$a_2=\sqrt{2}^{(a_1)}$$

etc...

 

[StatusX]

youre right, I'm wrong (those big letters are annoying, though). It seems like it would diverge for x>1, but it's only for x>e^(1/e). Which lead to the interesting paradox that x goes to 1 as a goes to infinity, and yet for x=1, a=1.

 

[dextercioby]

(those big letters are annoying, though).

They were meant to be annoying,not for the fact that u contradicted me when you were wrong,and i was right,but for doing that by assuming what i was thinking...

Daniel.

 

[saltydog]

Proof for convergence

Would anyone be willing to provide a reasonably short (if possible) proof for the convergence interval for the function? Or perhaps someone can cite an on-line reference for the proof. If I obtain one I can understand, I'll post it.

Thanks,
SD

 

[dextercioby]

The solution to the equation
$$x \upuparrows \infty = a$$

is $$a^{\frac{1}{a}}$$
,which is never bigger than $$e^{\frac{1}{e}}$$

Daniel.

 

[saltydog]

The solution to the equation
$$x \upuparrows \infty = a$$

is $$a^{\frac{1}{a}}$$
,which is never bigger than $$e^{\frac{1}{e}}$$

Daniel.

Thanks Daniel,
However, my understanding is that the hyperexponent function above converges only when x is in some interval. Can you please tell me how to incorporate math symbols into my posting so I can be more specific? I have Mathematica. Can I export data from there?

Thanks,
SD

 

[saltydog]

Thanks Daniel,
However, my understanding is that the hyperexponent function above converges only when x is in some interval. Can you please tell me how to incorporate math symbols into my posting so I can be more specific? I have Mathematica. Can I export data from there?

Thanks,
SD

Ok Daniel,
I checked out the FAQ and found LaTeX. Nice feature here!
SD

 

[Physics is Phun]

I still don't get how this works. I believe you that the answer is 2^1/2 but I still don't know why. Is is calcuable using a calcuator or only provable through algebra? Could somone post a sollution?

 

[StatusX]

I still don't get how this works. I believe you that the answer is 2^1/2 but I still don't know why. Is is calcuable using a calcuator or only provable through algebra? Could somone post a sollution?

If you mean how do you find that the answer is sqrt(2), look at my post above. If you mean how could it be that a number greater than 1 converges in an infinite tower, which at least I found strange, just think of it like this: Sqrt(2)^2 is 2, so sqrt(2)^x for any x less than 2 will be less than 2. This means that sqrt(2)^sqrt(2)<2, and so then sqrt(2)^(sqrt(2)^sqrt(2))<2, and so on.

 

[The Divine Zephyr]

Though I am still alittle quesitonable about sqrt 2 as an answer, I think I understand the reason the answer is correct.