25

Here's an expression I am struggling to evaluate: $$\LARGE {\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\:\cdot^{\:\cdot^{\:\cdot}}}}}} $$

The value turns out be $2$, but I don't understand how do we get it. Can anyone give the solution?

EDIT: The original problem is as follows:

If $y(x)= { x }^{ { x }^{ { x }^{ { x }^{. } } } }$, then evaluate $y(\sqrt { 2 })$.

2'5 9'2
  • 54,717
Das
  • 1,079

4 Answers4

53

Consider the sequence $\{0,1,\sqrt{2},\sqrt{2}^{\sqrt{2}},\sqrt{2}^{\sqrt{2}^{\sqrt{2}}},\ldots\}$. In other words the sequence with $a_0=0$ and $a_{n+1}=\sqrt{2}^{a_{n}}$.

Note that $a_0<2$. Assume that for some $n$, that $a_n<2$. Then $a_{n+1}=\sqrt{2}^{a_n}<\sqrt{2}^2=2$. So by induction, $a_n<2$ for all $n$.

Note that $a_0<a_1$. Suppose that for some $n$, that $a_{n-1}<a_n$. Then $\sqrt{2}^{a_{n-1}}<\sqrt{2}^{a_n}$, so $a_n<a_{n+1}$. So by induction, $a_n<a_{n+1}$ for all values of $n$. In other words, the sequence is increasing.

So we have that this sequence is increasing and bounded above. Therefore it converges to some limit $L$. That means $$\begin{align} a_{n+1}&=\sqrt{2}^{a_n}\\ \lim_{n\to\infty}a_{n+1}&=\lim_{n\to\infty}\sqrt{2}^{a_n}\\ L&=\sqrt{2}^L\\ \end{align}$$

By concavity, there are only two solutions to this equation, and fortunately it's easy to identify both: $L=2$ or $L=4$. But $L$ cannot be $4$, since the sequence was bounded by $2$.

So this establishes that $\lim_{n\to\infty}a_{n}$ exists and that its value is $2$.

2'5 9'2
  • 54,717
  • 3
    (+1) Very nice answer, but might it be worthwhile to point out that the sequence you are considering isn't arbitrary but rather is an interpretation (correct/only one) of the hand wavy question? I'm not sure how much this is obvious to the OP. – DRF Dec 19 '15 at 12:57
  • Alternate to the second induction paragraph: square root of two is greater than one. Therefore each term must be greater than the previous term. – WGroleau Dec 19 '15 at 15:23
  • @WGroleau I'm not sure that is straightforward. I think you are saying that "$x<\sqrt{2}^x$" is true because $\sqrt{2}>1$. But as revealed here, that's not true if $x\in[2,4]$. So to go that way, you have some more to show. – 2'5 9'2 Dec 19 '15 at 19:17
  • Could you not have just started with the line "Let the given expression be $x$. Then, we observe by the way the expression is constructed, that, $x=\sqrt{2}^x$. By concavity... etc. etc. – Deepak Gupta Dec 19 '15 at 19:25
  • @alex.jordan, No. I'm saying that if X & Y are both greater than 1, then X^Y is always greater than X. – WGroleau Dec 20 '15 at 01:47
  • @WGroleau But then you are using $X=\sqrt{2}$. So you only can conclude that $X^Y$ is greater than $\sqrt{2}$. This does not establish the sequence is increasing. Or if I still misunderstand you, please explain in full detail. – 2'5 9'2 Dec 20 '15 at 03:36
  • X > 1 → $X^X > X → $X^X > 1 → $X^{X^X}>$X^X —  Sorry, I don't know why I can’t make all the exponents behave. – WGroleau Dec 20 '15 at 04:10
  • @WGroleau Then that's the same inductive argument as the one in the answer, except you start it with a $0$th term of $1$. Your original comment sounded to me like you were bypassing a inductive chain of implications like this, and just immediately establishing $a_{n+1}>a_n$ somehow based on $\sqrt{2}>1$. So I'm not seeing why this is an alternative explanation. – 2'5 9'2 Dec 20 '15 at 04:38
  • It's a generalization. It's true for all X>1, not only for sort(2) – WGroleau Dec 20 '15 at 05:00
6

Really you wish to solve for $x$ in the expression below $$x^{x^{x^x…}}=2, $$ where $x$ exponentiates an infinite amount. One way to approach it would be to note $$\log_x( x^{x^{x^x…}})=\log_x(2)\rightarrow x^{x^{x^x…}}=\log_x(2). $$ But we know $x^{x^{x^x…}}=2 $, therefore, $$2=\log_x(2)\rightarrow x^2=2$$ and thus $$\boxed{x=\sqrt{2}}. $$

This is just a good way to show it. It is certainly not a rigorous proof because you could really perform the logarithm step as many times you like and get different solutions. But, it helps in basic understanding, in my opinion.

  • 2
    But if one didn't know $2$ was a solution. – mavavilj Dec 19 '15 at 07:17
  • I suppose you could set the expression equal to some number $n$. Perform the same as above with instead $x=\sqrt{2}$. You would end up with $\sqrt{2}=n^{1/n}$. The obvious solution here is then $n=2$. –  Dec 19 '15 at 07:23
  • 4
    This establishes that if the equation $x^{x^{x^⋰}}=2$ has a solution at all, the solution must be $\sqrt{2}$. But what if there is no actual solution? For instance, the equation $x^{x^{x^⋰}}=4$ has no solution. But this method would tell you that if it did, the solution would be $\sqrt{2}$. – 2'5 9'2 Dec 19 '15 at 07:33
  • 2
    Note that another solution to $\sqrt{2}=n^{1/n}$ is $\sqrt{2}=4^{1/4}$. – 2'5 9'2 Dec 19 '15 at 07:35
3

Let $x=\sqrt{2}^{\sqrt{2}^{\cdots}}$ then $x=\sqrt{2}^x$ or $x=2^{x/2}$ i.e., $x^2=2^x$

K_user
  • 2,438
0

If we have an infinite number of powers nested then we can say

$$v=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}=\sqrt{2}\,^v$$

Take the log to base 2.

$$lg(v)=v\,lg(\sqrt{2})$$

and

$$\frac{lg(v)}{v}=lg(\sqrt{2})=\frac{1}{2}lg(2)$$

And this gets us $v=2$

StephenG - Help Ukraine
  • 725
  • 2
    v=4 is also a solution. How do you ensure that there are no solutions other than the ones you have come up with (2 and 4, that is) – Deepak Gupta Dec 19 '15 at 19:10