# Why can't $\sqrt{2}^{\sqrt{2}{^\sqrt{2}{^\cdots}}}>2$?

So we have$$\sqrt{2}^{\sqrt{2}{^\sqrt{2}{^\cdots}}}=x\\\sqrt{2}^x=x$$where $x=2$ heuristically seems like a good solution. However, $x=4$ seems like an equally good solution. I was told in passing that $x$ was bounded at $2$, but I'm not sure how to show this.

Update

It would seem that the crux of this problem is whether the sequence $a_n$ converges or diverges, where $a_0=1$ and $a_{n+1}=\sqrt{2}^{a_n}$.

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Also, I want to ask about how to search for specific questions on this site. I'm almost certain this is a duplicate, but searching the LaTeX doesn't give any results. –  rnmartingale Nov 9 '12 at 1:50
Alas, searching (especially with embedded LaTeX) seems to be more of an art than a science, at least given the current state of the art in search engines. Wait a decade or so and perhaps the problem will go away. It's cold comfort, but I've had the same problem. –  Rick Decker Nov 9 '12 at 2:20
@RickDecker That's rather unfortunate. Well, at least there are enough people on M.SE to catch a duplicate if it happens. –  rnmartingale Nov 9 '12 at 3:44

Consider the corresponding sequence, where $a_0=1$ and $a_{n+1}=\sqrt2^{a_n}$, and use induction: $a_n\le 2$.

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So? This only shows that every finite step $a_n$ is bounded. Not $a_{\infty}$ –  fff Nov 9 '12 at 2:27
Is it sufficient to say we know there's an $a_n<2$, and by observing that $a_{n+1}=\sqrt{2}^{a_n}$, $a_{n+1}<\sqrt{2}^{2}=2$? –  rnmartingale Nov 9 '12 at 2:28
Along with fff I don't see how this shows anything about the limit as n approaches infinity. In fact, my hand calculator yields the following: 1.4^1.4 ~= 1.601, 1.601^1.4 ~= 1.93, 1.93^1.4 ~= 2.51. Also, (sqrt(2)^(sqrt(2))^sqrt(2)=2, and ((sqrt(2)^(sqrt(2))^sqrt(2))^sqrt(2) ~= 2.665. So, how in the world does there exist an upper bound? Did I do the calculations wrong? –  Doug Spoonwood Nov 9 '12 at 3:11
To the above commenters: Observe the function $x\mapsto \sqrt{2}^x$ is strictly increasing. If $x<2$ then $\sqrt{2}^x<\sqrt{2}^2=2$, thus with $a_0=1$ we have that $a_n<2$ for each $n$ by induction. Furthermore, if $a_n<2$ for every $n$ then $\displaystyle\lim_{n\to\infty}a_n\le 2$. –  anon Nov 9 '12 at 3:25
To conclude from what Berci wrote: One needs to begin by defining what the expression $\sqrt{2}^{\sqrt{2}{^\sqrt{2}{^\cdots}}}$ means, and just about the only sensible definition is as the limit of the sequence that Berci calls $a_1,a_2,\dots$ So, showing that each $a_n$ is below 2 tells us that their limit, if it exists, is at most 2. One also needs to argue that the sequence is increasing. From this, the completeness axiom gives us that the sequence converges (since it is increasing and bounded above), and it follows that its value is at most 2. –  Andres Caicedo Nov 9 '12 at 4:03

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Let us look at your sequence. We define our recurrence relation to be $$a_0=1$$ $$a_n=\sqrt{2}^{a_{n-1}}$$ Note that the number, (what you call $x$) is the limit$$lim_{n\rightarrow\infty}a_n$$. So how can we understand this situation and what is going on. Well let me describe to you a famous way of picturing the orbits of these types of recurrences.

So begin by drawing on the same set of axis the two functions $g(x)=\sqrt{2}^x$ , and $y=x$.

Now start with your starting value, $a_0=1$. Use this to label the point $(1,1)$. Draw the vertical line connecting $(1,1)$ to $(1,g(1))$. In words, you draw the vertical Line segment connecting $(1,1)$ to the graph of $g(x)=\sqrt{2}^x$. Now you draw the horizontal line that goes through the new point $(1,g(1))$ and see where it connects to the graph $y=x$. This new point you get is $(g(1),g(1))$. Now you take the vertical line (like we did for $(1,1)$) and see where it connects to the the graph $y=g(x)$, and we repeat the procedure forever. The picture you will get for this particular function is an infinite stair case whose corner points are $$\{(1,1);(1,g(1));(g(1)g(1));(g(1)g(g(1))); g(g(1));g(g(1));\cdots\}$$. You notice that this is converging to the point $(2,2)$.

Now some remarks in general. When you have any function, $f:\mathbb{R}\rightarrow\mathbb{R}$ (the domain does not necessarily need to br $\mathbb{R}$) and you want to find find the value of the limit $$a,f(a),f(f(a))$$, like we did here, the solution (if it exists) wil be a fixed point of the function. Now we can follow the iteration in similar way where we draw these line segments whose corners are $$\{(a,a);(a,f(a));(f(a),f(a));f(a),f(f(a));(f(f(a)),f(f(a)))\cdots\}$$. Our function above was special enough that the geometry of the initial point and the function gave us the that the limit is $(2,2).$ See the picture in the wikipedia artical

http://en.wikipedia.org/wiki/Fixed_point_%28mathematics%29#Attractive_fixed_points

The picture shows the same type of iteration except that the initial point is $x=1$, and the function doing the iteration is $y=cos(x)$. A difference here is that instead of a stair case, the "track\$ spirals around the fixed point.

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