what's the real value of $\sqrt2^{\sqrt2^{\large \sqrt2^{\sqrt2^{\unicode{x22F0}}}}}$? I heard this problem from my friend today, but unfortunately we could not figure out what's happening here.
$$x^{x^{\large x^{x^{\unicode{x22F0}}}}}=2$$
$$x^2=2$$
$$x=\sqrt2$$
but
$$x^{x^{\large  x^{x^{\unicode{x22F0}}}}}=4$$
$$x^4=4$$
$$x=\sqrt2$$
I have a feeling that this might be a duplicate, but I could not find this problem (basically I couldn't figure out what to search). So what's happening here, and what's the real value of
$$\sqrt2^{\sqrt2^{\large  \sqrt2^{\sqrt2^{\unicode{x22F0}}}}}$$
?
 A: Finding the value of an infinitely iterated function can be tricky. Usually, if $z$ is a fixed point of a function $f$ (that is, if $f(z)=z$), then it often happens that
$$\lim_{n\to\infty} f^n(x)=z, \forall x$$
...however, things tend to get a little bit complicated when $f$ has more than one fixed point, which is the case with your function
$$f(x)=\sqrt 2^x$$
that has fixed points
$$f(2)=2$$
$$f(4)=4$$
In this case, the value of $f^\infty(x)$ depends on $x$, which doesn't really make sense given the way you wrote it. Basically this means that its value depends on the number "on top" of your power tower. For example, if you start with $4$, and observe the sequence
$$a_0=4$$
$$a_{n+1}=\sqrt 2^{a^n}$$
then each $a_n=4$, and so
$$a_{\infty}=\lim_{n\to\infty}a_n=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}=4$$
However, if you take $a_0=1$,
$$a_{\infty}=\lim_{n\to\infty}a_n=\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}}=2$$
However, things are a little bit less complicated in this case, since $x=2$ turns out to be something called an "attracting fixed point". Basically, we end up with
$$f^\infty\left(x\right) = \left\{
     \begin{array}{lr}
       2 & : x \lt 4\\
       4 & : x = 4\\
       \infty & : x \gt 4\\
     \end{array}
   \right.\\$$
A: While some people in this thread may disagree, I believe the most obvious definition of a power tower $x^{x^{x^{\dots}}}$ is as the limit of the series $x, x^x, x^{x^x}, x^{x^{x^x}},\dots$
or, in other words, $\lim_{n\to \infty} a_n$ where $a_0 = x$ and $a_n = x^{a_{n-1}}$.
Now to answer your two questions:

*

*Why does $x=\sqrt{2}$ seemingly solve both of the following equations: $x^{x^{x^{\dots}}}=2$ and $x^{x^{x^{\dots}}}=4$ ?

The answer is that you need to be very careful when working with infinite sequences. Lots of properties that are normally true about every day numbers are no longer true about infinite things. For example, this reminds me of the following bogus proof that $1+2+4+8+\cdots = -1$

$S = 1+2+4+8+\cdots$
$S - 1 = 2 + 4 + 8 + \cdots$
$\frac{S-1}{2} = 1 + 2 + 4 + 8 + \cdots = S$
$S-1 = 2S \Rightarrow S = - 1$

The issue with the above proof is that it assumes a limit exists. If a limit were to exist, then it would be $-1$, but since no limit exists, that answer is moot.
Now back to your question. When you take $x^{x^{x^{\dots}}}=b$, plug in $b$ to get $x^b=b$, and realize you can solve for $x = \sqrt{2}$, what you are really showing is that if the limit $\lim_{n \to \infty} a_n = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}$ exists, then that limit is equal to one of those values $b$.


*What does $\lim_{n \to \infty} a_n = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\dots}}}$ truly equal?

Lemma 1: $a_n$ is an increasing series.
Proof by induction:
Since $\sqrt{2} > 1$, we have that $a_1 = \sqrt{2}^\sqrt{2} > \sqrt{2}^1 = a_0$. For the inductive step, assuming that $a_{n-1} > a_{n-2}$, it then follows that $a_n = \sqrt{2}^{a_{n-1}} > \sqrt{2}^{a_{n-2}} = a_{n-1}$.
Lemma 2: $a_n < 2$ for all $n$.
Proof by induction:
Obviously, $a_0 =\sqrt{2} < 2$. Since $a_{n-1} < 2$, it follows that $a_n = \sqrt{2}^{a_{n-1}} < \sqrt{2}^2 = 2$.
From these two lemmas, we have that $a_n$ is bounded above by $2$ and is increasing, which means $\lim_{n \to \infty} a_n$ exists. We know from above that if such a limit exists, it must be one of the fixed points 2 or 4. And since our series $a_n$ never gets near $4$, the limit must be $2$.
A: The problem here is that $4$ is not in the range (image) of the power tower function (which is only defined for $x\in [e^{−e},e^{1/e}]$). So, the assumption that the power tower equals 4 is false and thus gives a false "solution".
