Continued radical of powers of 4 equals 3 Can someone explain to me why
$$\sqrt{4 + \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \dots}}}} = 3???$$
I need an answer

Let $f_0(x)=0$ and recursively $$ f_{n+1}(x)=\sqrt{x+f_n(4x)}.$$
Next, let $$ f(x)=\lim_{n\to\infty}f_n(x).$$
Our goal is to find $f(4)$.
Claim 1. For $x\ge 0$ and $n\in\Bbb N$,
$$\tag1 f_{n}(x)\le f_{n+1}(x).$$
Proof.
This is trivial for $n=0$. Assume $(1)$ holds for $n$ and all $x\ge0$.
Then
$$f_{n+1}(x)=\sqrt{x+f_n(4x)}\le \sqrt{x+f_{n+1}(4x)} =f_{n+2}(x)$$
for all $x\ge 0$. Hence the claim follows by induction. $\square$
Claim 2. For all $n\in\Bbb N_0$ and all $x\ge 0$,
$$\tag2 f_n(x)\le\sqrt x+1.$$
Proof. The claim is certainly true for $n=0$. Assume $(2)$ holds for all $x\ge0$ and for some $n\in\Bbb N$. Then
$$f_{n+1}(x)=\sqrt{x+f_n(4x)}\le\sqrt{x+2\sqrt{x}+1} =\sqrt{(\sqrt x+1)^2}=\sqrt x+1,
$$
so $(2)$ holds for all $x\ge0$ also for $n+1$. By induction, the claim follows. $\square$
Combining claims 1 and 2, we see that $f_n(x)$ converges point-wise (so $f(x)$ is in fact defined for all $x\ge0$) and that
$$\tag3 \sqrt x\le f(x)\le \sqrt x+1. $$
In particular, $f(4)\le 3.$
We also note that
$$f(x)=\sqrt{x+f(4x)}. $$
Let $g(x)=\frac{f(x)}{\sqrt x+1}\in[0,1]$.
Then
$$ g(x)=\frac{\sqrt{x+g(4x)(\sqrt{4x}+1)}}{\sqrt x+1}=\sqrt{\frac{x+g(4x)(2\sqrt x+1)}{x+2\sqrt x+1}}$$
i.e.,
$ g(x)^2$ is a convex combination of $g(4x)$ and $1$ and hence is not further from $1$ than $g(4x)$ is. Then also $|g(x)-1|\le |g(4x)-1|$. By $(3)$, $\lim_{x\to\infty}g(x)=1$, so that we must have $g(x)=1$ for all $x>0$. We conclude
$$ f(x)=\begin{cases}0&x=0\\\sqrt x+1&x>0\end{cases}$$
