TITLE: Find the value of this infinitely nested radical (it appears to obtain multiple values)
QUESTION [5 upvotes]: Find the value of $$\sqrt{1-\sqrt{\frac{17}{16}-\sqrt{1-\sqrt{\frac{17}{16}-\cdots}}}}$$

This is not as simple as it looks for one reason - there are $2$ real solutions to the equation $$x=\sqrt{1-\sqrt{\frac{17}{16}-x}}\implies\begin{cases}x_1=0.5\\x_2\approx 0.073\end{cases}$$
You can see that for yourself on Wolfram Alpha.
How can we know the real value of this infinite radical?

REPLY [4 votes]: Let $f(x)=\sqrt{1-\sqrt{\frac{17}{16}-x}}$ on $\left[\frac1{16},\frac{17}{16}\right]$:
$\hspace{3cm}$
$f$ is monotonically increasing on $\left[\frac1{16},\frac{17}{16}\right]$. To be precise,
$$
f'(x)=\frac1{4f(x)\left(1-f(x)^2\right)}\tag{1}
$$
$B$, $C=\!\!\left(\frac12,\frac12\right)$, and $D=\!\!\left(\frac{17}{16},1\right)$ are the points where $f(x)=\frac89x+\frac1{18}$.
$A$ and $C$ are the points where $f(x)=x$.
Evaluating $f'$ at the two roots of $f(x)=x$, we get
$$
f'(A_x)\doteq3.4345\tag{2}
$$
and
$$
f'\left(\frac12\right)=\frac23\tag{3}
$$
This means that $A$ is an unstable fixed point of $f$ and $C$ is a stable fixed point of $f$. In fact, for $x\in\left[B_x,D_x\right]$, we have
$$
\frac23\le\frac{f(x)-\frac12}{x-\frac12}\le\frac89\tag{4}
$$
Furthermore, for $x\in\left(\frac1{16},B_x\right]$,
$$
f'(x)\gt\frac53\tag{5}
$$
$(4)$ and $(5)$ say that

if $x_0\in\left(A_x,\frac{17}{16}\right]$, then $x_{n+1}=f(x_n)$ will converge to $\frac12$.
if $x_0=A_x$, then $x_{n+1}=f(x_n)$ is the constant sequence $x_n=A_x$.
if $x_0\in\left[\frac1{16},A_x\right)$, then $x_{n+1}=f(x_n)$ will get smaller until at some point $x_n\lt\frac1{16}$.

where $A_x\doteq0.073182744516454277369$.
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