The value of $\sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+\cdots\sqrt{1-\sqrt{1+1}}}}}}$? How to find value of $\sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+\cdots\sqrt{1-\sqrt{1+1}}}}}}$ ? 
I've calculated it by MATLAB for some finite terms and I've got : $0.3001 - 0.4201i$, but I don't know how to find the value analytically! Would you mind helping me find it? Thanks
 A: The iteration of function $f(x) = \sqrt{1-\sqrt{1+x}}$ starting at $x=1$ approaches a 2-cycle of $.2229859448+.4133637969 i$ and $.2229859448-.4133637969 i$.
A: First of all let's assume the series is convergent. Looking for fixed points we have:
$$x=\sqrt{1-\sqrt{1+x}}$$
Now we will try to solve this equation. First squaring both sides:
$$1-x^2=\sqrt{1+x} \\
\left(\left( 1-x\right)\left( 1+x\right) \right)^2=1+x$$
Note that $x$ must be nonnegative, therefore:
$$(1-x)^2(1+x)-1=0 \\
\Rightarrow x^3-x^2-x=0$$
So $x=0$ is a solution. The other solutions are:
$$x^2-x-1=0 \Rightarrow x=\frac{1 \pm \sqrt{5}}{2}$$
where only $x=\frac{1 + \sqrt{5}}{2}$ is greater than or equal to zero and may look valid. But as people pointed out, one has to check if the answers actually fit into the initial equation. In this case $\frac{1 + \sqrt{5}}{2}$ doesn't, therefore the only fixed point we have found is $x=0$. 
But $x=0$ cannot be the convergence limit(it doesn't converge smoothly). Assume we deflect $x=0$ with the tiny amount of $\epsilon$(or rather starting with a tiny $x_1=\epsilon$). Putting it back into our initial equations and getting the next $x$:
$$x_2=\sqrt{1-\sqrt{1+\epsilon}}\approx \sqrt{1-\left( 1+\frac{\epsilon}{2}\right)} \approx \frac{i\sqrt{\epsilon}}{\sqrt{2}}$$
Now for $\epsilon < \frac{1}{2}$, $|x_2|>|x_1|$; ergo $x=0$ cannot be the convergence limit. We have proved that this infinite radicals doesn't have a single limit.
A: This $cannot$ have a positive real root,
because,
if $x$ is such a root,
then
$\sqrt{1+x} > 1$
so
$1-\sqrt{1+x} < 0$,
which means that
$\sqrt{1-\sqrt{1+x}}$
is complex, not real.
In other words,
finding a fixed point does not work -
there is $no$ fixed point.
The best that can be done
is to find a two-cycle
as GEdgar has done.
This involves solving
$x = f(f(x))$,
which is much more complicated.
Therefore Ali's work,
which I duplicated,
is wrong.
It solves $f(x) = x$,
but $f$ does not have a limit,
it has a two-cycle.
