A infinite and alternating square root of 2 I want to show 
$$
A=\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+...}}}}
$$
converges to a finite value. It's easy to see that
$$
\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}
$$
converges to 2 and obviously the value of A is bounded by 2. But it's not easy to see if A converges to a finite value or not. Any suggestion, idea, or comment is welcome, thanks!
 A: A bit of plotting shows that
$$ f(x) = \sqrt{2-\sqrt{2+x}} $$
is defined for $x\in[-2,2]$ and has a single fixpoint.
We can find the fixpoint by repeatedly rearranging and squaring to get rid of each of the square roots in turn, yielding
$$ (2-x^2)^2 = 2+x $$
This equation has four roots in $[-2,2]$ but three of them are spurious results of the squarings. Fortunately two of the spurious roots are the nice integers $-1$ and $2$, so we can divide them out; what is left is
$$ x^2+x-1=0 $$
which gives us the actual fixpoint as
$$ \frac{\sqrt 5-1}{2} $$
This fixpoint is attractive (namely, the derivative of $f$ there is absolutely less than $1$, by direct computation), so if we iterate $f$ starting at some place in $[-2,2]$ the iterates will converge towards it. So it makes sense to say that
$$ \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{\cdots}}}} = \frac{\sqrt 5-1}{2} $$
A: Clearly we have $$A=\sqrt{2-\sqrt{2+A}}$$
Now $A>0$ and $\sqrt{2+A}\le2\iff A\le2$ in fact $A<2$
WLOG $A=2\cos4y$ where $0<4y<90^\circ$
$\sqrt{2+A}=\sqrt{2(1+\cos4y)}=2\cos2y$ as $\cos2y>0$
$\sqrt{2-\sqrt{2+A}}=\sqrt{2-2\cos2y}=2\sin y$ as $\sin y>0$
So, we have $2\cos4y=2\sin y=2\cos(90^\circ-y)$
$\implies4y=360^\circ  m\pm(90^\circ-y)$ where $m$ is any integer
$+\implies y=72^\circ  m+18^\circ\implies4y=288^\circ m+72^\circ\equiv72^\circ,0,288^\circ,216^\circ,144^\circ\pmod{360^\circ}$
$-\implies4y=480^\circ m-120^\circ\equiv0,120^\circ,240^\circ\pmod{360^\circ}$
But $0<4y<90^\circ\implies4y=72^\circ$
