The value of $\sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+\cdots\sqrt{1-\sqrt{1+1}}}}}}$?

Question

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

Answer 1

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.

Answer 2

The iteration of function $f(x) = \sqrt{1-\sqrt{1+x}}$ starting at $x=1$ approaches a 2-cycle of $.2229859448+.4133637969 i$ and $.2229859448-.4133637969 i$.

Answer 3

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.

Answer 4

This question was already answered and one answer was accepted. This answer is just to add to the completeness for some later reader.

If one looks at the function $f(x) = \sqrt{1-\sqrt{1+\sqrt{1-\sqrt{1+x}}}}$ then this function has true fixpoints; the fixpoints are $\small t_0 = 0$, $ \small t_1 \approx 0.222985944830 + 0.413363796251 i$ and $ \small t_2 \approx 0.222985944830 - 0.413363796251 i$.

Around the fixpoint $t_0$ we get a real power series with a fractional cofaktor but without constant term due to wolframalpha, see my related question.

Developing around the complex fixpoints, say $t_1$ we get a usual (though with complex coefficients) power series, $ \small f_4(x+t_1)-t_1 = g(x) \approx 0.219471696356 x + $ $ \small (-0.102599142965 + 0.177579081091 i) x^2 + $ $ \small(-0.133743852861 - 0.220457785139 i) x^3 $ $ \small+ (0.375580757794 - 0.0217915275445 i) x^4 + O(x^5) $ where the absolute value of the coefficient of the linear term is smaller than 1 - which means, that this fixpoint $t_1$ is also attracting.

Thus infinite iteration of $f(x)$ beginning from any complex value (except 0) shall converge to that fixpoint $t_1$. This solves the first problem: is that OP's notation converging at all.

After that we can simply state, that any finite truncation of the OP's expression approximates one of the cycling fixpoints :

t1=0.222985944830 + 0.413363796251*I
x0=sqrt(1+t1)     :    %2164 = 1.1211469 + 0.18434863*I
x0=sqrt(1-x0)     :    %2165 = 0.22298594 - 0.41336380*I
x0=sqrt(1+x0)     :    %2166 = 1.1211469 - 0.18434863*I
x0=sqrt(1-x0)     :    %2167 = 0.22298594 + 0.41336380*I  \\ cycling occurs, approxi-
x0=sqrt(1+x0)     :    %2168 = 1.1211469 + 0.18434863*I   \\ mating the four-point
x0= ...           :     ...  =  ...                       \\  cycle