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How to find the real values of $x$ such that : $$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$

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I assume $\sqrt{z}$ always returns a positive value - that you wouldn't take $-2=\sqrt{4}?$ – Thomas Andrews Mar 19 '13 at 12:53
@Fredrik Meyer very bad idea :… – user67465 Mar 19 '13 at 12:56
The other part… – user67465 Mar 19 '13 at 12:57
You link to something unreadable and inaccessible... – Dennis Gulko Mar 19 '13 at 12:58
@Sasha I'm not seeing the difference. Looks the same to me. – Thomas Andrews Mar 19 '13 at 13:10
up vote 9 down vote accepted

Here's an approach leading to a "closed formula".

The two innermost square roots are defined only, if $x\in[-2,2]$, so we can write $x=2\cos\theta$ for some $\theta\in[0,\pi]$. My solution relies on the trig identities $$ \sqrt{\frac{1+\cos\alpha}2}=\cos\frac\alpha2 $$ and $$ \sqrt{\frac{1-\cos\alpha}2}=\sin\frac\alpha2 $$ that hold for all $\alpha\in[0,\pi]$ (outside this range we may get differing signs).

Thus $$ \sqrt{2+x}=\sqrt{2+2\cos\theta}=2\cos\frac\theta2, $$ and then $$ \sqrt{2-\sqrt{2+x}}=\sqrt{2-2\cos\frac\theta2}=2\sin\frac\theta4=2\cos(\frac\pi2-\frac\theta4). $$ Going on we get $$ 2\cos\theta=x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}=\sqrt{2+2\cos(\frac\pi2-\frac\theta4)}=2\cos(\frac\pi4-\frac\theta8). $$ In the interval $[0,\pi]$ cosine is injective, so we can conclude that $$ \theta=\frac\pi4-\frac\theta8\Leftrightarrow\theta=\frac{2\pi}9. $$ The only real solution is thus $$ x=2\cos\frac{2\pi}9\approx1.53209. $$

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@Fabian: Thanks for fixing the typo! – Jyrki Lahtonen Mar 19 '13 at 14:41

Elaborating some on what @Fredrik Meyer suggested, one can get: $$\begin{align*}x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}\hspace{5pt}&\Rightarrow\hspace{5pt} x^2=2+\sqrt{2-\sqrt{2+x}}\hspace{5pt}\Rightarrow\hspace{5pt} x^4-4x^2+4=2-\sqrt{2+x}\\ &\Rightarrow\hspace{5pt}x^8+16x^4+4-8x^6+4x^4-16x^2=2+x\\ &\Rightarrow\hspace{5pt} x^8-8x^6+20x^4-16x^2-x+2=0 \end{align*}$$ The last polynomial can be factored by noticing that it vanishes at $x=2$ and at $x=-1$ into $(x-2) (x+1) (x^3-3 x+1) (x^3+x^2-2 x-1)=0$ (using WA in the end).
Now you can find many roots - but beware: not all of them solve the initial problem: each transition above gives additional assumptions on $x$: first, $x\geq 0$. Then $x^2-2\geq0$ and after that $x^4-4x^2+2\leq0$. Do you see why?
All of those together imply that $\sqrt2\leq x\leq\sqrt{2+\sqrt2}$.
Using some real analysis (or WA :)), one can show that there exists a unique root of that polynomial on $(\sqrt2,\sqrt{2+\sqrt2})$.
NB: This is not the cleanest approach and not very elegant, but it works. I'm almost sure that I have seen much nicer way to solve it, but I can't think of one now.

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It's not hard to see that there can only be one real $x$ at most, since the RHS is decreasing in $x$. – Thomas Andrews Mar 19 '13 at 13:28

Here is a basic square free analysis approach. Clearly not the elegant approach Dennis Gulko has seen before. Consider the function $$f(x)=x-\sqrt{2+\sqrt{2-\sqrt{2+x}}}.$$ Its domain is $[-2,2]$, where it is continuous and increasing. Since $$ f(-2)=-2-\sqrt{2+\sqrt{2}}<0<f(2)=2\sqrt{2} $$ there exists (intermediate value theorem) a unique (since $f$ is increasing) zero $x_0$ for $f$. Here is a graph to confirm this claim. Using a calculator, one can check that $$ f(3/2)<0<f(8/5)\quad\Rightarrow\quad \frac{3}{2}=1.5<x_0<1.6=\frac{8}{5}. $$ If you want to go a little further further in the decimal expansion of $x_0$, you can use dichotomy. Or ask Wolfram.

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I would use the square-and-order method to turn it into a polynomial. The next step would be to find its real roots.

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Sorry for the typo, gratitude for the correction :) – András Hummer Mar 19 '13 at 14:15

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