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Solutions of following eqution $$x=\sqrt{2+{\sqrt{2-{\sqrt{2+{\sqrt{2-x}}}}}}}$$ is $\frac{1+\sqrt5}{2}$. This is solution of $$x=\sqrt{2+{\sqrt{2-x}}}$$

Does all of this type equation(repeating same shape) always have same solutions like this? Can you explain why?

user148928
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3 Answers3

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Note the following identities: $$2+2\sin(\theta) = 4\sin^2(\theta/2+\pi/4)$$ $$2-2\sin(\theta) = 4\sin^2(\pi/4-\theta/2)$$

Consider $$x=\sqrt{2+{\sqrt{2-{\sqrt{2+{\sqrt{2-x}}}}}}}$$

where $x = 2\sin(\theta)$.

Using the above identities, it must be that

$$2\sin(\theta)=\sqrt{2+{\sqrt{2-{\sqrt{2+2\sin(\pi/4-\theta/2)}}}}}$$

$$2\sin(\theta)=\sqrt{2+{\sqrt{2-2\sin(3\pi/8-\theta/4)}}}$$

$$2\sin(\theta)=\sqrt{2+2\sin(\theta/8+\pi/16)}$$

$$\sin(\theta) = \sin(\theta/16 + 9\pi/32)$$

$$\theta = 3\pi/10$$

Note that $$2\sin\left(\dfrac{3\pi}{10}\right) = \dfrac{1+\sqrt{5}}{2}.$$

I would conjecture that, yes, equations in that general form will have a solution in the form of the sine or cosine of some rational multiple of $\pi$.

It is worth noting that $\sin\!\left(\!\dfrac{p}{q}\pi\!\right)$ will be algebraic per this thread.

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Brad
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I think this will help in understanding it. Just see what is $ \sqrt{2+{\sqrt{2-\frac{(1+\sqrt5)}{2}}}} $. So we see that $\frac{(1+\sqrt5)}{2}$ is always a root but there will be many other solutions since the degree increases.

happymath
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  • More complex roots (at least if you take the multivalued version of the square root), but not more real solutions where you insist on the positive square root. – Robert Israel May 25 '14 at 05:07
  • @Robert why there are no more real solution? – user148928 May 25 '14 at 05:50
  • If $f(x) = \sqrt{2 + \sqrt{2 - x}}$, then $f$ is decreasing on $[-2,2]$, so $f(x) = x$ can only have one solution there. The fixed point at $x = p = (1+\sqrt{5})/2$ is attracting because $|f'(p)| < 1$. If the immediate basin of attraction of the fixed point is not the whole interval $(-2,2)$ it would have to be $(a,b)$ for a $2$-cycle ($f(a) = b$, $f(b) = a$). But $f(f(x)) = x$ has no other solutions in the interval $(-2,2)$, so that can't happen. Thus for every $x \in (-2,2)$ we have $f^{n}(x) \to p$ as $n \to \infty$, and in particular there are no solutions to $f^n(x) = x$ there. – Robert Israel May 25 '14 at 22:03
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If there is some handwaving here, we can continuously replace $x$ in each equation to get something like

$$x = \sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2- \cdots}}}}$$

so these two things are equal.

Now we just need to establish convergence but it is there.

Sandeep Silwal
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