A false nested radical of Ramanujan and a true nested radical

Question

On the wikipedia page I can see a nested radical by Ramanujan :

$$\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{\cdots}}}}}}}=\frac{2+\sqrt{5}+\sqrt{15-6\sqrt{5}}}{2}$$

Wolfram alpha says it's false

So I propose another one wich is true see here we have :

$$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{\cdots}}}}}}}=2\cos\Big(\frac{\pi}{9}\Big)$$

My question :

Can someone give me some steps to solve :

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

I know furthermore that is related to a cubic .

Any helps is highly appreciated .

Thanks a lot for all your contributions.

Ps: Can someone correct the wikipedia page and add the nested radical with $2$?

Answer 1

The octic factors as $$ (x-1)(x+2)(x^3 - 3x-1)(x^3 - x^2 - 2x + 1) $$

The two cubics have square discriminants, meaning cyclotomic roots.

The roots of $t^3 - 3 t + 1$ are $$ 2 \cos \frac{2 \pi}{9} \; \; , \; \; 2 \cos \frac{4 \pi}{9} \; \; , \; \;2 \cos \frac{8 \pi}{9} \; \; , \; \; $$ To get the roots of $x^3 - 3x+1,$ just negate these, giving $$ 2 \cos \frac{7 \pi}{9} \; \; , \; \; 2 \cos \frac{5 \pi}{9} \; \; , \; \;2 \cos \frac{ \pi}{9} \; \; . \; \; $$

The roots of $t^3 + t^2 - 2 t - 1$ are $$ 2 \cos \frac{2 \pi}{7} \; \; , \; \; 2 \cos \frac{4 \pi}{7} \; \; , \; \;2 \cos \frac{8 \pi}{7} \; \; , \; \; $$ To get the roots of $x^3 - x^2 - 2x+1,$ just negate these, giving $$ 2 \cos \frac{5 \pi}{7} \; \; , \; \; 2 \cos \frac{3 \pi}{7} \; \; , \; \;2 \cos \frac{ \pi}{7} \; \; . \; \; $$

The original equation with the square roots has just one real root, roughly 1.879.

$$ x^8 - 8x^6 + 20x^4 - 16x^2 + x + 2 $$

$$ -2.000000000000000000000000000, -1.532088886237956070404785301 , -1.246979603717467061050009768, -0.3472963553338606977034332535, 0.4450418679126288085778051290, 1.000000000000000000000000000, 1.801937735804838252472204639, 1.879385241571816768108218555$$ ?

Answer 2

Solving cyclic infinite nested square roots of 2 in the form $$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-...}}}}}}$$ Refer here Put $x = \sqrt{2+\sqrt{2+\sqrt{2-x}}}$ where $x = 2\cos\theta$

now the equation becomes easy to simplify

$2\cos\theta = \sqrt{2+\sqrt{2+\sqrt{2-2\cos\theta}}}$

= $\sqrt{2+\sqrt{2+{2\sin\frac{\theta}{2}}}}$ ...(from Half angle cosine formula for $\sin(\frac{\theta}{2}$)

= $\sqrt{2+\sqrt{2+{2\cos(\frac{\pi}{2}-\frac{\theta}{2})}}}$ = $\sqrt{2+{2\cos(\frac{\pi}{4}-\frac{\theta}{4})}}$ (from Half angle cosine formula for $\cos(\frac{\theta}{2}$)

= ${2\cos(\frac{\pi}{8}-\frac{\theta}{8})}$

now $\theta = \frac{\pi}{8}-\frac{\theta}{8}$ further simplification leads to

$\frac{9\theta}{8} = \frac{\pi}{8}$

$\therefore$ $\theta = \frac{\pi}{9}$ which is $20^\circ$

It is true that simplifying the cyclic infinite nested square roots of 2 $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-...}}}}}}$ is $2\cos\frac{\pi}{9}$

As for as Ramanujan's nested radical is concerned