Show that

\(\displaystyle \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}=2\sin \left(\frac{\pi}{18} \right)\)

Board index **‹** Computation of Series **‹** Nested Radicals Problem
## Nested Radicals Problem

Show that

\(\displaystyle \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}=2\sin \left(\frac{\pi}{18} \right)\)

I present a partial solution to the problem for now :

The nested radical repeats after twice positive signs producing a negative sign, i.e., [-, +, +, -]. One can then write is as :

\(\displaystyle x = \sqrt{2 - \sqrt{2 + \sqrt{2 + x}}}\)

Which gives the reducible octic equation

\(\displaystyle x^8 - 8x^6 + 20x^4 - 16x^2 - x + 2 = (x - 2)(x + 1)(x^3 - 3x + 1)(x^3 + x^2 - 2x - 1)\)

Some of the roots get outright ruled out, for example, \(\displaystyle 2\) or \(\displaystyle -1\). The hard part is to prove that the real roots of the fourth factor cannot be a solution which I think can be done by invoking some acceleration methods to the nested radical and then numerically checking each root.

Hence, the real roots of \(\displaystyle x^3 - 3x + 1\) are the only possible choice left. I think it wouldn't be hard to show that one root is \(\displaystyle 2\sin(\pi/18)\) and is the only value of the nested radical. The desired result should follow.

I present a partial solution to the problem for now :

The nested radical repeats after twice positive signs producing a negative sign, i.e., [-, +, +, -]. One can then write is as :

\(\displaystyle x = \sqrt{2 - \sqrt{2 + \sqrt{2 + x}}}\)

Which gives the reducible octic equation

\(\displaystyle x^8 - 8x^6 + 20x^4 - 16x^2 - x + 2 = (x - 2)(x + 1)(x^3 - 3x + 1)(x^3 + x^2 - 2x - 1)\)

Some of the roots get outright ruled out, for example, \(\displaystyle 2\) or \(\displaystyle -1\). The hard part is to prove that the real roots of the fourth factor cannot be a solution which I think can be done by invoking some acceleration methods to the nested radical and then numerically checking each root.

Hence, the real roots of \(\displaystyle x^3 - 3x + 1\) are the only possible choice left. I think it wouldn't be hard to show that one root is \(\displaystyle 2\sin(\pi/18)\) and is the only value of the nested radical. The desired result should follow.

There is no need. Two of the roots are negative and the other one hovers around $1.25$.

Substituting this value of the nested radical into the equation, we obtain:

$$ 1.55 \approx 2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}$$

$$0.45 \approx \sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}$$

$$0.2 \approx 2+\sqrt{2+\sqrt{2-\cdots}}$$

And we have a contradiction. Therefore, the value we are after does not belong to the solutions of $x^3 + x^2 = 2x +1$

**Moderators:** galactus, sos440, zaidalyafey

3 posts
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\(\displaystyle \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}}=2\sin \left(\frac{\pi}{18} \right)\)

The nested radical repeats after twice positive signs producing a negative sign, i.e., [-, +, +, -]. One can then write is as :

\(\displaystyle x = \sqrt{2 - \sqrt{2 + \sqrt{2 + x}}}\)

Which gives the reducible octic equation

\(\displaystyle x^8 - 8x^6 + 20x^4 - 16x^2 - x + 2 = (x - 2)(x + 1)(x^3 - 3x + 1)(x^3 + x^2 - 2x - 1)\)

Some of the roots get outright ruled out, for example, \(\displaystyle 2\) or \(\displaystyle -1\). The hard part is to prove that the real roots of the fourth factor cannot be a solution which I think can be done by invoking some acceleration methods to the nested radical and then numerically checking each root.

Hence, the real roots of \(\displaystyle x^3 - 3x + 1\) are the only possible choice left. I think it wouldn't be hard to show that one root is \(\displaystyle 2\sin(\pi/18)\) and is the only value of the nested radical. The desired result should follow.

mathbalarka wrote:

The nested radical repeats after twice positive signs producing a negative sign, i.e., [-, +, +, -]. One can then write is as :

\(\displaystyle x = \sqrt{2 - \sqrt{2 + \sqrt{2 + x}}}\)

Which gives the reducible octic equation

\(\displaystyle x^8 - 8x^6 + 20x^4 - 16x^2 - x + 2 = (x - 2)(x + 1)(x^3 - 3x + 1)(x^3 + x^2 - 2x - 1)\)

Some of the roots get outright ruled out, for example, \(\displaystyle 2\) or \(\displaystyle -1\). The hard part is to prove that the real roots of the fourth factor cannot be a solution which I think can be done by invoking some acceleration methods to the nested radical and then numerically checking each root.

Hence, the real roots of \(\displaystyle x^3 - 3x + 1\) are the only possible choice left. I think it wouldn't be hard to show that one root is \(\displaystyle 2\sin(\pi/18)\) and is the only value of the nested radical. The desired result should follow.

There is no need. Two of the roots are negative and the other one hovers around $1.25$.

Substituting this value of the nested radical into the equation, we obtain:

$$ 1.55 \approx 2-\sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}$$

$$0.45 \approx \sqrt{2+\sqrt{2+\sqrt{2-\cdots}}}$$

$$0.2 \approx 2+\sqrt{2+\sqrt{2-\cdots}}$$

And we have a contradiction. Therefore, the value we are after does not belong to the solutions of $x^3 + x^2 = 2x +1$

3 posts
• Page **1** of **1**