How is this nested radical expression for $\cos(2\pi /7)$ proven?

Question

On the Wolfram Mathworld page for the silver constant, the following identity is present, how is it proven?

$$2\cos\Big(\frac{2\pi}{7}\Big) = \frac{2+\sqrt[3]{\,7+7\sqrt[3]{\,7+...}}}{1+\sqrt[3]{\,7+7\sqrt[3]{\,7+...}}}$$

Link: http://mathworld.wolfram.com/SilverConstant.html

Answer 1

This is the same as

$$ 2\cos\left(\frac{2\pi}{7}\right)-1=\frac{1}{1+\sqrt[3]{7+7\sqrt[3]{7+\cdots}}}. $$

Write $T=2\cos\left(\frac{2\pi}{7}\right)=\zeta+\zeta^{-1}$ where $\zeta=e^{2\pi i/7}$ and $x=\sqrt[3]{7+7\sqrt[3]{7+\cdots}}$. Then observe

$$ \begin{array}{ll} T & = \zeta+\zeta^{-1} \\ T^2 & = \zeta^2+2+\zeta^{-2} \\ T^3 & = \zeta^3+3\zeta+3\zeta^{-1}+\zeta^{-3}. \end{array} $$

We know $0=\zeta^7-1=(\zeta-1)(\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1)$. Since $\zeta-1$ isn't $0$, that means the other factor is, and dividing by $\zeta^3$ yields the symmetric form

$$ \zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=0. $$

Therefore, $T^3+T^2-2T-1=0$. Next, observe $x^3=7+7x$ so $x^3-7x-7=0$. Define

$$ A=T-1, \quad B=1+x. $$

We want to show these are inverse of each other. Observe

$$ (A+1)^3+(A+1)^2-2(A+1)-1=0 \\ \implies f(A)=A^3+4A^2+3A-1=0 $$

$$ (B-1)^3-7(B-1)-7=0 \\ \implies g(B)=B^3-3B^2-4B-1=0. $$

In the case of $f(t)$, notice $f'(t)=3t^2+8t+3$ is always positive for $t>0$, and $f(0)=-1$, therefore $f$ has exactly one positive root (namely $t=A$). Now, $B$ is a root of $g(t)$, but $g(t)=-t^3f(1/t)$ is the opposite reciprocal polynomial, so $1/B$ must also be a (necessarily positve) root of $f(t)$. Conclude $A=1/B$, as desired.