Entry 7

Ramanujan found the nice trigonometric relation, $$\sqrt[3]{2\cos\big(\tfrac{2\pi}{7}\big)}+\sqrt[3]{2\cos\big(\tfrac{4\pi}{7}\big)} +\sqrt[3]{2\cos\big(\tfrac{6\pi}{7}\big)}  = -\sqrt[3]{-5+3\cdot7^{1/3}}= -0.904\dots$$ Equivalently, let $x_i$ be the three roots of the cubic $x^3+x^2-2x-1=0$. Then $$\sum_{k=1}^3 x_k^{1/3} = -\sqrt[3]{-5+3\cdot7^{1/3}} = -0.904\dots$$ The natural question to ask is, can this be generalized to quintics and $\cos\big(\tfrac{2\pi}{11}\big)$? It turns out it can. Noam Elkies found that if we let $y_i$ be the five roots of the quintic $y^5+6y^4-y^3-32y^2+16y-1=0$, then $$\sum_{k=1}^5 y_k^{1/5} = -\sqrt[5]{-274+5(-21\cdot11^{1/5}+13\cdot11^{2/5}+13\cdot11^{3/5})}=-0.093\dots$$ The $y_i$ can also be expressed as cosines $$y_k=-(z_k^2-1)^2-(z_k-1),\quad \text{and}\; z_k=2\cos\big(\tfrac{2\pi\,k}{11}\big)$$ Similarly, one can find a $7$th deg relation using $\cos\big(\tfrac{2\pi}{29}\big)$ and so on.