Entry 196

In Entry 194

$$2^{-1/4}G_{47} = x,\quad \text{where}\; x^5 - x^3 - 2x^2 - 2x - 1 = 0$$Ramanujan asked for the radical solution of this quintic. We give our version which partly uses the golden ratio \(\phi\)

$$\begin{align}y_1 &= \frac{1}{2\phi}\left(13+\frac{\sqrt{47(2+89\sqrt5)}}{5^{5/4}}\right)\\ y_2 &= \frac{1}{2\phi}\left(13-\frac{\sqrt{47(2+89\sqrt5)}}{5^{5/4}}\right)\\ y_3 &= \frac{\phi}{2}\left(13+\frac{\sqrt{47(-2+89\sqrt5)}}{5^{5/4}}\right)\\ y_4 &= \frac{\phi}{2}\left(13-\frac{\sqrt{47(-2+89\sqrt5)}}{5^{5/4}}\right) \end{align}$$

then the real root of quintic is

$$x = \frac{y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5}}{\sqrt5} = 1.73469134\dots$$

In fact, the \(y_n\) are the four roots of the quartic

$$3125y^4 - 40625y^3 - 521250y^2 + 55250y - 1= 0$$

where the leading term is a \(5\)th power \(3125 = 5^5\).