Entry 20

One of Ramanujan’s talents was spotting elegant special cases of general phenomena. Consider for example$$x=\small+\sqrt{a+\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a-\cdots}}}}}}}\tag1$$where the repeating pattern of the signs is \((+,+,+,-)\). Generally, \(x\) will be the root of a \(12\)th deg factor of a \(16\)th deg equation. But he found that $$x_1=\small+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}=2.7472\dots$$ Note that \(x\) is just a root of a quartic. Its other roots are given by the patterns \((+,+,-,+),\; (+,-,+,+),\; (-,+,+,+)\), respectively $$x_2=\small+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\cdots}}}}}}} = \frac{2-\sqrt 5 +\sqrt{15+6\sqrt 5}}{2}=2.5473\dots$$ $$x_3=\small+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\cdots}}}}}}} = \frac{2+\sqrt 5 -\sqrt{15-6\sqrt 5}}{2}=1.4888\dots$$ $$x_4=\small\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5\color{red}-\sqrt{5+\sqrt{5+\sqrt{5+\cdots}}}}}}} = \frac{2-\sqrt 5 -\sqrt{15+6\sqrt 5}}{2}=-2.7833\dots$$ This immediately implies that the four roots obey the system with \(a=5\), $$\begin{aligned}
x_1^2 &= x_2+a\\ x_2^2 &= x_3+a\\ x_3^2 &= x_4+a\\ x_4^2 &= x_1+a\\
\end{aligned}\tag2$$also studied by Ramanujan. In general, an infinitely nested radical with period length of 4 like \((1)\)$$x = \pm\sqrt{a\pm \sqrt{a\pm \sqrt{a\pm \sqrt{a\pm\dots}}}}$$and a system of 4 equations like \((2)\) can be expressed as$$x = (((x^2 - a)^2 - a)^2 - a)^2 - a\tag3$$Expanded out and factored, Ramanujan stated that \((3)\) was a product of 4 quartics, three of which had coefficients in the cubic,$$y^3+3y = 4(1+ay)$$However, this has a rational factor for the special cases when \(a=2,5\) so explains why the radical he found has no cubic irrationalities.