Entry 3

Ramanujan gave the beautiful continued fractions (with the second simplified by this author) $$5^{1/4}\sqrt{\phi}\,-\phi=\cfrac{e^{-2\pi/5}}{1 + \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1 + \cfrac{e^{-6\pi}}{1 + \ddots}}}}\tag1$$ $$\frac{\sqrt{5}}{1+\phi^{-1}\big({-\phi^5}+\sqrt{\phi^{10}+1}\big)^{1/5}}-\phi=\cfrac{e^{-2\pi/\sqrt{5}}}{1 + \cfrac{e^{-2\pi\sqrt{5}}}{1 + \cfrac{e^{-4\pi\sqrt{5}}}{1 + \cfrac{e^{-6\pi\sqrt{5}}}{1 + \ddots}}}}\tag2$$ $$\big({-\phi^5}+\sqrt{\phi^{10}+1}\big)^{1/5}=\cfrac{e^{-2\pi/(5\sqrt{5})}}{1 + \cfrac{e^{-2\pi/\sqrt{5}}}{1 + \cfrac{e^{-4\pi/\sqrt{5}}}{1 + \cfrac{e^{-6\pi/\sqrt{5}}}{1 + \ddots}}}}\tag3$$ with the golden ratio $\phi = \frac{1+\sqrt{5}}{2}$. In 1913, the British mathematician G.H. Hardy, after reading the letter Ramanujan sent to him (which included examples of these extraordinary continued fractions), remarked, “…the [theorems] defeated me completely; I had never seen anything in the least like them before.” He would have been even more amazed had he known that these were connected to geometry.