Square Root of 5 - Identities of Ramanujan

Identities of Ramanujan

The square root of 5 appears in various identities of Ramanujan involving continued fractions.

For example, this case of the Rogers–Ramanujan continued fraction:

$\cfrac{1}{1 + \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1 + \cfrac{e^{-6\pi}}{1 + \ddots}}}} = \left( \sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{\sqrt{5} + 1}{2} \right)e^{2\pi/5} = e^{2\pi/5}\left( \sqrt{\varphi\sqrt{5}} - \varphi \right).$

$\cfrac{1}{1 + \cfrac{e^{-2\pi\sqrt{5}}}{1 + \cfrac{e^{-4\pi\sqrt{5}}}{1 + \cfrac{e^{-6\pi\sqrt{5}}}{1 + \ddots}}}} = \left( {\sqrt{5} \over 1 + \left^{1/5}} - \varphi \right)e^{2\pi/\sqrt{5}}.$

$4\int_0^\infty\frac{xe^{-x\sqrt{5}}}{\cosh x}\,dx = \cfrac{1}{1 + \cfrac{1^2}{1 + \cfrac{1^2}{1 + \cfrac{2^2}{1 + \cfrac{2^2}{1 + \cfrac{3^2}{1 + \cfrac{3^2}{1 + \ddots}}}}}}}.$

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