Evaluation of R(e^{-2\pi/5})
In the last post we established the transformation formula \left[\left[\frac{\sqrt{5} + 1}{2}\right]^{5} + R^{5}(e^{-2\alpha})\right]\left[\left[\frac{\sqrt{5} + 1}{2}\right]^{5} + R^{5}(e^{-2\beta})\right] = 5\sqrt{5}\left[\frac{\sqrt{5} + 1}{2}\right]^{5}\tag{1} under the condition \alpha\beta = \pi^{2}/5.If we put \alpha = \pi then \beta = \pi/5 and since we already know the value of R(e^{-2\pi}) we can use equation (1) to evaluate R(e^{-2\pi/5}). But in order to do that we need to calculate R^{5}(e^{-2\pi}) first.
We have from an earlier post R(e^{-2\pi}) = \sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{\sqrt{5} + 1}{2}
