# Values of Rogers-Ramanujan Continued Fraction: Part 3

### 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 $$R(e^{-2\pi}) = \sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{\sqrt{5} + 1}{2}$$

# Values of Rogers-Ramanujan Continued Fraction: Part 2

Continuing our journey from the last post we will deduce further properties of the Rogers-Ramanujan Continued Fraction $R(q)$ which will help us to find out further values of $R(q)$. In this connection we first establish an identity concerning powers of $R(q)$.

### Identity Concerning $R^{5}(q)$

Using the identity $(3)$ from the last post, Ramanujan established another fundamental property of $R(q)$ namely: $$\frac{1}{R^{5}(q)} - 11 - R^{5}(q) = \frac{f^{6}(-q)}{qf^{6}(-q^{5})}\tag{1}$$

# Values of Rogers-Ramanujan Continued Fraction: Part 1

### A Wild Theorem by Ramanujan

In his letter dated 16th January 1913 to G. H. Hardy, Ramanujan presented the following wild theorem: $$\cfrac{1}{1 + \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1 + \cfrac{e^{-6\pi}}{1 + \cdots}}}} = \left(\sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{\sqrt{5} + 1}{2}\right)\sqrt[5]{e^{2\pi}}\tag{1}$$ The theorem looks so strange and surprising, coming out of nowhere that Hardy had to remark: "they must be true because, if they were not true, no one would have had the imagination to invent them." In this post we will prove the above theorem using elementary methods. The proof is essentially the one given by Watson who claimed that probably Ramanujan obtained the result in the same manner.