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 $$R(e^{-2\pi}) = \sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{\sqrt{5} + 1}{2}$$ Directly raising this to $5^{\text{th}}$ power is bit cumbersome and hence we follow another simpler approach. We note that the value $x = R(e^{-2\pi})$ is a root of the equation $$x^{2} + (\sqrt{5} + 1)x - 1 = 0\tag{2}$$ and it is the greater root of the equation. We propose to find the equation whose roots are $5$th powers of the roots of equation $(2)$. Let $a, b$ be roots of equation $(2)$ with $a > b$ so that $a = R(e^{-2\pi})$. Then $a + b = -(\sqrt{5} + 1), ab = -1$. It follows that $a^{5}b^{5} = -1$. What we require now is the sum $a^{5} + b^{5}$. To do this let $\Sigma_{i} = a^{i} + b^{i}$. Then we know that $\Sigma_{0} = 2, \Sigma_{1} = -(\sqrt{5} + 1)$.

From equation $(2)$ we can see that $\Sigma_{2} + (\sqrt{5} + 1)\Sigma_{1} - \Sigma_{0} = 0$ so that $$\Sigma_{2} = 2 + (\sqrt{5} + 1)^{2} = 8 + 2\sqrt{5}$$ Again $\Sigma_{3} + (\sqrt{5} + 1)\Sigma_{2} - \Sigma_{1} = 0$ and therefore \begin{align}\Sigma_{3} &= \Sigma_{1} - (\sqrt{5} + 1)\Sigma_{2}\notag\\ &= -(\sqrt{5} + 1) - (\sqrt{5} + 1)(8 + 2\sqrt{5})\notag\\ &= -(\sqrt{5} + 1)(9 + 2\sqrt{5}) = -(19 + 11\sqrt{5})\notag\end{align} Similarly \begin{align}\Sigma_{4} &= \Sigma_{2} - (\sqrt{5} + 1)\Sigma_{3}\notag\\ &= 8 + 2\sqrt{5} + (\sqrt{5} + 1)(19 + 11\sqrt{5})\notag\\ &= 82 + 32\sqrt{5}\notag\end{align} and \begin{align}\Sigma_{5} &= \Sigma_{3} - (\sqrt{5} + 1)\Sigma_{4}\notag\\ &= -(19 + 11\sqrt{5}) - (\sqrt{5} + 1)(82 + 32\sqrt{5})\notag\\ &= -(19 + 11\sqrt{5}) - (242 + 114\sqrt{5})\notag\\ &= -(261 + 125\sqrt{5})\notag\end{align} Hence it follows that $a^{5}, b^{5}$ are roots of the equation $$x^{2} + (261 + 125\sqrt{5})x - 1 = 0$$ and therefore \begin{align} R^{5}(e^{-2\pi}) &= a^{5} = \frac{-(261 + 125\sqrt{5}) + \sqrt{146250 + 65250\sqrt{5}}}{2}\notag\\ &= \sqrt{\frac{125(585 + 261\sqrt{5})}{2}} - \frac{261 + 125\sqrt{5}}{2}\notag\\ &= 15\sqrt{\frac{5(65 + 29\sqrt{5})}{2}} - \frac{261 + 125\sqrt{5}}{2}\notag\end{align} and therefore \begin{align}R^{5}(e^{-2\pi}) + \left(\frac{\sqrt{5} + 1}{2}\right)^{5} &= 15\sqrt{\frac{5(65 + 29\sqrt{5})}{2}} - \frac{261 + 125\sqrt{5}}{2} + \frac{11 + 5\sqrt{5}}{2}\notag\\ &= 15\sqrt{\frac{5(65 + 29\sqrt{5})}{2}} - 5(25 + 12\sqrt{5})\notag\\ &= 5\sqrt{5}\left(3\sqrt{\frac{65 + 29\sqrt{5}}{2}} - (12 + 5\sqrt{5})\right)\notag\end{align} Using the above calculation in equation $(1)$ and setting $\alpha = \pi, \beta = \pi/5$ we get \begin{align}\left(\frac{\sqrt{5} + 1}{2}\right)^{5} + R^{5}(e^{-2\pi/5}) &= \dfrac{\left(\dfrac{\sqrt{5} + 1}{2}\right)^{5}}{3\sqrt{\dfrac{65 + 29\sqrt{5}}{2}} - (12 + 5\sqrt{5})}\notag\\ &= \dfrac{\left(\dfrac{\sqrt{5} + 1}{2}\right)^{5}\left(3\sqrt{\dfrac{65 + 29\sqrt{5}}{2}} + (12 + 5\sqrt{5})\right)}{\dfrac{47 + 21\sqrt{5}}{2}}\notag\\ &= \left(\dfrac{\sqrt{5} + 1}{2}\right)^{5}\left(3\sqrt{85 - 38\sqrt{5}} + \frac{39 - 17\sqrt{5}}{2}\right)\notag\\ &= 3\sqrt{(85 - 38\sqrt{5})\cdot\left(\frac{11 + 5\sqrt{5}}{2}\right)^{2}}\notag\\ &\,\,\,\,\,\,\,\,\, + \frac{39 - 17\sqrt{5}}{2}\cdot\frac{11 + 5\sqrt{5}}{2}\notag\\ &= 3\sqrt{\frac{5 + \sqrt{5}}{2}} + 1 + 2\sqrt{5}\notag\end{align} so that \begin{align}R^{5}(e^{-2\pi/5}) &= 3\sqrt{\frac{5 + \sqrt{5}}{2}} + 1 + 2\sqrt{5} - \frac{11 + 5\sqrt{5}}{2}\notag\\ &= 3\sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{9 + \sqrt{5}}{2}\notag\end{align} and then $$R(e^{-2\pi/5}) = \left(3\sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{9 + \sqrt{5}}{2}\right)^{1/5}$$
In a similar manner using equation $(12)$ of last post we can show that $$S(e^{-\pi/5}) = \left(3\sqrt{\frac{5 - \sqrt{5}}{2}} + \frac{9 - \sqrt{5}}{2}\right)^{1/5}$$

Evaluation of $R(e^{-4\pi})$

We next evaluate $R(e^{-4\pi})$ using the values of some class invariants. The first such evaluation for $R(q)$ was done by Ramanathan which was later simplified by Bruce C. Berndt. We follow the approach suggested by Berndt.

Let $x = R(e^{-4\pi})$ so that $x$ satisfies the equation $$\frac{1}{x} - 1 - x = e^{4\pi/5}\cdot\frac{f(-e^{-4\pi/5})}{f(-e^{-20\pi})} = A$$ We next turn to the evaluation of constant $A$ defined above. Clearly we know that if $\eta(q) = q^{1/12}f(-q^{2})$ then $$\eta(e^{-\pi/\sqrt{n}}) = n^{1/4}\eta(e^{-\pi\sqrt{n}})$$ Putting $n = 25/4$ we get \begin{align}\eta(e^{-2\pi/5}) &= \sqrt{\frac{5}{2}}\eta(e^{-5\pi/2})\notag\\ \Rightarrow e^{-\pi/30}f(-e^{-4\pi/5}) &= \sqrt{\frac{5}{2}} e^{-5\pi/24}f(-e^{-5\pi})\notag\\ \Rightarrow f(-e^{-4\pi/5}) &= \sqrt{\frac{5}{2}} e^{-7\pi/40}f(-e^{-5\pi})\notag\end{align} and hence the constant $A$ defined above can be written as \begin{align}A &= \sqrt{\frac{5}{2}}e^{5\pi/8}\frac{f(-e^{-5\pi})}{f(-e^{-20\pi})} = \sqrt{\frac{5}{2}}e^{5\pi/8}\frac{f(-q)}{f(-q^{4})}\notag\\ &= \sqrt{\frac{5}{2}}e^{5\pi/8}\frac{(q;q)_{\infty}}{(q^{4};q^{4})_{\infty}}\notag\\ &= \sqrt{\frac{5}{2}}e^{5\pi/8}(q;q^{2})_{\infty}(q^{2};q^{4})_{\infty}\notag\\ &= \sqrt{\frac{5}{2}}e^{5\pi/8}\cdot 2^{1/4}q^{1/24}g(q)\cdot 2^{1/4}q^{1/12}g(q^{2})\notag\\ &= 5^{1/2}e^{5\pi/8}q^{1/8}g(q)g(q^{2})\notag\\ &= \sqrt{5}g_{25}g_{100}\notag\end{align} where $q = e^{-5\pi}$.

Using equations $(13)$ and $(14)$ of this post and setting $x = \sqrt[4]{5}$ we get \begin{align}A &= \sqrt{5}\frac{(1 + \sqrt{5})^{3/2}(3 + 2\sqrt[4]{5})^{1/2}}{2\sqrt{2}}\notag\\ &= \sqrt{5}\frac{(\sqrt{5} + 1)(\sqrt{5} - 1)(1 + \sqrt{5})^{1/2}(3 + 2\sqrt[4]{5})^{1/2}}{2\sqrt{2}(\sqrt{5} - 1)}\notag\\ &= \sqrt{5}\frac{\sqrt{2}}{\sqrt{5} - 1}\{(1 + \sqrt{5})(3 + 2\sqrt[4]{5})\}^{1/2}\notag\\ &= \frac{\sqrt{5}}{\sqrt{5} - 1}\{2(1 + x^{2})(3 + 2x)\}^{1/2}\notag\\ &= \frac{\sqrt{5}}{\sqrt{5} - 1}\{6 + 4x + 6x^{2} + 4x^{3}\}^{1/2}\notag\\ &= \frac{\sqrt{5}}{\sqrt{5} - 1}\{1 + 4x + 6x^{2} + 4x^{3} + x^{4}\}^{1/2}\notag\\ &= \frac{\sqrt{5}}{x^{2} - 1}\{(1 + x)^{4}\}^{1/2}\notag\\ &= \sqrt{5}\frac{x + 1}{x - 1}\notag\\ &= \sqrt{5}\frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}\notag\end{align} It now follows that $x = R(e^{-4\pi})$ is given by $$x = \sqrt{a^{2} + 1} - a$$ where $a$ is given by $$2a = 1 + A = 1 + \sqrt{5}\frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}$$

General Value of $R(e^{-\pi\sqrt{n}})$

If $q = e^{-\pi\sqrt{n}}$ where $n$ is a positive rational, then we know that the value of $k = k(q) = \theta_{2}^{2}(q)/\theta_{3}^{2}(q)$ is an algebraic number. Now let $l = k(q^{5})$ be the modulus corresponding to $q^{5}$ so that $l$ is also an algebraic number. Using notation of Ramanujan we set $\alpha = k^{2}, \beta = l^{2}$ so that $\beta$ is of degree $5$ over $\alpha$. From the transcription formulas in this post we have \begin{align}\frac{f^{6}(-q)}{qf^{6}(-q^{5})} &= \dfrac{2^{-1}z^{3}(q)(1 - \alpha)\alpha^{1/4}q^{-1/4}}{q\cdot 2^{-1}z^{3}(q^{5})(1 - \beta)\beta^{1/4}q^{-5/4}}\notag\\ &= m^{3}\frac{(1 - \alpha)\alpha^{1/4}}{(1 - \beta)\beta^{1/4}}\notag\end{align} Since the multiplier $m$ itself can be expressed as an algebraic function of $\alpha, \beta$ it follows from the above that the fraction $f^{6}(-q)/\{qf^{6}(-q^{5})\}$ is an algebraic number provided that $q = e^{-\pi\sqrt{n}}$, $n$ being a positive rational number. And therefore from the identity $$\frac{1}{R^{5}(q)} - 11 - R^{5}(q) = \frac{f^{6}(-q)}{qf^{6}(-q^{5})}$$ it follows that $R^{5}(q)$ and hence $R(q)$ is also an algebraic number.

Using sophisticated techniques of modular forms and class field theory it can be established that the fraction $f^{6}(-q)/\{qf^{6}(-q^{5})\}$ is an algebraic integer in which case the above identity for $R(q)$ implies that $R(q)$ is a unit whenever $q = e^{-\pi\sqrt{n}}$, $n$ being a positive rational. The same proposition holds for the values of $S(q) = -R(-q)$ and can be established in exactly the same manner as done for $R(q)$.

Following Dr. Bruce C. Berndt we will now establish some general formulas for $R(e^{-2\pi\sqrt{n}})$ in terms of the class invariants. Let us then suppose that $q = e^{-2\pi\sqrt{n}}$ and then we can define a constant $A$ by $$A = \frac{f(-q)}{q^{1/6}f(-q^{5})} = e^{\pi\sqrt{n}/3}\frac{f(-e^{-2\pi\sqrt{n}})}{f(-e^{-10\pi\sqrt{n}})}\tag{3}$$ so that $$\frac{1}{R^{5}(q)} - 11 - R^{5}(q) = A^{6}$$ and then $$R^{5}(e^{-2\pi\sqrt{n}}) = \sqrt{a^{2} + 1} - a\tag{4}$$ where $2a = A^{6} + 11$

To calculate the value of $A$ we need to make use of modular equations of degree $5$ established in an earlier post. We have from equations $(19)$ and $(20)$ of this post: $$m = \left(\frac{\beta}{\alpha}\right)^{1/4} + \left(\frac{1 - \beta}{1 - \alpha}\right)^{1/4} - \left(\frac{\beta(1 - \beta)}{\alpha(1 - \alpha)}\right)^{1/4}\tag{5}$$ $$\frac{5}{m} = \left(\frac{\alpha}{\beta}\right)^{1/4} + \left(\frac{1 - \alpha}{1 - \beta}\right)^{1/4} - \left(\frac{\alpha(1 - \alpha)}{\beta(1 - \beta)}\right)^{1/4}\tag{6}$$ From the above equations we can see that \begin{align} &\{\alpha(1 - \alpha)\}^{1/4}\left(m + \left(\frac{\beta(1 - \beta)}{\alpha(1 - \alpha)}\right)^{1/4}\right)\notag\\ &\,\,\,\,\,\,\,\,= \{\beta(1 - \beta)\}^{1/4}\left(\frac{5}{m} + \left(\frac{\alpha(1 - \alpha)}{\beta(1 - \beta)}\right)^{1/4}\right)\tag{7} \end{align} Now let $q$ correspond to $\alpha$ so that $q^{5}$ corresponds to $\beta$ and $m = \phi^{2}(q)/\phi^{2}(q^{5})$. We then have $$q^{-1/24}\chi(q) = 2^{1/6}\{\alpha(1 - \alpha)\}^{-1/24}$$ $$q^{-5/24}\chi(q^{5}) = 2^{1/6}\{\beta(1 - \beta)\}^{-1/24}$$ and thus the equation $(7)$ above can be transcribed into the following identity for theta functions $$\frac{2q^{1/4}}{\chi^{6}(q)}\left(\frac{\phi^{2}(q)}{\phi^{2}(q^{5})} + \frac{q\chi^{6}(q)}{\chi^{6}(q^{5})}\right) = \frac{2q^{5/4}}{\chi^{6}(q^{5})}\left(\frac{5\phi^{2}(q^{5})}{\phi^{2}(q)} + \frac{\chi^{6}(q^{5})}{q\chi^{6}(q)}\right)$$ which leads to $$\frac{\phi^{2}(q)}{\phi^{2}(q^{5})} - \frac{5q\phi^{2}(q^{5})\chi^{6}(q)}{\phi^{2}(q)\chi^{6}(q^{5})} = 1 - \frac{q\chi^{6}(q)}{\chi^{6}(q^{5})}\tag{8}$$
Noting that \begin{align}f(-q^{2}) &= (q^{2};q^{2})_{\infty}\notag\\ &= \frac{(-q;q^{2})_{\infty}(-q;q^{2})_{\infty}(q^{2};q^{2})_{\infty}}{(-q;q^{2})_{\infty}(-q;q^{2})_{\infty}}\notag\\ &= \frac{f(q, q)}{\chi^{2}(q)} = \frac{\phi(q)}{\chi^{2}(q)}\notag\end{align} we can write $\phi(q) = \chi^{2}(q)f(-q^{2})$ and then the equation $(8)$ is transformed into $$\frac{\chi^{4}(q)f^{2}(-q^{2})}{\chi^{4}(q^{5})f^{2}(-q^{10})} - \frac{5qf^{2}(-q^{10})\chi^{2}(q)}{f^{2}(-q^{2})\chi^{2}(q^{5})} = 1 - \frac{q\chi^{6}(q)}{\chi^{6}(q^{5})}\tag{9}$$
Next we know that the Ramanujan class invariant function $G(q)$ is given by $G(q) = 2^{-1/4}q^{-1/24}\chi(q)$ so that $\chi(q) = 2^{1/4}q^{1/24}G(q)$. Using this value of $\chi(q)$ in above equation we get $$\frac{G^{4}(q)f^{2}(-q^{2})}{q^{2/3}G^{4}(q^{5})f^{2}(-q^{10})} - \frac{5q^{2/3}G^{2}(q)f^{2}(-q^{10})}{G^{2}(q^{5})f^{2}(-q^{2})} = 1 - \frac{G^{6}(q)}{G^{6}(q^{5})}$$
Putting $q = e^{-\pi\sqrt{n}}$ in the above equation and noting that $G(q) = G_{n}, G(q^{5}) = G_{25n}$ and also observing that $A = q^{-1/3}f(-q^{2})/f(-q^{10})$, we finally get $$A^{2}\frac{G_{n}^{4}}{G_{25n}^{4}} - A^{-2}\frac{G_{n}^{2}}{G_{25n}^{2}} = 1 - \frac{G_{n}^{6}}{G_{25n}^{6}}$$ Setting $U = G_{25n}/G_{n}$ we can see that the above equation is transformed into $$\frac{A^{2}}{U} - 5\frac{U}{A^{2}} = U^{3} - \frac{1}{U^{3}}$$ If we replace $q$ by $-q$ in the equation $(9)$ and set $V = g_{25n}/g_{n}$ then we get $$\frac{A^{2}}{V} + 5\frac{V}{A^{2}} = V^{3} + \frac{1}{V^{3}}$$
We can now summarize our formulas as follows:
Let $n$ be a positive rational number and we define $$U = \frac{G_{25n}}{G_{n}},\,\, V = \frac{g_{25n}}{g_{n}}$$ Then the constant $A$ defined by $$A = e^{\pi\sqrt{n}/3}\frac{f(-e^{-2\pi\sqrt{n}})}{f(-e^{-10\pi\sqrt{n}})}$$ satisfies the following equations $$\frac{A^{2}}{U} - 5\frac{U}{A^{2}} = U^{3} - \frac{1}{U^{3}}$$ $$\frac{A^{2}}{V} + 5\frac{V}{A^{2}} = V^{3} + \frac{1}{V^{3}}$$ and the value of $R^{5}(e^{-2\pi\sqrt{n}})$ is given by $$R^{5}(e^{-2\pi\sqrt{n}}) = \sqrt{a^{2} + 1} - a$$ where $a$ is given by $$2a = A^{6} + 11$$ If we put $n = 1/5$ in the above formula we get $U = 1$ so that $A^{2} = \sqrt{5}$ and $a = (11 + 5\sqrt{5})/2$ and the value of $R^{5}(e^{-2\pi/\sqrt{5}})$ calculated using above formula matches the value calculated earlier.

There are analogous results for $S(q)$ which the reader can formulate and prove himself along the above lines. For more such formulas the reader should consult Dr. Bruce C. Berndt's papers and his Ramanujan's Notebooks.