What is the derivative of Rogers-Ramanujan Continued Fraction?

Question

We have many useful formulae about the derivatives of modular functions. For example, \begin{eqnarray} &&j'(\tau)=-\frac{E_6}{E_4} j(\tau), \\ &&\eta'(\tau)=\frac{1}{24}E_2 \eta(\tau), \\ &&E_2'(\tau)=\frac{1}{12}(E_2(\tau)^2-E_4(\tau)), \end{eqnarray} where $'=\frac{1}{2\pi i}\frac{d}{d\tau}.$ In order to calculate modular functions containing Rogers-Ramanujan continued fraction, I am looking for the derivative formulae about Rogers-Ramanujan continued fraction like these. Does anyone know any useful formulae?

Answer 1

If $|q|<1$ and $$ R(q)=\frac{q^{1/5}}{1+}\frac{q}{1+}\frac{q^2}{1+}\frac{q^3}{1+}\ldots\textrm{ :}\textrm{ (1)} $$ is the Rogers Ramanujan continued fraction, then $$ R'(q)=5^{-1}q^{-5/6}f(-q)^4R(q)\sqrt[6]{R(q)^{-5}-11-R(q)^5}\textrm{, }:(d1)$$ where $$ f(-q)=\prod^{\infty}_{n=1}(1-q^n)\textrm{, }q=e^{-\pi\sqrt{r}}\textrm{, }r>0 $$ is the Ramanujan eta function.

Moreover the Dedekind eta function is $$ \eta(z)=q^{1/24}\prod^{\infty}_{n=1}(1-q^n), $$ where $q=e(z)=e^{2\pi i z}$, $Im(z)>0$. Another relation (due to Ramanujan) is $$ R'(q)=\frac{f(-q)^5}{5qf(-q^5)}R(q)\textrm{, }:(d2) $$ Also $R(q)$ is function of the elliptic singular modulus $k_r$, $k'_r=\sqrt{1-k_r^2}$, hence $R(q)=F(k_r)$, where $F(x)$ is algebraic function solution of a six degree polynomial equation. Hence we can write (d1) as $$ \frac{dR(q)}{dk}=\frac{2^{1/3}}{5(k_rk'_r)^{2/3}}R(q)\sqrt[6]{R(q)^{-5}-11-R(q)^5}\textrm{ :}\textrm{ (2)} $$ We can now solve DE (2) and take the beatufull identity: $$ 2\pi\int^{+\infty}_{\sqrt{r}}\eta\left(it\right)^4dt=3\sqrt[3]{2k_{4r}}\cdot {}_2F_1\left(\frac{1}{3},\frac{1}{6};\frac{7}{6};k_{4r}^2\right)=5\int^{R(q^2)}_{0}\frac{dt}{t\sqrt[6]{t^{-5}-11-t^5}}, $$ where $q=e^{-\pi\sqrt{r}}$, $r>0$. The function $$ \Pi(r):=\sqrt[3]{2k_{4r}}\cdot {}_2F_1\left(\frac{1}{3},\frac{1}{6};\frac{7}{6};k_{4r}^2\right), $$ is Carty's function and is related to the famous Carty's problem (see Wikipedia).

Also $\Pi(r)$ satisfy the following functional equation $$ \Pi(r)+\Pi\left(\frac{1}{r}\right)=C_0\textrm{, }r>0 $$ where $C_0=2^{-4/3}\pi^{-1}\Gamma(1/3)^3\sqrt{3}$.

Answer 2

Another identity for the derivative of the Rogers-Ramanujan continued fraction can be found in this post