# What is the derivative of Rogers-Ramanujan Continued Fraction?

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?

REVISED.

If $$|q|<1$$ and $$R(q)=\frac{q^{1/5}}{1+}\frac{q}{1+}\frac{q^2}{1+}\frac{q^3}{1+}\ldots\tag 1$$ 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. Also if the Dedekind eta function is $$\eta(z)=q^{1/24}\prod^{\infty}_{n=1}(1-q^n),$$ where $$q=e(z)=e^{2\pi i z}$$, $$Im(z)>0$$ and if $$v(z)=R(q)$$, then $$v'(z)=\frac{2\pi i}{5}\eta(z)^4v(z)\sqrt[6]{v(z)^{-5}-11-v(z)^5}\textrm{, }:(d0)$$ 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 $$\frac{dR(q)}{dk}=\frac{2^{1/3}}{5(k_rk'_r)^{2/3}}R(q)\sqrt[6]{R(q)^{-5}-11-R(q)^5}\tag 2$$ 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 $$C_0=2^{-4/3}\pi^{-1}\Gamma(1/3)^3\sqrt{3}$$.

If $$q=e^{-\pi\sqrt{r}}$$, $$r>0$$ and $$u=k_r^{1/4}$$, $$v=k_{25r}^{1/4}$$, then $$R(q)^{-5}-11-R(q)^5=\frac{(1-u^8)(u-v^5)^3}{uv^2(1-u^3v)^3(1-v^8)}.\tag 3$$ Also if $$K(x)=\frac{\pi}{2} {}_2F_1\left(\frac{1}{2},\frac{1}{2};1,x^2\right)$$ is the complete elliptic integral of the first kind, then $$\frac{dR(q)}{dq}=\frac{2^{23/15}(k_r)^{5/12}(k'_r)^{5/3}}{5(k_{25r})^{1/12}(k'_{25r})^{1/3}}\frac{1}{\sqrt[5]{11+a_r+\sqrt{125+22a_r+a^2_r}}}\frac{K^2(k_r)}{\pi^2q\sqrt{M_5(r)}}, \tag 4$$ where $$a_r=\left(\frac{k'_r}{k'_{25r}}\right)^2\sqrt{\frac{k_r}{k_{25r}}}M_5(r)^{-3}$$ and $$M_5(r)=\frac{k_r^{1/4}\left(1-k_{25r}^{1/4}k_r^{3/4}\right)}{k_r^{1/4}-k_{25r}^{5/4}}.$$

• Thank you for your answer. Especially, (d2) seems to be very helpful for my study. Other relations were interesting to me, as well. Mar 13, 2017 at 5:37
• Note that $\frac{dR(q)}{dr}=\frac{dR(q)}{dq}\frac{dq}{dr}$, $q=e^{-\pi\sqrt{r}}$. The results can be generalized May 2, 2017 at 18:55

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

• Thank you for the information. Aug 4, 2017 at 11:43