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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?

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2 Answers 2

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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 $$ M_5(r)=\frac{k_r^{1/4}\left(1-k_{25r}^{1/4}k_r^{3/4}\right)}{k_r^{1/4}-k_{25r}^{5/4}}. $$

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  • $\begingroup$ Thank you for your answer. Especially, (d2) seems to be very helpful for my study. Other relations were interesting to me, as well. $\endgroup$
    – user356126
    Mar 13, 2017 at 5:37
  • $\begingroup$ Note that $\frac{dR(q)}{dr}=\frac{dR(q)}{dq}\frac{dq}{dr}$, $q=e^{-\pi\sqrt{r}}$. The results can be generalized $\endgroup$ May 2, 2017 at 18:55
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Another identity for the derivative of the Rogers-Ramanujan continued fraction can be found in this post

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  • $\begingroup$ Thank you for the information. $\endgroup$
    – user356126
    Aug 4, 2017 at 11:43

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