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The Rogers-Ramanujan cfrac is,

$$r = r(\tau)= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\ddots}}}$$

If $q = \exp(2\pi i \tau)$, then it is known that,

$$\frac{1}{r}-r =\frac{\eta(\tau/5)}{\eta(5\tau)}+1\tag1$$

$$\frac{1}{r^5}-r^5 =\left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11\tag2$$

with the Dedekind eta function, $\eta(\tau)$.

Q: Is there a similar simple identity known using ratios of the Jacobi theta functions $\vartheta_n(0,q)$?

$\color{brown}{Edit}$: In response to a comment, here are some details. Suppose we don't know $(2)$. One way to find such relations is to use known identities. Given the j-function $j(\tau)$, we have,

$$j(\tau)=-\frac{(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^3}{r^5(r^{10} + 11r^5 - 1)^5}\tag3$$

$$j(\tau)=\frac{(5x^2+10x+1)^3}{x^5}\tag4$$

where $x = \left(\frac{\eta(\tau)}{\sqrt{5}\,\eta(5\tau)}\right)^6$. Equate $(3),\,(4)$,

$$-\frac{(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^3}{r^5(r^{10} + 11r^5 - 1)^5} = \frac{(5x^2+10x+1)^3}{x^5}\tag5$$

and using symbolic software like Mathematica to factor $(5)$, we find one factor is given by,

$$\frac{1}{r^5}-r^5 =125x+11\tag6$$

and it only takes minor tweaking to make $(6)$ have the form of $(2)$. Thus, all we need is to express $j(\tau)$ not by an eta quotient $x$, but by a theta quotient $y=\left(\frac{\vartheta_n(0,q)}{\vartheta_n(0,q^5)}\right)^k$ so that,

$$j(\tau) = \frac{f_1(y)}{f_2(y)}\tag7$$

is a ratio of polynomials in $y$.

$\color{brown}{P.S.}$ The motivation for this is that the general quintic is solvable by the Rogers-Ramanujan cfrac via the eta quotient above as described in this post.

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Tito Piezas III
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  • The answer is in W.Duke's paper *continued fractions and modular functions*,equation (4.9) – Nicco Sep 07 '15 at 11:34
  • @Nicco: Thanks. By the way, can you ask a question on the forum how to solve the Brioschi quintic using elliptic functions? (I have an answer that may be useful to you and others interested, and is the reason for this theta function post.) – Tito Piezas III Sep 07 '15 at 11:44
  • @ Tito PiezasIII :I have asked the question here – Nicco Sep 07 '15 at 13:36
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    @Nicco: Thanks. Right now, I can do so using the complete elliptic integral $K(k)$ and the Dedekind eta function, via $R(q)$. I'm trying to adapt the method to the Jacobi theta. – Tito Piezas III Sep 07 '15 at 13:40
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    From another source, we have $$\frac{1}{r^5}-r^5 = \Big(\frac{ \vartheta_2(0,q),\vartheta_3(0,q), \vartheta_4(0,q) }{\vartheta_2(0,q^5),\vartheta_3(0,q^5), \vartheta_4(0,q^5) }\Big)^2+11$$ but I find this is cheating a little bit. – Tito Piezas III Sep 09 '15 at 03:34
  • @ Tito PiezasIII:it looks awesome,though – Nicco Sep 09 '15 at 05:12
  • The derivative of the theta function $$\theta_1(0,q)$$ is known to satisfy the identity $$\theta_1(0,q)\prime=\theta_2(0,q),\theta_3(0,q),\theta_4(0,q)$$ – Nicco Sep 09 '15 at 05:35
  • @Nicco: Very interesting. This would then imply $\theta_1(0,q)' = 2\eta^3(\tau)$. Substituting this into the identity, one recovers $$\frac{1}{r^5}-r^5 = \Big(\frac{\eta(\tau)}{\eta(5\tau)}\Big)^6+11$$ – Tito Piezas III Sep 09 '15 at 05:41
  • Another way to express $\eta$ in terms of Jacobi theta functions is eq. $(20)$ at Mathworld. This is not a nullvalue however. – ccorn Sep 10 '15 at 01:56
  • @ccorn: Or from this. – Tito Piezas III Sep 10 '15 at 02:12
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    Using basic identities, one can deduce a family of formulae for the RRCF like $$r(\tau) = \frac {\vartheta_1!\left(\frac{\pi}{5}\middle|\tau'\right)} {\vartheta_1!\left(\frac{2\pi}{5}\middle|\tau'\right)} = \frac {\vartheta_2(0\mid \tau'),\vartheta_3(0\mid\tau'),\vartheta_4(0\mid\tau')} {2,\vartheta_2!\left(\frac{\pi}{5}\middle|\tau'\right) \vartheta_3!\left(\frac{\pi}{5}\middle|\tau'\right) \vartheta_4!\left(\frac{\pi}{5}\middle|\tau'\right)}$$ where $\tau'=-(5\tau)^{-1}$. Unfortunately, the question requires thetanulls exclusively. Sigh. – ccorn Oct 23 '15 at 01:27
  • Could you concretize what you are asking for? The question is interesting, but currently it is not clear to me what a valid answer would be. It seems you might be satisfied with an algebraic equation for $r$ with coefficients in terms of thetanulls. You have already given some. I have inspected your Brioschi quintic post, but to adapt the method to thetanulls one can go directly via $\eta$, that is, the $2\eta^3$ as a product of thetanulls, as already mentioned. Do you expect something drastically simpler? Whatever reformulation we can come up with, it would at best be equivalent, I'm afraid. – ccorn Oct 25 '15 at 13:52
  • @ccorn: I have clarified the question. If you know of a formula for the j-function in terms of a theta quotient in just *one* thetanull $\frac{\vartheta_n(0,q)}{\vartheta_n(0,q^5)}$, then that can lead me to an answer. – Tito Piezas III Oct 25 '15 at 15:04
  • Thanks for the clarification. My gut feeling is that such a thing should exist. Thinking... – ccorn Oct 25 '15 at 16:34
  • @ccorn: Thanks to your prodding, I found what I was looking for. :) Kindly see answer below. – Tito Piezas III Oct 27 '15 at 06:36
  • I have asked a question which assumes $(6)$ and proves $(3),(4)$ based on it. I also provide an answer based on modular equations. See this question. – Paramanand Singh Jan 28 '21 at 01:58

5 Answers5

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Motivated by ccorns's comment, I decided to re-visit this question. After some effort, I found what I was looking for. The Rogers-Ramanujan cfrac is,

$$r = r(\tau)= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\ddots}}}$$

where $q = \exp(2\pi i \tau)$. It turns out we have the nice relations,

$$\begin{aligned}\frac{1}{r^5}-r^5 &=\,u^3+11\\ &= \frac{v(v-5)^2}{(v-1)^2}+11 \end{aligned}$$

where the three roots of the cubic in $u$ are eta quotients,

$$u_n = \left(\frac{\eta(\tau')}{\eta(5\tau')}\right)^2$$

with $\tau'=\tau+n$, while the three roots of the cubic in $v$ are theta quotients,

$$v_n = \left(\frac{\vartheta_{n+2}(0,p)}{\vartheta_{n+2}(0,p^5)}\right)^2$$

for the nome $p = e^{\pi i \tau}$ and for $n=0,1,2$.

Tito Piezas III
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    Nice indeed. If there were a 2nd volume of Whittaker & Watson, this one would make it to the legendary "Shew that..." exercises. – ccorn Oct 27 '15 at 19:17
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Given the Jacobi theta function $$ \vartheta_4(u,q)=1+2\sum^{\infty}_{n=1}(-1)^nq^{n^2}\cos(2nu) $$ I. Define the Rogers-Ramanujan as $$ R(q)=\frac{q^{1/5}}{1+}\frac{q}{1+}\frac{q^2}{1+}\frac{q^3}{1+}\ldots\textrm{, }\;|q|<1 $$ then $$ R(e^{-x})=e^{-x/5}\frac{\vartheta_4(3ix/4,\,e^{-5x/2})}{\vartheta_4(ix/4,\,e^{-5x/2})}\textrm{, }\;x>0 $$ II. Define the Ramanujan-Gordon-Gollniz as $$ H(q)=\frac{q^{1/2}}{1+q+}\frac{q^2}{1+q^3+}\frac{q^4}{1+q^5+}\frac{q^6}{1+q^7+}\ldots\textrm{, }\;|q|<1 $$ then $$ H(e^{-x})=e^{-x/2}\frac{\vartheta_4(3ix/2,\,e^{-4x})}{\vartheta_4(ix/2,\,e^{-4x})}\textrm{, }\;x>0 $$

Tito Piezas III
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Nikos Bagis
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  • +1 This is very nice. I tested it with Mathematica and they equate. Do you know if the cubic cfrac and the octic cfrac can be similarly expressed in terms of theta quotients? – Tito Piezas III Oct 29 '16 at 02:55
  • Can you look at this MO post about $x^5+y^5 = 1$? There are some functions $a,b,c,d$ there and the ratio $a/b$ can be expressed by Jacobi theta functions. Can you check if we can do so for the ratio $d/b$ as well? – Tito Piezas III Jan 21 '17 at 08:07
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I'm sorry about the octic cfrac. Yes it seems to be non trivial. In Berndt's book (see: B.C. Berndt. "Ramanujan Notebooks III". Springer Verlag, New York (1991), In chapter 19 Entry 1) one can find the expansion $$ M(q)=\tfrac{1}{2}\,q^{1/8}\frac{\sum^{\infty}_{n=-\infty}q^{n^2/2+n/2}}{\sum^{\infty}_{n=-\infty}q^{n^2}}=\frac{q^{1/8}}{1+}\frac{q}{1+q+}\frac{q^2}{1+q^2+}\frac{q^3}{1+q^3+}\ldots $$ Hence if $|q|<1$ and $$ \vartheta_3(z,q):=\sum^{\infty}_{n=-\infty}q^{n^2}e^{2niz} $$ For $x>0$ we have $$ M(e^{-x})=\tfrac{1}{2}\,e^{-x/8}\frac{\vartheta_3(ix/4,e^{-x/2})}{\vartheta_3(0,e^{-x})} $$ Equivalently, let $q = e^{2\pi i z}$, then $$M(q) = \tfrac{1}{2}\,q^{1/8} \frac{\vartheta_3\big(i\ln(q^{1/4}),q^{1/2}\big)}{\vartheta_3(0,q)}=\frac{\eta(z)\,\eta^2(4z)}{\eta^3(2z)}$$ where $\eta(z)$ is the Dedekind eta function.

Tito Piezas III
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Nikos Bagis
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For the cubic $V(q)$ cfrac is $$ V(q):=\frac{q^{1/3}}{1+}\frac{q+q^2}{1+}\frac{q^2+q^4}{1+}\frac{q^3+q^6}{1+}\ldots\textrm{, }|q|<1 $$ and $$ V(e^{-x})=e^{-x/3}\frac{\vartheta_4(ix/4,e^{-3x})}{\vartheta_4(0,e^{-3x})}\textrm{, }x>0 $$ See more formulas in https://arxiv.org/pdf/1107.2393v2.pdf

Nikos Bagis
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Set $T=\sqrt{1-8V(q)^3}$, $q=e^{-\pi\sqrt{r}}$, $r>0$, then $$ k_r^2=\frac{(1-T)(3+T)^3}{(1+T)(3-T)^3} $$ where $k_r$ is the elliptic singular modulus. (see https://arxiv.org/pdf/1008.1304v2.pdf)

Nikos Bagis
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