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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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  • $\begingroup$ The answer is in W.Duke's paper continued fractions and modular functions,equation (4.9) $\endgroup$ – Nicco Sep 7 '15 at 11:34
  • $\begingroup$ @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.) $\endgroup$ – Tito Piezas III Sep 7 '15 at 11:44
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    $\begingroup$ @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. $\endgroup$ – Tito Piezas III Sep 7 '15 at 13:40
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    $\begingroup$ 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. $\endgroup$ – Tito Piezas III Sep 9 '15 at 3:34
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    $\begingroup$ 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. $\endgroup$ – ccorn Oct 23 '15 at 1:27
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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 $n=0,1,2$.

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    $\begingroup$ Nice indeed. If there were a 2nd volume of Whittaker & Watson, this one would make it to the legendary "Shew that..." exercises. $\endgroup$ – 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 $$ H(e^{-x})=e^{-x/2}\frac{\vartheta_4(3ix/2,\,e^{-4x})}{\vartheta_4(ix/2,\,e^{-4x})}\textrm{, }\;x>0 $$

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  • $\begingroup$ +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? $\endgroup$ – Tito Piezas III Oct 29 '16 at 2:55
  • $\begingroup$ 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? $\endgroup$ – Tito Piezas III Jan 21 '17 at 8:07
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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

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  • $\begingroup$ +1 Great! Now all that is missing is Ramanujan's octic cfrac $\endgroup$ – Tito Piezas III Nov 5 '16 at 3:37
  • $\begingroup$ The octic cfrac of Ramanujan is just the "elementary" continued fraction of Euler and can be evaluated (with theta functions) in a very simply way. $\endgroup$ – Nikos Bagis Nov 5 '16 at 4:30
  • $\begingroup$ Kindly include so this list will be complete. $\endgroup$ – Tito Piezas III Nov 5 '16 at 4:32
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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.

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  • $\begingroup$ I've added some info to connect it to the $\eta(z)$ function. $\endgroup$ – Tito Piezas III Nov 6 '16 at 15:03
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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)

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