I have two questions about q-continued fractions, but a little intro first. Given Ramanujan's theta function,

$$f(a,b) = \sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}$$

then the following, $$A(q) = q^{1/8} \frac{f(-q,-q^3)}{f(-q^2,-q^2)}$$

$$B(q) = q^{1/5} \frac{f(-q,-q^4)}{f(-q^2,-q^3)}$$

$$C(q) = q^{1/3} \frac{f(-q,-q^5)}{f(-q^3,-q^3)}$$

$$D(q) = q^{1/2} \frac{f(-q,-q^7)}{f(-q^3,-q^5)}$$

$$E(q) = q^{1/1} \frac{f(-q,-q^{11})}{f(-q^5,-q^7)}$$

are q-continued fractions of degree $4,5,6,8,12$, respectively, namely,

$$A(q) = \cfrac{q^{1/8}}{1 + \cfrac{q}{1+q + \cfrac{q^2}{1+q^2 + \ddots}}},\;\;B(q) = \cfrac{q^{1/5}}{1 + \cfrac{q}{1 + \cfrac{q^2}{1 + \ddots}}}$$

$$C(q) = \cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \ddots}}},\;\;\;\;D(q) = \cfrac{q^{1/2}}{1 + q +\cfrac{q^2}{1+q^3 + \cfrac{q^4}{1+q^5 + \ddots}}}$$

$$E(q) = \cfrac{q(1-q)}{1-q^3 + \cfrac{q^3(1-q^2)(1-q^4)}{(1-q^3)(1+q^6)+\cfrac{q^3(1-q^8)(1-q^{10})}{(1-q^3)(1+q^{12}) + \ddots}}}$$

The first three are by Ramanujan, the fourth is the Ramanujan-Gollnitz-Gordon cfrac, while the last is by Naika, et al (using an identity by Ramanujan). Let $q = e^{2\pi i \tau}$ where $\tau = \sqrt{-n}$ and these can be simply expressed in terms of the Dedekind eta function $\eta(\tau)$ as,

$$\tfrac{1}{A^4(q)}+16A^4(q) = \left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8$$

$$\tfrac{1}{B(q)}-B(q) = \left(\tfrac{\eta(\tau/5)}{\eta(5\tau)}\right)+1$$

$$\tfrac{1}{C(q)}+4C^2(q) = \left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3$$

$$\tfrac{1}{D(q)}-D(q) = \big(\tfrac{1}{A(q^2)}\big)^2$$

$$E(q) = \;???$$

Question 1: Does anybody know how to express $E(q)$ in terms of $\eta(\tau)$? (It's SO frustrating not to complete this list. I believe there might be a simple relationship between orders 6 and 12, just like there is between 4 and 8.) This cfrac can be found in "On Continued Fraction of Order 12", but the authors do not address this point.

Question 2: Excluding these five and the Heine cfrac which gives $\eta(\tau)/\eta(2\tau)$, are there any other q-continued fractions which yield an algebraic value at imaginary arguments?


2 Answers 2


[Edited again to give a second identity relating $E$ to eta products]

Continued fraction or not, an expression $q^{\frac{(r-s)^2}{8(r+s)}} f(\pm q^r, \pm q^s)$ is a modular form of weight $1/2$ for all integers $r,s$ with $r+s>0$, because it is a sum $\sum_{n=-\infty}^\infty \pm q^{(cn+d)^2}$ with rational $c,d$ and periodic signs. Therefore the quotient of two such expressions is a modular function, and takes algebraic valus at quadratic imaginary values.

The quotient $$ E(q) = q - q^2 + q^6 - q^7 + q^8 - q^9 + q^{11} - 2q^{12} + 2q^{13} - 2q^{14} + 2q^{15} \cdots $$ looks like a modular unit $-$ its logarithmic derivative has small coefficients $-$ but not quite an eta product; instead it seems to be a quotient of Klein forms: $$ E(q) = q \prod_{n=1}^\infty (1-q^n)^{\chi(n)}, $$ where $\chi$ is the Dirichlet character of conductor $12$, given by

$$ \chi(n) = \cases{ +1,& if $n \equiv \pm 1 \bmod 12$; \cr -1,& if $n \equiv \pm 5 \bmod 12$; \cr 0,& otherwise. } $$ Two identities relating $E$ to $\eta$ products, similar to but somewhat more complicated than the ones you give for $A,B,C,D,$ are $$ \frac1{E(q)} - E(q) = \frac{\eta(2\tau)^2 \eta(6\tau)^4}{\eta(\tau)\eta(3\tau)\eta(12\tau)^4}, $$ and (a bit simpler) $$ \frac1{E(q)} + E(q) = \frac{\eta(4\tau)}{\eta(\tau)} \Bigl(\frac{\eta(3\tau)}{\eta(12\tau)}\Bigr)^3. $$

  • $\begingroup$ Thanks, Dr. Elkies. It is good to finally know a relation of $E(q)$ to an eta product! However, it may be possible to simplify this considering the other cfracs also have more complicated expressions. For example, in Duke's "Continued Fractions and Modular Functions", we have $\frac{1}{D^2(q)}+D^2(q) = \frac{\eta(4\tau)^2\eta(\tau)^4}{\eta(2\tau)^2\eta(8\tau)^4}+6$. I'll see if I can tweak your relation to find a reasonably simple one between $C(q)$ and $E(q)$. $\endgroup$ Oct 6, 2012 at 15:55
  • $\begingroup$ The $E^{-1} - E$ relation might not simplify much, but the one I just added for $E^{-1} + E$ might come closer to what you're hoping for. $\endgroup$ Oct 7, 2012 at 21:05
  • $\begingroup$ [It seems that meanwhile Somos also called attention to $E^{-1} + E$...] $\endgroup$ Oct 7, 2012 at 21:07

Based on Elkies' answer and an email by Michael Somos, we can give an alternative expression to my Question 1. If a sum is used, instead of a difference,



$$\frac{u(u-4)^3}{(u-1)^3} = \left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8$$

Or more simply, we can relate it to the cubic continued fraction $C(q)$ as,

$$u=\frac{1}{E(q)}+E(q) =\frac1{C(q)\,C(q^2)}$$

Since $\displaystyle C(q)=\frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}$, this implies,

$$u= \frac{\eta(4\tau)\,\eta^3(3\tau)}{\eta(\tau)\,\eta^3(12\tau)}$$

an eta quotient also mentioned by Elkies in his answer.


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