# A new $q$-continued fraction of order $12$

I think I may have discovered a $q$-continued fraction of order $12$ with a form different from that established by Mahadeva Naika.

Let $q=e^{2i \pi \tau}=\exp(2i \pi \tau)$, then,

\begin{aligned} N(q) &= q\,\dfrac{f(-q,-q^{11})}{f(-q^5,-q^7)} = q\prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\\[1.5mm] &= \dfrac{q(1-q)} {1+q^3-\dfrac{q^3(1+q^2)(1+q^4)} {1+q^9+\dfrac{q^6(1-q^5)(1-q^7)} {1+q^{15}-\dfrac{q^9(1+q^8)(1+q^{10})} {(1+q^{21})+\ddots }}}} \end{aligned} where $f(a,b)=\sum_{-\infty}^{\infty}a^{(n(n+1)/2}b^{n(n-1)/2}$ is Ramanujan's general theta function.

Does it look familiar? If so please let me know or better yet show me your methods on how you arrived at it.

• Is my reformatting correct? Also, please define $f$. Jul 5, 2015 at 21:38
• Yes thanks, your reformatting is correct sir, the function f(a,b) is the ramanujan theta function Jul 5, 2015 at 22:05
• @Nicco: I made some improvement to the details. It's a nice cfrac, by the way. Jul 6, 2015 at 3:08

I tested your cfrac with the order 12 discussed by Naika (which in turn is a special case of a general cfrac by Ramanujan) and labeled as $$D_1(q)$$ here,

$$D_1(q)= \dfrac{q(1-q)} {1-q^3+\dfrac{q^3(1-q^2)(1-q^4)} {(1-q^3)(1+q^6)+\dfrac{q^3(1-q^8)(1-q^{10})} {(1-q^3)(1+q^{12})+\dfrac{q^3(1-q^{14})(1-q^{16})} {(1-q^3)(1+q^{18})+\ddots }}}}$$

and as far as I can tell (numerically at least) your cfrac is indeed,

$$N(q) = D_1(q)$$

However, yours seem to converge faster. Also, since,

$$\frac{1}{D_1(q)}+D_1(q) = \frac{1}{C(q)C(q^2)}$$

where $$C(q)$$ is Ramanujan's cubic continued fraction, then we should be able to evaluate your cfrac as algebraic numbers. For example, I find,

\begin{align} \frac{1}{N(e^{-2\pi})}+N(e^{-2\pi}) &= \frac{4}{1-\sqrt{3\big(3+\sqrt3-3^{3/4}\sqrt{2+\sqrt3}\big)}}\\ &= \frac{\;8}{\;\sqrt{2}\,(3+ \sqrt{2}) - \sqrt[4]{3}\,(3+\sqrt{3})}\\[4pt] &= 536.4953904\dots\end{align}

• Thanks for such a satisfying answer Mr Piezas. In fact I have discovered many more continued fractions of this kind which I'll be posting here soon. If anyone has an answer to the question above may feel free to post it. Jul 6, 2015 at 12:29
• @Nicco: I think I've figured out the general form of your continued fraction as extended to other orders. However, I am not able to prove that it is equivalent to the one studied by Ramanujan and Naika, so I asked it in the forum for professional mathematicians. See here. I'm also curious how you found these variants in the first place. Jul 6, 2015 at 15:40