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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.

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

1 Answer 1

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(A partial answer.)

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}

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  • $\begingroup$ 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. $\endgroup$
    – Nicco
    Jul 6, 2015 at 12:29
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    $\begingroup$ @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. $\endgroup$ Jul 6, 2015 at 15:40

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