I believe Ramanujan would have loved this kind of identity. After deriving the identity, I wanted to share it with the mathematical community. If it's well known, please inform me and give me some links to it. Let $q=e^{2\pi\mathrm{i}\tau}$, then $$(1+q^{2}+q^{6}+q^{12}+q^{20}+q^{30}+\cdots)^{2}=\cfrac{1}{1+q-\cfrac{q(1+q)^2}{1+q^3+\cfrac{q^2(1-q^2)^2}{1+q^5-\cfrac{q^3(1+q^3)^2}{1+q^7+\ddots}}}}$$ for $|q|\lt1$. If possible, please provide more examples of this nature, available in the literature.

  • $\begingroup$ This is now the third continued fraction of this kind. The others: (1), (2). $\endgroup$ – ccorn Aug 7 '15 at 9:58
  • $\begingroup$ How did you derive it? You might find this problem interesting $\endgroup$ – Winther Aug 7 '15 at 9:59
  • $\begingroup$ Thanks for the links gentlemen,but all of those continued fractions were established by me.I will post my derivations soon,but as a self-taught mathematician it may be hard to follow my methods,being unrigorous perhaps. $\endgroup$ – Nicco Aug 7 '15 at 10:44
  • 2
    $\begingroup$ I second @Winther in that the derivation, or its main ingredients, ought to be given. Otherwise this question will probably share the same fate as the related questions linked above. $\endgroup$ – ccorn Aug 7 '15 at 10:46
  • 1
    $\begingroup$ The left-hand side does not change under the substitution $q\mapsto-q$, and the right-hand side does not need alternating signs then. I'd therefore prefer this form: $$(1+q^{2}+q^{6}+q^{12}+q^{20}+q^{30}+\cdots)^{2}=\cfrac{1}{1-q+{\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2}{1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\ddots}}}}}$$ $\endgroup$ – ccorn Aug 7 '15 at 10:51

To clarify, what you found is a q-continued fraction for the Jacobi theta function $\vartheta_2(0,q)$. Using ccorn's formulation,

$$\left(\frac{\vartheta_2(0,q)}{2\,q^{1/4}}\right)^2 =\Big(\sum_{n=0}^\infty q^{n(n+1)}\Big)^2 =\cfrac{1}{1-q+\cfrac{q(1\color{red}-q)^2}{1-q^3+\cfrac{q^2(1\color{red}-q^2)^2}{1-q^5+\cfrac{q^3(1\color{red}-q^3)^2}{1-q^7+\ddots}}}}\tag1$$

One can compare this to your other cfrac in this post,


They are beautifully similar, differing only in the $\pm$ within the square, though these two identities are not yet rigorously proven. (Update: The second was already established in 2005 by Michael Somos as the cfrac for sequence $A079006$ discussed in this answer.)

P.S. Ramanujan's octic cfrac can also express $(2)$, but I am unsure if there are q-cfracs for any of the $\vartheta_n(0,q)$. (I believe there are, but I'll have to go through my notes.)

  • $\begingroup$ Using ccorn's formulation ,I was able to show that the cfrac is a q-analogue of gauss's cfrac for pi,kindly see my answer here $\endgroup$ – Nicco Aug 20 '15 at 15:23
  • $\begingroup$ @Nicco: This is not really a Ramanujan theta function; Jacobi found it first. But you right, he would have liked your cfracs. :) $\endgroup$ – Tito Piezas III Aug 20 '15 at 15:32
  • $\begingroup$ @Nicco: What would be great is if you can find a $q$-cfrac for $\vartheta_3(0,q)$ as well. Then the ratio of two $q$-cfracs would also be a $q$-cfrac. $\endgroup$ – Tito Piezas III Aug 21 '15 at 11:55
  • $\begingroup$ @ Tito PiezasIII it has already been found by EIsenstein,see here $\endgroup$ – Nicco Aug 21 '15 at 12:35
  • $\begingroup$ @Nicco: Thanks. I now recall coming across Eisenstein's general cfrac before, and it must be somewhere in my notes. But too bad the ones for $\vartheta_3^k$ don't look similar to yours. Can you modify it so it will look similar? $\endgroup$ – Tito Piezas III Aug 21 '15 at 13:03

I refer to your claim with the sign of $q$ adjusted so that it reads $$\small(1+q^{2}+q^{6}+q^{12}+q^{20}+q^{30}+\cdots)^{2} =\cfrac{1}{1-q+{\cfrac{q\,(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2} {1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\cdots}}}}}$$ Given

  1. a related answer introducing a continued fraction formula by Ramanujan with parameters $a,b,q$ and making use of some Jacobi thetanull properties,
  2. plus another answer where the formula and the thetanull stuff is applied again with other settings of $a$ and $b$,
  3. and yet another one where we found and applied the insight that whenever $ab=q$, the result can be simplified along the same lines as before, this time using the two-argument Jacobi theta functions,

you know what comes next.

But it does not get boring: This one comes with another twist and a slight beautification.

As usual in posts like these, I write $q_n = \exp\frac{2\pi\mathrm{i}\tau}{n}$, thus $q_n^n=q$, and I consider expressions with $q_n$ implicity as functions of $\tau$.

I will again use Ramanujan's formula $$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty} = \cfrac{a+b}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)} {1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}}\tag{*}$$ If you look at formula $(2)$ in my previous answer, you may notice that, by pulling out the first factor of the $q$-Pochhammer symbol containing $q^{-1}$, the identity can be written in the fully symmetric form $$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty} = (a+b)\,\frac{(-a^2q^3;q^4)_\infty\,(-b^2q^3;q^4)_\infty} {(-a^2q;q^4)_\infty\,(-b^2q;q^4)_\infty} \qquad (ab=q) \tag{**}$$ Combining $(*)$ and $(**)$, you get the following formula, restricted to the case $ab=q$: $$\small\cfrac{1}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)} {1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}} = \frac{(-a^2q^3;q^4)_\infty\,(-b^2q^3;q^4)_\infty} {(-a^2q;q^4)_\infty\,(-b^2q;q^4)_\infty} \tag{***}$$ Note that the factor $(a+b)$ has been cancelled from both sides. This is important because now we are going to use it for $a+b=0$ by continuity.

Concretely, set $a=-\mathrm{i}q_2$, $b=\mathrm{i}q_2$, so $ab=q$ and $a/b=-1$. This yields $$\begin{align} \cfrac{1}{1-q+\cfrac{q\,(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2} {1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\cdots}}}} &\stackrel{(***)}{=} \frac{(q^4;q^4)_\infty^2}{(q^2;q^4)_\infty^2} \\ &\stackrel{(P1)}{=} (q^4;q^4)_\infty^2\,(-q^2;q^2)_\infty^2 \\ &\stackrel{(P2)}{=} (q^2;q^2)_\infty^2\,(-q^2;q^2)_\infty^4 \\ &\stackrel{(T1)}{=} \left(\frac{\vartheta_2(0\mid2\tau)}{2q_4}\right)^2 \\ &\stackrel{(T2)}{=} \left(\sum_{n=0}^\infty q^{n\,(n+1)}\right)^2 \end{align}$$ where $(\mathrm{P1})$ and $(\mathrm{P2})$ are $q$-Pochhammer symbol manipulation rules like $$\begin{align} (-q;q)_\infty\,(q;q^2)_\infty &= 1 \tag{P1} \\ (-q;q)_\infty\,(q;q)_\infty &= (q^2;q^2)_\infty \tag{P2} \end{align}$$ and $(\mathrm{T1})$ refers to the product representation of $\vartheta_2$: $$\vartheta_2(0\mid\tau) = 2q_8\,(-q;q)_\infty^2\,(q;q)_\infty \tag{T1}$$ while $(\mathrm{T2})$ describes the series representation $$\vartheta_2(0\mid\tau) = \sum_{k\in\mathbb{Z}} q_8^{(2k+1)^2} = 2q_8\sum_{n=0}^\infty q^{n\,(n+1)/2} \tag{T2}$$ which is linked with $(\mathrm{T1})$ by the triple product identity. Those ingredients are the same as for the other answers. Nothing new here.

That's it. Enjoy the formulae with more continued fractions of that sort.

  • $\begingroup$ This automatically solves the conjecture in this post $\endgroup$ – Nicco Oct 3 '15 at 18:47
  • $\begingroup$ @Nicco: Indeed. In that MO post, one would just have to flip the signs of both $a$ and $q$ to get the expressions used here. The condition $ab=q$ is preserved under that substitution. $\endgroup$ – ccorn Oct 3 '15 at 21:55
  • $\begingroup$ Ccorn:Surprisingly ,Bruce C Berndt once commented via e-mail that the symmetric form of the cfrac had not been previously observed. $\endgroup$ – Nicco Oct 3 '15 at 22:27
  • $\begingroup$ According to [ABBW85], in Ramanujan's 2nd notebook, chapter 16, entry 11, the sign of $b$ was flipped, making the expressions skew-symmetric. Entry 12 was then the cfrac mentioned in Tito's MO post. This leaves me with the impression that Ramanujan was aware of the possibility to specialize entry 11 to a simpler theta quotient, but sought something more general than just $ab=q$, and entry 12 was the result. I like entry 11 more though, particularly in the symmetric simplified form here for $ab=q$: The corresponding power series for the numerator and denominator converge to interesting series. $\endgroup$ – ccorn Oct 3 '15 at 22:38
  • $\begingroup$ @ ccorn :it's also of particular importance to note the equality $\frac{\sum_{n=-\infty}^{\infty} (a^{n}b^{n+1})^{2n+1}}{\sum_{n=-\infty}^{\infty} a^{n(2n+1)}b^{n(2n-1)}}=(a+b)\frac{(-a^5b^3;(ab)^4)_\infty\,(-a^3b^5;(ab)^4)_\infty} {(-ab^3;(ab)^4)_\infty\,(-a^3b;(ab)^4)_\infty}$ $\endgroup$ – Nicco Oct 12 '17 at 7:36

In a paper of me and Professor M.L. Glasser [1] we have generalized the continued fractions regarding this conversation. Namely

If $|q|<1$ and we set $$ (a;q)_{\infty}=\prod^{\infty}_{n=0}\left(1-aq^n\right),\tag 1 $$ then $$ \left(\frac{(-a;q)_{\infty}}{(a;q)_{\infty}}\right)^2=-1+\frac{2}{1-}\frac{2a}{1-q+}\frac{a^2(1+q)^2}{1-q^3+}\frac{a^2q(1+q^2)^2}{1-q^5+}\frac{a^2q^2(1+q^3)^2}{1-q^7+}\ldots,\tag 2 $$ $$ \sum^{\infty}_{n=0}\frac{q^n}{1-a^2q^{2n}}=\frac{1}{1-q+}\frac{-a^2(1-q)^2}{1-q^3+}\frac{-qa^2(1-q^2)^2}{1-q^5+}\frac{-q^2a^2(1-q^3)^2}{1-q^7+}\ldots,\tag 3 $$ where $a$ is a complex number.

Hence for example

1) If we set $a=iq$ in (3) and using [2] pg.17, pg.55: $$ cn(u)=\frac{2\pi}{Kk}\sum^{\infty}_{n=0}\frac{q^{n+1/2}}{1+q^{2n+1}}\cos((2n+1)z) $$ where $z=\frac{\pi}{2K}u$ and $cn(0)=1$, we get $$ \sum^{\infty}_{n=0}\frac{q^n}{1+q^{2n+1}}=\frac{1}{1-q+}\frac{q(1-q)^2}{1-q^3+}\frac{q^2(1-q^2)^2}{1-q^5+}\frac{q^2(1-q^3)^2}{1-q^7+}\ldots= $$ $$ =q^{-1/2}\frac{Kk}{2\pi}=\left(\frac{\theta_2(q)}{2q^{1/4}}\right)^2=\left(\sum^{\infty}_{n=0}q^{n(n+1)}\right)^2, $$ where $\theta_2(q)=\sum^{\infty}_{n=-\infty}q^{(n+1/2)^2}=\sqrt{2Kk/\pi}$, $q=e^{-\pi\sqrt{r}}$, $r>0$.

2) From [3] pg.37 we have $$ \theta_4(q)=\sum^{\infty}_{n=-\infty}(-1)^nq^{n^2}=\frac{(q;q)_{\infty}}{(-q,q)_{\infty}}. $$ Using (2) we arive to $$ \left(\sum^{\infty}_{n=-\infty}(-1)^nq^{n^2}\right)^{-2}=-1+\frac{2}{1-}\frac{2q}{1-q+}\frac{q^2(1+q)^2}{1-q^3+}\frac{q^3(1+q^2)^2}{1-q^5+}\frac{q^4(1+q^3)^2}{1-q^7+}\ldots $$

3) For $q\rightarrow q^2$ and $a=q$, we get $$ S_1=\left(\frac{(-q;q^2)_{\infty}}{(q;q^2)_{\infty}}\right)^2= -1+\frac{2}{1-}\frac{2q}{1-q^2+}\frac{q^2(1+q^2)^2}{1-q^6+}\frac{q^4(1+q^4)^2}{1-q^{10}+}\frac{q^6(1+q^6)^2}{1-q^{14}+}\ldots\tag 4 $$ we also get $$ S_2=\left(\frac{(q;q^2)_{\infty}}{(-q;q^2)_{\infty}}\right)^2=\frac{1}{S_1}= $$ $$ =-1+\frac{2}{1+}\frac{2q}{1-q^2+}\frac{q^2(1+q^2)^2}{1-q^6+}\frac{q^4(1+q^4)^2}{1-q^{10}+}\frac{q^6(1+q^6)^2}{1-q^{14}+}\frac{q^8(1+q^8)^2}{1-q^{18}+}\ldots\tag 5 $$ But $\chi(q)=(-q;q^2)_{\infty}$, $\psi(q)=\sum^{\infty}_{n=0}q^{n(n+1)/2}=\frac{(q^2;q^2)_{\infty}}{(q;q^2)_{\infty}}$, $f(-q)=(q;q)_{\infty}$ (note that $\chi(-q)=\frac{\theta_4(q)}{f(-q)}$). Moreover if $k'_r=\sqrt{1-k_r^2}$ is the complementary elliptic singular modulus, then from [3] chapter 16, Entry 24, pg. 39 we have the following cfrac expansions: $$ S_1=\frac{\chi(q)}{\chi(-q)}=\frac{f(q)}{f(-q)}=\frac{\psi(q)}{\psi(-q)}=\sqrt{\frac{\theta_3(q)}{\theta_4(q)}}=\frac{1}{\sqrt[4]{k'_r}} $$

Note. $$ k_r=\sqrt{m(q)}=4q^{1/2}\exp\left(-4\sum^{\infty}_{n=1}q^n\sum_{d|n}\frac{(-1)^{d+n/d}}{d}\right), $$
where $q=e^{-\pi\sqrt{r}}$, $r>0$.



[1]: N.D. Bagis and M.L. Glasser. "Evaluations of a Continued Fraction of Ramanujan". Rend. Sem. Mat. Univ. Padova, Vol. 133 (2015). (submited 2013)

[2]: J.V. Armitage W.F. Eberlein. 'Elliptic Functions'. Cambridge University Press, (2006).

[3]: Bruce C. Berndt. 'Ramanujan`s Notebooks Part III'. Springer Verlag, New York, (1991).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.