Articles of q series

$q$-series identity

I have to prove the following identity: $$\sum_{n\geq 0} (-1)^n(2n+1)q^{\frac{n(n+1)}{2}} = (q;q)_\infty^3$$ where $(a;q)_\infty = \prod_{i\geq 0}1-aq^i$ is the $q$-Pochhammer symbol. In my notes the identity is stated as a corollary of Jacobi triple product, whose statement is $$\sum_{n\in\mathbb Z} z^nq^{\frac{n(n+1)}{2}} = (-zq;q)_\infty(-1/z;q)_\infty(q;q)_\infty$$ What I have done so far is to change the range of […]

An expression for $U_{h,0}$ given $U_{n,k}=\frac{c^n}{c^n-1}(U_{n-1,k+1})-\frac{1}{c^n-1}(U_{n-1,k})$

Let $c\in\mathbb{R}\setminus\{ 1\}$, $c>0$. Let $U_i = \left\lbrace U_{i, 0}, U_{i, 1}, \dots \right\rbrace$, $U_i\in\mathbb{R}^\mathbb{N}$. We know that $U_{n+1,k}=\frac{c^{n+1}}{c^{n+1}-1}U_{n,k+1}-\frac{1}{c^{n+1}-1}U_{n,k}$. (As @TedShifrin pointed out, it can also be written $U_{n+1,k}=U_{n,k+1}+\frac{1}{c^{n+1}-1}\left(U_{n,k+1}-U_{n,k}\right)$) (obviously it implies that if $\lvert U_k \rvert=n$ then $\lvert U_{k+1}\rvert=n-1$ etc) Here is what I conjectured: $$\forall h\in\mathbb{N}, U_{h,0}=\sum\limits_{p=0}^h\frac{c^{\frac{p^2+p}{2}}}{\left(\prod\limits_{i=1}^p\left(c^i-1\right)\right)\prod\limits_{i=1}^{h-p}\left(1-c^i\right)}U_{0,p}$$ I tested it for some values […]

a q-continued fraction related to the octahedral group

Let $q=e^{2\pi i\tau}$. If $u(\tau)$ is Ramanujan’s octic continued fraction, $$u(\tau)=\cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$ is it true that the generator of the octahedral group is the continued fraction, $$\big(u(2\tau)\big)^2=\cfrac{2\,q^{1/2}}{1-q+\cfrac{q(1+q)^2}{1-q^3+\cfrac{q^2(1+q^2)^2}{1-q^5+\cfrac{q^3(1+q^3)^2}{1-q^7+\ddots}}}}$$ for $|q|\lt 1$?

Closed form of the integral ${\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx$

While doing some numerical experiments, I discovered a curious integral that appears to have a simple closed form: $${\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx\stackrel{\color{gray}?}=\frac{\pi^2}{6\sqrt3}\tag1$$ Could you suggest any ideas how to prove it? The infinite product in the integrand can be written using q-Pochhammer symbol: $$\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)=\left(e^{-24\!\;x};\,e^{-24\!\;x}\right)_\infty\tag2$$

Ramanujan theta function and its continued fraction

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 […]

Why $e^{\pi}-\pi \approx 20$, and $e^{2\pi}-24 \approx 2^9$?

This was inspired by this post. Let $q = e^{2\pi\,i\tau}$. Then, $$x := \left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24} = \frac{1}{q} – 24 + 276q – 2048q^2 + 11202q^3 – 49152q^4+ \cdots\tag1$$ where $\eta(\tau)$ is the Dedekind eta function. For example, let $\tau =\sqrt{-1}$ so $x = 2^9 =512 $ and $(1)$ “explains” why, $$x = 512 \approx e^{2\pi}-24 = […]

The ratio of jacobi theta functions

Let $q=e^{2\pi i\tau}$. If $\theta_2$ and $\theta_3$ are jacobi theta functions , is it true that the ratio of the two functions can be expressed as a continued fraction of the form $$ \frac{\theta_2(q^2)}{\theta_3(q^2)}=2q^{1/2}\prod_{n=1}^\infty \frac{(1+q^{4n})^2}{(1+q^{4n-2})^2}=\cfrac{2q^{1/2}}{1-q+\cfrac{q(1+q)^2}{1-q^3+\cfrac{q^2(1+q^2)^2}{1-q^5+\cfrac{q^3(1+q^3)^2}{1-q^7+\ddots}}}} $$ for $|q|\lt 1$?

the ratio of jacobi theta functions and a new conjectured q-continued fraction

Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$, define $$\begin{aligned}H(q)=\cfrac{2(1+q^2)}{1-q+\cfrac{(1+q)(1+q^3)}{1-q^3+\cfrac{2q^2(1+q^4)}{1-q^5+\cfrac{q^3(1+q)(1+q^5)}{1-q^7+\cfrac{q^4(1+q^2)(1+q^6)}{1-q^9+\cfrac{q^5(1+q^3)(1+q^7)}{1-q^{11}+\ddots}}}}}}\end{aligned}$$ Question: How do we show that $$H(q)\overset{\color{red}{?}}=\frac{2\,q^{1/2}\,\vartheta_3(0,q^2)}{\vartheta_2(0,q^2)}$$ where $\vartheta_n(0,q)$ are jacobi theta functions

Motivation for/history of Jacobi's triple product identity

I’m taking a short number theory course this summer. The first topic we covered was Jacobi’s triple product identity. I still have no sense of why this is important, how it arises, how it might have been discovered, etc. The proof we studied is a bit on the clever side for my taste, giving me […]

a conjecture of certain q-continued fractions

Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$, define, $$\begin{aligned}F(q)=\cfrac{1-q^2}{1-q^3+\cfrac{q^3(1-q)(1-q^5)}{1-q^9+\cfrac{q^6(1-q^4)(1-q^8)}{1-q^{15}+\cfrac{q^9(1-q^7)(1-q^{11})}{1-q^{21}+\ddots}}}}\overset{\color{red}{?}}=\prod_{n=1}^\infty\frac{(1-q^{12n-2})(1-q^{12n-10})}{(1-q^{12n-4})(1-q^{12n-8})}\\[1.5mm]&\end{aligned}$$ and $$\begin{aligned}G(q)=\cfrac{1-q^4}{1-q^3+\cfrac{q^3(1-\frac{1}{q})(1-q^7)}{1-q^9+\cfrac{q^6(1-q^2)(1-q^{10})}{1-q^{15}+\cfrac{q^9(1-q^5)(1-q^{13})}{1-q^{21}+\ddots}}}}\overset{\color{red}{?}}=\prod_{n=1}^\infty\frac{(1-q^{12n-4})(1-q^{12n-8})}{(1-q^{12n-2})(1-q^{12n-10})}\\[1.5mm]&\end{aligned}$$ Q: How do we prove rigorously that the two q-continued fractions are equal to the q-series? (if true, then the two q-continued fractions are reciprocals such that their product is unity, $F(q)\,G(q)=1$.)