Questions tagged [q-series]

Questions that are based on, use, or include the q-series in their content or solutions.

100 questions
28
votes
4 answers

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…
dfeuer
  • 8,689
  • 3
  • 31
  • 59
24
votes
2 answers

a new continued fraction for $\sqrt{2}$

In a q-continued fraction related to the octahedral group I defined a new q-continued fraction for the square of ramanujan's octic continued fraction which I discovered using certain three term relations and algebraic manipulations. Given…
Nicco
  • 2,763
  • 14
  • 32
22
votes
0 answers

A curious identity on $q$-binomial coefficients

Let's first recall some notations: The $q$-Pochhammer symbol is defined as $$(x)_n = (x;q)_n := \prod_{0\leq l\leq n-1}(1-q^l x).$$ The $q$-binomial coefficient (also known as the Gaussian binomial coefficient) is defined as $$\binom{n}{k}_q :=…
Henry
  • 2,865
  • 13
  • 32
18
votes
1 answer

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, $$\alpha(\tau) = \left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24} = \frac{1}{q} - 24 + 276q - 2048q^2 + 11202q^3 - 49152q^4+ \cdots\tag1$$ where $\eta(\tau)$ is the Dedekind eta…
Tito Piezas III
  • 47,981
  • 5
  • 96
  • 237
16
votes
3 answers

Combinatorial interpretation of this identity of Gauss?

Gauss came up with some bizarre identities, namely $$ \sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}=\prod_{k\geq 1}\frac{1-q^k}{1+q^k}. $$ How can this be interpreted combinatorially? It strikes me as being similar to many partition identities. Thanks.
Kally
  • 893
  • 4
  • 11
12
votes
1 answer

Show that $\prod\limits_{n=1}^\infty \frac{(1-q^{6n})(1-q^n)^2}{(1-q^{3n})(1-q^{2n})}=\sum\limits_{n=-\infty}^\infty q^{2n^2+n}-3q^{9(2n^2+n)+1}$.

Show that $\displaystyle \prod_{n=1}^\infty \frac{(1-q^{6n})(1-q^n)^2}{(1-q^{3n})(1-q^{2n})}=\sum_{n=-\infty}^\infty q^{2n^2+n}-3q^{9(2n^2+n)+1}$. I can't seem to be able to proceed with this question. I know that $\displaystyle \prod_{n=1}^\infty…
Icycarus
  • 1,112
  • 1
  • 7
  • 20
12
votes
3 answers

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 $q=e^{2\pi\mathrm{i}\tau}$,…
10
votes
2 answers

Show $1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$.

Show that $$1+\frac{8q}{1-q}+\frac{16q^2}{1+q^2}+\frac{24q^3}{1-q^3}+\dots=1+\frac{8q}{(1-q)^2}+\frac{8q^2}{(1+q^2)^2}+\frac{8q^3}{(1-q^3)^2}+\dots$$ where $|q|<1$ (q can be complex number). The hint is to convert the left side to a double…
10
votes
0 answers

The $q$-continued fraction for tribonacci constant and others

Let $q = e^{-2\pi}$. We are familiar with Ramanujan's beautiful continued fraction, $$\cfrac{q^{1/5}}{1 + \cfrac{q} {1 + \cfrac{q^2} {1 + \cfrac{q^3} {1+\ddots}}}} = {\sqrt{5+\sqrt{5}\over 2}-{1+\sqrt{5}\over 2}} = 5^{1/4}\sqrt{\phi}-\phi$$ Let $q…
9
votes
4 answers

Proving q-binomial identities

I was wondering if anyone could show me how to prove q-binomial identities? I do not have a single example in my notes, and I can't seem to find any online. For example, consider: ${a + 1 + b \brack b}_q = \sum\limits_{j=0}^{b} q^{(a+1)(b-j)}{a+j…
Nizbel99
  • 837
  • 1
  • 8
  • 23
9
votes
2 answers

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…
Nicco
  • 2,763
  • 14
  • 32
9
votes
2 answers

a conjecture of certain q-continued fractions

Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$,…
Nicco
  • 2,763
  • 14
  • 32
9
votes
2 answers

A $q$-continued fraction connected to the divisor function?

In this post, the following two continued fractions discussed by Nicco are given, $$A(q)= \left(\frac{\vartheta_2(0,q)}{2\,q^{1/4}}\right)^2=…
8
votes
2 answers

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…
Hippalectryon
  • 7,484
  • 2
  • 33
  • 61
8
votes
1 answer

Does the Rogers-Ramanujan continued fraction $R(q)$ satisfy this conjectured infinite series

Given the Rogers-Ramanujan continued fraction $R(q)= \cfrac{1}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$ where $q=\exp(2\pi i \tau)$, $|q|\lt1$ for the sake of brevity, let us introduce the following…
1
2 3 4 5 6 7