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

# Questions tagged [q-series]

100 questions

**28**

votes

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### 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

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votes

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### 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

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### 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

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votes

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### 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

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### 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

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### 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

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### 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 $q=e^{2\pi\mathrm{i}\tau}$,…

Nicco

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**10**

votes

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### 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…

Charlie Chang

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### 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…

Tito Piezas III

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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

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### 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

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### a conjecture of certain q-continued fractions

Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$,…

Nicco

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### 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=…

Tito Piezas III

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### 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

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### 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…

Nicco

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