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1
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0answers
26 views

Generating functions, Schur's identity

Let $S=\{n\in \mathbb{Z}_+ \mid n \equiv 1, 5 \,\,(\text{mod 6})\}.$ Let $a(n)$ be the number of partitions of $n$ into parts belonging to $S,$ and $b(n)$ be the number of partitions of $n$ into ...
2
votes
0answers
24 views

show $\sum_{j=0}^n (-1)^j {n \brack j}_q =0$ for n odd

I would like to show $\sum_{j=0}^n (-1)^j {n \brack j}_q =0$ for n odd, or preferably even more generally that $\sum_{j=0}^n (-1)^j {n \brack j }_q =\frac{1}{2}((-1)^n+1)(q;q)_{\frac{n}{2}}$. Using ...
0
votes
0answers
30 views

Weighted Q-binomial Coefficients

A possible identity popped up in a project for college, and if features q-binomial coefficient, which can be interpreted as the generating function for the number of Ferrer's boards fitting into a $k\...
14
votes
1answer
502 views

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, $$x := \left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24} = \frac{1}{q} - 24 + 276q - 2048q^2 + 11202q^3 - 49152q^4+ \cdots\tag1$$ ...
6
votes
2answers
202 views

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\...
1
vote
1answer
65 views

a conjecture of two equivalent q-continued fractions related to the reciprocal of the Göllnitz-Gordon continued fraction A111374-OEIS

Given the square of the nome $q=e^{2i\pi\tau}$ and ramanujan theta function $f(a,b)=\sum_{k=-\infty}^{\infty}a^{k(k+1)/2}b^{k(k-1)/2}$ with $|q|\lt1$, define, $$\begin{aligned}M(q)=\cfrac{1-q^3}{1-q^...
1
vote
0answers
31 views

How to prove that $\sum_{i=0}^n\frac{(a;q)_i}{(q;q)_i}\frac{(b;q)_{n-i}}{(q;q)_{n-i}}a^{n-i}=\frac{(ab;q)_n}{(q;q)_n}$?

By Cauchy identity, $${}_1\phi_0(a;—;q,z)=\sum_{n\geq0}\frac{(a;q)_n}{(q;q)_n}z^n=\frac{(az;q)_{\infty}}{(z;q)_\infty},\quad|z|<1,|q|<1,$$ we can obtain the $q-$analogue of $(1-z)^{-a}(1-z)^{-b}=...
1
vote
1answer
73 views

Generalizations of the pentagonal number theorem

Euler's pentagonal number theorem (see also the original paper and review by Jordan Bell) states $$ \prod_{n=1}^\infty (1 - q^n) = \sum_{k=-\infty}^{\infty} (-1)^{k} q^{(3k^2 - k)/2}, $$ where $k \, (...
8
votes
2answers
161 views

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^...
7
votes
2answers
256 views

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})}...
0
votes
0answers
19 views

A analytic representations of q- harmonic numbers

A formula, true or false ? $$\sum\limits_{m = n + 1}^\infty {\frac{{{}_q{H_m}}}{{{{\left( {m - n} \right)}_q}}}{q^{m - n}}} \mathop = \limits^? \sum\limits_{j = 1}^\infty {\frac{{{}_q{H_j}}}{{{j_q}}...
2
votes
2answers
32 views

A analytic representation of q- rational series

Using Mathematica, we can find $$\sum\limits_{n = 1}^\infty {\frac{{{{\left( {1 - q} \right)}^2}{q^n}}}{{\left( {1 - {q^n}} \right)\left( {1 - {q^{n + 1}}} \right)}}} = q,\;q \in \left( {0,1} \...
6
votes
2answers
181 views

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= \cfrac{1}{1-q+\cfrac{q(1\color{red}-q)^2}{1-q^3+\cfrac{q^...
8
votes
1answer
122 views

are these two continued fractions equivalent?

I would like to pose the following conjecture.Given $$\phi(q) =\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^3(1-q^2)^2}{1-q^5+\cfrac{q^5(1-q^3)^2}{1-q^7+\ddots}}}}$$ and $$\psi(q)=\cfrac{-q}{1-q+\...
3
votes
1answer
91 views

an interesting q-series and a certain continued-fraction

My aim is to find a rigorous proof of the following conjectured identity.Given $$1+q+q^2-q^4-q^5+q^7+q^8-q^{10}-q^{11}+\ddots=\cfrac{1}{1-q+\cfrac{(q^3)}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^...
7
votes
2answers
341 views

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 ...
18
votes
1answer
505 views

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 ...
4
votes
1answer
73 views

Some analogs of the pentagonal number theorem

There are the following analogs of the famous identity $$ \prod_{n\geqslant1}(1-q^n)=\sum_{n\in\mathbb Z}(-1)^nq^{\frac{3n^2-n}2}. $$ Let $v_2(n)$ denote the 2-adic valuation of $n$, that is, the ...
5
votes
2answers
247 views

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(...
3
votes
0answers
152 views

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 ...
3
votes
1answer
186 views

A new $q$-continued fraction of order $12$

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} ...
5
votes
0answers
62 views

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+\...
2
votes
0answers
59 views

q-Series no clue

I see in some table pages sum like $$\sum _{k=1}^{\infty } \frac{1}{\left(q^k+2\right)^2 \left(q^k+1\right)^2 \left(q^k+3\right)}\text{==}$$ $$\frac{2 \log (q) \left(-9 \psi _q^{(0)}\left(-\frac{\log ...
1
vote
0answers
37 views

Find simple proofs of the two $q$-series identities

When I read an article, I found the following two $q$-series identities very interesting $$ \sum_{k=-\infty}^{+\infty}(-1)^k{2n\brack n+2k}q^{2k^2}=(-q;q^2)_n, $$ $$ \sum_{k=-\infty}^{+\infty}(-1)^k{...
4
votes
0answers
199 views

Elementary proof of Ramanujan's “most beautiful identity”

Ramanujan presented many identities, Hardy chose one which for him represented the best of Ramanujan. There are many proofs for this identity. (for example, H. H. Chan’s proof, M. Hirschhorn's proof....
4
votes
0answers
100 views

Is there a formula for $\sum_{n=0}^{+\infty} q^{n^3}$?

When I studyied the representation of integers as sum of squares, I found that the most powerful tool is the Jacobi Triple Product, in fact this amazing identity allows us to find more useful ...
2
votes
1answer
48 views

double integrals on quantum calculus

I need references or book recommendations to find properties of double integrals on quantum calculus. Especially i need analogue of Fubini's theorem on q-calculus.
2
votes
0answers
78 views

improper integrals in q-calculus

In quantum calculus is this equality possible for improper integrals? $\lim_{x\to\infty}\int_0^xf(t)d_qt=\int_0^\infty f(x)d_qx$
2
votes
0answers
51 views

How to prove the $q$-series identity?

How to prove the following identity: $$\sum_{n\ge0}\frac{2q^{n^{2}+n}}{(q)_{n}^{2}(1+q^{n})}=\sum_{n\ge0}\frac{q^{n^{2}+n}}{(q)_{n}^{2}(1-q^{2n+2})}$$
1
vote
0answers
103 views

q-theta function and their properties

I want to compute the residue integral for the $q$-theta function, and derive its properties. First, I'll briefly explain the definition \begin{align} & \theta(a;q)=(a;q)(q/a;q)=\prod_{i=0}^{\...
1
vote
1answer
81 views

octagonal number theorem $q$-Pochhammer symbol expression

Setting the exponents of this analogue of the series in Euler's Pentagonal Number theorem to be the octagonal numbers: $$U(q)= \sum_{n\in\mathbb{Z}} (-1)^{n}q^{n(6n-4)/2}$$ in mpmath: ...
1
vote
0answers
42 views

a question on sum of q_binomials

I was trying to calculate something and at some point I get the following sum: \begin{equation} \sum_{t=0,t \text{ even}}^{s}{s+3n \brack s-t}\sum_{i = 0}^{t/2}q^{2i^2}{t/2+2n-i \brack t/2-i}{n \...
3
votes
0answers
119 views

Modular forms on the theta group

The theta group $\Gamma$ is the subgroup of $SL(2;\mathbb{Z})$ generated by $T=\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $S=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.$ It ...
8
votes
2answers
334 views

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,...
5
votes
0answers
91 views

About $\prod{\left(1-q^n\right)^{5}}$

Is there a result about the non-vanishing of coefficients of $$\prod_{n=1}^{+\infty}{\left(1-q^n\right)^{5}}=1-5q+5q^2+10q^3-15q^4-6q^5-5q^6+25q^7+15q^8-20q^9+\cdots \text{ ?}$$ Thanks !
4
votes
1answer
272 views

A question on a sum of $q$-binomial coefficients

I am trying to enumerate a certain quantity and at some point I get the following sum: \begin{equation} \sum_{i=0}^{m}{m \brack i}_q \sum_{j=0}^{n-m} q^{j(m-i)}{n-m \brack j}_q \sum_{k=0}^{r} {r\...
3
votes
1answer
110 views

$q$-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 ...
1
vote
1answer
127 views

Ramanujan Notebook Part 1 (1.16): $\sum q^{n^2} = (-q;q^2)_\infty^2(q^2;q^2)_\infty=\frac{(-q;-q)_\infty}{(q;-q)_\infty}$

I am having trouble with proving a statement in Ramanujan's Lost Notebook Part 1 (1.16). The statement is as follows: $\varphi(q)=f(q,q)=\sum_{n=-\infty}^\infty q^{n^2} = (-q;q^2)_\infty^2(q^2;q^2)_\...
4
votes
2answers
2k views

A problem with the geometric series and matrices?

Let $n$ be a positive integer. Let $A$ be a square matrix. Let $I$ be the identity matrix with the same size as $A$. I want to simplify $f_n(A) = I + A + A^2 + A^3 + A^4 + \cdots + A^n$ Now I know ...
6
votes
1answer
228 views

Combinatorial Identity $ \sum_{k=1}^n (-1)^{k-1} \cdot q^{\frac{k(k-1)}{2}} \cdot \frac{\prod_{i=n-k+1}^n(1-q^i)}{\prod_{i=1}^k(1-q^i)} = 1 $

I have to validate the following identity which is defined: $$ \sum_{k=1}^n (-1)^{k-1} \cdot q^{\frac{k(k-1)}{2}} \cdot \frac{\prod_{i=n-k+1}^n(1-q^i)}{\prod_{i=1}^k(1-q^i)} = 1 $$ where $0<q<1$....
2
votes
1answer
61 views

Q-series identities #2

Prove the following $$\frac{1}{(z;q)_{\infty}} = \sum^{\infty}_{k=0} \frac{z^k}{(q;q)_k}$$ I am looking for a proof that doesn't involve the q-binomial theorem . where $$(a;q)_k = \prod_{n=0}^{k-...
3
votes
1answer
212 views

q-series identities

Here is my first question in this site Prove the following $$\lim_{q \to 1 }\frac{(a)_{\infty}}{(aq^x)_{\infty}}=(1-a)^x$$ for $x$ an integer it was an easy task but it was generally for any ...
20
votes
3answers
1k views

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 ...
1
vote
2answers
62 views

A q-series related to adjoint representation of lie group

What is the infinite sum expansion in the degree of q? $$ \exp \left[\sum_{n=1}^\infty \frac{1}{n}\frac{2q^n }{1-q^n } \right] = \prod_{n=1}^\infty \exp \left[ \frac{1}{n}\frac{2q^n }{1-q^n } \right] ...
4
votes
1answer
523 views

How to evaluate this infinite product

How to evaluate this one $$\prod\limits_{n=1}^{\infty }{\left( 1-\frac{1}{{{2}^{n}}} \right)}$$
2
votes
2answers
238 views

$\log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr)+ \frac{1}{2} \log t$

Let $t$ be a real $>0$. Let $P(x) = \prod_{n=1}^\infty \dfrac{1}{1-x^n}$ $\log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr) + \frac{1}{2} \log t$ How to ...
7
votes
4answers
2k views

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 ...
15
votes
2answers
420 views

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 ...
4
votes
0answers
112 views

factoring infinite products of $q$-series with constant term equal to 1

I was thinking about the following infinite product: $$\prod_{n=0}^{\infty} \frac{ae^{-2n}+be^{-n}+c}{c}$$ The right way of generalizing it is to think in terms of $q$-Pochhammer symbols. If $r_{1}$ ...
2
votes
1answer
132 views

Closed form for $\sum_{m \geq 1} (-1)^m q^{m(m+1)/2 + m \Delta}$?

Is there a useful closed form for the following series ($|\Delta|$ is a small integer)? $$f(q,\Delta) =\sum_{m=1}^{\infty} (-1)^m q^{m(m+1)/2 + m \Delta}$$ It is a large-$n$ approximation of the ...