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14
votes
1answer
458 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
162 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 ...
1
vote
1answer
52 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, ...
1
vote
0answers
26 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
vote
1answer
67 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
143 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 ...
7
votes
2answers
248 views

a conjecture of certain q-continued fractions

Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$, define, ...
0
votes
0answers
16 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 ...
2
votes
2answers
28 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} ...
5
votes
2answers
166 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= ...
8
votes
1answer
105 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 ...
3
votes
1answer
83 views

an interesting q-series and a certain continued-fraction

My aim is to find a rigorous proof of the following conjectured identity.Given ...
4
votes
2answers
290 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
446 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 of the square of ramanujan's octic continued fraction which I discovered using certain three term ...
4
votes
1answer
68 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 ...
4
votes
2answers
218 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 $$ ...
2
votes
0answers
140 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
170 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} ...
4
votes
0answers
53 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
votes
0answers
58 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
36 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 $$ \sum_{k=-\infty}^{+\infty}(-1)^k{2n\brack n+2k}q^{2k^2}=(-q;q^2)_n, $$ $$ ...
3
votes
0answers
140 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 ...
4
votes
0answers
96 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
0answers
32 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
69 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
47 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
82 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} & ...
1
vote
1answer
79 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
39 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
109 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
330 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 ...
5
votes
0answers
86 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
250 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} ...
3
votes
1answer
103 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
122 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} = ...
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
219 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 ...
2
votes
1answer
55 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 = ...
3
votes
1answer
203 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 ...
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
478 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
230 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
405 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
131 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 ...
7
votes
1answer
101 views

Arithmetic/category theoretic information encoded in $q$-series reciprocals

According to the pentagonal number theorem: $$\prod_{n=1}^{\infty} (1-q^{n}) = \sum_{k=-\infty}^{\infty} (-1)^{k}q^{k(3k-1)/2}$$ Now the reciprocal of this has the partition numbers $p(k)$ in its ...
7
votes
1answer
183 views

How to prove the q-series identity?

How could I prove that $$(-q;q^2)_\infty (q;q)_\infty = 1 + 2 \sum_{i=1}^\infty (-1)^i q^{2 i^2}?$$ If that is too difficult is there a way to show $$(-q;q^2)_\infty (q;q)_\infty \equiv 1 \pmod 2?$$ ...
4
votes
1answer
115 views

Validity of a q-series theorem

Define the $q$-analog $(a;q)_n = \prod_{k=0}^n \left(1 - aq^k\right)$. I want to prove the identity $\frac{(q^2;q^2)_\infty}{(q;q)_\infty}=\frac{1}{(q;q^2)_\infty}$. I viewed the LHS this way: ...