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Questions tagged [q-series]

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

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38 views

Analyzing the convergence of a series

I have a series like this:$$\sum_{n=0}^∞\left(\frac{λδ_t}2\right)^n\frac1{n!}\sum_{j=0}^n(-1)^{n-j}C_n^j\exp\left(\frac{(n-2j)^2}{2δ^2}+\frac{(n-2j)γ}δ\right),$$where $C_n^j$ is the binominal ...
1
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2answers
289 views

Partial sum formula of following series

$$\sum_{m \ge 1} \frac{(xy)^m}{(2m)!(1-y^m)}, \quad\text{where }x,y \in \mathbb N$$ I have, to start, J.Jacquelin's answer.
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0answers
21 views

What is a closed form partial sum formula for the q-digamma function?

It is known that the partial sum of the digamma function can be expressed in closed form, but what of the q-digamma function? $$\sum_{x=1}^n\psi_q{\small(x)}\ =\ ?$$ $$q\in\mathbb N$$
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0answers
60 views

Double sum of Lambert series: Partial sum in closed form desired!

We desire the things stated in the title for: $\sum_{k=2}^m \sum_{n=1}^{k-1} {q^n\over {1-q^n}}$ Some things I've looked into that may be of some help: The first sum is just a truncated (partially ...
1
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2answers
146 views

How to find first 32 digits of $\displaystyle \prod_{n=1}^{\infty} (1-\gamma^n)$?

I need to find first 32 digits of $\displaystyle \prod_{n=1}^{\infty} (1-\gamma^n)$ but wolframalpha's brain is too narrow to contain the result, and I don't know any software and programming to find ...
4
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1answer
57 views

Ramanujan congruence mod 7

Hello I am trying to prove this congruence: $$P(7n+5)\equiv 0 \pmod{7}$$ In order to do that I have done the next thing: We have that $\displaystyle\sum_{n\geq0}\;P(n)q^{n}=\frac{1}{(q;q)_{\infty}}...
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0answers
24 views

For $|q|<1$, the function $\frac{(az;q)_\infty}{(z;q)_\infty}$ is analytic on $|z|<1$.

I want to prove that for $|q|<1$, the function $f(z):=\frac{(az;q)_\infty}{(z;q)_\infty}$ is analytic on the set $\{z:|z|<1\}$. My approach: We consider the sequence of functions $\{f_n\}$ ...
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2answers
113 views

Is this sum constant for n?

Hi I can prove that this sum is constant in $n\in \mathbb{N}$. However my proof is very long (a few pages with probability involved). Does anyone see a simple proof. The sum in question is (a q-series ...
1
vote
1answer
68 views

help with an infinite series, hypergeometric function

Hi I am doing some work in interacting particle systems. I have this sum $$ \sum_{k=0}^{\infty} r^{\frac{k}{2}(2m-1-k)} $$ where $m$ is some integer and $r>1$ is real. I don't how to work this ...
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110 views

Evaluate $ \frac{1}{(q)_\infty} \sum_{m \in \mathbb{Z}} q^{\frac{m^2}{2}} (-q^{-\frac{1}{2}}x)^m y^m(q^{1-m}y^{-1};q)_\infty $

This identity is taken from a physics paper [1] stated without proof, on page 43. $$ \frac{1}{(q)_\infty} \sum_{m \in \mathbb{Z}} q^{\frac{m^2}{2}} (-q^{-\frac{1}{2}}x)^m y^m(q^{1-m}y^{-1};q)_\infty =...
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0answers
80 views

Rogers-Ramanujan identity as a modular form

The Wikipedia article on the Rogers-Ramanujan identities remarks that $$q^{-1/60} G(q) = q^{-1/60} \sum_{n=0}^{\infty} \frac{q^{n^2}}{(1-q)...(1-q^n)} = q^{-1/60} \prod_{n=0}^{\infty} \frac{1}{(1 - q^{...
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1answer
143 views

Conjectured continued fraction for the Jacobi theta function $\vartheta_{4}(q)$

Given the Jacobi theta function $$\vartheta_{4}(q)=1+2\sum_{n=1}^{\infty}(-1)^nq^{n^2}$$ where $q=e^{2\pi i\tau}$, and $|q|\lt1$. It is conjectured that it has the following continued fraction $\...
4
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1answer
143 views

Conjectured continued fraction for the Generalized Rogers-Ramanujan continued fraction

Given the following Generalized Rogers-Ramanujan continued fraction,with $|q|\lt1$,which is equation (38) in mathworld $F(a,q)=1-\cfrac{aq}{1-\cfrac{aq^2}{1-\cfrac{aq^3}{1-\cfrac{aq^4}{1-\cfrac{aq^{5}...
1
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1answer
139 views

conjectured relations of mock theta functions of order $3$

Given the following mock theta functions of order $3$,found in this wikipedia article $f(q)=\sum_{n=0}^{\infty} \frac{q^{n^2}}{(-q;q)^2_{n}}$,$\phi(q)=\sum_{n=0}^{\infty} \frac{q^{n^2}}{(-q^2;q^2)_{n}...
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0answers
262 views

Identity between $q$-series

Fix some sequence $(a_k)$ such that $a_k=R_k+q^k$ and $R_k=\frac{1}{R(q^k,q)}$,where $R(x,q)$ is a Generalized Rogers-Ramanujan continued fraction for complex number $x$ and $q=\exp{(2\pi i\tau)}$ is ...
1
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1answer
88 views

q-exponential functions

We know that $$e_q(z)=\sum_{j\geq 0} \frac{z^j}{(q;q)_j}=\frac{1}{(z;q)_{\infty}}$$ where $(a;q)_{\infty}=\prod_{i=0}^{\infty}(1−aq^i)$ denotes the q-shifted factorial. The limit between the $q$-...
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0answers
108 views

q-shifted factorial

Is their any formula to simplify $$(a,ab;q)_n$$ where $(a;q)_n=\prod_{i=0}^{n-1}(1−aq^i)$ denotes the q-shifted factorial. P.S: $a$ and $b$ are complex numbers. I'm glade for your help.
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1answer
370 views

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 ...
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1answer
151 views

Convergence of series with q-Pochhammer symbol

I am working with the series $$\sum_{j=1}^{\infty}\rho \beta^{j-1} \prod_{n=0}^{j-2}(1+\rho \beta^n)$$ where the empty product is assumed as usual to be 1 and $\beta, \rho \in \mathbb R^+$, with $\...
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1answer
46 views

Separation in the q-shifted factorial

Is their any formula to separate the power in $$(q^{x+y};q)_n$$ where $(a;q)_n=\prod_{i=0}^{n-1}(1-aq^i)$ denotes the $q$-shifted factorial. P.S: $x,y$ and $n$ are all integers. I'm glade for your ...
1
vote
1answer
94 views

Differentiating q-Series Expression

Define the following (infinite) q-Pochhammer symbols: $$(a;q)_{\infty} := (a)_{\infty} := \prod_{k=0}^{\infty}(1-aq^{k})$$ $$(a_{1}, \ldots, a_{n})_{\infty} := (a_{1})_{\infty}\cdots(a_{n})_{\infty}$$ ...
3
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0answers
121 views

Parity of Ramanujan's Tau Function

I have been working problems out of Ram Murty's "Problems in the Theory of Modular Forms," which has been marvelous. His solutions are complete, but are way too slick for satisfaction (probably by his ...
3
votes
1answer
229 views

Relation between Ramanujan Theta Function and Jacobi Theta Function

In the theory of $q-$series, we have Ramanujan Theta function \begin{align}\label{rama-theta} f(a,b):=\sum_{n=-\infty}^{\infty} a^{\frac{n(n+1)}{2}}b^{\frac{n(n-1)}{2}} ,\qquad ...
6
votes
2answers
129 views

Proving $\frac 1{R(q)}-1-R(q)$ (Ramanujan's formula)

Question: How would you prove $$\dfrac 1{R(q)}-1-R(q)=\dfrac {f(-q^{1/5})}{q^{1/5} f(-q^5)}\tag{1}$$ where $f(-q)=(q;q^5)_\infty$ and $R(q)$ is defined as $$R(q)=q^{1/5}\dfrac {(q;q^5)_\infty (q^4;q^...
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0answers
44 views

What are the $q$ Power Series?

Question: What are the $q$ power series with $\pm1$ coefficients? This question arose when I saw this: $\ldots$ Thus, we may write$$f(-q^{1/5})=\sum\limits_{n=-\infty}^{\infty}(-1)^nq^{n(3n+1)/10}=...
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0answers
164 views

what is the expansion for $\log \sum q^{n^2}$?

what is the expansion for $\log \sum q^{n^2}$ ? Using the Jacobi triple product $$ (q;q)_\infty(z;q)_\infty (q/z ;q)_\infty = \sum_{k \in \mathbb{Z}} z^k q^{\binom{k}{2}} $$ if we set $q \to q^2$ ...
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0answers
42 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 ...
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0answers
60 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\...
17
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1answer
1k 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, $$\alpha(\tau) = \left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{24} = \frac{1}{q} - 24 + 276q - 2048q^2 + 11202q^3 - 49152q^4+ \cdots\...
6
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2answers
399 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
101 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^...
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0answers
52 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}=...
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1answer
132 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
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2answers
229 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^...
8
votes
2answers
303 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})}...
2
votes
2answers
36 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} \...
8
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2answers
249 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^...
3
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1answer
137 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^...
10
votes
3answers
597 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 ...
20
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2answers
743 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
94 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
531 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(...
4
votes
0answers
197 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 ...
4
votes
1answer
254 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} ...
9
votes
0answers
170 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
100 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
79 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{...
2
votes
0answers
218 views

q-Hermite polynomials

It is well known that the q-Hermite polynomials defined by $$H_n(\theta; q)= \sum\limits_{k=0}^n \frac{(q;q)_n}{(q;q)_k(q;q)_{n-k}}e^{i(n-2k)\theta}$$ are orthogonal in $\theta \in [0, \pi]$ with ...
7
votes
1answer
665 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....
5
votes
0answers
131 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 ...