Questions tagged [q-series]

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

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Need help in implementing q-SeriesToq-Product in Mathematica

In Mathematica Guidebook for symbolic computations (https://www.amazon.com/dp/0387950206/wolframresearch-20), in the Exercises, 30 (c)(p. 359), there is a question: I have no clue how to implement ...
Sangama's user avatar
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2 votes
1 answer
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Proving Gauss’s summation theorem for the $q$-binomial coefficients

I am following Warren P. Johnson's "An Introduction to q-analysis". We are supposed to prove $$\binom{n+1}{k+1}_q=\sum_{m=k}^{n}q^{m-k}\binom{m}{k}_q$$ Here is my attempt. We start off by ...
KanakD's user avatar
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An identity related to the series $\sum_{n\geq 0}p(5n+4)x^n$ in Ramanujan's lost notebook

While browsing through Ramanujan's original manuscript titled "The Lost Notebook" (the link is a PDF file with 379 scanned pages, so instead of a click it is preferable to download) I found ...
Paramanand Singh's user avatar
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7 votes
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Ramanujan's identity concerning a quotient of Dedekind's eta functions

In his paper On certain Arithmetical Functions (published in Transactions of the Cambridge Philosophical Society, XXII, No. 9, 1916, pp. 159-184) Ramanujan presents the following identities (as if ...
Paramanand Singh's user avatar
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3 votes
2 answers
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An identity related to $q$-series

While studying Ramanujan's theta functions, I encountered a q-series $(q;q)_\infty^2\phi(q)$. I calculated the first few terms of $(q;q)_\infty^2\phi(q)$ and observed that it seems to have the ...
Kevin's user avatar
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Closed form on variation of q-binomial theorem

The q-binomial theorem states that $$\sum_{n=0}^{\infty}\frac{\left(a;q\right)_n}{\left(q;q\right)_n}z^n =\frac{\left(az;q\right)_\infty}{\left(z;q\right)_\infty}$$. Is there a similar closed-form ...
Kenneth Goodenough's user avatar
3 votes
1 answer
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Numbers with a unique partition as a sum of two squares

The well known Ramanujan tau function $\tau(n)$ is defined as the nth Fourier coefficient of the modular discriminant $\displaystyle \Delta(q)=q\prod_{m=1}^\infty (1-q^m)^{24} = \sum_{n=1}^\infty \tau(...
Nicco's user avatar
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1 vote
1 answer
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A certain conjectured criterion for restricted partitions

Given the number of partitions of $n$ into distinct parts $q(n)$, with the following generating function $\displaystyle\prod_{m=1}^\infty (1+x^m) = \sum_{n=0}^\infty q(n)\,x^n\tag{1a}$ Which may be ...
Nicco's user avatar
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Exercise in the book "Basic Hypergeometric Series" of Gasper and Rahman

I am trying to solve Exercise 1.12.(iii) from the book "Basic Hypergeometric Series" of Gasper and Rahman (see the picture below). I am especially interested in the case where $c=ab$ and $n=...
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Ramanujan Identity Proof

Ramanujan defined a now famous q-series as $$\sum_{n=-\infty}^{\infty}q^{n^2} = \left(-q;q^2\right)^2_{\infty}\left(q^2;q^2\right)_{\infty}$$ I wanted to prove this identity but I wasn't sure where to ...
MokutekiJ's user avatar
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6 votes
2 answers
307 views

Conjectured identity for the ratio of Ramanujan theta functions

Following Ramanujan, we define theta functions as follows $$\chi(q):=\prod_{n = 1}^{\infty}\left(1+q^{2n-1}\right),\\\phi(q)=\sum_{n=-\infty}^{\infty}q^{n^2},\\\displaystyle \psi(q)=\sum_{n = 0}^{\...
Nicco's user avatar
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2 votes
1 answer
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Search for an algebraic proof of a near q-Vandermonde identity.

When trying to prove this q-binomial identity I had soon the idea that here we have a q-Vandermonde identity in disguise. I could transform the identity to \begin{align*} \color{blue}{(-1)^mq^{\binom{...
Markus Scheuer's user avatar
2 votes
1 answer
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Identity involving q-Pochhammer symbols -- "Normalization"

I am trying to prove the following identity involving q-Pochhammer symbols $$ \sum_{m=0}^{n} \dfrac{1}{(c^{-1};c^{-1})_m (c;c)_{n-m}}=1$$ where $n\in \mathbb{N}$, $c\in \mathbb{R}$ and $(a;q)_n=\prod_{...
Costantino Di Bello's user avatar
2 votes
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196 views

Asymptotics of ratio of q-Pochhammer symbols

In (9.23) of https://arxiv.org/abs/1908.08875 , they claim that it is "easy" to show the following limits $$\lim_{y\to 0}\frac{(yq^\frac{2-r+m}{2};q)_\infty}{(y^{-1}q^\frac{r+m}{2};q)_\infty}...
Blind Miner's user avatar
7 votes
2 answers
219 views

Combinatorial Interpretation of a partition identity

I am working on the book "Number Theory in the Spirit of Ramanujan" by Bruce Berndt. In Exercise $1.3.7$: He wants us to prove that $$ np\left(n\right) = \sum_{j = 0}^{n - 1}p\left(j\right)\...
ALNS's user avatar
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Question about the Cayley proof of the Jacobi triple product relation

In a stage of proving the Jacobi triple product identity, I have to prove this : For $|z|<1$ and $|q|<1$, we set $F_0(z,q)=1$ and $F_m(z,q)=\prod_{k=1}^m \frac{1}{1-q^kz}$. If $(E_m)_{m\in\...
Nicolas FRANCOIS's user avatar
26 votes
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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 ...
Henry's user avatar
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3 votes
1 answer
254 views

On $\mathrm{\sum_{x\in\Bbb Z}sech(x)=3.142242…}$

Inspired by This question, I started to wonder about simpler series. I have seen similar questions to the following, but none had this special case explicitly. It is related to the q-digamma ...
Тyma Gaidash's user avatar
2 votes
1 answer
199 views

Modularity of Euler $q$-series.

The Dedekind $\eta$ function is defined as a function on the upper half space $\mathbb{H}$ as $$\eta(\tau) = e^{\frac{\pi i \tau}{12}}\prod_{n>0}(1-e^{2\pi i n\tau})$$ or, using the circular ...
Mattia Coloma's user avatar
4 votes
3 answers
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Proving that odd partitions and distinct partitions are equal

I am working through The Theory of Partitions by George Andrews (I have the first paperback edition, published in 1998). Corollary 1.2 is a standard result that shows that the number of partitions of $...
seeker_after_truth's user avatar
1 vote
0 answers
66 views

Simplifying q-series expressions

Hi I am wondering if there is a general procedure (perhaps for a specific class) simplifying q-series of the form $(q^a;q^{na})_{\infty}$ where $n\in \mathbb{Z}^{+}$, into the product/qoutient of q-...
Broudy's user avatar
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Is this some sort of modular form? $f(\tau)=\eta(a\tau)\eta(b\tau)$

I recently found the following. Let $a,b\in\Bbb N$ with $24|(a+b)$ and consider the function $$f(\tau)=\eta(a\tau)\eta(b\tau).$$ We have the following symmetry relations: $$\begin{align} f(\tau+1)&...
clathratus's user avatar
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7 votes
2 answers
406 views

Relation between Rogers Ramanujan continued fraction and $j$-invariant

While going through this answer I found an interesting but slightly complicated relation between Rogers-Ramanujan continued fraction and the j-invariant. I would like to know an elementary proof of ...
Paramanand Singh's user avatar
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4 votes
1 answer
73 views

coefficient $[q^n]\sum_{m\ge1}\frac{q^{\alpha m}}{1-q^{\beta m}}$ where $0<\alpha<\beta$

Problem: I am looking for a finite-sum expression for the coefficient $c_n=c_n(\alpha,\beta)$, where $$C(\alpha,\beta;q)=\sum_{m\ge1}\frac{q^{\alpha m}}{1-q^{\beta m}}=\sum_{n\ge1}c_n(\alpha,\beta)q^n,...
clathratus's user avatar
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Symbolic calculation vs. desmos

Context: I am working on getting a recurrence of the form $$nc_n=\sum_{k=1}^nc_{n-k}R(k)\tag0$$ for the coefficients $c_n$ defined by $$f(q)=\vartheta_3^2(q)=\prod_{m\ge1}(1+q^{2m-1})^4(1-q^{2m})^2=\...
clathratus's user avatar
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2 votes
1 answer
105 views

Counting colored integer partitions with a strictness condition

For any $n \in \mathbb{N}$, let us first consider $P_2(n)$, the number of 2-colored integer partitions of $n$. Recall that these are enumerated by $(\lambda^{(1)},\lambda^{(2)})$, where $\lambda^{(1)}...
Astro Log's user avatar
2 votes
0 answers
64 views

Concerning the coefficient $[q^n]\sum_{n\ge1}\frac{(aq)^n}{(1-bq^n)^2}$.

Background: While trying to answer this question, I came up with a question of my own. Let $|a|,|b|,|q|<1$ and $$A(a,b;q)=\sum_{n\ge1}\frac{(aq)^n}{1-bq^n}.$$ One can show that $$A(a,b;q)=\sum_{n\...
clathratus's user avatar
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2 votes
1 answer
161 views

Modularity of Ramanujan-Sato series

The Ramanujan-Sato series $$j^*(\tau)=432\frac{\sqrt{ j(\tau)}+\sqrt{j(\tau)-1728}}{\sqrt{ j(\tau)}-\sqrt{j(\tau)-1728}}=432\frac{E_4(\tau)^{\frac32}+E_6(\tau)}{E_4(\tau)^{\frac32}-E_6(\tau)} \\ = \...
El Rafu's user avatar
  • 608
1 vote
1 answer
97 views

Is it true that $\frac 1q=\sum\limits_{\text{prime }p}^{\infty}\frac{1}{q^{p-1}-1}$ for all $q\notin[-1,1]$?

I believe I have proved that, for all $q\notin[-1,1]$, it follows $$\frac{1}{q}=\sum_{\text{prime }p}^\infty\frac{1}{q^{p-1}-1}$$ Of course this is a huge result, so I want to see if my proof is ...
Mr Pie's user avatar
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3 votes
1 answer
55 views

$q$-expansion of Klein's absolute invariant using infinite products

Given that $$j=\frac{1}{13824q^2}\left(2^8q^2\prod_{k\gt 0}(1+q^{2k})^{16}+\prod_{k\gt 0}(1+q^{2k-1})^{16}+\prod_{k\gt 0}(1-q^{2k-1})^{16}\right)^3,$$ how can I show that $$j=\frac{1}{1728q^2}(1+c_1 q^...
BIRA's user avatar
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3 votes
0 answers
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Mellin transform of $\sin(\pi x)\left(\frac{d}{dx}\log\left(x;j^{-1}\right)\right)$

Consider the smooth function : $$f(x)=\sin(\pi x)\left(\frac{d}{dx}\log\left(x;j^{-1}\right)\right)$$ $\left(z;q\right)$ being the Q-Pochhammer symbol, and $j$ is a positive integer. We want to ...
Mohammad Al Jamal's user avatar
10 votes
2 answers
797 views

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 ...
Charlie Chang's user avatar
2 votes
1 answer
369 views

What are specific proofs of Jacobi Triple Product Identity?

I am looking for the Special Proofs. Here is a reference from MSE. Motivation for/history of Jacobi's triple product identity I also know that a simple proof via Functional Equation from the book ...
user1062's user avatar
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1 vote
1 answer
261 views

An equivalent of Jacobi's two square theorem

Jacobi's two square theorem: The number of representations of $n$ as a sum of two squares is $4$ times the difference between the number of divisors of $n$ congruent to $1$ modulo $4$ and the ...
Redbox's user avatar
  • 819
3 votes
1 answer
152 views

Asymptotics of the quantum dilogarithm

Fadeev and Kashaev define the quantum dilogarithm by $$ \Psi(x) = \prod_{n=1}^\infty (1 - x q^n) $$ for $|q| < 1$. For $q = \exp(\epsilon)$, $\Re \epsilon < 0$, they say the asymptotic expansion ...
Calvin McPhail-Snyder's user avatar
2 votes
1 answer
412 views

how to prove these formulas about infinite product?

Recently , I read one paper titled 'Modular equations and approximations to π' by Ramanujan, in which there are some formulas for $q=\pi i \tau$( where $\tau=x+yi, y>0$, hence $|q|<1)$ : $$\...
Jacob.Lee's user avatar
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0 votes
1 answer
85 views

Reference request: $q$-gamma, $q$-polygamma, $q$-Pochhammer.

I'm trying to solve a problem related to a few difficult series and am using Mathematica to hammer out the difficult bits. The problem is that QPochhammer,...
Descartes Before the Horse's user avatar
13 votes
1 answer
374 views

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 ...
Icycarus's user avatar
  • 1,142
0 votes
0 answers
52 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 ...
artemiy's user avatar
1 vote
2 answers
403 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.
user3108815's user avatar
1 vote
0 answers
137 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 ...
user3108815's user avatar
1 vote
2 answers
172 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 ...
user avatar
4 votes
1 answer
109 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}}...
Liddo's user avatar
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2 votes
0 answers
49 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 $\{f_n\}$ ...
NoBinash's user avatar
  • 175
0 votes
2 answers
282 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 ...
Monty's user avatar
  • 2,230
1 vote
1 answer
91 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 ...
Monty's user avatar
  • 2,230
4 votes
0 answers
149 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 =...
cactus314's user avatar
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2 votes
0 answers
157 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^{...
user404188's user avatar
3 votes
1 answer
243 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 $\...
Nicco's user avatar
  • 2,813
6 votes
1 answer
276 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^{...
Nicco's user avatar
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