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Questions tagged [q-analogs]

Use this tag for questions pretaining to q-analogs of functions, for example q-Binomials, $q$-derivatives, the q-theta function, the q-Pochhammer symbol, etc.

2
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0answers
39 views

A closed form of $q$-analog finite sum

Let $[n]:=\frac{1-q^n}{1-q}$. I wish to find a closed form of the following q-sum $$I(p):=\sum\limits_{n_1+\cdots+n_p=n\atop n_1,\ldots,n_p\ge 1}\ \ \ \frac{1}{[n_1]\cdots [n_p]}=?$$ For example, we ...
5
votes
0answers
90 views

Irreducibility of q-factorial plus 1

Is it true that $[n]_q! + 1$ is an irreducible polynomial over $\mathbb{Z}$ for all positive integers $n$ ? I checked that this is true for $n$ up to $20$. Here $[n]_q! := 1 (1 + q) (1 + q + q^2) \...
1
vote
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.
0
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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? $$\sum_{x=1}^n\psi_q{\small(x)}\ =\ ?$$ $$q\in\mathbb N$$
0
votes
1answer
42 views

Can the following integral involving digamma be evaluated in closed form (special functions allowed but not preferred)?

$$\int_1^n \frac {\psi(x)}{x} dx \ s.t. \ n \in \mathbb N$$ EDIT: I've relocated the asking about the integral of the above except with the q-analog of the digamma to another question. This is so as ...
0
votes
0answers
11 views

Closed form of basic hypergeometric series: Adapt the answer given in linked question to solve my own variation?

I'm interested in two things: A computationally efficient (used here to mean the number of terms is bounded and not dependant on the size of the input) partial sum formulae of the expression below, ...
0
votes
0answers
37 views

What closed forms exist for this basic hypergeometric series?

I've run into: $$\sum_{x=1}^n {x^a\over 1-q^{x}}, \ s.t.\ q\in \mathbb N>1 \ or \ q\in (0, 1),\ a \in \mathbb N$$ I am interested mostly in the cases where $a = 1$ or $ a = 2$ Things I've done ...
1
vote
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 ...
2
votes
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 $\{f_n\}$ ...
10
votes
1answer
261 views

Bijection for $q$-binomial coefficient

Define the $q$-binomial (Gaussian) coefficient ${n+m\brack n}_q$ as the generating function for integer partitions (whose Ferrers diagrams are) fitting into a rectangle $n\times m$, i.e., for the set $...
0
votes
1answer
42 views

$q$-analogs confusion in some steps

I am understanding the proof of the general version of the Jacobi triple product, that is $$\prod_{k=1}^{\infty}(1+xq^{2k-2})=\sum_{k=0}^{\infty}\frac{q^{k(k-1)}}{(q^2)_k}x^k$$ In the proof to this ...
2
votes
1answer
117 views

Importance of $q$-analog

I am currently studying q-analog, but I was actually confused on what its actual purpose is. Like I see all these manipulations using $q$, but I have little idea on what they represent. Sure the ...
2
votes
1answer
60 views

Gauss identities of $q$-analog

How can we use the Jacobi-Triple Product Identity to show that in the area of q-analog series and theory: Jacobi- Triple Product Identity: $$\sum_{{n\in\mathbb{Z}}}x^nq^{n^2}=\prod_{k \ge 1}(1-q^{2k})...
2
votes
1answer
73 views

$q$-analog confusion

We have a very well known result by Gauss in the theory of $q$-analog. Which is given by: $$\sum_{r=0}^m (-1)^r\left[ \begin{array}{cc|c} m\\ r \end{array} \right]_q = \begin{cases} 0, & \...
1
vote
1answer
127 views

Properties of Gaussian binomial coefficient

Given $$\prod_{k=0}^{n-1}(1+q^kt)=\sum_{k=0}^{n}q^{k(k-1)/2}\binom{n}{k}_{q}t^k$$ How do I show that as $n \rightarrow \infty$ $$\prod_{k=0}^{\infty}(1+q^kt)=\sum_{k=0}^{\infty}\frac{q^{k(k-1)/2} \ t^...
2
votes
0answers
68 views

limit of the q-exponential function

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 $q$-...
4
votes
2answers
197 views

Combinatorial proof of the identity ${{n}\brack {m}}_{q}{{m}\brack {k}}_{q} = {{n}\brack {k}}_{q}{{n-k}\brack {m-k}}_{q}$

Problem Let $$ [n]_q = \sum_{0 \leq k < n} q^k = \frac{1-q^n}{1-q} $$ and $$[n]_q! = \prod_{0 \leq m < n} \sum_{0 \leq k < m} q^k. $$ Define $$ {{n}\brack{k}}_q = \frac{[n]_q!}{[k]_q![n-k]...
2
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0answers
52 views

q-analog of Stirling expansion.

It is well-known that $\ln\left(\Gamma(x)\right)=x\ln(x)-x+o(x)$. Now consider a integer $q>1$ and $\theta>1$ a real number. Can one write an asymptotic expansion (at least 2 terms) for $$\ln\...
8
votes
0answers
153 views

Enumerative interpretation of generalized $q$-hockey stick identity

Pascal's rule $$ \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k} \tag{1} $$ may be used recursively to obtain the hockey stick identity $$ \binom{n+1}{k+1}=\binom{n}{k}+\binom{n-1}{k}+\cdots+\binom{k}{...
0
votes
1answer
70 views

Stanley's Enumerative Combinatorics Problem 125 Page 140

The problem states: Find the number $f(n)$ of binary sequences $w = a_1a_2...a_k$ (where k is arbitrary) such that $a_1 = 1$, $a_k = 0$, and $inv(w) = n$. For instance, $f(4) = 5$, corresponding to ...
5
votes
1answer
221 views

Pascal's Identities for $q$-binomials

The recursive identity for building Pascal's triangle is $$ \binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}. \tag{$\circ$}$$ This has a simple combinatorial interpretation: every $k$-subset of $\{1,\...
1
vote
2answers
151 views

q-Binomial coefficients calculation

In the paper http://arminstraub.com/downloads/slides/2012qbinomials-illinois.pdf page 9 "D3" it is written a method of calculation of q-Binomial coefficients. For example for the $\binom{4}{2}_q$ the ...
1
vote
1answer
136 views

How to obtain the q-factorial from Bhargava's factorial?

For certain nice rings, Bhargava constructs a generalized factorial via p-orderings; the details of which can be found in Bhargava. In example 18, of section 10, he states that the generalized ...
1
vote
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 $q$-...
4
votes
2answers
498 views

Number of all subspaces of a vector space over a finite field

Let $V$ be an $n$-dimensional vector space over a finite field $\mathbb{F}_q$. We know that the number of $k$-dimensional subspaces of $V$ is given by the $q$-binomial coefficient $$\binom{n}{k}_q = \...
4
votes
1answer
88 views

Testing $q$-binomial theorem at $z=1/a$

The $q$-binomial theorem states that when $|q|<1$ $$ \frac{(az;q)_\infty}{(z;q)_\infty}={}_1 \phi_0 (a;\;;q;z)\equiv \sum_{n=0}^\infty \frac{(a;q)_n}{(q;q)_n}z^n \quad \text{for } |z|<1$$ with ...
1
vote
1answer
64 views

How do you differentiate $\ln(\exp(z))$

I thought it can be simplified to $\ln(e^z)$ I then thought $e^x$ differentiated is $e^x$ so I thought the $\exp$ part would stay the same I then thought $\ln x = \frac 1 x$ So I put it all ...
0
votes
0answers
57 views

q-exponentials, Convolution and Multiplication

I am doing a modeling study using q-exponential distribution. I come to this question that are q-exponential distributions closed under convolution and multiplication operations? Let say X1 and X2 are ...
1
vote
0answers
154 views

Recurrence relation of q-Stirling numbers of the second kind

I have seen in some papers the use of the recurrence fromula $S_q(n,k)= [k]_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1)$. However, I was wonderinghow we could prove this identity? I understand the proof of the ...
1
vote
1answer
189 views

Combinatorial interpretation of q-analog Stirling numbers of the second kind

I understand that combinatorially we can interpret the Stirling numbers of the second kind as the number of partitions of $n$ into $k$ parts. I was wondering what combinatorial interpretation we could ...
4
votes
1answer
587 views

Show that the Gaussian binomial coefficient is a symmetric polynomial

Deduce that ${{n} \brack {k}}_{q}$ is a symmetric polynomial of $q$, that is, if \begin{equation*} {{n}\brack {k}}_{q} = a_0 + a_1q + a_2q^2 + \ldots + a_Nq^N \end{equation*} with $a_N ...
1
vote
1answer
120 views

Proving a Newton generalized version of a binomial theorem identity

I am wanting to prove that $$\prod_{k=0}^{m-1}\frac{1}{1-zq^k} = \sum_{n \geq 0}{m+n-1 \brack n}_qz^n$$ which is found on this page. I've tried a few approaches. I was able to get a proof by taking ...
4
votes
0answers
266 views

Prove q-Vandermonde Identity using a lattice path counting argument

I have this problem, from Aigner's A Course in Enumeration: Vandermonde's formula for the $q$-binomial coefficients is $${{n+m}\brack {p}}_{q} = \sum_{k=0}^p q^{(m-k)(p-k)}{{m}\brack {k}}_{...
3
votes
3answers
424 views

Combinatorial proof of q-binomial recurrence relation

I have no idea how to proof combinatorially the following recurrence relation: $${n+1\choose k}_q = q^k {n \choose k}_q + {n \choose k-1}_q$$ Could anyone guide me through a combinatorial proof or ...
3
votes
1answer
99 views

q-analoge Van derMonde identity

I have the following problem from Van Lint and Wilson's book on combinatorics and have no idea how to solve it. The problem is the following: Determine the exponents $e_i$ (the may be functions of $m$...
1
vote
1answer
358 views

Q-multinomial coefficient combinatorial meaning

Could anyone please explain to me a combinatorial meaning for the expression ${n \choose{k_1,k_2,...,k_r}} _q$,where $_q$ means a q-multinomial coefficient. I understand what a normal multinomial ...
4
votes
1answer
181 views

Generalization of the q-Binomial Theorem

The following is the well-known q-Binomial Theorem: For all $n \geq 1$, we have: $$ \prod_{j=1}^{n}(1+xq^j) = \sum_{k = 0}^{n} q^{k(k+1)/2} {n \choose k}_q x^k $$ I am not too familiar with the ...
3
votes
1answer
230 views

q-Shifted factorials

The q-shifted factorial is defined as $(a;q)_n := (1-a)(1-aq)\ldots(1-aq^{n-1})$. It is supposed to be an analog of the Pochhammer symbol, or falling factorial: $x(x-1)\ldots(x-n+1)$. But the formulas ...
3
votes
2answers
185 views

$q$-digamma function evaluation

What is the value of $\psi_2^{(0)}(1)$, where $\psi_q^{(0)}(z)$ is the $q$-digamma function? My attempt: \begin{align*} \psi_2^{(0)}(z) &=\frac{1}{\Gamma_2(z)}\frac{d\Gamma_2(z)}{dz} \\&=\...
2
votes
0answers
53 views

Closed formula involving $q$-binomials

I was working on a combinatorial problem over finite fields, and the following quantity came up $$ \sum_{r=0}^k r\binom{n-k}{r}_q\binom{k}{k-r}_qq^{r^2},$$ where $k,n$ are integers such that $0<k&...
5
votes
0answers
92 views

An identity for $q-$Fibonacci numbers if $q$ is a root of unity.

In his proof of the Rogers-Ramanujan identities I. Schur introduced two $q-$analogues of the Fibonacci numbers ${F_n}({q})$ and ${G_n}({q})$, which satisfy ${F_0}({q})=0$, ${F_1}({q})=1$ and $${F_n}...
1
vote
1answer
57 views

How to prove ${}_2\phi_1(q^{-n},b;c;q,q)=\frac{(c/b;q)_n}{(c;q)_n}b^n.$

Firstly, we have already known the one of $q$-analogues of Vandermonde's formula, which is $${}_2\phi_1(q^{-n},b;c;q,cq^n/b)=\frac{(c/b;q)_n}{(c;q)_n}.$$ And there is a hint, when we change the order ...
1
vote
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}=...
8
votes
1answer
111 views

The value of $\sum_{n=0}^{\infty} \, \bigl(\prod_{i=0}^{n-1} q^n-q^i\bigr)^{-1}$

Let $q > 1$. What can we say about the value of $$\sum_{n=0}^{\infty} \, \bigl(\prod\limits_{i=0}^{n-1} q^n-q^i\bigr)^{-1} ~~?$$ The series clearly converges. Is there a closed form or something ...
3
votes
1answer
187 views

Barnes' double gamma function versus q-gamma function

According to wikipedia, the q-analog of the gamma function is closely related to a multiple gamma function defined by Barnes. Besides the fact that they are both generalizations of the Gamma function, ...
4
votes
1answer
177 views

Good article or book to q-analogs?

I want to learn more about the following Topics: $q$-analog calculus what can be done with the $q$-Pochhammer Symbol applications of $q$-analogs in combinatorics However, I have found very Little ...
7
votes
0answers
200 views

Infinite sum involving $q$-adic representations of whole numbers and $q$-factorial numbers

Let $q \in \mathbb{N}_{\geq 2}$. For $n \in \mathbb{N}_0$, let $$\mathrm{fac}_q(n) := \prod_{i=1}^n (1+q+\dots+q^{i-1}) = \prod_{i=1}^n \frac{q^i-1}{q-1}$$ be the $q$-factorial of $n$. In particular, ...
4
votes
0answers
98 views

Roots of derivative of q-expontial function

Let the q-deformation of the exponential function be defined by $$ e_q(z)=\sum_{n=0}^\infty{\frac{z^n}{[n]_q!}}. $$ Eq. (1.8) of this paper provides the product representation $$ e_q(z)=\prod_{k=0}^...
3
votes
1answer
125 views

Prove identity involving the Tsallis q-logarithm

The natural logarithm and the exponential can both be generalized to a called q-logarithms and q-exponentials.those functions are defined as follows: \begin{eqnarray} \log_q(x) &:=& \frac{x^{1-...
4
votes
1answer
547 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\...