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Questions tagged [q-analogs]

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8
votes
0answers
161 views

q-analog of $(d/dx) \binom{x}{k}$?

It is not hard to find easy formulas for the derivative of the function $\binom{x}{k}$, for instance it is not too hard to see (for $k$ an integer) that $\frac{d}{dx} \binom{x}{k} = \sum_{i=1}^k \...
2
votes
2answers
225 views

What partial sum formulae exist for this basic hypergeometric series?

I've run into: $$\sum_{x=1}^{\infty} {x^a\over 1-q^{x}}, \ s.t.\ q\in \mathbb N>1 \ or \ q\in (0, 1),\ a \in \mathbb N$$ I am interested mostly in the cases where $a = 1$ or $ a = 2$ Things I'...
12
votes
2answers
226 views

$q$ as a prime power and a root of unity

The number of points on the $(n-1)$-dimensional projective space $P^{n-1}(\mathbb{F}_q)$ over a finite field $\mathbb{F}_q$ is the $q$-integer $$[n]_q := \frac{q^n-1}{q-1}.$$ In analogy, the number of ...
15
votes
1answer
384 views

Schur-Weyl duality and q-symmetric functions

Disclaimer: I'm far from an expert on any of the topics of this question. I apologize in advance for any horrible mistakes and/or inaccuracies I have made and I hope that the spirit of the question ...
10
votes
4answers
577 views

A divisibility of q-binomial coefficients combinatorially

Let a and b be coprime positive integers. Then the number a+b divides the binomial coefficient ${a+b \choose a}$. I know how to prove this combinatorially - for example after choosing an ordered set ...
8
votes
0answers
148 views

For $q$-analogues of a known curious identity

In 2002 I published the folllowing curious combinatorial identity: $$(x+m+1)\sum_{i=0}^m(-1)^i\binom{x+y+i}{m-i}\binom{y+2i}i-\sum_{i=0}^m\binom{x+i}{m-i}(-4)^i=(x-m)\binom xm.$$ My original proof is ...
14
votes
1answer
422 views

Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?

The q-Vandermonde identity reads: $$ \binom{m + n}{k}_{\!\!q} =\sum_{j} \binom{m}{k - j}_{\!\!q} \binom{n}{j}_{\!\!q} q^{j(m-k+j)} $$ The q-binomial coefficients: $$ \binom{ a }{ b}_{\!\!q} $$ ...
30
votes
1answer
1k views

Mysterious symmetry - in search for a bijection

I have a mysterious symmetry that I have not managed to prove. First some definitions (see picture below) Fix a partition that fit in a staircase shape with $n$ rows. There are $Catalan(n)$ such ...
4
votes
1answer
264 views

$q$-analog of an integral from quantum field theory?

This question has been completely reformulated and a new property for the function $f_q$ has been added due to a series of helpful comments by fedja. Consider the integral from quantum field theory ...
6
votes
1answer
261 views

Inequality for functions on [0,1], continued

Let $0<a<1,\; \psi_a(x)=\displaystyle \prod_{j=0}^\infty (1-a^jx).$ For each $ k\in \mathbb{N},$ set $$f_k(a;x):=\frac{x^k}{(1-a)(1-a^2)\dots (1-a^k)}\,\psi_a(x).$$ Question. Is it true that, ...
3
votes
0answers
79 views

Does the Riemann characterization of the hypergeometric function have a q-analog?

This question is inspired by another recent one here, Characterization of the hypergeometric function. The latter is about the classical result of Riemann characterizing the hypergeometric functions ...
5
votes
0answers
215 views

Q-binomials at roots of unity

As the title says, given a general $q$-binomial $\binom{n}{k}_q$, is there some general result regarding its value at a root of unity, $q = \exp(2\pi i r/N)$?
5
votes
2answers
519 views

Some curious Hankel determinants

Let $f(n,q)=\prod_{j=1}^na(q^j)$ for a polynomial $a(q)$ and let $d(n)=\det(f(i+j,q))_{i,j=0}^n$ be its Hankel determinant. Computer experiments suggest that $$\lim_{q\to1}\frac{d(n)}{(q-1)^\binom{n+...
8
votes
1answer
131 views

Total positivity of $q$-Pascal matrix?

A matrix of real numbers is called totally positive if all its minors are non-negative. A well-known example is the Pascal matrix $(\binom{i}{j})$. Is it true that the minors of the $q$-Pascal matrix ...
2
votes
1answer
96 views

Does this q-analogue have a nice closed form? [closed]

Let $[n]_q=1+q+\cdots+q^{n-1}$. Is there a nice closed form of $\sum_{s=1}^i[s]_{q}$? One would expect that the answer will be some q-analog of $\frac{i(i+1)}{2}$, since $\sum_{s=1}^i s=\frac{i(i+1)...
13
votes
3answers
1k views

A “quantum” identity: in search of a proof -Part II

As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$ As a follow up on this ...
8
votes
3answers
363 views

A not quite theta not quite basic hypergeometric function

The study of matrix quantum group coactions on the noncommutative disk algebra turns up the following series, which is a $q$-deformation of the negative binomial series, for integer $t\ge 0$, complex $...
5
votes
0answers
195 views

Is a basic hypergeometric function ${}_2\phi_1(a, b; c; q, z)$ a meromorphic function in $z$?

Here a basic hypergeometric function is the analytic continuation of the basic hypergeometric series (or called the $q$-hypergeometric series) $$ {}_2\phi_1(a, b; c; q, z) = \sum^{\infty}_{n = 0} \...
5
votes
0answers
213 views

$q$-crystals - is there such a thing?

There are several important facts that I first heard about here on MO. One of the most enlightening of these is that $\mathscr D$-modules on a scheme $X$ may be viewed as sheaves on the groupoid of ...
3
votes
1answer
113 views

Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?

The formula $$ \small\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}=\frac{(-z,-q/z;q)_\infty}{\ln\frac{1}{q}}\int\limits_0^\infty\frac{\left(bt/z,pz/at;p\right)_\infty}{\left(-t,-...
11
votes
2answers
443 views

Does $q$-Catalan number count subspaces?

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\...
3
votes
1answer
200 views

What is the value of this sum involving q-binomials?

Let $n\ge 2r$ be positive integers. Is there a closed form for following finite summation involving in q-binomial coefficients $$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...
13
votes
2answers
613 views

A conjecture about algebraic values of $(-q;\,-q)_\infty/(q;\,q)_\infty$

Recall that $(a;\,q)_\infty$ is the $q$-Pochhammer symbol: $$(a;\,q)_\infty=\prod_{n=0}^\infty(1-a \, q^n).\tag1$$ Its important special case $(q;\,q)_\infty=\prod_{n=1}^\infty(1-q^n)$ is sometimes ...
4
votes
0answers
105 views

A $q-$binomial identity related to $q-$Narayana polynomials of type B

Denote by $ {n\brack {k}}$ a $q-$binomial coefficient. Let ${D_{n,k}}(t,q) = \sum\limits_{j = 0}^{n - k} {{q^{{j^2} + kj}}}{n\brack {j}}{n\brack {k+j}}t^j $ and ${R_n}(x,t,q) = \sum\limits_{k = 0}...
10
votes
3answers
622 views

Is this a q-count of Alternating Sign Matrices?

The number of Alternating Sign Matrices of size $n$ is well known to be $\prod_{k=0}^{n-1}\frac{(3k+1)!}{(n+k)!}$. Is it known whether the naive q-analog expression $$\prod_{k=0}^{n-1}\frac{[3k+1]_q!}{...
8
votes
1answer
326 views

q-analog of a combinatorial identity involving binomial coefficients

Using, e.g., properties of iterated finite differences it is easy to show that for any pair of integers $n$ and $m$ with $n>\!>m$ one has the identity $$ \sum_{k=0}^m(-1)^{k-m} {n-k\choose m}{m\...
6
votes
0answers
214 views

What is known about the $q$-analogue of the simplex?

I am interested in the field with one element. I am thus interested in combinatorial interpretations of the Gaussian binomial coefficients. Richard Stanley's "Enumerative combinatorics" mentions ...
12
votes
1answer
325 views

Multiplicative infinitesimals in q-analogs?

Risking to be downvoted, here is a very lightweight question. In various fields - say, algebraic geometry, nonstandard analysis, synthetic differential geometry - infinitely small quantities, i. e. ...
12
votes
1answer
726 views

Generating function for certain partitions (with a restriction on the Durfee square)

First of all my apologies if this question is well known or obvious: this is not in my area of research. Let $T(x)=\sum_{n=0}^\infty t_nx^n$, where $t_n$ is the number of partitions $\lambda$ of $n$ ...
8
votes
1answer
248 views

notation for $(a-b)(a-qb)\dots (a-q^{n-1}b)$

I wonder whether there is a notation for such thing, which I denote $[a;b]_q^n$ for a moment: $$ [a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n, $$ this last equation uses $q$-Pochhammer symbol ...
4
votes
2answers
213 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
362 views

q-Integer-valued polynomials

For $n \in \mathbb{Z}_{\geq 0}$, let $[n]_q := (1-q^n)/(1-q) = (1+q+...+q^{n-1})$ as is customary, with $[0]_q=0$. Let $R$ be the subring of $\mathbb{Q}(q)[x]$ consisting of all $f$ such that $f([n]...
7
votes
0answers
423 views

What is the $q$-analog of $\Gamma(z)\Gamma(1-z)=\frac\pi{\sin(\pi z)}$?

I would expect the $q$-Gamma function to have the property which would be the $q$-analog of the Euler reflection formula from my question title. More concretely: $\Gamma(z)$ has simple poles at ...
13
votes
2answers
1k views

Is there a $q$-L'Hospital's Rule?

Let $\binom{n}{j}_q$ be a $q$-binomial coefficient and $(x;q)_n = (1-x)(1-qx)\cdots(1-q^{n-1}x).$ Consider the sum $$f(n,m,r,k)= \sum\limits_{j = 0}^{2n} {( - 1)}^{ j}q^{mj^2+rj} \binom{2n}{j}_{q^k}$...
8
votes
1answer
715 views

Counting subspaces

We are given the finite vector space $V = V(n,p) = \mathbb{F}_p^n$ and two fixed subspaces $W_1, W_2 \subseteq V$ of dimensions $m_1$, $m_2$ respectively. Suppose that the dimension of $W_1 \cap W_2$ ...
22
votes
1answer
540 views

q-Catalan numbers from Grassmannians

In this question by $q$-Catalan numbers I mean the $q$-analog given by the formula $\frac{1}{[n+1]_q}\left[{2n\atop n}\right]_q$. The polynomial $\left[{2n\atop n}\right]_q$ represents the class of ...
13
votes
0answers
248 views

Some $q-$analogues of $ \sum\limits_{j = - k}^k {{{( - 1)}^{ j}}}\binom{n}{k-j}\binom{n}{k+j}=\binom{n}{k}.$

Let ${\left( {a;q} \right)_n}=\prod\limits_{j = 0}^{n - 1} {(1-{q^j}a} )$ and let $ {{n}\brack{k}}_q$ denote a $q-$binomial coefficient. I am interested in $q-$analogues of the identity $ \sum\...
12
votes
0answers
467 views

A $q$-analogue of Foulkes' character related to alternating permutations

My paper "Alternating permutations and symmetric functions" at http://math.mit.edu/~rstan/papers/altenum.pdf enumerates certain classes of alternating permutations, such as those whose inverse is ...
3
votes
1answer
262 views

Closed form for a simple hypergeometric $q$ series

I've run across an interesting hypergeometric $q$-series that I feel must have been found before: $\sum_{n=0}^{\infty}(-1)^n$$\frac{e^{n b y}}{\prod_{k=1}^{n}(e^{\pi k b^2}-e^{\pi k b^{-2}})} = \...
1
vote
1answer
162 views

Special values of continuous q - Hermite polynomials

The continuous $q-$Hermite polynomials are defined by $${H_{n + 1}}(x|q) = 2x{H_n}(x|q) +( {q^n}-1){H_{n - 1}}(x|q)$$ with initial values ${H_{ - 1}}(x|q) = 0$ and ${H_0}(x|q) = 1.$ Cf. e.g. http://...
4
votes
1answer
1k views

What does the $q$-Catalan Numbers count?

I had completed a paper describing the $q$-Catalan numbers, which is the $q$-analog of the Catalan numbers. The $n$-th Catalan numbers can be represented by: $$C_n=\frac{1}{n+1}{2n \choose n}$$ and ...
8
votes
1answer
435 views

A q,t-extension of Plancherel Measure thru Yang-Mills Theory ?

Buried in the physics paper by Nekrasov and Okounkov, a strange identity is proven: $$ \prod_{n > 0} (1 - q^n)^{\mu^2-1} = \sum_{\mathbf{k}} q^{|\mathbf{k}|} \prod_{\square \in k} \left( 1 - \frac{\...
4
votes
1answer
630 views

Are the following q-Genocchi numbers known?

The sequence of Genocchi numbers ${({G_{2n}})_{n \ge 0}}=$ $(0,1,1,3,17,155,2073,...)$ can be defined by the generating function $z\frac{{1 - {e^z}}}{{1 + {e^z}}} = \sum {{{( - 1)}^n}{G_{2n}}\frac{...
0
votes
0answers
372 views

Combinatorial Interpretation of an Extension of Gaussian Polynomials

It is well-known that the Gaussian polynomial (or Gaussian coefficient, q-binomial coefficient) $\binom{n}{k}_q$ counts the number of $k$-dimensional subspaces of an $n$-dimensional vector space over $...
12
votes
5answers
1k views

enumerative meaning of natural q-Catalan numbers

Define $[n]=(1-q^n)/(1-q)$ and $[n]!=[1][2][3] \cdots [n]$, so that $[2n]!/[n]![n+1]!$ is a polynomial in $q$ (the most algebraically natural $q$-analogue of the Catalan numbers); what enumerative ...
15
votes
2answers
872 views

Why are some q-analogues more canonical than others?

It is striking that some q-analogs of functions, operators, identities and especially whole theorems seem quite "canonical", e.g. the factorial and the q-Gamma function the basic hypergeometric ...
4
votes
1answer
493 views

q-analog of the matrix exponential

I am a fan of the Matrix exponential $\exp(X)$, defined for any complex matrix $X$ by \begin{equation*} \exp(X) := \sum_{k \ge 0} \frac{X^k}{k!}. \end{equation*} I have a fleeting acquaintance with ...
12
votes
1answer
516 views

$(q,x)$-analog of $n!$

While doing some work in geometric representation theory I have come across the following sequence of polynomials in two variables $(q,x)$ which I would like to denote by $n!_{q,x}$. For small $n$ ...
3
votes
1answer
306 views

Taylor expansion of a q-analog of the negative binomial distribution

Given $A,B \in \mathbb{Z}_+$ and $ 0 < t, q< 1$, I'd like to compute the coefficients $c_n(q,A,B)$ in the expansion of the product $$\prod_{i=0}^{A-1} \prod_{j=0}^{B-1} \frac{1}{1-t q^{i+j}} = \...
34
votes
2answers
2k views

Is there a “quantum” Riemann zeta function?

Occasionally I find myself in a situation where a naive, non-rigorous computation leads me to a divergent sum, like $\sum_{n=1}^\infty n$. In times like these, a standard approach is to guess the ...