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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.

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$q$-analog and $q$-series [closed]

Does a space of $q$-analogs of series exist? Where we define the following terms $q$-Analog of a Number The $q$-analog of a number $n$, denoted by $[n]_q$, is defined as: $$[n]_q = 1 + q + q^2 + \...
Siddartha11 's user avatar
6 votes
2 answers
210 views

A curious $q$-analog of $1-x$.

Let $$\frac{1}{\sum_{n \geq 0}q^\binom{n}{2}x^n}=\sum_{n \geq 0}u(n,q)x^n.$$ Computations suggest that $u(n,q)$ is a polynomial in $q$ with integer coefficients of degree $\binom{n}{2}.$ The first ...
Johann Cigler's user avatar
2 votes
1 answer
37 views

Prove $\sum_{j=0}^{n} q^{j^{2}}\binom{n}{j}_{q^{2}}$ generates the self-conjugate partitions with part at most $n$.

Prove $\sum_{j=0}^{n} q^{j^{2}}\binom{n}{j}_{q^{2}}$ generates the self-conjugate partitions with part at most $n$, and that it equals $(1+q)(1+q^{3})\cdot\cdot\cdot(1+q^{2n-1})$. For the first part, ...
JLGL's user avatar
  • 795
1 vote
1 answer
36 views

Prove the q-Vandermonde identity $\binom{m+n}{k}_{q} = \sum_{j} \binom{m}{k-j}_{q}\binom{n}{j}q^{(n-j)(k-j)}$ using q-commuting variables

I have seen a few questions on here surrounding the q-Vandermonde identity but in a different form. I've yet to find a proof that uses q-commuting variables. Does anybody have any suggestions on how ...
JLGL's user avatar
  • 795
3 votes
1 answer
96 views

Prove $\sum_{j=0}^{N} {N\brack j} \frac{z^{j}q^{j^{2}}}{(1-zq)(1-zq^{2})\cdot\cdot\cdot(1-zq^{j})}=\prod_{n=0}^{N} \frac{1}{1-zq^{n}}$

I am attempting exercise 101 from Integer Partitions by Andrews and Eriksson. That is, I want to prove $\sum_{j=0}^{N} {N\brack j} \frac{z^{j}q^{j^{2}}}{(1-zq)(1-zq^{2})\cdot\cdot\cdot(1-zq^{j})}=\...
JLGL's user avatar
  • 795
3 votes
0 answers
40 views

Applications of q-Lagrange inversion

I was reading a text on q,t-Catalan numbers and Diagonal Harmonics by Haglund, where they mention the following $q$-analogue of Lagrange Inversion, taken from Page 53: Let $e_n, h_n$ denote the ...
yeetcode's user avatar
  • 143
1 vote
0 answers
38 views

What is meant by "q-generalisation"?

I was reading Prof Gaurav Bhatnagnar, "How to prove Ramanujan q-continued fractions", on the first page he mentions: $$\text{the q-generalisation of } \\ 1+1+1 \cdots + 1 = n \\ \text{is } ...
Nishi's user avatar
  • 31
1 vote
0 answers
29 views

$q$-derivative satisfying linear approximation characterization

Suppose instead of defining the derivative as $$f'(x) = \lim_{h\to 0}\frac{f(x + h) - f(x)}{h}$$ we define it as $$f_q'(x) = \lim_{q\to 1}\frac{f(qx) - f(x)}{qx - x}.$$ Does this alternative ...
node196884's user avatar
1 vote
0 answers
106 views
+50

$q$-Pochhammer at root of unity

Are there any identities, papers/studies, posts, etc that go over $$(\ln\zeta_n^k;q)_{\infty} = \prod_{m=0}^{\infty}(1-\frac{2\pi i k q^m}{n})$$ which is sometimes called the $q$-Pochhammer or quantum ...
Mako's user avatar
  • 690
2 votes
1 answer
85 views

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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3 votes
0 answers
70 views

$(p,q)$-Weyl Algebra

In this Introduction to representation theory they define the '$q$-Weyl algebra by the primary defining relation $$xy = qyx$$ This seems appropriate in $q$-deformations based on the basic building ...
Mako's user avatar
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1 vote
0 answers
30 views

${[n]}_{q,q^{-1}}$ $q$-deformation

It seems that in some $q$-deformations the following definition of a $q$-number is used: $$(n)_q = \frac{q^n-q^{-n}}{q-q^{-1}}$$ If we define $${[n]}_q = \frac{1-q^n}{1-q}$$ as the 'conventional' $q$-...
Mako's user avatar
  • 690
4 votes
1 answer
101 views

Discriminant of numbers

In a previous post on a $q$-analog of number theory I present the fact that any $q$-number can be written as a unique product of cyclotomic polynomials, similar to the fundamental theorem of ...
Mako's user avatar
  • 690
1 vote
0 answers
75 views

Looking for a proper name for this vector subspaces

Let $K^n$ be the vector space of rows of length $n$ over $K$. A subspace $S$ is called coordinately complete if for every $i\ (1 \leq i \leq n)$ there is an element in $S$ with a nonzero $i$th ...
Nikolay Hodyunya's user avatar
12 votes
2 answers
422 views

$q$-analog of Number Theory [closed]

The main motivation behind this is to see whether the 'magic' of q-analogs can be felt in number theory. Obviously for q-analogs to be applied to number theory the parametrization in $q$ must yield a ...
Mako's user avatar
  • 690
4 votes
0 answers
63 views

q-derivative of complex valued functions

I am investigating q-differentiability of complex functions and I am trying to find a q-analog of the Cauchy Riemann conditions. Here is what I am trying to emulate for regular differentiability: $$\...
Corbin Smith's user avatar
7 votes
1 answer
185 views

Alternative q-Analog

Wikipedia gives the following definition of a q-analog: "In mathematics, a q-analog of a theorem, identity or expression is a generalization involving a new parameter $q$ that returns the ...
Mako's user avatar
  • 690
0 votes
0 answers
67 views

q-Leibniz Rule?

The famous Leibniz rule: $(uv)' = u'v+uv'$ can be generalized for any order derivative using the following formula $$(uv)^{(n)} = \sum_{k=0}^n{n\choose k}u^{(n-k)}v^{(k)}$$ This is a known formula ...
Mako's user avatar
  • 690
2 votes
1 answer
91 views

How do I construct a bijection to prove the $q$-binomial theorem?

\begin{align*} \sum_{k=0}^{n} q^{\binom{k}{2}} \binom{n}{k}_q x^k = \prod_{i=0}^{n-1} \left(1+xq^i\right),\quad \forall \ n\ge 1 \end{align*} This problem can be solved by mathematical induction, but ...
Always's user avatar
  • 304
6 votes
1 answer
184 views

Reference request: unusual expansion of product of binomial coefficients

There is a well-known formula for the product of the binomial coefficients: $$\binom{n}{a}\binom{n}{b}=\sum_{i=0}^{min(a,b)}\binom{a+b-i}{i,a-i,b-i}\binom{n}{a+b-i}$$ I'm interested in a different ...
Alvaro Martinez's user avatar
1 vote
0 answers
34 views

How to prove that $e_{q_1}(x)>e_{q_2}(x)$ for $q_1>q_2$?

I am struggling with a simple inequality with $q$-exponential functions. We call $q$-exponential function the following expansion serie : $$e_q(z) = \sum_{n=0}^\infty z^n \frac{(1-q)^n}{(1-q^n)(1-q^{n-...
NancyBoy's user avatar
  • 504
2 votes
1 answer
139 views

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
10 votes
2 answers
305 views

Prove $ \sum_{i=0}^{n}(-1)^{i}q^{(i+1)i/2}{n\choose i}_{q}{n+m-i\choose n}_{q}=1 $

Show that for any non-negative integers $n, m$ such that $n\le m$, we have $$ \sum_{i=0}^{n}(-1)^{i}q^{(i+1)i/2}{n\choose i}_{q}{n+m-i\choose n}_{q}=1 $$ where ${n\choose i}_{q}$ is the Gaussian ...
Bach's user avatar
  • 5,750
4 votes
1 answer
131 views

Is there a $q$-analog for the product of binomial coefficients?

The $q$-analog of the binomial coefficient $\binom{n}{k}$ may be defined as the coefficient of $x^k$ in $\prod_{i=0}^{n-1}(1+q^ix)$. Classical arithmetic identities tend to have $q$-analogs. I am ...
Alvaro Martinez's user avatar
2 votes
0 answers
49 views

q analogue of a number is a polynomial in $[2]_q$

$[m]_q = \frac{q^m−q^{-m}}{q-q^{-1}}$ is the q- analogue of the number $m\in\mathbb{Z}_{\geq0}$. $[0]_q=0$, $[1]_q=1$ and $[2]_q=q+q^{-1}$. I don't know how to prove Every $[m]_q$ can be expressed as ...
Learner's user avatar
  • 369
3 votes
1 answer
67 views

Question about Q-analogs

I am trying to prove the following Given $n \in \mathbb{N}$ we define $[n]_{q} = (1-q^{n})/(1-q)$. We also define $[n]_{q} ! = [n-1]_{q} ! \cdot [n]_{q}$, with $[1]_{q} ! =1$. Then I want to prove the ...
P.Luis's user avatar
  • 561
26 votes
0 answers
960 views

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
  • 3,136
3 votes
0 answers
69 views

Common divisor of Gaussian coefficient expressions

I have a question about common divisors of some expressions involving Gaussian coefficients, in particular in the case ${n \brack 1}_{q} = \frac{q^{n}-1}{q-1}$ where $q$ is a prime power. It is well ...
xxxxxxxxx's user avatar
  • 13.4k
-1 votes
1 answer
78 views

1st order q-differential equations

Can we solve 1st order q-differential equations using the usual methods of 1st order differential equations? For example, can we use integration factor method to solve this q-differential equations? $$...
LAMA S.M's user avatar
10 votes
0 answers
229 views

What is the inverse of the $q$-exponential?

The $q$-exponential function is given by the power series $$ e_q(x) = \sum_{n=0}^\infty \frac{x^n}{[n]!} $$ using the $q$-integers $[k]:=q^{k-1}+\cdots+q+1=(q^k-1)/(q-1)$ and $q$-factorials $[n]!=[n][...
anon's user avatar
  • 152k
2 votes
0 answers
47 views

A $\Bbb Z$-valued $\Bbb Q$-coefficient polynomial is a $\Bbb Z$-combination of $\binom{x}{k}$s: $q$-analog?

I am wondering what a $q$-analog of the following should be: Proposition. If $f(x)\in\Bbb Q[x]$ is a polynomial for which $f(\Bbb Z)\subseteq\mathbb{Z}$ (that is, $f(x)$ is an integer for all integers ...
anon's user avatar
  • 152k
2 votes
1 answer
97 views

On q-analogs of integer inequalities

I've been studying properties of q-analogs recently and had some difficulties on trying to prove some basic-looking inequalities. For $q \in (0, 1)$ and $n\in\mathbb{N}$ let's define q-integer $[n]_q =...
ascoli's user avatar
  • 43
6 votes
1 answer
335 views

Combinatorial interpretation of $\frac{1}{1-e^{x+1}}$ or $\frac{1}{1-e^{x-1}}$

After like 100 hours searching info to unravel a mystery about expressions on $e$ like $\frac{e^{6} A_{5}\left(e^{-1}\right)}{e^{6}\left(1-e^{-1}\right)^{6}} \approx 5 !$ talked by me in here, user ...
Alejandro Quinche's user avatar
1 vote
0 answers
52 views

Discriminant of q-analogs of positive integers

Computations suggest that the discriminant of the polynomial $[n]_{q^m}=1+q^{m}+\dots+q^{(n-1)m}$ with respect to $q$ is $ \pm {\left( {m{{(mn)}^{n - 2}}} \right)^m}.$ Could you please give me a ...
Johann Cigler's user avatar
2 votes
0 answers
51 views

$q$-analog generalization for limits not going to 1

I have been recently introduced to $q$-analogs, and find them quite interesting, specially for their connections going from Ramanujan expressions to very broad aplicable subjects. As I humbly ...
Alejandro Quinche's user avatar
1 vote
0 answers
46 views

Jacobi Quintuple Product Identity

I have to show the Jacobi Quintuple Product Identity $$\prod_{n = 1}^{\infty} (1-q^n)(1- \zeta q^{n-1})(1-\zeta^{-1} q^{n})(1-\zeta^{2} q^{2n-1})(1-\zeta^{-2}q^{2n-1}) = \sum_{n \in \mathbb{Z}} q^{\...
Yasmin's user avatar
  • 31
6 votes
0 answers
119 views

Linear-algebraic interpretation of $q$-multichoose

The $q$-analog $[n]$ of a whole number $n$ is $q^{n-1}+\cdots+q+1$, in which case the binomial and $q$-binomial are $$ \binom{n}{k}=\frac{n(n-1)\cdots}{k(k-1)\cdots} \qquad \left[\begin{matrix} n \\ k ...
anon's user avatar
  • 152k
3 votes
1 answer
182 views

The $q$ Multilinear Theorem

Let $R$ be the skew polynomial ring $k_\mathfrak{q}[x_1,\ldots,x_m]$ where $x_ix_j=qx_jx_i$ with $q\in k^*$ and for all $i<j$. The $q$ Multinomial Theorem states that $$(x_1+\ldots+x_m)^r=\Sigma_{...
user569423's user avatar
3 votes
0 answers
52 views

Divisor Polynomials

For an integer $n>0$, we may define a polynomial $$p_n(x) = \sum_{d\mid n} x^d.$$Have such polynomials been seriously studied and do they have any interesting properties? Looking at this more, I ...
Zach Hunter's user avatar
  • 1,828
11 votes
1 answer
194 views

Recurrence formula of the MacMahon $q$-analog of the Catalan numbers

Catalan number is defined by $C_{n}=\frac{1}{n+1}\binom{2n}{n}.$ Two natural $q$-analogs of Catalan numbers are (see Carlitz and Scoville, A note on weighted sequences, Fibonacci Quarterly, 13 (1975), ...
J. Yomcha's user avatar
  • 123
1 vote
1 answer
91 views

Is the Jackson integral of $e_q(x)$ , $e_q(x)$ itself?

Some info- Jackson integral (q-analog of standard integration) simply defined as -$$\int f(x){\mathrm d}_qx=(1-q)x\sum_{k=0}^\infty q^kf(q^kx)$$ q-exponential defined as $$e_q(x)=\sum_{n=0}^\infty \...
pjmathematician's user avatar
1 vote
0 answers
187 views

Calculate $\sum_{ k = 1 } ^ n \ln ( 1 - x ^ k )$

I am trying to calculate the following sum $$f_n ( x ) = \sum_{ k = 1 } ^ n \ln ( 1 - x ^ k ), \qquad x \geq 0.$$ I first tried to use the power expansion of $\ln(1-u) = - \sum_{ m = 1 } ^ { \infty } \...
Marie's user avatar
  • 53
2 votes
1 answer
213 views

What are the asymptotics of the $q$-binomial?

I have a rather basic question regarding the $q$-binomial $\begin{bmatrix}N \\ r \end{bmatrix}=\frac{(1-q^N)(1-q^{N-1} ) \dots (1-q^{N-r+1})}{ (1-q)(1-q^2)\dots(1-q^r) }$ as $N$ goes to infinity. ...
WLV's user avatar
  • 139
0 votes
1 answer
257 views

$q$-analogue of $\sum_{k=0}^n \, {n \choose k} = 2^n $

Is there a $q$-analogue of the formula $\sum_{k=0}^n \, {n \choose k} = 2^n $ in terms of the $q$-binomial coefficient ${n \choose k}_q$ and $(2^n)_q=(1+q)...(1+q^n)$?
Bipolar Minds's user avatar
1 vote
0 answers
326 views

Area under a lattice path, from q-binomial to q-multinomial

The $q$-binomial ${n+k \choose k}_q$ is a polynomial in $q$. The coefficient of each power $t \leq nk$ of $q$ gives the number of lattice paths from $(0,0)$ to $(n,k)$ which enclose an area $t$ ...
apg's user avatar
  • 2,797
1 vote
0 answers
56 views

An identity for generalized q-Fibonacci polynomials

Let $k$ be a positive integer and consider the generalized $q-$Fibonacci polynomials $F_n^{(k)}(x)$ which satisfy the recursion $F_n^{(k)}(x) = xF_{n - 1}^{(k)}(x) + q^{n-k} F_{n - k}^{(k)}(x)$ with ...
Johann Cigler's user avatar
2 votes
0 answers
86 views

Why is $q$ sometimes a complex number, but other times a prime power?

In the fields of representation theory and quantum algebra, we often start with some $\mathbf{C}$-algebra and study it's quantization as an algebra over $\mathbf{C}(q)$, using the algebra structure to ...
Mike Pierce's user avatar
1 vote
1 answer
52 views

Studying quantized algebras, why introduce $q^{1/2}$ instead of just $q$?

In the fields of representation theory and quantum algebra, we often study quantized versions of algebraic objects by regarding them as algebras over $\mathbf{C}(q)$, or some subring of $\mathbf{C}(q)$...
Mike Pierce's user avatar
5 votes
1 answer
429 views

Gaussian binomial coefficients, lattice paths, and vector spaces

The Gaussian binomial coefficient ${n+k \choose k}_q$ gives a probability generating function for the number of lattice paths from $(0,0)$ to $(n,k)$ enclosing an area $a$ in the upper-right quadrant ...
apg's user avatar
  • 2,797
4 votes
0 answers
531 views

Why is the "q-Pochhammer Symbol" referred to as such, despite not being a q-analog of the Pochhammer symbol?

The q-Pochhammer symbol (for $k>0$): $(a;q)_k = \prod\limits_{j=0}^{k-1}\left(1-aq^j\right)$ The Pochhammer symbol / rising factorial: $(a)_k = \prod\limits_{j=0}^{k-1}(a+j)$ The falling ...
H.v.M.'s user avatar
  • 165