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.

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

Read more about $q$-analogs in this Wikipedia article.

104 questions
81
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1 answer

Conjectured formula for the Fabius function

The Fabius function is the unique function ${\bf F}:\mathbb R\to[-1, 1]$ satisfying the following conditions: a functional–integral equation$\require{action} \require{enclose}{^{\texttip{\dagger}{a poet or philosopher could say "it knows and…
22
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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 coefficient) is defined as $$\binom{n}{k}_q :=…
Henry
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14
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Different notions of q-numbers

It seems that most of the literature dealing with q-analogs defines q-numbers according to $$[n]_q\equiv \frac{q^n-1}{q-1}.$$ Even Mathematica uses this definition: with the built-in function QGamma you obtain QGamma[n+1,q] / QGamma[n,q] = (q^n-1) /…
13
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4 answers

Intriguing polynomials coming from a combinatorial physics problem

For real $00 $ and integer $k\ge 0$, define $$[k, n]_q \equiv -\sum_{m=1}^{n} q^{m(k+1)} (q^{-n}; q)_m = -\sum_{m=1}^{n} q^{m(k+1)} \prod_{l=0}^{m-1} (1-q^{l-n})$$ where $(\cdot\; ; q)_n$ is a $q$-Pochhammer symbol. These…
13
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1 answer

Which families of groups have interesting formulas for the number of elements of given order?

Suppose that $G$ is a group and that $n$ is a positive integer diving the order of $G$. Let $f_n(G)$ be the number of elements satisfying $x^n = 1$ in $G$. According to a theorem of Frobenius, then we have $f_n(G) \equiv 0 \mod{n}$. Hence if we have…
11
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1 answer

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 $P_{n,m}$ of partitions with at most $n$ parts,…
10
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0 answers

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 $[n]_q! := 1 (1 + q) (1 + q + q^2) \cdots (1 + q + \cdots + q^{n-1})$ is the…
Penchez
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9
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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 $[n]!=[n][n-1]\cdots[2][1]$. What is the inverse function…
runway44
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9
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1 answer

Is there a gamma-like function for the q-factorial?

I'm looking at quantum calculus and just trying to understand what is going with this subject. Looking at the q-factorial made me wonder if this function could take all real or even complex numbers in the same way that $\Gamma (z)$ works as an…
9
votes
4 answers

Proving q-binomial identities

I was wondering if anyone could show me how to prove q-binomial identities? I do not have a single example in my notes, and I can't seem to find any online. For example, consider: ${a + 1 + b \brack b}_q = \sum\limits_{j=0}^{b} q^{(a+1)(b-j)}{a+j…
Nizbel99
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8
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3 answers

How to prove it? (one of the Rogers-Ramanujan identities)

Prove the following identity (one of the Rogers-Ramanujan identities) on formal power series by interpreting each side as a generating function for partitions: $$1+\sum_{k\geq1}\frac{z^k}{(1-z)(1-z^2)\cdots(1-z^k)}=\prod_{k\geq1}\frac{1}{1-z^k}$$
8
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1 answer

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 $$ \binom{n+1}{k+1}=\binom{n}{k}+\binom{n-1}{k}+\cdots+\binom{k}{k}. \tag{2} $$ The reason for the name is that if…
8
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1 answer

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 like that? Background: If $q$ is a prime power, then…
Martin Brandenburg
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8
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q-Analogue of the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$.

The stirling numbers of the second kind satisfy the formula $x^n=\sum_k\left\{n\atop k\right\}(x)_k$, where $(x)_k$ is the falling factorial. Consider the $q$-analog recursive definition of the stirling numbers, given by $$ \left\{n\atop…
7
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0 answers

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, $\mathrm{fac}_1(n) = n!$. Let $\alpha \in…
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