Use this tag for questions pretaining to q-analogs of functions, for example q-Binomials, $q$-derivatives, the q-theta function, the q-Pochammer symbol, etc.
5
votes
1answer
61 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 ...
1
vote
1answer
54 views
Combinatorial identity. Using echelon matrices.
Determine the exponents $e_i$ s.t. the following identity is correct.
$$\sum\limits_{i=0}^k q^{e_i} {\binom mi}_q {\binom{n}{k-i}}_q = {\binom{n+m}{k}}_q$$
Note: When $q=1$ the equation reduces to a ...
5
votes
3answers
101 views
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:
...
6
votes
1answer
182 views
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 ...
4
votes
0answers
60 views
A general Combinatorics problem (Coefficients of the q factorial)
I was solving a combinatorics problem when I encountered difficulties. The problem was:
$x_1 \in \{0,1\}$
$x_2 \in \{0,1,2\}$
.
.
$x_{n-1}\in\{0,1,2..,n-1\}$
We have to find the number of ways ...
3
votes
4answers
337 views
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 ...
5
votes
0answers
133 views
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 ...
3
votes
1answer
128 views
Reciprocity Law of the Gaussian (or $q$-Binomial) Coefficient
It is a standard exercise in combinatorics to show that the binomial coefficient satisfies the reciprocity law $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ for $n, k \geqslant 0$, which is the multiset ...
4
votes
0answers
53 views
Necessary and sufficient condition for $f(q^n)$ to be in $\mathbb{Z}[q,q^{-1}]$ when $f\in\mathbb{Q}(q)[x]$?
In this question, user bgins shows that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are in $\mathbb{Q}(q)$, the field of rational functions, such that ...
7
votes
2answers
217 views
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 ...
3
votes
2answers
188 views
Recurrence for $q$-analog for the Stirling numbers?
I read in some papers that the Stirling numbers (of the second kind) have a natural $q$-analog $S_q(n,k)$, which satisfy the recurrence
$$
S_q(n,k)=(k)_qS_q(n-1,k)+q^{k-1}S_q(n-1,k-1)
$$
with the ...
1
vote
1answer
150 views
Inferring Jacobi Triple product from $q$-binomial theorem?
Quite a while ago I asked a question about deducing the Jacobi triple product from the $q$-binomial theorem. In fact, from the $q$-binomial theorem
$$
...
2
votes
1answer
148 views
Is the following product of $q$-binomial coefficients a polynomial in $q$?
$$\frac{\binom{n}{j}_q\binom{n+1}{j}_q \cdots\binom{n+k-1}{j}_q}{\binom{j}{j}_q\binom{j+1}{j}_q\cdots\binom{j+k-1}{j}_q}$$ where $n,j,k$ are non-negative integers.
6
votes
1answer
219 views
Intermediate step in deducing Jacobi's triple product identity.
An intermediate step deduces Jacobi's triple product identity by taking the $q$-binomial theorem
$$
\prod_{i=1}^{m-1}(1+xq^i)=\sum_{j=0}^m\binom{m}{j}_q q^{\binom{j}{2}}x^j
$$
and deducing
$$
...
1
vote
2answers
243 views
Representing the $q$-binomial coefficient as a polynomial with coefficients in $\mathbb{Q}(q)$?
Trying a bit of combinatorics this winter break, and I don't understand a certain claim.
The claim is that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are ...
2
votes
1answer
119 views
Primitive roots as roots of a $q$-multinomial.
If $n$ is divisible by $m$, why is it the case that the $m$th primitive roots of unity are also roots of $\binom{n}{k}_q$ if and only if $m$ does not divide $k$?
I'm viewing $\binom{n}{k}_q$ as a ...
4
votes
1answer
152 views
Deriving Cauchy's identity from the $q$-binomial theorem?
Cauchy's identity states that
$$
\prod_{i\geq 0}\frac{1-axq^i}{1-xq^i}=\sum_{n\geq 0}\frac{(1-a)(1-aq)\cdots(1-aq^{n-1})}{(1-q)(1-q^2)\cdots(1-q^n)}x^n.
$$
Is it possible to somehow derive this ...
3
votes
1answer
130 views
What's the reasoning for this recurrence on $q$-multinomial coefficients?
I'm familiar with the recurrence for binomial coefficients based on Pascal's triangle. However, in general, there is the recurrence for $q$-multinomial coefficients given by
$$
...
2
votes
1answer
114 views
Closed form for $\sum_{m \geq 1} (-1)^m q^{m(m+1)/2 + m \Delta}$?
Is there a useful closed form for the following series ($|\Delta|$ is a small integer)?
$$f(q,\Delta) =\sum_{m=1}^{\infty} (-1)^m q^{m(m+1)/2 + m \Delta}$$
It is a large-$n$ approximation of the ...
13
votes
4answers
447 views
Intriguing polynomials coming from a combinatorial physics problem
For real $0<q<1$, integer $n >0 $ and integer $k\ge 0$, define
$$[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})$$
...
1
vote
1answer
105 views
Interpolation of a sequence of polynomials (viewed in terms of q-analogue powers)
I'm trying to find closed-form expressions for a sequence of coefficients, such that the index of the coefficient occurs as number such that I can later interpolate to fractional indexes as well. ...
7
votes
1answer
245 views
Is there a gamma-like function for the q-factorial?
I'm looking at quantum calculus and just trying to understand what is going with this subject. Looking at the q-factorial made me wonder if this function could take all real or even complex numbers in ...
6
votes
2answers
239 views
What is Eulerian?
I encountered an interesting function which is called "Eulerian" by the Wolfram's MathWorld:
$$\phi(q)=\prod_{k=1}^{\infty} (1-q^{k})$$
It is interesting because it seems that roots of any ...
