All Questions
Tagged with q-analogs q-identities
5 questions
12
votes
0
answers
631
views
$q$-analogue of the multinomial theorem?
The $q$-binomial theorem states that
$$
\prod_{k=0}^{n-1}(1+q^kt) = \sum_{k=0}^n q^{\binom k2}{n\brack k}_q t^k.
$$
This identity is a $q$-analogue of the binomial theorem
$$
(1+t)^n = \sum_{k=0}^n \...
9
votes
1
answer
420
views
notation for $(a-b)(a-qb)\dots (a-q^{n-1}b)$
I wonder whether there is a notation for such thing, which I denote $[a;b]_q^n$ for a moment:
$$
[a;b]_q^n:=(a-b)(a-qb)\dots (a-q^{n-1}b)=a^n(b/a;q)_n,
$$
this last equation uses $q$-Pochhammer symbol ...
9
votes
0
answers
192
views
For $q$-analogues of a known curious identity
In 2002 I published the folllowing curious combinatorial identity:
$$(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.$$
My original proof is ...
7
votes
1
answer
320
views
A curious $q$-series identity on a truncated Euler function
Recall that a $q$-Pochhammer symbol is defined as
$$
(x)_n = (x;q)_n := \prod_{l=0}^{n-1}(1-q^l x).
$$
I found the following curious $q$-series identity that seems to hold for any $n\geq 0$:
$$
(-1)^{...
4
votes
1
answer
672
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{...