All Questions
Tagged with q-analogs special-functions
8 questions
4
votes
1
answer
293
views
Double q-analog of Pochhammer
Has the function
$$(z;q_1,q_2)_\infty := \prod_{n_1,n_2=0}^\infty (1-z \, q_1^{n_1} q_2^{n_2}), \quad |q_1|,|q_2|<1$$
been studied in the math literature? For example, does it obey any difference ...
5
votes
1
answer
401
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 ...
3
votes
0
answers
106
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 ...
8
votes
3
answers
417
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
0
answers
388
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)
$$
{}_2\phi_1(a, b; c; q, z) = \sum^{\infty}_{n = 0} \...
3
votes
1
answer
1k
views
Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?
The formula
$$
\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,-...
3
votes
1
answer
348
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$$\frac{e^{n b y}}{\prod_{k=1}^{n}(e^{\pi k b^2}-e^{\pi k b^{-2}})} = \...
40
votes
1
answer
2k
views
Curious $q$-analogues
Consider the Fibonacci polynomials
$$F_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \right\rfloor }\binom{n-j}{j} x^{n - 2j} $$
and the Lucas polynomials
$$L_n (x) = \sum_{j = 0}^{\left\lfloor {n/2} \...