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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 ...
Nemo's user avatar
  • 5,624
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 ...
მამუკა ჯიბლაძე's user avatar
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 $...
Edwin Beggs's user avatar
  • 1,143
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} \...
Dong Wang's user avatar
  • 133
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}})} = \...
user40845's user avatar
  • 151
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} \...
Johann Cigler's user avatar