Series expansion of QPochhammer symbol

Mathematica has an internal function QPochhammer[a,q,n] which is given by $$\text{QPochhammer}[a,q,n]=\frac{\text{QPochhammer}[a,q]}{\text{QPochhammer}[a q^n,q]}=\frac{\prod_{k\geq 1}(1-a q^k)}{\prod_{k\geq 1}(1-a q^{n+k})}.$$

While I can series expand QPochhammer[q,q], I get the following when I try to expand QPochhammer[q,q,n].

In[1220]:= Series[QPochhammer[q, q, 2], {q, 0, 3}]

Out[1220]= SeriesData[q, 0, {
1, Derivative[0, 1, 0][QPochhammer][0, 0, 2] + Derivative[1, 0, 0][
QPochhammer][0, 0, 2],
Rational[1, 2] (
Derivative[0, 2, 0][QPochhammer][
0, 0, 2] + 2 Derivative[1, 1, 0][QPochhammer][
0, 0, 2] + Derivative[2, 0, 0][QPochhammer][0, 0, 2]),
Rational[1, 6] (
Derivative[0, 3, 0][QPochhammer][
0, 0, 2] + 3 Derivative[1, 2, 0][QPochhammer][
0, 0, 2] + 3 Derivative[2, 1, 0][QPochhammer][
0, 0, 2] + Derivative[3, 0, 0][QPochhammer][0, 0, 2])}, 0, 4, 1]


Is there a way to express this in terms of polynomials of q, and not have Derivative of QPochhammers?

• Which version are you using? Commented Nov 18, 2021 at 13:26

The series might not make fully sense as it is already a polynomial:

QPochhammer[q, q, 2] // FunctionExpand


gives:

(1 - q) (1 - q^2)


If I input your code I get:

Series[QPochhammer[q, q, 2], {q, 0, 3}]


Gives:

1-q-q^2+q^3+O[q]^4


in version 12.3.1.

• But it would be much more interesting to expand Series[QPochhammer[q, q, n], {n, Infinity, 1}] which is returned unevaluated. Commented Nov 18, 2021 at 14:26
• I think this might not be possible since it is alternating: DiscretePlot[QPochhammer[6, 6, n], {n, 0, 10}] Commented Nov 18, 2021 at 15:59
• I meant case 0 < q < 1, try 1/6 instead of 6. Commented Nov 18, 2021 at 16:04
• In closed form, QPochhammer[1/6, 1/6, n] / QPochhammer[1/6, 1/6] = 1 + 1/(5*6^n) + 6/(175*6^(2*n)) + ..., it would be good if Mathematica could generate this. Commented Nov 18, 2021 at 17:31
• @VaclavKotesovec Fixing minor issues, (Series[AsymptoticRSolveValue[y[n]==QPochhammer[1/6,1/6,n]/QPochhammer[1/6,1/6]//FunctionExpand,y[n],{n,\[Infinity],5}]/.n->Log[6,1/x]//Simplify,{x,0,5}]//Normal)/.x->6^-n Giving a result of $\frac{6^{10-5 n}}{378832015625}+\frac{6^{6-4 n}}{48724375}+\frac{6^{3-3 n}}{37625}+\frac{1}{175} 6^{1-2 n}+\frac{6^{-n}}{5}+1$
– Nugi
Commented Nov 19, 2021 at 7:37