Mathematica Asked on September 30, 2021
Consider the function
$$(a;q)_n=begin{cases}
1&n=0,
(1-a)(1-aq)cdots(1-aq^{n-1})&n=1,2,dots,
[(1-aq^{-1})(1-aq^{-2})cdots(1-aq^n)]^{-1}&n=-1,-2,dots
end{cases}$$
which in Mathematica is denoted as QPochhammer[a, q, n]
, and its infinite product cousin:
$$(a;q)_infty=prod_{k=0}^infty (1-aq^k)$$
which in Mathematica reads QPochhammer[a, q]
.
There exists the following relation among the finite product and infinite product expressions:
$$(a;q)_n=frac{(a;q)_infty}{(aq^n;q)_infty}$$
which means that we should have
QPochhammer[x, q]/QPochhammer[q^n x, q] == QPochhammer[x, q, n]
True
However, Mathematica does not produce any output for this input at all. Trying something like
QPochhammer[x, q]/QPochhammer[q^n x, q]//FullSimplify
does not return QPochhammer[x, q, n]
either. It seems that Mathematica is completely oblivious about this relation. Is there any way to get Mathematica to simplify these expressions properly? In general, I am interested in giving Mathematica a ratio of two infinite QPochhammer
functions and let it determine if a reduction to a finite version exists. Can this be done? Thanks for any suggestion!
It is even worse. While
FullSimplify[Product[(1 - q^k z), {k, 0, n - 1}]/Product[(1 - q^k z), {k, 1, n - 1}]]
yields 1-z
,
FullSimplify[QPochhammer[z, q, n]/QPochhammer[q z, q, n - 1]]
just returns the input. On the other hand,
FullSimplify[(1 - q) QGamma[x + 1, q]/QGamma[x, q]]
returns 1 - q^x
.
Answered by Tom Koornwinder on September 30, 2021
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