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`QPochhammer` function simplification?

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!

One Answer

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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