Difference between revisions of "Q-Pochhammer"

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$$(a;q)_{-n}=\dfrac{1} {(aq^{-n};q)_n} =\dfrac{1} {(1-aq^{-n})\ldots(1-aq^{-1})} = \dfrac{q^{\frac{n(n+1)}{2}}(-1)^n}{a^n (\frac{q}{a};q)_n}$$
 
$$(a;q)_{-n}=\dfrac{1} {(aq^{-n};q)_n} =\dfrac{1} {(1-aq^{-n})\ldots(1-aq^{-1})} = \dfrac{q^{\frac{n(n+1)}{2}}(-1)^n}{a^n (\frac{q}{a};q)_n}$$
 
<div align="center">
 
<gallery>
 
File:Qpochhammer(q,q)infty.png|Plot of $(q,q)_{\infty}$ for $q \in [-1,1]$.
 
File:Complex qpochhammer (q,q) infty.png|[[Domain coloring]] of [[analytic continuation]] of $(q,q)_{\infty}$ to the unit disk.
 
</gallery>
 
</div>
 
  
 
=Properties=
 
=Properties=

Revision as of 00:15, 5 May 2015

$$(a;q)_n=\dfrac{(a;q)_{\infty}}{(aq^n;q)_{\infty}}\stackrel{n \in \mathbb{Z}^+}{=} \displaystyle\prod_{j=0}^{n-1} (1-aq^j)$$ $$(a;q)_{\infty} = \displaystyle\prod_{j=0}^{\infty} (1-aq^j)$$

$$(a;q)_{-n}=\dfrac{1} {(aq^{-n};q)_n} =\dfrac{1} {(1-aq^{-n})\ldots(1-aq^{-1})} = \dfrac{q^{\frac{n(n+1)}{2}}(-1)^n}{a^n (\frac{q}{a};q)_n}$$

Properties

Relationship between q-factorial and q-pochhammer


$q$-calculus