Difference between revisions of "Q-Pochhammer"
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| − | $$(a;q)_n=\dfrac{(a;q)_{\infty}}{(aq^n;q)_{\infty}}\stackrel{n \in \mathbb{Z}^+}{=} \displaystyle\prod_{j=0}^{n-1} (1-aq^ | + | $$(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^k)$$ | $$(a;q)_{\infty} = \displaystyle\prod_{j=0}^{\infty} (1-aq^k)$$ | ||
$$(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}$$ | ||
Revision as of 02:18, 23 September 2014
$$(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^k)$$
$$(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}$$
