Difference between revisions of "Q-Pochhammer"
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Revision as of 18:55, 24 May 2016
$$(a;q)_n=\dfrac{(a;q)_{\infty}}{(aq^n;q)_{\infty}}\stackrel{n \in \mathbb{Z}^+}{=} \displaystyle\prod_{j=0}^{n-1} (1-aq^j)$$ $$(a;q)_{\infty} = \displaystyle\prod_{j=0}^{\infty} (1-aq^j)$$
$$(a;q)_{-n}=\dfrac{1} {(aq^{-n};q)_n} =\dfrac{1} {(1-aq^{-n})\ldots(1-aq^{-1})} = \dfrac{q^{\frac{n(n+1)}{2}}(-1)^n}{a^n (\frac{q}{a};q)_n}$$
Properties
Theorem: The following formula holds: $$(q;q)_{\infty} = \displaystyle\sum_{k=-\infty}^{\infty} (-1)^n q^{\frac{3k^2-k}{2}}.$$
Proof: █
Theorem: The following formula holds: $$(x;q)_{\infty} = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kq^{ {k \choose 2} }x^k}{(q;q)_k}.$$
Proof: █
Relationship between q-factorial and q-pochhammer Relationship between Euler phi and q-Pochhammer
