Difference between revisions of "Q-Pochhammer"

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=Properties=
 
=Properties=
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[[Relationship between q-factorial and q-pochhammer]]<br />
<strong>Theorem:</strong> The following formula holds:
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[[Relationship between Euler phi and q-Pochhammer]]<br />
$$(q;q)_{\infty} = \displaystyle\sum_{k=-\infty}^{\infty} (-1)^n q^{\frac{3k^2-k}{2}}.$$
 
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<strong>Proof:</strong> █
 
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<strong>Theorem:</strong> The following formula holds:
 
$$(x;q)_{\infty} = \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kq^{ {k \choose 2} }x^k}{(q;q)_k}.$$
 
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<strong>Proof:</strong> █
 
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{{:Relationship between q-factorial and q-pochhammer}}
 
{{:Relationship between Euler phi and q-Pochhammer}}
 
  
 
{{:q-calculus footer}}
 
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[[Category:SpecialFunction]]
 
[[Category:SpecialFunction]]

Revision as of 03:30, 22 June 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

Relationship between q-factorial and q-pochhammer
Relationship between Euler phi and q-Pochhammer

$q$-calculus