Difference between revisions of "Q-Binomial"
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(Created page with "$${n \brack m}_q = \dfrac{(q;q)_n}{(q;q)m(q;q)_{n-m}},$$ where $(q;q)_k$ is the q-Pochhammer symbol.") |
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$${n \brack m}_q = \dfrac{(q;q)_n}{(q;q)m(q;q)_{n-m}},$$ | $${n \brack m}_q = \dfrac{(q;q)_n}{(q;q)m(q;q)_{n-m}},$$ | ||
where $(q;q)_k$ is the [[q-Pochhammer symbol]]. | where $(q;q)_k$ is the [[q-Pochhammer symbol]]. | ||
| + | |||
| + | =Properties= | ||
| + | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px"> | ||
| + | <strong>Theorem:</strong> For $|x|<1,|q|<1$, | ||
| + | $$\displaystyle\sum_{k=0}^{\infty} \dfrac{(a;q)_k}{(q;q)_k} x^k = \dfrac{(ax;q)_{\infty}}{(x;q)_{\infty}},$$ | ||
| + | where $(a;q)_{\xi}$ is the [q-Pochhammer symbol | $q$-Pochhammer symbol]. | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> proof goes here █ | ||
| + | </div> | ||
| + | </div> | ||
Revision as of 18:32, 27 July 2014
$${n \brack m}_q = \dfrac{(q;q)_n}{(q;q)m(q;q)_{n-m}},$$ where $(q;q)_k$ is the q-Pochhammer symbol.
Properties
Theorem: For $|x|<1,|q|<1$, $$\displaystyle\sum_{k=0}^{\infty} \dfrac{(a;q)_k}{(q;q)_k} x^k = \dfrac{(ax;q)_{\infty}}{(x;q)_{\infty}},$$ where $(a;q)_{\xi}$ is the [q-Pochhammer symbol | $q$-Pochhammer symbol].
Proof: proof goes here █
