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 $|x|<1,|q|<1$, $$\displaystyle\sum_{k=0}^{\infty} \dfrac{(a;q)_k}{(q;q)_k} x^k = \dfrac{(ax;q)_{\infty}}{(x;q)_{\infty}}.$$

Proof: proof goes here █

@@ Line 5: / Line 5: @@
 <div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 <strong>Theorem:</strong> For $|x|<1,|q|<1$,
-$$\displaystyle\sum_{k=0}^{\infty} \dfrac{(a;q)_k}{(q;q)_k} x^k = \dfrac{(ax;q)_{\infty}}{(x;q)_{\infty}},$$
+$$\displaystyle\sum_{k=0}^{\infty} \dfrac{(a;q)_k}{(q;q)_k} x^k = \dfrac{(ax;q)_{\infty}}{(x;q)_{\infty}}.$$
-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>