Difference between revisions of "Q-Binomial"
From specialfunctionswiki
(→Properties) |
(→Properties) |
||
| Line 1: | Line 1: | ||
$${n \brack m}_q = \dfrac{(q;q)_n}{(q;q)m(q;q)_{n-m}},$$ | $${n \brack m}_q = \dfrac{(q;q)_n}{(q;q)m(q;q)_{n-m}},$$ | ||
where $(q;q)_k$ is the [[q-Pochhammer symbol]]. | where $(q;q)_k$ is the [[q-Pochhammer symbol]]. | ||
| + | |||
| + | The $q$-Binomial function is | ||
| + | $${n \brack k}_q = \dfrac{(q;q)_n}{(q;q)_k (q;q)_{n-k}}.$$ | ||
=Properties= | =Properties= | ||
| Line 10: | Line 13: | ||
</div> | </div> | ||
</div> | </div> | ||
| + | |||
| + | <div class="toccolours mw-collapsible mw-collapsed" style="width:800px"> | ||
| + | <strong>Corollary:</strong> | ||
| + | * $\displaystyle\sum_{k=0}^{\infty} \dfrac{x^k}{(q;q)_k} = \dfrac{1}{(x;q)_{\infty}}; |x|<1,|q|<1$ | ||
| + | * $\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kq^{k \choose 2}x^k}{(q;q)_k} =(x;q)_{\infty} ; |q|<1$ | ||
| + | * $\displaystyle\sum_{k=0}^N {N \brack k}_q (-1)^k q^{k \choose 2} x^k = (x;q)_N = (1-x)\ldots(1-xq^{N-1})$ | ||
| + | * $\displaystyle\sum_{k=0}^{\infty} {{N+k-1} \brack k}_q x^k = \dfrac{1}{(x;q)_N} = \dfrac{1}{(1-x)\ldots(1-xq^{N-1})} ;|x|<1$ | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> proof goes here █ | ||
| + | </div> | ||
| + | </div> | ||
| + | |||
| + | =References= | ||
| + | *Special Functions - G. Andrews, R. Askey, R. Roy | ||
Revision as of 18:41, 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.
The $q$-Binomial function is $${n \brack k}_q = \dfrac{(q;q)_n}{(q;q)_k (q;q)_{n-k}}.$$
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 █
Corollary:
- $\displaystyle\sum_{k=0}^{\infty} \dfrac{x^k}{(q;q)_k} = \dfrac{1}{(x;q)_{\infty}}; |x|<1,|q|<1$
- $\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kq^{k \choose 2}x^k}{(q;q)_k} =(x;q)_{\infty} ; |q|<1$
- $\displaystyle\sum_{k=0}^N {N \brack k}_q (-1)^k q^{k \choose 2} x^k = (x;q)_N = (1-x)\ldots(1-xq^{N-1})$
- $\displaystyle\sum_{k=0}^{\infty} {{N+k-1} \brack k}_q x^k = \dfrac{1}{(x;q)_N} = \dfrac{1}{(1-x)\ldots(1-xq^{N-1})} ;|x|<1$
Proof: proof goes here █
References
- Special Functions - G. Andrews, R. Askey, R. Roy
