Difference between revisions of "Q-Binomial"
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| − | $${n \brack | + | The $q$-Binomial function is |
| + | $${n \brack k}_q = \dfrac{(q;q)_n}{(q;q)_k (q;q)_{n-k}},$$ | ||
where $(q;q)_k$ is the [[q-Pochhammer symbol]]. | where $(q;q)_k$ is the [[q-Pochhammer symbol]]. | ||
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=Properties= | =Properties= | ||
Revision as of 18:41, 27 July 2014
The $q$-Binomial function is $${n \brack k}_q = \dfrac{(q;q)_n}{(q;q)_k (q;q)_{n-k}},$$ where $(q;q)_k$ is the q-Pochhammer symbol.
Properties
Theorem: For $|x|<1,|q|<1$, $$\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 █
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
