Q-Binomial

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The $q$-Binomial function is $${n \brack k}_q = \dfrac{[n]_q!}{[n-k]_q![k]_q!} = \dfrac{(q;q)_n}{(q;q)_k (q;q)_{n-k}},$$ where $[n]_q!$ denotes the $q$-factorial $(q;q)_k$ is the $q$-Pochhammer symbol.

Properties[edit]

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[edit]

  • Special Functions - G. Andrews, R. Askey, R. Roy
$q$-calculus