Difference between revisions of "Q-Binomial"

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$${n \brack m}_q = \dfrac{(q;q)_n}{(q;q)m(q;q)_{n-m}},$$
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The $q$-Binomial function is
where $(q;q)_k$ is the [[q-Pochhammer symbol]].
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$${n \brack k}_q = \dfrac{[n]_q!}{[n-k]_q![k]_q!} = \dfrac{(q;q)_n}{(q;q)_k (q;q)_{n-k}},$$
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where $[n]_q!$ denotes the [[q-factorial|$q$-factorial]] $(q;q)_k$ is the [[q-Pochhammer symbol|$q$-Pochhammer symbol]].
  
 
=Properties=
 
=Properties=
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> For $|x|<1,|q|<1$,
 
<strong>Theorem:</strong> For $|x|<1,|q|<1$,
$$\displaystyle\sum_{k=0}^{\infty} \dfrac{(a;q)_k}{(q;q)_k} x^k = \dfrac{(ax;q)_{\infty}}{(x;q)_{\infty}},$$
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$$\displaystyle\sum_{k=0}^{\infty} \dfrac{(a;q)_k}{(q;q)_k} x^k = \dfrac{(ax;q)_{\infty}}{(x;q)_{\infty}}.$$
where $(a;q)_{\xi}$ is the [q-Pochhammer symbol | $q$-Pochhammer symbol].
 
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> proof goes here █  
 
<strong>Proof:</strong> proof goes here █  
 
</div>
 
</div>
 
</div>
 
</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Corollary:</strong>
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* $\displaystyle\sum_{k=0}^{\infty} \dfrac{x^k}{(q;q)_k} = \dfrac{1}{(x;q)_{\infty}}; |x|<1,|q|<1$
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* $\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kq^{k \choose 2}x^k}{(q;q)_k} =(x;q)_{\infty} ; |q|<1$
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* $\displaystyle\sum_{k=0}^N {N \brack k}_q (-1)^k q^{k \choose 2} x^k = (x;q)_N = (1-x)\ldots(1-xq^{N-1})$
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* $\displaystyle\sum_{k=0}^{\infty} {{N+k-1} \brack k}_q x^k = \dfrac{1}{(x;q)_N} = \dfrac{1}{(1-x)\ldots(1-xq^{N-1})} ;|x|<1$
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> proof goes here █
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</div>
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</div>
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=References=
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*Special Functions - G. Andrews, R. Askey, R. Roy
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{{:q-calculus footer}}
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[[Category:SpecialFunction]]

Latest revision as of 18:56, 24 May 2016

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

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
$q$-calculus