Difference between revisions of "Q-Gamma"

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Let $0<q<1$. Define
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__NOTOC__
$$\Gamma_q(x) = \dfrac{(q;q)_{\infty}}{(q^x;q)_{\infty}}(1-q)^{1-x},$$
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Let $0<q<1$. Define the $q$-gamma function by the formula
where $(\cdot;\cdot)_{\infty}$ denotes the [[q-Pochhammer symbol]].
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$$\Gamma_q(z) = \dfrac{(q;q)_{\infty}}{(q^z;q)_{\infty}}(1-q)^{1-z},$$
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where $(\cdot;\cdot)_{\infty}$ denotes the [[q-shifted factorial]]. The function $\Gamma_q$ is a [[q-analogue | $q$-analogue]] of the [[gamma function]].
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<div align="center">
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<gallery>
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File:Qgamma,q=2plot.png|Graph of $\Gamma_2$.
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File:Complexqgammaq=2plot.png|[[Domain coloring]] of $\Gamma_2$.
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</gallery>
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</div>
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=Properties=
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[[q-Gamma at z+1]]<br />
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[[q-Gamma at 1]]<br />
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[[q-Gamma at 2]]<br />
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem ($q$-Bohr-Mollerup):</strong> Let $f$ be a function which satisfies
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$$f(x+1) = \dfrac{1-q^x}{1-q}f(x)$$
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for some $q \in (0,1)$,
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$$f(1)=1,$$
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and
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$\log f(x)$ is convex for $x>0$.
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Then $f(x) = \Gamma_q(x)$.
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> █
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</div>
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</div>
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<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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<strong>Theorem (Legendre Duplication Formula):</strong> $\Gamma_q(2x)\Gamma_{q^2}\left(\dfrac{1}{2}\right)=\Gamma_{q^2}(x)\Gamma_{q^2}\left( x +\dfrac{1}{2} \right)(1+q)^{2x+1}$
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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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<div class="toccolours mw-collapsible mw-collapsed">
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<strong>Theorem:</strong> ([[q-analog|$q$-analog]]) The following formula holds:
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$$\displaystyle\lim_{q \rightarrow 1^-} \Gamma_q(z) = \Gamma(z),$$
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where $\Gamma_q$ is the [[q-Gamma]] function and $\Gamma$ is the [[gamma]] function.
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<div class="mw-collapsible-content">
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<strong>Proof:</strong> █
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</div>
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</div>
  
 
=References=
 
=References=
Askey, Richard . The q-gamma and q-beta functions. Applicable Anal.  8  (1978/79),  no. 2, 125--141.
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* {{PaperReference|The q-Gamma and q-Beta functions|1978|Richard Askey|next=findme}}
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* {{PaperReference|The q-gamma function for q greater than 1|1980|Daniel S. Moak|next=Q-shifted factorial}}
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{{:q-calculus footer}}
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[[Category:SpecialFunction]]

Latest revision as of 00:13, 30 May 2017

Let $0<q<1$. Define the $q$-gamma function by the formula $$\Gamma_q(z) = \dfrac{(q;q)_{\infty}}{(q^z;q)_{\infty}}(1-q)^{1-z},$$ where $(\cdot;\cdot)_{\infty}$ denotes the q-shifted factorial. The function $\Gamma_q$ is a $q$-analogue of the gamma function.

Properties

q-Gamma at z+1
q-Gamma at 1
q-Gamma at 2

Theorem ($q$-Bohr-Mollerup): Let $f$ be a function which satisfies $$f(x+1) = \dfrac{1-q^x}{1-q}f(x)$$ for some $q \in (0,1)$, $$f(1)=1,$$ and $\log f(x)$ is convex for $x>0$. Then $f(x) = \Gamma_q(x)$.

Proof:

Theorem (Legendre Duplication Formula): $\Gamma_q(2x)\Gamma_{q^2}\left(\dfrac{1}{2}\right)=\Gamma_{q^2}(x)\Gamma_{q^2}\left( x +\dfrac{1}{2} \right)(1+q)^{2x+1}$

Proof: proof goes here █

Theorem: ($q$-analog) The following formula holds: $$\displaystyle\lim_{q \rightarrow 1^-} \Gamma_q(z) = \Gamma(z),$$ where $\Gamma_q$ is the q-Gamma function and $\Gamma$ is the gamma function.

Proof:

References

$q$-calculus