Difference between revisions of "Q-Gamma"

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(Properties)
(References)
 
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Let $0<q<1$. Define the $q$-gamma function by the formula
 
Let $0<q<1$. Define the $q$-gamma function by the formula
 
$$\Gamma_q(z) = \dfrac{(q;q)_{\infty}}{(q^z;q)_{\infty}}(1-q)^{1-z},$$
 
$$\Gamma_q(z) = \dfrac{(q;q)_{\infty}}{(q^z;q)_{\infty}}(1-q)^{1-z},$$
where $(\cdot;\cdot)_{\infty}$ denotes the [[q-Pochhammer symbol]]. The function $\Gamma_q$ is a [[q-analogue | $q$-analogue]] of the [[gamma function]].
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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]].
  
 
<div align="center">
 
<div align="center">
 
<gallery>
 
<gallery>
File:Qgamma.png|Graph of $\Gamma_q$ on $[-3,3]$.
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File:Qgamma,q=2plot.png|Graph of $\Gamma_2$.
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File:Complexqgammaq=2plot.png|[[Domain coloring]] of $\Gamma_2$.
 
</gallery>
 
</gallery>
 
</div>
 
</div>
  
 
=Properties=
 
=Properties=
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
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[[q-Gamma at z+1]]<br />
<strong>Proposition:</strong> $\Gamma_q(n+1)=1(1+q)\ldots(1+q+\ldots+q^{n-1})$
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[[q-Gamma at 1]]<br />
<div class="mw-collapsible-content">
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[[q-Gamma at 2]]<br />
<strong>Proof:</strong> proof goes here █
 
</div>
 
</div>
 
 
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Proposition:</strong> $\Gamma_q(z+1)=\dfrac{1-q^z}{1-q}\Gamma_q(z)$
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> proof goes here █
 
</div>
 
</div>
 
  
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
<strong>Proposition:</strong> $\Gamma_q(1)=\Gamma_q(2)=1$
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<strong>Theorem ($q$-Bohr-Mollerup):</strong> Let $f$ be a function which satisfies
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> proof goes here █
 
</div>
 
</div>
 
 
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem ($q$-analog of Bohr-Mollerup):</strong> Let $f$ be a function which satisfies
 
 
$$f(x+1) = \dfrac{1-q^x}{1-q}f(x)$$
 
$$f(x+1) = \dfrac{1-q^x}{1-q}f(x)$$
 
for some $q \in (0,1)$,  
 
for some $q \in (0,1)$,  
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Then $f(x) = \Gamma_q(x)$.
 
Then $f(x) = \Gamma_q(x)$.
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
<strong>Proof:</strong> proof goes here █  
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<strong>Proof:</strong> █  
 
</div>
 
</div>
 
</div>
 
</div>
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<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> proof goes here █  
 
<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> █
 
</div>
 
</div>
 
</div>
 
</div>
  
 
=References=
 
=References=
Askey, Richard . The q-gamma and q-beta functions. Applicable Anal.  8  (1978/79),  no. 2, 125--141.<br />
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* {{PaperReference|The q-Gamma and q-Beta functions|1978|Richard Askey|next=findme}}
[http://dlmf.nist.gov/5.18 DLMF entry on q-Gamma and q-Beta functions]
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* {{PaperReference|The q-gamma function for q greater than 1|1980|Daniel S. Moak|next=Q-shifted factorial}}
  
 
{{:q-calculus footer}}
 
{{: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