Difference between revisions of "Q-Gamma"

From specialfunctionswiki
Jump to: navigation, search
(Properties)
Line 2: Line 2:
 
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]].
+
where $(\cdot;\cdot)_{\infty}$ denotes the [[q-factorial]]. The function $\Gamma_q$ is a [[q-analogue | $q$-analogue]] of the [[gamma function]].
  
 
<div align="center">
 
<div align="center">
Line 46: Line 46:
  
 
=References=
 
=References=
Askey, Richard . The q-gamma and q-beta functions. Applicable Anal.  8  (1978/79),  no. 2, 125--141.<br />
+
* {{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]
 
  
 
{{:q-calculus footer}}
 
{{:q-calculus footer}}
  
 
[[Category:SpecialFunction]]
 
[[Category:SpecialFunction]]

Revision as of 20:04, 18 December 2016

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-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