Difference between revisions of "Q-Gamma"
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| − | <strong>Proposition:</strong> $\Gamma_q( | + | <strong>Proposition:</strong> $\Gamma_q(z+1)=\dfrac{1-q^z}{1-q}\Gamma_q(z)$ |
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Revision as of 15:39, 20 May 2015
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-Pochhammer symbol. The function $\Gamma_q$ is a $q$-analogue of the gamma function.
- Qgamma.png
Graph of $\Gamma_q$ on $[-3,3]$.
Properties
Proposition: $\Gamma_q(n+1)=1(1+q)\ldots(1+q+\ldots+q^{n-1})$
Proof: proof goes here █
Proposition: $\Gamma_q(z+1)=\dfrac{1-q^z}{1-q}\Gamma_q(z)$
Proof: proof goes here █
Proposition: $\Gamma_q(1)=\Gamma_q(2)=1$
Proof: proof goes here █
Theorem ($q$-analog of 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: proof goes here █
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 █
References
Askey, Richard . The q-gamma and q-beta functions. Applicable Anal. 8 (1978/79), no. 2, 125--141.
DLMF entry on q-Gamma and q-Beta functions
