Difference between revisions of "Q-Gamma"
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| − | <strong>Proposition:</strong> $\Gamma_q(1)=1$ | + | <strong>Proposition:</strong> $\Gamma_q(1)=\Gamma_q(2)=1$ |
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<strong>Proof:</strong> proof goes here █ | <strong>Proof:</strong> proof goes here █ | ||
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=References= | =References= | ||
Askey, Richard . The q-gamma and q-beta functions. Applicable Anal. 8 (1978/79), no. 2, 125--141. | Askey, Richard . The q-gamma and q-beta functions. Applicable Anal. 8 (1978/79), no. 2, 125--141. | ||
Revision as of 21:23, 2 October 2014
Let $0<q<1$. Define $$\Gamma_q(x) = \dfrac{(q;q)_{\infty}}{(q^x;q)_{\infty}}(1-q)^{1-x},$$ where $(\cdot;\cdot)_{\infty}$ denotes the q-Pochhammer symbol.
Properties
Proposition: $\Gamma_q(n+1)=1(1+q)\ldots(1+q+\ldots+q^{n-1})$
Proof: proof goes here █
Proposition: $\Gamma_q(x+1)=\dfrac{1-q^x}{1-q}\Gamma_q(x)$
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.
