Difference between revisions of "Generalized q-Bessel"
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<strong>Theorem:</strong> The following formula holds for $n \in \mathbb{Z}$: | <strong>Theorem:</strong> The following formula holds for $n \in \mathbb{Z}$: | ||
$$J_n(-x,a;q)=(-1)^nJ_n(x,a;q).$$ | $$J_n(-x,a;q)=(-1)^nJ_n(x,a;q).$$ | ||
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<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
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Revision as of 08:28, 18 May 2015
The generalized $q$-Bessel is defined by $$J_n(x,a;q) = \dfrac{(\frac{x}{2})^n}{(q;q)_n} \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^{\frac{ak}{2}(k+n)}}{(q^{n+1};q)_k} \dfrac{(\frac{x^2}{4})^k}{(q;q)_k}.$$
Properties
Theorem: The following formula holds for all $n \in \mathbb{Z}$: $$J_{-n}(x,a;q)=(-1)^{n(a+1)}J_n(x,a;q).$$
Proof: █
Theorem: The following formula holds for $n \in \mathbb{Z}$: $$J_n(-x,a;q)=(-1)^nJ_n(x,a;q).$$
Proof: █
Theorem: The following formula holds: $$J_n(x,a;q)=\dfrac{2}{x}(1-q^{n+1})J_{n+1}\left(q^{-\frac{a}{4}}x,a;q \right)+(-1)^{a+1}q^{\frac{(a+2)(n+1)}{2}} J_{n+2}(x,a;q).$$
Proof: █
