Continuous q-Hermite polynomials

${\displaystyle 2xH_{n}(x\mid q)=H_{n+1}(x\mid q)+(1-q^{n})H_{n-1}(x\mid q)}$

with the initial conditions

${\displaystyle H_{0}(x\mid q)=1,H_{-1}(x\mid q)=0}$

From the above, one can easily calculate:

${\displaystyle {\begin{aligned}H_{0}(x\mid q)&=1\\H_{1}(x\mid q)&=2x\\H_{2}(x\mid q)&=4x^{2}-(1-q)\\H_{3}(x\mid q)&=8x^{3}-2x(2-q-q^{2})\\H_{4}(x\mid q)&=16x^{4}-4x^{2}(3-q-q^{2}-q^{3})+(1-q-q^{3}+q^{4})\end{aligned}}}$

${\displaystyle \sum _{n=0}^{\infty }H_{n}(x\mid q){\frac {t^{n}}{(q;q)_{n}}}={\frac {1}{\left(te^{i\theta },te^{-i\theta };q\right)_{\infty }}}}$

where ${\displaystyle \textstyle x=\cos \theta }$.

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