Askey–Wilson polynomials

Remarks

Since $(a e^{i\theta}, a e^{-i\theta}; q)_k = \prod_{j=0}^{k-1} (1 - 2 a q^j x + a^2 q^{2j})$, each term is a polynomial in $x$, and $p_n$ is a polynomial of degree at most $n$ with leading coefficient $2^n (abcd\, q^{n-1}; q)_n$. The factor $a^{-n} (ab, ac, ad; q)_n$ cancels the denominators, and $p_n$ is symmetric in $a, b, c, d$.

Example: $n = 1$. The ${}_4\phi_3$ has two terms, and since $\frac{(1 - q^{-1})\, q}{1 - q} = -1$, $$p_1(x) = a^{-1} \left[ (1 - ab)(1 - ac)(1 - ad) - (1 - abcd)(1 - 2ax + a^2) \right] = 2(1 - abcd)\, x + (abc + abd + acd + bcd) - (a + b + c + d).$$

Orthogonality: if $0 < q < 1$ and $a, b, c, d$ are real with absolute values less than $1$, then for $m \ne n$ $$\int_{-1}^{1} p_m(x)\, p_n(x)\, w(x)\, \frac{dx}{\sqrt{1 - x^2}} = 0, \qquad w(\cos\theta) = \frac{(e^{2i\theta}, e^{-2i\theta}; q)_\infty}{(a e^{i\theta}, a e^{-i\theta}, b e^{i\theta}, b e^{-i\theta}, c e^{i\theta}, c e^{-i\theta}, d e^{i\theta}, d e^{-i\theta}; q)_\infty}.$$

Non-example: for parameters as in the orthogonality statement, $x^2$ is not $p_2$, since $p_2$ is orthogonal to the constant $p_0 = 1$ for the positive weight $w$, while $\int x^2 w(x) \frac{dx}{\sqrt{1-x^2}} > 0$.