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  "body": "\n## Definition\n\nLet $0 \u003c q \u003c 1$, let $a, b$ be real numbers, and let $n \\ge 0$ be an integer.\nWrite $(c; q)_k$ for the [[q-Pochhammer symbol|Y1X8ZZQ7VM]].\nAssume $aq^{j} \\ne 1$ for $j = 1, \\ldots, n$.\nThe *little $q$-Jacobi polynomial* of degree $n$ is\n$$p_n(x; a, b \\mid q) = {}_2\\phi_1\\!\\left( \\begin{matrix} q^{-n}, \\; abq^{n+1} \\\\ aq \\end{matrix} ; q, \\, qx \\right) = \\sum_{k=0}^{n} \\frac{(q^{-n}; q)_k \\, (abq^{n+1}; q)_k}{(aq; q)_k \\, (q; q)_k} (qx)^k,$$\nwhere ${}_2\\phi_1$ is the [[basic hypergeometric series|EB4N9RQR51]].\nThe series stops at $k = n$, because $(q^{-n}; q)_k = 0$ for $k \u003e n$.\n\n## Remarks\n\nOrthogonality: if $0 \u003c aq \u003c 1$ and $bq \u003c 1$, then for $m \\ne n$\n$$\\sum_{k=0}^{\\infty} \\frac{(bq; q)_k}{(q; q)_k} (aq)^k \\, p_m(q^k) \\, p_n(q^k) = 0,$$\nand the weights $\\frac{(bq; q)_k}{(q; q)_k} (aq)^k$ are positive.\n\nExample: for $n = 1$ the sum has two terms, and $(1 - q^{-1}) q / (1 - q) = -1$ gives\n$$p_1(x) = 1 - \\frac{1 - abq^2}{1 - aq} \\, x.$$\nCheck of orthogonality to $p_0 = 1$ when $b = 0$: the weights are $w_k = (aq)^k / (q; q)_k$, and the $q$-binomial theorem gives $\\sum_k w_k = 1/(aq; q)_\\infty$ and $\\sum_k w_k q^k = 1/(aq^2; q)_\\infty$.\nSo $\\sum_k w_k p_1(q^k) = \\frac{1}{(aq; q)_\\infty} - \\frac{1}{(1 - aq)(aq^2; q)_\\infty} = 0$, since $(aq; q)_\\infty = (1 - aq)(aq^2; q)_\\infty$.\n\nNon-example: for $b = 0$ the monomial $x$ is not the little $q$-Jacobi polynomial of degree $1$, since $x$ is not orthogonal to $p_0 = 1$ for these positive weights: $\\sum_k w_k q^k \u003e 0$.\n\nLimit: $\\lim_{q \\to 1} p_n(x; q^\\alpha, q^\\beta \\mid q) = P_n^{(\\alpha, \\beta)}(1 - 2x) / P_n^{(\\alpha, \\beta)}(1)$, where $P_n^{(\\alpha,\\beta)}$ is the [[Jacobi polynomial|TRD7DSF2KR]].\n",
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