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Gasper-Rahmanによるbibasic超幾何級数の和公式

次はGasper-Rahmanによって1990年に示された等式である.

Gasper-Rahman(1990)

$\begin{aligned} \sum_{k = - m}^{n} \frac{(1 - a d p^{k} q^{k}) (1 - b p^{k} / d q^{k})}{(1 - a d) (1 - b / d)} \frac{(a, b; p)_{k} (c, a d^{2} / b c; q)_{k}}{(d q, a d q / b; q)_{k} (a d p / c, b c p / d; p)_{k}} q^{k} \\ = \frac{(1 - a) (1 - b) (1 - c) (1 - a d^{2} / b c)}{d (1 - a d) (1 - b / d) (1 - c / d) (1 - a d / b c)} \\ \cdot (\frac{(a p, b p; p)_{n} (c q, a d^{2} q / b c; q)_{n}}{(d q, a d q / b; q)_{n} (a d p / c, b c p / d; p)_{n}} - \frac{(c / a d, d / b c; p)_{m + 1} (1 / d, b / a d; q)_{m + 1}}{(1 / c, b c / a d^{2}; q)_{m + 1} (1 / a, 1 / b; q)_{m + 1}}) \end{aligned}$

$\begin{array}{r} s_{k} := \frac{(a p, b p; p)_{k} (c q, a d^{2} q / b c; q)_{k}}{(d q, a d q / b; q)_{k} (a d p / c, b c p / d; p)_{k}} \end{array}$
とする.
$\begin{aligned} s_{k} - s_{k - 1} & = \frac{(a p, b p; p)_{k - 1} (c q, a d^{2} q / b c; q)_{k - 1}}{(d q, a d q / b; q)_{k} (a d p / c, b c p / d; p)_{k}} \\ \cdot ((1 - a p^{k}) (1 - b p^{k}) (1 - c q^{k}) (1 - a d^{2} q^{k} / b c) - (1 - d q^{k}) (1 - a d q^{k} / b) (1 - a d p^{k} / c) (1 - b c p^{k} / d)) \end{aligned}$
ここで,
$\begin{aligned} (1 - a p^{k}) (1 - b p^{k}) (1 - c q^{k}) (1 - a d^{2} q^{k} / b c) - (1 - d q^{k}) (1 - a d q^{k} / b) (1 - a d p^{k} / c) (1 - b c p^{k} / d) \\ = d (1 - c / d) (1 - a d / b c) (1 - a d p^{k} q^{k}) (1 - b p^{k} / d q^{k}) q^{k} \end{aligned}$
が成り立つ(これは $k = 0$ の場合を示してから $a \mapsto a p^{k}, b \mapsto b p^{k}, c \mapsto c q^{k}, d \mapsto d q^{k}$ とするとよい). よって,
$\begin{aligned} s_{k} - s_{k - 1} & = \frac{(a p, b p; p)_{k - 1} (c q, a d^{2} q / b c; q)_{k - 1}}{(d q, a d q / b; q)_{k} (a d p / c, b c p / d; p)_{k}} d (1 - c / d) (1 - a d / b c) (1 - a d p^{k} q^{k}) (1 - b p^{k} / d q^{k}) q^{k} \\ = \frac{d (1 - c / d) (1 - a d / b c) (1 - a d p^{k} q^{k}) (1 - b p^{k} / d q^{k})}{(1 - a) (1 - b) (1 - c) (1 - a d^{2} / b c)} \frac{(a, b; p)_{k} (c, a d^{2} / b c; q)_{k}}{(d q, a d q / b; q)_{k} (a d p / c, b c p / d; p)_{k}} q^{k} \\ = \frac{d (1 - a d) (1 - b / d) (1 - c / d) (1 - a d / b c)}{(1 - a) (1 - b) (1 - c) (1 - a d^{2} / b c)} \\ \cdot \frac{(1 - a d p^{k} q^{k}) (1 - b p^{k} / d q^{k})}{(1 - a d) (1 - b / d)} \frac{(a, b; p)_{k} (c, a d^{2} / b c; q)_{k}}{(d q, a d q / b; q)_{k} (a d p / c, b c p / d; p)_{k}} q^{k} \end{aligned}$
よって, これを足し合わせると,
$\begin{aligned} \sum_{k = - m}^{n} \frac{(1 - a d p^{k} q^{k}) (1 - b p^{k} / d q^{k})}{(1 - a d) (1 - b / d)} \frac{(a, b; p)_{k} (c, a d^{2} / b c; q)_{k}}{(d q, a d q / b; q)_{k} (a d p / c, b c p / d; p)_{k}} q^{k} \\ = \frac{(1 - a) (1 - b) (1 - c) (1 - a d^{2} / b c)}{d (1 - a d) (1 - b / d) (1 - c / d) (1 - a d / b c)} (s_{n} - s_{- m - 1}) \\ = \frac{(1 - a) (1 - b) (1 - c) (1 - a d^{2} / b c)}{d (1 - a d) (1 - b / d) (1 - c / d) (1 - a d / b c)} \\ \cdot (\frac{(a p, b p; p)_{n} (c q, a d^{2} q / b c; q)_{n}}{(d q, a d q / b; q)_{n} (a d p / c, b c p / d; p)_{n}} - \frac{(c / a d, d / b c; p)_{m + 1} (1 / d, b / a d; q)_{m + 1}}{(1 / c, b c / a d^{2}; q)_{m + 1} (1 / a, 1 / b; q)_{m + 1}}) \end{aligned}$
となって定理を得る.

特に $d = 1, m = 0$ として以下を得る. これはGasperによって1989年に示された公式である.

$\begin{aligned} \sum_{k = 0}^{n} \frac{(1 - a p^{k} q^{k}) (1 - b p^{k} q^{- k})}{(1 - a) (1 - b)} \frac{(a, b; p)_{k} (c, a / b c; q)_{k}}{(q, a q / b; q)_{k} (a p / c, b c p; p)_{k}} q^{k} & = \frac{(a p, b p; p)_{n} (c q, a q / b c; q)_{n}}{(q, a q / b; q)_{n} (a p / c, b c p; p)_{n}} \end{aligned}$

特に $b \to 0$ として以下を得る. これはGosperによる公式である.

$\begin{aligned} \sum_{k = 0}^{n} \frac{1 - a p^{k} q^{k}}{1 - a} \frac{(a; p)_{k} (c; q)_{k}}{(q; q)_{k} (a p / c; p)_{k}} c^{- k} & = \frac{(a p; p)_{n} (c q; q)_{n}}{(q; q)_{n} (a p / c; p)_{n}} c^{- n} \end{aligned}$

系1において $p = q$ とすると, 以下を得る.

$\begin{array}{r} \sum_{k = 0}^{n} \frac{1 - a q^{2 k}}{1 - a} \frac{(a, b, c, a / b c; q)_{k}}{(q, a q / b, a q / c, b c q; q)_{k}} q^{k} = \frac{(a q, b q, c q, a q / b c; q)_{n}}{(q, a q / b, a q / c, b c q; q)_{n}} \end{array}$