文献あり

Terminating q-Watson, q-Whippleの和公式

次はterminatingな場合の Watsonの $_{3} F_{2}$ 和公式の $q$ 類似である.

Andrews(1976)

非負整数 $N$ によって $b = q^{- N}$ と表されるとき,
$\begin{aligned} \sum_{0 \leq n} \frac{(a, b; q)_{n} (c; q^{2})_{n}}{(c, q; q)_{n} (a b q; q^{2})_{n}} q^{n} & = \frac{a^{N / 2} (a q, b q, c q / a, c q / b; q^{2})_{\infty}}{(q, a b q, c q, c q / a b; q^{2})_{\infty}} \end{aligned}$
が成り立つ.

$\begin{aligned} \sum_{0 \leq n} \frac{(a, b; q)_{n} (c; q^{2})_{n}}{(c, q; q)_{n} (a b q; q^{2})_{n}} q^{n} & =_{4} ϕ_{3} [\begin{array}{c} a, \sqrt{c}, - \sqrt{c}, b \\ c, \sqrt{a b q}, - \sqrt{a b q} \end{array}; q] \end{aligned}$
と書き直すと, Watsonの $_{8} ϕ_{7}$ 変換公式より,
$\begin{aligned} _{4} ϕ_{3} [\begin{array}{c} a, \sqrt{c}, - \sqrt{c}, b \\ c, \sqrt{a b q}, - \sqrt{a b q} \end{array}; q] & = \frac{(b q / \sqrt{c}, - b q / \sqrt{c}, - q, q / c; q)_{\infty}}{(- b q, q / \sqrt{c}, - q / \sqrt{c}, b q / c; q)_{\infty}} \\ \cdot_{8} ϕ_{7} [\begin{array}{c} - b, q \sqrt{- b}, - q \sqrt{- b}, \sqrt{b q / a}, - \sqrt{b q / a}, \sqrt{c}, - \sqrt{c}, b \\ \sqrt{- b}, - \sqrt{- b}, \sqrt{a b q}, - \sqrt{a b q}, b q / \sqrt{c}, - b q / \sqrt{c}, - q \end{array}; \frac{a q}{c}] \\ = \frac{(b^{2} q^{2} / c; q^{2})_{\infty} (- q, q / c; q)_{\infty}}{(q^{2} / c; q^{2})_{\infty} (- b q, b q / c; q)_{\infty}}_{4} ϕ_{3} [\begin{array}{c} b^{2}, - b q^{2}, b q / a, c \\ - b, a b q, b^{2} q^{2} / c \end{array}; q^{2}; \frac{a q}{c}] \end{aligned}$
である. ここで, $q$ -Dixonの和公式
$\begin{aligned} _{4} ϕ_{3} [\begin{array}{c} a, - \sqrt{a} q, b, c \\ - \sqrt{a}, a q / b, a q / c \end{array}; \frac{\sqrt{a} q}{b c}] & = \frac{(a q, \sqrt{a} q / b, \sqrt{a} q / c, a q / b c; q)_{\infty}}{(a q / b, a q / c, \sqrt{a} q, \sqrt{a} q / b c; q)_{\infty}} \end{aligned}$
より,
$\begin{aligned} _{4} ϕ_{3} [\begin{array}{c} b^{2}, - b q^{2}, b q / a, c \\ - b, a b q, b^{2} q^{2} / c \end{array}; q^{2}; \frac{a q}{c}] & = \frac{(b^{2} q^{2}, a q, b q / c, a b q / c; q^{2})_{\infty}}{(a b q, b^{2} q^{2} / c, b q^{2}, a q / c; q^{2})_{\infty}} \end{aligned}$
だから,
$\begin{aligned} \frac{(b^{2} q^{2} / c; q^{2})_{\infty} (- q, q / c; q)_{\infty}}{(q^{2} / c; q^{2})_{\infty} (- b q, b q / c; q)_{\infty}}_{4} ϕ_{3} [\begin{array}{c} b^{2}, - b q^{2}, b q / a, c \\ - b, a b q, b^{2} q^{2} / c \end{array}; q^{2}; \frac{a q}{c}] \\ = \frac{(b^{2} q^{2} / c; q^{2})_{\infty} (- q, q / c; q)_{\infty}}{(q^{2} / c; q^{2})_{\infty} (- b q, b q / c; q)_{\infty}} \frac{(b^{2} q^{2}, a q, b q^{2} / c, a b q / c; q^{2})_{\infty}}{(a b q, b^{2} q^{2} / c, b q^{2}, a q / c; q^{2})_{\infty}} \\ = \frac{(b q, b q^{2}, q / c; q^{2})_{\infty}}{(b^{2} q^{2}, b q / c, b q^{2} / c, q; q^{2})_{\infty}} \frac{(b^{2} q^{2}, a q, b q^{2} / c, a b q / c; q^{2})_{\infty}}{(a b q, b q^{2}, a q / c; q^{2})_{\infty}} \\ = \frac{(a q, b q, q / c, a b q / c; q^{2})_{\infty}}{(q, a b q, a q / c, b q / c; q^{2})_{\infty}} \end{aligned}$
ここで, $b = q^{- N}$ の $N$ が奇数ならば, $(b q; q^{2})_{\infty} = 0$ より $0$ となる. $N$ が偶数ならば,
$\begin{aligned} \frac{(q / c, a b q / c; q^{2})_{\infty}}{(a q / c, b q / c; q^{2})_{\infty}} \\ = \frac{(a q / c; q^{2})_{- N / 2}}{(q / c; q^{2})_{- N / 2}} \\ = a^{N / 2} \frac{(c q; q^{2})_{N / 2}}{(c q / a; q^{2})_{N / 2}} \\ = a^{N / 2} \frac{(c q / a, c q / b; q^{2})_{\infty}}{(c, c q / a b; q^{2})_{\infty}} \end{aligned}$
と書き換えられることから定理を得る.

次はterminatingな場合のWhippleの $_{3} F_{2}$ 和公式の $q$ 類似である.

Andrews(1976)

非負整数 $N$ によって $a = q^{- N}$ と表されるとき,
$\begin{aligned} \sum_{0 \leq n} \frac{(a, q / a; q)_{n} (c; q^{2})_{n}}{(e, c q / e; q)_{n} (q^{2}; q^{2})_{n}} q^{n} & = \frac{q^{N (N + 1) / 2} (a e, e q / a, a c q / e, c q^{2} / a e; q^{2})_{\infty}}{(e, c q / e; q)_{\infty}} \end{aligned}$
が成り立つ.

$\begin{aligned} \sum_{0 \leq n} \frac{(a, q / a; q)_{n} (c; q^{2})_{n}}{(e, c q / e; q)_{n} (q^{2}; q^{2})_{n}} q^{n} & =_{4} ϕ_{3} [\begin{array}{c} q / a, \sqrt{c}, - \sqrt{c}, a \\ c q / e, - q, e \end{array}; q] \end{aligned}$
と書き直すと, Watsonの $_{8} ϕ_{7}$ 変換公式より,
$\begin{aligned} _{4} ϕ_{3} [\begin{array}{c} q / a, \sqrt{c}, - \sqrt{c}, a \\ c q / e, - q, e \end{array}; q] & = \frac{(a e / \sqrt{c}, - a e / \sqrt{c}, - e, e / c; q)_{\infty}}{(- a e, e / \sqrt{c}, - e / \sqrt{c}, a e / c; q)_{\infty}} \\ \cdot_{8} ϕ_{7} [\begin{array}{c} - \frac{a e}{q}, q \sqrt{- \frac{a e}{q}}, - q \sqrt{- \frac{a e}{q}}, - a, \sqrt{c}, - \sqrt{c}, a \\ \sqrt{- \frac{a e}{q}}, - \sqrt{- \frac{a e}{q}}, - q, e, - a e / \sqrt{c}, a e / \sqrt{c}, - e \end{array}; \frac{e q}{a c}] \\ = \frac{(a^{2} e^{2} / c; q^{2})_{\infty} (- e, e / c; q)_{\infty}}{(e^{2} / c; q^{2})_{\infty} (- a e, a e / c; q)_{\infty}}_{4} ϕ_{3} [\begin{array}{c} \frac{a^{2} e^{2}}{q^{2}}, - a e q, a^{2}, c \\ - \frac{a e}{q}, e^{2}, \frac{a^{2} e^{2}}{c} \end{array}; q^{2}; \frac{e q}{a c}] \end{aligned}$
ここで, $q$ -Dixonの和公式より,
$\begin{aligned} _{4} ϕ_{3} [\begin{array}{c} \frac{a^{2} e^{2}}{q^{2}}, - a e q, a^{2}, c \\ - \frac{a e}{q}, e^{2}, \frac{a^{2} e^{2}}{c} \end{array}; q^{2}; \frac{e q}{a c}] & = \frac{(a^{2} e^{2}, e q / a, a e q / c, e^{2} / c; q^{2})_{\infty}}{(e^{2}, a^{2} e^{2} / c, a e q, e q / a c; q^{2})_{\infty}} \end{aligned}$
であるから,
$\begin{aligned} \frac{(a^{2} e^{2} / c; q^{2})_{\infty} (- e, e / c; q)_{\infty}}{(e^{2} / c; q^{2})_{\infty} (- a e, a e / c; q)_{\infty}}_{4} ϕ_{3} [\begin{array}{c} \frac{a^{2} e^{2}}{q^{2}}, - a e q, a^{2}, c \\ - \frac{a e}{q}, e^{2}, \frac{a^{2} e^{2}}{c} \end{array}; q^{2}; \frac{e q}{a c}] \\ = \frac{(a^{2} e^{2} / c; q^{2})_{\infty} (- e, e / c; q)_{\infty}}{(e^{2} / c; q^{2})_{\infty} (- a e, a e / c; q)_{\infty}} \cdot \frac{(a^{2} e^{2}, e q / a, a e q / c, e^{2} / c; q^{2})_{\infty}}{(e^{2}, a^{2} e^{2} / c, a e q, e q / a c; q^{2})_{\infty}} \\ = \frac{(e / c; q)_{\infty} (a e q / c; q^{2})_{\infty}}{(a e / c; q)_{\infty} (e q / a c; q^{2})_{\infty}} \cdot \frac{(a e, e q / a; q^{2})_{\infty}}{(e; q)_{\infty}} \end{aligned}$
ここで, $a = q^{- N}$ より
$\begin{aligned} \frac{(e / c; q)_{\infty} (a e q / c; q^{2})_{\infty}}{(a e / c; q)_{\infty} (e q / a c; q^{2})_{\infty}} & = \frac{(1 - e q^{1 - N} / c) (1 - e q^{3 - N} / c) \dots (1 - e q^{N - 1} / c)}{(1 - e q^{- N} / c) (1 - e q^{1 - N} / c) \dots (1 - e q^{- 1} / c)} \\ = q^{N (N + 1) / 2} \frac{(c q^{1 - N} / e; q^{2})_{N}}{(c q / e; q)_{N}} \\ = q^{N (N + 1) / 2} \frac{(c q / a e; q)_{\infty} (a c q / e; q^{2})_{\infty}}{(c q / e; q)_{\infty} (c q / a e; q^{2})_{\infty}} \\ = q^{N (N + 1) / 2} \frac{(a c q / e, c q^{2} / a e; q^{2})_{\infty}}{(c q / e; q)_{\infty}} \end{aligned}$
であるから, これを代入して定理を得る.