文献あり

Srivastavaによる二重q超幾何級数の展開公式

前回の記事の $q$ 類似を示す.

Srivastava(1984)

$\begin{aligned} \sum_{0 \leq m, n} \frac{(a_{1}, \dots, a_{r}; q)_{m} (c_{1}, \dots, c_{u}; q)_{n} (e; q)_{m + n}}{(b_{1}, \dots, b_{s}, q; q)_{m} (d_{1}, \dots, d_{v}, q; q)_{n} (f; q)_{m + n}} x^{n} y^{m} \\ = \sum_{0 \leq n} \frac{(a_{1}, \dots, a_{r}, c_{1}, \dots, c_{u}, e, f / e; q)_{n}}{(f q^{n - 1}; q)_{n} (f; q)_{2 n} (b_{1}, \dots, b_{s}, d_{1}, \dots, d_{v}; q)_{n}} q^{n (n - 1)} \frac{(e x y)^{n}}{(q; q)_{n}} \\ \cdot \sum_{0 \leq j} \frac{(a_{1} q^{n}, \dots, a_{r} q^{n}, e q^{n}; q)_{j}}{(b_{1} q^{n}, \dots, b_{s} q^{n}, f q^{2 n}, q; q)_{j}} x^{j} \\ \cdot \sum_{0 \leq k} \frac{(c_{1} q^{n}, \dots, c_{u} q^{n}, e q^{n}; q)_{k}}{(d_{1} q^{n}, \dots, d_{v} q^{n}, f q^{2 n}, q; q)_{k}} y^{k} \end{aligned}$

$α_{n} := \frac{(a_{1}, \dots, a_{r}; q)_{n}}{(b_{1}, \dots, b_{s}; q)_{n}}, β_{n} := \frac{(c_{1}, \dots, c_{u}; q)_{n}}{(d_{1}, \dots, d_{v}; q)_{n}}$ とすれば, 右辺は
$\begin{aligned} \sum_{0 \leq n} \frac{(1 - f q^{2 n - 1}) (f; q)_{n - 1} (f / e; q)_{n}}{(e; q)_{n}} e^{n} q^{n (n - 1)} \sum_{0 \leq j, k} \frac{(e; q)_{n + j} (e; q)_{n + k}}{(f; q)_{2 n + j} (f; q)_{2 n + k} (q; q)_{j} (q; q)_{k}} α_{n + j} β_{n + k} x^{n + j} y^{n + k} \\ = \sum_{0 \leq j, k} (e; q)_{j} (e; q)_{k} α_{j} β_{k} x^{j} y^{k} \sum_{0 \leq n} \frac{(1 - f q^{2 n - 1}) (f; q)_{n - 1} (f / e; q)_{n}}{(e; q)_{n} (f; q)_{n + j} (f; q)_{n + k} (q; q)_{j - n} (q; q)_{k - n}} e^{n} q^{n (n - 1)} \end{aligned}$
ここで, Rogersの $_{6} ϕ_{5}$ 和公式より,
$\begin{aligned} \sum_{0 \leq n} \frac{(1 - f q^{2 n - 1}) (f; q)_{n - 1} (f / e; q)_{n}}{(e; q)_{n} (f; q)_{n + j} (f; q)_{n + k} (q; q)_{j - n} (q; q)_{k - n}} e^{n} q^{n (n - 1)} \\ = \frac{1}{(f, q; q)_{j} (f, q; q)_{k}}_{6} ϕ_{5} [\begin{array}{c} f / q, \sqrt{f q}, - \sqrt{f q}, f / e, q^{- j}, q^{- k} \\ \sqrt{f / q}, - \sqrt{f / q}, e, f q^{j}, f q^{k} \end{array}; e q^{j + k}] \\ = \frac{1}{(f, q; q)_{j} (f, q; q)_{k}} \frac{(f; q)_{j} (f; q)_{k} (e; q)_{j + k}}{(e; q)_{j} (e; q)_{k} (f; q)_{j + k}} \\ = \frac{(e; q)_{j + k}}{(e, q; q)_{j} (e, q; q)_{k} (f; q)_{j + k}} \end{aligned}$
を代入して右辺を得る.

$e \mapsto 0$ とすると, 以下を得る.

$\begin{aligned} \sum_{0 \leq m, n} \frac{(a_{1}, \dots, a_{r}; q)_{m} (c_{1}, \dots, c_{u}; q)_{n}}{(b_{1}, \dots, b_{s}, q; q)_{m} (d_{1}, \dots, d_{v}, q; q)_{n} (f; q)_{m + n}} x^{n} y^{m} \\ = \sum_{0 \leq n} \frac{(a_{1}, \dots, a_{r}, c_{1}, \dots, c_{u}; q)_{n}}{(f q^{n - 1}; q)_{n} (f; q)_{2 n} (b_{1}, \dots, b_{s}, d_{1}, \dots, d_{v}; q)_{n}} q^{\frac{3}{2} n (n - 1)} \frac{(- f x y)^{n}}{(q; q)_{n}} \\ \cdot \sum_{0 \leq j} \frac{(a_{1} q^{n}, \dots, a_{r} q^{n}; q)_{j}}{(b_{1} q^{n}, \dots, b_{s} q^{n}, f q^{2 n}, q; q)_{j}} x^{j} \\ \cdot \sum_{0 \leq k} \frac{(c_{1} q^{n}, \dots, c_{u} q^{n}; q)_{k}}{(d_{1} q^{n}, \dots, d_{v} q^{n}, f q^{2 n}, q; q)_{k}} y^{k} \end{aligned}$

$f \mapsto 0$ とすると以下を得る.

$\begin{aligned} \sum_{0 \leq m, n} \frac{(a_{1}, \dots, a_{r}; q)_{m} (c_{1}, \dots, c_{u}; q)_{n} (e; q)_{m + n}}{(b_{1}, \dots, b_{s}, q; q)_{m} (d_{1}, \dots, d_{v}, q; q)_{n}} x^{n} y^{m} \\ = \sum_{0 \leq n} \frac{(a_{1}, \dots, a_{r}, c_{1}, \dots, c_{u}, e; q)_{n}}{(b_{1}, \dots, b_{s}, d_{1}, \dots, d_{v}; q)_{n}} q^{n (n - 1)} \frac{(e x y)^{n}}{(q; q)_{n}} \\ \cdot \sum_{0 \leq j} \frac{(a_{1} q^{n}, \dots, a_{r} q^{n}, e q^{n}; q)_{j}}{(b_{1} q^{n}, \dots, b_{s} q^{n}, q; q)_{j}} x^{j} \\ \cdot \sum_{0 \leq k} \frac{(c_{1} q^{n}, \dots, c_{u} q^{n}, e q^{n}; q)_{k}}{(d_{1} q^{n}, \dots, d_{v} q^{n}, q; q)_{k}} y^{k} \end{aligned}$