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Srivastavaによる二重超幾何級数の展開公式

次はSrivastavaによって1970年に示された公式である.

Srivastava(1970)

$\begin{aligned} \sum_{0 \leq m, n} \frac{(a_{1}, \dots, a_{r})_{m} (c_{1}, \dots, c_{u})_{n} (e)_{m + n}}{(b_{1}, \dots, b_{s}) m! (d_{1}, \dots, d_{v})_{n} n! (f)_{m + n}} x^{m} y^{n} \\ = \sum_{0 \leq n} \frac{(a_{1}, \dots, a_{r}, c_{1}, \dots, c_{u}, e, f - e)_{n}}{(f + n - 1)_{n} (f)_{2 n} (b_{1}, \dots, b_{s}, d_{1}, \dots, d_{v})_{n}} \frac{(x y)^{n}}{n!} \\ \cdot_{r} F_{s + 1} [\begin{array}{c} a_{1} + n, \dots, a_{r} + n, e + n \\ b_{1} + n, \dots, b_{s} + n, f + 2 n \end{array}; x]_{u} F_{v + 1} [\begin{array}{c} c_{1} + n, \dots, c_{u} + n, e + n \\ d_{1} + n, \dots, d_{v} + n, f + 2 n \end{array}; y] \end{aligned}$

$α_{n} = \frac{(a_{1}, \dots, a_{r})_{n}}{(b_{1}, \dots, b_{s})_{n}}, β_{n} = \frac{(c_{1}, \dots, c_{u})_{n}}{(d_{1}, \dots, d_{v})_{n}}$ とすると, 右辺は
$\begin{aligned} \sum_{0 \leq n} \frac{(f)_{n - 1} (f + 2 n - 1) (f - e)_{n}}{n! (e)_{n}} \sum_{0 \leq j, k} \frac{(e)_{n + j} (e)_{n + k}}{j! k! (f)_{2 n + j} (f)_{2 n + k}} α_{n + j} β_{n + k} x^{n + j} y^{n + k} \\ = \sum_{0 \leq j, k} (e)_{j} (e)_{k} α_{j} β_{k} x^{j} y^{k} \sum_{0 \leq n} \frac{(f)_{n - 1} (f + 2 n - 1) (f - e)_{n}}{(j - n)! (k - n)! n! (e)_{n} (f)_{n + j} (f)_{n + k}} (j \mapsto j - n, k \mapsto k - n) \end{aligned}$
ここで, Dougallの和公式より,
$\begin{aligned} \sum_{0 \leq n} \frac{(f)_{n - 1} (f + 2 n - 1) (f - e)_{n}}{(j - n)! (k - n)! n! (e)_{n} (f)_{n + j} (f)_{n + k}} \\ = \frac{1}{j! k! (f)_{j} (f)_{k}}_{5} F_{4} [\begin{array}{c} f - 1, 1 + \frac{f - 1}{2}, f - e, - j, - k \\ \frac{f - 1}{2}, e, f + j, f + k \end{array}; 1] \\ = \frac{1}{j! k! (f)_{j} (f)_{k}} \frac{(f)_{j} (f)_{k} (e)_{j + k}}{(e)_{j} (e)_{k} (f)_{j + k}} \\ = \frac{(e)_{j + k}}{j! k! (e)_{j} (e)_{k} (f)_{j + k}} \end{aligned}$
であるから, これを代入して定理を得る.

$x \mapsto \frac{x}{e}, y \mapsto \frac{y}{e}$ としてから $e \to \infty$ とすると以下を得る.

$\begin{aligned} \sum_{0 \leq m, n} \frac{(a_{1}, \dots, a_{r})_{m} (c_{1}, \dots, c_{u})_{n}}{(b_{1}, \dots, b_{s}) m! (d_{1}, \dots, d_{v})_{n} n! (f)_{m + n}} x^{m} y^{n} \\ = \sum_{0 \leq n} \frac{(a_{1}, \dots, a_{r}, c_{1}, \dots, c_{u})_{n}}{(f + n - 1)_{n} (f)_{2 n} (b_{1}, \dots, b_{s}, d_{1}, \dots, d_{v})_{n}} \frac{(- x y)^{n}}{n!} \\ \cdot_{r} F_{s + 1} [\begin{array}{c} a_{1} + n, \dots, a_{r} + n \\ b_{1} + n, \dots, b_{s} + n, f + 2 n \end{array}; x]_{u} F_{v + 1} [\begin{array}{c} c_{1} + n, \dots, c_{u} + n \\ d_{1} + n, \dots, d_{v} + n, f + 2 n \end{array}; y] \end{aligned}$

$x \mapsto f x, y \mapsto f y$ としてから $f \to \infty$ とすると以下を得る.

$\begin{aligned} \sum_{0 \leq m, n} \frac{(a_{1}, \dots, a_{r})_{m} (c_{1}, \dots, c_{u})_{n} (e)_{m + n}}{(b_{1}, \dots, b_{s}) m! (d_{1}, \dots, d_{v})_{n} n!} x^{m} y^{n} \\ = \sum_{0 \leq n} \frac{(e, a_{1}, \dots, a_{r}, c_{1}, \dots, c_{u})_{n}}{(e + n - 1)_{n} (e)_{2 n} (b_{1}, \dots, b_{s}, d_{1}, \dots, d_{v})_{n}} \frac{(x y)^{n}}{n!} \\ \cdot_{r + 1} F_{s} [\begin{array}{c} a_{1} + n, \dots, a_{r} + n, e + n \\ b_{1} + n, \dots, b_{s} + n \end{array}; x]_{u + 1} F_{v} [\begin{array}{c} c_{1} + n, \dots, c_{u} + n, e + n \\ d_{1} + n, \dots, d_{v} + n \end{array}; y] \end{aligned}$