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Baileyによる3F2の3次変換公式

Bailey(1928)

$b + c = 3 a + \frac{3}{2}$ のとき,
$\begin{aligned} (1 - x)^{- 3 a}_{3} F_{2} [\begin{array}{c} a, a + \frac{1}{3}, a + \frac{2}{3} \\ b, c \end{array}; - \frac{27 x}{4 (1 - x)^{3}}] \\ =_{3} F_{2} [\begin{array}{c} 3 a, \frac{1}{2} + b - c, \frac{1}{2} + c - b \\ b, c \end{array}; \frac{x}{4}] \end{aligned}$
が成り立つ.

$\begin{aligned} (1 - x)^{- 3 a}_{3} F_{2} [\begin{array}{c} a, a + \frac{1}{3}, a + \frac{2}{3} \\ b, c \end{array}; - \frac{27 x}{4 (1 - x)^{3}}] \\ = \sum_{0 \leq n} \frac{{(a, a + \frac{1}{3}, a + \frac{2}{3})}_{n}}{n! (b, c)_{n}} {(- \frac{27 x}{4})}^{n} \sum_{0 \leq k} \frac{(3 a + 3 n)_{k}}{k!} x^{k} \\ = \sum_{0 \leq n, k} \frac{(3 a)_{3 n + k}}{n! (b, c)_{n} k!} (- 1)^{n} 4^{- n} x^{n + k} \\ = \sum_{0 \leq n, k} \frac{(3 a)_{2 n + k}}{n! (b, c)_{n} (k - n)!} (- 1)^{n} 4^{- n} x^{k} \\ = \sum_{0 \leq k} x^{k} \sum_{0 \leq n} (- 1)^{n} 4^{- n} \frac{(3 a)_{2 n + k}}{n! (b, c)_{n} (k - n)!} \\ = \sum_{0 \leq k} \frac{(3 a)_{k}}{k!} x^{k}_{3} F_{2} [\begin{array}{c} \frac{3 a + k}{2}, \frac{3 a + k + 1}{2}, - k \\ b, c \end{array}; 1] \end{aligned}$
ここで, Saalschützの和公式から,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} \frac{3 a + k}{2}, \frac{3 a + k + 1}{2}, - k \\ b, c \end{array}; 1] & = \frac{{(b - \frac{3 a + k}{2}, b - \frac{3 a + k + 1}{2})}_{k}}{{(b, b - 3 a - k - \frac{1}{2})}_{k}} \\ = \frac{{(2 b - 3 a - k - 1)}_{2 k}}{(- 4)^{k} {(b, \frac{3}{2} + 3 a - b)}_{k}} \\ = \frac{{(2 b - 3 a - 1, 2 + 3 a - 2 b)}_{k}}{4^{k} {(b, c)}_{k}} \\ = \frac{{(\frac{1}{2} + b - c, \frac{1}{2} + c - b)}_{k}}{4^{k} {(b, c)}_{k}} \end{aligned}$
であるから定理が得られる.

Bailey(1928)

$b + c = 3 a + \frac{3}{2}$ のとき,
$\begin{aligned} (1 - x)^{- 3 a}_{3} F_{2} [\begin{array}{c} a, a + \frac{1}{3}, a + \frac{2}{3} \\ b, c \end{array}; \frac{27 x^{2}}{4 (1 - x)^{3}}] \\ =_{3} F_{2} [\begin{array}{c} 3 a, b - \frac{1}{2}, c - \frac{1}{2} \\ 2 b - 1, 2 c - 1 \end{array}; 4 x] \end{aligned}$
が成り立つ.

$\begin{aligned} (1 - x)^{- 3 a}_{3} F_{2} [\begin{array}{c} a, a + \frac{1}{3}, a + \frac{2}{3} \\ b, c \end{array}; \frac{27 x^{2}}{4 (1 - x)^{3}}] \\ = \sum_{0 \leq n} \frac{{(a, a + \frac{1}{3}, a + \frac{2}{3})}_{n}}{n! (b, c)_{n}} {(\frac{27 x^{2}}{4})}^{n} \sum_{0 \leq k} \frac{(3 a + 3 n)_{k}}{k!} x^{k} \\ = \sum_{0 \leq n, k} \frac{(3 a)_{3 n + k}}{n! (b, c)_{n} k!} 4^{- n} x^{2 n + k} \\ = \sum_{0 \leq n, k} \frac{(3 a)_{n + k}}{n! (b, c)_{n} (k - 2 n)!} 4^{- n} x^{k} \\ = \sum_{0 \leq k} \frac{(3 a)_{k}}{k!} x^{k}_{3} F_{2} [\begin{array}{c} 3 a + k, - \frac{k}{2}, \frac{1 - k}{2} \\ b, c \end{array}; 1] \end{aligned}$
ここで, Saalschutzの和公式から, $k = 2 j$ のとき,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 3 a + k, - \frac{k}{2}, \frac{1 - k}{2} \\ b, c \end{array}; 1] & = \frac{{(b - 3 a - 2 j, b + j - \frac{1}{2})}_{j}}{{(b, b - 3 a - j - \frac{1}{2})}_{j}} \\ = \frac{{(b + j - \frac{1}{2}, c + j - \frac{1}{2})}_{j}}{{(b, c)}_{j}} \\ = \frac{4^{k} {(b - \frac{1}{2}, c - \frac{1}{2})}_{k}}{{(2 b - 1, 2 c - 1)}_{k}} \end{aligned}$
全く同様に, $k = 2 j + 1$ の場合もSaalschutzの和公式から
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 3 a + k, - \frac{k}{2}, \frac{1 - k}{2} \\ b, c \end{array}; 1] & = \frac{4^{k} {(b - \frac{1}{2}, c - \frac{1}{2})}_{k}}{{(2 b - 1, 2 c - 1)}_{k}} \end{aligned}$
である. よって定理を得る.