文献あり

BaileyによるCayley-Orr型の定理

BaileyによっていくつかのCayley-Orr型の定理が示されている.

Bailey(1935)

$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ a + b + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} \frac{1}{2} + c - a, \frac{1}{2} + c - b \\ 2 c - a - b + \frac{1}{2} \end{array}; x] \\ = \sum_{0 \leq n} \frac{{(c + \frac{1}{2})}_{n}}{{(2 c - a - b + \frac{1}{2})}_{n}} x^{n} \sum_{k = 0}^{n} \frac{(2 a, 2 b, c)_{k}}{k! {(a + b + \frac{1}{2}, 2 c)}_{k}} \frac{{(c + \frac{1}{2} - a - b)}_{n - k}}{(n - k)!} \end{aligned}$

Bailey(1935)

$\begin{aligned} _{2} F_{1} [\begin{array}{c} b, c - b \\ c + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} a + \frac{1}{2}, c - a + \frac{1}{2} \\ c + \frac{1}{2} \end{array}; x] \\ = \sum_{0 \leq n} \frac{{(a + b + \frac{1}{2})}_{n}}{{(c + \frac{1}{2})}_{n}} x^{n} \sum_{k = 0}^{n} \frac{(2 a, 2 b, c)_{k}}{k! {(a + b + \frac{1}{2}, 2 c)}_{k}} \frac{{(c + \frac{1}{2} - a - b)}_{n - k}}{(n - k)!} \end{aligned}$

定理1, 定理2の証明

係数を比較して, それぞれ示すべき等式は,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} 2 a, 2 b, c, - n \\ 2 c, a + b + \frac{1}{2}, \frac{1}{2} - n + a + b - c \end{array}; 1] \\ = \frac{{(c + \frac{1}{2} - a, c + \frac{1}{2} - b)}_{n}}{{(c + \frac{1}{2}, c + \frac{1}{2} - a - b)}_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{1}{2} + a + b - 2 c - n, - n \\ \frac{1}{2} + a - c - n, \frac{1}{2} + b - c - n, a + b + \frac{1}{2} \end{array}; 1] \end{aligned}$
と,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} 2 a, 2 b, c, - n \\ 2 c, a + b + \frac{1}{2}, \frac{1}{2} - n + a + b - c \end{array}; 1] \\ = \frac{{(a + \frac{1}{2}, c + \frac{1}{2} - a)}_{n}}{{(a + b + \frac{1}{2}, c + \frac{1}{2} - a - b)}_{n}}_{4} F_{3} [\begin{array}{c} b, c - b, \frac{1}{2} - n - c, - n \\ c + \frac{1}{2}, \frac{1}{2} - a - n, \frac{1}{2} + a - c - n \end{array}; 1] \end{aligned}$
である. まず, Whippleの変換公式より,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, b, \frac{1}{2} + a + b - 2 c - n, - n \\ \frac{1}{2} + a - c - n, \frac{1}{2} + b - c - n, a + b + \frac{1}{2} \end{array}; 1] \\ = \frac{{(a + \frac{1}{2}, c + \frac{1}{2})}_{n}}{{(a + b + \frac{1}{2}, c + \frac{1}{2} - b)}_{n}}_{4} F_{3} [\begin{array}{c} b, c - b, \frac{1}{2} - n - c, - n \\ c + \frac{1}{2}, \frac{1}{2} - a - n, \frac{1}{2} + a - c - n \end{array}; 1] \end{aligned}$
であるから, 1つ目の式と2つ目の式は同値である. よって, 1つ目の式を示す. Whippleの変換公式より,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} 2 a, 2 b, c, - n \\ 2 c, a + b + \frac{1}{2}, \frac{1}{2} - n + a + b - c \end{array}; 1] \\ = \frac{{(2 c - 2 a, \frac{1}{2} + a - b + c)}_{n}}{{(2 c, \frac{1}{2} - a - b + c)}_{n}}_{4} F_{3} [\begin{array}{c} 2 a, \frac{1}{2} + a + b - c, \frac{1}{2} + a - b, - n \\ \frac{1}{2} + a - b + c, \frac{1}{2} + a + b, 1 - n + 2 a - 2 c \end{array}; 1] \end{aligned}$
次に WhippleのNearly-poised変換公式より,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} 2 a, \frac{1}{2} + a + b - c, \frac{1}{2} + a - b, - n \\ \frac{1}{2} + a - b + c, \frac{1}{2} + a + b, 1 - n + 2 a - 2 c \end{array}; 1] \\ = \frac{(2 c)_{n}}{(2 c - 2 a)_{n}}_{4} F_{3} [\begin{array}{c} 2 a, \frac{1}{2} + a + b - c, \frac{1}{2} + a - b, - n \\ \frac{1}{2} + a - b + c, \frac{1}{2} + a + b, 1 - n + 2 a - 2 c \end{array}; 1] \end{aligned}$
最後に, Whippleの $_{4} F_{3}$ 変換公式より,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} 2 a, \frac{1}{2} + a + b - c, \frac{1}{2} + a - b, - n \\ \frac{1}{2} + a - b + c, \frac{1}{2} + a + b, 1 - n + 2 a - 2 c \end{array}; 1] \\ = \frac{{(\frac{1}{2} + c - a, \frac{1}{2} + c - b)}_{n}}{{(c + \frac{1}{2}, c + \frac{1}{2} + a - b)}_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{1}{2} + a + b - 2 c - n, - n \\ \frac{1}{2} + a - c - n, \frac{1}{2} + b - c - n, a + b + \frac{1}{2} \end{array}; 1] \end{aligned}$
これらを合わせて, 示すべき等式が得られる.

Bailey(1935)

$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ a + b + \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} c - a, c - b \\ 2 c - a - b + \frac{1}{2} \end{array}; x] \\ = \sum_{0 \leq n} \frac{(c)_{n}}{{(2 c - a - b + \frac{1}{2})}_{n}} x^{n} \sum_{k = 0}^{n} \frac{{(2 a, 2 b, c + \frac{1}{2})}_{k}}{k! {(2 c, a + b + \frac{1}{2})}_{k}} \frac{(c - a - b)_{n - k}}{(n - k)!} \end{aligned}$

示すべき等式は,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} 2 a, 2 b, c + \frac{1}{2}, - n \\ 2 c, a + b + \frac{1}{2}, 1 - n + a + b - c \end{array}; 1] \\ = \frac{(c - a, c - b)_{n}}{(c, c - a - b)_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{1}{2} + a + b - 2 c - n, - n \\ a + b + \frac{1}{2}, 1 - n + a - c, 1 - n + b - c \end{array}; 1] \end{aligned}$
である. ここで,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} 2 a, 2 b, c - \frac{1}{2}, - n \\ 2 c - 1, a + b + \frac{1}{2}, 1 + a + b - c - n \end{array}; 1] \\ = \sum_{0 \leq k} \frac{{(2 a, 2 b, c + \frac{1}{2}, - n)}_{k}}{k! {(2 c, a + b + \frac{1}{2}, 1 + a + b - c - n)}_{k}} (1 - \frac{k}{2 k + 2 c - 1}) \\ =_{4} F_{3} [\begin{array}{c} 2 a, 2 b, c + \frac{1}{2}, - n \\ 2 c, a + b + \frac{1}{2}, 1 + a + b - c - n \end{array}; 1] \\ + \frac{a b n}{c (a + b + \frac{1}{2}) (1 + a + b - c - n)}_{4} F_{3} [\begin{array}{c} 2 a + 1, 2 b + 1, c + \frac{1}{2}, 1 - n \\ 2 c + 1, a + b + \frac{3}{2}, 2 + a + b - c - n \end{array}; 1] \end{aligned}$
よって, 定理1を用いると,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} 2 a, 2 b, c + \frac{1}{2}, - n \\ 2 c, a + b + \frac{1}{2}, 1 + a + b - c - n \end{array}; 1] \\ =_{4} F_{3} [\begin{array}{c} 2 a, 2 b, c - \frac{1}{2}, - n \\ 2 c - 1, a + b + \frac{1}{2}, 1 + a + b - c - n \end{array}; 1] \\ - \frac{a b n}{c (a + b + \frac{1}{2}) (1 + a + b - c - n)}_{4} F_{3} [\begin{array}{c} 2 a + 1, 2 b + 1, c + \frac{1}{2}, 1 - n \\ 2 c + 1, a + b + \frac{3}{2}, 2 + a + b - c - n \end{array}; 1] \\ = \frac{(c - a, c - b)_{n}}{(c, c - a - b)_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{3}{2} + a + b - 2 c - n, - n \\ 1 - n + a - c, 1 - n + b - c, a + b + \frac{1}{2} \end{array}; 1] \\ - \frac{a b n}{c (a + b + \frac{1}{2}) (1 + a + b - c - n)} \frac{{(c - a + \frac{1}{2}, c - b + \frac{1}{2})}_{n - 1}}{(c + 1, c - a - b)_{n - 1}}_{4} F_{3} [\begin{array}{c} a + \frac{1}{2}, b + \frac{1}{2}, \frac{3}{2} + a + b - 2 c - n, 1 - n \\ \frac{3}{2} - n + a - c, \frac{3}{2} - n + b - c, a + b + \frac{3}{2} \end{array}; 1] \end{aligned}$
ここで, Whippleの変換公式より,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a + \frac{1}{2}, b + \frac{1}{2}, \frac{3}{2} + a + b - 2 c - n, 1 - n \\ \frac{3}{2} - n + a - c, \frac{3}{2} - n + b - c, a + b + \frac{3}{2} \end{array}; 1] \\ = \frac{{(c - a, c - b)}_{n - 1}}{{(c - a + \frac{1}{2}, c - b + \frac{1}{2})}_{n - 1}}_{4} F_{3} [\begin{array}{c} a + 1, b + 1, \frac{3}{2} + a + b - 2 c - n, 1 - n \\ 2 - n + a - b, 2 - n + a - c, a + b + \frac{3}{2} \end{array}; 1] \end{aligned}$
を用いると,
$\begin{aligned} \frac{(c - a, c - b)_{n}}{(c, c - a - b)_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{3}{2} + a + b - 2 c - n, - n \\ 1 - n + a - c, 1 - n + b - c, a + b + \frac{1}{2} \end{array}; 1] \\ - \frac{a b n}{c (a + b + \frac{1}{2}) (1 + a + b - c - n)} \frac{{(c - a + \frac{1}{2}, c - b + \frac{1}{2})}_{n - 1}}{(c + 1, c - a - b)_{n - 1}}_{4} F_{3} [\begin{array}{c} a + \frac{1}{2}, b + \frac{1}{2}, \frac{3}{2} + a + b - 2 c - n, 1 - n \\ \frac{3}{2} - n + a - c, \frac{3}{2} - n + b - c, a + b + \frac{3}{2} \end{array}; 1] \\ = \frac{(c - a, c - b)_{n}}{(c, c - a - b)_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{3}{2} + a + b - 2 c - n, - n \\ 1 - n + a - c, 1 - n + b - c, a + b + \frac{1}{2} \end{array}; 1] \\ + \frac{a b n}{a + b + \frac{1}{2}} \frac{(c - a, c - b)_{n - 1}}{(c, c - a - b)_{n}}_{4} F_{3} [\begin{array}{c} a + 1, b + 1, \frac{3}{2} + a + b - 2 c - n, 1 - n \\ 2 - n + a - b, 2 - n + a - c, a + b + \frac{3}{2} \end{array}; 1] \\ = \frac{(c - a, c - b)_{n}}{(c, c - a - b)_{n}} \sum_{0 \leq k} \frac{{(a, b, \frac{3}{2} + a + b - 2 c - n, - n)}_{k}}{k! {(1 - n + a - c, 1 - n + b - c, a + b + \frac{1}{2})}_{k}} (1 - \frac{k}{k + \frac{1}{2} + a + b - 2 c - n}) \\ = \frac{(c - a, c - b)_{n}}{(c, c - a - b)_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{1}{2} + a + b - 2 c - n, - n \\ 1 - n + a - c, 1 - n + b - c, a + b + \frac{1}{2} \end{array}; 1] \end{aligned}$
となって示すべき等式が得られた.