文献あり

BaileyによるCayley-Orr型の定理2

Baileyによって1935年に示された結果のうち, 前の記事で書いていなかったものを書く.

Bailey(1935)

$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, b, \frac{c + d - 1}{2}, \frac{c + d}{2} \\ a + b, c, d \end{array}; 4 x (1 - x)] \\ = \sum_{0 \leq n} \frac{(c + d - 1)_{n}}{(a + b)_{n}} x^{n} \sum_{k = 0}^{n} \frac{(a, b)_{k}}{k! (c)_{k}} \frac{(a, b)_{n - k}}{(n - k)! (d)_{n - k}} \end{aligned}$

Whippleによる Nearly-poised $_{4} F_{3}$ の変換公式
$\begin{aligned} _{4} F_{3} [\begin{array}{c} - n, b, c, d \\ 1 - n - b, 1 - n - c, w \end{array}; 1] & = \frac{(w - d)_{n}}{(w)_{n}} \sum_{k = 0}^{n} \frac{(1 - n - b - c, 1 - n - w, d)_{k} (- n)_{2 k}}{k! (1 - n - b, 1 - n - c)_{k} (1 - n + d - w)_{2 k}} \end{aligned}$
において, $d \mapsto 1 - n - d$ として,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} - n, b, c, 1 - n - d \\ 1 - n - b, 1 - n - c, w \end{array}; 1] \\ = \frac{(w - 1 + n + d)_{n}}{(w)_{n}} \sum_{k = 0}^{n} \frac{(1 - n - b - c, 1 - n - w, 1 - n - d)_{k} (- n)_{2 k}}{k! (1 - n - b, 1 - n - c)_{k} (2 - d - w - 2 n)_{2 k}} \\ = \frac{(d + w - 1)_{2 n}}{(d + w - 1, w)_{n}} \sum_{k = 0}^{n} \frac{(1 - n - b - c, 1 - n - w, 1 - n - d)_{k} (- n)_{2 k}}{k! (1 - n - b, 1 - n - c)_{k} (2 - d - w - 2 n)_{2 k}} \end{aligned}$
が得られる. これを書き換えると,
$\begin{aligned} \frac{n! (d)_{n}}{(b, c)_{n}} \sum_{k = 0}^{n} \frac{(b, c)_{k}}{k! (w)_{k}} \frac{(b, c)_{n - k}}{(n - k)! (d)_{n - k}} \\ = \frac{n! (b + c, d)_{n}}{(d + w - 1, b, c)_{n}} \sum_{k = 0}^{n} \frac{(- 1)^{k} (d + w - 1)_{2 n - 2 k}}{k! (n - 2 k)!} \frac{(b, c)_{n - k}}{(b + c, w, d)_{n - k}} \end{aligned}$
だから,
$\begin{aligned} \sum_{0 \leq n} \frac{(d + w - 1)_{n}}{(b + c)_{n}} x^{n} \sum_{k = 0}^{n} \frac{(b, c)_{k}}{k! (w)_{k}} \frac{(b, c)_{n - k}}{(n - k)! (d)_{n - k}} \\ = \sum_{0 \leq n} x^{n} \sum_{k = 0}^{n} \frac{(- 1)^{k} (d + w - 1)_{2 n - 2 k}}{k! (n - 2 k)!} \frac{(b, c)_{n - k}}{(b + c, w, d)_{n - k}} \\ = \sum_{0 \leq n, k} x^{n + k} \frac{(- 1)^{k} (d + w - 1)_{2 n}}{k! (n - k)!} \frac{(b, c)_{n}}{(b + c, w, d)_{n}} \\ = \sum_{0 \leq n} x^{n} (1 - x)^{n} \frac{(d + w - 1)_{2 n}}{n!} \frac{(b, c)_{n}}{(b + c, w, d)_{n}} \\ =_{4} F_{3} [\begin{array}{c} b, c, \frac{d + w - 1}{2}, \frac{d + w}{2} \\ b + c, d, w \end{array}; 4 x (1 - x)] \end{aligned}$
となって示すべきことが得られた.

Bailey(1935)

$\begin{aligned} (1 - x)^{- a}_{4} F_{3} [\begin{array}{c} a, c - a, b, d \\ c, \frac{b + d}{2}, \frac{b + d + 1}{2} \end{array}; - \frac{x^{2}}{4 (1 - x)}] \\ = \sum_{0 \leq n} \frac{(c)_{n}}{(b + d)_{n}} x^{n} \sum_{k = 0}^{n} \frac{(a, b)_{k}}{k! (c)_{k}} \frac{(a, d)_{n - k}}{(n - k)! (d)_{n - k}} \end{aligned}$

Whippleによる Nearly-poised $_{4} F_{3}$ の変換公式
$\begin{aligned} _{4} F_{3} [\begin{array}{c} - n, b, c, d \\ 1 - n - b, 1 - n - c, w \end{array}; 1] & = \frac{(w - d)_{n}}{(w)_{n}} \sum_{k = 0}^{n} \frac{(1 - n - b - c, 1 - n - w, d)_{k} (- n)_{2 k}}{k! (1 - n - b, 1 - n - c)_{k} (1 - n + d - w)_{2 k}} \end{aligned}$
において, $c \mapsto 1 - n - c, w \mapsto 1 - n - w$ として,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} - n, b, 1 - n - c, d \\ 1 - n - b, c, 1 - n - w \end{array}; 1] & = \frac{(w + d)_{n}}{(w)_{n}} \sum_{k = 0}^{n} \frac{(c - b, w, d)_{k} (- n)_{2 k}}{k! (1 - n - b, c)_{k} (d + w)_{2 k}} \end{aligned}$
これを書き換えると,
$\begin{aligned} \frac{n! (c)_{n}}{(b, w)_{n}} \sum_{k = 0}^{n} \frac{(b, d)_{k}}{k! (c)_{k}} \frac{(b, w)_{n - k}}{(n - k)! (c)_{n - k}} & = \frac{n! (w + d)_{n}}{(b, w)_{n}} \sum_{k = 0}^{n} \frac{(- 1)^{k} (c - b, w, d)_{k} (b)_{n - k}}{k! (n - 2 k)! (c)_{k} (d + w)_{2 k}} \end{aligned}$
だから,
$\begin{aligned} \sum_{0 \leq n} \frac{(c)_{n}}{(d + w)_{n}} x^{n} \sum_{k = 0}^{n} \frac{(b, d)_{k}}{k! (c)_{k}} \frac{(b, w)_{n - k}}{(n - k)! (c)_{n - k}} \\ = \sum_{0 \leq n} x^{n} \sum_{k = 0}^{n} \frac{(- 1)^{k} (c - b, w, d)_{k} (b)_{n - k}}{k! (n - 2 k)! (c)_{k} (d + w)_{2 k}} \\ = \sum_{0 \leq n, k} x^{n + 2 k} \frac{(- 1)^{k} (c - b, w, d)_{k} (b)_{n + k}}{k! n! (c)_{k} (d + w)_{2 k}} \\ = \sum_{0 \leq k} x^{2 k} (1 - x)^{- k - b} \frac{(- 1)^{k} (b, c - b, w, d)_{k}}{k! (c)_{k} (d + w)_{2 k}} \\ = (1 - x)^{- b}_{4} F_{3} [\begin{array}{c} b, c - b, d, w \\ c, \frac{d + w}{2}, \frac{d + w + 1}{2} \end{array}; - \frac{x^{2}}{4 (1 - x)}] \end{aligned}$

Bailey(1935)

$c + c^{'} = a + a^{'} + b + b^{'}$ のとき,
$\begin{aligned} (1 - x)^{1 - c - c^{'}}_{4} F_{3} [\begin{array}{c} a, c - b, \frac{c + c^{'} - 1}{2}, \frac{c + c^{'}}{2} \\ a + a^{'}, a + b^{'}, c \end{array}; - \frac{4 x}{(1 - x)^{2}}] \\ = \sum_{0 \leq n} \frac{(c^{'}, c + c^{'} - 1)_{n}}{(a + a^{'}, a + b^{'})_{n}} x^{n} \sum_{k = 0}^{n} \frac{(a, b)_{k}}{k! (c)_{k}} \frac{(a^{'}, b^{'})_{n - k}}{(n - k)! (c^{'})_{n - k}} \end{aligned}$

Whippleの $_{4} F_{3}$ 変換公式
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, b, c, - n \\ d, e, f \end{array}; 1] \\ = \frac{(d - a, e - a)_{n}}{(d, e)_{n}}_{4} F_{3} [\begin{array}{c} a, f - b, f - c, - n \\ 1 - n + a - d, 1 - n + a - e, f \end{array}; 1], (a + b + c + 1 = d + e + f + n) \end{aligned}$
において, $c \mapsto c + n$ とすると,　 $a + b + c + 1 = d + e + f$ のとき,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, b, c + n, - n \\ d, e, f \end{array}; 1] \\ = \frac{(d - a, e - a)_{n}}{(d, e)_{n}}_{4} F_{3} [\begin{array}{c} a, f - b, f - c - n, - n \\ 1 - n + a - d, 1 - n + a - e, f \end{array}; 1] \end{aligned}$
が成り立つ. これを書き換えると,
$\begin{aligned} \frac{n!}{(c)_{n}} \sum_{k = 0}^{n} \frac{(- 1)^{k} (a, b)_{k} (c)_{n + k}}{k! (n - k)! (d, e, f)_{k}} \\ = \frac{n! (1 + c - f)_{n}}{(d, e)_{n}} \sum_{k = 0}^{n} \frac{(a, f - b)_{k}}{k! (f)_{k}} \frac{(d - a, e - a)_{n - k}}{(n - k)! (1 + c - f)_{n - k}} \end{aligned}$
よって,
$\begin{aligned} \sum_{0 \leq n} \frac{(c, 1 + c - f)_{n}}{(d, e)_{n}} x^{n} \sum_{k = 0}^{n} \frac{(a, f - b)_{k}}{k! (f)_{k}} \frac{(d - a, e - a)_{n - k}}{(n - k)! (1 + c - f)_{n - k}} \\ = \sum_{0 \leq n} x^{n} \sum_{k = 0}^{n} \frac{(- 1)^{k} (a, b)_{k} (c)_{n + k}}{k! (n - k)! (d, e, f)_{k}} \\ = \sum_{0 \leq n, k} x^{n + k} \frac{(- 1)^{k} (a, b)_{k} (c)_{n + 2 k}}{k! n! (d, e, f)_{k}} \\ = \sum_{0 \leq k} x^{k} (1 - x)^{- c - 2 k} \frac{(- 1)^{k} (a, b)_{k} (c)_{2 k}}{k! (d, e, f)_{k}} \\ = (1 - x)^{- c}_{4} F_{3} [\begin{array}{c} a, b, \frac{c}{2}, \frac{c + 1}{2} \\ d, e, f \end{array}; - \frac{4 x}{(1 - x)^{2}}] \end{aligned}$
$b \mapsto c - b, c \mapsto c + c^{'} - 1, d \mapsto a + a^{'}, e \mapsto a + b^{'}, f \mapsto c$ とすると,　条件は $c + c^{'} = a + a^{'} + b + b^{'}$ となって定理を得る.