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Cayley-Orrの定理

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Clausenの公式の一般化として, 以下のような定理が知られている.

Cayley-Orrの定理

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

係数を考えると, これは
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 2 a, 2 b, - n \\ 2 c, 1 - n + a + b - c \end{array}; 1] & = \frac{(c - a, c - b)_{n}}{(c, c - a - b)_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{1}{2} - c - n, - n \\ c + \frac{1}{2}, 1 - n + a - c, 1 - n + b - c \end{array}; 1] \end{aligned}$
と同値であること分かる.

$c = a + b$ とするとClausenの公式を得る.

$\begin{aligned} _{2} F_{1} {[\begin{array}{c} 2 a, 2 b \\ a + b + \frac{1}{2} \end{array}; x]}^{2} & =_{3} F_{2} [\begin{array}{c} 2 a, 2 b, a + b \\ 2 a + 2 b, a + b + \frac{1}{2} \end{array}; x] \end{aligned}$

次に, 定理1の類似として以下の2つの式が知られている.

Orrの定理

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

証明の前に補題を示しておく.

$\begin{aligned} _{3} F_{2} [\begin{array}{c} a, b, - n \\ e, f \end{array}; 1] & = \frac{(e - a, f - a)_{n}}{(e, f)_{n}}_{3} F_{2} [\begin{array}{c} 1 + a + b - e - f - n, a, - n \\ 1 - n + a - e, 1 - n + a - f \end{array}; 1] \end{aligned}$

Whippleの $_{4} F_{3}$ 変換公式
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, b, c, - n \\ d, e, f \end{array}; 1] & = \frac{(e - a, f - a)_{n}}{(e, f)_{n}}_{4} F_{3} [\begin{array}{c} a, d - b, d - c, - n \\ d, 1 - n + a - e, 1 - n + a - f \end{array}; 1] (1 + a + b + c = d + e + f + n) \end{aligned}$
において, $c, d$ 以外を固定して $c \to \infty$ とすると,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} a, b, - n \\ e, f \end{array}; 1] & = \frac{(e - a, f - a)_{n}}{(e, f)_{n}}_{4} F_{3} [\begin{array}{c} a, 1 + a + b - e - f - n, - n \\ 1 - n + a - e, 1 - n + a - f \end{array}; 1] \end{aligned}$
となって示される.

$\begin{aligned} _{3} F_{2} [\begin{array}{c} a, b, - n \\ 1 + a - b, w \end{array}; 1] & = \frac{(w - a)_{n}}{(w)_{n}}_{4} F_{3} [\begin{array}{c} \frac{1 + a}{2} - b, 1 + a - w, \frac{a}{2}, - n \\ 1 + a - b, \frac{1 - n + a - w}{2}, \frac{2 - n + a - w}{2} \end{array}; 1] \end{aligned}$

Whippleによる Nearly-poised $_{4} F_{3}$ の変換公式
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, b, c, - n \\ 1 + a - b, 1 + a - c, w \end{array}; 1] & = \frac{(w - a)_{n}}{(w)_{n}}_{5} F_{4} [\begin{array}{c} 1 + a - b - c, 1 + a - w, \frac{a}{2}, \frac{a + 1}{2}, - n \\ 1 + a - b, 1 + a - c, \frac{1 - n + a - w}{2}, \frac{2 - n + a - w}{2} \end{array}; 1] \end{aligned}$
において, $c = \frac{a + 1}{2}$ とすればよい.

定理1, 定理2の証明

係数を考えるとこれらはそれぞれ
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 2 a, 2 b, - n \\ 2 c, \frac{1}{2} - n + a + b - c \end{array}; 1] & = \frac{{(c + \frac{1}{2} - a, c + \frac{1}{2} - b)}_{n}}{{(c + \frac{1}{2}, c + \frac{1}{2} - a - b)}_{n}}_{4} F_{3} [\begin{array}{c} a, b, - n - c, - n \\ c, \frac{1}{2} - n + a - c, \frac{1}{2} - n + b - c \end{array}; 1] \\ _{3} F_{2} [\begin{array}{c} 2 a - 1, 2 b, - n \\ 2 c - 1, \frac{1}{2} - n + a + b - c \end{array}; 1] & = \frac{{(c + \frac{1}{2} - a, c - \frac{1}{2} - b)}_{n}}{{(c - \frac{1}{2}, c + \frac{1}{2} - a - b)}_{n}}_{4} F_{3} [\begin{array}{c} a, b, 1 - n - c, - n \\ c, \frac{1}{2} - n + a - c, \frac{3}{2} - n + b - c \end{array}; 1] \end{aligned}$
と同値であることが分かる. この2つの関係式は $1 + a + b + c = d + e + f + n$ の場合の Whippleの変換公式
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, b, c, - n \\ d, e, f \end{array}; 1] & = \frac{(e - a, f - a)_{n}}{(e, f)_{n}}_{4} F_{3} [\begin{array}{c} a, d - b, d - c, - n \\ d, 1 - n + a - e, 1 - n + a - f \end{array}; 1] \end{aligned}$
の特別な場合として,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, b, - n - c, - n \\ c, \frac{1}{2} - n + a - c, \frac{1}{2} - n + b - c \end{array}; 1] & = \frac{{(c + \frac{1}{2}, c - b)}_{n}}{{(c, c + \frac{1}{2} - b)}_{n}}_{4} F_{3} [\begin{array}{c} a + \frac{1}{2}, b, \frac{1}{2} - n - c, - n \\ c + \frac{1}{2}, \frac{1}{2} - n + a - c, 1 - n + b - c \end{array}; 1] \end{aligned}$
が成り立つことから同値であることが分かる. 次に, 定理2はそれぞれ,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 2 a, 2 b, - n \\ 2 c - 1, 1 - n + a + b - c \end{array}; 1] & = \frac{{(c - a, c - b)}_{n}}{{(c, c - a - b)}_{n}}_{4} F_{3} [\begin{array}{c} a, b, \frac{1}{2} - n - c, - n \\ c - \frac{1}{2}, 1 - n + a - c, 1 - n + b - c \end{array}; 1] \\ _{3} F_{2} [\begin{array}{c} 2 a + 1, 2 b, - n \\ 2 c, 1 - n + a + b - c \end{array}; 1] & = \frac{{(c - a, c - b)}_{n}}{{(c, c - a - b)}_{n}}_{4} F_{3} [\begin{array}{c} a + 1, b, 1 - n - c, - n \\ c + \frac{1}{2}, 1 - n + a - c, 1 - n + b - c \end{array}; 1] \end{aligned}$
と書き直せるので, 1つ目の式を $\frac{2 c - 1}{2 c - 2 a - 1}$ 倍, 2つ目の式を $\frac{2 a}{2 a - 2 c + 1}$ 倍して足し合わせると,
$\begin{aligned} \frac{(a)_{k}}{{(c + \frac{1}{2})}_{k}} & = \frac{2 c - 1}{2 c - 2 a - 1} \frac{(a)_{k}}{{(c - \frac{1}{2})}_{k}} + \frac{2 a}{2 a - 2 c + 1} \frac{(a + 1)_{k}}{{(c + \frac{1}{2})}_{k}} \\ \frac{(2 a)_{k}}{(2 c)_{k}} & = \frac{2 c - 1}{2 c - 2 a - 1} \frac{(2 a)_{k}}{{(2 c - 1)}_{k}} + \frac{2 a}{2 a - 2 c + 1} \frac{(2 a + 1)_{k}}{{(2 c)}_{k}} \end{aligned}$
であることから定理1が得られる. よって定理2の1つ目の式を示せば全ての式が示せることが分かった. つまり, 以下を示せば良い.
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 2 a, 2 b, - n \\ 2 c, \frac{1}{2} - n + a + b - c \end{array}; 1] & = \frac{{(c + \frac{1}{2} - a, c + \frac{1}{2} - b)}_{n}}{{(c + \frac{1}{2}, c + \frac{1}{2} - a - b)}_{n}}_{4} F_{3} [\begin{array}{c} a, b, - n - c, - n \\ c, \frac{1}{2} - n + a - c, \frac{1}{2} - n + b - c \end{array}; 1] \end{aligned}$
まず, 補題3より,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 2 a, 2 b, - n \\ 2 c, \frac{1}{2} - n + a + b - c \end{array}; 1] & = \frac{{(c + \frac{1}{2} + a - b, 2 c - 2 a)}_{n}}{{(2 c, c + \frac{1}{2} - a - b)}_{n}}_{3} F_{2} [\begin{array}{c} 2 a, \frac{1}{2} + a + b - c, - n \\ \frac{1}{2} + a - b + c, 1 - n + 2 a - 2 c \end{array}; 1] \end{aligned}$
が成り立つ.　次に, 補題4より,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} 2 a, \frac{1}{2} + a + b - c, - n \\ \frac{1}{2} + a - b + c, 1 - n + 2 a - 2 c \end{array}; 1] & = \frac{(2 c)_{n}}{(2 c - 2 a)_{n}}_{4} F_{3} [\begin{array}{c} a, c - b, 2 c + n, - n \\ c, c + \frac{1}{2}, \frac{1}{2} + a - b + c \end{array}; 1] \end{aligned}$
が成り立つ. 最後に, Whippleの変換公式より,
$\begin{aligned} _{4} F_{3} [\begin{array}{c} a, c - b, 2 c + n, - n \\ c, c + \frac{1}{2}, \frac{1}{2} + a - b + c \end{array}; 1] & = \frac{{(c - a + \frac{1}{2}, c - b + \frac{1}{2})}_{n}}{{(c + \frac{1}{2} + a - b, c + \frac{1}{2})}_{n}}_{4} F_{3} [\begin{array}{c} a, b, - c - n, - n \\ c, \frac{1}{2} - n + a - c, \frac{1}{2} - n + b - c \end{array}; 1] \end{aligned}$
であるから, これらを合わせて示すべき等式が得られる.

系として以下のような式が得られる.

$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ a + b - \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} a, b \\ a + b + \frac{1}{2} \end{array}; x] & =_{3} F_{2} [\begin{array}{c} 2 a, 2 b, a + b \\ 2 a + 2 b - 1, a + b + \frac{1}{2} \end{array}; x] \\ _{2} F_{1} [\begin{array}{c} a, b \\ a + b - \frac{1}{2} \end{array}; x]_{2} F_{1} [\begin{array}{c} a, b - 1 \\ a + b - \frac{1}{2} \end{array}; x] & =_{3} F_{2} [\begin{array}{c} 2 a, 2 b, a + b - 1 \\ 2 a + 2 b - 2, a + b - \frac{1}{2} \end{array}; x] \end{aligned}$