三重q超幾何級数とq超幾何級数の間の関係式

前の記事でSharmaによる公式の $q$ 類似を示したが, その三重類似も同じようにできることに気づいたので, それについて書く.

$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i}, q^{- n_{i}}; q)_{k_{i}}}{(q, b_{i} c_{i} q^{- n_{i}} / a; q)_{k_{i}}} q^{k_{i}}) \frac{(a q; q)_{k_{1} + k_{2} + k_{3}}}{(a q; q)_{k_{1} + k_{2}} (a q; q)_{k_{1} + k_{3}} (a q; q)_{k_{2} + k_{3}}} \\ = \prod_{i = 1}^{3} \frac{(a q / b_{i}, a q / c_{i}; q)_{n_{i}}}{(a q, a q / b_{i} c_{i}; q)_{n_{i}}} \\ \cdot_{12} ϕ_{11} [\begin{array}{c} a, \sqrt{a} q, - \sqrt{a} q, b_{1}, c_{1}, q^{- n_{1}}, b_{2}, c_{2}, q^{- n_{2}}, b_{3}, c_{3}, q^{- n_{3}} \\ \sqrt{a}, - \sqrt{a}, a q / b_{1}, a q / c_{1}, a q^{n_{1} + 1}, a q / b_{2}, a q / c_{2}, a q^{n_{2} + 1}, a q / b_{3}, a q / c_{3}, a q^{n_{3} + 1} \end{array}; \frac{a^{4} q^{n_{1} + n_{2} + n_{3} + 4}}{b_{1} b_{2} b_{3} c_{1} c_{2} c_{3}}] \end{aligned}$

Rogersの $_{6} ϕ_{5}$ 和公式より,
$\begin{aligned} \frac{(a q; q)_{k_{1}} (a q; q)_{k_{2}} (a q; q)_{k_{3}} (a q; q)_{k_{1} + k_{2} + k_{3}}}{(a q; q)_{k_{1} + k_{2}} (a q; q)_{k_{1} + k_{3}} (a q; q)_{k_{2} + k_{3}}} & =_{6} ϕ_{5} [\begin{array}{c} a, \sqrt{a} q, - \sqrt{a} q, q^{- k_{1}}, q^{- k_{2}}, q^{- k_{3}} \\ \sqrt{a}, - \sqrt{a}, a q^{k_{1} + 1}, a q^{k_{2} + 1}, a q^{k_{3} + 1} \end{array}; a q^{k_{1} + k_{2} + k_{3} + 1}] \end{aligned}$
であるから, 二重の場合と全く同様に,
$\begin{aligned} \sum_{0 \leq k} \frac{(b, c, q^{- n}; q)_{k}}{(q, a q, b c q^{- n}; q)_{k}} q^{k} \frac{(q^{- k}; q)_{j}}{(a q^{k + 1}; q)_{j}} q^{j k} & = \frac{(a q / b, a q / c; q)_{n}}{(a q, a q / b c; q)_{n}} \frac{(b, c, q^{- n}; q)_{j}}{(a q / b, a q / c, a q^{n + 1}; q)_{j}} {(\frac{a q^{n + 1}}{b c})}^{j} \end{aligned}$
を用いて定理を得る.

特別な場合として, 積で表される場合についても考えてみる.

$b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$ を適切に並び替えたとき $a q / b_{1}, a q / b_{2}, a q / b_{3}, a q / c_{1}, a q / c_{2}, a q / c_{3}$ に一致し, $b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = a^{3} q^{3}$ であるとき,
$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i}, q^{- n_{i}}; q)_{k_{i}}}{(q, b_{i} c_{i} q^{- n_{i}} / a; q)_{k_{i}}} q^{k_{i}}) \frac{(a q; q)_{k_{1} + k_{2} + k_{3}}}{(a q; q)_{k_{1} + k_{2}} (a q; q)_{k_{1} + k_{3}} (a q; q)_{k_{2} + k_{3}}} \\ = (\prod_{i = 1}^{3} \frac{(a q / b_{i}, a q / c_{i}; q)_{n_{i}}}{(a q / b_{i} c_{i}; q)_{n_{i}}}) \frac{(a q; q)_{n_{1} + n_{2} + n_{3}}}{(a q; q)_{n_{1} + n_{2}} (a q; q)_{n_{1} + n_{3}} (a q; q)_{n_{2} + n_{3}}} \end{aligned}$
が成り立つ.

定理1と Rogersの $_{6} ϕ_{5}$ 和公式より,
$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i}, q^{- n_{i}}; q)_{k_{i}}}{(q, b_{i} c_{i} q^{- n_{i}} / a; q)_{k_{i}}} q^{k_{i}}) \frac{(a q; q)_{k_{1} + k_{2} + k_{3}}}{(a q; q)_{k_{1} + k_{2}} (a q; q)_{k_{1} + k_{3}} (a q; q)_{k_{2} + k_{3}}} \\ = (\prod_{i = 1}^{3} \frac{(a q / b_{i}, a q / c_{i}; q)_{n_{i}}}{(a q, a q / b_{i} c_{i}; q)_{n_{i}}})_{6} ϕ_{5} [\begin{array}{c} a, \sqrt{a} q, - \sqrt{a} q, q^{- n_{1}}, q^{- n_{2}}, q^{- n_{3}} \\ \sqrt{a}, - \sqrt{a}, a q^{n_{1} + 1}, a q^{n_{2} + 1}, a q^{n_{3} + 1} \end{array}; a q^{n_{1} + n_{2} + n_{3} + 1}] \\ = (\prod_{i = 1}^{3} \frac{(a q / b_{i}, a q / c_{i}; q)_{n_{i}}}{(a q / b_{i} c_{i}; q)_{n_{i}}}) \frac{(a q; q)_{n_{1} + n_{2} + n_{3}}}{(a q; q)_{n_{1} + n_{2}} (a q; q)_{n_{1} + n_{3}} (a q; q)_{n_{2} + n_{3}}} \end{aligned}$

例えば, $a q = b_{1} c_{2} = b_{2} c_{3} = b_{3} c_{1}$ のとき, 上の条件が満たされる.

系1において, さらに, $n_{1}, n_{2}, n_{3} \to \infty$ とすると, 以下を得る.

$b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$ を適切に並び替えたとき $a q / b_{1}, a q / b_{2}, a q / b_{3}, a q / c_{1}, a q / c_{2}, a q / c_{3}$ に一致し, $b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} = a^{3} q^{3}$ であるとき,
$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i}; q)_{k_{i}}}{(q; q)_{k_{i}}} {(\frac{a q}{b_{i} c_{i}})}^{k_{i}}) \frac{(a q; q)_{k_{1} + k_{2} + k_{3}}}{(a q; q)_{k_{1} + k_{2}} (a q; q)_{k_{1} + k_{3}} (a q; q)_{k_{2} + k_{3}}} \\ = \frac{(b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}; q)_{\infty}}{(a q, a q, a q / b_{1} c_{1}, a q / b_{2} c_{2}, a q / b_{3} c_{3}; q)_{\infty}} \end{aligned}$
が成り立つ.

定理1において, $n_{1}, n_{2}, n_{3} \to \infty$ とすると以下を得る.

$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i}; q)_{k_{i}}}{(q; q)_{k_{i}}} {(\frac{a q}{b_{i} c_{i}})}^{k_{i}}) \frac{(a q; q)_{k_{1} + k_{2} + k_{3}}}{(a q; q)_{k_{1} + k_{2}} (a q; q)_{k_{1} + k_{3}} (a q; q)_{k_{2} + k_{3}}} \\ = \prod_{i = 1}^{3} \frac{(a q / b_{i}, a q / c_{i}; q)_{\infty}}{(a q, a q / b_{i} c_{i}; q)_{\infty}} \\ \cdot_{9} ϕ_{11} [\begin{array}{c} a, \sqrt{a} q, - \sqrt{a} q, b_{1}, c_{1}, b_{2}, c_{2}, b_{3}, c_{3} \\ \sqrt{a}, - \sqrt{a}, a q / b_{1}, a q / c_{1}, a q / b_{2}, a q / c_{2}, a q / b_{3}, a q / c_{3}, 0, 0, 0 \end{array}; \frac{a^{4} q^{4}}{b_{1} b_{2} b_{3} c_{1} c_{2} c_{3}}] \end{aligned}$

$q$ 類似においては, これをさらに特殊化して積で表される場合を見つけるのは難しそうである.

古典極限

定理1の古典極限を考えると以下を得る.

$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i}, - n_{i})_{k_{i}}}{(1, b_{i} + c_{i} - n_{i} - a)_{k_{i}}}) \frac{(1 + a)_{k_{1} + k_{2} + k_{3}}}{(1 + a)_{k_{1} + k_{2}} (1 + a)_{k_{1} + k_{3}} (1 + a)_{k_{2} + k_{3}}} \\ = \prod_{i = 1}^{3} \frac{(1 + a - b_{i}, 1 + a - c_{i})_{n_{i}}}{(1 + a, 1 + a - b_{i} - c_{i})_{n_{i}}} \\ \cdot_{11} F_{10} [\begin{array}{c} a, 1 + \frac{a}{2}, b_{1}, c_{1}, - n_{1}, b_{2}, c_{2}, - n_{2}, b_{3}, c_{3}, - n_{3} \\ \frac{a}{2}, 1 + a - b_{1}, 1 + a - c_{1}, 1 + a + n_{1}, 1 + a - b_{2}, 1 + a - c_{2}, 1 + a + n_{2}, 1 + a - b_{3}, 1 + a - c_{3}, 1 + a + n_{3} \end{array}; 1] \end{aligned}$

先ほどと同様に次の系が得られる.

$b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$ を適切に並び替えたとき $1 + a - b_{1}, 1 + a - b_{2}, 1 + a - b_{3}, 1 + a - c_{1}, 1 + a - c_{2}, 1 + a - c_{3}$ に一致するとき,
$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i}, - n_{i})_{k_{i}}}{(1, b_{i} + c_{i} - n_{i} - a)_{k_{i}}}) \frac{(1 + a)_{k_{1} + k_{2} + k_{3}}}{(1 + a)_{k_{1} + k_{2}} (1 + a)_{k_{1} + k_{3}} (1 + a)_{k_{2} + k_{3}}} \\ = (\prod_{i = 1}^{3} \frac{(1 + a - b_{i}, 1 + a - c_{i})_{n_{i}}}{(1 + a - b_{i} - c_{i})_{n_{i}}}) \frac{(1 + a)_{n_{1} + n_{2} + n_{3}}}{(1 + a)_{n_{1} + n_{2}} (1 + a)_{n_{1} + n_{3}} (1 + a)_{n_{2} + n_{3}}} \end{aligned}$
が成り立つ.

$Re (1 + a - b_{i} - c_{i}) > 0 (1 \leq i \leq 3)$ とする. $b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$ を適切に並び替えたとき $1 + a - b_{1}, 1 + a - b_{2}, 1 + a - b_{3}, 1 + a - c_{1}, 1 + a - c_{2}, 1 + a - c_{3}$ に一致するとき,
$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i})_{k_{i}}}{k_{i}!}) \frac{(1 + a)_{k_{1} + k_{2} + k_{3}}}{(1 + a)_{k_{1} + k_{2}} (1 + a)_{k_{1} + k_{3}} (1 + a)_{k_{2} + k_{3}}} \\ = \frac{Γ (1 + a)^{2} Γ (1 + a - b_{1} - c_{1}) Γ (1 + a - b_{2} - c_{2}) Γ (1 + a - b_{3} - c_{3})}{Γ (b_{1}) Γ (b_{2}) Γ (b_{3}) Γ (c_{1}) Γ (c_{2}) Γ (c_{3})} \end{aligned}$
が成り立つ.

定理2において, $n_{1}, n_{2}, n_{3} \to \infty$ とすると以下を得る.

$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i})_{k_{i}}}{k_{i}!}) \frac{(1 + a)_{k_{1} + k_{2} + k_{3}}}{(1 + a)_{k_{1} + k_{2}} (1 + a)_{k_{1} + k_{3}} (1 + a)_{k_{2} + k_{3}}} \\ = \prod_{i = 1}^{3} \frac{Γ (1 + a) Γ (1 + a - b_{i} - c_{i})}{Γ (1 + a - b_{i}) Γ (1 + a - c_{i})} \\ \cdot_{8} F_{7} [\begin{array}{c} a, 1 + \frac{a}{2}, b_{1}, c_{1}, b_{2}, c_{2}, b_{3}, c_{3} \\ \frac{a}{2}, 1 + a - b_{1}, 1 + a - c_{1}, 1 + a - b_{2}, 1 + a - c_{2}, 1 + a - b_{3}, 1 + a - c_{3} \end{array}; - 1] \end{aligned}$

この左辺が $_{4} F_{3}$ になる場合に積で表されることから, 以下の系が得られる.

$Re (1 + a - b_{i} - c_{i}) > 0 (1 \leq i \leq 3)$ とする. $b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$ の中から $4$ つの数を取り出して $1 + a - b_{1}, 1 + a - b_{2}, 1 + a - b_{3}, 1 + a - c_{1}, 1 + a - c_{2}, 1 + a - c_{3}$ のうち $4$ つの数に一致させることができるとき, 取り出されなかった $2$ つの数を $d_{1}, d_{2}$ とすると
$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i})_{k_{i}}}{k_{i}!}) \frac{(1 + a)_{k_{1} + k_{2} + k_{3}}}{(1 + a)_{k_{1} + k_{2}} (1 + a)_{k_{1} + k_{3}} (1 + a)_{k_{2} + k_{3}}} \\ = (\prod_{i = 1}^{3} \frac{Γ (1 + a) Γ (1 + a - b_{i} - c_{i})}{Γ (1 + a - b_{i}) Γ (1 + a - c_{i})}) \frac{Γ (1 + a - d_{1}) Γ (1 + a - d_{2})}{Γ (1 + a) Γ (1 + a - d_{1} - d_{2})} \end{aligned}$
が成り立つ.

これは系5の一般化である.

特別な場合として, 以下を得る.

$b_{2} + b_{3} = c_{1} + c_{3} = 1 + a$ または $b_{3} + c_{1} = b_{2} + c_{3} = 1 + a$ のとき,
$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} (\prod_{i = 1}^{3} \frac{(b_{i}, c_{i})_{k_{i}}}{k_{i}!}) \frac{(1 + a)_{k_{1} + k_{2} + k_{3}}}{(1 + a)_{k_{1} + k_{2}} (1 + a)_{k_{1} + k_{3}} (1 + a)_{k_{2} + k_{3}}} \\ = (\prod_{i = 1}^{3} \frac{Γ (1 + a) Γ (1 + a - b_{i} - c_{i})}{Γ (1 + a - b_{i}) Γ (1 + a - c_{i})}) \frac{Γ (1 + a - b_{1}) Γ (1 + a - c_{2})}{Γ (1 + a) Γ (1 + a - b_{1} - c_{2})} \end{aligned}$
が成り立つ.

上の証明の過程で, $n := n_{1}$ だけ極限を飛ばさずに残すと, Dougallの $_{5} F_{4}$ 和公式によって総和できるので, 上の和公式はもう少しだけ一般化される.

$b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$ の中から $4$ つの数を取り出して $1 + a - b_{1}, 1 + a - b_{2}, 1 + a - b_{3}, 1 + a - c_{1}, 1 + a - c_{2}, 1 + a - c_{3}$ のうち $4$ つの数に一致させることができるとき, 取り出されなかった $2$ つの数を $d_{1}, d_{2}$ とすると
$\begin{aligned} \sum_{0 \leq k_{1}, k_{2}, k_{3}} \frac{(b_{1}, c_{1}, - n)_{k_{1}}}{(1, b_{1} + c_{1} - n - a)_{k_{1}}} (\prod_{i = 2}^{3} \frac{(b_{i}, c_{i})_{k_{i}}}{k_{i}!}) \frac{(1 + a)_{k_{1} + k_{2} + k_{3}}}{(1 + a)_{k_{1} + k_{2}} (1 + a)_{k_{1} + k_{3}} (1 + a)_{k_{2} + k_{3}}} \\ = \frac{(1 + a - b_{1}, 1 + a - c_{1})_{n}}{(1 + a, 1 + a - b_{1} - c_{1})_{n}} (\prod_{i = 2}^{3} \frac{Γ (1 + a) Γ (1 + a - b_{i} - c_{i})}{Γ (1 + a - b_{i}) Γ (1 + a - c_{i})}) \frac{(1 + a, 1 + a - d_{1} - d_{2})_{n}}{(1 + a - d_{1}, 1 + a - d_{2})_{n}} \end{aligned}$
が成り立つ.