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Jainによる2F1の二次変換公式のq類似

以下は二次変換公式
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ 2 b \end{array}; 2 x] & = (1 - x)^{- a}_{2} F_{1} [\begin{array}{c} \frac{a}{2}, \frac{a + 1}{2} \\ b + \frac{1}{2} \end{array}; \frac{x^{2}}{(1 - x)^{2}}] \end{aligned}$
の $q$ 類似である.

Jain(1981)

$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a, b, - b \\ b^{2}, a x \end{array}; - x] & = \frac{(x; q)_{\infty}}{(a x; q)_{\infty}}_{2} ϕ_{1} [\begin{array}{c} a, a q \\ b^{2} q \end{array}; q^{2}; x^{2}] \end{aligned}$

$q$ -Vandermondeの和公式より,
$\begin{aligned} \sum_{0 \leq k} \frac{(q^{- n}, q^{1 - n}; q^{2})_{k}}{(b^{2} q, q^{2}; q^{2})_{k}} q^{2 k} & = \frac{(b^{2}; q^{2})_{n} q^{- n (n - 1) / 2}}{(b^{2}; q)_{n}} \end{aligned}$
であるから,
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a, b, - b \\ b^{2}, a x \end{array}; - x] \\ = \sum_{0 \leq n} \frac{(a; q)_{n}}{(q, a x; q)_{n}} (- x)^{n} q^{n (n - 1) / 2} \sum_{0 \leq k} \frac{(q^{- n}, q^{1 - n}; q^{2})_{k}}{(b^{2} q, q^{2}; q^{2})_{k}} q^{2 k} \\ = \sum_{0 \leq n, k} \frac{(a; q)_{n}}{(a x; q)_{n}} (- x)^{n} q^{n (n - 1) / 2} \frac{1}{(b^{2} q, q^{2}; q^{2})_{k} (q; q)_{n - 2 k}} q^{k (2 k - 2 n + 1)} \\ = \sum_{0 \leq n, k} \frac{(a; q)_{n + 2 k}}{(a x; q)_{n + 2 k}} (- x)^{n} q^{n (n - 1) / 2} \frac{1}{(b^{2} q, q^{2}; q^{2})_{k} (q; q)_{n}} x^{2 k} \\ = \sum_{0 \leq k} \frac{(a; q)_{2 k}}{(a x; q)_{2 k} (b^{2} q, q^{2}; q^{2})_{k}} x^{2 k}_{1} ϕ_{1} [\begin{array}{c} a q^{2 k} \\ a x q^{2 k} \end{array}; x] \end{aligned}$
ここで, Heineの和公式の系として
$\begin{aligned} _{1} ϕ_{1} [\begin{array}{c} a q^{2 k} \\ a x q^{2 k} \end{array}; x] & = \frac{(x; q)_{\infty}}{(a x q^{2 k}; q)_{\infty}} \end{aligned}$
と表されるから, これを代入することによって定理を得る.

以下は二次変換公式
$\begin{aligned} _{2} F_{1} [\begin{array}{c} 2 a, a + b \\ 2 a + 2 b \end{array}; x] & = (1 - x)^{- a}_{2} F_{1} [\begin{array}{c} a, b \\ a + b + \frac{1}{2} \end{array}; - \frac{x^{2}}{4 (1 - x)}] \end{aligned}$
の $q$ 類似である.

Jain(1981)

$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a^{2}, a b, - a b \\ a^{2} b^{2}, - a^{2} x \end{array}; x] & = \frac{(a^{2} x^{2}; q^{2})_{\infty}}{(- a^{2} x, x; q)_{\infty}}_{2} ϕ_{2} [\begin{array}{c} a^{2}, b^{2} \\ a^{2} b^{2} q, a^{2} x^{2} \end{array}; q^{2}; a^{2} x^{2} q] \end{aligned}$

定理1
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a, b, - b \\ b^{2}, a x \end{array}; - x] & = \frac{(x; q)_{\infty}}{(a x; q)_{\infty}}_{2} ϕ_{1} [\begin{array}{c} a, a q \\ b^{2} q \end{array}; q^{2}; x^{2}] \end{aligned}$
の右辺に Jacksonの $_{2} ϕ_{2}$ 変換公式
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a, b \\ c \end{array}; x] & = \frac{(a x; q)_{\infty}}{(x; q)_{\infty}}_{2} ϕ_{2} [\begin{array}{c} a, c / b \\ c, a x \end{array}; b x] \end{aligned}$
を適用すると
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a, b, - b \\ b^{2}, a x \end{array}; - x] & = \frac{(x; q)_{\infty}}{(a x; q)_{\infty}} \frac{(a x^{2}; q^{2})_{\infty}}{(x^{2}; q^{2})_{\infty}}_{2} ϕ_{2} [\begin{array}{c} a, b^{2} / a \\ b^{2} q, a x^{2} \end{array}; q^{2}; x^{2}] \\ = \frac{(a x^{2}; q^{2})_{\infty}}{(a x, - x; q)_{\infty}}_{2} ϕ_{2} [\begin{array}{c} a, b^{2} / a \\ b^{2} q, a x^{2} \end{array}; q^{2}; x^{2}] \end{aligned}$
ここで, $a \mapsto a^{2}, b \mapsto a b, x \mapsto - x$ と置き換えることによって定理を得る.

以下は二次変換公式
$\begin{aligned} _{2} F_{1} [\begin{array}{c} 2 a, 2 b \\ a + b + \frac{1}{2} \end{array}; x] & =_{2} F_{1} [\begin{array}{c} a, b \\ a + b + \frac{1}{2} \end{array}; 4 x (1 - x)] \end{aligned}$
の有限和の場合の $q$ 類似である.

$a, b$ のいずれかが非負整数 $N$ を用いて $q^{- N}$ と表されるとき,
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} a^{2}, b^{2}, x \\ a b \sqrt{q}, - a b \sqrt{q} \end{array}; q] & =_{3} ϕ_{2} [\begin{array}{c} a^{2}, b^{2}, x^{2} \\ a^{2} b^{2} q, 0 \end{array}; q^{2}; q^{2}] \end{aligned}$
が成り立つ.

$a, b$ のいずれかが非負整数 $N$ を用いて $q^{- N}$ と書けるとき, Singhの二次変換公式より,
$\begin{aligned} _{4} ϕ_{3} [\begin{array}{c} a^{2}, b^{2}, c, d \\ a b \sqrt{q}, - a b \sqrt{q}, - c d \end{array}; q] & =_{4} ϕ_{3} [\begin{array}{c} a^{2}, b^{2}, c^{2}, d^{2} \\ a^{2} b^{2} q, - c d, - c d q \end{array}; q^{2}; q^{2}] \end{aligned}$
である. $d = 0$ として, $c \mapsto x$ とすれば定理を得る.