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Gasper-Rahmanの二次変換公式

Gasper-Rahmanの1986年の論文でいくつかの二次変換公式が示されている.

Gasper-Rahman(1986)

$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a, b \\ a q / b \end{array}; \frac{x q}{b^{2}}] & = \frac{(x q / b; a x^{2} q / b^{2}; q)_{\infty}}{(a x q / b, x^{2} q / b^{2}; q)_{\infty}} \\ \cdot_{8} ϕ_{7} [\begin{array}{c} a x / b, q \sqrt{a x / b}, - q \sqrt{a x / b}, x, \sqrt{a}, - \sqrt{a}, \sqrt{a q}, - \sqrt{a q} \\ \sqrt{a x / b}, - \sqrt{a x / b}, a q / b, \sqrt{a} x q / b, - \sqrt{x} q / b, \sqrt{a q} x / b, - \sqrt{a q} x / b \end{array}; \frac{x q}{b^{2}}] \end{aligned}$

BaileyのNearly-Poised $_{5} ϕ_{4}$ の変換公式
$\begin{aligned} _{5} ϕ_{4} [\begin{array}{c} a, b, c, d, q^{- n} \\ a q / b, a q / c, a q / d, a^{2} q^{- n} / w^{2} \end{array}; q] \\ = \frac{(w q / a, w^{2} q / a; q)_{n}}{(w q, w^{2} q / a^{2}; q)_{n}}_{12} ϕ_{11} [\begin{array}{c} w, \sqrt{w} q, - \sqrt{w} q, w b / a, w c / a, w d / a, \sqrt{a}, - \sqrt{a}, \sqrt{a q}, - \sqrt{a q}, w^{2} q^{n + 1} / a, q^{- n} \\ \sqrt{w}, - \sqrt{w}, a q / b, a q / c, a q / d, w q / \sqrt{a}, - w q / \sqrt{a}, w \sqrt{q / a}, - w \sqrt{q / a}, a q^{- n} / w, w q^{n + 1} \end{array}; q] (w = a^{2} q / b c d) \end{aligned}$
において, $c \mapsto c q^{- n}, d \mapsto d q^{n}$ とすると,
$\begin{aligned} _{5} ϕ_{4} [\begin{array}{c} a, b, c q^{- n}, d q^{n}, q^{- n} \\ a q / b, a q^{n + 1} / c, a q^{1 - n} / d, a^{2} q^{- n} / w^{2} \end{array}; q] \\ = \frac{(w q / a, w^{2} q / a; q)_{n}}{(w q, w^{2} q / a^{2}; q)_{n}}_{12} ϕ_{11} [\begin{array}{c} w, \sqrt{w} q, - \sqrt{w} q, w b / a, w c q^{- n} / a, w d q^{n} / a, \sqrt{a}, - \sqrt{a}, \sqrt{a q}, - \sqrt{a q}, w^{2} q^{n + 1} / a, q^{- n} \\ \sqrt{w}, - \sqrt{w}, a q / b, a q^{n + 1} / c, a q^{1 - n} / d, w q / \sqrt{a}, - w q / \sqrt{a}, w \sqrt{q / a}, - w \sqrt{q / a}, a q^{- n} / w, w q^{n + 1} \end{array}; q] (w = a^{2} q / b c d) \end{aligned}$
である. $n \to \infty$ とすると,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a, b \\ a q / b \end{array}; \frac{w^{2} c d}{a^{3}}] \\ = \frac{(w q / a, w^{2} q / a; q)_{\infty}}{(w q, w^{2} q / a^{2}; q)_{\infty}}_{8} ϕ_{7} [\begin{array}{c} w, \sqrt{w} q, - \sqrt{w} q, w b / a, \sqrt{a}, - \sqrt{a}, \sqrt{a q}, - \sqrt{a q} \\ \sqrt{w}, - \sqrt{w}, a q / b, w q / \sqrt{a}, - w q / \sqrt{a}, w \sqrt{q / a}, - w \sqrt{q / a} \end{array}; \frac{w^{2} c d}{a^{3}}] (w = a^{2} q / b c d) \end{aligned}$
ここで, $d = a q / c x$ とすると, $w = a x / b$ であり,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a, b \\ a q / b \end{array}; \frac{x q}{b^{2}}] \\ = \frac{(x q / b, a x^{2} q / b^{2}; q)_{\infty}}{(a x q / b, x^{2} q / b^{2}; q)_{\infty}}_{8} ϕ_{7} [\begin{array}{c} a x / b, q \sqrt{a x / b}, - q \sqrt{a x / b}, x, \sqrt{a}, - \sqrt{a}, \sqrt{a q}, - \sqrt{a q} \\ \sqrt{a x / b}, - \sqrt{a x / b}, a q / b, \sqrt{a} x q / b, - \sqrt{a} x q / b, \sqrt{a q} x / b, - \sqrt{a q} x / b \end{array}; \frac{x q}{b^{2}}] \end{aligned}$
となって定理を得る.

上の定理において, $a \mapsto q^{a}, b \mapsto q^{b}$ として, $q \to 1$ とすれば,
$\begin{aligned} _{2} F_{1} [\begin{array}{c} a, b \\ 1 + a - b \end{array}; x] & = (1 + x)^{- a}_{2} F_{1} [\begin{array}{c} \frac{a}{2}, \frac{a + 1}{2} \\ 1 + a - b \end{array}; \frac{4 x}{(1 + x)^{2}}] \end{aligned}$
となるので, 定理1は実は $_{2} F_{1}$ の二次変換公式の $q$ 類似になっていることが分かる.

Gasper-Rahman(1986)

$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a^{2}, b^{2} \\ a^{2} q^{2} / b^{2} \end{array}; q^{2}, \frac{x^{2} q^{2}}{b^{4}}] \\ = \frac{(a q / b^{2}, x^{2} q / b^{2}; q)_{\infty}}{(a x^{2} q / b^{2}, a^{2} q / b^{2}; q)_{\infty}} \frac{(a^{2} x^{2} q / b^{2}, a^{2} x^{2} q^{2} / b^{4}; q^{2})_{\infty}}{(x^{2} q / b^{2}, x^{2} q^{2} / b^{4}; q^{2})_{\infty}} \\ \cdot_{8} ϕ_{7} [\begin{array}{c} a x^{2} / b^{2}, q \sqrt{a x^{2} / b^{2}}, - q \sqrt{a x^{2} / b^{2}}, a, x, - x, x \sqrt{q} / b, - x \sqrt{q} / b \\ \sqrt{a x^{2} / b^{2}}, - \sqrt{a x^{2} / b^{2}}, x^{2} q / b^{2}, a x q / b^{2}, - a x q / b^{2}, a x \sqrt{q} / b, - a x \sqrt{q} / b \end{array}; \frac{a q}{b^{2}}] \end{aligned}$

一つ補題を用意する.

Gasper-Rahman(1986)

$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} x^{2}, y^{2}, a^{2} \\ a^{2} b^{2}, x^{2} y^{2} b^{2} \end{array}; b^{2} q] \\ = \frac{(- a, a b^{2}; q)_{\infty} (b^{2}; q^{2})_{\infty}}{(- 1, b^{2}; q)_{\infty} (a^{2} b^{2}; q^{2})_{\infty}}_{5} ϕ_{4} [\begin{array}{c} a, b x, - b x, b y, - b y \\ - q, a b^{2}, b x y, - b x y \end{array}; q] \\ + \frac{(a, - a b^{2}; q)_{\infty} (b^{2}; q^{2})_{\infty}}{(- 1, b^{2}; q)_{\infty} (a^{2} b^{2}; q^{2})_{\infty}}_{5} ϕ_{4} [\begin{array}{c} - a, b x, - b x, b y, - b y \\ - q, - a b^{2}, b x y, - b x y \end{array}; q] \end{aligned}$

Non-terminating $q$ -Saalschützの和公式
$\begin{aligned} \frac{(e, f; q)_{\infty}}{(a, b, c; q)_{\infty}}_{3} ϕ_{2} [\begin{array}{c} a, b, c \\ e, f \end{array}; q] - \frac{q}{e} \frac{(q^{2} / e, f q / e; q)_{\infty}}{(a q / e, b q / e, c q / e; q)_{\infty}}_{3} ϕ_{2} [\begin{array}{c} a q / e, b q / e, c q / e \\ q^{2} / e, f q / e \end{array}; q] \\ = \frac{(e, q / e, f / a, f / b, f / c; q)_{\infty}}{(a, b, c, a q / e, b q / e, c q / e; q)_{\infty}} (a b c q = e f) \end{aligned}$
において, $e = - q, b = - c, a \mapsto a q^{r}$ とすると,

$\begin{aligned} \frac{(- q, a b^{2} q^{r}; q)_{\infty}}{(a q^{r}, b, - b; q)_{\infty}}_{3} ϕ_{2} [\begin{array}{c} a q^{r}, b, - b \\ - q, a b^{2} q^{r} \end{array}; q] + \frac{(- q, - a b^{2} q^{r}; q)_{\infty}}{(- a q^{r}, - b, b; q)_{\infty}}_{3} ϕ_{2} [\begin{array}{c} - a q^{r}, - b, b \\ - q, - a b^{2} q^{r} \end{array}; q] \\ = \frac{(- q, - 1, b^{2}, a b q^{r}, - a b q^{r}; q)_{\infty}}{(a q^{r}, b, - b, - a q^{r}, - b, b; q)_{\infty}} \end{aligned}$
両辺に
$\begin{array}{r} \frac{(x^{2}, y^{2}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}; q^{2})_{r}} b^{2 r} q^{r} \end{array}$
を掛けて足し合わせると,
$\begin{aligned} \frac{(- q, - 1, b^{2}; q)_{\infty} (a^{2} b^{2}; q^{2})_{\infty}}{(a^{2}, b^{2}, b^{2}; q^{2})_{\infty}}_{3} ϕ_{2} [\begin{array}{c} x^{2}, y^{2}, a^{2} \\ a^{2} b^{2}, x^{2} y^{2} b^{2} \end{array}; b^{2} q] \\ = \sum_{0 \leq r} \frac{(x^{2}, y^{2}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}; q^{2})_{r}} b^{2 r} q^{r} \frac{(- q, a b^{2} q^{r}; q)_{\infty}}{(a q^{r}, b, - b; q)_{\infty}}_{3} ϕ_{2} [\begin{array}{c} a q^{r}, b, - b \\ - q, a b^{2} q^{r} \end{array}; q] \\ + \sum_{0 \leq r} \frac{(x^{2}, y^{2}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}; q^{2})_{r}} b^{2 r} q^{r} \frac{(- q, - a b^{2} q^{r}; q)_{\infty}}{(- a q^{r}, - b, b; q)_{\infty}}_{3} ϕ_{2} [\begin{array}{c} - a q^{r}, - b, b \\ - q, - a b^{2} q^{r} \end{array}; q] \end{aligned}$
ここで,
$\begin{aligned} \sum_{0 \leq r} \frac{(x^{2}, y^{2}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}; q^{2})_{r}} b^{2 r} q^{r} \frac{(- q, a b^{2} q^{r}; q)_{\infty}}{(a q^{r}, b, - b; q)_{\infty}}_{3} ϕ_{2} [\begin{array}{c} a q^{r}, b, - b \\ - q, a b^{2} q^{r} \end{array}; q] \\ = \sum_{0 \leq r} \frac{(x^{2}, y^{2}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}; q^{2})_{r}} b^{2 r} q^{r} \frac{(- q, a b^{2} q^{r}; q)_{\infty}}{(a q^{r}, b, - b; q)_{\infty}} \sum_{0 \leq j} \frac{(a q^{r}, b, - b; q)_{j}}{(- q, q, a b^{2} q^{r}; q)_{j}} q^{j} \\ = \frac{(- q, a b^{2}; q)_{\infty}}{(a, b, - b; q)_{\infty}} \sum_{0 \leq r, j} \frac{(x^{2}, y^{2}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}; q^{2})_{r}} \frac{(a; q)_{r + j} (b, - b; q)_{j}}{(a b^{2}; q)_{r + j} (- q, q; q)_{j}} b^{2 r} q^{r + j} \\ = \frac{(- q, a b^{2}; q)_{\infty}}{(a, b, - b; q)_{\infty}} \sum_{0 \leq r, j} \frac{(x^{2}, y^{2}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}; q^{2})_{r}} \frac{(a; q)_{j} (b^{2}; q^{2})_{j - r}}{(a b^{2}; q)_{j} (q^{2}; q^{2})_{j - r}} b^{2 r} q^{j} \\ = \frac{(- q, a b^{2}; q)_{\infty}}{(a, b, - b; q)_{\infty}} \sum_{0 \leq j} \frac{(a; q)_{j} (b^{2}; q^{2})_{j}}{(a b^{2}; q)_{j} (q^{2}; q^{2})_{j}} q^{j} \sum_{0 \leq r} \frac{(x^{2}, y^{2}, q^{- 2 r}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}, q^{2 - 2 r} / b^{2}; q^{2})_{r}} q^{2 r} \\ = \frac{(- q, a b^{2}; q)_{\infty}}{(a, b, - b; q)_{\infty}} \sum_{0 \leq j} \frac{(a; q)_{j} (b^{2}; q^{2})_{j}}{(a b^{2}; q)_{j} (q^{2}; q^{2})_{j}} q^{j} \frac{(x^{2} b^{2}, y^{2} b^{2}; q^{2})_{j}}{(x^{2} y^{2} b^{2}, b^{2}; q)_{j}} \\ = \frac{(- q, a b^{2}; q)_{\infty}}{(a, b, - b; q)_{\infty}}_{5} ϕ_{4} [\begin{array}{c} a, b x, - b x, b y, - b y \\ - q, a b^{2}, b x y, - b x y \end{array}; q] \end{aligned}$
ここで, 最後から2行目の等号は $q$ -Saalschützの和公式による. $a \mapsto - a$ とすれば,
$\begin{aligned} \sum_{0 \leq r} \frac{(x^{2}, y^{2}; q^{2})_{r}}{(q^{2}, x^{2} y^{2} b^{2}; q^{2})_{r}} b^{2 r} q^{r} \frac{(- q, - a b^{2} q^{r}; q)_{\infty}}{(- a q^{r}, - b, b; q)_{\infty}}_{3} ϕ_{2} [\begin{array}{c} - a q^{r}, - b, b \\ - q, - a b^{2} q^{r} \end{array}; q] \\ = \frac{(- q, - a b^{2}; q)_{\infty}}{(- a, b, - b; q)_{\infty}}_{5} ϕ_{4} [\begin{array}{c} - a, b x, - b x, b y, - b y \\ - q, - a b^{2}, b x y, - b x y \end{array}; q] \end{aligned}$
も得られるから,
$\begin{aligned} _{3} ϕ_{2} [\begin{array}{c} x^{2}, y^{2}, a^{2} \\ a^{2} b^{2}, x^{2} y^{2} b^{2} \end{array}; b^{2} q] \\ = \frac{(- a, a b^{2}; q)_{\infty} (b^{2}; q^{2})_{\infty}}{(- 1, b^{2}; q)_{\infty} (a^{2} b^{2}; q^{2})_{\infty}}_{5} ϕ_{4} [\begin{array}{c} a, b x, - b x, b y, - b y \\ - q, a b^{2}, b x y, - b x y \end{array}; q] \\ + \frac{(a, - a b^{2}; q)_{\infty} (b^{2}; q^{2})_{\infty}}{(- 1, b^{2}; q)_{\infty} (a^{2} b^{2}; q^{2})_{\infty}}_{5} ϕ_{4} [\begin{array}{c} - a, b x, - b x, b y, - b y \\ - q, - a b^{2}, b x y, - b x y \end{array}; q] \end{aligned}$
を得る.

定理2の証明

補題3において, $y = a b$ とすると,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a^{2}, x^{2} \\ a^{2} b^{4} x^{2} \end{array}; b^{2} q] \\ = \frac{(- a, a b^{2}; q)_{\infty} (b^{2}; q^{2})_{\infty}}{(- 1, b^{2}; q)_{\infty} (a^{2} b^{2}; q^{2})_{\infty}}_{4} ϕ_{3} [\begin{array}{c} a, b x, - b x, - a b^{2} \\ - q, a b^{2} x, - a b^{2} x \end{array}; q] \\ + \frac{(a, - a b^{2}; q)_{\infty} (b^{2}; q^{2})_{\infty}}{(- 1, b^{2}; q)_{\infty} (a^{2} b^{2}; q^{2})_{\infty}}_{4} ϕ_{3} [\begin{array}{c} - a, b x, - b x, a b^{2} \\ - q, a b^{2} x, - a b^{2} x \end{array}; q] \end{aligned}$
ここで, non-terminating $q$ -Whippleの変換公式より,
$\begin{aligned} _{8} ϕ_{7} [\begin{array}{c} a b^{2} x^{2} / q, q \sqrt{a b^{2} x^{2} / q}, - q \sqrt{a b^{2} x^{2} / q}, a, x, - x, b x, - b x \\ \sqrt{a b^{2} x^{2} / q}, - \sqrt{a b^{2} x^{2} / q}, b^{2} x^{2}, a b^{2} x, - a b^{2} x, a b x, - a b x \end{array}; a b^{2}] \\ = \frac{(a b^{2} x^{2}, b x, - b x, - a; q)_{\infty}}{(b^{2} x^{2}, a b x, - a b x, - 1; q)_{\infty}} (_{4} ϕ_{3} [\begin{array}{c} a b^{2}, a, b x, - b x \\ a b^{2} x, - a b^{2} x, - q \end{array}; q] + \frac{(a, - a b^{2}; q)_{\infty}}{(- a, a b^{2}; q)_{\infty}}_{4} ϕ_{3} [\begin{array}{c} a b^{2}, - a, b x, - b x \\ a b^{2} x, - a b^{2} x, - q \end{array}; q]) \end{aligned}$
であるから,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a^{2}, x^{2} \\ a^{2} b^{4} x^{2} \end{array}; q^{2}, b^{2} q] \\ = \frac{(- a, a b^{2}; q)_{\infty} (b^{2}; q^{2})_{\infty}}{(- 1, b^{2}; q)_{\infty} (a^{2} b^{2}; q^{2})_{\infty}} \frac{(b^{2} x^{2}, a b x, - a b x, - 1; q)_{\infty}}{(a b^{2} x^{2}, b x, - b x, - a; q)_{\infty}} \\ \cdot_{8} ϕ_{7} [\begin{array}{c} a b^{2} x^{2} / q, q \sqrt{a b^{2} x^{2} / q}, - q \sqrt{a b^{2} x^{2} / q}, a, x, - x, b x, - b x \\ \sqrt{a b^{2} x^{2} / q}, - \sqrt{a b^{2} x^{2} / q}, b^{2} x^{2}, a b^{2} x, - a b^{2} x, a b x, - a b x \end{array}; a b^{2}] \\ = \frac{(a b^{2}, b^{2} x^{2}; q)_{\infty} (a^{2} b^{2} x^{2}, b^{2}; q^{2})_{\infty}}{(b^{2}, a b^{2} x^{2}; q)_{\infty} (b^{2} x^{2}, a^{2} b^{2}; q^{2})_{\infty}} \\ \cdot_{8} ϕ_{7} [\begin{array}{c} a b^{2} x^{2} / q, q \sqrt{a b^{2} x^{2} / q}, - q \sqrt{a b^{2} x^{2} / q}, a, x, - x, b x, - b x \\ \sqrt{a b^{2} x^{2} / q}, - \sqrt{a b^{2} x^{2} / q}, b^{2} x^{2}, a b^{2} x, - a b^{2} x, a b x, - a b x \end{array}; a b^{2}] \end{aligned}$
ここで, Heineの変換公式より,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a^{2}, x^{2} \\ a^{2} b^{4} x^{2} \end{array}; q^{2}, b^{2} q] & = \frac{(a^{2} b^{2} q, b^{4} x^{2}; q^{2})_{\infty}}{(b^{2} q, a^{2} b^{4} x^{2}; q^{2})_{\infty}}_{2} ϕ_{1} [\begin{array}{c} a^{2}, q / b^{2} \\ a^{2} b^{2} q \end{array}; q^{2}, b^{4} x^{2}] \end{aligned}$
であるから,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a^{2}, q / b^{2} \\ a^{2} b^{2} q \end{array}; q^{2}, b^{4} x^{2}] \\ = \frac{(a b^{2}, b^{2} x^{2}; q)_{\infty} (a^{2} b^{2} x^{2}, a^{2} b^{4} x^{2}; q^{2})_{\infty}}{(a^{2} b^{2}, a b^{2} x^{2}; q)_{\infty} (b^{2} x^{2}, b^{4} x^{2}; q^{2})_{\infty}} \\ \cdot_{8} ϕ_{7} [\begin{array}{c} a b^{2} x^{2} / q, q \sqrt{a b^{2} x^{2} / q}, - q \sqrt{a b^{2} x^{2} / q}, a, x, - x, b x, - b x \\ \sqrt{a b^{2} x^{2} / q}, - \sqrt{a b^{2} x^{2} / q}, b^{2} x^{2}, a b^{2} x, - a b^{2} x, a b x, - a b x \end{array}; a b^{2}] \end{aligned}$
ここで, $b \mapsto \sqrt{q} / b$ とすると,
$\begin{aligned} _{2} ϕ_{1} [\begin{array}{c} a^{2}, b^{2} \\ a^{2} q^{2} / b^{2} \end{array}; q^{2}, \frac{x^{2} q^{2}}{b^{4}}] \\ = \frac{(a q / b^{2}, x^{2} q / b^{2}; q)_{\infty} (a^{2} x^{2} q / b^{2}, a^{2} x^{2} q^{2} / b^{4}; q^{2})_{\infty}}{(a^{2} q / b^{2}, a x^{2} q / b^{2}; q)_{\infty} (x^{2} q / b^{2}, x^{2} q^{2} / b^{4}; q^{2})_{\infty}} \\ \cdot_{8} ϕ_{7} [\begin{array}{c} a x^{2} / b^{2}, q \sqrt{a x^{2} / b^{2}}, - q \sqrt{a x^{2} / b^{2}}, a, x, - x, x \sqrt{q} / b, - x \sqrt{q} / b \\ \sqrt{a x^{2} / b^{2}}, - \sqrt{a x^{2} / b^{2}}, x^{2} q / b^{2}, a x q / b^{2}, - a x q / b^{2}, a x \sqrt{q} / b, - a x \sqrt{q} / b \end{array}; \frac{a q}{b^{2}}] \end{aligned}$

定理2においては, $a \mapsto q^{a}, b \mapsto q^{b}$ として, $q \to 1$ としても右辺が上手く表せなさそうであるから, これ自体が古典的な $_{2} F_{1}$ の二次変換公式の $q$ 類似になっているわけではないかもしれない.

参考文献

[1]

George Gasper, Mizan Rahman, Positivity of the Poisson kernel for the continuous q-Jacobi polynomials and some quadratic transformation formulas for basic hypergeometric series., SIAM J. Math. Anal., 1986, 970-999

投稿日：5月7日