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q-ガウスの連分数

204

q-ガウスの連分数

$\frac{_{2} ϕ_{1} [\begin{matrix} a q, b \\ c q \end{matrix}; q, z]}{(1 - c)_{2} ϕ_{1} [\begin{matrix} a, b \\ c \end{matrix}; q, z]} = K_{n = 0}^{\infty} \frac{t_{n}}{1 - c q^{n}}$
$\begin{array}{r} {\begin{array}{l} t_{0} & = & 1 \\ t_{2 n - 1} & = & q^{n - 1} (c q^{n - 1} - a) (1 - b q^{n - 1}) z \\ t_{2 n} & = & q^{n - 1} (c q^{n} - b) (1 - a q^{n}) z \end{array} \end{array}$

q-超幾何関数の比が連分数で表わされるというものであり、ガウスの連分数のq-類似に相当する。
次の補題2を用いて、ガウスの連分数と同様の処理をすることで導く。

$_{2} ϕ_{1} [\begin{matrix} a, b \\ c \end{matrix}; q, z] -_{2} ϕ_{1} [\begin{matrix} a q, b \\ c q \end{matrix}; q, z] = \frac{(c - a) (1 - b) z}{(1 - c) (1 - c q)}_{2} ϕ_{1} [\begin{matrix} a q, b q \\ c q^{2} \end{matrix}; q, z]$

$\begin{array}{rcl} (左辺) & = & \sum_{n = 0}^{\infty} (\frac{(a, b; q)_{n}}{(c, q; q)_{n}} - \frac{(a q, b; q)_{n}}{(c q, q; q)_{n}}) z^{n} \\ = & \sum_{n = 0}^{\infty} \frac{(a q; q)_{n - 1} (b; q)_{n}}{(c; q)_{n + 1} (q; q)_{n}} {(1 - a) (1 - c q^{n}) - (1 - c) (1 - a q^{n})} z^{n} \\ = & (c - a) \sum_{n = 0}^{\infty} \frac{(a q; q)_{n - 1} (b; q)_{n}}{(c; q)_{n + 1} (q; q)_{n}} (1 - q^{n}) z^{n} \\ = & (c - a) \sum_{n = 1}^{\infty} \frac{(a q; q)_{n - 1} (b; q)_{n}}{(c; q)_{n + 1} (q; q)_{n - 1}} z^{n} \\ = & \frac{(c - a) (1 - b) z}{(1 - c) (1 - c q)} \sum_{n = 0}^{\infty} \frac{(a q, b q; q)_{n}}{(c q^{2}, q; q)_{n}} z^{n} \\ = & \frac{(c - a) (1 - b) z}{(1 - c) (1 - c q)}_{2} ϕ_{1} [\begin{array}{c} a q, b q \\ c q^{2} \end{array}; q, z] \\ = & (右辺) \end{array}$

$\frac{(- x, - y; q)_{\infty} - (x, y; q)_{\infty}}{(- x, - y; q)_{\infty} + (x, y; q)_{\infty}} = \frac{x + y}{1 - q + K_{n = 1}^{\infty} \frac{q^{n - 1} (x + q^{n} y) (y + q^{n} x)}{1 - q^{2 n + 1}}}$

$\begin{array}{rcl} (左辺) & = & \frac{\frac{1}{2} (\frac{(- x; q)_{\infty}}{(y; q)_{\infty}} - \frac{(x; q)_{\infty}}{(- y; q)_{\infty}})}{\frac{1}{2} (\frac{(- x; q)_{\infty}}{(y; q)_{\infty}} + \frac{(x; q)_{\infty}}{(- y; q)_{\infty}})} \\ = & \frac{\sum_{n = 0}^{\infty} \frac{(- \frac{x}{y}; q)_{2 n + 1}}{(q; q)_{2 n + 1}} y^{2 n + 1}}{\sum_{n = 0}^{\infty} \frac{(- \frac{x}{y}; q)_{2 n}}{(q; q)_{2 n}} y^{2 n}} (∵ q - 二項定理) \\ = & \frac{\frac{1 + \frac{x}{y}}{1 - q}_{2} ϕ_{1} [\begin{array}{c} - \frac{q^{2} x}{y}, - \frac{q x}{y} \\ q^{3} \end{array}; q^{2}, y^{2}]}{_{2} ϕ_{1} [\begin{array}{c} - \frac{x}{y}, - \frac{q x}{y} \\ q \end{array}; q^{2}, y^{2}]} \\ = & \frac{x + y}{1 - q + K_{n = 1}^{\infty} \frac{q^{n - 1} (x + q^{n} y) (y + q^{n} x)}{1 - q^{2 n + 1}}} \\ = & (右辺) \end{array}$

次回は、今回の定理を用いて次式を示す。
$\begin{array}{rcl} \frac{\sqrt[8]{2} - 1}{\sqrt[8]{2} + 1} & = & K_{n = 0}^{\infty} \frac{1}{2 \sinh ((2 n + 1) π)} \\ = & \frac{1}{2 \sinh π + \frac{1}{2 \sinh (3 π) + \frac{1}{2 \sinh (5 π) + ⋱}}} \end{array}$
$\begin{array}{rcl} \frac{\sqrt[4]{2} - 1}{\sqrt[4]{2} + 1} & = & K_{n = 0}^{\infty} \frac{\cosh^{2} (n π)}{\sinh ((2 n + 1) π)} \\ = & \frac{1}{\sinh π + \frac{\cosh^{2} π}{\sinh (3 π) + \frac{\cosh^{2} (2 π)}{\sinh (5 π) + ⋱}}} \end{array}$