文献あり

Partial theta functionに関するRamanujanの公式2

前の記事で, 次の定理1を用いていくつかのRamanujanによる公式を示した.

Andrews(1981)

$\begin{aligned} \sum_{0 \leq n} \frac{(B, - A b q; q)_{n}}{(- a q, - b q; q)_{n}} q^{n} & = - a^{- 1} \frac{(B, - A b q; q)_{\infty}}{(- a q, - b q; q)_{\infty}} \sum_{0 \leq n} \frac{(1 / A; q)_{n}}{(- B / a; q)_{n + 1}} {(\frac{A b q}{a})}^{n} \\ + (1 + b) \sum_{0 \leq n} \frac{(- 1 / a; q)_{n + 1} (- A B q / a; q)_{n}}{(- B / a, A b q / a; q)_{n + 1}} (- b)^{n} \end{aligned}$

今回もいくつかRamanujanによる公式を示す.

$\begin{aligned} (1 + \frac{1}{a}) \sum_{0 \leq n} \frac{(q; q^{2})_{n} q^{2 n + 1}}{(- a q, - q / a; q^{2})_{n + 1}} \\ = \sum_{0 \leq n} (- a)^{n} q^{\frac{1}{2} n (n + 1)} - \frac{(q; q^{2})_{\infty}}{(- a q, - q / a; q^{2})_{\infty}} \sum_{0 \leq n} a^{3 n} q^{3 n^{2} + n} (1 - a^{2} q^{4 n + 2}) \end{aligned}$

定理1において, $q \mapsto q^{2}, a \mapsto q / a, A = - a^{- 1} q^{- 2}, B = 0, b \mapsto a q$ として
$\begin{aligned} (1 + \frac{1}{a}) \sum_{0 \leq n} \frac{(q; q^{2})_{n} q^{2 n + 1}}{(- a q, - q / a; q^{2})_{n + 1}} \\ = - \frac{(q; q^{2})_{\infty}}{(- a q, - q / a; q^{2})_{\infty}} \sum_{0 \leq n} (- a; q^{2})_{n + 1} (- a)^{n} + \sum_{0 \leq n} \frac{(- a q; q^{2})_{n} (- a q)^{n}}{(- a q^{2}; q^{2})_{n}} \end{aligned}$
ここで, Rogers-Fineの恒等式
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q)_{n}}{(b; q)_{n}} t^{n} & = \sum_{0 \leq n} \frac{(a, a t q / b; q)_{n} (b t)^{n} q^{n^{2} - n} (1 - a t q^{2 n})}{(b; q)_{n} (t; q)_{n + 1}} \end{aligned}$
において, $q \mapsto q^{2}, a \mapsto - a q, b = - a q^{2}, t = - a q$ とすると
$\begin{aligned} \sum_{0 \leq n} \frac{(- a q; q^{2})_{n} (- a q)^{n}}{(- a q^{2}; q^{2})_{n}} & = \sum_{0 \leq n} a^{2 n} q^{2 n^{2} + n} (1 - a q^{2 n + 1}) \\ = \sum_{0 \leq n} (- a)^{n} q^{\frac{1}{2} n (n + 1)} \end{aligned}$
を得る. また, Rogers-Fineの恒等式において, $q \mapsto q^{2}, a \mapsto - a q^{2}, b = 0, t = - a$ とすると,
$\begin{aligned} \sum_{0 \leq n} (- a; q^{2})_{n + 1} (- a)^{n} & = \sum_{0 \leq n} a^{3 n} q^{3 n^{2} + n} (1 - a^{2} q^{4 n + 2}) \end{aligned}$
を得る. よって, これらを代入して示される.

$\begin{aligned} (1 + \frac{1}{a}) \sum_{0 \leq n} \frac{(- q; q)_{2 n} q^{2 n + 1}}{(a q, q / a; q^{2})_{n + 1}} = \frac{1}{(q, a q, q / a; q^{2})_{\infty}} \sum_{0 \leq n} (- a)^{n} q^{\frac{1}{2} n (n + 1)} - \sum_{0 \leq n} (- a)^{n} q^{n (n + 1)} \end{aligned}$

定理1において, $q \mapsto q^{2}$ として $a \mapsto - q / a, B = - q, b = - a q, A = - 1 / a q$ として整理すると,
$\begin{aligned} (1 + \frac{1}{a}) \sum_{0 \leq n} \frac{(- q; q)_{2 n} q^{2 n + 1}}{(a q, q / a; q^{2})_{n + 1}} \\ = \frac{(1 + a) (- q; q)_{\infty}}{(a q, q / a; q^{2})_{\infty}} \sum_{0 \leq n} \frac{(- a q; q^{2})_{n} (- a q)^{n}}{(- a; q^{2})_{n + 1}} - \sum_{0 \leq n} \frac{(a q, q; q^{2})_{n} (a q)^{n}}{(- a q^{2}; q^{2})_{n} (- a q; q^{2})_{n + 1}} \end{aligned}$
ここで, 1つ目の項には定理2の証明過程で示した式
$\begin{aligned} \sum_{0 \leq n} \frac{(- a q; q^{2})_{n} (- a q)^{n}}{(- a q^{2}; q^{2})_{n}} & = \sum_{0 \leq n} (- a)^{n} q^{\frac{1}{2} n (n + 1)} \end{aligned}$
を用いることができる.
$\begin{aligned} \sum_{0 \leq n} \frac{(a q, q; q^{2})_{n} (a q)^{n}}{(- a q^{2}; q^{2})_{n} (- a q; q^{2})_{n + 1}} \\ = \frac{1}{1 + a q} \sum_{0 \leq n} \frac{(a q; q^{2})_{n} (q; q)_{2 n} (a q)^{n}}{(q^{2}; q^{2})_{n} (- a q; q)_{2 n}} \end{aligned}$
ここで, AndrewsによるHeineの変換公式の類似
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q^{2})_{n} (b; q)_{2 n}}{(q^{2}; q^{2})_{n} (c; q)_{2 n}} t^{n} & = \frac{(b; q)_{\infty} (a t; q^{2})_{\infty}}{(c; q)_{\infty} (t; q^{2})_{\infty}} \sum_{0 \leq n} \frac{(c / b; q)_{n} (t; q^{2})_{n}}{(q; q)_{n} (a t; q^{2})_{n}} b^{n} \end{aligned}$
において, $a \mapsto a q, b = q, c = - a q^{2}, t = a q$ とすると,
$\begin{aligned} \frac{1}{1 + a q} \sum_{0 \leq n} \frac{(a q; q^{2})_{n} (q; q)_{2 n} (a q)^{n}}{(q^{2}; q^{2})_{n} (- a q; q)_{2 n}} \\ = \frac{(q; q)_{\infty} (a^{2} q^{2}; q^{2})_{\infty}}{(- a q; q)_{\infty} (a q; q^{2})_{\infty}} \sum_{0 \leq n} \frac{(- a q; q)_{n} (a q; q^{2})_{n}}{(q; q)_{n} (a^{2} q^{2}; q^{2})_{n}} q^{n} \\ = (q; q)_{\infty} (a q^{2}; q^{2})_{\infty} \sum_{0 \leq n} \frac{(\sqrt{a q}, - \sqrt{a q}; q)_{n}}{(q, a q; q)_{n}} q^{n} \end{aligned}$
ここで, Heineの変換公式
$\begin{aligned} \sum_{0 \leq n} \frac{(a, b; q)_{n}}{(q, c; q)_{n}} t^{n} & = \frac{(b, a t; q)_{\infty}}{(c, t; q)_{\infty}} \sum_{0 \leq n} \frac{(c / b, t; q)_{n}}{(q, a t; q)_{n}} b^{n} \end{aligned}$
において, $a \mapsto \sqrt{a q}, b = - \sqrt{a q}, c = a q, t = q$ として,
$\begin{aligned} \sum_{0 \leq n} \frac{(\sqrt{a q}, - \sqrt{a q}; q)_{n}}{(q, a q; q)_{n}} q^{n} & = \frac{(a q; q^{2})_{\infty}}{(q, a q; q)_{\infty}} \sum_{0 \leq n} \frac{(\sqrt{a q}; q)_{n} (a q)^{\frac{n}{2}}}{(- \sqrt{a q}; q)_{n + 1}} \end{aligned}$
であり, Rogers-Fineの恒等式
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q)_{n}}{(b; q)_{n}} t^{n} & = \sum_{0 \leq n} \frac{(a, a t q / b; q)_{n} (b t)^{n} q^{n^{2} - n} (1 - a t q^{2 n})}{(b; q)_{n} (t; q)_{n + 1}} \end{aligned}$
において, $a \mapsto \sqrt{a q}, b = - q \sqrt{a q}, t = \sqrt{a q}$ とすると,
$\begin{aligned} \sum_{0 \leq n} \frac{(\sqrt{a q}; q)_{n} (a q)^{\frac{n}{2}}}{(- \sqrt{a q}; q)_{n + 1}} & = \sum_{0 \leq n} (- a)^{n} q^{n (n + 1)} \end{aligned}$
を得る. よって, これらを合わせると,
$\begin{array}{r} \sum_{0 \leq n} \frac{(a q, q; q^{2})_{n} (a q)^{n}}{(- a q^{2}; q^{2})_{n} (- a q; q^{2})_{n + 1}} = \sum_{0 \leq n} (- a)^{n} q^{n (n + 1)} \end{array}$
であるからこれを代入して定理を得る.

$\begin{aligned} \sum_{0 \leq n} \frac{(q; q^{2})_{n} q^{2 n}}{(- a q^{2}, - q^{2} / a; q^{2})_{n}} \\ = (1 + a) \sum_{0 \leq n} (- a)^{n} q^{\frac{1}{2} n (n + 1)} - \frac{a (q; q^{2})_{\infty}}{(- a q^{2}, - q^{2} / a; q^{2})_{\infty}} \sum_{0 \leq n} a^{3 n} q^{3 n^{2} + 2 n} (1 - a q^{2 n + 1}) \end{aligned}$

定理1において, $q \mapsto q^{2}, a \mapsto 1 / a, A = - 1 / a q, B = 0, b = a$ として,
$\begin{aligned} \sum_{0 \leq n} \frac{(q; q^{2})_{n} q^{2 n}}{(- a q^{2}, - q^{2} / a; q^{2})_{n}} \\ = - \frac{a (q^{2}; q^{2})_{\infty}}{(- a q^{2}, - q^{2} / a; q^{2})_{\infty}} \sum_{0 \leq n} (- a q; q^{2})_{n} (- a q)^{n} + (1 + a) \sum_{0 \leq n} \frac{(- a; q^{2})_{n + 1} (- a)^{n}}{(- a q; q^{2})_{n + 1}} \end{aligned}$
ここで, Rogers-Fineの恒等式
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q)_{n}}{(b; q)_{n}} t^{n} & = \sum_{0 \leq n} \frac{(a, a t q / b; q)_{n} (b t)^{n} q^{n^{2} - n} (1 - a t q^{2 n})}{(b; q)_{n} (t; q)_{n + 1}} \end{aligned}$
において, $q \mapsto q^{2}, a \mapsto - a q, t = - a q, b \to 0$ として,
$\begin{array}{r} \sum_{0 \leq n} (- a q; q^{2})_{n} (- a q)^{n} = \sum_{0 \leq n} a^{3 n} q^{3 n^{2} + 2 n} (1 - a q^{2 n + 1}) \end{array}$
を得る. また, Rogers-Fineの恒等式において, $a \mapsto - a q^{2}, b = - a q^{3}, t = - a$ として,
$\begin{aligned} \sum_{0 \leq n} \frac{(- a; q^{2})_{n + 1} (- a)^{n}}{(- a q; q^{2})_{n + 1}} & = \sum_{0 \leq n} a^{2 n} q^{2 n^{2} + n} (1 - a q^{2 n + 1}) \\ = \sum_{0 \leq n} (- a)^{n} q^{\frac{1}{2} n (n + 1)} \end{aligned}$
となるから, これらを代入して定理を得る.

$\begin{aligned} \sum_{0 \leq n} \frac{(q; q^{2})_{n} q^{n}}{(- a q, - q / a; q)_{n}} = (1 + a) \sum_{0 \leq n} (- a)^{n} q^{\frac{1}{2} n (n + 1)} - \frac{a (q; q^{2})_{\infty}}{(- a q, - q / a; q)_{\infty}} \sum_{0 \leq n} (- 1)^{n} a^{2 n} q^{n (n + 1)} \end{aligned}$

定理の等式を $q \mapsto q^{2}$ とした等式
$\begin{aligned} \sum_{0 \leq n} \frac{(q, - q; q^{2})_{n} q^{2 n}}{(- a q^{2}, - q^{2} / a; q^{2})_{n}} \\ = (1 + a) \sum_{0 \leq n} (- a)^{n} q^{n (n + 1)} - \frac{a (q, - q; q^{2})_{\infty}}{(- a q^{2}, - q^{2} / a; q^{2})_{\infty}} \sum_{0 \leq n} (- 1)^{n} a^{2 n} q^{2 n (n + 1)} \end{aligned}$
を示せば良い. 定理1において, $q \mapsto q^{2}, a \mapsto 1 / a, b = a, A = 1 / a q, B = q$ とすると,
$\begin{aligned} \sum_{0 \leq n} \frac{(q, - q; q^{2})_{n} q^{2 n}}{(- a q^{2}, - q^{2} / a; q^{2})_{n}} \\ = - \frac{a (q, - q; q^{2})_{\infty}}{(- a q^{2}, - q^{2} / a; q^{2})_{\infty}} \sum_{0 \leq n} \frac{(a q; q^{2})_{n} (a q)^{n}}{(- a q; q^{2})_{n + 1}} + (1 + a) \sum_{0 \leq n} \frac{(- a; q^{2})_{n + 1} (- q^{2}; q^{2})_{n} (- a)^{n}}{(- a q, a q; q^{2})_{n + 1}} \end{aligned}$
Rogers-Fineの恒等式
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q)_{n}}{(b; q)_{n}} t^{n} & = \sum_{0 \leq n} \frac{(a, a t q / b; q)_{n} (b t)^{n} q^{n^{2} - n} (1 - a t q^{2 n})}{(b; q)_{n} (t; q)_{n + 1}} \end{aligned}$
において, $q \mapsto q^{2}, a \mapsto a q, b = - a q^{3}, t = a q$ として,
$\begin{aligned} \sum_{0 \leq n} \frac{(a q; q^{2})_{n} (a q)^{n}}{(- a q; q^{2})_{n + 1}} & = \sum_{0 \leq n} (- 1)^{n} a^{2 n} q^{2 n (n + 1)} \end{aligned}$
を得る. また, AndrewsによるHeineの変換公式の類似
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q^{2})_{n} (b; q)_{2 n}}{(q^{2}; q^{2})_{n} (c; q)_{2 n}} t^{n} & = \frac{(b; q)_{\infty} (a t; q^{2})_{\infty}}{(c; q)_{\infty} (t; q^{2})_{\infty}} \sum_{0 \leq n} \frac{(c / b; q)_{n} (t; q^{2})_{n}}{(q; q)_{n} (a t; q^{2})_{n}} b^{n} \end{aligned}$
において, $q \mapsto q^{2}, a \mapsto a^{2} q^{2}, b = - a, c = a^{2} q^{2}, t = q^{4}$ とすると,
$\begin{aligned} \sum_{0 \leq n} \frac{(- a; q^{2})_{n + 1} (- q^{2}; q^{2})_{n} (- a)^{n}}{(- a q, a q; q^{2})_{n + 1}} \\ = \frac{1 + a}{1 - a^{2} q^{2}} \sum_{0 \leq n} \frac{(- a q^{2}; q^{2})_{n} (q^{4}; q^{4})_{n} (- a)^{n}}{(q^{2}; q^{2})_{n} (a^{2} q^{6}; q^{4})_{n}} \\ = \frac{(1 + a) (a^{2} q^{2}; q^{2})_{\infty} (q^{4}; q^{4})_{\infty}}{(1 - a^{2} q^{2}) (- a; q^{2})_{\infty} (a^{2} q^{6}; q^{4})_{\infty}} \sum_{0 \leq n} \frac{(a^{2} q^{2}; q^{4})_{n} (- a; q^{2})_{2 n} q^{4 n}}{(q^{4}; q^{4})_{n} (a^{2} q^{2}; q^{4})_{2 n}} \\ = (a q^{2}; q^{2})_{\infty} (q^{4}; q^{4})_{\infty} \sum_{0 \leq n} \frac{(- a, - a q^{2}; q^{4})_{n} q^{4 n}}{(q^{4}, a^{2} q^{4}; q^{4})_{n}} \end{aligned}$
ここで, Heineの変換公式
$\begin{aligned} \sum_{0 \leq n} \frac{(a, b; q)_{n}}{(q, c; q)_{n}} t^{n} & = \frac{(b, a t; q)_{\infty}}{(c, t; q)_{\infty}} \sum_{0 \leq n} \frac{(c / b, t; q)_{n}}{(q, a t; q)_{n}} b^{n} \end{aligned}$
において, $q \mapsto q^{4}, a \mapsto - a, b = - a q^{2}, c = a^{2} q^{4}, t = q^{4}$ とすると,
$\begin{aligned} \sum_{0 \leq n} \frac{(- a, - a q^{2}; q^{4})_{n} q^{4 n}}{(q^{4}, a^{2} q^{4}; q^{4})_{n}} \\ = \frac{(- a q^{2}, - a q^{4}; q^{4})_{\infty}}{(a^{2} q^{4}, q^{4}; q^{4})_{\infty}} \sum_{0 \leq n} \frac{(- a q^{2}; q^{4})_{n} (- a q^{2})^{n}}{(- a q^{4}; q^{4})_{n}} \end{aligned}$
ここで, Rogers-Fineの恒等式
$\begin{aligned} \sum_{0 \leq n} \frac{(a; q)_{n}}{(b; q)_{n}} t^{n} & = \sum_{0 \leq n} \frac{(a, a t q / b; q)_{n} (b t)^{n} q^{n^{2} - n} (1 - a t q^{2 n})}{(b; q)_{n} (t; q)_{n + 1}} \end{aligned}$
において, $q \mapsto q^{4}, a \mapsto a q^{2}, b = - a q^{4}, t = - a q^{2}$ とすると,
$\begin{aligned} \sum_{0 \leq n} \frac{(- a q^{2}; q^{4})_{n} (- a q^{2})^{n}}{(- a q^{4}; q^{4})_{n}} & = \sum_{0 \leq n} a^{2 n} q^{4 n^{2} + 2 n} (1 - a q^{4 n + 2}) \\ = \sum_{0 \leq n} (- a)^{n} q^{n (n + 1)} \end{aligned}$
である. これらより,
$\begin{array}{r} \sum_{0 \leq n} \frac{(- a; q^{2})_{n + 1} (- q^{2}; q^{2})_{n} (- a)^{n}}{(- a q, a q; q^{2})_{n + 1}} = \sum_{0 \leq n} (- a)^{n} q^{n (n + 1)} \end{array}$
であるからこれらを代入して定理を得る.