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RamanujanによるJackson-Slaterの恒等式の類似

$\begin{aligned} \sum_{0 \leq n} \frac{q^{n}}{(q; q)_{2 n}} & = \frac{(- q^{3}, - q^{5}, q^{8}; q^{8})_{\infty}}{(q; q)_{\infty}} \\ \sum_{0 \leq n} \frac{q^{n}}{(q; q)_{2 n + 1}} & = \frac{(- q, - q^{7}, q^{8}; q^{8})_{\infty}}{(q; q)_{\infty}} \end{aligned}$

まず, $q$ 二項定理より,
$\begin{aligned} \sum_{0 \leq n} \frac{q^{n}}{(q^{2}; q^{2})_{n}} & = \frac{1}{(q; q^{2})_{\infty}} \end{aligned}$
である. よって, Jacobiの三重積より
$\begin{aligned} \sum_{0 \leq n} \frac{q^{2 n}}{(q^{2}; q^{2})_{2 n}} & = \frac{1}{2} \sum_{0 \leq n} \frac{(1 + (- 1)^{n}) q^{n}}{(q^{2}; q^{2})_{n}} \\ = \frac{1}{2} (\frac{1}{(q; q^{2})_{\infty}} + \frac{1}{(- q; q^{2})_{\infty}}) \\ = \frac{1}{2} (q^{2}; q^{2})_{\infty} ((- q, - q^{3}, q^{4}; q^{4})_{\infty} + (q, q^{3}, q^{4}; q^{4})_{\infty}) \\ = \frac{1}{2 (q^{2}; q^{2})_{\infty}} (\sum_{0 \leq n} q^{2 n^{2} + n} + \sum_{n \in Z} (- 1)^{n} q^{2 n^{2} + n}) \\ = \frac{1}{(q^{2}; q^{2})_{\infty}} \sum_{0 \leq n} q^{8 n^{2} + 2 n} \\ = \frac{(- q^{6}, - q^{10}, q^{16}; q^{16})_{\infty}}{(q^{2}; q^{2})_{\infty}} \end{aligned}$
である. $q^{2} \mapsto q$ として1つ目の等式を得る. 2つ目の等式も同様にJacobiの三重積より
$\begin{aligned} \sum_{0 \leq n} \frac{q^{2 n + 1}}{(q^{2}; q^{2})_{2 n + 1}} & = \frac{1}{2} \sum_{0 \leq n} \frac{(1 - (- 1)^{n}) q^{n}}{(q^{2}; q^{2})_{n}} \\ = \frac{1}{2} (\frac{1}{(q; q^{2})_{\infty}} - \frac{1}{(- q; q^{2})_{\infty}}) \\ = \frac{1}{2} (q^{2}; q^{2})_{\infty} ((- q, - q^{3}, q^{4}; q^{4})_{\infty} - (q, q^{3}, q^{4}; q^{4})_{\infty}) \\ = \frac{1}{2 (q^{2}; q^{2})_{\infty}} (\sum_{0 \leq n} q^{2 n^{2} + n} - \sum_{n \in Z} (- 1)^{n} q^{2 n^{2} + n}) \\ = \frac{q}{(q^{2}; q^{2})_{\infty}} \sum_{0 \leq n} q^{8 n^{2} - 6 n} \\ = \frac{q (- q^{2}, - q^{14}, q^{16}; q^{16})_{\infty}}{(q^{2}; q^{2})_{\infty}} \end{aligned}$
であるから両辺を $q$ で割って $q^{2} \mapsto q$ として得られる.

$\begin{aligned} \sum_{0 \leq n} \frac{q^{n}}{(- q; q)_{2 n}} & = \sum_{0 \leq n} q^{12 n^{2} + n} (1 - q^{22 n + 11}) \\ + q \sum_{0 \leq n} q^{12 n^{2} + 7 n} (1 - q^{10 n + 5}) \\ \sum_{0 \leq n} \frac{q^{n}}{(- q; q)_{2 n + 1}} & = \sum_{0 \leq n} q^{12 n^{2} + 5 n} (1 - q^{14 n + 7}) \\ + q^{2} \sum_{0 \leq n} q^{12 n^{2} + 11 n} (1 - q^{2 n + 1}) \end{aligned}$

前の記事で示した等式
$\begin{aligned} \sum_{0 \leq n} \frac{q^{n}}{(- a q, - q / a; q)_{n}} & = (1 + a) \sum_{0 \leq n} a^{3 n} q^{\frac{1}{2} n (3 n + 1)} (1 - a^{2} q^{2 n + 1}) \\ - \frac{1}{(- a q, - q / a; q)_{\infty}} \sum_{0 \leq n} (- 1)^{n} a^{2 n + 1} q^{\frac{1}{2} n (n + 1)} \end{aligned}$
において, $a = i$ として
$\begin{aligned} h (q) & := \sum_{0 \leq n} \frac{q^{n}}{(- q^{2}; q^{2})_{n}} \\ = Re ((1 + i) \sum_{0 \leq n} i^{3 n} q^{\frac{1}{2} n (3 n + 1)} (1 + q^{2 n + 1})) \\ = \sum_{0 \leq n} (- 1)^{n} q^{n (6 n + 1)} (1 + q^{4 n + 1}) + \sum_{0 \leq n} (- 1)^{n} q^{(2 n + 1) (3 n + 2)} (1 + q^{4 n + 3}) \end{aligned}$
を得る. これより,
$\begin{aligned} \sum_{0 \leq n} \frac{q^{2 n}}{(- q^{2}; q^{2})_{2 n}} \\ = \frac{1}{2} \sum_{0 \leq n} \frac{(1 + (- 1)^{n}) q^{n}}{(- q^{2}; q^{2})_{n}} \\ = \frac{1}{2} (h (q) + h (- q)) \\ = Even part of (\sum_{0 \leq n} (- 1)^{n} q^{n (6 n + 1)} (1 + q^{4 n + 1}) + \sum_{0 \leq n} (- 1)^{n} q^{(2 n + 1) (3 n + 2)} (1 + q^{4 n + 3})) \\ = \sum_{0 \leq n} q^{24 n^{2} + 2 n} - \sum_{0 \leq n} q^{24 n^{2} + 34 n + 12} \\ + \sum_{0 \leq n} q^{24 n^{2} + 14 n + 2} - \sum_{0 \leq n} q^{24 n^{2} + 46 n + 22} \end{aligned}$
ここで, $q^{2} \mapsto q$ として1つ目の等式が得られる.
$\begin{aligned} \sum_{0 \leq n} \frac{q^{2 n + 1}}{(- q^{2}; q^{2})_{2 n + 1}} \\ = \frac{1}{2} \sum_{0 \leq n} \frac{(1 - (- 1)^{n}) q^{n}}{(- q^{2}; q^{2})_{n}} \\ = \frac{1}{2} (h (q) - h (- q)) \\ = Odd part of (\sum_{0 \leq n} (- 1)^{n} q^{n (6 n + 1)} (1 + q^{4 n + 1}) + \sum_{0 \leq n} (- 1)^{n} q^{(2 n + 1) (3 n + 2)} (1 + q^{4 n + 3})) \\ = - \sum_{0 \leq n} q^{24 n^{2} + 26 n + 7} + \sum_{0 \leq n} q^{24 n^{2} + 10 n + 1} \\ - \sum_{0 \leq n} q^{24 n^{2} + 38 n + 15} + \sum_{0 \leq n} q^{24 n^{2} + 22 n + 5} \end{aligned}$
ここで, 両辺を $q$ で割って $q^{2} \mapsto q$ とすれば2つ目の等式が得られる.