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Cooper-Hirschhornの恒等式

$[a; q]_{\infty} = (a, q / a; q)_{\infty}, [a_{1}, \dots, a_{r}; q]_{\infty} := [a_{1}; q]_{\infty} \dots [a_{r}; q]_{\infty}$ とする. 無限積の三項関係式は以下のように表される.

$A^{2} = b c d e$ のとき,
$\begin{aligned} [A / b, A / c, A / d, A / e; q]_{\infty} - [b, c, d, e; q]_{\infty} & = b [A, A / b c, A / b d, A / b e; q]_{\infty} \end{aligned}$
が成り立つ.

この等式の応用として, 以下の恒等式を示す.

Cooper-Hirschhorn(2001)

$\begin{aligned} \sum_{n \in Z} q^{n^{2}} + \sum_{n \in Z} q^{2 n^{2}} & = 2 \frac{(q^{3}, q^{5}, q^{8}; q^{8})_{\infty}}{(q, q^{4}, q^{7}; q^{8})_{\infty}} \\ \sum_{n \in Z} q^{n^{2}} - \sum_{n \in Z} q^{2 n^{2}} & = 2 q \frac{(q, q^{7}, q^{8}; q^{8})_{\infty}}{(q^{3}, q^{4}, q^{5}; q^{8})_{\infty}} \end{aligned}$

二つの等式,
$\begin{aligned} \sum_{n \in Z} q^{n^{2}} & = \frac{(q^{3}, q^{5}, q^{8}; q^{8})_{\infty}}{(q, q^{4}, q^{7}; q^{8})_{\infty}} + q \frac{(q, q^{7}, q^{8}; q^{8})_{\infty}}{(q^{3}, q^{4}, q^{5}; q^{8})_{\infty}} \\ \sum_{n \in Z} q^{2 n^{2}} & = \frac{(q^{3}, q^{5}, q^{8}; q^{8})_{\infty}}{(q, q^{4}, q^{7}; q^{8})_{\infty}} - q \frac{(q, q^{7}, q^{8}; q^{8})_{\infty}}{(q^{3}, q^{4}, q^{5}; q^{8})_{\infty}} \end{aligned}$
を示せばよい. Jacobiの三重積より,
$\begin{aligned} \sum_{n \in Z} q^{n^{2}} & = (- q, - q, q^{2}; q^{2})_{\infty} \\ = \frac{(q^{2}; q^{2})_{\infty} (q^{2}, q^{2}; q^{4})_{\infty}}{(q, q; q^{2})_{\infty}} \\ = \frac{(q^{2}, q^{2}, q^{2}, q^{4}, q^{6}, q^{6}, q^{6}, q^{8}; q^{8})_{\infty}}{(q, q, q^{3}, q^{3}, q^{5}, q^{5}, q^{7}, q^{7}; q^{8})_{\infty}} \end{aligned}$
よって1つ目の等式は
$\begin{array}{r} \frac{(q^{2}, q^{2}, q^{2}, q^{4}, q^{6}, q^{6}, q^{6}, q^{8}; q^{8})_{\infty}}{(q, q, q^{3}, q^{3}, q^{5}, q^{5}, q^{7}, q^{7}; q^{8})_{\infty}} = \frac{(q^{3}, q^{5}, q^{8}; q^{8})_{\infty}}{(q, q^{4}, q^{7}; q^{8})_{\infty}} + q \frac{(q, q^{7}, q^{8}; q^{8})_{\infty}}{(q^{3}, q^{4}, q^{5}; q^{8})_{\infty}} \end{array}$
つまり,
$\begin{array}{r} (q^{2}, q^{2}, q^{2}, q^{4}, q^{4}, q^{6}, q^{6}, q^{6}; q^{8})_{\infty} = (q, q^{3}, q^{3}, q^{3}, q^{5}, q^{5}, q^{5}, q^{7}; q^{8})_{\infty} + q (q, q, q, q^{3}, q^{5}, q^{7}, q^{7}, q^{7}; q^{8})_{\infty} \end{array}$
を示せば良い. これは
$\begin{array}{r} [q^{2}, q^{2}, q^{2}, q^{4}; q^{8}]_{\infty} = [q, q^{3}, q^{3}, q^{3}; q^{8}]_{\infty} + q [q, q, q, q^{3}; q^{8}]_{\infty} \end{array}$
と表され, 定理1において, $q \mapsto q^{8}, A = q^{5}, b = q, c = d = e = q^{3}$ として従う. 2つ目の等式は Jacobiの三重積より,
$\begin{aligned} \sum_{n \in Z} q^{2 n^{2}} & = \frac{(q^{4}; q^{4})_{\infty} (q^{4}, q^{4}; q^{8})_{\infty}}{(q^{2}, q^{2}; q^{4})_{\infty}} \\ = \frac{(q^{4}, q^{4}, q^{4}, q^{8}; q^{8})_{\infty}}{(q^{2}, q^{2}, q^{6}, q^{6}; q^{8})_{\infty}} \end{aligned}$
であるから,
$\begin{array}{r} \frac{(q^{4}, q^{4}, q^{4}, q^{8}; q^{8})_{\infty}}{(q^{2}, q^{2}, q^{6}, q^{6}; q^{8})_{\infty}} = \frac{(q^{3}, q^{5}, q^{8}; q^{8})_{\infty}}{(q, q^{4}, q^{7}; q^{8})_{\infty}} - q \frac{(q, q^{7}, q^{8}; q^{8})_{\infty}}{(q^{3}, q^{4}, q^{5}; q^{8})_{\infty}} \end{array}$
つまり,
$\begin{array}{r} (q, q^{3}, q^{4}, q^{4}, q^{4}, q^{4}, q^{5}, q^{7}; q^{8})_{\infty} = (q^{2}, q^{2}, q^{3}, q^{3}, q^{5}, q^{5}, q^{6}, q^{6}; q^{8})_{\infty} - q (q, q, q^{2}, q^{2}, q^{6}, q^{6}, q^{7}, q^{7}; q^{8})_{\infty} \end{array}$
を示せば良い. これは
$\begin{array}{r} [q, q^{3}, q^{4}, q^{4}; q^{8}]_{\infty} = [q^{2}, q^{2}, q^{3}, q^{3}; q^{8}]_{\infty} - q [q, q, q^{2}, q^{2}; q^{8}]_{\infty} \end{array}$
と表され, 定理1において, $A = q^{6}, b = q, c = q^{3}, d = e = q^{4}$ として従う. よって定理が示された.