ヤコビの三重積

ヤコビの三重積 (ヤコビのさんじゅうせき、Jacobi triple product)とは、次の恒等式をいう。 ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n v = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ + 2 π i v ) ( 1 + e ( 2 m − 1 ) π i τ − 2 π i v ) {\displaystyle \sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }inv}}=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}} 但し、 Im ⁡ τ > 0 {\displaystyle \operatorname {Im} {\tau }>0} とする。この恒等式はヤコビによるテータ関数の研究から生まれたものであるが、 q = e π i τ , z = e 2 π i v {\displaystyle q=e^{{\pi }i{\tau }},z=e^{2{\pi }iv}} と置くことにより ∑ n = − ∞ ∞ q n 2 z n = ∏ m = 1 ∞ ( 1 − q 2 m ) ( 1 + q 2 m − 1 z ) ( 1 + q 2 m − 1 z − 1 ) {\displaystyle \sum _{n=-\infty }^{\infty }{q^{n^{2}}z^{n}}=\prod _{m=1}^{\infty }{\left(1-q^{2m}\right)\left(1+q^{2m-1}z\right)\left(1+q^{2m-1}z^{-1}\right)}} 或いは、 q = e π i τ , z = e − π i τ + 2 π i v {\displaystyle q=e^{{\pi }i{\tau }},z=e^{-{\pi }i{\tau }+2{\pi }iv}} と置くことにより ∑ n = − ∞ ∞ q n ( n + 1 ) z n = ∏ m = 1 ∞ ( 1 − q 2 m ) ( 1 + q 2 m z ) ( 1 + q 2 m − 2 z − 1 ) {\displaystyle \sum _{n=-\infty }^{\infty }{q^{n(n+1)}z^{n}}=\prod _{m=1}^{\infty }{\left(1-q^{2m}\right)\left(1+q^{2m}z\right)\left(1+q^{2m-2}z^{-1}\right)}} となり、数論にも適する形になる。カール・グスタフ・ヤコブ・ヤコビが1829年の著書 Jacobi (1829)で示した。 左辺を ϑ ( v , τ ) {\displaystyle \vartheta (v,\tau )} 、右辺を Θ ( v , τ ) {\displaystyle \Theta (v,\tau )} と置き、まず、右辺が疑二重周期を持つことを示す。 Θ ( v + 1 , τ ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ + 2 π i ( v + 1 ) ) ( 1 + e ( 2 m − 1 ) π i τ − 2 π i ( v + 1 ) ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ + 2 π i v ) ( 1 + e ( 2 m − 1 ) π i τ − 2 π i v ) = Θ ( v , τ ) {\displaystyle {\begin{aligned}\Theta (v+1,\tau )&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }i(v+1)}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }i(v+1)}\right)}\\&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}\\&=\Theta (v,\tau )\end{aligned}}} Θ ( v + τ , τ ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m + 1 ) π i τ + 2 π i v ) ( 1 + e ( 2 m − 3 ) π i τ − 2 π i v ) = 1 + e − π i τ − 2 π i v 1 + e π i τ + 2 π i v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ + 2 π i v ) ( 1 + e ( 2 m − 1 ) π i τ − 2 π i v ) = e − π i τ − 2 π i v + 1 1 + e π i τ + 2 π i v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ + 2 π i v ) ( 1 + e ( 2 m − 1 ) π i τ − 2 π i v ) = e − π i τ − 2 π i v Θ ( v , τ ) {\displaystyle {\begin{aligned}\Theta (v+\tau ,\tau )&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m+1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-3){\pi }i{\tau }-2{\pi }iv}\right)}\\&={\frac {1+e^{-{\pi }i{\tau }-2{\pi }iv}}{1+e^{{\pi }i{\tau }+2{\pi }iv}}}\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}\\&={\frac {e^{-{\pi }i{\tau }-2{\pi }iv}+1}{1+e^{{\pi }i{\tau }+2{\pi }iv}}}\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}\\&=e^{-{\pi }i{\tau }-2{\pi }iv}\Theta (v,\tau )\end{aligned}}} Im ⁡ τ > 0 {\displaystyle \operatorname {Im} {\tau }>0} により | e 2 m π i τ | < 1 {\displaystyle |e^{2m{\pi }i{\tau }}|<1} であるから、右辺の零点は ( 1 + e ( 2 m − 1 ) π i τ + 2 π i v ) ( 1 + e ( 2 m − 1 ) π i τ − 2 π i v ) = 0 {\displaystyle {\begin{aligned}\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)&=0\\\end{aligned}}} ( 1 + 2 e ( 2 m − 1 ) π i τ cos ⁡ 2 π v + e 2 ( 2 m − 1 ) π i τ ) = 0 {\displaystyle {\begin{aligned}\left(1+2e^{(2m-1){\pi }i{\tau }}\cos {2{\pi }v}+e^{2(2m-1){\pi }i{\tau }}\right)&=0\\\end{aligned}}} cos ⁡ 2 π v = − e ( 2 m − 1 ) π i τ + e − ( 2 m − 1 ) π i τ 2 cos ⁡ 2 π v = e ( 2 m − 1 ) π i τ + π i + e − ( 2 m − 1 ) π i τ − π i 2 2 π v = ( ( 2 m − 1 ) π τ + π ) ± 2 π n v = 1 + τ 2 + n ′ + m ′ τ {\displaystyle {\begin{aligned}\cos {2{\pi }v}&=-{\frac {e^{(2m-1){\pi }i{\tau }}+e^{-(2m-1){\pi }i{\tau }}}{2}}\\\cos {2{\pi }v}&={\frac {e^{(2m-1){\pi }i{\tau }+{\pi }i}+e^{-(2m-1){\pi }i{\tau }-{\pi }i}}{2}}\\2{\pi }v&=\left((2m-1){\pi }{\tau }+{\pi }\right)\pm 2{\pi }n\\v&={\frac {1+\tau }{2}}+n'+m'{\tau }\end{aligned}}} に限られる。一方、左辺は ϑ ( v + 1 ; τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n ( v + 1 ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n v = ϑ ( v ; τ ) {\displaystyle {\begin{aligned}\vartheta (v+1;\tau )&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }in(v+1)}}\\&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }inv}}\\&=\vartheta (v;\tau )\\\end{aligned}}} ϑ ( v + τ ; τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n ( v + τ ) = ∑ n = − ∞ ∞ e π i τ ( n + 1 ) 2 − π i τ + 2 π i ( n + 1 ) v − 2 π i v = ∑ n = − ∞ ∞ e π i τ n 2 − π i τ + 2 π i n v − 2 π i v = e − π i τ e − 2 π i v ϑ ( v ; τ ) {\displaystyle {\begin{aligned}\vartheta (v+\tau ;\tau )&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }in(v+\tau )}}\\&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }(n+1)^{2}-{\pi }i{\tau }+2{\pi }i(n+1)v-2{\pi }iv}}\\&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}-{\pi }i{\tau }+2{\pi }inv-2{\pi }iv}}\\&=e^{-{\pi }i\tau }e^{-2{\pi }i{v}}\vartheta (v;\tau )\\\end{aligned}}} ϑ 3 ( 1 + τ 2 ; τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n ( 1 + τ 2 ) = ∑ n = − ∞ ∞ ( − 1 ) n e π i τ ( n + 1 2 ) 2 − 1 4 π i τ = ∑ n = 0 ∞ ( − 1 ) n e π i τ ( n + 1 2 ) 2 − 1 4 π i τ + ∑ n = 0 ∞ ( − 1 ) n + 1 e π i τ ( − n − 1 + 1 2 ) 2 − 1 4 π i τ = 0 {\displaystyle {\begin{aligned}\vartheta _{3}({\frac {1+\tau }{2}};\tau )&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }in({\frac {1+\tau }{2}})}}\\&=\sum _{n=-\infty }^{\infty }{(-1)^{n}e^{{\pi }i{\tau }(n+{\frac {1}{2}})^{2}-{\frac {1}{4}}{\pi }i{\tau }}}\\&=\sum _{n=0}^{\infty }{(-1)^{n}e^{{\pi }i{\tau }(n+{\frac {1}{2}})^{2}-{\frac {1}{4}}{\pi }i{\tau }}}+\sum _{n=0}^{\infty }{(-1)^{n+1}e^{{\pi }i{\tau }(-n-1+{\frac {1}{2}})^{2}-{\frac {1}{4}}{\pi }i{\tau }}}\\&=0\end{aligned}}} であるから、右辺と同じ準二重周期を持ち、少なくとも右辺が零点を持つところに悉く零点を持つ。従って、リウヴィルの定理により、 c ( τ , v ) = ϑ ( v , τ ) Θ ( v , τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ + 2 π i v ) ( 1 + e ( 2 m − 1 ) π i τ − 2 π i v ) {\displaystyle c(\tau ,v)={\frac {\vartheta (v,\tau )}{\Theta (v,\tau )}}={\frac {\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }inv}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}}}} は v {\displaystyle v} に依存しない。 c ( 1 2 , τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + π i n ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ + π i ) ( 1 + e ( 2 m − 1 ) π i τ − π i ) = ∑ n = − ∞ ∞ ( − 1 ) n e π i τ n 2 ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 − e ( 2 m − 1 ) π i τ ) ( 1 − e ( 2 m − 1 ) π i τ ) {\displaystyle {\begin{aligned}c\left(\textstyle {\frac {1}{2}},\tau \right)&={\frac {\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+{\pi }in}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+{\pi }i}\right)\left(1+e^{(2m-1){\pi }i{\tau }-{\pi }i}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-1)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{(2m-1){\pi }i{\tau }}\right)\left(1-e^{(2m-1){\pi }i{\tau }}\right)}}}\\\end{aligned}}} c ( 1 4 , τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + π i n / 2 ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ + π i / 2 ) ( 1 + e ( 2 m − 1 ) π i τ − π i / 2 ) = ∑ n = − ∞ ∞ ( − i ) n e π i τ n 2 ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + i e ( 2 m − 1 ) π i τ ) ( 1 − i e ( 2 m − 1 ) π i τ ) = ∑ n = − ∞ ∞ ( − i ) n e π i τ n 2 ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 4 m − 2 ) π i τ ) = ∑ n = − ∞ ∞ ( − i ) n e π i τ n 2 ∏ m = 1 ∞ ( 1 − e 4 m π i τ ) ( 1 − e ( 4 m − 2 ) π i τ ) ( 1 + e ( 4 m − 2 ) π i τ ) = ∑ n = − ∞ ∞ ( − i ) n e π i τ n 2 ∏ m = 1 ∞ ( 1 − e 4 m π i τ ) ( 1 − e ( 8 m − 4 ) π i τ ) = ∑ n = − ∞ ∞ ( − i ) n e π i τ n 2 ∏ m = 1 ∞ ( 1 − e 8 m π i τ ) ( 1 − e ( 8 m − 4 ) π i τ ) ( 1 − e ( 8 m − 4 ) π i τ ) {\displaystyle {\begin{aligned}c\left(\textstyle {\frac {1}{4}},\tau \right)&={\frac {\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+{\pi }in/2}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+{\pi }i/2}\right)\left(1+e^{(2m-1){\pi }i{\tau }-{\pi }i/2}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+ie^{(2m-1){\pi }i{\tau }}\right)\left(1-ie^{(2m-1){\pi }i{\tau }}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(4m-2){\pi }i{\tau }}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{4m{\pi }i{\tau }}\right)\left(1-e^{(4m-2){\pi }i{\tau }}\right)\left(1+e^{(4m-2){\pi }i{\tau }}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{4m{\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{8m{\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)}}}\\\end{aligned}}} 分子の級数においてnが奇数の項は正負で打ち消しあうから2nをnに置き換える。 c ( 1 4 , τ ) = ∑ n = − ∞ ∞ ( − 1 ) n e 4 π i τ n 2 ∏ m = 1 ∞ ( 1 − e 8 m π i τ ) ( 1 − e ( 8 m − 4 ) π i τ ) ( 1 − e ( 8 m − 4 ) π i τ ) = c ( 1 2 , 4 τ ) {\displaystyle {\begin{aligned}c\left(\textstyle {\frac {1}{4}},\tau \right)&={\frac {\sum _{n=-\infty }^{\infty }{(-1)^{n}e^{4{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{8m{\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)}}}\\&=c\left(\textstyle {\frac {1}{2}},4\tau \right)\end{aligned}}} c ( v , τ ) {\displaystyle c(v,\tau )} は v {\displaystyle v} に依存しないから c ( v , 4 τ ) = c ( v , τ ) = lim n → ∞ c ( v , 4 − n τ ) = lim τ ′ → 0 c ( v , τ ′ ) = c ( v , 0 ) {\displaystyle c\left(v,4\tau \right)=c\left(v,\tau \right)=\lim _{n\to \infty }c\left(v,4^{-n}\tau \right)=\lim _{\tau '\to 0}c\left(v,\tau '\right)=c(v,0)} であり、 c ( v , τ ) {\displaystyle c(v,\tau )} は τ {\displaystyle \tau } にも依存しない定数である。 τ → + i ∞ {\displaystyle \tau \to {+i}\infty } として c ( v , τ ) = 1 {\displaystyle c(v,\tau )=1} を得る。結局、両辺は等しい。 ヤコビの三重積はラマヌジャンの和公式の特殊な場合である。ラマヌジャンの和公式 ∑ n = − ∞ ∞ ( a ; q ) n ( b ; q ) n z n = ( a z ; q ) ∞ ( q ; q ) ∞ ( q a z ; q ) ∞ ( b a ; q ) ∞ ( z ; q ) ∞ ( b ; q ) ∞ ( b a z ; q ) ∞ ( q a ; q ) ∞ ( | q | < 1 , | b / a | < | z | < 1 ) {\displaystyle \sum _{n=-\infty }^{\infty }{\frac {(a;q)_{n}}{(b;q)_{n}}}z^{n}={\frac {(az;q)_{\infty }(q;q)_{\infty }\left({\frac {q}{az}};q\right)_{\infty }\left({\frac {b}{a}};q\right)_{\infty }}{(z;q)_{\infty }(b;q)_{\infty }\left({\frac {b}{az}};q\right)_{\infty }\left({\frac {q}{a}};q\right)_{\infty }}}\qquad (|q|<1,|b/a|<|z|<1)} はq二項定理から導かれる。ラマヌジャンの和公式に b = 0 {\displaystyle b=0} を代入すると ∑ n = − ∞ ∞ ( a ; q ) n z n = ( a z ; q ) ∞ ( q ; q ) ∞ ( q / a z ; q ) ∞ ( z ; q ) ∞ ( q / a ; q ) ∞ {\displaystyle \sum _{n=-\infty }^{\infty }(a;q)_{n}z^{n}={\frac {(az;q)_{\infty }(q;q)_{\infty }\left(q/az;q\right)_{\infty }}{(z;q)_{\infty }\left(q/a;q\right)_{\infty }}}} となり、 q {\displaystyle q} を q 2 {\displaystyle q^{2}} と書き、 z {\displaystyle z} を − q z / a {\displaystyle -qz/a} と書けば ∑ n = − ∞ ∞ ( a ; q 2 ) n ( − q z a ) n = ( − q z ; q 2 ) ∞ ( q 2 ; q 2 ) ∞ ( − q / z ; q 2 ) ∞ ( − q z / a ; q 2 ) ∞ ( q 2 / a ; q 2 ) ∞ {\displaystyle \sum _{n=-\infty }^{\infty }(a;q^{2})_{n}\left(-{\frac {qz}{a}}\right)^{n}={\frac {(-qz;q^{2})_{\infty }(q^{2};q^{2})_{\infty }\left(-q/z;q^{2}\right)_{\infty }}{(-qz/a;q^{2})_{\infty }\left(q^{2}/a;q^{2}\right)_{\infty }}}} となる。qポッホハマー記号の変換式 ( a q − n + 1 ; q ) n = ( − a ) n q − n ( n − 1 ) / 2 ( 1 a ; q ) n {\displaystyle \left(aq^{-n+1};q\right)_{n}=\left(-a\right)^{n}q^{-n(n-1)/2}\left({\frac {1}{a}};q\right)_{n}} により、左辺は ∑ n = − ∞ ∞ ( a ; q 2 ) n ( − q z a ) n = ∑ n = − ∞ ∞ ( a q 2 n − 2 q − 2 n + 2 ; q 2 ) n ( − q z a ) n = ∑ n = − ∞ ∞ ( − a q 2 n − 2 ) n q − 2 n ( n − 1 ) / 2 ( 1 / a q 2 n − 2 ; q ) n ( − q z a ) n = ∑ n = − ∞ ∞ q n 2 z n ( 1 / a q 2 n − 2 ; q ) n {\displaystyle {\begin{aligned}\sum _{n=-\infty }^{\infty }(a;q^{2})_{n}\left(-{\frac {qz}{a}}\right)^{n}&=\sum _{n=-\infty }^{\infty }(aq^{2n-2}q^{-2n+2};q^{2})_{n}\left(-{\frac {qz}{a}}\right)^{n}\\&=\sum _{n=-\infty }^{\infty }(-aq^{2n-2})^{n}q^{-2n(n-1)/2}(1/aq^{2n-2};q)_{n}\left(-{\frac {qz}{a}}\right)^{n}\\&=\sum _{n=-\infty }^{\infty }q^{n^{2}}z^{n}(1/aq^{2n-2};q)_{n}\\\end{aligned}}} であるから、 a → ∞ {\displaystyle a\to \infty } の極限を取れば ∑ n = − ∞ ∞ q n 2 z n = ( − q z ; q 2 ) ∞ ( q 2 ; q 2 ) ∞ ( − q / z ; q 2 ) ∞ {\displaystyle \sum _{n=-\infty }^{\infty }q^{n^{2}}z^{n}=(-qz;q^{2})_{\infty }(q^{2};q^{2})_{\infty }\left(-q/z;q^{2}\right)_{\infty }} となり、qポッホハマー記号を展開して ∑ n = − ∞ ∞ q n 2 z n = ∏ m = 1 ∞ ( 1 − q 2 m ) ( 1 + q 2 m − 1 / z ) ( 1 + q 2 m − 1 z ) {\displaystyle \sum _{n=-\infty }^{\infty }q^{n^{2}}z^{n}=\prod _{m=1}^{\infty }(1-q^{2m})(1+q^{2m-1}/z)(1+q^{2m-1}z)} を得る。 Jacobi (1829)の原証明は冪級数の操作のみを用いている。まず Q ( x ) = ( 1 − x 2 ) ( 1 − x 4 ) ( 1 − x 6 ) ⋯ , R ( x , z ) = ( 1 + x z ) ( 1 + x 3 z ) ( 1 + x 5 z ) ⋯ , S ( x , z ) = 1 ( 1 − x z ) ( 1 − x 2 z ) ( 1 − x 3 z ) ⋯ {\displaystyle {\begin{aligned}Q(x)=&(1-x^{2})(1-x^{4})(1-x^{6})\cdots ,\\R(x,z)=&(1+xz)(1+x^{3}z)(1+x^{5}z)\cdots ,\\S(x,z)=&{\frac {1}{(1-xz)(1-x^{2}z)(1-x^{3}z)\cdots }}\end{aligned}}} とおく。ヤコビの三重積は Q ( x ) R ( x , z − 1 ) R ( x , z ) {\displaystyle Q(x)R(x,z^{-1})R(x,z)} であらわされる。 まず R ( x , z ) {\displaystyle R(x,z)} について、 R ( x , z x 2 ) = ( 1 + x 3 z ) ( 1 + x 5 z ) ( 1 + x 7 z ) ⋯ {\displaystyle R(x,zx^{2})=(1+x^{3}z)(1+x^{5}z)(1+x^{7}z)\cdots } より R ( x , z ) = ( 1 + x z ) R ( x , z x 2 ) {\displaystyle R(x,z)=(1+xz)R(x,zx^{2})} が成り立つ。そこで R ( x , z ) = 1 + a 1 ( x ) z + a 2 ( x ) z 2 + ⋯ {\displaystyle R(x,z)=1+a_{1}(x)z+a_{2}(x)z^{2}+\cdots } とおくと ( 1 + a 1 ( x ) z + a 2 ( x ) z 2 + ⋯ ) = ( 1 + x z ) ( 1 + a 1 ( x ) x 2 z + a 2 ( x ) x 4 z 2 + ⋯ ) {\displaystyle (1+a_{1}(x)z+a_{2}(x)z^{2}+\cdots )=(1+xz)(1+a_{1}(x)x^{2}z+a_{2}(x)x^{4}z^{2}+\cdots )} より a n ( x ) = a n ( x ) x 2 n + a n − 1 ( x ) x 2 n − 1 , {\displaystyle a_{n}(x)=a_{n}(x)x^{2n}+a_{n-1}(x)x^{2n-1},} つまり a n ( x ) = a n − 1 ( x ) x 2 n − 1 1 − x 2 n , a 0 ( x ) = 1 {\displaystyle a_{n}(x)={\frac {a_{n-1}(x)x^{2n-1}}{1-x^{2n}}},a_{0}(x)=1} が成り立つ。この漸化式を解くと a n ( x ) = x n 2 ( 1 − x 2 ) ( 1 − x 4 ) ⋯ ( 1 − x 2 n ) {\displaystyle a_{n}(x)={\frac {x^{n^{2}}}{(1-x^{2})(1-x^{4})\cdots (1-x^{2n})}}} が得られる。 次に S ( x , z ) {\displaystyle S(x,z)} について、 S ( x , z x ) = 1 ( 1 − x 2 z ) ( 1 − x 3 z ) ( 1 − x 4 z ) ⋯ = ( 1 − x z ) S ( x , z ) {\displaystyle S(x,zx)={\frac {1}{(1-x^{2}z)(1-x^{3}z)(1-x^{4}z)\cdots }}=(1-xz)S(x,z)} が成り立つ。そこで S ( x , z ) = 1 + b 1 ( x ) z 1 − x z + b 2 ( x ) z 2 ( 1 − x z ) ( 1 − x 2 z ) + ⋯ {\displaystyle S(x,z)=1+{\frac {b_{1}(x)z}{1-xz}}+{\frac {b_{2}(x)z^{2}}{(1-xz)(1-x^{2}z)}}+\cdots } とおくと S ( x , z ) = S ( x , z x ) 1 − x z = 1 1 − x z ( 1 + b 1 ( x ) x z 1 − x 2 z + b 2 ( x ) x 2 z 2 ( 1 − x 2 z ) ( 1 − x 3 z ) + ⋯ ) = 1 1 − x z + b 1 ( x ) x z ( 1 − x z ) ( 1 − x 2 z ) + b 2 ( x ) x 2 z 2 ( 1 − x z ) ( 1 − x 2 z ) ( 1 − x 3 z ) ⋯ {\displaystyle {\begin{aligned}S(x,z)=&{\frac {S(x,zx)}{1-xz}}\\=&{\frac {1}{1-xz}}\left(1+{\frac {b_{1}(x)xz}{1-x^{2}z}}+{\frac {b_{2}(x)x^{2}z^{2}}{(1-x^{2}z)(1-x^{3}z)}}+\cdots \right)\\=&{\frac {1}{1-xz}}+{\frac {b_{1}(x)xz}{(1-xz)(1-x^{2}z)}}+{\frac {b_{2}(x)x^{2}z^{2}}{(1-xz)(1-x^{2}z)(1-x^{3}z)}}\cdots \end{aligned}}} であるが b m ( x ) x m z m ( 1 − x z ) ( 1 − x 2 z ) ⋯ ( 1 − x m + 1 z ) = b m ( x ) x m z m ( 1 − x z ) ( 1 − x 2 z ) ⋯ ( 1 − x m z ) + b m ( x ) x 2 m + 1 z m + 1 ( 1 − x z ) ( 1 − x 2 z ) ⋯ ( 1 − x m + 1 z ) {\displaystyle {\frac {b_{m}(x)x^{m}z^{m}}{(1-xz)(1-x^{2}z)\cdots (1-x^{m+1}z)}}={\frac {b_{m}(x)x^{m}z^{m}}{(1-xz)(1-x^{2}z)\cdots (1-x^{m}z)}}+{\frac {b_{m}(x)x^{2m+1}z^{m+1}}{(1-xz)(1-x^{2}z)\cdots (1-x^{m+1}z)}}} より b n ( x ) = b n ( x ) x n + b n − 1 x 2 n − 1 , {\displaystyle b_{n}(x)=b_{n}(x)x^{n}+b_{n-1}x^{2n-1},} つまり b n ( x ) = b n − 1 ( x ) x 2 n − 1 1 − x n , b 0 ( x ) = 1 {\displaystyle b_{n}(x)={\frac {b_{n-1}(x)x^{2n-1}}{1-x^{n}}},b_{0}(x)=1} が成り立つ。この漸化式を解くと b n ( x ) = x n 2 ( 1 − x ) ( 1 − x 2 ) ⋯ ( 1 − x n ) {\displaystyle b_{n}(x)={\frac {x^{n^{2}}}{(1-x)(1-x^{2})\cdots (1-x^{n})}}} が得られる。 さて、ヤコビの三重積の冪級数展開を得たいが、代わりに R ( x , z ) R ( x , z − 1 ) {\displaystyle R(x,z)R(x,z^{-1})} の冪級数展開について考える。 R ( x , z ) R ( x , z − 1 ) = ( 1 + a 1 ( x ) z + a 2 ( x ) z 2 + ⋯ ) ( 1 + a 1 ( x ) z − 1 + a 2 ( x ) z − 2 + ⋯ ) {\displaystyle R(x,z)R(x,z^{-1})=(1+a_{1}(x)z+a_{2}(x)z^{2}+\cdots )(1+a_{1}(x)z^{-1}+a_{2}(x)z^{-2}+\cdots )} より R ( x , z ) R ( x , z − 1 ) {\displaystyle R(x,z)R(x,z^{-1})} を冪級数展開したときの z m {\displaystyle z^{m}} および z − m {\displaystyle z^{-m}} の係数は共に a m ( x ) + a m + 1 ( x ) a 1 ( x ) + a m + 2 ( x ) a 2 ( x ) + ⋯ = a m ( x ) ( 1 + x 2 m + 2 ( 1 − x 2 ) ( 1 − x 2 m + 2 ) + x 4 m + 8 ( 1 − x 2 ) ( 1 − x 4 ) ( 1 − x 2 m + 2 ) ( 1 − x 2 m + 4 ) + ⋯ ) = a m ( x ) ∑ k = 0 ∞ x 2 k 2 ( 1 − x 2 ) ( 1 − x 4 ) ⋯ ( 1 − x 2 k ) × x 2 k m ( 1 − x 2 m + 2 ) ( 1 − x 2 m + 4 ) ⋯ ( 1 − x 2 ( m + k ) ) {\displaystyle {\begin{aligned}&a_{m}(x)+a_{m+1}(x)a_{1}(x)+a_{m+2}(x)a_{2}(x)+\cdots \\=&a_{m}(x)\left(1+{\frac {x^{2m+2}}{(1-x^{2})(1-x^{2m+2})}}+{\frac {x^{4m+8}}{(1-x^{2})(1-x^{4})(1-x^{2m+2})(1-x^{2m+4})}}+\cdots \right)\\=&a_{m}(x)\sum _{k=0}^{\infty }{\frac {x^{2k^{2}}}{(1-x^{2})(1-x^{4})\cdots (1-x^{2k})}}\times {\frac {x^{2km}}{(1-x^{2m+2})(1-x^{2m+4})\cdots (1-x^{2(m+k)})}}\\\end{aligned}}} に一致するが、これは、上記の S ( x , z ) {\displaystyle S(x,z)} の展開より = a m ( x ) ∑ k = 0 ∞ b k ( x 2 ) x 2 k m ( 1 − x 2 m + 2 ) ( 1 − x 2 m + 4 ) ⋯ ( 1 − x 2 ( m + k ) ) = a m ( x ) S ( x 2 , x 2 m ) = x m 2 ( 1 − x 2 ) ( 1 − x 4 ) ⋯ ( 1 − x 2 m ) × ( 1 − x 2 m + 2 ) ( 1 − x 2 m + 4 ) ⋯ = x m 2 ( 1 − x 2 ) ( 1 − x 4 ) ⋯ = x m 2 Q ( x ) {\displaystyle {\begin{aligned}=&a_{m}(x)\sum _{k=0}^{\infty }{\frac {b_{k}(x^{2})x^{2km}}{(1-x^{2m+2})(1-x^{2m+4})\cdots (1-x^{2(m+k)})}}\\=&a_{m}(x)S(x^{2},x^{2m})\\=&{\frac {x^{m^{2}}}{(1-x^{2})(1-x^{4})\cdots (1-x^{2m})\times (1-x^{2m+2})(1-x^{2m+4})\cdots }}\\=&{\frac {x^{m^{2}}}{(1-x^{2})(1-x^{4})\cdots }}\\=&{\frac {x^{m^{2}}}{Q(x)}}\end{aligned}}} に一致する。よって Q ( x ) R ( x , z ) R ( x , z − 1 ) = ∑ m = − ∞ ∞ x m 2 z m {\displaystyle Q(x)R(x,z)R(x,z^{-1})=\sum _{m=-\infty }^{\infty }x^{m^{2}}z^{m}} が成り立つ。 ワトソンの五重積 オイラーの五角数定理 Jacobi, C. G. J. (1829) (Latin), Fundamenta nova theoriae functionum ellipticarum, Königsberg: Borntraeger, ISBN 978-1-108-05200-9, Reprinted by Cambridge University Press 2012, https://archive.org/details/fundamentanovat00jacogoog  Andrews, G. E. (1965), A simple proof of Jacobi's triple product identity, American Mathematical Society, https://www.ams.org/journals/proc/1965-016-02/S0002-9939-1965-0171725-X/  Carlitz, L (1962), A note on the Jacobi theta formula, American Mathematical Society, https://projecteuclid.org/euclid.bams/1183524930  Wright, E. M. (1965), An Enumerative Proof of An Identity of Jacobi, London Mathematical Society, https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s1-40.1.55