ヤコビの三重積

左辺を${\displaystyle \vartheta (v,\tau )}$、右辺を${\displaystyle \Theta (v,\tau )}$と置き、まず、右辺が疑二重周期を持つことを示す。

${\displaystyle {\begin{aligned}\Theta (v+1,\tau )&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }i(v+1)}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }i(v+1)}\right)}\\&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}\\&=\Theta (v,\tau )\end{aligned}}}$

${\displaystyle {\begin{aligned}\Theta (v+\tau ,\tau )&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m+1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-3){\pi }i{\tau }-2{\pi }iv}\right)}\\&={\frac {1+e^{-{\pi }i{\tau }-2{\pi }iv}}{1+e^{{\pi }i{\tau }+2{\pi }iv}}}\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}\\&={\frac {e^{-{\pi }i{\tau }-2{\pi }iv}+1}{1+e^{{\pi }i{\tau }+2{\pi }iv}}}\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}\\&=e^{-{\pi }i{\tau }-2{\pi }iv}\Theta (v,\tau )\end{aligned}}}$

${\displaystyle \operatorname {Im} {\tau }>0}$により${\displaystyle |e^{2m{\pi }i{\tau }}|<1}$であるから、右辺の零点は

${\displaystyle {\begin{aligned}\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)&=0\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\left(1+2e^{(2m-1){\pi }i{\tau }}\cos {2{\pi }v}+e^{2(2m-1){\pi }i{\tau }}\right)&=0\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\cos {2{\pi }v}&=-{\frac {e^{(2m-1){\pi }i{\tau }}+e^{-(2m-1){\pi }i{\tau }}}{2}}\\\cos {2{\pi }v}&={\frac {e^{(2m-1){\pi }i{\tau }+{\pi }i}+e^{-(2m-1){\pi }i{\tau }-{\pi }i}}{2}}\\2{\pi }v&=\left((2m-1){\pi }{\tau }+{\pi }\right)\pm 2{\pi }n\\v&={\frac {1+\tau }{2}}+n'+m'{\tau }\end{aligned}}}$

に限られる。一方、左辺は

${\displaystyle {\begin{aligned}\vartheta (v+1;\tau )&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }in(v+1)}}\\&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }inv}}\\&=\vartheta (v;\tau )\\\end{aligned}}}$

${\displaystyle {\begin{aligned}\vartheta (v+\tau$ ;\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}}}

${\displaystyle {\begin{aligned}\vartheta _{3}({\frac {1+\tau }{2}};\tau )&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }in({\frac {1+\tau }{2}})}}\\&=\sum _{n=-\infty }^{\infty }{(-1)^{n}e^{{\pi }i{\tau }(n+{\frac {1}{2}})^{2}-{\frac {1}{4}}{\pi }i{\tau }}}\\&=\sum _{n=0}^{\infty }{(-1)^{n}e^{{\pi }i{\tau }(n+{\frac {1}{2}})^{2}-{\frac {1}{4}}{\pi }i{\tau }}}+\sum _{n=0}^{\infty }{(-1)^{n+1}e^{{\pi }i{\tau }(-n-1+{\frac {1}{2}})^{2}-{\frac {1}{4}}{\pi }i{\tau }}}\\&=0\end{aligned}}}$

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、

${\displaystyle c(\tau ,v)={\frac {\vartheta (v,\tau )}{\Theta (v,\tau )}}={\frac {\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+2{\pi }inv}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }-2{\pi }iv}\right)}}}}$

は${\displaystyle v}$に依存しない。

${\displaystyle {\begin{aligned}c\left(\textstyle {\frac {1}{2}},\tau \right)&={\frac {\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+{\pi }in}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+{\pi }i}\right)\left(1+e^{(2m-1){\pi }i{\tau }-{\pi }i}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-1)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{(2m-1){\pi }i{\tau }}\right)\left(1-e^{(2m-1){\pi }i{\tau }}\right)}}}\\\end{aligned}}}$

${\displaystyle {\begin{aligned}c\left(\textstyle {\frac {1}{4}},\tau \right)&={\frac {\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}+{\pi }in/2}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }+{\pi }i/2}\right)\left(1+e^{(2m-1){\pi }i{\tau }-{\pi }i/2}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+ie^{(2m-1){\pi }i{\tau }}\right)\left(1-ie^{(2m-1){\pi }i{\tau }}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(4m-2){\pi }i{\tau }}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{4m{\pi }i{\tau }}\right)\left(1-e^{(4m-2){\pi }i{\tau }}\right)\left(1+e^{(4m-2){\pi }i{\tau }}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{4m{\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)}}}\\&={\frac {\sum _{n=-\infty }^{\infty }{(-i)^{n}e^{{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{8m{\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)}}}\\\end{aligned}}}$

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。

${\displaystyle {\begin{aligned}c\left(\textstyle {\frac {1}{4}},\tau \right)&={\frac {\sum _{n=-\infty }^{\infty }{(-1)^{n}e^{4{\pi }i{\tau }n^{2}}}}{\prod _{m=1}^{\infty }{\left(1-e^{8m{\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)\left(1-e^{(8m-4){\pi }i{\tau }}\right)}}}\\&=c\left(\textstyle {\frac {1}{2}},4\tau \right)\end{aligned}}}$

${\displaystyle c(v,\tau )}$は${\displaystyle v}$に依存しないから

${\displaystyle c\left(v,4\tau \right)=c\left(v,\tau \right)=\lim _{n\to \infty }c\left(v,4^{-n}\tau \right)=\lim _{\tau '\to 0}c\left(v,\tau '\right)=c(v,0)}$

であり、${\displaystyle c(v,\tau )}$は${\displaystyle \tau }$にも依存しない定数である。${\displaystyle \tau \to {+i}\infty }$として${\displaystyle c(v,\tau )=1}$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

${\displaystyle \sum _{n=-\infty }^{\infty }{\frac {(a;q)_{n}}{(b;q)_{n}}}z^{n}={\frac {(az;q)_{\infty }(q;q)_{\infty }\left({\frac {q}{az}};q\right)_{\infty }\left({\frac {b}{a}};q\right)_{\infty }}{(z;q)_{\infty }(b;q)_{\infty }\left({\frac {b}{az}};q\right)_{\infty }\left({\frac {q}{a}};q\right)_{\infty }}}\qquad (|q|<1,|b/a|<|z|<1)}$

はq二項定理から導かれる。ラマヌジャンの和公式に${\displaystyle b=0}$を代入すると

${\displaystyle \sum _{n=-\infty }^{\infty }(a;q)_{n}z^{n}={\frac {(az;q)_{\infty }(q;q)_{\infty }\left(q/az;q\right)_{\infty }}{(z;q)_{\infty }\left(q/a;q\right)_{\infty }}}}$

となり、${\displaystyle q}$を${\displaystyle q^{2}}$と書き、${\displaystyle z}$を${\displaystyle -qz/a}$と書けば

${\displaystyle \sum _{n=-\infty }^{\infty }(a;q^{2})_{n}\left(-{\frac {qz}{a}}\right)^{n}={\frac {(-qz;q^{2})_{\infty }(q^{2};q^{2})_{\infty }\left(-q/z;q^{2}\right)_{\infty }}{(-qz/a;q^{2})_{\infty }\left(q^{2}/a;q^{2}\right)_{\infty }}}}$

となる。qポッホハマー記号の変換式

${\displaystyle \left(aq^{-n+1};q\right)_{n}=\left(-a\right)^{n}q^{-n(n-1)/2}\left({\frac {1}{a}};q\right)_{n}}$

により、左辺は

${\displaystyle {\begin{aligned}\sum _{n=-\infty }^{\infty }(a;q^{2})_{n}\left(-{\frac {qz}{a}}\right)^{n}&=\sum _{n=-\infty }^{\infty }(aq^{2n-2}q^{-2n+2};q^{2})_{n}\left(-{\frac {qz}{a}}\right)^{n}\\&=\sum _{n=-\infty }^{\infty }(-aq^{2n-2})^{n}q^{-2n(n-1)/2}(1/aq^{2n-2};q)_{n}\left(-{\frac {qz}{a}}\right)^{n}\\&=\sum _{n=-\infty }^{\infty }q^{n^{2}}z^{n}(1/aq^{2n-2};q)_{n}\\\end{aligned}}}$

であるから、${\displaystyle a\to \infty }$の極限を取れば

${\displaystyle \sum _{n=-\infty }^{\infty }q^{n^{2}}z^{n}=(-qz;q^{2})_{\infty }(q^{2};q^{2})_{\infty }\left(-q/z;q^{2}\right)_{\infty }}$

となり、qポッホハマー記号を展開して

${\displaystyle \sum _{n=-\infty }^{\infty }q^{n^{2}}z^{n}=\prod _{m=1}^{\infty }(1-q^{2m})(1+q^{2m-1}/z)(1+q^{2m-1}z)}$

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

${\displaystyle {\begin{aligned}Q(x)=&(1-x^{2})(1-x^{4})(1-x^{6})\cdots ,\\R(x,z)=&(1+xz)(1+x^{3}z)(1+x^{5}z)\cdots ,\\S(x,z)=&{\frac {1}{(1-xz)(1-x^{2}z)(1-x^{3}z)\cdots }}\end{aligned}}}$

とおく。ヤコビの三重積は ${\displaystyle Q(x)R(x,z^{-1})R(x,z)}$ であらわされる。

まず ${\displaystyle R(x,z)}$ について、

${\displaystyle R(x,zx^{2})=(1+x^{3}z)(1+x^{5}z)(1+x^{7}z)\cdots }$

より

${\displaystyle R(x,z)=(1+xz)R(x,zx^{2})}$

が成り立つ。そこで

${\displaystyle R(x,z)=1+a_{1}(x)z+a_{2}(x)z^{2}+\cdots }$

とおくと

${\displaystyle (1+a_{1}(x)z+a_{2}(x)z^{2}+\cdots )=(1+xz)(1+a_{1}(x)x^{2}z+a_{2}(x)x^{4}z^{2}+\cdots )}$

より

${\displaystyle a_{n}(x)=a_{n}(x)x^{2n}+a_{n-1}(x)x^{2n-1},}$

つまり

${\displaystyle a_{n}(x)={\frac {a_{n-1}(x)x^{2n-1}}{1-x^{2n}}},a_{0}(x)=1}$

が成り立つ。この漸化式を解くと

${\displaystyle a_{n}(x)={\frac {x^{n^{2}}}{(1-x^{2})(1-x^{4})\cdots (1-x^{2n})}}}$

が得られる。

次に ${\displaystyle S(x,z)}$ について、

${\displaystyle S(x,zx)={\frac {1}{(1-x^{2}z)(1-x^{3}z)(1-x^{4}z)\cdots }}=(1-xz)S(x,z)}$

が成り立つ。そこで

${\displaystyle S(x,z)=1+{\frac {b_{1}(x)z}{1-xz}}+{\frac {b_{2}(x)z^{2}}{(1-xz)(1-x^{2}z)}}+\cdots }$

とおくと

${\displaystyle {\begin{aligned}S(x,z)=&{\frac {S(x,zx)}{1-xz}}\\=&{\frac {1}{1-xz}}\left(1+{\frac {b_{1}(x)xz}{1-x^{2}z}}+{\frac {b_{2}(x)x^{2}z^{2}}{(1-x^{2}z)(1-x^{3}z)}}+\cdots \right)\\=&{\frac {1}{1-xz}}+{\frac {b_{1}(x)xz}{(1-xz)(1-x^{2}z)}}+{\frac {b_{2}(x)x^{2}z^{2}}{(1-xz)(1-x^{2}z)(1-x^{3}z)}}\cdots \end{aligned}}}$

であるが

${\displaystyle {\frac {b_{m}(x)x^{m}z^{m}}{(1-xz)(1-x^{2}z)\cdots (1-x^{m+1}z)}}={\frac {b_{m}(x)x^{m}z^{m}}{(1-xz)(1-x^{2}z)\cdots (1-x^{m}z)}}+{\frac {b_{m}(x)x^{2m+1}z^{m+1}}{(1-xz)(1-x^{2}z)\cdots (1-x^{m+1}z)}}}$

より

${\displaystyle b_{n}(x)=b_{n}(x)x^{n}+b_{n-1}x^{2n-1},}$

つまり

${\displaystyle b_{n}(x)={\frac {b_{n-1}(x)x^{2n-1}}{1-x^{n}}},b_{0}(x)=1}$

が成り立つ。この漸化式を解くと

${\displaystyle b_{n}(x)={\frac {x^{n^{2}}}{(1-x)(1-x^{2})\cdots (1-x^{n})}}}$

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 ${\displaystyle R(x,z)R(x,z^{-1})}$ の冪級数展開について考える。

${\displaystyle R(x,z)R(x,z^{-1})=(1+a_{1}(x)z+a_{2}(x)z^{2}+\cdots )(1+a_{1}(x)z^{-1}+a_{2}(x)z^{-2}+\cdots )}$

より ${\displaystyle R(x,z)R(x,z^{-1})}$ を冪級数展開したときの ${\displaystyle z^{m}}$ および ${\displaystyle z^{-m}}$ の係数は共に

${\displaystyle {\begin{aligned}&a_{m}(x)+a_{m+1}(x)a_{1}(x)+a_{m+2}(x)a_{2}(x)+\cdots \\=&a_{m}(x)\left(1+{\frac {x^{2m+2}}{(1-x^{2})(1-x^{2m+2})}}+{\frac {x^{4m+8}}{(1-x^{2})(1-x^{4})(1-x^{2m+2})(1-x^{2m+4})}}+\cdots \right)\\=&a_{m}(x)\sum _{k=0}^{\infty }{\frac {x^{2k^{2}}}{(1-x^{2})(1-x^{4})\cdots (1-x^{2k})}}\times {\frac {x^{2km}}{(1-x^{2m+2})(1-x^{2m+4})\cdots (1-x^{2(m+k)})}}\\\end{aligned}}}$

に一致するが、これは、上記の ${\displaystyle S(x,z)}$ の展開より

${\displaystyle {\begin{aligned}=&a_{m}(x)\sum _{k=0}^{\infty }{\frac {b_{k}(x^{2})x^{2km}}{(1-x^{2m+2})(1-x^{2m+4})\cdots (1-x^{2(m+k)})}}\\=&a_{m}(x)S(x^{2},x^{2m})\\=&{\frac {x^{m^{2}}}{(1-x^{2})(1-x^{4})\cdots (1-x^{2m})\times (1-x^{2m+2})(1-x^{2m+4})\cdots }}\\=&{\frac {x^{m^{2}}}{(1-x^{2})(1-x^{4})\cdots }}\\=&{\frac {x^{m^{2}}}{Q(x)}}\end{aligned}}}$

に一致する。よって

${\displaystyle Q(x)R(x,z)R(x,z^{-1})=\sum _{m=-\infty }^{\infty }x^{m^{2}}z^{m}}$

が成り立つ。