ヤコビの三重積
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}}}
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
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}}}
( 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}}}