テータ関数
θ 3 ( 0 , τ ) 2 = ∑ m = − ∞ ∞ e π i τ m 2 ∑ n = − ∞ ∞ e π i τ n 2 = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( m 2 + n 2 ) = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( m + n ) 2 + ( m − n ) 2 ) / 2 = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( 2 m ) 2 + ( 2 n ) 2 ) / 2 + ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( 2 m + 1 ) 2 + ( 2 n + 1 ) 2 ) / 2 ( ⌊ m + n 2 ⌋ ↦ m , ⌊ m − n 2 ⌋ ↦ n ) = θ 3 ( 0 , 2 τ ) 2 + θ 2 ( 0 , 2 τ ) 2 {\displaystyle {\begin{aligned}\vartheta _{3}\left(0,\tau \right)^{2}&=\sum _{m=-\infty }^{\infty }e^{\pi {i}\tau {m^{2}}}\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {n^{2}}}=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {(m^{2}+n^{2})}}\\&=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((m+n)^{2}+(m-n)^{2}\right)/2}\\&=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((2m)^{2}+(2n)^{2}\right)/2}+\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((2m+1)^{2}+(2n+1)^{2}\right)/2}\qquad (\left\lfloor \textstyle {\frac {m+n}{2}}\right\rfloor \mapsto {m},\left\lfloor \textstyle {\frac {m-n}{2}}\right\rfloor \mapsto {n})\\&=\vartheta _{3}\left(0,2\tau \right)^{2}+\vartheta _{2}\left(0,2\tau \right)^{2}\end{aligned}}}
θ 1 ( v , τ ) := − θ 11 ( v , τ ) = − ∑ n = − ∞ ∞ e π i τ ( n + 1 2 ) 2 + 2 π i ( n + 1 2 ) ( v + 1 2 ) = 2 ∑ n = 0 ∞ ( − 1 ) n q ( n + 1 2 ) 2 sin  ( 2 n + 1 ) π v , θ 2 ( v , τ ) := θ 10 ( v , τ ) = ∑ n = − ∞ ∞ e π i τ ( n + 1 2 ) 2 + 2 π i ( n + 1 2 ) v = 2 ∑ n = 0 ∞ q ( n + 1 2 ) 2 cos  ( 2 n + 1 ) π v , θ 3 ( v , τ ) := θ 00 ( v , τ ) = ∑ n = − ∞ ∞ e i π τ n 2 + 2 π i n v = 1 + 2 ∑ n = 1 ∞ q n 2 cos  2 n π v , θ 4 ( v , τ ) := θ 01 ( v , τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n ( v + 1 2 ) = 1 + 2 ∑ n = 1 ∞ ( − 1 ) n q n 2 cos  2 n π v , {\displaystyle {\begin{aligned}\vartheta _{1}(v,\tau )&:=-\vartheta _{11}(v,\tau )=-\sum _{n=-\infty }^{\infty }e^{\pi i\tau \left(n+{\frac {1}{2}}\right)^{2}+2\pi i\left(n+{\frac {1}{2}}\right)\left(v+{\frac {1}{2}}\right)}\\&=2\sum _{n=0}^{\infty }{(-1)^{n}q^{{\left(n+{\frac {1}{2}}\right)}^{2}}\sin(2n+1)\pi v},\\\vartheta _{2}(v,\tau )&:=\vartheta _{10}(v,\tau )=\sum _{n=-\infty }^{\infty }e^{\pi i\tau \left(n+{\frac {1}{2}}\right)^{2}+2\pi i\left(n+{\frac {1}{2}}\right)v}\\&=2\sum _{n=0}^{\infty }q^{{\left(n+{\frac {1}{2}}\right)}^{2}}\cos(2n+1)\pi v,\\\vartheta _{3}(v,\tau )&:=\vartheta _{00}(v,\tau )=\sum _{n=-\infty }^{\infty }e^{i\pi \tau n^{2}+2\pi inv}\\&=1+2\sum _{n=1}^{\infty }q^{n^{2}}\cos 2n\pi v,\\\vartheta _{4}(v,\tau )&:=\vartheta _{01}(v,\tau )=\sum _{n=-\infty }^{\infty }e^{\pi i\tau n^{2}+2\pi in\left(v+{\frac {1}{2}}\right)}\\&=1+2\sum _{n=1}^{\infty }(-1)^{n}q^{n^{2}}\cos 2n\pi v,\\\end{aligned}}}
ヤコビの三重積の公式により、