テータ関数
であるが、 n ± m {\displaystyle n{\pm }m} は共に偶数か共に奇数であるから、 N = ⌊ n + m 2 ⌋ , M = ⌊ n − m 2 ⌋ {\displaystyle N=\lfloor {\tfrac {n+m}{2}}\rfloor ,M=\lfloor {\tfrac {n-m}{2}}\rfloor } とすれば
θ 3 ( x + y , τ ) θ 3 ( x − y , τ ) = ∑ N = − ∞ ∞ ∑ M = − ∞ ∞ e 2 π i τ N 2 + 4 π i N x + 2 π i τ M 2 + 4 π i M y ( n ± m even ) + ∑ N = − ∞ ∞ ∑ M = − ∞ ∞ e 2 π i τ ( N + 1 2 ) 2 + 4 π i ( N + 1 2 ) x + 2 π i τ ( M + 1 2 ) 2 + 4 π i ( M + 1 2 ) y ( n ± m odd ) = θ 3 ( 2 x , 2 τ ) θ 3 ( 2 y , 2 τ ) + θ 2 ( 2 x , 2 τ ) θ 2 ( 2 y , 2 τ ) {\displaystyle {\begin{aligned}\vartheta _{3}\left(x+y,\tau \right)\vartheta _{3}\left(x-y,\tau \right)&=\sum _{N=-\infty }^{\infty }\sum _{M=-\infty }^{\infty }e^{2{\pi }i{\tau }N^{2}+4{\pi }iNx+2{\pi }i{\tau }M^{2}+4{\pi }iMy}\qquad (n{\pm }m\;{\mbox{even}})\\&\quad +\sum _{N=-\infty }^{\infty }\sum _{M=-\infty }^{\infty }e^{2{\pi }i{\tau }\left(N+{\tfrac {1}{2}}\right)^{2}+4{\pi }i\left(N+{\tfrac {1}{2}}\right)x+2{\pi }i{\tau }\left(M+{\tfrac {1}{2}}\right)^{2}+4{\pi }i\left(M+{\tfrac {1}{2}}\right)y}\qquad (n{\pm }m\;{\mbox{odd}})\\&=\vartheta _{3}\left(2x,2\tau \right)\vartheta _{3}\left(2y,2\tau \right)+\vartheta _{2}\left(2x,2\tau \right)\vartheta _{2}\left(2y,2\tau \right)\end{aligned}}}
θ 2 ( v + τ ; τ ) = e − π i τ e − 2 π i v θ 2 ( v ; τ ) {\displaystyle \vartheta _{2}(v+\tau ;\tau )=e^{-{\pi }i\tau }e^{-2{\pi }i{v}}\vartheta _{2}(v;\tau )}
θ 3 ( v + τ ; τ ) = e − π i τ e − 2 π i v θ 3 ( v ; τ ) {\displaystyle \vartheta _{3}(v+\tau ;\tau )=e^{-{\pi }i\tau }e^{-2{\pi }i{v}}\vartheta _{3}(v;\tau )}
θ 4 ( v + τ ; τ ) = − e − π i τ e − 2 π i v θ 4 ( v ; τ ) {\displaystyle \vartheta _{4}(v+\tau ;\tau )=-e^{-{\pi }i\tau }e^{-2{\pi }i{v}}\vartheta _{4}(v;\tau )}