テータ関数
となる。ここで x ↦ x + 1 2 τ {\displaystyle x\mapsto {x+{\tfrac {1}{2}}\tau }} とすれば
x = y = z とすれば
θ 3 ( v + τ 2 , τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n ( v + τ / 2 ) = ∑ n = − ∞ ∞ e π i τ ( n + 1 / 2 ) 2 − π i τ / 4 + 2 π i ( n + 1 / 2 ) v − π i v = e − π i τ / 4 e − π i v θ 2 ( v , τ ) {\displaystyle \vartheta _{3}\left(v+{\frac {\tau }{2}},\tau \right)=\sum _{n=-\infty }^{\infty }e^{\pi {i\tau }n^{2}+2\pi {in}(v+\tau /2)}=\sum _{n=-\infty }^{\infty }e^{\pi {i\tau }(n+1/2)^{2}-\pi {i\tau }/4+2\pi {i(n+1/2)v}-\pi {iv}}=e^{-\pi {i\tau }/4}e^{-\pi {iv}}\vartheta _{2}(v,\tau )}
擬二重周期と併せて
θ 1 ( v ± 1 2 , τ ) = ± θ 2 ( v ; τ ) {\displaystyle \vartheta _{1}\left(v\pm {\frac {1}{2}},\tau \right)=\pm \vartheta _{2}(v;\tau )}
θ 2 ( v ± 1 2 , τ ) = ∓ θ 1 ( v ; τ ) {\displaystyle \vartheta _{2}\left(v\pm {\frac {1}{2}},\tau \right)=\mp \vartheta _{1}(v;\tau )}
θ 3 ( v ± 1 2 , τ ) = θ 4 ( v ; τ ) {\displaystyle \vartheta _{3}\left(v\pm {\frac {1}{2}},\tau \right)=\vartheta _{4}(v;\tau )}