テータ関数
θ 1 ( x + y , τ ) θ 1 ( x − y , τ ) θ 3 2 ( 0 , τ ) = θ 1 2 ( x , τ ) θ 3 2 ( y , τ ) − θ 3 2 ( x , τ ) θ 1 2 ( y , τ ) = θ 4 2 ( x , τ ) θ 2 2 ( y , τ ) − θ 2 2 ( x , τ ) θ 4 2 ( y , τ ) θ 2 ( x + y , τ ) θ 2 ( x − y , τ ) θ 3 2 ( 0 , τ ) = θ 2 2 ( x , τ ) θ 3 2 ( y , τ ) − θ 4 2 ( x , τ ) θ 1 2 ( y , τ ) = θ 3 2 ( x , τ ) θ 2 2 ( y , τ ) − θ 1 2 ( x , τ ) θ 4 2 ( y , τ ) θ 3 ( x + y , τ ) θ 3 ( x − y , τ ) θ 3 2 ( 0 , τ ) = θ 3 2 ( x , τ ) θ 3 2 ( y , τ ) + θ 1 2 ( x , τ ) θ 1 2 ( y , τ ) = θ 2 2 ( x , τ ) θ 2 2 ( y , τ ) + θ 4 2 ( x , τ ) θ 4 2 ( y , τ ) θ 4 ( x + y , τ ) θ 4 ( x − y , τ ) θ 3 2 ( 0 , τ ) = θ 4 2 ( x , τ ) θ 3 2 ( y , τ ) + θ 2 2 ( x , τ ) θ 1 2 ( y , τ ) = θ 1 2 ( x , τ ) θ 2 2 ( y , τ ) + θ 3 2 ( x , τ ) θ 4 2 ( y , τ ) {\displaystyle {\begin{aligned}&\vartheta _{1}\left(x+y,\tau \right)\vartheta _{1}\left(x-y,\tau \right)\vartheta _{3}^{\;2}(0,\tau )=\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)-\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)-\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)\\&\vartheta _{2}\left(x+y,\tau \right)\vartheta _{2}\left(x-y,\tau \right)\vartheta _{3}^{\;2}(0,\tau )=\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)-\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)-\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)\\&\vartheta _{3}\left(x+y,\tau \right)\vartheta _{3}\left(x-y,\tau \right)\vartheta _{3}^{\;2}(0,\tau )=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)+\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)+\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)\\&\vartheta _{4}\left(x+y,\tau \right)\vartheta _{4}\left(x-y,\tau \right)\vartheta _{3}^{\;2}(0,\tau )=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)+\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)+\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)\\\end{aligned}}}
θ 4 ( 0 , τ ) 2 = ∑ m = − ∞ ∞ ( − 1 ) m e π i τ m 2 ∑ n = − ∞ ∞ ( − 1 ) n e π i τ n 2 = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ ( − 1 ) m + n e π i τ ( m 2 + n 2 ) = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ ( − 1 ) 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 _{4}\left(0,\tau \right)^{2}&=\sum _{m=-\infty }^{\infty }(-1)^{m}e^{\pi {i}\tau {m^{2}}}\sum _{n=-\infty }^{\infty }(-1)^{n}e^{\pi {i}\tau {n^{2}}}=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }(-1)^{m+n}e^{\pi {i}\tau {(m^{2}+n^{2})}}\\&=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }(-1)^{m+n}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 1 ( v , τ ) = ∂ ∂ v log  θ 1 ( v , τ ) = π cos  π v sin  π v − 2 π i ∑ m = 1 ∞ e 2 m π i τ e 2 π i v 1 − e 2 m π i τ e 2 π i v + 2 π i ∑ m = 1 ∞ e 2 m π i τ e − 2 π i v 1 − e 2 m π i τ e − 2 π i v = π cot  π v − 2 π i ∑ m = 1 ∞ ( ∑ n = 1 ∞ e 2 m n π i τ e 2 π i n v ) + 2 π i ∑ m = 1 ∞ ( ∑ n = 1 ∞ e 2 m n π i τ e − 2 π i n v ) = π cot  π v − 2 π i ∑ n = 1 ∞ ( ∑ m = 1 ∞ e 2 m n π i τ ) ( e 2 π i n v − e − 2 π i n v ) = π cot  π v + 4 π ∑ n = 1 ∞ e 2 n π i τ 1 − e 2 n π i τ sin  2 π n v {\displaystyle {\begin{aligned}{\frac {\vartheta _{1}'(v,\tau )}{\vartheta _{1}(v,\tau )}}&={\frac {\partial }{\partial {v}}}\log \vartheta _{1}(v,\tau )\\&={\frac {\pi \cos \pi {v}}{\sin \pi {v}}}-2\pi {i}\sum _{m=1}^{\infty }{\frac {e^{2m\pi {i\tau }}e^{2\pi {iv}}}{1-e^{2m\pi {i\tau }}e^{2\pi {iv}}}}+2\pi {i}\sum _{m=1}^{\infty }{\frac {e^{2m\pi {i\tau }}e^{-2\pi {iv}}}{1-e^{2m\pi {i\tau }}e^{-2\pi {iv}}}}\\&=\pi \cot \pi {v}-2\pi {i}\sum _{m=1}^{\infty }\left(\sum _{n=1}^{\infty }e^{2mn\pi {i\tau }}e^{2\pi {inv}}\right)+2\pi {i}\sum _{m=1}^{\infty }\left(\sum _{n=1}^{\infty }e^{2mn\pi {i\tau }}e^{-2\pi {inv}}\right)\\&=\pi \cot \pi {v}-2\pi {i}\sum _{n=1}^{\infty }\left(\sum _{m=1}^{\infty }e^{2mn\pi {i\tau }}\right)\left(e^{2\pi {inv}}-e^{-2\pi {inv}}\right)\\&=\pi \cot \pi {v}+4\pi \sum _{n=1}^{\infty }{\frac {e^{2n\pi {i\tau }}}{1-e^{2n\pi {i\tau }}}}\sin 2\pi {nv}\\\end{aligned}}}