テータ関数

テータ関数（テータかんすう、英: theta function）は、 ϑ ( z , τ ) := ∑ n = − ∞ ∞ e π i n 2 τ + 2 π i n z . {\displaystyle \vartheta (z,\tau ):=\sum _{n=-\infty }^{\infty }e^{\pi in^{2}\tau +2\pi inz}.} で定義される関数のことである。それ以外にも、指標付きのテータ関数 ϑ a b ( z , τ ) {\displaystyle \vartheta _{ab}(z,\tau )} 、ヤコビのテータ関数、楕円テータ関数 ϑ i ( z , τ ) {\displaystyle \vartheta _{i}(z,\tau )} と呼ばれる一連のテータ関数が存在する。 指標付きのテータ関数や楕円テータ関数は、その定義にいくつかの流儀があり、同じ記号を使いながら違ったものを指していることがあるので注意が必要である。 これらの関数は、z の関数と見た場合には擬二重周期を持ち楕円関数に関係し、τ の関数と見た場合はモジュラー形式に関係する。 テータ関数は次のように定義される関数のことを指す。 ϑ ( z , τ ) := ∑ n = − ∞ ∞ e π i n 2 τ + 2 π i n z . {\displaystyle \vartheta (z,\tau ):=\sum _{n=-\infty }^{\infty }e^{\pi in^{2}\tau +2\pi inz}.} テータ関数を z の関数と見た場合、周期 1 の周期関数である。 ϑ ( z + 1 , τ ) = ϑ ( z , τ ) . {\displaystyle \vartheta (z+1,\tau )=\vartheta (z,\tau ).} 一般には以下の等式を満たす。 ϑ ( z + m τ + n , τ ) = e − π i m 2 τ − 2 π i m z ϑ ( z , τ ) . {\displaystyle \vartheta (z+m\tau +n,\tau )=e^{-\pi im^{2}\tau -2\pi imz}\vartheta (z,\tau ).} ヤコビのテータ関数は狭義の意味では次の関数のことを指す。 Θ ( u ) := ( 2 k ′ K π ) 1 / 2 exp ⁡ ( ∫ 0 u d t   Z ( t ) ) , Θ 1 ( u ) := Θ ( u + K ) . {\displaystyle {\begin{aligned}\Theta (u)&:=\left({\frac {2k'K}{\pi }}\right)^{1/2}\exp \left(\int _{0}^{u}\mathrm {d} t~Z(t)\right),\\\Theta _{1}(u)&:=\Theta (u+K).\end{aligned}}} ただし、 k ′ := 1 − k 2 {\displaystyle k':={\sqrt {1-k^{2}\,}}} は補母数、 K = K ( k ) {\displaystyle K=K(k)} は 第1種完全楕円積分、 Z ( u ) {\displaystyle Z(u)} はヤコビのツェータ関数 Z ( u ) := E ( u ) − E ( k ) u K ( k ) , E ( u ) := ∫ 0 u d t d n 2 t = ∫ 0 s n u d t 1 − k 2 t 2 1 − t 2 = ∫ 0 a m u d θ 1 − k 2 sin 2 ⁡ θ , {\displaystyle {\begin{aligned}Z(u)&:={\mathcal {E}}(u)-{\frac {E(k)u}{K(k)}},\\{\mathcal {E}}(u)&:=\int _{0}^{u}\mathrm {d} t\,\mathrm {dn} ^{2}t=\int _{0}^{\mathrm {sn} u}\mathrm {d} t{\sqrt {\frac {1-k^{2}t^{2}}{1-t^{2}}}}=\int _{0}^{\mathrm {am} u}\mathrm {d} \theta {\sqrt {1-k^{2}\sin ^{2}\theta }},\end{aligned}}} E ( u ) {\displaystyle {\mathcal {E}}(u)} はヤコビのイプシロン関数、 E ( k ) {\displaystyle E(k)} は第2種完全楕円積分、 s n u = s n ( u , k ) {\displaystyle \mathrm {sn} \,u=\mathrm {sn} (u,k)} , d n u = d n ( u , k ) {\displaystyle dn\,u=dn(u,k)} は ヤコビの楕円関数、 a m u = a m ( u , k ) {\displaystyle \mathrm {am} \,u=\mathrm {am} (u,k)} は振幅関数である。 また、ヤコビのエータ関数 H ( u ) := − i exp ⁡ ( ( 2 u + i K ′ ) π i / ( 4 K ) ) Θ ( u + i K ′ ) , i := − 1 , H 1 ( u ) := H ( u + K ) , {\displaystyle {\begin{aligned}H(u)&:=-i\exp \left((2u+iK')\pi i/(4K)\right)\Theta (u+iK'),\quad i:={\sqrt {-1\,}},\\H_{1}(u)&:=H(u+K),\end{aligned}}} を含めて、 Θ ( u ) {\displaystyle \Theta (u)} , Θ 1 ( u ) {\displaystyle \Theta _{1}(u)} , H ( u ) {\displaystyle H(u)} , H 1 ( u ) {\displaystyle H_{1}(u)} のことをヤコビのテータ関数と呼ぶこともある。ただし、 K ′ := K ( k ′ ) {\displaystyle K':=K(k')} である。ヤコビのテータ関数は、後述の楕円テータ関数と以下の関係で結ばれている。 Θ ( u ) = ϑ 0 ( u / ( 2 ω 1 ) ) , Θ 1 ( u ) = ϑ 3 ( u / ( 2 ω 1 ) ) , H ( u ) = ϑ 1 ( u / ( 2 ω 1 ) , H 1 ( u ) = ϑ 2 ( u / ( 2 ω 1 ) ) , {\displaystyle {\begin{aligned}\Theta (u)&=\vartheta _{0}(u/(2\omega _{1})),\quad \Theta _{1}(u)=\vartheta _{3}(u/(2\omega _{1})),\\H(u)&=\vartheta _{1}(u/(2\omega _{1}),\quad H_{1}(u)=\vartheta _{2}(u/(2\omega _{1})),\end{aligned}}} ただし、 ω 1 {\displaystyle \omega _{1}} は、楕円関数の基本周期の半分で、 τ = ω 3 / ω 1 {\displaystyle \tau =\omega _{3}/\omega _{1}} である（ 2 ω 1 {\displaystyle 2\omega _{1}} , 2 ω 3 {\displaystyle 2\omega _{3}} が楕円関数の基本周期に相当する）。 物理の教科書では後述の ϑ i ( z , τ ) {\displaystyle \vartheta _{i}(z,\tau )} をヤコビのテータ関数と呼んでいるが、やや不正確な言い方である。 以下のように定義された、添え字を 2 つ持つテータ関数のことを指標付きのテータ関数と呼ぶ。 ϑ a , b ( z , τ ) := ∑ n = − ∞ ∞ e π i ( n + a ) 2 τ + 2 π i ( n + a ) ( z + b ) , a , b ∈ R . {\displaystyle \vartheta _{a,b}(z,\tau ):=\sum _{n=-\infty }^{\infty }e^{\pi i(n+a)^{2}\tau +2\pi i(n+a)(z+b)},\quad a,b\in \mathbb {R} .} なお、指標付きのテータ関数の定義には 2 つの流儀があって統一的に用いられていないため、文献を読むときには注意しなければならない 。 この記事で使われているのは、Mumford 2006 で使われているのと同じ定義である。 楕円テータ関数（だえんテータかんすう、英: elliptic theta function）は、以下のように定義された関数である。 ただし、 I m τ > 0 {\displaystyle \mathrm {Im} \,\tau >0} , q := e π i τ {\displaystyle q:=e^{\pi i\tau }} である。 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}}} 楕円テータ関数にも定義に 2 つの流儀があり、注意が必要である。 フルヴィッツ・クーランの「楕円関数論」の定義では添え字が 1 から 4 ではなく、 0 から 3 である。 その場合は ϑ 1 ( v , τ ) {\displaystyle \vartheta _{1}(v,\tau )} , ϑ 2 ( v , τ ) {\displaystyle \vartheta _{2}(v,\tau )} , ϑ 3 ( v , τ ) {\displaystyle \vartheta _{3}(v,\tau )} の定義は変わらず、 ϑ 0 ( v , τ ) := ϑ 4 ( v , τ ) {\displaystyle \vartheta _{0}(v,\tau ):=\vartheta _{4}(v,\tau )} で定義される。 文脈から v あるいは τ が明らかな場合は ϑ i ( v ) {\displaystyle \vartheta _{i}(v)} あるいは ϑ i ( τ ) {\displaystyle \vartheta _{i}(\tau )} と書き、更に ϑ i = ϑ i ( 0 , τ ) {\displaystyle \vartheta _{i}=\vartheta _{i}(0,\tau )} と書く。Mathematica では、 π v {\displaystyle \pi v} のことを v と書いている。 テータ関数は擬二重周期を持つ。 ϑ 1 ( v + 1 ; τ ) = − ∑ n = − ∞ ∞ e π i τ ( n + 1 2 ) 2 + 2 π i ( n + 1 2 ) ( v + 1 2 ) + 2 π i ( n + 1 2 ) = − ∑ n = − ∞ ∞ e π i τ ( n + 1 2 ) 2 + 2 π i ( n + 1 2 ) ( v + 1 2 ) + π i = ∑ n = − ∞ ∞ e π i τ ( n + 1 2 ) 2 + 2 π i ( n + 1 2 ) ( v + 1 2 ) = − ϑ 1 ( v ; τ ) {\displaystyle {\begin{aligned}\vartheta _{1}(v+1;\tau )&=-\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }\left(n+{\frac {1}{2}}\right)^{2}+2{\pi }i(n+{\frac {1}{2}})(v+{\frac {1}{2}})+2{\pi }i(n+{\frac {1}{2}})}}\\&=-\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }\left(n+{\frac {1}{2}}\right)^{2}+2{\pi }i(n+{\frac {1}{2}})(v+{\frac {1}{2}})+{\pi }i}}\\&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }\left(n+{\frac {1}{2}}\right)^{2}+2{\pi }i(n+{\frac {1}{2}})(v+{\frac {1}{2}})}}\\&=-\vartheta _{1}(v;\tau )\\\end{aligned}}} ϑ 2 ( v + 1 ; τ ) = − ϑ 2 ( v ; τ ) {\displaystyle \vartheta _{2}(v+1;\tau )=-\vartheta _{2}(v;\tau )} ϑ 3 ( v + 1 ; τ ) = ϑ 3 ( v ; τ ) {\displaystyle \vartheta _{3}(v+1;\tau )=\vartheta _{3}(v;\tau )} ϑ 4 ( v + 1 ; τ ) = ϑ 4 ( v ; τ ) {\displaystyle \vartheta _{4}(v+1;\tau )=\vartheta _{4}(v;\tau )} ϑ 1 ( v + τ ; τ ) = − ∑ n = − ∞ ∞ e π i τ ( n + 1 2 ) 2 + 2 π i ( n + 1 2 ) ( v + 1 2 ) + 2 π i ( n + 1 2 ) τ = − ∑ n = − ∞ ∞ e π i τ ( n + 1 + 1 2 ) 2 + 2 π i ( n + 1 + 1 2 ) ( v + 1 2 ) − π i τ − 2 π i ( v + 1 2 ) = − ∑ n = − ∞ ∞ e π i τ ( n + 1 2 ) 2 + 2 π i ( n + 1 2 ) ( v + 1 2 ) − π i τ − 2 π i ( v + 1 2 ) = ∑ n = − ∞ ∞ e π i τ ( n + 1 2 ) 2 + 2 π i ( n + 1 2 ) ( v + 1 2 ) − π i τ − 2 π i v = − e − π i τ e − 2 π i v ϑ 1 ( v ; τ ) {\displaystyle {\begin{aligned}\vartheta _{1}(v+\tau ;\tau )&=-\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }\left(n+{\frac {1}{2}}\right)^{2}+2{\pi }i(n+{\frac {1}{2}})(v+{\frac {1}{2}})+2{\pi }i(n+{\frac {1}{2}})\tau }}\\&=-\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }\left(n+1+{\frac {1}{2}}\right)^{2}+2{\pi }i(n+1+{\frac {1}{2}})(v+{\frac {1}{2}})-{\pi }i{\tau }-2{\pi }i(v+{\frac {1}{2}})}}\\&=-\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }\left(n+{\frac {1}{2}}\right)^{2}+2{\pi }i(n+{\frac {1}{2}})(v+{\frac {1}{2}})-{\pi }i{\tau }-2{\pi }i(v+{\frac {1}{2}})}}\\&=\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }\left(n+{\frac {1}{2}}\right)^{2}+2{\pi }i(n+{\frac {1}{2}})(v+{\frac {1}{2}})-{\pi }i{\tau }-2{\pi }iv}}\\&=-e^{-{\pi }i\tau }e^{-2{\pi }i{v}}\vartheta _{1}(v;\tau )\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 )} ヤコビの三重積の公式により、 ϑ 1 ( v ; τ ) = − ∑ n = − ∞ ∞ e π i τ ( n + 1 / 2 ) 2 e 2 π i ( n + 1 / 2 ) ( v + 1 / 2 ) = − i e π i τ / 4 e π i v ∑ n = − ∞ ∞ e π i τ n 2 e π i τ n + 2 π i v n + π i n = − i e π i τ / 4 e π i v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 − e 2 m π i τ e 2 π i v ) ( 1 − e ( 2 m − 2 ) π i τ e − 2 π i v ) = − i e π i τ / 4 e π i v ( 1 − e − 2 π i v ) ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 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 e π i τ / 4 sin ⁡ π v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 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 e π i τ / 4 sin ⁡ π v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 − 2 e 2 m π i τ cos ⁡ 2 π v + e 4 m π i τ ) {\displaystyle {\begin{aligned}\vartheta _{1}(v;\tau )&=-\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }\left(n+1/2\right)^{2}}e^{2{\pi }i(n+1/2)(v+1/2)}}\\&=-ie^{{\pi }i{\tau }/4}e^{{\pi }iv}\sum _{n=-\infty }^{\infty }{e^{{\pi }i{\tau }n^{2}}e^{{\pi }i{\tau }n+2{\pi }ivn+{\pi }in}}\\&=-ie^{{\pi }i{\tau }/4}e^{{\pi }iv}\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{2m{\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1-e^{(2m-2){\pi }i{\tau }}e^{-2{\pi }iv}\right)}\\&=-ie^{{\pi }i{\tau }/4}e^{{\pi }iv}(1-e^{-2{\pi }iv})\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{2m{\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1-e^{2m{\pi }i{\tau }}e^{-2{\pi }iv}\right)}\\&=2e^{{\pi }i{\tau }/4}\sin {\pi }v\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{2m{\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1-e^{2m{\pi }i{\tau }}e^{-2{\pi }iv}\right)}\\&=2e^{{\pi }i{\tau }/4}\sin {\pi }v\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-2e^{2m{\pi }i{\tau }}\cos {2{\pi }v}+e^{4m{\pi }i{\tau }}\right)}\\\end{aligned}}} ϑ 2 ( v ; τ ) = 2 e π i τ / 4 cos ⁡ π v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 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 e π i τ / 4 cos ⁡ π v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + 2 e 2 m π i τ cos ⁡ 2 π v + e 4 m π i τ ) {\displaystyle {\begin{aligned}\vartheta _{2}(v;\tau )&=2e^{{\pi }i{\tau }/4}\cos {\pi }v\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{2m{\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1+e^{2m{\pi }i{\tau }}e^{-2{\pi }iv}\right)}\\&=2e^{{\pi }i{\tau }/4}\cos {\pi }v\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+2e^{2m{\pi }i{\tau }}\cos {2{\pi }v}+e^{4m{\pi }i{\tau }}\right)}\\\end{aligned}}} ϑ 3 ( v ; τ ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ e 2 π i v ) ( 1 + e ( 2 m − 1 ) π i τ e − 2 π i v ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + 2 e ( 2 m − 1 ) π i τ cos ⁡ 2 π v + e 2 ( 2 m − 1 ) π i τ ) {\displaystyle {\begin{aligned}\vartheta _{3}(v;\tau )&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1+e^{(2m-1){\pi }i{\tau }}e^{-2{\pi }iv}\right)}\\&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+2e^{(2m-1){\pi }i{\tau }}\cos {2{\pi }v}+e^{2(2m-1){\pi }i{\tau }}\right)}\\\end{aligned}}} ϑ 4 ( v ; τ ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 − e ( 2 m − 1 ) π i τ e 2 π i v ) ( 1 − e ( 2 m − 1 ) π i τ e − 2 π i v ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 − 2 e ( 2 m − 1 ) π i τ cos ⁡ 2 π v + e 2 ( 2 m − 1 ) π i τ ) {\displaystyle {\begin{aligned}\vartheta _{4}(v;\tau )&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{(2m-1){\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1-e^{(2m-1){\pi }i{\tau }}e^{-2{\pi }iv}\right)}\\&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-2e^{(2m-1){\pi }i{\tau }}\cos {2{\pi }v}+e^{2(2m-1){\pi }i{\tau }}\right)}\\\end{aligned}}} | e 2 m π i τ | < 1 {\displaystyle |e^{2m{\pi }i{\tau }}|<1} であるから ϑ 3 ( v ; τ ) {\displaystyle \vartheta _{3}(v;\tau )} の零点は 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 = 2 n ′ + 1 2 + 2 m ′ + 1 2 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 {2n'+1}{2}}+{\frac {2m'+1}{2}}\tau \end{aligned}}} である。他の関数の零点も同様にして求められる。 ϑ 1 ( v ; τ ) = 0 ⇔ v = n + m τ ϑ 2 ( v ; τ ) = 0 ⇔ v = 2 n + 1 2 + m τ ϑ 3 ( v ; τ ) = 0 ⇔ v = 2 n + 1 2 + 2 m + 1 2 τ ϑ 4 ( v ; τ ) = 0 ⇔ v = n + 2 m + 1 2 τ {\displaystyle {\begin{aligned}&\vartheta _{1}(v;\tau )=0\;\Leftrightarrow \;v=n+m\tau \\&\vartheta _{2}(v;\tau )=0\;\Leftrightarrow \;v={\frac {2n+1}{2}}+m\tau \\&\vartheta _{3}(v;\tau )=0\;\Leftrightarrow \;v={\frac {2n+1}{2}}+{\frac {2m+1}{2}}\tau \\&\vartheta _{4}(v;\tau )=0\;\Leftrightarrow \;v=n+{\frac {2m+1}{2}}\tau \\\end{aligned}}} v = 0 のときのテータ関数の値をテータ定数（英: theta constant）あるいはテータ零値（独: Thetanullwerte）という。これは定数といいながら実は τ の関数である。 ϑ 2 = ϑ 2 ( 0 ; τ ) = 2 e π i τ / 4 ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e 2 m π i τ ) 2 {\displaystyle {\begin{aligned}\vartheta _{2}=\vartheta _{2}(0;\tau )&=2e^{{\pi }i{\tau }/4}\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{2m{\pi }i{\tau }}\right)^{2}}\\\end{aligned}}} ϑ 3 = ϑ 3 ( 0 ; τ ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 + e ( 2 m − 1 ) π i τ ) 2 {\displaystyle {\begin{aligned}\vartheta _{3}=\vartheta _{3}(0;\tau )&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1+e^{(2m-1){\pi }i{\tau }}\right)^{2}}\\\end{aligned}}} ϑ 4 = ϑ 4 ( 0 ; τ ) = ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 − e ( 2 m − 1 ) π i τ ) 2 {\displaystyle {\begin{aligned}\vartheta _{4}=\vartheta _{4}(0;\tau )&=\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{(2m-1){\pi }i{\tau }}\right)^{2}}\\\end{aligned}}} ϑ 1 = ϑ 1 ( 0 ; τ ) = 0 {\displaystyle \vartheta _{1}=\vartheta _{1}(0;\tau )=0} であるから、代わりに導関数を用いる。 ϑ 1 ′ = [ d d v ϑ 1 ( v ; τ ) ] v = 0 = 2 e π i τ / 4 π cos ⁡ ( 0 ) ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) 3 + 2 e π i τ / 4 sin ⁡ ( 0 ) d d v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 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 e π i τ / 4 ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) 3 {\displaystyle {\begin{aligned}\vartheta _{1}'&=\left[{\frac {d}{dv}}\vartheta _{1}(v;\tau )\right]_{v=0}\\&=2e^{{\pi }i{\tau }/4}\pi \cos(0)\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)^{3}}+2e^{{\pi }i{\tau }/4}\sin(0){\frac {d}{dv}}\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{2m{\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1-e^{2m{\pi }i{\tau }}e^{-2{\pi }iv}\right)}\\&=2{\pi }e^{{\pi }i{\tau }/4}\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)^{3}}\\\end{aligned}}} c = π ϑ 2 ϑ 3 ϑ 4 / ϑ 1 ′ {\displaystyle c=\pi \vartheta _{2}\vartheta _{3}\vartheta _{4}/\vartheta _{1}'} とすると c = ∏ m = 1 ∞ ( 1 + e 2 m π i τ ) 2 ( 1 + e ( 2 m − 1 ) π i τ ) 2 ( 1 − e ( 2 m − 1 ) π i τ ) 2 = ∏ m = 1 ∞ ( 1 + e 2 m π i τ ) 2 ( 1 − e 2 ( 2 m − 1 ) π i τ ) 2 {\displaystyle {\begin{aligned}c&=\prod _{m=1}^{\infty }{\left(1+e^{2m{\pi }i{\tau }}\right)^{2}\left(1+e^{(2m-1){\pi }i{\tau }}\right)^{2}\left(1-e^{(2m-1){\pi }i{\tau }}\right)^{2}}\\&=\prod _{m=1}^{\infty }{\left(1+e^{2m{\pi }i{\tau }}\right)^{2}\left(1-e^{2(2m-1){\pi }i{\tau }}\right)^{2}}\\\end{aligned}}} となるが、オイラーの分割恒等式により、 ∏ m − 1 ∞ ( 1 + e 2 m π i τ ) = ∏ m − 1 ∞ ( 1 − e 2 ( 2 m − 1 ) π i τ ) − 1 {\displaystyle \prod _{m-1}^{\infty }\left(1+e^{2m{\pi }i{\tau }}\right)=\prod _{m-1}^{\infty }\left(1-e^{2(2m-1){\pi }i{\tau }}\right)^{-1}} であるから c = 1 であり、故に ϑ 1 ′ = π ϑ 2 ϑ 3 ϑ 4 {\displaystyle \vartheta _{1}'=\pi \vartheta _{2}\vartheta _{3}\vartheta _{4}} である。 テータ関数の間で次の恒等式が成立する。 ϑ 3 ( v + 1 2 , τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n ( v + 1 / 2 ) = ϑ 4 ( v , τ ) {\displaystyle \vartheta _{3}\left(v+{\frac {1}{2}},\tau \right)=\sum _{n=-\infty }^{\infty }e^{\pi {i\tau }n^{2}+2\pi {in}(v+1/2)}=\vartheta _{4}(v,\tau )} ϑ 2 ( v + 1 2 , τ ) = ∑ n = − ∞ ∞ e π i τ ( n + 1 / 2 ) 2 + 2 π i ( n + 1 / 2 ) ( v + 1 / 2 ) = − ϑ 1 ( v , τ ) {\displaystyle \vartheta _{2}\left(v+{\frac {1}{2}},\tau \right)=\sum _{n=-\infty }^{\infty }e^{\pi {i\tau }(n+1/2)^{2}+2\pi {i(n+1/2)}(v+1/2)}=-\vartheta _{1}(v,\tau )} 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 )} ϑ 4 ( v ± 1 2 , τ ) = ϑ 3 ( v ; τ ) {\displaystyle \vartheta _{4}\left(v\pm {\frac {1}{2}},\tau \right)=\vartheta _{3}(v;\tau )} ϑ 1 ( v ± τ 2 , τ ) = ± i e − π i τ / 4 e ∓ π i v ϑ 4 ( v ; τ ) {\displaystyle \vartheta _{1}\left(v\pm {\frac {\tau }{2}},\tau \right)=\pm {i}e^{-\pi {i\tau }/4}e^{\mp \pi {i}v}\vartheta _{4}(v;\tau )} ϑ 2 ( v ± τ 2 , τ ) = e − π i τ / 4 e ∓ π i v ϑ 3 ( v ; τ ) {\displaystyle \vartheta _{2}\left(v\pm {\frac {\tau }{2}},\tau \right)=e^{-\pi {i\tau }/4}e^{\mp \pi {i}v}\vartheta _{3}(v;\tau )} ϑ 3 ( v ± τ 2 , τ ) = e − π i τ / 4 e ∓ π i v ϑ 2 ( v ; τ ) {\displaystyle \vartheta _{3}\left(v\pm {\frac {\tau }{2}},\tau \right)=e^{-\pi {i\tau }/4}e^{\mp \pi {i}v}\vartheta _{2}(v;\tau )} ϑ 4 ( v ± τ 2 , τ ) = ± i e − π i τ / 4 e ∓ π i v ϑ 1 ( v ; τ ) {\displaystyle \vartheta _{4}\left(v\pm {\frac {\tau }{2}},\tau \right)=\pm {i}e^{-\pi {i\tau }/4}e^{\mp \pi {i}v}\vartheta _{1}(v;\tau )} 次の恒等式はヤコビの虚数変換式という。 ϑ 1 ( v τ , − 1 τ ) = i e − i π / 4 τ 1 / 2 e π i v 2 / τ ϑ 1 ( v , τ ) , ϑ 2 ( v τ , − 1 τ ) = e − i π / 4 τ 1 / 2 e π i v 2 / τ ϑ 4 ( v , τ ) , ϑ 3 ( v τ , − 1 τ ) = e − i π / 4 τ 1 / 2 e π i v 2 / τ ϑ 3 ( v , τ ) , ϑ 4 ( v τ , − 1 τ ) = e − i π / 4 τ 1 / 2 e π i v 2 / τ ϑ 2 ( v , τ ) . {\displaystyle {\begin{aligned}\vartheta _{1}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)&=ie^{-i\pi /4}\tau ^{1/2}e^{\pi iv^{2}/\tau }\vartheta _{1}\left(v,\tau \right),\\\vartheta _{2}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)&=e^{-i\pi /4}\tau ^{1/2}e^{\pi iv^{2}/\tau }\vartheta _{4}\left(v,\tau \right),\\\vartheta _{3}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)&=e^{-i\pi /4}\tau ^{1/2}e^{\pi iv^{2}/\tau }\vartheta _{3}\left(v,\tau \right),\\\vartheta _{4}\left({\frac {v}{\tau }},-{\frac {1}{\tau }}\right)&=e^{-i\pi /4}\tau ^{1/2}e^{\pi iv^{2}/\tau }\vartheta _{2}\left(v,\tau \right).\end{aligned}}} 他に τ を変換するものとして ϑ 3 ( v , τ + 1 ) = ∑ n = − ∞ ∞ e π i ( τ + 1 ) n 2 + 2 π i n v = ∑ n = − ∞ ∞ ( − 1 ) n e π i τ n 2 + 2 π i n v = ϑ 4 ( v , τ ) {\displaystyle \vartheta _{3}\left(v,\tau +1\right)=\sum _{n=-\infty }^{\infty }e^{\pi {i}(\tau +1)n^{2}+2\pi {i}nv}=\sum _{n=-\infty }^{\infty }(-1)^{n}e^{\pi {i}\tau {n^{2}}+2\pi {i}nv}=\vartheta _{4}(v,\tau )} ϑ 3 ( v , τ ) = ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n v = ∑ n = − ∞ ∞ e π i τ ( 2 n ) 2 + 2 π i ( 2 n ) v + ∑ n = − ∞ ∞ e π i τ ( 2 n + 1 ) 2 + 2 π i ( 2 n + 1 ) v ( n ↦ 2 n , 2 n + 1 ) = ϑ 3 ( 2 v , 4 τ ) + ϑ 2 ( 2 v , 4 τ ) {\displaystyle {\begin{aligned}\vartheta _{3}\left(v,\tau \right)&=\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {n^{2}}+2\pi {i}nv}\\&=\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {(2n)^{2}}+2\pi {i}(2n)v}+\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {(2n+1)^{2}}+2\pi {i}(2n+1)v}\qquad ({n}\mapsto {2n,2n+1})\\&=\vartheta _{3}\left(2v,4\tau \right)+\vartheta _{2}\left(2v,4\tau \right)\end{aligned}}} ϑ 4 ( v , τ ) = ∑ n = − ∞ ∞ ( − 1 ) n e π i τ n 2 + 2 π i n v = ∑ n = − ∞ ∞ e π i τ ( 2 n ) 2 + 2 π i ( 2 n ) v − ∑ n = − ∞ ∞ e π i τ ( 2 n + 1 ) 2 + 2 π i ( 2 n + 1 ) v ( n ↦ 2 n , 2 n + 1 ) = ϑ 3 ( 2 v , 4 τ ) − ϑ 2 ( 2 v , 4 τ ) {\displaystyle {\begin{aligned}\vartheta _{4}\left(v,\tau \right)&=\sum _{n=-\infty }^{\infty }(-1)^{n}e^{\pi {i}\tau {n^{2}}+2\pi {i}nv}\\&=\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {(2n)^{2}}+2\pi {i}(2n)v}-\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {(2n+1)^{2}}+2\pi {i}(2n+1)v}\qquad ({n}\mapsto {2n,2n+1})\\&=\vartheta _{3}\left(2v,4\tau \right)-\vartheta _{2}\left(2v,4\tau \right)\end{aligned}}} 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 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 2 ( 0 , τ ) 2 = ∑ m = − ∞ ∞ e π i τ ( m + 1 / 2 ) 2 ∑ n = − ∞ ∞ e π i τ ( n + 1 / 2 ) 2 = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( m + 1 / 2 ) 2 + ( n + 1 / 2 ) 2 ) = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( m + n + 1 ) 2 + ( m − n ) 2 ) / 2 = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( 2 m + 1 ) 2 + ( 2 n ) 2 ) / 2 + ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ e π i τ ( ( 2 m + 2 ) 2 + ( 2 n + 1 ) 2 ) / 2 ( ⌊ m + n 2 ⌋ ↦ m , ⌊ m − n 2 ⌋ ↦ n ) = 2 ϑ 3 ( 0 , 2 τ ) ϑ 2 ( 0 , 2 τ ) {\displaystyle {\begin{aligned}\vartheta _{2}\left(0,\tau \right)^{2}&=\sum _{m=-\infty }^{\infty }e^{\pi {i}\tau {(m+1/2)^{2}}}\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {(n+1/2)^{2}}}=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau {\left((m+1/2)^{2}+(n+1/2)^{2}\right)}}\\&=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((m+n+1)^{2}+(m-n)^{2}\right)/2}\\&=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((2m+1)^{2}+(2n)^{2}\right)/2}+\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }e^{\pi {i}\tau \left((2m+2)^{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})\\&=2\vartheta _{3}\left(0,2\tau \right)\vartheta _{2}\left(0,2\tau \right)\end{aligned}}} これにより ϑ 3 ( 0 , τ ) 4 − ϑ 4 ( 0 , τ ) 4 = ( ϑ 3 ( 0 , 2 τ ) 2 + ϑ 2 ( 0 , 2 τ ) 2 ) 2 − ( ϑ 3 ( 0 , 2 τ ) 2 − ϑ 2 ( 0 , 2 τ ) 2 ) 2 = 4 ϑ 3 ( 0 , 2 τ ) 2 ϑ 2 ( 0 , 2 τ ) 2 = ϑ 2 ( 0 , τ ) 4 {\displaystyle {\begin{aligned}\vartheta _{3}\left(0,\tau \right)^{4}-\vartheta _{4}\left(0,\tau \right)^{4}&=\left(\vartheta _{3}\left(0,2\tau \right)^{2}+\vartheta _{2}\left(0,2\tau \right)^{2}\right)^{2}-\left(\vartheta _{3}\left(0,2\tau \right)^{2}-\vartheta _{2}\left(0,2\tau \right)^{2}\right)^{2}\\&=4\vartheta _{3}\left(0,2\tau \right)^{2}\vartheta _{2}\left(0,2\tau \right)^{2}\\&=\vartheta _{2}\left(0,\tau \right)^{4}\end{aligned}}} 次の恒等式はランデンの公式 (Landen's formula) という。 ϑ 4 ( 0 , 2 τ ) ϑ 4 ( 2 v , 2 τ ) = ϑ 3 ( v , τ ) ϑ 4 ( v , τ ) {\displaystyle \vartheta _{4}(0,2\tau )\vartheta _{4}(2v,2\tau )=\vartheta _{3}(v,\tau )\vartheta _{4}(v,\tau )} ϑ 4 ( 0 , 2 τ ) ϑ 1 ( 2 v , 2 τ ) = ϑ 2 ( v , τ ) ϑ 1 ( v , τ ) {\displaystyle \vartheta _{4}(0,2\tau )\vartheta _{1}(2v,2\tau )=\vartheta _{2}(v,\tau )\vartheta _{1}(v,\tau )} 第一式の右辺を展開すれば ϑ 3 ( v , τ ) ϑ 4 ( v , τ ) = ( ∑ n = − ∞ ∞ e π i τ n 2 + 2 π i n v ) ( ∑ m = − ∞ ∞ ( − 1 ) m e π i τ m 2 + 2 π i m v ) = ∑ n = − ∞ ∞ ∑ m = − ∞ ∞ ( − 1 ) m e π i τ ( n 2 + m 2 ) + 2 π i ( n + m ) v {\displaystyle {\begin{aligned}\vartheta _{3}(v,\tau )\vartheta _{4}(v,\tau )&=\left(\sum _{n=-\infty }^{\infty }e^{\pi {i\tau }n^{2}+2\pi {inv}}\right)\left(\sum _{m=-\infty }^{\infty }(-1)^{m}e^{\pi {i\tau }m^{2}+2\pi {imv}}\right)\\&=\sum _{n=-\infty }^{\infty }\sum _{m=-\infty }^{\infty }(-1)^{m}e^{\pi {i\tau }(n^{2}+m^{2})+2\pi {i(n+m)v}}\\\end{aligned}}} となるが、 n ± m {\displaystyle {n}\pm {m}} が奇数の項は n ≷ m {\displaystyle {n}\gtrless {m}} で打ち消し合うから ϑ 3 ( v , τ ) ϑ 4 ( v , τ ) = ∑ n ′ = − ∞ ∞ ∑ m ′ = − ∞ ∞ ( − 1 ) n ′ − m ′ e 2 π i τ ( n ′ 2 + m ′ 2 ) + 4 π i n ′ v ( n ↦ n ′ + m ′ , m ↦ n ′ − m ′ ) = ( ∑ n ′ = − ∞ ∞ ( − 1 ) n ′ e 2 π i τ n ′ 2 + 4 π i n ′ v ) ( ∑ m ′ = − ∞ ∞ ( − 1 ) m ′ e 2 π i τ m ′ 2 ) = ϑ 4 ( 2 v , 2 τ ) ϑ 4 ( 0 , 2 τ ) {\displaystyle {\begin{aligned}\vartheta _{3}(v,\tau )\vartheta _{4}(v,\tau )&=\sum _{n'=-\infty }^{\infty }\sum _{m'=-\infty }^{\infty }(-1)^{n'-m'}e^{2\pi {i\tau }(n'^{2}+m'^{2})+4\pi {in'v}}\qquad ({n}\mapsto {n'+m'},{m}\mapsto {n'-m'})\\&=\left(\sum _{n'=-\infty }^{\infty }(-1)^{n'}e^{2\pi {i\tau }n'^{2}+4\pi {in'v}}\right)\left(\sum _{m'=-\infty }^{\infty }(-1)^{m'}e^{2\pi {i\tau }m'^{2}}\right)\\&=\vartheta _{4}(2v,2\tau )\vartheta _{4}(0,2\tau )\end{aligned}}} となり、左辺を得る。第二式は第一式に v ′ = v + τ 2 {\displaystyle v'=v+\textstyle {\frac {\tau }{2}}} を代入して得られる。 例えば ϑ 3 ( x + y , τ ) ϑ 3 ( x − y , τ ) = ∑ n = − ∞ ∞ ∑ m = − ∞ ∞ e π i τ n 2 + 2 π i n ( x + y ) + π i τ m 2 + 2 π i m ( x − y ) = ∑ n = − ∞ ∞ ∑ m = − ∞ ∞ e 2 π i τ ( n + m 2 ) 2 + 4 π i ( n + m 2 ) x + 2 π i τ ( n − m 2 ) 2 + 4 π i ( n − m 2 ) y {\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^{{\pi }i{\tau }n^{2}+2{\pi }in(x+y)+{\pi }i{\tau }m^{2}+2{\pi }im(x-y)}\\&=\sum _{n=-\infty }^{\infty }\sum _{m=-\infty }^{\infty }e^{2{\pi }i{\tau }\left({\tfrac {n+m}{2}}\right)^{2}+4{\pi }i\left({\tfrac {n+m}{2}}\right)x+2{\pi }i{\tau }\left({\tfrac {n-m}{2}}\right)^{2}+4{\pi }i\left({\tfrac {n-m}{2}}\right)y}\\\end{aligned}}} 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}}} となる。ここで x ↦ x + 1 2 τ {\displaystyle x\mapsto {x+{\tfrac {1}{2}}\tau }} とすれば ϑ 2 ( x + y , τ ) ϑ 2 ( x − y , τ ) = ϑ 2 ( 2 x , 2 τ ) ϑ 3 ( 2 y , 2 τ ) + ϑ 3 ( 2 x , 2 τ ) ϑ 2 ( 2 y , 2 τ ) {\displaystyle \vartheta _{2}\left(x+y,\tau \right)\vartheta _{2}\left(x-y,\tau \right)=\vartheta _{2}\left(2x,2\tau \right)\vartheta _{3}\left(2y,2\tau \right)+\vartheta _{3}\left(2x,2\tau \right)\vartheta _{2}\left(2y,2\tau \right)} となり、 x ↦ x + 1 2 {\displaystyle x\mapsto {x+{\tfrac {1}{2}}}} とすれば ϑ 4 ( x + y , τ ) ϑ 4 ( x − y , τ ) = 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 \vartheta _{4}\left(x+y,\tau \right)\vartheta _{4}\left(x-y,\tau \right)=\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)} となる。これらにより ϑ 3 ( x + y , τ ) ϑ 3 ( x − y , τ ) ϑ 3 2 ( 0 , τ ) = ( ϑ 3 ( 2 x , 2 τ ) ϑ 3 ( 2 y , 2 τ ) + ϑ 2 ( 2 x , 2 τ ) ϑ 2 ( 2 y , 2 τ ) ) ( ϑ 3 2 ( 0 , 2 τ ) + ϑ 2 2 ( 0 , 2 τ ) ) = ( ϑ 3 ( 2 x , 2 τ ) ϑ 2 ( 0 , 2 τ ) + ϑ 2 ( 2 x , 2 τ ) ϑ 3 ( 0 , 2 τ ) ) ( ϑ 2 ( 2 y , 2 τ ) ϑ 3 ( 0 , 2 τ ) + ϑ 3 ( 2 y , 2 τ ) ϑ 2 ( 0 , 2 τ ) ) + ( ϑ 3 ( 2 x , 2 τ ) ϑ 3 ( 0 , 2 τ ) − ϑ 2 ( 2 x , 2 τ ) ϑ 2 ( 0 , 2 τ ) ) ( ϑ 3 ( 2 y , 2 τ ) ϑ 3 ( 0 , 2 τ ) − ϑ 2 ( 2 y , 2 τ ) ϑ 2 ( 0 , 2 τ ) ) = ϑ 2 2 ( x , τ ) ϑ 2 2 ( y , τ ) + ϑ 4 2 ( x , τ ) ϑ 4 2 ( y , τ ) {\displaystyle {\begin{aligned}\vartheta _{3}\left(x+y,\tau \right)\vartheta _{3}\left(x-y,\tau \right)\vartheta _{3}^{\;2}(0,\tau )&=\left(\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)\right)\left(\vartheta _{3}^{\;2}\left(0,2\tau \right)+\vartheta _{2}^{\;2}\left(0,2\tau \right)\right)\\&=\left(\vartheta _{3}\left(2x,2\tau \right)\vartheta _{2}\left(0,2\tau \right)+\vartheta _{2}\left(2x,2\tau \right)\vartheta _{3}\left(0,2\tau \right)\right)\left(\vartheta _{2}\left(2y,2\tau \right)\vartheta _{3}\left(0,2\tau \right)+\vartheta _{3}\left(2y,2\tau \right)\vartheta _{2}\left(0,2\tau \right)\right)\\&\quad +\left(\vartheta _{3}\left(2x,2\tau \right)\vartheta _{3}\left(0,2\tau \right)-\vartheta _{2}\left(2x,2\tau \right)\vartheta _{2}\left(0,2\tau \right)\right)\left(\vartheta _{3}\left(2y,2\tau \right)\vartheta _{3}\left(0,2\tau \right)-\vartheta _{2}\left(2y,2\tau \right)\vartheta _{2}\left(0,2\tau \right)\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)\end{aligned}}} が得られ、同様にして数十もの恒等式が得られる。 ϑ 1 ( x + y , τ ) ϑ 1 ( x − y , τ ) ϑ 2 2 ( 0 , τ ) = ϑ 1 2 ( x , τ ) ϑ 2 2 ( y , τ ) − ϑ 2 2 ( x , τ ) ϑ 1 2 ( y , τ ) = ϑ 4 2 ( x , τ ) ϑ 3 2 ( y , τ ) − ϑ 3 2 ( x , τ ) ϑ 4 2 ( y , τ ) ϑ 2 ( x + y , τ ) ϑ 2 ( x − y , τ ) ϑ 2 2 ( 0 , τ ) = ϑ 2 2 ( x , τ ) ϑ 2 2 ( y , τ ) − ϑ 1 2 ( x , τ ) ϑ 1 2 ( y , τ ) = ϑ 3 2 ( x , τ ) ϑ 3 2 ( y , τ ) − ϑ 4 2 ( x , τ ) ϑ 4 2 ( y , τ ) ϑ 3 ( x + y , τ ) ϑ 3 ( x − y , τ ) ϑ 2 2 ( 0 , τ ) = ϑ 3 2 ( x , τ ) ϑ 2 2 ( y , τ ) + ϑ 4 2 ( x , τ ) ϑ 1 2 ( y , τ ) = ϑ 1 2 ( x , τ ) ϑ 4 2 ( y , τ ) + ϑ 2 2 ( x , τ ) ϑ 3 2 ( y , τ ) ϑ 4 ( x + y , τ ) ϑ 4 ( x − y , τ ) ϑ 2 2 ( 0 , τ ) = ϑ 4 2 ( x , τ ) ϑ 2 2 ( y , τ ) + ϑ 3 2 ( x , τ ) ϑ 1 2 ( y , τ ) = ϑ 2 2 ( x , τ ) ϑ 4 2 ( y , τ ) + ϑ 1 2 ( x , τ ) ϑ 3 2 ( y , τ ) {\displaystyle {\begin{aligned}&\vartheta _{1}\left(x+y,\tau \right)\vartheta _{1}\left(x-y,\tau \right)\vartheta _{2}^{\;2}(0,\tau )=\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)-\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)-\vartheta _{3}^{\;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 _{2}^{\;2}(0,\tau )=\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)-\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)-\vartheta _{4}^{\;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 _{2}^{\;2}(0,\tau )=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)+\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)+\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)\\&\vartheta _{4}\left(x+y,\tau \right)\vartheta _{4}\left(x-y,\tau \right)\vartheta _{2}^{\;2}(0,\tau )=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)+\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)+\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)\\\end{aligned}}} 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 1 ( x + y , τ ) ϑ 1 ( x − y , τ ) ϑ 4 2 ( 0 , τ ) = ϑ 1 2 ( x , τ ) ϑ 4 2 ( y , τ ) − ϑ 4 2 ( x , τ ) ϑ 1 2 ( y , τ ) = ϑ 3 2 ( x , τ ) ϑ 2 2 ( y , τ ) − ϑ 2 2 ( x , τ ) ϑ 3 2 ( y , τ ) ϑ 2 ( x + y , τ ) ϑ 2 ( x − y , τ ) ϑ 4 2 ( 0 , τ ) = ϑ 2 2 ( x , τ ) ϑ 4 2 ( y , τ ) − ϑ 3 2 ( x , τ ) ϑ 1 2 ( y , τ ) = ϑ 4 2 ( x , τ ) ϑ 2 2 ( y , τ ) − ϑ 1 2 ( x , τ ) ϑ 3 2 ( y , τ ) ϑ 3 ( x + y , τ ) ϑ 3 ( x − y , τ ) ϑ 4 2 ( 0 , τ ) = ϑ 3 2 ( x , τ ) ϑ 4 2 ( y , τ ) − ϑ 2 2 ( x , τ ) ϑ 1 2 ( y , τ ) = ϑ 4 2 ( x , τ ) ϑ 3 2 ( y , τ ) − ϑ 1 2 ( x , τ ) ϑ 2 2 ( y , τ ) ϑ 4 ( x + y , τ ) ϑ 4 ( x − y , τ ) ϑ 4 2 ( 0 , τ ) = ϑ 4 2 ( x , τ ) ϑ 4 2 ( y , τ ) − ϑ 1 2 ( x , τ ) ϑ 1 2 ( y , τ ) = ϑ 3 2 ( x , τ ) ϑ 3 2 ( y , τ ) − ϑ 2 2 ( x , τ ) ϑ 2 2 ( y , τ ) {\displaystyle {\begin{aligned}&\vartheta _{1}\left(x+y,\tau \right)\vartheta _{1}\left(x-y,\tau \right)\vartheta _{4}^{\;2}(0,\tau )=\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;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 _{2}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)\\&\vartheta _{2}\left(x+y,\tau \right)\vartheta _{2}\left(x-y,\tau \right)\vartheta _{4}^{\;2}(0,\tau )=\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;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 _{1}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)\\&\vartheta _{3}\left(x+y,\tau \right)\vartheta _{3}\left(x-y,\tau \right)\vartheta _{4}^{\;2}(0,\tau )=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)-\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)-\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)\\&\vartheta _{4}\left(x+y,\tau \right)\vartheta _{4}\left(x-y,\tau \right)\vartheta _{4}^{\;2}(0,\tau )=\vartheta _{4}^{\;2}\left(x,\tau \right)\vartheta _{4}^{\;2}\left(y,\tau \right)-\vartheta _{1}^{\;2}\left(x,\tau \right)\vartheta _{1}^{\;2}\left(y,\tau \right)=\vartheta _{3}^{\;2}\left(x,\tau \right)\vartheta _{3}^{\;2}\left(y,\tau \right)-\vartheta _{2}^{\;2}\left(x,\tau \right)\vartheta _{2}^{\;2}\left(y,\tau \right)\\\end{aligned}}} ϑ 1 ( x + y , τ ) ϑ 2 ( x − y , τ ) ϑ 3 ( 0 , τ ) ϑ 4 ( 0 , τ ) = ϑ 1 ( x , τ ) ϑ 2 ( x , τ ) ϑ 3 ( y , τ ) ϑ 4 ( y , τ ) + ϑ 3 ( x , τ ) ϑ 4 ( x , τ ) ϑ 1 ( y , τ ) ϑ 2 ( y , τ ) ϑ 1 ( x + y , τ ) ϑ 3 ( x − y , τ ) ϑ 2 ( 0 , τ ) ϑ 4 ( 0 , τ ) = ϑ 1 ( x , τ ) ϑ 3 ( x , τ ) ϑ 2 ( y , τ ) ϑ 4 ( y , τ ) + ϑ 2 ( x , τ ) ϑ 4 ( x , τ ) ϑ 1 ( y , τ ) ϑ 3 ( y , τ ) ϑ 1 ( x + y , τ ) ϑ 4 ( x − y , τ ) ϑ 3 ( 0 , τ ) ϑ 2 ( 0 , τ ) = ϑ 1 ( x , τ ) ϑ 4 ( x , τ ) ϑ 3 ( y , τ ) ϑ 2 ( y , τ ) + ϑ 3 ( x , τ ) ϑ 2 ( x , τ ) ϑ 1 ( y , τ ) ϑ 4 ( y , τ ) {\displaystyle {\begin{aligned}&\vartheta _{1}\left(x+y,\tau \right)\vartheta _{2}\left(x-y,\tau \right)\vartheta _{3}(0,\tau )\vartheta _{4}(0,\tau )=\vartheta _{1}\left(x,\tau \right)\vartheta _{2}\left(x,\tau \right)\vartheta _{3}(y,\tau )\vartheta _{4}(y,\tau )+\vartheta _{3}\left(x,\tau \right)\vartheta _{4}\left(x,\tau \right)\vartheta _{1}(y,\tau )\vartheta _{2}(y,\tau )\\&\vartheta _{1}\left(x+y,\tau \right)\vartheta _{3}\left(x-y,\tau \right)\vartheta _{2}(0,\tau )\vartheta _{4}(0,\tau )=\vartheta _{1}\left(x,\tau \right)\vartheta _{3}\left(x,\tau \right)\vartheta _{2}(y,\tau )\vartheta _{4}(y,\tau )+\vartheta _{2}\left(x,\tau \right)\vartheta _{4}\left(x,\tau \right)\vartheta _{1}(y,\tau )\vartheta _{3}(y,\tau )\\&\vartheta _{1}\left(x+y,\tau \right)\vartheta _{4}\left(x-y,\tau \right)\vartheta _{3}(0,\tau )\vartheta _{2}(0,\tau )=\vartheta _{1}\left(x,\tau \right)\vartheta _{4}\left(x,\tau \right)\vartheta _{3}(y,\tau )\vartheta _{2}(y,\tau )+\vartheta _{3}\left(x,\tau \right)\vartheta _{2}\left(x,\tau \right)\vartheta _{1}(y,\tau )\vartheta _{4}(y,\tau )\\\end{aligned}}} x = y = z とすれば ϑ 1 ( 2 z , τ ) ϑ 2 ( 0 , τ ) ϑ 3 ( 0 , τ ) ϑ 4 ( 0 , τ ) = 2 ϑ 1 ( z , τ ) ϑ 2 ( z , τ ) ϑ 3 ( z , τ ) ϑ 4 ( z , τ ) ϑ 2 ( 2 z , τ ) ϑ 2 3 ( 0 , τ ) = ϑ 2 4 ( z , τ ) − ϑ 1 4 ( z , τ ) = ϑ 3 4 ( z , τ ) − ϑ 4 4 ( z , τ ) ϑ 3 ( 2 z , τ ) ϑ 3 3 ( 0 , τ ) = ϑ 3 4 ( z , τ ) + ϑ 1 4 ( z , τ ) = ϑ 2 4 ( z , τ ) + ϑ 4 4 ( z , τ ) ϑ 4 ( 2 z , τ ) ϑ 4 3 ( 0 , τ ) = ϑ 4 4 ( z , τ ) − ϑ 1 4 ( z , τ ) = ϑ 3 4 ( z , τ ) − ϑ 2 4 ( z , τ ) {\displaystyle {\begin{aligned}&\vartheta _{1}\left(2z,\tau \right)\vartheta _{2}(0,\tau )\vartheta _{3}(0,\tau )\vartheta _{4}(0,\tau )=2\vartheta _{1}\left(z,\tau \right)\vartheta _{2}(z,\tau )\vartheta _{3}(z,\tau )\vartheta _{4}(z,\tau )\\&\vartheta _{2}\left(2z,\tau \right)\vartheta _{2}^{\;3}(0,\tau )=\vartheta _{2}^{\;4}\left(z,\tau \right)-\vartheta _{1}^{\;4}\left(z,\tau \right)=\vartheta _{3}^{\;4}\left(z,\tau \right)-\vartheta _{4}^{\;4}\left(z,\tau \right)\\&\vartheta _{3}\left(2z,\tau \right)\vartheta _{3}^{\;3}(0,\tau )=\vartheta _{3}^{\;4}\left(z,\tau \right)+\vartheta _{1}^{\;4}\left(z,\tau \right)=\vartheta _{2}^{\;4}\left(z,\tau \right)+\vartheta _{4}^{\;4}\left(z,\tau \right)\\&\vartheta _{4}\left(2z,\tau \right)\vartheta _{4}^{\;3}(0,\tau )=\vartheta _{4}^{\;4}\left(z,\tau \right)-\vartheta _{1}^{\;4}\left(z,\tau \right)=\vartheta _{3}^{\;4}\left(z,\tau \right)-\vartheta _{2}^{\;4}\left(z,\tau \right)\\\end{aligned}}} などが得られ、更に z = 0 とすれば ϑ 3 4 ( 0 , τ ) = ϑ 2 4 ( 0 , τ ) + ϑ 4 4 ( 0 , τ ) {\displaystyle \vartheta _{3}^{\;4}(0,\tau )=\vartheta _{2}^{\;4}\left(0,\tau \right)+\vartheta _{4}^{\;4}\left(0,\tau \right)} が得られる。 無限乗積表示 ϑ 1 ( v , τ ) = 2 e π i τ / 4 sin ⁡ π v ∏ m = 1 ∞ ( 1 − e 2 m π i τ ) ( 1 − e 2 m π i τ e 2 π i v ) ( 1 − e 2 m π i τ e − 2 π i v ) {\displaystyle \vartheta _{1}(v,\tau )=2e^{{\pi }i{\tau }/4}\sin {\pi }v\prod _{m=1}^{\infty }{\left(1-e^{2m{\pi }i{\tau }}\right)\left(1-e^{2m{\pi }i{\tau }}e^{2{\pi }iv}\right)\left(1-e^{2m{\pi }i{\tau }}e^{-2{\pi }iv}\right)}} の対数微分により 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}}} である。同様に ϑ 2 ′ ( v , τ ) ϑ 2 ( v , τ ) = − π tan ⁡ π v + 4 π ∑ n = 1 ∞ ( − 1 ) n e 2 n π i τ 1 − e 2 n π i τ sin ⁡ 2 π n v ϑ 3 ′ ( v , τ ) ϑ 3 ( v , τ ) = 4 π ∑ n = 1 ∞ ( − 1 ) n e n π i τ 1 − e 2 n π i τ sin ⁡ 2 π n v ϑ 4 ′ ( v , τ ) ϑ 4 ( v , τ ) = 4 π ∑ n = 1 ∞ e n π i τ 1 − e 2 n π i τ sin ⁡ 2 π n v {\displaystyle {\begin{aligned}{\frac {\vartheta _{2}'(v,\tau )}{\vartheta _{2}(v,\tau )}}&=-\pi \tan \pi {v}+4\pi \sum _{n=1}^{\infty }{\frac {{(-1)^{n}}e^{2n\pi {i\tau }}}{1-e^{2n\pi {i\tau }}}}\sin 2\pi {nv}\\{\frac {\vartheta _{3}'(v,\tau )}{\vartheta _{3}(v,\tau )}}&=4\pi \sum _{n=1}^{\infty }{\frac {{(-1)^{n}}e^{n\pi {i\tau }}}{1-e^{2n\pi {i\tau }}}}\sin 2\pi {nv}\\{\frac {\vartheta _{4}'(v,\tau )}{\vartheta _{4}(v,\tau )}}&=4\pi \sum _{n=1}^{\infty }{\frac {e^{n\pi {i\tau }}}{1-e^{2n\pi {i\tau }}}}\sin 2\pi {nv}\\\end{aligned}}} である。