テータ関数
&=\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{align}
\begin{align}\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{align}
\begin{align}\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{align}
\begin{align}\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{align}
次の恒等式はランデンの公式 (Landen's formula) という。
\vartheta_4(0,2\tau)\vartheta_4(2v,2\tau)=\vartheta_3(v,\tau)\vartheta_4(v,\tau)
\vartheta_4(0,2\tau)\vartheta_1(2v,2\tau)=\vartheta_2(v,\tau)\vartheta_1(v,\tau)
\begin{align}\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{align}
となるが、{n}\pm{m} が奇数の項は {n}\gtrless{m} で打ち消し合うから
\begin{align}\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{align}
となり、左辺を得る。第二式は第一式に v'=v+\textstyle\frac{\tau}{2} を代入して得られる。
\begin{align}\vartheta_3\left(x+y,\tau\right)\vartheta_3\left(x-y,\tau\right)
&=\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}e^\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{align}
であるが、n{\pm}m は共に偶数か共に奇数であるから、N=\lfloor\tfrac{n+m}{2}\rfloor,M=\lfloor\tfrac{n-m}{2}\rfloor とすれば
\begin{align}\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{align}
となる。ここで x\mapsto{x+\tfrac{1}{2}\tau} とすれば
\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\mapsto{x+\tfrac{1}{2 とすれば
\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
\begin{align}\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{align}
が得られ、同様にして数十もの恒等式が得られる。
\begin{align}
&\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{align}
\begin{align}
&\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{align}
\begin{align}
&\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{align}
\begin{align}
&\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{align}
x  y  z とすれば
\begin{align}
&\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)\\