14.6.3 $\theta$ 函数

可以用 $\theta$ 函数 (theta functions) 来计算雅可比函数:

$$\begin{array}{}\text{(14.113a)}& {\vartheta }_1\left(z,q\right)=2q^{\frac{1}{4}}\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{q}^{n\left(n+1\right)}\sin \left(2n+1\right)z,\end{array}$$$$\begin{array}{}\text{(14.113b)}& {\vartheta }_2\left(z,q\right)=2q^{\frac{1}{4}}\sum _{n=0}^{\mathrm{\infty }}{q}^{n\left(n+1\right)}\cos \left(2n+1\right)z,\end{array}$$$$\begin{array}{}\text{(14.113c)}& {\vartheta }_3\left(z,q\right)=1+2\sum _{n=1}^{\mathrm{\infty }}{q}^{n^2}\cos 2nz,\end{array}$$$$\begin{array}{}\text{(14.113d)}& {\vartheta }_4\left(z,q\right)=1+2\sum _{n=1}^{\mathrm{\infty }}{(-1)}^n{q}^{n^2}\cos 2nz.\end{array}$$

如果 $\left|q\right|<1$ ( $q$ 是复数),则级数 $\left(14.113\mathrm{a}\right)\sim \left(14.113\mathrm{d}\right)$ 对于每个复变量 $z$ 都是收敛的. 当 $q$ 是常数的情形,用下述简单的记号:

$$\begin{array}{}\text{(14.114)}& {\vartheta }_k\left(z\right)\text{:=}{\vartheta }_k\left(\pi z,q\right)\left(k=1,2,3,4\right).\end{array}$$

此时, 雅可比函数有表达式:

$$\begin{array}{}\text{(14.115a)}& \mathrm{sn}z=2K\frac{{\vartheta }_4\left(0\right)}{{\vartheta }_1^{\prime }\left(0\right)}\frac{{\vartheta }_1\left(\frac{z}{2K}\right)}{{\vartheta }_4\left(\frac{z}{2K}\right)},\end{array}$$$$\begin{array}{}\text{(14.115b)}& \mathrm{cn}z=\frac{{\vartheta }_4\left(0\right)}{{\vartheta }_2\left(0\right)}\frac{{\vartheta }_2\left(\frac{z}{2K}\right)}{{\vartheta }_4\left(\frac{z}{2K}\right)},\end{array}$$$$\begin{array}{}\text{(14.115c)}& \mathrm{dn}z=\frac{{\vartheta }_4\left(0\right)}{{\vartheta }_3\left(0\right)}\frac{{\vartheta }_3\left(\frac{z}{2K}\right)}{{\vartheta }_4\left(\frac{z}{2K}\right)},\end{array}$$

其中

$$\begin{array}{}\text{(14.115d)}& q=\exp \left(-\pi \frac{K^{\prime }}{K}\right),k={\left(\frac{{\vartheta }_2\left(0\right)}{{\vartheta }_3\left(0\right)}\right)}^2,\end{array}$$

以及 $K,K^{\prime }$ 如 (14.109) 所述.