ν-阶贝塞尔方程
\[z^2u''(z) + zu'(z) + (z^2 - \nu^2)u(z) = 0, \quad \nu \neq \frac{m}{2}, \quad m \in \mathbb{Z} \]\[p(z) = \frac{1}{z}, \quad q(z) = 1 - \frac{\nu^2}{z^2} \]\(x_0 = 0\) — 方程的正则奇点
设 \(u(z)\) 展开的最低次幂为 \(s\) 次 \(u(z) = \sum_{k=0}^{\infty} a_k(z - x_0)^{s+k}\) 代入方程
\[\sum_{k=0}^{\infty} (s + k)(s + k - 1)a_k z^{s+k} + \sum_{k=0}^{\infty} (s + k)a_k z^{s+k} + (z^2 - \nu^2)\sum_{k=0}^{\infty} a_k z^{s+k} = 0 \]指标方程 — \(s(s - 1) + s - \nu^2 = 0\) \(\rightarrow s_1 = \nu \quad s_2 = -\nu\)
\[u_1(z) = \sum_{k=0}^{\infty} a_k(z - x_0)^{\nu+k} \qquad u_2(z) = \sum_{k=0}^{\infty} b_k(z - x_0)^{-\nu+k} \]\(s = \nu\)
\[[(k + \nu)(k + \nu - 1) + (k + \nu) - \nu^2] a_k + a_{k-2} = 0 \]\[a_k = -\frac{1}{k(k + 2\nu)} a_{k-2} \]\[[(\nu + 1)^2 - \nu^2] a_1 = 0 \rightarrow a_1 = 0 \]\[u_1(z) = \sum_{n=0}^{\infty} a_{2n} z^{\nu+2n} \]收敛半径 — \(R^2 = \lim_{k \to \infty} \frac{a_{k-2}}{a_k} = \infty\)
令 \(a_0 = \frac{1}{2^\nu \Gamma(\nu + 1)} \rightarrow u_1(z) \equiv J_\nu(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \Gamma(n + \nu + 1)} \left(\frac{z}{2}\right)^{2n+\nu}\)
上式称为ν-阶贝塞尔函数
\(s = -\nu\)，同理
\[u_2(z) \equiv J_{-\nu}(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \Gamma(n - \nu + 1)} \left(\frac{z}{2}\right)^{2n-\nu} \]上式称为(-ν)-阶贝塞尔函数
ν-阶诺伊曼函数（由上述贝塞尔函数的一种比较好用而特殊的线性组合）
\[N_\nu(z) \equiv \cot \nu\pi J_\nu(z) - \csc \nu\pi J_{-\nu}(z) \] 标签：4.5,infty,frac,求解,sum,贝塞尔,quad,nu From： https://www.cnblogs.com/RES-HON/p/18546418