Bessel's equation

Example

Let $\nu \in \C$. Bessel's equation of order $\nu$ is $$z^2 y'' + z y' + (z^2 - \nu^2)\, y = 0.$$ In normalised form, $p(z) = \frac{1}{z}$ and $q(z) = 1 - \frac{\nu^2}{z^2}$. Then $z p(z) = 1$ and $z^2 q(z) = z^2 - \nu^2$ are analytic at $0$, so $0$ is a regular singular point. The indicial equation is $r (r - 1) + r - \nu^2 = r^2 - \nu^2 = 0$, with exponents $r = \pm \nu$. If $\nu$ is not a negative integer, the Frobenius method with $r = \nu$ gives the recurrence $\left( (\nu + k)^2 - \nu^2 \right) a_k = - a_{k-2}$, that is, $k (k + 2 \nu)\, a_k = - a_{k-2}$, with $a_1 = 0$. With $a_0 = \frac{1}{2^\nu \Gamma(\nu + 1)}$ it gives the Bessel function of the first kind $$J_\nu(z) = \sum_{m \ge 0} \frac{(-1)^m}{m!\, \Gamma(m + \nu + 1)} \left( \frac{z}{2} \right)^{2m + \nu},$$ where $\Gamma$ is the gamma function. The series converges for all $z$, since the coefficients of the equation are analytic on $\C \setminus \set{0}$. At $\infty$, the function $z^2 q(z) = z^2 - \nu^2$ is not bounded, so $\infty$ is an irregular singular point.