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  "body": "\n## Definition\n\nLet $p_1, \\ldots, p_n$ be meromorphic functions near $z_0 \\in \\C$, and consider the [[linear equation|0QQRP17Z7S]]\n$$y^{(n)} + p_1(z)\\, y^{(n-1)} + \\cdots + p_n(z)\\, y = 0.$$\nThe point $z_0$ is a *regular singular point* of the equation if:\n\n1. $z_0$ is a [[singular point|QK2J038VB4]] of the equation, and\n2. for each $k = 1, \\ldots, n$, the function $(z - z_0)^k\\, p_k(z)$ is [[analytic|081WKEDK6C]] at $z_0$.\n\nThe point $\\infty$ is a regular singular point if $w = 0$ is one for the equation obtained by the substitution $z = 1/w$.\n\n## Remarks\n\nFor $n = 2$, written $y'' + p(z) y' + q(z) y = 0$, condition 2 says that $(z - z_0) p(z)$ and $(z - z_0)^2 q(z)$ are analytic at $z_0$.\nFor $n = 2$, the point $\\infty$ is ordinary or regular singular if and only if $z\\, p(z)$ and $z^2 q(z)$ are analytic at $\\infty$.\n\nExample: the [[Cauchy–Euler equation|03F7N3915D]] $z^2 y'' + a z y' + b y = 0$ with $(a, b) \\ne (0, 0)$ has $p = \\frac{a}{z}$ and $q = \\frac{b}{z^2}$.\nSo $0$ is singular, while $z p = a$ and $z^2 q = b$ are constant; hence $0$ is a regular singular point.\n\nNon-example: for $z^2 y'' + y' = 0$, one has $p = \\frac{1}{z^2}$, and $z p = \\frac{1}{z}$ is not analytic at $0$.\nSo $0$ is singular but not regular singular; it is an irregular singular point.\n\nFuchs proved a characterisation by growth.\nA singular point is regular singular if and only if every solution grows at most like a power of $\\abs{z - z_0}^{-1}$ as $z \\to z_0$ in each sector with vertex $z_0$.\n",
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