# Minimum Radius of convergence of a power series in Differential equations

If $x_0$ is an ordinary point of the differential equation $(1)$, we can always ﬁnd two linearly independent solutions in the form of a power series centered at $x_0$.
A power series solution converges at least on some interval deﬁned by $|x-x_0|< R$ , where $R$ is the distance from $x_0$ to the closest singular point.
A solution of the form is said to be a solution about the ordinary point $x_0$. The distance $R$ in Theorem 6.2.1 is the minimum value or lower bound for the radius of convergence.

This passage is taken from my Differential equation book and I just want to focus on the last small but confusing adjective MINIMUM VALUE used in the last quoted sentence , now it is true that if we want to center a given power series solution of a DE at a point $x_0$ we want $x_0$ to be an ordinary points , fine till here .

And we dont want any ordinary points in our radius of convergence which explains $x-x_0 < R$ , BUT ! shouldn't $R$ be called the maximum radius of convergence ? How is it considered minimum ? These definitions just keep me confused

Under certain circunstances it may happen that the actual radius of convergence of the solution is greater than that minimum value. Consider the equation $$(1-x)\,y''+y'+y=0.$$ $x=0$ is a regular (or ordinary) point, and $x=1$ is a singular point. Any power series solution centered at $0$ has a radius of convergence at least $1$. But $y(x)=0$ is a solution, whose power series around $x=0$ has a radius of convergence $\infty$.
• Ups! For a trivial example consider the solution $y=0$. A less trivial example is $(1-x)\,y''+x\,y'-y=0$ and $y=x$. Mar 10 '16 at 18:21