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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

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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 $\infty$.

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  • $\begingroup$ I understood your reasoning but isn't this particular DE example wrong ? If you substitute y=1 you get 1=0 ? $\endgroup$
    – Mike Harb
    Mar 10 '16 at 17:47
  • $\begingroup$ Ups! For a trivial example consider the solution $y=0$. A less trivial example is $(1-x)\,y''+x\,y'-y=0$ and $y=x$. $\endgroup$ Mar 10 '16 at 18:21

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