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### Radius of Convergence

#### Discussion

Consider the linear homogenous variable coefficient initial value problem $$ \begin{align} xy'' + y' + y &= 0 \\ y(0) &= -1 \\ y'(0) &= 1 \end{align} $$ Suppose we try to solve for $y''(0)$ as we did in section 1. If we plug $x=0$ and the initial values into the problem we find $$ 0 + 1 + -1 = 0 $$ which is certainly true, but gives us no hint about the value of $y''(0)$. The trouble is that the coefficient of $y''$ (which is $x$) is 0 at $x=0$. A point where the coefficient of $y''$ is $0$ is called a singular point of the differential equation. More generally, we can define singular points as follows.
*Definition:* Let $p(x)$, $q(x)$ and $r(x)$ be analytic functions.
A point $x_0$ is a **singular point** for the differential
equation $p(x)y'' + q(x)y' + r(x)y=0$ if either $q(x)/p(x)$ or $r(x)/p(x)$ is
undefined at $x=x_0$. A point which is not a singular point is an
**ordinary point**.

A function is analytic in an interval if its Taylor series converges to the function in that interval. Don't let that definition bother you. Any polynomial or rational function is analytic, as are the exponential, log, and trigonometric functions, away from their singularities. I will not try to trick you by using non-analytic functions in this class. It should be noted that if $p(x)$, $q(x)$ and $r(x)$ are all polynomials, then $x_0$ will only be a singular point if $p(x_0)=0$.

Example: $x^2 y'' + xy' = 0$ has a singular point at $x=0$ since $x/x^2=1/x$ is undefined at $x=0$.

Example: $2y''+y/(x-1)=0$ has a singular point at $x=1$ since $(1/(x-1))/2=1/(2x-2)$ is undefined at $x=1$.

The techniques we learned in the last section will work for ordinary points but will not work for singular points. We will study solutions about singular points in the next several sections after this one.

So far, we have not worried about where a power series solution to a differential equation is valid in this class. We have just gone through the manipulations to find the solution and then assumed what we found was actually a solution. But of course, the manipulations we went through are only valid inside the radius of convergence of the power series. So now we will take up the question of how large the radius of convergence of our series solution will be. Of course, we already have some examples to build on from the labs. Fortunately, we don't have to try to remember all the old tests for deciding on the radius of convergence of a power series from calculus. Instead, we have the following theorem.

*Theorem:* Suppose
$\displaystyle y(x) = \sum_{n=0}^{\infty}a_n(x-x_0)^n$
is the series solution to the
differential equation $p(x)y'' + q(x)y' + r(x)y = 0$ where $p(x)$, $q(x)$
and
$r(x)$ are all polynomials and $x_0$ is an ordinary point.
Then
the radius of convergence for $y(x)$ is at least as large as the distance from
$x_0$ to the nearest singular point of the equation.

NOTE: We must consider complex $x$, even though the equation is real.

#### Paradigm

What is the radius of convergence of the series solution to $(x+1)y'' + y' + (x+1)y = 0$ about the point $x_0 = 0$?
*Step 1:*
Check the series is expanded about an ordinary point (so the
theorem applies).

Since all the coefficients are polynomials and $p(0)=0+1=1\ne0$, 0 is an ordinary point and the above theorem applies.

*Step 2:* Compute all singular points.

The only singular point is $-1$.

*Step 3:* Find the distances from the singular points to center of
expansion.

$-1$ is a distance of 1 away from 0

*Step 4:* The radius of convergence is at least as large as the
smallest
distance from a singular point to the center of expansion of the series.

The radius of convergence is at least 1.

EXAMPLE: What is the radius of convergence of the series solution to $ (x^2 - 2x - 3)y'' + (x+1)y' + 3y = 0$ about the point $x_0 = 0$?

Step 1: Since all coefficients are polynomial and $p(0)=0+0-3=-3\ne 0$, 0 is an ordinary point and the above theorem applies.

Step 2: The two singular points are $-1$ and $3$.

Step 3: $-1$ is 1 unit from 0 and 3 is 3 units from 0.

Step 4: $-1$ is the singular point closest to 0 and it is a distance 1 away. The radius of convergence is at least 1.

EXAMPLE: What is the radius of convergence of the series solution to $(x^3 + x^2 + x + 1)y'' + 2xy' + (x-1)y = 0$ about the point $x_0 = 1$?

Step 1: Since all coefficients are polynomial and $p(1)=1+1+1+1=4\ne0$, 1 is an ordinary point and the above theorem applies.

Step 2: The three singular points are $-1$, $i$ and $-i$.

Step 3: $-1$ is 2 away from 1 while $\pm i$ are $\sqrt2<2$ away from 1.

Step 4: The radius of convergence is at least $\sqrt2$.

This is a very powerful theorem because we can decide where a power series solution will be valid even before we have found the series. The hypothesis can be weakened somewhat ($p$, $q$, $r$ analytic instead of polynomial), but this leads to more technical questions than we need to consider. The easiest mistake to make is to forget to consider complex singular points, but complex analysis is used in the proof and so complex points must be considered. In fact, if you go back to the labs and consider the Taylor series expansion for $1/(1+x^2 )$ about $x_0=0$ you will see that the Taylor series had a radius of convergence of 1, corresponding to the singularities at $\pm i$ which are a distance 1 from 0. The other point to remember is that this just gives a lower bound on the radius of convergence. It is possible for the radius of convergence to be larger than this test indicates. But that is fairly rare in practice; this test works out pretty accurately.

If you have any problems with this page, please contact bennett@math.ksu.edu.

©2010, 2014 Andrew G. Bennett