]> 30.4 Power Series and Radius of Convergence

## 30.4 Power Series and Radius of Convergence

Suppose we have a power series in the variable $x$ .

If it converges for some value of $x$ , it will converge (by the comparison test) for any smaller value of $x$ .

Thus the series will converge up to some maximum value of $x$ , for which the ratio of successive terms becomes 1.

The maximum value of $| x |$ at which the series converges is called its radius of convergence.

It is obvious that the same series represents a convergent and infinitely differentiable function for all values of $x$ whose absolute value is strictly less than its radius of convergence.

On the other hand, there is generally a value of $x$ in the complex plane at a distance given by the radius of convergence from the origin, at which the series is singular.

Thus, the radius of convergence of a series represents the distance in the complex plane from the expansion point to the nearest singularity of the function expanded.

For example, the geometric series in $x$ (the series for $( 1 − x ) − 1$ ) blows up at $x = 1$ and 1 is its radius of convergence, and this behavior is typical of all power series.

This same function, $( 1 − x ) − 1$ can be expanded in a power series about argument -1. Since the distance between -1 and the singular point at $x = 1$ is 2, this series will have radius of convergence 2

$( 1 − x ) − 1 = 1 2 + ( x + 1 ) 4 + ( x + 1 ) 2 8 + ( x + 1 ) 3 16 + ⋯$

Exercises:

30.6 Prove this statement by subtracting $1 2$ from the right hand side and dividing the result by $x + 1 2$ , and rearranging the resulting statement.

30.7 Figure out the comparable series for the same function expanded in powers of $x + 3$ .

30.8 What is the radius of convergence of the exponential series expanded about the origin?

Another nice feature about power series is that if you start with the function $f$ , you can deduce its series expansion about the point $z$ by Taylor's theorem. $f$ will have the expansion

$f ( x ) = ∑ f ( k ) ( z ) ( x − z ) k k !$

where $f ( k ) ( z )$ is the k-th derivative of $f$ at argument $z$ , and the sum is from $k = 0$ on.

Exercise 30.9 Find the series expansion for $( 1 − x ) − 1$ expanded about $x = − 3$ by using Taylor's theorem, that is, by computing its derivatives there.