]> 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.