Home  18.013A  Chapter 30 


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 $\leftx\right$ 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 ${(1x)}^{1}$ ) blows up at $x=1$ and 1 is its radius of convergence, and this behavior is typical of all power series.
This same function, ${(1x)}^{1}$ 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
Exercises:
30.6 Prove this statement by subtracting $\frac{1}{2}$ from the right hand side and dividing the result by $\frac{x+1}{2}$ , 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
where ${f}^{(k)}(z)$ is the kth derivative of $f$ at argument $z$ , and the sum is from $k=0$ on.
Exercise 30.9 Find the series expansion for ${(1x)}^{1}$ expanded about $x=3$ by using Taylor's theorem, that is, by computing its derivatives there.
