Radius of convergence

From Wikipedia, the free encyclopedia

In mathematics, the radius of convergence of a power series is a non-negative quantity, either a real number or $\infty,$ that represents a range (within the radius) in which the function will converge.

1 Definition
2 Finding the radius of convergence
3 Clarity and simplicity result from complexity
- 3.1 A simple example
- 3.2 A more complicated example
4 Convergence on the "circle of convergence"
5 A graphical example
6 Abscissa of convergence of a Dirichlet series
7 References
8 External links

[edit] Definition

For a power series f defined as:

$f(z) = \sum_{n=0}^\infty c_n (z-a)^n,$

where

a is a constant, the center of the circle of convergence,

c_n is the n^th complex coefficient (note that real numbers are a very common special case of complex numbers), and

z is a variable.

The radius of convergence r is a nonnegative real number or $\infty$ , such that the series converges if

$|z-a| <r \!$

and diverges if

$|z-a| >r. \!$

In other words, the series converges if z is close enough to the center and diverges if it is too far away. The radius of convergence specifies how close is close enough. The radius of convergence is infinite if the series converges for all complex numbers z.

[edit] Finding the radius of convergence

The radius of convergence can be found by applying the root test to the terms of the series. The root test uses the number

$C = \limsup_{n\rightarrow\infty}\sqrt[n]{|f_n|}$

where f_n is the nth term c_n(z − a)ⁿ ("lim sup" denotes the limit superior). The root test states that the series converges if |C| < 1 and diverges if |C| > 1. It follows that the power series converges if the distance from z to the center a is less than

$r = \frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|c_n|}}$

and diverges if the distance exceeds that number. Note that r = 1/0 is interpreted as an infinite radius, meaning that f is an entire function.

The limit involved in the ratio test is usually easier to compute, but the limit may fail to exist, in which case the root test should be used. The ratio test uses the limit

$L = \lim_{n\rightarrow\infty}\left|\frac{f_{n+1}}{f_n}\right|.$

In the case of a power series, this can be used to find that

$r = \lim_{n\rightarrow\infty} \left| \frac{c_n}{c_{n+1}} \right|.$

[edit] Clarity and simplicity result from complexity

One of the best examples of clarity and simplicity following from thinking about complex numbers where confusion would result from thinking about real numbers is this theorem of complex analysis:

The radius of convergence is always equal to the distance from the center to the nearest point where the function f has a (non-removable) singularity; if no such point exists then the radius of convergence is infinite.

The nearest point means the nearest point in the complex plane, not necessarily on the real line, even if the center and all coefficients are real. See holomorphic functions are analytic; the result stated above is a by-product of the proof found in that article.

[edit] A simple example

The arctangent function of trigonometry can be expanded in a power series familiar to calculus students:

$\arctan(z)=z-\frac{z^3}{3}+\frac{z^5}{5}-\frac{z^7}{7}+\cdots .$

It is easy to apply the ratio test in this case to find that the radius of convergence is 1. But we can also view the matter thus:

$\frac{d}{dz}\arctan(z)=\frac{1}{1+z^2}$

and a zero appears in the denominator when z² = − 1, i.e., when z = i or − i. The center in this power series is at 0. The distance from 0 to either of these two singularities is 1. That is therefore the radius of convergence.

[edit] A more complicated example

Consider this power series:

$\frac{z}{e^z-1}=\sum_{n=0}^\infty B_n z^n$

where the coefficients B_n are the Bernoulli numbers. It may be cumbersome to try to apply the ratio test to find the radius of convergence of this series. But the theorem of complex analysis stated above quickly solves the problem. At z = 0, there is in effect no singularity since the singularity is removable. The only non-removable singularities are therefore located at the other points where the denominator is zero. We solve

e z - 1 = 0

by recalling that if z = x + iy and e^iy = cos(y) + i sin(y) then

e z = e x e i y = e x (cos(y) + i sin(y)),

and then take x and y to be real. Since y is real, the absolute value of cos(y) + i sin(y) is necessarily 1. Therefore, the absolute value of e^z can be 1 only if e^x is 1; since x is real, that happens only if x = 0. Therefore we need cos(y) + i sin(y) = 1. Since y is real, that happens only if cos(y) = 1 and sin(y) = 0, so that y is an integral multiple of 2π. Since the real part x is 0 and the imaginary part y is a nonzero integral multiple of 2π, the solution of our equation is

z = a nonzero integral multiple of 2πi.

The singularity nearest the center (the center is 0 in this case) is at 2πi or − 2πi. The distance from the center to either of those points is 2π. That is therefore the radius of convergence.

[edit] Convergence on the "circle of convergence"

The circle of convergence of a power series is the set of points in the complex plane at distance $r$ from the point the series is expanded around, where $r$ is the radius of convergence. It is of course a circle. A power series may diverge at every point on the circle of convergence, or diverge on some points of the circle and converge at other points, or, the power series may converge at every point of the circle.

Example 1: The power series for the function $f (z) = (1 - z) - 1,$ expanded around $z = 0,$ has radius of convergence 1 and diverges at every point on the circle of convergence.

Example 2: The power series for $g (z) = ln(1 - z)$ has radius of convergence $r = 1$ expanded around $z = 0,$ and diverges for $z = 1$ but converges for all other points on the circle of convergence. $f (z)$ in Example 1 is the derivative of the negative of $g (z).$

A graph of the functions explained in the text : approximations in red and blue, circle of convergence in white

Example 3: The power series

$\sum_{n=2}^\infty \frac{1}{(-n)(n-1)} z^n$

has radius of convergence 1 and converges everywhere on the circle. If $h (z)$ is the function represented by this series, then the derivative of $h (z)$ is $g (z)$ in Example 2.