Definitions

# circle of convergence

In mathematics, the radius of convergence of a power series is a non-negative quantity, either a real number or $scriptstyle infty,$ that represents a range (within the radius) in which the series will converge. If the series converges, it is the Taylor series for the function to which it converges inside its radius of convergence.

## Definition

For a power series f defined as:

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

where

a is a constant, the center of the disk of convergence,
cn is the nth complex coefficient, and
z is a variable.

The radius of convergence r is a nonnegative real number or $scriptstyle infty$ 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. On the boundary, that is, where |z - a| = r, the behavior of the power series may be complicated, and the series may converge for some values of z and diverge for others. The radius of convergence is infinite if the series converges for all complex numbers z.

## 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

$C = limsup_\left\{nrightarrowinfty\right\}sqrt\left[n\right]$
= limsup_{nrightarrowinfty}sqrt[n]{|c_n|z-a|>

"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

$r = frac\left\{1\right\}\left\{limsup_\left\{nrightarrowinfty\right\}sqrt\left[n\right]$>}

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

The limit involved in the ratio test is usually easier to compute, and when that limit exists, it shows that

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

## Radii of convergence in complex analysis

A power series with a positive radius of convergence can be made into a holomorphic function by taking its argument to be a complex variable. The radius of convergence can be characterized by the following theorem:
The radius of convergence of a power series f centered on a point a is equal to the distance from a to the nearest point where f cannot be defined in a way that makes it holomorphic.
The set of all points whose distance to a is strictly less than the radius of convergence is called the disk of convergence.

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. For example, the function

$f\left(z\right)=frac\left\{1\right\}\left\{1+z^2\right\}$
has no real roots. Its Taylor series about 0 is given by
$sum_\left\{n=0\right\}^infty \left(-1\right)^n z^\left\{2n\right\}$
The root test shows that its radius of convergence is 1. In accordance with this, the function $f\left(z\right)$ has singularities at ±i, which are at a distance 1 from 0.

For a proof of this theorem, see holomorphic functions are analytic.

### A simple example

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

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

It is easy to apply the root test in this case to find that the radius of convergence is 1.

### A more complicated example

Consider this power series:

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

where the rational numbers Bn 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

$e^z = e^x e^\left\{iy\right\} = e^x\left(cos\left(y\right)+isin\left(y\right)\right),,$

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 z is pure imaginary and 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. Consequently the singular points of this function occur at

z = a nonzero integral multiple of 2πi.

The singularities nearest 0, which is the center of the power series expansion, are at ±2πi. The distance from the center to either of those points is 2π, so the radius of convergence is 2π.

## Convergence on the boundary

If the power series is expanded around the point a and the radius of convergence is r, then the set of all points z such that |z − a| = r is a circle called the boundary of the disk of convergence. A power series may diverge at every point on the boundary, or diverge on some points and converge at other points, or converge at all the points on the boundary. Furthermore, even if the series converges on the boundary, it may not converge absolutely.

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

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 boundary. ƒ(z) in Example 1 is the derivative of the negative of g(z).

Example 3: The power series

$sum_\left\{n=1\right\}^infty frac\left\{1\right\}\left\{n^2\right\} z^n$

has radius of convergence 1 and converges everywhere on the boundary. If h(z) is the function represented by this series, then the derivative of h(z) is g(z) divided by z in Example 2. It turns out that h(z) is the dilogarithm function.

Example 4: The power series

$P\left(z\right) = sum_\left\{n=1\right\}^infty frac\left\{\left(-1\right)^\left\{n-1\right\}\right\}\left\{2^ncdot n\right\}\left(z^\left\{2^\left\{n-1\right\}\right\} + cdots + z^\left\{2^n-1\right\}\right)$
has radius of convergence 1 and converges uniformly on the boundary {|z|=1}, but does not converge absolutely on the boundary.

## Comments on rate of convergence

If we expand the function

$f\left(x\right)=sin x = sum^\left\{infin\right\}_\left\{n=0\right\} frac\left\{\left(-1\right)^n\right\}\left\{\left(2n+1\right)!\right\} x^\left\{2n+1\right\} = x - frac\left\{x^3\right\}\left\{3!\right\} + frac\left\{x^5\right\}\left\{5!\right\} - cdotstext\left\{ for all \right\} x$

around the point x = 0, we find out that the radius of convergence of this series is $scriptstyleinfty$ meaning that this series converges for all complex numbers. However, in applications, one is often interested in the precision of a numerical answer. Both the number of terms and the value at which the series is to be evaluated affect the accuracy of the answer. For example, if we want to calculate ƒ(0.1) = sin(0.1) accurate up to five decimal places, we only need the first two terms of the series. However, if we want the same precision for x = 1, we must evaluate and sum the first five terms of the series. For ƒ(10), one requires the first 18 terms of the series, and for ƒ(100), we need to evaluate the first 141 terms.

So the fastest convergence of a power series expansion is at the center, and as one moves away from the center of convergence, the rate of convergence slows down until you reach the boundary (if it exists) and cross over, in which case the series will diverge.

## A graphical example

Consider the function 1/(z2 + 1).

This function has poles at z = $scriptstyle pm$i.

As seen in the first example, the radius of convergence this function is 1 as the distance from 0 to each of those poles is 1.

Then the Taylor series of this function around z = 0 will only converge if |z| < 1, as depicted on the example on the right.

## Abscissa of convergence of a Dirichlet series

An analogous concept is the '''abscissa of convergence of a Dirichlet series

$sum_\left\{n=1\right\}^infty \left\{a_n over n^s\right\}.$

Such a series converges if the real part of s is less than a particular number depending on the coefficients an: the abscissa of convergence.