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

For a power series f defined as:

$f(z)\; =\; \backslash sum\_\{n=0\}^\backslash infty\; f\_n\; =\; \backslash sum\_\{n=0\}^\backslash infty\; c\_n\; (z-a)^n,$

where

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

c_n is the nth complex coefficient (note that real numbers are a very common special case of complex numbers),

z is a variable and

f_n is the nth term of the series

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

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

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.

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 is defined as:

$C\; =\; \backslash limsup\_\{n\backslash rightarrow\backslash infty\}\backslash sqrt*\{|f\_n|\}$

and in the case of a power series, this can be used to find that:

$r\; =\; \backslash limsup\_\{n\backslash rightarrow\backslash infty\}\; \backslash frac\{1\}\{\backslash sqrt*\{|c\_n|\}\}$

lim sup denotes the limit superior and

Note that 1/0 is interpreted as an infinite radius, meaning that f is an entire function.

The ratio test is usually easier to compute, but the limit may be infinite (i.e. non-existant limit), in which case the root test should be used. The ratio test is defined as:

$L\; =\; \backslash lim\_\{n\backslash rightarrow\backslash infty\}\backslash left|\backslash frac\{f\_\{n+1\}\}\{f\_n\}\backslash right|$

and in the case of a power series, this can be used to find that:

$r\; =\; \backslash lim\_\{n\backslash rightarrow\backslash infty\}\; |\; \backslash frac\{c\_n\}\{c\_\{n+1\}\}\; |$.

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.

A simple example

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

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

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:

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

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.

A more complicated example

Consider this power series:

$\backslash frac\{z\}\{e^z-1\}=\backslash sum\_\{n=0\}^\backslash infty\; B\_n\; z^n$

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

$e^z\; =\; e^x\; e^\{iy\}\; =\; e^x(\backslash cos(y)+i\backslash sin(y)),$

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.

External links

• What is radius of convergence?

Calculus | Complex analysis | Mathematical series

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