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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:

${\displaystyle f(z)=\sum _{n=0}^{\infty }f_{n}=\sum _{n=0}^{\infty }c_{n}(z-a)^{n},}$

where

a is a constant (sometimes called the center of the series since the radius of convergence "centers" around a, like a circle centers around its center),

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

z is a variable and

f_n represents the nth term of the series

The radius of convergence is defined such that the series converges if

${\displaystyle |z-a|<r\!}$

and diverges if

${\displaystyle |z-a|>r\!}$

where

r is the radius of convergence, which may be a real number or ∞.

In other words, the series converges if z is close enough to the center. 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:

${\displaystyle C=\limsup _{n\rightarrow \infty }{\sqrt{|f_{n}|}}}$

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

${\displaystyle r=\limsup _{n\rightarrow \infty }{\frac {1}{\sqrt{|c_{n}|}}}}$

where

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:

${\displaystyle L=\lim _{n\rightarrow \infty }\left|{\frac {f_{n+1}}{f_{n}}}\right|}$

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

${\displaystyle r=\lim _{n\rightarrow \infty }|{\frac {c_{n}}{c_{n+1}}}|}$.

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 warm-up example

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

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

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:

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

and a zero appears in the denominator when z = − 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.

(This famous example also immediately gives us a method for calculating the value of ${\displaystyle \pi }$. It is an interesting application of Abel's theorem. In view of Leibniz' test (described in the entry alternating series) the series

${\displaystyle 1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+\cdots .}$

converges. So Abel's theorem tells us that the sum of this series must equal

${\displaystyle \lim _{z\uparrow 1}\arctan(z)={\frac {\pi }{4}}}$.

In view of Leibniz' theorem we can also easily determine how many terms of this series we need to use to find ${\displaystyle \pi }$ to within any required accuracy. For a slightly different explanation of this calculation see the entry Leibniz formula for pi.)

A gaudier example

Consider this power series:

${\displaystyle {\frac {z}{e^{z}-1}}=\sum _{n=0}^{\infty }B_{n}z^{n}}$

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 where the denominator is zero. We solve

${\displaystyle e^{z}-1=0}$

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

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

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 can be 1 only if e 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