Editing Radius of convergence (section)

== Radius 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_of_a_function|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''.

[[File:TaylorComplexConv.png|thumb|300px|Radius of convergence (white) and Taylor approximations (blue) for <math>\frac{1}{1+z^2}</math>.]]

''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

: <math>f(z)=\frac 1 {1+z^2}</math>

has no singularities on the real line, since <math>1+z^2</math> has no real roots. Its Taylor series about 0 is given by

:<math>\sum_{n=0}^\infty (-1)^n z^{2n}.</math>

The root test shows that its radius of convergence is 1. In accordance with this, the function ''f''(''z'') has singularities at&nbsp;±''i'', which are at a distance 1 from&nbsp;0.

For a proof of this theorem, see [[analyticity of holomorphic functions]].

===A simple example===
The arctangent function of [[trigonometry]] can be expanded in a power series:

:<math>\arctan(z)=z-\frac{z^3} 3 + \frac{z^5} 5 -\frac{z^7} 7 +\cdots .</math>

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:

:<math>\frac z {e^z-1}=\sum_{n=0}^\infty \frac{B_n}{n!} z^n </math>

where the rational numbers ''B''<sub>''n''</sub> 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 [[removable singularity|the singularity is removable]].  The only non-removable singularities are therefore located at the ''other'' points where the denominator is zero.  We solve

:<math>e^z - 1 = 0</math>

by recalling that if {{math|1=''z'' = ''x'' + ''iy''}} and {{math|1=''e''{{i sup|''iy''}} = cos(''y'') + ''i'' sin(''y'')}} then

:<math>e^z = e^x e^{iy} = e^x(\cos(y)+i\sin(y)),</math>

and then take ''x'' and ''y'' to be real.  Since ''y'' is real, the absolute value of {{math|cos(''y'') + ''i'' sin(''y'')}} is necessarily 1.  Therefore, the absolute value of ''e''{{i sup|''z''}} can be 1 only if ''e''{{i sup|''x''}} is 1; since ''x'' is real, that happens only if ''x'' = 0.  Therefore ''z'' is purely imaginary and {{math|1=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 integer multiple of&nbsp;2{{pi}}.  Consequently the singular points of this function occur at

: ''z'' = a nonzero integer multiple of&nbsp;2{{pi}}''i''.

The singularities nearest 0, which is the center of the power series expansion, are at ±2{{pi}}''i''.  The distance from the center to either of those points is 2{{pi}}, so the radius of convergence is&nbsp;2{{pi}}.