Jump to content
Main menu
Main menu
move to sidebar
hide
Navigation
Main page
Recent changes
Random page
Help about MediaWiki
Special pages
Niidae Wiki
Search
Search
Appearance
Create account
Log in
Personal tools
Create account
Log in
Pages for logged out editors
learn more
Contributions
Talk
Editing
Radius of convergence
(section)
Page
Discussion
English
Read
Edit
View history
Tools
Tools
move to sidebar
hide
Actions
Read
Edit
View history
General
What links here
Related changes
Page information
Appearance
move to sidebar
hide
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
== Convergence on the boundary == If the power series is expanded around the point ''a'' and the radius of convergence is {{math|''r''}}, then the set of all points {{math|''z''}} such that {{math|1={{mabs|''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 everywhere on the boundary (even uniformly), it does not necessarily converge absolutely. Example 1: The power series for the function {{math|1=''f''(''z'') = 1/(1 − ''z'')}}, expanded around {{math|1=''z'' = 0}}, which is simply :<math> \sum_{n=0}^\infty z^n,</math> has radius of convergence 1 and diverges at every point on the boundary. Example 2: The power series for {{math|1=''g''(''z'') = −ln(1 − ''z'')}}, expanded around {{math|1=''z'' = 0}}, which is :<math> \sum_{n=1}^\infty \frac{1}{n} z^n,</math> has radius of convergence 1, and diverges for {{math|1=''z'' = 1}} but converges for all other points on the boundary. The function {{math|''f''(''z'')}} of Example 1 is the [[derivative]] of {{math|''g''(''z'')}}. Example 3: The power series :<math> \sum_{n=1}^\infty \frac 1 {n^2} z^n </math> has radius of convergence 1 and converges everywhere on the boundary absolutely. If {{math|''h''}} is the function represented by this series on the unit disk, then the derivative of ''h''(''z'') is equal to ''g''(''z'')/''z'' with ''g'' of Example 2. It turns out that {{math|''h''(''z'')}} is the [[dilogarithm]] function. Example 4: The power series :<math>\sum_{i=1}^\infty a_i z^i \text{ where } a_i = \frac{(-1)^{n-1}}{2^nn}\text{ for } n = \lfloor\log_2(i)\rfloor+1\text{, the unique integer with }2^{n-1}\le i < 2^n,</math> has radius of convergence 1 and converges [[uniform convergence|uniformly]] on the entire boundary {{math|1={{mabs|''z''}} = 1}}, but does not [[Absolute convergence|converge absolutely]] on the boundary.<ref>{{cite journal |url=https://eudml.org/doc/215384|title=O szeregu potęgowym, który jest zbieżny na całem swem kole zbieżności jednostajnie, ale nie bezwzględnie|journal=Prace Matematyczno-Fizyczne|year=1918|volume=29|issue=1|pages=263–266|last1=Sierpiński|first1=W.|author-link=Wacław Sierpiński}}</ref>
Summary:
Please note that all contributions to Niidae Wiki may be edited, altered, or removed by other contributors. If you do not want your writing to be edited mercilessly, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource (see
Encyclopedia:Copyrights
for details).
Do not submit copyrighted work without permission!
Cancel
Editing help
(opens in new window)
Search
Search
Editing
Radius of convergence
(section)
Add topic