My Math Forum  

Go Back   My Math Forum > College Math Forum > Complex Analysis

Complex Analysis Complex Analysis Math Forum

Thanks Tree1Thanks
  • 1 Post By Country Boy
LinkBack Thread Tools Display Modes
March 9th, 2018, 12:04 PM   #1
Senior Member
Joined: Nov 2011

Posts: 250
Thanks: 3

Simple English

Can one explain to me what is Magnus Gustaf (Gösta) Mittag-Leffler theorem in Simple English?
Thanks for trying to help me.
shaharhada is offline  
March 9th, 2018, 12:14 PM   #2
Global Moderator
Joined: Dec 2006

Posts: 21,105
Thanks: 2324

Are you referring to this theorem?
skipjack is offline  
March 9th, 2018, 12:39 PM   #3
Senior Member
Joined: Nov 2011

Posts: 250
Thanks: 3

Please, one explain me in Simple English the theorem!!!
shaharhada is offline  
April 24th, 2018, 11:56 AM   #4
Math Team
Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

The statement of the "Mittag-Lefler theorem" at Wikipedia is
"Let D be an open subset of the set of all complex numbers and E a closed discrete subset of D. For each point a in E, let $\displaystyle p_a(x)$ be a polynomial in 1/(z- a). There exist a meromorphic function, f, on D such that for each a in E, the function $\displaystyle f(z)- p_a(z)$ has only a removable discontinuity at a."

Now, the problem is that we do not know what you would consider "simple English".

Do you know what "complex numbers" are? I am going to assume you do since otherwise I can't possibly make sense of this to you. A meromorphic function is a function that is analytic (infinitely differentiable) everywhere except for poles at an isolated set of points. The Mittag-Lefler theorem says that, given such a polynomial, we can find a function, f, that is analytic everywhere except at a discrete set of points such that $\displaystyle f(x)-p_a(z)$ has only a removable discontinuity at z= a. A function, f, has a "removable" discontinuity at z= a if the limit f(z), as z goes a exist but f(a) either does not exist or is not equal to that limit. We can "remove" the limit by redefining f(a) to be equal to that limit. For example, the function $\displaystyle f(z)= \frac{z^2- a}{z- a}$ is the same as $\displaystyle z+ a$ for all z except a. The limit as z goes to a is a+ a= 2a but the function is not defined at z= a. We can "remove" that discontinuity by defining f(a)= 2a. (Then the function is identical to f(z)= z+ a so is continuous at z= a.)
Thanks from topsquark
Country Boy is offline  

  My Math Forum > College Math Forum > Complex Analysis

english, simple

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Simple(6,74) English(7,74) Gematria(8,74): GOD=7_4 (July 4) Brad Watson, Miami New Users 2 January 2nd, 2013 09:05 AM
English Gematria Gematria New Users 14 September 16th, 2012 03:20 PM
Test on english mathbalarka New Users 7 June 20th, 2012 06:35 PM
english Ghulam hussain New Users 2 January 21st, 2011 08:07 AM

Copyright © 2019 My Math Forum. All rights reserved.