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 March 9th, 2018, 11:04 AM #1 Senior Member   Joined: Nov 2011 Posts: 194 Thanks: 2 Simple English Can one explain to me what is Magnus Gustaf (Gösta) Mittag-Leffler theorem in Simple English? Thanks for trying to help me.
 March 9th, 2018, 11:14 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,956 Thanks: 1602 Are you referring to this theorem?
 March 9th, 2018, 11:39 AM #3 Senior Member   Joined: Nov 2011 Posts: 194 Thanks: 2 Please, one explain me in Simple English the theorem!!!
 April 24th, 2018, 10:56 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 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

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