Mittag-Leffler and Principle of the Argument

Transcription

1 Mittag-Leffler and Principle of the Argument Thursday, November 21, :54 PM Homework 3 due Friday, November 22 at 5 PM. Homework 4 will be posted tonight, due Wednesday, December 11 at 5 PM. We'll next show how to use contour integration techniques to evaluate the numerical values of sums of the form: where f(n) is a rational function (ratio of polynomials) where the degree of the denominator is at least one more than the degree of the numerator. Prototypical example: (One only needs to write these infinite sums in terms of the specific limits above when the degree of the denominator is exactly one higher than the degree of the numerator, because then the series is not absolutely convergent: The basis for doing these calculation is to recognize some important Mittag-Leffler representations. This is just a generalization of partial fraction representations applied to meromorphic functions (functions which are analytic everywhere on the complex plane except for a finite or countable sequence of poles). New Section 2 Page 1

2 Why is this true? First of all the right hand side is an absolutely convergent series for any for integers n. By modifying the argument we just made, the series on the RHS can be shown to be uniformly convergent over any region of the form: for all, with R, fixed parameters. New Section 2 Page 2

3 Proof of uniform convergence on this region works by modifying the above argument using the Weierstrass M-test by cutting off the sums at rather than. On such regions, the series defines an analytic function because it's a uniformly convergent limit of analytic functions. Since are arbitrary positive numbers, the series on the RHS defines an analytic function everywhere in the complex plane except at the integers on the real axis, which are poles. So why is The argument is: Both the LHS and RHS have singularities at the same locations, namely the integers on the real axis (and nowhere else). Not only are the singularities identical, but the principal parts are the same, namely, they are double poles with the principal part: Clearly these are the principal parts of the RHS. For the LHS, first consider the singularity at z=0. We know that: For general Notice that has period 1. New Section 2 Page 3

4 So therefore by periodicity, the principal parts of are just translations of the principal part at the origin, so matches up with the RHS. Next, one observes that because the singularities match exactly, the difference LHS - RHS is an entire function (no singularities on the finite complex plane). Then show that LHS-RHS is a bounded function (direct estimates on the functions, and using periodicity of both LHS and RHS). Therefore by Liouville theorem, LHS-RHS is a constant. Then show the the constant is 0 by actually noting that both the LHS and RHS go to 0 as. Now if we integrate both sides of the Mittag-Leffler representation for, we formally get: Have to write this more carefully: New Section 2 Page 4

5 Why does this sum converge?: One can also show, through similar techniques that: So how can we make use of them? Suppose we want to compute a sum of the form: Suppose that and f has no singularities on the integers. Then consider the following contour integral: New Section 2 Page 5

6 apply residue calculus: From the MLE, we know that New Section 2 Page 6

7 So: So then equating this to the limit on the RHS: This formula can also be shown to be true when the degree of the denominator is only one more than the degree of the numerator. It spoils the integral inequality to show the contour integral goes to zero. So instead, subtract off the part of f(z) that decays like just and show that it doesn't contribute to the sum. A parallel argument shows that: New Section 2 Page 7

8 Principle of the argument and Rouche's theorem These have to do with root finding of analytic functions in the complex plane. When you work with Laplace transforms of differential equations,for example, then the locations of the zeros of the denominator of the Laplace transform tell you about the stability of the system. Principle of the argument: If f(z) is a meromorphic function on and inside a simple closed contour C with positive orientation, and is analytic on C, then: Note that Nth order poles count N times, so really P is the sum of the orders of the poles inside C. Similarly Z is the sum of the orders of the zeros of f(z) inside C. New Section 2 Page 8

9 Often we know where the poles are, or that there aren't any, in which case the principle of the argument counts zeros. Then one can use this to help hunt for places where f(z) vanishes, and in particular rule out places where it does not. Proof of the principle of the argument: Clearly by residue calculus: At a zero of order m, The same argument shows that poles of f will give residues that are equal to minus their order. New Section 2 Page 9

10 Another useful variation on the principle of the argument: Define: as the change in the argument of a function f(z) as it traverses a contour C. That is, if we parameterize a contour C by, and write in polar form with modulus and argument, where this polar representation varies continuously with the parameter t as the contour is traversed. Then One can show this result is independent of any continuous choice of parameterization. In fact, is called the winding number of the function f(z) around the closed contour C. Geometrically, it indicates how many times the image of the contour under the mapping f winds around the origin as the contour is traversed. So the extended statement of the principle of the argument is: New Section 2 Page 10

11 That is, for a function f(z) to have a zero (or pole) at a point, the function f(z) must map sufficiently smalll loops around to loops around the origin. This is essentially saying that meromorphic functions can't fold domains. To prove the extended principle of the argument, just compute the contour integral parametrically: New Section 2 Page 11

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