-
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).
Mittag-Leffler and Principle of the ArgumentThursday, November
21, 20131:54 PM
New Section 2 Page 1
-
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.
Proof of uniform convergence on this region works by modifying
the
New Section 2 Page 2
-
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
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:
•
The argument is:
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
-
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
-
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
-
apply residue calculus:
From the MLE, we know that
New Section 2 Page 6
-
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
-
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
-
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
-
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
-
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
-
New Section 2 Page 12