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CONTOUR INTEGRALS AND THEIR APPLICATIONSPowerPoint Presentation

CONTOUR INTEGRALS AND THEIR APPLICATIONS

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### CONTOUR INTEGRALS AND THEIR APPLICATIONS

### ARYABHATA

### BEZOUT

### INEQUALITY CONSTRAINTS

### UPPER LENGTH BOUNDS

### LOWER LENGTH BOUNDS

### CONTOUR INTEGRAL

### CONTOUR CONSTRUCTION

### CONCLUSIONS AND EXTENSIONS

Wayne Lawton

Department of Mathematics

National University of Singapore

S14-04-04, [email protected]

http://math.nus.edu.sg/~matwml

where a and b are integers. Clearly this equation admits a solution if and only if a and b have no common factors other than 1, -1 (are relatively prime) and then Euclid’s algorithm gives a solution. Furthermore, if (x,y) is a solution then the set of solutions is the infinite set

characterized the set { (x, y) } of integer solutions of the equation

Van der Warden, Geometry and Algebra in Ancient Civilizations, Springer-Verlag, New York, 1984.

Clearly this equation has a solution iff

and

have no

common roots and then Euclid’s algorithm gives a solution.

Bezout identities in general rings arise in numerous areas of mathematics and its application to science and engineering:

investigated the polynomial version of this equation

Algebraic Polynomials: control, Quillen-Suslin Theorem

Laurent Polynomials: wavelet, splines, Swan’s Theorem

H_infinity: the Corona Theorem

Entire Functions: distributional solutions of systems of PDE’s

Matrix Rings: control, signal processing

E. Bezout, Theorie Generale des Equations Algebriques, Paris, 1769.

are

on the unit circle

then

LP

on

and

Proof. Let LP

real on

with

that is real on

with

Choose a LP

then choose

W.Lawton & C.Micchelli, Construction of conjugate quadrature filters with specified zeros, Numerical Algorithms, 14:4 (1997) 383-399

W.Lawton & C.Micchelli, Bezout identities with inequality constraints, Vietnam J. Math. 28:2(2000) 97-126

There exists

with

Furthermore, for fixed

and for fixed L

Proof: Uses resultants.

For any positive integer n, there exist LP

with

and

Proof: See VJM paper.

Question: Are there better ways to obtain bounds that ‘bridge the gap’ between the upper and lower bounds

representation for the Bezout identity is given by

Theorem Let

are a disjoint contours and the

interior

of

contains all roots of

and

excludes all roots of

then for

where

are LP, real on T, and satisfy the Bezout identity.

Proof Follows from the residue calculus.

on T,

hence if

are

-invariant contours then it suffices

to consider these quantities inside of the unit disk D.

For k=1,2 let

union of open disks of radius

centered at zeros of

in D

and

be the disk of this radius centered at 0.

if

else

Theorem

The contour integral method provide sharper bounds for

and therefore for B than the resultant method but

sharper bounds are required to ‘bridge the gap’.

Contour integrals for BI with n > terms are given by

where

encloses all zeros of T except for those of

Residue current integrals give multivariate versions

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