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CONTOUR INTEGRALS AND THEIR APPLICATIONS. Wayne Lawton Department of Mathematics National University of Singapore S14-04-04, [email protected] http://math.nus.edu.sg/~matwml. ARYABHATA.

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### CONTOUR INTEGRALS AND THEIR APPLICATIONS

Wayne Lawton

Department of Mathematics

National University of Singapore

S14-04-04, [email protected]

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

### ARYABHATA

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.

### BEZOUT

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

### INEQUALITY CONSTRAINTS

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

### UPPER LENGTH BOUNDS

There exists

with

Furthermore, for fixed

and for fixed L

Proof: Uses resultants.

For any positive integer n, there exist LP

### LOWER LENGTH BOUNDS

with

and

Proof: See VJM paper.

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

### CONTOUR INTEGRAL

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.

### SOLUTION BOUNDS

Lemma

where

a contour that is disjoint from

and whose (annular) interior contains T

on T,

### CONTOUR CONSTRUCTION

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

### CONCLUSIONS AND EXTENSIONS

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