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

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  1. CONTOUR INTEGRALS AND THEIR APPLICATIONS Wayne Lawton Department of Mathematics National University of Singapore S14-04-04, matwml@nus.edu.sg http://math.nus.edu.sg/~matwml

  2. 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.

  3. 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.

  4. Theorem If RPLP 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

  5. Theorem UPPER LENGTH BOUNDS There exists with Furthermore, for fixed and for fixed L Proof: Uses resultants.

  6. Theorem 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

  7. representation for the Bezout identity is given by 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.

  8. SOLUTION BOUNDS Lemma where a contour that is disjoint from and whose (annular) interior contains T

  9. Since 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

  10. The contour integral method provide sharper bounds for 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

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