BEZOUT IDENTITIES WITH INEQUALITY CONSTRAINTS Wayne M. Lawton Department of Mathematics National University of Singapore Lower Kent Ridge Road Singapore Tel (65) Fax (65)

Bezout Identities where products belong to a ring R with identity 1 Origin in diophantine problems, where p’s are coprime integers, gave rise to the Euclidean Algorithm and the Chinese Remainder Theorem Related to Corona Theorem, transcendental numbers, wavelets, deconvolution, interpolation, and control

Interpolation Concepts Lattice Subroups Restriction Operator Inclusion Operator Convolution Rings

Interpolation Method Interpolatory Filter The Interpolation Operator defined by satisfies

Desired Properties Accuracy: ifis a low-degree polynomial then Positivity: if then is positive-definite Remark: positivity is required for interpolation of statistical autocovariance functions, occurs if and only if the interpolatory filter p is positive-definite

Z-Transform Construct an isomorphism onto the ring of Laurent polynomials, from a basis for and define the torus group A Laurent polynomial is determined by the trigonometric polynomial defined by its values on the torus group. Denote

Stability Group Torus group acts as a group of transformations on Define the stability group Then

Equivalent Properties Accuracy:zero, to specified order, on Positivity: is nonnegative-valued on Interpolatory ( Poisson ) ( Bochner )

Filter Design Approach Step 3. Define is coprime Step 1. Construct on Step 2. Solve Bezout on

Step 1 Step 1.3 Define are distinct Step 1.1 Select monomial Step 1.2 Construct

Step 2 Use Theorem 1 (Lawton-Micchelli) to solve Bezout on

are coprime and If Theorem 1 onthen there exist on that solve the Bezout identity

Proof of Theorem 1 where Step 1. Use Quillen-Suslin to compute invertible matrix whose first row equals and belong to the subring Step 2. Form the invertible matrix of Laurent polynomials that are real-valued on

Proof of Theorem 1 Use this fact to define vectors having continuous entries that are positive, real-valued, respectively Step 3. Sincehave no common zeros in is strictly positive on

Proof of Theorem 1 so the entries of Step 3. Approximate are positive on by Then the Bezout identity with positive constraints is solved by entries of