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**ON MULTIVARIATE POLYNOMIAL INTERPOLATION**

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**Subject The purpose of this work is to provide the problem of**

Hermite-Lagrange multivariate polynomial interpolation and especially the computation of the inverse of a two variable polynomial matrix. We start the presentation stating some basic definitions for univariate case.

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**Interpolation polynomial in one variable**

Problem-Definitions Let be distinct points on which the values of , are known. Find a polynomial of degree which takes the same values as at the same points. Essentially we are looking for a polynomial which satisfies the below interpolation conditions for The points are called interpolation points and the interpolation polynomial of degree .

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**Existence and Uniqueness if interpolation polynomial**

Theorem: For any set of distinct points and the values , there is only one polynomial , which is satisfying for

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**Hermite-Lagrange polynomial interpolation in two variables**

Definitions and Hermite’s interpolation problem : We consider the Hermite’s interpolation problem for polynomials in two variables. The interpolation points are located in several circles centered at the origin and the interpolation matches preassigned data of function’s values and consecutive normal derivatives. When no derivative values are interpolated the problem is reduced to a Lagrange’s interpolation problem. The total degree of a polynomial , in two variables is defined by

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**Hermite-Lagrange polynomial interpolation in two variables**

A polynomial of total degree is of the form An interpolation problem is defined to be poised if it has a unique solution. Unlike the polynomial interpolation problem in one variable case, the Hermite or Lagrange interpolation problem in the multivariate case is not always poised. Normal Derivative: , For positive number, denote the circle of radius , centered at the origin by

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**Hermite-Lagrange polynomial interpolation in two variables**

Let denote the integer part of . Interpolation problem: Let (total degree) be a positive integer. Let and let be positive integers such that Denote by distinct points on the circle where ,

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**Hermite-Lagrange polynomial interpolation in two variables**

Then the interpolation problem has a unique solution for any preassigned data if the following interpolation conditions are satisfied, where If for all , then the problem becomes Lagrange on circles. If there is only one circle which we choose to be the unit and the problem becomes Hermite interpolation problem on the unit circle.

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**Hermite-Lagrange polynomial interpolation in two variables**

The most natural choice of interpolation points on the circle is the equidistant points, that is where , states that the equidistant points on different circles can differ by a rotation. Theorem: The Hermite interpolation problem based on the equidistant points is poised.

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**Hermite-Lagrange polynomial interpolation in two variables**

Example: Consider the two variable function It is enough to select as positive integers such that . The only solutions to this problem are (Lagrange interpolation problem on 2 circles) and (Hermite interpolation problem on the unit circle) . If we consider the first case then we have the interpolation points for the first circle of radius ½ that gives

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**Hermite-Lagrange polynomial interpolation in two variables**

and for the second circle of radius 1 that gives The values of at the above equidistant gives Let the two-variable polynomial of order two with the same values at the interpolation points with

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**Hermite-Lagrange polynomial interpolation in two variables**

Then by solving the system of equations we obtain that and thus

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**On the computation of the determinant of 2-D polynomial matrix**

Let be a two-variable polynomial matrix. For the evaluation of the determinant of the matrix we give the below algorithm. Algorithm: Step 1: Compute the upper bound n for the total degree of the determinant of Let Then Therefore is

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**On the computation of the determinant of 2-D polynomial matrix**

Step 2: Find the solution of equation, If , there is only one circle and the problem becomes the Hermite interpolation problem on the unit circle. If , for all , the problem becomes the Lagrange interpolation problem on circles. In all other cases we select λ circles , with radius Step 3: Determine the n interpolations points, where and is a number independent of .

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**On the computation of the determinant of 2-D polynomial matrix**

Step 4: Apply the points on the interpolation conditions for where are preassigned data of the matrix at the points Example: Consider the polynomial matrix Let denote the determinant of .

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**On the computation of the determinant of 2-D polynomial matrix**

Step 1: Compute the total degree n of Let Then it is easy to determine the maximum degree in variable x (or y) of , Therefore is of total degree , i.e. Step 2: Find the solution of equation for or equivalently

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**On the computation of the determinant of 2-D polynomial matrix**

Case 1: Let Then and thus we have the Lagrange interpolation problem in circles, , for and . We choose , , Step 3: Determine the interpolations points. Let the denote distinct points on the circles where and We choose equidistant points, that is,

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**On the computation of the determinant of 2-D polynomial matrix**

Step 4: Construct the interpolation conditions. Because of the Lagrange’s interpolation, for all the points on the circles. That is, we only need to interpolate the determinant’s values and not it’s derivative values. Therefore the interpolation conditions become To obtain the data we substitute the interpolation points on the matrix and for the each point we compute the determinant of the matrix. A system of 15 equations with 15 unknown follows from interpolation conditions. Using Mathematica the solution gives

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**On the computation of the determinant of 2-D polynomial matrix**

There are more 2 cases we can interpolate. The first one is where and In that case we have interpolation points on two circles. For the second circle ( ) we have to interpolate not only determinant’s values but also determinant’s first derivative values. In order to evaluate these values we use the following formula where comes from taking partial derivatives in terms of x (or y) from the elements of the i-th series of

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**On the computation of the determinant of 2-D polynomial matrix**

The second case is , that is, interpolation points on the unit circle on which we have to interpolate determinant’s values, determinant’s first and second derivative values. The second derivative values can arise by modifying the previous formula. Both the two cases give the same interpolation polynomial as case 1. For approaching the determinant of a two polynomial matrix we can also use methods based on Discrete Fourier Transform.

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**On the computation of the determinant of 2-D polynomial matrix**

Algorithm: Step 1: Compute , the maximum degree of x and y respectively, in the determinant of the matrix Step 2: Calculate the number of the interpolation numbers from Step 3: Determine the interpolation points, ; , , , Step 4: Apply the points on the matrix and compute the determinant at each point.

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**On the computation of the determinant of 2-D polynomial matrix**

Step 5: Use the inverse DFT in order to obtain the coefficients where , Step 6: Compute the polynomial-determinant from the formula

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**Inversion of a 2-D polynomial matrix via interpolation**

Algorithm: Step 1: Interpolate the determinant of the matrix using Hermite- Lagrange interpolation. Step 2: Interpolate the Step 2.1: Evaluate the on the interpolation points we used for the determinant’s interpolation using the formula Step 2.2: Construct the interpolation conditions for each element of We use the same form of polynomials as we used on determinant’s interpolation. Step 3: Compute the inverse from the following formula

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