2 Subject The purpose of this work is to provide the problem of Hermite-Lagrange multivariate polynomial interpolation andespecially the computation of the inverse of a two variablepolynomial matrix.We start the presentation stating some basic definitionsfor univariate case.
3 Interpolation polynomial in one variable Problem-DefinitionsLet be distinct points on which the values of, are known. Find a polynomial ofdegree which takes the same values as at the samepoints.Essentially we are looking for a polynomial whichsatisfies the below interpolation conditionsforThe points are called interpolation points andthe interpolation polynomial of degree .
4 Existence and Uniqueness if interpolation polynomial Theorem: For any set of distinct pointsand the values , there is only onepolynomial , which is satisfyingfor
5 Hermite-Lagrange polynomial interpolation in two variables Definitions and Hermite’s interpolation problem :We consider the Hermite’s interpolation problem forpolynomials in two variables. The interpolation points are locatedin several circles centered at the origin and the interpolationmatches preassigned data of function’s values and consecutivenormal derivatives.When no derivative values are interpolated the problem isreduced to a Lagrange’s interpolation problem.The total degree of a polynomial , in two variablesis defined by
6 Hermite-Lagrange polynomial interpolation in two variables A polynomial of total degree is of the formAn interpolation problem is defined to be poised if ithas a unique solution. Unlike the polynomial interpolationproblem in one variable case, the Hermite or Lagrangeinterpolation problem in the multivariate case is notalways poised.Normal Derivative:,For positive number, denote the circle of radius , centered atthe origin by
7 Hermite-Lagrange polynomial interpolation in two variables Let denote the integer part of .Interpolation problem:Let (total degree) be a positive integer. Letand let be positive integers such thatDenote by distinct points on the circlewhere ,
8 Hermite-Lagrange polynomial interpolation in two variables Then the interpolation problem has a unique solution for anypreassigned data if the following interpolation conditionsare satisfied,whereIf for all , then the problem becomes Lagrange oncircles. If there is only one circle which wechoose to be the unit and the problem becomes Hermiteinterpolation problem on the unit circle.
9 Hermite-Lagrange polynomial interpolation in two variables The most natural choice of interpolation points on the circleis the equidistant points, that iswhere , states that the equidistant points on different circlescan differ by a rotation.Theorem: The Hermite interpolation problem based on theequidistant points is poised.
10 Hermite-Lagrange polynomial interpolation in two variables Example: Consider the two variable functionIt is enough to select as positive integers such that. The only solutions tothis problem are (Lagrange interpolation problem on2 circles) and (Hermite interpolation problem on the unitcircle) . If we consider the first case then we have the interpolationpointsfor the first circle of radius ½ that gives
11 Hermite-Lagrange polynomial interpolation in two variables andfor the second circle of radius 1 that givesThe values of at the above equidistant givesLet the two-variable polynomial of order twowith the same values at the interpolation points with
12 Hermite-Lagrange polynomial interpolation in two variables Then by solving the system of equationswe obtain thatand thus
13 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 thebelow algorithm.Algorithm:Step 1: Compute the upper bound n for the total degree of thedeterminant of LetThen Therefore is
14 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 becomesthe Hermite interpolation problem on the unit circle.If , for all , the problem becomes the Lagrangeinterpolation problem on circles.In all other cases we select λ circles , withradiusStep 3: Determine the n interpolations points,where and is a number independent of .
15 On the computation of the determinant of 2-D polynomial matrix Step 4: Apply the points on the interpolation conditionsfor where arepreassigned data of the matrix at the pointsExample: Consider the polynomial matrixLet denote the determinant of .
16 On the computation of the determinant of 2-D polynomial matrix Step 1: Compute the total degree n of LetThen 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 equationforor equivalently
17 On the computation of the determinant of 2-D polynomial matrix Case 1: LetThen and thus we have the Lagrange interpolationproblem in circles, , for and. We choose , ,Step 3: Determine the interpolations points.Let the denote distinct points on the circleswhere and We chooseequidistant points, that is,
18 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 pointson the circles. That is, we only need to interpolate thedeterminant’s values and not it’s derivative values. Therefore theinterpolation conditions becomeTo obtain the data we substitute the interpolation points onthe matrix and for the each point we compute the determinant ofthe matrix.A system of 15 equations with 15 unknown follows frominterpolation conditions. Using Mathematica the solution gives
19 On the computation of the determinant of 2-D polynomial matrix There are more 2 cases we can interpolate. The first one iswhere and In that case we haveinterpolation points on two circles. For the second circle ( )we have to interpolate not only determinant’s values but alsodeterminant’s first derivative values. In order to evaluate thesevalues we use the following formulawhere comes from taking partial derivatives in terms of x(or y) from the elements of the i-th series of
20 On the computation of the determinant of 2-D polynomial matrix The second case is , that is, interpolation points on theunit circle on which we have to interpolate determinant’s values,determinant’s first and second derivative values. The secondderivative values can arise by modifying the previous formula.Both the two cases give the same interpolation polynomial ascase 1.For approaching the determinant of a two polynomial matrixwe can also use methods based on Discrete FourierTransform.
21 On the computation of the determinant of 2-D polynomial matrix Algorithm:Step 1: Compute , the maximum degree of x and yrespectively, in the determinant of the matrixStep 2: Calculate the number of the interpolation numbersfromStep 3: Determine the interpolation points,; ,, ,Step 4: Apply the points on the matrix and compute thedeterminant at each point.
22 On the computation of the determinant of 2-D polynomial matrix Step 5: Use the inverse DFT in order to obtain thecoefficientswhere ,Step 6: Compute the polynomial-determinant from theformula
23 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 theStep 2.1: Evaluate the on the interpolation points we used for the determinant’s interpolation using the formulaStep 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