1. ON MULTIVARIATE POLYNOMIAL INTERPOLATION

2. 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.

3. 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 .

4. 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

5. 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

6. 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

7. 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 ,

8. 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.

9. 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.

10. 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

11. 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

12. Hermite-Lagrange polynomial interpolation in two variables
Then by solving the system of equations
we obtain that
and 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 the
below algorithm.
Algorithm:
Step 1: Compute the upper bound n for the total degree of the
determinant of Let
Then 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 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 .

15. 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 .

16. 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

17. 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,

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 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

19. 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

20. 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.

21. 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.

22. 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

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 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