on multivariate polynomial interpolation. subject the purpose of this work is to provide the problem...

ON MULTIVARIATE

POLYNOMIAL

INTERPOLATION

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.

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 andthe interpolation polynomial of degree .

1n 0, 1,..., nx x x

0 1( ), ( ),..., ( )nf x f x f xf ( )np xf

( )np x

( ) ( )n i ip x f x 0,1,...,i n0, 1,..., nx x x

( )np x n

n

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

1n 0, 1,..., nx x x

( ) np x P( ) ( )i if x p x 0,1,...,i n

0 1( ), ( ),..., ( )nf x f x f x

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

( , )p x yn

max{ : ( , ), , 0,1... }i jn i j x y p x y i j k


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

( , )p x y n

,0 0

( , )n k

j k j

j kk j

p x y c x y

cos sinp p p

r x y

( , ) ( cos , sin )x y r r

2 2 2( ) {( , ) : }S r x y x y r r


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 ,

[ ]t t

n 1 20 ... 1r r r

1, 2 ,...,

1 2 ... [ / 2] 1n

, ,{( , ) : 0 2 }l j l jx y j m ( )lS r

[( 1) / 2]m n 1 l


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.

, ,j l kf

, , , ,,k

l j l j j l kk

Px y f

r

0 1,1 ,0 2lk l j m

l 1l

[ / 2] 1n 1


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.

( )lS r

, , , , ,

(2 )( , ) ( cos , sin ), ,0 2

2 1l

l j l j l l j l l j l j

j ax y r r j m

m

la


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

2 2,g x y x y

1, 2 ,..., 1 2 ... [ / 2] 1 [3 / 2] 1 2n

1 21, 1

1 2

1, 1,1 2 1 2

, cos , sin ,0 22 2 1 2 2 1j j

j jx y j

m m

1 1 3 1 3,0 , , , ,2 4 4 4 4


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 .

2, 2,2 2

, cos ,sin ,0 22 1 2 1j jj j

x y jm m

1 3 1 3{1,0}, , , ,

2 2 2 2

,g x y

1 1 1 1 1, , ,1, ,4 8 8 2 2

2 200 01 10 11 02 20,q x y a a y a x a xy a y a x

,g x y


Then by solving the system of equations

we obtain that

and thus

, ,k k k kq x y g x y

11 02 01 20 10 00a 0,a 1,a 0,a 1,a 0,a 0

2 2,q x y x y

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

, ,A x y x y

( , )p x y ,A x y

,max{deg ( , ) / , 1, 2}x i jl a x y i j

,max{deg ( , ) / , 1, 2}y i jk a x y i j

n l k ,p x y

,0 0

( , )n k

j k jj k

k j

p x y p x y


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 .

1 2 ... [ / 2] 1n 1

1l l[ / 2] 1n

( )lS r 1 l

1 20 ... 1r r r

, ,

2 2, cos , sin

2 1 2 1l l

l j l j l l

j a j ax y r r

m m

0 2j m la j


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 .

, , , ,,k

l j l j j l kk

px y f

r

0 1,1 ,0 2lk l j m , ,j l kf

,A x y , ,,l j l jx y

1,

1

x xyx y

y xy

( , )p x y A


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

( , )p x y

,max{deg ( ) / , 1,2} 1x i jl a i j

,max{deg ( ) / , 1,2} 1y i jk a i j

4n4

,0 0

( , )k

j k j

j kk j

p x y p x y

( , )p x y

maxdeg , 2x p x y maxdeg , 2y p x y

,p x y

1 2 ... [ / 2] 1n 4n

1 2 ... 3


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,

( )lS r

1 2 30 1r r r

, ,{( , ) : 0 2 }l j l jx y j m

, , , ,( , ) ( cos , sin ),l j l j l l j l l jx y r r ,

2,

2 1l j

j

m

[( 1) / 2] [5/ 2] 2m n

1,2,3l 1 1/ 4r

2 1/ 2r 3 1r

1 2 3 3

1 2 3 1 [ / 2] 1 [4 / 2] 1 3n

1 3l

0 2j m


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 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

, , , ,, ,k

l j l j j l kk

Px y f

r

0k

, , ,( , )l j l j l jP x y f,l jf

2 2( , ) det ( , )p x y x y xy x y A x y


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 .

1 2 3 1 1 2 2

2 2

1

det ( , ) det ( , )n

ii

A x y A x yx

( , )iA x y

( , )A x y


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.

1 3


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.

1M 2M

,A x y

R

1 2( 1) ( 1)R M M

1 1 2 2,x u r y u r ( ) jri j iu r W

2

1i

I

MiW e

1 10,1,...,r M 2 20,1,...,r M 1,2i

,A x y

1 2, 1 1 2 2det ,r rp A u r u r


Step 5: Use the inverse DFT in order to obtain the

coefficients

where , .

Step 6: Compute the polynomial-determinant from the

formula

1 2,l lp

1 2

1 1 2 2

1 2 1 2

1 2

, ,1 2

1M M

r l r ll l r r

r r

p p W WR

0,1,..., , 1, 2i il M i 1 2( 1) ( 1)R M M

1 2

0 0

( , )i j

M Mi j

l li j

p x y p x y

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

adjA

adjA

1

, , , , , ,( , ) ( , ) det ( )l j l j l j l j l j l jadjA x y A x y A x y

adjA

1 1

detA adjA

A

on multivariate polynomial interpolation. subject the purpose of this work is to provide the problem...

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