On Multivariate Hermite Interpolation - CiteSeerX

On Multivariate Hermite Interpolation by Thomas Sauer Mathematical Institute University Erlangen{Nuremberg 91054 Erlangen Germany [email protected]

Yuan Xuy Department of Mathematics University of Oregon Eugene, Oregon 97403 USA [email protected]

Abstract. We study the problem of Hermite interpolation by polynomials

in several variables. A very general de nition of Hermite interpolation is adopted which consists of interpolation of consecutive chains of directional derivatives. We discuss the structure and some aspects of poisedness of the Hermite interpolation problem; using the notion of blockwise structure which we introduced in [10], we establish an interpolation formula analogous to that of Newton in one variable and use it to derive an integral remainder formula for a regular Hermite interpolation problem. For Hermite interpolation of degree n of a function f , the remainder formula is a sum of integrals of certain (n + 1){st directional derivatives of f multiplied by simplex spline functions.

Keywords: Hermite interpolation, blockwise structure, poisedness, regular Hermite interpolation, simplex spline, remainder formula.

1991 Mathematics Subject Classi cation: Primary: 41A05, 41A10, 65D05, 65D10

Version December 22, 1995

ySupported

by National Science Foundation Grant No. 9302721 and the Alexander von Humboldt Foundation

1 Introduction Let d be the space of all polynomials in d variables, and let dn be the subspace of polynomials of total degree at most n. For a positive integer N , let XN be a set of pairwise distinct points in Rd and assume that the cardinality N of XN is equal to dimdn. The Lagrange interpolation problem associated to XN asks whether there is a unique polynomial p in dn which matches preassigned data on XN . If the answer is yes, the interpolation problem is said to be poised; we denote the interpolating polynomial by Ln f , assuming that the data to be interpolated are the values of a function f at the interpolation points. In a recent paper [10], we studied the Lagrange interpolation problem using a nite dierence approach which oers formulae very much comparable to the classical univariate Newton interpolation formula. The starting point in [10] is that, in analogy to univariate the points of XN are grouped in blocks whose cardinality is equal   dinterpolation, d to dim k =k?1 ; we refer to the corresponding interpolation scheme as interpolation in blocks. The blockwise structure allowed us to de ne a nite dierence from which we derived a Newton formula for Lagrange interpolation and an integral remainder formula for f ? Lnf . The remainder formula is given in terms of a sum of integrals of n{fold directional derivatives of f multiplied by simplex splines. For d = 1, it reduces to the well known integral remainder formula of univariate interpolation. The purpose of this paper is to extend these results to Hermite interpolation. For one variable, the Hermite interpolation problem means to nd a polynomial which matches, on a set of distinct points, values of a function and its consecutive derivatives. For several variables, however, the de nition is less clear; in the literature it is often de ned as interpolation of all partial derivatives up to a certain order (cf. [7]). One way of looking at the univariate Hermite interpolation problem is to treat it as the result of the collapsing of points in a Lagrange interpolation problem; for example, if two points collapse, the Lagrange interpolation conditions at the two points turn into a Hermite condition of order one which matches both the value of function and the value of its rst order derivative at the \double" point. However, in several variables, even if we assume that points collapse only along lines (cf. [6, 4, 5]), they can collapse from many dierent directions; this suggests that, in general, we need to deal with directional derivatives instead of just partial derivatives. Moreover, even when we interpolate partial derivatives, instead of interpolating all derivatives of certain orders, we may need to interpolate only some of the derivatives in order to keep the number of interpolation conditions to agree with dimdn (see, for example, [1]); for instance, we may interpolate only one partial derivative of rst order and one of second order at some points. It is natural, therefore, to consider Hermite interpolation at a point as interpolation of chains of directional derivatives of consecutive order; more precisely, for given directions y1; : : :; yk in Rd, a typical example is to interpolate Dy ; Dy Dy ; : : : ; Dyk


Dy : In such a generality, the Hermite interpolation problem in several variables has not been studied before. There are clearly many questions that have to be answered; even the proper notations have to be developed in order to de ne the problem in precise terms. We 1

2

1

1

1

mention, however, that under the more restricted de nitions, the Hermite interpolation problem in several variables has been studied and used in the eld of nite elements. We refer to [7, page 171{182] for extensive references. More speci c comments on the related results will be given in the proper context in the paper. In the following we outline the content of the paper. In Section 2, we introduce the notation of trees to de ne and discuss the general notion of Hermite interpolation. Our basic approach is to give the Hermite interpolation problem a blockwise structure, which amounts to a change of basis in dn. For Lagrange interpolation, this amounts to switch from Lagrange fundamental polynomials to Newton fundamental polynomials. The block structure and the related matter are discussed in Section 3. One of the diculties in multivariate interpolation lies in the fact that the interpolation problem may not be poised. A Lagrange interpolation problem is poised if the interpolation points are not all on an algebraic surface of degree n ? 1; since such points form a set of measure zero, Lagrange interpolation problem is almost poised, meaning that it is poised for almost all choices of points. For an Hermite interpolation problem, almost poisedness means that the problem is poised for almost all choices of interpolation points and directional vectors in the derivative conditions. There are examples, however, of Hermite interpolation problems that are never poised, no matter what points and vectors we choose. We discuss questions in this respect in Section 4. In Section 5, we de ne a nite dierence associated to Hermite interpolation and use it to write the interpolating polynomial in its Newton form. In Section 6, we derive an integral remainder formula for all blockwise Hermite interpolation problems satisfying an additional regularity condition which, roughly speaking, is necessary due to the commutativity of directional derivatives. The remainder formula is new even for very simple interpolation schemes; it extends the one we proved for Lagrange interpolation in [10]. From the remainder formula, one can derive estimates on the error of interpolation in terms of the derivatives of functions. To illustrate the various de nitions and results, numerous examples are presented throughout the paper. In the remain of this section, we give the basic notation that will be used throughout the paper. We will use standard multiindex notation. For each multiindex  = (1; : : :; d) 2 Nd0 we write jj = 1 + : : : + d, and ! = 1!


d!; we also de ne the multinomials jj! := jj! = jj! :  ! 1!


d ! For x 2 Rd we write x = (1; : : : ; d) and for  2 Nd0 we denote by

x = 1


dd ; 1

a monomial of total degree jj. A natural basis for dn is formed by the monomials fx : 0  jj  ng. Let rnd denote the number of all monomials x which have total degree exactly n. It follows that ! ! n + d n + d ? 1 d d and dimn = n : rn = n 2

2 Hermite Interpolation Polynomials In this section we give the de nition of Hermite interpolation problems. As already mentioned in the introduction, by Hermite interpolation at a point we mean interpolating a chain of directional derivatives and any chain of directional derivatives subordinate to it. An example of Hermite interpolation conditions at one point is as follows.

Example 2.1. Let x0 be a point in Rd, y1; y2; y3 and z be nonzero vectors in Rd. The

following is an example of Hermite interpolation conditions, f (x0); Dy f (x0); Dy Dy f (x0); Dy Dy Dy f (x0); Dz Dy f (x0); where there is no further assumption imposed on yi and z. 1

2

1

3

2

1

1

2

To de ne Hermite interpolation in such a generality, we need to nd a way to describe the assignment of derivatives without being overwhelmed by the notation. One way of doing so is by using the notion of a tree from graph theory. Indeed, in the above example, we can take x0; y1; y2; y3, and z as vertices of a tree, and describe the interpolation condition by specifying the maximal chains (branches) in the tree. The precise de nition of this concept is given below. Let x 2 Rd and m 2 N0 . We consider an index set

E = (1j"11; : : :; "1rd j : : : j"m1 ; : : :; "mrmd ); where "ki = 0 or 1, and a diagram Tx = Tx;E de ned by Tx = (xjy11; : : : ; yr1d j : : : jy1m; : : :; yrmmd ); 1

1

(2.1) (2.2)

where yik 2 Rd and yik = 0 if "ki = 0. We say that E is of tree structure, if for each "ki = 1, k > 1, there exists an j = j (i), uniquely determined, such that "jk?1 = 1, where "jk?1 is called the predecessor of "ki . Moreover, the edges of the tree connect only a vertex and its predecessor. If E is of tree structure, we call Tx a tree and call x root and the yik vertices of the tree; moreover, we denote by jTxj the number of nonzero vertices, including the root, in Tx. We consider an edge joining a vertex and its predecessor to be directed from the predecessor to the vertex. We remark that our de nition is more restrictive than the ordinary de nition of a tree in graph theory, where it usually means a directed graph without loop. To give a tree explicitly one usually needs to specify the set of edges; for our purpose, though, it is much easier to specify maximal chains. A sequence  = (i1; : : :ik ) is called a chain in the tree structure E , if "1i = : : : = "kik = 1 where for each j , 1  j  k ? 1, "jij is the predecessor of "jij+1 ; it is called a maximal chain if its nal vertex "kik is not the predecessor of another element in E . If  = (i1; : : : ik ) is a chain and j  k, then we say that 0 = (i1; : : :; ij ) is subordinate to , which we denote by 0  . Clearly, a tree structure can be uniquely determined by specifying all maximal chains; moreover, 1

+1

3

every vertex in a tree Tx is connected to the root by exactly one chain. Let us look at an example.

Example 2.2. Let E = f1j1; 0j1; 1; 0j1; 0; 0; 0g and Tx = fx0jy1; 0jz; y2; 0jy3; 0; 0; 0g. The 0

tree structure is de ned by giving the maximal chains

1 = (1; 2; 1); 2 = (1; 1): The tree structure is most easily seen by its graph, which is depicted in Fig. 1.

(a)

2

(b)

Figure 1: All later examples of trees in this paper will be presented by their graphs as in Fig. 1. In drawing the graph of the tree Tx we use \" for the root and \" for other vertices in the graph. For examples in R2, the arrows which connect the vertices may depict the actual directions of the vectors they represent, as in (a) of Fig. 1; in general, however, they merely stand for the connectivity in the tree, as in part (b) of Fig. 1. To distinguish the two possibilities, we will align the vertices in the latter case vertically according to the levels in Tx; i.e., vertices y1j ; : : : ; yrjjd which form level j of Tx are aligned in a column, which we also emphasize by gray vertical bars as in part (b) of Fig. 1. The vertices corresponding to "kj = 0 are not used in the tree, they are depicted by unconnected circles. Let  = (i1; : : :ik ) be a chain in Tx; we write it as  2 Tx from now on. We de ne y

= yi1 : : :yikk and () = k 1

and a dierential operator

Dy() = Dyi : : :Dyik ; which is a multiple directional derivative, or a chain of directional derivatives. We suppress the superscripts of the directions yikk in order to keep the notation simple. Moreover, for convenience, we interpret  for which () = 0 as 0 and Dy0 as the identity operator I . Hermite interpolation at a point x can now be described by a tree structure in the 1

0

4

following way: whenever a chain of directional derivatives is interpolated, all subordinate chains of directional derivatives are interpolated as well. As an example, we can describe the Hermite interpolation conditions given in Example 2.1 by Dy() f (x0); for all  2 Tx; where the tree Tx is given as in Example 2.2; the derivative interpolation conditions correspond to the chains in Tx, where the maximal chains are (1; 1) and (1; 2; 1) whose subordinate chains are (1) and (1), (1; 2). Now we are ready to give the de nition of the Hermite interpolation polynomials that we will study in this paper. Let M 2 N0 and let x1; : : : ; xM be pairwise distinct points in Rd. For 1  k  M , let mk 2 N0 such that M X mk = dimdn (2.3) k=1

for some n 2 N0 . For each k, 1  k  M , we associate to xk a tree, denoted by Tk , whose root is xk and jTkj = mk .

De nition 2.3. Let x1; : : :; xM and T1; : : : ; TM be as above. The problem of Hermite interpolation is to nd a polynomial P 2 dn such that for any f : Rd 7! R, f 2 C m(Rd), where m = maxfmk : 1  k  M g, Dy() P (xk ) = Dy() f (xk );  2 Tk ; 1  k  M: (2.4) If this problem has a unique solution, we say that the Hermite interpolation problem is poised, and we denote the interpolating polynomial by Hn(f ;
). Since this de nition of the Hermite interpolation problem is rather general, we illustrate it by a few examples before we proceed. Instead of explicitly writing down the interpolation conditions, we will state the examples by showing the graph of their trees at each node, as done in Example 2.2. In our rst example, however, we give the interpolation conditions explicitly.

Example 2.4. Let x1; x2 2 R2 and let e1 = (1; 0), e2 = (0; 1). Of course, De = @=@1 1

and De = @=@2. The interpolation conditions are f (x1); De f (x1); De +e De f (x1); De De f (x1) f (x2); De ?e f (x2) : This situation is depicted in Fig. 2 together with the structure of the trees T1; T2. 2 Example 2.5. For two points x1; x2 on R2 and 6 interpolation conditions, there can be a number of interpolation problems. In Fig. 3 we depict the graph of some of them, where we do not specify the directional vectors of the derivatives. Note that the graph in Fig. 3(a) corresponds to Example 2.4. 2 2

1

1

1

2

2

5

1

2

1

T1

T2

Figure 2:

(a)

(b)

(c)

(d)

Figure 3:

Example 2.6. Uniform Hermite interpolation problem. If at each point the interpolation

conditions are the same, the Hermite interpolation problem is called uniform Hermitian. If, in addition, the interpolation conditions consist of all partial derivatives of order up to m, then the Hermite interpolation problem is called uniform Hermitian of type total degree (cf. [7]). In the latter case, the interpolation conditions (2.4) are usually written as @ P (x ) = @ f (x );  2 Nd ; jj  m; 1  k  M ; (2.5) 0 @1


@dd k @1


@dd k the corresponding tree, T , which is the same for every node, is a complete directed graph. In this particular case, all directional vectors are parallel to coordinate axes. We depict the cases d = 2, m = 1; 2; 3 in Fig. 4. Let us indicate how the case m = 3 in Fig. 4 ts into the notation of trees as in (1.1). To describe the tree structure as in (1.1), we need to impose an order on the derivative conditions at each level and specify the maximal chains; dierent orders will lead to dierent ways of writing down the tree. The most canonical way seems to be as follows: o n T = x0je1; e2je1; e1; e2je1; e1; e1; e2 1

1

with the maximal chains (1; 1; 1), (2; 2; 2), (2; 3; 3) and (2; 3; 4). We depict this choice in Fig. 5 (a); the alignment of the vertices is shown in part (b). 2 6

m=1

m=2

m=3

Figure 4:

(a)

(b)

Figure 5:

Example 2.7. Partially uniform Hermite interpolation problem. If we only take a subset of the possible partial derivatives in (2.5), then we will speak of a partially uniform Hermite interpolation problem. To be precise, let fi1; : : :; ipg be a subset of f1; 2; : : : ; dg.

Then the interpolation conditions of equation (2.5) are restricted as follows: @ f (x );  2 Np ; jj  m; 1  k  M: @ P (x ) = 0 @i


@ipp k @i


@ipp k One of the simplest examples is to interpolate @=@1 at each point. 2 As mentioned above, the tree depicted in Fig. 5 is only one choice among several options; in other words, there are several, though equivalent, ways to formulate the same interpolation problem. In general, since the tree structure does not require specifying the directional vectors, the way we formulate a problem depends not only on the tree structure, but also on the directions we select. To illustrate this point, let us rst consider a simple example. 1 1

1 1

7

Example 2.8. Equivalent interpolation problems. Consider two interpolation problems of rst order at a point x0 2 R2. In the rst one we interpolate the partials De and De ; 1

2

in the second one we interpolate Dy and Dy , where y1 = (11; 12) and y2 = (21; 22) are two linearly independent vectors both dierent from e1 and e2. The graphs are depicted in Fig. 6. Clearly, the two problems have the same tree structure. For f 2 C 1(Rd), let 1

2

(a)

(b)

Figure 6:

p 2 11 be the unique solution of the interpolation problem p(x0) = f (x0);

De p(x0) = De f (x0); 1

1

De p(x0) = De f (x0): 2

2

Then it follows readily that

Dy p(x0) = 11De p(x0) + 12De p(x0) = 11De f (x0) + 12De f (x0) = Dy f (x0); 1

1

2

1

2

1

and, similarly, Dy p(x0) = Dy f (x0). In other words, the solution of the rst interpolation problem is also a solution of the second one, and, as long as y1 and y2 are linearly independent, the solution of the second problem is the solution of the rst one as well. 2 2

2

This example suggests the following de nition. De nition 2.9. Suppose the Hermite interpolation problem with points x1; : : :; xM and trees T1; : : :; TM and another one with points x^1; : : :; x^M and trees T^1; : : :; T^M , respectively, are both poised. Let their corresponding interpolation polynomials be denoted by Hn (f;
) and H^ n (f;
), respectively. We say that the two problems are equivalent if Hn (f; x) = H^ n(f; x); f 2 C 1(Rd); x 2 Rd: It is easily seen that two interpolation problems are equivalent if, and only if, the interpolation operators commute, i.e.,   Hn H^ n (f;
);
= H^ n (Hn (f;
);
) ; f 2 C 1(Rd): The notion of equivalent interpolation problems will be of interest later in Section 6. Example 2.10. Two Hermite interpolation problems can be equivalent even though they have dierent tree structures, as shown by a simple example in Fig. 7 2 8

(a)

(b)

Figure 7:

3 Blockwise Structure The Hermite interpolation problem de ned in the previous section is very general; in order to approach it, we make an additional assumption on the structure of the interpolation scheme, which is the notion of blockwise interpolation that we introduced in [10] for the study of the Lagrange interpolation problem. To describe this notion we need further notation. Let x1; : : :; xM and T1; : : :; TM be as in De nition 2.3. Since each point xk has mk = jTk j interpolation conditions attached to it and the conditions are Hermitian, we can consider xk as a point of multiplicity mk , or as a point having mk identical copies. We denote the copies of xk by xjk;i such that

xjk;i = xk ; for "jk;i = 1; where "jk;i is from the index set Ek correspondingPto Tk as in (2.1) and (2.2); we note that "0k;1 = 1 according to (2.2). Recalling that mk = dimdn, we denote by Xn the collection of all nodes with their copies,

Xn = fxjk;i : "jk;i = 1; 1  k  mk g;

(3.1)

the cardinality of Xn is dimdn. For each k we say that xjk;i is a copy of xk in level j . Our main assumption on the Hermite interpolation problem is that Xn can be put into a proper block form; in other words, that the points and conditions can be indexed in a consistent way by using multi-indices. If a set of points is given in the form fxgjjn, we say that it is in block form and call the sets fxgjj=k , 0  k  n, blocks.

De nition 3.1. The Hermite interpolation problem de ned in (2.4) is said to be of blockwise structure, if Xn can be rearranged into the block form, i.e., if Xn = fx : j j  n; 2 Nd0 g; 9

such that for each k, copies of xk from the same level are in the same block and the order of the levels is preserved in the order of blocks. Precisely, two copies of xk , say xjk;i and xlk;i , are in the same block if and only if j = l. If j < l, then xjk;i is in the n1-th block and xlk;i is in the n2-th block for some n1 < n2. We call an interpolation problem with blockwise structure a blockwise Hermite interpolation problem. Remark 3.2. It is worth to note that every univariate Hermite interpolation problem has a blockwise structure. Indeed, for a univariate Hermite interpolation problem which interpolates up to (i1 ? 1)-th derivatives on xi, x1 <


< xN , with polynomials of degree n, the block structure merely means to write xi as multiple points and order them as x1 =


= xi < x2 =


= xi <


< xN =


= xiN ; where it is necessary that i1 + : : : + iN = n + 1, since the interpolation polynomials are of degree n. 1

2

1

2

1

2

Blockwise structure is a property possessed solely by the interpolation scheme, it has nothing to do with the actual choice of nodes or directional derivatives. Many Hermite interpolation problems have blockwise structure, although not all of them. Again we illustrate this concept by examples and state these examples by giving their graphs. Blockwise structure means to arrange the trees for all nodes into a joint diagram such that each tree preserves its structure in the diagram: vertices from the same level of each tree appear in the same level of the joint diagram, vertices from lower levels also appear in the lower levels of the joint diagram. In depicting the graph of the joint diagram, we align the vertices according to the blocks, which are the levels of the joint diagram, and emphasize the alignment by gray vertical bars again.

Example 3.3. The examples (a), (b), (c) in Example 2.5 can be put into blocks; the corresponding joint diagrams of the tree structures are depicted in Fig. 8.

(a)

(b)

2

(c)

Figure 8: For a given Hermite interpolation problem, there may be several dierent ways of putting it into blocks; in other words, the blockwise structure of an interpolation scheme is not unique. The following example depicts the non{uniqueness. 10

Example 3.4. Let x1; x2; x3 be points in R2, and y1; y2; z 2 R2 be directional vectors.

We consider the Hermite interpolation problem de ned by interpolating f (x1); Dy f (x1); Dy f (x1); f (x2); Dz f (x2); f (x3); this interpolation scheme can be put into blocks in several dierent ways which are depicted in the Fig. 9. 2 1

(a)

2

(b)

(c)

Figure 9: Let us remark that, by computing the general Vandermonde determinant (see below), it is not hard to show that (d) in Example 2.5 does not give rise to a poised interpolation problem for any choice of points. On the other hand we also have

Example 3.5. The interpolation scheme of (d) in Example 2.5 cannot be put into blocks. 2

Actually, for d = 2, the case (d) in Example 2.5 is a prototype of uniform Hermite interpolation of rst total degree, which means m = 1 in Example 2.6, and of two nodes. In general, in order that a uniform Hermite interpolation problem of rst total degree is poised, it is necessary that the number of nodes, M , satis es the equation 3M = (n + 1)(n + 2)=2 = dim2n for some n 2 N0, see (2.5). The case (d) in Example 2.5 corresponds to M = 2 and n = 2. It has been pointed out (cf. [7, p. 32]) that the uniform Hermite interpolation problem is not poised for M  6, which consists of the two cases M = 2; n = 2 and M = 5; n = 4, and is almost poised for M  7 (cf. [7, p. 117]). The case M = 7 corresponds to n = 5, which leads to interpolation at 7 points with 21 conditions. It can be easily seen that for M = 2; 5 the uniform Hermite interpolation of the rst total degree cannot be put into blocks, while the case M = 7 is given in the next example.

Example 3.6. The blockwise diagram of the uniform Hermite interpolation problem of

rst total degree with d = 2, M = 7, is depicted in Fig. 10. 2 From now on we assume that the Hermite interpolation problem under consideration has already been put into blocks. For Xn in block form it is more convenient to identify 11

Figure 10: a point by using multiindices. If x = xji;k and  = (i1; : : :; ij ) 2 Tk is a chain with ij = i j to the root, then we which, we recall, means that  is a chain connecting the vertex yi;k de ne  = (); Dy = Dy() ; i.e. y = y : This slight abuse of symbols should not cause any confusion. The advantage of this notation, however, lies in the fact that a Hermite interpolation problem can now be completely described by the pairs (x ; Dy ), j j  n. Combining De nition 2.3 and De nition 3.1, the de nition of blockwise Hermite interpolation can be restated as follows.

De nition 3.7. Let x1; : : :; xM and T1; : : : ; TM be as in De nition 2.3, and let Xn in (3.1)

be of blockwise structure as in De nition 3.1. Then the problem of blockwise Hermite interpolation is to nd a polynomial P 2 dn such that for any f : Rd 7! R, f 2 C m(Rd),

Dy P (x ) = Dy f (x );

j j  n:

(3.2)

If the problem has a unique solution, we say that the interpolation problem is poised and denote the polynomial solving (3.2) by Hn (f ;n
). If for each k, 0o k  n, the Hermite interpolation problem corresponding to Xk = (x ; Dy ) : j j  k is poised, then we say that the interpolation problem is poised in block. For  2 Nd0 we let  denote the monomials in d,

(x) = x;

 2 Nd0 :

The generalized Vandermonde matrix associated to the Hermite interpolation problem (3.2) is de ned by   V (Xn) = Dy (x ) jjn;j jn ; 12

where the rows and columns are arranged according to the lexicographical order in f : jj = kg and 0  k  n. Since every interpolation problem can be rewritten as a linear system of equations in terms of coecients of polynomials, the following proposition is obvious. Proposition 3.8. The Hermite interpolation problem (3.2) is poised if, and only if, det V (Xn) 6= 0. For a poised Hermite interpolation problem, the fundamental interpolating polynomials, denoted by q 2 dn, jj  n, are uniquely de ned by the conditions

Dy q (x ) =  ; ; if j j; j j  n;

(3.3)

where  ; =  ; : : : d; d and  i; i is the Kronecker delta. The reason that they are called fundamental lies in the fact X  Hn (f ; x) = Dy f (x )q (x): 1

1

j jn

Actually, these polynomials can be given explicitly as follows: for 2 Nd0 , j j  n, let V (Xnjx) be matrices de ned by   

V (Xnjx) = Dy (x ) jjn;j jn ; x = xx; ; ifif 6= = . Then the fundamental polynomials of Hermite interpolation are given by V (Xnjx) ; j j  n; 2 Nd : q (x) = det 0 det V (Xn) Since each of the fundamental polynomial is the solution of the Hermite interpolation problem with zero-one conditions, the set of q, jj  n, forms a basis for dn. It is very dicult, however, to compute the polynomials q practically by working with the determinants. For our study below it is much more convenient to represent Hn (f ;
) in another form, which is the analogy of the Newton interpolation formula in one variable. To do so, we need to change the basis of dn from fq : jj  ng to another one. The elements of the new basis are denoted by p, where jj  n; we call p Newton fundamental polynomials. The polynomial p 2 dj j is uniquely de ned by the conditions

Dy p (x ) =  ; ; if j j  j j:

(3.4)

For each level k there are rkd many Newton fundamental polynomials which are of degree exactly k. Each of them is the solution of a Hermite interpolation problem which consists of the rst k blocks of the original interpolation problem with zero-one conditions. Since we assume that the interpolation problem is poised in block, each matrix V (Xk ), 0  k  n, has nonzero determinant. The polynomials p are related to q. Indeed, if 2 Nd0 , 13

j j = n, then p = q . For later use, we write down the Newton fundamental polynomials explicitly as V (Xk jx) ; j j = k; 0  k  n; 2 Nd : p (x) = det (3.5) 0 det V (X ) k

The Newton formula for Hermite interpolation polynomials will be given in Section 5.

4 Some aspects of poisedness In this section we deal with some respects of structural questions of poisedness, more precisely, almost poisedness. The main dierence is that, loosely spoken, the poisedness of a Hermite interpolation problem depends on the exact choice of the points and directions, while almost poisedness depends only on the tree structure of the interpolation problem. To make it more precise, let us again have a look at the Vandermonde matrix of the Hermite interpolation problem   V (Xn) = (Dy (
))(x ) jj;j jn :

The Vandermonde determinant, det V (Xn), is then a polynomial in the variables x and y , jj  n. This polynomial can be identically zero, which means that the Hermite interpolation problem can never be solved, or it can vanish only on a set of measure zero, which means that the interpolation problem is solvable for almost all choices of x and y . This leads to the following de nition: De nition 4.1. The Hermite interpolation problem (3.2) is called almost poised if there exists a collection of points x and directions y, jj  n, such that the associated Vandermonde matrix V (Xn) is nonsingular. A Hermite interpolation problem which is not almost poised will be called never poised. We note that even when all directional vectors y are xed in a Hermite interpolation problem, for example, when they are taken as coordinate vectors so that the Hermite conditions involve only partial derivatives, the notion of almost poisedness still applies; the only dierence is that in this case only the points to be interpolated at are free parameters. Actually, the almost poisedness in this sense is used in [7] for the partial uniform or uniform Hermite interpolation of total degree and, more generally, for Birkho interpolation. Clearly, the almost poisedness in the De nition 4.1 is more general in the sense that it is less restricted. We shall not try to formally distinguish the case when directional vectors are xed, but we will point it out whenever the results warrant. Before we discuss the necessary and/or sucient conditions for Hermite interpolation problems to be almost poised, let us look at one of the simplest example of a never poised problem. Example 4.2. (Case (d) in Example 2.5.) Consider two points x1 6= x2 2 R2 and directions y1;1; y1;2; y2;1; y2;2. The Hermite interpolation problem which interpolates f (x1); Dy ; f (x1); Dy ; f (x1); f (x2); Dy ; f (x2); Dy ; f (x2) 11

12

21

14

22

2

is never poised in 22.

It is not hard to verify that the problem in this example is never poised by computing its Vandermonde determinant, for example. We do not give a proof here because this con guration violates a necessary condition for almost poisedness which we are going to state in the following. To derive our necessary condition for the poisedness of Hermite interpolation, we use what is known as Bezier representation of polynomials which is primarily used in CAGD. Since it may be alien to readers from other elds, we will rst recapitulate the basic idea of the Bezier representation of a multivariate polynomial. We refer to [3] for an extensive survey and additional references. Let v0; : : :; vk 2 Rd, k  d, be given and assume that they are Din general position; i.e., E the vectors vj ? v0, j = 1; : : : ; k, are linearly independent. Let v0; : : :; vk denote the ane hull of v0; : : :; vk ; i.e., D0 E v ; : : : ; vk = fc0v0 + : : : + ck vkjci 2 R; c1 +


+ ck = 1g: E D Then each point x 2 v0; : : : ; vk can be uniquely described by its barycentric coordinates u = u(x) = (u0(x); : : :; uk (x)) as follows:

x=

k X i=1

k X

ui(x)vi;

i=1

ui(x)  1:

i h In particular, the points in v0; : : : ; vk , which denotes the k{dimensional simplex spanned by v0; : : :; vk , are characterized by the additional condition ui(x)  0,h i = 0; : : :i; k. For P 2 dn , let p = P j[v ;:::;vk ] denote the restriction of P on the simplex v0; : : :; vk . Then we can use barycentric coordinates to write p as ! X j  j p(x) = p B(x); B(x) =  u(x) =  !j

j
! ! u0(x)


uk (x)k ; 0 k jj=n 0

0

where  = (0; : : : ; k ) 2 N0k+1 . This is the so{called Bernstein{Bezier representation of p. The coecients with respect to this representation, p, provide convenient access to geometric properties of the polynomial p. In particular, it follows that

p(vj ) = pnej ; and

Dvi ?vj p(x) =

X jj=n?1

j = 0; : : : ; n;

(Ei ? Ej ) p B(x);

(4.1) 0  i; j  k;

(4.2)

where the formal shift operators Ej , j = 0; : : : ; k, are de ned as Ej p = p+ej . Note that (4.2) is a rst order dierence applied to the coecients p , jj = n. Iterating (4.2) and invoking (4.1), one obtains 15

Lemma 4.3. Let 1  k  d and let 2 Nk0+1 such that j j  n and j = 0 for some xed 1  j  k. Then Dv ?vj p(vj ) = Dv ?vj


Dv kk ?vj p(vj ) = (E0 ? Ej )


(Ek ? Ej ) k p(n?j j)ej : (4.3) 0

0 0

To state the necessary condition in its most general form we have to introduce a quantity which measures to what extent an interpolation problem contains locally a subproblem which is of type total degree. Precisely, for each k = 1; : : : ; M , let k be the maximal number such that the tree Tk , associated to xk , contains a subtree of order k which, upon properly choosing directions yik , contains all partial derivatives of order  k . In other words, k is the maximal number such that the rst k + 1 levels of the tree Tk , root included, are equivalent (in the sense of De nition 2.9) to a tree which is of type total degree with order k . For example, for Lagrange interpolation at a point xk we have k = 0, but we also have k = 0 if we interpolate only Dyj for j = 0; : : :; mk , at xk . In general, k = 0, k = 1; : : : ; M , for problems that are partially uniform but not uniform of total degree. Theorem 4.4. Let a Hermite interpolation problem with points xk and trees Tk , k = 1; : : : ; M , be almost poised in dn. Then j + k < n for any 1  j; k  M . Proof : Without loss of generality let us assume that there exists a poised Hermite interpolation problem such that 1 + 2  n. Let P 2 dn denote the unique solution of the problem P (x1) = 1; and Dy() P (xj ) = 0; otherwise: In particular, using the de nition of 1 and 2, we have @ jj P (x ) = 0; 1  jj   ; i = 1; 2: P (x1) = 1; P (x2) = 0; @x (4.4) i i  Let p denote the restriction of P to the linear segment joining x1; x2, denoted by [x1; x2], and write p in its Bezier representation ! n X n p(x) = pi Bi(x); Bi(x) = i ui(1 ? u)n?i; i=0 where u = u1(x) and 1 ? u = u0(x) are the barycentric coordinates of x with respect to [x1; x2], i.e., x = (1 ? u)x1 + ux2. Using (4.4) and applying Lemma 4.3 with = (0; k), where 0  k  1, we obtain that (E2 ? E1)k p(n?k)e = k;0; 0  k  1; 1

which implies inductively that p0 =


= p = 1. Similarly, choosing = (k; 0), we conclude that pn? =


= pn = 0. However, under the assumption 1 + 2  n or, equivalently, 1  n ? 2, we end up with the contradiction that p has to have value 0 and 1 at the same time. 2 1

2

1

16

Clearly, Theorem 4.4 can be applied to the case when directional vectors in the Hermite conditions are xed. The fact that the interpolation problem in the Example 4.2 is never poised follows easily from the theorem: indeed, in this case 1 = 2 = 1 and n = 2. Another example is as follows:

Example 4.5. Consider a Hermite interpolation problem at x1; x2; x3; x4; x5 2 R2 in 24,

which has Lagrange interpolation condition at x1; x2; x3 and uniform Hermite condition of total degree 2 at x4; x5; the total number of interpolation condition is 15 = dim24. In this case, we have 1 = 2 = 3 = 0 and 4 = 5 = 2. Since 4 + 5 = 4, the interpolation problem is never poised in 24 by Theorem 4.4. 2 The necessary condition in the Theorem 4.4 seems to be new even when all directional derivatives are xed as partial derivatives. In particular, for uniform Hermite interpolation of total degree, we can apply Theorem 4.4 to derive the following result: Corollary 4.6. If a uniform Hermite interpolation problem of nonzero total degree at M points in Rd, M  2 and d  2, is almost poised in dn, then hdi ( n + 2)( n + 4)


( n + 2 2 ) h d+1 i h d+1 i : (4.5) M  2d (n + 2 2 + 1)(n + 2 2 + 3)


(n + 2d ? 1)

Proof : Suppose that a uniform Hermite interpolation problem at M points of total degree r is poised in dn. By Theorem 4.4, it is necessary that 2r < n, or equivalently, 1  r  n?2 1 : Moreover, the number of interpolation   conditions has to equal to the

dimension of dn, which means that M r+d d = n+n d . Therefore, it follows that n+d 1)


(n + d)  (n + 1)


(n + d) M = r+d d = ((nr + + 1)


(r + d) ((n ? 1)=2 + 1)


((n ? 1)=2 + d) ; d

where the last expression is the same as the right hand side of (4.5). 2 In particular, since the right hand side of (4.5) can be easily seen to be a monotone function in n, we have Yd + 3 M  jj + 1 > d + 1; j =1

where the last inequality follows easily by induction for d  2. Therefore, as a consequence of Corollary 4.6, we have given a simple proof of the following result due to G.G. Lorentz and R.A. Lorentz (see [7, p. 27]). Corollary 4.7. If a uniform Hermite interpolation problem of nonzero total degree at at least two points in Rd, d  2, is almost poised, then it contains more than d + 1 points. 17

In other words, this corollary states that the only poised uniform Hermite interpolation problem of total degree in Rd, d  2, with fewer than d + 1 points is the Lagrange one. The condition M > d + 1 in Corollary 4.7 is sharp if we consider the class of all uniform Hermite interpolation problems of total degree. On the other hand, for each n, the necessary condition in Corollary 4.6 gives a much stronger result for d > 2. Indeed, for d = 3 and n  4, it states that + 2 = 16 > 5; M  8 44 + 5 3 while Corollary 4.7 gives only M > 4. We explain the application of Corollary 4.6 further in the following example.

Example 4.8. If a uniform Hermite interpolation problem   of type total degree with order r at M points in Rd is poised, then the condition M r+d d = n+n d has to be satis ed. For d = 3; 4; 5, solutions of this equation are found in [7, p. 47] for small M . In particular, for d = 3 two of the solutions are M = 6; n = 7; r = 3; and M = 7; n = 19; r = 9; for which the poisedness of the corresponding interpolation problems was left open. For d = 3, condition (4.5) reads + 2; M  8 nn + 5 the case n = 19 and M = 6 clearly violates this inequality, thus, not poised. The case n = 7 and M = 6 satis es the inequality with equal sign, hence, its poisedness can not be decided by the necessary condition (4.5). Similarly, we can conclude from the condition (4.5) that the two other undecided cases in [7, p. 47] d = 4 : M = 6; n = 6; r = 3; d = 5 : M = 12; n = 5; r = 2; are never poised. 2 Sucient conditions for almost poisedness are usually more dicult to establish. It turns out that the argument similar to the one used in the proof of Theorem 4.4 also leads to a sucient condition for those Hermite interpolation which contains only trees of total degree; it applies whenever the number of genuine Hermite interpolation points is small enough. Theorem 4.9. Let a Hermite interpolation problem be given which consists of trees T1; : : : ; TM of total degree such that only k  d + 1 of the trees, say T1; : : :; Tk , are of nonzero order nj > 0, j = 1; : : : ; k. Then the Hermite interpolation problem is almost poised in dn if and only if

1  i; j  k:

ni + nj < n; 18

(4.6)

Before we prove this theorem, we rst show that Lagrange interpolation is always almost poised in any polynomial subspace. This extends yet another theorem of Lorentz who proved that Lagrange interpolation is almost poised for any polynomial subspace decided by a lower set [7, p. 61]. Lemma 4.10. Let p1 ; : : :; pn be linearly independent polynomials in dn. Then the Lagrange interpolation problem of n points in Rd is almost poised in the span of p1 ; : : :; pn . Proof : We use induction on n; the case n = 1 is trivial. Suppose the claim has been proved for n  1 and let p1; : : :; pn+1 be linearly independent. By induction hypothesis, there exist q1; : : : ; qn in the span of p1; : : :; pn and points x1; : : :; xn such that qi(xj ) = ij ; i; j = 1; : : : ; n: Because of linear independence, the polynomial n X qn+1 = pn+1 ? pn+1 (xi)qi i=1

does not vanish identically; hence, there is some xn+1 such that qn+1(xn+1 ) 6= 0. Now we set  qn+1(x) = qqn+1(x(x) ) and qi = qi(x) ? qi(xn+1 )qn+1(x); i = 1; : : : ; n: n+1 n+1 We observe that qi(xj ) = ij , 0  i; j  n + 1, therefore, these polynomials are the Lagrange fundamental polynomials with respect to x1; : : :; xn+1 in the span of p1; : : :; pn+1 . In particular, the interpolation problem is poised. 2 Proof of Theorem 4.9 : Clearly the necessity of (4.6) follows from Theorem 4.4. To prove the suciency, we choose x1; : : :; xk to be the points associated to T1; : : :; Tk and a set of additional points xk+1; : : : ; xd+1 such that these points are in general position; i.e., the simplex
= [xj : j = 1; : : : ; d + 1] has dimension d. We use the Bernstein-Bezier representation of an interpolation polynomial p 2 dn. Using Lemma 4.3 and induction on j j again, we see that the interpolation conditions determined by Tj exactly de ne the values of the coecients p , j  n ? nj , j = 1; : : : ; k. Moreover, by means of (4.6), the index sets f : j  n ? nj ; j j  ng, j = 1; : : : ; k, do not intersect each other. In addition, we observe that the polynomials B , j < n ? nj , j = 1; : : : ; k, annihilate all interpolation conditions determined by T1; : : :; Tk ; thus, they form a basis for the space of polynomials that annihilate these interpolation conditions. Since, by Lemma 4.10, Lagrange interpolation is almost poised in any polynomial subspace, hence, in particular, in the span of B , j < n ? nj , j = 1; : : :; k. Therefore we can nd additional points xd+2; : : : ; xM such that the original Hermite interpolation problem is poised. 2 We remark that Theorem 4.9 can be applied to the case when directional vectors in the Hermite conditions are xed. 19

The sort of Hermite interpolation problems in Theorem 4.9 may be used in connection with nite element methods; for example, a triangular nite element scheme yields a global function u on a triangulated domain by de ning it locally in each triangle as an interpolation polynomial, where one usually interpolates derivatives up to a certain order at the vertices of triangles and determine the remaining degrees of freedom by additional constraints. If the additional constraints are given in terms of Lagrange interpolation conditions, then Theorem 4.9 characterizes the solvability of all interpolation of this sort. A typical example is given as follows: Example 4.11. Let x1; x2; x3 2 R2 be distinct points and assume that they are in general position, which means that they are vertices of a non-degenerate triangle. Consider a Hermite interpolation problem in 23 which interpolates the values of f , @x@ and @y@ at x1; x2; x3 as well as interpolates the value of f at an additional point x0. By Theorem 4.9 this interpolation problem is almost poised. In Fig. 11 we depict the graph of the interpolation problem together with its tree structure.

Figure 11: Actually, by computing the Vandermonde determinant of the interpolation problem, it is found that the problem is poised as long as x0 is not on one of the edges of the triangle (cf. [7, pp. 76 and 154]). In particular, the point x0 can be located at the center of the triangle, i.e., x0 = (x +1 +x2 + x3)=3. 2 Clearly, many other type of nite elements can be veri ed this way. One of them is as follows. Example 4.12. Let x1; x2; x3 2 R2 be vertices of a triangle. Consider a Hermite interpolation problem in 25 which interpolates the values of f and its partial derivatives up to second order at x1; x2; x3, and interpolates f at three additional points inside the triangle. This problem has 6 points and 3  6 + 3 = 21 = dim25 conditions. Again, by Theorem 4.9, the interpolation problem is almost poised. Moreover, since the \free" Bernstein{Bezier basis polynomials in the proof of Theorem 4.9 all vanish on the boundary of the triangle, this interpolation problem is poised as long as the additional points for Lagrange interpolation do not lie on the edges of the triangle. 2 20

5 Finite dierence and Newton formula Suppose that the Hermite interpolation problem de ned in De nition 3.7 is poised in block. We will de ne a nite dierence associated to it. To simplify the notation, we introduce vectors h i xn = [x]jjn and x = x0 ; x1; : : : : De nition 5.1. Let x and y be as in the blockwise Hermite interpolation problem. The nite dierence in Rd, denoted by n [x0; : : : ; xn?1; x]f; x 2 Rd; is de ned recursively as 0[x]f := f (x) (5.1) 0 n 0 n ? 1 n+1 [x ; : : : ; x ; x]f := n [X x ; : : :; x ; x]f ? Dy n [x0; : : :; xn?1 ; x ]f
p (x): (5.2) j j=n

The recursive de nition works in the following way: to compute the (n + 1)-st dierence, one needs the n-th dierence as a function in the parameter x as well as the values  of the n-th dierence and its Dy derivatives at the point x for all j j = n. For the Lagrange interpolation problem, this has been introduced in [10] and it is easy to see that in the case of a Lagrange interpolation problem, i.e.,  = 0, j j  n, the two notions actually coincide. The nite dierence is vital to our further development because of the following analogy of the Newton formula for the Hermite interpolation polynomials. Theorem 5.2. Let the Hermite interpolation problem de ned in De nition 3.7 be poised in block. Then X  Hn (f ; x) = (5.3) Dy j j[x0; : : :; xj j?1; x ]f
p (x): j jn

Proof : By the de nition of the Hermite interpolation problem, we need to verify the conditions

X j jn

Dy j j[x0; : : :; xj j?1; x ]f
Dy p (x) = Dy f (x); jj  n;

(5.4)

since (5.4) implies that the left-hand side of (5.3) satis es the interpolation conditions from (3.2). By the assumption that the Hermite interpolation polynomial is unique, this will conclude the proof. Because of the de nition of p , it suces to verify (5.4) for (x; Dy ), jj = n. We shall use the following formula: X  n [x0; : : :; xn?1 ; x]f = f (x) ? (5.5) Dy j j[x0; : : : ; xj j?1; x ]f
p (x); j jn?1

21

which follows from (5.1) and (5.2) by iteration. Applying Dy on (5.5) and evaluating at x = x leads to Dy n [x0; : : : ; xn?1; x]f X  = Dy f (x) ? Dy j j[x0; : : :; xj j?1; x ]f
Dy p (x): (5.6) j jn?1

From the de nition of p , i.e., p (x) = ; if j j = n, we then have X  0 Dy n [x0; : : : ; xn?1; x]f = Dy n [x ; : : :; xn?1 ; x ]f
Dy p (x): j j=n

(5.7)

Together, equations (5.6) and (5.7) verify (5.4) for (x; Dy ). 2 The following theorem gives a remainder formula for the Hermite interpolation polynomials in terms of the nite dierence. Theorem 5.3. Let the Hermite interpolation problem de ned in De nition 3.7 be poised in block. Then f (x) ? Hn (f ; x) = n+1 [x0; : : : ; xn; x]f: (5.8)

Proof : Equation (5.8) follows immediately from (5.5) upon replacing n by n + 1 and using (5.3). 2 It is worth to point out that one can also take (5.8) as the de nition of the nite dierence n+1 [x0; : : :; xn ; x]f . One then needs to prove that both (5.2) and (5.3) are consequences of such a de nition, which can be done easily. We choose to de ne the nite dierence as in De nition 5.1 for two reasons: rst, the proof of the integral remainder formula in Section 6.2 will use induction based on the recurrence relation (5.1) and (5.2). Moreover, for Lagrange interpolation based on the lattice points inside a standard simplex, the nite dierence coincides with the classical forward dierence which is usually de ned recursively (cf. [11]). For the purpose of deriving the integral remainder formula, the recurrence relation of the nite dierence as de ned in (5.1) and (5.2) is very useful. For computational purposes, however, this recurrence relation may be dicult to use, because to compute the (n + 1){st dierence we have to compute directional derivatives of n{th dierences which is not numerically ecient. For this reason we introduce a modi ed nite dierence

n [x0; : : : xn?1; x; y]f; where y = (y1; : : :; y ); de ned by the recurrence

0[x; y]f = Dy f (x); (5.9) 0 n 0 n ? 1

n+1[x ; : : :; x ; x; y]f = n[xX; : : :; x ; x; y]f ? n [x0; : : :; xn?1 ; x; y]f
Dy p(x): (5.10) jj=n

22

The relation between the two dierences is then as follows

Proposition 5.4.

n[x0; : : :; xn?1; x; y]f = Dy n [x0; : : :; xn?1; x]f; therefore,

Hn (f ; x) =

X j jn

j j

h

x0 ;


; xj j?1; x ; y

y = (y1 ; : : :; y );

i

f
p (x)

(5.11) (5.12)

for any Hermite interpolation problem that is poised in block. Proof : Using (5.2) and (5.8) it can be easily veri ed that Dy n [x0; : : :; xn?1; x]f satis es the recurrence relations (5.9) and (5.10) which de ne n[x0; : : :; xn?1; x; y]f . The second statement is then obvious. 2 The advantage of the dierence n [x0; : : : ; xn?1; x; y]f lies in the fact that the dierentiations are performed on the function f and the polynomials p(x), where the latter ones can be easily handled numerically. Moreover, if we want to compute the Hermite interpolation polynomial as in (5.3), the recurrence relation for the dierence only involves the data to be interpolated as well as evaluation of polynomials and their derivatives. To compute the (n + 1){st dierence we have to compute all the n{th dierences

n [x0; : : :; xn?1 ; x ; y ]f for all j j = n rst, but this can be done by an eective triangular scheme similar to the case of the Lagrange interpolation (cf. [9]). Since we are mainly concerned with the remainder formula for the rest of this paper, we will not elaborate this respect any further.

6 Remainder Formula In this section we establish the integral representation of the nite dierence in terms of simplex splines from which we derive a remainder formula for the Hermite interpolation problem. For this purpose, however, we need to impose a regularity condition on the directions of the higher order derivatives in the Hermite interpolation problem. This condition is de ned and discussed in the rst subsection, Section 6.1. The remainder formulae will be stated and explained in Section 6.2; the proof itself will be given in Section 6.3. The remainder formula we will derive expresses f ? Hn f as a sum of integrals of (n +1)-st directional derivatives of f against simplex splines. Since the terms appearing in the sum are de ned through the notion of path which is simple in concept but complicated in notation, several examples are given in Section 6.4 to illustrate the formula. The regular condition imposed on Hermite interpolation is essential in deriving the general remainder formula, which allows a simple and perceptive geometric explanation; in case that this condition is not satis ed, the remainder formula takes a somewhat strange form, which we will explain through one of the examples in Section 6.4. Finally, we discuss the error estimation formula derived from the remainder formula in Section 6.5. From now on we always assume that the Hermite interpolation problem is poised in block; i.e., the points and derivatives are arranged as x and y , respectively, such that the 23

order of the underlying trees is preserved. More precisely, let us recall that if j j  j j  n such that x = x , then either  <  if j j < j j or  =  if j j = j j.

6.1 Regular Hermite interpolation

The notion of Hermite interpolation introduced in the previous section is fairly general and it is dicult to establish an integral remainder formula for all cases. In fact, we will have to restrict ourselves to a special (but still quite general) class of interpolation problems which we will name as regular Hermite interpolation. De nition 6.1. Let Tx be a tree. For each  = (i1; : : : ; ik ) 2 Tx and each integer j , 1  j  k, let j = (i1; : : :; ij?1; ij+1; ik ). The tree Tx is called regular if for each chain  and each j , there exists a chain 0 in Tx such that 0 y  = y j : A Hermite interpolation problem is called regular if all its trees T1; : : : ; TM are regular. Before we discuss the notion of regularity in depth, let us rst point out that regularity only applies to directional derivatives of second and higher order. This means that, as long as the Hermite interpolation problem only consists of Lagrange and rst order Hermite conditions, it is automatically regular. Let us also remark that regularity could be seen as a \commutativity" condition. Indeed, for k = (i1; : : : ; ik?0 1)   = (i1; : : : ; ik ), we can trivially choose 0 = (i1; : : :; ik?1) as a chain 0 such that y = yk as required in De nition 6.1. In fact this follows from the tree structure of the Hermite interpolation problem which automatically interpolates all subordinate chains of directions whenever a chain of directions is interpolated. On the other hand, if we look at the dierential operator Dy() , we note that Dy() = Dyi


Dyik = Dyi


Dyi k ; for any permutation  of the numbers 1; : : : ; k; directional derivatives clearly commute as long as they are applied to functions that are suciently smooth. Hence, it is reasonable to assume that for any such permutation of the directions there exists a chain 0 2 Tx such that 0 y  = yi


yi k? : In other words, the commutativity of the directions should carry over to the existence of the respective chains which is equivalent to regularity. There is also a second argument justifying the assumption of regularity, which is related to the ideal generated by an interpolation scheme. It is well known that in the case of Lagrange interpolation, where the interpolating polynomial is denoted by Ln f , all polynomials p 2 d such that Ln p = 0 generate an ideal, say I  d; i.e., for any q 2 d and p 2 I we have p
q 2 I . Let us look how this property reads in the case of Hermite interpolation. By de nition, a polynomial p 2 d satis es Hn p = 0 if, and only if, Dy() p(xi) = 0;  2 Ti; i = 1; : : : ; M: 1

(1)

(1)

24

(

1)

( )

The polynomials satisfying these conditions generate an ideal if for any q 2 d we have, applying the Leibnitz rule, X 0 = Dy() (p
q) (xi) = Dy p(xi)Dy q(xi);  2 Ti; i = 1; : : : ; M: 1

y1 +y2 =y

2

This implies, since q is arbitrary, that in order to generate an ideal, the polynomial p must satisfy Dy0 p(xi) = 0;  2 Ti; i = 1; : : : ; M; whenever the directional vectors in y0 are a subset of the directional vectors in y . For a regular Hermite problem, the above conditions are automatically satis ed. In other words, if a Hermite interpolation problem is regular, then H = fp 2 d : Hn p = 0g is an ideal in d. If the interpolation problem is not regular, then the requirement that H is an ideal will impose additional interpolation conditions; hence, the interpolation problem will be over-determined. Let us look at some examples of regular and irregular trees next. Example 6.2. Several examples of bivariate regular trees are given in Fig. 12. Regularity

(a)

(b)

(c)

Figure 12: is very sensitive to the position of the directional vectors: for example, the regularity in the rst example will be lost if the direction of the second vector is slightly changed. More precisely, the Hermite condition f (x); Dy f (x); Dy Dy f (x) is regular if, and only if, y1 is parallel to y2. In this sense, regularity requires much more than a structural property. In particular, consider the two interpolation problems in Fig. 13 which have the same tree structure; the rst one is the uniform interpolation problem of total degree 3 which interpolates all partial derivatives, its tree is depicted in (a); the second one is also a uniform interpolation problem with tree depicted in (b). Clearly, the two interpolation problems are equivalent in the sense of De nition 2.9. However, the example (a) shows a regular tree, and example (b) shows a highly irregular tree. 2 For the poisedness of Hermite interpolation problems, we see that an interpolation problem can be almost poised (which means that it has the structural prerequisites for 1

1

25

2

(a)

(b)

Figure 13: being poised), or never poised (which means that the problem will not be poised no matter what points and directional vectors we choose). Regularity has a similar structural property which is explained in the next example. Example 6.3. Three dierent types of irregular trees are shown in Fig. 14, where we depict the vectors according to the actual geometric positions in the plane as well as according to the respective block structures. The tree in (a) has proper structure and is

(a)

(b)

(c)

Figure 14: only irregular because one direction is wrong; here irregularity is in a sense \removable"; i.e., if we switch to the dotted direction then the problem turns into a regular one. On the other hand, the tree (b) has one vertex to which there are attached more directions than to its predecessor; the tree is irregular and cannot be turned into a regular one by choosing an appropriate direction. We remark that even if irregularity can be removed by adjusting its directional vectors, it is irregular for almost all directions, which is in sharp 26

contrast to almost poisedness. Another example is given in (c) where the graph looks similar to (a), except that we interpolate one more second order mixed partial derivative at the point and interpolate the function value at an additional point. Once again, the problem can be changed to a regular one if we replace the \bad" direction by the dotted one. On the other hand, in the sense of De nition 2.9, all irregular problems are equivalent to the regular one with dotted direction. 2 For convenience we reformulate the de nition of regularity in a form which we will use in the sequel. For any chain of directions y = (y1; : : :; y ) we de ne yb k

= (y1; : : :; yk?1; yk+1; : : : ; y );

1  k  :

(6.1)

An immediate consequence of De nition 6.1 is Lemma 6.4. Let the Hermite interpolation problem be poised in block and regular. Then for any j j  n and any j j  j j   ?1  k = 1; : : : ;  : (6.2) Dyb

k p (x ) = 0; Another question one may ask is whether there is a possible connection between poisedness and regularity. The answer is negative in both directions. Recall that if a Hermite interpolation problem is poised, then the Vandermonde determinant  i h V (Xn) = Dy (
) (x ) jj;j jn ; is a non-vanishing polynomial in x and y ;1; : : :; y ; , j j  n; it will remain nonzero in a small \neighborhood" of the poised problem; i.e., any suciently small distortion of the interpolation problem will not aect poisedness. On the other hand, regular problems are very singular; i.e., a slight change of a single direction may spoil the whole regularity. Thus, regularity cannot be necessary for poisedness. Conversely, the interpolation problem in Example 4.2 is regular but even never poised. Nevertheless, for a regular Hermite interpolation problem there is a necessary condition for poisedness in blocks which we give below. This condition generalizes the necessary condition given in [10] for Lagrange interpolation. Proposition 6.5. Let the Hermite interpolation problem be poised in block and regular. Then for any jj < n the simplex spanned by x and the points ( x ; j j = jj + 1; x^ = x +x y; ; xx 6=  ; = x has full dimension d; i.e.,

vold [x; x^ : j j = jj + 1] > 0: 27

Proof : Fix an  with jj < n and assume that the simplex spanned by x and x^ , j j = jj + 1 has dimension < d. Then there exists a nonzero linear polynomial ` 2 d1 which vanishes at all these points. This implies that 0 = `(x) = `(x );

j j = jj + 1;

and

0 = Dy ; `(x); j j = jj + 1; x = x: Now, choose some such that j j  jj + 1 and consider  X   Dy (`
p) (x ) = `(x )Dy p (x ) + Dyb i ?1p(x )Dy ;i `(x ): i=1

(6.3) (6.4) (6.5)

We prove that the right hand side of (6.5) vanishes, for which we need to distinguish several cases. If j j  jj, then by Lemma 6.4, by the de nition of p and because of the fact that `(x) = 0, we always have 0 = Dby i ?1p(x ) = `(x )Dy p (x ); hence the right hand side of (6.5) vanishes. If j j = jj + 1, then by (6.3) we have `(x ) = 0 and therefore (6.5) reduces to  X Dy (`
p) (x ) = Dyb i ?1 p(x )Dy ;i `(x ): i=1

Considering the case x 6= x rst, we only have to note that regularity implies that for any i = 1; : : : ;  there exists some i 6=  such that j ij  jj, x i = x , and yb i = y i . Thus, Dyb i ?1p(x ) = Dy i?1p(x i ) = 0; i = 1; : : : ;  ; hence, the right hand side of (6.5) vanishes again. The remaining case is j j = jj + 1 and x = x. Of course, as long as yb i 6= y, we can apply the same argument as before to show that Dyb i ?1p(x ) = 0: On the other hand, if yb i = y, i.e.,

Dby i ?1p(x ) = 1; then y ;i = y ; (because of regularity and the tree structure of the directions to be interpolated); we can use (6.4) to conclude that in this case the right hand side of (6.5) vanishes as well. 28

Thus, the polynomial `
p 2 djj+1 annihilates all interpolation conditions of level  jj +1, which contradicts the assumption of poisedness in block. Therefore the simplex has to have dimension d. 2

Remark 6.6. The geometric meaning of the condition in Proposition 6.5 is explained in the pictures in Fig. 15, where only a segment of the tree is depicted.. The requirement

(a)

(b)

Figure 15: states that in order to obtain a poised Hermite interpolation problem the convex hull of the \circles" depicted in (b) must have nonzero d{dimensional volume, in this case, nonzero area in R2. The following example illustrates the necessary condition given in Proposition 6.5. Example 6.7. Consider two points x1; x2 2 R2 and assume that the trees associated to these points are T1 = fx1jy1; 0jy2; 0; 0g and T2 = fx2jz1; z2g, as depicted in case (b) of Example 2.5; its tree structure is depicted in Example 3.3. For convenience, we reproduce these two pictures here. For the sake of simplicity let us choose x1 = (0; 0), x1 = (0; 1), y2

Figure 16:

y1 = e1, y2 = (1; 2) and z1 = e1, z2 = e2. Then the poisedness and the regularity of 29

the interpolation problem only depends on y2. It can beeasily veri ed that the blockwise arrangement of the interpolation conditions x; Dy and the Newton fundamental polynomials p are as follows: 2 3 2 2 2(2 ? 1) 3 ( x 2 ; (e )) 77 66 2 77 66 (x2; ()) 6 7 66 12 ? 2 12 777 ; (x2; (e1)) 77 ; [p]jj2 = 66 1 [(x; y)]jj2 = 6 (x1; ()) 75 64 1 75 64 (x1; (e1)) 1 2 (x1; (e1; y)) 2 1 where the columns of these formulae correspond to elements whose index satis es jj = i, i = 0; 1; 2, and where y = () means the interpolation condition being the point evaluation. Clearly, the Newton fundamental polynomials are only de ned if 1 6= 0; indeed, if we evaluate the Vandermonde determinant we obtain det V (X2) = 21; i.e., the Hermite interpolation problem is poised if, and only if, 1 6= 0. Let us have a look at what this means in terms of the necessary condition for poisedness which we derived in Proposition 6.5. For this purpose we choose  = (1; 0), i.e., x is the level one copy of x1 and consider the points on next level, fx^ : j j = 2g = fx2; x2; x1 + yg as de ned in Proposition 6.5. The simplex spanned by these points and x = x1 now has dimension 2 if, and only if, y is not parallel to x2 ? x1, i.e., if, and only if, 1 6= 0. Hence, in this particular case the necessary condition for poisedness is also a sucient one. 2 Finally, we brie y go back to the observation that Hermite interpolation conditions may arise as the nodes of a Lagrange interpolation problem collapse, which we mentioned in the introduction. If a Hermite interpolation problem is poised, then its Vandermonde determinant is not zero. Let a node x be a double node with a Hermite interpolation condition of the rst order attached to it, say, interpolating f (x ) and Dy f (x ); then by considering the modi ed interpolation problem which replaces the Hermite condition by Lagrange interpolation at the points x and x + yh, the Hermite condition at x can be considered as the result of h ! 0 in the modi ed interpolation problem in an obvious way. Upon performing one column operation, the Vandermonde determinant with respect to the Lagrange interpolation problem will contain a column
hy (x ) := (x + hy) ? (x ) in place of the column of (x + hy), which will converge, when divided by h, to the Vandermonde determinant with respect to the Hermite problem. If the node x is attached with a Hermite interpolation condition of the second order or higher, we can use the higher order dierence operator, which for y = (y1; : : :; y ) is de ned by
y f (x) := f (x + y) ? f (x);
y f (x) :=
y
yb? 1f (x); where yb i = (y1; : : :; yi?1; yi+1; : : :; y ) for i = 1; : : : ; . For f 2 C  (Rd) 1
 f (x); Dy f (x) = hlim x 2 Rd: !0 h hy However, it is not hard to see that the above limiting procedure, or collapsing of points, works only when the Hermite interpolation problem is regular. 2

1

1

+

30

6.2 Remainder Formulae

The remainder formula will be given with simplex splines acting as kernels of integrals. First we recall the de nition of simplex splines, following the fundamental paper of Micchelli [8]. Let v0; : : :; vn be vectors in Rd and let Sn := f = (0; : : :; n) : i  0; 0 +


+ n = 1g be the n{dimensional unit simplex. The simplex spline M (
jv0; : : :; vn) is de ned by the condition Z Z 0; : : : ; v n) dx = (n ? d)! f (0v0 +


+ nvn)d; f 2 C (Rd): (6.6) f ( x ) M ( x j v d R

Sn

As long as the convex hull of the knots, [v0; : : : ; vn], has dimension d, the simplex spline M (
jv0; : : : ; vn) is a nonnegative piecewise polynomial of degree n ? d in Rd, supported on [v0; : : :; vn]. The order of dierentiability of simplex splines depends strongly on the position of the knots; if, for example, the knots are in general position, which means that any subset of d + 1 of the knots spans a proper simplex, then the simplex spline has maximal order of dierentiability n ? d ? 1. Let us recall the notion of a path which we introduced in [10]. An n{path  is a vector of multiindices with ascending lengths, i.e., n o  2 n := (0; : : :; n ) : i 2 Nd0 ; jij = i; i = 0; : : : ; n : We remark that from now on we use the symbol  to denote paths in n ; it should not be confused with chains in a certain tree. In particular, k , k = 0; : : :; n, now denotes the k{th entry of  = (0; : : :; n ), i.e., a multiindex in Nd0 of length k. For each pair of multiindices jj = j j ? 1 < n and y = (y ;1; : : :; y ; ), we de ne a dierential operator 8  > if x 6= x <  Dy (6.7) D; = > XD ?1 if x = x  : i=1 by i and a direction ( x ? x if x 6= x : (6.8) z; = y ; if x = x Having this in hand we associate to each path  2 n a directional derivative of order n Dzn = Dzn? ;n


Dz ; ; (6.9) and a number (x; y) = Dn? ;n pn? (xn )


D ; p (x ): (6.10) Finally, by x we denote as in [10] the set of points that the path passes through; i.e., x = fx ; : : :; xn g: (6.11) 1

1

0

1

0

31

1

0

1

0

1

Then the remainder formula reads as Theorem 6.8. Let the Hermite interpolation problem be poised in block and regular. Then for n 2 N and x 2 Rd we have Z X   0 n pn (x)(x ; y ) Rd Dx?xn Dzn f (t)M (tjx; x)dt: (6.12) n+1 [x ; : : :; x ; x]f = 2n

Together with Theorem 5.3 this readily gives Corollary 6.9. For n 2 N and x 2 Rd Z X   pn (x)(x ; y ) Rd Dx?xn Dzn f (t)M (tjx; x)dt: f (x) ? Hn (f; x) = 2n

(6.13)

Before we turn to the proof of Theorem 6.8, let us rst give a geometric interpretation of the remainder formula (6.12). The basic ingredient of (6.12) that depends on the function f is the integral Z D Dn f (t)M (tjx; x)dt; d x?xn z R

associated to a path  2 n . Clearly, two things depend on the path here: the vector of directions z and the collection of knots x. Looking at the directions rst, we see from (6.8) that whenever the path goes from a point x to a dierent point x on the next level (i.e., j j = jj + 1 and x 6= x ), the respective direction is just the dierence of the two points, which is the same as in the case of Lagrange interpolation. If the two points are copies of the same point, however, we cannot use the dierence between them since it is zero; instead, we take the direction attached to the end of the chain at the point x ; i.e., we take y ; which is the last direction in the chain y . The rule for the knots in the simplex spline is strikingly simple: we only have to collect those points whose indices occur on the path. In particular, there is an interesting comparison to the univariate case: whenever a point appears successively in the path (i.e., whenever we use the direction y ; ), the respective point x appears as a multiple knot of the simplex spline. The dierence to the univariate case lies in the fact that some knot can appear as a multiple knot in the simplex spline without the associated direction appearing in the dierential operator in (6.12). This can happen if the path passes points like xk = xk 6= xk , see Fig. 17 below. In such a case we have xk as a a multiple knot for the simplex spline, although the direction associated to xk (since the point has a lower level copy, xk , we interpolate derivatives here) does not show up under the integral. In particular, if the interpolation problem is put into block with \jumps" (see, for example, Fig. 10), the above discussion implies that some direction may not appear under the integral in the remainder formula. Of course, the remainder formula still depends on the missing direction, since the formula will be in uenced by it via the dierential operator D. +2

+2

32

+1

Figure 17:

Remark 6.10. It is easily seen that the number of terms ink+the remainder formula is d ? 1 generally very large. Since the cardinality of f : jj = kg is d?1 , our set n contains a total number of n k + d ? 1! Y d?1 k=0

paths. On the other hand, in many cases there will not be so many terms in the remainder formula, as can be seen, for instance, in Example 6.17. The reason is that it might happen that D; p(x ) = 0 for some jj + 1 = j j  n; (6.14) which implies that (x; y) = 0 whenever jj =  and j j = . Such a condition will remove a whole subtree of paths from the remainder formula. In fact, the equality (6.14) may happen in a generic situation, namely, when a Hermite interpolation problem of degree n contains the interpolation of an n{th order directional derivative at a certain point x0. Clearly, if this problem is poised in block, then any level k contains some copies x = x0, j j = k, of x0 with associated directional vectors y such that  = k. Since for  such a of length k the operator Dy is a homogeneous dierential operator of order k which annihilates all polynomials of degree  k ? 1, we have

D; p(x ) = Dy p(x ) = 0; whenever jj = k ? 1 such that x = 6 x0 and j j = k such that x = x0, k = 1; : : :; n. In other words: if xk = x0 for some 0  k  n, then (x; y ) = 6 0 only if xj = x0 for all 0  j < k. The most extreme case in this respect are Taylor polynomials of degree n, which will be discussed in Example 6.16 of Section 6.4. Remark 6.11. Let us brie y point out that, in addition to the geometric notion of a path, there is another interpretation of formula (6.12). Let us rewrite it into Z X   0 n pn (x)(x ; y ) Rd Dx?xn Dzn f (t)M (tjx; x)dt n+1 [x ; : : : ; x ; x]f = 2n Z X d X pn (x)(x; y) Rd (i ? n ;i) @@ Dzn f (t)M (tjx; x)dt = i i=1 2n 33

=

d XX 2n i=1

pn (x)(i ? n ;i)

(x; y)

Z Rd

@ Dn f (t)M (tjx; x)dt: @i z

The interesting feature of that representation lies in the fact that the polynomials

p (i ? n ;i) ; jj = n; i = 1; : : : ; d; n o generate the ideal H = p 2 d : p 6= 0; Hn (p; x)  0 of nonzero polynomials which annihilate all the interpolation conditions. It is, however, worthwhile to remark that we have implicitly used regularity twice in this simple statement: we need regularity to assure that the above polynomials indeed annihilate the interpolation conditions and regularity is also necessary for H to be an ideal at the rst place. This again underlines the importance of the notion of regularity. Finally, let us have a brief look at two extreme cases: rst, assume that all the points are distinct, i.e., we deal with a Lagrange interpolation problem. Then the nite dierence and the remainder formula reduce to the ones we gave in [10]. On the other hand, if all points coincide (i.e., x = x0 for all jj  n) and if we take, for convenience, partial derivatives instead of general directions, then formula (6.13) reduces to a remainder formula for the Taylor polynomial of degree n, which we will work out explicitly in Example 6.16 of Section 6.4.

6.3 Proof of the remainder formula

To prove (6.12) we have to establish several preliminary results rst. Lemma 6.12. Let y = (y1; : : :; y ) and x0 2 Rd. Then for any polynomial p 2 d  X x 2 Rd: (6.15) Dy ((x ? x0)p(x)) = (x ? x0)Dy p(x) + yiDyb?i 1 p(x); i=1

Proof : The case  = 0 is trivial; we proceed by induction on , assuming that the result has been proved for some   0. Using the notation introduced in (6.1) we write y = (yb +1 ; y+1 ) and

obtain by the induction hypothesis   Dy+1 ((x ? x0)p(x)) = Dy Dyb ((x ? x0)p(x)) !  X  ? 1  = Dy (x ? x0)Dby p(x) + yiDbyi; p(x) i=1  X  +1  = (x ? x0)Dy p(x) + y+1Dyb p(x) + yiDy Dby?i;1 p(x) i=1 X +1 = (x ? x0)Dy+1 p(x) + yiDybi p(x) i=1 +1

+1

+1

+1

+1

+1

34

+1

+1

2

Next we derive a formula for derivatives of simplex splines which may be of independent interest. We want to compare multiple directional derivatives of a simplex spline function M (tjv; x), v = fv0; : : :; vmg, vi 2 Rd, as a function of t 2 Rd as well as a function of x 2 Rd. To distinguish between directional derivatives with respect to t and with respect to x we will denote these dierentiations as Dy (@t) and Dy (@x), respectively. More precisely, let rt and rx denote the gradient with respect to t or x, respectively, and let y = (y1 ; : : :; y ), yi 2 Rd, i = 1; : : : ;  , then Dy (@t) = (y1
rt)


(y
rt) ; Dy (@x) = (y1
rx)


(y
rx) : The following lemma shows that the simplex spline M (tjv; x) is almost symmetric with respect to these operations; the lemma seems to be known as a folklore in the spline community. Since we cannot locate a reference, we oer a simple proof here for selfcontainedness. Proposition 6.13. Let v = fv0; : : : ; vmg, m  0, be a set of knots in Rd and let y = (y1 ; : : :; y ). Then d Dy (@x)M (tjv; x) = (?1) ! Dy (@t)M (tjv; x; | :{z: :; x}); x 2 R : +1

Proof : First we x 0 < k  , y 2 Rd and h > 0. With help of the derivative formula

for simplex splines we then compute M (tjv; x; | :{z: :; x}) ? M (tjv; x| + hy; :{z: :; x + hy}) k 0k 1 kX ?1 B@M (tjv; x; : : :; x; x + hy; : : :; x + hy) ? M (tjv; x; : : :; x; x + hy; : : :; x + hy)CA = | {z } | | {z } | {z } {z } =

i=0 kX ?1

k?i

i+1

Dhy (@t)M (tjv; x; | :{z: :; x}; x| + hy; :{z: :; x + hy})

i=0 kX ?1

= h

k?i?1

i

i=0

k?i

i+1

Dy (@t)M (tjv; x; | :{z: :; x}; x| + hy; :{z: :; x + hy}): k?i

i+1

Thus,

1 0 1B C Dy (@x)M (tjv; xk) = hlim @M (tjv; x| + hy; :{z: :; x + hy}) ? M (tjv; x; | :{z: :; x})A !0 h k k 0 1 k ? 1 1 Bh X D (@t)M (tjv; x; : : :; x; x + hy; : : :; x + hy)C = hlim ? @ y | {z } | {z }A !0 h i=0

k?i

35

i+1

= ?

kX ?1

lim Dy (@t)M (tjv; x; | :{z: :; x}; x| + hy; :{z: :; x + hy})

i=0 h!0

k?i

i+1

= ?kDy (@t)M (tjv; x; | :{z: :; x}): k+1

This already includes the case  = 1; to complete the proof we use induction on , which gives   Dy (@x)M (tjv; x) = Dy (@x) Dby? 1 (@t)M (tjv; x) 0 1  ? 1  ? 1 A = Dy (@x) @(?1) ( ? 1)! Dyb (@t)M (tjv; x; | :{z: :; x})  ) : = (?1) ! Dy (@t)M (tjv; x; : : :; x | {z } +1

2

Let us remark that, strictly speaking, the above limiting argument holds only in the distributional sense; since we will always use simplex splines this way, i.e., as a kernel functions in an integral, this does not cause any problems. Lemma 6.14. Let  2 n?1 and let y = (y1; : : :; y ),  > 0. Then Z  Dy (@x) Rd Dx?x Dx f (t)M (tjx; x)dt Z = ! Rd Dy (@t)Dx?x Dx f (t)M (tjx; x; | :{z: :; x})dt +1 Z   + ! d Dy (@t)Dx f (t)M (tjx ; x; | :{z: :; x})dt: R 

Proof : Applying the same idea as in the proof of Lemma 6.12, we obtain Z Dy (@x) d Dx?x Dx f (t)M (tjx; x)dt R =

Z

 Z X   D D  f (t)Dby?i 1 (@x)M (tjx; x)dt: D D  f (t)Dy (@x)M (tjx ; x)dt + d yi x R Rd x?x x i=1

For the rst term on the right hand side we obtain by Proposition 6.13 and integration by parts Z Dx?x Dx f (t)Dy (@x)M (tjx; x)dt d R Z = d Dx?x Dx f (t) (?1) ! Dy (@t)M (tjx; x; | :{z: :; x})dt R +1 Z = ! d Dy (@t)Dx?x Dx f (t)M (tjx; x; | :{z: :; x})dt: R +1

36

Similarly, the second term transforms into  Z X D D  f (t)Dyb?i 1 (@x)M (tjx; x)dt d yi x R i=1  Z X Dyi Dx f (t)(?1)?1( ? 1)! Dby?i 1 (@t)M (tjx; x; = d | :{z: :; x})dt i=1 R  Z  X D?1 (@t)Dyi Dx f (t)M (tjx; x; = ( ? 1)! d ybi | :{z: :; x})dt R i=1   Z X = ( ? 1)! Dy (@t)Dx f (t)M (tjx; x; d | :{z: :; x})dt i=1 R  Z = ! d Dy (@t)Dx f (t)M (tjx; x; | :{z: :; x})dt: R 

2

Lemma 6.15. Assume that the Hermite interpolation problem is poised in block and regular. Then for any j j = k ? 1, k = 1; : : : ; n, we have X  p (x) (x ? x ) = Dy (p (
)(
? x )) (x)p(x): (6.16) jj=k

Proof : Since the Hermite interpolation problem is poised in block, it suces to show

that the left hand side and the right hand side of (6.16) coincide in all interpolation conditions. For j j = k it is easily seen that applying Dy , j j = k, on both sides of (6.16) and then inserting x indeed gives identical values. For j j  k ? 1, after applying Dy and evaluating at x , the left hand side reads as  X  (6.17) (x ? x )Dy p (x ) + yj Dyb j p (x ): j =1

Here is the point where regularity enters the scene for the rst time. Because of Lemma  6.4 the sum in (6.17) vanishes completely, while (x ? x )Dy p (x ) also has value zero: if 6= then Dy p (x ) = 0 by the de nition of Newton fundamental polynomials, and the case = is obvious. On the other hand, applying the same operations to the right hand side of (6.16) clearly gives zero value. 2

Proof of Theorem 6.8 : We use induction on n. The case n = 0 is Lagrange interpolation using the constant function for which the remainder formula clearly holds true. Suppose the theorem is already proved for an n  0, i.e., we have Z X n [x0; : : :; xn?1 ; x]f = pn? (x)(x; y) d Dx?xn? Dzn?1 f (t)M (tjx; x)dt: R 2n?1

1

1

(6.18)

37

In order to use the recurrence relation in the inductional step n ! n +1, we rst   (5.2)  0 n ? 1 have to derive a closed formula for Dy n [x ; : : :; x ;
]f (x). From (6.18) we obtain, for any y = (y1; : : : ; y ), by the Leibnitz rule of dierentiation and by Lemma 6.14

Dy n[x0; : : :; xn?1; x]f 1 0 Z X = Dy (@x) @ pn? (x)(x; y ) d Dx?xn? Dzn?1 f (t)M (tjx; x)dtA R 2n? Z X X     = Dy pn? (x)(x ; y )Dy (@x) d Dx?xn? Dzn?1 f (t)M (tjx; x)dt R 2n? y +y =y X X  = Dy pn? (x)(x; y )2!  2n? y +y =y Z  Rd Dy Dx?xn? Dzn?1 f (t)M (tjx; x; | :{z: :; x})dt  +1 Z    + d Dy Dz f (t)M (tjx ; x; | :{z: :; x})dt ; R 1

1

1

1

1

1

1

1 1

1

1 1

1

2 2

1

2

2

2 2

1

2

2 2

2

with the convention that for 2 = 0 the last integral is zero. In particular, for jj = n,   0  Dy n [x ; : : :; xn?1;
]f (x) X X  Dy pn? (x)(x ; y)2!  = 2n? y +y =y Z   ; : : :; x )dt n?1 D y Dx ?xn? Dz f (t)M (tjx ; x d |  {z } R  +1 Z    + Rd Dy Dz f (t)M (tjx ; x| ; :{z: : ; x})dt : 1 1

1

1

1

2

2 2

1

2

2 2

2

Now it's again the assumption of regularity which signi cantly simpli es the above sum: Lemma 6.4 states that, whenever 1 <  ? 1, regularity implies

Dy pn? (x) = 0: 1 1

1

Therefore,   0  Dy n[x ; : : :; xn?1;
]f (x) Z X     Dy pn? (x)(x ; y ) d Dx ?xn? Dzn?1 f (t)M (tjx; x)dt = R 2n? Z X  ?1 Dybi ?1pn? (x)(x; y) d Dy;i Dx?xn? Dzn?1 f (t)M (tjx; x; x)dt + R i=1 Z X  ?1  Dybi ?1pn? (x)(x; y) Rd Dy;i Dzn?1 f (t)M (tjx; x)dt : + i=1 1

1

1

1

1

1

38

Using regularity once more we observe that the second sum in the above expansion vanishes for all n?1 except a single one which satis es xn? = x; but then the integral contains a zero derivative and therefore vanishes. Hence, the second sum disappears completely, giving   0  Dy n [x ; : : : ; xn?1;
]f (x) Z X     = Dy pn? (x)(x ; y ) Rd Dx?xn? Dzn?1 f (t)M (tjx; x)dt 2n? Z X  ?1  Dybi ?1pn? (x)(x; y ) Rd Dy;i Dzn?1 f (t)M (tjx; x)dt : (6.19) + i=1 1

1

1

1

1

Rewriting (6.16) as !   X  X   pn? (x) x ? xn? = p(x) (x ? xn? )Dy pn? (x) + y;iDbyi pn? (x) 1

1

1

jj=n

1

1

i=1

(6.20)

and incorporating the linearity of directional derivatives we obtain  X X pn? (x)Dx?xn? = p(x)Dy pn? (x)Dx?xn? + Dybi pn? (x)Dy;i : (6.21) 1

1

1

jj=n

1

1

i=1

Inserting (6.21) into (6.18) yields n [x0; : : : ; xn?1; x]f Z X = pn? (x)(x; y ) d Dx?xn? Dzn?1 f (t)M (tjx; x)dt R 2n? Z   X X   (x ; y ) = p(x) Dy pn? (x) d Dx?xn? Dzn?1 f (t)M (tjx; x)dt R 2n? jj=n Z   X (6.22) + Dybi ?1 pn? (x) Rd Dy;i Dzn?1 f (t)M (tjx; x)dt : i=1 1

1

1

1

1

1

1

Combining (6.19) and (6.22) with the recurrence relation (5.2) then gives n+1 [x0; : : :; xn; x]f  X   0 = n[x0; : : :; xn?1; x]f ? Dy n [x ; : : :; xn?1 ;
] (x)f
p(x) jj=n Z   X X   p(x) Dy pn? (x) d Dx?xn? Dzn?1 f (t)M (tjx; x)dt = (x ; y ) R 2n? jj=n Z   X + Dybi ?1pn? (x) d Dy;i Dzn?1 f (t)M (tjx; x)dt R i=1 Z   X X   ? p(x) (x ; y ) Dy pn? (x) d Dx?xn? Dzn?1 f (t)M (tjx; x)dt 1

1

1

1

jj=n

1

2n?1

39

R

1

Z  Dybi ?1pn? (x) d Dy;i Dzn?1 f (t)M (tjx; x)dt R i=1 X = (x ; y) p(x)  2n? jj=n Z    Dy pn? (x) Rd Dx?xn? Dzn?1 f (t) (M (tjx; x) ? M (tjx; x)) dt Z   X + Dybi ?1pn? (x) d Dy;i Dzn?1 f (t) (M (tjx; x) ? M (tjx; x)) dt : R i=1 + X

 X

1

1

1

1

1

Since

M (tjx; x)dt ? M (tjx; x) = ?Dx?x M (tjx; x; x); we apply integration by parts to obtain n+1 [x0; : : : ; xn; x]f Z  X X (x; y ) p (x) Dy pn? (x) d Dx?xn? Dzn?1 f (t)M (tjx; x; x)dt = R 2n? jj=n Z   X (6.23) + Dybi ?1 pn? (x) Rd Dy;i Dzn?1 f (t)M (tjx; x; x)dt : i=1 1

1

1

1

Notice that the rst part of the above sum vanishes whenever x = xn? and let us have a closer look at the second term, namely, Z  X  ? 1  Dybi pn? (x) d Dy;i Dzn?1 f (t)M (tjx; x; x)dt: R i=1 1

1

Since the interpolation problem is regular, we again stress Lemma 6.4 to conclude that for any i = 1; : : : ; , there exist multiindices i, j ij < jj, such that y i = ybi and x i = x. On the other hand, by the de nition of Newton fundamental polynomials,

j j = jj ? 1;

Dy i?1p (x);

vanishes whenever x 6= x and takes value one if = i for some 1  i  . Now, we replace n by  in (6.23) to obtain

n+1 [x0; : : :; xn ; x]f Z  n X   (x ; y )pn (x) Dyn pn? (x) Rd Dxn ?xn? Dzn?1 f (t)M (tjx; x; x)dt = 2n Z   X  ? 1  (6.24) + Dybi n pn? (xn ) d Dy;i Dzn?1 f (t)M (tjx; xn ; x)dt : R i=1 1

1

1

Applying the de nitions of Dz and (x; y ) from (6.9) and (6.10) respectively, formula (6.24) turns into (6.12).

2

40

6.4 Examples of remainder formulae

In this section we illustrate and explain the remainder formula (6.12) by applying it to several particular examples. All the problems we consider can be veri ed to be almost poised. Whenever we give a general remainder formula without specializing the points and directions explicitly, then we assume that the points and directions are chosen such that the Hermite interpolation problem is poised. Our rst example considers an extremal case of Hermite interpolation, namely, the Taylor polynomial, where we have only a single point with a tree of partial derivatives of total degree. Example 6.16. Taylor polynomial. For j j  n we choose

x = x0

and

y

= (e| 1
{z

e1}


e| d
{z

ed}): 1

d

This implies that

@ j j f (x ): Dy f (x ) = @x 0 Moreover, the Newton fundamental polynomials p (x) are easily seen to be (x ? x0) = !.  Since they satisfy Dy p (x ) = ; , the nite dierence boils down to @ j j f (x ); Dy j j[x0; : : : ; xj j?1]f = Dy f (x ) = @x 0

and therefore the Hermite interpolation polynomial reads as X 1 @ j j Hn (f ; x) = f (x0 ) (x ? x0 ) ; ! @x j jn

which is the n{th Taylor polynomial of f at the point x0. To nd the explicit formula for the remainder we rst note that the dierential operator from (6.7) takes the form

D; =

d X @ j j?1 ; i @x ?ei i=1

and, therefore,

jj + 1 = j j  n;

(

k + ei; 1  i  d; Dk ;k pk (xk ) = k0+1; ;i ; if k+1 = otherwise. +1

Hence,

+1



(x; y ) =

(

n!;  2 n ; 0;  26 n ;

41

where n is the set of all pathes going along the coordinate vectors; i.e., n o n =  2 n : k = k?1 + eik ; 1  ik  d; k = 1; : : : ; n : For convenience we also introduce the shorthand notation n () = f 2 n : n = g ;

n() = n() \ n;

which enables us to compute the remainder term from (6.12) as

f (x) ? Hn (f; x) = n+1 [x0; : : :; xn; x]f Z X d X X @ n f (t)M (tjx; x)dt   = p(x) (x ; y ) Rd (i ? ;i) @@ @x i  i=1 jj=n 2n () 0 1 Z n+1 d X XX @ f (t)M (tjx; x)dt: (i ? ;i) p(x) @ = (x; y)A d @x +ei R

2n ()

jj=n i=1

Taking into account that the cardinality of n () for jj = n is jn()j = independently of  2 n(), we have (x; y ) = !, this turns into

jj 

and that,

f (x) ? Hn (f; x) = n+1 [x0; : : : ; xn; x]f d (x ? x )+ei jj! Z XX @ n+1 f (t)M (tjx ; : : : ; x ; x)dt 0  ! = 0 0  ! Rd @x+ei jj=n i=1 d (x ? x ) Z @ n+1 X X 0 = n! i f (t)M (tjx0 ; : : :; x0; x)dt j j=n+1 i=1 ( ? e )! Rd @x d ! (x ? x ) Z @ n+1 X X 0 i = n! f (t)M (tjx0; : : : ; x0; x)dt ! Rd @x j j=n+1 i=1 X (x ? x0) Z @ n+1 f (t) ((n + 1)! M (tjx0; : : :; x0; x)) dt: = ! Rd @x j j=n+1 This is the remainder formula for the multivariate Taylor polynomial which can also be obtained from the univariate formula (see, e.g., [2, p. 70]). Recall that in each of the spline functions x0 appears as a knot of multiplicity n. 2 All examples below are given for the case d = 2; for convenience, we write the Newton fundamental polynomials p as

pnk = p;

 = (k; n ? k) 2 N2 ; jj = n:

In the second example we consider a Hermite interpolation problem at three points which involves only rst order derivatives and therefore is trivially regular. 42

Example 6.17. Three points. Let x1; x2; x3 2 R2 be the interpolation points, and let y1; y2; z 2 R2 be the directional vectors. We consider the Hermite interpolation problem with respect to f (x1); (Dz f )(x1); f (x2); (Dy f )(x2); (Dy f )(x2); f (x3): The respective tree structures are given as follows: T1 = fx2jz; 0g; T2 = fx3jy1; y2g; T3 = fx3g: The nonzero conditions imposed on the Newton fundamental polynomials are       p11(x2) = Dz p12 (x1) = Dy p21 (x2) = Dy p22 (x2) = p23(x3) = 1: This situation and the respective joint diagram for the blockwise structure are depicted in Fig. 18. The remainder formula in Theorem 6.8 is now a sum over six paths, where 1

2

1

2

z x3

y1

x1

z x1

y2 x3

x2

y2 x2 y1

(a)

(b)

Figure 18: each path corresponds to one term in the formula. To illustrate the paths and the formula geometrically, we give the remainder formula explicitly below and depict aside each term in the summation its corresponding path in the joint diagram:

f (x) ? H2(f; x) = 3[x0; x1; x2; x]f Z 2 1 = p1(x)p1(x2) Dx?x Dy Dx ?x f (t)M (tjx1; x2; x2; x)dt R Z +p22(x)p11(x2) R Dx?x Dy Dx ?x f (t)M (tjx1; x2; x2; x)dt Z 2 1 +p3(x)p1(x3) Dx?x Dx ?x Dx ?x f (t)M (tjx1; x2; x3; x)dt R Z +p21(x)(Dy p12)(x2) Dx?x Dx ?x Dz f (t)M (tjx1; x1; x2; x)dt R Z 2 1 +p2(x)(Dy p2)(x2) R Dx?x Dx ?x Dz f (t)M (tjx1; x1; x2; x)dt 2

2

2

2

1

2

2

2

1

2

2

3

3

2

1

2

1

2

2

1

2

2

1

2

2

1

43

+p23(x)p12(x3)

Z R

2

Dx?x Dx ?x Dz f (t)M (tjx1; x2; x3; x)dt 3

3

1

The graph behind each formula shows the path corresponding to the term; for example, the third one follows the path fx1; x1; x2g where the repeated nodes x1; x1 correspond to the directional derivative along the direction z, the third order directional derivative is taken along the path and the same nodes appear in the simplex spline. 2 Our next example explains the in uence of regularity on the remainder formula. Example 6.18. Two points. Let x1; x2 2 R2 be the interpolation points, and let y1; y2; z1; z2 2 R2 be directional vectors. We consider a Hermite interpolation problem at these two points with the conditions f (x1); (Dy f )(x1); (Dy Dy f )(x1); f (x2); (Dz f )(x2); (Dz f )(x2): The tree structure are given by T1 = fx1jy1; 0jy2; 0; 0g; T2 = fx2jz1; z2g; and the joint diagram for the blockwise structure is depicted as in Fig. 19. The blockwise 1

2

1

1

2

y2 y1

z2

x1

x2 x1

z1

z1 y1

x2

y2

z2

Figure 19: structure implies that the Newton fundamental polynomials of the rst degree satisfy the following conditions, p11(x1) = p11 (x2) = 0; (Dy p11)(x1) = 1 and p12 (x1) = (Dy p12)(x1) = 0; p12(x2) = 1: The fundamental polynomials p2i , i = 1; 2; 3, satisfy the zero-one interpolation conditions with (Dy Dy p21 )(x1) = 1; (Dz p22)(x2) = 1; (Dz p23)(x2) = 1; respectively. In Example 6.7, these polynomials are given explicitly for a particular choice of points and directions. It is readily seen that this interpolation problem is regular if, and only if, y1 is parallel to y2; i.e., if, and only if, y2 is a multiple of y1. When the problem is not regular, we cannot apply Theorem 6.8 to it. However, for such a simple and low{degree problem, we can still derive the explicit remainder formula directly; a tedious computation leads to the following: 1

2

1

1

1

2

44

f (x) ? H2(f; x) = 3[x0; x1; x2; x]f Z 2 = 2p1(x) Dx?x Dy Dy f (t)M (tjx1; x1; x1; x)dt R Z +p22(x)(Dz p11)(x2) R Dx?x Dx ?x Dy f (t)M (tjx1; x1; x2; x)dt Z 2 1 +p3(x)(Dz p1)(x2) Dx?x Dx ?x Dy f (t)M (tjx1; x1; x2; x)dt R Z +p22(x) Dx?x Dz Dx ?x f (t)M (tjx1; x2; x2; x)dt R Z +p23(x) R Dx?x Dz Dx ?x f (t)M (tjx1; x2; x2; x)dt Z +p21(x)(Dy p12)(x1) R Dx ?x Dy Dx ?x f (t)M (tjx1; x2; x1; x)dt: 1

2

2

1

1

2

2

2

2

2

2

1

2

1

2

2

2

1

2

2

2

1

2

2

1

1

2

2

1

1

1

2

1

Again we have depicted the graphs aside each term, except the last one, to indicate the corresponding path. The last term is of a dierent character which contains no x in its integral part; actually, it is a term due to irregularity. Indeed, if the interpolation problem is regular, then y2 = cy1. By the de nition of p12 , it follows that (Dcy p12)(x1) = c(Dy p12 )(x1) = 0 and the last term does not show up in the remainder formula. Therefore, if the problem is regular, then the remainder formula contains ve terms as above. In terms of paths, it means that one path does not lead to a term in the remainder formula. The disappearance 2

1

Figure 20: of this path, which is depicted in Fig. 20, is due to the fact that (Dy p12(x1)) = 0. It is also interesting to note that along this path the dierential operator, which would take the form Dx?x Dx ?x Dx ?x , seems to have no proper interpretation; indeed, since the point x1 appears twice in the path, there should be a directional derivative involved in the dierential operator corresponding to it, yet it is lacking. This is exactly the phenomenon we explained in the paragraph following Theorem 6.8. 2 Example 6.19. In the above Example 6.18 we now choose the particular selection of points and directions as considered in Example 6.7, where the polynomials are explicitly computed. If the interpolation is regular, which is equivalent to 2 = 0 in Example 6.7, then the remainder formula takes the form 2

1

1

2

2

1

45

f (x) ? H2(f; x) = 3[x0; x1; x2; x]f Z = 12+ 1 12 R Dx?x Dy De f (t)M (tjx1; x1; x1; x)dt 1 Z +12 Dx?x Dx ?x De f (t)M (tjx1; x1; x2; x)dt ZR +12 Dx?x De Dx ?x f (t)M (tjx1; x2; x2; x)dt R Z +2(2 ? 1) R Dx?x De Dx ?x f (t)M (tjx1; x2; x2; x)dt: 1

1

2

2

2

2

2

2

2

1

1

1

2

1

2

2

2

1

In this case we have only four terms in the remainder formula since (Dz p11)(x1) = (De 2)(x1) = 0; making one more of the terms in the previous example disappear. Moreover, it is interesting to observe that in the case 1 = ?1 even the rst term in the remainder formula vanishes; we are then left with a remainder formula that contains only three terms. This example also illustrates Remark 6.10. We note, however, that the terms missing in the remainder formula for this interpolation problem are missing for entirely dierent reasons. In the rst case the term does not appear because of the structure of the interpolation problem; while in the second case the term vanishes because the corresponding fundamental polynomial vanishes at the point accidentally. 2 1

1

6.5 Error estimates

From the remainder formula (6.13), we can derive an error estimation formula for a regular Hermite interpolation problem. To do so, we recall the notation n () = f 2 n : n = g ; jj = n; and rewrite (6.13) as Z X   f (x) ? Hn (f; x) = pn (x)(x ; y ) d Dx?xn Dzn f (t)M (tjx; x)dt R 2n Z d X XX (i ? ;i) p(x) = (x; y ) d Dei Dzn f (t)M (tjx; x)dt: jj=n i=1

Using the expansion

Dzn f (t) = and the fact that

R

2n ()

X

n ) (zn? ;n ) n


(z ; ) @ @
f

(t@ n 2f1;:::;dgn 1

0

1

1

1

Z Rd

M (tjx; x)dt = (n +1 1)! ; 46

we obtain the error estimation formula as follows. Theorem 6.20. Let f 2 C n+1(Rd) and let  Rd be a convex domain which contains all nodes of interpolation. Then the following inequality holds for all x 2 : d XX jf (x) ? Hn (f; x)j  k(fnk+n+11)!;

j(i ? ;i) p(x)j C;j (Xn); (6.25) jj=n i=1 where and

n+1 @ f (x) kf kn+1; = sup max x2 j j=n+1 @x X X C;j (Xn) = j(x; y )j (zn? ;n ) n


(z ; ) ; 2n ()

1

2f1;:::;dgn

0

1

1

for jj = n; j = 1; : : : ; d, are constants independent of x. Let us rst remark that in the case d = 1 the error estimate (6.25) boils down to the well{known formula (n+1) k k f jf (x) ? Hn (f; x)j  (n + 1)! (x ? x1)k


(x ? xM )kM ; 1

(6.26)

where we assume that Hn (f ) interpolates derivatives of f of order up to ki ? 1 at xi, i = 1; : : : ; M . There are other error estimate formulae in the literature for the Hermite interpolation problem; most of them deal with special con gurations of points. The one that applies to the general interpolation scheme that we are aware of is found in [1], which we learned from [12], where the error estimation formula is derived using the Taylor expansion. The idea is to use the fact that Hermite interpolation polynomial is a projection on dn, which allows one to write Hn (f ) ? f = Hn (f ? Tn) ? (f ? Tn) for any polynomial Tn in dn , in particular, the Taylor polynomial. The error estimate formula so derived takes the form X jHn (f; x) ? f (x)j  k(fnk+n+11)!; jq(x)j ; jjn   which uses all fundamental polynomials q de ned in (3.3), a sum of n+d d = O(nd) terms. For the case d = 1 and Lagrange interpolation, such an estimate depends on the term n X Y Ln(x) = jqk (x)j; qk (x) = xx ?? xxk ; k k=0 i6=k i which is usually called the Lebesgue function; it is known that the uniform norm of Ln is unbounded as n ! 1 for any choice of points. As a contrast, the error estimate (6.26) 47

contains only one term that depends on x, which is clearly  bounded under mild conditions on xi. In general, the estimate (6.25) contains d  n+d?d?1 1 = O(nd?1 ) terms that depend on x, which correspond to the fundamental polynomials of the highest degree. Actually, if we recall Remark 6.11, the polynomials (j ? ;j )p(x);

jj = n; j = 1; : : : ; d; are of degree n + 1 and generate the ideal H of all nonzero polynomials that annihilate

the Hermite interpolation problem. On the other hand, another set of polynomials that generate the ideal are

'(x) = x ? Hn ((
); x) ;

jj = n + 1:

Thus, by a change of basis one could rewrite (6.25) into a formula which only contains n+d?1 terms depending on x. d?1

48

References [1] P.G. Ciarlet and P.A. Raviart. General Lagrange and Hermite interpolation in Rn with applications to nite element methods. Arch. Ratioal Mech. Anal., 46 (1972), 178{199. [2] P.J. Davis. Interpolation & Approximation. Dover Books, 2. edition, 1975. [3] C. de Boor. B { form basics. In G. Farin, editor, Geometric Modelling: Algorithms and new trends. SIAM, 1987. [4] C. de Boor and A. Ron. On multivariate polynomial interpolation. Constr. Approx., 6 (1990), 287{302. [5] C. de Boor and A. Ron. Computational aspects of polynomial interpolation in several variables. Math. Comp., 58 (1992), 705{727. [6] M. Gasca and J.I. Maeztu. On Lagrange and Hermite interpolation in Rk. Numer. Math., 39 (1982), 1{14. [7] R. A. Lorentz. Multivariate Birkho Interpolation. Number 1516 in Lecture Notes in Mathematics. Springer Verlag, 1992. [8] C. A. Micchelli. On a numerically ecient method of computing multivariate B{ splines. In W. Schempp and K. Zeller, editors, Multivariate Approximation Theory, pages 211{248. Birkhauser, Basel, 1979. [9] Th. Sauer. Computational aspects of multivariate polynomial interpolation. Advances in Comp. Math., (1995). to appear. [10] Th. Sauer and Y. Xu. On multivariate Lagrange interpolation. Math. Comp., (1995). to appear. [11] Th. Sauer and Y. Xu. A case study in multivariate Lagrange interpolation. to appear in NATO-ASI Proceedings, Maratea, 1995. [12] S. Waldron. A multivariate form of Hardy's inequality and Lp-error bounds for multivariate Lagrange interpolation. Preprint, 1994.

49