Lagrange Interpolation. Xue-Zhang Liang and Chun-Mei Lu. Abstract. This paper develops a theorem based on a problem proposed by the rst author in 1965, ...
Properly Posed Set of Nodes for Bivariate Lagrange Interpolation Xue-Zhang Liang and Chun-Mei Lu
Abstract.
This paper develops a theorem based on a problem proposed by the rst author in 1965, and proves some theorems concerning the geometrical structure of certain properly poised set of nodes for bivariate Lagrange interpolation.
x1. Introduction
In this paper we discuss the bivariate Lagrange interpolation problem in the two-dimensional complex plane C 2 (or in the two-dimensional real plane IR2). Let n be a nonnegative integer. n denotes the space of all bivariate polynomials of total degree n
8 9 < X = i j n = : aij x y jaij 2 C ; : 0i+j n
Let qi = (xi ; yi ) 2 C 2; i = 1; 2; : : : ; k, be k distinct points, where k = 21 (n + 1)(n + 2). We consider the following Lagrange interpolation problem: Problem 1. Given any set of complex numbers ffi 2 C ji = 1; 2; : : : ; kg: We seek a polynomial P (x; y) 2 n satisfying P (qi ) = fi ; i = 1; 2; : : : ; k: (1) We call the set < = fqi gki=1 a properly poised set of nodes (or PPSN, for short) for n if there exists a unique solution for equations (1). It is clear that the condition k = dimn = 21 (n + 1)(n + 2) is necessary for < being a PPSN for n. But the condition is not sucient. This situation is quite di erent from the univariate case. As well known for univariate polynomial interpolation, we have the following trivial observation Approximation Theory IX, Volume 1: Theoretical Aspects Charles K. Chui and Larry L. Schumaker (eds.), pp. 189{196. Copyright oc 1998 by Vanderbilt University Press, Nashville, TN. ISBN 0-8265-1325-5. All rights of reproduction in any form reserved.
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Observation. Let x1 ; x2 ; : : : ; xn+1 be n +1 distinct points in C 1 . Then given n +1 complex numbers f1 ; f2; : : : ; fn+1, there exists a unique polynomial P (x) of degree n such that P (xi ) = fi ; i = 1; 2; : : : ; n + 1: This observation is fundamental to the numerical analyst. But it cannot be simply generalized to the multivariate case. Therefore, the study of properly poised sets of nodes (namely PPSN) constitutes the rst problem of multivariate polynomial interpolation. There are two approaches to this research: (1) to nd the properly poised sets of interpolation conditions for a given space of interpolating polynomials; (2) to nd the properly poised space of interpolating polynomials for a given set of interpolation conditions. C. de Boor and A. Ron have obtained some results (see [3]) on the second approach. Our research work (see [6, 7, 8]) was focused on (1). In 1965, X. Z. Liang gave the following lemma and theorem: Lemma 1. fqi gki=1 (k = 21 (n + 1)(n + 2)) is a PPSN for n , if and only if fqi gki=1 is not contained in any curve in n . (We call P (x; y) = 0 a curve in n if P (x; y) 2 n and if P (x; y) is not identically zero.) Theorem 1. If fqi gki=1 is a PPSN for n , and if none of these points lie on an irreducible curve Q(x; y) = 0 of degree l (l = 1 or l = 2; l = 1 means a straight line; l = 2 means a conic), then fqi gki=1 with the (n + 3)l ? 1 points being distinct and selected freely on the irreducible curve must constitute a PPSN for n+l . The purpose of this paper is to generalize the above theorem to the case l 3, and nally deduce the geometrical structure of PPSN. To this end, we must introduce a few new concepts.
x2. Lagrange Interpolation along an Algebraic Curve without Multiple Factors First we introduce the following De nition 1. Let n; l be two natural numbers,
n+2 n+2?l l, k = ((2n+2));? (2 ); ifif nn l ? 1. 2 Suppose that Q(x; y) = 0 is an algebraic curve without multiple factors (or ACWMF) of degree l and fqi gki=1 are distinct points on Q(x; y ) = 0. If the assumptions
P (x; y) 2 n and P (qi ) = 0; i = 1; 2; : : : ; k;
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imply
P (x; y) 0 on the curve Q(x; y) = 0, then we call fqi gki=1 a properly poised set of nodes for the interpolation of degree n along the ACWMF Q(x; y) = 0 of degree l, and write fqi gki=1 2 In(Q): The following theorem gives a general method to produce a PPSN for the interpolation of degree n along an ACWMF. It also shows the existence of PPSN for the interpolation. Theorem 2. Suppose Q(x; y) = 0 is an ACWMF of degree l, and Q(x; y) is factorized as Q(x; y) = Q1 (x; y) : : : Qm (x; y); where Qi (x; y)(i = 1; 2; : : : ; m) are distinct irreducible polynomials of degree li, respectively, and l1 + l2 + : : : + lm = l. Let n be a nonnegative integer, + m ? (n + 1)(n + 2)=2; if n l ? 1, r = nl m + l(l ? 3)=2; if n l. Suppose that nli + 1 distinct points are freely chosen on each factor curve Qi (i = 1; 2; : : : ; m). Then we can properly delete altogether r points from their aggregate, such that the remaining points constitute a PPSN for the interpolation of degree n along curve Q. Proof: The proof of this theorem for the case n l k? 1 is simple. Hence, we only give a proof for the case n l . Let T = fqigi=1 be the points chosen on the curve Q, where k = nl + m. After deleting some r points from them, the number of remaining points is exactly equal to the number of points in a PPSN for the interpolation of degree n along Q. If P (x; y) 2 n is the solution of system of equations
P (qi ) = 0; i = 1; 2; : : : ; k; (3) then the nth algebraic curve P and li th algebraic curve Qi intersect at least at nli +1 distinct points (i = 1; 2; : : : ; m). Due to Bezout's Theorem (see [9]), there exists R(x; y) 2 n?l such that P (x; y) = Qi (x; y):::Qm (x; y)R(x; y) = Q(x; y)R(x; y ): (4) Conversely, any P (x; y) 2 n expressed by Equation (4) is a solution of Equation (3). Let fT1(x; y); : : : ; Ts1 (x; y)g, fR1(x; y); : : : ; Rs2 (x; y)g be a basis for n , n?l, respectively, s1 = 21 (n + 1)(n + 2), s2 = 12 (n ? l + 1)(n ? l + 2). From Equation (4), there exists in the solution space only s2 linealy independent polynomials of degree n, say Pi (x; y) = Q(x; y)Ri (x; y); i = 1; 2; : : : ; s2 ;
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which pass through T . This shows the rank of the coecient matrix [Tj (qi )], (i = 1; 2; : : : ; k; j = 1; 2; : : : ; s1 ) of equation (3) must be s1 ? s2 = nl ? 21 (l2 ? 3l): So there exists a nonsingular submatrix of order s1 ? s2 and the s1 ? s2 points related to the submatrix belong to In(Q). This completes the proof. In particular, we have Theorem 3. Suppose Q(x; y) = 0 is a lth (l 1) irreducible algebraic curve in C 2. And suppose that nl + 1 distinct points are freely chosen in the curve Q. Then we can properly delete r = 12 (l ? 1)(l ? 2) points from them, such that the remaining points belong to In(Q). Let us recall (see [4]) the following Cramer Strange Proposition. Suppose the curve P (x; y) = 0 of degree n and2 the curve Q(x; y) = 0 of degree n meet exactly at n2 distinct points fqi gni=1. Then any curve P0 of degree n which passes through 21 n(n +3) ? 1 of these points must pass through the other 21 (n ? 1)(n ? 2) points of intersection. From this proposition and Theorem 2 we can get the following Theorem 4. Suppose the curve P (x; y) = 0 of degree n and2 the curve Q(x; y) = 0 of degree n meet exactly at n2 distinct points2fqi gni=1. On the curve Q(x; y) = 0 we choose freely a point q0 beyond fqi gni=1. Then q0 with 2 the 21 n(n + 3) ? 1 points being distinct and selected freely in fqi gni=1 must constitute a PPSN for the interpolation of degree n along curve Q. Proof: We let fqi gki=1(k = 21 n(n + 3) ? 1) be the points taken from fqi gni=12 . Then the number of points in fqigki=0 is 21 n(n + 3), which is exactly equal to the number of points in a PPSN for the interpolation of degree n along curve Q. Next we show fqi gki=0 2 In(Q). In fact, if
P (qi ) = 0; i = 1; : : : ; k; then by the Cramer Strange Proposition, the curve P must pass through the other 12 (n ? 1)(n ? 2) points of intersection, so that
P (qi ) = 0; i = 1; : : : ; n2: Noticing that P (q0 ) = 0, then the nth algebraic curve P and the irreducible curve Q intersect at n2 + 1 points. Since Q(x; y) is irreducible, it follows from Bezout's Theorem, that there exists a constant c such that P (x; y) = cQ(x; y) This proves the theorem.
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Furthermore, we prove the following Theorem 5. Suppose the curve Q(x; y) = 0 of degree k and the straight line P (x; y) = 0 meet exactly at k distinct points fqi gki=1, and suppose Tn 2 In(Q) (n k ? 2) and Tn \ fqi gki=1 = ;. Then we have Tn [ fqigki=1 2 In+1(Q): Proof: We suppose that P (x; y) = ax ? y + c and qi = (xi ; yi ); i = 1; : : : ; k with x1 ; x2 ; : : : ; xk are pairwise distinct.(Otherwise, we can make a coordinate rotation transform). Then we have Q(x; y) = (ax ? y + c)Q1 (x; y) + Q(x; ax + c); with deg Q1 k ? 1, deg Q(x; ax + c) k and Q(xi ; axi + c) = 0; i = 1; 2; : : : ; k: So there must exist a constant c1(6= 0) such that Q(x; ax + c) = c1(x ? x1)(x ? x2 ):::(x ? xk ): On the other hand, for any given bivariate polynomial F (x; y) with deg F at most n + 1 and satisfying F (q) = 0 for all q in Tn; F (qi ) = 0; i = 1; 2; : : : ; k; we have F (x; y) = (ax ? y + c)F1(x; y) + F (x; ax + c); where deg F1 n, deg F (x; ax + c) n + 1 and F (xi ; axi + c) = F (xi ; yi ) ? (axi ? yi + c)F1(xi ; yi ) = 0; i = 1; 2; : : : ; k: It is clear that there exists a univariate polynomial G(x) of degree n ? k +1 such that F (x; ax + c) = G(x)(x ? x1)(x ? x2 ):::(x ? xk ): Therefore, we have F (x; y) = (ax ? y + c)F1(x; y) + G(x)[Q(x; y) ? (ax ? y + c)Q1 (x; y)]=c1 = (ax ? y + c)[F1(x; y) ? G(x)Q1 (x; y)=c1 ] + G(x)Q(x; y)=c1 = (ax ? y + c)H (x; y) + G(x)Q(x; y)=c1 with deg H n: Since F (q) = 0 for all q in Tn, we have H (q) = 0 for all q in Tn. Since Tn 2 In(Q), we have H (x; y) = R(x; y)Q(x; y ); F (x; y) = ((ax ? y + c)R(x; y) + G(x)=c1 )Q(x; y) which completes the proof.
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x3. The Lagrange Interpolation in C 2 (or IR2) In this section we study the structure of PPSN for Lagrange interpolation in C 2 (or IR2). The following theorem relates the interpolation problem in C 2 to the problem of interpolation along an ACWMF. Theorem 6. (Recursive Construction Theorem). Let < be a PPSN for n , and the number of points in < be 21 (n + 1)(n + 2). If none of these points lie on an ACWMF Q(x; y) = 0 of degree l, then for any T 2 In+l(Q), T [ < must be a PPSN for n+l. Proof: The number of points in T [ < is 1 (n + 1)(n + 2) + (n + l)l ? 1 (l2 ? 3l) = 1 (n + l + 1)(n + l + 2); 2 2 2 which is exactly equal to the dimension of n+l(C 2). If T [ < is not a PPSN for n+l, then there must exist a curve P (x; y) 2 n+l such that P passes through all of these points in T [
