Baltzer Journals
December 22, 1995
Regular Points for Lagrange Interpolation on the Unit Disk Thomas Sauer1; and Yuan Xu2;y 1Mathematical Institute, University Erlangen{Nuremberg, 90537 Erlangen, Germany
E-mail:
[email protected]
2
Department of Mathematics, University of Oregon, Eugene, Oregon 97403, USA
E-mail:
[email protected]
A set of points on the unit disk of the Euclidean plane is given, which admits unique Lagrange interpolation. The points have rotational symmetry and they form an example of natural lattices of Chung and Yao [2]. Properties of Lagrange interpolation with respect to these points are studied. Keywords: Lagrange interpolation, two variables, unit disk. Subject classi cation: AMS(MOS) Primary: 41A05, 65D10
1 Introduction
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For n 0 let dn be the space of polynomials of degree n in d variables. Let N = dim dn = n+d d . If XN = fx1; : : :; xN g is a set of distinct points in Rd and if for any f : Rd 7! R, there exists a unique polynomial in dn such that P (xk ) = f (xk ), xk 2 XN , then the Lagrange interpolation problem with respect to XN is said to be poised. It is well-known that the Lagrange interpolation problem in d variables is poised if the points are not on an algebraic surface of degree n. But these conditions are dicult to verify. In [2], Chung and Yao introduced a geometric condition on XN that ensures the poisedness of the Lagrange interpolation problem. In particular, they de ned natural lattices which satisfy their geometric condition. Let H1; : : :; Hn+d be a collection of hyperplanes such that any distinct d hyperplanes chosen among them intersect at one point and di erent choices give di erent points; the set of all intersection points is called the nth order natural lattice in Rd generated by H1 ; : : :; Hn+d . They called these con gurations of points \natural lattice", probably because of the fact that the product of certain linear polynomials vanishes on all points except one, which allows one to write down the Lagrange interpolation polynomial in a way very similar to the univariate case. Interpolation based on similar con guration of points is also studied in [3]. However, we are not aware of any set of regular points which have been constructed explicitly in this way; here by regular we mean points that enjoy certain symmetry and locate inside a regular geometric region which has the same symmetry properties. The purpose of this paper is to provide a remarkable example in which the points are Work done when visiting the University of Oregon at Eugene, Oregon y Supported by National Science Foundation under Grant No. 9302721
T. Sauer, Y. Xu / Regular Points for Interpolation
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all located inside the unit disk on the Euclidean plane. All points can be constructed in a symmetric way and they can be given in compact formulae, which yields very simple formulae for the Lagrange interpolation polynomial and the error of interpolation. Since, in contrast to interpolation in one variable, there are few interesting examples of points in two variables which admit unique polynomial interpolation, we believe that the points constructed in the paper are of interest.
2 Construction of points Let m 2 N and let x0; : : :; x2m be 2m + 1 points, equidistributed on the unit circle. Let `j R2 be the line passing through the points xj and xj +m , j = 0; : : :; 2m, where we make use of the convention that x2m+1+j = xj , j = 0; : : :; 2m. We depict this situation for m = 2; 3 in Fig. 1.
Figure 1: Clearly, any two lines `j and `k , j 6= k, intersect at a point
xj;k = `j \ `k ;
0 j; k 2m;
inside the unit circle and, of course, xj;k = xk;j . To prove that these points indeed admit unique polynomial interpolation of degree n = 2m ? 1, we incorporate the following observation:
Proposition 1
The lines `0; : : :; `2m intersect at m(2m + 1) points which form regular (2m + 1)-gons on the circles with radii cos m rk = 2mk+1 ; k = 1; : : :; m: (1) cos 2m+1
Proof
It is convenient to use the standard coordinates for the unit circle such that 2k ; ; sin xk = (xk ; yk ) = cos 2m2k k = 0; : : :; 2m: + 1 2m + 1
(2)
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T. Sauer, Y. Xu / Regular Points for Interpolation
Then the line `k is the set of solutions of the equation 0 = `k (x) = cos (22km++m1) x + sin (22km++m1) y ? cos 2mm+ 1 ; (3) Note that we use the symbol `k to denote both the line and the ane function which vanishes there. Since the meaning will be clear from the context, this should not cause any confusion. In particular, if we set k = 0 in (3), we obtain the line `0 as the solution of cos 2mm+ 1 x + sin 2mm+ 1 y ? cos 2mm+ 1 = 0:
(4)
m) , gives Multiplying (3) by sin 2mm+1 and subtracting (4), multiplied by sin (22km++1 m cos (k + m) sin k ; x = 2 cos sin 2m2k +1 2m + 1 2m + 1 2m + 1 which leads to cos 2mm+1 (k + m) x= (5) cos 2m + 1 : cos 2mk+1 m) {multiple of (4) In the same manner we multiply (3) by cos 2mm+1 and subtract the cos (22km++1 to obtain m sin (k + m) sin k ; sin 2m2k y = 2 cos +1 2m + 1 2m + 1 2m + 1 and, nally, cos 2mm+1 (k + m) y= (6) sin 2m + 1 : cos 2mk+1 De ning rk , k = 1; : : :; 2m, as in (1) we thus have (k + m) (k + m) x0;k = rk cos 2m + 1 ; sin 2m + 1 and, in particular, kx0;kk22 = x20;k + y02;k = rk2: We note that 0 < r1 < r2 < < rm = 1: Therefore, all intersections of `0 and `k , k = 1; : : :; 2m, lie on one of the circles with radius rk , j = 1; : : :; m. Moreover, it is easily seen that exactly two points, x0;m+k and x0;m+1?k , 0 k m, are on the circle of radius rk , and that these two points are distinct. Thus, the line `0 and the lines `j , j = 1; : : :; 2m, intersect in 2m distinct points which lie on m circles with radius r1; : : :; rm. Since `k can be obtained by applying a rotation of 2m2k+1 to `0, any point xj;k , j 6= k, is the result of applying a rotation of 2m2k+1 to xj ?k;0 . Formally,
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2k ? sin 2m2k+1 x 2m+1 xj;k = cos j ?k;0 2k sin 2m+1 cos 2m2k+1 m) ; sin (k + j + m) : = rj ?k cos (k +2mj + +1 2m + 1
(7)
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Again, for any k = 0; : : :; 2m, all the points xj;k , j 6= k, are distinct and lie on the circles with radii rk , k = 1; : : :; m. To con rm that these are actually m(2m + 1) distinct points, just note that there are 2m points on each of the 2m + 1 lines. Since each point is the intersection of exactly two lines and is counted twice, we end up with a total sum of m(2m + 1) points as claimed. In the following Fig. 2 we depict those circles for the cases m = 2; 3. The pictures suggest
Figure 2: that the regular (2m + 1)-gons on the circles are rotated in two di erent angles only. In fact, the intersections can be explicitly computed as 2j 2 j rk cos 2m + 1 ; sin 2m + 1 ; j = 0; : : :; 2m; if k + m is even and rk cos (22jm++1)1 ; sin (22jm++1)1 ; j = 0; : : :; 2m; if k + m is odd, k = 1; : : :; m. The distribution of points is as shown in Fig. 3. It seems to be worthwhile to remark that for n = 2m ? 2 there also exists a collection of points which admits unique Lagrange interpolation. The easiest way to obtain these con gurations is to consider the intersections of the lines `0; : : :; `2m?1 as de ned above in the case of n = 2m ? 1. Although this case is still symmetric along the line perpendicular to `2m, it no longer has rotational symmetry, because of which we will not consider them in any detail. However, we depict the cases m = 2; 3 in Fig. 4. Also we note that in this con guration there are 2m ? 1 (nonsymmetrical) points on each of the m circles of radius r1; : : :; rm.
3 Properties of Lagrange interpolation Because of Proposition 1, the set of intersection points Xm = fxj;k : 0 j < k 2mg ;
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T. Sauer, Y. Xu / Regular Points for Interpolation
Figure 3:
Figure 4: forms a natural lattice set for polynomials of degree n = 2m ? 1 in the sense of Chung and Yao [2]. Indeed, this is an immediate consequence of the fact that polynomials
Pj;k (x) =
nY +1 =0 6 6= = s
s
j; s
k
`s (x) ; `s(xj;k )
0 j < k 2m;
are of degree n and satisfy
Pj;k (xr;s) = jr ks ; 0 j < k 2m; 0 r < s 2m: Hence, the interpolation polynomial Ln (f; ) with respect to some function f : R2 ! R is given as
Ln(f; x) =
X
j
