Rev. Anal. Num´er. Th´eor. Approx., vol. 34 (2005) no. 1, pp. 47–53 ictp.acad.ro/jnaat
BOUNDS FOR THE REMAINDER IN THE BIVARIATE SHEPARD INTERPOLATION OF LIDSTONE TYPE
TEODORA C ˘ATINAS¸∗
Abstract. We study the bivariate Shepard-Lidstone interpolation operator and obtain new estimates for the remainder. Some numerical examples are provided.
MSC 2000. 41A63, 41A80.
Keywords. Bivariate Shepard-Lidstone interpolation, remainder.
1. INTRODUCTION
As it is pointed out in [7] and [15], interpolation at nodes having no ex- ploitable pattern is referred to as the case of scattered data and there are two important methods of interpolation in this case: the method of Shepard and the interpolation by radial basis functions.
Consider −∞ < a < b < ∞ and −∞ < c < d < ∞ and let ∆ : a = x0 < x1 < . . . < xN+1 = b and ∆0 : c = y0 < y1 < . . . < yM+1 = d denote uniform partitions of the intervals [a, b] and [c, d] with stepsizes h = (b−a)/(N+ 1) andl= (d−c)/(M+ 1),respectively.Further, letρ= ∆×∆0 be a rectangular partition of [a, b]×[c, d].For the univariate functionf and the bivariate functiongand each positive integerrwe denote byDrf = drf /dxr, Drxg=∂rg/∂xr and Dyrg=∂rg/∂yr.
According to [1] and [2], for a fixed ∆ denote the setLm(∆) ={h∈C[a, b] : h is a polynomial of degree at most 2m −1 in each subinterval [xi, xi+1], 0≤i≤N}.
Definition 1. [2]For a given functionf ∈C2m−2[a, b]we say thatL∆mf is the Lidstone interpolant of f if L∆mf ∈Lm(∆) with
D2k(L∆mf)(xi) =f(2k)(xi), 0≤k≤m−1, 0≤i≤N+ 1.
According to [2], forf ∈C2m−2[a, b] the Lidstone interpolantL∆mf uniquely exists and on the subinterval [xi, xi+1],0≤i≤N, can be explicitly expressed
∗“Babe¸s-Bolyai” University, Faculty of Mathematics and Computer Science, Department of Applied Mathematics, 400084 Cluj-Napoca, Romania, e-mail:
as (1)
(L∆mf)|[xi,xi+1](x) =
m−1
X
k=0
hΛk xi+1h−xf(2k)(xi) + Λk x−xh if(2k)(xi+1)ih2k, where Λk is the Lidstone polynomial of degree 2k+ 1, k ∈N on the interval [0,1].
We have the interpolation formula
f =L∆mf+R∆mf, whereR∆mf denotes the remainder.
For a fixed rectangular partition ρ= ∆×∆0 of [a, b]×[c, d] the set Lm(ρ) is defined as Lm(ρ) =Lm(∆)⊗Lm(∆0) (see, e.g., [1] and [2]).
Definition 2. [2] For a given function f ∈ C2m−2,2m−2([a, b]×[c, d]) we say thatLρmf is the two-dimensional Lidstone interpolant off ifLρmf ∈Lm(ρ) with
D2µx Dy2ν(Lρmf)(xi, yj) =f(2µ,2ν)(xi, yj), 0≤i≤N+ 1, 0≤j≤M + 1, 0≤µ, ν ≤m−1.
According to [2], forf ∈C2m−2,2m−2([a, b]×[c, d]),the Lidstone interpolant Lρmf uniquely exists and can be explicitly expressed as
(2) (Lρmf)(x, y) =
N+1
P
i=0 m−1
P
µ=0 M+1
P
j=0 m−1
P
ν=0
rm,i,µ(x)rm,j,ν(y)f(2µ,2ν)(xi, yj),
where rm,i,j, 0≤i≤N + 1, 0≤j ≤m−1, are the basic elements of Lm(ρ) satisfying
(3) D2υrm,i,j(xµ) =δiµδ2υ,j, 0≤µ≤N + 1, 0≤υ≤m−1.
Lemma 3. [2] If f ∈C2m−2,2m−2([a, b]×[c, d]), then
(Lρmf)(x, y) = (L∆mL∆m0f)(x, y) = (L∆m0L∆mf)(x, y).
Corollary 4. [2] For a function f ∈ C2m−2,2m−2([a, b]× [c, d]), from Lemma 3, we have that
f −Lρmf = (f −L∆mf) +L∆m(f−L∆m0f) (4)
= (f −L∆mf) + [L∆m(f −L∆m0f)−(f −L∆m0f)] + (f −L∆m0f).
With the previous assumptions we denote by L∆,im f the restriction of the Lidstone interpolation polynomialL∆mfto the subinterval [xi, xi+1],0≤i≤N, given by (1), and in analogous way we obtain the expression of L∆m0,if, the restriction ofL∆m0f,to the subinterval [yi, yi+1]⊆[c, d],0≤i≤N. We denote
by SL the univariate combined Shepard-Lidstone operator, introduced by us in [4]:
(SLf)(x) =
N
X
i=0
Ai(x)(L∆,im f)(x), withAi, i= 0, ..., N, given by
(5) Ai(x, y) =
N
Q
j=0 j6=i
rjµ(x, y) N
P
k=0 N
Q
j=0 j6=k
rjµ(x, y)
and
(6) PN
i=0
Ai = 1.
The univariate Shepard-Lidstone interpolation formula is
(7) f =SLf +RLf.
We considerf ∈C2m−2,2m−2([a, b]×[c, d]) and the set of Lidstone functio- nals
ΛiLi=nf(xi, yi), f(xi+1, yi+1), . . . , f(2m−2,2m−2)
(xi, yi), f(2m−2,2m−2)(xi+1, yi+1)o,
regarding each subrectangle [xi, xi+1]×[yi, yi+1],0 ≤ i ≤ N, with ΛiLi = 4m, 0≤i≤N. We denote byLρ,imf the restriction of the polynomial given by (2) to the subrectangle [xi, xi+1]×[yi, yi+1],0≤i≤N.This 2m−1 polynomial, in each variable, solves the interpolation problem corresponding to the set ΛiLi, 0≤i≤N and it uniquely exists.
We have
(Lρ,imf)(2ν,2ν)(xk, yk) =f(2ν,2ν)(xk, yk), 0≤i≤N; 0≤ν ≤m−1;k=i, i+ 1.
The bivariate Shepard operator of Lidstone type SLi, introduced by us in [5], is given by
(8) (SLif)(x, y) = PN
i=0
Ai(x, y)(Lρ,imf)(x, y).
We obtain the bivariate Shepard-Lidstone interpolation formula,
(9) f =SLif+RLif,
whereSLifis given by (8) andRLif denotes the remainder of the interpolation formula.
Next, we give an error estimation using the modulus of smoothness of order k.For a functiong defined on [a, b] we have
ωk(g;δ) = sup{|∆khg(x)| :|h| ≤δ, x, x+kh∈[a, b]},
withδ ∈[0,(b−a)/k] and
∆khg(x) =
k
X
i=0
(−1)k+i kig(x+ih).
We consider the norm in the setC(X) of continuous functions defined onX by kfkC(X)= max
x∈X |f(x)|. We recall first a result from [17]:
Theorem 5. [17] Let L be a bounded operator and let L(P) =P for every P ∈ Pk−1. Then for every bounded function f : [a, b] → R the following inequality is fulfilled
kf −L(f)kC[a,b]≤(1 +kLk[a,b])Wkωkf;b−ak , where Wk is Whitney’s constant.
We apply this result for the operators L∆m and L∆m0. For f ∈ C[a, b] and g∈C[c, d] we have
f −L∆m(f)
C[a,b]≤(1 +kL∆mk[a,b])W2mω2mf;b−a2m, (10)
g−L∆m0(g)C[c,d]≤(1 +kL∆m0k[c,d])W2mω2m
g;d−c2m.
Now we give an estimation of the remainderRLf from (7), in terms of the modulus of smoothness.
Theorem 6. If f ∈C2m−2[a, b], then (11) kRLfkC[a,b]≤(1 +L∆m
[a,b])W2mω2m
f;b−a2m. Proof. We have
(RLf)(x) =f(x)−PN
i=0
Ai(x)(L∆,im f)(x)
= PN
i=0
Ai(x)f(x)−PN
i=0
Ai(x)(L∆,im f)(x)
= PN
i=0
Ai(x)[f(x)−(L∆,im f)(x)], and taking into account (10) and that
N
P
i=0
|Ai(x)|= 1,
relation (11) follows.
The next result provides an estimation of the error in formula (9).
Theorem 7. If f ∈C2m−2,2m−2([a, b]×[c, d]), then
RLif
C[a,b]≤ 1 +L∆m
C[a,b]
W2m max
y∈[c,d]ω2mf(·, y);b−a2m + 1 +L∆m
C[a,b]
W2m max
y∈[c,d]ω2m (f −L∆m0f)(·, y);b−a2m + 1 +kL∆m0kC[c,d]W2m max
x∈[a,b]ω2m
f(x,·);d−c2m,
where Wk is Whitney’s constant.
Proof. Taking into account (8) and (6) we get (RLif)(x, y) =f(x, y)−(SLif)(x, y)
=f(x, y)− PN
i=0
Ai(x, y)(Lρ,imf)(x, y)
=
N
P
i=0
Ai(x, y)f(x, y)− PN
i=0
Ai(x, y)(Lρ,imf)(x, y)
=
N
P
i=0
Ai(x, y)[f(x, y)−(Lρ,imf)(x, y)].
Next applying the results (4) given by Corollary 4 and (6) it follows that (RLif)(x, y) =
N
P
i=0
Ai(x, y){(f−L∆,im f)(x, y)
+ [L∆,im (f−L∆m0,if)(x, y)−(f−L∆m0,if)(x, y)]
+ (f−L∆m0,if)(x, y)}
=PN
i=0
Ai(x, y)(f −L∆,im f)(x, y) + PN
i=0
Ai(x, y)[L∆,im (f −L∆m0,if)(x, y)−(f−L∆m0,if)(x, y)]
+
N
P
i=0
Ai(x, y)(f −L∆m0,if)(x, y)
=
"
f(x, y)PN
i=0
Ai(x, y)− PN
i=0
Ai(x, y)(L∆,im f)(x, y)
#
− PN
i=0
Ai(x, y)[(f−L∆m0,if)(x, y)−L∆,im (f−L∆m0,if)(x, y)]
+
"
f(x, y)PN
i=0
Ai(x, y)− PN
i=0
Ai(x, y)(L∆m0,if)(x, y)
#
and, finally, (RLif)(x, y) =
"
f(x, y)− PN
i=0
Ai(x, y)(L∆,im f)(x, y)
#
− PN
i=0
Ai(x, y)[(f−L∆m0,if)(x, y)−L∆,im (f−L∆m0,if)(x, y)]
+
"
f(x, y)− PN
i=0
Ai(x, y)(L∆m0,if)(x, y)
# .
Applying Theorem 6 three times the conclusion follows.
Example 1. Letf : [−2,2]×[−2,2]→R, f(x, y) =xe−(x2+y2)
and consider the nodes z1 = (−1,−1), z2 = (−0.5,−0.5), z3 = (−0.3,−0.1), z4 = (0,0), z5 = (0.5,0.8), z6 = (1,1). In Figure 1 we plot the graphics of f and SLif for µ = 1. In Figure 2 we plot the error (in absolute value) for Shepard interpolation regarding these data, and also, the error for Shepard interpolation of Lidstone type; we notice that in both cases, the maximum value is 0.5.
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−2
−1 0 1
−0.52
−0.4
−0.3
−0.2
−0.1 0 0.1 0.2 0.3 0.4 0.5
(a) Graph off.
−1 −0.5 0 0.5 1
−1
−0.5 0.5 0
1
−0.4
−0.3
−0.2
−0.1 0 0.1 0.2 0.3
(b) The interpolant SLi(2)f, µ = 1.
Fig. 1. Graph off andSLi(2)f.
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−1
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Received by the editors: March 10, 2004.