Northeast . Math. J . 16 (2) (2000) , 243 - 252
Received date : Oct . 27 , 1999.
Foundation item: The NNSF (19571035) of China

.
Lagrange Interpolation on a Sphere

LIANG Xue-zhang (梁学章) 　FENG Ren-zhong (冯仁忠) 　CUI Li-hong (崔利宏)
( Institute of Mathematics , Jilin University , Changchun , 130023)

Abstract : In this paper , we provide a Circumference2Superposition Process to construct the
properly posed set of nodes for interpolation on a sphere. As an application of the process , we give
a concrete algorithm to do Lagrange interpolation on a sphere. Finally , we also give a kind of
piecewise interpolation on a sphere , a kind of spline interpolation on a sphere and their error
estimates.
Key words : Lagrange interpolation on a sphere , properly posed set of nodes for interpolation ,
spline interpolation
1991 MR subject classif ication : 41A10 , 41A30
CLC number : O241. 5
Document code : A
Article ID : 100021778 (2000) 0220243210

§1. Introduction
　 Let n be a nonnegative integer and S = { ( x , y , z ) ∈R3 | x2 + y2 + z 2 = 1} be the unit
sphere in R3 . P(2)
n and P(3)
n denote the space of all bivariate polynomials of total degree ≤n
and the space of all t rivariate polynomials of total degree ≤n respectively , i. e.
P(2)
n = ∑ 0 ≤i+ j ≤n
aij x iyj | aij ∈ R ,
P(3)
n = ∑ 0 ≤i+ j+ k ≤n
aijkx iyjz k | aijk ∈ R .
Obviously
dim P(2)
n =
n + 2
2
, 　dim P(3)
n =
n + 3
3
.
We consider the following Lagrange interpolation problem on the sphere S .
　 Problem　Suppose m = ( n + 1) 2 , and Qi = ( x i , yi , z i ) , i = 1 , ?, m , are m distinct
point s on the unit sphere S . Given any set of real numbers { f i ∈R| i = 1 , ?, m} , find a
polynomial p ( x , y , z ) ∈P(3)
n satisfying p ( Qi) = f i , i = 1 , ?, m .
　 Def inition 1 　We call the set of points { Qi } m
i = 1 a properly posed set of nodes
( abbreviation PPSN) of degree n f or interpolation on the sphere S if there alw ays exists a
sol ution to the problem f or each set of real numbers { f i} m
i = 1 .
　 Def inition 2 　Let Qi ∈R3 , i = 1 , ?, m′, be m′distinct points , w here m′=
n + 3
3
.
If f or any given set of numbers { f i ∈ R | i = 1 , ?, m′} , we can alw ays f ind a unique
polynomial p ( x , y , z ) ∈P(3)
n satisf ying
p ( Qi) = f i , 　i = 1 , ?, m′,
then we call the set { Qi} m′
i = 1 a PPSN f or P(3)
n .
　 The study on the Lagrange interpolation on a sphere is very significant for practical
application (see [1 ] and [6 ]) . In this paper , by means of remaking the Line2Superposition
Process and the Conic2Superposition Process of const ructing the PPSN for bivariate Lagrange
interpolation described in [ 2 —5 ] , we give a Circumference2Superposition Process of
const ructing the PPSN for interpolation on the unit sphere. Then based on the result , we
give a concrete algorithm of generating Lagrange interpolation polynomial on the unit sphere.
This algorithm is also used to const ruct t rivariate interpolation polynomial inside a ball.
Finally , we also provide a kind of piecewise inter2polation on the sphere , a kind of spline
interpolation on the sphere and their error estimates.
　 Let the perpendicular coordinates and the spherical polar coordinates of any point M in R3
be respectively denoted by
M = ( x , y , z ) , M = ( rsinθcosφ, rsinθsinφ, rcosθ) ,
where 0 ≤φ ≤2π, 0 ≤θ≤π. In the description of the paper , we will use the following
t ransform of rotation axes
x y z
=
cosθcosφ sinφ - sinθcosφ
- cosθsinφ cosφ sinθsinφ
sinθ 0 cosθ
ax
ay
az
. (1)
§2. The Properly Posed Set of Nodes for
Lagrange Interpolation on a Sphere
　 We use Pn ( C) and Pn ( S ) to respectively denote the space of all bivariate polynomials of
total degree ≤n on a circumference C and the space of all t rivariate polynomials of total
degree ≤n on the unit sphere S . It is easy to see that
dim Pn ( C) =
n + 2
2
-
n 2
= 2 n + 1 ,
dim Pn ( S ) =
n + 3
3
-
n + 1
3
= ( n + 1) 2 .
With respect to the PPSN for interpolation on the sphere , we have the following theorems.
4 4 2 NORTHEAST. MATH. J . 　　　　　　　　　　　　　VOL. 16
　 Theorem 1 　Let the set of points { Qi} m
i = 1 be a PPSN of degree n f or interpolation on the
unit sphere S , w here m = ( n + 1) 2 . If there exists a polynomial p ( x , y , z ) ∈ P(3)
n
satisf ing
p ( Qi) = 0 , 　i = 1 , ?, m ,
then there must exist a polynomial q ( x , y , z ) ∈P(3)
n - 2 such that
p ( x , y , z ) = q ( x , y , z ) ( x2 + y2 + z 2 - 1) . (2)
　 Proof 　Let r =
n + 3
3
and the set of point s Θ= { Qm + 1 , ?, Qr} inside the unit ball be a
PPSN of the space P(3)
n - 2 . Then , for any given set of numbers { f 1 , ?, f r} , we can const ruct
a corresponding polynomial p ( x , y , z ) ∈P(3)
n to satisfy p ( Qi ) = f i , i = 1 , ?, r. The
concrete const ructing process is as follows.
　 Since { Qi } m
i = 1 is a PPSN of degree n for interpolation on the sphere S , there exist s a
polynomial p
≈ ( x , y , z ) such that
p
≈ ( Qi) = f i , 　i = 1 , ?, m .
Let R ( x , y , z ) = 0 denote the unit sphere. Because Θ is a PPSN of the space P(3)
n - 2 inside
the ball , R ( Qi) ≠0 and there exist s a unique polynomial q ( x , y , z ) ∈P(3)
n - 2 determined by
the following interpolation condition :
q ( Qi) = ( f i - p
≈ ( Qi) ) / R ( Qi) , i = m + 1 , ?, r.
Now , we const ruct a polynomial
ap ( x , y , z ) = p
≈ ( x , y , z ) + ( x 2 + y2 + z 2 - 1) q ( x , y , z ) .
It is not hard to verify that ap ( Qi) = f i , i = 1 , ?, r. Hence , this shows that { Qi} r
i = 1 is a
PPSN of the space P(3)
n .
　 Now , we set p
≈ ( Qi) = f i = 0 , i = 1 , ?, m and f i = p ( Qi) , i = m + 1 , ?, r. According
to the above const ructing process , in order to prove (2) , we only need to take p
≈ ( x , y , z ) ≡
0 and
ap ( x , y , z ) = ( x2 + y2 + z 2 - 1) q ( x , y , z ) ∈ P(3)
n ,
which satisfies the conditions
ap ( Qi) = p ( Qi) , 　i = 1 , ?, r.
Thus , ap ( X ) and p ( X ) satisfy the same interpolation condition. According to the
interpolation uniqueness in the space P(3)
n , we have
p ( x , y , z ) = ap ( x , y , z ) = q ( x , y , z ) ( x2 + y2 + z 2 - 1) .
The proof is finished.
　 Theorem 2 ( The Circumference2Superposition Process) 　Let m = ( n + 1) 2 and { Qi } m
i = 1
be a PPSN of the space Pn ( S ) . Planeπ: xcosα+ ycosβ+ zcosγ- ρ= 0 (0 <ρ< 1) does not
pass through these m points and intersect the unit sphere S at a ci rcumf erence C.
A rbit rarit y select 2 n + 3 points { Qi } ( n + 2)
2
i = m + 1 on the ci rcumf erence C. Then these 2 n + 3
points together with those m points f orm a PPSN of the space Pn + 1 ( S ) .
　 Proof 　We may suppose the coordinates of the center of the circumference C are (ρcosα,
5 4 2 NO. 2 　　　　　　L IA N G X . Z. et al . 　LAGRANGE INTERPOLATION ON A SPHERE
ρcosβ,ρcosγ) = (ρsinθcosφ,ρsinθsinφ,ρcosθ) . Using the t ransform (1) , we obtain the
equation of the circumference C under a new coordinate system Oaxayaz :
ax 2 + ay2 = 1 - ρ2 = r2 , 　az = ρ.
Let R denote the set of 2 n + 3 point s { Qi } ( n + 2)
2
i = m + 1 on the circumference C. Set Φ = R∪
{ Qi } m
i = 1 . Since R = { Qi } ( n + 2)
2
i = m + 1 are 2n + 3 distinct point s on the circumference C ,
according to [ 2 —3 ] , these 2n + 3 point s must form a PPSN of the space Pn + 1 ( C) under
the new coordinate system. Then for any given set of numbers
{ f i | i = 1 , ?, ( n + 2) 2} ,
there exist s a polynomial ap′( ax , ay) ∈P( 2)
n + 1 satisfying
ap′( Qi) = f i , 　i = m + 1 , ?, ( n + 2) 2 .
Set ap ( x , y , z ) = ap′( ax , ay) . Then ap ( x , y , z ) ∈P( 3)
n + 1 and satisfies
ap ( Qi) = f i , 　i = m + 1 , ?, ( n + 2) 2 .
Now , we const ruct a polynomial
p ( x , y , z ) = ap ( x , y , z ) + ( xcosα + ycosβ+ zcosγ - p) p
≈ ( x , y , z ) ,
such that p
≈ ( x , y , z ) ∈P( 3)
n and
p ( Qi) = ap ( Qi) + ( x icosα + yicosβ+ z icosγ - ρ) p
≈ ( Qi) , i = 1 , ?, m ,
i. e. ,
p
≈ ( Qi) =
p ( Qi) - ap ( Qi)
x icosα + yicosβ+ z icosγ - ρ, 　i = m + 1 , ?, ( n + 2) 2 . (3)
Since { Qi} m
i = 1 is a PPSN of Pn ( S ) , we can find a polynomial p
≈ ( x , y , z ) satisfying ( 3) .
This indicates that there exist s a polynomial p ( x , y , z ) ∈P( 3)
n + 1 such that
p ( Qi) = f i , 　i = 1 , ?, ( n + 2) 2 .
This proves that Φ is a PPSN of Pn + 1 ( S ) .
　 The above discussion on the PPSN for interpolation on a sphere may be used to study the
interpolation problem inside a three2dimensional ball.
　 Theorem 3 ( The Sphere2Superposition Process) 　S uppose the set of points { Qi} m′
i = 1 with
m′=
n + 3
3
inside the unit ball is a PPSN of the space P(3)
n and { Qi } r′
i = m′+ 1 with r′=
n + 5
3
is a PPSN of the space Pn + 2 ( S ) . Then { Qi} r′
i = 1 must f orm a PPSN of the sapce
P(3)
n + 2 .
　 Proof 　Let Ψ = { Qi} r′
i = m′+ 1 , Ω= Ψ ∪{ Q} m′
i = 1 . For any given set of numbers { f i} r′
i = 1 ,
since { Qi} r′
i = m′+ 1 is a PPSN of the space Pn + 2 ( S ) , there exist s a polynomial ap ( x , y , z ) ∈
Pn + 2 ( S ) satisfying
ap ( Qi) = f i , 　i = m′+ 1 , ?, r′.
Now , we const ruct a polynomial p ( x , y , z ) ∈P(3)
n + 2 as follows :
p ( x , y , z ) = ap ( x , y , z ) + ( x 2 + y2 + z 2 - 1) p
≈ ( x , y , z ) ,
6 4 2 NORTHEAST. MATH. J . 　　　　　　　　　　　　　VOL. 16
where p
≈ ( x , y , z ) ∈P(3)
n and satisfies the following interpolation condition :
p ( Qi) = ap ( Qi) + ( x2
i + y2
i + z 2
i - 1) p
≈ ( Qi) , 　i = 1 , ?, m′,
i. e. ,
p
≈ ( Qi) = ( p ( Qi) - ap ( Qi) ) / ( x2
i + y2
i + z 2
i - 1) . (4)
Since { Qi} m′
i = 1 is inside the unit ball and is a PPSN of P(3)
n , it follows that p
≈ ( x , y , z ) can be
uniquely determined by (4) . This shows that there exist s a polynomial p ( x , y , z ) ∈P(3)
n + 2
satisfying
p ( Qi) = f i , 　i = 1 , ?, r′.
Noting that r′=
n + 5
3
equals dim P(3)
n + 2 and { f i } r′
i = 1 can be f reely selected , we conclude
that Ω is a PPSN of the space P(3)
n + 2 .
§3. An Algorithm of Lagrange Interpolation on a Shpere
　 Based on the Circumference2Superposition Process in Section 2 , we give a concrete
algorithm to const ruct Lagrange interpolation polynomial on the unit sphere. Let the
equation of the unit sphere in R3 be x2 + y2 + z 2 - 1 = 0. Take n + 1 distinct planes
πi ( x , y , z ) = xcosαi + ycosβi + zcosγi - ρi = 0 , 　0 < ρi < 1 , i = 0 , ?, n ,
which intersect the unit ball at n + 1 distinct circumferences Ci , i = 0 , ?, n. On the
circumference Ci , we select 2 ( n - i ) + 1 distinct point s ( x ( i)
0 , y ( i)
0 , z ( i)
0 ) , ?, ( x ( i)
2 ( n - i) ,
y ( i)
2 ( n - i) , z ( i)
2 ( n - i) ) , none of which lies on i circumferences C0 , ?, Ci - 1 previous to Ci . Then
these ∑
n
i =0
(2 ( n - i) + 1) = ( n + 1) 2 point s form a PPSN of the space Pn ( S ) .
　 It might as well assume that the coordinate of the center of the circumference Ci under the
coordinate system Ox yz is
(cosαi , cosβi , cosγi) = ( sinθicosφi , sinθisinφi , cosθi) .
Making use of the t ransform (1) , we get the equation of the circumference Ci under the
coordinate system Oaxayaz :
( ax ( i) ) 2 + ( ay ( i) ) 2 = 1 - ρ2
i = r2
i , 　az ( i) = ρi .
Let f ( x , y , z ) ∈C( R3) be an interpolated function and define
f 0 ( x , y , z ) = f ( x , y , z ) , gi ( ax ( i) , ay ( i) , az ( i) ) = f i ( x , y , z ) , i = 0 , ?, n.
We int roduce polar coordinates
( ax ( i) , ay ( i) ) = ( risinθ( i) , risinθ( i) ) , 　i = 0 , ?, n.
Then the polar coordinates of 2 ( n - i) + 1 point s selected on the circumference Ci are
( ax ( i)
j , ay ( i)
j ) = ( ricosθ( i)
j , ri sinθ( i)
j ) , j = 0 , ?,2 ( n - i) , 　i = 0 , ?, n.
Set
ω( i)
k (θ) = ∏
2 ( n- i)
j = 0
sin
θ- θ( i)
j
2
sin
θ- θ( i)
k
2
, 　k = 0 , ?,2 ( n - i) .
7 4 2 NO. 2 　　　　　　L IA N G X . Z. et al . 　LAGRANGE INTERPOLATION ON A SPHERE
Then we can get an interpolation polynomial pn ( x , y , z ) in the space Pn ( S ) corresponding
to f ( x , y , z ) , whose expression is determined by
pn ( x , y , z ) = ∑
n
i = 0
ci ( ax ( i) , ay ( i) ) ·∏
i - 1
j =0
πj ( x , y , z ) ,
where ci ( ax ( i) , ay ( i) ) is calculated by the following formulas :
Ti (θ) = ∑
2 ( n - i)
k = 0
gi ( ax ( i)
k , ay ( i)
k ,ρi)
ω( i)
k (θ)
ω( i)
k (θ( i)
k )
= a0 + ∑
2 ( n- i)
k = 1
a ( i)
k cos kθ+ b( i)
k sin kθ ,
ci ( ax ( i) , ay ( i) ) = a0 + Re ∑
2 ( n - i)
k =1
( a ( i)
k - jb( j)
k )
ax ( i) + jay ( i)
ri
k
,
f i +1 ( x , y) =
f i ( x , y , z ) - ci ( ax ( i) , ay ( i) )
πi ( x , y , z )
, 　i = 0 ,1 , ?, n.
where j is the unit of imaginary number.
　 The calculation order is f 0 →g0 →T0 →c0 →f 1 →g1 →T1 →c1 →?→f n →gn →Tn →cn →
pn .
§4. Piecewise Interpolation and Spl ine Interpolation
　 We know that any point A selected on the unit sphere can form a PPSN of P0 ( S ) .
According to Theorem 2 in Section 2 , three point s additionally selected on the sphere
together with the point A form a PPSN of P1 ( S ) if the plane determined by these three
point s does not pass through the point A . It is easy to see that for any given set of
interpolation conditions , the t rivariate polynomial of total degree ≤1 satisfying these
interpolation conditions is uniquely determined by these four point s. In this section , we first
give a piecewise interpolation method and it s error estimate , then give a spline interpolation
method and it s error estimate.
　 Method 1 ( The Piecewise Interpolation Method)
　 We make an inscribed polyhedron of the unit ball such that it s each face is t riangle piece ,
all it s vertexes are on the sphere and all interior angles of each t riangle piece are all more than
some fixed small angle α0 but not more than
π
2 . Passing by two vertexes of each edge of the
polyhedron , we make a geodesic. These geodesics together with all vertexes of the
polyhedron form a t riangulation for the sphere.
　 Suppose t riangle ABC is one of these t riangle pieces , and G is the circumcenter of the
t riangle ABC (see Figure 1) . The plane passing through the t riangle ABC divides the unit
ball into two spherical caps. Use Σ to denote the less spherical cap and Φ to denote the
intersection set of the unit sphere and Σ. Passing through the point G , we draw a line
which is vertical to the plane ABC and intersect s Φ at the point D. Again passing by point s
A , C and the ball center O , we generate a plane OA C and make it intersect Φ at the
8 4 2 NORTHEAST. MATH. J . 　　　　　　　　　　　　　VOL. 16
geodesic A C ; likewise do geodesic BC and AB . So , we obtain a subfield DABC on the
sphere S and write it Λ.
　 Let f ( x , y , z ) ∈C2 ( R3 ) be a interpolated function. Then , for the set of numbers
{ f ( A ) , f ( B) , f ( C) , f ( D) } , we know that these four point s can uniquely determine an
interpolation polynomial p ( x , y , z ) of total degree ≤1 such that
p ( A ) = f ( A ) , p ( B) = f ( B) , p ( C) = f ( C) , p ( D) = f ( D) .
If p ( x , y , z ) is confined on the subfield Λ, then we have the following error estimate.
　 Theorem 4 　Let the subf iel dΛ ( see Figure 1) be determined by Method 1 , f ( x , y , z ) ∈
C2 ( R3) be interpolated f unction and p ( x , y , z ) be the interpolation polynomial described
above. We use ( x i , yi , z i) , i = 1 ,2 ,3 ,4 , to denote respectively thei r coordinates of points ,
A , B , C , D and w rite
d = max
( x , y , z) ∈S
min
( x′, y′, z′) ∈Γ | x - x′| 2 +| y - y′| 2 +| z - z′| 2 ,
w hereΓ denotes the vertex set of the inscribed polyhedron of the unit ball . Then we have
| f ( x , y , z ) - p ( x , y , z ) | = O
d2
sinα0
.
　 Proof 　For the convenience of description , we use R to denote the circumradius of the
t riangle ABC and a , b , c to denote respectively the lengths of edges BC , A C , AB ;
additionally write the thickness of the spherical cap Σ by h. By the sine law , we have
2 RsinB = b , 　2 Rsin A = a , 　2 Rsin C = c
and easily calculate
h = 1 - 1 - R2 =
R2
1 + 1 - R2
< R2 < R .
Let ε( x , y , z ) = f ( x , y , z ) - p ( x , y , z ) . According to Kincaid’s technique ( see [ 2 ]) ,
ε( x , y , z ) has the following expression :
ε( x , y , z ) = -
1
2 ∑
4
i = 1
( x i - x)
5
5 x
+ ( yi - y)
5
5 y
+ ( z - z i)
5
5 z
2
·
·f (θ1 ( x i - x ) + x , θ1 ( yi - y) + y , θ1 ( z i - z ) + z ) ·
·li ( x , y , z ) , 　0 ≤θ1 ≤1 ,
where l i ( x , y , z ) , i = 1 , 2 , 3 , 4 , are Lagrange’s basic interpolation polynomials. We will
9 4 2 NO. 2 　　　　　　L IA N G X . Z. et al . 　LAGRANGE INTERPOLATION ON A SPHERE
indicate that l i ( x , y , z ) , i = 1 ,2 ,3 ,4 , are uniformly bounded. In fact ,
| l4 ( x , y , z ) | =
V ( ( x , y , z ) - ABC)
V D - ABC
< 1 ,
| l3 ( x , y , z ) | =
V ( ( x , y , z ) - ABD)
V D - ABC
≤
VΣ
V D - ABC
≤
π( h2 - h3/ 3)
abc ·h
12 R
≤12πhR
abc
≤12πR3
abc
=
12πR3
8 R3sin A sinB sin C
,
where V D - ABC denotes the volume of the simplex D - ABC and VΣ denotes the volume of
the spherical cap Σ. From the confinement to t riangle ABC in Method 1 , we know
α0 ≤max{ ∠A , ∠B , ∠C} <
π
2 .
It might as well assume that the least angle of three angles ∠A , ∠B , ∠C is ∠C. It is not
hard to draw the conclusion that ∠C ≤
π 3
and
1
sin A sinB
≤4. Therefore we have
| l3 ( x , y , z ) | ≤ 6π
sinα0
. (5)
Similarly ,
| l1 ( x , y , z ) | ≤ 6π
sinα0
, 　| l2 ( x , y , z ) | ≤ 6π
sinα0
. (6)
From (5) and (6) , it follows that
| ε( x , y , z ) | ≤ 1
2 ·M ·d2 ∑
4
i =1
| l i ( x , y , z ) | = O
d2
sinα0
,
where M is a constant . So Theorem 4 is proved.
　 Next , we give a spline interpolation method of degree 1 in C0 ( R) on the sphere S .
　 Method 2 ( The C0 Spline Interpolation Method)
　 The t riangulation way for the sphere in Method 2 is similar to that in Method 1 , but we
make a little revision. As Method 1 we give an inscribed polyhedron of the sphere S and take
any two neighboring t riangles q1 q2 q3 , q2 q3 q5 ( see Figure 2 ) . Point s q4 , q6 are their
respective fourth point s selected by Method 1. Let f ( x , y , z ) ∈C2 ( R3 ) be interpolated
function on the sphere S . According to the above analysis , we know that point s q1 , q2 , q3 ,
0 5 2 NORTHEAST. MATH. J . 　　　　　　　　　　　　　VOL. 16
q4 can uniquely determine a polynomial p ( x , y , z ) satisfying
p ( q1) = f ( q1) , p ( q2) = f ( q2) , p ( q3) = f ( q3) , p ( q4) = f ( q4) ,
and the other four point s q2 , q3 , q5 , q6 can also uniquely determine a polynomial p′( x , y , z )
satisfying
p′( q2) = f ( q2) , p′( q3) = f ( q3) , p′( q5) = f ( q5) , p′( q6) = f ( q6) .
Now the problem is how we make a new interface line between the subfield q1 q2 q3 q4 and
subfield q2 q3 q5 q6 on the sphere instead of the geodesic passing through two point s q2 , q3 to
ensure that these interpolation polynomials are continuous on the interface. Our processing is
as follows.
　 Suppose p ( x , y , z ) =α1 x +β1 y + γ1 z + d1 , p′( x , y , z ) = α2 x +β2 y + γ2 + d2 are
determined by Method 1. To ensure that p ( x , y , z ) and p′( x , y , z ) are continuous on the
interface line , we only make their values on the interface be equal , i. e. ,
(α1 - α2) x + (β1 - β2) y + (γ1 - γ2) z + d1 - d2 = 0 , 　( x , y , z ) ∈ R3 . (7)
We can see that the equation (7) is a plane passing through point s q2 , q3 . Then the circle
passing through point s q2 , q3 is determined by the following equations :
(α1 - α2) x + (β1 - β2) y + (γ1 - γ2) z + ( d1 - d2) = 0 ,
x 2 + y2 + z 2 = 1.
(8)
The circle (8) is divided into two part s by point s q2 , q3 . We take the less part as the new
interface line between the subfield q1 q2 q3 q4 and subfield q2 q3 q5 q6 on the sphere instead of
the geodesic. By this way , we modify all geodesics in Method 1. In the end of this section ,
we give the error estimate of the C0 spline interpolation.
　 Theorem 5 　S uppose f ( x , y , z ) ∈C2 ( R3) is an interpolated f unction and the sphere S
has been t riangulated by the above method . If the ci rcumci rcle diameter of each t riangle
piece of the inscribed polyhedron is not more than 3 , then the error beween f ( x , y , z ) and
the interpolation polynomial p ( x , y , z ) over each subf iel d satisf ies the f ollowing
relationshi p :
| f ( x , y , z ) - p ( x , y , z ) | = O
d2
sinα0
,
w here d is as that in Theorem 4.
　 Proof 　Arbit rarily take two neighboring modified subfield q1 q2 q3 q4 and q2 q3 q5 q6 ( see
Figure 2) and let polynomial p ( x , y , z ) be determined by the interpolation conditions
p ( q1) = f ( q1) , p ( q2 ) = f ( q2 ) , p ( q3 ) = f ( q3 ) , p ( q4 ) = f ( q4 ) . Analysing the
disturbance of the interface lines of the subfield q1 q2 q3 q4 , we find that these interface lines
are not beyond the three balls whose diameters are respectively line segment s q1 q2 , q2 q3 ,
q3 q1 and whose centers are respectively the cent res of line segment s q1 q2 , q2 q3 , q3 q1 . We
can prove that the spherical cap whose vertex is point q4 and whose thickness is 4 h can
cantains these three balls provided that the circumcircle diameter of the t riangle q1 q2 q3 is not
more than 3. We denote the spherical cap by Σ′. According to Kincaid’s technique , we
have
1 5 2 NO. 2 　　　　　　L IA N G X . Z. et al . 　LAGRANGE INTERPOLATION ON A SPHERE
| ε( x , y , z ) | =| f ( x , y , z ) - p ( x , y , z ) |
≤M ·∑
4
i =1
| l i ( x , y , z ) | .
We are able to prove that l i ( x , y , z ) , i = 1 , 2 , 3 , 4 , are uniformly bounded. In fact , we
have
| l4 ( x , y , z ) | =
V ( ( x , y , z ) - q1 q2 q3
)
V q4 - q1 q2 q3
< 3 ,
| l1 ( x , y , z ) | =
V ( ( x , y , z ) - q2 q3 q4
)
V q4 - q1 q2 q3
≤
VΣ′
V q4 - q1 q2 q3
=
π
3
(4 h) 2 (3 - 4 h)
abch
12 R
≤192πRh
abc
≤ 24π
sin A sinB sin C
≤96π
sinα0
,
where VΣ′denotes the volume of Σ′. Likewise ,
| l2 ( x , y , z ) | ≤ 96π
sinα0
, 　| l3 ( x , y , z ) | ≤ 96π
sinα0
.
　 So , we get
| ε( x , y , z ) | = O
d2
sinα0
.
Theorem 5 is this proved.
References
[1 ] Lawson , C. L. , C12surface interpolation for scattered data on a sphere , Rocky Mountain J . Math. , 14
(1) (1984) , 177 —202.
[2 ] Wang , R. H. and Liang , X. Z. , Approximation of Functions in Several Variables (in Chinese) , Science
Press , Beijing , 1988.
[3 ] Liang , X. Z. , Lagrange representation of multivariate interpolation , Sci . China , A4 (1989) , 385 —
396.
[4 ] Liang , X. Z. and Li , L. Q. , On bivariate osciltory interpolation , J . Comput . A ppl . Math. , 38
(1991) , 271 —282.
[5 ] Liang , X. Z. and Lü, C. M. , Properly posed set of nodes for bivariate Lagrange interpolation ,
Approximation Theory IX , Vol. 2 : Computational Aspect , Vanderbilt University Press , Nashville , TN ,
1998 , 189 —196.
[6 ] Kurt Jetter , Joachim St¨ockles and Joseph D. Ward , Error estimates for scattered data interpolation on
spheres , Math. Comput . , 68 (266) (1999) , 733 —734.
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