ORDERING OF SOME SUBSETS FROM
IR
dUSED IN THE
MULTIDIMENSIONAL INTERPOLATION THEORY
by
Crainic Nicolae
Abstract: The ordering of the elements of a subset of INd, d ≥1, with the order relations defined at the beginning of this article and implicitly obtaining some expressions for them, corresponding to the defined order relations, is useful for the multidimensional interpolation theory, especially when it is necessary to exemplify some general theoretical notions. This is the reason for presenting expressions for ordered subsets from IN2, IN3 and in general from INd.
Next we present the most usual order relations encountered in the interpolation theory and used throughout this article.
1. We say that i=(i1,i2,...,id)∈INd is in the relation “«” with d
d IN
j j
j ∈
=( 1, 2,..., )
j and we write i«
j
, wheni
1<
j
1, ori
1=
j
1 andi
2<
j
2,…, ori
1=
j
1,…,i
d−1=
j
d−1andi
d≤
j
d.2. We say that i=(i1,i2,...,id)∈INd is in the relation
"
p
"
withd
d IN
j j
j ∈
=( 1, 2,..., )
j and we write
i
p
j
, when i < j or i = j and i«j
. Let be the set S={
i
=
(
i
1,
i
2)
/
0
≤
i
1≤
n
,
0
≤
i
2≤
m
j,
j
=
0
,
n
)
}
from2 IN , denoted by
n
m m n
S
; ,...,0 , wheren and 0≤
m
0≤
m1…≤
m
n are given natural numbers. It is obvious that we would rather write out the set ( ; ,...,,
0 mn
m n
S
«) than the set)
,
(
; ,...,0 mn
p
m n
S
whose last terms are difficult to be written. Thus we have:( ; ,...,
,
0 mn
m n
S
«)=UU
n
j m
k
j k j 0 0
)} , {(
= =
=
U
n
j
j
m j j
j 0
)} , ( ),..., 1 , ( ), 0 , {(
=
=
If in the set
n
m m n
S
; ,...,0 we take
m
0=n,m
j=n
−
j
,j
=
1
,
n
, and respectively m0 =m1 =...=mn =n2, n=n1, we obtain some of the most usual forms of some subsets from S⊂
IN2, that is the triangleT
n2, respectively the rectangleR
n2, wheren
=
(
n
1,
n
2)
∈
IN
02. More concrete, these sets are{
IN n}
Tn = i∈ / i ≤2 0 2
,
{
02/
0
,
1
,
2
}
2 , 2
2
1
=
∈
≤
≤
=
=
R
IN
i
n
j
R
n n ni
j jand they are depicted in Figure 1 a and 1 b.
0 1 2 n-1 n 0 1 2
0 0
1 1
n-2
2 n-1
n n2
n1
[image:2.595.136.475.212.436.2](a)
(b)
Figure 1
It is quite obvious that the grid of points from Figure 1
ais made out
of the parallel lines
2
k
∆
=
j
k
j
k
n
k
j
,
0
,
)}
,
{(
0
=
−
=
U
.
It follows the next explicit expression of triangle
T
n2 ordered by relation"
"
p
) , ( 2
p
n
T =
UU
n
k k
j
j
k
j
0 0
)}
,
{(
= =
−
=
U
n
k
k
k
k
0
)}
0
,
(
),...,
1
,
1
(
),
,
0
{(
={(0,0),(0,1),(1,0),...,(0,k),(1,k−1),...,(k,0),...,(0,n),(1,n−1),...,(n,0)}. (2)
Moreover, if we consider the relation (1) and the fact that
T
n2=S
n;n,n−1,...,0, we have:(
T
n2,«)=UU
n
j j n
k
k j 0 0
)} , {(
= −
=
=
U
n
j
j n j j
j 0
)} ,
( ),..., 1 , ( ), 0 , {(
= −
=
)} 0 , ( ),..., 1 , 1 ( ), 0 , 1 ( ),..., 1 , 1 ( ),..., 1 , 1 ( ), 0 , 1 ( ), , 0 ( ),..., 1 , 0 ( ), 0 , 0
{( n n− n− n− n
= .
(3) From the same relation (1) and the fact that
T
n2=2 2 2 1;n ,n ,...,n
n
S
, we obtain(
R
n2,«)=U
10
2)} , ( ),..., 1 , ( ), 0 , {(
n
j
n j j
j
=
=
UU
1 20 0
)} , {(
n
j n
k
k j
= =
=
= {(0,0),(0,1),...,(0,n2),(1,0),(1,1),...,(1,n2),...,(n1,0),(n1,1),...,(n1,n2)}. (4)
An explicit expression of
(
R
n2,
p
)
is more difficult because it depends on the position of the integers n1and n2 with respect to each other. But for n1 =n2 we have:) , ( 2
,n p n
R =(Tn2,p)
U
{(1,n),(2,n−1),...,(n,1),(2,n),(3,n−1),...,(n,2),...,(n,n)}. We keep also in mind the following expressions of the ordered subsets from 2IN
:(
T
n2\T
k2,p)={(
0
,
).(
1
,
1
),...,
(
,
0
)}
1
j
j
j
nk j
−
+=
U
=U U
n
k j
j
i
i
j
i
1 0
)}
,
{(
+
= =
−
=
=
)} 0 , ( ),..., 1 , 1 ( ), , 0 ( ),..., 0 , 2 ( ),..., 1 , 1 ( ), 2 , 0 ( ), 0 , 1 ( ),..., , 1 ( ), 1 , 0
{( k+ k k+ k+ k+ k+ n n− n
,
(Rn2,n \Tn2,p)=
,
{(
1
,
n
),
(
2
,
n
−
1
),...,
(
n
,
1
)(
2
,
n
),
(
3
,
n
−
1
),...,
(
n
,
2
),...,
(
n
,
n
)}
and (Rn2,n \Tn2,«)=)}
,
(
),...,
1
,
(
),
0
,
(
),...,
,
3
(
),
1
,
3
(
),
2
,
3
(
),
,
2
(
),
1
,
2
(
),
,
1
In general, extensions of the two dimensional case are used as examples in 3
, d ≥
INd , among which we exemplify the simplex (Figure 2, in
IN
03){
IN n}
Tnd = i∈ 0d / i ≤ , respectively the hyper-parallelipiped (Figure 4, in
IN
03){
IN
i
n
j
d
}
R
R
j jd d
n n n d
d 0
/
0
,
1
,
,..., ,2
1
=
∈
≤
≤
=
=
i
n .
Proposition: IfS is an arbitrary set from
IN
d, and ii S
n ∈
=max , then
d n
T
S
⊂
.Proof: For any i∈S, with i ≤ i
i∈S
max =
n
, we havei
∈
T
nd.Next we will build successively an expression for the simplex
(
d,
p
)
n
[image:4.595.151.459.298.588.2]T
.Figure 2
2
k
∆
=j
k
j
k
n
k
j
,
0
,
)}
,
{(
0
=
−
=
U
,their traversal order being (
∆
2k,p)=(2
k
∆
,«), and of each of them, from left to right, in descendant order. It follows that∆
2k given in the previous relation is (∆
2k,p). As each element of the set∆
2k has the sum of indexes k-constant it follows that (∆
2k,p)=(2
k
∆
,«). From the same relation we have that2
i k−
∆
=U
i k
j
j
i
k
j
−=0
−
−
)}
,
{(
,i
=
0
,
k
.If we translate every line
∆
2k−i,i
=
1
,
k
in thei
=
1
,
k
x-coordinate plane,obviously parallel with the initial
yOz
plane, we obtain the ordered triangle∆
3kfrom 30
IN
, that is:(
∆
3k,p)=U
ki i k i
0 2
,
= ∆ −
=
UU
k
i i k
j
j
i
k
j
i
0 0
)}
,
,
{(
= −
=
−
−
=
=
−
−
−
−
−
=
=U
ki
i
k
i
i
k
i
i
k
i
i
k
i
0
)}
0
,
,
(
),
,
0
,
(
),...,
1
,
1
,
(
),
,
0
,
{(
)}
0
,
0
,
(
),...,
0
,
1
,
1
(
),...,
2
,
1
,
1
(
),
1
,
0
,
1
(
),
0
,
,
0
(
),...,
1
,
1
,
0
(
),
,
0
,
0
{(
k
k
−
k
k
−
k
−
k
−
k
=
,depicted in Figure 2, whose vertices are the points of coordinates
(
k
,
0
,
0
),
(
0
,
k
,
0
)
and)
,
0
,
0
(
k
. To denote the plane of which∆
2k is a part of after translation, we added the iindex to the line
∆
2k thus obtaining2
i k−
∆
. Actuallyi
becomes the x-coordinate for all points from the translated i x-coordinate plane. The traversal order of the lines ∆2i,k−i is ∆20,k, ∆21,k−1 ,… , ∆2k,0. It follows that:)
,
(
T
n3p
=U
n
k k
0 3
= ∆ =
UUU
n
k k
i i k
j
j
i
k
j
i
0 0 0
)}
,
,
{(
= = −
=
−
−
(5)which is the same to
(
T
n3,
p
)
, taking into consideration the traversal order of theAn explicit form of the ordered set
(
T
n3,
p
)
is:)
,
(
T
n3p
=),..., 0 , 0 , 2 ( ), 0 , 1 , 1 ( ), 1 , 0 , 1 ( ), 0 , 2 , 0 ( ), 1 , 1 , 0 ( ), 2 , 0 , 0 ( ), 0 , 0 , 1 ( ), 0 , 1 , 0 ( ), 1 , 0 , 0 ( ), 0 , 0 , 0 {( ), 0 , , 0 ( ),..., 1 , 1 , 0 ( ), , 0 , 0
( n n− n
(
1
,
0
,
n
−
1
),
(
1
,
1
,
n
−
2
),...,
(
1
,
n
−
1
,
0
),...,
(
n
,
0
,
0
))}
. Proceeding similarly to the previous situation, we obtain(
∆
3k−t,p)=U
t k i i t k i −
=0∆ −−
2
, =
U U
t k i i t k j
j
i
t
k
j
i
− = − − =−
−
−
0 0)}
,
,
{(
, respectively(∆3t,k−t,p)=
U U
t k i i t k j
j
i
t
k
j
i
t
− = − − =−
−
−
0 0)}
,
,
,
{(
,from where we obtain that 4
k
∆
=U
k t t k t 0 3 , =
∆
−=
UU U
k t t k i i t k j
j
i
t
k
j
i
t
0 0 0
)}
,
,
,
{(
= − = − − =−
−
−
, respectively)
,
(
T
n4p
=U
n k k 0 4 =
∆
=
UUU U
n k k t t k i i t k j
j
i
t
k
j
i
t
0 0 0 0
)}
,
,
,
{(
= = − = − − =−
−
−
.Now the generalization to the d-dimensional space becomes obvious, namely:
d k
∆
U U U
kU
i i k i i i k i i i k i d d d d
i
i
i
k
i
i
i
0 0 ) ( 0 ) ... ( 0 1 2 1 1 2 1 1 1 2 2 1 3 2 1 1)}
...
(
,
,...,
,
{(
...
= − = + − = + + − = − − − −+
+
+
−
=
, (6) respectively)
,
(
dp
n
T
=U
UU U U
U
n k k i i k i i i k i i i k i d d n k d k d d
i
i
i
k
i
i
i
0 0 0 ) ( 0 ) ... ( 0 1 2 1 1 2 1 0 1 1 2 2 1 3 2 1 1
)}
...
(
,
,...,
,
{(
...
= = − = + − = + + − = − − = − −+
+
+
−
=
∆
(7)Theorem: An arbitrary set
(
S
,
p
)
fromIN
d can be written in the form)
,
(
S
p
=U
n
t d kt
1 '
=
∆
, (8)
where i
i S
n
k ∈
=max , dk d kt ⊂∆
∆'
are given by (6).
Example:
S
=
{(
0
,
0
),
(
0
,
2
),
(
1
,
1
),
(
1
,
2
)}
={(
0
,
0
)}
∪
{(
0
,
2
),
(
1
,
1
)}
∪
{(
1
,
2
)}
= =∆'2 ∪1
k ∆ ∪
2 '
2
k
2 '
3
k
∆ =
U
31 2 '
=
∆
t t
k , where
2 '
1
k
∆ =
∆
20, '2 2k
∆ =
{(
0
,
2
),
(
1
,
1
)}
⊂
∆
22, and 2' 3
k
∆ =
{(
1
,
2
)}
⊂
∆
23. [image:7.595.155.453.325.597.2]In what follows, we will build in successive steps another expression for the simplex
T
nd using partially the relations"
p
"
and “«”. Eventually we obtain a total order relation defined onT
nd, which we denote by “∠”.The grid of points from the zero x-coordinate plane, Figure 3 (plane yOz), according to relation (.3), is the triangle (Tn2,p)=
UU
n
k k
j
j
k
j
0 0
)}
,
{(
= =
−
, made out of the
parallel lines
∆
2k=j
k
j
k
n
kj
,
0
,
)}
,
{(
0
=
−
=
U
, their traversal order being∆
20,∆
21 , …,
∆
2n, and of each of them, from left to right, in descending order. It follows that) , ( 2 p
i n
T− =
j
k
j
i
n
in
k k
j
,
0
,
)}
,
{(
0 0
=
−
−
= =
UU
.If we translate each triangle ( 2 ,p)
i n
T − , with
i
=
1
,
n
, in thei
=
1
,
n
x-coordinate plane, obviously parallel to the initial plane of the triangle, it results the ordered tetrahedron,
(
T
n3 ∠)=U
n
i i n i
T 0
2 ,
= −
from Figure 3., which is different both from
(
T
n3,
p
)
and (T
n3,«). We added the indexi
to the triangle Tn2−i to indicate the x-coordinate of the plane of which Tn2−i is part of after translation. In facti
becomes x-coordinate for all points from thei
x-coordinate translated plane. The traversal order of the triangles Ti2,n−i is 2 ,, 0n
T 2 , 1 , 1n−
T … ,Tn2,0. It follows that:
,
(
T
n3 ∠)=UUU
n
i i n
k k
j
j
k
j
i
0 0 0
)}
,
,
{(
= −
= =
−
.
An explicit form of the ordered set
(
3,
n
T
∠) is:,
(
T
n3 ∠)=), 0 , , 0 ( ),..., 1 , 1 , 0 ( ), , 0 , 0 ( ),..., 0 , 2 , 0 ( ), 1 , 1 , 0 ( ), 2 , 0 , 0 ( ), 0 , 1 , 0 ( ), 1 , 0 , 0 ( ), 0 , 0 , 0
{( n n− n
))} 0 , 0 , ( ,..., 0 , 1 , 1 ( ...,
),..., 2 , 1 , 1 ( ), 1 , 0 , 1 ( ),..., 0 , 2 , 1 ( ), 1 , 1 , 1 ( ), 2 , 0 , 1 ( ), 0 , 1 , 1 ( ), 1 , 0 , 1 ( ), 0 , 0 , 1 (
n n
n n
−
− −
,
(
T
nd ∠)=U U U
U
U
U
ni i n
i
i i n
i
i i n
i
i i n
k
k
i
d d
d
d
d
d
i
k
i
i
i
0 0 ) (
0
) ... (
0
) ... (
0 0
1 1
2 1 1
1
2
2 1
3
3 1
2
2 1
1
)}
,
,...,
,
{(
...
= −
= + −
=
+ + −
=
+ + −
= = − −
−
−
−
−
−
.(9)
An explicit expression for the simplex
(
T
nd,
«) can be obtained starting from the triangle (T
n2,«)=UU
n
j j n
k
k j 0 0
)} , {(
= −
=
given by relation (3), lying in the plane yOz from Figure 3. The elements of the triangle (
T
n2,«) are the grid of points lying in the mentioned plane, whose explicit form is obtained if we traverseT
n2from bottom to top and from left to right. We can also write that(
T
n2−i,«)=U U
i nj j i n
k
k j
−
= − −
=
0 0
)} , {( .
Translating each triangle ( 2 ,p)
i n
T − ,
i
=
1
,
n
in thei
=
1
,
n
x-coordinate plane, we obtain the ordered tetrahedron,
(
T
n3 «)=U
n
i i n i
T 0
2 ,
= −
.
The iindex added to Tn2−i, indicates the plane of which Tn2−i is part of after translation and becomes the x-coordinate for all points from this plane. The traversal order of triangles Ti2,n−i is also T02,n, ,
2 1 , 1n−
T … ,Tn2,0. It follows that:
,
(
T
n3 «)=UU U
n
i i n
j j i n
k
k
j
i
0 0 0
)}
,
,
{(
= −
= − −
=
Eventually, for the d-dimensional space we obtain that:
,
(
dn
T
«)=U U U
U
U
n
i i n
i
i i n
i
i i n
i
i i n
i
d d
d
d
d
d
i
i
i
i
0 0 ) (
0
) ... (
0
) ... (
0
1 2 1 1
1
2
2 1
3
2 1
1
1 1
)}
,
,...,
,
{(
...
= −
= + −
=
+ + −
=
+ + −
= −
−
−
−
(10)
Figure 4
The grid of points from the zero x-coordinate plane, Figure 4 (plane yOz), according to relation (4), is the rectangle (
R
n2,«)=( 2,2 1n
n
R ,«)=
UU
1 20 0
)} , {(
n
j n
k
k j
= =
, its traversal order being from bottom to top and from left to right. If we “multiply” the rectangle
R
n2 by successive translations, in the i=1,n3 x-coordinate planes, obviously parallel to the initial plane of the rectangle, we obtain the ordered parallelepiped(
R
n3,«)=U
32 1 0
2 ) , ( ,
n
i
n n i
R
=
from Figure 4. The
i
index added toR
n2 indicates the plane of which this is a part of after translation and becomes x-coordinate for all points from thei
x-coordinate plane. The traversal order of the rectanglesR
n2 is R02,(n1,n2), ,2 ) , ( , 1 n1n2
R …, 2,( , ) 2 1 3 n n
n
R . It
follows that:
( 3, , 2 1 3n n
n
R ,«)=
UUU
3 1 20 0 0
)}
,
,
{(
n
i n
j n
k
k
j
i
= = =Generalizing to the d-dimensional space we obtain: (Rnd1,n2,...,nd,«)=
U U U U
1 1 2 2 1 1 0 0 0 0
2
1
,
,...,
)}
{(
...
n i n i n i n i d d d d di
i
i
= = = = − − (11)Remark: An expression for ( , ,..., , ) 2 1 p d n n n d
R can be obtained applying the relation (8) from the previous theorem.
As we pointed out in the beginning, the notions presented in this article are useful in the field of multidimensional interpolation. We illustrate this by two examples.
Example1:
The totally ordered set S=( nd n n d R , ,...,
2
1 ,«), given by relation (11), generates the following multidimensional polinom
)
(
z
P
=∑
∈S d i d i i
x
x
x
a
i i...
2 1 21 =
∑∑ ∑
= = = 1 1 2 2 2 1 2 1 0 0 0
2 1 ,..., ,
...
...
n i n i n i i d i i i i i d d d dx
x
x
a
,and in this situation
=
∂
∂
∂
+ +)
(
...
2 1 ... 1 1z
P
x
x
d d α α α α∑ ∑ ∑
= = = − − −−
−
−
=
1 1 1 2 2 2 2 2 1 1 21
(
)!
...
!
...
)!
(
!
)!
(
!
...
1 22 2 2 1 1 1 ,..., , n i n i n i i d i i d d d i i i d d d d d
d
i
x
x
x
i
i
i
i
i
a
α α α α α αα
α
α
. Example 2:If S=Tn2, then
∑
∈
=
=
S i iy
x
a
y
x
P
P
i iz
)
(
,
)
1 2(
, and= ∂ ∂∂ + ) ( 2 1 2 1 z P y xα α
α α
∑
∈ − − − − S i i y x i i i i a i i 2 2 1 1 )! ( ! )! ( ! 2 2 2 1 11 α α
α
α
, for any α =(α
1,α
2)∈S withα
≤
i.If S=(Tn2,p) and taking into consideration the relation (2), then
P
(
x
,
y
)
j k j n k k j y x a − = =
∑∑
= 0 0 j -kj, , 1 2 ( ) 2
1
z P y xα α
=
∑ ∑
+ =
−
=
− − −
−
−
−
−
−
n
k k
j
j k j j
k
j
x
y
j
k
j
k
j
j
a
2 1
2
1
2 1
)!
(
)!
(
)!
(
!
2 1
, α
α α
α
α α
α
α
If S = (
T
n2,«) and taking into account the relation (3), then=
)
,
(
x
y
P
j kn
j j n
k
y
x
a
∑∑
=0 −=0 j,k , and= ∂
∂∂
+
) , ( 2 1
2 1
y x P y xα α
α α
2 1
1 2
(
)!
!
)!
(
!
2 1
α α
α α
α
α
− −
= −
=
∑ ∑
−
−
j kn
j j n
k
y
x
k
k
j
j
a
j,k .REFERENCES
[1] Lorentz, G. G., Solvability of multivariate interpolation, J. reine angew. Math., 398 (1989), 101-104.
[2] Rudolf A. Lorentz, Multivariate Birkhoff Interpolation, Springer – Verlag Berlin Heidelberg 1992.
[3] Bloom, T. and Calvi, J.P., A continuity property of multivariate Lagrange interpolation, Mathematics of computation, Volume 66, Number 220, 1997.
[4] Wang, R. H., Piecewise algebraic curves and its branches, The Seventh Conf. Of Chinese Mathematical Society, Beijing, May 18-21, 1995.