A decomposition method to construct cubature formulae of degree 3

0.14 MB·English

by  Zhaoliang Meng

A decomposition method to construct cubature formulae of
degree 3∗
Zhaoliang Meng†
School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, China
Zhongxuan Luo
School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024, China
3 School of Software, Dalian University of Technology, Dalian, 116620, China
1
0
2

n
a Abstract
J
5 Numerical integration formulas in n-dimensional Euclidean space of degree three are
2 discussed. For the integrals with permutation symmetry we present a method to construct
itsthird-degreeintegrationformulaswith2nrealpoints. Wepresentadecompositionmethod

]
A and only need to deal with n one-dimensional moment problems independently.
N
Keywords: Numerical integration; Degree three; Cubature formulae; Decomposi-
.
h tion method; One-dimensional moment problem
t
a
m
1 Introduction
[

1 Let Πn = R[x ,...,x ] be the space of polynomials in n real variables and L be a square
v 1 n
8 positive linear functional defined on Πn such as those given by L(f)= Rnf(x)W(x)dx, where
8 W is a nonnegative weight function with finite moments of all order. RLet Πn be the space of
d
9
polynomials of degree at most d. Here we discuss numerical integration formulas of the form
5
.
1 L(f)≈ a f(u(k)), (1)
0 k
3 Xk
1
: where a are constants and u(k) are points in the spaces. The formulas are called degree of d if
v k
i they are exact for integrations of any polynomials of x of degree at most d but not d+1.
X
In this paper, we only deal with the construction of third-degree cubature formula. It
r
a looks like a simple problem, however it remains to be solved. The well known result is due to
Stroud [1–3]. He presented a method to construct numerical integration formulas of degree 3 for
centrally symmetrical region and recently Xiu [5] also considered the similar numerical formulas
for integrals as
b b
L(f) = ... w(x )...,w(x )f(x ,x ...,x )dx ...dx .
1 n 1 2 n 1 n
Z Z
a a
In [5], Xiu assumed that every single integral is symmetrical, which means his result naturally
belongs to centrally symmetrical case. Recently, the authors [12] extended Stroud’s results and
∗TheprojectissupportedbyNNSFofChina(Nos. 61033012,11171052,11271060,61272371) andalsosupported
by“the FundamentalResearch Fundsfor theCentral Universities”.
†Corresponding author. E-mail: m zh [email protected]
1
presented formulas of degree 3 of 2n points or 2n+1 points for integrals as
b1 bn
L(f)= ... w (x )...w (x )f(x ,...,x ) dx ...dx ,
1 1 n n 1 n 1 n
Z Z
a1 an
w (x )≥ 0, x ∈[a ,b ], i = 1,...,n.
i i i i i
Besides, manyscholars employed theinvarianttheory methodtodealwithsymmetricalcase and
we can refer to [7–9] and the references therein. As far as we know, 2n is the minimum of the
integration points except some special regions(see [11]), and for centrally symmetrical region
Mysovskikh [6] had shown this point. However, for the general integration case, it remains
unknown how to construct the formulas of degree 3 with 2n points. In the two-dimensional
case, third degree integration formulas with 4 real points was given in [13, 14] for any regions.
But it is difficult to extend it to higher dimension. For other related work, we can refer to
[15–17] and the reference therein.
This paper will extend the results in [12] for the integrals of product regions to those with
permutation symmetry. Firstwe presentacondition which is satisfied by theintegral. Andthen
we prove that under this condition, the construction problem of cubature formulae with degree
three can be transformed into two smaller sub-cubature problems. Finally, for the construction
of cubature rules of the integrals with permutation symmetry can be decomposed into n one-
dimensional moment problems.
This paper is organized as follows. The construction of cubature formulas of degree 3 are
presented in section 2. And section 3 will present two examples to illustrate the construction
process. Finally, section 4 will make a conclusion.
2 The construction of third-degree formulas
Assume that L has the following property:
There exist n linearly independent polynomial l (x ,...,x ),i = 1,2,...,n
i 1 n
(P) such that all l l (i = 1,...,n−1) are the orthogonal polynomials of degree
i n
two with respect to L.
Let
T : l (x ,...,x ) −→ t , i = 1,2,...,n (2)
i 1 n i
be a linear transformation and L be transformed into L′. Then by the assumption all t t (i =
i n
1,2...,n−1) are the orthogonal polynomial of degree two with respect to L′. Here we do not
requirethat all t t (i = 1,2...,n−1) can constitute a basis of orthogonal polynomials of degree
i n
2 with respect to L′. We can also assume that the third-degree formula of L′ has the following
form
v(1) = (v ,v ,v ...,v ,0) ω
1,1 1,2 1,3 1,n−1 1
v(2) = (v ,v ,v ...,v ,0) ω
2,1 2,2 2,3 2,n−1 2
······························ (3)
v(N) = (v ,v ,v ...,v ,0) ω
N,1 N,2 N,3 N,n−1 N
v(N+j) =(0,0,0...,0,v ) ω , j = 1,2.
N+j,n N+j
To enforce polynomial exactness of degree 3, it suffices to require (3) to be exact for
1, t , t ,...,t , t t , t t t i,j,k = 1,2,...,n.
1 2 n i j i j k
2
Then we have
ω +ω +...+ω = L′(1) (4a)
1 2 N+2
ω v +ω v +...+ω v = L′(t ), i= 1,2,...,n−1
1 1,i 2 2,i N N,i i (4b)
(cid:26) ωN+1vN+1,n+ωN+2vN+2,n = L′(tn)
ω v v +ω v v +...+ω v v = L′(t t ), i,j = 1,2,...,n−1
1 1,i 1,j 2 2,i 2,j N N,i N,j i j
(4c)
(cid:26) ωN+1vN2+1,n+ωN+2vN2+2,n = L′(t2n)
ω v v v +...+ω v v v = L′(t t t ), i,j,k = 1,2,...,n−1
1 1,i 1,j 1,k N N,i N,j N,k i j k
(4d)
(cid:26) ωN+1vN3+1,n+ωN+2vN3+2,n = L′(t3n)
and the equation (4a) can be rewritten as
ω +ω +...+ω = ξ ,
1 2 N 1
ω +ω = ξ
N+1 N+2 2
ξ +ξ = L′(1).
1 2
Hence we can rewrite the equations (4) as
ω +ω +...+ω = ξ ,
1 2 N 1
 ω1v1,i+ω2v2,i+...+ωNvN,i = L′(ti), i= 1,2,...,n−1 (5a)
 ω v v +ω v v +...+ω v v = L′(t t ), i,j = 1,2,...,n−1
 1 1,i 1,j 2 2,i 2,j N N,i N,j i j
ω v v v +...+ω v v v = L′(t t t ), i,j,k = 1,2,...,n−1
1 1,i 1,j 1,k N N,i N,j N,k i j k


 ω +ω = ξ
N+1 N+2 2
 ωN+1vN+1,n+ωN+2vN+2,n = L′(tn)
(5b)
 ωN+1vN2+1,n+ωN+2vN2+2,n = L′(t2n)
ω v3 +ω v3 = L′(t3)
N+1 N+1,n N+2 N+2,n n

 ξ +ξ = L′(1). (5c)
1 2
Once ξ is determined by (5c), then (5a) and (5b) become one n − 1 dimensional and the
i
other one-dimensional moment problems respectively. If these two lower dimensional moment
problems can be solved, then we can get a cubature formula of degree 3 with respect to the
original integration problem. Generally speaking, the one-dimensional moment problem can be
easily solved, butitis difficultto besolved for then−1dimensionalmomentproblem. However,
if the n−1 dimensional problem can be divided into one n−2 dimensional moment problem
and the other one-dimensional moment problem and further the n − 2 dimensional moment
problem can continue this process, then the original integration problem can be turned into n
one-dimensional moment problems.
In what follows, we will prove that if L is a integral functional with permutation symmetry,
then the construction problem of third-degree cubature formulae can be turned into n one-
dimensional moment problems. In fact, we usually encounter this kind of integral functional,
for example, the integration over the simplex, the square, the ball or the positive sector of the
ball, that is {(x ,x ,...,x )|x ≥ 0,i = 1,2,...,n;x2+x2+...,+x2 ≤ 1}.
1 2 n i 1 2 n
We first prove that the integral functional with permutation symmetry must meet the prop-
erty (P). Thus the original cubature problem can be turned into two sub-cubature problems.
Theorem 1. Let n≥ 2 and
l (x ,...,x ) = x −x , i = 1,2,...,n−1
i 1 n i n
l (x ,...,x ) = x +x +...+x +c
n 1 n 1 2 n n
where
L(x3+(n−3)x2x −(n−2)x x x )
c = − 1 1 2 1 2 3 .
n L(x2−x x )
1 1 2
3
If L is permutation symmetrical, then l l (i = 1,2,...,n−1) is orthogonal to the polynomials
i n
of degree ≤ 1.
Proof. Take l l as an example. We first prove L(x2−x x ) 6= 0 to confirm the existence of
1 n 1 1 2
l . In fact, by the symmetry and the positivity,
n
1 1
L(x2−x x ) = L(x2−x x −x x +x2) = L((x −x )2) > 0.
1 1 2 2 1 1 2 1 2 2 2 1 2
Let us exam the orthogonality of l l . Assume n ≥ 3. By the symmetry, we have
1 n
L(l l ) = L x2−x2 +(x −x )(x +...+x +c ) = 0,
1 n 1 n 1 n 2 n−1 n
L(x l l ) = L(cid:0) x2x −x2x +(x −x )(x x +...+x (cid:1) x +c x ) = 0, 2 ≤ i≤ n−1
i 1 n 1 i n i 1 n 2 i n−1 i n i
(cid:0) (cid:1)
and for i = 1(the same to i = n)
L(x l l )= L x3−x2x +(x −x )(x x +...+x x +c x )
1 1 n 1 n 1 1 n 2 1 n−1 1 n 1
= L x3+(n−3)(cid:0)x2x −(n−2)x x x +c (x2−x x ) = 0. (cid:1)
1 1 2 1 2 3 n 1 1 2
(cid:0) (cid:1)
It is easy to verify that the result holds when n = 2. This completes the proof.
Let L and L1 be two linear functionals defined on Πn−1 and Π1, whose moments are
1
determined by
L (t t t ) = L(l l l ), L (t t ) = L(l l ), L (t ) = L(l ), L (1) = ξ(0), (6)
1 i j k i j k 1 i j i j 1 i i 1 1
L1(t3)= L(l3), L1(t2) = L(l2), L1(t ) = L(l ), L1(1) = ξ(0) (7)
n n n n n n 2
ξ(0)+ξ(0) =L(1) and 1 ≤i,j,k ≤ n−1
1 2
respectively. Thustheconstructionproblemofthird-degreeformulaswithrespecttoL isturned
into two smaller problems, one of which is the construction with respect to L and the other of
1
which is the construction with respect to L1. It is easy to compute
L1(t ) = L(l ) = n·L(x )+c L(1),
n n 1 n
L1(t 2)= nL(x2)+n(n−1)L(x x )+2nc ·L(x )+c2L(1)
n 1 1 2 n 1 n
n n
L1(t 3)= nL(x3)+6 L(x2x )+6 L(x x x )
n 1 (cid:18)2(cid:19) 1 2 (cid:18)3(cid:19) 1 2 3
+3c nL(x2)+n(n−1)L(x x ) +3nc2L(x )+c3L(1)
n 1 1 2 n 1 n
= nL((cid:0)x3)+3n(n−1)L(x2x )+n((cid:1)n−1)(n−2)L(x x x )
1 1 2 1 2 3
+3c nL(x2)+n(n−1)L(x x ) +3nc2L(x )+c3L(1).
n 1 1 2 n 1 n
(cid:0) (cid:1)
Next we will show this decomposition process can continue.
Let us consider the cubature formula with respect to L . Obviously, L is also permutation
1 1
symmetrical, which allows us to employ theorem 1 continuously. Define
(1)
l (t ,...,t ) = t −t , i = 1,2,...,n−2,
i 1 n−1 i n−1
(1)
l (t ,...,t ) = t +t +...+t +c ,
n−1 1 n 1 2 n−1 n−1
where
L (t3+(n−4)t2t −(n−3)t t t )
c = − 1 1 1 2 1 2 3 ,
n−1 L (t2−t t )
1 1 1 2
4
(1) (1)
then l l ,n = 1,2,...,n−2 are the orthogonal polynomials of degree two with respect to
i n−1
L . Noticing t = l (x ,x ,...,x ), we have
1 i i 1 2 n
(1)
l (t ,...,t ) = t −t = x −x , i= 1,2,...,n−2,
i 1 n−1 i n−1 i n−1
(1)
l (t ,...,t ) = t +t +...+t +c
n−1 1 n 1 2 n−1 n−1
= x +x +...+x −(n−1)x +c .
1 2 n−1 n n−1
Again let L and L2 be two linear functionals defined on Πn−2 and Π1, whose moments are
2
determined by
L (τ τ τ ) = L (l(1)l(1)l(1)), L (τ τ ) = L (l(1)l(1)), L (τ ) = L (l(1)), L (1) = ξ(1),
2 i j k 1 i j k 2 i j 1 i j 2 i 1 i 2 1
L2(τ3 )= L ((l(1) )3), L2(τ2 )= L ((l(1) )2), L2(τ ) = L (l(1)), L2(1) = ξ(1)
n−1 1 n−1 n−1 1 n−1 n−1 1 n 2
(1) (1) (0)
ξ +ξ = ξ and 1 ≤ i,j,k ≤ n−2
1 2 1
respectively. It is easy to compute
L2(τ )= L (l(1) ) =L(l(1) )+c L (1)−L(1) ,
n−1 1 n−1 n−1 n−1 1
L2(τ 2) = L ((l(1) )2) = L [(t +t (cid:0)+...+t )+(cid:1)c ]2
n−1 1 n−1 1 1 2 n−1 n−1
= L((l(1) )2)+c2 (cid:0) L (1)−L(1) , (cid:1)
n−1 n−1 1
L2(τ 3) = L((l(1) )3)+c3 (cid:0)L (1)−L(1)(cid:1).
n−1 n−1 n−1 1
(cid:0) (cid:1)
Assume that L is a linear functional defined on Πn−k for every k(0 ≤ k < n) and satisfies
k
property (P). Then a cubature problem of degree 3 with respect to L can be divided into
k
two smaller cubature problems—one with respect to L and the other with respect to Lk+1.
k+1
Moreover L also satisfies the property (P) and then this process can continue and will end
k+1
when k = n−1. Finally, an n-dimensional cubature problem can be transformed into n one-
dimensional cubature problems.
Theorem 2. Let
(k)
l = x +x +...+x −(n−k)x +c
n−k 1 2 n−k n−k+1 n−k
and Lk be a linear functional defined on Π1 according to the above process, then the correspond-
ing moments are
Lk+1(1) = ξ(k),
2
Lk+1(τ) = L(l(k) )+c L (1)−L(1)
n−k n−k k
= c L (1), (cid:0) (cid:1)
n−k k
Lk+1(τ2) = L((l(k) )2)+c2 L (1)−L(1)
n−k n−k k
= (n−k)(n−k+1)L(cid:0) (x2−x x )+(cid:1)c2 L (1),
1 1 2 n−k k
Lk+1(τ3) = L((l(k) )3)+c3 L (1)−L(1)
n−k n−k k
= (n−k)(n−k+1)((cid:0)n−k+2)L(−(cid:1)x3+3x2x −2x x x )+c3 L (1)
1 1 2 1 2 3 n−k k
where
L(x3−3x2x +2x x x )
c = − 1 1 2 1 2 3 , k = 1,2,...,n−2. (8)
n−k L(x2−x x )
1 1 2
5
Proof. It remains to prove Eq.(8). It follows from theorem 1 that
L (t3+(n−3−k)t2t −(n−2−k)t t t )
c = − k 1 1 2 1 2 3 .
n−k L (t2−t t )
k 1 1 2
According to the definition of L , we have
k
L (t3+(n−3−k)t2t −(n−2−k)t t t )
k 1 1 2 1 2 3
= L(t3+(n−3−k)t2t −(n−2−k)t t t )
1 1 2 1 2 3
= L (x −x )3+(n−3−k)(x −x )2(x −x )
1 n−k 1 n−k 2 n−k
(cid:16)
−(n−2−k)(x −x )(x −x )(x −x )
1 n−k 2 n−k 3 n−k
(cid:17)
= L x3−3x2x +2x x x ,
1 1 2 1 2 3
(cid:16) (cid:17)
L (t2−t t ) = L((x −x )2−(x −x )(x −x ))
k 1 1 2 1 n−k 1 n−k 2 n−k
= L(x2−x x ),
1 1 2
where the permutation symmetry is used. This completes the proof.
Remark1. InfactL isalsoaone-dimensional integration functional. The corresponding
n−1
moment can be calculated by
L (1) = ξ(n−2),
n−1 1
L (τ) = L(x −x ) = 0,
n−1 1 2
(9)
L (τ2) = L((x −x )2) =2L(x2−x x ),
n−1 1 2 1 1 2
L (τ3) = L((x −x )3) =0.
n−1 1 2
For convenience, in what follows let Ln = L .
n−1
Suppose that
nk
Lk(g) ≈ w g(t ),k = 1,2,...,n (10)
i,k i,k
Xi=1
is exact for any g ∈Π1. And let vi,k = (v(1),v(2),...,v(n)) be the solution of
3 i,k i,k i,k
x +x +...+x +c = t
1 2 n n i,1
 xn−1−xn = 0
 ························ for k = 1, (11)



x −x = 0
2 n
 x1−xn = 0


 x +x +...+x +c = 0
1 2 n n
 x1+x2+...+xn−1−(n−1)xn +cn−1 = 0
 ························ for k = n, (12)



x +x −2x +c = 0
1 2 3 2
 x1−x2 = ti,n



6
and
x +x +...+x +c = 0
1 2 n n
 x1+x2+...+xn−1−(n−1)xn +cn−1 = 0
 ························


 x1+x2+...+xn−k+1−(n−k+1)xn−k+2+cn−k+1 = ti,k for k = 2,3,...,n−1,


x −x =0
n−k n−k+1
 ························


 x1−xn−k+1 = 0

 (13)
Hence the final cubature formula can be written as
n nk
L(f)≈ w f(vi,k)
i,k
Xk=1Xi=1
which is exact for any f ∈ Πn. It is clear that the solution of Eq.(11) is
3
t −c
i,1 n
(η ,η ,...,η ), η =
i i i i
n
and the solution of Eq.(13) is
c −c −δ t
n−1 n 2,k i,2
x = ,
n
 n
c −c
 xn−1 = xn+ n−n2 −1n−1 = xn

 .........


 xn−k+3 = xn−k+4
c −c −t t
n−k+2 n−k+3 i,k i,k
x = x + = x −
 n−k+2 n−k+3 n−k+2 n−k+3 n−k+2

 t −c
 x1 = x2 = ... = xn−k+1 = xn−k+2+ i,nk−kn+−k1+1




where δ = 1 if k = 2 and δ = 0 if k 6= 2, and the solution of Eq.(12) is
2,k 2,k
c −c
n−1 n
x = x = ... = x =
n n−1 3
 n
c +t
 n−1 i,n
 x = x −
 2 3 2
t −c
i,n n−1
x = x +t = x +
1 2 i,n 3
 2




Collecting the above discussion, we have
Theorem 3. Assume that L is permutation symmetrical. Then there must exist a cubature
formula
n mk m1
L(f) ≈ w f(α ,...,α ,β ,γ,γ,...,γ)+ w f(α ,...,α ) := C(f)
i,k i,k i,k i,k i,1 i,1 i,1
Xk=2Xi=1 Xi=1
n−k+1 k−2
| {z } | {z }
which is exact for every polynomial of degree ≤ 3. In the formula, α’s and β’s can be computed
7
by
c −c
n−1 n
γ =
 n
t
 i,k
 β = γ− , 2 ≤ k ≤ n−1;
 i,k
 n−k+2
 α = β + ti,k −cn−k, 2 ≤ k ≤ n−1;
 i,k i,k
 n−k+1
 ti,1−cn
α = ,
i,1
n
 βi,n = γ− ti,n2+c2

 αi,n = βi,n+ti,n




where t ’s and w ’s are the nodes and weights of the quadrature formula (10) with respect to
i,k i,k
Lk.
The proof is a direct result of the computation and is omitted.
Remark 2. For the quadrature problem of the one-dimensional moment, it is well known
that the number of the nodes n = 2 in the general case. Hence the total number of the nodes of
k
the cubature formula with respect to L is 2n generally and 2n is usually the minimum among
all the cubature formula of degree 3 except one case of the integration over the n-dimensional
simplex [11]. For more knowledge of the problem of the one-dimensional moment, we can refer
to the appendix of [4].
Remark 3. For convenience, we present the relations of ξ’s as follows
L(1) −→ ξ(0) −→ ξ(1) −→ ... −→ ξ(n−2)
1 1 1
+ + +
(0) (1) (n−2)
ξ ξ ξ
2 2 2
According to the previous discussion, it is clear that
Lk(1) = ξ(k−1) for 1 ≤ k ≤ n−1 and Ln(1) = L (1) = ξ(n−2).
2 n−1 1
3 Numerical Examples
• Firstly take the integration over the n-dimensional simplex as an example. Define
L(f)= f(x ,x ,...,x )dx dx ...dx
1 2 n 1 2 n
Z
Tn
where
T = {(x ,x ,...,x )|x +x +...+x ≤ 1, x ≥ 0, i= 1,2,...,n}.
n 1 2 n 1 2 n i
It is well known that
α !α !...α !
L(xα1xα2...xαn) = 1 2 n .
1 2 n (n+α +α +...+α )!
1 2 n
Then by a simple computation, we have
2 n+2
c = − ,i = 2,3,...,n−1, c = − ,
i n
n+3 n+3
1
γ = .
n+3
8
Here if we take
1 1
(i) (n−2)
ξ = t · , ξ = t · , i= 0,1,...,n−2,
2 i+1 n·n! 1 n n·n!
and n t = n, then the moments of Lk+1(1 ≤ k ≤ n−2) are
i=1 i
P
1
Lk+1(1) = t ,
k+1
n·n!
k
2(n− t )
Lk+1(τ) = c L (1) = − i=1 k ,
n−k k
(n+P3)n·n!
k
(n−k)(n−k+1) 4(n− t )
Lk+1(τ2) = L((l(k) )2)+c2 L (1)−L(1) = + i=1 k ,
n−k n−k k (n+2)! n(n+P3)2 ·n!
(cid:0) (cid:1)
Lk+1(τ3) = (n−k)(n−k+1)(n−k+2)L(−x3+3x2x −2x x x )+c3 L (1)
1 1 2 1 2 3 n−k k
k
−2(n−k)(n−k+1)(n−k+2) 8(n− t )
= − i=1 k
(n+3)! n(n+P3)3·n!
and
t
L1(1) = 1 ,
n·n!
−2
L1(τ) = ,
(n+1)!(n+3)
n2+5n+8
L1(τ2) = ,
(n+3)!(n+3)
2(n+2)(n+4)
L1(τ3) = −
(n+3)2(n+3)!
and
t
Ln(1) = n ,
n·n!
Ln(τ) = 0,
2
Ln(τ2) = ,
(n+2)!
Ln(τ3) = 0.
By taking different values for ξs, we can get different cubature formulae. For n = 3 and n = 4,
if we take all t = 1, then we can get formulas as showed in Tables 1 and 2.
i
x x x weight
1 2 3
0.34240723692377 0.34240723692377 0.34240723692377 0.01469064053612
0.14125289379518 0.14125289379518 0.14125289379518 0.04086491501944
0.41353088165296 0.41353088165296 0.00627157002742 0.01887111233337
0.12380973765487 0.12380973765487 0.58571385802358 0.03668444322218
0.60719461208592 0.05947205458075 0.16666666666667 0.02777777777778
0.05947205458075 0.60719461208592 0.16666666666667 0.02777777777778
Table 1: Nodes and weights for T
3
In Table 1, the first point is outside of the region. To avoid it, we can take
93 378 108
t = , t = , t = ,
1 2 3
85 391 115
9
x x x x
1 2 3 4
0.27145760185760 0.27145760185760 0.27145760185760 0.27145760185760
0.12024746726682 0.12024746726682 0.12024746726682 0.12024746726682
0.30652570925957 0.30652570925957 0.30652570925957 -0.06243427063585
0.11154151763119 0.11154151763119 0.11154151763119 0.52251830424930
0.37131176827505 0.37131176827505 -0.02833782226438 0.14285714285714
0.09266869570542 0.09266869570542 0.52894832287488 0.14285714285714
0.54391317546145 0.02751539596712 0.14285714285714 0.14285714285714
0.02751539596712 0.54391317546145 0.14285714285714 0.14285714285714
w = 0.00254167472911, w = 0.00787499193755, w = 0.00294495824332,
1 2 3
w = 0.00747170842335, w = 0.00365639117145, w = 0.00676027549522,
4 5 6
w = 0.00520833333333, w = 0.00520833333333.
7 8
Table 2: Nodes and weights for T
4
x x x weight
1 2 3
0.33333333333333 0.33333333333333 0.33333333333333 0.01875000000000
0.14285714285714 0.14285714285714 0.14285714285714 0.04203431372549
0.41666666666667 0.41666666666667 0.00000000000000 0.01875000000000
0.12037037037037 0.12037037037037 0.59259259259259 0.03495843989770
0.61593041596355 0.05073625070311 0.16666666666667 0.02608695652174
0.05073625070311 0.61593041596355 0.16666666666667 0.02608695652174
Table 3: Nodes and weights for T
3
and the corresponding cubature formula is listed in Table 3. In the formula, the first and third
nodes are on the boundary of the region T .
3
If we take
94 76
t = , t = 1, t = ,
1 2 3
85 85
then all the nodes are inside the region, see Table 4.
x x x weight
1 2 3
0.33237874197689 0.33237874197689 0.33237874197689 0.01927056497746
0.14303319621635 0.14303319621635 0.14303319621635 0.04216734351927
0.41247250927755 0.41247250927755 0.00838831477823 0.02043165185637
0.12085734331926 0.12085734331926 0.59161864669481 0.03512390369919
0.61593041596355 0.04371016618787 0.16666666666667 0.02516339869281
0.04371016618787 0.61593041596355 0.16666666666667 0.02516339869281
Table 4: Nodes and weights for T
3
In Table 2, all the weights are positive. However, we find there exist three points outside of
the region. If we take
104 3577 9947 49
t = , t = , t = , t = .
1 2 3 4
75 2775 8880 60
and add one more point with weight −49/80, then the correspondingformula is showed in Table
5.
In Table 5, there are 5 points on the boundary. To make all the nodes inside the region, we
can take
7 187 179522 5
t = , t = , t = , t =
1 2 3 4
5 145 160283 6
and add one more node with weight −618391/961698 and the corresponding formula is shown
in Table 6.
10