Method of Undetermined Coefficients

5.3.1 Method of Undetermined Coefficients

We have previously seen how to derive quadrature rules using Lagrange interpolation or operator series. We now outline another general technique, the method of undetermined coefficients, for determining quadrature formulas of maximum order with both free and prescribed nodes.

Let L be a linear functional and consider approximation formulas of the form

Lf ≈ ˜Lf =

a i f (x i ) +

b j f (z j ), (5.3.1)

i =1

j =1

where the x i are p given nodes, while the z j are q free nodes. The latter are to be determined together with the weight factors a i ,b j . The altogether p + 2q parameters in the formula

566 Chapter 5. Numerical Integration are to be determined, if possible, so that the formula becomes exact for all polynomials of

degree less than N = p + 2q. We introduce the two node polynomials

s(x) = (x − z 1 ) · · · (x − z q ) (5.3.2) of degree p and q, respectively.

r(x) = (x − x 1 ) · · · (x − x p ),

be a basis of the space of polynomials of degree less than N. We assume that the quantities Lφ k , k = 1 : p + 2q are known. Then we obtain the nonlinear system

Let φ 1 ,φ 2 ,...,φ N

φ k (x i )a i + φ k (z j )b j = Lφ k , k = 1, 2, . . . , p + 2q, (5.3.3)

i =1

j =1

for the p + 2q parameters. This system is nonlinear in z j , but of a very special type. Note that the free nodes z j appear in a symmetric fashion; the system (5.3.3) is invariant

with respect to permutations of the free nodes together with their weights. We therefore first ask for their elementary symmetric functions, i.e., for the coefficients g j of the node polynomial

s(x) =φ q +1 (x) −

that has the free nodes z 1 ,...,z q as zeros. We change the basis to the set φ 1 (x), . . . , φ q (x), s(x)φ 1 (x), . . . , s(x)φ p +q (x).

In the system (5.3.3), the equations for k = 1 : q will not be changed, but the equations for k = 1 + q : p + 2q become

φ k ′ (x i )s(x i )a i +

φ k ′ (z j )s(z j )b j = L(sφ k ′ ),

1≤k ′ ≤ p + q. (5.3.5)

i =1

j =1

Here the second sum disappears since s(z j ) = 0 for all j. (This is the nice feature of this treatment.) Further, by (5.3.4),

L(sφ k ′ ) = L(φ k ′ φ q +1 ) − L(φ k ′ φ j )g j , 1≤k ′ ≤ p + q. (5.3.6)

j =1

We thus obtain the following linear system for the computation of the p + q quantities, g j , and A i = s(x i )a i :

L(φ k ′ φ j )g j + φ k ′ (x i )A i = L(φ k ′ φ q +1 ), k ′ = 1 : p + q. (5.3.7)

j =1

i =1

The weights of the fixed nodes are a i =A i /s(x i ) . The free nodes z j are then determined by finding the q roots of the polynomial s(x). Methods for computing roots of a polynomial are given in Sec. 6.5. Finally, with a i and z j known, the weights b j are obtained by the solution of the first q equations of the system (5.3.3) which are linear in b j .

5.3. Quadrature Rules with Free Nodes 567 The remainder term Rf = (Lf − ˜Lf ) of the method, exact for all polynomials of

degree less than N = p + 2q, is of the form Rf

f (N ) (ξ ), c = R(f − P N N ≈c N N = R(x )/N !, where c N is called the error constant. Note that R(x N ) = R(φ N +1 ) , where φ (N +1) is any

monic polynomial of degree N, since x N −φ (N +1) is a polynomial of degree less than N. Hence, for the determination of the error constant we compute the difference between the right-hand and the left-hand sides of

φ k (x i )a i + φ k (z j )b j + N!c N = Lφ N +1 , N = p + 2q, (5.3.8)

i =1 j =1

and divide by (N)!. If, for example, a certain kind of symmetry is present, then it can happen that c p +2q = 0. The formula is then more accurate than expected, and we take

N = p + 2q + 1 instead. The case that also c p +2q+1 = 0 may usually be ignored. It can occur if several of the given nodes are located, where free nodes would have been placed.

From a pure mathematical point of view all bases are equivalent, but equation (5.3.3) may be better conditioned with some bases than with others, and this turns out to be an important issue when p + 2q is large. We mention three different situations.

(i) The most straightforward choice is to set [a, b] = [0, 1] and use the monomial basis φ k (x)

=x k −1 , x ∈ (0, b) (b may be infinite). For this choice the condition number of (5.3.3) increases exponentially with p + 2q. Then the free nodes and corresponding

weights may become rather inaccurate when p + 2q is large. It is usually found, however, that unless the condition number is so big that the solution breaks down

completely, the computed solution will satisfy equation (5.3.3) with a small residual. This is what really matters for the application of formula (5.3.1).

(ii) Take [a, b] = [−1, 1], and assume that the weight function w(x) and the given nodes x i are symmetrical with respect to the origin. Then the weights a i and b i , and the free

nodes z j will also be symmetrically located, and with the monomial basis it holds that L(φ k (x)) = 0, when k is even. If p = 2p ′ is even, the number of parameters will be

reduced to p ′ + q by the transformation x = ξ,ξ ∈ [0, b 2 √

]. Note that w(x) will be replaced by w( ξ )/ ξ . If p is odd, one node is at the origin, and one can proceed in an analogous way. This should also reduce the condition number approximately to its square root, and it is possible to derive in a numerically stable way formulas with about twice as high an order of accuracy as in the unsymmetric case.

(iii) Taking φ k to be the orthogonal polynomials for the given weight function will give

a much better conditioned system for determining the weights. This case will be considered in detail in Sec. 5.3.5.

Example 5.3.1.

0 f (x) dx . Set p = 0, q = 3 and choose the monomial basis φ i (x)

Consider the linear functional L(f ) = 1

=x i −1 . Introducing the node polynomial s(x)

3 = (x − z 2

1 )(x −z 2 )(x −z 3 ) =x −s 3 x −s 2 x −s 1 ,

568 Chapter 5. Numerical Integration the linear system (5.3.6) becomes

s 3 1/6 The exact solution is s 1 = 1/20, s 2 = −3/5, and s 3 = 3/2. The free nodes thus are the

zeros of s(x) = x √ 3 − 3x 2 / 2 + 3x/5 − 1/20, which are z 2 = 1/2 and z 1,3 = 1/2 ± 3/20.

The weights b 1 ,b 2 ,b 3 are then found by solving (5.3.3) for k = 1 : 3.

The matrix of the above system is a Hankel matrix. The reader should verify that when p> 0 the matrix becomes a kind of combination of a Hankel matrix and a Vandermonde matrix.