Gauss quadrature

Gerard MEURANT

October, 2008

1 Quadrature rules

2 The Gauss rule

3 The Gauss–Radau rule

4 The Gauss–Lobatto rule

5 Computation of the Gauss rules

6 The anti–Gauss rule

7 Nonsymmetric Gauss quadrature rules

8 The block Gauss quadrature rules

Quadrature rules

Given a measure α on the interval [a, b] and a function f , aquadrature rule is a relation∫ b

af (λ) dα =

N∑j=1

wj f (tj) + R[f ]

R[f ] is the remainder which is usually not known exactlyThe real numbers tj are the nodes and wj the weights

The rule is said to be of exact degree d if R[p] = 0 for allpolynomials p of degree d and there are some polynomials q ofdegree d + 1 for which R[q] 6= 0

I Quadrature rules of degree N − 1 can be obtained byinterpolation

I Such quadrature rules are called interpolatory

I Newton–Cotes formulas are defined by taking the nodes to beequally spaced

I A popular choice for the nodes is the zeros of the Chebyshevpolynomial of degree N. This is called the Fejer quadraturerule

I Another interesting choice is the set of extrema of theChebyshev polynomial of degree N − 1. This gives theClenshaw–Curtis quadrature rule

TheoremLet k be an integer, 0 ≤ k ≤ N. The quadrature rule has degreed = N − 1 + k if and only if it is interpolatory and∫ b

a

N∏j=1

(λ− tj)p(x) dα = 0, ∀p polynomial of degree ≤ k − 1.

see Gautschi

If the measure is positive, k = N is maximal for interpolatoryquadrature since if k = N + 1 the condition in the last theoremwould give that the polynomial

N∏j=1

(λ− tj)

is orthogonal to itself which is impossible

Gauss quadrature rules

The optimal quadrature rule of degree 2N − 1 is called a GaussquadratureIt was introduced by C.F. Gauss at the beginning of the nineteenthcentury

The general formula for a Riemann–Stieltjes integral is

I [f ] =

∫ b

af (λ) dα(λ) =

N∑j=1

wj f (tj) +M∑

k=1

vk f (zk) + R[f ], (1)

where the weights [wj ]Nj=1, [vk ]Mk=1 and the nodes [tj ]

Nj=1 are

unknowns and the nodes [zk ]Mk=1 are prescribed

see Davis and Rabinowitz; Gautschi; Golub and Welsch

I If M = 0, this is the Gauss rule with no prescribed nodes

I If M = 1 and z1 = a or z1 = b we have the Gauss–Radau rule

I If M = 2 and z1 = a, z2 = b, this is the Gauss–Lobatto rule

The term R[f ] is the remainder which generally cannot beexplicitly computedIf the measure α is a positive non–decreasing function

R[f ] =f (2N+M)(η)

(2N + M)!

∫ b

a

M∏k=1

(λ−zk)

N∏j=1

(λ− tj)

2

dα(λ), a < η < b

(2)Note that for the Gauss rule, the remainder R[f ] has the sign of

f (2N)(η)see Stoer and Bulirsch

The Gauss rule

How do we compute the nodes tj and the weights wj?

I One way to compute the nodes and weights is to usef (λ) = λi , i = 1, . . . , 2N − 1 and to solve the non linearequations expressing the fact that the quadrature rule is exact

I Use of the orthogonal polynomials associated with themeasure α

∫ b

api (λ)pj(λ) dα(λ) = δi ,j

P(λ) = [p0(λ) p1(λ) · · · pN−1(λ)]T , eN = (0 0 · · · 0 1)T

λP(λ) = JNP(λ) + γNpN(λ)eN

JN =

ω1 γ1

γ1 ω2 γ2

. . .. . .

. . .

γN−2 ωN−1 γN−1

γN−1 ωN

JN is a Jacobi matrix, its eigenvalues are real and simple

TheoremThe eigenvalues of JN (the so–called Ritz values θ

(N)j which are

also the zeros of pN) are the nodes tj of the Gauss quadrature rule.The weights wj are the squares of the first elements of thenormalized eigenvectors of JN

Proof.The monic polynomial

∏Nj=1(λ− tj) is orthogonal to all

polynomials of degree less than or equal to N − 1. Therefore, (upto a multiplicative constant) it is the orthogonal polynomialassociated to α and the nodes of the quadrature rule are the zerosof the orthogonal polynomial, that is the eigenvalues of JN

The vector P(tj) is an unnormalized eigenvector of JN

corresponding to the eigenvalue tjIf q is an eigenvector with norm 1, we have P(tj) = ωq with ascalar ω. From the Christoffel–Darboux relation

wjP(tj)TP(tj) = 1, j = 1, . . . ,N

ThenwjP(tj)

TP(tj) = wjω2‖q‖2 = wjω

2 = 1

Hence, wj = 1/ω2. To find ω we can pick any component of theeigenvector q, for instance, the first one which is different fromzero ω = p0(tj)/q1 = 1/q1. Then, the weight is given by

wj = q21

If the integral of the measure is not 1

wj = q21µ0 = q2

1

∫ b

adα(λ)

The knowledge of the Jacobi matrix and of the first moment allowsto compute the nodes and weights of the Gauss quadrature rule

Golub and Welsch showed how the squares of the first componentsof the eigenvectors can be computed without having to computethe other components with a QR–like method

I [f ] =

∫ b

af (λ) dα(λ) =

N∑j=1

wGj f (tG

j ) + RG [f ]

with

RG [f ] =f (2N)(η)

(2N)!

∫ b

a

N∏j=1

(λ− tGj )

2

dα(λ)

The monic polynomial∏N

j=1(tGj − λ) which is the determinant χN

of JN − λI can be written as γ1 · · · γN−1pN(λ)

TheoremSuppose f is such that f (2n)(ξ) > 0, ∀n, ∀ξ, a < ξ < b, and let

LG [f ] =N∑

j=1

wGj f (tG

j )

The Gauss rule is exact for polynomials of degree less than orequal to 2N − 1 and

LG [f ] ≤ I [f ]

Moreover ∀N, ∃η ∈ [a, b] such that

I [f ]− LG [f ] = (γ1 · · · γN−1)2 f (2N)(η)

(2N)!

The Gauss–Radau rule

To obtain the Gauss–Radau rule, we have to extend the matrix JN

in such a way that it has one prescribed eigenvalue z1 = a or b

Assume z1 = a. We wish to construct pN+1 such that pN+1(a) = 0

0 = γN+1pN+1(a) = (a− ωN+1)pN(a)− γNpN−1(a)

This gives

ωN+1 = a− γNpN−1(a)

pN(a)

Note that(JN − aI )P(a) = −γNpN(a)eN

Let δ(a) = [δ1(a), · · · , δN(a)]T with

δl(a) = −γNpl−1(a)

pN(a)l = 1, . . . ,N

This gives ωN+1 = a + δN(a) and δ(a) satisfies

(JN − aI )δ(a) = γ2NeN

I we generate γN

I we solve the tridiagonal system for δ(a), this gives δN(a)

I we compute ωN+1 = a + δN(a)

JN+1 =

(JN γNeN

γN(eN)T ωN+1

)gives the nodes and the weights of the Gauss–Radau quadrature

rule

TheoremSuppose f is such that f (2n+1)(ξ) < 0, ∀n, ∀ξ, a < ξ < b. Let

UGR [f ] =N∑

j=1

waj f (ta

j ) + va1 f (a)

waj , va

1 , taj being the weights and nodes computed with z1 = a and

let LGR

LGR [f ] =N∑

j=1

wbj f (tb

j ) + vb1 f (b)

wbj , vb

1 , tbj being the weights and nodes computed with z1 = b.

The Gauss–Radau rule is exact for polynomials of degree less thanor equal to 2N and we have

LGR [f ] ≤ I [f ] ≤ UGR [f ]

Theorem (end)

Moreover ∀N ∃ ηU , ηL ∈ [a, b] such that

I [f ]− UGR [f ] =f (2N+1)(ηU)

(2N + 1)!

∫ b

a(λ− a)

N∏j=1

(λ− taj )

2

dα(λ)

I [f ]− LGR [f ] =f (2N+1)(ηL)

(2N + 1)!

∫ b

a(λ− b)

N∏j=1

(λ− tbj )

2

dα(λ)

The Gauss–Lobatto rule

We would like to have

pN+1(a) = pN+1(b) = 0

Using the recurrence relation(pN(a) pN−1(a)pN(b) pN−1(b)

)(ωN+1

γN

)=

(a pN(a)b pN(b)

)Let

δl = − pl−1(a)

γNpN(a), µl = − pl−1(b)

γNpN(b), l = 1, . . . ,N

then(JN − aI )δ = eN , (JN − bI )µ = eN

(1 −δN

1 −µN

)(ωN+1

γ2N

)=

(ab

)

I we solve the tridiagonal systems for δ and µ, this gives δN andµN

I we compute ωN+1 and γN

JN+1 =

(JN γNeN

γN(eN)T ωN+1

)

TheoremSuppose f is such that f (2n)(ξ) > 0, ∀n, ∀ξ, a < ξ < b and let

UGL[f ] =N∑

j=1

wGLj f (tGL

j ) + vGL1 f (a) + vGL

2 f (b)

tGLj , wGL

j , vGL1 and vGL

2 being the nodes and weights computedwith a and b as prescribed nodes. The Gauss–Lobatto rule is exactfor polynomials of degree less than or equal to 2N + 1 and

I [f ] ≤ UGL[f ]

Moreover ∀N ∃ η ∈ [a, b] such that

I [f ]−UGL[f ] =f (2N+2)(η)

(2N + 2)!

∫ b

a(λ−a)(λ−b)

N∏j=1

(λ− tGLj )

2

dα(λ)

Computation of the Gauss rulesThe weights wi are given by the squares of the first components ofthe eigenvectors wi = (z i

1)2 = ((e1)T z i )2

Theorem

N∑l=1

wl f (tl) = (e1)T f (JN)e1

Proof.N∑

l=1

wl f (tl) =N∑

l=1

(e1)T z l f (tl)(zl)T e1

= (e1)T

(N∑

l=1

z l f (tl)(zl)T

)e1

= (e1)TZN f (ΘN)ZTN e1

= (e1)T f (JN)e1

The anti–Gauss rule

A usual way of obtaining an estimate of I [f ]− LNG [f ] is to use

another quadrature rule Q[f ] of degree greater than 2N − 1 and toestimate the error as Q[f ]− LN

G [f ]Laurie proposed to construct a quadrature rule with N + 1 nodescalled an anti–Gauss rule

HN+1[f ] =N+1∑j=1

$j f (ϑj),

such thatI [p]− HN+1[p] = −(I [p]− LN

G [p])

for all polynomials of degree 2N + 1. Then, the error of the Gaussrule can be estimated as

1

2(HN+1[f ]− LN

G [f ])

HN+1[p] = 2I [p]− LNG [p]

for all polynomials p of degree 2N + 1. Hence, HN+1 is a Gaussrule with N + 1 nodes for the functional I(·) = 2I [·]− LN

G [·]We have

I [pq] = I(pq)

for p a polynomial of degree N − 1 and q a polynomial of degreeN and

I(p2N) = 2I (p2

N)

where pj are the orthogonal polynomials associated to IUsing the Stieltjes formulas for the coefficients we obtain theJacobi matrix

JN+1 =

ω1 γ1

γ1 ω2 γ2

. . .. . .

. . .

γN−2 ωN−1 γN−1

γN−1 ωN

√2γN√

2γN ωN+1

The anti–Gauss nodes ϑj , j = 2, . . . ,N are inside the integrationintervalHowever, the first and the last nodes can eventually be outside ofthe integration intervalActually, in some cases, the matrix JN+1 can be indefinite even ifJN is positive definite

One can construct a quadrature rule SN+1[f ] such that

I [p]− SN+1[p] = −γ(I [p]− LNG [p])

for all polynomials of degree 2N + 1. The parameter γ is positiveand less than or equal to 1

JN+1 =

ω1 γ1

γ1 ω2 γ2

. . .. . .

. . .

γN−2 ωN−1 γN−1

γN−1 ωN γN√

1 + γγN√

1 + γ ωN+1

The error of the Gauss rule can be estimated as

1

1 + γ(SN+1[f ]− LN

G [f ])

Nonsymmetric Gauss quadrature rules

We consider the case where the measure α can be written as

α(λ) =l∑

k=1

αkδk , λl ≤ λ < λl+1, l = 1, . . . ,N − 1

where αk 6= δk and αkδk ≥ 0We assume that there exists two sequences of mutually orthogonal(sometimes called bi–orthogonal) polynomials p and q such that

γjpj(λ) = (λ− ωj)pj−1(λ)− βj−1pj−2(λ), p−1(λ) ≡ 0, p0(λ) ≡ 1

βjqj(λ) = (λ− ωj)qj−1(λ)− γj−1qj−2(λ), q−1(λ) ≡ 0, q0(λ) ≡ 1

with 〈pi , qj〉 = 0, i 6= j

LetP(λ)T = [p0(λ) p1(λ) · · · pN−1(λ)]

Q(λ)T = [q0(λ) q1(λ) · · · qN−1(λ)]

and

JN =

ω1 γ1

β1 ω2 γ2

. . .. . .

. . .

βN−2 ωN−1 γN−1

βN−1 ωN

In matrix form

λP(λ) = JNP(λ) + γNpN(λ)eN

λQ(λ) = JTN Q(λ) + βNqN(λ)eN

Proposition

pj(λ) =βj · · ·β1

γj · · · γ1qj(λ)

Hence, qN is a multiple of pN and the polynomials have the sameroots which are also the common real eigenvalues of JN and JT

N

We define the quadrature rule as∫ b

af (λ) dα(λ) =

N∑j=1

f (θj)sj tj + R[f ]

where θj is an eigenvalue of JN , sj is the first component of theeigenvector uj of JN corresponding to θj and tj is the firstcomponent of the eigenvector vj of JT

N corresponding to the sameeigenvalue, normalized such that vT

j uj = 1

TheoremAssume that γjβj 6= 0, then the nonsymmetric Gauss quadraturerule is exact for polynomials of degree less than or equal to 2N − 1

The remainder is characterized as

R[f ] =f (2N)(η)

(2N)!

∫ b

apN(λ)2 dα(λ)

The extension of the Gauss–Radau and Gauss–Lobatto rules to thenonsymmetric case is almost identical to the symmetric case

The block Gauss quadrature rules

The integral∫ ba f (λ)dα(λ) is now a 2× 2 symmetric matrix. The

most general quadrature formula is of the form∫ b

af (λ)dα(λ) =

N∑j=1

Wj f (Tj)Wj + R[f ]

where Wj and Tj are symmetric 2× 2 matrices. This can bereduced to

2N∑j=1

f (tj)uj(uj)T

where tj is a scalar and uj is a vector with two components

There exist orthogonal matrix polynomials related to α such that

λpj−1(λ) = pj(λ)Γj + pj−1(λ)Ωj + pj−2(λ)ΓTj−1

p0(λ) ≡ I2, p−1(λ) ≡ 0

This can be written as

λ[p0(λ), . . . , pN−1(λ)] = [p0(λ), . . . , pN−1(λ)]JN+[0, . . . , 0, pN(λ)ΓN ]

where

JN =

Ω1 ΓT

1

Γ1 Ω2 ΓT2

. . .. . .

. . .

ΓN−2 ΩN−1 ΓTN−1

ΓN−1 ΩN

is a symmetric block tridiagonal matrix of order 2N

The nodes tj are the zeros of the determinant of the matrixorthogonal polynomials that is the eigenvalues of JN and ui is thevector consisting of the two first components of the correspondingeigenvectorHowever, the eigenvalues may have a multiplicity larger than 1Let θi , i = 1, . . . , l be the set of distinct eigenvalues and ni theirmultiplicities. The quadrature rule is then

l∑i=1

ni∑j=1

(w ji )(w

ji )

T

f (θi )

The block quadrature rule is exact for polynomials of degree lessthan or equal to 2N − 1 but the proof is rather involved

The block Gauss–Radau rule

We would like a to be a double eigenvalue of JN+1

JN+1P(a) = aP(a)− [0, . . . , 0, pN+1(a)ΓN+1]T

apN(a)− pN(a)ΩN+1 − pN−1(a)ΓTN = 0

If pN(a) is non singular

ΩN+1 = aI2 − pN(a)−1pN−1(a)ΓTN

But

(JN − aI )

−p0(a)TpN(a)−T

...−pN−1(a)

TpN(a)−T

=

0...

ΓTN

I We first solve

(JN − aI )

δ0(a)...

δN−1(a)

=

0...

ΓTN

I We compute

ΩN+1 = aI2 + δN−1(a)TΓT

N

The block Gauss–Lobatto rule

The generalization of the Gauss–Lobatto construction to the blockcase is a little more difficultWe would like to have a and b as double eigenvalues of the matrixJN+1

It gives (I2 p−1

N (a)pN−1(a)

I2 p−1N (b)pN−1(b)

)(ΩN+1

ΓTN

)=

(aI2bI2

)Let δ(λ) be the solution of

(JN − λI )δ(λ) = (0 . . . 0 I2)T

Then, as before

δN−1(λ) = −pN−1(λ)TpN(λ)−TΓ−TN

Solving the 4× 4 linear system we obtain

ΓTNΓN = (b − a)(δN−1(a)− δN−1(b))−1

Thus, ΓN is given as a Cholesky factorization of the right handside matrix which is positive definite because δN−1(a) is a diagonalblock of the inverse of (JN − aI )−1 which is positive definite and−δN−1(b) is the negative of a diagonal block of (JN − bI )−1 whichis negative definite

From ΓN , we compute

ΩN+1 = aI2 + ΓNδN−1(a)ΓTN

Computation of the block Gauss rules

Theorem

2N∑i=1

f (ti )uiuTi = eT f (JN)e

where eT = (I2 0 . . . 0)

P.J. Davis and P. Rabinowitz, Methods of numericalintegration, Second Edition, Academic Press, (1984)

W. Gautschi, Orthogonal polynomials: computation andapproximation, Oxford University Press, (2004)

G.H. Golub and G. Meurant, Matrices, moments andquadrature, in Numerical Analysis 1993, D.F. Griffiths andG.A. Watson eds., Pitman Research Notes in Mathematics,v 303, (1994), pp 105–156

G.H. Golub and J.H. Welsch, Calculation of Gaussquadrature rules, Math. Comp., v 23, (1969), pp 221–230

D.P. Laurie, Anti–Gaussian quadrature formulas,Math. Comp., v 65 n 214, (1996), pp 739–747

J. Stoer and R. Bulirsch, Introduction to numericalanalysis, second edition, Springer Verlag, (1983)