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8/8/2019 Gauss quadrature 2
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Gauss quadrature
Gerard MEURANT
October, 2008
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Quadrature rules
Given a measure on the interval [a, b] and a function f, aquadrature rule is a relation
b
a
f() d =N
j=1
wjf(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] = 0
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Quadrature rules of degree N
1 can be obtained by
interpolation Such quadrature rules are called interpolatory
NewtonCotes formulas are defined by taking the nodes to beequally spaced
A popular choice for the nodes is the zeros of the Chebyshevpolynomial of degree N. This is called the Fejer quadraturerule
Another interesting choice is the set of extrema of the
Chebyshev polynomial of degree N 1. This gives theClenshawCurtis quadrature rule
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TheoremLet k be an integer, 0 k N. The quadrature rule has degreed = N
1 + k if and only if it is interpolatory and
ba
Nj=1
( tj)p(x) d = 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
Nj=1
( tj)
is orthogonal to itself which is impossible
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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 RiemannStieltjes integral is
I[f] =
ba
f() d() =N
j=1
wjf(tj) +Mk=1
vkf(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
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If M = 0, this is the Gauss rule with no prescribed nodes
IfM = 1 and z1
= a or z1
= b we have the GaussRadau rule
If M = 2 and z1 = a, z2 = b, this is the GaussLobatto rule
The term R[f] is the remainder which generally cannot beexplicitly computedIf the measure is a positive nondecreasing function
R[f] =f(2N+M)()
(2N + M)!
ba
Mk=1
(zk)
Nj=1
( tj)
2
d(), a < < b
(2)Note that for the Gauss rule, the remainder R[f] has the sign off(2N)()see Stoer and Bulirsch
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The Gauss rule
How do we compute the nodes tj and the weights wj?
One way to compute the nodes and weights is to usef() = i, i = 1, . . . , 2N 1 and to solve the non linearequations expressing the fact that the quadrature rule is exact
Use of the orthogonal polynomials associated with themeasure
ba pi()pj() d() = i
,j
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P() = [p0() p1() pN1()]T, eN = (0 0 0 1)T
P() = JNP() + NpN()eN
JN =
1 11 2 2
. . .. . .
. . .
N2 N1 N1N1 N
JN is a Jacobi matrix, its eigenvalues are real and simple
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TheoremThe eigenvalues of JN (the socalled Ritz values (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( 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 and the nodes of the quadrature rule are the zerosof the orthogonal polynomial, that is the eigenvalues of JN
Th P( ) i li d i f J
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The vector P(tj) is an unnormalized eigenvector of JNcorresponding to the eigenvalue tjIf q is an eigenvector with norm 1, we have P(tj) = q with ascalar . From the ChristoffelDarboux relation
wjP(tj)TP(tj) = 1, j = 1, . . . , N
ThenwjP(tj)
TP(tj) = wj2q2 = wj2 = 1
Hence, wj = 1/2. To find we can pick any component of theeigenvector q, for instance, the first one which is different fromzero = p0(tj)/q1 = 1/q1. Then, the weight is given by
wj = q21
If the integral of the measure is not 1
wj = q210 = q
21
b
a
d()
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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 QRlike method
I[f] =ba f() d() =
Nj=1 w
G
j f(t
G
j ) + RG[f]
with
RG[f] =f(2N)()
(2N)!
b
a
N
j=1
(
tGj )2
d()
The monic polynomialN
j=1(tGj ) which is the determinant N
of JN I can be written as 1 N1pN()
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The GaussRadau rule
To obtain the GaussRadau rule, we have to extend the matrix JNin 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 = N+1pN+1(a) = (a
N+1)pN(a)
NpN1(a)
This gives
N+1 = a NpN1(a)pN(a)
Note that (JN aI)P(a) = NpN(a)eN
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Let (a) = [1(a), , N(a)]T with
l(a) =
N
pl1(a)
pN(a)
l = 1, . . . , N
This gives N+1 = a + N(a) and (a) satisfies
(JN aI)(a) = 2NeN
we generate N we solve the tridiagonal system for (a), this gives N(a)
we compute N+1 = a + N(a)
JN+1 =
JN NeN
N(eN)T N+1
gives the nodes and the weights of the GaussRadau quadrature
rule
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TheoremSuppose f is such that f(2n+1)() < 0, n, , a < < b. Let
UGR[f] =N
j=1
waj f(taj ) + va1 f(a)
waj , va1 , t
aj being the weights and nodes computed with z1 = a and
let LGR
LGR[f] =N
j=1
wbj f(tbj ) + v
b1 f(b)
wbj , vb1 , t
bj being the weights and nodes computed with z1 = b.
The GaussRadau rule is exact for polynomials of degree less thanor equal to 2N and we have
LGR[f] I[f] UGR[f]
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Theorem (end)MoreoverN U, L [a, b] such that
I[f]
UGR[f] =
f(2N+1)(U)
(2N + 1)!
b
a
(
a)
N
j=1
(
taj )
2
d()
I[f] LGR[f] = f(2N+1)(L)
(2N + 1)!
ba
( b)
N
j=1( tbj )
2
d()
G
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The GaussLobatto rule
We would like to have
pN+1(a) = pN+1(b) = 0
Using the recurrence relation
pN(a) pN1(a)pN(b) pN1(b)
N+1N
=a pN(a)
b pN(b)
Let
l =
pl1(a)
NpN(a)
, l =
pl1(b)
NpN(b)
, l = 1, . . . , N
then(JN aI) = eN, (JN bI) = eN
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1 N1
N
N+12N =
a
b
we solve the tridiagonal systems for and , this gives N andN
we compute N+1 and N
JN+1 =
JN NeNN(e
N)T N+1
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TheoremSuppose f is such that f(2n)() > 0, n, , a < < b and let
UGL[f] =
Nj=1
wGLj f(tGLj ) + vGL1 f(a) + vGL2 f(b)
tGLj , wGLj , v
GL1 and v
GL2 being the nodes and weights computed
with a and b as prescribed nodes. The GaussLobatto rule is exactfor polynomials of degree less than or equal to 2N + 1 and
I[f] UGL[f]
Moreover
N
[a, b] such that
I[f]UGL[f] = f(2N+2)()
(2N + 2)!
ba
(a)(b) Nj=1
( tGLj )2
d()
C t ti f th G l
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Computation of the Gauss rulesThe weights wi are given by the squares of the first components ofthe eigenvectors wi = (z
i1)2 = ((e1)Tzi)2
TheoremNl=1
wlf(tl) = (e1)Tf(JN)e
1
Proof.Nl=1
wlf(tl) =Nl=1
(e1)Tzlf(tl)(zl)Te1
= (e1)T N
l=1
zlf(tl)(zl)T
e1
= (e1)TZNf(N)ZTNe
1
= (e1)Tf(JN)e1
Th ti G l
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The antiGauss rule
A usual way of obtaining an estimate of I[f] LNG[f] is to useanother quadrature rule Q[f] of degree greater than 2N 1 and toestimate the error as Q[f] L
NG[f]
Laurie proposed to construct a quadrature rule with N + 1 nodescalled an antiGauss rule
HN+1[f] =N+1
j=1
j
f(j
),
such thatI[p] HN+1[p] = (I[p] LNG[p])
for all polynomials of degree 2N+ 1. Then, the error of the Gaussrule can be estimated as
1
2(HN+1[f] LNG[f])
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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[] LNG[]We have
I[pq] =
I(pq)
for p a polynomial of degree N 1 and q a polynomial of degreeN and
I(p2N) = 2I(p2N)where pj are the orthogonal polynomials associated to
IUsing the Stieltjes formulas for the coefficients we obtain theJacobi matrix
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JN+1 =
1 1
1 2 2. . .. . .
. . .
N2 N1 N1N1 N
2N
2N
N+1
The antiGauss nodes j, j = 2, . . . , N are inside the integrationintervalHowever, the first and the last nodes can eventually be outside of
the integration intervalActually, in some cases, the matrix JN+1 can be indefinite even ifJN is positive definite
O t t d t l SN+1[f ] h th t
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One can construct a quadrature rule SN+1[f] such that
I[p] SN+1[p] = (I[p] LNG[p])
for all polynomials of degree 2N + 1. The parameter is positiveand less than or equal to 1
JN+1 =
1 11 2 2
. . . . . . . . .
N2 N1 N1N1 N N
1 +
N
1 + N+1
The error of the Gauss rule can be estimated as
1
1 + (SN+1[f] LNG[f])
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Proposition
pj() =
j
1
j 1 qj()
Hence, qN is a multiple of pN and the polynomials have the sameroots which are also the common real eigenvalues of JN and J
TN
We define the quadrature rule asba
f() d() =N
j=1
f(j)sjtj + R[f]
where j is an eigenvalue of JN, sj is the first component of theeigenvector uj of JN corresponding to j and tj is the firstcomponent of the eigenvector vj of J
TN corresponding to the same
eigenvalue, normalized such that vTj uj = 1
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TheoremAssume that jj = 0, then the nonsymmetric Gauss quadraturerule is exact for polynomials of degree less than or equal to 2N 1The remainder is characterized as
R[f] = f(2N)
()(2N)!
ba
pN()2 d()
The extension of the GaussRadau and GaussLobatto rules to thenonsymmetric case is almost identical to the symmetric case
The block Gauss quadrature rules
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The block Gauss quadrature rules
The integralba f()d() is now a 2 2 symmetric matrix. Themost general quadrature formula is of the form
b
a
f()d() =N
j=1Wjf(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
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There exist orthogonal matrix polynomials related to such that
pj1() = pj()j + pj1()j + pj2()Tj1
p0() I2, p1() 0This can be written as
[p0(), . . . , pN1()] = [p0(), . . . , pN1()]JN+[0, . . . , 0, pN()N]
where
JN =
1 T1
1 2 T2
. . .. . .
. . .
N2 N1 TN1
N1 N
is a symmetric block tridiagonal matrix of order 2N
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The nodes tj are the zeros of the determinant of the matrixorthogonal polynomials that is the eigenvalues of JN and ui is the
vector consisting of the two first components of the correspondingeigenvectorHowever, the eigenvalues may have a multiplicity larger than 1Let i, i = 1, . . . , l be the set of distinct eigenvalues and ni theirmultiplicities. The quadrature rule is then
li=1
ni
j=1
(wji)(wji)
T
f(i)
The block quadrature rule is exact for polynomials of degree lessthan or equal to 2N 1 but the proof is rather involved
The block GaussRadau rule
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We would like a to be a double eigenvalue of JN+1
JN+1P(a) = aP(a) [0, . . . , 0, pN+1(a)N+1]T
apN(a) pN(a)N+1 pN1(a)TN = 0If p
N(a) is non singular
N+1 = aI2 pN(a)1pN1(a)TNBut
(JN aI)
p0(a)TpN(a)
T
...pN1(a)TpN(a)T
= 0
...TN
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We first solve
(JN aI)
0(a)...
N1(a)
=
0...
TN
We computeN+1 = aI2 + N1(a)
TTN
The block GaussLobatto rule
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The generalization of the GaussLobatto construction to the blockcase is a little more difficult
We would like to have a and b as double eigenvalues of the matrixJN+1It gives
I2 p
1N (a)pN1(a)
I2 p
1
N (b)pN
1(b)
N+1
T
N =
aI2
bI2
Let () be the solution of
(JN I)() = (0 . . . 0 I2)T
Then, as before
N1() = pN1()TpN()TTN
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Solving the 4 4 linear system we obtain
TNN = (b a)(N1(a) N1(b))
1
Thus, N is given as a Cholesky factorization of the right handside matrix which is positive definite because N1(a) is a diagonalblock of the inverse of (JN
aI)1 which is positive definite and
N1(b) is the negative of a diagonal block of (JN bI)1 whichis negative definite
From N, we compute
N+1
= aI2
+ N
N
1(a)T
N
Computation of the block Gauss rules
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Theorem
2Ni=1
f(ti)uiuTi = eTf(JN)e
where eT = (I2 0 . . . 0)
P.J. Davis and P. Rabinowitz, Methods of numericali t ti S d Editi A d i P (1984)
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integration, 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 105156
G.H. Golub and J.H. Welsch, Calculation of Gaussquadrature rules, Math. Comp., v 23, (1969), pp 221230
D.P. Laurie, AntiGaussian quadrature formulas,Math. Comp., v 65 n 214, (1996), pp 739747
J. Stoer and R. Bulirsch, Introduction to numericalanalysis, second edition, Springer Verlag, (1983)
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