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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