Orthogonal Polynomials and Polynomial Approximations

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Transcript of Orthogonal Polynomials and Polynomial Approximations


    Orthogonal Polynomials and PolynomialApproximations

    3.1. Orthogonal polynomials

    The orthogonal polynomials play the most important role in spectralmethods, so it is useful to understand some general properties of the or-thogonal polynomials.

    Given an interval (a, b) and a weight function (x) which is positive on(a, b) and w L1(a, b), we define the weighted Sobolev space L2(a, b) by

    L2(a, b) ={f :

    baf2(x)(x)dx < +

    }. (3.1.1)

    It is obvious that (, ) defined by

    (f, g) := b


    is an inner product on L2(a, b). Hence, fL2 = (f, f)12 . Hereafter, the

    subscript will be omitted from the notation when (x) 1.Two functions f and g are said to be orthogonal in L2(a, b) if

    (f, g) = 0.

    A sequence of polynomials {pn}n=0 with deg(pn) = n is said to be or-thogonal in L2(a, b) if

    (pi, pj) = 0 for i 6= j. (3.1.2)Since the orthogonality is not altered by a multiplicative constant, we maynormalize the polynomial pn so that the coefficient of xn is one, i.e.

    pn(x) = xn + a(n)n1x

    n1 + + a(n)0 .Such a polynomial is said to be monic.

    3.1.1. Existence and uniqueness. Our immediate goal is to establishthe existence of a sequence of orthogonal polynomials. Although we could,in principle, determine the coefficients a(n)j of pn in the natural basis by usingthe orthogonality conditions (3.1.2), it is computationally advantageous toexpress pn in terms of lower-order orthogonal polynomials.

    Let us denotePn := span

    {1, x, x2, , xn

    }. (3.1.3)



    Then, if {pk}k=0 is a sequence of polynomials such that pk is exactly ofdegree k, it is obvious by dimension argument that

    Pn := span {p0, p1, , pn} . (3.1.4)A direct consequence of this result is the following:

    Lemma 3.1. If the sequence of polynomials {pk}k=0 is orthogonal, thenthe polynomial pn+1 is orthogonal to any polynomial q of degree n or less.

    Proof. This can be established by writing

    q = bnpn + bn1pn1 + + b0p0and observing

    (pn+1, q) = bn(pn+1, pn) + bn1(pn+1, pn1) + + b0(pn+1, p0) = 0,the last equality following from the orthogonality of the polynomials.

    Theorem 3.1. For any given positive weight function (x) L1(a, b),there exists a unique set of monic orthogonal polynomials {pn}. More pre-cisely, {pn} can be constructed as follows:

    p0 = 1, p1 = x 1 with 1 = b



    andpn+1 = (x n+1)pn n+1pn1, n 1, (3.1.5)


    n+1 = b




    n+1 = b



    Proof. We shall first establish the existence of monic orthogonal poly-nomials. We begin by computing the first two. Since p0 is monic and ofdegree zero,

    p0(x) 1.Since p1 is monic and of degree one, it must have the form

    p1(x) = x 1.To determine 1, we use orthogonality:

    0 = (p1, p0) = b

    a(x)xdx 1


    Since the weight function is positive in the interval of integration, it followsthat

    1 = b




    In general we will seek pn+1 in the form

    pn+1 = xpn n+1pn n+1pn1 n+1pn2 .

    As in the construction of p1, we will use orthogonality to determine thecoefficients above. To determine n+1, write

    0 = (pn+1, pn) = (xpn, pn) n+1(pn, pn) n+1(pn1, pn) .

    By orthogonality, we have baxp2ndx n+1

    bap2ndx = 0,

    which gives

    n+1 = b



    For n+1, using the fact that (pn+1, pn1) = 0 gives

    n+1 = b



    Next, we have

    n+1 = b


    bap2n2dx = 0

    since xpn2 is of degree n 1, and thus orthogonal to pn. Likewise, thecoefficients of pn3, pn4, ... are all zero.

    The above procedure leads to a sequence of monic orthogonal polyno-mials. We now show that such a sequence is unique.

    Assume that {qn}n=0 is another sequence of orthogonal (monic) polyno-mials, i.e. the coefficient of xn in qn is 1, and b

    aqi(x)qj(x)dx = 0, i 6= j.

    Since pn, given by (3.1.5), is also monic, we obtain that

    deg(pn+1 qn+1

    ) n.

    Now using Lemma 3.1 gives

    (pn+1, pn+1 qn+1) = 0, (qn+1, pn+1 qn+1) = 0.

    The above results yield (pn+1 qn+1, pn+1 qn+1) = 0. This implies thatpn+1 qn+1 0 for all n 0.

    The above theorem reveals a remarkable property of the orthogonalpolynomials, namely, all orthogonal polynomials can be constructed witha three-term recurrence relation (3.1.5).


    3.1.2. Zeros of orthogonal polynomials. The zeros of the orthogo-nal polynomials play an important role in the implementations of the spec-tral methods. The main result concerning the zeros of orthogonal polyno-mials is the following:

    Theorem 3.2. The zeros of pn+1 are all real, simple, and lie in theinterval (a, b).

    Proof. Since (1, pn+1) = 0, there exists at least one zero of pn+1 in(a, b). Let x0, x1, , xk in (a, b) be the zeros of odd multiplicity of pn+1;i.e., x0, x1, , xk are the points where pn+1 changes sign. If k = n, weare through, since {xi}ni=0 are the n + 1 simple zeros of pn+1. If k < n, weconsider the polynomial

    q(x) = (x x0)(x x1) (x xk).Since deg(q) = k + 1 < n+ 1, by orthogonality

    (pn+1, q) = 0.

    On the other hand pn+1(x)q(x) cannot change sign on (a, b) since each signchange in pn+1(x) is canceled by a corresponding sign change in q(x). Itfollows that

    (pn+1, q) 6= 0,which is a contradiction.

    ??? Give a procedure to compute zeroes (from an eigenvalue problem)???

    3.1.3. Gauss type quadratures. We wish to create quadrature for-mulae of the type b



    f(xj)j .

    If the choice of nodes x0, x1, , xN are made a priori, then in generalthe above formula can only be exact for polynomials of degree N . Moreprecisely, setting

    j = b

    ahj(x)(x)dx, with hj(x) =

    i6=j(x xi)i6=j(xj xi)


    being the Lagrange polynomial associated with the nodes {xj}Nj=0, we have bap(x)(x)dx =


    p(xj)j , p PN . (3.1.7)

    However, if we are free to choose nodes {xk}Nk=0, we can expect the quad-rature formulae of the above form to be exact for polynomials of degree 2N + 1.

    We assume in this subsection that {pn}n=0 is a sequence of orthogonalpolynomials, associated with a weight function in the interval (a, b).


    Theorem 3.3. (Gauss quadrature) Let x0, x1, , xN be the rootsof pN+1 and we define j (j = 0, 1, . . . , N) by (3.1.6). Then j > 0 forj = 0, 1, , N and b

    ap(x)(x)dx =


    p(xj)j , for all p P2N+1. (3.1.8)

    ???? Add formulae for j ???

    Proof. For any p PN , we can write p(x) =N

    j=0 p(xj)hj(x). Hence, bap(x)(x)dx =


    p(xj) b

    ahj(x)(x)dx =


    p(xj)j .

    Now for any p P2N+1, we can write p = rpN+1 + s where r, s PN . SincePN+1(xj) = 0, we have p(xj) = s(xj) for j = 0, 1, . . . , N . Since PN+1 isorthogonal to r, and s PN , we find b

    ap(x)(x)dx =




    s(xj)j =N


    p(xj)j .

    It remains to prove that wj > 0 for j = 0, 1, . . . , N . To this end, we takep(x) = h2k(x) in the above relation to find

    0 < b

    ah2k(x)(x)dx =


    h2k(xj)j = k.

    The Gauss quadrature is optimal in the sense that it is not possible tofind {xj , j}Nj=0 such that (3.1.8) holds for all p P2N+2. However, it isdifficult to enforce any boundary condition since the end points a and b arenot among the Gauss nodes. Therefore, we need generalized Gauss quadra-tures which are suitable for enforcing boundary conditions. More precisely,to enforce the Dirichlet boundary condition at both end points, we need theso called Gauss-Lobatto quadrature below, and to enforce the boundary con-dition at one end point, we should use the Gauss-Radau quadrature. Othergeneralized Gauss quadratures enforcing derivative boundary conditions canalso be constructed similarly, see [9, 4].

    Assuming we would like to include the left endpoint a in the quadrature.We choose N = pN+1(a)/pN (a) and set

    qN (x) =PN+1(x) + NpN (x)

    x a.


    It is obvious that qN PN , and for any rN1 PN1, we have baqN (x)rN1(x)(x)(x a)dx = b

    a(pN+1(x) + NpN (x))rN1(x)(x)dx = 0.


    Hence, {qn} is a set of orthogonal polynomials with weight (x)(x a).

    Theorem 3.4. (Gauss-Radau quadrature) Let x0, x1, , xN be theroots of (x a)qN and j (j = 0, 1, . . . , N) defined by (3.1.7). Then j > 0for j = 0, 1, , N and b

    ap(x)(x)dx =


    p(xj)j , for all p P2N . (3.1.10)

    Proof. The proof is similar to that of Theorem 3.3. Obviously, bap(x)(x)dx =


    p(xj) b

    ahj(x)(x)dx =


    p(xj)j , p PN .

    (3.1.11)Now for any p P2N , we write p = r(x a)qN + s with r PN1, s PN ,and p(xj) = s(xj) for j = 0, 1, . . . , N . Therefore, thanks to (3.1.9), we have b

    ap(x)(x)dx =




    s(xj)j =N


    p(xj)j .

    Again, by taking p(x) = h2k(x) in the above relation, we conclude thatwk > 0 for k = 0, 1, . . . , N .

    A second Gauss-Radau quadrature can be constructed if we want toinclude the right endpoint b instead of the leftend point a.

    We now consider the Gauss-Lobatto quadrature whose nodes include thetwo endpoints. We choose N and N such that

    pN+1(x) + NpN (x) + NpN1(x) = 0, x = a, b,

    and set

    zN1(x) =pN+1(x) + NpN (x) + NpN1(x)

    (x a)(b x).

    Hence, zN1 PN1 and for any rN2 PN2,