12.0 Properties of orthogonal polynomials

Transcription

1 12.0 Properties of orthogonal polynomials In this section we study orthogonal polynomials to use them for the construction of quadrature formulas investigate projections on polynomial spaces and their convergence when the degree of the polynomials is increased 96

2 12.1 Orthogonal polynomials form a basis Theorem. [cf. Th. 12.1] Let B be a Hilbert space with P n B. Let {φ i, i = 0,..., n} be a set of polynomials with the properties Then, φ i P i and (φ i, φ j ) = 0 for i j. the functions {φ i } are linearly independent and if ψ P k is orthogonal to all polynomials in P k 1 then there exists a c 0 such that ψ = cφ k. 97

3 12.2 Orthogonal polynomials have real, distinct roots Theorem. [cf. Th. 12.2] Let 0 φ k P k be orthogonal to all elements in P k 1. exactly k distinct and real roots in [a, b] Then, φ k has Note, [a, b] is the interval which defines the inner product. The roots of the orthogonal polynomials depend only on k, [a, b] and the weight function w(x). 98

4 12.3 Quadrature (1) The basic idea of quadrature (integration) methods for computing is b a f(x)dx The approximation of the exact integral of f is taken as the exact integral of an approximation to f. As an approximation a polynomial which interpolates f in the points {x i, i = 0, 1,..., k} [a, b] is taken. 99

5 12.4 Quadrature (2) Let p(x) = k f(x i )L(x) be the interpolation polynomial of f at the interpolation points x i = a + c i (b a) and L the corresponding Lagrange polynomial. Then we make the ansats i=0 b a f(x)dx b a p(x)dx = h 1 0 p(a + θ)dθ = h k f(a + c i h) L(θ)dθ } 0 {{} i=0 1 =:b i with h = b a. 100

6 12.5 Order of a quadrature formula (3) Definition. Let q be the biggest integer such that the quadrature formula integrates f(x) = p(x) P q 1 exactly, then q is called the oder of the method. Evidently by construction q k + 1. Some methods: Method k f evals c 0 c 1 c 2 order q Rectangle rule Midpoint rule Trapezoidal rule Simpson s rule Gauss

7 12.6 Gauss quadrature (1) By placing the interpolation points, the order can be changed. What is the largest order for a given k? Theorem. property Let the x i s be the zeros of a polynomial φ s P s then the b a w(x)φ s (x)p(x)dx = 0 p P m 1 holds if and only if the corresponding quadrature method (Gauss method) has order s + m. 3 Example: a = 1, b = 1, w(x) = 1, c 0 = 5, c 3 1 = 0, c 2 = 5 defines the 3 stage Gauss-method, which has order 6. The c i s are zeros of the Legendre polynomials. 102

8 12.7 Gauss quadrature (2) The Gauss method has maximal order, i.e. same k and higher order. there is no method with the For preserving monotonicity the following theorem is important: Theorem. The coefficients b i for the Gauss quadrature method are positive. 103

9 Characterization of orthogonal polynomials Orthogonal polynomials are given by a 3-term recursion. The following theorem gives an alternative characterization: Theorem. The function φ k+1 C[a, b] satisfies the orthogonality condition b a w(x)φ k+1 (x)p(x)dx = 0 p P k if and only if there exists u such that w(x)φ k+1 (x) = u (k+1) (x) x [a, b] u (i) (a) = u (i) (b) = 0 i = 0, 1,..., k 104

10 12.8 Special orthogonal polynomials Definition. The polynomial defined by the orthogonality relation 1 1 (1 x) α (1 + x) β φ i (x)φ j (x) = 0 i j α 1, β 1 are called Jacobi polynomials This gives u(x) = (1 x) α+k+1 (1 + x) β+k+1. Example: α = β = 0 (1 x) k+1 (1 + x) k+1. By differentiation we get the so-called Legendre polynomials φ 0 (x) = 1 φ 1 (x) = dx d (1 + x)(1 x) = 2x φ 2 (x) = d2 dx 2 (1 + x)2 (1 x) 2 = 12x 2 4. For α = β = 1/2 we get the Chebychev polynomials. 105

11 12.9 Chebychev operator R n (1) Let the weight function w be w(x) := 1 1 x 2 We the corresponding norm: f 2 w := 1 1 f(x) 2 dx. Let R n : C[ 1, 1] P n be defined as the best approximation f R n f w = min p P n f p w. 106

12 12.10 Chebychev operator R n (2) We call R n the Chebychev operator. It is a linear projection. The orthogonal polynomials concerning this norm are the Chebychev polynomials φ j : n (φ j, f) (R n f)(x) = (φ j, φ j ) φ j(x) j=0 If f P n+1 then R n f is even the best approximation in the. -norm. How is R n f related to the best minmax approximation p for general f? The answer is given by R n and Theorem 3.1: f R n f (1 + R n ) f p. 107

13 12.11 Chebychev operator R n (3) Theorem. [Th. 12.7] R n = 1 π 1 2n π sin(n θ) 0 sin( 1 2 θ) dx = n ( ) 1 jπ j tan. 2n + 1 π j=1 108