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MATH 304 Linear Algebra Lecture 38: Orthogonal Polynomials. Problem

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MATH 304 Linear Algebra Lecture 38: Orthogonal Polynomials. Problem MATH 304 Linear algebra Lecture 38: Orthogonal polynomials. Problem. Approximate the function f (x) = ex on the interval [ 1, 1] by a quadratic polynomial. − The best approximation would be a polynomial p(x) that minimizes the distance relative to the uniform norm: f p = max f (x) p(x) . k − k∞ x 1 | − | | |≤ However there is no analytic way to find such a polynomial. Another approach is to find a “least squares” approximation that minimizes the integral norm 1 1/2 2 f p 2 = f (x) p(x) dx . k − k 1 | − | Z− The norm 2 is induced by the inner product k · k 1 g, h = g(x)h(x) dx. h i 1 Z− Therefore f p 2 is minimal if p is the k − k orthogonal projection of the function f on the subspace 3 of quadratic polynomials. P Suppose that p0, p1, p2 is an orthogonal basis for 3. Then P f , p0 f , p1 f , p2 p(x) = h i p0(x) + h i p1(x) + h i p2(x). p0, p0 p1, p1 p2, p2 h i h i h i Orthogonal polynomials : the vector space of all polynomials with real P 2 n coefficients: p(x) = a0 + a1x + a2x + + anx . ··· Basis for : 1, x, x 2,..., x n,... P Suppose that is endowed with an inner product. P Definition. Orthogonal polynomials (relative to the inner product) are polynomials p0, p1, p2,... such that deg pn = n (p0 is a nonzero constant) and pn, pm = 0 for n = m. h i 6 Orthogonal polynomials can be obtained by applying the Gram-Schmidt orthogonalization process to the basis 1, x, x 2,... : p0(x)=1, x, p0 p1(x) = x h i p0(x), − p0, p0 h 2 i 2 2 x , p0 x , p1 p2(x) = x h ip0(x) h ip1(x), − p0, p0 − p1, p1 .................................................h i h i n n n x , p0 x , pn 1 pn(x) = x h ip0(x) h − i pn 1(x), − p0, p0 −···− pn 1, pn 1 − .................................................h i h − − i Then p0, p1, p2,... are orthogonal polynomials. Theorem (a) Orthogonal polynomials always exist. (b) The orthogonal polynomial of a fixed degree is unique up to scaling. (c) A polynomial p = 0 is an orthogonal 6 polynomial if and only if p, q = 0 for any h i polynomial q with deg q < deg p. (d) A polynomial p = 0 is an orthogonal 6 k polynomial if and only if p, x = 0 for any h i 0 k < deg p. ≤ 1 Example. p, q = p(x)q(x) dx. h i 1 Z− Note that x n, x m =0 if m + n is odd. h i Hence p2k (x) contains only even powers of x while p2k+1(x) contains only odd powers of x. p0(x)=1, p1(x) = x, 2 2 x , 1 2 1 p2(x) = x h i = x , − 1, 1 − 3 h 3 i 3 x , x 3 3 p3(x) = x h ix = x x. − x, x − 5 h i p0, p1, p2,... are called the Legendre polynomials. Instead of normalization, the orthogonal polynomials are subject to standardization. The standardization for the Legendre polynomials is Pn(1) = 1. In particular, P0(x)=1, P1(x) = x, 1 2 1 3 P2(x) = (3x 1), P3(x) = (5x 3x). 2 − 2 − Problem. Find P4(x). 4 3 2 Let P4(x) = a4x + a3x + a2x + a1x + a0. k We know that P4(1) = 1 and P4, x = 0 for h i 0 k 3. ≤ ≤ P4(1) = a4 + a3 + a2 + a1 + a0, 2 2 2 2 P4, 1 = 5a4 + 3a2 + 2a0, P4, x = 5a3 + 3a1, h i2 2 2 2 h i3 2 2 P4, x = a4 + a2 + a0, P4, x = a3 + a1. h i 7 5 3 h i 7 5 a4 + a3 + a2 + a1 + a0 = 1 2 2 5a4 + 3a2 + 2a0 = 0 2 2 5a3 + 3a1 = 0 2 2 2 7a4 + 5a2 + 3a0 = 0 2a 2a 7 3 + 5 1 = 0 2 2 5a3 + 3a1 = 0 2 2 = a1 = a3 = 0 ( 7a3 + 5a1 = 0 ⇒ 35 a4 + a2 + a0 = 1 a4 = 8 2 2 30 5a4 + 3a2 + 2a0 = 0 a2 = 8 2 2 2 ⇐⇒ −3 7a4 + 5a2 + 3a0 = 0 a0 = 8 1 4 2 Thus P4(x) = (35x 30x + 3). 8 − Legendre polynomials Problem. Find a quadratic polynomial that is the best least squares fit to the function f (x) = x on | | the interval [ 1, 1]. − The best least squares fit is a polynomial p(x) that minimizes the distance relative to the integral norm 1 1/2 f p = f (x) p(x) 2 dx k − k 1 | − | Z− over all polynomials of degree 2. The norm f p is minimal if p is the orthogonal k − k projection of the function f on the subspace 3 of P polynomials of degree at most 2. The Legendre polynomials P0, P1, P2 form an orthogonal basis for 3. Therefore P f , P0 f , P1 f , P2 p(x)= h i P0(x)+ h i P1(x)+ h i P2(x). P0, P0 P1, P1 P2, P2 h i h i h i 1 1 f , P0 = x dx = 2 x dx = 1, h i −1 | | 0 Z 1 Z f , P1 = x x dx = 0, h i −1 | | Z 1 2 1 3x 1 2 1 f , P2 = x − dx = x(3x 1) dx = , h i −1 | | 2 0 − 4 Z Z 2 1 1 3x 2 1 2 P0, P0 = dx = 2, P2, P2 = − dx = . h i −1 h i −1 2 5 Z Z 2 In general, Pn, Pn = . h i 2n + 1 Problem. Find a quadratic polynomial that is the best least squares fit to the function f (x) = x on | | the interval [ 1, 1]. − 1 5 Solution: p(x) = P0(x) + P2(x) 2 8 1 5 3 = + (3x 2 1) = (5x 2 + 1). 2 16 − 16 Recurrent formula for the Legendre polynomials: (n + 1)Pn+1 = (2n + 1)xPn(x) nPn 1(x). − − For example, 4P4(x) = 7xP3(x) 3P2(x). − Definition. Chebyshev polynomials T0, T1, T2,... are orthogonal polynomials relative to the inner product 1 p(x)q(x) p, q dx = 2 , h i 1 √1 x Z− − with the standardization Tn(1) = 1. Remark. “T” is like in “Tschebyscheff”. Change of variable in the integral: x = cos φ. π p(cos φ) q(cos φ) p, q = cos′ φ dφ h i − 0 1 cos2 φ Zπ − = p(cospφ) q(cos φ) dφ. Z0 Theorem. Tn(cos φ) = cos nφ. π Tn, Tm = Tn(cos φ)Tm(cos φ) dφ h i 0 π Z = cos(nφ) cos(mφ) dφ =0 if n = m. 0 6 Z Recurrent formula: Tn+1(x) = 2xTn(x) Tn 1(x). − − T0(x)=1, T1(x) = x, 2 T2(x) = 2x 1, 3 − T3(x) = 4x 3x, 4 − 2 T4(x) = 8x 8x +1, ... − That is, cos 2φ = 2 cos2 φ 1, 3 − cos 3φ = 4 cos φ 3 cos φ, 4 − 2 cos 4φ = 8 cos φ 8 cos φ +1, ... − Chebyshev polynomials.
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