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# TAMU MATH 304 - Lect4-04web

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MATH 304Linear AlgebraLecture 38:Orthogonal polynomials.Problem. Approximate the function f (x) = exonthe interval [−1, 1] by a quadratic polynomial.The best approximation would be a polynomial p(x)that minimizes the distance relative to the uniformnorm:kf − pk∞= max|x |≤1|f (x) − p(x)|.However there is no analytic way to find such apolynomial. Another approach is to find a “leastsquares” approximation that minimizes the integralnormkf − pk2=Z1−1|f (x) − p(x)|2dx1/2.The norm k · k2is induced by the inner producthg, hi =Z1−1g(x)h(x) dx.Therefore kf − pk2is minimal if p is theorthogonal projection of the function f on thesubspace P3of quadratic polynomials.Suppose that p0, p1, p2is an orthogonal basis forP3. Thenp(x) =hf , p0ihp0, p0ip0(x) +hf , p1ihp1, p1ip1(x) +hf , p2ihp2, p2ip2(x).Orthogonal polynomialsP: the vector space of all polynomials with realcoefficients: p(x) = a0+ a1x + a2x2+ ··· + anxn.Basis for P: 1, x, x2, . . . , xn, . . .Suppose that P is endowed with an inner product.Definition. Orthogonal polynomials (relative tothe inner product) are polynomials p0, p1, p2, . . .such that deg pn= n (p0is a nonzero constant)and hpn, pmi = 0 for n 6= m.Orthogonal polynomials can be obtained by applyingthe Gram-Schmidt orthogonalization processto the basis 1, x, x2, . . . :p0(x) = 1,p1(x) = x −hx, p0ihp0, p0ip0(x),p2(x) = x2−hx2, p0ihp0, p0ip0(x) −hx2, p1ihp1, p1ip1(x),. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .pn(x) = xn−hxn, p0ihp0, p0ip0(x) − ··· −hxn, pn−1ihpn−1, pn−1ipn−1(x),. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Then p0, p1, p2, . . . are orthogonal polynomials.Theorem (a) Orthogonal polynomials always exist.(b) The orthogonal polynomial of a fixed degree is unique upto scaling.(c) A polynomial p 6= 0 is an orthogonal polynomial if a ndonly if hp, qi = 0 for any polynomial q with deg q < deg p.(d) A polynomial p 6= 0 is an orthogonal polynomial if andonly if hp, xki = 0 for any 0 ≤ k < deg p.Proof of statement (b): Suppose that P and R are twoorthogonal polynomials of the same degree n. ThenP(x) = anxn+ an−1xn−1+ ··· + a1x + a0andR(x) = bnxn+ bn−1xn−1+ ··· + b1x + b0, where an, bn6= 0.Consider a polynomial Q(x) = bnP(x) − anR(x). Byconstruction, deg Q < n. It follows from statement (c) thathP, Qi = hR, Qi = 0. ThenhQ, Qi = hbnP − anR, Qi = bnhP, Qi − anhR, Qi = 0,which means that Q = 0. Thus R(x) = (a−1nbn) P(x).Example. hp, qi =Z1−1p(x)q(x) dx.Note that hxn, xmi = 0 if m + n is odd.Hence p2k(x) contains only even powers of x whilep2k+1(x) contains only odd powers of x.p0(x) = 1,p1(x) = x,p2(x) = x2−hx2, 1ih1, 1i= x2−13,p3(x) = x3−hx3, xihx, xix = x3−35x.p0, p1, p2, . . . are called the Legendre polynomials.Instead of normalization, the orthogonalpolynomials are subject to standardization.The standardization for the Legendre polynomials isPn(1) = 1. In particular, P0(x) = 1, P1(x) = x,P2(x) =12(3x2− 1), P3(x) =12(5x3− 3x).Problem. Find P4(x).Let P4(x) = a4x4+ a3x3+ a2x2+ a1x + a0.We know that P4(1) = 1 and hP4, xki = 0 for0 ≤ k ≤ 3.P4(1) = a4+ a3+ a2+ a1+ a0,hP4, 1i =25a4+23a2+ 2a0, hP4, xi =25a3+23a1,hP4, x2i =27a4+25a2+23a0, hP4, x3i =27a3+25a1.a4+ a3+ a2+ a1+ a0= 125a4+23a2+ 2a0= 025a3+23a1= 027a4+25a2+23a0= 027a3+25a1= 0(25a3+23a1= 027a3+25a1= 0=⇒ a1= a3= 0a4+ a2+ a0= 125a4+23a2+ 2a0= 027a4+25a2+23a0= 0⇐⇒a4=358a2= −308a0=38Thus P4(x) =18(35x4− 30x2+ 3).Legendre polynomialsProblem. Find a quadratic polynomial that is thebest least squares fit to the function f (x) = |x| onthe interval [−1, 1].The best least squares fit is a polynomial p(x) thatminimizes the distance relative to the integral normkf − pk =Z1−1|f (x) − p(x)|2dx1/2over all polynomials of degree 2.The norm kf − pk is minimal if p is the orthogonalprojection of the function f on the subspace P3ofpolynomials of degree at most 2.The Legendre polynomials P0, P1, P2form an orthogonal basisfor P3. Thereforep(x) =hf , P0ihP0, P0iP0(x) +hf , P1ihP1, P1iP1(x) +hf , P2ihP2, P2iP2(x).hf , P0i =Z1−1|x|dx = 2Z10x dx = 1,hf , P1i =Z1−1|x|x dx = 0,hf , P2i =Z1−1|x|3x2− 12dx =Z10x(3x2− 1) dx =14,hP0, P0i =Z1−1dx = 2, hP2, P2i =Z1−13x2− 122dx =25.In general, hPn, Pni =22n + 1.Problem. Find a quadratic polynomial that is thebest least squares fit to the function f (x) = |x| onthe interval [−1, 1].Solution: p(x) =12P0(x) +58P2(x)=12+516(3x2− 1) =316(5x2+ 1).Recurrent formula for the Legendre polynomials:(n + 1)Pn+1(x) = (2n + 1)xPn(x) − nPn−1(x).For example, 4P4(x) = 7xP3(x) − 3P2(x).Legendre polynomialsDefinition. Chebyshev polynomials T0, T1, T2, . . .are orthogonal polynomials relative to the innerproducthp, qi =Z1−1p(x)q(x)√1 − x2dx,with the standardization Tn(1) = 1.Remark. “T” is like in “Tschebyscheff”.Change of variable in the integral: x = cos φ.hp, qi = −Zπ0p(cos φ) q(cos φ)p1 − cos2φcos′φ dφ=Zπ0p(cos φ) q(cos φ) dφ.Theorem. Tn(cos φ) = cos nφ.hTn, Tmi =Zπ0Tn(cos φ)Tm(cos φ) dφ=Zπ0cos(n φ) cos(mφ) dφ = 0 if n 6= m.Recurrent formula: Tn+1(x) = 2xTn(x) − Tn−1(x).T0(x) = 1, T1(x) = x,T2(x) = 2x2− 1,T3(x) = 4x3− 3x,T4(x) = 8x4− 8x2+ 1, . . .That is, cos 2φ = 2 cos2φ − 1,cos 3φ = 4 cos3φ − 3 cos φ,cos 4φ = 8 cos4φ − 8 cos2φ + 1, . . .Chebyshev

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