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Math 304 Answers to Selected Problems 1 Section 5.5 6. Let {u1 , u2 , u3 } be an orthonormal basis for an inner product space V and let u = u1 + 2u2 + 2u3 and v = u1 + 7u3 Determine the value of each of the following. (a) hu, vi (b) kuk and kvk (c) The angle θ between u and v Answer: (a) By Corollary 5.5.3, hu, vi = 1(1) + 2(0) + 2(7) = 15 (b) By Corollary 5.5.4, √ √ kuk = 12 + 22 + 22 = 9 = 3 √ √ √ 12 + 7 2 = 50 = 5 2 kvk = (c) The angle between u and v can be computed using the following formula: cos θ = hu, vi kukkvk Thus, cos θ = Thus, θ = π . 4 1 15 1 √ =√ 3·5 2 2 9. The set S= 1 √ , cos x, cos 2x, cos 3x, cos 4x 2 is an orthonormal set of vectors in C[−π, π] with inner product defined by (2). (a) Use trigonometric identities to write the function sin4 x as a linear combination of elements of S. (b) Use part (a) and Theorem 5.5.2 to find the values of the following integrals: Rπ (i) −π sin4 x cos x dx Rπ (ii) −π sin4 x cos 2x dx Rπ (iii) −π sin4 x cos 3x dx Rπ (iv) −π sin4 x cos 4x dx Answer: (a) We will use the trigonometric identity sin2 x = 1 2 − 21 cos 2x. 2 1 1 − cos 2x 2 2 1 1 1 − cos 2x + cos2 2x 4 2 4 1 1 1 1 1 − cos 2x + + cos 4x 4 2 4 2 2 3 1 1 − cos 2x + cos 4x 8 2 8 3 1 1 1 √ √ − cos 2x + cos 4x 2 8 4 2 2 4 sin x = = = = = (b) Since S is an orthonormal set, Theorem 5.5.2 applies. Using part (a) and the Theorem, we have that 2 1 4 sin x, √ 2 4 sin x, cos x 4 sin x, cos 2x 4 sin x, cos 3x 4 sin x, cos 4x = = = = = 3 √ 4 2 0 1 − 2 0 1 8 We can use these to evaluate the integrals. Z π (i) sin4 x cos x dx = π sin4 x, cos x = 0 −π (ii) Z π Z π π sin4 x cos 2x dx = π sin4 x, cos 2x = − 2 −π sin4 x cos 3x dx = π sin4 x, cos 3x = 0 (iii) −π Z (iv) π π sin4 x cos 4x dx = π sin4 x, cos 4x = 8 −π 21. Let A= 1 2 1 2 1 2 1 2 − 12 − 12 1 2 1 2 (a) Show that the column vectors of A form an orthonormal set in R4 . 3 (b) Solve the least squares problem Ax = b for each of the following choices of b. (i) b = (4, 0, 0, 0)T (ii) b = (1, 2, 3, 4)T (iii) b = (1, 1, 2, 2)T Answer: (a) Let a1 and a2 denote the first and second column vectors of A, respectively. To show that the column vectors of A form an orthonormal set in R4 , we just need to check that ha1 , a2 i = 0, ka1 k = 1, and ka2 k = 1. 1 1 1 1 1 1 1 1 − + − + + =0 ha1 , a2 i = 2 2 2 2 2 2 2 2 s 2 2 2 2 1 1 1 1 ka1 k = + + + =1 2 2 2 2 s 2 2 2 2 −1 1 1 −1 ka2 k = + + + =1 2 2 2 2 Thus, the column vectors of A form an orthonormal set in R4 . (b) By Theorem 5.5.6, since the column vectors of A form an orthonormal set, the solution to the least squares problem is x̂ = AT b. 1 4 1 1 1 2 2 2 2 0 2 (i) x̂ = 0 = −2 1 1 1 1 −2 −2 2 2 0 1 1 1 1 1 2 2 2 2 2 = 5 (ii) x̂ = 3 2 − 21 − 12 12 12 4 1 1 1 1 1 2 2 2 2 1 = 3 (iii) x̂ = 2 1 − 21 − 12 12 12 2 4 27. Given the vector space C[−1, 1] with inner product Z 1 hf, gi = f (x)g(x) dx −1 and norm kf k = (hf, f i)1/2 (a) Show that the vectors 1 and x are orthogonal. (b) Compute k1k and kxk. (c) Find the least squares approximation to x1/3 on [−1, 1] by a linear function `(x) = c1 1 + c2 x. (d) Sketch the graphs x1/3 and `(x) on [−1, 1]. Answer: (a) The vectors 1 and x are orthogonal if h1, xi = 0. Z 1 h1, xi = x dx = 0 −1 The integral equals 0 because the function x is an odd function. (b) First, we compute h1, 1i: Z 1 h1, 1i = 1 dx = 2 −1 p √ Thus, k1k = h1, 1i = 2. We compute hx, xi: 1 1 3 2 hx, xi = x dx = x = 3 3 −1 −1 r p 2 Thus, kxk = hx, xi = . 3 (c) To find the least squares approximation to x1/3 on [−1, 1] by a linear function, we project x1/3 onto the subspace Span(1, x). The formula to compute the projection requires an orthonormal basis Z 1 2 5 for the subspace. Since 1 and x are orthogonal, we can find an orthonormal basis by k1k and kxk. Thus, the ortho( by dividing √ ) 3x 1 . The projection of x1/3 onto this normal basis is √ , √ 2 2 subspace is * √ +√ 1 1 3x 3x √ + x1/3 , √ √ x1/3 , √ 2 2 2 2 We can compute each of these inner products: 4/3 1 Z 1 1/3 3x x 1/3 1 √ dx = √ x ,√ = =0 2 2 4 2 −1 −1 * " √ #1 √ + Z 1 √ 4/3 7/3 3x 3x 3x 3 √ √ x1/3 , √ = dx = = 2 2 7 2 −1 −1 √ 6 3 √ 7 2 Thus, the projection of x1/3 onto Span(1, x) is * √ +√ √ 1 3x 3x 6 3 1 1/3 1/3 √ + x , √ √ = √ x ,√ 2 2 2 2 7 2 9 x = 7 √ ! 3x √ 2 Thus, the least squares approximation to x1/3 by a linear function 9 is `(x) = x. 7 6