MATH ASSIGNMENT 06 SOLUTIONS

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1 MATH ASSIGNMENT 06 SOLUTIONS 7.14 Find the polynomial of degree 10 that best fits the function b(t) = cos(4t) at 50 equally spaced points t between 0 and 1. Set up the matrix A and right-hand side vector b and determine the polynomial coefficients in two different ways: (a) By using the Matlab command x = A\b (which uses a QR decomposition). (b) By solving the normal equations A T Ax = A T b. This can be done in Matlab by typing x = (A *A)\(A *b). Print the results to 16 digits (using format long e) and comment on the diffferences you see. [Note you can compute the condition number of A or of A T A using the Matlab function cond.] Solution. The following Matlab code can be used to complete the exercise: t = linspace(0, 1, 50).'; A = vander(t); A = A(:,end-10:end); b = cos(4*t); xqr = A\b; xne = (A'*A)\(A'*b); format long e table(xqr, xne, 'Var', {'xqr', 'xne'}) % Display the relative difference in the coefficients table(abs(xqr-xne)./abs(xqr), 'Var', {'RelDiff'}) The above generates the followinng output: ans = xqr e e e e e e e e e e e+00 xne e e e e e e e e e e e+00 1

2 ans = RelDiff e e e e e e e e e e e-09 The difference between coefficients is explained by the squaring of the condition number of A when solving the normal equations. Once squared, the condition of A is so large as to ensure that one can only expect about two correct digits in the solution to the normal equations In the inclined plane experiment of Galileo, described in Example 7.6.2, if one knows the horizontal speed v of the ball when it leaves the table and enters free fall, then one can determine the time T that it takes for the ball to fall to the ground due to the force of gravity and therefore the horizontal distance vt that it travels before hitting the ground. Assume that the acceleration of the ball on the inclined plane is constant in the direction of the incline: a(t) = C, where C will depend on gravity and on the angle of the inclined plane. Then the velocity v(t) will be proportional to t: v(t) = Ct, since v(0) = 0. The distance s(t) that the ball has moved along the inclined plane is therefore s(t) = 1 2 Ct2, since s(0) = 0. The ball reaches the bottom of the inclined plane when s(t) = h/ sin(θ), where θ is the angle of incline. (a) Write down an expression in terms of C, h, and θ for the time t h at which the ball reaches the bottom of the inclined plane and for its velocity v(t h ) at that time. Show that the velocity is proportional to h. Solution. As noted above, the ball reaches the bottom of the inclined plane when s(t) = h, which implies that t sin(θ) h is a solution to A little algebra establishes that 1 2 Ct2 = h sin(θ). t h = 2 h C sin(θ) 2

3 and thus the velocity at t h is given by 2 2 v(t h ) = C h = h C sin(θ) sin(θ) which is proportional to h. (b) Upon hitting the table, the ball is deflected so that what was its velocity along the incline now becomes its horizontal velocity as it flies off the table. From this observation and part (a), one would expect Galileo s measured distances to be proportional to h. Modify the Matlab code in Example to do a least squares fit of the given distances d in table 7.5 to a function of the form c 0 + c 1 h. Determine the best coefficients c0 and c 1 and the residual norm 4 i=1 (d i (c 0 + c 1 h))2. Solution. The following Matlab code can be used to complete this exercise: % Set up data h = [0.282; 0.564; 0.752; 0.940]; d = [0.752; 1.102; 1.248; 1.410]; % Form the 4-by-2 matrix A and solve for the coefficient vector c. A = zeros(size(h,1), 2); A(:,1) = sqrt(h); A(:,2) = 1; c = A\d % Plot the data points. plot(h, d, 'b*'); title('least Squares Fit'); xlabel('release Height'); ylabel('horizontal Distance'); % Plot the best fit function of the form f(x) = c(1)*sqrt(h) + c(2) pts = linspace(min(h), max(h), 101); f c(1)*sqrt(x)+c(2); hold on; plot(pts, f(pts), 'r'); axis tight; hold off % Compute residual resid = norm(d - f(h)) The following output is generated: c = resid =

4 The following plot is generated: 7.16 Consider the following least squares approach for ranking sports teams. Suppose we have four college football teams, called simply T1, T2, T3, and T4. These four teams play each other with the following outcomes: T1 beats T2 by 4 points: 21 to 17. T3 beats T1 by 9 points: 27 to 18. T1 beats T4 by 6 points: 16 to 10. T3 beats T4 by 3 points: 10 to 7. T2 beats T4 by 7 points: 17 to 10. To determine ranking points r 1,..., r 4 for each team, we do a least squares fit to the overdetermined linear system: r 1 r 2 = 4 r 3 r 1 = 9 r 1 r 4 = 6 r 3 r 4 = 3 r 2 r 4 = 7. 4

5 This system does not have a unique least squares solution, however, since if (r 1,..., r 4 ) T is one solution and we add to it any constant vector, such as the vector (1, 1, 1, 1) T, then we obtain another vector for which the residual is exactly the same. Show that if (r 1,..., r 4 ) T solves the least squares problem above then so does (r 1 +c,..., r 4 +c) T for any constant c. To make the solution unique, we can fix the total number of ranking points, say, at 20. To do this, we add the following equation to those listed above: r 1 + r 2 + r 3 + r 4 = 20. Note that this equation will be satisfied exactly since it will not affect how well the other equalities can be approximated. Use Matlab to determine the values r 1, r 2, r 3, r 4 that most closely satisfy these equations, and based on your results, rank the four teams. Solution. The system generated by the ranking problem has coefficient matrix A given by A = With e being the vector of all ones, note that Ae = = Therefore, for any r and any constant c we have that b A(r + ce) = b Ar cae = b Ar. That is, if r solves the least squares poblem, r + ce also solves the least squares problem for any choice of constant c. We can therefore impose the constraint that r 1 + r 2 + r 3 + r 4 = K for some choice of K, say K = 20. The least squares problem to solve, then, is min b Ar A = The following Matlab code solves this problem: b = A = zeros(6,4); b = zeros(6,1); 5

6 % T1 beats T2 by 4 A(1, [1 2]) = [1-1]; b(1) = 4; % T3 beats T1 by 9 A(2, [3 1]) = [1-1]; b(2) = 9; % T1 beats T4 by 6 A(3, [1 4]) = [1-1]; b(3) = 6; % T3 beats T4 by 3 A(4, [3 4]) = [1-1]; b(4) = 3; % T2 beats T4 by 7 A(5, [2 4]) = [1-1]; b(5) = 7; % Fix number of ranking points at 20. A(6, :) = 1; b(6) = 20; r = A\b And it generates this output: r = Based on these results, we arrive at the ranking T3, T1, T2, T4. (2) Let u, v be two n-vectors and let P = uv T. (a) Under what conditions is P a projector? Solution. P is a projector if P 2 = P. That is, uv T uv = uv T must be the case. This occurs when v T u = 1. (b) Under what conditions is P an orthogonal projector? Solution. In order for P to be an orthogonal projector, P T = P must be the case. This occurs only when v = u and u T u = 1. (3) Let P be a nonzero projector. Show that P 2 1 with equality if and only if P is an orthogonal projector. Solution. Whenever P is a projector, it follows that whenever y col(p ), y = P y and so P y 2 = y 2 and thus P 2 = max y 0 P y 2 y 2 1. Now, suppose that P 2 = 1. We must show that this implies that P is orthogonal. Recall that the definition of orthogonal projector is that null(p ) = col(p ). So, let x null(p ) and let y = P x x. As P is a projector, P y = P x P x = 0 and so y null(p ). As a result, x T y = 0 and so x + y 2 2 = x y

7 Now then x 2 2 x y 2 2 = x + y 2 2 = P x 2 2 x 2 2. It follows, then that y = 0 and so P x = x and thus x col(p ). We have established that null(p ) col(p ). On the other hand, let z col(p ) so that P z = z. Also, let x null(p ) and y null(p ) such that z = x + y. Thus P z = P x + P y = P y = y with the last equality following from the fact that null(p ) col(p ). As z = y, it follows that col(p ) null(p ) and so null(p ) = col(p ). Therefore, P is an orthogonal projector. Now, when P is orthogonal, P x 2 2 = x T P T P x = x T P x x 2 P x 2 by Cauchy-Schwarz. It follows then, for all x, P x 2 x 2 and so P 2 1. Since for any projector P 2 1, it follows that P 2 = 1 whenever P is an orthogonal projector. 7