Math 5334: Homework 2 Solutions

Transcription

1 Math 5334: Homework 2 Solutions Victoria E. Howle February 20, 2008 Problem 2.15 a. The condition number of golub(n) grows exponentially. x = []; y = []; % For golub matrices size n = 1 to 10, we will run 10 tests of % the condition number for each size n and average the results. for j = 1:10 for k = 1:10 x(k) = cond(golub(j)); end y(j) = mean(x); end % Plot the resulting condition numbers (on a log scale). semilogy(1:10, y); % The result looks exponential. We can see that condition number % grows approximately like 64^n hold on n = 1:0.1:10; semilogy(n,64.^n, : ); legend( cond(golub(n)), 64^n ); xlabel( n ) ylabel( log(cond(golub(n))) averaged over 10 tests ); title( HW2: Problem 2.15 ) Texas Tech University, Department of Mathematics & Statistics 1

2 b. We can see with lugui(golub(n)) that with diagonal pivoting (i.e., always using the diagonal element as the pivot, therefore, no pivoting) the pivots all turn out to be one. This makes the resulting L and U factors both have all ones on the diagonal. This is unusual behavior since we might expect large pivots with such a badly conditioned matrix. But note that bad pivots implies badly conditioner, but badly conditioned may or may not mean bad pivots. c. The determinant of a triangular matrix is equal to the product of its diagonal elements. For golub matrices, this means that det(l) = det(u) = 1 since they are triangular matrices with ones on the diagonal. And since the determinant of a product of matrices is equal to the product of the determinants, det(golub(n)) = 1. 2

3 Problem 2.19 The given system of equations in matrix form is: 2 1 x x x = x n x n n = 100; % (a) a = -1*ones(n-1,1); d = 2*ones(n,1); A = diag(a,-1) + diag(d) + diag(a,1); rhs = (1:n) ; % solve with lutx: [L,U,p] = lutx(a); % i.e., LUx = b(p) % so to solve we want x = U^{-1} L^{-1} b(p): x1 = U \ (L \ rhs(p)); % solve with bslashtx x2 = bslashtx(a,rhs); n 1 n % (b) % one simple form of the call to spdiags needs the vector a and b to be the % same length. So we will append an nth value -1 to the vector a: c = a; c(n) = -1; A = spdiags([c d c], [-1 0 1], n, n); x3 = A\rhs; % (c) x4 = tridisolve(a,d,a,rhs); % (d) cnum = condest(a); disp(sprintf( condest(a) = %e, cnum)); condest(a) = e+03 3

4 Problem 3.3 x = [ ] ; y = [ ] ; u = (-1:0.01:1) ; v1 = piecelin(x,y,u); v2 = polyinterp(x,y,u); v3 = splinetx(x,y,u); v4 = pchiptx(x,y,u); % plot the data points plot(x,y, o ) % plot the 4 interpolations on the same graph hold on plot(u,v1) plot(u,v2, r ) plot(u,v3, g ) plot(u,v4, k ) % label the four plots and label the graph legend( (x,y), piecelin, polyinterp, splinetx, pchiptx ) title( HW2, problem 3.3 ); xlabel( u ); ylabel( v = P(u) ); % evaluate each interpolant at x = -0.3 >> piecelin(x,y,-0.3) 4

5 ans = >> polyinterp(x,y,-0.3) ans = >> splinetx(x,y,-0.3) ans = >> pchiptx(x,y,-0.3) ans = I prefer the value at x = 0.3 given by pchiptx. The point x = 0.3 lies in the interval between x 3 = 0.65 and x 4 = 0.10, but the values of P ( 0.3) given by polyinterp and splinetx lie significantly below y 3 and y 4. In fact, the given y values of data are monotonically increasing with x. So we have no reason to suspect that the true function actually oscillates as polyinterp and splinetx do. That leaves pchiptx and piecewise linear, which give very similar answers for P ( 0.3). Since pchiptx also produces a smooth function, I prefer pchiptx for this data. To estimate the coefficients of the low-degree polynomial used to generate the date, we can try using a Vandermonde matrix. In general, it is a bad idea to compute using the Vandermonde matrix. But since cond(vander(x)) isn t too large ( 222), we will try it. format long c = vander(x)\y c = This is an inaccurate computation, so we will guess the actual coefficients to be these coefficients rounded to the nearest integer. That gives the polynomial: y = 16x 5 20x 3 + 5x 5

6 which appears again in problem 3.8 as the power form of the Chebyshev polynomial of degree five. We can check if our guess is right by evaluating this polynomial at the 6 given data points: testy = 16*x.^5-20*x.^3 + 5*x testy = which equals our given y data values to the four digits of accuracy given. Problem 3.4 The pchiptx interpolation seems to make a smoother, nicer looking, version of a hand. The splinetx version has some unrealistic looking sharp turns. For the same reasons, I d would think that the hand picture in the book was done with pchiptx. 6

7 Problem 3.7 Suppose we have two polynomials P (x) and Q(x) of degree less than n that agree at n distinct points. Let and P (x) = c 1 x n 1 + c 2 x n c n 1 x + c n Q(x) = a 1 x n 1 + a 2 x n a n 1 x + a n, where the coefficients c k and a k are not yet known. Then, if P and Q agree at n distinct points x k, k = 1,... n, we have n separate equations and P (x k ) = c 1 x n 1 k Q(x k ) = a 1 x n 1 k + c 2 x n 2 k + a 2 x n 2 k + + c n 1 x k + c n + + a n 1 x k + a n for k = 1,..., n. Since P (x k ) = Q(x k ), if we subtract each Q equation from each P equation, we get n equations (c 1 a 1 )x n 1 k + (c 2 a 2 )x n 2 k + + (c n 1 a n 1 )x k + (c n a n ) = 0. We can set up a Vandermonde matrix system to solve for the n unknown coefficients (c k a k ). (Vandermonde matrices are often not a good idea for computation, but they are very handy for analyzing properties of polynomials.) x n 1 1 x n 2 x n 1 2 x n 2. x n 1 n 1 x 2 1 x 1 1 c 1 a x 2 2 x 2 1 c 2 a = 0. x n 2 n x 2 n x n 1 c n a n 0 Vandermonde matrices are nonsingular for distinct x k values. Therefore, the only possible solution to this system is c k a k = 0 for each k = 1,..., n. That is, P (x) = Q(x). (There are other possible proofs; this is just one.) Problem 3.8 See prob3 8.m for the Matlab code. All six representations are plotted here. We can see that they represent the same function since the plots are identical. 7

8 Problem 3.13 The only thing we are changing in this problem is which boundary conditions are used in forming d 1 and d n in pchip and spline. perpchip In pchiptx.m, the internal d values are determined with a particular formula: If δ k 1 and δ k have opposite signs (i.e., we are at a local max or min): d k = 0 Otherwise, we use the formula: w 1 = 2h k + h k 1 w 2 = h k + 2h k 1 d k = w 1+w 2 w 1 + w 2 δ k 1 δ k We now use this same formula for d 1 and d n, taking advantage of the periodicity assumptions δ k = δ k+n 1, etc. That is, for d 1, Check if δ 0 and δ 1 have opposite signs. We don t actually have δ 0, but periodicity tells us that δ 0 = δ n 1, which we do have. So we check if δ n 1 and δ 1 have opposite signs. If they do, then: d 1 = 0 8

9 and by periodicity, we also set d n = d 1 Otherwise, we use the formula (using k = 1): w 1 = 2h 1 + h 0 w 2 = h 1 + 2h 0 d 1 = w 1+w 2 w 1 + w 2 δ 0 δ 1 where again by periodicity, this becomes: w 1 = 2h 1 + h n 1 w 2 = h 1 + 2h n 1 d 1 = w 1+w 2 w 1 + w 2 δ n 1 δ 1 See perpchip vh.m for a Matlab implementation. perspline In splinetx.m, the internal d values are determined with the formula: h k d k 1 + 2(h k 1 + h k )d k + h k 1 d k+1 = 3(h k δ k 1 + h k 1 δ k ). This is a formula for the kth row of a matrix system Ad = r which we then solve for the internal d values. For periodic boundary conditions, we use this same formula (plus our periodicity assumptions) for d 1 and d 1. In particular, for k = 1 h 1 d 0 + 2(h 0 + h 1 )d 1 + h 0 d 2 = 3(h 1 δ 0 + h 0 δ 1 ). We again use our periodicity assumptions to replace the values (such as h 0 and δ 0 ) that we do not have. This gives us: h 1 d n 1 + 2(h n 1 + h 1 )d 1 + h n 1 d 2 = 3(h 1 δ n 1 + h n 1 δ 1 ). We can do the same process with the nth equation giving: h n d n 1 + 2(h n 1 + h n )d n + h n 1 d 2 = 3(h n δ n 1 + h n 1 δ n ). where we have also used the periodicity assumption to see that d n+1 = d 2. See perspline vh.m for a Matlab implementation of these BCs. x = 0:pi/4:2*pi; x = x ; % using column vectors y = cos(x); 9

10 u = 0:pi/50:2*pi; u = u ; v1 = perpchip_vh(x,y,u); v2 = perspline_vh(x,y,u); plot(x,y, o ) hold on plot(u,v1) plot(u,v2, r ) legend( data, perpchip\_vh, perspline\_vh ) title( HW2, problem 3.13 ) xlabel( u ) ylabel( v = P(u) ) In the following two figures from interp2gui we can see that the nonperiodic versions of pchip and spline have more trouble with overshooting when used for 2d closed-curve interpolation. 10

11 Problem 3.18 The Vandermonde matrix for the given t values is extremely badly conditioned: t = 1900:10:2000; V = vander(t); cond(v) 11

12 ans = e+48 Yikes. This is not a matrix we would want to use in a linear solve. We would expect the answer (from backslash or any other solver) to be all error. We can get a much better behaved polynomial with centering and scaling. For example, the next two figures show a cubic polynomial fit (not interpolating) to the original data, then the polynomial we get after centering and scaling. 12

13 We see even more extreme differences if we look at the degree 10 polynomials without, then with, centering and scaling: Matlab s centering and scaling is using the change of variable described in (c). To get the coefficients of Q(s) in terms of those of P (t), let s look at a smaller degree problem (say a degree 3 polynomial). P (t) = c 1 t 3 + c 2 t 2 + c 3 t + c 4 13

14 and Q(s) = a 1 t 3 + a 2 t 2 + a 3 t + a 4 = a 1 ( t µ σ )3 + a 2 ( t µ σ )2 + a 3 ( t µ σ ) + a 4 = ( a 1 σ 3 )t3 + ( 3a 1µ σ 3 + a 2 σ 2 )t2 + ( 3a 1µ 2 σ 3 2a 2µ σ 2 + a 3 σ )t + ( a 1µ 3 σ 3 + a 2µ 2 σ 2 a 3µ σ + a 4) giving us c 1 = a 1 σ 3 c 2 = 3a 1µ σ 3 + a 2 σ 2 c 3 = 3a 1µ 2 σ 3 2a 2µ σ 2 + a 3 σ c 4 = a 1µ 3 σ 3 + a 2µ 2 σ 2 a 3µ σ + a 4 We could do the same thing with degree 10. That would be a job for Mathematica or the Matlab symbolic toolbox (I don t have either). The transformation to s = t µ σ makes a substantial difference in the condition number of the Vandermonde matrix: t = 1900:10:2000; mu = mean(t) mu = 1950 sigma = std(t) sigma = s = (t - mu)/sigma; cond(vander(t)) ans = e+48 cond(vander(s)) ans = e+04 This is much more promising condition number. It is not necessarily the best we can do. If we take other values around this σ value, we can get a lower condition number. For example if we let sigtest take on values around σ we can get a smaller condition number. 14

15 sigest = 10:1:50; cc = []; for k = 1:41 cc(k) = cond(vander((t - mu)/sigest(k))); end min(cc) e+04 which is a smaller condition number than we got with the given σ and µ. But they are reasonable choices and lead to a much better conditioned Vandermonde matrix. 15