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### Hw7Solution08f

Course: M 426, Fall 2009
School: Delaware
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Word Count: 881

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426/CISC Math 410 08F, All Sections R.J. Braun Homework 7 Solutions, Hints and Answers Problem 3.2.1. Here we use the normal equation approach. AT A = 8 -6 -6 6 , AT b = -10 6 . Solving AT Ax = AT b gives x = [-2 - 1]T . 3.2.2. We can compute all the quantities we need to get the pseudoinverse, even if there is only one column in A. Here AT A = 12 + (-2)2 + 32 = 14. Then A+ = (AT A)-1 AT = 1 [1 - 2 3] . 14...

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426/CISC Math 410 08F, All Sections R.J. Braun Homework 7 Solutions, Hints and Answers Problem 3.2.1. Here we use the normal equation approach. AT A = 8 -6 -6 6 , AT b = -10 6 . Solving AT Ax = AT b gives x = [-2 - 1]T . 3.2.2. We can compute all the quantities we need to get the pseudoinverse, even if there is only one column in A. Here AT A = 12 + (-2)2 + 32 = 14. Then A+ = (AT A)-1 AT = 1 [1 - 2 3] . 14 3.2.9. Let P and Q be orthogonal matrices; then for either of them, the inverse is the same as the transpose. If the product P Q is orthogonal then (P Q)T (P Q) = (P Q)(P Q)T = I. Let's try it. (P Q)T (P Q) = QT P T P Q = QT IQ = QT Q = I; we can do this if both P and Q are orthogonal. Also (P Q)(P Q)T = P QQT P T = P IP T = P P T = I. Because the order of the products doesn't matter and we get the identity matrix as a result, the product P Q of orthogonal matrices is still an orthogonal matrix. 3.2.10. (a) For an orthogonal matrix, Q-1 = QT ; that is, QQT = QT Q = I. Let P = QT , and now try the same approach as in the last problem. P P T = QT (QT )T = QT Q = I, and P T P = (QT )T QT = QQT = I; thus, QT is also an orthogonal matrix. (b) Now 2 (Q) = ||Q-1 ||2 ||Q||2 = ||QT ||2 ||Q||2 . Recall that ||Q|| = max||x||2 =1 ||Qx||2 is the definition of the 2-norm of the matrix. Then for an orthogonal matrix, ||Qx||2 = (Qx)T (Qx) = xT QT Qx = xT Ix = xT x = ||x||2 . 2 2 Because in the definition of the norm we must have ||x||2 = 1, then ||Q||2 = 1. Now ||QT ||2 = 1 because QT is also an orthogonal matrix. Then 2 (Q) = (1)(1) = 1. 3.3.4. The QR factorization can be written as follows. ^ A = QR = Q Q0 ^ R 0 . 1 ^ ^ Note that A is m-by-n, Q is m-by-n and R0 is n-by-n; the others fill out the dimensions with Q0 being m-by-(m - n) and the zero matrix is (m - n)-by-n. Because of that zero matrix, ^^ we can reconstruct A with the compressed QR facorization: A = QR. Now use this in the formula for the pseudoinverse. A+ = (AT A)-1 AT ^^ ^^ ^^ = [(QR)T QR]-1 [QR]T , ^ ^ ^^ ^ ^ = [RT QT QR]-1 RT QT , ^ ^ ^ ^ = [RT R]-1 RT QT , ^ ^ ^ ^ = R-1 [RT ]-1 RT QT , ^ ^ ^ ^ = R-1 I QT = R-1 QT . 3.3.6. (b) Use the definition of the norm. ||A||2 = = = = ||x||2 =1 ||x||2 =1 ||x||2 =1 ||x||2 =1 max ||Ax||2 max ||QRx||2 max ||Q(Rx)||2 max ||Rx||2 = ||R||2 . (a) Count the number of flops in the qrfact function; use m = n. Note that there are no flops in lines 9 to 13, but there is important there: information the sizes of everything are established. For example, v and z are n - k + 1 long. Let s = n - k + 1 for convenience. Line 14: There are 3 flops before the semicolon and there are s - 1 flops (multiplication by -1 after it, for a total of s - 1 + 3 = s + 2. Line 15: This line is non-trivial to count! Start with the parentheses. The last term is v'*v, which is a dot product; the result is a scalar that takes 2s-1 flops. The other term in parentheses is v*v', which is an outer product giving an s-by-s matrix as output; this matrix takes s2 flops. (You have to write out the matrix to see this.) Now the division sign takes s2 flops, as does the multiplication by 2, and the substraction from the s-by-s identity matrix. Thus the total flops count is 4s2 + 2s - 1 for this line. Line 16: This is multiplication of two matrices of size s-by-s; there are s2 locations and each one takes a dot product requiring 2s - 1 flops, so there are s2 (2s - 1) flops required here. Line 17: Same as line 16, so s2 (2s - 1) flops here too. That's it for the flops. Now we need to total them up, for each k we have (s + 2) + (4s2 + 2s - 1) + [s2 (2s - 1)] + [s2 (2s - 1)] = 4s3 - 2s2 + 3s + 1, s = n - k + 1. 2 Now summing over the loop gives n n 4s3 - 2s2 + 3s - 3 = k=1 k=1 n 4(n - k + 1)3 - 2(n - k + 1)2 + 3(n - k + 1) + 1, 4n3 - 4k 3 + O(n2 ) + O(k 2 ), k=1 n n = 4 n3 - k=1 k=1 k3 , 4(n4 - n4 /4), = 3n4 . Thus,...

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Delaware - M - 426
Math 426/CISC 410 08F, All Sections R.J. Braun Homework 5 Solutions, Hints and AnswersProblem 2.4.1(d). Take limits. 1 n+ n = lim 1 + = 1; lim n n n n this implies that n + n n. Likewise, n+2 n 2 lim = lim 1 + = 1; n n n n this implies that n
Delaware - M - 426
Math 426/CISC 410 08F, All sections R.J. Braun Homework 4 Solutions, Hints and AnswersProblem 1.6.2. Note that log is denoting the natural log here. Here is one way to solve this problem. See the web page for a published version of this problem. % S
Delaware - M - 428
UNIVERSITY OF DELAWARE Mathematical Sciences Department Math 428/Cisc 411 Algorithmic &amp; Numerical Solution of Differential Equations Spring 2008 Honors Sections Instructor: Dr. Richard J. Braun, Ewing 509, (302) 831-1869, braun@math.udel.edu. Text:
Delaware - M - 426
Beginning Matlab ExercisesR. J. Braun Department of Mathematical Sciences University of Delaware Version of September 5, 20081IntroductionThis collection of exercises is intended to help you start learning Matlab. Matlab is a huge package with
Delaware - M - 426
University of Delaware Department of Mathematical Sciences Math 426/CISC 410 Intro to Numerical Analysis and Algorithmic Computing 08F R. J. Braun Your name: Project 3: SVD Image Compression In class, we used the SVD to compress an image. We treated
Delaware - M - 428
University of Delaware Department of Mathematical Sciences Math 428 Algorithmic and Numerical Solution of Differential Equations 08S R. J. Braun Project 2: FitzHugh-Nagumo Equations Due date: 5/1/08, 3:30 pm. YOUR NAME: In this project you will use M
Delaware - M - 426
Delaware - M - 428
Math 428 Numerical and Algorithmic Solution of Dierential Equations 08S R. J. Braun Department of Mathematical Sciences University of Delaware Honors Project 1. Due 4/17/08, 3:30pm. In this project you will reproduce some of the eorts to compute 10 d
Delaware - M - 428
UNIVERSITY OF DELAWARE Mathematical Sciences Department Math 428/Cisc 411 Algorithmic &amp; Numerical Solution of Differential Equations Spring 2008 Instructor: Dr. Richard J. Braun, Ewing 509, (302) 831-1869, braun@math.udel.edu. Text: &quot;Numerical Analys
Delaware - M - 426
Math 426/CISC 410 08F, All Sections R.J. Braun Homework 9 Solutions, Hints and AnswersProblem 4.3.1. The ratios below lead one to the hypothesis that = |1 /2 | where 1 is the smallest eigenvalue and 2 is the next largest. I used the absolute value
Delaware - M - 426
Math 426/CISC 410 08F, All Sections R.J. Braun Homework 6 Solutions, Hints and AnswersProblem 2.7.1. We assume that the relative change in b is eps; that is, |b|/|b| =eps. This script will do the job. % Script: Prob2_7_1.m % Problem 2.7.1: verify re
Delaware - M - 426
Math 426/CISC 410 08F, All Sections R.J. Braun Homework 10 Solutions, Hints and AnswersProblem 4.6.2. (a) Assume that A is n-by-n and nonsingular. The SVD is A = U SV ; here S is a diagonal matrix with the singular values on the diagonal, and U and
Delaware - M - 353
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Delaware - M - 428
Math 428 Numerical and Algorithmic Solution of Dierential Equations 08S Project 4. Due Friday, 5/16/05, 4:30pm. R. J. Braun Your name: In this project you will examine the numerical solution of the boundary value problem u + eu = 0, u(0) = u(1) = 0,
Delaware - M - 353
University of Delaware Department of Mathematical Sciences Math 353 Engineering Mathematics III 05F R. J. Braun Your name: Due date: 11/30/05, 5pm. Project 2: Solving a Nonlinear ODE IVP In this project you will study the solution of a nonlinear seco
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University of Delaware Department of Mathematical Sciences Math 353 Engineering Mathematics III 05F R. J. Braun Homework 8: Due Tuesday, 10/11/05, 4:00pm1. Read 3.3-5. Reading 3.1,2 is optional; do so if you're not sure about all this matrix-vector
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University of Delaware Department of Mathematical Sciences Math 353 Engineering Mathematics III 05F R. J. Braun Homework 7: Due Monday, 10/3/05, 5:00pm 1. Read 2.2, 2.3, 2.4. 2. 2.2, Problems 2.2.3(a,d), 2.2.11. 3. Apply the method of bisection 3 tim
Delaware - M - 426
qriter_testThe approximation to the smallest eigenvalue:ans = -0.059675578871905 1.564191203698703 1.987691412628296 2.349297531449158 3.409078620563635 -1.263166795003825 0.386479838574264 0.760280531365414 0.8920698056836
Delaware - M - 426
type Hw1_07f.m% Script for Hw 1, 07f. % Problem 3; display the output for second part of 11.2 of the Beginning% Matlab Problemsg = [1 2 3 4; 5 6 7 8; 9 10 11 12]h = [3 3 4 4; 5 5 6 6; 7 7 8 8]h &gt;= gg = hbigger = (g &gt;= h)g(bigger)g([0 0 0
Delaware - M - 426
type TestESeq% Script TestESeq.m% try approximation for exp(1) approx = zeros(1,15);format short e;for k = 1:15 approx(k) = (1+10^(-k)^(10^k);endabserror = abs(exp(1)-approx)%% make a nice table, like that from the basic matlab problem
Delaware - MATH - 611
A=[2 1 1 0; 4 3 3 1; 8 7 9 5; 6 7 9 8 ][L,U] = lufact(A);LUL*U - Ab = [1;2;3;4]y = L\bx = U\yA\bA = [ 1e-12 1; 1 1A = [ 1e-12 1; 1 1 ]norm(A)cond(A)
Delaware - MATH - 611
t = linspace(0,2*pi,200);A=[1 3;4 2]/4;for i=1:200, v=A*[cos(t(i);sin(t(i)]; plot(v(1),v(2),'o'), hold on, endginput(1)norm(ans)norm(A)
Delaware - MATH - 611
A = magic(8)A=A+A';eig(A)lam = poweriter(A);plot(lam,'-o')plot(520-lam,'-o')shgsemilogy(abs(520-lam),'-o')shg[x,y]=ginput(2)diff(log(y) / diff(x)exp(ans)eigeig(A)(ans(end-1)/ans(end)^2eig(A)lam=inviter(A,160);plot(lam,'-o')shglam=i
Delaware - MATH - 611
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Delaware - MATH - 611
m=5;A = toeplitz([1;-ones(m-1,1)],[1 zeros(1,m-1)]); A(:,m)=1[L,U]=lu(A)m=50;A=[2 1 1 0; 4 3 3 1; 8 7 9 5; 6 7 9 8 ];A = toeplitz([1;-ones(m-1,1)],[1 zeros(1,m-1)]); A(:,m)=1;x = rand(50,1);b = A*x;[L,U]=lu(A);x1 = U\(L\b);norm(x-x1)/norm(x
Delaware - MATH - 611
A = magic(5)A = A(:,1:3)[Q,R] = gs(A)Q'*Qans-eye(3)Q*R - A[Q,R]=qr(A,0)[Q,R]=qr(A)N = 40;N = 400;x = linspace(-1,1,N+1)';A = [ x.^0 x.^1 x.^2 x.^3 x.^4 ];[Q,R]=qr(A,0);plot(x,Q)
Delaware - MATH - 611
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