Numerical Methods. Midterm 2. May 14, Please give details of your calculation. A direct answer without explanation is not counted.

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1 Numerical Methods Midterm 2 May 4, 208 Please give details of your calculation A direct answer without explanation is not counted Your answers must be in English You can bring notes and the textbook Other books or electronic devices are not allowed Please read problem statements carefully Problem (45%) Consider the following 2 2 matrix A = [ ] (a) (0%) Calculate Hint: Let A = max x = Ax Ax 2 = x A Ax and replace the constraint with cos 2 θ + sin 2 θ = (b) (0%) Calculate A = max x = A x Hint: Make sure your calculation of A is correct as that is going to be used in (d)-(e)

2 (c) (5%) What is the condition number of A? (d) (0%) Consider If b is slightly changed to b = A b + δb = [ ] [ ] 07, 07 then what is the new solution x + δx? What are x and (e) (0%) If A is slightly changed to Solution A + δa = A A δb b? [ ] 3 22, 22 3 then what is the new solution x + δx? What are x and A A δa A? (a) The optimization problem is [ ] [ ] A A = [ ] 3 2 = 2 3 max Ax 2 subject to x 2 + x 2 2 = Let x = cos θ and x 2 = sin θ Then max x, x 2 3x 2 24x x 2 + 3x 2 2 max θ max θ 3 24 sin θ cos θ 3 2 sin 2θ 2

3 The solution is sin 2θ = Thus A = 25 = 5 (b) The inverse matrix is A = = 5 [ [ ] ] We calculate [ 3 2 ] [ ] 3 2 (A ) A = 25 = 25 The optimization problem becomes Let [ ] max x, x 2 25 (3x2 24x x 2 + 3x 2 2) subject to x 2 + x 2 2 = x = cos θ and x 2 = sin θ Then The solution is max θ max θ (3 24 sin θ cos θ) 25 (3 2 sin 2θ) 25 sin 2θ = Thus A = 25 = 25 3

4 (c) From (a) and (b), the condition number of A is A A = 5 = 5 (d) Given [ ] [ ] 3 2 b = 2 3 [ ] =, [ ] 03 δb = 03 The new solution is x + δx = A (b + δb) [ ] [ ] = [ ] = 5 35 [ ] 07 =, 07 [ ] 03 δx = 03 We have x = ( 03)2 + ( 03) = 03 A A δb b = 5 = ( 03) 2 ( )

5 (e) The new solution is x + δx = (A + δa) b [ ] [ ] 3 22 = 22 3 = [ ] [ ] = [ ] [ ] 5/4 =, 5/4 [ ] /4 δx = /4 We have x = ( /4)2 + ( /4) = /4 Next, we solve Given We know that A A δa A A + δa = [ ] [ ] 3 22 δa = 22 3 [ ] 0 02 = 02 0 [ 3 ] To calculate δa, we follow the similar derivation as in problem (a) The optimization problem is max δax 2 subject to x 2 + x 2 2 = Let x = cos θ and x 2 = sin θ 5

6 Then max 004x x 2 2 = 004 x, x 2 Thus δa = 004 = 02 We have A A δa A = = /5 Here, we show the / x + δx value x + δx = ( /4) 2 + ( /4) 2 (5/4) 2 + (5/4) 2 = /5 Problem 2 (20%) (a) (0%) Consider the following definition Is it a norm or not? v = min i v i (b) (0%) Consider v = number of v i 0 Is it a norm or not? Solution The norm of a vector should satisfy (I) v 0 with equality iff v = 0 (II) αv = α v for any α (III) u + v u + v 6

7 (a) No Condition (I) does not holds Because For example, v = [ ] 0 v 0 st v = 0 (b) No Condition (II) does not holds Consider [ ] 0 α = 0, v = We have [ [ ] [ ] αv = 0 ] 0 = 0 = α v = = 0 Problem 3 (25%) Consider v R n, w R n that are stored as regular vectors We would like to construct the following rankone matrix vw and store it as a sparse matrix in the compressed column format Write a code to do this task Some requirements (a) Known zeros in the resulting matrix must not be stored (b) The complexity of the algorithm should be no more than the order of the number of non-zeros in the output matrix Hint: To achieve (b), you might need to first transform v to a sparse vector You are allowed to have extra O(n) arrays We assume each array starts with rather than 0 Solution The solution procedure (a) v is first transformed to a sparse array (b) For every non-zero w j, we store w j v to the output matrix 7

8 Algorithm : procedure ([APtr, AInd, AVal] = outerprod(v, w, n)) 2: k = ; 3: ind = ; 4: for i = : n do 5: if v(i) 0 then 6: vind(k) = i; 7: vval(k) = v(i); 8: k = k + ; 9: end if 0: end for : for j = : n do 2: APtr(j) = ind; 3: if w(j) 0 then 4: for i = : k- do 5: AVal(ind) = vval(i) * w(j); 6: AInd(ind) = vind(i); 7: ind = ind + ; 8: end for 9: end if 20: end for 2: APtr(n+) = ind; 22: end procedure Common mistake: >> v = [2 0 ]; 2 >> w = [3 0 2]; 3 >> outerprod (v,w,3) 4 5 APtr = AInd = AVal = Listing : Program output 8

9 Problem 4 (0%) Consider A = 2 2, b = In our lecture we show that the Jacobi method diverges by using 0 as the initial 0 solution What if we switch to use Gauss-Seidel method? Solution Using Gauss-Seidel method, in the first iteration, we have In the second iteration, we have x = (5 2(0) 2(0))/ = 5 x 2 = (5 2(5) 2(0))/ = 5 x 3 = (5 2(5) 2( 5))/ = 5 x = (5 2( 5) 2(5))/ = 5 x 2 = (5 2(5) 2(5))/ = 5 x 3 = (5 2(5) 2( 5))/ = 25 As we observe, the sequence of x, x 2, and x 3 still diverges 9

Numerical methods, midterm test I (2018/19 autumn, group A) Solutions

Numerical methods, midterm test I (2018/19 autumn, group A Solutions x Problem 1 (6p We are going to approximate the limit 3/2 x lim x 1 x 1 by substituting x = 099 into the fraction in the present form