Chapter 4 No. 4.0 Answer True or False to the following. Give reasons for your answers.

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1 MATH 434/534 Theoretical Assignment 3 Solution Chapter 4 No 40 Answer True or False to the following Give reasons for your answers If a backward stable algorithm is applied to a computational problem, the solution will be accurate False See Conditioning, Stability, and Accuracy on page 60 When a (backward) stable algorithm is applied to a well-conditioned problem, the computed solution should be near the exact solution However, if a (backward) stable algorithm is applied to an illcondition problem, accuracy is not guaranteed (b) A backward stable algorithm produces a good approximation to an exact solution False The same reason as the previous problem (c) Well-conditioning is a good property of an algorithm False The conditioning of a problem is a property of the problem itself It reals with how the solution of the problem will change when the input data is pertubed (d) Cancellation is always bad False (e) If the zeros of a polynomial are all distinct, then they must be well-conditioned False Consider the Wilkinson polynomial of degree 20, p(x) = (x )(x 2) (x 20) = x 20 20x 9 + The roots of p(x) are, 2,, 20 However, the Wilkinson polynomial is ill-conditioned Read The Wilkinson Polynomial on page 59 (f) An efficient algorithm is necessarily a stable algorithm False The efficiency of an algorithm depends on the amount of computer time consumed in its implementation Read Efficiency of an Algorithm on page 52 (g) Backward erros relates the errors to the data of the problem False The conditioning of a problem relates the errors to the data of the problems (h) A backward stable algorithm applied to a well-condtioned problem produces an accurate solution True See Conditioning, Stability, and Accuracy on page 60 (i) Stability analysis of an algorithm is performed by means of pertubation analysis False A pertubation analysis is a study that one can detect whether or not a given problem is good when small perturbations in the data will cause a large or small change in the solution (j) A symmetric matrix must be well-conditioned

2 False Consider the Hilbert martix: A = n 3 n 4 n+ n= 2n For n = 0, Cond 2 (A) = ; Cond (A) = ; Cond (A) = Therefore, A is ill-conditioned (k) If the determine of a matrix A is small, then it must be close to a singular matrix True In this case close to singular means that a matrix is theoreically nonsingular, but a small perturbation will make the matrix singular For example, consider a 0 0 diagonal matrix all of whose entries are equal to 000 Take a diagonal matrix of order 0 with all the diagonal entries equal to 000 Its determinant is : 0 30, very small Bt it is perfectly a nonsingular matrix, its condition number is (l) One must perform a large amount of computation to obtain a large round-off error?? No 4 Show that the floating point computations of the sum, product, and division of two numbers are backward stable Let x and y be floating point numbers First consider the case of the sum of x and y fl(x + y) = (x + y)( + δ) = x( + δ) + y( + δ) = x + y Since δ µ, both x and y are close to x and y, respectively Therefore, the sum of two floating point numbers is backward stable Next, we look at the case of the product of x and y In a similar way, we can see that fl(xy) = (xy)( + δ) = (x + δ)(y + δ) = x y Similarly, since δ µ, x and y are close to x and y Therefore, the product of two floating point number is backward stable Finally, we look at the case of the division of x and y where y 0 fl(x/y) = (x/y)( + δ) = x( + δ) y = x y Again, since δ µ, x and y are close to x and y Therefore, the division of two floating point number is backward stable No 42 Are the following floating point computaitons backward stable? Give reasons for your answer 2

3 in each case fl(x + ) (b) fl(x(y + z)) From No4-, it is backward stable (b) Since the inner product of two vectors, x T y = x y + x 2 y x n y n and the computed innder product of two vectors fl(x T y = x y + x 2 y x n y n ) is backward stable from the example 48 on page 55 Taking these two vectors x = (x, x) and y = (y, z), we have fl(xy + xz) = fl(x(y + z)) Therefore, fl(x(y + z)) is backward stable Let x, y and z be floating point numbers Then consider the following floating point computation: fl(x(y + z)) = {x fl(y + z)} ( + δ ) = x(y + z)( + δ 2 )( + δ ) = x(y + z)( + δ δ2 + δ + δ2) x(y + z)( + δ 3 ) where δ 3 = δ + δ 2 Since δ and δ 2 are small, δ δ 2 is neglected No 43 Show that the roots of the following polynomials are ill-conditioned and give reasons for your answers x 3 3x 2 + 3x (c) (x )(x 099)(x 2) Consider the problem of solving x 3 3x 2 + 3x = 0 The solutions of this equation are x =,, Let p(x) = x 3 3x 2 +3x Now assume that the coefficient of the second term is perturbed by 0 5 Let the pertubed polynomial be f(x) = x x 2 +3x+ The roots of this perturbed polynomial f(x) = x x 2 +3x+ are x = , x 2 = i and x 3 = i The relative errors in x, x 2 and x 3 are , and , respectively The relative error in the data is e 006 Therefore, a small perturbation in the data causes the large error between the roots of p(x) and the roots of the perturbed polynomial f(x) (b) Let p(x) = (x )(x 099)(x 2) = x 3 399x x 98 The roots of p(x) are x =, x = 099, and x = 2 Now perturbing the coefficient of the second term is pertubed by 0 5, then let f(x) = x 3 ( )x x 98 The roots of this perturbed polynomial of f(x) are x = , x 2 = and x 3 = The relative errors in x, x 2 and x 3 are , and , respectively The relative error in the data is e 006 Therefore, a small perturbation in the data causes the large error between the roots of p(x) and the roots of the perturbed polynomial f(x) 3

4 No 44 Work out the flop-counts for the following simple matrix operations (i) Multiplicaiton of matrices A and B of orders n m and m p, respectively (vi) Computation of the matrix A = uvt, where u and v are m column vectors u T v (vii) Computation of the matrix B = A uv T, where A and B are two n n matrices and u and v are two column vectors (i) Let A i be the ith row vector of a matrix A such that A i = (a i, a i2,, a im ), and B i be the ith column vector of a matrix B such that B i = (b i, b 2i,, b mi ) T A n m B m p = = A A 2 A n ( ) B B 2 B p A B A B 2 A B p A 2 B A 2 B 2 A 2 B p A n B A n B 2 A n B p Let C = AB Then each entry of C can be expressed as C ij = A i B j C ij = A i B j = (a i, a i2,, a im ) To compute C ij, multiplications and additions are carried out m times and m, respectively So, the flop-count to compute C ij is 2m Since the size of C is n p, that is, there are n p entries, the total flop-counts for the multiplication of A n m and B ( m p) is np(2m ) (vi) First consider the flop-count for computing the outer product uv T uv T = u v u v 2 u v m u 2 v u 2 v 2 u 2 v m b j b 2j b mj u m v u m v 2 u m v m The flop-count for computibg uv T is m 2 Now consider the flop-count for computing the inner product u T v u u T u 2 ( ) v = v v 2 v m u m = u v + u 2 v u m v m 4

5 So, the flop-count for computing the inner product u T v is 2m Since each entry of A = uvt u T v, A ij, is expressed as A ij = u T v u iv j (another m 2 flops-count), the total flop-counts are m 2 + 2m + m 2 = 2m 2 + 2m + (vii) From part (vi), the flop-count for computing uv T are n 2 Let a ij and b ij be an entry of A and B, respectivly Since we have b ij = a ij u i v j for each entry of B, that is, B = a u v a 2 u v 2 a n u v n a 2 u 2 v a 22 u 2 v 2 a 2n u 2 v n a n u n v a n2 u n v 2 a nn u n v n Therefore, the total flop-count is n 2 + n 2 = 2n 2 No 45 Develop an algorithm to compute the following matrix products Your algorithm should take advantage of the special structure of the matrices in each case Give flop-count and show storage requirement in each case A nad B are both lower triangular matrices (b) A is arbitrary and B is lower triangular Let A and B be both n n lower triangular matrices, that is, a 0 0 b 0 0 a 2 a 22 0 A = and B = b 2 b 22 0 a n a nn a nn b n b nn b nn Let a matrix C be the matrix such that C = AB Since the product of two lower triangular matrices is a triangular matrix, a matrix C is a lower triangular matrix Letting c ij is the entry of a matrix C The following is the algorithm for this computation: For i, j =, 2,, n If i < j, c ij = 0 (entries of the lower part of C) If i = j, c ij = a ii b ii (diagonal entries of C) If i > j, c ij = i k=j a ik b kij (diagonal entries of C) The flop-count for computations of the nth row of C is n k= (2k ) = n 2 The flopcount for computations of the (n )th row of C is n k= (2k ) = (n )2 The flopcount for computations of the (n 2)th row of C is n 2 k= (2k ) = (n 2)2, and so on On this computation the flop-count for computations of the (n l)th row of C is n l k= (2k ) = (n l)2 where l n Therefore, the total flop-count is n l= (n l) 2 = n(n + )(2n + ) 6 5

6 (b) Let A be an n n arbitrary matrixand and B be n n lower triangular matrix such that b 0 0 b 2 b 22 0 A = c and B = b n b nn b nn Let a matrix C be the matrix such that C = AB Letting c ij is the entry of a matrix C The following is the algorithm for this computation: For i, j =, 2,, n c ij = n k=j a ik b kj Since each entry of the first column of C takes 2n flops, the flop-count for computations of the st column of C is n (2n ) = n(2n ) Each entry of the second column takes 2n 3 flops Thus, the flop-count for computations of the second column of C is n (2n 3) = n(2n 3) The each entry of the third column of C takes 2n 5 flops Similarly we can find the flop-count for computations of the third column, that is, n (2n 5) = n(2n 5) On this computation the flop-count for computations of the lth column of C is n(2n 2l + ) Therefore, the total flop-count is n n n(2n 2l + ) = (2n 2 2nl + n) = n 3 l= l= No 46 A square matrix A = (a ij ) is said to be a band matrix of bandwidth 2k + if a ij = 0 whenever i j > k Develop an algorithm to compute the product C = AB, where A is arbitrary and B is a band matrix of bandwidth 3, taking advantage of the structure of the matrix B Overwrite A with AB and give flop-count Let A and B be n n matrices Setting bandwidth 2k + = 3, we have k = By the definition of a band matrix, a matrix B is a tridiagonal such as b b b 2 b 22 b 23 0 B = 0 b 32 bn,n b n,n 0 0 b n,n b nn Now let c ij be an entry of a matrix C Since C = AB, we consider a a 2 a 3 a b b n a 2 a 22 a 23 a 2n b 2 b 22 b 23 0 C = AB = a 3 a 32 a 33 a 3n 0 b 32 bn,n b n,n a n a n2 a n3 a nn 0 0 b n,n b nn 6

7 First cosidner the frist column of C Since just only first two entries of B are nonzero, the first column of C can be expressed as 2 c i = a ik b k k= Similarly, the nth column of C is expressed as c in = n k=n a ik b kn For the column indecies j = 2, 3,, n, we have c ij = j+ k=j a ik b kj Therefore, the following is the algorithm for this computation: For i =, 2,, n For j = c in = n k=n a ik b kn For j = 2, 3,, n c ij = j+ k=j a ikb kj For j = n c in = n k=n a ik b kn Each element of first column and the nth column takee 3 flops, that is, the total flops to compute these two colums is 6n flops Since it takes 5 flops to compute the other entries, the total flops to compute entries which does not belonging to the first and the nth columns is 5n(n 2) Therefore, the total flop-count for computing C is 6n + 5n(n 2) = 5n 2 4n 7