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Denote by ek = x - Xk the error and r= b - Ax the residual after the kth iteration of the Gauss-Seidel method. The operators rest and prol introduced above are linear, and so can be represented by matrices R ER(-1)×(2-1) and PE R (2-1)×(-1), respectively, i.e. rest(v) = Rv and prol(z) = Pz. Assuming ek = PRek, show that where AR = RAP = R (C-1)×(C-1) AR(Rek) = Rrk,

Denote by ek = x - Xk the error and r= b - Ax the residual after the kth iteration of the Gauss-Seidel method. The operators rest and prol introduced above are linear, and so can be represented by matrices R ER(-1)×(2-1) and PE R (2-1)×(-1), respectively, i.e. rest(v) = Rv and prol(z) = Pz. Assuming ek = PRek, show that where AR = RAP = R (C-1)×(C-1) AR(Rek) = Rrk,

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question 2.1 Denote by ek = X — X the error and r = b - Ax the residual after the kth iteration of the Gauss-Seidel method. The operators rest and prol introduced above are linear, and so can be represented by matrices R = R(²-1)×(27−1) and P = R (2€-1)×(€−1), respectively, i.e. rest(v) = Rv and prol(z) = Pz. Assuming ek = PRe, show that where AR = RAP = R(-1)×(C-1) AR(Rek) = Rrk,
Transcribed Image Text:Question 2.1 Denote by ek = X — X the error and r = b - Ax the residual after the kth iteration of the Gauss-Seidel method. The operators rest and prol introduced above are linear, and so can be represented by matrices R = R(²-1)×(27−1) and P = R (2€-1)×(€−1), respectively, i.e. rest(v) = Rv and prol(z) = Pz. Assuming ek = PRe, show that where AR = RAP = R(-1)×(C-1) AR(Rek) = Rrk,
In this question, we want to develop a version of iterative refinement (as seen in the Week 5 computer lab) for iterative methods for solving Ax = b. The main steps in iterative refinement for Gaussian elimination were as follows: 1. Solve Ax = b with Gaussian elimination, to obtain computed solution 2. Compute the residual r = b - AX 3. Solve Ae = r with Gaussian elimination, to obtain computed solution ê 4. Update = x + ê. In step 4, we are using the fact that the error e = x - in the computed solution satisfies Ae = Ax - Ax = b - Ax = r. The solve in step 3 can be done cheaply, in O(n²) cost, since we can re-use the LU factorisation that was already computed in step 1. So the additional cost of adding steps 2-4 is negligible, but they significantly increase the accuracy of the computed solution. For iterative methods, such as the Jacobi and Gauss-Seidel methods, both solves in steps 1 and 3 would have (²) cost, and so the "refinement" steps 2-4 would substantially increase the cost. To reduce the cost, we want to solve a smaller system in step 3. Let's see how that's done. For the remainder of this question, assume A E R (2-1)×(2-1), for some / EN. R-1 by For EN, define a restriction operator rest : R²-1 for all v € R2€-1. Similarly, define a prolongation operator prol: Re-1 for all z E R²-¹ (rest(v)), = = (v2-1 + 2v₂ + V2+1), Vi= 1,..., 1, R2-1 by (prol(z)); Z1/2 if i is even, (z(i-1)/2 + z(i+1)/2) if i is odd, i ‡ 1, i ‡ 2l − 1, (Z(i+1)/2) if i = 1, (Z(1-1)/2) if i = 2l - 1, Vi= 1,..., 2t - 1,
Transcribed Image Text:In this question, we want to develop a version of iterative refinement (as seen in the Week 5 computer lab) for iterative methods for solving Ax = b. The main steps in iterative refinement for Gaussian elimination were as follows: 1. Solve Ax = b with Gaussian elimination, to obtain computed solution 2. Compute the residual r = b - AX 3. Solve Ae = r with Gaussian elimination, to obtain computed solution ê 4. Update = x + ê. In step 4, we are using the fact that the error e = x - in the computed solution satisfies Ae = Ax - Ax = b - Ax = r. The solve in step 3 can be done cheaply, in O(n²) cost, since we can re-use the LU factorisation that was already computed in step 1. So the additional cost of adding steps 2-4 is negligible, but they significantly increase the accuracy of the computed solution. For iterative methods, such as the Jacobi and Gauss-Seidel methods, both solves in steps 1 and 3 would have (²) cost, and so the "refinement" steps 2-4 would substantially increase the cost. To reduce the cost, we want to solve a smaller system in step 3. Let's see how that's done. For the remainder of this question, assume A E R (2-1)×(2-1), for some / EN. R-1 by For EN, define a restriction operator rest : R²-1 for all v € R2€-1. Similarly, define a prolongation operator prol: Re-1 for all z E R²-¹ (rest(v)), = = (v2-1 + 2v₂ + V2+1), Vi= 1,..., 1, R2-1 by (prol(z)); Z1/2 if i is even, (z(i-1)/2 + z(i+1)/2) if i is odd, i ‡ 1, i ‡ 2l − 1, (Z(i+1)/2) if i = 1, (Z(1-1)/2) if i = 2l - 1, Vi= 1,..., 2t - 1,
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