# Lecture 6 - 6 Conditioning and Stability A computing...

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6 Conditioning and Stability A computing problem is well-posed if 1. a solution exists (e.g., we want to rule out situations that lead to division by zero), 2. the computed solution is unique , 3. the solution depends continuously on the data, i.e., a small change in the data should result in a small change in the answer. This phenomenon is referred to as stability of the problem. Example Consider the following three different recursion algorithms to compute x n = ( 1 3 ) n : 1. x 0 = 1, x n = 1 3 x n - 1 for n 1, 2. y 0 = 1, y 1 = 1 3 , y n +1 = 4 3 y n - 1 3 y n - 1 for n 1, 3. z 0 = 1, z 1 = 1 3 , z n +1 = 10 3 z n - z n - 1 for n 1. The validity of the latter two approaches can be proved by induction. We illustrate these algorithms with the Maple worksheet 477 577 stability.mws . Use of slightly perturbed initial values shows us that the first algorithm yields stable errors throughout. The second algorithm has stable errors, but unstable relative errors. And the third algorithm is unstable in either sense. 6.1 The Condition Number of a Matrix Consider solution of the linear system A x = b , with exact answer x and computed answer ˜ x . Thus, we expect an error e = x - ˜ x . Since x is not known to us in general we often judge the accuracy of the solution by looking at the residual r = b - A ˜ x = A x - A ˜ x = A e and hope that a small residual guarantees a small error. Example We consider A x = b with A = 1 . 01 0 . 99 0 . 99 1 . 01 , b = 2 2 , and exact solution x = [1 , 1] T . 1. (a) Let’s assume we computed a solution of ˜ x = [1 . 01 , 1 . 01] T . Then the error e = x - ˜ x = - 0 . 01 - 0 . 01 53

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is small, and the residual r = b - A x = 2 2 - 2 . 02 2 . 02 = - 0 . 02 - 0 . 02 is also small. Everything looks good. 2. (b) Now, let’s assume that we computed a solution of ˜ x = [2 , 0] T . This “solutions” is obviously not a good one. Its error is e = - 1 1 , which is quite large. However, the residual is r = 2 2 - 2 . 02 1 . 98 = - 0 . 02 0 . 02 , which is still small. This is not good. This shows that the residual is not a reliable
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