Conditioning and Stability A problem is well conditioned if a small change in the input creates a small change in the output (solution). A computation is backward stable if it produces the exact solution to a nearby problem. There is room for debate in this deﬁnition, but it does succinctly capture the idea. I like better: A computation is backward stable if the result is close to the exact solution to a nearby problem. In any case, we say that a method is backward stable for a set of problems if it is backward stable for each problem in the set. If the problem we are trying to solve has a unique solution, then we can formulate it as “evaluate f ( x )”, where x represents the input and f ( x ) represents the solution (output). Let’s represent our input space by D and our output space by R . Then our computed result can be represented by ¯ f : D -→ R , where the exact result is represented by f : D -→ R . That is, ¯ f ( x ) is our computed approximation to f ( x ). A problem is
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