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I would appreciate a definition clarification.

if a numerical method is "unstable", does it mean that if we introduce a small random error in one of the steps, the error would be magnified greatly after further steps? is this true for all unstable algorithms or are there some where the random error is never made significant, say wrt the error of the method itself?

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    More or less, yes. A tiny error committed in the course of the algorithm's preceedings gets easily magnified as the algorithm marches on. But one must always keep separate the notion of an "ill-conditioned problem" and an "unstable algorithm".2011-12-31

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