среда, 29 июня 2011 г.

Piecewise Chebyshev approximation, Part 4: Clarifying conditions

Some posts ago I've said that for optimality of piecewise approximation we must have equal deviations at each interval. Now is the time to clarify this condition. Let's look at the following problem: we have  defined at . The has the minimum point at . Now let's try to approximate it using two intervals. Deviations will be equal only if minimum point is deleted, and deviations will be 0.5 each. Otherwise, one deviation will be 0.5, but another deviation will be 0.4(9). So we must change our previous condition  to , where .

Some notes about previous problem with one extremum. There must be a condition I called "big extrema condition". Formulation, given in previos post is applicable only is extremum is "pretty big". It means that  and doesn't change it's sign at least at and . If this "big extrema condition" is not satisfied, depending on the situation , or both points doensn't exist.

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