Piecewise Chebyshev approximation, Part 3: Moving along with analytical solution

Let's go further with our piecewise approximation problem. Now we'll take a function that has only one extremum at and is differentiable twice in the point of extremum. Continuality conditions will be discussed later. We couldn't find really analytical solution for this case.


We can divide into three intervals: two of them containing monotonic parts of function -- ,  and one containing extremum -- . Also, I must say that interval with extremum is containing only one or two (is this case equal up to some epsilon) parts of approximation partition. Appoximating at  by parabola (using Taylor series, deviations must be pretty small) we can find and . Let's assume that is extrema point, so , . Summing all of above we can write a following formulation: , where and are defined above. Look at  expression. We can't really say if it will be 1 or 2, so this relation is nondeterministic, so we cannot simply invert it to get relation (also we can't do this analytically because of occurences of in formulation above). So we only have nondeterministic solution for inverse problem.
Two words about continuality: must be continuos at least at interval.