
Re: Computationally efficient method of assessing one measure of variation of a function
Posted:
Feb 23, 2013 10:37 PM


On Fri, 22 Feb 2013, pepstein5@gmail.com wrote:
> Let N be a positive integer. Let f be a function from the nonnegative > integers <= N to the reals. Let d > 0. What is a computationally > efficient way of finding the largest possible k such that there exists M > >=0, M + k <=N such that abs(f(x)  f(y)) <= d for all x, y such that x > and y are both >= M and <= M + k?
What's the ranges of the variables? Reals, integers, positive integers?
> I'm also interested in continuous analogies. For example, suppose f is > a continuous function defined on a closed interval. How do we find the > length of the longest interval I in the domain of f such that abs(f(x)  > f(y)) <= d whenever x and y both lie in I.
For continuous f:[a,b] > R and d > 0, you want to find the largest interval I subset [a,b] with the lenght of the interval f(I) <= d?
That is highly dependent upon how f varies. If f is constant, then I = [a,b].
If f(x) = x, then I = [a, a+d] if a+d <= b is one maximal interval among possibly uncountable many. If b < a + d, then I = [a,b].
If f(x) = rx, r > 0, then I = [a, a+d/r] if a+d/r <= b is one maximal interval among possibly uncountable many. If b < a + d/r, then I = [a,b].
In general, if len f([a,b]) <= d, [a,b] is the maximum interval. It seems to me that the maximal intervals would be associated where f" is the smallest.

