Date: Feb 23, 2013 10:37 PM
Author: William Elliot
Subject: Re: Computationally efficient method of assessing one measure of<br> variation of a function
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.