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The main objective of interval arithmetic is to provide a simple way of calculating upper and lower bounds of a function's range in one or more variables. These endpoints are not necessarily the true supremum or infimum of a range since the precise calculation of those values can be difficult or impossible; the bounds only need to contain the function's range as a subset.
13934 and other numbers x such that x ≥ 13934 would be an upper bound for S. The set S = {42} has 42 as both an upper bound and a lower bound; all other numbers are either an upper bound or a lower bound for that S. Every subset of the natural numbers has a lower bound since the natural numbers have a least element (0 or 1, depending on ...
There is a corresponding greatest-lower-bound property; an ordered set possesses the greatest-lower-bound property if and only if it also possesses the least-upper-bound property; the least-upper-bound of the set of lower bounds of a set is the greatest-lower-bound, and the greatest-lower-bound of the set of upper bounds of a set is the least ...
A lattice is an abstract structure studied in the mathematical subdisciplines of order theory and abstract algebra.It consists of a partially ordered set in which every pair of elements has a unique supremum (also called a least upper bound or join) and a unique infimum (also called a greatest lower bound or meet).
where i is the index of summation; a i is an indexed variable representing each term of the sum; m is the lower bound of summation, and n is the upper bound of summation. The "i = m" under the summation symbol means that the index i starts out equal to m. The index, i, is incremented by one for each successive term, stopping when i = n. [b]
More generally, one may define upper bound and least upper bound for any subset of a partially ordered set X, with “real number” replaced by “element of X ”. In this case, we say that X has the least-upper-bound property if every non-empty subset of X with an upper bound has a least upper bound in X.
The new solution (¯,) is used to update the lower bound. If the gap between the best upper and lower bound is less than then the procedure terminates and the value of ¯ is determined by solving the primal residual problem fixing ¯. Otherwise, the procedure continues on to the next iteration.
Following the rules of this construction, would have to be an upper bound of , contradicting property 2 of all sequences of nested intervals. In two steps, it has been shown that is an upper bound of and that a lower upper bound