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The finite form of Jensen's inequality is a special case of this result. Consider the real numbers x 1, …, x n ∈ I and let := + + + denote their arithmetic mean.Then (x 1, …, x n) majorizes the n-tuple (a, a, …, a), since the arithmetic mean of the i largest numbers of (x 1, …, x n) is at least as large as the arithmetic mean a of all the n numbers, for every i ∈ {1, …, n − 1}.
Bennett's inequality, an upper bound on the probability that the sum of independent random variables deviates from its expected value by more than any specified amount Bhatia–Davis inequality , an upper bound on the variance of any bounded probability distribution
When Helping Hurts uses the Bible and the Great Commission to state that the church's mission should be to help the poor and the desolate. Corbett and Fikkert state that the definition of poverty will change depending on who is defining it, with the poor defining it through the psychological and social scope while more wealthy churches emphasize the lack of material things or a geographical ...
For instance, to solve the inequality 4x < 2x + 1 ≤ 3x + 2, it is not possible to isolate x in any one part of the inequality through addition or subtraction. Instead, the inequalities must be solved independently, yielding x < 1 / 2 and x ≥ −1 respectively, which can be combined into the final solution −1 ≤ x < 1 / 2 .
In the general case, constraint problems can be much harder, and may not be expressible in some of these simpler systems. "Real life" examples include automated planning, [6] [7] lexical disambiguation, [8] [9] musicology, [10] product configuration [11] and resource allocation. [12] The existence of a solution to a CSP can be viewed as a ...
Following Antman (1983, p. 283), the definition of a variational inequality is the following one.. Given a Banach space, a subset of , and a functional : from to the dual space of the space , the variational inequality problem is the problem of solving for the variable belonging to the following inequality:
Reducing and re-arranging the coefficients by adding multiples of as necessary, we can assume < (in fact, this is the unique such satisfying the equation and inequalities). Similarly we take u , v {\displaystyle u,v} satisfying N − k = u a + v b {\displaystyle N-k=ua+vb} and 0 ≤ u < b {\displaystyle 0\leq u<b} .
Since all the inequalities are in the same form (all less-than or all greater-than), we can examine the coefficient signs for each variable. Eliminating x would yield 2*2 = 4 inequalities on the remaining variables, and so would eliminating y. Eliminating z would yield only 3*1 = 3 inequalities so we use that instead.