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A facet of a polytope is the set of its points which satisfy an essential defining inequality of the polytope with equality. If the polytope is d-dimensional, then its facets are (d − 1)-dimensional. For any graph G, the facets of MP(G) are given by the following inequalities: [1]: 275–279 x ≥ 0 E
Kraft's inequality limits the lengths of codewords in a prefix code: if one takes an exponential of the length of each valid codeword, the resulting set of values must look like a probability mass function, that is, it must have total measure less than or equal to one.
In the context of metric measure spaces, the definition of a Poincaré inequality is slightly different.One definition is: a metric measure space supports a (q,p)-Poincare inequality for some , < if there are constants C and λ ≥ 1 so that for each ball B in the space, ‖ ‖ () ‖ ‖ ().
For graphs of constant arboricity, such as planar graphs (or in general graphs from any non-trivial minor-closed graph family), this algorithm takes O (m) time, which is optimal since it is linear in the size of the input. [18] If one desires only a single triangle, or an assurance that the graph is triangle-free, faster algorithms are possible.
Hardy's inequality is an inequality in mathematics, named after G. H. Hardy.. Its discrete version states that if ,,, … is a sequence of non-negative real numbers, then for every real number p > 1 one has
Two-dimensional linear inequalities are expressions in two variables of the form: + < +, where the inequalities may either be strict or not. The solution set of such an inequality can be graphically represented by a half-plane (all the points on one "side" of a fixed line) in the Euclidean plane. [2]
The bounds these inequalities give on a finite sample are less tight than those the Chebyshev inequality gives for a distribution. To illustrate this let the sample size N = 100 and let k = 3. Chebyshev's inequality states that at most approximately 11.11% of the distribution will lie at least three standard deviations away from the mean.
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}.