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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]
Jensen's inequality generalizes the statement that a secant line of a convex function lies above its graph. Visualizing convexity and Jensen's inequality. In mathematics, Jensen's inequality, named after the Danish mathematician Johan Jensen, relates the value of a convex function of an integral to the integral of the convex function.
Several graph-theoretic concepts are related to each other via complementation: The complement of an edgeless graph is a complete graph and vice versa. Any induced subgraph of the complement graph of a graph G is the complement of the corresponding induced subgraph in G. An independent set in a graph is a clique in the complement graph and vice ...
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Occasionally, chained notation is used with inequalities in different directions, in which case the meaning is the logical conjunction of the inequalities between adjacent terms. For example, the defining condition of a zigzag poset is written as a 1 < a 2 > a 3 < a 4 > a 5 < a 6 > ... .
In mathematical analysis, the Minkowski inequality establishes that the L p spaces are normed vector spaces.Let be a measure space, let < and let and be elements of (). Then + is in (), and we have the triangle inequality
A pictorial representation of a simple linear program with two variables and six inequalities. The set of feasible solutions is depicted in yellow and forms a polygon , a 2-dimensional polytope . The optimum of the linear cost function is where the red line intersects the polygon.
In case there are several permutations with this property, let σ denote one with the highest number of integers from {, …,} satisfying = (). We will now prove by contradiction , that σ {\displaystyle \sigma } has to keep the order of y 1 , … , y n {\displaystyle y_{1},\ldots ,y_{n}} (then we are done with the upper bound in ( 1 ), because ...