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Bernoulli's inequality can be proved for case 2, in which is a non-negative integer and , using mathematical induction in the following form: we prove the inequality for r ∈ { 0 , 1 } {\displaystyle r\in \{0,1\}} ,
Thus we can find a graph with at least e − cr(G) edges and n vertices with no crossings, and is thus a planar graph. But from Euler's formula we must then have e − cr(G) ≤ 3n, and the claim follows. (In fact we have e − cr(G) ≤ 3n − 6 for n ≥ 3). To obtain the actual crossing number inequality, we now use a probabilistic argument.
The first of these quadratic inequalities requires r to range in the region beyond the value of the positive root of the quadratic equation r 2 + r − 1 = 0, i.e. r > φ − 1 where φ is the golden ratio. The second quadratic inequality requires r to range between 0 and the positive root of the quadratic equation r 2 − r − 1 = 0, i.e. 0 ...
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.
Proof [2]. Since + =, =. A graph = on the -plane is thus also a graph =. From sketching a visual representation of the integrals of the area between this curve and the axes, and the area in the rectangle bounded by the lines =, =, =, =, and the fact that is always increasing for increasing and vice versa, we can see that upper bounds the area of the rectangle below the curve (with equality ...