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The real absolute value function is an example of a continuous function that achieves a global minimum where the derivative does not exist. The subdifferential of | x | at x = 0 is the interval [−1, 1]. [14] The complex absolute value function is continuous everywhere but complex differentiable nowhere because it violates the Cauchy–Riemann ...
Cauchy–Schwarz inequality (Modified Schwarz inequality for 2-positive maps [27]) — For a 2-positive map between C*-algebras, for all , in its domain, () ‖ ‖ (), ‖ ‖ ‖ ‖ ‖ ‖. Another generalization is a refinement obtained by interpolating between both sides of the Cauchy–Schwarz inequality:
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
Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space L p (μ), and also to establish that L q (μ) is the dual space of L p (μ) for p ∈ [1, ∞). Hölder's inequality (in a slightly different form) was first found by Leonard James Rogers .
The standard absolute value on the integers. The standard absolute value on the complex numbers.; The p-adic absolute value on the rational numbers.; If R is the field of rational functions over a field F and () is a fixed irreducible polynomial over F, then the following defines an absolute value on R: for () in R define | | to be , where () = () and ((), ()) = = ((), ()).
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 ...
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 .
The second inequality is the elementary inequality between and . The last inequality follows by applying reverse Fatou lemma , i.e. applying the Fatou lemma to the non-negative functions g − f n {\displaystyle g-f_{n}} , and again (up to sign) cancelling the finite ∫ X g d μ {\displaystyle \int _{X}g\,d\mu } term.