Sentence examples for is a convex function in from inspiring English sources

It can be shown that the Fisher information is a convex function in both the input and the experiment configuration parameters.

Suppose that is twice continuously differentiable, that is a convex and monotone function in for all and that is a convex function in for all, for all.

The left-hand side of (16) is a convex function in the entries of diagonal matrix D. For (D=operatorname{diag}(x_{1},ldots,x_{n},x_{n})) denote the left-hand side by (f(x_{1},ldots,x_{n},x_{n})).

Therefore, f l (T ab ) is a convex function in the range Tmin1 ≤ T < Td 1. Then we will show that f r (T ab ) is a quasi-convex function in the range T > Td 1, where we will use the following lemma.

Furthermore, it is a convex function in S and its global minimum is (21) S = S ∼ ○ α (W T W ) + + 2 γ S 0 2 γ S ∼ + α (W T W ) − + α diag (W T W J J K. can be proved using a similar idea to that in (Ding et al., 2006) by validating (i) L (S ) ≤ Z (S, S ∼ ), (ii) L (S ) = Z (S, S ) (iii) Z (S, S ∼ ) is convex with respect to S. The formal proof is provided in the Supplementary Material.

Let (u D rightarrowmathbb{R}) be a convex function in the class (C^{3}).

Let g be a convex function in U and let h ( z ) = g ( z ) + n α z g ′ ( z ), for z ∈ U, where α > 0 and n is a positive integer.

where is a convex function, is interval in, and.

In this case, the cost function of NPCS is a convex function, which is linear in p k. Thus, following [44], there exists a Nash equilibrium.

One of the best-known inequalities for convex functions is the following Hermite-Hadamard inequality: If (f Isubseteq mathbb{R}rightarrow mathbb{R}) (I is an interval) is a convex function and (a, b in I) with (a< b), then begin{aligned} f biggl( frac{a+b}{2} biggr) leq frac{1}{b-a} int_{a}^{b}f ( x ),dxleq frac{f ( a ) +f ( b ) }{2}.

Let φ(x) = x+, ψ(x) = -x-, then φ is a convex function satisfying the condition in Theorem 2.1 and ψ is a concave function satisfying the condition in Theorem 2.2 By Theorems 2.1 and 2.2 we have - ln E [ exp ( - X + ) ] ≥ ( - ln E [ exp ( - X ) ] ) +, - ln E [ exp ( - ( - X - ) ) ] ≤ - ( - ln E [ exp ( - X ) ] ) -.