Euler's one step method is undoubtedly the simplest method for approximating the solution to an ordinary differential equation. It goes something like this: Given a first order initial value problem y = f (x, y), x ∈ (a, b) y(a) = y a we observe that defining x j = a + h(j − 1), for j = 1, 2, · · · , (N + 1) where h = (b − a)/N we have y(x k+1) − y(x k) h ≈ f (x k , y(x k)). Therefore we can wr...

and Applied Analysis 3 Theneuron activation functions,g p (y p (⋅)) (p = 1, . . . , n), are assumed to be nondecreasing, bounded, and globally Lipschitz; that is, l − p ≤ g p (ξ p ) − g p (ξ q ) ξ p − ξ q ≤ l + p , ∀ξ p , ξ q ∈ R, ξ p ̸ = ξ q , (5) where l− p and l+ p are constant values. For simplicity, in stability analysis of the network (1), the equilibrium point y∗ = [y∗ 1 , . . . , y ∗ n ]...

Solution: Since Λ(y) has finitely many (3) possible values, all values of η between any adjacent pair lead to the same threshold test. Thus, for η > 1, Λ(y) > η, leads to the decision ĥ = 0 if and only if (iff) Λ(y) = 1, i.e., iff −1 ≤ y < 0. For η = 1, the rule is the same, Λ(y) > η iff Λ(y) = 1, but here there is a ‘don’t care’ case Λ(y) = 1 where 0 ≤ y ≤ 1 leads to ĥ = 1 simply because of th...

and Applied Analysis 3 Equation (8) can be expressed as ̇ e 1 = − 2y 1 x 1 [a (x 2 − x 1 ) + x 4 ] − (1 + x 2 1 ) × [a 1 (y 2 − y 1 ) + y 4 ] − u 1 + 3x 2 4 (rx 4 + x 2 x 3 ) , ̇ e 2 = − 2y 2 x 2 (dx 1 + cx 2 − x 1 x 3 ) − (1 + x 2 2 ) × (b 1 y 1 − y 2 − y 1 y 3 ) − u 2 + 3x 2 1 [a (x 2 − x 1 ) + x 4 ] , ̇ e 3 = − 2y 3 x 3 (−bx 3 + x 1 x 2 ) − (1 + x 2 3 ) × (−c 1 y 3 + y 1 y 2 ) − u 3 + 3x 2 2 (d...

In this paper. (1) We determine the complex-valued solutions of the following variant of Van Vleck's functional equation $$int_{S}f(sigma(y)xt)dmu(t)-int_{S}f(xyt)dmu(t) = 2f(x)f(y), ;x,yin S,$$ where $S$ is a semigroup, $sigma$ is an involutive morphism of $S$, and $mu$ is a complex measure that is linear combinations of Dirac measures $(delta_{z_{i}})_{iin I}$, such that for all $iin I$, $z_{...

1. Proof Proposition 1. Ψ y ν (x) is a surrogate function of ψ ν (x). Proof. A function g y (x) is said to a surrogate function of f (x) provided ⇢ f (x) ≤ g y (x), ∀x, y ∈ [0, ∞) f (y)=g y (y), ∀y ∈ [0, ∞). (1) Let us consider the following function: f ν (x)=(1− exp(−νx))/ν. (2) Since f 00 ν (x) < 0, this function is strictly concave, ∀x, y, f ν (x) ≤ f ν (y)+(x − y)f 0 ν (y), (3) with equalit...

y=1 φ(y) = 0, θ ∈ {1, 2, . . .}, i.e., φ(y) = 0, y ∈ {1, 2 . . .}. Consequently, if g : IR → IR is an unbiased estimator of θ on the basis of Y , it is “essentially unique”: if g̃ : IR → IR is another unbiased estimator, then it must hold by the completeness of {U(1, . . . , θ), θ ∈ {1, 2 . . .}} that g̃(Y ) = g(Y ) Pθ-a.s. θ ∈ {1, 2, . . .}, whence g̃(y) = g(y), y ∈ {1, 2, . . .}. Next, Eθ[g(Y )]...

The functional equation f(y − x) − g(xy) = h (1/x− 1/y) is solved for general solution. The result is then applied to show that the three functional equations f(xy) = f(x)+f(y), f(y−x)−f(xy) = f(1/x−1/y) and f(y−x)−f(x)−f(y) = f(1/x−1/y) are equivalent. Finally, twice differentiable solution functions of the functional equation f(y − x) − g1(x) − g2(y) = h (1/x− 1/y) are determined.