Sentence examples for ls let us from inspiring English sources

We first consider the following situation: α GS > 0, α LS > 0, α GS ≠ α LS α GS > 0, α LS > 0, α GS ≠ α LS Let us first see that in this case the spectrum of K cannot contain K i for i ∈ {1, 2, 4, 6}.

Calling D C ll) the 4 × 4 matrix B C (l) ⊗ B C (l), let us introduce now the following cost function: E = ∑ l = 1 l max d l - D C l g l 2 + λΦ g 1, …, g l max (11).

By Lp,q and W p, q l let us denoted, respectively, the (p, q -integrable function space and Sobolev space with mixed norms, where 1 ≤ p, q -integrable0].

Theorem 5 For n ≥ 0, we have D ˆ n, ξ = ξ n ∑ l = 0 n S 1 ( n, l ) ( − 1 ) l B l. Let us consider the generating function of the n th twisted Daehee numbers of the second kind as follows: ∑ n = 0 ∞ D ˆ n, ξ t n n !

We agree that in the sequel one and the same symbol { κ n } denotes different sequences in l 2. Let us choose the model boundary value problem L = L ( q ˜ ( x ), T ), so that ω ˜ = ω (for example, one can take q ˜ ( x ) ≡ 2 ω / T ).

Again, for the purpose of correctly modeling the game among the L users, let us define the following potential: T ( c 1, …, c L ) = - ∑ l = 1 L d l ( c l ) † I + ∑ h = 1, h ≠ l L ∑ k = - N + 1 N - 1 G ( h, l ) J k c h c h † J k T + G ( l, l ) ∑ k = - N + 1, k ≠ 0 N - 1 J k c l c l † J k T d l ( c l ), (23).

u satisfies and d u d t ∈ L ∞ ( 0, T, L 2 . Let us consider the set C = { v ∈ X : v ( t ) ∈ C, ∂ v ∂ t ≥ 0 a.e.

Before presenting necessary conditions so that (hbox {ARS}_{supseteq {L}_0, supseteq {R}_0} ({s}_0,{s}_1,{c}_0,{c}_1)) and ({AR}_{{L}_0,{R}_0} ({L}, {S})) are not empty, let us use some additional notations as follows.

Given y ∈ L 2 ( R ), let us show that φ ˆ ( ⋅, y ) is measurable.

For L > 0, let us introduce in the space Y 0 the equivalent norm defined as ∥ y ∥ ∗ = sup t ∈ J ( e − L t ∥ y ( t ) ∥ ), since for any ψ ∈ L 1 ( J, X ), lim L → + ∞ sup t ∈ J ∫ 0 t e − L ( t − s ) ψ ( s ) d s = 0, we can take the appropriate L to satisfy M 2 C 1 ∗ sup t ∈ J ∫ 0 t e − L ( t − s ) ( t − s ) − 1 − q γ μ ( s ) d s ≤ 1 2, (3.1).

With (28), when b = -1, then y = M-1(r + s A) becomes y = x + M-1e + M-1q, where y = [y1, y2,..., y l ] T. Let us denote x + M-1e = v p + v n, where v p = v p 1, v p 2,..., p p l T is the vector whose elements are non-negative, and v n = v n 1, v n 2,..., v n l T is the vector whose elements are less than zero.