Sentence examples for d for some from inspiring English sources

Since {x n - x0} is bounded, we have limn→∞||x n - x0|| = d for some a constant d.

Let ( X, κ ) be a finite collection of digital m-simplices, 0 ≤ m ≤ d, for some nonnegative integer d.

Definition 2.5 Let ( X, κ ) be a finite collection of digital m-simplexes, 0 ≤ m ≤ d for some nonnegative integer d.

Assume that g ¯ Y is a subword of f ¯ Y, say, f ¯ Y = c g ¯ Y d for some c, d ∈ Y ∗.

It follows that lim n → ∞ w ( x n, x n + 1 ) = d for some d ≥ 0. Now we claim that (2.11) holds.

Proof Let r l = p m N and for some r, l ∈ N and prime p. Since l does not divide p, we can write r = p m d for some d ∈ N. Therefore, we have (5).

where J is the duality mapping on E, {α n } is a sequence in [0, 1] and { r j, n } n = 1 ∞ ⊂ [ d, ∞ ) for some d > 0 (j = 1, 2,..., m). If lim infn→∞(1 - α n ) > 0, then {x n } converges strongly to p ∈ 픉, where p = Π 픉 f x 0. Proof We split the proof into five steps.

where J is the duality mapping on E, {α n } is a sequence in [0, 1] and { r j, n } n = 1 ∞ ⊂ [ d, ∞ ) for some d > 0 (j = 1, 2,..., m). If lim infn→∞(1 - α n ) > 0, then {x n } converges strongly to p ∈ 픉, where p = Π 픉 f x 0. Remark 3.3. Corollary 3.2 extends and improves the result of Li et al. [26].

If C ( d ′ ) ∩ C ( d ) ≠ ∅, then there is K such that K ∈ C j ( d ) and K ∈ C p ( d ′ ) for some j and p. Consequently, ⋃ i = 0 j C i ( d ) ⊆ ⋃ i = p q ( d ′ ) C i ( d ′ ) and ⋃ i = j q ( d ) C i ( d ) ⊆ ⋃ i = p q ( d ′ ) C i ( d ′ ), by definition of C ( d ′ ).

A function f ∈ K s if it satisfies the following inequality Re z 2 f ′ ( z ) g ( z ) g ( - z ) < 0 ( z ∈ D ). for some function g ∈ S*(1/2).

In the context of the DG category of A -modules M o d ∞ for some A ∞ algebra A, it follows from the definition that the length filtration in (2.11) equals that of the F filtration on the bar constructions F Hom M o d ∞ (M, N ) = F Hom (B M, B N ).