Sentence examples for mapping for x from inspiring English sources

From [11, Lemma 2.6], (M x, y,cdot)) is a nondecreasing mapping for (x; yin X).

Step 2. In this step, we prove that F is a contraction mapping for (x,yin B_{r}).

Let's consider the mapping for x ″ k, 1 and assume x k, 1, 1 ″ occupies OFDM symbol 1 at subcarrier K 1, x k, 1, 2 ″ occupies OFDM symbol 2 at subcarrier K2,..., and x k, 1, N c ″ occupies OFDM symbol N c at subcarrier K N c.

Let (X, d) be a complete cone metric space and ϕ : P → P be a non-vanishing, sub-additive cone integrable mapping on each [a, b] ⊂ P such that for each ε ≫ 0, and the mapping for (x ≥ 0), has a continuous inverse at zero.

In other words, x and y should map to the same vertex s in S. Moreover the result of the leaf removal from T x I should result in a different LCA mapping for x and y.

Essentially, the returned edge (pa y), y) implies that, in a scenario where g maps to s and g is a transfer node with (g, x)∈Ξ, the best possible mapping for x (i.e. one for which cΘ g, s) is minimized) is y.

ϕ: X × X → R is a monotone and R-semi-continuous mapping; for each x ∈ X, f y) + ϕ x, y) is an R-convex function related to the variant y; for any N = {x 0, x 1} ∈ 〈X〉, lim inf t → 0 + f ( x t ) ≥ f ( x 0 ), where x t = φ N ((1 - t)e 0 + te 1), t ∈ [0, 1].

Proof Since f is an onto mapping, for each x 0 ∈ X, there exists f x 1 = x 0. Continuing this process, we can define { x n } by x n = f x n + 1, n = 0, 1, 2, …  .

Otherwise, for any g x ∈ F x 0, one has g x ≥ g x 0. As F has a g-approximative multivalued map, for x 1 ∈ X, there exists g x 1 ∈ F x 0 with g x 1 ≥ g x 0 and p ( g x 0, g x 1 ) = p ( F x 0, g x 0 ).

The present study represents part of our efforts to generate a genetic map for X. tropicalis using microsatellites as markers.

The Integromics platform produces a correlation map for X and Y, identifying the variables that display strong correlations (Fig.  5).