Sentence examples for set q on from inspiring English sources

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Given a quartet topology set Q on a taxon set S = { s1, s2,..., s n }, Q is complete if Q contains exactly one quartet topology for every quartet of S. In this paper, we assume Q is complete.

By the equicontinuity of the set Q on compact intervals, we have ω 0 T ( Q n ) = 0, for  n = 0, 1, …,  and  T ≥ 0. Set z n ( t ) = μ ( Q n ( t ) ).

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In this paper, our aim is to study the behavior of the solution set Q and the conditions on these operators S, E, F, M, N, η 1, η 2, A 1, A 2 under which the function Q is continuous or Lipschitz continuous with respect to the parameter ∈ Ω × Λ.

As we found that the former method always gives better performance in our algorithm, we thus let: (5) N (i, j ) = N p i - N p j = (N p i, q 1 - N p j, q 1, …, N p i, q m - N p j, q m ) which is the neural response of the pair (p i, p j) on the templates set q 1...q m.

This algorithm takes a set Q of weighted quartet topologies on X as input and, using a modification of Neighbor-Net, produces a set of weighted splits S on X that is circular, and thus can be represented by an outer-planar split network.

□ We say a set Q of compatible quartets is redundant if for some q∈ Q, Q∖{ q}⊩ q; otherwise, we say that Q is irredundant.

SNP-PRAGE yielded 2 significant gene sets (q-values: 0.01, 0.03) based on q-value 0.05 as cut off, while Z-statistic method did not yield any significant gene sets.

maps U r onto the set q ( U r ) on the right half-plane Re > 0. The boundary ∂ q ( U r ) is tangent to the imaginary axis at q ( z 1 ) and at q ( z 2 ) because ∂ p ( U r ) is tangent to the sector − π β / 2 < arg w < π α / 2 at p ( z 1 ) and at p ( z 2 ).

The Stechkin problem on the best approximation of the operator A by linear bounded operators on set Q consists in evaluating quantity (9) and finding extremal operators (if any exists) delivering an infimum in the right hand part of (9).

This system is considered on the set Q T 0 = ] 0, 1 0, T 0 [, where Q T 0 is the domain of our generalized solution (see [10]).

The function Ω = Ω ( δ, A, Q ) : = sup { ∥ A f ∥ Y : f ∈ Q, ∥ f ∥ X ⩽ δ }, δ ⩾ 0, (8). is called the modulus of continuity of the operator A on the set Q.

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