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However, it is commonly believed that satisfying the first and the second criteria simultaneously is not feasible, and a tradeoff between them is always definite.
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The definite-valued properties have also been characterized somewhat differently (Bub and Clifton 1996; for an improved version, see Bub, Clifton and Goldstein 2000), that is, in terms of the quantum state |φ⟩ plus a "privileged observable" R, which is privileged in the sense that it represents a property that is always definite-valued (see also Dieks 2005, 2007).
A variant on this approach is that of Bub (1997), on which some observable R is chosen to to be always-definite; the set of definite observables is then expanded to the maximal set that avoids a KS obstruction.
V ( X ) is always positive definite, and therefore the negative definite of its derivative should be examined; it means W ( X ) in equation (17) should always be positive definite so that V ˙ ( X ) would be negative definite.
Go to Step 1. Step 4 of Algorithm 1 can ensure that (B_{k}) is always positive definite.
In a parametric setting, it is shown that the Fisher information matrix about the unknown parameters of a GRSS sample minus that of an SRS sample of the same size is always positive definite.
Symbol schemes that are notational can be compared in their workings to the way digital instruments of measurement work: for any measurement indicated by the instrument there is always a definite answer to the question, What is the measurement?
△ [ p ( n ) △ u ( n − 1 ) ] − L ( n ) u ( n ) + ∇ W ( n, u ( n ) ) = 0, where p ( n ) and L ( n ) are N × N real symmetric matrices for all n ∈ Z, and p ( n ) is always positive definite.
For a Laplacian matrix,
The shrinkage estimator provides a more accurate estimate which is always positive-definite such that all of the eigenvalues in Step 3 become positive.
According to ref. [24], we obtain that the matrix P ( r ) is the unique solution of ARE (3) which is always symmetric and positive-definite for r > 0. Using the following properties: A D ( r ) = r D ( r ) A, D ( r ) A T = r A T D ( r ), C D ( r ) = C, D ( r ) C T = C T, we get D − 1 ( r ) A = r A D − 1 ( r ), A T D − 1 ( r ) = r D − 1 ( r ) A T, C = C D − 1 ( r ), C T = D − 1 ( r ) C T. (4).
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com