On Saturday, a member of Mr. Berlusconi's legal team, Piersilvio Cipollotti, called the decision a "most positive solution, given the circumstances".
Here the max indicates the selection of the most positive solution which is the best score chosen by taking the smallest P value from the t -test, SAM, or CATT results or taking the largest score from the results of information gain.
(2) Secondly, normalize the gene scores in each individual resource by dividing each of them by the Euclidean norm of all the gene scores from the same resource, as in formula (2 ); for i = 1,2,…, n and j = 1,2,…, m, (2) U i j = X i j ∑ i = 1 n X i j 2. (3) Thirdly, obtain the most positive solution (U j +) and the most negative solution (U j −) for each type of genetic resources.
Secondly, normalize the gene scores in each individual resource by dividing each of them by the Euclidean norm of all the gene scores from the same resource, as in formula (2); for i = 1,2,…, n and j = 1,2,…, m, (2) U i j = X i j ∑ i = 1 n X i j 2. Thirdly, obtain the most positive solution (U j +) and the most negative solution (U j −) for each type of genetic resources.
It is possible that (y=Y_{1}(x)) and (y=Y_{2}(x)) have two solutions, but at most one positive solution.
Then there exists at most one positive solution to the equation (kappa_{theta}=alpha) with respect to θ.
We now prove that the equation (F(x_{1})=0) has at most one positive solution on the interval ((0, frac {r_{1}}{a_{1}})).
In Section 4, assuming that Ω = ( p, q ) is an interval, we find that (1.1) has at most one positive solution when γ 2 ( c + d ) ≤ 2. In this section, we establish the existence and nonexistence of positive solutions to (1.1).
By equation (3.23) and the fact ϑ ( 1 ; 0, 0 ; 0 ⋅ a ) = 0, we can deduce that ϑ ( 1 ; 0, 0, λ a ) = k π p, k ≥ 1, has at most one positive solution and one negative solution, denoted by λ k N ( a ) and λ − k N ( a ) respectively, if they exist.
For any k ≥ 1, it follows from equations (3.18) and (3.22) that ϑ ( 1 ; 0, − π p 2, λ a ) = k π p − π p 2. has at most one positive solution and one negative solution, denoted by λ k D ( a ) and λ − k D ( a ), respectively, if they exist.
It follows from equations (3.35) and (3.36) that for any k ≥ 1, the equation θ ( 1 ; 0, − π p 2, λ a, λ b ) = k π p − π p 2. has at most one positive solution and one negative solution, denoted, respectively, by Λ k D ( a, b ) and Λ − k D ( a, b ), if they exist.
