By Jean-Paul Gauthier

This paintings provides a normal conception in addition to optimistic technique so as to resolve "observation problems," specifically, these difficulties that pertain to reconstructing the total information regarding a dynamical method at the foundation of partial saw info. A common technique to manage strategies at the foundation of the observations is additionally constructed. Illustrative yet useful purposes within the chemical and petroleum industries are proven.

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**Example text**

20) 1 ∂x ∂x Proof. Let us chose coordinates x 0 , . . ,i Ker(d x j ). Then, ∂ , and for 0 ≤ j ≤ n − 2, ∂ x n−1 Dn−1 = {0}, Dn−2 = Span L ∂ ∂ x n−1 j L fu h = 0 = L fu L ∂ ∂ x n−1 j−1 L fu h + L j−1 ∂ ∂ x n−1 But, L ∂ ∂ x n−1 j−1 L f u h = 0, , fu L f u h. P1: FBH CB385-Book CB385-Gauthier May 31, 2001 17:54 Char Count= 0 2. Normal Form for a Uniform Canonical Flag 23 therefore, ∂ ∂ x n−1 , f u ∈ Dn−3 . An obvious induction shows that ∂ , f u ∈ Dk−2 , ∂xk which implies ∂h = 0 for i ≥ 1, ∂xi ∂ fi = 0 for j ≥ i + 2.

As we said, the statements of our results are the same for Sr and S 0,r . Our proofs also are the same because the dependence of h in the control plays no role in them. In the following, we shall have to consider subspaces of Sr , F r , H r , or S 0,r , F r , H 0,r , which will be denoted by the letters S, F, H , possibly with some additional indices, and which will have the following two properties: (A1 ) S, F, H are subspaces of Sr , F r , H r , or S 0,r , F r , H 0,r . They are Banach spaces for a stronger norm than the one from the overspaces Sr , F r , .

2 for Bˆ 2 (k) imply that the TX f ( p)i , P1: FBH CB385-Book CB385-Gauthier June 21, 2001 11:11 Char Count= 0 The Case d y > du 48 0 ≤ i ≤ ρ, are linearly independent. Hence, Codim Bˆ 6 (k, ρ), J k S × R (k−1)du ≥ Codim µ−1 (0), J k S× X T X × R (k−1)du + 1 − n. 2 implies immediately that µ is a submersion. Hence Codim(µ−1 (0), J k H × X ×U G) = kd y, (resp. Codim(µ−1 (0), J k H 0 × X ×U G) = kd y ). Using 4, Codim µ−1 (0), J k S× X T X × R (k−1)du = kd y + n + (ρ − 1)du , Codim Bˆ 6 (k, ρ), J k S × R (k−1)du ≥ kd y + n + (ρ − 1)du − n + 1, 5.