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Extra info for Algebre, solutions developpees des exercices, 2eme partie, algebre lineaire [Algebra]

Sample text

1) there holds φ ∞ ≤ C| ln l| h for l > 0 small. 27) Proof. We need an estimate on the value of c. 28) B because B ∆φP Z = B ∆P Zφ = 0 and B P Z = 0. 8) we get that P Z = O(1) and | B hP Z| = O( B |h|) = O( h ∗ ). 8). We have that for any j = 0, 1, 2 + ρ2 eU φZj · aj = O φ ∞ |y|≤1/δρ B ρl−1 φ =O = O(l φ ∞ 3 |y|≤l 2 /δρ 8 (1 + |y|2 )2 8 (1 + |y|2 )2 + O( φ ∞ δ −1 |y − lδ −1 ρ−1 aj | + |y − lδ −1 ρ−1 aj |2 2 δ −2 3 |y|≥l 2 /δρ 8 ) (1 + |y|2 )2 ∞) and for any k = j − eUj φZk · ak = O ρ2 φ ∞ |y|≤1/ ρ B +O( 2 ρ2 φ +O( φ ∞ ∞) ρl−1 φ =O 3 |y|≥l 2 / ρ ∞ 3 |y|≤l 2 8 |y + l −1 ρ−1 (aj − ak )| (1 + |y|2 )2 1 + |y + l −1 ρ−1 (aj − ak )|2 / ρ 8 (1 + |y|2 )2 8 ) + O( 2 ρ2 φ (1 + |y|2 )2 ∞) = O(l φ ∞) as l → 0, uniformly on φ.

S. gives a contribution ρ2n e(Un ) + P zn − zn − B 16π 16π 2 H(0, 0) φn = ρ 3 3 n +O δn2 ρ2n ln2 (δn ρn ) = O(δn ρn ) → 0 + e(Un ) (H(x, 0) − H(0, 0)) φn B as n → +∞. S. 25) we get that 2 eP Un + e−P Un − e(Un ) ρ2n + B j=0 2 = Bj,n j=0 + 16π 3 B 8 Φj,n zn ( n ρn y + ln aj ) (1 + |y|2 )2 2 j=0 − e(Un )j φn P zn + O(ln2 ln2 ln ) φn P zn = ρ2n Bj,n = 8F ln ln R2 8 Φj,n H( n ρn y + ln aj , 0) + O(ln2 ln2 ln ) (1 + |y|2 )2 8(|y|2 − 1) + 16πF H(0, 0) (1 + |y|2 )3 R2 8(1 − |y|2 ) + o(1) → 0 (1 + |y|2 )3 2 2 as n → +∞, by means of Lebesgue Theorem and Φj,n → F 1−|y| 1+|y|2 in Cloc (R ).

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