Dirk Ferus's Analysis II [Lecture notes] PDF

By Dirk Ferus

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Extra info for Analysis II [Lecture notes]

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An ) k=1 j=1 n =:av jk n bkj avjk . ,n = (det(a1 , . . , aj−1 , ek , aj+1 , . . ,n , so ist DA det(B) gerade die Summe der Diagonalelemente der Matrix B adj(A), die sogenannte Spur dieser Matrix: DA det(B) = Spur(B adj(A)). Wir merken noch an: 60 1. Ist A = E, so ist nach (22) evjk = det(e1 , . . , ej−1 , ek , ej+1 , . . , en ) = δjk , also adj(E) = E und DE det(B) = Spur(B). (23) 2. Allgemein gilt nach (22) n n avik akj = det(a1 , . . , ai−1 , k=1 akj ek , ai+1 , . . , an ) k=1 = det(a1 , .

K=0 Aus n+m n+m Sn+m − Sn = A −1 (A − B) k A −1 A−1 (A − B) ≤ k A−1 k=n+1 k=n+1 folgt mit (28), dass die Sn eine Cauchyfolge bilden. Weil L(V, V ) endlich-dimensional, also ein Banachraum ist, existiert S := lim Sn . Aus Sn A = E + A−1 (A − B) + . . + (A−1 (A − B))n Sn A(A−1 (A − B)) = A−1 (A − B) + . . + (A−1 (A − B))n + (A−1 (A − B))n+1 folgt durch Subtraktion: Sn B = E − (A−1 (A − B))n+1 . h. S = B −1 . Versch¨arft man (27) zu 1 A−B ≤ , (29) 2 A−1 so folgt mit der Dreiecksungleichung und (26) n A−1 (A − B) Sn ≤ k A−1 ≤ k=0 1 1− 1 2 A−1 = 2 A−1 .

49 (13) Beispiel 108. Sei f : R2 → R, (x, y) → 1+3x+4y+5xy 2 . Dann ist f in (0, 0) differenzierbar mit D(0,0) f (u, v) = 3u + 4v. Es ist n¨ amlich f (x, y) = 1 + 3(x − 0) + 4(y − 0) + 5xy 2 , und weil 5xy 2 in (x − 0) und (y − 0) kubisch ist“, geht der Rest f¨ ur (x, y) → (0, 0) gegen ” null: x 5xy 2 =5 y 2 ≤ 5y 2 . 2 2 2 x +y x + y2 Durch Nachrechnen k¨ onnen Sie best¨atigen, dass f (x, y) = 32 + 23(x − 1) + 24(y − 2) + 20(x − 1)(y − 2) + 5(y − 2)2 + 5(x − 1)(y − 2)2 . =:R(x,y) ur (x, y) → (1, 2) wieder gegen null.

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Analysis II [Lecture notes] by Dirk Ferus


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