Analysis I - IV - download pdf or read online

By Hilgert J.

Der vorliegende textual content ist eine vorlaufige Ausarbeitung meiner Vorlesungen research I-IV (Wintersemester 2004/2005 { Sommersemester 2006) an der Universitat Paderborn.

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Ii) F¨ ur a = 0 ist das wieder klar. F¨ ur a > 0 gilt −a = −|a| < 0 < a = |a| und f¨ ur a < 0 gilt |a| = −a > 0 > a = −|a|. (iii) Wenn a ≥ 0, dann folgt aus a ≤ b die Ungleichung |a| = a ≤ b. Wenn a < 0, dann folgt aus −b ≤ a erst −a ≤ b und dann die Ungleichung |a| = −a ≤ b. Umgekehrt folgt aus |a| ≤ b die Ungleichung −b ≤ −|a| und dann −b ≤ −|a| ≤ a ≤ |a| ≤ b. 2. POSITIVITAT 27 (iii’) Dies folgt sofort aus (iii). 5(ii)) erh¨alt man −(|a| + |b|) ≤ a + b ≤ |a| + |b|. Mit (iii) folgt |a + b| ≤ |a| + |b|.

X 40 KAPITEL 2. 5 : Die Menge {f ∈ F ]a, ∞[, R | ∃ lim f (x)} x→∞ ¨ ist ein Untervektorraum von F(]a, ∞[, R) (vgl. 9 : Sei M ⊆ R nach oben unbeschr¨ ankt und f : M → R eine Funktion. Setze N := { x1 | x ∈ M \ {0}} und g : N → R, t → f ( 1t ). Dann gilt ]0, δ[ ∩N = ∅ f¨ ur jedes δ > 0 und lim f (x) = y ⇔ x→∞ lim g(t) = y. 5 und den Definitionen. Wenn limx→∞ f (x) = y, dann gibt es zu (∗) > 0 ein x0 ∈ R mit |f (x) − y| < f¨ ur x ∈ ]x0 , ∞ ∩ M .

14 : Man fertige eine qualitative Skizze1 der folgenden Funktionen an: (i) f : R → R; f (x) = 1 . (x−121)2 +1 (ii) f : R → R; f (x) = xn f¨ ur n = 1, 2, 3, 4, 5 (in ein Schaubild). (iii) f : D → R; f (x) = 1 1 x . Welches ist die gr¨ oßte Teilmenge D ⊆ R, f¨ ur die f definiert ist? 15 : Es sei f: D x → → R 1 x − 1 x . (i) Man bestimme den gr¨ oßtm¨ oglichen Definitionsbereich D ⊆ R von f . (ii) Man fertige eine qualitative Skizze von f an. ur alle n ∈ Z und alle r ∈ [0, 1[ mit (n, r) = (0, 0) gilt: (iii) Man beweise, daß f¨ f 1 n+r = r.

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Analysis I - IV by Hilgert J.


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