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11 Homework/Mathe II/Homework 2.typ
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74
11 Homework/Mathe II/Homework 2.typ
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#set page(
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header: align(right, [
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#set text(size: 9pt)
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#set par(leading: .3em)
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#pad(y: -.5cm, [
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Mathe II \
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Hausaufgabe 02 \
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Jan Meyer \
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664237
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])
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])
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)
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== Problem 1: Integration rationaler Funktionen
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=== (a) $f(x) = (x + 2) / (x^3 - 3x^2 - x + 3)$
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Faktorisierung des Nenners:
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$ x^3 - 3x^2 - x + 3 = (x - 1)(x + 1)(x - 3) $
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Partialbruchzerlegung:
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$ (x + 2) / ((x - 1)(x + 1)(x - 3)) = A / (x - 1) + B / (x + 1) + C / (x - 3) $
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$ A = -3/4, B = 1/8, C = 5/8 $
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Stammfunktion:
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$ F(x) = -3/4 ln|x - 1| + 1/8 ln|x + 1| + 5/8 ln|x - 3| + C $
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---
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=== (b) $f(x) = x^6 / (x^4 + 3x^2 + 2)$
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Polynomdivision:
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$ x^6 / (x^4 + 3x^2 + 2) = x^2 - 3 + (7x^2 + 6) / ((x^2 + 1)(x^2 + 2)) $
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Partialbruchzerlegung des Restes:
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$ (7x^2 + 6) / ((x^2 + 1)(x^2 + 2)) = 8 / (x^2 + 2) - 1 / (x^2 + 1) $
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Stammfunktion:
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$ F(x) = 1/3 x^3 - 3x - arctan(x) + 4 sqrt(2) arctan(x / sqrt(2)) + C $
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#pagebreak()
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== Problem 3: Eigenwerte
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=== (a) Berechnung der Eigenwerte
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#let id = "I"
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#let det = "det"
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==== (i) $A_1 = mat(3, -1, 0; 3, -1, 0; 4, -1, -2)$
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Charakteristisches Polynom (Entwicklung nach der 3. Spalte):
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$ p(lambda) = (-2-lambda) det mat(3-lambda, -1; 3, -1-lambda) $
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$ p(lambda) = (-2-lambda) (lambda^2 - 2lambda) = -lambda(lambda - 2)(lambda + 2) $
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*Eigenwerte:* $lambda_1 = 0, lambda_2 = 2, lambda_3 = -2$
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==== (ii) $A_2 = mat(1, 0, -1; 1, 0, 2; 1, 0, 1)$
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Charakteristisches Polynom (Entwicklung nach der 2. Spalte):
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$ p(lambda) = (-lambda) det mat(1-lambda, -1; 1, 1-lambda) $
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$ p(lambda) = -lambda (lambda^2 - 2lambda + 2) $
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Nullstellen via $p q$-Formel: $lambda = 1 plus.minus sqrt(-1)$
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*Eigenwerte:* $lambda_1 = 0, lambda_2 = 1 + i, lambda_3 = 1 - i$
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---
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=== (b) Beweis: Eigenwerte von $A B$ und $B A$
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Sei $lambda != 0$ ein Eigenwert von $A B$. Dann existiert ein Eigenvektor $v != 0$, sodass:
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$ (A B) v = lambda v $
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Multiplikation von links mit $B$:
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$ B(A B v) = B(lambda v) $
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$ (B A)(B v) = lambda (B v) $
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Da $lambda != 0$ und $v != 0$, ist $B v != 0$.
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Somit ist $B v$ ein Eigenvektor von $B A$ zum Eigenwert $lambda$. $square$
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