mathe doof
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== Exercise Sheet 1 - Basics
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=== Exercise 1 @Exercise[1, 1]
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#block(
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fill: luma(230),
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inset: 8pt,
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radius: 4pt,
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[
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Determine whether the following sequences converge. Calculate the limit in case of convergence.
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#set enum(numbering: "(a)")
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+ $ a_n = (2024 (1+n+n^2))/(n(n+2023)) $
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+ $ a_n = sqrt(n^2+n dot b_1+b_2)-n; quad b_1,b_2 in RR $
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+ $ a_n = (n^4-2)/(n^2+4) + (n^3(3-n^2))/(n^3+1) $
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])
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#set enum(numbering: "(a)")
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+ $
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a_n &= (2024 (1+n+n^2))/(n(n+2023)) \
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&= ((2024 (1+n+n^2))/n^2)/(1+2023/n) \
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&= (2024 (1/n^2+1/n+1/))/(1+2023/n) \
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&= (2024/n^2+2024/n+2024)/(1+2023/n) \
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=> lim_(n -> infinity) a_n &= lim_(n -> infinity) (2024/n^2+2024/n+2024)/(1+2023/n) \
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&= lim_(n -> infinity) 2024/1 = 2024 checkmark \
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$
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+ $
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a_n &= sqrt(n^2+n dot b_1+b_2)-n; quad b_1,b_2 in RR \
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&= ((sqrt(n^2+n dot b_1+b_2)-n) (sqrt(n^2+n dot b_1+b_2)+n))/(sqrt(n^2+n dot b_1+b_2)+n) \
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&= (n^2+n dot b_1+b_2 - n^2)/(sqrt(n^2+n dot b_1+b_2)+n) \
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&= (n dot b_1+b_2)/(sqrt(n^2+n dot b_1+b_2)+n) \
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&= (b_1+b_2/n)/((sqrt(n^2+n dot b_1+b_2)+n)/n) \
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&= (b_1+b_2/n)/((sqrt(n^2+n dot b_1+b_2))/n+1) \
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&= (b_1+b_2/n)/((sqrt(n^2+n dot b_1+b_2))/sqrt(n^2)+1) \
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&= (b_1+b_2/n)/(sqrt(1+b_1/n+b_2/(n^2))+1) \
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=> lim_(n -> infinity) a_n &= lim_(n -> infinity) (b_1+b_2/n)/sqrt(1+b_1/n+b_2/(n^2))+1 \
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&= b_1/(sqrt(1) + 1) = b_1/2 checkmark
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$
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+ $
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a_n &= (n^4-2)/(n^2+4) + (n^3(3-n^2)) / (n^3+1) \
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&= ((n^4-2)(n^3+1) + (n^3(3-n^2))(n^2+4)) / ((n^2+4)(n^3+1)) \
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&= (n^7+n^4-2n^3-2 + 3n^5+12n^3-n^7-4n^5) / ((n^2+4)(n^3+1)) \
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&= (-n^5+n^4+10n^3-2) / (n^5+n^2+4n^3+4) \
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&= (-1+1/n+10/(n^2)-2/(n^5)) / (1+1/(n^3)+4/(n^2)+4/(n^5)) \
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=> lim_(n -> infinity) a_n
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&= lim_(n -> infinity) (-1+1/n+10/(n^2)-2/(n^5)) / (1+1/(n^3)+4/(n^2)+4/(n^5)) \
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&= lim_(n -> infinity) -1/1 = 1 checkmark
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$
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#include "1.typ"
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@ -330,7 +330,7 @@ $ <v1-partint>
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An dieser Stelle stimmt die Lösung in @Vorlesung leider wieder nicht. Ich habe
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das Ergebnis wie in <v1-partint> maschinell überprüfen lassen, und es war
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korrekt. Lauf @Vorlesung wäre das Ergebnis:
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korrekt. Laut @Vorlesung wäre das Ergebnis:
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$ A = integral x sin(x) d x = x(-cos(x)) - integral sin(x) d x = cos(x)(1-x)
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" "#emoji.crossmark $
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