Added a build script
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141
src/Analysis_rewrite.typ
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141
src/Analysis_rewrite.typ
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#set page(
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paper: "a4",
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margin: (
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bottom: 10mm,
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top: 5mm,
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left: 5mm,
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right: 5mm
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),
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flipped:true,
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numbering: "— 1 —",
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number-align: center
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)
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#set text(
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size: 8pt,
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)
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#place(top+center, scope: "parent", float: true, heading(
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[Analysis 1 (IE)]
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))
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#let subHeading(it: content, fill: color) = {
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box(
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align(
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top+center,
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text(
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it,
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size: 10pt,
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weight: "regular",
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style: "italic",
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)
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),
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fill: fill,
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width: 100%,
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inset: 1mm,
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height: auto
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)
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}
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#let SeperatorLine = line(length: 100%, stroke: (paint: black, thickness: 0.3mm))
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#let MathAlignLeft(e) = {
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align(left, block(e))
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}
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#let colorFolgen = color.hsl(202.05deg, 92.13%, 75.1%)
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#let colorReihen = color.hsl(280deg, 92.13%, 75.1%)
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#let colorAbleitung = color.hsl(356.92deg, 92.13%, 75.1%)
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#let colorIntegral = color.hsl(34.87deg, 92.13%, 75.1%)
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#columns(5, gutter: 2mm)[
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#subHeading(fill: colorFolgen, it: [Folgen])
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$ lim_(x -> infinity) a_n $
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*Beschränkt:* $exists k in RR$ sodass $abs(a_n) <= k$
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- Beweiße: durch Induktion
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- Beweiße: Hat min. ein konvergent Teilefolge
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- (Beweiße: Ungleichung $abs(a_n) <= k$)
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*Monoton fallend/steigended*
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- Beweise: Induktion
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#grid(columns: (1fr, 1fr),
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gutter: 1mm,
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row-gutter: 2mm,
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align(top+center, [*Fallend*]), align(top+center, [*Fallend*]),
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[$ a_(n+1) <= a_(n) $],
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[$ a_(n+1) >= a_(n) $],
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[$ a_(n+1)/a_(n) > 1 $],
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[$ a_(n+1)/a_(n) < 1 $],
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)
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*Konvergentz Allgemein*
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$ lim_(n -> infinity) a_n = a $
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$forall epsilon > 0 space exists n_epsilon in NN$ sodass \
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- Konvergent $-> a$: $a_n in [a - epsilon, a + epsilon] $
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- Divergent $-> infinity$: $a_n in [epsilon, infinity) $
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- Divergent $-> infinity$: $a_n in (-infinity, epsilon) $
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$space forall n > n_epsilon$
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*Konvergentz Häufungspunkte*
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- $a_n -> a <=>$ Alle Teilfolgen $-> a$
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*Konvergenz Beweißen*
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- Monoton UND Beschränkt $=>$ Konvergenz
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NICHT Umgekehert
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- (Cauchyfolge \
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$forall epsilon > 0 space exists n_epsilon in NN space$ sodass \
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$forall m,n >= n_epsilon : abs(a_n - a_m) < epsilon$ \
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Cauchyfolge $=>$ Konvergenz)
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*Konvergent Grenzwert finden*
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- Von Bekannten Ausdrücken aufbauen
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- Fixpunk Gleichung: $a = f(a)$ \
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für $a_(n+1) = f(a_n)$
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- Bernoulli-Ungleichung Folgen der Art $(a_n)^n$: \
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$(1 + a)^n >= 1 + n a$
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#subHeading(fill: colorFolgen, it: [Konvergent Folge Regeln])
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#grid(
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columns: (auto, auto),
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align: bottom,
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gutter: 2mm,
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[$ lim_(n->infinity) (a_n + b_n) = a + b $],
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grid.cell(
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rowspan: 2,
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[$ lim_(n->infinity) (a_n / b_n) = a / b $ für ($b != 0$)],
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),
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MathAlignLeft($ lim_(n->infinity) (a_n dot b_n) = a dot b $),
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MathAlignLeft($ lim_(n->infinity) sqrt(a_n) = sqrt(a) $),
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MathAlignLeft($ lim_(n->infinity) abs(a_n) = abs(a) $),
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MathAlignLeft($ lim_(n->infinity) c dot a_n = c dot lim_(n->infinity) a_n $),
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)
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#subHeading(fill: colorFolgen, it: [Bekannte Folgen])
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#grid(
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columns: (auto, auto, auto),
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column-gutter: 4mm,
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row-gutter: 2mm,
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align: bottom,
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MathAlignLeft($ lim_(n->infinity) 1/n = 0 $),
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MathAlignLeft($ lim_(n->infinity) q^n = 0 $),
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MathAlignLeft($ lim_(n->infinity) q^n = 0 $),
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grid.cell(colspan: 2, MathAlignLeft($ lim_(n->infinity) sqrt(n) = + infinity $)), [],
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grid.cell(colspan: 2, MathAlignLeft($ lim_(n->infinity) k = k, k in RR $)), [],
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grid.cell(colspan: 2, MathAlignLeft($ exp(x) = e^x = lim_(n->infinity) (1 + x/n)^n $))
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)
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#subHeading(fill: colorReihen, it: [Reihen])
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#subHeading(fill: colorReihen, it: [Potenzreihen])
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#subHeading(fill: colorAbleitung, it: [Funktionen])
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#subHeading(fill: colorAbleitung, it: [Ableitung])
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#colbreak()
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]
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