From 563e354acddbcf465b81bf1366833eee9707781d Mon Sep 17 00:00:00 2001 From: Manuel Thalmann Date: Thu, 2 Jun 2022 18:27:54 +0200 Subject: [PATCH] Add notes on Analysis II --- .../AN2 - Analysis 2/Horner Method.png | Bin 0 -> 30471 bytes .../Semester 2/AN2 - Analysis 2/PartInteg.png | Bin 0 -> 11466 bytes .../AN2 - Analysis 2/ProductInt.png | Bin 0 -> 22483 bytes .../AN2 - Analysis 2/Zusammenfassung SEP.md | 284 ++++++++++++++++++ 4 files changed, 284 insertions(+) create mode 100644 Notes/Semester 2/AN2 - Analysis 2/Horner Method.png create mode 100644 Notes/Semester 2/AN2 - Analysis 2/PartInteg.png create mode 100644 Notes/Semester 2/AN2 - Analysis 2/ProductInt.png create mode 100644 Notes/Semester 2/AN2 - Analysis 2/Zusammenfassung SEP.md diff --git a/Notes/Semester 2/AN2 - Analysis 2/Horner Method.png b/Notes/Semester 2/AN2 - Analysis 2/Horner Method.png new file mode 100644 index 0000000000000000000000000000000000000000..59b5b60b582abb57a666ada082ccce9a4c3318ad GIT binary patch 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zwz-787VS=$?=!Liitko5H6oxzS>$#k^+>P{ulqT8;_-2Te#E`v3p{ literal 0 HcmV?d00001 diff --git a/Notes/Semester 2/AN2 - Analysis 2/Zusammenfassung SEP.md b/Notes/Semester 2/AN2 - Analysis 2/Zusammenfassung SEP.md new file mode 100644 index 0000000..6f6fcbe --- /dev/null +++ b/Notes/Semester 2/AN2 - Analysis 2/Zusammenfassung SEP.md @@ -0,0 +1,284 @@ + + + + +# Zusammenfassung Analysis 2 +## Nullstellen durch Horner-Schema +Das Horner-Schema erlaubt es, Nullstellen leicht zu bestimmen. + +Der Vorgang wird anhand der folgenden Gleichung aufgezeigt: + +$$f(x) = 3x^3 - 15x + 12$$ + +Eine Nullstelle dieser Formel ist die $1$. Dies lässt sich wie folgt mit dem Horner-Schema prüfen: + +![](Horner%20Method.png) + +**1. Nullstelle aufschreiben** + +Im Bild rot markiert + +**2. Die Koeffizienten aller $x$-Potenzen aufschreiben** + +Zu sehen im Bild in der obersten Reihe. +Entspricht eine der Koeffizienten $0$ (wie in diesem Beispiel $x^2$), so **muss** im Horner-Schema eine $0$ eingetragen werden. + +**4. Startwert notieren** +Die Berechnung wird gestartet indem man den ersten Koeffizient (in diesem Fall $3$) in die unterste Reihe überträgt. + +**5. Berechnung** + +In diesem Schritt multipliziert man die Eingabe-Zahl mit der bekannten Nullstelle (im Bild durch graue Pfeile markiert) und addiert das Ergebnis mit dem nachfolgenden Koeffizienten. + +Ist das letzte Resultat $0$, so handelt es sich auch wirklich um eine Nullstelle. + +> **Note:** +> Anhand der Resultat-Reihe im Horner-Schema lässt sich das übrigbleibende Polynom ablesen. +> +> Rechnet man die Nullstelle $1$ aus der Rechnung $3x^3 - 15x + 12$ heraus, so würde das übrigbleibende Polynom folgendermassen lauten: +> $$3x^2 + 3x - 12$$ +> +> Verdeutlicht: +> $$3x^3 - 15x + 12 = (x - 1) \cdot (3x^2 + 3x - 12)$$ + +## Stammfunktion +Umkehrung von Ableitungen gemäss der Seite über Ableitungen[^Derivation]. + +Alle Ableitungsregeln können unter [Ableitungen][Derivation] nachgeschlagen werden. + +## Integrale +Integrale erlauben es, die Fläche unter Funktionen zu berechnen: + +

+ +**Beispiel anhand der Funktion $f(x) = x^3 + 5$** + +Integrale haben folgende Erscheinungsform: + +$$\int_{a}^{b}{f(x)}dx$$ + +Die Zeichen haben folgende Bedeutung: + - $\int$: Integrations-Zeichen + - $a$: Die Untergrenze (Punkt ab dem integriert werden soll) + - $b$: Die Obergrenze (Punkt bis zu dem integriert werden soll) + - $f(x)$: Zu integrierende Funktion + - $dx$: Bezeichnet, dass $x$ integriert wird, indem unendlich kleine Rechtecke aufsummiert werden. + +### Integration von Produkten +Produkte benötigen zum Teil spezielle Vorgehensweisen um sie zu integrieren. + +Zwei gängige Wege dazu sind im Folgenden Beschrieben. + +#### Integration durch Substitution +Die Integration durch Substitution basiert auf der folgenden Regel: + +$$\int_a^b{f(u(x)) \cdot u'(x)}dx = \int_{u(a)}^{u(b)}{f(u)}du$$ + +Die Integration durch Substitution wird hier anhand des folgenden Beispiels gezeigt: + +$$\int_0^{\sqrt{\frac{\pi}{2}}}{\cos(x^2) \cdot x}dx$$ + +**1. Funktionen bestimmen** + +Im Folgenden werden die Haupt- und die Unterfunktion bestimmt. + + - $u(x) = x^2$ + - $g(x) = \cos(u)$ + - $f(x) = g(u(x))$ + +**2. Substitutions-Gleichung für $dx$** + +Diese kann mit folgender Regel ermittelt werden: + +$$\frac{du}{dx} = u'(x) \Rightarrow dx = \frac{du}{u'(x)}$$ + +Das ergibt in diesem Beispiel folgendes: + +$$\frac{du}{dx} = u'(x) = 2x \Rightarrow dx = \frac{du}{2x}$$ + +**3. Substitution** + +Nun müssen die errechneten Werte in die folgende Formel eingesetzt werden: + +$$\int_{u(a)}^{u(b)}{f(u)}du$$ + +Wobei $du$ durch unser errechnetes $dx$ ersetzt werden muss. + +Wichtig ist hierbei, dass sich $x$ komplett wegkürzen lassen **muss**. + +$$\int_0^{\sqrt{\frac{\pi}{2}}}{\cos(x^2) \cdot x}dx = +\int_{u(0)}^{u(\sqrt{\frac{\pi}{2}})}{\cos(u) \cdot x \cdot \frac{du}{2x}}$$ + +Vereinfacht ergibt das folgendes: + +$$\int_0^{\frac{\pi}{2}}{\frac{1}{2} \cdot \cos(u)}du = \left[\frac{1}{2} \cdot \sin(u)\right]_0^{\frac{\pi}{2}}$$ + +**4. Resultat berechnen** + +$$\left[\frac{1}{2} \cdot \sin(u)\right]_0^{\frac{\pi}{2}} = \left(\frac{1}{2} \cdot 1\right) - \left(\frac{1}{2} \cdot 0\right) = \frac{1}{2}$$ + +Bei Integralen, die keine Grenzen definiert haben, lässt sich das Ergebnis nicht eindeutig bestimmen. + +Hätte das ursprüngliche Integral keine Grenzen, wäre das Ergebnis folgendes: + +$$\left[\frac{1}{2} \cdot \sin(u)\right] = \frac{1}{2} \cdot \sin(u) + C$$ + +**5. Rücksubstitution** + +Im Falle eines Integrals ohne Grenzen muss die Variable $u$ rück-substituiert werden: + +$$\frac{1}{2} \cdot \sin(u) + C = \frac{1}{2} \cdot \sin(x^2) + C$$ + +#### Partielle Integration +Die Partielle Integration beruht auf folgender Regel: + +$$\int_a^b{u'(x) \cdot v(x)dx} = \left[u(x) \cdot v(x)\right]_a^b - \int_a^b{u(x) \cdot v'(x)}dx$$ + +Die Partielle Integration wird anhand des folgenden Beispiels erklärt: + +$$\int_0^\pi{sin(x) \cdot x}dx$$ + +Folgende Schritte müssen für die partielle Integration durchgeführt werden: + +**1. Unterfunktionen bestimmen** + +Einem Teil der Funktion $v$ und einem anderen $u'$ zuordnen. Sollte die partielle Integration nicht funktionieren, kann in diesem Schritt die Zuordnung von $v$ und $u'$ vertauscht werden. + + - $u'(x) = sin(x)$ + - $v(x) = x$ + +**2. Stammfunktion von $u'$ und Ableitung von $v$ bestimmen** + + - $u(x) = -\cos(x)$ + - $v'(x) = 1$ + +**3. Resultat berechnen** + +Daraus ergibt sich folgende Gleichung: + +$$\int_0^\pi{u'(x) \cdot v(x)dx} = \left[u(x) \cdot v(x)\right]_0^\pi - \int_0^\pi{u(x) \cdot v'(x)}dx$$ +$$\left[-\cos(x) \cdot x\right]_0^\pi - \int_0^\pi{-\cos(x) \cdot 1}dx$$ +$$((1 \cdot \pi) - (1 \cdot 0)) - \int_0^\pi{-\cos(x)}dx$$ +$$(\pi - 0) - \left[\sin(x)\right]_0^\pi$$ +$$(\pi - 0) - (\sin(0) - \sin(\pi))$$ +$$(\pi - 0) - (0 - 0) = \pi$$ + +> **Note:** +> Die Partielle Integration kann auch für einfache Operationen verwendet werden, indem eine Multiplikation mit $1$ durchgeführt wird. +> +> Beispiel: +> +> $$\int{ln(x)}dx = \int{ln(x) \cdot 1}dx$$ + +#### Partialbruchzerlegung +Für die Partialbruchzerlegung muss der zu integrierende Bruch **vollständig gekürzt** sein. + +Die Partialbruchzerlegung wird anhand der folgenden Aufgabe erklärt: + +$$\int{\frac{x + 1}{x^3 + 5x^2 + 8x - 4}}dx$$ + +Folgende Schritte müssen durchgeführt werden: + +**1. Nullstellen des Nenners bestimmen** + +Durch erraten: (Eine der Nullstellen ist $1$) + +Linearfaktor abspalten mit Horner-Schema: + +![](PartInteg.png) + +Verbleibendes Polynom: $x^2 - 4x + 4 = (x - 2) \cdot (x - 2)$ + +Die verbleibende, **doppelte** Nullstelle ist also $x = 2$. + +**2. Jeder Nullstelle eine Summe von Nullstellen zuordnen** + + - Für einfache Nullstellen: + $$\rightarrow \frac{A}{x - x_1}$$ + - Für doppelte Nullstellen: + $$\rightarrow \frac{A_1}{x - x_1} + \frac{A_2}{(x - x_1)^2}$$ + - Für $r$-fache Nullstellen: + $$\rightarrow \frac{A_1}{x - x_1} + \frac{A_2}{(x - x_1)^2} + ... + \frac{A_r}{(x - x_1)^r}$$ + +Das ergibt im Fall des Beispiels folgendes: + +$$\frac{x + 1}{x^3 - 5x^2 + 8x - 4} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2}$$ + +**3. Brüche gleichnamig machen** + +$$\frac{x + 1}{(x - 1)(x - 2)^2} = \frac{A(x - 2)^2}{(x + 1)(x - 2)^2} + \frac{B(x - 1)(x - 2)}{(x + 1)(x - 2)^2} + \frac{C(x - 1)}{(x + 1)(x - 2)^2}$$ + +Da die Brüche nun gleichnamig sind, können sämtliche Nenner weggekürzt werden: + +$$x + 1 = A(x - 2)^2 + B(x - 1)(x - 2) + C(x - 1)$$ + +**4. Koeffizienten mit Hilfe von LGS ausrechnen** + +Gleichung ausmultiplizieren und nach Potenz zerlegen: + +$$(A + B)x^2 + (C - 4A - 3B)x + (4A + 2B - C)$$ + + - $x + 1$ beinhaltet kein $x^2$. Also muss $(A + B)$ zwingend $0$ ergeben: + $$A + B = 0$$ + - $x + 1$ beinhaltet $1x$. Also muss $C - 4A - 3B$ $1$ ergeben: + $$C - 4A - 3B = 1$$ + - $x + 1$ beinhaltet eine Konstante $1$. Also muss $4A + 2B - C$ $1$ ergeben: + $$4A + 2B - C = 1$$ + +Das Lösen des Gleichungssystems ergibt folgende Resultate: + + - $A = 2$ + - $B = -2$ + - $C = 3$ + +Daraus ergibt sich folgendes: + +$$\frac{x + 1}{(x - 1)(x - 2)^2} = \frac{2}{x - 1} + \frac{-2}{x - 2} + \frac{3}{(x - 2)^2}$$ + +**5. Integration der Teilbrüche** + +Für Nenner ohne Exponent: + +$$\int{\frac{1}{x - x_0}}dx = \int{\frac{1}{u}}du = \ln(|u|) + c = \ln(|x - x_0|) + c$$ + +Für Nenner mit Exponent: +$$\begin{split} + \int{\frac{1}{(x - x_0)^r}}dx = \int{u^{-r}}du & = \frac{u^{-r + 1}}{-r + 1} + c \\ + & = \frac{(x - x_0)^{-r + 1}}{1 - r} + c \\ + & = \frac{1}{(1 - r)(x - x_o)^{r - 1}} + c +\end{split}$$ + +Für das aktuelle Beispiel ergibt das folgendes: + +$$\begin{split} + \int{\frac{2}{x - 1} + \frac{2}{x - 2} + \frac{3}{(x - 2)^2}}dx & = \int{\frac{2}{x - 1}} - \int{\frac{2}{x - 2}} + \int{\frac{3}{(x - 1)^2}} \\ + & = 2 \cdot \int{\frac{1}{x - 1} - 2 \cdot \int{\frac{1}{x - 2}}} + 3 \cdot \int{\frac{1}{(x - 2)^2}} \\ + & = 2 \cdot \ln(|x - 1|) - 2 \cdot \ln(|x - 2|) - 3 \cdot \frac{1}{x - 2} + c +\end{split}$$ + +#### Leitfaden +![](ProductInt.png) + +[^Derivation]: [Ableitungen][Derivation] + +[Derivation]: ../.../../../Semester%201/AN1%20-%20Analysis%201/Ableitungen.md \ No newline at end of file