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| \documentclass{../../../lecture} | |||
| \begin{document} | |||
| \begin{satz}[Reihenentwicklung Sinus / Cosinus] | |||
| Für alle $x \in \R$ gilt (absolut konvergente | |||
| Potenzreihendarstellung) | |||
| \[ | |||
| \cos(x) = \sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k}}{(2k)!} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \ldots | |||
| .\] und | |||
| \[ | |||
| \sin(x) = \sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \ldots | |||
| .\] | |||
| \end{satz} | |||
| \begin{proof} | |||
| Die absolute Konvergenz folgt als Teilreihe der Exponentialreihe (als Majorante) | |||
| Es gilt für $m \in \N_0$ | |||
| \[ | |||
| i^{n} = \begin{cases} | |||
| 1 & n = 4m \\ | |||
| i & n = 4m+1 \\ | |||
| -1 & n = 4m+2 \\ | |||
| -i & n = 4m+3 | |||
| \end{cases} | |||
| .\] Es folgt | |||
| \begin{align*} | |||
| e^{ix} &= \sum_{n=0}^{\infty} \frac{(ix)^{n}}{n!} = \sum_{n=0}^{\infty} i^{n} \frac{x^{n}}{n!} \\ | |||
| &= \underbrace{\sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k}}{(2k)!}}_{\cos(x)} + i \underbrace{\sum_{k=0}^{\infty} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!}}_{\sin(x)} | |||
| .\end{align*} | |||
| \end{proof} | |||
| \begin{satz}[Restgliedabschätzung Sinus / Cosinus] | |||
| Für $n \in \N_0$ gilt | |||
| \[ | |||
| \cos(x) = \sum_{k=0}^{n} (-1)^{k} \frac{x^{2k}}{(2k)!} + R_{2n+2}(x) | |||
| .\] und | |||
| \[ | |||
| \sin(x)= \sum_{k=0}^{n} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!} + R_{2n+3}(x) | |||
| .\] | |||
| mit | |||
| \[ | |||
| |R_{2n+2}(x)| \le \frac{|x|^{2n+2}}{(2n+2)!} \text{ für } |x| \le 2n+3 | |||
| .\] bzw. | |||
| \[ | |||
| |R_{2n+3}(x)| \le \frac{|x|^{2n+3}}{(2n+3)!} \text{ für } |x| \le 2n+4 | |||
| .\] | |||
| \end{satz} | |||
| \begin{proof} | |||
| Es gilt | |||
| \begin{align*} | |||
| R_{2n+2}(x) &= \sum_{k=n+1}^{\infty} (-1)^{k} \frac{x^{2k}}{(2k)!} \\ | |||
| &= (-1)^{n+1} \frac{x^{2n+2}}{(2n+2)!} | |||
| \left( \sum_{k=n+1}^{\infty} (-1)^{k-(n+1)} | |||
| \frac{x^{2(k - (n+1))}}{(2k)! \frac{1}{(2n+2)!}}\right) \\ | |||
| &= (-1)^{n+1} \frac{x^{2n+2}}{(2n+2)!} | |||
| \left( \sum_{k=0}^{\infty} (-1)^{k} | |||
| \frac{x^{2k}(2n+2)!}{(2k+2n+2)!} \right) | |||
| .\end{align*} | |||
| Für $k \in \N$ setze | |||
| \begin{align*} | |||
| a_k :&= \frac{x^{2k}(2n+2)!}{(2k+2n+2)!} | |||
| = \frac{x^{2k}}{(2n+3)(2n+4) \ldots (2k + 2n + 2)} \\ | |||
| a_{k-1} &= \frac{x^{2k-2}(2n+2)!}{(2k+2n)!} | |||
| \intertext{damit} | |||
| a_k &= a_{k-1} \cdot \frac{x^{2}}{(2k+2n+1)(2k+2n+2)} | |||
| .\end{align*} | |||
| Es gilt für $|x| \le 2n+3, k\ge 1$ | |||
| \[ | |||
| \frac{x^{2}}{(2k+2n+1)(2k+2n+2)} \le \frac{(2n+3)^{2}}{(2n+3)(2n+4)} < 1 | |||
| .\] $\implies$ | |||
| \[ | |||
| a_k \le \frac{(2n+3)^{k}}{(2n+4)^{k}} a_0 \quad a_0 = \frac{1}{(2n+2)!} | |||
| .\] $\stackrel{\text{Leibniz}}{\implies}$ | |||
| \[ | |||
| \sum_{k=0}^{\infty} (-1)^{k} a_k | |||
| .\] konvergent mit | |||
| \[ | |||
| 0 < \underbrace{\underbrace{1 - a_1}_{> 0} + \underbrace{a_2 - a_3}_{> 0} | |||
| + \underbrace{a_4 - \ldots}_{> 0}}_{< 1} < 1 | |||
| .\] $\implies$ | |||
| \[ | |||
| |R_{2n+2}(x)| \le \frac{|x|^{2n+2}}{(2n+2)!} \text{ für } |x| \le 2n+3 | |||
| .\] Genauso für $R_{2n+3}(x)$ (Sinus). | |||
| \end{proof} | |||
| \begin{lemma} | |||
| Sinus und Cosinus Funktionen haben das folgende Verhalten | |||
| \[ | |||
| \lim_{x \to 0} \frac{\sin(x)}{x} = 1 | |||
| .\] | |||
| \[ | |||
| \lim_{x \to 0} \frac{\cos(x)-1}{x} = 0 | |||
| .\] | |||
| \end{lemma} | |||
| \begin{proof} | |||
| \begin{align*} | |||
| \left| \frac{\sin(x)}{x} - 1 \right| | |||
| &= \left| \underbrace{1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!}}_{\frac{\sin(x)}{x}} - \ldots - 1\right| \\ | |||
| &= \left| x \sum_{k=1}^{\infty} (-1)^{k}\frac{x^{2k-1}}{(2k+1)!} \right| \\ | |||
| &\stackrel{|x| < 1}{\le |x|} \cdot \left| \sum_{k=1}^{\infty} \frac{1}{(2k+1)!} \right| | |||
| \le |x| \cdot e | |||
| .\end{align*} $\implies$ | |||
| \[ | |||
| \underbrace{\left| \frac{\sin(x)}{x} -1 \right|}_{\to 0} | |||
| \le \underbrace{|x| \cdot e}_{\to 0} | |||
| .\] | |||
| genauso für $\lim_{x \to 0} \frac{\cos(x) - 1}{x}$. | |||
| \end{proof} | |||
| \subsection{Die Zahl $\pi$} | |||
| Ziel: Analytische Definition von $\pi \in \R$. | |||
| \begin{satz}[und Definition] | |||
| Die Funktion $\cos\colon [0,2] \to \R$ hat genau eine Nullstelle | |||
| im Intervall $[0,2]$, welche mit $\frac{\pi}{2}$ bezeichnet | |||
| wird ($\pi := 2 \frac{\pi}{2}$ ). | |||
| \end{satz} | |||
| \begin{proof} | |||
| in 4 Schritten. | |||
| Schritt 1 / Lemma 1: $\cos(2) \le -\frac{1}{3}$. \\ | |||
| Restgliedabschätzung liefert ($|x| \le 5$ ). | |||
| \[ | |||
| \cos(x) = 1 - \frac{x^2}{2} + R_4(x) \text{ mit } |R_4(x)| \le \frac{|x|^{4}}{24} | |||
| .\] $\implies$ | |||
| \[ | |||
| \cos(2) = 1 - 2 + \underbrace{R_4(2)}_{\le \frac{16}{24} = \frac{2}{3}} \le -1 + \frac{2}{3} = -\frac{1}{3} | |||
| .\] | |||
| Schritt 2 / Lemma 2: $\sin(x) > 0$ $\forall x \in \; ]0, 2[$\\ | |||
| Es gilt | |||
| \begin{align*} | |||
| \sin(x) = x + R_3(x) = x (1 + \frac{R_3(x)}{x}) | |||
| \left| \frac{R_3(x)}{x} \right| \le \frac{|x|^2}{6} | |||
| \stackrel{0 < x \le 2}{\le} \frac{4}{6} = \frac{2}{3} | |||
| \intertext{$\implies$} | |||
| 1 + \frac{R_3(x)}{x} \ge \frac{1}{3} | |||
| .\end{align*} | |||
| Schritt 3 / Lemma 3: $\cos: [0,2] \to \R$ ist streng monoton fallend.\\ | |||
| Sei $0 \le y < x \le 2$. Dann gilt | |||
| \begin{align*} | |||
| \cos(x) - \cos(y) \stackrel{\text{Additionstheorem}}{=} | |||
| - 2 \underbrace{\sin\left( \frac{x+y}{2} \right)}_{> 0} | |||
| \underbrace{\sin\left( \frac{x-y}{2} \right)}_{> 0} < 0 | |||
| .\end{align*} | |||
| Schritt 4 (Beweis der Definition von $\pi$ ) | |||
| $\cos(0) = 1$ (nach Definition). | |||
| \[ | |||
| \cos(2) \le - \frac{1}{3} \stackrel{\text{Zwischenwertsatz}}{\implies} | |||
| \exists x_0 \in [0,2] \text{ mit } \cos(x_0) = 0 | |||
| .\] Nach Lemma 3 ist $x_0$ eindeutig. | |||
| \end{proof} | |||
| \begin{korrolar}[Spezielle Werte von $\exp$] | |||
| Es gilt: $e^{i \frac{\pi}{2}} = i$, | |||
| $e^{i \pi} = -1$, $e^{i \frac{3\pi}{2}} = -i$, $e^{2\pi i} = 1$ | |||
| \end{korrolar} | |||
| \begin{proof} | |||
| Übung. | |||
| \end{proof} | |||
| \begin{korrolar}[Eigenschaften Sinus / Cosinus] | |||
| $\forall x \in \R$ gilt: | |||
| \begin{enumerate}[(i)] | |||
| \item $\cos(x + 2\pi) = \cos(x) \quad \sin(x+2\pi) = \sin(x)$ \\ | |||
| $2 \pi$: Periodizität | |||
| \item $\cos(x + \pi) = - \cos(x) \quad \sin(x+ \pi) = - \sin(x)$ | |||
| \item $\cos(x) = \sin(\frac{\pi}{2} - x) \quad \sin(x) = \cos(\frac{\pi}{2} - x)$ | |||
| \item Nullstellen von $\sin / \cos$.\\ | |||
| $\{x \in \R | \sin x = 0\} = \{x = k\pi | k \in \Z\} $ \\ | |||
| $\{x \in \R | \cos x = 0\} = \{x = \left(k+\frac{1}{2}\right)\pi | k \in \Z\} $ \\ | |||
| \end{enumerate} | |||
| \end{korrolar} | |||
| \begin{proof} | |||
| folgt aus den Additionstheoremen, der Definition von $\frac{\pi}{2}$, | |||
| den speziellen Werten von $\exp$ und folgender Tabelle | |||
| \begin{tabular}{l|l|l|l|l|l} | |||
| x & 0 & $\frac{\pi}{2}$ & $\pi$ & $\frac{3}{2} \pi$ & $2 \pi$ \\ \hline | |||
| $\cos x$ & 1 & 0 & $-1$ & 0 & 1 \\ \hline | |||
| $\sin x$ & 0 & 1 & 0 & $-1$ & 0 \\ | |||
| \end{tabular}. | |||
| \end{proof} | |||
| \begin{korrolar}[$e^{z} = 1$] | |||
| Es gilt $\{z \in \mathbb{C} | e^{z} = 1\} = \{i 2 \pi k | k \in \Z\} $ | |||
| \end{korrolar} | |||
| \begin{proof} | |||
| ohne Beweis. | |||
| \end{proof} | |||
| \begin{definition}[Tangens, Cotangens] | |||
| \begin{enumerate}[(i)] | |||
| \item Die Tangensfunktion | |||
| \begin{align*} | |||
| &\tan: \R \setminus \{x = (k + \frac{1}{2}) \pi | k \in \Z\} | |||
| \to \R | |||
| \intertext{ist definiert durch} | |||
| &\tan x := \frac{\sin x}{\cos x} | |||
| .\end{align*} | |||
| \item Die Cotangensfunktion | |||
| \begin{align*} | |||
| &\cot: \R \setminus \{x = k \pi | k \in \Z\} \to \R | |||
| \intertext{ist definiert durch} | |||
| &\cot x := \frac{\cos(x)}{\sin(x)} | |||
| .\end{align*} | |||
| \end{enumerate} | |||
| \end{definition} | |||
| \begin{figure}[htpb] | |||
| \centering | |||
| \begin{tikzpicture} | |||
| \begin{axis}% | |||
| [grid=both, | |||
| minor tick num=4, | |||
| grid style={line width=.1pt, draw=gray!10}, | |||
| major grid style={line width=.2pt,draw=gray!50}, | |||
| axis lines=middle, | |||
| enlargelimits={abs=0.2}, | |||
| ymax=5, | |||
| ymin=-5 | |||
| ] | |||
| \addplot[domain=-3:3,samples=50,smooth,red] {tan(deg(x))}; | |||
| \end{axis} | |||
| \end{tikzpicture} | |||
| \caption{$\tan(x)$} | |||
| \end{figure} | |||
| \begin{figure}[htpb] | |||
| \centering | |||
| \begin{tikzpicture} | |||
| \begin{axis}% | |||
| [grid=both, | |||
| minor tick num=4, | |||
| grid style={line width=.1pt, draw=gray!10}, | |||
| major grid style={line width=.2pt,draw=gray!50}, | |||
| axis lines=middle, | |||
| enlargelimits={abs=0.2}, | |||
| ymax=5, | |||
| ymin=-5 | |||
| ] | |||
| \addplot[domain=-3:3,samples=50,smooth,red] {cot(deg(x))}; | |||
| \end{axis} | |||
| \end{tikzpicture} | |||
| \caption{$\cot(x)$} | |||
| \end{figure} | |||
| \begin{definition}[Arcusfunktionen (Umkehrfunktionen der Trigonometrischen | |||
| Funktionen)] | |||
| \begin{enumerate}[(i)] | |||
| \item $\cos\colon [0, \pi] \to [-1, 1]$ ist streng monoton fallend | |||
| und bijektiv. Die Umkehrfunktion heißt | |||
| Arcus-Cosinus. | |||
| \[ | |||
| \arccos: [-1,1] \to [0, \pi] | |||
| .\] | |||
| \item $\sin\colon \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \to [-1, 1]$ | |||
| ist streng monoton wachsend und bijektiv. Die Umkehrfunktion | |||
| heißt Arcus-Sinus. | |||
| \[ | |||
| \arcsin: [-1,1] \to \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] | |||
| .\] | |||
| \item $\tan\colon \; ] - \frac{\pi}{2}, \frac{\pi}{2} [ \to \R$ ist streng | |||
| monoton wachsend und bijektiv. Die Umkehrfunktion heißt | |||
| Arcus-Tangens. | |||
| \[ | |||
| \arctan: \R \to ] - \frac{\pi}{2}, \frac{\pi}{2} [ | |||
| .\] | |||
| \end{enumerate} | |||
| \end{definition} | |||
| \begin{satz}[Polarkoordinaten] | |||
| Jedes $z \in \mathbb{C}$ lässt sich schreiben als | |||
| $z = r\cdot e^{i \varphi}$, $\varphi \in \R$ und | |||
| $r = |z| \in [0, \infty[$. | |||
| Für $z \neq 0$ ist $\varphi$ bis auf ein ganzzahliges Vielfaches von | |||
| $2\pi$ eindeutig bestimmt. | |||
| \end{satz} | |||
| \begin{proof} | |||
| Rannacher. | |||
| \end{proof} | |||
| \end{document} | |||
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| \documentclass[uebung]{../../../lecture} | |||
| \title{Übungsblatt 9 Analysis 1} | |||
| \author{Leon Burgard, Christian Merten, Mittwoch Übungsgruppe} | |||
| \begin{document} | |||
| % punkte tabelle | |||
| \begin{tabular}{|c|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|@{}m{0cm}@{}} | |||
| \hline | |||
| Aufgabe & \centering A1 & \centering A2 & \centering A3 & \centering A4 | |||
| & \centering A5 & \centering A6 & \centering A7 & \centering A8 | |||
| & \centering $\sum$ & \\[5mm] \hline | |||
| Punkte & & & & & & & & & & \\[5mm] \hline | |||
| \end{tabular} | |||
| \begin{aufgabe}[Vollständige Induktion] | |||
| \begin{enumerate}[(a)] | |||
| \item Beh.: | |||
| \[ | |||
| \sum_{k=1}^{n} k^{2} = \frac{1}{6} n (n+1) (2n+1) | |||
| .\] | |||
| \begin{proof} | |||
| durch vollständige Induktion | |||
| I.A.: $n=1 $ | |||
| \[ | |||
| \sum_{k=1}^{1} k^2 = 1 = \frac{1}{6} \cdot 2 \cdot 3 | |||
| .\] | |||
| I.S.: $n \to n+1$. Es existiere ein festes aber beliebiges | |||
| $n \in \N$ mit $\sum_{k=1}^{n} k^2 = \frac{1}{6}n(n+1)(2n+1)$. | |||
| \begin{align*} | |||
| \frac{1}{6}(n+1)(n+2)(2(n+1)+1) | |||
| &= \frac{1}{6} (n^2 + 3n + 2)(2n +3) \\ | |||
| &= \frac{1}{6} (2n^{3} + 3n^2 + n + 6n^2 + 12n + 6) \\ | |||
| &= \frac{1}{6} (2n^{3} + 3n^2 +n) + n^2 + 2n + 1 \\ | |||
| &= \frac{1}{6}n(2n^{3} + 3n + 1) + (n+1)^{2} \\ | |||
| &= \frac{1}{6}n (n (2n+3) +1) + (n+1)^2 \\ | |||
| &= \frac{1}{6}n (n+1)(2n+1) + (n+1)^2 \\ | |||
| &\stackrel{\text{I.V.}}{=} \sum_{k=1}^{n} k^2 + (n+1)^2 \\ | |||
| &= \sum_{k=1}^{n+1} k^2 | |||
| .\end{align*} | |||
| \end{proof} | |||
| \item Beh.: | |||
| \[ | |||
| \sum_{k=1}^{n} (3k+2)^2 = \frac{1}{2} n \left( 6n^2 + 21n + 23 \right) | |||
| .\] | |||
| \begin{proof} | |||
| \begin{align*} | |||
| \sum_{k=1}^{n} (3k+2)^2 &= \sum_{k=1}^{n} (9k^2 + 12 k + 4) \\ | |||
| &= 9 \sum_{k=1}^{n} k^2 | |||
| + 12 \sum_{k=1}^{n} k | |||
| + \sum_{k=1}^{n} 4 \\ | |||
| &\stackrel{\text{(a) und kl. Gauß}}{=} | |||
| \frac{3}{2} n(n+1)(2n+1) + 6n (n+1) + 4n \\ | |||
| &= \frac{1}{2} n \left( 3(n+1)(2n+1) + 12n + 12 + 8 \right) \\ | |||
| &= \frac{1}{2} n \left( 6n^2 + 21n + 23 \right) | |||
| .\end{align*} | |||
| \end{proof} | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \begin{enumerate}[(a)] | |||
| \item Beh.: $a_n := \sqrt[n]{n F^{n}}$. $\lim_{n \to \infty} a_n = F$ | |||
| \begin{proof} $\lim_{n \to \infty} a_n | |||
| = \lim_{n \to \infty} \sqrt[n]{n} \cdot F = F$ | |||
| \end{proof} | |||
| \item Beh.: $b_n := \sum_{k=0}^{n} \left(\frac{\rho -1}{\rho}\right)^{k}$, $\rho \in [2,100]. \lim_{n \to \infty} b_n = \rho$ | |||
| \begin{proof} | |||
| Mit $q := \frac{\rho - 1}{\rho}$ folgt $0 < q < 1$ $\forall \rho > 1 $. \\ | |||
| $\implies$ | |||
| \begin{align*} | |||
| \lim_{n \to \infty} b_n \stackrel{\text{geometrische Reihe}}{=} \frac{1}{1- q} = \frac{1}{1 - \frac{\rho - 1}{\rho}} = \frac{1}{\frac{\rho - (\rho - 1)}{\rho}} = \rho | |||
| .\end{align*} | |||
| \end{proof} | |||
| \item Beh.: $c_n := \sum_{k=0}^{n} \frac{s^{k}}{k!}$. $\lim_{n \to \infty} c_n = e^{S}$ | |||
| \begin{proof} | |||
| Mit $e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!}$ folgt direkt | |||
| $\lim_{n \to \infty} c_n = e^{S}$ | |||
| \end{proof} | |||
| \item Beh.: $d_n := \frac{3 - Fn^{5}}{\frac{n^{5}}{E} + n} | |||
| \cdot \frac{R - GSTn}{\frac{U}{n} + Gn}$. $\lim_{n \to \infty} d_n = FEST$ | |||
| \begin{proof} | |||
| \begin{align*} | |||
| d_n &= \frac{3 - Fn^{5}}{\frac{n^{5}}{E} + n} | |||
| \cdot \frac{R - GSTn}{\frac{U}{n} + Gn} | |||
| = \frac{\frac{3}{n^{5}} - F}{\frac{1}{E} + \frac{1}{n^{4}}} \cdot \frac{\frac{R}{n} - GST}{\frac{U}{n^2} + G} | |||
| \intertext{$\implies$} | |||
| \lim_{n \to \infty} d_n &= \frac{F}{\frac{1}{E}} \cdot \frac{GST}{G} = FEST | |||
| .\end{align*} | |||
| \end{proof} | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \begin{enumerate}[(a)] | |||
| \item | |||
| \begin{enumerate}[(1)] | |||
| \item $\sum_{k=0}^{\infty} k$ konvergiert nicht, | |||
| da $k$ keine Nullfolge. Die Folge der Partialsummen ist: $s_n = \sum_{k=0}^{n} k \stackrel{\text{kl. Gauß}}{=} \frac{n(n+1)}{2}$. | |||
| \item $\sum_{m=1}^{\infty} \frac{1}{m(m+1)} = \sum_{m=1}^{\infty} \frac{1}{k^2+k} < \sum_{m=1}^{\infty} \frac{1}{k^2}$ $\forall k \in \N$ | |||
| ist konvergent, da $\sum_{k=1}^{\infty} \frac{1}{k^2}$ | |||
| konvergente Majorante. Die Folge der Partialsummen ist | |||
| $s_n := \sum_{m=1}^{n} \frac{1}{m(m+1)}$. | |||
| \end{enumerate} | |||
| \item | |||
| \begin{enumerate}[(i)] | |||
| \item | |||
| \begin{align*} | |||
| \sum_{k=2}^{\infty} \frac{4\cdot 2^{k+1}}{3^{k}} | |||
| = \sum_{k=2}^{\infty} \frac{4 \cdot 2 \cdot 2^{k}}{3^{k}} | |||
| = 8 \sum_{k=2}^{\infty} \left(\frac{2}{3}\right)^{k} | |||
| = 8 \left( \sum_{k=0}^{\infty} \left( \frac{2}{3} \right)^{k} | |||
| - \sum_{k=0}^{1} \left( \frac{2}{3} \right)^{k}\right) | |||
| &= 8 \left( 3 - \frac{5}{3} \right) = \frac{32}{3} | |||
| .\end{align*} | |||
| \item | |||
| \begin{align*} | |||
| \sum_{k=0}^{\infty} (-1)^{k} \frac{1}{\sqrt{3^{k+1}} - \sqrt{3^{k}} } = \sum_{k=0}^{\infty} (-1)^{k} \frac{1}{\sqrt{3^{k}}(\sqrt{3} - 1)} | |||
| &\qquad \;= \frac{1}{\sqrt{3} - 1} \sum_{k=0}^{\infty} \left(-\frac{1}{\sqrt{3} }\right)^{k} \\ | |||
| &\stackrel{\text{Geometr. Reihe}}{=} \frac{1}{\sqrt{3} - 1} \cdot \frac{1}{1 + \frac{1}{\sqrt{3} }} \\ | |||
| &\qquad \;= \frac{1}{\sqrt{3} - \frac{1}{\sqrt{3} }} \\ | |||
| &\qquad \;= \frac{\sqrt{3} }{2} | |||
| .\end{align*} | |||
| \end{enumerate} | |||
| \item | |||
| \begin{enumerate}[(i)] | |||
| \item $\sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k}$ | |||
| ist nicht absolut konvergent, | |||
| da $\sum_{k=1}^{\infty} \frac{1}{k}$ divergiert. | |||
| \item $\sum_{k=1}^{\infty} \frac{2^{k}}{k!} = e^2 - 1$ | |||
| konvergiert absolut. | |||
| \end{enumerate} | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \begin{enumerate}[(a)] | |||
| \item Sei $h\colon \R \to \R$ eine beschränkte und | |||
| $u\colon \R \to \R$ definiert durch $u(x) := x \cdot h(x)$. | |||
| Beh.: $u$ im Punkt $x_0 = 0$ stetig. | |||
| \begin{proof} | |||
| Da $h$ beschränkt $\implies$ $\exists C \in \R$, s.d. | |||
| $|h(x)| \le C$ $\forall x \in \R$. Also gilt | |||
| $|u(x)| \le x \cdot C$ $\forall x \in \R$. | |||
| Damit folgt | |||
| \begin{align*} | |||
| 0 \le \lim_{x \nearrow 0} |u(x)| | |||
| \le \lim_{x \nearrow 0} x \cdot C = 0 | |||
| \intertext{und} | |||
| 0 \le \lim_{x \searrow 0} |u(x)| | |||
| \le \lim_{x \searrow 0} x \cdot C = 0 | |||
| \intertext{$\implies$} | |||
| \lim_{x \nearrow 0} u(x) = 0 = \lim_{x \searrow 0} u(x) | |||
| .\end{align*} | |||
| $\implies f$ stetig in $x_0$. | |||
| \end{proof} | |||
| \item Beh.: | |||
| \[ | |||
| f(x) := \begin{cases} | |||
| 1 & x \in \Q \\ | |||
| -1 & x \in \R \setminus \Q | |||
| \end{cases} | |||
| .\] ist unstetig auf ganz $\R$ aber $|f(x)|$ ist stetig auf $\R$. | |||
| \begin{proof} | |||
| $f(x)$ ist unstetig analog zur Dirichlet Funktion und | |||
| $|f(x)| = 1$ ist offensichtlich stetig. | |||
| \end{proof} | |||
| \item Beh.: Es gibt keine Funktion die im Punkt $x_0 = 0$ stetig | |||
| und in allen anderen Punkten unstetig ist. | |||
| \begin{proof} | |||
| Sei $f\colon \R \to \R$ stetig in $x_0 = 0$ und $\epsilon > 0$ | |||
| beliebig. Dann $\exists \delta > 0$, s.d. | |||
| $\forall x \in \R\colon |x| < \delta $ | |||
| $|f(x) - f(0)| < \frac{\epsilon}{2}$. Wähle | |||
| $a := \frac{\delta }{2}$. | |||
| Zz.: $f$ ist stetig in $a$. | |||
| Wähle $\delta' := \frac{\delta}{2}$. Sei $x' \in \R$ | |||
| mit $|x' - a| < \frac{\delta }{2}$. Dann | |||
| gilt $|f(0) - f(x')| < \frac{\epsilon}{2}$. Mit | |||
| $|f(0) - f(a)| < \frac{\epsilon}{2}$ folgt | |||
| \begin{align*} | |||
| |f(a) - f(x')| &= |f(a) - f(0) + f(0) - f(x')| \\ | |||
| &\le |f(a) - f(0)| + |f(0) - f(x')| \\ | |||
| &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon | |||
| .\end{align*} $\implies f$ stetig in $a$. | |||
| \end{proof} | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \begin{enumerate}[(a)] | |||
| \item | |||
| Sei $(a_n)_{n \in \N}$ Folge in $\R^{+}$. | |||
| Beh (i) .: | |||
| \[ | |||
| \frac{a_{n+1}}{a_n} \xrightarrow{n \to \infty} a \implies | |||
| \sqrt[n]{a_n} \xrightarrow{n \to \infty} a | |||
| .\] | |||
| \begin{proof} | |||
| Sei $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = a$. Damit gilt | |||
| \[ | |||
| \liminf_{n \to \infty} \frac{a_{n+1}}{a_n} = a = \limsup_{n \to \infty} \frac{a_{n+1}}{a_n} | |||
| .\] Mit Blatt 7 folgt damit: | |||
| \[ | |||
| a = \liminf_{n \to \infty} \frac{a_{n+1}}{a_n} | |||
| \le \liminf_{n \to \infty} \sqrt[n]{a_n} | |||
| \le \limsup_{n \to \infty} \sqrt[n]{a_n} | |||
| \le \limsup_{n \to \infty} \frac{a_{n+1}}{a_n} = a | |||
| .\] Also $\liminf_{n \to \infty} \sqrt[n]{a_n} = a = \limsup_{n \to \infty} \sqrt[n]{a_n} $ \\ | |||
| $\implies \lim_{n \to \infty} \sqrt[n]{a_n} = a$. | |||
| \end{proof} | |||
| Beh (ii) .: | |||
| \[ | |||
| \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \xrightarrow{n \to \infty} | |||
| \infty \implies \sqrt[n]{a_n} \xrightarrow{n \to \infty} \infty | |||
| .\] | |||
| \begin{proof} | |||
| Sei $\frac{a_{n+1}}{a_n} \xrightarrow{n \to \infty} \infty$. | |||
| Dann existiert eine streng monoton wachsende, nach oben | |||
| unbeschränkte Teilfolge | |||
| $(a_{n_k})_{k \in\N}$ von $(a_n)_{n\in\N}$. | |||
| Sei nun $q > 1$ beliebig. Dann $\exists k_0 \in \N$, s.d. | |||
| $\forall k > k_0\colon \frac{a_{n_k}}{a_{n_{k-1}}} > q$. Damit | |||
| folgt: | |||
| \begin{align*} | |||
| &a_{n_k} > q \cdot a_{n_{k-1}} > q^{2} \cdot a_{n_{k - 2}} | |||
| > \ldots > q^{k - k_0} a_{n_{k_0}} \\ | |||
| \implies& \sqrt[k]{a_{n_k}} > q^{1 - \frac{k_0}{k}} \sqrt[k]{a_{n_{k_0}}} | |||
| .\end{align*} Für $k \to \infty$ folgt | |||
| \[ | |||
| \limsup_{k \to \infty} \sqrt[k]{a_{n_k}} > q | |||
| .\] Da $q > 1$ beliebig groß folgt damit | |||
| \[ | |||
| \limsup_{n \to \infty} \sqrt[n]{a_n} = \infty | |||
| .\] | |||
| \end{proof} | |||
| \item | |||
| \begin{enumerate}[(i)] | |||
| \item $a_n := \sqrt[n]{n!}$. Mit | |||
| $\frac{(n+1)!}{n!} = n+1 \xrightarrow{n \to \infty} \infty$ | |||
| folgt mit (a ii) $a_n \xrightarrow{n \to \infty} \infty$. | |||
| \item $b_n := \sqrt[n]{\frac{n^{n}}{n!}}$ | |||
| \[ | |||
| \frac{(n+1)^{n+1}}{(n+1)!} \cdot \frac{n!}{n^{n}} | |||
| = \frac{(n+1)^{n}}{n^{n}} | |||
| = \left( \frac{n+1}{n} \right) ^{n} | |||
| = \left( 1 + \frac{1}{n} \right)^{n} | |||
| \xrightarrow{n \to \infty} e | |||
| .\] | |||
| Mit (a i) folgt direkt $\lim_{n \to \infty} b_n = e$. | |||
| \item $c_n := \frac{n^{n}}{n!} = | |||
| \sqrt[n]{\left( \frac{n^{n}}{n!} \right)^{n}}$. | |||
| \begin{align*} | |||
| \left( \frac{(n+1)^{n+1}}{(n+1)!} \right)^{n+1} | |||
| \cdot \left( \frac{n!}{n^{n}} \right)^{n} | |||
| &= \frac{(n+1)^{(n+1)(n+1)}}{((n+1)!)^{n+1}} | |||
| \cdot \frac{(n!)^{n}}{n^{n^2}} \\ | |||
| &= \frac{(n+1)^{n^2 + 2n + 1}}{(n+1)!(n+1)^{n}} | |||
| \cdot \frac{1}{n^{n^2}} \\ | |||
| &= \frac{(n+1)^{n^2 + n + 1}}{(n+1)! \cdot n^{n^2}} \\ | |||
| &> \frac{(n+1)^{n^2 + n + 1}}{(n+1)^{n} \cdot (n+1)^{n^2}} \\ | |||
| &= \frac{(n+1)^{n^2 + n + 1}}{(n+1)^{n + n^2}} \\ | |||
| &= n+1 \xrightarrow{n \to \infty} \infty | |||
| .\end{align*} | |||
| Mit (a ii) folgt damit $c_n \xrightarrow{n \to \infty} \infty$. | |||
| \end{enumerate} | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} Ergebnisse | |||
| \begin{tabular}{m{1.5cm}|m{3cm}|m{3cm}|m{3cm}|m{3.5cm}@{}m{0pt}@{}} | |||
| Aufgabe & Beschränkt nach unten & Beschränkt nach oben & Monoton? & Konvergent? & \\[2mm] \hline | |||
| (a) & Ja, durch $\frac{1}{2}$ & Ja, durch $1$ & Ja, streng monoton wachsend & Ja, da monoton und beschränkt & \\[5mm] \hline | |||
| (b) & Ja, durch $1$ & Ja, durch $2$ & Nein & Ja, nach Quotientenkriterium für Folgen & \\[2mm] | |||
| \end{tabular} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \begin{enumerate}[(a)] | |||
| \item | |||
| \begin{enumerate}[(i)] | |||
| \item ist konvergent nach Leibniz Kriterium, da | |||
| $\frac{1}{\ln(k)}$ monoton fallende Nullfolge. | |||
| \item ist divergent, da $(-1)^{k} \frac{(k+1)^{k}-k^{k}}{(k+1)^{k}}$ | |||
| keine Nullfolge ist. | |||
| \item $\sum_{k=2}^{\infty} 2^{k}\cdot \frac{1}{2^{k}\cdot \ln^2(2^{k})}$ ist konvergent und damit ist (iii) nach Verdichtungskriterium konvergent. | |||
| \end{enumerate} | |||
| \item | |||
| Der Konvergenzradius $\rho$ ist $\frac{1}{4}$, da | |||
| Häufungspunkte von $\sqrt[k]{|a_k|}$ bei $\pi$ und $4$ vorliegen. | |||
| Wegen $4 > \pi$ ist damit | |||
| $\limsup_{k \to \infty} \sqrt[k]{a_k} = 4$. | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \begin{aufgabe} | |||
| \begin{enumerate}[(a)] | |||
| \item $\lim_{x \to \infty} \frac{2x + 3}{5x + 1} = \frac{2}{5}$ | |||
| \item $\lim_{x \to \infty} \sqrt{4x^2-2x+3} -2x = -\frac{1}{2}$ | |||
| \item $\lim_{x \to \infty} 2^{-x} = 0$ | |||
| \item $\lim_{x \to \infty} \frac{x+\sin(x)}{x} = 1$ | |||
| \item $\lim_{x \to 1} \frac{x^2 - x}{x^2 - 1} = \frac{1}{2}$ | |||
| \item $\frac{x^2 + x}{x^2 -1} \to \infty$ für $x \searrow 1$. | |||
| \end{enumerate} | |||
| \end{aufgabe} | |||
| \end{document} | |||