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\documentclass{../../../lecture}
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\begin{document}
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\begin{satz}[Reihenentwicklung Sinus / Cosinus]
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Für alle $x \in \R$ gilt (absolut konvergente
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Potenzreihendarstellung)
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\[
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\cos(x) = \sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k}}{(2k)!} = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \ldots
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.\] und
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\[
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\sin(x) = \sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \ldots
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.\]
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\end{satz}
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\begin{proof}
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Die absolute Konvergenz folgt als Teilreihe der Exponentialreihe (als Majorante)
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Es gilt für $m \in \N_0$
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\[
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i^{n} = \begin{cases}
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1 & n = 4m \\
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i & n = 4m+1 \\
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-1 & n = 4m+2 \\
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-i & n = 4m+3
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\end{cases}
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.\] Es folgt
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\begin{align*}
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e^{ix} &= \sum_{n=0}^{\infty} \frac{(ix)^{n}}{n!} = \sum_{n=0}^{\infty} i^{n} \frac{x^{n}}{n!} \\
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&= \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)}
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.\end{align*}
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\end{proof}
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\begin{satz}[Restgliedabschätzung Sinus / Cosinus]
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Für $n \in \N_0$ gilt
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\[
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\cos(x) = \sum_{k=0}^{n} (-1)^{k} \frac{x^{2k}}{(2k)!} + R_{2n+2}(x)
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.\] und
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\[
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\sin(x)= \sum_{k=0}^{n} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!} + R_{2n+3}(x)
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.\]
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mit
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\[
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|R_{2n+2}(x)| \le \frac{|x|^{2n+2}}{(2n+2)!} \text{ für } |x| \le 2n+3
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.\] bzw.
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\[
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|R_{2n+3}(x)| \le \frac{|x|^{2n+3}}{(2n+3)!} \text{ für } |x| \le 2n+4
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.\]
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\end{satz}
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\begin{proof}
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Es gilt
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\begin{align*}
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R_{2n+2}(x) &= \sum_{k=n+1}^{\infty} (-1)^{k} \frac{x^{2k}}{(2k)!} \\
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&= (-1)^{n+1} \frac{x^{2n+2}}{(2n+2)!}
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\left( \sum_{k=n+1}^{\infty} (-1)^{k-(n+1)}
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\frac{x^{2(k - (n+1))}}{(2k)! \frac{1}{(2n+2)!}}\right) \\
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&= (-1)^{n+1} \frac{x^{2n+2}}{(2n+2)!}
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\left( \sum_{k=0}^{\infty} (-1)^{k}
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\frac{x^{2k}(2n+2)!}{(2k+2n+2)!} \right)
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.\end{align*}
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Für $k \in \N$ setze
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\begin{align*}
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a_k :&= \frac{x^{2k}(2n+2)!}{(2k+2n+2)!}
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= \frac{x^{2k}}{(2n+3)(2n+4) \ldots (2k + 2n + 2)} \\
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a_{k-1} &= \frac{x^{2k-2}(2n+2)!}{(2k+2n)!}
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\intertext{damit}
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a_k &= a_{k-1} \cdot \frac{x^{2}}{(2k+2n+1)(2k+2n+2)}
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.\end{align*}
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Es gilt für $|x| \le 2n+3, k\ge 1$
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\[
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\frac{x^{2}}{(2k+2n+1)(2k+2n+2)} \le \frac{(2n+3)^{2}}{(2n+3)(2n+4)} < 1
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.\] $\implies$
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\[
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a_k \le \frac{(2n+3)^{k}}{(2n+4)^{k}} a_0 \quad a_0 = \frac{1}{(2n+2)!}
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.\] $\stackrel{\text{Leibniz}}{\implies}$
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\[
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\sum_{k=0}^{\infty} (-1)^{k} a_k
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.\] konvergent mit
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\[
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0 < \underbrace{\underbrace{1 - a_1}_{> 0} + \underbrace{a_2 - a_3}_{> 0}
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+ \underbrace{a_4 - \ldots}_{> 0}}_{< 1} < 1
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.\] $\implies$
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\[
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|R_{2n+2}(x)| \le \frac{|x|^{2n+2}}{(2n+2)!} \text{ für } |x| \le 2n+3
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.\] Genauso für $R_{2n+3}(x)$ (Sinus).
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\end{proof}
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\begin{lemma}
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Sinus und Cosinus Funktionen haben das folgende Verhalten
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\[
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\lim_{x \to 0} \frac{\sin(x)}{x} = 1
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.\]
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\[
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\lim_{x \to 0} \frac{\cos(x)-1}{x} = 0
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.\]
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\end{lemma}
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\begin{proof}
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\begin{align*}
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\left| \frac{\sin(x)}{x} - 1 \right|
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&= \left| \underbrace{1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!}}_{\frac{\sin(x)}{x}} - \ldots - 1\right| \\
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&= \left| x \sum_{k=1}^{\infty} (-1)^{k}\frac{x^{2k-1}}{(2k+1)!} \right| \\
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&\stackrel{|x| < 1}{\le |x|} \cdot \left| \sum_{k=1}^{\infty} \frac{1}{(2k+1)!} \right|
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\le |x| \cdot e
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.\end{align*} $\implies$
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\[
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\underbrace{\left| \frac{\sin(x)}{x} -1 \right|}_{\to 0}
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\le \underbrace{|x| \cdot e}_{\to 0}
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.\]
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genauso für $\lim_{x \to 0} \frac{\cos(x) - 1}{x}$.
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\end{proof}
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\subsection{Die Zahl $\pi$}
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Ziel: Analytische Definition von $\pi \in \R$.
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\begin{satz}[und Definition]
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Die Funktion $\cos\colon [0,2] \to \R$ hat genau eine Nullstelle
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im Intervall $[0,2]$, welche mit $\frac{\pi}{2}$ bezeichnet
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wird ($\pi := 2 \frac{\pi}{2}$ ).
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\end{satz}
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\begin{proof}
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in 4 Schritten.
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Schritt 1 / Lemma 1: $\cos(2) \le -\frac{1}{3}$. \\
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Restgliedabschätzung liefert ($|x| \le 5$ ).
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\[
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\cos(x) = 1 - \frac{x^2}{2} + R_4(x) \text{ mit } |R_4(x)| \le \frac{|x|^{4}}{24}
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.\] $\implies$
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\[
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\cos(2) = 1 - 2 + \underbrace{R_4(2)}_{\le \frac{16}{24} = \frac{2}{3}} \le -1 + \frac{2}{3} = -\frac{1}{3}
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.\]
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Schritt 2 / Lemma 2: $\sin(x) > 0$ $\forall x \in \; ]0, 2[$\\
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Es gilt
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\begin{align*}
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\sin(x) = x + R_3(x) = x (1 + \frac{R_3(x)}{x})
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\left| \frac{R_3(x)}{x} \right| \le \frac{|x|^2}{6}
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\stackrel{0 < x \le 2}{\le} \frac{4}{6} = \frac{2}{3}
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\intertext{$\implies$}
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1 + \frac{R_3(x)}{x} \ge \frac{1}{3}
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.\end{align*}
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Schritt 3 / Lemma 3: $\cos: [0,2] \to \R$ ist streng monoton fallend.\\
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Sei $0 \le y < x \le 2$. Dann gilt
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\begin{align*}
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\cos(x) - \cos(y) \stackrel{\text{Additionstheorem}}{=}
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- 2 \underbrace{\sin\left( \frac{x+y}{2} \right)}_{> 0}
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\underbrace{\sin\left( \frac{x-y}{2} \right)}_{> 0} < 0
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.\end{align*}
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Schritt 4 (Beweis der Definition von $\pi$ )
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$\cos(0) = 1$ (nach Definition).
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\[
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\cos(2) \le - \frac{1}{3} \stackrel{\text{Zwischenwertsatz}}{\implies}
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\exists x_0 \in [0,2] \text{ mit } \cos(x_0) = 0
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.\] Nach Lemma 3 ist $x_0$ eindeutig.
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\end{proof}
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\begin{korrolar}[Spezielle Werte von $\exp$]
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Es gilt: $e^{i \frac{\pi}{2}} = i$,
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$e^{i \pi} = -1$, $e^{i \frac{3\pi}{2}} = -i$, $e^{2\pi i} = 1$
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\end{korrolar}
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\begin{proof}
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Übung.
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\end{proof}
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\begin{korrolar}[Eigenschaften Sinus / Cosinus]
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$\forall x \in \R$ gilt:
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\begin{enumerate}[(i)]
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\item $\cos(x + 2\pi) = \cos(x) \quad \sin(x+2\pi) = \sin(x)$ \\
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$2 \pi$: Periodizität
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\item $\cos(x + \pi) = - \cos(x) \quad \sin(x+ \pi) = - \sin(x)$
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\item $\cos(x) = \sin(\frac{\pi}{2} - x) \quad \sin(x) = \cos(\frac{\pi}{2} - x)$
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\item Nullstellen von $\sin / \cos$.\\
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$\{x \in \R | \sin x = 0\} = \{x = k\pi | k \in \Z\} $ \\
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$\{x \in \R | \cos x = 0\} = \{x = \left(k+\frac{1}{2}\right)\pi | k \in \Z\} $ \\
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\end{enumerate}
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\end{korrolar}
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\begin{proof}
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folgt aus den Additionstheoremen, der Definition von $\frac{\pi}{2}$,
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den speziellen Werten von $\exp$ und folgender Tabelle
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\begin{tabular}{l|l|l|l|l|l}
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x & 0 & $\frac{\pi}{2}$ & $\pi$ & $\frac{3}{2} \pi$ & $2 \pi$ \\ \hline
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$\cos x$ & 1 & 0 & $-1$ & 0 & 1 \\ \hline
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$\sin x$ & 0 & 1 & 0 & $-1$ & 0 \\
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\end{tabular}.
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\end{proof}
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\begin{korrolar}[$e^{z} = 1$]
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Es gilt $\{z \in \mathbb{C} | e^{z} = 1\} = \{i 2 \pi k | k \in \Z\} $
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\end{korrolar}
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\begin{proof}
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ohne Beweis.
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\end{proof}
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\begin{definition}[Tangens, Cotangens]
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\begin{enumerate}[(i)]
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\item Die Tangensfunktion
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\begin{align*}
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&\tan: \R \setminus \{x = (k + \frac{1}{2}) \pi | k \in \Z\}
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\to \R
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\intertext{ist definiert durch}
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&\tan x := \frac{\sin x}{\cos x}
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.\end{align*}
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\item Die Cotangensfunktion
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\begin{align*}
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&\cot: \R \setminus \{x = k \pi | k \in \Z\} \to \R
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\intertext{ist definiert durch}
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&\cot x := \frac{\cos(x)}{\sin(x)}
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.\end{align*}
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\end{enumerate}
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\end{definition}
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\begin{figure}[htpb]
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\centering
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\begin{tikzpicture}
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\begin{axis}%
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[grid=both,
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minor tick num=4,
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grid style={line width=.1pt, draw=gray!10},
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major grid style={line width=.2pt,draw=gray!50},
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axis lines=middle,
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enlargelimits={abs=0.2},
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ymax=5,
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ymin=-5
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]
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\addplot[domain=-3:3,samples=50,smooth,red] {tan(deg(x))};
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\end{axis}
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\end{tikzpicture}
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\caption{$\tan(x)$}
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\end{figure}
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\begin{figure}[htpb]
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\centering
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\begin{tikzpicture}
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\begin{axis}%
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[grid=both,
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minor tick num=4,
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grid style={line width=.1pt, draw=gray!10},
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major grid style={line width=.2pt,draw=gray!50},
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axis lines=middle,
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enlargelimits={abs=0.2},
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ymax=5,
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ymin=-5
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]
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\addplot[domain=-3:3,samples=50,smooth,red] {cot(deg(x))};
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\end{axis}
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\end{tikzpicture}
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\caption{$\cot(x)$}
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\end{figure}
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\begin{definition}[Arcusfunktionen (Umkehrfunktionen der Trigonometrischen
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Funktionen)]
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\begin{enumerate}[(i)]
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\item $\cos\colon [0, \pi] \to [-1, 1]$ ist streng monoton fallend
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und bijektiv. Die Umkehrfunktion heißt
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Arcus-Cosinus.
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\[
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\arccos: [-1,1] \to [0, \pi]
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.\]
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\item $\sin\colon \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \to [-1, 1]$
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ist streng monoton wachsend und bijektiv. Die Umkehrfunktion
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heißt Arcus-Sinus.
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\[
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\arcsin: [-1,1] \to \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]
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.\]
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\item $\tan\colon \; ] - \frac{\pi}{2}, \frac{\pi}{2} [ \to \R$ ist streng
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monoton wachsend und bijektiv. Die Umkehrfunktion heißt
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Arcus-Tangens.
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\[
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\arctan: \R \to ] - \frac{\pi}{2}, \frac{\pi}{2} [
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.\]
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\end{enumerate}
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\end{definition}
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\begin{satz}[Polarkoordinaten]
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Jedes $z \in \mathbb{C}$ lässt sich schreiben als
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$z = r\cdot e^{i \varphi}$, $\varphi \in \R$ und
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$r = |z| \in [0, \infty[$.
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Für $z \neq 0$ ist $\varphi$ bis auf ein ganzzahliges Vielfaches von
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$2\pi$ eindeutig bestimmt.
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\end{satz}
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\begin{proof}
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Rannacher.
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\end{proof}
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\end{document}
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\documentclass[uebung]{../../../lecture}
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\title{Übungsblatt 9 Analysis 1}
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\author{Leon Burgard, Christian Merten, Mittwoch Übungsgruppe}
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\begin{document}
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% punkte tabelle
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\begin{tabular}{|c|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|m{1cm}|@{}m{0cm}@{}}
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\hline
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Aufgabe & \centering A1 & \centering A2 & \centering A3 & \centering A4
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& \centering A5 & \centering A6 & \centering A7 & \centering A8
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& \centering $\sum$ & \\[5mm] \hline
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Punkte & & & & & & & & & & \\[5mm] \hline
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\end{tabular}
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\begin{aufgabe}[Vollständige Induktion]
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\begin{enumerate}[(a)]
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\item Beh.:
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\[
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\sum_{k=1}^{n} k^{2} = \frac{1}{6} n (n+1) (2n+1)
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.\]
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\begin{proof}
|
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durch vollständige Induktion
|
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I.A.: $n=1 $
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\[
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\sum_{k=1}^{1} k^2 = 1 = \frac{1}{6} \cdot 2 \cdot 3
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.\]
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I.S.: $n \to n+1$. Es existiere ein festes aber beliebiges
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$n \in \N$ mit $\sum_{k=1}^{n} k^2 = \frac{1}{6}n(n+1)(2n+1)$.
|
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\begin{align*}
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\frac{1}{6}(n+1)(n+2)(2(n+1)+1)
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&= \frac{1}{6} (n^2 + 3n + 2)(2n +3) \\
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&= \frac{1}{6} (2n^{3} + 3n^2 + n + 6n^2 + 12n + 6) \\
|
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&= \frac{1}{6} (2n^{3} + 3n^2 +n) + n^2 + 2n + 1 \\
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||||
&= \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 \\
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&\stackrel{\text{I.V.}}{=} \sum_{k=1}^{n} k^2 + (n+1)^2 \\
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&= \sum_{k=1}^{n+1} k^2
|
||||
.\end{align*}
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\end{proof}
|
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\item Beh.:
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||||
\[
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\sum_{k=1}^{n} (3k+2)^2 = \frac{1}{2} n \left( 6n^2 + 21n + 23 \right)
|
||||
.\]
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||||
\begin{proof}
|
||||
\begin{align*}
|
||||
\sum_{k=1}^{n} (3k+2)^2 &= \sum_{k=1}^{n} (9k^2 + 12 k + 4) \\
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||||
&= 9 \sum_{k=1}^{n} k^2
|
||||
+ 12 \sum_{k=1}^{n} k
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||||
+ \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) \\
|
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&= \frac{1}{2} n \left( 6n^2 + 21n + 23 \right)
|
||||
.\end{align*}
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\end{aufgabe}
|
||||
|
||||
\begin{aufgabe}
|
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\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}
|
||||
Reference in New Issue
Block a user