2 измененных файлов: 318 добавлений и 0 удалений
-
Двоичные данныеws2019/ana/uebungen/ana9.pdf
-
+318 -0ws2019/ana/uebungen/ana9.tex
Двоичные данные
ws2019/ana/uebungen/ana9.pdf
Просмотреть файл
+ 318
- 0
ws2019/ana/uebungen/ana9.tex
Просмотреть файл
| @@ -0,0 +1,318 @@ | |||||
| \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} | |||||