77 lines
2.4 KiB
TeX
77 lines
2.4 KiB
TeX
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\documentclass[12pt,a4paper,german]{article}
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\usepackage{url}
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%\usepackage{graphics}
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\usepackage{times}
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\usepackage[T1]{fontenc}
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\usepackage{ngerman}
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\usepackage{float}
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\usepackage{diagbox}
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\usepackage[utf8]{inputenc}
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\usepackage{geometry}
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\usepackage{amsfonts}
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\usepackage{amsmath}
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\usepackage{csquotes}
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\usepackage{graphicx}
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\usepackage{epsfig}
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\usepackage{paralist}
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\geometry{left=2.0cm,textwidth=17cm,top=3.5cm,textheight=23cm}
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%%%%%%%%%% Fill out the the definitions %%%%%%%%%
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\def \name {Valentin Brandl} %
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\def \matrikel {108018274494} %
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\def \pname {Marvin Herrmann} %
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\def \pmatrikel {108018265436} %
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\def \qname {Pascal Brackmann}
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\def \qmatrikel {108017113834} %
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\def \gruppe {2 (Mi 10-12 Andre)}
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\def \uebung {3} %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% DO NOT MODIFY THIS HEADER
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\newcommand{\hwsol}{
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\vspace*{-2cm}
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\noindent \matrikel \quad \name \hfill \"Ubungsgruppe: \gruppe \\
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\noindent \pmatrikel \quad \pname \\
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\noindent \qmatrikel \quad \qname \\
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\begin{center}{\Large \bf L\"osung f\"ur \"Ubung \# \uebung}\end{center}
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}
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\begin{document}
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%Import header
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\hwsol
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\section*{Aufgabe 3.3}
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\begin{enumerate}[1.]
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\item
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\begin{eqnarray*}
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\ln n &= o(e^{\sqrt{\ln n}}) \\
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\text{Anwendung des Satz von L'Hospital:} \\
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\lim\limits_{x \to x_0} \frac{f(x)}{g(x)} &= \lim\limits_{x \to x_0} \frac{f'(x)}{g'(x)} \\
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\Rightarrow &\lim\limits_{n \to \infty} \left( \frac{\frac{1}{n}}{e^{\sqrt{\ln(n)}}
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\frac{1}{2n\sqrt{\ln(n)}}} \right) \\
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&= \lim\limits_{n \to \infty} \left( \frac{\frac{1}{n}n}{e^{\sqrt{\ln(n)}}\frac{n}{2n\sqrt{\ln(n)}}}
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\right) \\
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&= \lim\limits_{n \to \infty} \left( \frac{1}{\frac{e^{\sqrt{\ln(n)}}}{2\sqrt{\ln(n)}}} \right) \\
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&= \lim\limits_{n \to \infty} \frac{1}{\infty} \\
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\end{eqnarray*}
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da $e^{\sqrt{\ln(n)}}$ schneller wächst als $2\sqrt{\ln(n)}$
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$\Rightarrow \ln n = o(e^{\sqrt{\ln n}})$
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\item
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\begin{eqnarray*}
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e^{\sqrt{\ln n}} &= o(n^{\varepsilon}) \text{ für alle } \varepsilon \in \mathbb{R}, \varepsilon > 0 \\
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\lim\limits_{n \to \infty} \frac{e^{\sqrt{\ln n}}}{e^{\ln(n^{\varepsilon})}} =
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\lim\limits_{n \to \infty} \frac{e^{\sqrt{\ln n}}}{e^{\varepsilon \cdot \ln n}} = 0
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\end{eqnarray*}
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Gilt für alle $\varepsilon > 0$, da $\ln n$ schneller wächst als $\sqrt{\ln n}$.
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$\Rightarrow e^{\sqrt{\ln n}} = o(n^{\varepsilon})$
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\end{enumerate}
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\end{document}
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