Add dima u03
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school/di-ma/uebung/03/03_1.tex
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89
school/di-ma/uebung/03/03_1.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 \gruppe {2 (Mi 10-12 Andre)}
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\def \qname {Pascal Brackmann}
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\def \qmatrikel {108017113834} %
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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 3.1}
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\begin{enumerate}[1.]
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\item Getränke $\widehat{=}$ Bälle (unterscheidbar); Tablets $\widehat{=}$ Urnen (nicht unterscheidbar)
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In jeder Urne mindestens ein Ball $\Rightarrow$ surjektiv.
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$\Rightarrow$ $S_{n,m}$ mit $n = 8$, $m = 4$
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\begin{tabular}{c|cccccccccc}
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$n/m$ & 0 & 1 & 2 & 3 & 4 \\\hline
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0 & 1 &&&&\\
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1 & 0 & 1 & & & \\
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2 & 0 & 1 & 1 & & \\
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3 & 0 &1 &3 &1 & \\
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4 & 0 &1 &7 &6 &1 \\
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5 & 0 &1 &15 &25 &10 \\
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6 & 0 &1 &31 &90 &65 \\
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7 & 0 &1 &63 &301 &350 \\
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8 & 0 &1 &127 &966 &\underline{1701} \\
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\end{tabular}
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\begin{eqnarray*}
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S_{8,4} = 1701
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\end{eqnarray*}
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\item
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\begin{align}
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S_{8,4} &+& S_{8,3} &+& S_{8,2} &+& S_{8,1} &=& \\
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1701 &+& 966 &+& 127 &+& 1 &=& 2795
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\end{align}
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\item Freunde $\widehat{=}$ Urnen (unterscheidbar); Getränke $\widehat{=}$ Bällen (unterscheidbar).
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Mindestens ein Ball pro Urne $\Rightarrow$ surjektiv.
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$\Rightarrow$ $m! * S_{n,m}$ mit $n = 8$, $m = 4$.
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\begin{eqnarray*}
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4! * S_{8,4} &= 24 * 1701 \\
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&= 40824
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\end{eqnarray*}
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\end{enumerate}
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\end{document}
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64
school/di-ma/uebung/03/03_2.tex
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64
school/di-ma/uebung/03/03_2.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 3.2}
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\begin{itemize}
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\item Hilfsdiagonale von $(1,0)$ nach $(12,11)$ legen und damit ein relatives \enquote{Unterkoordinatensystem}
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erzeugen, in dem von Punkt $(1,0)$ gestartet wird.
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\item Analog zur korrekten Klammerung, muss der erste Schritt nach rechts (öffnende Klammer) und der letzte Schritt
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nach oben (schließende Klammer) gehen
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\item Wir betrachten die Diagonale von $(1,0)$ nach $(12,11)$. Hier gibt es $C_{11}$ gültige wege.
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\item Weiter betrachten wir $(2,1)$ nach $(11,10)$ (was analog ist zu $(2,1)$ nach $(12,11)$, da es nur einen
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gültigen Weg von $(11,10)$ nach $(12,10)$ gibt) $= C_{10}$ Wege, die man von $(3,0)$ aus \underline{nicht} gehen
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kann.
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\item $\Rightarrow C_{11} - C_{10} = 58786 - 16796 = 41990$
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\end{itemize}
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\end{document}
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76
school/di-ma/uebung/03/03_3.tex
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76
school/di-ma/uebung/03/03_3.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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75
school/di-ma/uebung/03/03_4.tex
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75
school/di-ma/uebung/03/03_4.tex
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@ -0,0 +1,75 @@
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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{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 3.4}
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% \begin{itemize}
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% \item $o$ impliziert $O$
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% \item $\omega$ impliziert $\Omega$
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% \item $f = \Theta(g) \Rightarrow f = \Omega(g) \text{ und } f = O(g)$
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% \item $f = O(g) \Rightarrow g = \Omega(f)$
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% \item $f = o(g) \Rightarrow g = \omega(f)$
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% \end{itemize}
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\begin{tabular}{|l||l|l|l|l|l|l|l|l|}
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\hline
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\backslashbox{f}{g} & $4^n$ & $e^{2n}$ & $\sum\limits_{k=1}^n k$ & $n \ln n$ & $2^{(2^{n+1})}$ & $51^{\ln n}$ & $2^{(2^n)}$ &
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$2^{\sqrt{\ln n}}$ \\\hline
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$4^n$ & $\Theta$ & $o$ & $o$ & $\omega$ & $O$ & $\omega$ & $O$ & $\omega$ \\\hline
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$e^{2n}$ & $\omega$ & $\Theta$ & $o$ & $\omega$ & $\omega$ & $\omega$ & $\omega$ & $\omega$ \\\hline
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$\sum\limits_{k=1}^n k$ & $\omega$ & $\omega$ & $\Theta$ & $\omega$ & $\omega$ & $\omega$ & $\omega$ & $\omega$ \\\hline
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$n \ln n$ & $o$ & $o$ & $o$ & $\Theta$ & $o$ & $o$ & $o$ & $\omega$ \\\hline
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$2^{(2^n)}$ & $\Omega$ & $o$ & $o$ & $\omega$ & $\Theta$ & $\Omega$ & $\Omega$ & $\omega$ \\\hline
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$51^{\ln n}$ & $o$ & $o$ & $o$ & $\omega$ & $O$ & $\Theta$ & $O$ & $\omega$ \\\hline
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$2^{(2n)}$ & $\Omega$ & $o$ & $o$ & $\omega$ & $O$ & $\Omega$ & $\Theta$ & $\omega$ \\\hline
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$2^{\sqrt{\ln n}}$ & $o$ & $o$ & $o$ & $o$ & $o$ & $o$ & $o$ & $\Theta$ \\\hline
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\end{tabular}
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\end{document}
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41
school/di-ma/uebung/03/st.py
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41
school/di-ma/uebung/03/st.py
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#!/usr/bin/env python
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def p(n,k):
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if k == 0:
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return 0
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elif n == k or k == 1:
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return 1
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elif n < k:
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return 0
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else:
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return p(n-1,k-1) + p(n-k,k)
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def s(n,k):
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if n == k:
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return 1
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elif k == 0:
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return 0
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elif n < k:
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return 0
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else:
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return s(n-1,k-1) + (n-1)*s(n-1,k)
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def S(n, k):
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if n == k:
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return " S_{" + n + "," + k + "} "
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# return 1
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elif k == 1:
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return " S_{" + n + "," + k + "} "
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# return 1
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elif k == 0:
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return " S_{" + n + "," + k + "} "
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# return 0
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elif n < k:
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return " S_{" + n + "," + k + "} "
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# return 0
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else:
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return " S_{" + str(n-1) + "," + str(k-1) + "} + " + str(k) + " * S_{" + str(n-1) + "," + str(k) + "} "
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# return S(n-1,k-1) + k * S(n-1,k)
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print(S(5,3))
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# print(p(10,4))
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22
school/di-ma/uebung/03/tmp
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22
school/di-ma/uebung/03/tmp
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12 → \item Hilfsdiagonale von $(1,0)$ nach $(12,11)$ legen und damit ein
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↪relatives \enquote{Unterkoordinatensystem}¬
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11 → → erzeugen, in dem von Punkt $(1,0)$ gestartet wird.¬
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10 ¬
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9 → \item Analog zur korrekten Klammerung, muss der erste Schritt nach rechts
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↪(öffnende Klammer) und der letzte Schritt¬
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8 → → nach oben (schließende Klammer) gehen¬
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7 ¬
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6 → \item Wir betrachten die Diagonale von $(1,0)$ nach $(12,11)$. Hier gibt es
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↪$C_{11}$ gültige wege.¬
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5 ¬
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4 → \item Weiter betrachten wir $(2,1)$ nach $(11,10)$ (was analog ist zu $(2,
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↪1)$ nach $(12,11)$, da es nur einen¬
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3 → → gültigen Weg von $(11,10)$ nach $(12,10)$ gibt) $= C_{10}$ Wege, die man
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↪von $(3,0)$ aus \underline{nicht} gehen¬
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2 → → kann.¬
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1 ¬
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59 → \item $\Rightarrow C_{11} - C_{10} = 58786 - 16796 = 41990$¬
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1 ¬
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2 ¬
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3 \end{itemize}¬
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