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MATH1-ALA: Abgaben hinzugefügt
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\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
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\usepackage[T1]{fontenc}
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\usepackage[utf8]{inputenc}
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\usepackage[ngerman]{babel}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{paralist}
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\usepackage{gauss}
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\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
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\usepackage{tikz}
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\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
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\usepackage{polynom}
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\polyset{style=C, div=:,vars=x}
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\pagenumbering{arabic}
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\def\thesection{\arabic{section})}
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\def\thesubsection{\alph{subsection})}
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\def\thesubsubsection{(\roman{subsubsection})}
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\makeatletter
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\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
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\hskip -\arraycolsep
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\let\@ifnextchar\new@ifnextchar
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\array{#1}}
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\makeatother
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\begin{document}
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\author{Jim Martens}
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\title{Hausaufgaben zum 11. April}
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\maketitle
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\section{} %1
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\begin{equation*}
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\frac{3}{x+5} \geq 3
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\end{equation*}
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1. Fall $x > -5$:\\
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\begin{alignat*}{3}
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&& \frac{3}{x+5} &\geq & 3 \\
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\Leftrightarrow && 3 &\geq & 3(x+5) \\
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\Leftrightarrow && 3 &\geq & 3x + 15 \\
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\Leftrightarrow && -12 &\geq & 3x \\
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\Leftrightarrow && -4 &\geq & x
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\end{alignat*}
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\\
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2. Fall $x < -5$:\\
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\begin{alignat*}{3}
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&& \frac{3}{x+5} &\geq & 3 \\
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\Leftrightarrow && 3 &\leq & 3(x+5) \\
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\Leftrightarrow && 3 &\leq & 3x + 15 \\
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\Leftrightarrow && -12 &\leq & 3x \\
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\Leftrightarrow && -4 &\leq & x
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\end{alignat*}
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\\
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$L = [-4]$
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\section{} %2
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\begin{equation*}
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|3x-4| \geq 2
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\end{equation*}
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1. Fall $x \geq \frac{4}{3}$:\\
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\begin{alignat*}{3}
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&& |3x-4| &\geq & 2 \\
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\Leftrightarrow && 3x-4 &\geq & 2 \\
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\Leftrightarrow && 3x &\geq & 6 \\
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\Leftrightarrow && x &\geq & 2
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\end{alignat*}
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\\
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2. Fall $x < \frac{4}{3}$:\\
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\begin{alignat*}{3}
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&& |3x-4| &\geq & 2 \\
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\Leftrightarrow && -(3x-4) &\geq & 2 \\
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\Leftrightarrow && -3x + 4 &\geq & 2 \\
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\Leftrightarrow && -3x &\geq & -2 \\
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\Leftrightarrow && x &\leq & \frac{2}{3}
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\end{alignat*}
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\\
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$L = (-\infty,\frac{2}{3}] \cup [2,\infty)$
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\section{} %3
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\subsection{} %a
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\begin{alignat*}{3}
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&& |a_{n} - a| &=& |\frac{2n-1}{n+3} - 2| \\
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\Leftrightarrow && &=& |\frac{2n-1}{n+3} - \frac{2(n+3)}{n+3}| \\
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\Leftrightarrow && &=& |\frac{2n-1 - 2n - 6}{n+3}| \\
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\Leftrightarrow && &=& |\frac{-7}{n+3}| \\
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\Leftrightarrow && &=& \frac{7}{n+3}
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\end{alignat*}
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\subsection{} %b
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Es sei $\varepsilon > 0$. Aufgrund von a) gilt:\\
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\begin{alignat*}{3}
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&& |a_{n} - a| &<& \varepsilon \label{eq:1}\tag{1}\\
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\Leftrightarrow && |a_{n} - a| = |\frac{-7}{n+3}| = \frac{7}{n+3} &<& \varepsilon \\
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\Leftrightarrow && -7 &<& \varepsilon (n+3) \\
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\Leftrightarrow && \frac{-7}{\varepsilon} &<& n+3 \\
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\Leftrightarrow && \frac{-7}{\varepsilon} - 3 &<& n
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\end{alignat*}
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\\
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Wählt man $N > \frac{-7}{\varepsilon} - 3$, so ergibt sich aus \eqref{eq:1}, dass $|a_{n} - a| < \varepsilon$ für alle $n \geq N$ gilt. Das zeigt $(a_{n}) \rightarrow a = 2$.
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\subsection{} %c
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Es sei $\varepsilon = \frac{1}{10}$:\\
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\begin{alignat*}{3}
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&& \frac{-7}{\frac{1}{10}} - 3 &<& n \\
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\Leftrightarrow && -70 - 3 &<& n \\
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\Leftrightarrow && -73 &<& n
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\end{alignat*}
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Wählt man $N = -72$, so ergibt sich aus \eqref{eq:1}, dass $|a_{n} - a| < \varepsilon$ für alle $n \geq N$ gilt.\\
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\\
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Es sei $\varepsilon = \frac{1}{100}$:\\
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\begin{alignat*}{3}
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&& \frac{-7}{\frac{1}{100}} - 3 &<& n \\
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\Leftrightarrow && -700 - 3 &<& n \\
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\Leftrightarrow && -703 &<& n
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\end{alignat*}
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Wählt man $N = -702$, so ergibt sich aus \eqref{eq:1}, dass $|a_{n} - a| < \varepsilon$ für alle $n \geq N$ gilt.\\
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\\
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Es sei $\varepsilon = \frac{1}{100000}$:\\
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\begin{alignat*}{3}
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&& \frac{-7}{\frac{1}{100000}} - 3 &<& n \\
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\Leftrightarrow && -700000 - 3 &<& n \\
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\Leftrightarrow && -700003 &<& n
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\end{alignat*}
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Wählt man $N = -700002$, so ergibt sich aus \eqref{eq:1}, dass $|a_{n} - a| < \varepsilon$ für alle $n \geq N$ gilt.
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\section{} %4
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\textbf{Behauptung:} Die folgende Aussage gilt für alle $n \in \mathbb{N}$:\\
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\begin{equation*}
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0 \leq a_{n} < \frac{1}{2} \label{eq:2}\tag{2}
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\end{equation*}\\
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Die Folge ($a_{n}$) sei rekursiv definiert durch \\
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\begin{alignat*}{2}
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a_{1} &=& \frac{2}{5} \label{eq:3}\tag{3}\\
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a_{n+1} &=& a_{n}^{2} + \frac{1}{4} \label{eq:4}\tag{4}
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\end{alignat*}
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\textbf{Beweis:} Durch vollständige Induktion.\\
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Mit $A(n)$ sei die Aussage \eqref{eq:2} bezeichnet.\\\\
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\underline{Induktionsanfang:} \\
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$A(1)$ ist richtig, da die Aussage \eqref{eq:2} für \eqref{eq:3} wie folgt gilt:
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\begin{alignat*}{2}
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0 \leq \frac{2}{5} = \frac{4}{10} < \frac{1}{2} = \frac{5}{10}
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\end{alignat*}\\
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\underline{Induktionsannahme:}\\
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Die Aussage \eqref{eq:2} gilt für ein beliebig fest gewähltes $n \in \mathbb{N}$.\\\\
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\underline{Zu zeigen:}\\
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$A(n+1)$ gilt, d. h. Folgendes gilt für die Aussage \eqref{eq:4}:\\
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\begin{equation*}
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0 \leq a_{n+1} < \frac{1}{2} \label{eq:5}\tag{5}
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\end{equation*}
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\underline{Induktionsschluss:}\\
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Aus \eqref{eq:5} folgt für $0 \leq a_{n+1}$ Folgendes:\\
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\begin{alignat*}{3}
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&& 0 &\leq & a_{n}^{2} + \frac{1}{4}
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\end{alignat*}
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Diese Aussage gilt, da $\frac{1}{4}$ auf triviale Weise die Aussage erfüllt und $a_{n}^{2}$ immer positiv oder gleich Null sein muss, da eine beliebige Zahl zum Quadrat immer größer gleich Null ist.\\
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Für $a_{n+1} < \frac{1}{2}$ ergibt sich Folgendes:\\
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\begin{alignat*}{5}
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&& a_{n}^{2} + \frac{1}{4} &<& \frac{1}{2} && \;&|& -\frac{1}{4} \\
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\Leftrightarrow && a_{n}^{2} &<& \frac{1}{4} && && \\
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\Leftrightarrow && a_{n} \cdot a_{n} &<& \frac{1}{4} && &&
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\end{alignat*}
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Aufgrund der Induktionsannahme gilt $a_{n} < \frac{1}{2}$. Daher ist das Quadrat von $a_{n}$ auf jeden Fall kleiner als $\frac{1}{4}$.\\
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Nach dem Induktionsprinzip folgt aus dem Induktionsanfang und dem Induktionsschluss die Behauptung. \hfill $\Box$\\
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Es ist zu zeigen, dass $a_{n+1} \geq a_{n}$ für alle $n \in \mathbb{N}$ gilt. Es ergibt sich Folgendes:\\
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\begin{alignat*}{5}
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&& a_{n+1} &\geq & a_{n} && && \\
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\Leftrightarrow && a_{n}^{2} + \frac{1}{4} &\geq & a_{n} && \;&|& -a_{n} \\
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\Leftrightarrow && a_{n}^{2} - a_{n} + \frac{1}{4} &\geq & 0 && \;&|& \text{Binomische Formel erzeugen} \\
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\Leftrightarrow && (a_{n} - \frac{1}{2})^{2} &\geq & 0 && &&
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\end{alignat*}
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Diese Aussage gilt, da ein Quadrat einer beliebigen Zahl immer größer gleich Null ist.
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\end{document}
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@ -0,0 +1,253 @@
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\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
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\usepackage[T1]{fontenc}
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\usepackage[utf8]{inputenc}
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\usepackage[ngerman]{babel}
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\usepackage{amsmath}
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\usepackage{amsfonts}
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\usepackage{amssymb}
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\usepackage{paralist}
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\usepackage{gauss}
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\usepackage{pgfplots}
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\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
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\usepackage{tikz}
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\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
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\usepackage{polynom}
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\polyset{style=C, div=:,vars=x}
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\pgfplotsset{compat=1.8}
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\pagenumbering{arabic}
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\def\thesection{\arabic{section})}
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\def\thesubsection{\alph{subsection})}
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\def\thesubsubsection{(\roman{subsubsection})}
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\makeatletter
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\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
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\hskip -\arraycolsep
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\let\@ifnextchar\new@ifnextchar
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\array{#1}}
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\makeatother
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\begin{document}
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\author{Jim Martens (6420323)}
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\title{Hausaufgaben zum 27. Juni}
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\maketitle
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\section{} %1
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\subsubsection{} %i
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\begin{alignat*}{2}
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f(x,y) &=& 2x^{2}y^{2} - 3xy + 4x + 2 \\
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f_{x} &=& 4xy^{2} - 3y + 4 \\
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f_{y} &=& 4x^{2}y - 3x
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\end{alignat*}
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\subsubsection{} %ii
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\begin{alignat*}{2}
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f(x,y) &=& \cos(x^{2}y) \cdot e^{xy} \\
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f_{x} &=& -\sin(x^{2}y) \cdot 2xy + \cos(x^{2}y) \cdot e^{xy} \cdot y \\
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f_{y} &=& -\sin(x^{2}y) \cdot x^{2} + \cos(x^{2}y) \cdot e^{xy} \cdot x \\
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\end{alignat*}
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\subsubsection{} %iii
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\begin{alignat*}{2}
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f(x,y) &=& \frac{\sin x + \cos y}{x^{2} + y^{2}} \\
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f_{x} &=& \frac{\cos x \cdot (x^{2} + y^{2}) - (\sin x + \cos y) \cdot 2x}{(x^{2} + y^{2})^{2}} \\
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f_{y} &=& \frac{-\sin y \cdot (x^{2} + y^{2}) - (\sin x + \cos y) \cdot 2y}{(x^{2} + y^{2})^{2}}
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\end{alignat*}
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\subsubsection{} %iv
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\begin{alignat*}{2}
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f(x,y) &=& \sqrt{1 - x^{2} - y^{2}} = (1 - x^{2} - y^{2})^{\frac{1}{2}} \\
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f_{x} &=& \frac{1}{2}(1 - x^{2} - y^{2})^{-\frac{1}{2}} \cdot (-2x) \\
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f_{y} &=& \frac{1}{2}(1 - x^{2} - y^{2})^{-\frac{1}{2}} \cdot (-2y)
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\end{alignat*}
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\section{} %2
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\begin{alignat*}{2}
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f(x,y) &=& x^{2}y^{3} + y \cdot e^{x^{2}y} \\
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f_{x} &=& 2xy^{3} + y \cdot e^{x^{2}y} \cdot 2xy\\
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&=& 2xy^{3} + e^{x^{2}y} \cdot 2xy^{2} \\
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f_{y} &=& 3x^{2}y^{2} + e^{x^{2}y} + y \cdot e^{x^{2}y} \cdot x^{2} \\
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f_{xx} &=& 2y^{3} + e^{x^{2}y} \cdot 2xy \cdot 2xy^{2} + e^{x^{2}y} \cdot 2y^{2}\\
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&=& 2y^{3} + e^{x^{2}y} \cdot 4x^{2}y^{3} + e^{x^{2}y} \cdot 2y^{2} \\
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f_{yx} &=& 6xy^{2} + e^{x^{2}y} \cdot x^{2} \cdot 2xy^{2} + e^{x^{2}y} \cdot 4xy\\
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&=& 6xy^{2} + e^{x^{2}y} \cdot 2x^{3}y^{2} + e^{x^{2}y} \cdot 4xy \\
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f_{xy} &=& 6xy^{2} + e^{x^{2}y} \cdot 2xy + y \cdot (e^{x^{2}y} \cdot 2xy \cdot x^{2} + e^{x^{2}y} \cdot 2x)\\
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&=& 6xy^{2} + e^{x^{2}y} \cdot 2x^{3}y^{2} + e^{x^{2}y} \cdot 4xy \\
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f_{yy} &=& 6x^{2}y + e^{x^{2}y} \cdot x^{2} + e^{x^{2}y} \cdot x^{2} + y \cdot e^{x^{2}y} \cdot x^{2} \cdot x^{2} \\
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&=& 6x^{2}y + e^{x^{2}y} \cdot 2x^{2} + e^{x^{2}y} \cdot x^{4}y
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\end{alignat*}
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\section{} %3
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\setcounter{subsubsection}{0}
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\subsubsection{} %i
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\begin{alignat*}{2}
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f(x,y) &=& 2x^{2} + y^{2} - 2xy -2x -4y + 5 \\
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I f_{x} &=& 4x - 2y - 2 = 0 \\
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II f_{y} &=& 2y - 2x - 4 = 0 \\
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f_{xx} &=& 4 \\
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f_{xy} &=& - 2 \\
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f_{yx} &=& - 2 \\
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f_{yy} &=& 2 \\
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I + II &=& 2x -6 = 0 \\
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&\Leftrightarrow & 2x = 6 \\
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&\Leftrightarrow & x = 3 \\
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\intertext{in II einsetzen}
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&\Rightarrow & 2y - 2 \cdot 3 -4 = 0 \\
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&\Leftrightarrow & 2y - 6 - 4 = 0 \\
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&\Leftrightarrow & 2y - 10 = 0 \\
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&\Leftrightarrow & 2y = 10 \\
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&\Leftrightarrow & y = 5 \\
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\intertext{Einsetzen von beiden Werten in I}
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&\Rightarrow & 4 \cdot 3 - 2 \cdot 5 - 2 = 0 \\
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&\Leftrightarrow & 12 - 10 - 2 = 0 \\
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&\Leftrightarrow & 0 = 0 \\
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\intertext{Die einzige kritische Stelle befindet sich an (3,5). Aufstellen der Hesse-Matrix}
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H &=& \begin{pmatrix} 4 & -2 \\
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-2 & 2 \end{pmatrix} \\
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\bigtriangleup_{1} &=& 4 > 0 \\
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\bigtriangleup_{2} &=& 4 > 0 \\
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\intertext{Die Hesse-Matrix ist positiv definit und damit befindet sich an der kritischen Stelle ein lokales Minimum.}
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\end{alignat*}
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\subsubsection{} %ii
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\begin{alignat*}{2}
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f(x,y) &=& x^{2} + 2y^{2} - 3xy -x -y +7 \\
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I f_{x} &=& 2x - 3y - 1 = 0 \\
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II f_{y} &=& 4y - 3x - 1 = 0 \\
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f_{xx} &=& 2 \\
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f_{xy} &=& -3 \\
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f_{yx} &=& -3 \\
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f_{yy} &=& 4 \\
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II + I &=& -x + y -2 = 0 \\
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&\Leftrightarrow & y = x + 2 \\
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\intertext{Einsetzen in I}
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&\Rightarrow & 2x - 3(x+2) - 1 = 0 \\
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&\Leftrightarrow & 2x - 3x - 6 - 1 = 0 \\
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&\Leftrightarrow & -x - 7 = 0 \\
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&\Leftrightarrow & x = -7 \\
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\intertext{Einsetzen in I}
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&\Rightarrow & 2 \cdot (-7) - 3y - 1 = 0 \\
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&\Leftrightarrow & -14 - 3y - 1 = 0 \\
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&\Leftrightarrow & 3y = -15 \\
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&\Leftrightarrow & y = -5 \\
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\intertext{Einsetzen beider Werte in II}
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&\Rightarrow & 4 \cdot (-5) - 3 \cdot (-7) - 1 = 0 \\
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&\Leftrightarrow & -20 + 21 - 1 = 0 \\
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&\Leftrightarrow & 0 = 0 \\
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\intertext{Die einzige kritische Stelle befindet sich an (-7, -5). Aufstellen der Hesse-Matrix}
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H &=& \begin{pmatrix} 2 & -3 \\
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-3 & 4 \end{pmatrix} \\
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\bigtriangleup_{1} &=& 2 > 0 \\
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\bigtriangleup_{2} &=& -1 < 0 \\
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\intertext{Die Hesse-Matrix ist damit weder positiv noch negativ definit und daher infinit. Daher liegt an der kritischen Stelle kein lokales Extremum vor.}
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\end{alignat*}
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\subsubsection{} %iii
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\begin{alignat*}{2}
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f(x,y) &=& 2x^{3} + y^{3} - 12x -27y +2 \\
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I f_{x} &=& 6x^{2} -12 = 0\\
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II f_{y} &=& 3y^{2} - 27 = 0\\
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f_{xx} &=& 12x \\
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f_{xy} &=& 0 \\
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f_{yx} &=& 0 \\
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f_{yy} &=& 6y \\
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I &\Rightarrow & 6x^{2} - 12 = 0 \\
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&\Leftrightarrow & 6x^{2} = 12 \\
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&\Leftrightarrow & x^{2} = 2 \\
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&\Rightarrow & x_{1} = \sqrt{2} \\
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&\Rightarrow & x_{2} = -\sqrt{2} \\
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I + II &=& 6x^{2} + 3y^{2} - 39 = 0 \\
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\intertext{Einsetzen von den x-Werten und berechnen von y}
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x_{1} &\Rightarrow & 6 \cdot 2 + 3y^{2} -39 = 0 \\
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&\Leftrightarrow & 3y^{2} - 27 = 0 \\
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&\Leftrightarrow & 3y^{2} = 27 \\
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&\Leftrightarrow & y^{2} = 9 \\
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&\Leftrightarrow & y \pm 3 \\
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x_{2} &\Rightarrow & 6 \cdot 2 + 3y^{2} - 39 = 0 \\
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&\Leftrightarrow & 3y^{2} = 27 \\
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&\Leftrightarrow & y \pm 3 \\
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\intertext{Es gibt also vier kritische Stellen: $(\sqrt{2}, 3), (\sqrt{2}, -3), (-\sqrt{2}, 3)$ und $(-\sqrt{2},-3)$. Aufstellen der Hesse-Matrix}
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H &=& \begin{pmatrix}12x & 0 \\
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0 & 6y\end{pmatrix} \\
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\intertext{Berechnen der Definitheit für erste kritische Stelle:}
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\bigtriangleup_{1} &=& 12 \cdot \sqrt{2} > 0 \\
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\bigtriangleup_{2} &=& 12 \cdot \sqrt{2} \cdot 18 \\
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&=& 216 \cdot \sqrt{2} > 0 \\
|
||||
\intertext{Die Hesse-Matrix ist für die erste kritische Stelle positiv definit. An der ersten kritischen Stelle liegt also ein lokales Minimum vor. Berechnen der Definitheit für die zweite kritische Stelle:}
|
||||
\bigtriangleup_{1} &=& 12 \cdot \sqrt{2} > 0 \\
|
||||
\bigtriangleup_{2} &=& 12 \cdot \sqrt{2} \cdot (-18) \\
|
||||
&=& -216 \cdot \sqrt{2} < 0 \\
|
||||
\intertext{Die Hesse-Matrix ist für die zweite kritische Stelle infinit. An der zweiten kritischen Stelle liegt also kein lokales Extremum vor. Berechnen der Definitheit für die dritte kritische Stelle:}
|
||||
\bigtriangleup_{1} &=& -12 \cdot \sqrt{2} < 0 \\
|
||||
\bigtriangleup_{2} &=& -12 \cdot \sqrt{2} \cdot 18 \\
|
||||
&=& -216 \cdot \sqrt{2} < 0 \\
|
||||
\intertext{Die Hesse-Matrix ist für die dritte kritische Stelle infinit. An der dritten kritischen Stelle liegt also kein lokales Extremum vor. Berechnen der Definitheit für die vierte kritische Stelle:}
|
||||
\bigtriangleup_{1} &=& -12 \cdot \sqrt{2} < 0 \\
|
||||
\bigtriangleup_{2} &=& -12 \cdot \sqrt{2} \cdot (-18) \\
|
||||
&=& 216 \cdot \sqrt{2} > 0 \\
|
||||
\intertext{Die Hesse-Matrix ist für die vierte kritische Stelle negativ definit. An der vierten kritischen Stelle liegt also ein lokales Maximum vor.}
|
||||
\end{alignat*}
|
||||
\section{} %4
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
C(x,y) &=& 0.01x^{2} + 0.02xy + 0.16y^{2} + 5x + 6y + 120 \\
|
||||
&=& \frac{1}{100}x^{2} + \frac{1}{50}xy + \frac{4}{25}y^{2} + 5x + 6y + 120 \\
|
||||
\intertext{Aufstellen der Gewinnfunktion}
|
||||
G(x,y) &=& 12x + 28y - C(x,y) \\
|
||||
&=& 12x + 28y - \frac{1}{100}x^{2} - \frac{1}{50}xy - \frac{4}{25}y^{2} - 5x - 6y - 120 \\
|
||||
&=& -\frac{1}{100}x^{2} - \frac{4}{25}y^{2} - \frac{1}{50}xy + 7x + 22y - 120 \\
|
||||
I\, G_{x} &=& -\frac{1}{50}x - \frac{1}{50}y + 7 = 0 \\
|
||||
II\, G_{y} &=& -\frac{8}{25}y - \frac{1}{50}x + 22 = 0 \\
|
||||
G_{xx} &=& -\frac{1}{50} \\
|
||||
G_{xy} &=& 0 \\
|
||||
G_{yx} &=& 0 \\
|
||||
G_{yy} &=& -\frac{8}{25} \\
|
||||
I &\Rightarrow & -\frac{1}{50}x - \frac{1}{50}y + 7 = 0 \\
|
||||
&\Leftrightarrow & \frac{1}{50}x = 7 - \frac{1}{50}y \\
|
||||
&\Leftrightarrow & x = 350 - y \\
|
||||
\intertext{Einsetzen in II}
|
||||
II &\Rightarrow & -\frac{8}{25}y - \frac{1}{50} (350-y) + 22 = 0 \\
|
||||
&\Leftrightarrow & \frac{8}{25}y = 22 - \frac{1}{50} (350 - y) \\
|
||||
&\Leftrightarrow & \frac{8}{25}y = 22 - 7 + \frac{1}{50}y \\
|
||||
&\Leftrightarrow & \frac{15}{50}y = 15 \\
|
||||
&\Leftrightarrow & \frac{3}{10}y = 15 \\
|
||||
&\Leftrightarrow & y = 50 \\
|
||||
\intertext{Einsetzen in I}
|
||||
&\Rightarrow & x = 350 - 50 = 300
|
||||
\intertext{Die einige kritische Stelle von G(x,y) befindet sich an $\left(300, 50\right)$. Aufstellen der Hesse-Matrix:}
|
||||
H &=& \begin{pmatrix}-\frac{1}{50} & 0 \\
|
||||
0 & -\frac{8}{25} \end{pmatrix} \\
|
||||
\bigtriangleup_{1} &=& -\frac{1}{50} < 0 \\
|
||||
\bigtriangleup_{2} &=& \frac{4}{625} > 0 \\
|
||||
\intertext{Die Hesse-Matrix ist an der kritischen Stelle negativ definit. An der kritischen Stelle befindet sich daher ein Maximum. Der höchste Gewinn ist demnach mit 300 Einheiten des Gutes A und 50 Einheiten des Gutes B zu erreichen.}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
G(x,y) &=& -\frac{1}{100}x^{2} - \frac{4}{25}y^{2} - \frac{1}{50}xy + 7x + 22y - 120 \\
|
||||
n(x,y) &=& x + 2y = 320 \\
|
||||
&\Leftrightarrow & x= 320 - 2y \\
|
||||
\intertext{Einsetzen in G(x,y)}
|
||||
G(y) &=& -\frac{1}{100} \cdot (320-2y)^{2} - \frac{4}{25}y^{2} - \frac{1}{50} \cdot (320 - 2y)y + 7(320-2y) + 22y - 120 \\
|
||||
&=& -\frac{1}{100} \cdot (102400 -1280y + 4y^{2}) - \frac{4}{25}y^{2} - \frac{1}{50} \cdot (320y -2y^{2}) + 2240 - 14y + 22y - 120 \\
|
||||
&=& -1024 + \frac{128}{10}y - \frac{1}{25}y^{2} - \frac{4}{25}y^{2} - \frac{32}{5}y + \frac{1}{25}y^{2} + 8y + 2120 \\
|
||||
&=& \frac{64}{5}y - \frac{5}{25}y^{2} - \frac{32}{5} + \frac{1}{25}y^{2} + 1096 + 8y \\
|
||||
&=& -\frac{4}{25}y^{2} + \frac{72}{5}y + 1096 \\
|
||||
G'(y) &=& -\frac{8}{25}y + \frac{72}{5} = 0 \\
|
||||
&\Leftrightarrow & \frac{8}{25}y = \frac{72}{5} \\
|
||||
&\Leftrightarrow & y = 45 \\
|
||||
G''(y) &=& -\frac{8}{25} < 0 \\
|
||||
\intertext{Unter der Nebenbedingung n(x,y) gibt es ein lokales Maximum für 45 Einheiten von Gut B.
|
||||
Einsetzen von y in die Nebenbedingung:}
|
||||
n(x) &=& x + 2 \cdot 45 = 320 \\
|
||||
&\Leftrightarrow & x + 90 = 320 \\
|
||||
&\Leftrightarrow & x = 230
|
||||
\intertext{Die optimalen Mengen des Outputs liegen bei 230 Einheiten von Gut A und 45 Einheiten von Gut B.}
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
Berechnen des maximalen Gewinns für Fall a)\\
|
||||
\begin{alignat*}{2}
|
||||
G(300,50) &=& -\frac{1}{100} \cdot 300^{2} - \frac{4}{25} \cdot 50^{2} - \frac{1}{50} \cdot 300 \cdot 50 + 7 \cdot 300 + 22 \cdot 50 - 120 \\
|
||||
&=& -900 - \frac{4}{25} \cdot 2500 - 300 + 2100 + 1100 - 120 \\
|
||||
&=& 1880 - 400 \\
|
||||
&=& 1480 \\
|
||||
\intertext{Der maximale Gewinn im Fall a) beträgt 1480 Geldeinheiten.}
|
||||
\end{alignat*}\\
|
||||
Berechnen des maximalen Gewinns für Fall b)\\
|
||||
\begin{alignat*}{2}
|
||||
G(230,45) &=& -\frac{1}{100} \cdot 230^{2} - \frac{4}{25} \cdot 45^{2} - \frac{1}{50} \cdot 230 \cdot 45 + 7 \cdot 230 + 22 \cdot 45 - 120 \\
|
||||
&=& -529 - \frac{4}{25} \cdot 2025 - \frac{1}{50} \cdot 10350 + 1610 + 990 - 120 \\
|
||||
&=& 1951 - 324 - 207 \\
|
||||
&=& 1420
|
||||
\intertext{Der maximale Gewinn im Fall b) beträgt 1420 Geldeinheiten.}
|
||||
\end{alignat*}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,222 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pgfplotsset{compat=1.8}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 4. Juli}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
f(x,y,z) &=& 2x^{2} + y^{2} + 4z^{2} - 2yz - 2x - 6y + 8 \\
|
||||
I f_{x} &=& 4x - 2 = 0\\
|
||||
II f_{y} &=& 2y - 7z - 6 = 0\\
|
||||
III f_{z} &=& 8z - 2y = 0 \\
|
||||
I &\Rightarrow & 4x - 2 = 0 \\
|
||||
&\Leftrightarrow & 4x = 2 \\
|
||||
&\Leftrightarrow & x = \frac{1}{2} \\
|
||||
III &\Rightarrow & 8z - 2y = 0 \\
|
||||
&\Leftrightarrow & 8z = 2y \\
|
||||
&\Leftrightarrow & 4z = y \\
|
||||
II &\Rightarrow & 2 \cdot 4z - 7z - 6 = 0 \\
|
||||
&\Leftrightarrow & 8z - 7z = 6 \\
|
||||
&\Leftrightarrow & z = 6 \\
|
||||
III &\Rightarrow & 4 \cdot 6 = y \\
|
||||
&\Leftrightarrow & 24 = y \\
|
||||
\intertext{Die einzige kritische Stelle befindet sich an ($\frac{1}{2}, 24, 6$).}
|
||||
f_{xx} &=& 4 \\
|
||||
f_{yx} &=& 0 \\
|
||||
f_{zx} &=& 0 \\
|
||||
f_{xy} &=& 0 \\
|
||||
f_{yy} &=& 2 \\
|
||||
f_{zy} &=& -7 \\
|
||||
f_{xz} &=& 0 \\
|
||||
f_{yz} &=& -2 \\
|
||||
f_{zz} &=& 8 \\
|
||||
\intertext{Aufstellen der Hesse-Matrix}
|
||||
H &=& \begin{pmatrix}4 & 0 & 0 \\
|
||||
0 & 2 & -7 \\
|
||||
0 & -2 & 8\end{pmatrix} \\
|
||||
\bigtriangleup_{1} &=& 4 > 0\\
|
||||
\bigtriangleup_{2} &=& 8 > 0\\
|
||||
\bigtriangleup_{3} &=& 64 - 56 = 8 > 0\\
|
||||
\intertext{Die Hesse-Matrix ist positiv definit. Daher liegt an der kritischen Stelle ein Minimum vor.}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
grad\,f(1,1,1) &=& (4 - 2, 2 - 7 - 6, 8 - 2) \\
|
||||
&=& (2, -11, 6) \\
|
||||
||grad\,f(1,1,1)|| &=& \sqrt{2^{2} + 11^{2} + 6^{2}} \\
|
||||
&=& \sqrt{4 + 121 + 36} \\
|
||||
&=& \sqrt{161} \\
|
||||
\intertext{In Richtung von (2, -11, 6) steigt die Temperatur von (1,1,1) aus am stärksten an. In Richtung (-2, 11, -6) sinkt die Temperatur am stärksten. Die Größe des stärksten Anstiegs beträgt $\sqrt{161}$.}
|
||||
\end{alignat*}
|
||||
\section{} %2
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
A &=& \begin{pmatrix}-i & -1 \\
|
||||
3 & i \end{pmatrix} \\
|
||||
B &=& \begin{pmatrix} i \\
|
||||
1 + i\end{pmatrix} \\
|
||||
C &=& \begin{pmatrix}-i & i \end{pmatrix} \\
|
||||
AB &=& \begin{pmatrix}-i \cdot i + (1+i)\cdot (-1) \\
|
||||
3i + i(1+i) \end{pmatrix} \\
|
||||
&=& \begin{pmatrix}-i^{2} - 1 - i \\
|
||||
3i + i + i^{2} \end{pmatrix} \\
|
||||
&=& \begin{pmatrix}- 1 - i \\
|
||||
4i - 1 \end{pmatrix} \\
|
||||
\intertext{AC existiert nicht, da A mehr Spalten hat, als C Zeilen hat.}
|
||||
BC &=& \begin{pmatrix}i \cdot (-i) & i \cdot i \\
|
||||
(1+i) \cdot (-i) & (1+i)i \end{pmatrix} \\
|
||||
&=& \begin{pmatrix}-i^{2} & i^{2} \\
|
||||
-i - i^{2} & i + i^{2} \end{pmatrix} \\
|
||||
&=& \begin{pmatrix}1 & -1 \\
|
||||
-i +1 & i -1 \end{pmatrix} \\
|
||||
CB &=& \begin{pmatrix}-i \cdot i + i(1+i) \end{pmatrix} \\
|
||||
&=& \begin{pmatrix}-i^{2} + i + i^{2} \end{pmatrix} \\
|
||||
&=& \begin{pmatrix}1 + i -1 \end{pmatrix} \\
|
||||
&=& \begin{pmatrix}i \end{pmatrix}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
\overline{z} &=& \frac{3 + 2i}{4 - 3i} \\
|
||||
&=& \frac{(3 + 2i)(4+3i)}{(4 - 3i)(4+3i)} \\
|
||||
&=& \frac{12 + 9i + 8i + 6i^{2}}{16 + 12i - 12i - 9i^{2}} \\
|
||||
&=& \frac{12 - 6 + 17i}{16 + 9} \\
|
||||
&=& \frac{6 + 17i}{25} \\
|
||||
&=& \frac{6}{25} + \frac{17}{25}i \\
|
||||
z &=& \frac{6}{25} - \frac{17}{25}i \\
|
||||
a &=& \frac{6}{25} \\
|
||||
b &=& \frac{17}{25}
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\begin{alignat*}{2}
|
||||
z_{1} &=& -1 - i = (-1, -1) \\
|
||||
z_{2} &=& \sqrt{2} \cdot \cos \frac{\pi}{4} + \sqrt{2} \cdot i \cdot \sin \frac{\pi}{4} \\
|
||||
&=& \sqrt{2} \cdot \cos \left(\frac{1}{2} \cdot \frac{\pi}{2} \right) + \sqrt{2} \cdot \sin \left(\frac{1}{2} \cdot \frac{\pi}{2} \right) \\
|
||||
&=& 1 + i = (1, 1) \\
|
||||
z_{3} &=& (-1 -i)(1+ i) \\
|
||||
&=& -1 - i - i - i^{2} \\
|
||||
&=& -1 + 1 - 2i = 0 - 2i = (0, -2) \\
|
||||
z_{4} &=& -1 + i = (-1, 1)
|
||||
\end{alignat*}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=-5,ymax=5,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$\Re$},
|
||||
ylabel={$\Im$},
|
||||
xmin=-5,xmax=5
|
||||
]
|
||||
|
||||
\node at (axis cs: -1,-1) {$z_{1}$};
|
||||
\node at (axis cs: 1,1) {$z_{2}$};
|
||||
\node at (axis cs: 0,-2) {$z_{3}$};
|
||||
\node at (axis cs: -1,1) {$z_{4}$};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\subsection{} %d
|
||||
\subsubsection{} %i
|
||||
In dieser Teilmenge sind alle komplexen Zahlen enthalten, die sich auf der Geraden befinden, die durch die Punkte (1,1) und (2,0) geht.
|
||||
\subsubsection{} %ii
|
||||
In der Teilmenge sind alle komplexen Zahlen enthalten, die sich auf der Kreislinie eines Kreises mit dem Radius 1 um den Punkt (1,1) befinden.
|
||||
\section{} %3
|
||||
\begin{alignat*}{2}
|
||||
f(x,y) &=& -\frac{1}{5}x^{2} - xy - \frac{25}{10}y^{2} + 48x + 235y - 88 \\
|
||||
g(x,y) &=& \frac{1}{5}x + y = 40
|
||||
\end{alignat*}
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
g_{x} &=& \frac{1}{5} \\
|
||||
g_{y} &=& 1 \\
|
||||
\begin{pmatrix}\frac{\partial g}{\partial x} (x,y) & \frac{\partial g}{\partial y} (x,y)\end{pmatrix} &=& \begin{pmatrix}\frac{1}{5} & 1 \end{pmatrix} \\
|
||||
\intertext{Der Rang dieser Matrix ist für alle $x,y$ gleich 2. Damit ist die Regularitätsbedingung erfüllt.}
|
||||
L(x,y,\lambda) &=& -\frac{1}{5}x^{2} - xy - \frac{25}{10}y^{2} + 48x + 235y - 88 + \lambda(\frac{1}{5}x + y - 40) \\
|
||||
I L_{x} &=& - \frac{2}{5}x - y + 48 + \lambda \cdot \frac{1}{5} = 0 \\
|
||||
II L_{y} &=& - x - 5y + 235 + \lambda = 0 \\
|
||||
III L_{\lambda} &=& \frac{1}{5}x + y - 40 = 0 \\
|
||||
II &\Rightarrow & \lambda = x + 5y - 235 \\
|
||||
I &\Rightarrow & - \frac{2}{5}x - y + 48 + (x + 5y - 235) \cdot \frac{1}{5} = 0\\
|
||||
&\Leftrightarrow & - \frac{2}{5}x - y + 48 + \frac{1}{5}x + y - 47 = 0 \\
|
||||
\intertext{Beachten der dritten Gleichung}
|
||||
&\Leftrightarrow & - \frac{1}{5}x - \frac{1}{5}x - y + 48 - 7 = 0\\
|
||||
&\Leftrightarrow & - \frac{1}{5}x + 1=\frac{1}{5}x + y - 40 \\
|
||||
\intertext{Beachten der dritten Gleichung}
|
||||
&\Leftrightarrow & - \frac{1}{5}x + 1 = 0 \\
|
||||
&\Leftrightarrow & \frac{1}{5}x = 1 \\
|
||||
&\Leftrightarrow & x = 5 \\
|
||||
\intertext{Einsetzen in I}
|
||||
&\Rightarrow & - \frac{1}{5} \cdot 5 - \frac{1}{5} \cdot 5 - y + 48 - 7 = 0 \\
|
||||
&\Leftrightarrow & - 1 - 1 - y + 41 = 0 \\
|
||||
&\Leftrightarrow & -y + 39 = 0 \\
|
||||
&\Leftrightarrow & y = 39 \\
|
||||
\intertext{Einsetzen in III}
|
||||
\lambda &=& \frac{1}{5} \cdot 5 + 39 - 40 \\
|
||||
&=& 1 + 39 - 40 \\
|
||||
&=& 0
|
||||
\intertext{Die einzige kritische Stelle befindet sich an (5, 39).}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
L_{xx} &=& - \frac{2}{5} \\
|
||||
L_{yx} &=& - 1 \\
|
||||
L_{xy} &=& - 1 \\
|
||||
L_{yy} &=& - 5 \\
|
||||
\overline{H} &=& \begin{pmatrix}0 & \frac{1}{5} & 1 \\
|
||||
\frac{1}{5} & -\frac{2}{5} & -1\\
|
||||
1 & -1 & -5 \end{pmatrix} \\
|
||||
det \, \overline{H} &=& -\frac{1}{5} - \frac{1}{5} + \frac{2}{5} + \frac{1}{5} \\
|
||||
&=& \frac{1}{5} > 0 \\
|
||||
\intertext{An der kritischen Stelle liegt ein Maximum vor.}
|
||||
\end{alignat*}
|
||||
\section{} %4
|
||||
\begin{alignat*}{2}
|
||||
f(x,y) &=& -\frac{1}{5}x^{2} - xy - \frac{25}{10}y^{2} + 48x + 235y - 88 \\
|
||||
g(x,y) &=& \frac{1}{5}x + y -40 = 0 \\
|
||||
&\Leftrightarrow & y = -\frac{1}{5}x + 40 \\
|
||||
\intertext{Einsetzen in f(x,y)}
|
||||
f(x) &=& -\frac{1}{5}x^{2} - x \cdot (-\frac{1}{5}x + 40) - \frac{25}{10} \cdot (-\frac{1}{5}x + 40)^{2} + 48x + 235 \cdot (-\frac{1}{5}x + 40) - 88 \\
|
||||
&=& -\frac{1}{5}x^{2} + \frac{1}{5}x^{2} - 40x - \frac{25}{10} \cdot (\frac{1}{25}x^{2} -16x + 1600) + 48x -47x + 9400 - 88 \\
|
||||
&=& - 40x - \frac{1}{10}x^{2} + 40x - 4000 + 48x -47x + 9400 - 88 \\
|
||||
&=& - \frac{1}{10}x^{2} + x + 5312 \\
|
||||
f'(x) &=& -\frac{1}{5}x + 1 = 0 \\
|
||||
&\Leftrightarrow & \frac{1}{5}x = 1 \\
|
||||
&\Leftrightarrow & x = 5 \\
|
||||
f''(x) &=& -\frac{1}{5} < 0 \\
|
||||
\intertext{Unter der Nebenbedingung g(x,y) gibt es ein lokales Maximum für x = 5.
|
||||
Einsetzen von x in die Nebenbedingung:}
|
||||
g(y) &=& \frac{1}{5} \cdot 5 + y -40 = 0 \\
|
||||
&\Leftrightarrow & 1 + y - 40 = 0 \\
|
||||
&\Leftrightarrow & y - 39 = 0 \\
|
||||
&\Leftrightarrow & y = 39 \\
|
||||
\intertext{An der Stelle (5, 39) ist der Gewinn maximal.}
|
||||
\end{alignat*}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,102 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pgfplotsset{compat=1.8}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 11. Juli}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\begin{alignat*}{2}
|
||||
\iint\limits_{I} f(x,y)\, d(x,y) &=& \iint\limits_{I} (2x^{2}y)\, d(x,y) \\
|
||||
\intertext{Version 1}
|
||||
&=& \int\limits_{1}^{2} \left( \int\limits_{-1}^{3} (2x^{2}y) \, dy \right) \, dx \\
|
||||
&=& \int\limits_{1}^{2} \left( 2x^{2} \int\limits_{-1}^{3} y \, dy \right)\, dx \\
|
||||
&=& \int\limits_{1}^{2} \left( 2x^{2} \left[\frac{1}{2}y^{2}\right]_{-1}^{3} \right)\, dx \\
|
||||
&=& \int\limits_{1}^{2} \left( 2x^{2} \left[\frac{9}{2} - \frac{1}{2}\right] \right)\, dx \\
|
||||
&=& \int\limits_{1}^{2} \left( 2x^{2} \cdot 4 \right)\, dx \\
|
||||
&=& 8 \int\limits_{1}^{2} x^{2}\, dx \\
|
||||
&=& 8 \left[\frac{1}{3}x^{3} \right]_{1}^{2} \\
|
||||
&=& 8 \left[\frac{8}{3} - \frac{1}{3} \right] \\
|
||||
&=& 8 \cdot \frac{7}{3} = \frac{56}{3} \\
|
||||
\intertext{Version 2}
|
||||
&=& \int\limits_{-1}^{3} \left( \int\limits_{1}^{2} (2x^{2}y) \, dx \right) \, dy \\
|
||||
&=& \int\limits_{-1}^{3} \left( 2y\int\limits_{1}^{2} x^{2} \, dx \right) \, dy \\
|
||||
&=& \int\limits_{-1}^{3} \left( 2y\left[\frac{1}{3}x^{3} \right]_{1}^{2} \right) \, dy \\
|
||||
&=& \int\limits_{-1}^{3} \left( 2y\left[\frac{8}{3} - \frac{1}{3} \right] \right) \, dy \\
|
||||
&=& \int\limits_{-1}^{3} \left( 2y \cdot \frac{7}{3} \right) \, dy \\
|
||||
&=& \frac{14}{3}\int\limits_{-1}^{3} y \, dy \\
|
||||
&=& \frac{14}{3}\left[\frac{1}{2}y^{2}\right]_{-1}^{3} \\
|
||||
&=& \frac{14}{3}\left[\frac{9}{2} - \frac{1}{2}\right] \\
|
||||
&=& \frac{14}{3} \cdot 4 = \frac{56}{3}
|
||||
\end{alignat*}
|
||||
\section{} %2
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
\iint\limits_{G} f(x,y) \, d(x,y) &=& \iint\limits_{G} (xy^{2}) \, d(x,y) \\
|
||||
&=& \int\limits_{0}^{1} \left(\int\limits_{0}^{3x} xy^{2}\, dy \right) \, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(x\int\limits_{0}^{3x} y^{2}\, dy \right) \, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(x \left[\frac{1}{3}y^{3} \right]_{0}^{3x} \right) \, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(x \left[\frac{1}{3} \cdot 27x^{3} \right] \right) \, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(x \cdot 9x^{3} \right) \, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(9x^{4} \right) \, dx \\
|
||||
&=& 9\int\limits_{0}^{1} x^{4} \, dx \\
|
||||
&=& 9\left[\frac{1}{5}x^{5} \right]_{0}^{1} \\
|
||||
&=& 9 \cdot \frac{1}{5} = \frac{9}{5}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
\iint\limits_{G} f(x,y) \, d(x,y) &=& \iint\limits_{G} (xy^{2}) \, d(x,y) \\
|
||||
&=& \int\limits_{0}^{1} \left(\int\limits_{3x}^{3} xy^{2} \, dy \right)\, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(x\int\limits_{3x}^{3} y^{2} \, dy \right)\, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(x \left[\frac{1}{3}y^{3} \right]_{3x}^{3} \right)\, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(x \left[\frac{27}{3} - \frac{27}{3}x^{3} \right] \right)\, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(9x - 9x^{4} \right)\, dx \\
|
||||
&=& \int\limits_{0}^{1} \left(9(x - x^{4}) \right)\, dx \\
|
||||
&=& 9\int\limits_{0}^{1} x - x^{4}\, dx \\
|
||||
&=& 9\int\limits_{0}^{1} x \, dx - 9\int\limits_{0}^{1} x^{4}\, dx \\
|
||||
&=& 9 \left[\frac{1}{2}x^{2} \right]_{0}^{1} - 9\left[\frac{1}{5}x^{5} \right]_{0}^{1} \\
|
||||
&=& 9 \cdot \frac{1}{2} - 9 \cdot \frac{1}{5} \\
|
||||
&=& \frac{9}{2} - \frac{9}{5} \\
|
||||
&=& \frac{45}{10} - \frac{18}{10} = \frac{27}{10}
|
||||
\end{alignat*}
|
||||
\section{} %3
|
||||
\subsection{} %a
|
||||
Klarerweise gilt $f_{4}(n) = O(f_{5}(n))$. Ebenfalls gilt $f_{3}(n) = O(f_{1}(n))$. Außerdem ist klar, dass $f_{1}(n) = O(f_{4}(n))$ gilt.
|
||||
Damit ergibt sich die Reihenfolge $f_{3}, f_{1}, f_{4}, f_{5}$.
|
||||
Es müssen noch $f_{2}$ und $f_{6}$ eingeordnet werden. $f_{2}(n) = O(f_{1}(n))$ gilt ebenso wie $f_{3}(n) = O(f_{2}(n))$. $f_{2}$ kann demnach zwischen $f_{3}$ und $f_{1}$ eingeordnet werden, womit sich die Reihenfolge $f_{3}, f_{2}, f_{1}, f_{4}, f_{5}$ ergibt.
|
||||
|
||||
Abschließend muss noch $f_{6}(n)$ eingeordnet werden. $n^{2}$ kommt als Faktor auch in $f_{1}$ vor. Es bleibt daher die Frage, ob $\sqrt{n}$ schneller wächst als $\log_{2}(n)$. Dem ist so, da Wurzelfunktionen allgemein schneller wachsen als Logarithmusfunktionen. Daher gilt $f_{6}(n) = O(f_{1}(n))$. Gleichzeitig gilt, dass $\sqrt{2n}$ langsamer wächst als $f_{6}(n)$, womit auch $f_{2}(n) = O(f_{6}(n))$ gilt. Die fertige Reihenfolge ist daher $f_{3}, f_{2}, f_{6}, f_{1}, f_{4}, f_{5}$.
|
||||
\subsection{} %b
|
||||
\section{} %4
|
||||
\subsection{} %a
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Es gilt $n = \lfloor x \rfloor, x \geq 1$ für beide Funktionen:}
|
||||
f(n) &=& n \\
|
||||
g(n) &=& n^{1+ \lceil \sin (x) \rceil}
|
||||
\end{alignat*}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,216 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 18. April}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsubsection{} %(i)
|
||||
\begin{alignat*}{3}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left(\frac{-3n^{4}+2n^{2}+n+1}{-7n^{4}+25} \right) &\Rightarrow & \frac{-3n^{4}+2n^{2}+n+1}{-7n^{4}+25} && \\
|
||||
\intertext{Ausklammern von $n^{4}$}
|
||||
&\Leftrightarrow & \frac{-3 + \frac{2}{n^{2}} + \frac{1}{n^{3}} + \frac{1}{n^{4}}}{-7 + \frac{25}{n^{4}}} &\rightarrow & \frac{3}{7}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %(ii)
|
||||
\begin{alignat*}{3}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left(\frac{-3n^{4}+2n^{2}+n+1}{-7n^{5}+25} \right) &\Rightarrow & \frac{-3n^{4}+2n^{2}+n+1}{-7n^{5}+25} && \\
|
||||
\intertext{Ausklammern von $n^{4}$ im Zähler und $n^{5}$ im Nenner}
|
||||
&\Leftrightarrow & \frac{1}{n} \cdot \frac{-3 + \frac{2}{n^{2}} + \frac{1}{n^{3}} + \frac{1}{n^{4}}}{-7 + \frac{25}{n^{5}}} &\rightarrow & 0
|
||||
\end{alignat*}
|
||||
\subsubsection{} %(iii)
|
||||
\begin{alignat*}{3}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left(\frac{-3n^{5}+2n^{2}+n+1}{-7n^{4}+25} \right) &\Rightarrow & \frac{-3n^{5}+2n^{2}+n+1}{-7n^{4}+25} && \\
|
||||
\intertext{Ausklammern von $n^{5}$ im Zähler und $n^{4}$ im Nenner}
|
||||
&\Leftrightarrow & n \cdot \frac{-3 + \frac{2}{n^{3}} + \frac{1}{n^{4}} + \frac{1}{n^{5}}}{-7 + \frac{25}{n^{4}}} &\rightarrow & \infty
|
||||
\end{alignat*}
|
||||
\subsubsection{} %(iv)
|
||||
\begin{alignat*}{3}
|
||||
\hspace{-2.5cm}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left( \frac{6n^{3}+2n-3}{9n^{2}+2} - \frac{2n^{3}+5n^{2}+7}{3n^{2}+3} \right) &\Rightarrow & \frac{6n^{3}+2n-3}{9n^{2}+2} - \frac{2n^{3}+5n^{2}+7}{3n^{2}+3} && \\
|
||||
\intertext{Auf gleichen Nenner bringen}
|
||||
\hspace{-2.5cm}
|
||||
&\Leftrightarrow & \frac{(6n^{3}+2n-3)(3n^{2}+3) - (2n^{3}+5n^{2}+7)(9n^{2}+2)}{(9n^{2}+2)(3n^{2}+3)} && \\
|
||||
\intertext{Klammern auflösen und zusammenfassen}
|
||||
\hspace{-2.5cm}
|
||||
&\Leftrightarrow & \frac{-45n^{4} + 20n^{3}-82n^{2}+6n-23}{27n^{4}+33n^{2}+6} && \\
|
||||
\intertext{Ausklammern von $n^{4}$}
|
||||
\hspace{-2.5cm}
|
||||
&\Leftrightarrow & \frac{-45 + \frac{20}{n}-\frac{82}{n^{2}}+\frac{6}{n^{3}}-\frac{23}{n^{4}}}{27+\frac{33}{n^{2}}+\frac{6}{n^{4}}} &\rightarrow & \frac{-45}{27} = \frac{-5}{3}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %(v)
|
||||
\begin{alignat*}{3}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left( \frac{\sqrt{9n^{4}+n^{2}+1}-2n^{2}+3}{\sqrt{2n^{2}+1} \cdot \sqrt{2n^{2}+n+1}} \right) &\Rightarrow & \frac{\sqrt{9n^{4}+n^{2}+1}-2n^{2}+3}{\sqrt{2n^{2}+1} \cdot \sqrt{2n^{2}+n+1}} && \\
|
||||
\intertext{Anwendung der Wurzelgesetze}
|
||||
&\Leftrightarrow &\frac{\sqrt{9n^{4}+n^{2}+1}-2n^{2}+3}{\sqrt{(2n^{2}+1) \cdot (2n^{2}+n+1)}} && \\
|
||||
\intertext{Zusammenfassen}
|
||||
&\Leftrightarrow &\frac{\sqrt{9n^{4}+n^{2}+1}-2n^{2}+3}{\sqrt{4n^{4}+2n^{3}+4n^{2} + 1}} && \\
|
||||
\intertext{$n^{2}$ ausklammern}
|
||||
&\Leftrightarrow &\frac{\sqrt{9+\frac{1}{n^{2}}+\frac{1}{n^{4}}}-2+\frac{3}{n^{2}}}{\sqrt{4+\frac{2}{n}+\frac{4}{n^{2}} + \frac{1}{n^{4}}}} &\rightarrow & \frac{7}{2}
|
||||
\end{alignat*}
|
||||
\section{} %2
|
||||
\subsection{} %a
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{3}
|
||||
a_{0} &=& 1 &&&\\
|
||||
a_{1} &=& \frac{2}{5} &&& \\
|
||||
a_{2} &=& \left(\frac{2}{5}\right)^{2} = \frac{4}{25} &&& \\
|
||||
a_{3} &=& \left(\frac{2}{5}\right)^{3} = \frac{8}{125} &&& \\
|
||||
a_{4} &=& \left(\frac{2}{5}\right)^{4} = \frac{16}{625} &&& \\
|
||||
s_{0} &=& a_{0} &=&& 1 \\
|
||||
s_{1} &=& a_{0} + a_{1} = 1 + \frac{2}{5} = \frac{7}{5} &=&& 1.4 \\
|
||||
s_{2} &=& a_{0} + a_{1} + a_{2} = \frac{7}{5} + \frac{4}{25} = \frac{39}{25} &=&& 1.56 \\
|
||||
s_{3} &=& a_{0} + a_{1} + a_{2} + a_{3} = \frac{39}{25} + \frac{8}{125} = \frac{203}{125} &=&& 1.624 \\
|
||||
s_{4}&=& a_{0} + a_{1} + a_{2} + a_{3} + a_{4} = \frac{203}{125} + \frac{16}{625} = \frac{1031}{625} &=&& 1.6496
|
||||
\end{alignat*}\\
|
||||
Bestimmung des Grenzwertes mithilfe der Geometrischen Summenformel:\\
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left( \sum\limits_{i=0}^{n} \left(\frac{2}{5}\right)^{i} \right) &=& \underset{n \rightarrow \infty}{\text{lim}} \left(\frac{1-\left(\frac{2}{5}\right)^{n+1}}{1-\left(\frac{2}{5}\right)} \right) \\
|
||||
&=& \frac{1}{\frac{3}{5}} = \frac{5}{3}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{3}
|
||||
a_{0} &=& 1 &&& \\
|
||||
a_{1} &=& \frac{5}{2}&&& \\
|
||||
a_{2} &=& \left(\frac{5}{2}\right)^{2} = \frac{25}{4} &&&\\
|
||||
a_{3} &=& \left(\frac{5}{2}\right)^{3} = \frac{125}{8} &&& \\
|
||||
a_{4} &=& \left(\frac{5}{2}\right)^{4} = \frac{625}{16} &&& \\
|
||||
s_{0} &=& a_{0} &=& 1 \\
|
||||
s_{1} &=& a_{0} + a_{1} = 1 + \frac{5}{2} = \frac{7}{2} &=&& 3.5 \\
|
||||
s_{2} &=& a_{0} + a_{1} + a_{2} = \frac{7}{2} + \frac{25}{4} = \frac{39}{4} &=&& 9.75 \\
|
||||
s_{3} &=& a_{0} + a_{1} + a_{2} + a_{3} = \frac{39}{4} + \frac{125}{8} = \frac{203}{8} &=&& 25.375 \\
|
||||
s_{4}&=& a_{0} + a_{1} + a_{2} + a_{3} + a_{4} = \frac{203}{8} + \frac{625}{16} = \frac{1031}{16} &=&& 64,4375
|
||||
\end{alignat*}\\
|
||||
Die Reihe divergiert, da geometrische Reihen immer divergieren, wenn der Betrag von q größer als $1$ ist. Dies ist mit $\frac{5}{2}$ der Fall.
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{3}
|
||||
a_{0} &=& 1 &&&\\
|
||||
a_{1} &=& -\frac{2}{5} &&& \\
|
||||
a_{2} &=& \left(-\frac{2}{5}\right)^{2} = \frac{4}{25} &&& \\
|
||||
a_{3} &=& \left(-\frac{2}{5}\right)^{3} = -\frac{8}{125} &&& \\
|
||||
a_{4} &=& \left(-\frac{2}{5}\right)^{4} = \frac{16}{625} &&& \\
|
||||
s_{0} &=& a_{0} &=&& 1 \\
|
||||
s_{1} &=& a_{0} + a_{1} = 1 - \frac{2}{5} = \frac{3}{5} &=&& 0.6 \\
|
||||
s_{2} &=& a_{0} + a_{1} + a_{2} = \frac{3}{5} + \frac{4}{25} = \frac{19}{25} &=&& 0.76 \\
|
||||
s_{3} &=& a_{0} + a_{1} + a_{2} + a_{3} = \frac{19}{25} - \frac{8}{125} = \frac{87}{125} &=&& 0.696 \\
|
||||
s_{4}&=& a_{0} + a_{1} + a_{2} + a_{3} + a_{4} = \frac{87}{125} + \frac{16}{625} = \frac{451}{625} &=&& 0.7216
|
||||
\end{alignat*} \\
|
||||
Bestimmung des Grenzwertes mithilfe der Geometrischen Summenformel:\\
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left( \sum\limits_{i=0}^{n} \left(-\frac{2}{5}\right)^{i} \right) &=& \underset{n \rightarrow \infty}{\text{lim}} \left(\frac{1-\left(-\frac{2}{5}\right)^{n+1}}{1+\left(\frac{2}{5}\right)} \right) \\
|
||||
&=& \frac{1}{\frac{7}{5}} = \frac{5}{7}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\subsubsection{} %i
|
||||
Bestimmung des Grenzwertes mithilfe der Geometrischen Summenformel:\\
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left( \sum\limits_{i=0}^{n} \left(-\frac{3}{10}\right)^{i} \right) &=& \underset{n \rightarrow \infty}{\text{lim}} \left(\frac{1-\left(-\frac{3}{10}\right)^{n+1}}{1+\left(\frac{3}{10}\right)} \right) \\
|
||||
&=& \frac{1}{\frac{13}{10}} = \frac{10}{13}
|
||||
\end{alignat*}
|
||||
Die Reihe konvergiert gegen den Wert $\frac{10}{13} \approx 0.769$.
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{3}
|
||||
&& \underset{n \rightarrow \infty}{\text{lim}} \left( \sum\limits_{i=0}^{n} x^{i} \right) &=& \frac{5}{8} \\
|
||||
\Rightarrow && \frac{1}{1-x} &=& \frac{5}{8} \\
|
||||
\overset{\cdot (1-x)}{\Leftrightarrow} && 1 &=& \frac{5}{8} - \frac{5}{8}x \\
|
||||
\overset{-\frac{5}{8}}{\Leftrightarrow} && \frac{3}{8} &=& -\frac{5}{8}x \\
|
||||
\overset{\cdot -\frac{8}{5}}{\Leftrightarrow} && -\frac{3}{5} &=& x
|
||||
\end{alignat*}
|
||||
$x$ ist gleich $-\frac{3}{5}$.
|
||||
\section{} %3
|
||||
\subsection{}
|
||||
\subsubsection{} %(i)
|
||||
Die Reihe konvergiert, da der Betrag von $q = \frac{7}{9}$ kleiner als $1$ ist.\\
|
||||
Berechnung des Grenzwertes mithilfe der Geometrischen Summenformel:\\
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left( \sum\limits_{i=0}^{n} \left(\frac{7}{9}\right)^{i} \right) &=& \underset{n \rightarrow \infty}{\text{lim}} \left(\frac{1-\left(\frac{7}{9}\right)^{n+1}}{1-\left(\frac{7}{9}\right)} \right) \\
|
||||
&=& \frac{1}{\frac{2}{9}} = \frac{9}{2}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %(ii)
|
||||
Die Reihe konvergiert, da der Betrag von $q = -\frac{7}{9}$ kleiner als $1$ ist.\\
|
||||
Berechnung des Grenzwertes mithilfe der Geometrischen Summenformel:\\
|
||||
\begin{alignat*}{2}
|
||||
\sum\limits_{i=1}^{\infty} \left(-\frac{7}{9}\right)^{i} &=& \sum\limits_{i=0}^{\infty} \left(-\frac{7}{9}\right)^{i} - \left(-\frac{7}{9}\right)^{0} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \sum\limits_{i=0}^{n} \left(-\frac{7}{9}\right)^{i} - \left(-\frac{7}{9}\right)^{0} \right) \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \sum\limits_{i=0}^{n} \left(-\frac{7}{9}\right)^{i} \right) - 1 \\
|
||||
&=& \frac{1}{1 + \frac{7}{9}} - 1 = \frac{1}{\frac{16}{9}} \\
|
||||
&=& \frac{9}{16} - 1 = -\frac{7}{16}
|
||||
\end{alignat*}
|
||||
Der Grenzwert ist $-\frac{7}{16}$.
|
||||
\subsubsection{} %(iii)
|
||||
\begin{alignat*}{2}
|
||||
\sum\limits_{i=2}^{\infty} (-1)^{i} \cdot \left( \frac{7}{9} \right)^{i+1} &=& \sum\limits_{i=2}^{\infty} (-1)^{i} \cdot \left( \frac{7}{9} \right)^{i} \cdot \left( \frac{7}{9} \right) \\
|
||||
&=& \sum\limits_{i=2}^{\infty} \left( -\frac{7}{9} \right)^{i} \cdot \left( \frac{7}{9} \right) \\
|
||||
&=& \left( \frac{7}{9} \right) \cdot \sum\limits_{i=2}^{\infty} \left( -\frac{7}{9} \right)^{i} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \frac{7}{9} \cdot \sum\limits_{i=2}^{n} \left( -\frac{7}{9} \right)^{i} \right) \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \frac{7}{9} \cdot \left( \sum\limits_{i=0}^{n} \left( -\frac{7}{9} \right)^{i} - \left( -\frac{7}{9} \right)^{0} - \left( -\frac{7}{9} \right)^{1} \right) \right) \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \frac{7}{9} \cdot \left( \sum\limits_{i=0}^{n} \left( -\frac{7}{9} \right)^{i} - 1 + \frac{7}{9} \right) \right) \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \frac{7}{9} \cdot \left( \frac{1- \left(-\frac{7}{9} \right)^{n+1}}{1 - \left(-\frac{7}{9} \right)} - 1 + \frac{7}{9} \right) \right) \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \frac{7}{9} \cdot \left( \frac{1- \left(-\frac{7}{9} \right)^{n+1}}{1 + \frac{7}{9}} - 1 + \frac{7}{9} \right) \right)\\
|
||||
&\rightarrow & \frac{7}{9} \cdot \left( \frac{1}{1 + \frac{7}{9}} - 1 + \frac{7}{9} \right) \\
|
||||
&=& \frac{7}{9} \cdot \left( \frac{9}{16} - 1 + \frac{7}{9} \right) \\
|
||||
&=& \frac{7}{9} \cdot \left( -\frac{7}{16} + \frac{7}{9} \right) \\
|
||||
&=& \frac{7}{9} \cdot \frac{49}{144} \\
|
||||
&=& \frac{343}{1296} \approx 0.26
|
||||
\end{alignat*}\\
|
||||
Der Grenzwert beträgt $\frac{343}{1296}$.
|
||||
\subsubsection{} %(iv)
|
||||
|
||||
\begin{alignat*}{3}
|
||||
\sum\limits_{i=2}^{\infty} \frac{1}{(i+1)i} &=& \sum\limits_{i=2}^{\infty} \left( \frac{1}{i} - \frac{1}{i+1} \right) \\
|
||||
&=& \sum\limits_{i=2}^{\infty} \frac{1}{i} - \sum\limits_{i=2}^{\infty} \frac{1}{i+1} \\
|
||||
&=& \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n} - \frac{1}{3} - \frac{1}{4} - ... - \frac{1}{n+1} \\
|
||||
&\Rightarrow & \underset{n \rightarrow \infty}{\text{lim}} \left(\frac{1}{2} - \frac{1}{n+1} \right) = \frac{1}{2}
|
||||
\end{alignat*}\\
|
||||
Der Grenzwert ist $\frac{1}{2}$.
|
||||
\section{} %4
|
||||
\subsection{}
|
||||
\subsubsection{} %(i)
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right)^{n+3} &=& \underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right)^{n} \cdot \underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right)^{3} \\
|
||||
&=& e \cdot 1 = e
|
||||
\end{alignat*}\\
|
||||
\subsubsection{} %(ii)
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right)^{3n} &=& \underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right)^{n} \cdot \underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right)^{n} \cdot \underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right)^{n} \\
|
||||
&=& e \cdot e \cdot e \\
|
||||
&=& e^{3}
|
||||
\end{alignat*}\\
|
||||
\subsubsection{} %(iii)
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right)^{3} &=& \underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right) \cdot \underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right) \cdot \underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{n} \right) \\
|
||||
&=& 1 \cdot 1 \cdot 1 = 1
|
||||
\end{alignat*}
|
||||
\subsubsection{} %(iv)
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \left(1 + \frac{1}{3n} \right)^{3n} &=& \underset{n \rightarrow \infty}{\text{lim}} \left( \left(1 + \frac{1}{3n} \right)^{3} \right)^{n} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \left(1 + \frac{1}{3n} \right)^{2} \cdot \left(1 + \frac{1}{3n} \right) \right)^{n} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \left(1 + \frac{2}{3n} + \frac{1}{9n^{2}} \right) \cdot \left(1 + \frac{1}{3n} \right) \right)^{n} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( 1 + \frac{1}{3n} + \frac{2}{3n} + \frac{2}{9n^{2}} + \frac{1}{9n^{2}} + \frac{1}{27n^{3}} \right)^{n} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( 1 + \frac{3}{3n} + \frac{3}{9n^{2}} + \frac{1}{27n^{3}} \right)^{n} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( 1 + \frac{1}{n} + \frac{1}{3n^{2}} + \frac{1}{27n^{3}} \right)^{n} \\
|
||||
&\rightarrow & 1
|
||||
\end{alignat*}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,105 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 25. April}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsection{} %a
|
||||
\vspace{11cm}
|
||||
Die Unstetigkeitsstellen sind $x=2$ und $x=6$.
|
||||
\subsection{} %b
|
||||
\vspace{4cm}
|
||||
Die Funktion $g(x)$ ist periodisch. Betrachtet man die Periode für $x=0$ bis $x=1$, so ist ersichtlich, dass sowohl $x=0$ als auch $x=1$ Unstetigkeitsstellen sind. Nähert man sich an die beiden Stellen von links an, so stimmt der Grenzwert nicht mit dem Funktionswert überein. Nähert man sich von rechts an, stimmt er überein.\\
|
||||
Stetigkeit erfordert jedoch, dass der Grenzwert existiert und gleich dem Funktionswert ist, unabhängig von der Folge mit der man sich annähert.
|
||||
Daher ist $g(x)$ in diesen beiden Stellen unstetig. Aufgrund der Periodizität der Funktion ist $g(x)$ an allen Stellen $x \in \mathbb{Z}$ unstetig.\\
|
||||
\\
|
||||
Betrachtet man hingegen eine andere Stelle in der Periode, so stimmen Grenzwert und Funktionswert überein, unabhängig davon ob man sich von rechts oder links annähert. Daher ist $g(x)$ in allen Stellen der Periode mit $x \neq 0$ und $x \neq 1$ stetig. Aufgrund der Periodizität der Funktion ist $g(x)$ an allen Stellen $x \not\in \mathbb{Z}$ stetig.
|
||||
\section{} %2
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} a_{n} &=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\sqrt{3n^{2}-2n+5}-\sqrt{n}}{\sqrt{n^{2}-n+1}+4n} \\
|
||||
\intertext{Ausklammern von $n^{2}$ unterhalb der Wurzeln}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\sqrt{n^{2} (3 - \frac{2}{n} + \frac{5}{n^{2}})} - \sqrt{n^{2} \cdot \frac{1}{n}}}{\sqrt{n^{2} (1 -\frac{1}{n} + \frac{1}{n^{2}})} + 4n} \\
|
||||
\intertext{Wurzelgesetze anwenden}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\sqrt{n^{2}} \cdot \sqrt{3 - \frac{2}{n} + \frac{5}{n^{2}}} - \sqrt{n^{2}} \cdot \sqrt{\frac{1}{n}}}{\sqrt{n^{2}} \cdot \sqrt{1 -\frac{1}{n} + \frac{1}{n^{2}}} + 4n} \\
|
||||
\intertext{Wurzel auflösen}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{n \cdot \sqrt{3 - \frac{2}{n} + \frac{5}{n^{2}}} - n \cdot \sqrt{\frac{1}{n}}}{n \cdot \sqrt{1 -\frac{1}{n} + \frac{1}{n^{2}}} + 4n} \\
|
||||
\intertext{$n$ ausklammern}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{n \cdot (\sqrt{3 - \frac{2}{n} + \frac{5}{n^{2}}} - \sqrt{\frac{1}{n}})}{n \cdot (\sqrt{1 -\frac{1}{n} + \frac{1}{n^{2}}} + 4)} \\
|
||||
\intertext{$n$ kürzen}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\sqrt{3 - \frac{2}{n} + \frac{5}{n^{2}}} - \sqrt{\frac{1}{n}}}{\sqrt{1 -\frac{1}{n} + \frac{1}{n^{2}}} + 4} \\
|
||||
\intertext{lim mit Wurzelfunktion vertauschen, da Wurzelfunktion stetig}
|
||||
&=& \frac{\sqrt{\underset{n \rightarrow \infty}{\text{lim}} (3 - \frac{2}{n} + \frac{5}{n^{2}})} - \sqrt{\underset{n \rightarrow \infty}{\text{lim}}(\frac{1}{n})}}{\sqrt{\underset{n \rightarrow \infty}{\text{lim}} (1 -\frac{1}{n} + \frac{1}{n^{2}})} + 4} \\
|
||||
\intertext{limes anwenden und Nullfolgen entfernen}
|
||||
&=& \frac{\sqrt{3} - \sqrt{0}}{\sqrt{1} + 4} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \frac{\sqrt{3}}{5}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
&& \underset{n \rightarrow \infty}{\text{lim}} \left( \text{cos} \left( \frac{\sqrt{10n^{2}-n}-n}{2n+3} \right) \right) \\
|
||||
\intertext{Ausklammern von $n^{2}$ unter der Wurzel}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \text{cos} \left( \frac{\sqrt{n^{2} (10-\frac{1}{n})}-n}{2n+3} \right) \right) \\
|
||||
\intertext{Wurzelgesetze anwenden}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \text{cos} \left( \frac{\sqrt{n^{2}} \cdot \sqrt{10-\frac{1}{n}}-n}{2n+3} \right) \right) \\
|
||||
\intertext{Wurzel auflösen}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \text{cos} \left( \frac{n \cdot \sqrt{10-\frac{1}{n}}-n}{2n+3} \right) \right) \\
|
||||
\intertext{Ausklammern von $n$ in Zähler und Nenner}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \text{cos} \left( \frac{n \cdot (\sqrt{10 - \frac{1}{n}} - 1)}{n \cdot (2 + \frac{3}{n})} \right) \right) \\
|
||||
\intertext{Kürzen von $n$}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \left( \text{cos} \left( \frac{\sqrt{10 - \frac{1}{n}} - 1}{2 + \frac{3}{n}} \right) \right) \\
|
||||
\intertext{cos mit lim vertauschen, da Cosinusfunktion stetig}
|
||||
&=& \text{cos} \left( \underset{n \rightarrow \infty}{\text{lim}} \left( \frac{\sqrt{10 - \frac{1}{n}} - 1}{2 + \frac{3}{n}} \right) \right) \\
|
||||
\intertext{lim in Wurzel ziehen, da Wurzelfunktion stetig}
|
||||
&=& \text{cos} \left( \frac{\sqrt{\underset{n \rightarrow \infty}{\text{lim}} (10 - \frac{1}{n})} - \underset{n \rightarrow \infty}{\text{lim}} (1) }{\underset{n \rightarrow \infty}{\text{lim}} (2 + \frac{3}{n})} \right) \\
|
||||
&=& \text{cos} \left( \frac{\sqrt{10} - 1}{2} \right)\\
|
||||
&\approx & 0.47
|
||||
\end{alignat*}
|
||||
\section{} %3
|
||||
$g \circ f$ kann auch so geschrieben werden $g(f(x))$. Vereinfacht gesagt, liefert $g$ den Funktionswert an der Stelle, die dem Funktionswert von $f$ an der Stelle $x$ entspricht.\\
|
||||
Das berücksichtigend wissen wir, dass $f$ an der Stelle $x_{0}$ stetig ist. Der Funktionswert für diese Stelle ist $f(x_{0}) = y_{0}$. Wir wissen ferner, dass $g$ an der Stelle $y_{0}$ stetig ist.\\
|
||||
\\
|
||||
Da der Funktionswert von $f$ an der Stelle $x_{0}$ der Stelle entspricht, an der $g$ bekanntermaßen stetig ist, werden hier zwei stetige Funktionen nacheinander ausgeführt. Und die Nacheinanderausführung von zwei stetigen Funktionen ist selbst wiederum stetig.\\
|
||||
Es ist somit ersichtlich, dass $g(f(x_{0}))$ den soeben beschrieben Fall darstellt und damit klarstellt, dass $g \circ f$ ebenfalls an der Stelle $x_{0}$ stetig ist.
|
||||
\section{} %4
|
||||
$g(x)$:\\
|
||||
\begin{alignat*}{2}
|
||||
\underset{x \rightarrow 0}{\text{lim}} g(x) &\Rightarrow & \underset{x \rightarrow 0}{\text{lim}} (x^{2} \cdot 1 ) \geq \underset{x \rightarrow 0}{\text{lim}} \left(x^{2} \cdot \text{sin} \left(\frac{1}{x}\right) \right) \geq \underset{x \rightarrow 0}{\text{lim}} (x^{2} \cdot -1) \\
|
||||
&\Rightarrow & 0 \geq \underset{x \rightarrow 0}{\text{lim}} \left(x^{2} \cdot \text{sin} \left(\frac{1}{x} \right) \right) \geq 0
|
||||
\end{alignat*}
|
||||
Daraus folgt, dass $g(x)$ für alle $x \in \mathbb{R}$ stetig ist. \\
|
||||
|
||||
$h(x)$:\\
|
||||
\begin{alignat*}{2}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} h(x_{n}) &=& \underset{n \rightarrow \infty}{\text{lim}} \text{sin} \left( \frac{1}{x_{n}} \right) \\
|
||||
\intertext{$x_{n}$ sei $\frac{1}{2\pi n}, n \in \mathbb{N}$}
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \text{sin} \left(\frac{1}{\frac{1}{2 \pi n}}\right) \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \text{sin} (2 \pi n) \\
|
||||
&=& \text{sin} \left( \underset{n \rightarrow \infty}{\text{lim}} (2 \pi n) \right) = 0
|
||||
\end{alignat*}
|
||||
$h(x)$ ist stetig für alle $x = x_{n} = \frac{1}{2\pi n}$. Die Funktion ist nicht stetig für andere $x$.
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,284 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 2. Mai}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsection{} %a
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& 7x^{5} + 3x^{3} + x + 1 \\
|
||||
f'(x) &=& 35x^{4} + 9x^{2} + 1
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& (3x^{7} - 4x^{3} + x^{2} - 3x + 1)^{8} \\
|
||||
f'(x) &=& 8 \cdot (3x^{7} - 4x^{3} + x^{2} - 3x + 1)^{7} \cdot (21x^{6} - 12x^{2} + 2x - 3)
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& (3x^{4} + 2x) \cdot \sqrt{x^{2} + 1} \\
|
||||
f'(x) &=& (12x^{3} + 2) \cdot \sqrt{x^{2} + 1} + (3x^{4} + 2x) \cdot \frac{2x}{2 \cdot \sqrt{x^{2} + 1}}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iv
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& (x^{3} + 1) \cdot \ln (x^{4} + 3x^{2} + 1) \\
|
||||
f'(x) &=& 3x \cdot \ln (x^{4} + 3x^{2} + 1) + (x^{3} + 1) \cdot \frac{4x^{3} + 6x}{x^{4} + 3x^{2} + 1}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %v
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& e^{x^{3} + x^{2} + 1} \cdot \sqrt{x} \\
|
||||
f'(x) &=& e^{x^{3} + x^{2} + 1} \cdot (3x^{2} + 2x) \cdot \sqrt{x} + e^{x^{3} + x^{2} + 1} \cdot \frac{1}{2 \cdot \sqrt{x}}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %vi
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \sqrt{x^{4} + 1} \cdot \ln x \\
|
||||
f'(x) &=& \frac{4x^{3}}{2 \cdot \sqrt{x^{4} + 1}} \cdot \ln x + \sqrt{x^{4} + 1} \cdot \frac{1}{x}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
q(x) &=& \frac{5x^{2} + 1}{x - 3} \\
|
||||
q'(x) &=& \frac{10x \cdot (x-3) - (5x^{2} + 1) \cdot 1}{(x-3)^{2}} \\
|
||||
\intertext{Ausmultiplizieren}
|
||||
&=& \frac{10x^{2} - 30x - (5x^{2} + 1)}{x^{2} - 6x + 9}\\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \frac{5x^{2} - 30x - 1}{x^{2} - 6x + 9}\\
|
||||
q''(x) &=& \frac{(10x-30) \cdot (x^{2} - 6x + 9) - (5x^{2} - 30x - 1) \cdot (2x - 6)}{(x^{2} - 6x + 9)^{2}} \\
|
||||
\intertext{Ausmultiplizieren}
|
||||
&=& \frac{10x^{3} - 60x^{2} + 90x - 30x^{2} + 180x - 270 - (10x^{3} - 30x^{2} - 60x^{2} + 180x -x^{2} - 9)}{x^{4} - 6x^{3} + 9x^{2} - 6x^{3} + 36x^{2} - 54x + 9x^{2} - 54x + 81} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \frac{10x^{3} - 90x^{2} + 270x - 270 -10x^{3} + 91x^{2} - 180x + 9}{x^{4} - 12x^{3} + 54x^{2} - 108x + 81} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \frac{x^{2} + 90x - 261}{x^{4} - 12x^{3} + 54x^{2} - 108x + 81} \\
|
||||
q'''(x) &=& \frac{(2x + 90) \cdot (x^{4} - 12x^{3} + 54x^{2} - 108x + 81) - (x^{2} + 90x - 261) \cdot (4x^{3} - 36x^{2} + 108x - 108)}{(x^{4} - 12x^{3} + 54x^{2} - 108x + 81)^{2}} \\
|
||||
\intertext{Ausmultiplizieren und Zusammenfassen}
|
||||
&=& \frac{-2x^{5} - 258x^{4} + 3204x^{3} - 14364x^{2} +28350x - 20898}{x^{8} - 24x^{7} + 252x^{6} -1404x^{5} +5670x^{4} - 13716x^{3} + 20412x^{2} -8748x - 2187}
|
||||
\end{alignat*}
|
||||
\section{} %2
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \left| 3 - \frac{1}{2}x \right| \\
|
||||
\intertext{Sei $x_{n} = 6 + \frac{1}{n} : n \in \mathbb{N}$}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \frac{f(x_{n}) - f(x_{0})}{x_{n} - x_{0}} &=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\left| 3 - \frac{1}{2} \cdot (6 + \frac{1}{n}) \right| - \left| 3 - \frac{1}{2} \cdot 6 \right|}{6 + \frac{1}{n} - 6} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\left| 3 - 3 - \frac{1}{2n} \right| - \left| 3 - 3 \right|}{\frac{1}{n}} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\left|-\frac{1}{2n} \right|}{\frac{1}{n}} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\frac{1}{2n}}{\frac{1}{n}} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{n}{2n} = \frac{1}{2} \\
|
||||
\intertext{Sei $x_{n} = 6 - \frac{1}{n} : n \in \mathbb{N}$}
|
||||
\underset{n \rightarrow \infty}{\text{lim}} \frac{f(x_{n}) - f(x_{0})}{x_{n} - x_{0}} &=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\left| 3 - \frac{1}{2} \cdot (6 - \frac{1}{n}) \right| - \left| 3 - \frac{1}{2} \cdot 6 \right|}{6 - \frac{1}{n} - 6} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\left| 3 - 3 + \frac{1}{2n} \right| - \left| 3 - 3 \right|}{-\frac{1}{n}} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\left|\frac{1}{2n} \right|}{-\frac{1}{n}} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} \frac{\frac{1}{2n}}{-\frac{1}{n}} \\
|
||||
&=& \underset{n \rightarrow \infty}{\text{lim}} -\frac{n}{2n} = -\frac{1}{2}
|
||||
\end{alignat*}
|
||||
Der Grenzwert existiert an der Stelle $x _{0} = 6$ nicht. Daher ist die Funktion $f$ an der Stelle $x_{0} = 6$ nicht differenzierbar.
|
||||
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=-10,ymax=10,
|
||||
x=1em,
|
||||
y=1em,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
]
|
||||
\addplot[no marks, black, -] expression[domain=-10:6,samples=100]{3 - (1/2)*x} node[pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=6:10,samples=100]{-1*(3 - (1/2)*x)} node[pos=0.65,anchor=north]{};
|
||||
|
||||
\draw (16em, 10em) circle (2pt);
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\section{} %3
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& (x^{4}+1)^{x+2} \\
|
||||
&=& e^{ln\left((x^{4}+1)^{x+2}\right)} \\
|
||||
&=& e^{(x+2) \cdot \ln(x^{4}+1)} \\
|
||||
f'(x) &=& e^{(x+2) \cdot \ln(x^{4}+1)} \cdot \left((x+2) \cdot \ln(x^{4}+1)\right)' \\
|
||||
&=& e^{(x+2) \cdot \ln(x^{4}+1)} \cdot \left((x+2)' \cdot \ln(x^{4}+1) + (x+2) \cdot \ln(x^{4}+1)'\right) \\
|
||||
&=& e^{(x+2) \cdot \ln(x^{4}+1)} \cdot \left(1 \cdot \ln(x^{4}+1) + (x+2) \cdot \frac{1}{x^{4}+1} \cdot (x^{4}+1)'\right) \\
|
||||
&=& (x^{4}+1)^{x+2} \cdot \left(\ln(x^{4}+1) + (x+2) \cdot \frac{4x^{3}}{x^{4}+1}\right)
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& x^{\frac{1}{2}} \\
|
||||
&=& e^{\ln \left(x^{\frac{1}{2}} \right)} \\
|
||||
&=& e^{\frac{1}{2} \cdot \ln (x)} \\
|
||||
f'(x) &=& e^{\frac{1}{2} \cdot \ln (x)} \cdot \left(\frac{1}{2} \cdot \ln (x) \right)' \\
|
||||
&=& e^{\frac{1}{2} \cdot \ln (x)} \cdot \frac{1}{2} \cdot \frac{1}{x} \cdot (x)' \\
|
||||
&=& e^{\frac{1}{2} \cdot \ln (x)} \cdot \frac{1}{2x} \\
|
||||
&=& x^{\frac{1}{2}} \cdot \frac{1}{2x} \\
|
||||
&=& \frac{1}{2} \cdot \frac{x^{\frac{1}{2}}}{x} \\
|
||||
&=& \frac{1}{2} \cdot x^{-\frac{1}{2}} \\
|
||||
g(x) &=& \left( \frac{1}{2}\right)^{x} \\
|
||||
&=& e^{\ln \left( \left( \frac{1}{2}\right)^{x} \right)} \\
|
||||
&=& e^{x \cdot \ln \left( \frac{1}{2}\right)} \\
|
||||
g'(x) &=& e^{x \cdot \ln \left( \frac{1}{2}\right)} \cdot \left(x \cdot \ln \left( \frac{1}{2}\right) \right)' \\
|
||||
&=& e^{x \cdot \ln \left( \frac{1}{2}\right)} \cdot (x)' \cdot \ln \left( \frac{1}{2}\right) \\
|
||||
&=& e^{x \cdot \ln \left( \frac{1}{2}\right)} \cdot 1 \cdot \ln \left( \frac{1}{2}\right) \\
|
||||
&=& \left( \frac{1}{2}\right)^{x} \cdot \ln \left( \frac{1}{2}\right)
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
g(x) &=& (x^{2}+1)^{4x+1} \\
|
||||
&=& e^{\ln \left( (x^{2}+1)^{4x+1}\right)} \\
|
||||
&=& e^{(4x+1) \cdot \ln (x^{2}+1)} \\
|
||||
g'(x) &=& e^{(4x+1) \cdot \ln (x^{2}+1)} \cdot \left((4x+1) \cdot \ln (x^{2}+1) \right)' \\
|
||||
&=& e^{(4x+1) \cdot \ln (x^{2}+1)} \cdot \left((4x+1)' \cdot \ln (x^{2}+1) + (4x+1) \cdot \ln (x^{2}+1)' \right) \\
|
||||
&=& e^{(4x+1) \cdot \ln (x^{2}+1)} \cdot \left(4 \cdot \ln (x^{2}+1) + (4x+1) \cdot \frac{1}{x^{2}+1} \cdot (x^{2}+1)' \right) \\
|
||||
&=& e^{(4x+1) \cdot \ln (x^{2}+1)} \cdot \left(4 \cdot \ln (x^{2}+1) + (4x+1) \cdot \frac{2x}{x^{2}+1} \right) \\
|
||||
&=& (x^{2}+1)^{4x+1} \cdot \left(4 \cdot \ln (x^{2}+1) + \frac{8x^{2} + 2x}{x^{2}+1} \right)
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
h(x) = (x-3)^{3x^{4}+5} \\
|
||||
&=& e^{\ln \left((x-3)^{3x^{4}+5} \right)} \\
|
||||
&=& e^{(3x^{4}+5) \cdot \ln (x-3)} \\
|
||||
h'(x) &=& e^{(3x^{4}+5) \cdot \ln (x-3)} \cdot \left((3x^{4}+5) \cdot \ln (x-3) \right)' \\
|
||||
&=& e^{(3x^{4}+5) \cdot \ln (x-3)} \cdot \left((3x^{4}+5)' \cdot \ln (x-3) + (3x^{4}+5) \cdot \ln (x-3)' \right) \\
|
||||
&=& e^{(3x^{4}+5) \cdot \ln (x-3)} \cdot \left(12x^{3} \cdot \ln (x-3) + (3x^{4}+5) \cdot \frac{1}{x-3} \cdot (x-3)' \right) \\
|
||||
&=& e^{(3x^{4}+5) \cdot \ln (x-3)} \cdot \left(12x^{3} \cdot \ln (x-3) + (3x^{4}+5) \cdot \frac{1}{x-3} \right) \\
|
||||
&=& (x-3)^{3x^{4}+5} \cdot \left(12x^{3} \cdot \ln (x-3) + \frac{3x^{4}+5}{x-3} \right)
|
||||
\end{alignat*}
|
||||
\section{} %4
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
g(p) &=& 10^{5} \cdot \left( \frac{1}{p} - \frac{3}{p^{2}} \right) \\
|
||||
&=& 10^{5} \cdot \left( p^{-1} - 3p^{-2} \right) \\
|
||||
g'(p) &=& 10^{5} \cdot \left( p^{-1} - 3p^{-2} \right)' \\
|
||||
&=& 10^{5} \cdot \left((p^{-1})' - (3p^{-2})' \right) \\
|
||||
&=& 10^{5} \cdot \left(-p^{-2} - (-6p^{-3}) \right) \\
|
||||
&=& 10^{5} \cdot \left(-p^{-2} + 6p^{-3} \right) \\
|
||||
&=& 10^{5} \cdot \left(-\frac{1}{p^{2}} + \frac{6}{p^{3}} \right) \\
|
||||
g''(p) &=& 10^{5} \cdot \left(-p^{-2} + 6p^{-3} \right)' \\
|
||||
&=& 10^{5} \cdot \left((-p^{-2})' + (6p^{-3})' \right) \\
|
||||
&=& 10^{5} \cdot \left(2p^{-3} - 18p^{-4} \right) \\
|
||||
&=& 10^{5} \cdot \left(\frac{2}{p^{3}} - \frac{18}{p^{4}} \right)
|
||||
\end{alignat*}
|
||||
\begin{enumerate}
|
||||
\item Bestimmung der Nullstellen der 1. Ableitung:
|
||||
\begin{alignat*}{2}
|
||||
-\frac{1}{p^{2}} + \frac{6}{p^{3}} &=& 0 \\
|
||||
-\frac{1}{p^{2}} &=& - \frac{6}{p^{3}} \\
|
||||
\frac{1}{p^{2}} &=& \frac{6}{p^{3}} \\
|
||||
\frac{p^{3}}{p^{2}} &=& 6 \\
|
||||
p &=& 6 \\
|
||||
\intertext{Einsetzen in $g'(p)$}
|
||||
0 &=& 10^{5} \cdot \left(-\frac{1}{6^{2}} + \frac{6}{6^{3}} \right) \\
|
||||
&=& 10^{5} \cdot \left(-\frac{1}{36} + \frac{6}{216} \right) \\
|
||||
&=& 10^{5} \cdot \left(-\frac{1}{36} + \frac{1}{36} \right) \\
|
||||
&=& 10^{5} \cdot 0 = 0
|
||||
\end{alignat*} \\
|
||||
\item Einsetzen von $p=6$ in $g''(p)$:
|
||||
\begin{alignat*}{2}
|
||||
g''(6) &=& 10^{5} \cdot \left(\frac{2}{6^{3}} - \frac{18}{6^{4}} \right) \\
|
||||
&=& 10^{5} \cdot \left(\frac{2}{216} - \frac{18}{1296} \right) \\
|
||||
&=& 10^{5} \cdot \left(\frac{2}{216} - \frac{3}{216} \right) \\
|
||||
&=& 10^{5} \cdot \left(- \frac{1}{216} \right) = -\frac{10^{5}}{216}
|
||||
\end{alignat*}
|
||||
Daraus ergibt sich, dass $p=6$ ein lokales Maximum der Funktion $g(p)$ ist. Es bleibt noch festzustellen, dass es auch das globale Maximum ist. \\
|
||||
\item Feststellung des globalen Maximums:\\
|
||||
Da es nur eine Nullstelle für die erste Ableitung gibt, kann es insofern nur einen Extrempunkt geben. Da der Funktionswert der 2. Ableitung für diese Stelle negativ ist, liegt an der Stelle ein Maximum. Der Funktionswert von $g(p)$ für die Stelle wird wie folgt berechnet:
|
||||
\begin{alignat*}{2}
|
||||
g(p) &=& 10^{5} \cdot \left(\frac{1}{p} - \frac{3}{p^{2}} \right) \\
|
||||
g(6) &=& 10^{5} \cdot \left( \frac{1}{6} - \frac{3}{3^{2}} \right) \\
|
||||
&=& 10^{5} \cdot \left( \frac{1}{6} - \frac{3}{9} \right) \\
|
||||
&=& 10^{5} \cdot \left( \frac{1}{6} - \frac{2}{6} \right) \\
|
||||
&=& 10^{5} \cdot \left(- \frac{1}{6} \right) \\
|
||||
&=& -\frac{10^{5}}{6} = \frac{5000}{3} \\
|
||||
&=& \frac{4998}{3} + \frac{2}{3} \\
|
||||
&=& 1666 + \frac{2}{3}
|
||||
\end{alignat*} \\\\\\
|
||||
Zuletzt muss noch geprüft werden, ob die Grenzen des Definitionsbereiches einen höheren Wert aufweisen.
|
||||
\begin{alignat*}{2}
|
||||
g(p) &=& 10^{5} \cdot \left(\frac{1}{p} - \frac{3}{p^{2}} \right) \\
|
||||
g(3) &=& 10^{5} \cdot \left(\frac{1}{3} - \frac{3}{3^{2}} \right) \\
|
||||
&=& 10^{5} \cdot \left(\frac{1}{3} - \frac{3}{9} \right) \\
|
||||
&=& 10^{5} \cdot \left(\frac{1}{3} - \frac{1}{3} \right) \\
|
||||
&=& 10^{5} \cdot 0 = 0 \\
|
||||
g(100) &=& 10^{5} \cdot \left(\frac{1}{100} - \frac{3}{100^{2}} \right) \\
|
||||
&=& 10^{5} \cdot \left(\frac{1}{100} - \frac{3}{10000} \right) \\
|
||||
&=& 10^{5} \cdot \left(\frac{100}{10000} - \frac{3}{10000} \right) \\
|
||||
&=& 10^{5} \cdot \frac{97}{10000} \\
|
||||
&=& 970
|
||||
\end{alignat*}
|
||||
Die Funktion hat nur einen Extrempunkt und die beiden Definitionsgrenzen weisen einen niedrigeren Funktionswert auf, als der zuvor bestimmte lokale Extrempunkt mit $p = 6$. Somit ist der einzige lokale Extrempunkt auch der globale Extrempunkt. Da es sich bei dem Extrempunkt um ein Maximum handelt, ist er somit das globale Maximum.
|
||||
\end{enumerate}
|
||||
\subsection{} %b
|
||||
\subsubsection{} %i
|
||||
Bestimmung der Nullstellen von $f, f'$ und $f''$:\\
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& -2x^{3} - x + 25 \\
|
||||
f'(x) &=& -6x^{2} - 1 \\
|
||||
f''(x) &=& -18x \\
|
||||
f'''(x) &=& -18 \\
|
||||
f(2) &=& -2 \cdot 2^{3} - 2 + 25 \\
|
||||
&=& -2 \cdot 8 +23 \\
|
||||
&=& -16 + 23 = 7 \\
|
||||
f(3) &=& -2 \cdot 3^{3} - 3 + 25 \\
|
||||
&=& -2 \cdot 27 + 22 \\
|
||||
&=& -54 + 22 \\
|
||||
&=& -32 \\
|
||||
f'(0) &=& -6 \cdot 0^{2} - 1 \\
|
||||
&=& 0 -1 = -1 \\
|
||||
f''(0) &=& -18 \cdot 0 = 0
|
||||
\end{alignat*}
|
||||
Die Nullstelle von $f(x)$ liegt zwischen $x=2$ und $x=3$. $f'(x)$ hat keine Nullstelle. Der höchste erreichbare Wert ist $-1$.
|
||||
$f''(x)$ hat eine Nullstelle, welche nachweislich bei $x=0$ liegt.
|
||||
$f'''(x)$ hat offensichtlich keine Nullstelle.\\
|
||||
\\
|
||||
$f(x) > 0$ für $x \leq 2$ und $f(x) < 0$ für $x \geq 3$. $f'(x) < 0$ für alle $x \in \mathbb{R} \wedge x \geq -5 \wedge x \leq 5$. $f''(x) < 0$ für $x < 0$ und $f''(x) > 0$ für $x > 0$. $f'''(x) > 0$ für alle $x \in \mathbb{R} \wedge x \geq -5 \wedge x \leq 5$. \\
|
||||
Es gibt keine Maxima oder Minima. Allerdings gibt es einen Wendepunkt bei $x=0$, da $f''(0) = 0$ und $f'''(0) \neq 0$. \\
|
||||
Da es keine Nullstellen der ersten Ableitung gibt und $f(2) > f(3)$ gilt, ist $f(x)$ auf dem gesamten Intervall streng monoton fallend. Der höchste Punkt ist demnach am Beginn des Definitionsbereiches, der niedrigste am Ende.
|
||||
Daraus ergibt sich:\\
|
||||
\begin{alignat*}{2}
|
||||
f(-5) &=& -2 \cdot (-5)^{3} - (-5) + 25 \\
|
||||
&=& -2 \cdot (-125) + 30 \\
|
||||
&=& 250 + 30 = 280 \\
|
||||
f(5) &=& -2 \cdot 5^{3} - 5 + 25 \\
|
||||
&=& -2 \cdot 125 + 20 \\
|
||||
&=& -250 + 20 = -230
|
||||
\end{alignat*}
|
||||
$f(x)$ nimmt das globale Maximum bei $x=-5$ und das globale Minimum bei $x=5$ an.
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
g(x) &=& x^{3} - 6x^{2} + 3x + 8 \\
|
||||
g'(x) &=& 3x^{2} - 12x + 3 \\
|
||||
g''(x) &=& 6x - 12 \\
|
||||
g'''(x) &=& 6
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
h(x) &=& e^{2x-3} - e^{x+2} \\
|
||||
h'(x) &=& e^{2x-3} \cdot 2 - e^{x+2} \\
|
||||
h''(x) &=& e^{2x-3} \cdot 4 - e^{x+2} \\
|
||||
h'''(x) &=& e^{2x-3} \cdot 8 - e^{x+2}
|
||||
\end{alignat*}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,349 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 16. Mai}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \frac{1}{\sqrt[4]{x^{5}}} \cdot \sqrt[3]{\sqrt{x^{7}}} \\
|
||||
\intertext{Wurzeln umformen}
|
||||
&=& \frac{1}{(x^{5})^{\frac{1}{4}}} \cdot (\sqrt{x^{7}})^{\frac{1}{3}} \\
|
||||
\intertext{Wurzeln umformen}
|
||||
&=& \frac{1}{(x^{5})^{\frac{1}{4}}} \cdot ((x^{7})^{\frac{1}{2}})^{\frac{1}{3}} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \frac{1}{x^{\frac{5}{4}}} \cdot x^{\frac{7}{6}} \\
|
||||
\intertext{Umformen}
|
||||
&=& x^{-\frac{5}{4}} \cdot x^{\frac{7}{6}} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& x^{\frac{7}{6} - \frac{5}{4}} \\
|
||||
\intertext{Auf selben Nenner bringen}
|
||||
\intertext{Zusammenfassen}
|
||||
&=& x^{\frac{14 - 15}{12}} \\
|
||||
&=& x^{-\frac{1}{12}} \\
|
||||
f'(x) &=& -\frac{1}{12}x^{-\frac{13}{12}}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \sin (x^{2}) \\
|
||||
f'(x) &=& \cos (x^{2}) \cdot 2x
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \sin ^{2} x \\
|
||||
&=& \sin x \cdot \sin x \\
|
||||
f'(x) &=& \cos x \cdot \sin x + \sin x \cdot \cos x
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iv
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \sin x \cdot \cos x \\
|
||||
f'(x) &=& \cos x \cdot \cos x + \sin x \cdot (- \sin x) \\
|
||||
&=& \cos x \cdot \cos x - \sin x \cdot + \sin x)
|
||||
\end{alignat*}
|
||||
\subsubsection{} %v
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \arcsin (\sqrt{x}) \\
|
||||
f'(x) &=& \frac{1}{\sqrt{1 - \sqrt{x}}} \cdot (\sqrt{x})' \\
|
||||
&=& \frac{1}{\sqrt{1 - \sqrt{x}}} \cdot \frac{1}{2 \sqrt{x}}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %vi
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& (x^{3}-1)^{\arctan x} \\
|
||||
&=& e^{\ln \left((x^{3}-1)^{\arctan x} \right)} \\
|
||||
&=& e^{\left(\arctan x \right) \cdot \ln \left(x^{3}-1 \right)} \\
|
||||
f'(x) &=& e^{\left(\arctan x \right) \cdot \ln \left(x^{3}-1 \right)} \cdot \left( \left(\arctan x \right) \cdot \ln \left(x^{3}-1 \right)\right)' \\
|
||||
&=& e^{\left(\arctan x \right) \cdot \ln \left(x^{3}-1 \right)} \cdot \left( \left(\arctan x \right)' \cdot \ln \left(x^{3}-1 \right) + \left(\arctan x \right) \cdot \ln \left(x^{3}-1 \right)' \right) \\
|
||||
&=& e^{\left(\arctan x \right) \cdot \ln \left(x^{3}-1 \right)} \cdot \left( \frac{1}{\sqrt{1 - x}} \cdot \ln \left(x^{3}-1 \right) + \left(\arctan x \right) \cdot \frac{1}{x^{3}-1} \cdot (x^{3}-1)' \right) \\
|
||||
&=& e^{\left(\arctan x \right) \cdot \ln \left(x^{3}-1 \right)} \cdot \left( \frac{1}{\sqrt{1 - x}} \cdot \ln \left(x^{3}-1 \right) + \left(\arctan x \right) \cdot \frac{3x^{2}}{x^{3}-1}\right) \\
|
||||
&=& (x^{3}-1)^{\arctan x} \cdot \left( \frac{1}{\sqrt{1 - x}} \cdot \ln \left(x^{3}-1 \right) + \left(\arctan x \right) \cdot \frac{3x^{2}}{x^{3}-1}\right)
|
||||
\end{alignat*}
|
||||
\section{} %2
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \frac{2x}{1+x^{2}} \\
|
||||
f'(x) &=& \frac{2 \cdot (1+x^{2}) - 2x \cdot 2x}{\left(1+x^{2} \right)^{2}} \\
|
||||
&=& \frac{2 \cdot (1+x^{2}) - 4x^{2}}{\left(1+x^{2} \right)^{2}} \\
|
||||
f''(x) &=& \frac{\left(2 \cdot (1+x^{2}) - 4x^{2} \right)' \cdot \left(1+x^{2} \right)^{2} - ((2 \cdot (1+x^{2}) - 4x^{2}) \cdot \left(\left(1+x^{2} \right)^{2} \right)'}{\left(1+x^{2} \right)^{4}} \\
|
||||
&=& \frac{(4x-8x) \cdot \left(1+x^{2} \right)^{2} - (2 \cdot (1+x^{2}) - 4x^{2}) \cdot 2 \cdot (1+x^{2}) \cdot 2x}{\left(1+x^{2} \right)^{4}} \\
|
||||
\intertext{$(1+x^{2})$ ausklammern und kürzen}
|
||||
&=& \frac{-4x \cdot \left(1+x^{2} \right) - (2 \cdot (1+x^{2}) - 4x^{2}) \cdot 2 \cdot 2x}{\left(1+x^{2} \right)^{3}} \\
|
||||
&=& \frac{4x \cdot \left( -\left(1+x^{2} \right) - (2 \cdot (1+x^{2}) - 4x^{2})\right)}{\left(1+x^{2} \right)^{3}} \\
|
||||
&=& -\frac{4x \cdot \left(1+x^{2} + (2 \cdot (1+x^{2}) - 4x^{2}\right)}{\left(1+x^{2} \right)^{3}} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& -\frac{4x \cdot \left(3-x^{2}\right)}{\left(1+x^{2} \right)^{3}}
|
||||
\end{alignat*}
|
||||
\begin{enumerate}
|
||||
\item $x \in \mathbb{R}$
|
||||
\item
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& 0 \\
|
||||
\intertext{$f(x)$ mit Zähler ersetzen, da der über Nullstelle bestimmt}
|
||||
2x &=& 0 \\
|
||||
\intertext{geteilt durch $2$}
|
||||
x &=& 0 \\
|
||||
f'(x) &=& 0 \\
|
||||
\intertext{$f'(x)$ mit Zähler ersetzen, da der über Nullstelle bestimmt}
|
||||
2 - 2x^{2} &=& 0 \\
|
||||
\intertext{$2$ subtrahieren}
|
||||
-2x^{2} &=& -2 \\
|
||||
\intertext{geteilt durch $-2$}
|
||||
x^{2} &=& 1 \\
|
||||
\intertext{Wurzel ziehen}
|
||||
x_{1} &=& 1 \\
|
||||
x_{2} &=& -1 \\
|
||||
f''(x) &=& 0 \\
|
||||
\intertext{$f''(x)$ mit Zähler ersetzen, da der über Nullstelle bestimmt}
|
||||
-4x \cdot \left(3-x^{2}\right) &=& 0 \\
|
||||
&\Rightarrow & x_{1} = 0, x_{2} = \sqrt{3}, x_{3} = - \sqrt{3}
|
||||
\end{alignat*}
|
||||
\item Es gibt keine Randpunkte des Definitionsbereiches.
|
||||
\item
|
||||
\begin{alignat*}{2}
|
||||
f(-1) &=& \frac{2\cdot (-1)}{1 + (-1)^{2}} \\
|
||||
&=& \frac{-2}{2} = -1 \\
|
||||
f(1) &=& \frac{2 \cdot 1}{1 + 1^{2}} \\
|
||||
&=& \frac{2}{2} = 1 \\
|
||||
f'(-2) &=& \frac{2 - 2 \cdot (-2)^{2}}{\left(1+ (-2)^{2} \right)^{2}} \\
|
||||
&=& \frac{2 - 2 \cdot 4}{\left(1+ 4 \right)^{2}} \\
|
||||
&=& \frac{-6}{\left(5 \right)^{2}} \\
|
||||
&=& \frac{-6}{25} \\
|
||||
f'(0) &=& \frac{2 - 2 \cdot 0^{2}}{\left(1+ 0^{2} \right)^{2}} \\
|
||||
&=& \frac{2}{1} = 2 \\
|
||||
f'(2) &=& \frac{2 - 2 \cdot 2^{2}}{\left(1+ 2^{2} \right)^{2}} \\
|
||||
&=& \frac{2 - 8}{\left(5 \right)^{2}} \\
|
||||
&=& \frac{-6}{25}
|
||||
\end{alignat*}\\
|
||||
\begin{alignat*}{2}
|
||||
f''(-2) &=& -\frac{4 \cdot (-2) \cdot \left(3- (-2)^{2}\right)}{\left(1+ (-2)^{2} \right)^{3}} \\
|
||||
&=& -\frac{-8 \cdot \left(3- 4 \right)}{\left(1+ 4 \right)^{3}} \\
|
||||
&=& -\frac{-8 \cdot -1}{5^{3}} \\
|
||||
&=& -\frac{8}{125} \\
|
||||
f''(-1) &=& -\frac{4 \cdot (-1) \cdot \left(3- (-1)^{2}\right)}{\left(1+ (-1)^{2} \right)^{3}} \\
|
||||
&=& -\frac{-4 \cdot \left(3- 1\right)}{\left(1+ 1 \right)^{3}} \\
|
||||
&=& -\frac{-4 \cdot 2}{2^{3}} \\
|
||||
&=& -\frac{-8}{8} \\
|
||||
&=& 1 \\
|
||||
f''(1) &=& -\frac{4 \cdot 1 \cdot \left(3- 1^{2}\right)}{\left(1+ 1^{2} \right)^{3}} \\
|
||||
&=& -\frac{4 \cdot 2}{2^{3}} \\
|
||||
&=& -\frac{8}{8} \\
|
||||
&=& -1 \\
|
||||
f''(2) &=& -\frac{4 \cdot 2 \cdot \left(3- 2^{2}\right)}{\left(1+ 2^{2} \right)^{3}} \\
|
||||
&=& -\frac{8 \cdot \left(3- 4\right)}{\left(1+ 4 \right)^{3}} \\
|
||||
&=& -\frac{8 \cdot -1}{5^{3}} \\
|
||||
&=& -\frac{-8}{125} \\
|
||||
&=& \frac{8}{125}
|
||||
\end{alignat*}\\
|
||||
$f(x)$ ist negativ für $x < 0$ und positiv für $x > 0$. $f$ ist streng monoton fallend für $x < -1$, streng monoton steigend für $-1 < x < 1$ und streng monoton fallend für $x > 1$. $f$ ist streng konkav für $x < -\sqrt{3}$, streng konvex für $- \sqrt{3} < x < 0$, streng konkav für $0 < x < \sqrt{3}$ und streng konvex für $x > \sqrt{3}$.
|
||||
\label{enum:1}
|
||||
\item Aus \ref{enum:1}\text{.} ergibt sich: $f$ hat ein Minimum an $x=-1$, ein Maximum an $x=1$, Wendepunkte an $x=-\sqrt{3}, x=0$ und $x=\sqrt{3}$.
|
||||
\item Mithilfe von Satz 22 zeige ich, dass $f$ für $x \rightarrow \infty$ eine Asymptote $g(x) =ax +b$ besitzt und berechne diese:\\
|
||||
\begin{alignat*}{2}
|
||||
a = \lim_{x \rightarrow \infty} \left( \frac{f(x)}{x} \right) &=& \lim_{x \rightarrow \infty} \left(\frac{2x}{x+x^{3}} \right) \\
|
||||
&=& \lim_{x \rightarrow \infty} \left(\frac{(2) \cdot x}{(\frac{1}{x^{2}}+1) \cdot x^{3}} \right) \\
|
||||
&=& \lim_{x \rightarrow \infty} \left(\frac{2}{(\frac{1}{x^{2}}+1) \cdot x^{2}} \right) \\
|
||||
\intertext{Der Nenner geht gegen unendlich, der Zähler ist konstant, daher geht der Ausdruck gegen Null.}
|
||||
&=& 0\\
|
||||
\\
|
||||
b = \lim_{x \rightarrow \infty} \left( f(x) - ax \right) &=& \lim_{x \rightarrow \infty} \left( \frac{2x}{1+x^{2}} - 0 \right) \\
|
||||
&=& \lim_{x \rightarrow \infty} \left( \frac{(2) \cdot x}{(\frac{1}{x^{2}}+1) \cdot x^{2}} \right) \\
|
||||
\lim_{x \rightarrow \infty} \left( \frac{2}{(\frac{1}{x^{2}}+1) \cdot x} \right) \\
|
||||
\intertext{Der Nenner geht gegen unendlich, der Zähler ist konstant, daher geht der Ausdruck gegen Null.}
|
||||
&=& 0
|
||||
\end{alignat*}\\
|
||||
Also besitzt $f$ für $x \rightarrow \infty$ die Asymptote $g(x)=0$. Eine ähnliche Überlegung ergibt, dass $f$ für $x \rightarrow -\infty$ ebenfalls die Asymptote $g(x)=0$ besitzt.\\
|
||||
Es gibt einen Schnittpunkt von $f$ mit der Asymptote bei $x=0$, da\\
|
||||
\begin{alignat*}{2}
|
||||
\frac{2x}{1+x^{2}} &=& 0 \\
|
||||
\intertext{Mit $1+x^{2}$ multiplizieren geht, da der Term nie gleich 0 sein kann.}
|
||||
2x &=& 0 \\
|
||||
x &=& 0
|
||||
\end{alignat*}\\
|
||||
gilt.
|
||||
\item
|
||||
\begin{alignat*}{2}
|
||||
f(0) &=& \frac{2 \cdot 0}{1 + 0^{2}} \\
|
||||
&=& \frac{0}{1} = 0\\
|
||||
f(-1) &=& \frac{2 \cdot -1}{1 + (-1)^{2}} \\
|
||||
&=& \frac{-2}{2} = -1 \\
|
||||
f(1) &=& \frac{2 \cdot 1}{1 + 1^{2}} \\
|
||||
&=& \frac{2}{2} = 1 \\
|
||||
f(-\sqrt{3}) &=& \frac{2 \cdot (-\sqrt{3})}{1 + (-\sqrt{3})^{2}} \\
|
||||
&=& \frac{-2\sqrt{3}}{1 + 3} \\
|
||||
&=& \frac{-2\sqrt{3}}{4} \\
|
||||
&=& -\frac{\sqrt{3}}{2} \\
|
||||
f(\sqrt{3}) &=& \frac{2 \cdot \sqrt{3}}{1 + \sqrt{3}^{2}} \\
|
||||
&=& \frac{2\sqrt{3}}{1 + 3} \\
|
||||
&=& \frac{2\sqrt{3}}{4} \\
|
||||
&=& \frac{\sqrt{3}}{2}
|
||||
\end{alignat*}
|
||||
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=-5,ymax=5,
|
||||
x=1em,
|
||||
y=1em,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
]
|
||||
\addplot[no marks, black, -] expression[domain=-10:10,samples=100]{(2*x)/(1+(x*x))} node[pos=0.65,anchor=north]{};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\end{enumerate}
|
||||
\section{} %3
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& 4x^{3} - 10x +5 \\
|
||||
f'(x) &=& 12x^{2} - 10 \\
|
||||
\intertext{Die Berechnung der Näherungswerte erfolgt durch folgende Formel:}
|
||||
x_{n+1} &=& x_{n} - \frac{f(x_{n})}{f'(x_{n})} \\
|
||||
\intertext{Exemplarische Berechnung von $f(x)$ für $x=2$:}
|
||||
f(2) &=& 4 \cdot 2^{3} - 10 \cdot 2 +5 \\
|
||||
&=& 4 \cdot 8 - 20 + 5 \\
|
||||
&=& 32 -15 \\
|
||||
&=& 17 \\
|
||||
\intertext{Exemplarische Berechnung von $f'(x)$ für $x=2$:}
|
||||
f'(2) &=& 12 \cdot 2^{2} -10 \\
|
||||
&=& 48 - 10 \\
|
||||
&=& 38 \\
|
||||
\intertext{Berechnung der Näherungswerte:}
|
||||
x_{1} &=& 2 - \frac{17}{38} \\
|
||||
&=& \frac{76 -17}{38} \\
|
||||
&=& \frac{59}{38} \\
|
||||
x_{2} &=& \frac{59}{38} - \frac{f(\frac{59}{38})}{f'(\frac{59}{38})} \\
|
||||
&\approx & 1.317784436 \\
|
||||
x_{3} &\approx & 1.317784436 - \frac{f(1.317784436)}{f'(1.317784436)} \\
|
||||
&\approx & 1.227756731 \\
|
||||
x_{4} &\approx & 1.227756731 - \frac{f(1.227756731)}{f'(1.227756731)} \\
|
||||
&\approx & 1.212272195 \\
|
||||
x_{5} &\approx & 1.212272195 - \frac{f(1.212272195)}{f'(1.212272195)} \\
|
||||
&\approx & 1.211811475 \\
|
||||
x_{6} &\approx & 1.211811475 - \frac{f(1.211811475)}{f'(1.211811475)} \\
|
||||
&\approx & 1.21181107 \\
|
||||
x_{7} &=& 1.21181107 - \frac{f(1.21181107)}{f'(1.21181107)} \\
|
||||
&=& 1.21181107
|
||||
\end{alignat*}
|
||||
\section{} %4
|
||||
Die Länge der Schnur berechnet sie wie folgt:\\
|
||||
\begin{alignat*}{2}
|
||||
L &=& 2x + y \\
|
||||
\intertext{Nach y umgestellt ergibt sich}
|
||||
y &=& L - 2x \\
|
||||
\intertext{Eingesetzt in die Formel für den Flächeninhalt von Rechtecken ergibt sich:}
|
||||
A(x) &=& x \cdot (L-2x) \\
|
||||
&=& -2x^{2} + Lx \\
|
||||
A'(x) &=& -4x + L \\
|
||||
A''(x) &=& -4 \\
|
||||
\intertext{Gleichsetzen von $A'(x)$ mit Null}
|
||||
0 &=& -4x + L \\
|
||||
4x &=& L \\
|
||||
x &=& \frac{L}{4} \\
|
||||
\intertext{Einsetzen in zweite Ableitung}
|
||||
A''(\frac{L}{4}) &=& -4 \Rightarrow \text{$< 0$, daher Maximum}
|
||||
\end{alignat*}
|
||||
Daraus folgt, dass die beiden gleichlangen Seiten des Rechtecks, die von der Schnur begrenzt werden, jeweils ein Viertel der Schnur ausmachen. Die dritte von der Schnur begrenzte Seite ist folglich halb so lang wie die Schnur.
|
||||
\section{} %5
|
||||
\subsection{} %a
|
||||
Die Oberfläche ($A$) und das Volumen ($V$) der Dose sei durch folgende Formeln gegeben:\\
|
||||
\begin{alignat*}{2}
|
||||
A &=& 2\pi r^{2} + 2\pi rh \\
|
||||
V &=& 1000 cm^{3} = \pi r^{2}h \\
|
||||
\intertext{Umstellen von $V$ nach $h$}
|
||||
h &=& \frac{1000 cm^{3}}{\pi r^{2}} \\
|
||||
\intertext{Einsetzen in $A$}
|
||||
A(r) &=& 2 \pi r^{2} + 2 \pi r \cdot \frac{1000 cm^{3}}{\pi r^{2}} \\
|
||||
\intertext{Kürzen}
|
||||
&=& 2 \pi r^{2} + 2 \cdot \frac{1000 cm^{3}}{r} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& 2 \pi r^{2} + 2000 cm^{3} \cdot r^{-1} \\
|
||||
A'(r) &=& 4 \pi r - 2000 cm^{3} \cdot r^{-2} \\
|
||||
A''(r) &=& 4 \pi + 4000 cm^{3} \cdot r^{-3} \\
|
||||
\intertext{Erste Ableitung mit Null gleichsetzen}
|
||||
0 &=& 4 \pi r - 2000 cm^{3} \cdot r^{-2} \\
|
||||
2000 cm^{3} \cdot r^{-2} &=& 4 \pi r \\
|
||||
2000 cm^{3} &=& 4 \pi r^{3} \\
|
||||
\frac{2000 cm^{3}}{4 \pi} &=& r^{3} \\
|
||||
\frac{500 cm^{3}}{\pi} &=& r^{3} \\
|
||||
\intertext{Dritte Wurzel ziehen}
|
||||
\sqrt[3]{\frac{500 cm^{3}}{\pi}} &=& r \\
|
||||
\intertext{Einsetzen in zweite Ableitung}
|
||||
A''\left(\sqrt[3]{\frac{500 cm^{3}}{\pi}} \right) &=& 4 \pi + 4000 cm^{3} \cdot \left(\sqrt[3]{\frac{500 cm^{3}}{\pi}}\right)^{-3} \\
|
||||
\intertext{Umformen}
|
||||
&=& 4 \pi + 4000 cm^{3} \cdot \frac{1}{\left(\sqrt[3]{\frac{500 cm^{3}}{\pi}} \right)^{3}} \\
|
||||
\intertext{Auflösen der Wurzel}
|
||||
&=& 4 \pi + 4000 cm^{3} \cdot \frac{1}{\frac{500 cm^{3}}{\pi}} \\
|
||||
\intertext{Auflösen Doppelbruch}
|
||||
&=& 4 \pi + 4000 cm^{3} \cdot \frac{\pi}{500 cm^{3}} \\
|
||||
\intertext{Kürzen}
|
||||
&=& 4 \pi + 8 \cdot \pi \\
|
||||
&=& 12 \pi \Rightarrow \text{$>0$, daher Minimum} \\
|
||||
\intertext{Einsetzen des Radius in Formel für $h$}
|
||||
h &=& \frac{1000 cm^{3}}{\pi \cdot \left(\sqrt[3]{\frac{500 cm^{3}}{\pi}} \right)^{2}} \\
|
||||
\intertext{Umformen}
|
||||
&=& \frac{1000 cm^{3}}{\pi \cdot \left(\sqrt[3]{500 cm^{3}\pi^{-1}} \right)^{2}} \\
|
||||
\intertext{Umformen}
|
||||
&=& \frac{1000 cm^{3}}{\pi \cdot \left(\left(500 cm^{3}\pi^{-1}\right)^{\frac{1}{3}} \right)^{2}} \\
|
||||
\intertext{innere Klammer auflösen}
|
||||
&=& \frac{1000 cm^{3}}{\pi \cdot \left((500 cm^{3})^{\frac{1}{3}} \cdot \pi^{-\frac{1}{3}} \right)^{2}} \\
|
||||
\intertext{Klammer auflösen}
|
||||
&=& \frac{1000 cm^{3}}{\pi \cdot (500 cm^{3})^{\frac{2}{3}} \cdot \pi^{-\frac{2}{3}}} \\
|
||||
\intertext{Zusammenfassen und Bruch auseinanderziehen}
|
||||
&=& \frac{1000 cm^{3}}{(500 cm^{3})^{\frac{2}{3}} } \cdot \frac{1}{\pi^{\frac{1}{3}}}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
Die Oberfläche $A$ wird in diesem Fall ähnlich berechnet. Allerdings muss die Grundfläche nur einmal eingerechnet werden:\\
|
||||
\begin{alignat*}{2}
|
||||
A &=& \pi r^{2} + 2 \pi r h \\
|
||||
\intertext{Die Berechnung von r ist analog. Lediglich der Faktor vor dem $\pi r^{2}$ ist anders. Daraus folgt für $A(r)$:}
|
||||
A(r) &=& \pi r^{2} + 2000 cm^{3} \cdot r^{-1} \\
|
||||
A'(r) &=& 2 \pi r - 2000 cm^{3} \cdot r^{-2} \\
|
||||
\intertext{Aufgrund der Änderung ergibt sich nach umstellen nach r folgendes:}
|
||||
\frac{2000 cm^{3}}{2 \pi} &=& r^{3} \\
|
||||
\intertext{Durch Kürzen ergibt sich:}
|
||||
\frac{1000 cm^{3}}{\pi} &=& r^{3} \\
|
||||
\intertext{Ziehen der dritten Wurzel:}
|
||||
\sqrt[3]{\frac{1000 cm^{3}}{\pi}} &=& r \\
|
||||
\intertext{Beim Einsetzen in die zweite Ableitung verändert sich nicht viel:}
|
||||
A''\left(\sqrt[3]{\frac{1000 cm^{3}}{\pi}} \right) = 2 \pi + 4000 cm^{3} \cdot \left(\sqrt[3]{\frac{1000 cm^{3}}{\pi}} \right)^{-3} \\
|
||||
&=& 2 \pi + 4000 cm^{3} \cdot \frac{1}{ \left(\sqrt[3]{\frac{1000 cm^{3}}{\pi}} \right)^{3}} \\
|
||||
&=& 2 \pi + 4000 cm^{3} \cdot \frac{1}{ \frac{1000 cm^{3}}{\pi}} \\
|
||||
&=& 2 \pi + 4000 cm^{3} \cdot \frac{\pi}{ 1000 cm^{3}} \\
|
||||
&=& 2 \pi + 4 \pi \\
|
||||
&=& 6 \pi \Rightarrow \text{$>0$, daher Minimum}\\
|
||||
\intertext{Durch Einsetzen in Gleichung von $h$ ergibt sich nun:}
|
||||
h &=& \frac{1000 cm^{3}}{\pi \cdot \left(\sqrt[3]{\frac{1000 cm^{3}}{\pi}} \right)^{2}} \\
|
||||
&=& \frac{1000 cm^{3}}{\pi \cdot \left(\left(\frac{1000 cm^{3}}{\pi}\right)^{\frac{1}{3}} \right)^{2}} \\
|
||||
&=& \frac{1000 cm^{3}}{\pi \cdot \left(\left(1000 cm^{3} \cdot \pi^{-1} \right)^{\frac{1}{3}} \right)^{2}} \\
|
||||
&=& \frac{1000 cm^{3}}{\pi \cdot \left((1000 cm^{3})^{\frac{1}{3}} \cdot \pi^{-\frac{1}{3}}\right)^{2}} \\
|
||||
&=& \frac{1000 cm^{3}}{\pi \cdot (1000 cm^{3})^{\frac{2}{3}} \cdot \pi^{-\frac{2}{3}}} \\
|
||||
&=& \frac{1000 cm^{3}}{(1000 cm^{3})^{\frac{2}{3}} \cdot \pi^{\frac{1}{3}}} \\
|
||||
&=& \frac{1000 cm^{3}}{(1000 cm^{3})^{\frac{2}{3}}} \cdot \frac{1}{\pi^{\frac{1}{3}}} \\
|
||||
\intertext{Kürzen}
|
||||
&=& (1000 cm^{3})^{-\frac{1}{3}} \cdot \frac{1}{\pi^{\frac{1}{3}}}
|
||||
\end{alignat*}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,392 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 30. Mai}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsection{} %a
|
||||
Berechnen von $\lim\limits_{n \rightarrow \infty} O_{n}$. Es gilt:\\
|
||||
\begin{alignat*}{2}
|
||||
O_{n} &=&& \frac{b-a}{n} \sum\limits_{i=1}^{n} f(x_{i}) = \frac{3}{n} \sum\limits_{i=1}^{n} f\left( \frac{3i}{n}\right) = \frac{3}{n} \sum\limits_{i=1}^{n} \left( \frac{3i}{n}\right)^{3} = \frac{3}{n^{4}} \sum\limits_{i=1}^{n} 27 \cdot i^{3}\\
|
||||
&=&& \frac{81}{n^{4}} \sum\limits_{i=1}^{n} i^{3} = \frac{81}{n^{4}} \cdot \frac{n^{2}(n+1)^{2}}{4} = \frac{81 \cdot n^{2}(n+1)^{2}}{4 \cdot n^{4}} = \frac{81 \cdot (n+1)^{2}}{4 \cdot n^{2}} \\
|
||||
&=&& \frac{81 \cdot (n^{2} + 2n + 1)}{4 \cdot n^{2}} = \frac{81n^{2} + 162n + 81}{4n^{2}}
|
||||
\end{alignat*}\\
|
||||
Also gilt $\lim\limits_{n \rightarrow \infty} O_{n} = \lim\limits_{n \rightarrow \infty} \left(\frac{81n^{2} + 162n + 81}{4n^{2}} \right) = \frac{81}{4}$. Der gesuchte Flächeninhalt hat den Wert $\frac{81}{4}$.
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
\int\limits_{a}^{b} f(x) \,dx = \int\limits_{0}^{3} x^{3} \,dx &=& \left[\frac{1}{4}x^{4}\right]_{0}^{3} \\
|
||||
&=& \frac{1}{4} \cdot 3^{4} - \frac{1}{4} \cdot 0^{4} = \frac{1}{4} \cdot 81 - 0 = \frac{81}{4}
|
||||
\end{alignat*}
|
||||
\section{} %2
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
\int\limits_{1}^{3} (x^{2} -x-6) \,dx &=& \left[\frac{1}{3}x^{3} - \frac{1}{2}x^{2} - 6x \right]_{1}^{3} \\
|
||||
&=& \left(\frac{1}{3} \cdot 3^{3} - \frac{1}{2} \cdot 3^{2} - 6 \cdot 3\right) - \left(\frac{1}{3} \cdot 1^{3} - \frac{1}{2} \cdot 1^{2} - 6 \cdot 1\right) \\
|
||||
&=& \left(9 - \frac{9}{2} - 18\right) - \left(\frac{1}{3} - \frac{1}{2} - 6\right) = \left(\frac{18 - 9 - 36}{2}\right) - \left(\frac{2 - 3 - 36}{6}\right) \\
|
||||
&=& \left(\frac{-27}{2}\right) - \left(\frac{-37}{6}\right) = \frac{-27}{2} + \frac{37}{6} = \frac{-81 + 37}{6} \\
|
||||
&=& -\frac{44}{6} = -\frac{22}{3}
|
||||
\end{alignat*}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=-7,ymax=1,
|
||||
x=1em,
|
||||
y=1em,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=0,xmax=4
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=1:3,samples=100]{x*x -x -6} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 2.75,-3) {f};
|
||||
\node at (axis cs: 1.75,-2.5) {A};
|
||||
\draw[>=stealth] (axis cs:1,0) -- (axis cs:1,-6) node [pos=0.65,anchor=north]{};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
\int\limits_{1}^{3} x^{\frac{1}{3}} \,dx &=& \left[\frac{3}{4}x^{\frac{4}{3}} \right]_{1}^{3} \\
|
||||
&=& \left(\frac{3}{4} \cdot 3^{\frac{4}{3}}\right) - \left(\frac{3}{4} \cdot 1^{\frac{4}{3}}\right) = 3^{\frac{4}{3}} \cdot \frac{3}{4} - \frac{3}{4} \\
|
||||
&=& \left(3^{\frac{4}{3}} - 1\right) \cdot \frac{3}{4}
|
||||
\end{alignat*}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=3,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=0,xmax=4
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=1:3,samples=100]{x^(1/3)} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 2,1.5) {f};
|
||||
\node at (axis cs: 2,0.5) {A};
|
||||
\draw[>=stealth] (axis cs:1,0) -- (axis cs:1,1) node [pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:3,0) -- (axis cs:3,1.44224957) node [pos=0.65,anchor=north]{};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
\int\limits_{1}^{3} \frac{1}{1+x^{2}} \,dx &=& \left[\arctan x \right]_{1}^{3} \\
|
||||
&=& \arctan 3 - \arctan 1
|
||||
\end{alignat*}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=2,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=0,xmax=4
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=1:3,samples=100]{1/(1+x*x)} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 2,0.5) {f};
|
||||
\node at (axis cs: 1.2,0.2) {A};
|
||||
\draw[>=stealth] (axis cs:1,0) -- (axis cs:1,0.5) node [pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:3,0) -- (axis cs:3,0.1) node [pos=0.65,anchor=north]{};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\subsubsection{} %iv
|
||||
\begin{alignat*}{2}
|
||||
\int\limits_{1}^{3} \ln x \,dx &=& \left[x \cdot \ln x - x \right]_{1}^{3} \\
|
||||
&=& \left(3 \cdot \ln 3 - 3 \right) - \left(1 \cdot \ln 1 - 1 \right) = 3 \cdot \ln 3 - 3 + 1 \\
|
||||
&=& 3 \cdot \ln 3 - 2
|
||||
\end{alignat*}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=2,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=0,xmax=4
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=1:3,samples=100]{ln x} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 2,1) {f};
|
||||
\node at (axis cs: 2.2,0.4) {A};
|
||||
%\draw[>=stealth] (axis cs:1,0) -- (axis cs:1,0.5) node [pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:3,0) -- (axis cs:3,1.098612289) node [pos=0.65,anchor=north]{};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\subsubsection{} %v
|
||||
\begin{alignat*}{2}
|
||||
\int\limits_{1}^{3} e^{-x} \,dx &=& \left[-e^{-x} \right]_{1}^{3} \\
|
||||
&=& \left(-e^{-3} \right) - \left( -e^{-1}\right) = -e^{-3} + e^{-1}
|
||||
\end{alignat*}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=2,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=0,xmax=4
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=1:3,samples=100]{e^(-x)} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 2,0.5) {f};
|
||||
\node at (axis cs: 1.15,0.12) {A};
|
||||
\draw[>=stealth] (axis cs:1,0) -- (axis cs:1,0.367879441) node [pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:3,0) -- (axis cs:3,0.049787068) node [pos=0.65,anchor=north]{};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\section{} %3
|
||||
\setcounter{subsubsection}{0}
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
\int (x^{4} + 2x^{3} -x +5) \,dx &=& \frac{1}{5}x^{5} + \frac{1}{4}x^{4} - \frac{1}{2}x^{2} + 5x
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
\int \frac{1}{\sqrt{x^{3}}} \,dx &=& \int \frac{1}{x^{\frac{3}{2}}} \,dx = \int x^{\frac{2}{3}} \,dx \\
|
||||
&=& \frac{3}{5}x^{\frac{5}{3}}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
\int x \cdot \sin(3x) \,dx &=& \frac{-\cos(3x)}{3} \cdot x + \frac{\sin(3x)}{9} \\
|
||||
\intertext{Probe:}
|
||||
\left(\frac{-\cos(3x)}{3} \cdot x + \frac{\sin(3x)}{9}\right)' &=& \left(\frac{-\cos(3x)}{3}\right)' \cdot x + \frac{-\cos(3x)}{3} \cdot 1 + \left(\frac{\sin(3x)}{9}\right)' \\
|
||||
&=& \frac{\sin(3x) \cdot 3}{3} \cdot x + \frac{-\cos(3x)}{3} + \frac{\cos(3x) \cdot 3}{9} \\
|
||||
&=& \sin(3x) \cdot x + \frac{-\cos(3x)}{3} + \frac{\cos(3x)}{3} \\
|
||||
&=& \sin(3x) \cdot x
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iv
|
||||
\begin{alignat*}{2}
|
||||
\int x^{3} \cdot \ln x \,dx &=& \frac{1}{4}x^{4} \cdot \ln x - \int \frac{1}{4}x^{4} \cdot \frac{1}{x}\,dx \\
|
||||
&=& \frac{1}{4}x^{4} \cdot \ln x - \int \frac{1}{4}x^{3}\,dx \\
|
||||
&=& \frac{1}{4}x^{4} \cdot \ln x - \frac{1}{16}x^{4} \\
|
||||
\intertext{Probe:}
|
||||
\left(\frac{1}{4}x^{4} \cdot \ln x - \frac{1}{4}x^{4}\right)' &=& (\frac{1}{4}x^{4} \cdot \ln x)' - \left(\frac{1}{16}x^{4}\right)' \\
|
||||
&=& x^{3} \cdot \ln x + \frac{1}{4}x^{4} \cdot \frac{1}{x} - \frac{1}{4}x^{3} \\
|
||||
&=& x^{3} \cdot \ln x + \frac{1}{4}x^{3} - \frac{1}{4}x^{3}\\
|
||||
&=& x^{3} \cdot \ln x
|
||||
\end{alignat*}
|
||||
\subsubsection{} %v
|
||||
\begin{alignat*}{2}
|
||||
\int x^{2}e^{x} \,dx &=& x^{2}e^{x} - \int 2x \cdot e^{x}\,dx \\
|
||||
&=& x^{2}e^{x} - 2 \cdot \int x \cdot e^{x}\,dx \\
|
||||
&=& x^{2}e^{x} - 2 \cdot \left(x \cdot e^{x} - \int 1 \cdot e^{x}\,dx \right) \\
|
||||
&=& x^{2}e^{x} - 2 \cdot (x \cdot e^{x} - e^{x}) \\
|
||||
&=& x^{2}e^{x} - 2x \cdot e^{x} +2e^{x} \\
|
||||
\intertext{Probe:}
|
||||
(x^{2}e^{x} - 2x \cdot e^{x} +2e^{x})' &=& (x^{2}e^{x})' - (2x \cdot e^{x})' + 2e^{x} \\
|
||||
&=& (2x \cdot e^{x} + x^{2}e^{x}) - (2e^{x} + 2x \cdot e^{x}) + 2e^{x} \\
|
||||
&=& 2x \cdot e^{x} + x^{2}e^{x} - 2e^{x} - 2x \cdot e^{x} + 2e^{x} \\
|
||||
&=& x^{2}e^{x}
|
||||
\end{alignat*}
|
||||
\section{} %4
|
||||
\setcounter{subsubsection}{0}
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
t &=& \sqrt{2x+5} \\
|
||||
\frac{dt}{dx} &=& \frac{2}{2 \cdot \sqrt{2x+5}} \\
|
||||
\sqrt{2x+5}\,dt &=& dx \\
|
||||
dx &=& t\,dt\\
|
||||
\int \cos(\sqrt{2x+5}) \,dx &=& \int \cos(t) \cdot t\,dt \\
|
||||
&=& \int \cos(t) \cdot t\,dt \\
|
||||
&=& \sin(t) \cdot t - \int \sin(t) \cdot 1 \,dt \\
|
||||
&=& \sin(t) \cdot t + \cos(t) \\
|
||||
\int \cos(\sqrt{2x+5}) \,dx &=& \sqrt{2x+5} \cdot \sin(\sqrt{2x+5}) + \cos(\sqrt{2x+5})
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{3}
|
||||
\intertext{Probe:}
|
||||
\left(\sqrt{2x+5} \cdot \sin(\sqrt{2x+5}) + \cos(\sqrt{2x+5})\right)' &=& (\sqrt{2x+5} \cdot \sin(\sqrt{2x+5}))' + (\cos(\sqrt{2x+5}))' \\
|
||||
\intertext{Ableiten}
|
||||
&=\begin{split}\frac{2 \cdot \sin(\sqrt{2x+5})}{2 \cdot \sqrt{2x+5}}\\ + \sqrt{2x+5} \cdot \frac{\cos(\sqrt{2x+5}) \cdot 2}{2 \cdot \sqrt{2x+5}} \\ + (-\sin(\sqrt{2x+5}) \cdot (\sqrt{2x+5})')\end{split}\\
|
||||
\intertext{Zusammenfassen und Ableiten}
|
||||
&= \begin{split}\frac{\sin(\sqrt{2x+5})}{\sqrt{2x+5}} + \cos(\sqrt{2x+5})\\ - \frac{\sin(\sqrt{2x+5}) \cdot 2}{2 \cdot \sqrt{2x+5}}\end{split} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&= \begin{split}\frac{\sin(\sqrt{2x+5})}{\sqrt{2x+5}} + \cos(\sqrt{2x+5})\\ - \frac{\sin(\sqrt{2x+5})}{\sqrt{2x+5}}\end{split} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \cos(\sqrt{2x+5})
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
t &=& \sqrt[3]{x} \\
|
||||
\frac{dt}{dx} &=& \frac{1}{3 \cdot \left(\sqrt[3]{x}\right)^{2}} \\
|
||||
dx &=& 3 \cdot \left(\sqrt[3]{x}\right)^{2}\,dt \\
|
||||
dx &=& 3 \cdot t^{2}\,dt\\
|
||||
\int \sin(\sqrt[3]{x})\,dx &=& \int \sin(t) \cdot 3t^{2}\,dt\\
|
||||
&=& 3 \int \sin(t) \cdot t^{2}\,dt \\
|
||||
&=& 3\left(-\cos(t) \cdot t^{2} - \int -\cos(t) \cdot 2t \right) \\
|
||||
3\left(-\cos(t) \cdot t^{2} + 2\int \cos(t) \cdot t \right) \\
|
||||
&=& 3\left(-\cos(t) \cdot t^{2} + 2\left( \sin(t) \cdot t - \int \sin(t) \cdot 1 \right)\right) \\
|
||||
&=& 3\left(-\cos(t) \cdot t^{2} + 2\left( \sin(t) \cdot t + \cos(t) \right)\right) \\
|
||||
&=& 3\left(-\cos(t) \cdot t^{2} + \sin(t) \cdot 2t + 2 \cdot \cos(t)\right) \\
|
||||
&=& -3t^{2} \cdot \cos(t) + 6t \cdot \sin(t) + 6 \cdot \cos(t) \\
|
||||
\int \sin(\sqrt[3]{x})\,dx &=& -3 \cdot \sqrt[3]{x}^{2} \cdot \cos(\sqrt[3]{x}) + 6 \cdot \sqrt[3]{x} \cdot \sin(\sqrt[3]{x}) + 6 \cdot \cos(\sqrt[3]{x}) \\
|
||||
&=& -3 \cdot x^{\frac{2}{3}} \cdot \cos(x^{\frac{1}{3}}) + 6 \cdot x^{\frac{1}{3}} \cdot \sin(x^{\frac{1}{3}}) + 6 \cdot \cos(x^{\frac{1}{3}})
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Probe:}
|
||||
&& \left(-3 \cdot x^{\frac{2}{3}} \cdot \cos(x^{\frac{1}{3}}) + 6 \cdot x^{\frac{1}{3}} \cdot \sin(x^{\frac{1}{3}}) + 6 \cdot \cos(x^{\frac{1}{3}})\right)' \\
|
||||
&=& \left(-3 \cdot x^{\frac{2}{3}} \cdot \cos(x^{\frac{1}{3}})\right)' + 6\left(\left(x^{\frac{1}{3}} \cdot \sin(x^{\frac{1}{3}})\right)' + \left(\cos(x^{\frac{1}{3}})\right)'\right) \\
|
||||
&=& -3\left(\frac{2}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) + x^{\frac{2}{3}} \cdot (-\sin(x^{\frac{1}{3}})) \cdot \frac{1}{3}x^{-\frac{2}{3}}\right)+ 6\left(\left(x^{\frac{1}{3}} \cdot \sin(x^{\frac{1}{3}})\right)' + \left(\cos(x^{\frac{1}{3}})\right)'\right) \\
|
||||
&=& -3\left(\frac{2}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) - \frac{1}{3} \cdot \sin(x^{\frac{1}{3}})\right)+ 6\left(\left(x^{\frac{1}{3}} \cdot \sin(x^{\frac{1}{3}})\right)' + \left(\cos(x^{\frac{1}{3}})\right)'\right) \\
|
||||
\intertext{Ableiten}
|
||||
&= \begin{split}-3\left(\frac{2}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) - \frac{1}{3} \cdot \sin(x^{\frac{1}{3}})\right)\\ + 6\left(\left(\frac{1}{3}x^{-\frac{2}{3}} \cdot \sin(x^{\frac{1}{3}}) + x^{\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) \cdot \frac{1}{3}x^{-\frac{2}{3}} \right) + \left(\cos(x^{\frac{1}{3}})\right)'\right) \end{split} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&= \begin{split}-3\left(\frac{2}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) - \frac{1}{3} \cdot \sin(x^{\frac{1}{3}})\right)\\ + 6\left(\frac{1}{3}x^{-\frac{2}{3}} \cdot \sin(x^{\frac{1}{3}}) + \frac{1}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) + \left(\cos(x^{\frac{1}{3}})\right)'\right) \end{split} \\
|
||||
\intertext{Ableiten}
|
||||
&= \begin{split}-3\left(\frac{2}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) - \frac{1}{3} \cdot \sin(x^{\frac{1}{3}})\right)\\ + 6\left(\frac{1}{3}x^{-\frac{2}{3}} \cdot \sin(x^{\frac{1}{3}}) + \frac{1}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) - \sin(x^{\frac{1}{3}}) \cdot \frac{1}{3}x^{-\frac{2}{3}} \right) \end{split} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&= \begin{split}-3\left(\frac{2}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) - \frac{1}{3} \cdot \sin(x^{\frac{1}{3}})\right)\\ + 6\left(\frac{1}{3}x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}})\right) \end{split} \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& -2x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) + \sin(x^{\frac{1}{3}}) + 2x^{-\frac{1}{3}} \cdot \cos(x^{\frac{1}{3}}) \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \sin(x^{\frac{1}{3}})
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
t &=& \sqrt{\frac{2}{7}x+3} \\
|
||||
\frac{dt}{dx} &=& \frac{2}{14 \cdot \sqrt{\frac{2}{7}x+3}} \\
|
||||
\frac{dt}{dx} &=& \frac{1}{7 \cdot \sqrt{\frac{2}{7}x+3}} \\
|
||||
dx &=& 7 \cdot \sqrt{\frac{2}{7}x+3}\, dt \\
|
||||
&=& 7t\\
|
||||
\int e^{\sqrt{\frac{2}{7}x+3}}\,dx &=& \int e^{t} \cdot 7t\,dt \\
|
||||
&=& e^{t} \cdot 7t - \int e^{t} \cdot 7\,dt \\
|
||||
&=& e^{t} \cdot 7t - 7\int e^{t}\,dt \\
|
||||
&=& e^{t} \cdot 7t - 7 \cdot e^{t} \\
|
||||
&=& e^{t}(7t - 7) \\
|
||||
&=& e^{\sqrt{\frac{2}{7}x+3}}\left(7 \cdot \sqrt{\frac{2}{7}x+3} - 7\right)
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Probe:}
|
||||
&& \left(e^{\sqrt{\frac{2}{7}x+3}}\left(7 \cdot \sqrt{\frac{2}{7}x+3} - 7\right) \right)' \\
|
||||
&=& \left(e^{\sqrt{\frac{2}{7}x+3}}\right)' \cdot \left(7 \cdot \sqrt{\frac{2}{7}x+3} - 7\right) + \left(e^{\sqrt{\frac{2}{7}x+3}}\right) \cdot \left(7 \cdot \sqrt{\frac{2}{7}x+3} - 7\right)' \\
|
||||
&=& e^{\sqrt{\frac{2}{7}x+3}} \cdot \frac{2}{14 \cdot \sqrt{\frac{2}{7}x+3}} \cdot \left(7 \cdot \sqrt{\frac{2}{7}x+3} - 7\right) + \left(e^{\sqrt{\frac{2}{7}x+3}}\right) \cdot \left(7 \cdot \left(\frac{2}{7}x+3\right)^{\frac{1}{2}} - 7\right)' \\
|
||||
&=& e^{\sqrt{\frac{2}{7}x+3}} \cdot \frac{1}{7 \cdot \sqrt{\frac{2}{7}x+3}} \cdot \left(7 \cdot \sqrt{\frac{2}{7}x+3} - 7\right) + \left(e^{\sqrt{\frac{2}{7}x+3}}\right) \cdot \left(7 \cdot \left(\frac{2}{7}x+3\right)^{\frac{1}{2}}\right)'\\
|
||||
&=& e^{\sqrt{\frac{2}{7}x+3}} \cdot \frac{1}{7 \cdot \sqrt{\frac{2}{7}x+3}} \cdot \left(7 \cdot \sqrt{\frac{2}{7}x+3} - 7\right) + \left(e^{\sqrt{\frac{2}{7}x+3}}\right) \cdot \left(\frac{7}{2} \cdot \left(\frac{2}{7}x+3\right)^{-\frac{1}{2}} \cdot \frac{2}{7}\right) \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& e^{\sqrt{\frac{2}{7}x+3}} \cdot \left(1 - \frac{1}{\sqrt{\frac{2}{7}x+3}}\right) + \left(e^{\sqrt{\frac{2}{7}x+3}}\right) \cdot \left(\frac{2}{7}x+3\right)^{-\frac{1}{2}} \\
|
||||
\intertext{Ausklammern}
|
||||
&=& e^{\sqrt{\frac{2}{7}x+3}} \left(\left(1 - \frac{1}{\sqrt{\frac{2}{7}x+3}}\right) + \frac{1}{\sqrt{\frac{2}{7}x+3}} \right) \\
|
||||
\intertext{Klammern auflösen}
|
||||
&=& e^{\sqrt{\frac{2}{7}x+3}} \left(1 - \frac{1}{\sqrt{\frac{2}{7}x+3}} + \frac{1}{\sqrt{\frac{2}{7}x+3}}\right) \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& e^{\sqrt{\frac{2}{7}x+3}}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iv
|
||||
\begin{alignat*}{2}
|
||||
t &=& \ln(2x+1) \\
|
||||
e^{t} &=& 2x+1 \\
|
||||
x &=& \frac{e^{t}-1}{2} \\
|
||||
\frac{dx}{dt} &=& \frac{1}{2}\left(e^{t}\right) \\
|
||||
dx &=& \frac{1}{2} \cdot e^{t}\,dt \\
|
||||
\int \left(\ln (2x+1)\right)^{2}\,dx &=& \int t^{2} \cdot \frac{1}{2} \cdot e^{t}\,dt \\
|
||||
&=& \frac{1}{2} \int t^{2} \cdot e^{t} \\
|
||||
&=& \frac{1}{2}\left(t^{2} \cdot e^{t} - \int 2t \cdot e^{t} \right) \\
|
||||
&=& \frac{1}{2}\left(t^{2} \cdot e^{t} - 2\int t \cdot e^{t} \right) \\
|
||||
&=& \frac{1}{2}\left(t^{2} \cdot e^{t} - 2\left(t \cdot e^{t} - \int 1 \cdot e^{t} \right)\right) \\
|
||||
&=& \frac{1}{2}\left(t^{2} \cdot e^{t} - 2t \cdot e^{t} + 2 \cdot e^{t} \right) \\
|
||||
&=& \frac{1}{2}t^{2} \cdot e^{t} - t \cdot e^{t} + e^{t} \\
|
||||
&=& \frac{1}{2} \left(\ln(2x+1)\right)^{2} \cdot e^{\ln(2x+1)} - \ln(2x+1) \cdot e^{\ln(2x+1)} + e^{\ln(2x+1)} \\
|
||||
&=& \left(\frac{1}{2} \left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right)e^{\ln(2x+1)} \\
|
||||
&=& \left(\frac{1}{2} \left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right) \cdot (2x+1)
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Probe:}
|
||||
&& \left(\left(\frac{1}{2} \left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right)\cdot (2x+1)\right)' \\
|
||||
&=& \left(\frac{1}{2} \left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right)' \cdot (2x+1) + \left(\frac{1}{2}\left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right) \cdot (2x+1)' \\
|
||||
&=& \left(\frac{2 \cdot \ln(2x+1)}{2x+1} - \frac{2}{2x+1}\right) \cdot (2x+1) + \left(\frac{1}{2}\left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right) \cdot 2 \\
|
||||
&=& \left(\frac{2 \cdot \ln(2x+1) - 2}{2x+1}\right) \cdot (2x+1) + \left(\frac{1}{2}\left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right) \cdot 2 \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \frac{2 \cdot \ln(2x+1) \cdot (2x+1) - 2 \cdot (2x+1)}{2x+1} + \left(\frac{1}{2}\left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right) \cdot 2 \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& 2 \cdot \ln(2x+1) - 2 + \left(\frac{1}{2}\left(\ln(2x+1)\right)^{2} - \ln(2x+1) + 1\right) \cdot 2 \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& 2 \cdot \ln(2x+1) - 2 + \left(\ln(2x+1)\right)^{2} - 2 \cdot \ln(2x+1) + 2 \\
|
||||
\intertext{Zusammenfassen}
|
||||
&=& \left(\ln(2x+1)\right)^{2}
|
||||
\end{alignat*}
|
||||
\section{} %5
|
||||
\setcounter{subsubsection}{0}
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& x^{3} - 12x^{2} + 36x + 1\\
|
||||
f'(x) &=& 3x^{2} -24x + 36 \\
|
||||
f''(x) &=& 6x - 24 \\
|
||||
f'''(x) &=& 6 \\
|
||||
\intertext{Berechnung Nullstellen erste Ableitung}
|
||||
f'(x) &=& 0 \\
|
||||
0 &=& 3x^{2} - 24x + 36 \\
|
||||
0 &=& x^{2} - 8x + 12 \\
|
||||
\intertext{pq-Formel}
|
||||
x_{1,2} &=& 4 \pm \sqrt{4^{2} -12} \\
|
||||
&=& 4 \pm \sqrt{16 - 12} \\
|
||||
&=& 4 \pm \sqrt{4} \\
|
||||
x_{1} &=& 4 + 2 = 6 \\
|
||||
x_{2} &=& 4-2 = 2 \\
|
||||
\intertext{Einsetzen in zweite Ableitung}
|
||||
f''(2) &=& 6 \cdot 2 - 24 \\
|
||||
&=& -12 \Rightarrow \text{ Maximum} \\
|
||||
f''(6) &=& 6 \cdot 6 - 24 \\
|
||||
&=& 12 \Rightarrow \text{ Minimum} \\
|
||||
\intertext{Einsetzen in Funktion}
|
||||
f(2) &=& 2^{3} - 12 \cdot 2^{2} + 36 \cdot 2 + 1 \\
|
||||
&=& 8 - 48 + 72 + 1 \\
|
||||
&=& 33 \\
|
||||
f(6) &=& 6^{3} - 12 \cdot 6^{2} + 36 \cdot 6 + 1 \\
|
||||
&=& 216 - 12 \cdot 36 + 216 + 1 \\
|
||||
&=& 216 - 432 + 216 + 1 \\
|
||||
&=& 1 \\
|
||||
\intertext{Überprüfen der Intervallgrenzen - nur $f(0)$ noch nötig, da $f(6)$ bereits berechnet}
|
||||
f(0) &=& 0^{3} - 12 \cdot 0^{2} + 36 \cdot 0 + 1 \\
|
||||
&=& 1
|
||||
\end{alignat*}
|
||||
Die Tageshöchsttemperatur ist demnach $33\,^{\circ} \mathrm{C}$.
|
||||
\subsubsection{} %ii
|
||||
Wie in i) berechnet, ist die Tagestiefsttemperatur $1\,^{\circ} \mathrm{C}$.
|
||||
\subsubsection{} %iii
|
||||
Der Mittelwert wird klassisch so berechnet: Addieren aller Einzelwerte und Teilen der Summe durch deren Anzahl. In diesem Fall entspricht der Flächeninhalt zwischen dem Integral und der x-Achse der Summe aller einzelnen Temperaturen. Nun muss das Ergebnis noch mal $\frac{1}{6}$ genommen werden.\\
|
||||
\begin{alignat*}{2}
|
||||
\frac{1}{6} \cdot \int\limits_{0}^{6} (x^{3} - 12x^{2} + 36x +1) \,dx &=& \frac{1}{6} \cdot \left[\frac{1}{4}x^{4} - 4x^{3} + 18x^{2} + x \right]_{0}^{6} \\
|
||||
&=& \frac{1}{6} \cdot \left(\left(\frac{1}{4} \cdot 6^{4} - 4 \cdot 6^{3} + 18 \cdot 6^{2} + 6 \right) - \left(\frac{1}{4} \cdot 0^{4} - 4 \cdot 0^{3} + 18 \cdot 0^{2} + 0 \right)\right) \\
|
||||
&=& \frac{1}{6} \cdot \left(\frac{1}{4} \cdot 1296 - 4 \cdot 216 + 18 \cdot 36 + 6 \right)\\
|
||||
&=& \frac{1}{6} \cdot \left(324 - 864 + 648 + 6 \right) \\
|
||||
&=& \frac{1}{6} \cdot 114 = 19
|
||||
\end{alignat*}\\
|
||||
Die Durchschnittstemperatur des Tages ist demnach $19\,^{\circ} \mathrm{C}$.
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,525 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pgfplotsset{compat=1.8}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 6. Juni}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
\int \frac{x + 1}{x^{2} - x +6}\,dx &=& \int \frac{x}{x^{2} - x + 6}\,dx + \int \frac{1}{x^{2} -x + 6}\,dx \\
|
||||
&=& \frac{1}{2}\int \frac{2x - 1 + 1}{x^{2} - x + 6}\,dx + \int \frac{1}{x^{2} -x + 6}\,dx \\
|
||||
&=& \frac{1}{2}\int \frac{2x - 1}{x^{2} - x + 6}\,dx + \frac{1}{2}\int \frac{1}{x^{2} - x + 6}\,dx + \int \frac{1}{x^{2} -x + 6}\,dx \\
|
||||
&=& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2}\int \frac{1}{x^{2} - x + 6}\,dx + \int \frac{1}{x^{2} -x + 6}\,dx
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
x^{2} - x + 6 &=& x^{2} - x + \frac{1}{4} + 6 - \frac{1}{4} \\
|
||||
&=& \left( x - \frac{1}{2}\right)^{2} + 6 - \frac{1}{4}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Setze $c = 6 - \frac{1}{4}$. Substitution: $t = x - \frac{1}{2}$, $\frac{dt}{dx} = 1$ und $dx = dt$.}
|
||||
&& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2}\int \frac{1}{x^{2} - x + 6}\,dx + \int \frac{1}{x^{2} -x + 6}\,dx\\
|
||||
&=& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2}\int \frac{1}{\left( x - \frac{1}{2}\right)^{2} + c}\,dx + \int \frac{1}{\left( x - \frac{1}{2}\right)^{2} + c}\,dx \\
|
||||
&=& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2}\int \frac{1}{t^{2} + c}\,dt + \int \frac{1}{t^{2} + c}\,dt
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Bestimmung Nullstellen des Nennerpolynoms}
|
||||
0 &=& x^{2} - x + 6 \\
|
||||
x_{1,2} &=& \frac{1}{2} \pm \sqrt{\left(\frac{1}{2}\right)^{2} - 6} \\
|
||||
&=& \frac{1}{2} \pm \sqrt{- \frac{23}{4}}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Eine negative Wurzel kann nicht gelöst werden. Demnach gibt es keine Nullstellen. Setzen von $r = \sqrt{c}$. Dann ist $c = r^{2}$.}
|
||||
&& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2}\int \frac{1}{x^{2} - x + 6}\,dx + \int \frac{1}{x^{2} -x + 6}\,dx\\
|
||||
&=& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2}\int \frac{1}{t^{2} + r^{2}}\,dt + \int \frac{1}{t^{2} + r^{2}}\,dt \\
|
||||
&=& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2} \cdot \frac{1}{r} \cdot \arctan \left(\frac{t}{r} \right) + \frac{1}{r} \cdot \arctan \left(\frac{t}{r} \right) \\
|
||||
&=& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2} \cdot \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{t}{\sqrt{c}} \right) + \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{t}{\sqrt{c}} \right) \\
|
||||
\intertext{Resubsitution}
|
||||
&& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2}\int \frac{1}{x^{2} - x + 6}\,dx + \int \frac{1}{x^{2} -x + 6}\,dx \\
|
||||
&=& \frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2} \cdot \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right) + \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right)
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Probe:}
|
||||
&& \left(\frac{1}{2} \cdot \ln |x^{2} - x + 6| + \frac{1}{2} \cdot \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right) + \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right)\right)' \\
|
||||
&=& \left(\frac{1}{2} \cdot \ln |x^{2} - x + 6|\right)' + \left(\frac{1}{2} \cdot \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right)\right)' + \left(\frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right)\right)' \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x - 1}{x^{2} - x + 6} + \frac{1}{2} \cdot \frac{1}{\sqrt{c}} \cdot \frac{1}{\left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right)^{2} + 1} \cdot \left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right)' + \frac{1}{\sqrt{c}} \cdot \frac{1}{\left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right)^{2} + 1} \cdot \left(\frac{x - \frac{1}{2}}{\sqrt{c}} \right)' \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x - 1}{x^{2} - x + 6} + \frac{1}{2} \cdot \frac{1}{\sqrt{c}} \cdot \frac{1}{\sqrt{c}} \cdot \frac{1}{\frac{\left(x - \frac{1}{2}\right)^{2}}{c} + 1} \cdot \left(x - \frac{1}{2}\right)' + \frac{1}{\sqrt{c}} \cdot \frac{1}{\sqrt{c}} \cdot \frac{1}{\frac{\left(x - \frac{1}{2}\right)^{2}}{c} + 1} \cdot \left(x - \frac{1}{2}\right)' \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x - 1}{x^{2} - x + 6} + \frac{1}{2} \cdot \frac{1}{c} \cdot \frac{1}{\frac{\left(x - \frac{1}{2}\right)^{2}}{c} + 1} + \frac{1}{c} \cdot \frac{1}{\frac{\left(x - \frac{1}{2}\right)^{2}}{c} + 1} \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x - 1}{x^{2} - x + 6} + \frac{1}{2} \cdot \frac{1}{c\left(\frac{\left(x - \frac{1}{2}\right)^{2}}{c} + 1\right)} + \frac{1}{c\left(\frac{\left(x - \frac{1}{2}\right)^{2}}{c} + 1\right)} \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x - 1}{x^{2} - x + 6} + \frac{1}{2} \cdot \frac{1}{\left(x - \frac{1}{2}\right)^{2} + c} + \frac{1}{\left(x - \frac{1}{2}\right)^{2} + c} \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x - 1}{x^{2} - x + 6} + \frac{1}{2} \cdot \frac{1}{x^{2} - x + \frac{1}{4} + 6 - \frac{1}{4}} + \frac{1}{x^{2} - x + \frac{1}{4} + 6 - \frac{1}{4}} \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x - 1}{x^{2} - x + 6} + \frac{1}{2} \cdot \frac{1}{x^{2} - x + 6} + \frac{1}{x^{2} - x + 6} \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x - 1 + 1}{x^{2} - x + 6} + \frac{1}{x^{2} - x + 6} \\
|
||||
&=& \frac{1}{2} \cdot \frac{2x}{x^{2} - x + 6} + \frac{1}{x^{2} - x + 6} \\
|
||||
&=& \frac{x}{x^{2} - x + 6} + \frac{1}{x^{2} - x + 6} \\
|
||||
&=& \frac{x + 1}{x^{2} - x + 6}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
0 &=& x^{2} - 4x + 4 \\
|
||||
x_{1,2} &=& 2 \pm \sqrt{2^{2} - 4} \\
|
||||
&=& 2 \pm \sqrt{0} \\
|
||||
x_{1} = x_{2} &=& 2 \\
|
||||
x^{2} - 4x + 4 &=& (x-2)(x-2) = (x-2)^{2}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\frac{2x + 1}{x^{2} -4x +4} = \frac{2x + 1}{(x-2)^{2}} &=& \frac{A}{(x-2)^{2}} + \frac{B}{x-2} \\
|
||||
2x + 1 &=& \frac{A(x-2)^{2}}{(x-2)^{2}} + \frac{B(x-2)^{2}}{x-2} \\
|
||||
&=& A + B(x-2) \\
|
||||
&=& A + Bx -2B \\
|
||||
&=& Bx - 2B + A
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Vergleichen der Koeffizienten}
|
||||
B &=& 2 \\
|
||||
-2B + A &=& 1 \\
|
||||
-2 \cdot 2 + A &=& 1 \\
|
||||
-4 + A &=& 1 \\
|
||||
A &=& 5
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Partialbruchzerlegung:}
|
||||
\frac{2x+1}{x^{2}-4x+4} &=& \frac{5}{(x-2)^{2}} + \frac{2}{x-2}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Integrieren}
|
||||
&& \int \frac{2x+1}{x^{2}-4x+4}\,dx \\
|
||||
&=& \int \left(\frac{5}{(x-2)^{2}} + \frac{2}{x-2}\right)\,dx \\
|
||||
&=& \int \frac{5}{(x-2)^{2}}\,dx + \int \frac{2}{x-2}\,dx \\
|
||||
&=& 5 \int (x-2)^{-2}\,dx + 2 \int \frac{1}{x-2}\,dx \\
|
||||
&=& -5 (x-2)^{-1} + 2 \cdot \ln |x-2|
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Probe}
|
||||
&& \left(-5 (x-2)^{-1} + 2 \cdot \ln |x-2|\right)' \\
|
||||
&=& \left(-5 (x-2)^{-1}\right)' + \left(2 \cdot \ln |x-2|\right)' \\
|
||||
&=& 5(x-2)^{-2} + \frac{2}{x-2} \\
|
||||
&=& \frac{5}{(x-2)^{2}} + \frac{2}{x-2} \\
|
||||
&=& \frac{5 + 2(x-2)}{(x-2)^{2}} \\
|
||||
&=& \frac{5 + 2x -4)}{(x-2)^{2}} \\
|
||||
&=& \frac{2x + 1}{x^{2} -4x +4}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
\int \frac{4x + 1}{x^{2} + 4x +8}\,dx &=& \int \frac{4x}{x^{2} + 4x + 8}\,dx + \int \frac{1}{x^{2} + 4x + 8}\,dx \\
|
||||
&=& 2\int \frac{2x +4 - 4}{x^{2} + 4x + 8}\,dx + \int \frac{1}{x^{2} + 4x + 8}\,dx \\
|
||||
&=& 2\int \frac{2x + 4}{x^{2} + 4x + 8}\,dx - 8\int \frac{1}{x^{2} + 4x + 8}\,dx + \int \frac{1}{x^{2} + 4x + 8}\,dx \\
|
||||
&=& 2 \cdot \ln |x^{2} + 4x + 8| - 8\int \frac{1}{x^{2} + 4x + 8}\,dx + \int \frac{1}{x^{2} + 4x + 8}\,dx
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
x^{2} + 4x + 8 &=& x^{2} + 4x + \frac{16}{4} + 8 - \frac{16}{4} \\
|
||||
&=& (x +2)^{2} + 4
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Setze $c = 4$. Substitution: $t = x + 2$, $\frac{dt}{dx} = 1$ und $dx = dt$.}
|
||||
&& 2 \cdot \ln |x^{2} + 4x + 8| - 8\int \frac{1}{x^{2} + 4x + 8}\,dx + \int \frac{1}{x^{2} + 4x + 8}\,dx\\
|
||||
&=& 2 \cdot \ln |x^{2} + 4x + 8| -8\int \frac{1}{(x+2)^{2} + c}\,dx + \int \frac{1}{(x+2)^{2} + c}\,dx \\
|
||||
&=& 2 \cdot \ln |x^{2} + 4x + 8| -8\int \frac{1}{t^{2} + c}\,dt + \int \frac{1}{t^{2} + c}\,dt
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Bestimmung Nullstellen des Nennerpolynoms}
|
||||
0 &=& x^{2} + 4x + 8 \\
|
||||
x_{1,2} &=& -2 \pm \sqrt{(2)^{2} - 8} \\
|
||||
&=& -2 \pm \sqrt{-4}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Eine negative Wurzel kann nicht gelöst werden. Demnach gibt es keine Nullstellen. Setzen von $r = \sqrt{c}$. Dann ist $c = r^{2}$.}
|
||||
&& 2 \cdot \ln |x^{2} + 4x + 8| - 8\int \frac{1}{x^{2} + 4x + 8}\,dx + \int \frac{1}{x^{2} + 4x + 8}\,dx\\
|
||||
&=& 2 \cdot \ln |x^{2} + 4x + 8| -8\int \frac{1}{t^{2} + r^{2}}\,dt + \int \frac{1}{t^{2} + r^{2}}\,dt \\
|
||||
&=& 2 \cdot \ln |x^{2} + 4x + 8| -8 \cdot \frac{1}{r} \cdot \arctan \left(\frac{t}{r} \right) + \frac{1}{r} \cdot \arctan \left(\frac{t}{r} \right) \\
|
||||
&=& 2 \cdot \ln |x^{2} + 4x + 8| -8 \cdot \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{t}{\sqrt{c}} \right) + \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{t}{\sqrt{c}} \right) \\
|
||||
\intertext{Resubsitution}
|
||||
&& 2 \cdot \ln |x^{2} + 4x + 8| - 8\int \frac{1}{x^{2} + 4x + 8}\,dx + \int \frac{1}{x^{2} + 4x + 8}\,dx \\
|
||||
&=& 2 \cdot \ln |x^{2} + 4x + 8| -8 \cdot \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x+2}{\sqrt{c}} \right) + \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x+2}{\sqrt{c}} \right)
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Probe:}
|
||||
&& \left(2 \cdot \ln |x^{2} + 4x + 8| -8 \cdot \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x+2}{\sqrt{c}} \right) + \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x+2}{\sqrt{c}} \right)\right)' \\
|
||||
&=& \left(2 \cdot \ln |x^{2} + 4x + 8|\right)' + \left(-6 \cdot \frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x+2}{\sqrt{c}} \right)\right)' + \left(\frac{1}{\sqrt{c}} \cdot \arctan \left(\frac{x+2}{\sqrt{c}} \right)\right)' \\
|
||||
&=& 2 \cdot \frac{2x + 4}{x^{2} + 4x + 8} - 8 \cdot \frac{1}{\sqrt{c}} \cdot \frac{1}{\left(\frac{x + 2}{\sqrt{c}} \right)^{2} + 1} \cdot \left(\frac{x + 2}{\sqrt{c}} \right)' + \frac{1}{\sqrt{c}} \cdot \frac{1}{\left(\frac{x + 2}{\sqrt{c}} \right)^{2} + 1} \cdot \left(\frac{x + 2}{\sqrt{c}} \right)' \\
|
||||
&=& 2 \cdot \frac{2x + 4}{x^{2} + 4x + 8} - 8 \cdot \frac{1}{\sqrt{c}} \cdot \frac{1}{\sqrt{c}} \cdot \frac{1}{\frac{(x + 2)^{2}}{c} + 1} \cdot (x + 2)' + \frac{1}{\sqrt{c}} \cdot \frac{1}{\sqrt{c}} \cdot \frac{1}{\frac{(x + 2)^{2}}{c} + 1} \cdot (x + 2)' \\
|
||||
&=& 2 \cdot \frac{2x + 4}{x^{2} + 4x + 8} - 8 \cdot \frac{1}{c} \cdot \frac{1}{\frac{(x + 2)^{2}}{c} + 1} + \frac{1}{c} \cdot \frac{1}{\frac{(x + 2)^{2}}{c} + 1} \\
|
||||
&=& 2 \cdot \frac{2x + 4}{x^{2} + 4x + 8} - 8 \cdot \frac{1}{c\left(\frac{(x + 2)^{2}}{c} + 1\right)} + \frac{1}{c\left(\frac{(x + 2)^{2}}{c} + 1\right)} \\
|
||||
&=& 2 \cdot \frac{2x + 4}{x^{2} + 4x + 8} - 8 \cdot \frac{1}{(x + 2)^{2} + c} + \frac{1}{(x + 2)^{2} + c} \\
|
||||
&=& 2 \cdot \frac{2x + 4}{x^{2} + 4x + 8} - 8 \cdot \frac{1}{x^{2} + 4x + 4 + 4} + \frac{1}{x^{2} + 4x + 4 + 4} \\
|
||||
&=& 2 \cdot \frac{2x + 4}{x^{2} + 4x + 8} - 2\cdot \frac{4}{x^{2} + 4x + 8} + \frac{1}{x^{2} + 4x + 8} \\
|
||||
&=& 2 \cdot \frac{2x + 4 - 4}{x^{2} + 4x + 8} + \frac{1}{x^{2} + 4x + 8} \\
|
||||
&=& 2 \cdot \frac{2x}{x^{2} + 4x + 8} + \frac{1}{x^{2} + 4x + 8} \\
|
||||
&=& \frac{4x + 1}{x^{2} + 4x + 8}
|
||||
\end{alignat*}
|
||||
\section{} %2
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& e^{-x} \\
|
||||
f'(x) &=& -e^{-x} \\
|
||||
f''(x) &=& e^{-x} \\
|
||||
f'''(x) &=& -e^{-x} \\
|
||||
\intertext{Wendepunkte bestimmen}
|
||||
f''(x) &=& 0 \\
|
||||
e^{-x} &=& 0 \\
|
||||
-x &=& \ln 0 \Rightarrow \text{$\ln 0$ ist nicht definiert, daher kann es keine Wendepunkte für $e^{-x}$ geben.}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
g(x) &=& \frac{1}{1+x} \\
|
||||
g'(x) &=& -(x+1)^{-2} \\
|
||||
g''(x) &=& 2(x+1)^{-3} \\
|
||||
g'''(x) &=& -6(x+1)^{-4} \\
|
||||
\intertext{Wendepunkte bestimmen}
|
||||
g''(x) &=& 0 \\
|
||||
2(x+1)^{-3} &=& 0 \\
|
||||
\frac{2}{(x+1)^{3}} &=& 0 \Rightarrow \text{Die Funktion wird niemals $0$. Daher kann es keine Wendepunkte geben.}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
h(x) &=& \frac{1}{1+x^{2}} \\
|
||||
h'(x) &=& -(1+x^{2})^{-2} \cdot 2x \\
|
||||
h''(x) &=& 2(1+x^{2})^{-3} \cdot 2x \cdot 2x -(1+x^{2})^{-2} \cdot 2 \\
|
||||
&=& 2(1+x^{2})^{-3} \cdot 4x^{2} - 2(1+x^{2})^{-2} \\
|
||||
h'''(x) &=& -6(1+x^{2})^{-4} \cdot 2x \cdot 4x^{2} + 2(1+x^{2})^{-3} \cdot 8x + 4(1+x^{2})^{-3} \cdot 2x \\
|
||||
&=& -6(1+x^{2})^{-4} \cdot 8x^{3} + 2(1+x^{2})^{-3} \cdot 8x + 4(1+x^{2})^{-3} \cdot 2x \\
|
||||
\intertext{Wendepunkte bestimmen}
|
||||
h''(x) &=& 0 \\
|
||||
2(1+x^{2})^{-3} \cdot 4x^{2} - 2(1+x^{2})^{-2} &=& 0 \\
|
||||
\intertext{Es ergibt sich das Ergebnis:}
|
||||
x = \frac{1}{\sqrt{3}}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Probe}
|
||||
2(1+\left(\frac{1}{\sqrt{3}}\right)^{2})^{-3} \cdot 4\left(\frac{1}{\sqrt{3}}\right)^{2} - 2(1+\left(\frac{1}{\sqrt{3}}\right)^{2})^{-2} &=& 0 \\
|
||||
2(1+\frac{1}{3})^{-3} \cdot 4 \cdot \frac{1}{3} - 2(1+\frac{1}{3})^{-2} &=& 0 \\
|
||||
\frac{2 \cdot \frac{4}{3}}{\left(\frac{4}{3}\right)^{3}} - \frac{2}{\left(\frac{4}{3}\right)^{2}} &=& 0 \\
|
||||
\frac{2 \cdot \frac{4}{3}}{\frac{64}{27}} - \frac{2}{\frac{16}{9}} &=& 0 \\
|
||||
2 \cdot \frac{4}{3} \cdot \frac{27}{64} - 2 \cdot \frac{9}{16} &=& 0 \\
|
||||
\frac{4}{3} \cdot \frac{54}{64} - \frac{18}{16} &=& 0 \\
|
||||
\frac{4}{3} \cdot \frac{27}{32} - \frac{9}{8} &=& 0 \\
|
||||
\intertext{27 mit 3 kürzen und 32 mit 4 kürzen}
|
||||
1 \cdot \frac{9}{8} - \frac{9}{8} &=& 0 \\
|
||||
\frac{9 - 9}{8} &=& 0 \\
|
||||
0 &=& 0
|
||||
\end{alignat*}
|
||||
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=5,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=0,xmax=5
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=0:5,samples=100]{e^(-x)} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 1,0.8) {f};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=5,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=0,xmax=5
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=0:5,samples=100]{1/(1+x)} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 1,0.8) {g};
|
||||
\end{axis}
|
||||
\end{tikzpicture}\\
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=5,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=0,xmax=5
|
||||
]
|
||||
\addplot[no marks, black, -] expression[domain=0:5,samples=100]{1/(1+x^2)} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 2,0.8) {h};
|
||||
\draw (axis cs:0.577350269,0.75) circle (2pt);
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Integrieren von f}
|
||||
\int\limits_{0}^{b} e^{-x}\,dx &=& \left[ -e^{-x}\right]_{0}^{b} \\
|
||||
&=& -e^{-b} + e^{0} \\
|
||||
&=& -\frac{1}{e^{b}} + 1 \rightarrow 1 \text{ für } b \rightarrow \infty \\
|
||||
&\Longrightarrow & \lim\limits_{b \rightarrow \infty} \int\limits_{0}^{b} e^{-x}\,dx = 1
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Integrieren von g}
|
||||
\int\limits_{0}^{b} \frac{1}{1+x}\,dx &=& \left[\ln |1+x|\right]_{0}^{b} \\
|
||||
&=& \ln |1 + b| - \ln |1 + 0| \\
|
||||
&=& \ln |1 + b| \rightarrow \infty \text{ für } b \rightarrow \infty \\
|
||||
&\Longrightarrow & \lim\limits_{b \rightarrow \infty} \int\limits_{0}^{b} \frac{1}{1+x}\,dx = \infty
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Integrieren von h}
|
||||
\int\limits_{0}^{b} \frac{1}{1+x^{2}} \,dx &=& \left[ \arctan x \right]_{0}^{b} \\
|
||||
&=& \arctan b - \arctan 0 \\
|
||||
&=& \arctan b \rightarrow \frac{\pi}{2} \text{ für } b \rightarrow \infty \\
|
||||
&\Longrightarrow & \lim\limits_{b \rightarrow \infty} \int\limits_{0}^{b} \frac{1}{1+x^{2}}\,dx = \frac{\pi}{2}
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=2,
|
||||
x=2cm,
|
||||
y=2cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$y$},
|
||||
xmin=-1,xmax=1
|
||||
]
|
||||
\addplot[no marks, black, -] expression[domain=-1:1,samples=100]{1/sqrt(1-x^2)} node[pos=0.65,anchor=north]{};
|
||||
\node at (axis cs: 2,0.8) {f};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Integrieren von f}
|
||||
\int\limits_{-1}^{1} \frac{1}{\sqrt{1-x^{2}}}\,dx &=& \left[\arcsin x\right]_{-1}^{1} \\
|
||||
&=& \arcsin 1 - \arcsin (-1) \\
|
||||
&=& \frac{\pi}{2} + \frac{\pi}{2}\\
|
||||
&=& \pi
|
||||
\end{alignat*}
|
||||
\section{} %3
|
||||
\begin{alignat*}{2}
|
||||
\intertext{$n=4$}
|
||||
\int\limits_{0}^{1} \sin x \,dx &\approx & \frac{1}{8}\left(\sin(0) + 2\sin\left(\frac{1}{4}\right) + 2\sin\left(\frac{1}{2}\right) + 2\sin\left(\frac{3}{4}\right) + \sin(1)\right) \\
|
||||
&\approx & 0.4573009376
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{$n=5$}
|
||||
\int\limits_{0}^{1} \sin x \,dx &\approx & \frac{1}{10}\left(\sin(0) + 2\sin\left(\frac{1}{5}\right) +
|
||||
2\sin\left(\frac{2}{5}\right) + 2\sin\left(\frac{3}{5}\right) + 2\sin\left(\frac{4}{5}\right) + \sin(1)\right) \\
|
||||
&\approx & 0.4581643460
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{$n=10$}
|
||||
\int\limits_{0}^{1} \sin x \,dx &\approx \begin{split} \frac{1}{20}\left(\sin(0) + 2\sin\left(\frac{1}{10}\right) +
|
||||
2\sin\left(\frac{1}{5}\right) + 2\sin\left(\frac{3}{10}\right) + 2\sin\left(\frac{2}{5}\right) +
|
||||
2\sin\left(\frac{1}{2}\right)\right.\\ +
|
||||
\left. 2\sin\left(\frac{3}{5}\right) + 2\sin\left(\frac{7}{10}\right) + 2\sin\left(\frac{4}{5}\right) +
|
||||
2\sin\left(\frac{9}{10}\right) + \sin(1)\right)\end{split} \\
|
||||
&\approx & 0.4593145489
|
||||
\end{alignat*}
|
||||
\section{} %4
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
f(1) &=& 10 \cdot e^{-\frac{2}{5}} \\
|
||||
&\approx & 6.7032 \\
|
||||
f(2) &=& 10 \cdot 2 \cdot e^{-frac{2}{5}\cdot 2} \\
|
||||
&=& 20 \cdot e^{-\frac{4}{5}} \\
|
||||
&\approx & 8.9866 \\
|
||||
f(6) &=& 10 \cdot 6 \cdot e^{-\frac{2}{5}\cdot 6}\\
|
||||
&=& 60 \cdot e^{-\frac{12}{5}} \\
|
||||
&\approx & 5.4431 \\
|
||||
f(12) &=& 10 \cdot 12 \cdot e^{-\frac{2}{5}\cdot 12} \\
|
||||
&=& 120 \cdot e^{-\frac{24}{5}} \\
|
||||
&\approx & 0.9876 \\
|
||||
f(24) &=& 10 \cdot 24 \cdot e^{-\frac{2}{5}\cdot 24} \\
|
||||
&=& 240 \cdot e^{-\frac{48}{5}} \\
|
||||
&=& 0.0163
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
f'(t) &=& 10 \cdot e^{-\frac{2}{5}t} - 10t \cdot e^{-\frac{2}{5}t} \cdot \frac{2}{5} \\
|
||||
&=& 10 \cdot e^{-\frac{2}{5}t} - \frac{20}{5}t \cdot e^{-\frac{2}{5}t} \\
|
||||
&=& 10 \cdot e^{-\frac{2}{5}t} - 4t \cdot e^{-\frac{2}{5}t} \\
|
||||
&=& (10 - 4t) \cdot e^{-\frac{2}{5}t} \\
|
||||
&=& \frac{10-4t}{e^{\frac{2}{5}t}} \\
|
||||
f''(x) &=& \left((10 - 4t) \cdot e^{-\frac{2}{5}t}\right)' \\
|
||||
&=& (10 - 4t)' \cdot e^{-\frac{2}{5}t} + (10 - 4t) \cdot \left(e^{-\frac{2}{5}t}\right)' \\
|
||||
&=& -4 \cdot e^{-\frac{2}{5}t} + (10 - 4t) \cdot e^{-\frac{2}{5}t} \cdot \left(-\frac{2}{5}\right) \\
|
||||
\intertext{Berechnung der Nullstelle(n) von $f'(x)$}
|
||||
f'(x) &=& 0 \\
|
||||
\intertext{$f'(x)$ mit Zähler ersetzen, da der über Nullstelle bestimmt}
|
||||
10-4t &=& 0 \\
|
||||
10 &=& 4t \\
|
||||
\frac{5}{2} &=& t \\
|
||||
\intertext{Einsetzen in $f(x)$}
|
||||
f\left(\frac{5}{2}\right) &=& 10 \cdot \frac{5}{2} \cdot e^{-\frac{2}{5} \cdot \frac{5}{2}} \\
|
||||
&=& 25 \cdot e^{-1} \\
|
||||
&\approx & 9.1970
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\begin{alignat*}{2}
|
||||
\int 10t \cdot e^{-\frac{2}{5}t}\,dx &=& -\frac{5}{2}\cdot e^{-\frac{2}{5}t} \cdot 10t - \int -\frac{5}{2} \cdot e^{-\frac{2}{5}t} \cdot 10\,dx \\
|
||||
&=& -\frac{5}{2}\cdot e^{-\frac{2}{5}t} \cdot 10t + 25 \int e^{-\frac{2}{5}t}\,dx \\
|
||||
&=& -\frac{5}{2}\cdot e^{-\frac{2}{5}t} \cdot 10t - \frac{5}{2} \cdot e^{-\frac{2}{5}t} \cdot 25 \\
|
||||
&=& -\frac{5}{2}\cdot e^{-\frac{2}{5}t} \cdot \left( 10t + 25\right) \\
|
||||
\frac{1}{6} \int\limits_{0}^{6} 10t \cdot e^{-\frac{2}{5}t}\,dx &=& \frac{1}{6} \cdot \left[-\frac{5}{2}\cdot e^{-\frac{2}{5}t} \cdot (10t + 25)\right]_{0}^{6} \\
|
||||
&=& \frac{1}{6}\left(-\frac{5}{2}\cdot e^{-\frac{2}{5} \cdot 6} \cdot ( 10 \cdot 6 + 25) - \left(-\frac{5}{2}\cdot e^{-\frac{2}{5} \cdot 0} \cdot ( 10 \cdot 0 + 25)\right)\right) \\
|
||||
&=& \frac{1}{6}\left(-\frac{5}{2}\cdot e^{-\frac{12}{5}} \cdot 85 + \frac{5}{2}\cdot e^{0} \cdot 25\right) \\
|
||||
&=& \frac{1}{6}\left(-\frac{425}{2}\cdot e^{-\frac{12}{5}} + \frac{125}{2}\right) \\
|
||||
&\approx & 7.2037
|
||||
\end{alignat*}
|
||||
\subsection{} %d
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Die Stammfunktion wurde bereits in c) berechnet. Daher setze ich direkt die Werte entsprechend ein.}
|
||||
\frac{1}{6} \int\limits_{6}^{12} 10t \cdot e^{-\frac{2}{5}t}\,dx &=& \frac{1}{6} \cdot \left[-\frac{5}{2}\cdot e^{-\frac{2}{5}t} \cdot (10t + 25)\right]_{6}^{12} \\
|
||||
&=& \frac{1}{6}\left(-\frac{5}{2}\cdot e^{-\frac{2}{5} \cdot 12} \cdot ( 10 \cdot 12 + 25) - \left(-\frac{5}{2}\cdot e^{-\frac{2}{5} \cdot 6} \cdot ( 10 \cdot 6 + 25)\right)\right) \\
|
||||
&=& \frac{1}{6}\left(-\frac{5}{2}\cdot e^{-\frac{24}{5}} \cdot 145 + \frac{5}{2}\cdot e^{-\frac{12}{5}} \cdot 85\right) \\
|
||||
&=& \frac{1}{6}\left(-\frac{725}{2}\cdot e^{-\frac{24}{5}} + \frac{425}{2}\cdot e^{-\frac{12}{5}}\right) \\
|
||||
&\approx & 2.7157
|
||||
\end{alignat*}
|
||||
\subsection{} %e
|
||||
\begin{alignat*}{2}
|
||||
f'(t) &=& \frac{10-4t}{e^{\frac{2}{5}t}} \\
|
||||
f''(t) &=& -4 \cdot e^{-\frac{2}{5}t} + (10 - 4t) \cdot e^{-\frac{2}{5}t} \cdot \left(-\frac{2}{5}\right) \\
|
||||
&=& e^{-\frac{2}{5}t} \cdot \left(-4 + (10 - 4t) \cdot \left(-\frac{2}{5}\right) \right) \\
|
||||
&=& e^{-\frac{2}{5}t} \cdot \left(-4 - 4 + \frac{8}{5}t \right) \\
|
||||
&=& e^{-\frac{2}{5}t} \cdot \left(-8 + \frac{8}{5}t \right) \\
|
||||
&=& \frac{-8 + \frac{8}{5}t}{ e^{-\frac{2}{5}t}} \\
|
||||
f'''(t) &=& \left(e^{-\frac{2}{5}t} \cdot \left(-8 + \frac{8}{5}t \right)\right)' \\
|
||||
&=& e^{-\frac{2}{5}t} \cdot \left(-\frac{2}{5} \right) \cdot \left(-8 + \frac{8}{5}t \right) + e^{-\frac{2}{5}t} \cdot \left(-8 + \frac{8}{5}t \right)' \\
|
||||
&=& e^{-\frac{2}{5}t} \cdot \left(\frac{16}{5} - \frac{16}{25}t \right) + e^{-\frac{2}{5}t} \cdot \left(\frac{8}{5}\right) \\
|
||||
&=& e^{-\frac{2}{5}t} \cdot \left(\frac{16}{5} - \frac{16}{25}t + \frac{8}{5}\right) \\
|
||||
&=& e^{-\frac{2}{5}t} \cdot \left(\frac{24}{5} - \frac{16}{25}t\right) \\
|
||||
\intertext{Nullstelle(n) von $f''(x)$ berechnen}
|
||||
f''(x) &=& 0 \\
|
||||
\intertext{$f''(x)$ mit dem Zähler ersetzen, da der für die Nullstelle zuständig ist}
|
||||
-8 + \frac{8}{5}t &=& 0 \\
|
||||
\frac{8}{5} &=& 8 \\
|
||||
t &=& 5 \\
|
||||
f'''(5) &=& e^{-\frac{2}{5} \cdot 5} \cdot \left(\frac{24}{5} - \frac{16}{25} \cdot 5\right) \\
|
||||
&=& e^{-2} \cdot \left(\frac{24}{5} - \frac{16}{5}\right) \\
|
||||
&=& e^{-2} \cdot \frac{8}{5} > 0 \Rightarrow \text{Minimum} \\
|
||||
f(5) &=& 10 \cdot 5 \cdot e^{-\frac{2}{5} \cdot 5} \\
|
||||
&=& 50 \cdot e^{-2} \\
|
||||
&\approx & 6.7668
|
||||
\end{alignat*}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=10,
|
||||
x=1em,
|
||||
y=1em,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$t$},
|
||||
ylabel={$f(t)$},
|
||||
xmin=0,xmax=24
|
||||
]
|
||||
\addplot[no marks, black, -] expression[domain=0:24,samples=100]{10*x*e^(-(2/5)*x)} node[pos=0.65,anchor=north]{};
|
||||
\draw (axis cs:5,6.766764162) circle (2pt);
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\section{} %5
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
h(x) &=& (x^{2}+1)^{\cos(x)} \\
|
||||
&=& e^{\ln \left((x^{2}+1)^{\cos(x)}\right)}\\
|
||||
&=& e^{\cos(x) \cdot \ln (x^{2}+1)} \\
|
||||
h'(x) &=& e^{\cos(x) \cdot \ln (x^{2}+1)} \cdot \left(\cos(x) \cdot \ln (x^{2}+1)\right)' \\
|
||||
&=& e^{\cos(x) \cdot \ln (x^{2}+1)} \cdot \left((\cos(x))' \cdot \ln (x^{2}+1) + \cos(x) \cdot (\ln (x^{2}+1))'\right) \\
|
||||
&=& e^{\cos(x) \cdot \ln (x^{2}+1)} \cdot \left(-\sin(x) \cdot \ln (x^{2}+1) + \cos(x) \cdot \frac{(x^{2}+1)'}{x^{2}+1}\right) \\
|
||||
&=& e^{\cos(x) \cdot \ln (x^{2}+1)} \cdot \left(-\sin(x) \cdot \ln (x^{2}+1) + \cos(x) \cdot \frac{2x}{x^{2}+1}\right) \\
|
||||
&=& (x^{2}+1)^{\cos(x)} \cdot \left(-\sin(x) \cdot \ln (x^{2}+1) + \cos(x) \cdot \frac{2x}{x^{2}+1}\right)
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
t &=& \sqrt{\frac{x}{4} + 3} \\
|
||||
t^{2} &=& \frac{x}{4} + 3 \\
|
||||
t^{2} - 3 &=& \frac{x}{4} \\
|
||||
4t^{2} - 12 &=& x \\
|
||||
8t &=& \frac{dx}{dt} \\
|
||||
8t\,dt &=& dx
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\int \sin \left(\sqrt{\frac{x}{4} + 3}\right) &\Rightarrow & \int \sin (t) \cdot 8t\,dt \\
|
||||
\int \sin (t) \cdot 8t\,dt &=& -\cos(t) \cdot 8t - \int (-\cos (t) \cdot 8)\,dt \\
|
||||
&=& -\cos(t) \cdot 8t + 8\int \cos(t)\,dt \\
|
||||
&=& -\cos(t) \cdot 8t + 8 \cdot \sin(t) \\
|
||||
\int \sin \left(\sqrt{\frac{x}{4} + 3}\right) &=& -\cos \left(\sqrt{\frac{x}{4} + 3} \right) \cdot 8 \cdot \left(\sqrt{\frac{x}{4} + 3} \right) + 8 \cdot \sin \left(\sqrt{\frac{x}{4} + 3}\right)
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\begin{alignat*}{2}
|
||||
g(x) = x^{2}-x-6 &=& 0 \\
|
||||
x_{1,2} &=& \frac{1}{2} \pm \sqrt{\frac{1}{4} + 6 } \\
|
||||
&=& \frac{1}{2} \pm \sqrt{\frac{25}{4}} \\
|
||||
&=& \frac{1}{2} \pm \frac{5}{2} \\
|
||||
x_{1} &=& \frac{6}{2} = 3 \\
|
||||
x_{2} &=& -\frac{4}{2} = -2 \\
|
||||
g(x) = x^{2}-x-6 &=& (x-3)(x+2)
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\frac{3x+2}{x^{2}-x-6} = \frac{1}{(x-3)(x+2)} &=& \frac{A}{x-3} + \frac{B}{x+2} \\
|
||||
&=& \frac{A(x+2) + B(x-3)}{(x-3)(x+2)} \\
|
||||
&=& \frac{Ax + Bx + 2A - 3B}{(x-3)(x+2)} \\
|
||||
&=& \frac{(A+B)x + 2A - 3B}{(x-3)(x+2)} \\
|
||||
A + B &=& 3 \\
|
||||
2A - 3B &=& 2 \\
|
||||
A &=& 3 - B \\
|
||||
2(3-B) - 3B &=& 2 \\
|
||||
6 - 2B - 3B &=& 2 \\
|
||||
-5B &=& -4 \\
|
||||
5B &=& 4 \\
|
||||
B &=& \frac{4}{5}\\
|
||||
A &=& 3 - \frac{4}{5} \\
|
||||
&=& \frac{11}{5} \\
|
||||
\frac{3x+2}{x^{2}-x-6} &=& \frac{11}{5} \cdot \frac{1}{x-3} + \frac{4}{5} \cdot \frac{1}{x+2}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\int \frac{3x+2}{x^{2}-x-6}\,dx &=& \int \frac{11}{5} \cdot \frac{1}{x-3}\,dx + \int \frac{4}{5} \cdot \frac{1}{x+2}\,dx \\
|
||||
&=& \frac{11}{5} \int \frac{1}{x-3}\,dx + \frac{4}{5} \int \frac{1}{x+2}\,dx \\
|
||||
&=& \frac{11}{5} \cdot \ln |x-3| + \frac{4}{5} \cdot \ln |x+2|
|
||||
\end{alignat*}
|
||||
\subsection{} %d
|
||||
\begin{alignat*}{2}
|
||||
g(x) = x^{2} + 8x + 16 &=& (x+4)^{2} \\
|
||||
\frac{x+1}{(x+4)^{2}} &=& \frac{A}{(x+4)^{2}} + \frac{B}{x+4} \\
|
||||
&=& \frac{A + B(x+4)}{(x+4)^{2}} \\
|
||||
&=& \frac{A + Bx + 4B}{(x+4)^{2}} \\
|
||||
&=& \frac{Bx + A + 4B}{(x+4)^{2}} \\
|
||||
\Rightarrow B &=&1 \\
|
||||
A + 4B &=& 1 \\
|
||||
A &=& -4B + 1 \\
|
||||
A &=& -4 + 1 \\
|
||||
A &=& -3 \\
|
||||
\frac{x+1}{x^{2}+8x+16} &=& -\frac{3}{(x+4)^{2}} + \frac{1}{x+4}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\int \frac{x+1}{x^{2}+8x+16} \,dx &=& \int -\frac{3}{(x+4)^{2}}\,dx + \int \frac{1}{x+4}\,dx \\
|
||||
&=& -3\int (x+4)^{-2}\,dx + \int \frac{1}{x+4}\,dx \\
|
||||
&=& -3\left(-(x+4)^{-1}\right)+ \ln |x+4| \\
|
||||
&=& 3(x+4)^{-1} + \ln |x+4|
|
||||
\end{alignat*}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,297 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pgfplotsset{compat=1.8}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 13. Juni}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Anwenden der Limes-Version des Wurzelkriteriums}
|
||||
\lim\limits_{i \rightarrow \infty} \sqrt[i]{\left| \frac{i}{2^{i}}\right|} &=& \lim\limits_{i \rightarrow \infty} \frac{\sqrt[i]{\left| i \right|}}{\sqrt[i]{\left| 2^{i}\right|}} \\
|
||||
&=& \lim\limits_{i \rightarrow \infty} \frac{\sqrt[i]{|i|}}{2} \\
|
||||
\intertext{Anwenden, dass $\sqrt[i]{|i|} \rightarrow 1$ für $i \rightarrow \infty$ gilt}
|
||||
&=& \frac{1}{2} < 1 \Rightarrow \text{ Konvergenz}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Anwenden der Limes-Version des Quotientenkriteriums}
|
||||
\lim\limits_{i \rightarrow \infty} \left|\frac{\frac{(-1)^{i+1} \cdot (i+1)!}{(i+1)^{i+1}}}{\frac{(-1)^{i} \cdot i!}{i^{i}}} \right| &=& \lim\limits_{i \rightarrow \infty} \left| \frac{(-1)^{i+1} \cdot (i+1)! \cdot i^{i}}{(i+1)^{i+1} \cdot (-1)^{i} \cdot i!}\right| \\
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left| \frac{(-1)^{i+1} \cdot (i+1)! \cdot i^{i}}{(-1)^{i} \cdot i! \cdot (i+1)^{i+1}}\right| \\
|
||||
\intertext{Kürzen}
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left| \frac{(-1) \cdot (i+1) \cdot i^{i}}{(i+1)^{i+1}}\right| \\
|
||||
\intertext{Kürzen}
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left| \frac{(-1) \cdot i^{i}}{(i+1)^{i}}\right| \\
|
||||
\intertext{Ausklammern von $i^{i}$ im Nenner, Nullfolgen werden durch $...$ symbolisiert}
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left| \frac{(-1) \cdot i^{i}}{(1 + ...) \cdot i^{i}}\right| \\
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left| \frac{-1}{(1 + ...)}\right| \\
|
||||
&=& \left|\frac{-1}{1}\right| = 1 \Rightarrow \text{ Keine Aussage möglich}
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Es sei $x \in \mathbb{R}$ eine beliebig fest gewählte Zahl.}
|
||||
\lim\limits_{i \rightarrow \infty} \left|\frac{(i+1)^{2}2^{i+1}x^{i+1}}{i^{2}2^{i}x^{i}} \right| &=& \lim\limits_{i \rightarrow \infty} \left|\frac{(i+1)^{2} \cdot 2x}{i^{2}} \right| \\
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left|\frac{(i+1)^{2}}{i^{2}} \cdot 2x \right| \\
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left(\left|\frac{(i+1)^{2}}{i^{2}}\right| \cdot |2x| \right) \\
|
||||
\intertext{Anwenden, dass $x$ fest gewählt ist}
|
||||
&=& 2|x| \cdot \lim\limits_{i \rightarrow \infty} \left|\frac{(i+1)^{2}}{i^{2}}\right| \\
|
||||
&=& 2|x| \cdot \lim\limits_{i \rightarrow \infty} \left|\frac{i^{2} + 2i + 1}{i^{2}}\right| \\
|
||||
\intertext{Ausklammern von $i^{2}$ im Zähler und Nenner, anschließend kürzen}
|
||||
&=& 2|x| \cdot \lim\limits_{i \rightarrow \infty} \left|\frac{1 + \frac{2}{i} + \frac{1}{i^{2}}}{1}\right| \\
|
||||
&=& 2|x| \cdot 1 = 2|x|
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
2|x| < 1 &\Longleftrightarrow & |x| < \frac{1}{2} \\
|
||||
2|x| > 1 &\Longleftrightarrow & |x| > \frac{1}{2}
|
||||
\end{alignat*}
|
||||
Demzufolge liegt Konvergenz vor, falls $|x| < \frac{1}{2}$ gilt; Divergenz liegt vor, falls $|x| > \frac{1}{2}$ gilt. Also gilt $R = \frac{1}{2}$.
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Es sei $x \in \mathbb{R}$ eine beliebig fest gewählte Zahl.}
|
||||
\lim\limits_{i \rightarrow \infty} \sqrt[i]{\left| i^{2} \cdot 2^{i} \cdot x^{i} \right|} &=& \lim\limits_{i \rightarrow \infty} \left( \sqrt[i]{\left| i^{2} \right|} \cdot \sqrt[i]{\left| 2^{i} \right|} \cdot \sqrt[i]{\left| x^{i} \right|} \right) \\
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left( \sqrt[i]{\left| i^{2} \right|} \cdot |2| \cdot |x| \right) \\
|
||||
\intertext{Anwenden, dass $x$ fest gewählt ist}
|
||||
&=& 2 \cdot |x| \cdot \lim\limits_{i \rightarrow \infty} \sqrt[i]{\left| i^{2} \right|} \\
|
||||
&=& 2 \cdot |x| \cdot \lim\limits_{i \rightarrow \infty} \sqrt[i]{|i|} \cdot \lim\limits_{i \rightarrow \infty} \sqrt[i]{|i|} \\
|
||||
\intertext{Anwenden, dass $\sqrt[i]{|i|} \rightarrow 1$ für $i \rightarrow \infty$ gilt}
|
||||
&=& 2 \cdot |x| \cdot 1 \cdot 1 = 2|x|
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
2|x| < 1 &\Longleftrightarrow & |x| < \frac{1}{2} \\
|
||||
2|x| > 1 &\Longleftrightarrow & |x| > \frac{1}{2}
|
||||
\end{alignat*}
|
||||
Demzufolge liegt Konvergenz vor, falls $|x| < \frac{1}{2}$ gilt; Divergenz liegt vor, falls $|x| > \frac{1}{2}$ gilt. Also gilt $R = \frac{1}{2}$.
|
||||
\section{} %2
|
||||
\setcounter{subsubsection}{0}
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
\lim\limits_{i \rightarrow \infty} \left|\frac{\frac{-1}{2^{i+2}}}{\frac{-1}{2^{i+1}}} \right| &=& \lim\limits_{i \rightarrow \infty} \left|\frac{(-1) \cdot 2^{i+1}}{2^{i+2} \cdot (-1)} \right| \\
|
||||
\intertext{Kürzen}
|
||||
&=& \lim\limits_{i \rightarrow \infty} \left|\frac{1}{2} \right| \\
|
||||
&=& \frac{1}{2} < 1 \Rightarrow \text{ Konvergenz}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\intertext{Berechnen des Grenzwertes}
|
||||
\sum\limits_{i=1}^{\infty} \frac{-1}{2^{i+1}} &=& - \sum\limits_{i=1}^{\infty} \frac{1}{2^{i+1}} \\
|
||||
&=& - \sum\limits_{i=1}^{\infty} \frac{1^{i+1}}{2^{i+1}} \\
|
||||
&=& - \sum\limits_{i=1}^{\infty} \left(\frac{1}{2}\right)^{i+1} \\
|
||||
&=& - \sum\limits_{i=2}^{\infty} \left(\frac{1}{2}\right)^{i} \\
|
||||
&=& -\left( \sum\limits_{i=0}^{\infty} \left(\frac{1}{2}\right)^{i} - \left(\frac{1}{2}\right)^{0} - \left(\frac{1}{2}\right)^{1} \right) \\
|
||||
&=& -\left( \frac{1}{1 - \frac{1}{2}} - 1 - \frac{1}{2} \right) \\
|
||||
&=& -\left( 2 - 1 - \frac{1}{2} \right) \\
|
||||
&=& -\left( \frac{1}{2} \right)
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{2}
|
||||
\sum\limits_{i=1}^{\infty} \frac{(-1)^{i} \cdot i}{2(i+1)} &=& \sum\limits_{i=0}^{\infty} \frac{(-1)^{i} \cdot i}{2(i+1)} - \frac{(-1)^{0} \cdot 0}{2(0+1)} \\
|
||||
&=& \sum\limits_{i=0}^{\infty} \frac{(-1)^{i} \cdot i}{2(i+1)} - 0 \\
|
||||
&=& \sum\limits_{i=0}^{\infty} (-1)^{i} \cdot \frac{i}{2(i+1)} \\
|
||||
\intertext{Die Glieder der Reihe werden betragsmäßig immer größer und bilden daher keine Nullfolge. Deswegen divergiert die Reihe.}
|
||||
a_{0} &=& 0 \\
|
||||
a_{1} &=& \frac{1}{4} \\
|
||||
a_{2} &=& \frac{1}{3} = \frac{8}{24} \\
|
||||
a_{3} &=& \frac{3}{8} = \frac{9}{24} = \frac{15}{40}\\
|
||||
a_{4} &=& \frac{2}{5} = \frac{16}{40}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
\sum\limits_{i=1}^{\infty} \frac{1}{2(i+1)} &=& \frac{1}{2} \cdot \sum\limits_{i=2}^{\infty} \frac{1}{i} \\
|
||||
&=& \frac{1}{2} \cdot \left( \sum\limits_{i=1}^{\infty} \frac{1}{i} - 1\right)
|
||||
\intertext{Die Harmonische Reihe divergiert. Damit divergiert auch jede Reihe, die durch Hinzufügen, Weglassen oder Abändern endlich vieler Glieder entsteht.
|
||||
Daher divergiert diese Reihe.}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iv
|
||||
\begin{alignat*}{2}
|
||||
\sum\limits_{i=1}^{\infty} \frac{(-1)^{i+1}}{2^{i}} &=& \sum\limits_{i=0}^{\infty} \frac{(-1)^{i+1}}{2^{i}} - \frac{(-1)^{0+1}}{2^{0}} \\
|
||||
&=& \sum\limits_{i=0}^{\infty} \frac{(-1)^{i+1}}{2^{i}} - \frac{-1}{1} \\
|
||||
&=& \sum\limits_{i=0}^{\infty} \frac{(-1)^{i+1}}{2^{i}} + 1
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\lim\limits_{i \rightarrow \infty} \sqrt[i]{\left|\frac{(-1)^{i+1}}{2^{i}} \right|} &=& \lim\limits_{i \rightarrow \infty} \frac{\sqrt[i]{\left|(-1)^{i+1}\right|}}{\sqrt[i]{|2^{i}|}} \\
|
||||
&=& \lim\limits_{i \rightarrow \infty} \frac{\sqrt[i]{|(-1)^{i}| \cdot |(-1)|}}{2} \\
|
||||
&=& \lim\limits_{i \rightarrow \infty} \frac{\sqrt[i]{|(-1)^{i}|} \cdot \sqrt[i]{|-1|}}{2} \\
|
||||
&=& \frac{1}{2} < 1 \Rightarrow \text{ Konvergenz}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
\sum\limits_{i=0}^{\infty} \frac{(-1)^{i+1}}{2^{i}} + 1 &=& \sum\limits_{i=0}^{\infty} (-1) \cdot \frac{(-1)^{i}}{2^{i}} + 1 \\
|
||||
&=& \sum\limits_{i=0}^{\infty} (-1) \cdot \left(-\frac{1}{2}\right)^{i} + 1 \\
|
||||
&=& -\sum\limits_{i=0}^{\infty} \left(-\frac{1}{2}\right)^{i} + 1 \\
|
||||
&=& -\left(\frac{1}{1 + \frac{1}{2}} \right) + 1 \\
|
||||
&=& -\left(\frac{1}{\frac{3}{2}} \right) + 1 \\
|
||||
&=& -\frac{2}{3} + 1 \\
|
||||
&=& \frac{1}{3}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %v
|
||||
\begin{alignat*}{2}
|
||||
\sum\limits_{i=0}^{\infty} \frac{(-1)^{i}}{2i+1} &=& \sum\limits_{i=0}^{\infty} (-1)^{i} \cdot \frac{1}{2i+1} \\
|
||||
\intertext{Nach dem Leibnitz-Kriterium konvergiert diese Reihe, da die Reihenglieder $a_{i} = \frac{1}{2i+1}$ eine monotone Nullfolge bilden.}
|
||||
s_{0} &=& 1 \\
|
||||
s_{1} &=& 1 - \frac{1}{3} = \frac{2}{3}\\
|
||||
s_{2} &=& 1 - \frac{1}{3} + \frac{1}{5} = \frac{13}{15} \approx 0.867 \\
|
||||
s_{3} &=& \frac{13}{15} - \frac{1}{7} = \frac{91}{105} - \frac{15}{105} = \frac{76}{105} \approx 0.724
|
||||
\intertext{Der Grenzwert könnte $\approx 0.7853$ sein.}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %vi
|
||||
\begin{alignat*}{2}
|
||||
\sum\limits_{i=1}^{\infty} \frac{(-1)^{i}}{2i} &=& \sum\limits_{i=1}^{\infty} (-1)^{i} \cdot \frac{1}{2i} \\
|
||||
\intertext{Nach dem Leibnitz-Kriterium ist diese Reihe konvergent.}
|
||||
s_{0} &=& -\frac{1}{2} \\
|
||||
s_{1} &=& -\frac{1}{2} + \frac{1}{4} = -\frac{1}{4} \\
|
||||
s_{2} &=& -\frac{1}{4} - \frac{1}{6} = -\frac{10}{24} = -\frac{5}{12} \approx -0.4166667 \\
|
||||
s_{3} &=& -\frac{5}{12} + \frac{1}{8} = -\frac{40}{96} + \frac{12}{96} = -\frac{28}{96} \approx -0.2916667
|
||||
\intertext{Der Grenzwert könnte $\approx -0.3541667$ sein.}
|
||||
\end{alignat*}
|
||||
\section{} %3
|
||||
\subsection{} %a
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=10,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$f(x)$},
|
||||
xmin=0,xmax=10
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=0:10,samples=100]{1/x} node[pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=1:2,samples]{1} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:1,1) -- (axis cs:1,0) node [pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:2,1) -- (axis cs:2,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=2:3,samples]{1/2} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:3,0.5) -- (axis cs:3,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=3:4,samples]{1/3} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:4,0.3333333333333) -- (axis cs:4,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=4:5,samples]{1/4} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:5,0.25) -- (axis cs:5,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=5:6,samples]{1/5} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:6,0.2) -- (axis cs:6,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=6:7,samples]{1/6} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:7,0.1666666667) -- (axis cs:7,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=7:8,samples]{1/7} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:8,0.142857142857) -- (axis cs:8,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=8:9,samples]{1/8} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:9,0.125) -- (axis cs:9,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=9:10,samples]{1/9} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:10,0.111111111) -- (axis cs:10,0) node [pos=0.65,anchor=north]{};
|
||||
|
||||
\node at (axis cs: 1.5,1.25) {1};
|
||||
\node at (axis cs: 2.5,0.75) {$\frac{1}{2}$};
|
||||
\node at (axis cs: 3.5,0.5833333) {$\frac{1}{3}$};
|
||||
\node at (axis cs: 4.5,0.5) {$\frac{1}{4}$};
|
||||
\node at (axis cs: 5.5,0.45) {$\frac{1}{5}$};
|
||||
\node at (axis cs: 6.5,0.4166667) {$\frac{1}{6}$};
|
||||
\node at (axis cs: 7.5,0.392857142857) {$\frac{1}{7}$};
|
||||
\node at (axis cs: 8.5,0.375) {$\frac{1}{8}$};
|
||||
\node at (axis cs: 9.5,0.36111111) {$\frac{1}{9}$};
|
||||
\end{axis}
|
||||
\end{tikzpicture}\\
|
||||
Anhand der Skizze erkennt man:\\
|
||||
\begin{alignat*}{3}
|
||||
H_{n} &\geq & \int\limits_{1}^{n+1} \frac{1}{x}\,dx &=& [\ln x]_{1}^{n+1} \\
|
||||
&\geq & &=& \ln (n+1) - \ln 1 \\
|
||||
&\geq & &=& \ln (n+1) \\
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
Der Logarithmus $\ln$ ist divergent. Da die harmonische Reihe nach entsprechender Formel größer oder gleich dem natürlichen Logarithmus ist, wächst somit die harmonische Reihe ebenfalls über alle Schranken.
|
||||
\section{} %4
|
||||
\subsection{} %a
|
||||
Aus Aufgabe 3 ist bekannt, dass Folgendes gilt:\\
|
||||
\begin{alignat*}{2}
|
||||
\ln(n+1) &\leq & H_{n} \; (n = 1,2,...)\\
|
||||
\intertext{Da der natürliche Logarithmus monoton wachsend ist, gilt somit auch dies:}
|
||||
\ln(n) &\leq & H_{n} \; (n = 1,2,...)
|
||||
\end{alignat*}
|
||||
\begin{tikzpicture}[>=stealth]
|
||||
\begin{axis}[
|
||||
ymin=0,ymax=10,
|
||||
x=1cm,
|
||||
y=1cm,
|
||||
axis x line=middle,
|
||||
axis y line=middle,
|
||||
axis line style=->,
|
||||
xlabel={$x$},
|
||||
ylabel={$f(x)$},
|
||||
xmin=0,xmax=9
|
||||
]
|
||||
|
||||
\addplot[no marks, black, -] expression[domain=0:9,samples=100]{1/x} node[pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=1:2,samples=100]{1/2} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:1,0.5) -- (axis cs:1,0) node [pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:2,0.5) -- (axis cs:2,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=2:3,samples=100]{1/3} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:3,0.3333333333333) -- (axis cs:3,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=3:4,samples=100]{1/4} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:4,0.25) -- (axis cs:4,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=4:5,samples=100]{1/5} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:5,0.2) -- (axis cs:5,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=5:6,samples=100]{1/6} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:6,0.1666666667) -- (axis cs:6,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=6:7,samples=100]{1/7} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:7,0.142857142857) -- (axis cs:7,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=7:8,samples=100]{1/8} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:8,0.125) -- (axis cs:8,0) node [pos=0.65,anchor=north]{};
|
||||
\addplot[no marks, black, -] expression[domain=8:9,samples=100]{1/9} node[pos=0.65,anchor=north]{};
|
||||
\draw[>=stealth] (axis cs:9,0.111111111) -- (axis cs:9,0) node [pos=0.65,anchor=north]{};
|
||||
%\addplot[no marks, black, -] expression[domain=9:10,samples=100]{1/10} node[pos=0.65,anchor=north]{};
|
||||
%\draw[>=stealth] (axis cs:10,0.1) -- (axis cs:10,0) node [pos=0.65,anchor=north]{};
|
||||
|
||||
\node at (axis cs: 1.5,0.75) {$\frac{1}{2}$};
|
||||
\node at (axis cs: 2.5,0.5833333) {$\frac{1}{3}$};
|
||||
\node at (axis cs: 3.5,0.5) {$\frac{1}{4}$};
|
||||
\node at (axis cs: 4.5,0.45) {$\frac{1}{5}$};
|
||||
\node at (axis cs: 5.5,0.4166667) {$\frac{1}{6}$};
|
||||
\node at (axis cs: 6.5,0.392857142857) {$\frac{1}{7}$};
|
||||
\node at (axis cs: 7.5,0.375) {$\frac{1}{8}$};
|
||||
\node at (axis cs: 8.5,0.36111111) {$\frac{1}{9}$};
|
||||
%\node at (axis cs: 9.5,0.35) {$\frac{1}{10}$};
|
||||
\end{axis}
|
||||
\end{tikzpicture}\\
|
||||
Anhand der Skizze erkennt man:\\
|
||||
\begin{alignat*}{3}
|
||||
H_{n} - 1 &\leq & \int\limits_{1}^{n} \frac{1}{x}\,dx &=& [\ln x]_{1}^{n} \\
|
||||
&\leq & &=& \ln (n) - \ln 1 \\
|
||||
&\leq & &=& \ln (n)
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{3}
|
||||
\intertext{Daraus ergibt sich:}
|
||||
H_{n} - 1 &\leq & \ln(n) &\leq & H_{n}
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{3}
|
||||
\intertext{Aus a) ergibt sich:}
|
||||
\lim\limits_{n \rightarrow \infty} \frac{H_{n}-1}{H_{n}} &\leq & \lim\limits_{n \rightarrow \infty} \frac{\ln(n)}{H_{n}} &\leq & \lim\limits_{n \rightarrow \infty} \frac{H_{n}}{H_{n}} \\
|
||||
\lim\limits_{n \rightarrow \infty} \frac{H_{n}}{H_{n}} - \frac{1}{H_{n}} &\leq & \lim\limits_{n \rightarrow \infty} \frac{\ln(n)}{H_{n}} &\leq & 1 \\
|
||||
1 - 0 &\leq & \lim\limits_{n \rightarrow \infty} \frac{\ln(n)}{H_{n}} &\leq & 1 \\
|
||||
1 &\leq & \lim\limits_{n \rightarrow \infty} \frac{\ln(n)}{H_{n}} &\leq & 1
|
||||
\intertext{Nach dem Einschließungssatz muss der Grenzwert von $\lim\limits_{n \rightarrow \infty} \frac{\ln(n)}{H_{n}}$ $1$ sein.}
|
||||
\end{alignat*}
|
||||
\end{document}
|
||||
|
|
@ -0,0 +1,185 @@
|
|||
\documentclass[10pt,a4paper,oneside,ngerman,numbers=noenddot]{scrartcl}
|
||||
\usepackage[T1]{fontenc}
|
||||
\usepackage[utf8]{inputenc}
|
||||
\usepackage[ngerman]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{amsfonts}
|
||||
\usepackage{amssymb}
|
||||
\usepackage{paralist}
|
||||
\usepackage{gauss}
|
||||
\usepackage{pgfplots}
|
||||
\usepackage[locale=DE,exponent-product=\cdot,detect-all]{siunitx}
|
||||
\usepackage{tikz}
|
||||
\usetikzlibrary{matrix,fadings,calc,positioning,decorations.pathreplacing,arrows,decorations.markings}
|
||||
\usepackage{polynom}
|
||||
\polyset{style=C, div=:,vars=x}
|
||||
\pgfplotsset{compat=1.8}
|
||||
\pagenumbering{arabic}
|
||||
\def\thesection{\arabic{section})}
|
||||
\def\thesubsection{\alph{subsection})}
|
||||
\def\thesubsubsection{(\roman{subsubsection})}
|
||||
\makeatletter
|
||||
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
|
||||
\hskip -\arraycolsep
|
||||
\let\@ifnextchar\new@ifnextchar
|
||||
\array{#1}}
|
||||
\makeatother
|
||||
|
||||
\begin{document}
|
||||
\author{Jim Martens (6420323)}
|
||||
\title{Hausaufgaben zum 20. Juni}
|
||||
\maketitle
|
||||
\section{} %1
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
T_{8}(x) &=&& 1 - \frac{1}{2}x^{2} + \frac{1}{24}x^{4} - \frac{1}{720}x^{6} + \frac{1}{40320}x^{8} \\
|
||||
T_{9}(x) &=&& 1 - \frac{1}{2}x^{2} + \frac{1}{24}x^{4} - \frac{1}{720}x^{6} + \frac{1}{40320}x^{8} \\
|
||||
T_{10}(x) &=&& 1 - \frac{1}{2}x^{2} + \frac{1}{24}x^{4} - \frac{1}{720}x^{6} + \frac{1}{40320}x^{8} - \frac{1}{10!}x^{10}\\
|
||||
T_{11}(x) &=&& 1 - \frac{1}{2}x^{2} + \frac{1}{24}x^{4} - \frac{1}{720}x^{6} + \frac{1}{40320}x^{8} - \frac{1}{10!}x^{10} \\
|
||||
T_{12}(x) &=&& 1 - \frac{1}{2}x^{2} + \frac{1}{24}x^{4} - \frac{1}{720}x^{6} + \frac{1}{40320}x^{8} - \frac{1}{10!}x^{10} + \frac{1}{12!}x^{12}\\
|
||||
T_{13}(x) &=&& 1 - \frac{1}{2}x^{2} + \frac{1}{24}x^{4} - \frac{1}{720}x^{6} + \frac{1}{40320}x^{8} - \frac{1}{10!}x^{10} + \frac{1}{12!}x^{12}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
T_{9}(1) &=& 1 - \frac{1}{2} \cdot 1^{2} + \frac{1}{24} \cdot 1^{4} - \frac{1}{720} \cdot 1^{6} + \frac{1}{40320} \cdot 1^{8} \\
|
||||
&\approx & 0.5403026 \\
|
||||
T_{11}(1) &=& 1 - \frac{1}{2} \cdot 1^{2} + \frac{1}{24} \cdot 1^{4} - \frac{1}{720} \cdot 1^{6} + \frac{1}{40320} \cdot 1^{8} - \frac{1}{10!} \cdot 1^{10} \\
|
||||
&\approx & 0.5403023 \\
|
||||
T_{13}(1) &=& 1 - \frac{1}{2} \cdot 1^{2} + \frac{1}{24} \cdot 1^{4} - \frac{1}{720} \cdot 1^{6} + \frac{1}{40320} \cdot 1^{8} - \frac{1}{10!} \cdot 1^{10} + \frac{1}{12!} \cdot 1^{12} \\
|
||||
&\approx & 0.5403023
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \sqrt{1+x} = (1+x)^{\frac{1}{2}} \\
|
||||
f'(x) &=& \frac{1}{2} \cdot (1+x)^{-\frac{1}{2}} \\
|
||||
f''(x) &=& -\frac{1}{4} \cdot (1+x)^{-\frac{3}{2}} \\
|
||||
f'''(x) &=& \frac{3}{8} \cdot (1+x)^{-\frac{5}{2}} \\
|
||||
f^{(4)}(x) &=& -\frac{15}{16} \cdot (1+x)^{-\frac{7}{2}}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
T_{0}(x) &=&& 1 \\
|
||||
T_{1}(x) &=&& 1 + \frac{1}{2}x \\
|
||||
T_{2}(x) &=&& 1 + \frac{1}{2}x - \frac{1}{8}x^{2}\\
|
||||
T_{3}(x) &=&& 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{3}{48}x^{3} \\
|
||||
T_{4}(x) &=&& 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{3}{48}x^{3} - \frac{15}{384}x^{4}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
g(x) &=& \sqrt[3]{1+x} = (1+x)^{\frac{1}{3}} \\
|
||||
g'(x) &=& \frac{1}{3} \cdot (1+x)^{-\frac{2}{3}} \\
|
||||
g''(x) &=& -\frac{2}{9} \cdot (1+x)^{-\frac{5}{3}} \\
|
||||
g'''(x) &=& \frac{10}{27} \cdot (1+x)^{-\frac{8}{3}} \\
|
||||
g^{(4)}(x) &=& -\frac{80}{81} \cdot (1+x)^{-\frac{11}{3}}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
T_{0}(x) &=&& 1 \\
|
||||
T_{1}(x) &=&& 1 + \frac{1}{3}x \\
|
||||
T_{2}(x) &=&& 1 + \frac{1}{3}x - \frac{1}{9}x^{2} \\
|
||||
T_{3}(x) &=&& 1 + \frac{1}{3}x - \frac{1}{9}x^{2} + \frac{5}{81}x^{3} \\
|
||||
T_{4}(x) &=&& 1 + \frac{1}{3}x - \frac{1}{9}x^{2} + \frac{5}{81}x^{3} - \frac{10}{243}x^{4}
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& e^{x} \cdot \sin x \\
|
||||
f'(x) &=& e^{x} \cdot \cos x \\
|
||||
f''(x) &=& -e^{x} \cdot \sin x \\
|
||||
f'''(x) &=& -e^{x} \cdot \cos x \\
|
||||
f^{(4)}(x) &=& e^{x} \cdot \sin x \\
|
||||
f^{(5)}(x) &=& e^{x} \cdot \cos x
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
T_{5}(x) &=& 0 + x - 0x^{2} - \frac{1}{6}x^{3} + 0x^{4} + \frac{1}{120}x^{5} \\
|
||||
&=& x - \frac{1}{6}x^{3} + \frac{1}{120}x^{5}
|
||||
\end{alignat*}
|
||||
\section{} %2
|
||||
\subsubsection{} %i
|
||||
\begin{alignat*}{2}
|
||||
\lim\limits_{x \rightarrow 1} \left(\frac{x^{3}-3x^{2}+x+2}{x^{2}-5x+6} \right) &=& \frac{1-3+1+2}{1-5+6} \\
|
||||
&=& \frac{1}{2}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %ii
|
||||
\begin{alignat*}{3}
|
||||
\lim\limits_{x \rightarrow 2} \left(\frac{x^{3}-3x^{2}+x+2}{x^{2}-5x+6} \right) &=& \lim\limits_{x \rightarrow 2} \left(\frac{3x^{2} - 6x + 1}{2x - 5} \right) &=& \frac{3 \cdot 2^{2} - 6 \cdot 2 + 1}{2 \cdot 2 - 5} \\
|
||||
&& &=& \frac{12 - 12 + 1}{4 - 5} \\
|
||||
&& &=& \frac{1}{-1} \\
|
||||
&& &=& -1
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iii
|
||||
\begin{alignat*}{2}
|
||||
\lim\limits_{x \rightarrow 0} (1+3x)^{\frac{1}{2x}}
|
||||
\end{alignat*}
|
||||
\subsubsection{} %iv
|
||||
\begin{alignat*}{2}
|
||||
\lim\limits_{x \rightarrow 0} \left(\frac{1}{e^{x} - 1} - \frac{1}{\sin x} \right)
|
||||
\end{alignat*}
|
||||
\section{} %3
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& 3^{x} \\
|
||||
f'(x) &=& 3^{x} \cdot \ln 3 \\
|
||||
t'(x) = f'(2) &=& 9 \cdot \ln 3 \\
|
||||
&\approx & 9.88751 \\
|
||||
\intertext{Die Steigung der Tangente beträgt $9 \cdot \ln 3$ oder rund $9.88751$.}
|
||||
t(x) &=& \ln (3) \cdot 9x + b \\
|
||||
\intertext{Bestimmen des Schnittpunkts mit der y-Achse}
|
||||
b &=& t(x) - \ln (3) \cdot 9x \\
|
||||
&=& t(2) - \ln (3) \cdot 18 \\
|
||||
&=& 9 - \ln (3) \cdot 18 \\
|
||||
&\approx & -10.77502 \\
|
||||
t(x) &=& \ln 3 \cdot 9x + 9 - \ln (3) \cdot 18 \\
|
||||
\intertext{Bestimmen des Schnittpunkts mit der x-Achse}
|
||||
0 &=& \ln (3) \cdot 9x + 9 - \ln (3) \cdot 18 \\
|
||||
\ln (3) \cdot 18 - 9 &=& \ln (3) \cdot 9x \\
|
||||
\frac{\ln (3) \cdot 18 - 9}{\ln 3} &=& 9x \\
|
||||
\frac{\ln (3) \cdot 18 - 9}{\ln (3) \cdot 9} &=& x \\
|
||||
x &\approx & 1.08976
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
f(x) &=& \sqrt[7]{x+2} \\
|
||||
&=& (x+2)^{\frac{1}{7}} \\
|
||||
f'(x) &=& \frac{1}{7} \cdot (x+2)^{-\frac{6}{7}} \\
|
||||
f''(x) &=& -\frac{6}{49} \cdot (x+2)^{-\frac{13}{7}}
|
||||
\end{alignat*}
|
||||
\begin{alignat*}{2}
|
||||
T_{0} &=&& 2^{\frac{1}{7}} \\
|
||||
T_{1} &=&& 2^{\frac{1}{7}} + \frac{1}{7} \cdot 2^{-\frac{6}{7}}x \\
|
||||
T_{2} &=&& 2^{\frac{1}{7}} + \frac{1}{7} \cdot 2^{-\frac{6}{7}}x - \frac{3}{49} \cdot 2^{-\frac{13}{7}}
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\begin{alignat*}{2}
|
||||
x_{n} &=& \frac{1}{2\pi n} \\
|
||||
\lim\limits_{n\rightarrow \infty} f(x_{n}) = \lim\limits_{n\rightarrow \infty} \cos \left(\frac{1}{x_{n}} \right) &=& \lim\limits_{n\rightarrow \infty} \cos \left(\frac{1}{\frac{1}{2 \pi n}} \right) \\
|
||||
&=& \lim\limits_{n\rightarrow \infty} \cos (2 \pi n) = 1
|
||||
\end{alignat*} \\
|
||||
Der mögliche Grenzwert 1 stimmt nicht mit dem Funktionswert überein. Daher ist $h(x)$ im Punkt $x_{0}=0$ nicht stetig.
|
||||
\subsection{} %d
|
||||
\setcounter{subsubsection}{0}
|
||||
\subsubsection{} %i
|
||||
$\mathcal{B}$ enthält alle Folgen, die gegen eine reelle Zahl konvergieren, als auch solche, die zwischen zwei Werten oszillieren.
|
||||
Lediglich uneigentlich konvergente Folgen sind nicht enthalten.
|
||||
\subsubsection{} %ii
|
||||
Aus $i)$ ergibt sich, dass nicht jede Folge in $\mathcal{B}$ konvergiert. Oszillierende Folgen konvergieren nicht, allerdings erfüllen sie die Bedingungen von $\mathcal{B}$.
|
||||
\section{} %4
|
||||
\subsection{} %a
|
||||
\begin{alignat*}{2}
|
||||
\lim\limits_{x \rightarrow \infty} \left(\frac{a^{x}}{x^{n}}\right) &=& \lim\limits_{x \rightarrow \infty} \left(\frac{a^{x} \cdot \ln a}{n \cdot x^{n-1}}\right) \\
|
||||
\intertext{Nach n Ableitungen}
|
||||
&=& \lim\limits_{x \rightarrow \infty} \left(\frac{a^{x} \cdot (\ln a)^{n}}{n! \cdot x^{0}}\right) \\
|
||||
\intertext{Zähler geht gegen unendlich, Nenner gegen konstante Zahl $n!$. Daher existiert der Grenzwert nicht.}
|
||||
&=& \lim\limits_{x \rightarrow \infty} \left(\frac{a^{x} \cdot (\ln a)^{n}}{n!}\right) = \infty
|
||||
\end{alignat*}
|
||||
\subsection{} %b
|
||||
\begin{alignat*}{2}
|
||||
\lim\limits_{x \rightarrow \infty} \left(\frac{x^{r}}{\ln^{k} x} \right) &=& \lim\limits_{x \rightarrow \infty} \left(\frac{rx^{r-1}}{k \cdot \ln^{k-1} x \cdot \frac{1}{x}} \right) \\
|
||||
&=& \lim\limits_{x \rightarrow \infty} \left(\frac{rx^{r}}{k \cdot \ln^{k-1} x} \right) \\
|
||||
\intertext{Nach der k-ten Ableitung sieht es so aus}
|
||||
&=& \lim\limits_{x \rightarrow \infty} \left(\frac{r^{k} \cdot x^{r}}{k! \cdot \ln x} \right) \\
|
||||
\intertext{Herausziehen der Konstanten}
|
||||
&=& \frac{r^{k}}{k!} \cdot \lim\limits_{x \rightarrow \infty} \left(\frac{x^{r}}{\ln x} \right) \\
|
||||
\intertext{Nach Satz 27 im Skript, existiert der Grenzwert $\lim\limits_{x \rightarrow \infty} \left(\frac{x^{r}}{\ln x} \right)$ nicht bzw. ist unendlich. Unendlich mal eine Konstante ist immer noch unendlich.}
|
||||
&=& \infty
|
||||
\end{alignat*}
|
||||
\subsection{} %c
|
||||
\setcounter{subsubsection}{0}
|
||||
\subsubsection{} %i
|
||||
\subsubsection{} %ii
|
||||
\end{document}
|
||||
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Reference in New Issue