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Aufgabe15_Vorlage.pdf
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Pearson0001.pgf
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Pearson0001.pgf
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\begin{tikzpicture}
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\usetikzlibrary{plotmarks}
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\draw[color=gray, thin] (-3,0) -- (3,0);
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%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
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\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
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%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
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%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
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\draw[fill=blue!50,color=blue!50, ultra thick] (-1.5, 0) circle (.075);
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\draw[fill=blue!50,color=blue!50, ultra thick] (1.5, 0) circle (.075);
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\node[color=blue!50,] at (-1.5,0.35) {\textbf{a}};
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\node[color=blue!50] at (1.5,0.35) {\textbf{b}};
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\end{tikzpicture}
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Pearson0001.tikz
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Pearson0001.tikz
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++++\begin{tikzpicture}
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\usetikzlibrary{plotmarks}
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\draw[color=gray, thick] (-3,0) -- (3,0);
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%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
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\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
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%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
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%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
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\draw[color=blue!50, ultra thick] (-1.5, 0) circle (.075);
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\draw[color=blue!50, ultra thick] (1.5, 0) circle (.075);
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\node[color=blue!50] at (-1.5,0.35) {A};
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\node[color=blue!50] at (1.5,0.35) {B};
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\end{tikzpicture}
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Pearson0002.pgf
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Pearson0002.pgf
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\begin{tikzpicture}
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\usetikzlibrary{plotmarks}
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\draw[color=gray, thin] (-3,0) -- (3,0);
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%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
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\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
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%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
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%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
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\draw[color=blue!50, ultra thick] (-1.5, 0) circle (.075);
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\draw[color=blue!50, ultra thick] (1.5, 0) circle (.075);
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\node[color=blue!50,] at (-1.5,0.35) {\textbf{a}};
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\node[color=blue!50] at (1.5,0.35) {\textbf{b}};
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\end{tikzpicture}
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Pearson0003.pgf
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Pearson0003.pgf
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\begin{tikzpicture}
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\usetikzlibrary{plotmarks}
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\draw[color=gray, thin] (-3,0) -- (3,0);
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%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
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\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
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%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
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%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
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\draw[fill=blue!50,color=blue!50, ultra thick] (-1.5, 0) circle (.075);
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\draw[color=blue!50, ultra thick] (1.5, 0) circle (.075);
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\node[color=blue!50] at (-1.5,0.35) {\textbf{a}};
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\node[color=blue!50] at (1.5,0.35) {\textbf{b}};
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\end{tikzpicture}
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Pearson0004.pgf
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Pearson0004.pgf
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\begin{tikzpicture}
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\usetikzlibrary{plotmarks}
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\draw[color=gray, thin] (-3,0) -- (3,0);
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%http://tex.stackexchange.com/questions/64567/how-to-draw-circle-square-and-triangle-marks-in-tikz-picture
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\draw[color=blue!50, ultra thick] (-1.4625,0) -- (1.4625,0);
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%\node[mark size=2.5pt,color=blue!50, fill=none] at (-1.5,0) {\pgfuseplotmark{*}};
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%\node[mark size=2.5pt,color=blue!50] at (1.5,0) {\pgfuseplotmark{*}};
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\draw[color=blue!50, ultra thick] (-1.5, 0) circle (.075);
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\draw[fill=blue!50,color=blue!50, ultra thick] (1.5, 0) circle (.075);
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\node[color=blue!50] at (-1.5,0.35) {\textbf{a}};
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\node[color=blue!50] at (1.5,0.35) {\textbf{b}};
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\end{tikzpicture}
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VektorenInEbeneVeranschaulichung01.eps
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VektorenInEbeneVeranschaulichung01.eps
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VektorenInEbeneVeranschaulichung01.pst
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VektorenInEbeneVeranschaulichung01.pst
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\begin{pspicture}(-1,-1)(4.9,4.9)
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%\psgrid[griddots=10,gridlabels=0pt, subgriddiv=0]
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%%%%%%%%%%%%%%%%%%%%%%%%Koor%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psaxes[Ox=0,Dx=2,Oy=0,Dy=2,linewidth=0.1pt]{->}(0,0)(0,0)(4.5,4.5)%Schnittpunkt, x0, y0 x1,y1
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\rput[l](1.7,-0.8){$1.\ Koordinate$}
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\rput[c](4.8,0){$x$}
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\rput[l](0.4,4){$2.\ Koordinate$}
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\rput[c](0,4.8){$y$}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Vektor / Gerade%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psline[linecolor=red, linewidth=0.1pt](0,0)(3,3)
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\rput[c](3.0,-0.3){$a_1$}
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\rput[c](-0.3,3){$a_2$}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%Hlfslinien%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psline[linecolor=blue, linestyle=dashed, linewidth=0.1pt](3,-0.2)(3,3)
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\psline[linecolor=blue, linestyle=dashed, linewidth=0.1pt](-0.2,3)(3,3)
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Pfeil%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psline{->}(3.5,2.5)(3.1,2.9)
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Kreuz %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psline[linecolor=green] (2.9,3.1)(3.1,2.9)
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\psline[linecolor=green] (2.9,2.9)(3.1,3.1)
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Beschriftung am Pfeil %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\rput[l](3.3,2.3){Punkt}
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\rput[l](3.2,1.9){$(a_1, a_2)$}
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\end{pspicture}
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VektorenInEbeneVeranschaulichung02.eps
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VektorenInEbeneVeranschaulichung02.eps
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VektorenInEbeneVeranschaulichung02.pst
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VektorenInEbeneVeranschaulichung02.pst
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\begin{pspicture}(-1,-1)(4.9,4.9)
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%\psgrid[griddots=10,gridlabels=0pt, subgriddiv=0]
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Koordinaten%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psaxes[Ox=0,Dx=2,Oy=0,Dy=2,linewidth=0.1pt]{->}(0,0)(-0.0,-0.0)(4.5,4.5)%Schnittpunkt, x0, y0 x1,y1
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\rput[c](4.8,0){$x$}
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\rput[c](0,4.8){$y$}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Vektor%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psline[linecolor=red, linewidth=0.1pt]{->}(0,0)(3,3)
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Hilfslinien%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psline[linecolor=blue, linestyle=dashed, linewidth=0.1pt](3,-0.2)(3,3)
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\psline[linecolor=blue, linestyle=dashed, linewidth=0.1pt](-0.2,3)(3,3)
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\rput[c](3.0,-0.3){$a_1$}
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\rput[c](-0.3,3){$a_2$}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%Pfeil mit Beschriftung%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\psline{->}(3.5,2.5)(3.1,2.9)
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\rput[l](1.8,2.3){$Vektor\ ("`Pfeil"')$}
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\rput[l](2.5,1.9){$\vec{a}$}
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\end{pspicture}
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VeranschaulichungVonVektorenImRaum01.eps
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VeranschaulichungVonVektorenImRaum01.eps
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VeranschaulichungVonVektorenImRaum01.pdf
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VeranschaulichungVonVektorenImRaum01.pdf
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VeranschaulichungVonVektorenImRaum02.eps
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VeranschaulichungVonVektorenImRaum02.pdf
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VeranschaulichungVonVektorenImRaum02.pdf
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definitions.tex
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%!TEX root=main.tex
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\usepackage{geometry}
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\geometry{
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%a4paper,
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%total={170mm,257mm},
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left=20mm,
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top=20mm,
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right=40mm,
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marginparwidth=30mm
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}
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%\usepackage[top=2cm, bottom=1.3cm, left=10mm, right=0.5cm, heightrounded,
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%marginparwidth=30mm, marginparsep=3mm]{geometry}
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%\usepackage{pythontex}
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\usepackage[ngerman]{babel}
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\usepackage[utf8]{inputenc}
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%\usepackage{microtype}
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%\usepackage[sfmath,slantedGreeks]{kpfonts}
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%%
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% Just some sample text
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\usepackage{lipsum}
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\usepackage{wrapfig}
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%%
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% For nicely typeset tabular material
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\usepackage{booktabs}
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%\usepackage[utf8]{inputenc}
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%%
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% For graphics / images
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\usepackage{graphicx}
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%\setkeys{Gin}{width=\linewidth,totalheight=\textheight,keepaspectratio}
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%\graphicspath{{graphics/}}
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\usepackage{mathtools}
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%\usepackage[inline]{asymptote}
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% The fancyvrb package lets us customize the formatting of verbatim
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% environments. We use a slightly smaller font.
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%\usepackage{fancyvrb}
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%\fvset{fontsize=\normalsize}
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\usepackage{cmbright}
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%%
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% Prints argument within hanging parentheses (i.e., parentheses that take
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% up no horizontal space). Useful in tabular environments.
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\newcommand{\hangp}[1]{\makebox[0pt][r]{(}#1\makebox[0pt][l]{)}}
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%%
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% Prints an asterisk that takes up no horizontal space.
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% Useful in tabular environments.
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\newcommand{\hangstar}{\makebox[0pt][l]{*}}
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%%
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% Prints a trailing space in a smart way.
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\usepackage{xspace}
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%\usepackage[makeroom]{cancel}
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%%
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% Some shortcuts for Tufte's book titles. The lowercase commands will
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% produce the initials of the book title in italics. The all-caps commands
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% will print out the full title of the book in italics.
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%\newcommand{\vdqi}{\textit{VDQI}\xspace}
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%\newcommand{\ei}{\textit{EI}\xspace}
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%\newcommand{\ve}{\textit{VE}\xspace}
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%\newcommand{\be}{\textit{BE}\xspace}
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%\newcommand{\VDQI}{\textit{The Visual Display of Quantitative Information}\xspace}
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%\newcommand{\EI}{\textit{Envisioning Information}\xspace}
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%\newcommand{\VE}{\textit{Visual Explanations}\xspace}
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%\newcommand{\BE}{\textit{Beautiful Evidence}\xspace}
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%\newcommand{\TL}{Tufte-\LaTeX\xspace}
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%20180930 Fonts LX
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%\usepackage{lxfonts}
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%\usepackage[sfdefault,scaled=1.5]{FiraSans}
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%\usepackage[sfdefault,lining,scaled=1.5]{FiraSans} %% option 'sfdefault' activates Fira Sans as the default text font
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%\usepackage[fakebold, scaled=1.5]{firamath-otf}
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%\renewcommand*\oldstylenums[1]{{\firaoldstyle #1}}
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\usepackage[defaultfam,tabular,lining]{montserrat} %% Option 'defaultfam'
|
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%% only if the base font of the document is to be sans serif
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\usepackage[T1]{fontenc}
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\renewcommand*\oldstylenums[1]{{\fontfamily{Montserrat-TOsF}\selectfont #1}}
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%\usepackage{newtxsf}
|
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%\setkomafont{subsection}{\usefont{T1}{fvm}{m}{n}}
|
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\setkomafont{section}{\usefont{T1}{fvs}{b}{n}\Large}
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\setkomafont{subsection}{\usefont{T1}{fvs}{b}{n}}
|
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\setkomafont{subsubsection}{\usefont{T1}{fvs}{b}{n}}
|
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\setcounter{secnumdepth}{3}
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%20180930 Fonts LX
|
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\usepackage{enumitem}
|
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|
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% Prints the month name (e.g., January) and the year (e.g., 2008)
|
||||||
|
%\newcommand{\monthyear}{%
|
||||||
|
% \ifcase\month\or January\or February\or March\or April\or May\or June\or
|
||||||
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% July\or August\or September\or October\or November\or
|
||||||
|
% December\fi\space\number\year
|
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%}
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%wenn eps --> aus gnuplot
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|
%\usepackage{graphicx}
|
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|
%\usepackage{epstopdf}
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%\epstopdfsetup{update} % only regenerate pdf files when eps file is newer
|
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|
% lua aus gnuplot
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||||||
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%\usepackage{gnuplot-lua-tikz}
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||||||
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||||||
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% Prints an epigraph and speaker in sans serif, all-caps type.
|
||||||
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%\newcommand{\openepigraph}[2]{%
|
||||||
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% %\sffamily\fontsize{14}{16}\selectfont
|
||||||
|
% \begin{fullwidth}
|
||||||
|
% \sffamily\large
|
||||||
|
% \begin{doublespace}
|
||||||
|
% \noindent\allcaps{#1}\\% epigraph
|
||||||
|
% \noindent\allcaps{#2}% author
|
||||||
|
% \end{doublespace}
|
||||||
|
% \end{fullwidth}
|
||||||
|
%}
|
||||||
|
|
||||||
|
% Inserts a blank page
|
||||||
|
\newcommand{\blankpage}{\newpage\hbox{}\thispagestyle{empty}\newpage}
|
||||||
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|
||||||
|
\usepackage{units}
|
||||||
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|
||||||
|
% Typesets the font size, leading, and measure in the form of 10/12x26 pc.
|
||||||
|
%\newcommand{\measure}[3]{#1/#2$\times$\unit[#3]{pc}}
|
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%
|
||||||
|
%% Macros for typesetting the documentation
|
||||||
|
%\newcommand{\hlred}[1]{\textcolor{Maroon}{#1}}% prints in red
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||||||
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|
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||||||
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|
||||||
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|
||||||
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|
||||||
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|
||||||
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|
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|
||||||
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|
||||||
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||||||
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||||||
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|
||||||
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||||||
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|
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|
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|
||||||
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|
||||||
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|
||||||
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|
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|
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|
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|
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|
||||||
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|
||||||
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|
||||||
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|
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|
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|
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|
||||||
124
folgen01.tex
Normal file
124
folgen01.tex
Normal file
@@ -0,0 +1,124 @@
|
|||||||
|
%!TEX root=main.tex
|
||||||
|
\section{Folgen}
|
||||||
|
|
||||||
|
\paragraph*{Beispiel 1:}
|
||||||
|
|
||||||
|
Monatlicher Umsatz eines Shops: \\
|
||||||
|
\begin{center}
|
||||||
|
\begin{tabular}{|l|cccccc|}
|
||||||
|
\hline
|
||||||
|
\rule{0pt}{12pt}{\cellcolor[rgb]{1,0.647,0}}Monat &1 & 2 & 3 & 4 & 5 & 6 \\
|
||||||
|
\hline
|
||||||
|
\rule{0pt}{12pt}{\cellcolor[rgb]{1,0.647,0}}Umsatz in T\texteuro& 1 & 2 & 3 & 4 & 5 & 6 \\
|
||||||
|
\hline
|
||||||
|
\end{tabular}
|
||||||
|
|
||||||
|
\end{center}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{center}
|
||||||
|
\begin{tikzpicture}[scale=1.25]
|
||||||
|
\definecolor{orange1}{rgb}{1,0.647,0}
|
||||||
|
\begin{axis}[%
|
||||||
|
xlabel={Monat},
|
||||||
|
ylabel={Umsatz [T\texteuro]},
|
||||||
|
%clickable coords={(xy): \thisrow{label}},%
|
||||||
|
scatter/classes={%
|
||||||
|
a={mark=square*,orange1}}]
|
||||||
|
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|
||||||
|
scatter src=explicit symbolic]%
|
||||||
|
table[meta=label] {
|
||||||
|
x y label
|
||||||
|
1 1 a
|
||||||
|
2 2 a
|
||||||
|
3 3 a
|
||||||
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4 4 a
|
||||||
|
5 5 a
|
||||||
|
6 6 a
|
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|
};
|
||||||
|
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|
||||||
|
\end{tikzpicture}
|
||||||
|
\end{center}
|
||||||
|
|
||||||
|
\paragraph*{Beispiel 2:}
|
||||||
|
|
||||||
|
Rückzahlung eines Kredites (monatlich 200 \texteuro): \\
|
||||||
|
\begin{center}
|
||||||
|
\begin{tabular}{|l|cccccc|}
|
||||||
|
\hline
|
||||||
|
\rule{0pt}{12pt}{\cellcolor[rgb]{0.98,0.1,0}}Monat &1 & 2 & 3 & 4 & 5 & 6 \\
|
||||||
|
\hline
|
||||||
|
\rule{0pt}{12pt}{\cellcolor[rgb]{0.98,0.1,0}}Schuld in T\texteuro& 7,0 & 6,8 & 6,6 & 6,4 & 6,2 & 6,0 \\
|
||||||
|
\hline
|
||||||
|
\end{tabular}
|
||||||
|
|
||||||
|
\end{center}
|
||||||
|
|
||||||
|
|
||||||
|
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|
||||||
|
\begin{tikzpicture}[scale=1.25]
|
||||||
|
\definecolor{red1}{rgb}{0.98,0.1,0}
|
||||||
|
\begin{axis}[%
|
||||||
|
xlabel={Monat},
|
||||||
|
ylabel={Schuld [T\texteuro]},
|
||||||
|
%clickable coords={(xy): \thisrow{label}},%
|
||||||
|
scatter/classes={%
|
||||||
|
a={mark=square*,red1}}]
|
||||||
|
\addplot[scatter,only marks,%
|
||||||
|
scatter src=explicit symbolic]%
|
||||||
|
table[meta=label] {
|
||||||
|
x y label
|
||||||
|
1 7.0 a
|
||||||
|
2 6.8 a
|
||||||
|
3 6.6 a
|
||||||
|
4 6.4 a
|
||||||
|
5 6.2 a
|
||||||
|
6 6.0 a
|
||||||
|
};
|
||||||
|
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|
||||||
|
\end{tikzpicture}
|
||||||
|
\end{center}
|
||||||
|
|
||||||
|
\fancybox[Definition \glqq undendliche Folge\grqq]{Eine unendliche Folge is eine Abbildung, die jeder natürlichen Zahl $n$ (also $n \in \mathbb{ N}$ bzw. $n \in \mathbb{N}_0$) eine reelle Zahl $a_n$ zuordnet.
|
||||||
|
|
||||||
|
1 - 4:52 weiter
|
||||||
|
}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\vfill
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
x-Achse = Abzisse
|
||||||
|
y-Achse = Ordinate
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
%\def\hcenter#1{\hfil#1\hfil}
|
||||||
|
%\begin{table}[]
|
||||||
|
% \begin{tabu}{
|
||||||
|
% >{\columncolor[HTML]{9A0000}}X[l]|X[c]|X[c]|X[c]|X[c]|X[c]|X[c]|}
|
||||||
|
% {\color[HTML]{FFFFFF} \textbf{Monat}} & {\color[HTML]{9A0000} \textbf{1}} & {\color[HTML]{9A0000} \textbf{2}} & {\color[HTML]{9A0000} \textbf{3}} & {\color[HTML]{9A0000} \textbf{4}} & {\color[HTML]{9A0000} \textbf{5}} & {\color[HTML]{9A0000} \textbf{6}}\\
|
||||||
|
% \rule{10pt}{20pt}{\color[HTML]{FFFFFF} \textbf{Umsatz in T\texteuro}} & \textbf{1} & \textbf{2} & \textbf{3} & \textbf{4} & \textbf{5} & \textbf{6}
|
||||||
|
% \end{tabu}
|
||||||
|
%\end{table}
|
||||||
|
|
||||||
|
|
||||||
|
% Muster
|
||||||
|
%\begin{tabular}{|>{\columncolor{blue!40}}r|rrrrr|}
|
||||||
|
% \hline
|
||||||
|
% \rowcolor[gray]{.8} \textbf{No.} & {\bf 134} & {\bf 135} & {\bf 136} & {\bf 137} & {\bf 138} \\
|
||||||
|
% \hline
|
||||||
|
% \textbf{Milch } & 0.00 & 0.05 & 0.00 & 0.04 & 0.00 \\
|
||||||
|
% \textbf{Käse } & 49.57 & 49.15 & 49.80 & 49.68 & 50.18 \\
|
||||||
|
% \textbf{Zucker } & 0.01 & 0.00 & 0.00 & 0.00 & 0.00 \\
|
||||||
|
% {\bf Apfel } & 0.00 & 0.06 & 0.00 & 0.01 & 0.01 \\
|
||||||
|
% {\bf Wurst } & 46.14 & 46.56 & 46.32 & 46.48 & 46.31\\\hline
|
||||||
|
% {\bf Total } & {\bf 97.13} & {\bf 97.23} & {\bf 97.53} & {\bf 97.65} & {\bf 98.04} \\\hline
|
||||||
|
%\end{tabular}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\newpage
|
||||||
24
gnuplot01.gp
Normal file
24
gnuplot01.gp
Normal file
@@ -0,0 +1,24 @@
|
|||||||
|
set terminal png
|
||||||
|
set output "gnuplot01.png"
|
||||||
|
set grid
|
||||||
|
set samples 100000
|
||||||
|
unset border
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-7:10]
|
||||||
|
set xrange [-8:8]
|
||||||
|
set size 10./10.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
set ytics axis
|
||||||
|
set xtics axis
|
||||||
|
#set object circle at 3,3 size 0.2
|
||||||
|
set label "(2,0.33)" at 2.1,0.77 tc rgb "#FF0000"
|
||||||
|
plot 1/(x**2 -1) lt rgb "#006300" lw 2 notitle, '-' w p pt 7 ps 1.5 lc rgb "#FF0000" notitle
|
||||||
|
2.0 0.33
|
||||||
|
e
|
||||||
|
|
||||||
|
#"<echo '1 2'" with points pt 7 ps 2 #pt = filled circle ps = size
|
||||||
BIN
gnuplot01.png
Normal file
BIN
gnuplot01.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 5.5 KiB |
20
gnuplot01a.gp
Normal file
20
gnuplot01a.gp
Normal file
@@ -0,0 +1,20 @@
|
|||||||
|
set terminal png
|
||||||
|
set output "gnuplot01a.png"
|
||||||
|
set grid
|
||||||
|
unset border
|
||||||
|
set samples 100000
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-7:10]
|
||||||
|
set xrange [-8:8]
|
||||||
|
set size 10./10.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
set ytics axis
|
||||||
|
set xtics axis
|
||||||
|
set arrow from 3,-5 to -0.9,-5 lt rgb "#FF0000" lw 2
|
||||||
|
set label "von rechts" at 0.9,-4.5 tc rgb "#FF0000"
|
||||||
|
plot 1/(x**2 -1) lt rgb "#000FCF" lw 2 notitle
|
||||||
BIN
gnuplot01a.pdf
Normal file
BIN
gnuplot01a.pdf
Normal file
Binary file not shown.
BIN
gnuplot01a.png
Normal file
BIN
gnuplot01a.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 5.5 KiB |
21
gnuplot01b.gp
Normal file
21
gnuplot01b.gp
Normal file
@@ -0,0 +1,21 @@
|
|||||||
|
set terminal png
|
||||||
|
set output "gnuplot01b.png"
|
||||||
|
set grid
|
||||||
|
set samples 100000
|
||||||
|
unset border
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-7:10]
|
||||||
|
set xrange [-8:8]
|
||||||
|
set size 10./10.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
set ytics axis
|
||||||
|
set xtics axis
|
||||||
|
set arrow from -4.2,6 to -1.1,6 lt rgb "#FF0000" lw 2
|
||||||
|
set label "von links" at -3.5,6.5 tc rgb "#FF0000"
|
||||||
|
;set style line 1 lt 2 lw 2 pt 3 ps 0.5
|
||||||
|
plot 1/(x**2 -1) lt rgb "#006300" lw 2 notitle
|
||||||
BIN
gnuplot01b.png
Normal file
BIN
gnuplot01b.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 5.4 KiB |
BIN
gnuplot02.png
Normal file
BIN
gnuplot02.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 5.5 KiB |
17
gnuplot02a.gp
Normal file
17
gnuplot02a.gp
Normal file
@@ -0,0 +1,17 @@
|
|||||||
|
set terminal png
|
||||||
|
set output "gnuplot02a.png"
|
||||||
|
set size 10./10.
|
||||||
|
set nogrid
|
||||||
|
unset border
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-8:8]
|
||||||
|
set xrange [-8:8]
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
;set xtics axis out scale 0.5
|
||||||
|
set ytics axis
|
||||||
|
plot 1/x lt rgb "#FF00FF" notitle
|
||||||
BIN
gnuplot02a.png
Normal file
BIN
gnuplot02a.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 3.9 KiB |
21
gnuplot02b.gp
Normal file
21
gnuplot02b.gp
Normal file
@@ -0,0 +1,21 @@
|
|||||||
|
set terminal png
|
||||||
|
set output "gnuplot01b.png"
|
||||||
|
set grid
|
||||||
|
set samples 100000
|
||||||
|
unset border
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-7:10]
|
||||||
|
set xrange [-8:8]
|
||||||
|
set size 10./10.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
set ytics axis
|
||||||
|
set xtics axis
|
||||||
|
set arrow from -4.2,6 to -1.1,6 lt rgb "#FF0000" lw 2
|
||||||
|
set label "von links" at -3.5,6.5 tc rgb "#FF0000"
|
||||||
|
;set style line 1 lt 2 lw 2 pt 3 ps 0.5
|
||||||
|
plot 1/(x**2 -1) lt rgb "#006300" lw 2 notitle
|
||||||
BIN
gnuplot02b.pdf
Normal file
BIN
gnuplot02b.pdf
Normal file
Binary file not shown.
17
gnuplot03a.gp
Normal file
17
gnuplot03a.gp
Normal file
@@ -0,0 +1,17 @@
|
|||||||
|
set terminal png
|
||||||
|
set output "gnuplot03a.png"
|
||||||
|
set nogrid
|
||||||
|
unset border
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-1:10]
|
||||||
|
set xrange [-8:3]
|
||||||
|
set size 10./10.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
;set xtics axis out scale 0.5
|
||||||
|
set ytics axis
|
||||||
|
plot exp(x) lt rgb "#0000FF" notitle
|
||||||
18
gnuplot03a.gp.tex
Normal file
18
gnuplot03a.gp.tex
Normal file
@@ -0,0 +1,18 @@
|
|||||||
|
\centering
|
||||||
|
\begin{gnuplot}[terminal=pdf,terminaloptions={font ",14" linewidth 2}]
|
||||||
|
set nogrid
|
||||||
|
unset border
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-1:10]
|
||||||
|
set xrange [-8:3]
|
||||||
|
set size 10./10.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
;set xtics axis out scale 0.5
|
||||||
|
set ytics axis
|
||||||
|
plot exp(x) lt rgb "#0000FF"
|
||||||
|
\end{gnuplot}
|
||||||
BIN
gnuplot03a.png
Normal file
BIN
gnuplot03a.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 3.2 KiB |
17
gnuplot04.gp
Normal file
17
gnuplot04.gp
Normal file
@@ -0,0 +1,17 @@
|
|||||||
|
\begin{gnuplot}[terminal=pdf,terminaloptions={font ",10" linewidth 2}]
|
||||||
|
unset border
|
||||||
|
set nogrid
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-12:-1]
|
||||||
|
set xrange [-4:4]
|
||||||
|
set size 5/5.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
set xtics axis out scale 0.5
|
||||||
|
set ytics axis
|
||||||
|
plot(sqrt(4-x)/(sqrt(12+x)-4)) lt rgb "#FFAA00"
|
||||||
|
\end{gnuplot}
|
||||||
17
gnuplot05.gp
Normal file
17
gnuplot05.gp
Normal file
@@ -0,0 +1,17 @@
|
|||||||
|
\begin{gnuplot}[terminal=pdf,terminaloptions={font ",10" linewidth 2}]
|
||||||
|
unset border
|
||||||
|
set nogrid
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-15:-4]
|
||||||
|
set xrange [-4:4]
|
||||||
|
set size 5/5.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
set xtics axis out scale 0.5
|
||||||
|
set ytics axis
|
||||||
|
plot(-x**2 +2*x -8) lt rgb "#FFAA00"
|
||||||
|
\end{gnuplot}
|
||||||
BIN
gnuplot06a.png
Normal file
BIN
gnuplot06a.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 4.2 KiB |
18
gnuplot07a.gp
Normal file
18
gnuplot07a.gp
Normal file
@@ -0,0 +1,18 @@
|
|||||||
|
set terminal png
|
||||||
|
set output "gnuplot07a.png"
|
||||||
|
unset border
|
||||||
|
set nogrid
|
||||||
|
set samples 100000
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-6:6]
|
||||||
|
set xrange [-10:30]
|
||||||
|
set size 5/5.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
set xtics axis out scale 0.5
|
||||||
|
set ytics axis
|
||||||
|
plot(7/(9-x)) lt rgb "#CBAAC0" lw 2 notitle
|
||||||
BIN
gnuplot07a.png
Normal file
BIN
gnuplot07a.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 4.1 KiB |
18
gnuplot08a.gp
Normal file
18
gnuplot08a.gp
Normal file
@@ -0,0 +1,18 @@
|
|||||||
|
set terminal png
|
||||||
|
set output "gnuplot08a.png"
|
||||||
|
unset border
|
||||||
|
set nogrid
|
||||||
|
set samples 100000
|
||||||
|
set lmargin at screen 0
|
||||||
|
set rmargin at screen 1
|
||||||
|
set bmargin at screen 0
|
||||||
|
set tmargin at screen 1
|
||||||
|
set yrange [-1:6]
|
||||||
|
set xrange [-1:20]
|
||||||
|
set size 5/5.
|
||||||
|
set key center top reverse Left
|
||||||
|
set xzeroaxis
|
||||||
|
set yzeroaxis
|
||||||
|
set xtics axis out scale 0.5
|
||||||
|
set ytics axis
|
||||||
|
plot(sin(x)**(1/log(x))) lt rgb "#CBAAC0" lw 2 notitle
|
||||||
77
limit001.pgf
Normal file
77
limit001.pgf
Normal file
@@ -0,0 +1,77 @@
|
|||||||
|
\begin{tikzpicture}[scale=1.5]
|
||||||
|
|
||||||
|
\pgfplotsset{compat=1.11}
|
||||||
|
|
||||||
|
\definecolor{FireBrick}{rgb}{0.7, 0.13, 0.13}
|
||||||
|
|
||||||
|
\definecolor{NewBlue}{rgb}{0.27, 0.45, 0.76}
|
||||||
|
|
||||||
|
\tikzfading[name=arrowfading, top color=transparent!0, bottom color=transparent!95]
|
||||||
|
\tikzset{arrowfill/.style={#1,general shadow={fill=black, shadow yshift=-0.8ex, path fading=arrowfading}}}
|
||||||
|
\tikzset{arrowstyle/.style n args={3}{draw=#2,arrowfill={#3}, single arrow,minimum height=#1, single arrow,
|
||||||
|
single arrow head extend=.3cm,}}
|
||||||
|
|
||||||
|
%\NewDocumentCommand{\tikzfancyarrow}{O{2cm} O{FireBrick} O{top color=orange!20!red, bottom color=red} m}{
|
||||||
|
%\tikz[baseline=-0.5ex]\node [arrowstyle={#1}{#2}{#3}] {#4};
|
||||||
|
%}
|
||||||
|
|
||||||
|
|
||||||
|
%\node [
|
||||||
|
% fill=blue!50, draw,
|
||||||
|
% single arrow, single arrow head indent=0ex,
|
||||||
|
% rotate=0,
|
||||||
|
% font=\sffamily
|
||||||
|
%] at (1,1.5)
|
||||||
|
%{\rotatebox{0}{ \qquad}};
|
||||||
|
|
||||||
|
|
||||||
|
%\draw[color=gray!10,step=2mm,help lines] (-0.7,0) grid (72mm,58mm);
|
||||||
|
%\draw[color=gray!70,step=10mm,xshift=4mm,yshift=-1mm] (-0.5,0) grid (70mm,60mm);
|
||||||
|
|
||||||
|
\begin{axis}[
|
||||||
|
thin,
|
||||||
|
axis x line=center,
|
||||||
|
axis y line=center,
|
||||||
|
%xtick={-2,...,2},
|
||||||
|
ytick={-1,...,6},
|
||||||
|
grid = both,
|
||||||
|
minor x tick num=5,
|
||||||
|
minor y tick num=5,
|
||||||
|
%minor x tick style={line width=0pt},
|
||||||
|
major x grid style={black, line width=0.2pt},
|
||||||
|
major y grid style={black, line width=0.2pt},
|
||||||
|
minor x grid style={gray, line width=0.1pt},
|
||||||
|
%minor y tick num={5},
|
||||||
|
%minor y tick style={line width=0pt},
|
||||||
|
%yminorgrids,
|
||||||
|
%minor y grid style={black, line width=0.1pt},
|
||||||
|
xlabel style={below right},
|
||||||
|
ylabel style={above left},
|
||||||
|
minor tick length=0pt,
|
||||||
|
xmin=-2.9,
|
||||||
|
xmax=2.9,
|
||||||
|
ymin=-1.2,
|
||||||
|
style={>=latex},
|
||||||
|
yticklabel style = {font=\footnotesize,xshift=0.5ex},
|
||||||
|
xticklabel style = {font=\footnotesize,yshift=0.5ex},
|
||||||
|
% grid=both,
|
||||||
|
% grid style={line width=.1pt, draw=gray!10},
|
||||||
|
% major grid style={line width=.2pt,draw=gray!50},
|
||||||
|
ymax=5.5]
|
||||||
|
\addplot [mark=none,domain=-3.8:3.8, color=NewBlue, line width=0.75mm,smooth] {x^2};
|
||||||
|
\end{axis}
|
||||||
|
|
||||||
|
\draw[-{Triangle[scale=3,length=5,width=6]}, color=NewBlue, line width=3mm, fill=white] (2,1.5) to (0.8,1.5);
|
||||||
|
%http://latexcolor.com
|
||||||
|
|
||||||
|
|
||||||
|
\draw[-{Triangle[scale=3,length=5,width=6]}, color=NewBlue, line width=3mm] (4.8,1.5) to (6,1.5);
|
||||||
|
|
||||||
|
|
||||||
|
\draw[draw=NewBlue, fill=white, line width=0.75mm] (-1,1.2) rectangle (0.5,1.8) node[pos=.5] {\textbf{\textcolor{black}{$-\infty$}}};
|
||||||
|
|
||||||
|
|
||||||
|
\draw[draw=NewBlue, fill=white, line width=0.75mm] (6.3,1.2) rectangle (7.8,1.8) node[pos=.5] {\textbf{\textcolor{black}{$+\infty$}}};
|
||||||
|
%\tikzfancyarrow[3cm]{} arrow
|
||||||
|
|
||||||
|
\end{tikzpicture}
|
||||||
53
limit01.pgf
Normal file
53
limit01.pgf
Normal file
@@ -0,0 +1,53 @@
|
|||||||
|
\begin{tikzpicture}[scale=0.8]
|
||||||
|
|
||||||
|
\pgfplotsset{compat=1.11}
|
||||||
|
|
||||||
|
\definecolor{FireBrick}{rgb}{0.7, 0.13, 0.13}
|
||||||
|
|
||||||
|
\definecolor{NewBlue}{rgb}{0.27, 0.45, 0.76}
|
||||||
|
|
||||||
|
\tikzfading[name=arrowfading, top color=transparent!0, bottom color=transparent!95]
|
||||||
|
\tikzset{arrowfill/.style={#1,general shadow={fill=black, shadow yshift=-0.8ex, path fading=arrowfading}}}
|
||||||
|
\tikzset{arrowstyle/.style n args={3}{draw=#2,arrowfill={#3}, single arrow,minimum height=#1, single arrow,
|
||||||
|
single arrow head extend=.3cm,}}
|
||||||
|
|
||||||
|
%\NewDocumentCommand{\tikzfancyarrow}{O{2cm} O{FireBrick} O{top color=orange!20!red, bottom color=red} m}{
|
||||||
|
%\tikz[baseline=-0.5ex]\node [arrowstyle={#1}{#2}{#3}] {#4};
|
||||||
|
%}
|
||||||
|
|
||||||
|
|
||||||
|
%\node [
|
||||||
|
% fill=blue!50, draw,
|
||||||
|
% single arrow, single arrow head indent=0ex,
|
||||||
|
% rotate=0,
|
||||||
|
% font=\sffamily
|
||||||
|
%] at (1,1.5)
|
||||||
|
%{\rotatebox{0}{ \qquad}};
|
||||||
|
|
||||||
|
|
||||||
|
%\draw[color=gray!10,step=2mm,help lines] (-0.7,0) grid (72mm,58mm);
|
||||||
|
%\draw[color=gray!70,step=10mm,xshift=4mm,yshift=-1mm] (-0.5,0) grid (70mm,60mm);
|
||||||
|
\begin{axis}[
|
||||||
|
x=1cm,y=1cm,
|
||||||
|
axis x line=center,
|
||||||
|
axis y line=center,
|
||||||
|
%axis lines=middle,
|
||||||
|
ymajorgrids=true,
|
||||||
|
xmajorgrids=true,
|
||||||
|
xmin=-5,
|
||||||
|
xmax=5,
|
||||||
|
ymin=-6,
|
||||||
|
ymax=6,
|
||||||
|
xtick={-5,-4,...,5},
|
||||||
|
ytick={-5,-4,...,6},]
|
||||||
|
\addplot [mark=none,domain=-4.8:-1.05, color=NewBlue, line width=0.5mm,step=10000, smooth, tension=0.2] {1/(x^2-1)};
|
||||||
|
\addplot [mark=none,domain=-0.95:0.95, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
|
||||||
|
\addplot [mark=none,domain=1.05:4.8, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
|
||||||
|
|
||||||
|
%\clip(-17.083986586441775,-20.54798056618339) rectangle (4.103328404466779,7.349328615703948);
|
||||||
|
\end{axis}
|
||||||
|
\filldraw[red](7,6.33) circle (0.75mm) node[above,right, yshift=4]{(2,0.33)};
|
||||||
|
|
||||||
|
%\node[] (A) at ( 1,3) {\textbf{$y^2=x^2$}};
|
||||||
|
|
||||||
|
\end{tikzpicture}
|
||||||
715
limit01.tex
Normal file
715
limit01.tex
Normal file
@@ -0,0 +1,715 @@
|
|||||||
|
%!TEX root=main.tex
|
||||||
|
\section{Grenzwerte}
|
||||||
|
|
||||||
|
\subsection{Der Limes ~\cite{studimup.de}}
|
||||||
|
Mit dem Limes können Grenzwerte angegeben werden. Der Limes beschreibt, was passiert, wenn man für eine Variable Werte einsetzt, die einem bestimmten Wert immer näherkommen. Dabei steht unter dem „lim“ die Variable und gegen welche Zahl sie geht (also welchem Wert die Variable immer näher kommt). Nach dem „lim“ steht dann die Funktion, worin dann die Werte für x eingesetzt werden, zum Beispiel:
|
||||||
|
$\lim _ { x \rightarrow \infty } \frac { 1 } { x }$
|
||||||
|
|
||||||
|
Diese Schreibweise bedeutet, dass man für $x$ in die Funktion $1/x$ Werte einsetzt, die immer näher an unendlich herankommen. Man kann ja keinen unendlichen Wert einsetzen, aber man kann mit dem Limes „gucken“ was für unendlich herauskommen würde. Man spricht dann „Limes gegen unendlich“. Das geht natürlich auch mit allen anderen Werten, nicht nur für unendlich.
|
||||||
|
|
||||||
|
\subsection{Grenzwerte im Unendlichen}
|
||||||
|
Grenzwerte im Unendlichen beschreiben, was mit der Funktion passiert, also an welchen Wert sich die Funktion immer mehr annähert, wenn $x$ gegen unendlich läuft (das heißt, wenn $x$ immer größer wird bis unendlich). Dabei kann $x$ gegen $+\infty$ und $-\infty$ laufen, also immer kleiner oder größer werden. Es sieht dann in mathematischer Schreibweise folgendermaßen aus:
|
||||||
|
|
||||||
|
$\lim _ { x \rightarrow \infty } f ( x ) \quad$ und $\quad \lim _ { x \rightarrow - \infty } f ( x )$
|
||||||
|
|
||||||
|
\begin{center}
|
||||||
|
\input{limit001.pgf}
|
||||||
|
\end{center}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\subsection{Merkblatt}\label{lbl:MerkblattGrenzwert}
|
||||||
|
|
||||||
|
\subsubsection{Wichtige Grenzwerte}
|
||||||
|
|
||||||
|
% \setlength\extrarowheight{10pt}
|
||||||
|
%\begin{tabular}{|C{2cm}|C{2cm}|}
|
||||||
|
% \hline
|
||||||
|
% & \\
|
||||||
|
% \hline
|
||||||
|
% $\frac { 1 } { \infty }$ & $0$ \\
|
||||||
|
% \hline
|
||||||
|
% $\frac { 1 } { \pm 0 }$ & $\pm \infty$\\
|
||||||
|
% \hline
|
||||||
|
% & \\
|
||||||
|
% \hline
|
||||||
|
% & \\
|
||||||
|
% \hline
|
||||||
|
% & \\
|
||||||
|
% \hline
|
||||||
|
% & \\
|
||||||
|
% \hline
|
||||||
|
% & \\
|
||||||
|
% \hline
|
||||||
|
%\end{tabular}
|
||||||
|
%\begin{tabular}{l}
|
||||||
|
|
||||||
|
\begin{TAB}(r,7mm,8mm)[10pt]{|c|l|}{|c|c|c|c|c|c|}% (rows,min,max)[tabcolsep]{columns}{rows}
|
||||||
|
$\frac { 1 } { \infty }$ & $0$ \\
|
||||||
|
$\frac { 1 } { \pm 0 }$ & $\pm \infty$ \\
|
||||||
|
$\mathrm { e } ^ { \infty }$ & $\infty$\\
|
||||||
|
$q ^ { \infty }$ & $\text {falls } | q | < 1$\\
|
||||||
|
$\ln ( \infty )$&$\infty$\\
|
||||||
|
$\ln \left( 0 ^ { + } \right) $&$ - \infty$
|
||||||
|
\end{TAB}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\begin{longtable}{|l|l|l|}
|
||||||
|
$\left[ \frac { 0 } { 0 } \right] , \left[ \frac { \infty } { \infty } \right]$ & \textbf{Zähler} und \textbf{Nenner} einzeln \textbf{ableiten}. & $\lim _ { x \rightarrow 0 } \frac { \sin ( x ) } { x } \frac { \left[ \frac { 0 } { 0 } \right] } { = } \lim _ { x \rightarrow 0 } \frac { \cos ( x ) } { 1 } = 1$\\\hline
|
||||||
|
& & $\lim _ { x \rightarrow \infty } \frac { 3 x } { e ^ { 2 x } } \stackrel { \left[ \frac { \infty } { \infty } \right] } { = } \lim _ { x \rightarrow \infty } \frac { 3 } { 2 e ^ { 2 x } } = 0$ \\
|
||||||
|
\\
|
||||||
|
$[ 0 \cdot \infty ]$ & \textbf{Umformen zu} $\left[ \frac { 0 } { 0 } \right]$ \textbf{oder} $\left[ \frac { \infty } { \infty } \right]$: & \\
|
||||||
|
|
||||||
|
& \textbf{Bruch vorhanden?}: Bruch \glqq zusammensetzen\grqq\ & $\lim _ { x \rightarrow \infty } \frac { 1 } { x } \cdot \ln ( x ) \stackrel { [ 0 \cdot \infty ] } { = }$ \\
|
||||||
|
|
||||||
|
& & $\lim _ { x \to \infty } \frac { \ln ( x ) } { x } \stackrel { \left[ \frac { \infty } { \infty } \right] } { = } \ldots$ \\ \\
|
||||||
|
& \textbf{Sonst}: Bruch \glqq erzeugen\grqq mit $a \cdot b = \frac { a } { 1 / b } = \frac { b } { 1 / a }$& $\lim _ { x \rightarrow 0 ^ +} x \cdot \ln ( x ) \mathop =\limits^{\left[ 0 \cdot \infty \right]} $\\
|
||||||
|
& & $\lim _ { x \rightarrow 0 ^ { + } } \frac { \ln ( x ) } { 1 / x } \mathop = \limits^{\left[ \frac{\infty }{\infty } \right]}\ldots$\\ \\
|
||||||
|
$ \left[ 0 ^ 0 \right] , \left[ 1 ^ \infty \right],$ & \textbf{Umformen zu} $\left[ 0 \cdot \infty \right] : \quad a ^ b = e ^ { b \cdot \ln ( a ) }$ & $\lim _ { x \rightarrow 0 ^ { + }
|
||||||
|
} ( 1 - x ) ^ { \frac { 1 } { x } } \stackrel { \left[ 1 ^ { \infty } \right] } { = }$\\
|
||||||
|
$\left[ \infty ^ 0 \right]$& & $\lim _ { x \rightarrow 0 ^ { + } } e ^ { \frac { 1 } { x } \cdot \ln ( 1 - x ) } \stackrel { [ 0 \cdot \infty ] } { = } \ldots$\\
|
||||||
|
$[ \infty - \infty ]$ & \textbf{Umformen}: &\\
|
||||||
|
&\textbf{Bruch vorhanden?}: Hauptnenner bilden & $\lim _ { x \rightarrow 0 ^ { + } } \frac { 1 } { x } - \frac { 1 } { \ln ( 1 + x ) } \stackrel { [ \infty - \infty ] } { = }$\\
|
||||||
|
& & $\stackrel { [ \infty - \infty ] } { = }\lim _ { x \rightarrow 0 ^ { + } } \frac { \ln ( 1 + x ) - x } { x \cdot \ln ( 1 + x ) } \stackrel { \left[ \frac { 0 } { 0 } \right] } { = } \ldots$\\
|
||||||
|
& \textbf{Quadratwurzel vorhanden?:} $a - b = \frac { a ^ { 2 } - b ^ { 2 } } { a + b }$ & $\lim _ { x \rightarrow \infty } x - \sqrt { x ^ { 2 } + 1 } \stackrel { [ \infty - \infty ] } { = }$\\
|
||||||
|
|
||||||
|
\end{longtable}
|
||||||
|
|
||||||
|
\vfill
|
||||||
|
\pagebreak
|
||||||
|
|
||||||
|
|
||||||
|
\subsection{Wichtige Grenzwerte}
|
||||||
|
Empfehlung: Die Zahl die gegen das $x$ läuft in die Funktion einsetzen. Im einfachsten Fall kommt sofort das Endergebnis heraus.
|
||||||
|
\begin{enumerate}
|
||||||
|
\item Einfach
|
||||||
|
z.B. $\mathop {\lim }\limits_{x \to - 3} {x^2} = 9$
|
||||||
|
\item $\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0$
|
||||||
|
\item $\mathop {\lim }\limits_{x \to {2^ + }} \frac{{\ln \left( {x - 1} \right)}}{{x - 2}} = 1$
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
\textbf{Beispiel}: Funktion $\mathop {\lim }\limits_{x \to - 2} \frac{1}{{{x^2} - 1}}$
|
||||||
|
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
% \pgfplotsset{compat=1.13}
|
||||||
|
% \begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=0.6]
|
||||||
|
% \definecolor{ffqqqq}{rgb}{1,0,0}
|
||||||
|
% \definecolor{qqwuqq}{rgb}{0,0.39215686274509803,0}
|
||||||
|
% \begin{axis}[
|
||||||
|
% x=1cm,y=1cm,
|
||||||
|
% axis lines=middle,
|
||||||
|
% ymajorgrids=true,
|
||||||
|
% xmajorgrids=true,
|
||||||
|
% xmin=-7.72,
|
||||||
|
% xmax=7.7,
|
||||||
|
% ymin=-5.5,
|
||||||
|
% ymax=5.5,
|
||||||
|
% xtick={-7,-6,...,7},
|
||||||
|
% ytick={-5,-4,...,5},]
|
||||||
|
% \clip(-7.72,-5.48) rectangle (7.72,5.48);
|
||||||
|
% \draw[line width=0.75pt,color=qqwuqq,smooth,samples=400,domain=-7.72:7.719999999999998] plot(\x,{1/((\x)^(2)-1)});
|
||||||
|
% \begin{scriptsize}
|
||||||
|
% %\draw[color=qqwuqq] (-7.54,-0.05) node {$f$};
|
||||||
|
% \draw [fill=ffqqqq] (2,0.3333333333333333) circle (2.5pt);
|
||||||
|
% \draw[color=ffqqqq] (2.8,0.77) node {$(2, 0.33)$};
|
||||||
|
% \end{scriptsize}
|
||||||
|
% \end{axis}
|
||||||
|
% \end{tikzpicture}
|
||||||
|
\centering
|
||||||
|
%\includegraphics[scale=.7]{gnuplot01.png}
|
||||||
|
\input{limit01.pgf}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
|
||||||
|
%\begin{figure}[h]
|
||||||
|
%\pgfplotsset{compat=1.15}
|
||||||
|
%\begin{tikzpicture}[>=stealth, baseline=0,scale=0.75,x=1cm,y=1cm]
|
||||||
|
%\definecolor{ffqqqq}{rgb}{1,0,0}
|
||||||
|
%\definecolor{qqwuqq}{rgb}{0,0.39215686274509803,0}
|
||||||
|
%\begin{axis}[
|
||||||
|
% x=1cm,y=1cm,
|
||||||
|
% axis lines=middle,
|
||||||
|
% ymajorgrids=true,
|
||||||
|
% xmajorgrids=true,
|
||||||
|
% xmin=-7.72,
|
||||||
|
% xmax=7.7,
|
||||||
|
% ymin=-5.5,
|
||||||
|
% ymax=5.8,
|
||||||
|
% xtick={-7,-6,...,7},
|
||||||
|
% ytick={-5,-4,...,5},
|
||||||
|
% restrict y to domain=-5.5:5.8,
|
||||||
|
% restrict y to domain=-7.72:7.7]
|
||||||
|
% \clip(-7.72,-5.48) rectangle (7.72,5.48);
|
||||||
|
% \draw[line width=0.8pt,color=qqwuqq,smooth,samples=500,domain=-7:7] plot(\x,{1/((\x)^(2)-1)});
|
||||||
|
% \begin{scriptsize}
|
||||||
|
% \draw [fill=ffqqqq] (2,0.3333333333333333) circle (2.5pt);
|
||||||
|
% \draw[color=ffqqqq] (2.8,0.77) node {$(2, 0.33)$};
|
||||||
|
% \end{scriptsize}
|
||||||
|
%\end{axis}
|
||||||
|
%\end{tikzpicture}
|
||||||
|
%\end{figure}
|
||||||
|
|
||||||
|
\marginpar{bekommt man allgemein bei dieser Methode ein eindeutiges Ergebnis, so ist die Aufgabe gelöst}Hier kann man auch erst alle $x$ mit dem Grenzwert füllen und berechnen. Hier in diesem Fall ergibt sich ein eindeutiges Ergebnis: $\mathop {\lim }\limits_{x \to - 2} \frac{1}{{{x^2} - 1}} = \frac{1}{3}$
|
||||||
|
|
||||||
|
Schaut man sich die Funktion an der Stelle $-1$ an, so kann man hier von zwei Seiten diesen Wert untersuchen. Wenn man von links kommt so schreibt man $\mathop {\lim }\limits_{x \to {1^ - }} \frac{1}{{{x^2} - 1}}$. Hier geht Funktion gegen $+\infty$. Von rechts $\mathop {\lim }\limits_{x \to {1^ + }} \frac{1}{{{x^2} - 1}}$ geht die Funktion gegen $-\infty$.
|
||||||
|
|
||||||
|
\definecolor{ffqqqq}{rgb}{1,0,0}
|
||||||
|
\definecolor{qqwuqq}{rgb}{0,0.39215686274509803,0}
|
||||||
|
|
||||||
|
%%%%%%%%%%%%%%Test%%%%%%%%%%%%%%
|
||||||
|
%\begin{figure}[htb]
|
||||||
|
% \centering
|
||||||
|
% \begin{minipage}[t]{.45\linewidth}
|
||||||
|
% \centering
|
||||||
|
% \pgfplotsset{compat=1.13}
|
||||||
|
% \begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=0.5]
|
||||||
|
% \definecolor{ffqqqq}{rgb}{1,0,0}
|
||||||
|
% \definecolor{qqwuqq}{rgb}{0,0.39215686274509803,0}
|
||||||
|
% \definecolor{ffzzqq}{rgb}{1,0.2,0}
|
||||||
|
% \begin{axis}[
|
||||||
|
% x=1cm,y=1cm,
|
||||||
|
% axis lines=middle,
|
||||||
|
% ymajorgrids=true,
|
||||||
|
% xmajorgrids=true,
|
||||||
|
% xmin=-7.72,
|
||||||
|
% xmax=7.7,
|
||||||
|
% ymin=-5.5,
|
||||||
|
% ymax=5.5,
|
||||||
|
% xtick={-7,-6,...,7},
|
||||||
|
% ytick={-5,-4,...,5},]
|
||||||
|
% \clip(-7.72,-5.48) rectangle (7.72,5.48);
|
||||||
|
% \draw[line width=0.75pt,color=qqwuqq,smooth,samples=400,domain=-7.72:7.719999999999998] plot(\x,{1/((\x)^(2)-1)});
|
||||||
|
% \begin{scriptsize}
|
||||||
|
% %\draw[color=qqwuqq] (-7.54,-0.05) node {$f$};
|
||||||
|
% \draw [->,line width=2pt,color=ffzzqq] (-5,4) -- (-1.32,4);
|
||||||
|
% %\draw [fill=ffqqqq] (2,0.3333333333333333) circle (2.5pt);
|
||||||
|
% %\draw[color=ffqqqq] (2.8,0.77) node {$(2, 0.33)$};
|
||||||
|
% \draw[color=ffzzqq] (-3.211316965163661,4.440165320759923) node {von links};
|
||||||
|
% \end{scriptsize}
|
||||||
|
% \end{axis}
|
||||||
|
% \end{tikzpicture}
|
||||||
|
% \end{minipage}%
|
||||||
|
% \hfill%
|
||||||
|
% \begin{minipage}[t]{.45\linewidth}
|
||||||
|
% \centering
|
||||||
|
% \pgfplotsset{compat=1.13}
|
||||||
|
% \begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1cm,y=1cm,scale=0.5]
|
||||||
|
% \definecolor{ffqqqq}{rgb}{1,0,0}
|
||||||
|
% \definecolor{qqwuqq}{rgb}{0,0.39215686274509803,0}
|
||||||
|
% \definecolor{ffzzqq}{rgb}{1,0.2,0}
|
||||||
|
% \begin{axis}[
|
||||||
|
% x=1cm,y=1cm,
|
||||||
|
% axis lines=middle,
|
||||||
|
% ymajorgrids=true,
|
||||||
|
% xmajorgrids=true,
|
||||||
|
% xmin=-7.72,
|
||||||
|
% xmax=7.7,
|
||||||
|
% ymin=-5.5,
|
||||||
|
% ymax=5.5,
|
||||||
|
% xtick={-7,-6,...,7},
|
||||||
|
% ytick={-5,-4,...,5},]
|
||||||
|
% \clip(-7.72,-5.48) rectangle (7.72,5.48);
|
||||||
|
% \draw[line width=0.75pt,color=qqwuqq,smooth,samples=400,domain=-7.72:7.719999999999998] plot(\x,{1/((\x)^(2)-1)});
|
||||||
|
% \draw [<-,line width=2pt,color=ffzzqq] (-0.7673053253313482,-5.056552230598019) -- (2.9126946746686526,-5.056552230598019);
|
||||||
|
% \begin{scriptsize}
|
||||||
|
% \draw[color=ffzzqq] (1.5,-4.6911412980822496) node {von rechts};
|
||||||
|
% \end{scriptsize}
|
||||||
|
% \end{axis}
|
||||||
|
% \end{tikzpicture}
|
||||||
|
% \end{minipage}
|
||||||
|
%% \caption{Bildtitel}
|
||||||
|
%\end{figure}
|
||||||
|
\begin{figure}[htb]
|
||||||
|
\centering
|
||||||
|
\begin{minipage}[t]{.45\linewidth}
|
||||||
|
\centering
|
||||||
|
\input{limit01a.pgf}
|
||||||
|
\caption*{An der Stelle $-2 \rightarrow 2^+$}
|
||||||
|
\end{minipage}%
|
||||||
|
\hfill%
|
||||||
|
\begin{minipage}[t]{.45\linewidth}
|
||||||
|
\centering
|
||||||
|
\includegraphics[scale=0.45]{gnuplot01b.png}
|
||||||
|
\caption*{An der Stelle $-2 \rightarrow 2^-$}
|
||||||
|
\end{minipage}%
|
||||||
|
\end{figure}
|
||||||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||||
|
\pagebreak
|
||||||
|
|
||||||
|
|
||||||
|
\begin{itemize}
|
||||||
|
\item $\frac{1}{{ \pm 0}} \to \pm \infty $ hiermit ist gemeint, je mehr der Nenner gegen $0$ geht, da eine Division durch Null nicht möglich ist.
|
||||||
|
\item $\frac{1}{\pm \infty} \to 0$
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
%\resizebox{!}{.15\paperheight}{\input{gnuplot02.gp}}
|
||||||
|
%\resizebox{!}{.15\paperheight}{\includegraphics{gnuplot01a.png}}
|
||||||
|
\includegraphics[scale=0.5]{gnuplot02a.png}
|
||||||
|
\caption*{$\frac{1}{x}$}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{itemize}
|
||||||
|
\item $e^{\infty} \to \infty $
|
||||||
|
\item $e^{- \infty} \to 0$
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
%TODO
|
||||||
|
%\resizebox{!}{.15\paperheight}{\input{gnuplot03a.gp}}
|
||||||
|
\includegraphics[scale=0.5]{gnuplot03a.png}
|
||||||
|
\caption*{$e^x$}
|
||||||
|
\end{figure}
|
||||||
|
\pagebreak
|
||||||
|
|
||||||
|
%http://tutorial.math.lamar.edu/Classes/CalcI/InverseFunctions.aspx
|
||||||
|
\marginpar{die Logarithmus-Funktion ist nur im positiven Bereich. Man kann sich nur von rechts der $0$ annähern.}
|
||||||
|
\begin{itemize}
|
||||||
|
\item $\ln (\infty) \to \infty $
|
||||||
|
\item $\ln (0^+) \to -\infty$
|
||||||
|
\end{itemize}
|
||||||
|
\begin{figure}[ht]
|
||||||
|
\centering
|
||||||
|
%\resizebox{!}{.15\paperheight}{\input{gnuplot03.gp}}
|
||||||
|
\includegraphics[scale=0.5]{gnuplot04a.png}
|
||||||
|
\caption*{$\ln x$}%
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
\begin{itemize}
|
||||||
|
\item $q^\infty = 0$, falls $\left|q\right|<1$
|
||||||
|
\item $\ln (0^+) \to -\infty$
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
\vfill
|
||||||
|
|
||||||
|
\pagebreak
|
||||||
|
\subsection{Regeln von L'Hospital}
|
||||||
|
Wenn man Grenzwerte wie $\lim _ { x \rightarrow x _ { 0 } } f ( x )$ bestimmen soll, schaut man zuerst einmal, was man denn erhalten würde, wenn man $x_0$ einfach einsetzt.
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
Kommt bei der Berechnung von Grenzwerten einer der nachfolgenden Sonderfälle heraus, so kann man diese gegebenenfalls mit der Regeln von L'Hospital lösen.
|
||||||
|
\begin{description}
|
||||||
|
\item [1.]\tikz[na]\node [coordinate] (n1) {}; $\frac{0}{0}$
|
||||||
|
\item [2.]\tikz[na]\node [coordinate] (n2) {}; $\frac{\infty}{\infty}$
|
||||||
|
\end{description}
|
||||||
|
Diese beiden Punkte können mit der Regel von L'Hospital gelöst werden.
|
||||||
|
\textbf{Beispiel:} $\mathop {\lim }\limits_{x \to 2} {\frac{x^2+x-6}{x^2-3x+2}}$ wird hier nun die $2$ eingesetzt, so ergibt sich:
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 2} {\frac{2^2+2-6}{2^2-3\cdot 2+2}} = \mathop {\lim }\limits_{x \to 2} {\frac{4+2-6}{4-6+2}} = \frac{0}{0}$. Da das Ergebnis $\frac{0}{0}$ beträgt wird nun die Regel von L'Hospital angewandt. Hierzu werden der Zähler und der Nenner separat differenziert. Somit ergibt sich $\mathop {\lim }\limits_{x \to 2} {\frac{2x+1}{2x-3}}=\frac{5}{1}=5$.
|
||||||
|
|
||||||
|
\begin{description}
|
||||||
|
\item [3.]\tikz[na]\node [coordinate] (n3) {}; $0\cdot \infty$ \qquad \tikz[na]\node [coordinate] (n31){}; Bruch vorhanden: $\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln \left( x \right)=\mathop {\lim }\limits_{x \to \infty } \frac{\ln \left( x \right)}{x}$
|
||||||
|
|
||||||
|
|
||||||
|
\hspace{1.35cm} \tikz[na]\node [coordinate] (n32){}; sonst: $\mathop {\lim }\limits_{x \to {0^ + }} x \cdot \ln \left( x \right)=\mathop {\lim }\limits_{x \to {0^ + }} \frac{\ln \left( x \right)}{\frac{1}{x}}$
|
||||||
|
%\begin{description}
|
||||||
|
%\item[]
|
||||||
|
|
||||||
|
% \item[]
|
||||||
|
%\end{description}
|
||||||
|
\end{description}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{description}
|
||||||
|
\item [4.]\tikz[na]\node [coordinate] (n41) {};$0^0$\tikz[na]\node [coordinate] (n4) {};
|
||||||
|
\item [5.]\tikz[na]\node [coordinate] (n51) {};$\infty^0$\tikz[na]\node [coordinate] (n5) {};
|
||||||
|
|
||||||
|
|
||||||
|
%$\mathop {\lim }\limits_{x \to {0^ + }} x \cdot \ln \left( x \right)=\mathop {\lim }\limits_{x \to {0^ + }} \frac{\ln \left( x \right)}{\frac{1}{x}}$
|
||||||
|
\item [6.]\tikz[na]\node [coordinate] (n61) {};$1^\infty$\tikz[na]\node [coordinate] (n6) {};
|
||||||
|
\item [7.]\tikz[na]\node [coordinate] (n7){};$\infty-\infty$ Bruch vorhanden: Hauptnenner bilden
|
||||||
|
|
||||||
|
$\sqrt[2]{\ldots} a-b=\frac{a^2-b^2}{a+b} $
|
||||||
|
|
||||||
|
$\sqrt[3]{\ldots} a-b =\frac{a^3-b^3}{a^2+ab+b^2}$
|
||||||
|
\end{description}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{tikzpicture}[overlay]
|
||||||
|
%\node [xshift=0cm,yshift=0cm] at (current page.south west)
|
||||||
|
%{
|
||||||
|
\path[-latex, color=blue!40, line width=0.5mm, opacity=0.5] (n3) edge [bend left=70] (n1);
|
||||||
|
\path[-latex, color=red!40, line width=0.5mm, opacity=0.5] (n3) edge [bend left=60] (n2);
|
||||||
|
|
||||||
|
\path[-latex, color=green!40!black, line width=0.25mm, opacity=0.5] (n3) edge [bend right=20] (n31);
|
||||||
|
|
||||||
|
\path[-latex, color=green!80!black, line width=0.25mm, opacity=0.5] (n3) edge [bend left=20] (n32);
|
||||||
|
|
||||||
|
\path[-latex, color=red!80!black, line width=0.25mm, opacity=0.5] (n41) edge [bend left=70] (n3);
|
||||||
|
|
||||||
|
\path[-latex, color=red!60!black, line width=0.25mm, opacity=0.5] (n51) edge [bend left=70] (n3);
|
||||||
|
|
||||||
|
\path[-latex, color=red!40!black, line width=0.25mm, opacity=0.5] (n61) edge [bend left=70] (n3);
|
||||||
|
%\path (n6) -| node[coordinate] (n4) {} (n1);
|
||||||
|
%\draw[thick,decorate,decoration={brace,amplitude=5pt}]
|
||||||
|
%(n1) -- (n3);
|
||||||
|
%\node[right=4pt] at ($(n1)!0.5!(n3)$) {One and two are cool};
|
||||||
|
|
||||||
|
\node [inner sep=3pt, fit=(n4) (n5) (n6) ] (bracemarks) {};
|
||||||
|
\draw[thick,decorate,decoration={brace,amplitude=5pt}]
|
||||||
|
(bracemarks.north east) -- (bracemarks.south east) node[midway, right=6pt] {$a^b=e^{b\cdot\ln(a)} \Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} {x^{\tikz\node[draw,shape=circle,anchor=base, color=red] {$2x$} ;}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{2x \cdot \ln \left( x \right)}}$};
|
||||||
|
%}
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
\pagebreak
|
||||||
|
\subsubsection{Beispiele zu $\frac{0}{0}$ und $\frac{\infty}{\infty}$}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{enumerate}
|
||||||
|
\item $\mathop {\lim }\limits_{x \to 2} {\frac{x^3-6x^2+12x-8}{x^2-4x+4}}=\frac{0}{0}$ Anwenden der Regel von L'Hospital.
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 2} {\frac{3x^2-12x+12}{2x-4}}=\frac{0}{0}$ somit muss hier die L'Hospitalsche Regel noch einmal angewandt werden
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 2} {\frac{6x-12}{2}}=\frac{0}{2} = \underline{\underline{0}}$
|
||||||
|
|
||||||
|
\item $\mathop {\lim }\limits_{x \to \infty}\frac{-2x^3+3x-1}{4\sqrt{5}+x^2+1}$
|
||||||
|
\marginpar{In der Unendlichkeit überleben immer nur die stärksten Terme}
|
||||||
|
|
||||||
|
Ausklammern der höchsten (des am stärksten wachsenden Term) Potenz in Nenner und Zähler. Die beiden höchsten Potenzen sind hier $x^3$ und $x^{\frac{5}{2}}$. Die am stärksten wachsende ist in diesem Fall $x^3$.
|
||||||
|
|
||||||
|
Somit $\mathop {\lim }\limits_{x \to \infty}\frac{x^3\left(-2+\frac{3}{x^2}-\frac{1}{x^3}\right)}{{x^3}\left( {4{x^{ - \frac{1}{2}}} + \frac{1}{x} + \frac{1}{{{x^3}}}} \right)}=\mathop {\lim }\limits_{x \to \infty}\frac{x^3\left(-2+\frac{3}{x^2}-\frac{1}{x^3}\right)}{{x^3}\left( {{\frac{4}{{{x^{\frac{1}{2}}}}}} + \frac{1}{x} + \frac{1}{{{x^3}}}} \right)} $\marginpar{$\frac{a^m}{a^n}=a^{m-n}$}
|
||||||
|
Alle Terme in der eine Konstante durch $x$ geteilt wird gehen nach $0$. Somit
|
||||||
|
$\mathop {\lim }\limits_{x \to \infty}\frac{x^3\left(-2+\cccancelto[red]{\frac{3}{x^2}}{0}-\cccancelto[red]{\frac{1}{x^3}}{0}\right)}{{x^3}\left( \bccancelto[orange]{\frac{ 4 }{x^{\frac{1}{2}}}}{0} + \bccancelto[orange]{\frac{1}{x}}{0} + \bccancelto[orange]{\frac{1}{x^3}}{0} \right)}$
|
||||||
|
|
||||||
|
Als Ergebnis erhält man hier $\frac{-2}{0}$ was in diesem Fall bedeutet das das Ergebnis nach unendlich geht, hier ist es $-\infty$
|
||||||
|
|
||||||
|
\item $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {{x^3}} \right)}}{{4{x^6}}}$ wird die $0$ eingesetzt so ergibt sich ein $\frac{0}{0}$ Ergebnis:
|
||||||
|
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos \left( {{0^3}} \right)}}{{4 \cdot {0^6}}} = \frac{{1 - 1}}{0} = \frac{0}{0}$
|
||||||
|
|
||||||
|
Anwendung von L'Hospital: \marginpar{Kettenregel:
|
||||||
|
|
||||||
|
$f\left( x \right) = g\left( {h\left( x \right)} \right) \to f'\left( x \right) = g'\left(h\left( x \right) \right) \cdot h'\left( x \right)$ }
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^3}} \right) \cdot \cccancelto[blue]{3}{}{\cccancelto[red]{x^2}{}}}}{\bccancelto[blue]{24}{8} \cdot {x^{\bccancelto[red]{5}{3}}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {{x^3}} \right)}}{{8{x^3}}}$
|
||||||
|
|
||||||
|
Dies ergibt wieder $\frac{0}{0} \rightarrow \sin(0) = 0$ und $8\cdot 0^3 = 0$ somit wird wieder differenziert:
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 0} \frac{\cos \left( x^3 \right) \cdot \cccancelto[green!50!black]{3}{}\cccancelto[orange]{x^2}{}}{\bccancelto[green!50!black]{24}{8}\bccancelto[orange]{x^2}{}}=\frac{1}{8}$
|
||||||
|
|
||||||
|
\item \marginpar{\[\frac{0}{{\sqrt {16} - 4}} = \frac{0}{0}\]}$\lim \limits_{x \to 4^-}\frac{\sqrt{4-x}}{\sqrt{12+x}-4} = \lim \limits_{x \to 4^-} \frac{{{{\left( {4 - x} \right)}^{\frac{1}{2}}}}}{{{{\left( {12 + x} \right)}^{\frac{1}{2}}} - 4}}$
|
||||||
|
|
||||||
|
Nach dieser Umstellung kann man die L'Hospitalsche Regel anwenden:
|
||||||
|
|
||||||
|
Somit: $\mathop {\lim }\limits_{x \to 4^-}\frac{{\frac{1}{2}{{\left( {4 - x} \right)}^{ - \frac{1}{2}}} \cdot - 1}}{{\frac{1}{2}{{\left( {12 + x} \right)}^{ - \frac{1}{2}}}}} = \mathop {\lim }\limits_{x \to 4^-}\frac{{ - \frac{1}{2}{{\left( {4 - x} \right)}^{ - \frac{1}{2}}}}}{{\frac{1}{2}{{\left( {12 + x} \right)}^{ - \frac{1}{2}}}}} = \mathop {\lim }\limits_{x \to 4^-}\frac{{\frac{1}{2}{{\left( {12 + x} \right)}^{\frac{1}{2}}}}}{{ - \frac{1}{2}{{\left( {4 - x} \right)}^{\frac{1}{2}}}}} = \mathop {\lim }\limits_{x \to 4^-}\frac{{\cccancelto[black]{\frac{1}{2}}{}{{\left( {12 + x} \right)}^{\frac{1}{2}}}}}{{ - \bccancelto[black]{\frac{1}{2}}{}{{\left( {4 - x} \right)}^{\frac{1}{2}}}}}$
|
||||||
|
|
||||||
|
|
||||||
|
$ = - \frac{{{{\left( {12 + x} \right)}^{\frac{1}{2}}}}}{{{{\left( {4 - x} \right)}^{\frac{1}{2}}}}} = \frac{-4}{0}$
|
||||||
|
|
||||||
|
Die Division durch $0$ bedeutet in diesem Fall das das Ergebnis gegen unendlich geht, hier $-\infty$.
|
||||||
|
|
||||||
|
\item $\mathop {\lim }\limits_{x \to 0}\frac{e^x-2x-e^{-x}}{x-\sin{x}}$
|
||||||
|
|
||||||
|
\marginpar{Man kann die Regel solange anwenden, solange sich Änderungen ergeben, wen nicht sollte man ausklammern etc... ausprobieren}Einsetzen der $0$ ergibt
|
||||||
|
$\frac{{{e^0} - 2 \cdot 0 - {e^{ - 0}}}}{{0 - 0}} = \frac{0}{0}$
|
||||||
|
Somit muss wieder L'Hospital angewandt werden:
|
||||||
|
|
||||||
|
Ableiten von Nenner und Zähler: $\mathop {\lim }\limits_{x \to 0}\frac{e^x-2+e^{-x}}{1-\cos{x}}$
|
||||||
|
|
||||||
|
Einsetzen von $0$ ergibt: $\frac{{{e^0} - 2 + {e^{ - 0}}}}{{1 - \cos (0)}} = \frac{{1 - 2 + 1}}{{1 - 1}} = \frac{0}{0}$
|
||||||
|
|
||||||
|
Da noch immer keine Lösung erhalten wurde, wird nun wieder differenziert:
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 0}\frac{e^x-2+e^{-x}}{1-\cos{x}}=\mathop{\lim }\limits_{x \to 0}\frac{e^x - e^{ - x}}{\sin (x)}$
|
||||||
|
|
||||||
|
Und wieder:
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos (x)}} = \frac{2}{1} = 2$
|
||||||
|
|
||||||
|
\item $\mathop {\lim }\limits_{x \to \infty } \frac{{{2^x} - {2^{ - x}}}}{{{2^x} + {2^{ - x}}}}$
|
||||||
|
|
||||||
|
Hier wird ist nun ein Fall bei der die Regel von L'Hospital zwar zur Anwendung kommen kann, jedoch nichts bewirkt.
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to \infty } \frac{2^x - 2^{ - x}}{2^x + 2^{ - x}}=\mathop { \lim }\limits_{ x\to \infty }\frac { 2^{ x }\ln { ( } 2)-\left( -\ln { ( } 2)\cdot 2^{ -x } \right) }{ { 2 }^x\ln(2)-2^{-x}\ln(2) } =\mathop { \lim }\limits_{ x\to \infty }\frac { 2^{ x }\ln { \left(2\right)} +\ln { \left(2\right)}\cdot 2^{ -x } }{ { 2 }^{ x }\ln { \left(2\right)}-2^{ -x }\ln { \left(2\right)} } \\=\mathop {\lim }\limits_{ x\to \infty }\frac { \ln { \left( 2 \right) } \left( 2^{ x }+2^{ -x } \right) }{ \ln { \left( 2 \right) } \left( { 2 }^{ x }-2^{ -x } \right) } =\mathop {\lim }\limits_{ x\to \infty }\frac { \left( 2^{ x }+2^{ -x } \right) }{\left( { 2 }^{ x }-2^{ -x } \right) } $
|
||||||
|
|
||||||
|
Beim nächsten Differenzieren würde sich wiederum nur das Operationszeichen umkehren, also + wird wieder - etc. In diesem Fall wird ausgeklammert:
|
||||||
|
|
||||||
|
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\cccancelto[red]{\ln (2)}{}\left( {{2^x} + {2^{ - x}}} \right)}}{{\bccancelto[red]{\ln (2)}{}\left( {{2^x} - {2^{ - x}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {{2^x} + {2^{ - x}}} \right)}}{{\left( {{2^x} - {2^{ - x}}} \right)}} $
|
||||||
|
|
||||||
|
Nun wird $2^x$ ausgeklammert.
|
||||||
|
|
||||||
|
|
||||||
|
Es gilt $\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}} \Rightarrow \frac{{{2^{ - x}}}}{{{2^x}}} = {2^{ - x - x}} = {2^{ - 2x}}$
|
||||||
|
|
||||||
|
und somit:
|
||||||
|
|
||||||
|
$ = \mathop {\lim }\limits_{x \to \infty } \frac{{{2^x}\left( {1 + {2^{ - 2x}}} \right)}}{{{2^x}\left( {1 - {2^{ - 2x}}} \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{1 + \cccancelto[blue]{2^{ - 2x}}{0}}{1 - \bccancelto[blue]{2^{ - 2x}}{0}}=\frac{1}{1}=1$
|
||||||
|
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
\pagebreak
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
%}\frac{\left(4-x\right)^{-\frac{1}{2}}}{\left(12+x\right)^{-\frac{1}{2}}}} = \lim \limits_{x \to 4^-}\frac{{{{\left( {12 + x} \right)}^{\frac{1}{2}}}}}{{{{\left( {4 - x} \right)}^{\frac{1}{2}}}}}$ \marginpar{wieder Anwendung der Kettenregel}
|
||||||
|
%fsdfsdfsdfsd
|
||||||
|
%\marginpar{
|
||||||
|
%\begin{wrapfigure}{l}{}
|
||||||
|
%\centering
|
||||||
|
%\resizebox{!}{.99\marginparwidth}{\input{gnuplot04.gp}}
|
||||||
|
%\caption{$\frac{\sqrt{4-x}}{\sqrt{12+x}-4}$}%
|
||||||
|
%\end{wrapfigure}
|
||||||
|
%}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
%$\left\{
|
||||||
|
%\begin{tabular}{p{.8\textwidth}}
|
||||||
|
%\begin{itemize}
|
||||||
|
%\item Second line
|
||||||
|
%\item Third line, which is quite long and seemingly tedious in the extreme
|
||||||
|
%\item Fourth line, which isn't as long as the third
|
||||||
|
%\end{itemize}
|
||||||
|
%\end{tabular}
|
||||||
|
%\right.$
|
||||||
|
|
||||||
|
%$\xlimes{ 2 }{ \frac{ x^2 }{ x } }$
|
||||||
|
|
||||||
|
\subsubsection{Beispiele zu $0\cdot\infty$}
|
||||||
|
|
||||||
|
\begin{enumerate}
|
||||||
|
\item $\mathop {\lim }\limits_{x \to \infty } \frac{1}{x}\ln \left( {2x} \right)$
|
||||||
|
|
||||||
|
Hier tritt beim Einsetzen von $\infty$ geht $\frac{1}{\infty}$ gegen $0$ und $\ln\left(2x\right)$ gegen $\infty$.
|
||||||
|
|
||||||
|
Umwandeln der Funktion zu einem Bruch, damit man den Fall $\frac{\infty}{\infty}$ bzw. $\frac{0}{0}$ erhält und somit die Regel von L'Hospital anwenden kann. Somit
|
||||||
|
\marginpar{$f(x)=\ln{x}$
|
||||||
|
|
||||||
|
$f'(x)=\frac{1}{x}$}
|
||||||
|
$\mathop {\lim }\limits_{x \to \infty } \frac{\ln \left( 2x\right)}{x}\Rightarrow \frac{\infty}{\infty}\Rightarrow \mathop {\lim }\limits_{x\to \infty }\frac { \frac { 1 }{ \cccancelto[red]{2}{}x } \cccancelto[red]{2}{} }{ 1 } = \mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0 $
|
||||||
|
|
||||||
|
|
||||||
|
\item $\mathop {\lim }\limits_{ x\to { 0 }^{ + } }x\cdot \ln { \left( 2x \right) } $
|
||||||
|
|
||||||
|
Hier ist $x=0$ und $\ln{\left(2\cdot 0^+\right)} = -\infty$ \marginpar{siehe wichtige Grenzwerte\ref{lbl:MerkblattGrenzwert}}
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{ x\to { 0 }^{ + } }\frac { \ln { \left( 2x \right) } }{ \frac { 1 }{ x } } = \frac{\infty}{\infty} \Rightarrow $\marginpar{$\frac{1}{x}=x^{-1}=-x^{-2}=-\frac{1}{x2}$} $\frac{\frac{1}{\cccancelto[red]{2}{}x}\cccancelto[red]{2}{}}{-\frac{1}{x^2}}=\frac{x^2}{-x}=x=0$
|
||||||
|
|
||||||
|
\marginpar{$\frac{1}{\frac{1}{x}}=\frac{1}{x^{-1}}$}
|
||||||
|
\item $\mathop {\lim }\limits_{x\to\infty}X^3\cdot e^{-2x}\Rightarrow $ \marginpar{$\infty^3=\infty$ und $e^{-2\infty}=0$}
|
||||||
|
|
||||||
|
Da: $e^{-2x}=\frac{1}{e^{2x}}\Rightarrow \mathop {\lim }\limits_{x\to\infty}\frac{x^3}{e^{2x}}=\mathop {\lim }\limits_{x\to\infty}\frac{3x^2}{e^{2x}}\Rightarrow \frac{\infty}{\infty}\Rightarrow \mathop {\lim }\limits_{x\to\infty}\frac{6x}{2e^{2x}}= \mathop {\lim }\limits_{x\to\infty}\frac{6}{8e^{2x}}=0$ \marginpar{$\frac{6}{8e^{2\cdot\infty}}=0$}
|
||||||
|
|
||||||
|
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
\subsubsection{Beispiele zu $0^0$, $\infty^0$ und $1^\infty$}
|
||||||
|
|
||||||
|
\begin{description}
|
||||||
|
|
||||||
|
\item[1.] $\mathop {\lim }\limits_{x\to 0^{+}} \tikz[na]\node [coordinate, xshift=1mm,yshift=1mm] (n11){}; x^{-3 \tikz[na]\node [coordinate,yshift=1mm] (n21) {};x}=$ $0^{-3\cdot 0}\Rightarrow 0^0$ Umformen durch das Hinzufügen der Eulerschen Zahl $e$ als Basis, der alte Exponent hier $-\tikz[na]\node [coordinate, xshift=1mm,yshift=1mm] (n22) {}; 3x$ bleibt und wird um den natürlichen Logarithmus der eigentlichen Basis ($\tikz[na]\node [coordinate, xshift=1mm,yshift=1mm] (n12) {}; x$) erweitert:
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x\to 0^+}e^{-\tikz[na]\node [coordinate, xshift=1mm,yshift=-1mm] (n23) {};3x\ln\left(\tikz[na]\node [coordinate,yshift=-1mm] (n13) {}; x\right)}$
|
||||||
|
|
||||||
|
Jetzt nur der Exponent betrachtet:
|
||||||
|
|
||||||
|
\begin{equation*}\mathop{\lim }\limits_{ x\to 0^+ } x\cdot \left(\ln\left(\tikz[na]\node [coordinate,yshift=-2mm, xshift=-1mm] (n31) {};x\right)\right)\Rightarrow 0 \cdot \infty \Rightarrow \mathop{\lim }\limits_{ x\to 0^+ } \frac{\ln\left(x\right)}{\frac{1}{x}} \mathop \Rightarrow \limits^{\frac{\infty }{\infty }} \mathop = \limits^{L'H} \mathop{\lim }\limits_{ x\to 0^+ }\frac{\frac{1}{x}}{-\frac{1}{x^2}}=-\frac{x^{\cccancelto[red]{2}{}}}{\cccancelto[red]{x}{}}=x= {\color{red}\tikz[na]\node [coordinate, xshift=1mm,yshift=2mm] (n32) {};0}
|
||||||
|
\end{equation*}%\tikz[baseline, remember picture ]{\node[fill=blue!20,anchor=base] (t1) {$0$};}
|
||||||
|
|
||||||
|
|
||||||
|
\vspace{5mm}
|
||||||
|
|
||||||
|
Nun wird dieses Ergebnis in die Ausgangs-Exponentialgleichung eingesetzt:
|
||||||
|
|
||||||
|
\begin{equation*}
|
||||||
|
\mathop {\lim }\limits_{x \to {0^ + }} e^{-3 {\color{red}\ln\tikz[na]\node [coordinate, yshift=1.5mm] (n33) {};\left(x\right)}} = e^{-3 \cdot {\color{red}\tikz[na]\node [coordinate, yshift=1mm] (n34) {};0}} = e^0 = 1
|
||||||
|
\end{equation*}
|
||||||
|
|
||||||
|
\item[2.] $\mathop {\lim }\limits_{x \to 0 + } \sin {\left( x \right)^{\frac{1}{{\ln \left( x \right)}}}}\Rightarrow \sin
|
||||||
|
\left(
|
||||||
|
0
|
||||||
|
\right)=0 \text{ und } \frac{1}{\ln
|
||||||
|
\left(
|
||||||
|
0^+
|
||||||
|
\right)}=\frac{1}{-\infty}=0$
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 0 + } e^{\frac{1}{\ln\left(x\right)}\cdot \ln\left(\sin\left(x\right)\right)}=\mathop {\lim }\limits_{x \to 0 + } e^{\frac{\ln
|
||||||
|
\left(
|
||||||
|
\sin
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)
|
||||||
|
\right)}{\ln
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}}$
|
||||||
|
\marginpar{Man kann einzelne Teile aus dem kritischen Fall $\frac{0}{0}$ rauslösen, wenn man erstens nichts verändert und man zweitens keinen neuen kritischen Fall erzeugt. Da hier der $\cos\left(0\right)=1$ ist, kann dieser rausgelöst werden. }
|
||||||
|
|
||||||
|
\textbf{Nebenrechnung}:
|
||||||
|
$\mathop {\lim }\limits_{x \to 0 + }\frac{\ln
|
||||||
|
\left(
|
||||||
|
\sin
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)
|
||||||
|
\right)}{\ln
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}=\mathop {\lim }\limits_{x \to 0 + }\frac{\frac{1}{\sin
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}\cdot\cos
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}{\frac{1}{x}}=\frac{\frac{\cos
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}{\sin
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}}{\frac{1}{x}}=\frac{x\cdot \cos\left(x\right)}{\sin\left(x\right)}$
|
||||||
|
|
||||||
|
|
||||||
|
$\underbrace {\mathop {\lim }\limits_{x \to 0 + }\cos
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}_{=1}\cdot\mathop {\lim }\limits_{x \to 0 + }\frac{x}{\sin
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}\mathop = \limits^{\frac{0}{0}}=\mathop {\lim }\limits_{x \to 0 + }\frac{1}{\cos
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}=1$
|
||||||
|
|
||||||
|
Dieses Ergebnis wird nun wieder in die umgeformte Ausgangsgleichung eingesetzt:
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to 0 + } e^{\frac{\ln
|
||||||
|
\left(
|
||||||
|
\sin
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)
|
||||||
|
\right)}{\ln
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}}=e^1=\underline{\underline{e}}$
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\end{description}
|
||||||
|
|
||||||
|
\begin{tikzpicture}[overlay]
|
||||||
|
\path[-latex, color=blue!40, line width=0.5mm, opacity=0.5] (n11) edge [bend left=70] (n12);
|
||||||
|
\path[-latex, color=red!40, line width=0.5mm, opacity=0.5] (n21) edge [bend left=70] (n22);
|
||||||
|
\path[-latex, color=red!40, line width=0.5mm, opacity=0.5] (n22) edge [bend left=70] (n23);
|
||||||
|
\path[-latex, color=blue!40, line width=0.5mm, opacity=0.5] (n12) edge [bend left=70] (n13);
|
||||||
|
|
||||||
|
\path[-latex, color=red!80, line width=0.2mm, opacity=0.5](n31) edge [bend right] (n33);
|
||||||
|
\path[-latex, color=red!80, line width=0.2mm, opacity=0.5] (n32) edge [bend angle =-45, bend left] (n34);
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
\newpage
|
||||||
|
|
||||||
|
\begin{description}
|
||||||
|
\item[3.] $\mathop {\lim }\limits_{x \to 1} \sqrt[{1 - x}]{x}=\mathop {\lim }\limits_{x \to 1}x^{\frac{1}{1-x}}=\mathop {\lim }\limits_{x \to 1}e^{\frac{1}{1-x}\cdot \ln\left(x\right)}=e^{-\tikz[na]\node [coordinate, yshift=1mm] (n42) {};1}$
|
||||||
|
|
||||||
|
|
||||||
|
\textbf{Nebenrechnung:} $\mathop {\lim }\limits_{x \to 1}\frac{\ln
|
||||||
|
\left(
|
||||||
|
x
|
||||||
|
\right)}{1-x}= \mathop = \limits^{\frac{\infty }{\infty }} = \mathop {\lim }\limits_{x \to 1} \frac{\frac{1}{x}}{-1}=-\tikz[na]\node [coordinate, yshift=1.5mm] (n41) {};1$
|
||||||
|
|
||||||
|
\item[4.]\marginpar{\raggedright{\color{red}$\frac{1}{x} = {x^{ - 1}} \left( {{x^{ - 1}}} \right)' = - {x^{ - 2}} = - \frac{1}{{{x^2}}}$
|
||||||
|
|
||||||
|
\color{blue}$1+\frac{a}{x}=\left(1+ax^{-1}\right)'=-ax^{-2}=-\frac{a}{x^2} $}
|
||||||
|
}
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to \infty}\left(1+\frac{a}{x}\right)^x =\mathop {\lim }\limits_{x \to \infty}e^{x\cdot\ln\left(1+\frac{a}{x}\right)}=e^{\tikz[na]\node [coordinate, xshift=1mm, yshift=1mm] (n52){};a}$
|
||||||
|
|
||||||
|
\textbf{Nebenrechnung:} $\mathop {\lim }\limits_{x \to \infty}x\cdot\ln
|
||||||
|
\left(
|
||||||
|
1+\frac{a}{x}
|
||||||
|
\right)=\mathop {\lim }\limits_{x \to \infty } \frac{{1 + \frac{a}{x}}}{{\frac{1}{x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{1}{{1 + \frac{a}{}}} \cdot \left( { - \frac{a}{{{x^2}}}} \right)}}{{ - \frac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - \frac{a}{{{x^2} \cdot 1 + \frac{a}{x}}}}}{{ - \frac{1}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to \infty } - \frac{a}{\cccancelto[red]{x^2}{} \cdot 1 + \frac{a}{x}} \cdot -\cccancelto[red]{x^2}{} = \mathop {\lim }\limits_{x \to \infty } \frac{a}{{1 + \frac{a}{x}}}$ Einsetzen von $\infty$ in $x$ $\frac{a}{{1 + \frac{a}{\infty }}} = \frac{a}{{1 + 0}} = \tikz[na]\node [coordinate, yshift=1.5mm,xshift=1mm] (n51){};a$
|
||||||
|
|
||||||
|
\item[5.] $\mathop {\lim }\limits_{x \to \infty }
|
||||||
|
\left(10+x\right)^{\frac{2}{x}}=\mathop {\lim }\limits_{x \to \infty } e^{\frac{2}{x}\cdot\ln\left(10+x\right)}=e^0=1$
|
||||||
|
|
||||||
|
\textbf{Nebenrechnung:} $\mathop {\lim }\limits_{x \to \infty }\frac{2\cdot
|
||||||
|
\ln\left(
|
||||||
|
10+x
|
||||||
|
\right)}{x}=\frac{2\cdot\frac{1}{10+x}\cdot 1}{1}=\frac{\frac{2}{10+x}}{1}=\frac{2\cdot\frac{1}{\infty}}{1}=\frac{0}{1}=0$
|
||||||
|
|
||||||
|
\item[6.]$\mathop {\lim }\limits_{x \to \infty } \left( e^{5x}-4x \right)^{\frac{1}{x}}=\mathop {\lim }\limits_{x \to \infty }e^{\frac{1}{x}\ln\left(e^{5x}-4x\right)} = e^{\textbf{\textcolor{orange}{5}}}$
|
||||||
|
|
||||||
|
|
||||||
|
Nebenrechnung: $\mathop {\lim }\limits_{x \to \infty }\frac{1}{x}\ln\left(e^{5x}-4x\right)=\mathop {\lim }\limits_{x \to \infty }\frac{\ln\left(e^{5x}-4x\right)}{x}=\mathop {\lim }\limits_{x \to \infty }\frac{1}{e^{5x}-4x}\cdot 5e^{5x}-4=\mathop {\lim }\limits_{x \to \infty}\frac{5e^{5x}-4}{e^{5x}-4x}=\mathop {\lim }\limits_{x \to \infty}{\frac{25e^{5x}}{5e^{5x}-4}}=\frac{125e^{5x}}{25e^{5x}}=\textbf{\textcolor{orange}{5}}$
|
||||||
|
|
||||||
|
\end{description}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\begin{tikzpicture}[overlay]
|
||||||
|
\path[-latex, color=green!40!black, line width=0.5mm, opacity=0.5] (n41) edge [bend left=-70] (n42);
|
||||||
|
\path[-latex, color=yellow!40!black, line width=0.5mm, opacity=0.5] (n51) edge [bend left=-50] (n52);
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
\subsubsection{Beispiele zu $\infty - \infty$}
|
||||||
|
|
||||||
|
\begin{enumerate}
|
||||||
|
\marginpar{Im ersten Term wird im Zähler die 0 postiv, da $1^+$ und somit geht der Term gegen $\infty$}\item $\mathop {\lim }\limits_{x \to 1^+}\frac{1}{x-1}-\frac{1}{\ln\left(x\right)}$
|
||||||
|
|
||||||
|
Bilden eines Hauptnenners: $\frac{\ln\left(x\right)-\left(x-1\right)}{\left(x-1\right)\ln\left(x\right)}$ Dadurch entsteht der Fall $\frac{0}{0}$
|
||||||
|
|
||||||
|
|
||||||
|
\marginpar{$\left(u\cdot v\right)'=u'\cdot v+u\cdot v'$}$\mathop {\lim }\limits_{x \to 1^+}\frac{\ln\left(x\right)-\left(x-1\right)}{\left(x-1\right)\ln\left(x\right)}=\mathop {\lim }\limits_{x \to 1^+}\frac{\frac{1}{x}-1}{1\cdot\ln\left(x\right)+\left(x-1\right)\cdot\frac{1}{x}}=\mathop {\lim }\limits_{x \to 1^+}\frac{-\frac{1}{x^2}}{\frac{1}{x} + \left( { - \frac{1}{{{x^2}}}} \right)\left( {x - 1} \right) + \frac{1}{x}}$
|
||||||
|
|
||||||
|
Einsetzen der $1$
|
||||||
|
|
||||||
|
$\frac{-1}{1-1\cdot 0 +1}=-\frac{1}{2}$
|
||||||
|
|
||||||
|
\marginpar{Über dritte binomische Formel: $a-b=\frac{a^2-b^2}{a+b}$ also $\left(a-b\right)\left(a+b\right)=a^2-b^2$}
|
||||||
|
\item $\mathop {\lim }\limits_{x \to \infty} x-\sqrt{x^2+1}=\mathop {\lim }\limits_{x \to \infty}\underbrace x_{\textcolor{red}{a}} - \underbrace {\sqrt {{x^2} - 1} }_{\textcolor{red}{b}}=\mathop {\lim }\limits_{x \to \infty}\frac{x^2-\left(x^2+1\right)}{x+\sqrt{x^2+1}}=\mathop {\lim }\limits_{x \to \infty}\frac{x^2-\left(x^2+1\right)}{x+\sqrt{x^2-1}}=\mathop {\lim }\limits_{x \to \infty}\frac{-1}{x+\sqrt{x^2-1}}=-\frac{1}{\infty}=0$
|
||||||
|
|
||||||
|
\item \marginpar{Zum Auflösen der Wurzel $a - b = \frac{{{a^3} - {b^3}}}{{{a^2} + ab + {b^3}}}$}$\mathop {\lim }\limits_{x \to \infty } x - \sqrt[3]{{{x^3} - {x^2}}}=\mathop {\lim }\limits_{x \to \infty } \underbrace x_a - \underbrace {\sqrt[3]{{{x^3} - {x^2}}}}_b=\mathop {\lim }\limits_{x \to \infty }\frac{{{x^3} - {{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^3}}}{{{x^2} + x\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right) + {{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^2}}} =\mathop {\lim }\limits_{x \to \infty }\frac{{{x^3} - \left( {{x^3} - {x^2}} \right)}}{{{x^2} + x\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right) + {{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^2}}}$
|
||||||
|
|
||||||
|
Hier muss nun der am stärksten wachsende Term gesucht werden:
|
||||||
|
Zweiter Term des Nenners: $x\left( \sqrt[3]{x^3 - x^2} \right)$ hier würde beim Einsetzen von $\infty$ der Term $x\left( \sqrt[3]{x^3} \right)$ übrig, welcher sich dann wiederum auf $x\cdot x$ also $x^2$ reduziert.
|
||||||
|
|
||||||
|
Der dritte Term ${{{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^2}}$ reduziert sich bei Betrachtung in der Unendlichkeit auf $\left( \sqrt[3]{x^3} \right)^2$. Da die dritte Wurzel aus $x^3 = x$ ist, bleibt $x^2$ übrig. Somit wachsen alle Terme im Nenner so schnell wie $x^2$.
|
||||||
|
|
||||||
|
Im Zähler bleibt da $x^3 - \left( x^3 - x^2 \right)$ nur $-x^2$ übrig, so also auch hier wächst alles so schnell wie $x^2$.
|
||||||
|
|
||||||
|
Das bedeutet das man jetzt den Faktor $x^2$ ausklammert:
|
||||||
|
|
||||||
|
\begin{itemize}
|
||||||
|
\item Term 2 des Nenners:
|
||||||
|
$\frac{x\left(\sqrt[3]{x^3 - x^2} \right)}{x^2} = \frac{\sqrt[3]{x^3 - x^2}}{x}$
|
||||||
|
|
||||||
|
Um nun die Nenner $x$ in die Wurzel zu bekommen muss dieser hoch 3 genommen werden, somit ergibt sich $\frac{\sqrt[3]{x^3 - x^2}}{x^3} = \sqrt[3]{\frac{x^3 - x^2}{x^3}}$
|
||||||
|
|
||||||
|
Nebenrechnung: $\frac{x^3 - x^2}{x^3} = \frac{x^2\left( x - 1 \right)}{x^2 \cdot x} = \frac{x - 1}{x} = 1 - \frac{1}{x}$
|
||||||
|
|
||||||
|
$=\sqrt[3]{1 - \frac{1}{x}}$
|
||||||
|
|
||||||
|
\item Term 3 des Nenners: $\frac{\left( \sqrt[3]{x^3 - x^2} \right)^2}{x^2} = {\left( {\frac{{\sqrt[3]{{{x^3} - {x^2}}}}}{x}} \right)^2} = {\left( {\sqrt[3]{{\frac{{{x^3} - {x^2}}}{{{x^3}}}}}} \right)^2} = {\left( {\sqrt[3]{{1 - \frac{1}{x}}}} \right)^2}$
|
||||||
|
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
Die nun in die Ausgangsfunktion eimsetzen:
|
||||||
|
|
||||||
|
$\mathop {\lim }\limits_{x \to \infty} \frac{x^2}{x^2\left(1+\sqrt[3]{1+\frac{1}{x}}+\left(\sqrt[3]{1+\frac{1}{x}}\right)^2\right)}=\mathop {\lim }\limits_{x \to \infty} \frac{1}{1+\sqrt[3]{1-\frac{1}{x}}+\left(\sqrt[3]{1-\frac{1}{x}}\right)^2}$
|
||||||
|
|
||||||
|
Da sich im Zähler nun kein $x$ mehr befindet, wird jetzt im Nenner $\infty$ eingesetzt. Alle $\frac{1}{x}$ werden $0$. Somit $\frac{1}{1+1+1}=\frac{1}{3}$
|
||||||
|
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
%\includegraphics[scale=0.45]{gnuplot08a.png}
|
||||||
|
|
||||||
|
%\begin{pycode}
|
||||||
|
%print ('Hello, \LaTeX')
|
||||||
|
%\end{pycode}
|
||||||
|
|
||||||
|
%https://www.studimup.de/abitur/analysis/grenzwerte/
|
||||||
52
limit01a.pgf
Normal file
52
limit01a.pgf
Normal file
@@ -0,0 +1,52 @@
|
|||||||
|
\begin{tikzpicture}[scale=0.7]
|
||||||
|
|
||||||
|
\pgfplotsset{compat=1.11}
|
||||||
|
|
||||||
|
\definecolor{FireBrick}{rgb}{0.7, 0.13, 0.13}
|
||||||
|
|
||||||
|
\definecolor{NewBlue}{rgb}{0.27, 0.45, 0.76}
|
||||||
|
|
||||||
|
\tikzfading[name=arrowfading, top color=transparent!0, bottom color=transparent!95]
|
||||||
|
\tikzset{arrowfill/.style={#1,general shadow={fill=black, shadow yshift=-0.8ex, path fading=arrowfading}}}
|
||||||
|
\tikzset{arrowstyle/.style n args={3}{draw=#2,arrowfill={#3}, single arrow,minimum height=#1, single arrow,
|
||||||
|
single arrow head extend=.3cm,}}
|
||||||
|
|
||||||
|
%\NewDocumentCommand{\tikzfancyarrow}{O{2cm} O{FireBrick} O{top color=orange!20!red, bottom color=red} m}{
|
||||||
|
%\tikz[baseline=-0.5ex]\node [arrowstyle={#1}{#2}{#3}] {#4};
|
||||||
|
%}
|
||||||
|
|
||||||
|
|
||||||
|
%\node [
|
||||||
|
% fill=blue!50, draw,
|
||||||
|
% single arrow, single arrow head indent=0ex,
|
||||||
|
% rotate=0,
|
||||||
|
% font=\sffamily
|
||||||
|
%] at (1,1.5)
|
||||||
|
%{\rotatebox{0}{ \qquad}};
|
||||||
|
|
||||||
|
|
||||||
|
%\draw[color=gray!10,step=2mm,help lines] (-0.7,0) grid (72mm,58mm);
|
||||||
|
%\draw[color=gray!70,step=10mm,xshift=4mm,yshift=-1mm] (-0.5,0) grid (70mm,60mm);
|
||||||
|
\begin{axis}[
|
||||||
|
x=1cm,y=1cm,
|
||||||
|
axis x line=center,
|
||||||
|
axis y line=center,
|
||||||
|
%axis lines=middle,
|
||||||
|
ymajorgrids=true,
|
||||||
|
xmajorgrids=true,
|
||||||
|
xmin=-5,
|
||||||
|
xmax=5,
|
||||||
|
ymin=-6,
|
||||||
|
ymax=5,
|
||||||
|
xtick={-4,-3,...,4},
|
||||||
|
ytick={-5,-4,...,4},]
|
||||||
|
\addplot [mark=none,domain=-4.8:-1.05, color=NewBlue, line width=0.5mm,step=10000, smooth, tension=0.2] {1/(x^2-1)};
|
||||||
|
\addplot [mark=none,domain=-0.95:0.95, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
|
||||||
|
\addplot [mark=none,domain=1.05:4.8, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
|
||||||
|
|
||||||
|
%\clip(-17.083986586441775,-20.54798056618339) rectangle (4.103328404466779,7.349328615703948);
|
||||||
|
\end{axis}
|
||||||
|
\draw[-latex,red, line width=0.75mm](6.15,2.5) node[above, right, yshift=0.5mm] {von rechts} -- (4.15,2.5) ;
|
||||||
|
%\node[] (A) at ( 1,3) {\textbf{$y^2=x^2$}};
|
||||||
|
|
||||||
|
\end{tikzpicture}
|
||||||
72
limit01b.pgf
Normal file
72
limit01b.pgf
Normal file
@@ -0,0 +1,72 @@
|
|||||||
|
%!tikz editor 1.0
|
||||||
|
\documentclass{article}
|
||||||
|
\usepackage{tikz}
|
||||||
|
\usepackage[graphics, active, tightpage]{preview}
|
||||||
|
\PreviewEnvironment{tikzpicture}
|
||||||
|
|
||||||
|
%!tikz preamble begin
|
||||||
|
\usepackage{pgf,tikz,pgfplots}
|
||||||
|
\usetikzlibrary{fadings,shapes.arrows,shadows}
|
||||||
|
\usetikzlibrary{arrows.meta}
|
||||||
|
%!tikz preamble end
|
||||||
|
|
||||||
|
|
||||||
|
\begin{document}
|
||||||
|
%!tikz source begin
|
||||||
|
\begin{tikzpicture}
|
||||||
|
\begin{tikzpicture}[scale=0.7]
|
||||||
|
|
||||||
|
\pgfplotsset{compat=1.11}
|
||||||
|
|
||||||
|
\definecolor{FireBrick}{rgb}{0.7, 0.13, 0.13}
|
||||||
|
|
||||||
|
\definecolor{NewBlue}{rgb}{0.27, 0.45, 0.76}
|
||||||
|
|
||||||
|
\tikzfading[name=arrowfading, top color=transparent!0, bottom color=transparent!95]
|
||||||
|
\tikzset{arrowfill/.style={#1,general shadow={fill=black, shadow yshift=-0.8ex, path fading=arrowfading}}}
|
||||||
|
\tikzset{arrowstyle/.style n args={3}{draw=#2,arrowfill={#3}, single arrow,minimum height=#1, single arrow,
|
||||||
|
single arrow head extend=.3cm,}}
|
||||||
|
|
||||||
|
%\NewDocumentCommand{\tikzfancyarrow}{O{2cm} O{FireBrick} O{top color=orange!20!red, bottom color=red} m}{
|
||||||
|
%\tikz[baseline=-0.5ex]\node [arrowstyle={#1}{#2}{#3}] {#4};
|
||||||
|
%}
|
||||||
|
|
||||||
|
|
||||||
|
%\node [
|
||||||
|
% fill=blue!50, draw,
|
||||||
|
% single arrow, single arrow head indent=0ex,
|
||||||
|
% rotate=0,
|
||||||
|
% font=\sffamily
|
||||||
|
%] at (1,1.5)
|
||||||
|
%{\rotatebox{0}{ \qquad}};
|
||||||
|
|
||||||
|
|
||||||
|
%\draw[color=gray!10,step=2mm,help lines] (-0.7,0) grid (72mm,58mm);
|
||||||
|
%\draw[color=gray!70,step=10mm,xshift=4mm,yshift=-1mm] (-0.5,0) grid (70mm,60mm);
|
||||||
|
\begin{axis}[
|
||||||
|
x=1cm,y=1cm,
|
||||||
|
axis x line=center,
|
||||||
|
axis y line=center,
|
||||||
|
%axis lines=middle,
|
||||||
|
ymajorgrids=true,
|
||||||
|
xmajorgrids=true,
|
||||||
|
xmin=-5,
|
||||||
|
xmax=5,
|
||||||
|
ymin=-6,
|
||||||
|
ymax=5,
|
||||||
|
xtick={-4,-3,...,4},
|
||||||
|
ytick={-5,-4,...,4},]
|
||||||
|
\addplot [mark=none,domain=-4.8:-1.05, color=NewBlue, line width=0.5mm,step=10000, smooth, tension=0.2] {1/(x^2-1)};
|
||||||
|
\addplot [mark=none,domain=-0.95:0.95, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
|
||||||
|
\addplot [mark=none,domain=1.05:4.8, color=NewBlue, line width=0.5mm,step=10000] {1/(x^2-1)};
|
||||||
|
|
||||||
|
%\clip(-17.083986586441775,-20.54798056618339) rectangle (4.103328404466779,7.349328615703948);
|
||||||
|
\end{axis}
|
||||||
|
\draw[-latex,red, line width=0.75mm](3.9,2.5) node[above, left, yshift=0.5mm] {von links} -- (5.9,2.5) ;
|
||||||
|
%\node[] (A) at ( 1,3) {\textbf{$y^2=x^2$}};
|
||||||
|
|
||||||
|
\end{tikzpicture}
|
||||||
|
\end{tikzpicture}
|
||||||
|
%!tikz source end
|
||||||
|
|
||||||
|
\end{document}
|
||||||
169
main.tex
Normal file
169
main.tex
Normal file
@@ -0,0 +1,169 @@
|
|||||||
|
|
||||||
|
%!TEX root=main.tex
|
||||||
|
%\documentclass{tufte-book}
|
||||||
|
\documentclass[fleqn, a4paper,12pt]{scrbook}
|
||||||
|
|
||||||
|
%\hypersetup{colorlinks}% uncomment this line if you prefer colored hyperlinks (e.g., for onscreen viewing)
|
||||||
|
|
||||||
|
%%
|
||||||
|
% Book metadata
|
||||||
|
\title{Formeln und Notizen}
|
||||||
|
%\author[]{}
|
||||||
|
%\publisher{Ich}
|
||||||
|
|
||||||
|
%%
|
||||||
|
% If they're installed, use Bergamo and Chantilly from www.fontsite.com.
|
||||||
|
% They're clones of Bembo and Gill Sans, respectively.
|
||||||
|
%\IfFileExists{bergamo.sty}{\usepackage[osf]{bergamo}}{}% Bembo
|
||||||
|
%\IfFileExists{chantill.sty}{\usepackage{chantill}}{}% Gill Sans
|
||||||
|
|
||||||
|
\input{definitions.tex}
|
||||||
|
|
||||||
|
|
||||||
|
%%% Local Variables:
|
||||||
|
%%% TeX-master: "master"
|
||||||
|
%%% End:
|
||||||
|
|
||||||
|
|
||||||
|
\begin{document}
|
||||||
|
\tikzstyle{every picture}+=[remember picture]
|
||||||
|
% % Front matter
|
||||||
|
% \frontmatter
|
||||||
|
%
|
||||||
|
% % r.1 blank page
|
||||||
|
% \blankpage
|
||||||
|
%
|
||||||
|
% % v.2 epigraphs
|
||||||
|
%
|
||||||
|
%
|
||||||
|
% % r.3 full title page
|
||||||
|
% \maketitle
|
||||||
|
%
|
||||||
|
%
|
||||||
|
% % v.4 copyright page
|
||||||
|
% \newpage
|
||||||
|
% \begin{fullwidth}
|
||||||
|
% ~\vfill
|
||||||
|
% \thispagestyle{empty}
|
||||||
|
% \setlength{\parindent}{0pt}
|
||||||
|
% \setlength{\parskip}{\baselineskip}
|
||||||
|
% Copyright \copyright\ \the\year\ \thanklessauthor
|
||||||
|
%
|
||||||
|
% \par\smallcaps{Published by \thanklesspublisher}
|
||||||
|
%
|
||||||
|
% \par\smallcaps{tufte-latex.googlecode.com}
|
||||||
|
%
|
||||||
|
% \par Licensed under the Apache License, Version 2.0 (the ``License''); you may not
|
||||||
|
% use this file except in compliance with the License. You may obtain a copy
|
||||||
|
% of the License at \url{http://www.apache.org/licenses/LICENSE-2.0}. Unless
|
||||||
|
% required by applicable law or agreed to in writing, software distributed
|
||||||
|
% under the License is distributed on an \smallcaps{``AS IS'' BASIS, WITHOUT
|
||||||
|
% WARRANTIES OR CONDITIONS OF ANY KIND}, either express or implied. See the
|
||||||
|
% License for the specific language governing permissions and limitations
|
||||||
|
% under the License.\index{license}
|
||||||
|
%
|
||||||
|
% \par\textit{First printing, \monthyear}
|
||||||
|
% \end{fullwidth}
|
||||||
|
|
||||||
|
% r.5 contents
|
||||||
|
|
||||||
|
|
||||||
|
%\tableofcontents
|
||||||
|
|
||||||
|
%\listoffigures
|
||||||
|
|
||||||
|
% \listoftables
|
||||||
|
|
||||||
|
% r.7 dedication
|
||||||
|
\cleardoublepage
|
||||||
|
%~\vfill
|
||||||
|
%\begin{doublespace}
|
||||||
|
%\noindent\fontsize{18}{22}\selectfont\itshape
|
||||||
|
%\nohyphenation
|
||||||
|
%Dedicated to those who appreciate \LaTeX{}
|
||||||
|
%and the work of \mbox{Edward R.~Tufte}
|
||||||
|
%and \mbox{Donald E.~Knuth}.
|
||||||
|
%\end{doublespace}
|
||||||
|
%\vfill
|
||||||
|
%\vfill
|
||||||
|
|
||||||
|
|
||||||
|
% r.9 introduction
|
||||||
|
\cleardoublepage
|
||||||
|
%\chapter*{Introduction}
|
||||||
|
%
|
||||||
|
%
|
||||||
|
%
|
||||||
|
%This sample book discusses the design of Edward Tufte's
|
||||||
|
%books\cite{Tufte2001,Tufte1990,Tufte1997,Tufte2006}
|
||||||
|
%and the use of the \doccls{tufte-book} and \doccls{tufte-handout} document classes.
|
||||||
|
%
|
||||||
|
%
|
||||||
|
%%%
|
||||||
|
%% Start the main matter (normal chapters)
|
||||||
|
\mainmatter
|
||||||
|
|
||||||
|
|
||||||
|
% \begin{asydef}
|
||||||
|
% // Global Asymptote definitions can be put here.
|
||||||
|
% import three;
|
||||||
|
% usepackage("bm");
|
||||||
|
%texpreamble("\def\V#1{\bm{#1}}");
|
||||||
|
% // One can globally override the default toolbar Settings here:
|
||||||
|
% // settings.toolbar=true;
|
||||||
|
% \end{asydef}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
% \begin{figure}
|
||||||
|
% \begin{framebox}
|
||||||
|
%
|
||||||
|
% \begin{asy}
|
||||||
|
% label("Hallo Welt");
|
||||||
|
%
|
||||||
|
% unitsize(3cm);
|
||||||
|
% size(4cm,4cm);
|
||||||
|
% import graph;
|
||||||
|
%
|
||||||
|
% real f(real x) {
|
||||||
|
% return sqrt(2*x - x^3);
|
||||||
|
% }
|
||||||
|
% draw((-4,0) -- (3,0), arrow=Arrow(HookHead));
|
||||||
|
% draw((0,-.1) -- (0,2), arrow=Arrow(HookHead));
|
||||||
|
% path g = graph(f, -3, 1.4142);
|
||||||
|
% draw(g);
|
||||||
|
% \end{asy}
|
||||||
|
% \end{framebox}
|
||||||
|
% \end{figure}
|
||||||
|
|
||||||
|
\input{pearson_Funktionen_und_ihre_Graphen.tex}
|
||||||
|
|
||||||
|
% \input{integral01.tex}
|
||||||
|
|
||||||
|
%\input{calculus_one.tex}
|
||||||
|
|
||||||
|
% \input{formeln.tex}
|
||||||
|
|
||||||
|
%\input{FHTW.tex}
|
||||||
|
|
||||||
|
%\section{Umrechnung rad - Degree und umgekehrt}
|
||||||
|
|
||||||
|
|
||||||
|
%\begin{marginfigure}
|
||||||
|
|
||||||
|
%\input{trigon01.tikz.tex}
|
||||||
|
|
||||||
|
%\end{marginfigure}
|
||||||
|
|
||||||
|
\backmatter
|
||||||
|
|
||||||
|
% \bibliography{sample-handout}
|
||||||
|
% \bibliographystyle{plainnat}
|
||||||
|
\bibliographystyle{plain}
|
||||||
|
\bibliography{formelbib}
|
||||||
|
|
||||||
|
|
||||||
|
\printindex
|
||||||
|
|
||||||
|
\end{document}
|
||||||
46
pearson0201.pgf
Normal file
46
pearson0201.pgf
Normal file
@@ -0,0 +1,46 @@
|
|||||||
|
\begin{tikzpicture}[scale=0.75]
|
||||||
|
\pgfplotsset{compat=1.11}
|
||||||
|
|
||||||
|
\definecolor{FireBrick}{rgb}{0.7, 0.13, 0.13}
|
||||||
|
|
||||||
|
\definecolor{NewBlue}{rgb}{0.27, 0.45, 0.76}
|
||||||
|
|
||||||
|
\tikzfading[name=arrowfading, top color=transparent!0, bottom color=transparent!95]
|
||||||
|
\tikzset{arrowfill/.style={#1,general shadow={fill=black, shadow yshift=-0.8ex, path fading=arrowfading}}}
|
||||||
|
\tikzset{arrowstyle/.style n args={3}{draw=#2,arrowfill={#3}, single arrow,minimum height=#1, single arrow,
|
||||||
|
single arrow head extend=.3cm,}}
|
||||||
|
|
||||||
|
%\NewDocumentCommand{\tikzfancyarrow}{O{2cm} O{FireBrick} O{top color=orange!20!red, bottom color=red} m}{
|
||||||
|
%\tikz[baseline=-0.5ex]\node [arrowstyle={#1}{#2}{#3}] {#4};
|
||||||
|
%}
|
||||||
|
|
||||||
|
|
||||||
|
%\node [
|
||||||
|
% fill=blue!50, draw,
|
||||||
|
% single arrow, single arrow head indent=0ex,
|
||||||
|
% rotate=0,
|
||||||
|
% font=\sffamily
|
||||||
|
%] at (1,1.5)
|
||||||
|
%{\rotatebox{0}{ \qquad}};
|
||||||
|
|
||||||
|
|
||||||
|
%\draw[color=gray!10,step=2mm,help lines] (-0.7,0) grid (72mm,58mm);
|
||||||
|
%\draw[color=gray!70,step=10mm,xshift=4mm,yshift=-1mm] (-0.5,0) grid (70mm,60mm);
|
||||||
|
\begin{axis}[
|
||||||
|
x=1cm,y=1cm,
|
||||||
|
axis lines=middle,
|
||||||
|
ymajorgrids=true,
|
||||||
|
xmajorgrids=true,
|
||||||
|
xmin=-5,
|
||||||
|
xmax=5,
|
||||||
|
ymin=-5,
|
||||||
|
ymax=5,
|
||||||
|
xtick={-5,-4,...,5},
|
||||||
|
ytick={-5,-4,...,5},]
|
||||||
|
%\clip(-17.083986586441775,-20.54798056618339) rectangle (4.103328404466779,7.349328615703948);
|
||||||
|
\draw [line width=1pt,color=orange,domain=-5:5] plot(\x,{(-0--1*\x)/1});
|
||||||
|
\draw [line width=1pt,color=orange,domain=-5:5] plot(\x,{(-0-1*\x)/1});
|
||||||
|
\end{axis}
|
||||||
|
|
||||||
|
\node[] (A) at ( 1,3) {\textbf{$y^2=x^2$}};
|
||||||
|
\end{tikzpicture}
|
||||||
807
pearson_Funktionen_und_ihre_Graphen.tex
Normal file
807
pearson_Funktionen_und_ihre_Graphen.tex
Normal file
@@ -0,0 +1,807 @@
|
|||||||
|
\chapter{Pearson Brückenkurs Mathematik}
|
||||||
|
\section{Beispiele zu Mengen}
|
||||||
|
\subsection{Mengenoperationen}
|
||||||
|
|
||||||
|
\subsubsection{Vereinigung $A\cup B$}
|
||||||
|
|
||||||
|
Die Vereinigungsmenge, ist die Menge aller Elemente, die sowohl zu $A$ oder zu $B$ oder zu beiden Mengen gehören. (Der gesamte Inhalt der Mengen $A$ und $B$)
|
||||||
|
|
||||||
|
\begin{tikzpicture}
|
||||||
|
% Set the colors and patterns
|
||||||
|
\begin{scope}
|
||||||
|
% First circle (set A)
|
||||||
|
\fill[pattern=north west lines, pattern color=blue!50] (-1,0) circle (1.5);
|
||||||
|
% Second circle (set B)
|
||||||
|
\fill[pattern=north east lines, pattern color=red!50] (1,0) circle (1.5);
|
||||||
|
\end{scope}
|
||||||
|
|
||||||
|
% Draw the circles' borders
|
||||||
|
\draw (-1,0) circle (1.5) node {$A$};
|
||||||
|
\draw (1,0) circle (1.5) node {$B$};
|
||||||
|
|
||||||
|
% Add a label for the union
|
||||||
|
\node at (0,-2) {$A \cup B$};
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
\subsubsection{Schnittmenge $A \cap B$}
|
||||||
|
Die Schnittmenge ist die Menge aller Elemente, die sowohl zu $A$ als auch zu $B$ gehören. (Gemeinsamkeiten)
|
||||||
|
|
||||||
|
\begin{tikzpicture}
|
||||||
|
% Set the colors and patterns
|
||||||
|
\begin{scope}
|
||||||
|
% First circle (set A) - light color outside intersection
|
||||||
|
\fill[pattern=north west lines, pattern color=blue!30] (-1,0) circle (1.5);
|
||||||
|
% Second circle (set B) - light color outside intersection
|
||||||
|
\fill[pattern=north east lines, pattern color=red!30] (1,0) circle (1.5);
|
||||||
|
|
||||||
|
% Intersection area color - darker to highlight the overlap
|
||||||
|
\begin{scope}
|
||||||
|
\clip (-1,0) circle (1.5);
|
||||||
|
\fill[blue!50, opacity=0.5] (1,0) circle (1.5);
|
||||||
|
\end{scope}
|
||||||
|
\begin{scope}
|
||||||
|
\clip (1,0) circle (1.5);
|
||||||
|
\fill[red!50, opacity=0.5] (-1,0) circle (1.5);
|
||||||
|
\end{scope}
|
||||||
|
\end{scope}
|
||||||
|
|
||||||
|
% Draw the circles' borders and labels in the middle of each circle
|
||||||
|
\draw (-1,0) circle (1.5) node {$A$};
|
||||||
|
\draw (1,0) circle (1.5) node {$B$};
|
||||||
|
|
||||||
|
% Add a label for the intersection
|
||||||
|
\node at (0,-2) {$A \cap B$};
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
|
||||||
|
\subsubsection{Differenzmenge $A\setminus B$}
|
||||||
|
\begin{tikzpicture}
|
||||||
|
% Set the colors and patterns
|
||||||
|
\begin{scope}
|
||||||
|
% First circle (set A) - lighter outside B
|
||||||
|
\fill[pattern=north west lines, pattern color=blue!50] (-1,0) circle (1.5);
|
||||||
|
|
||||||
|
% Second circle (set B) - no fill to exclude it from A
|
||||||
|
\fill[pattern=north east lines, pattern color=red!30] (1,0) circle (1.5);
|
||||||
|
|
||||||
|
% Exclude the intersection (make it white)
|
||||||
|
\begin{scope}
|
||||||
|
\clip (1,0) circle (1.5);
|
||||||
|
\fill[white] (-1,0) circle (1.5);
|
||||||
|
\end{scope}
|
||||||
|
\end{scope}
|
||||||
|
|
||||||
|
% Draw the circles' borders and labels in the middle of each circle
|
||||||
|
\draw (-1,0) circle (1.5) node {$A$};
|
||||||
|
\draw (1,0) circle (1.5) node {$B$};
|
||||||
|
|
||||||
|
% Add a label for the difference
|
||||||
|
\node at (0,-2) {$A \setminus B$};
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
\paragraph{Beispiel 1}\mbox{}\\
|
||||||
|
|
||||||
|
Gegeben sind die Mengen $A:=[1,5), B:=\{2,3,4\}$ und $C:=\{z \in \mathbb{Z} \mid-1 \leq z<3\}$.
|
||||||
|
|
||||||
|
Bestimme:
|
||||||
|
\begin{enumerate}[label=\alph*)]
|
||||||
|
\item $A \cup B$
|
||||||
|
\item $A \cap C$
|
||||||
|
\item $C \backslash B$
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
\begin{enumerate}[label=\alph*)]
|
||||||
|
\item Das "'`$\cup$"' - Symbol steht für die Vereinigung der beiden Mengen $A$ und $B$.
|
||||||
|
|
||||||
|
|
||||||
|
Bei der Vereinigung nimmt man die Elemente, die in $A$ oder $B$ liegen - gewissermaßen also einfach alle Elemente aus beiden Mengen.
|
||||||
|
|
||||||
|
|
||||||
|
In $A \cup B$ liegen also alle Elemente aus $A$ und $B$...Wie schreibt man das ordentlich auf? Am besten wäre es ja, wenn wir Elemente nur einmal auflisten \ldots
|
||||||
|
|
||||||
|
Das Ziel ist also $A \cup B$ so kompakt wie möglich aufzuschreiben. Man kann einfach
|
||||||
|
|
||||||
|
$$
|
||||||
|
A \cup B=[1,5) \cup\{2,3,4\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
schreiben.
|
||||||
|
|
||||||
|
Hier werden aber Elemente doppelt aufgezählt, weil die $3$ beispielsweise in $[1,5)$ und in $\{2,3,4\}$ auftauchen - und das ist nicht Sinn der Übung.
|
||||||
|
|
||||||
|
Um keine Elemente doppelt aufzuzählen muss man zum Beispiel schauen, welche Elemente aus $B$ schon in $A$ liegen. Die kann man dann beim Aufschreiben weg lassen.
|
||||||
|
$A$ ist ein Intervall und enthält also alle reellen Zahlen zwischen $1$ und $5$ (welche selbst nicht mit drin liegt). $2$, $3$ und $4$ sind nun aber alles Zahlen die dort drin sind.
|
||||||
|
|
||||||
|
Somit kommen durch $B$ gar keine neuen Elemente zu $A$ hinzu, da $B$ eine Teilmenge ist. Also ist die Vereinigung von $A$ und $B$ einfach wieder $$[1,5)$$.
|
||||||
|
|
||||||
|
\textbf{Grafische Lösung}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{tikzpicture}
|
||||||
|
% Draw the number line
|
||||||
|
\draw[->] (0,0) -- (6,0) node[anchor=north] {x}; % arrow at the end
|
||||||
|
|
||||||
|
% Draw ticks and labels
|
||||||
|
\foreach \x in {0,1,2,3,4,5}
|
||||||
|
\draw (\x,0.1) -- (\x,-0.1) node[below] {\x};
|
||||||
|
|
||||||
|
% Highlight the interval [1,5)
|
||||||
|
\draw[thick] (1,0) -- (5,0); % draw the line between 1 and 5
|
||||||
|
|
||||||
|
% Solid circle at 1 (inclusive)
|
||||||
|
\filldraw (1,0) circle (2pt);
|
||||||
|
|
||||||
|
% Open circle at 5 (exclusive)
|
||||||
|
\draw[thick] (5,0) circle (2pt);
|
||||||
|
|
||||||
|
% Label the interval A
|
||||||
|
\node[above] at (3,0.3) {$A = [1,5)$};
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\begin{tikzpicture}
|
||||||
|
% Draw the number line
|
||||||
|
\draw[->] (0,0) -- (6,0) node[anchor=north] {x}; % arrow at the end
|
||||||
|
|
||||||
|
% Draw ticks and labels
|
||||||
|
\foreach \x in {0,1,2,3,4,5}
|
||||||
|
\draw (\x,0.1) -- (\x,-0.1) node[below] {\x};
|
||||||
|
|
||||||
|
% Highlight the elements 2, 3, 4
|
||||||
|
\filldraw (2,0) circle (2pt); % Solid circle at 2
|
||||||
|
\filldraw (3,0) circle (2pt); % Solid circle at 3
|
||||||
|
\filldraw (4,0) circle (2pt); % Solid circle at 4
|
||||||
|
|
||||||
|
% Label the set B
|
||||||
|
\node[above] at (3,0.3) {$B = \{2, 3, 4\}$};
|
||||||
|
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{tikzpicture}
|
||||||
|
% Draw the number line
|
||||||
|
\draw[->] (0,0) -- (6,0) node[anchor=north] {x}; % arrow at the end
|
||||||
|
|
||||||
|
% Draw ticks and labels
|
||||||
|
\foreach \x in {0,1,2,3,4,5}
|
||||||
|
\draw (\x,0.1) -- (\x,-0.1) node[below] {\x};
|
||||||
|
|
||||||
|
% Highlight the set A in blue
|
||||||
|
\draw[thick, blue] (1,0.7) -- (5,0.7); % Line for set A from 1 to 5
|
||||||
|
\filldraw[blue] (1,0.7) circle (2pt); % Solid circle at 1 (A)
|
||||||
|
\draw[blue, thick] (5,0.7) circle (2pt); % Open circle at 5 (A)
|
||||||
|
\node[above, blue] at (3,0.8) {$A = [1, 5)$};
|
||||||
|
|
||||||
|
% Highlight the set B in red
|
||||||
|
\foreach \x in {2,3,4}
|
||||||
|
\filldraw[red] (\x,0) circle (2pt); % Solid circles for B
|
||||||
|
\node[above, red] at (3,0) {$B = \{2, 3, 4\}$};
|
||||||
|
|
||||||
|
% Highlight the union A ∪ B in green
|
||||||
|
\draw[thick, green] (1,-0.7) -- (5,-0.7); % Line for A ∪ B
|
||||||
|
\filldraw[green] (1,-0.7) circle (2pt); % Solid circle at 1 (A ∪ B)
|
||||||
|
\draw[green, thick] (5,-0.7) circle (2pt); % Open circle at 5 (A ∪ B)
|
||||||
|
\node[below, green] at (3,-0.9) {$A \cup B = [1, 5)$};
|
||||||
|
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
\item Gesucht sind die Elemente, die sowohl in $A$ als auch in $C$ enthalten sind.
|
||||||
|
|
||||||
|
$$
|
||||||
|
C=\{z \in \mathbb{Z} \mid-1 \leq z<3\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Die Menge $C$ ist einfach geschrieben $\{-1,0,1,2\}$.
|
||||||
|
|
||||||
|
Welche Elemente davon liegen auch in $A=[1,5)$ ?
|
||||||
|
|
||||||
|
Geht man mal die Reihe nach durch:
|
||||||
|
$$
|
||||||
|
\begin{array}{rrll}
|
||||||
|
& \fcolorbox{red}{white}{-1}<1 & \text { also nicht in } A \\
|
||||||
|
& \fcolorbox{red}{white}{0}<1 & \text { also nicht in } A \\
|
||||||
|
1 & \leq \fcolorbox{red}{white}{1}<5 & \text { also in } A \\
|
||||||
|
1 & \leq \fcolorbox{red}{white}{2}<5 & \text { also in } A
|
||||||
|
\end{array}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Somit liegen nur 1 und 2 in beiden Mengen. Die Schnittmenge lautet:
|
||||||
|
|
||||||
|
$$
|
||||||
|
A \cap C=\{1,2\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
|
||||||
|
\textbf{Grafische Lösung}
|
||||||
|
|
||||||
|
\begin{tikzpicture}
|
||||||
|
% Draw the number line
|
||||||
|
\draw[->] (-2,0) -- (6,0) node[anchor=north] {x}; % arrow at the end
|
||||||
|
|
||||||
|
% Draw ticks and labels
|
||||||
|
\foreach \x in {-1,0,1,2,3,4,5}
|
||||||
|
\draw (\x,0.1) -- (\x,-0.1) node[below] {\x};
|
||||||
|
|
||||||
|
% Highlight the set A in blue
|
||||||
|
\draw[thick, blue] (1,0.7) -- (5,0.7); % Line for set A from 1 to 5
|
||||||
|
\filldraw[blue] (1,0.7) circle (2pt); % Solid circle at 1 (A)
|
||||||
|
\draw[blue, thick] (5,0.7) circle (2pt); % Open circle at 5 (A)
|
||||||
|
\node[above, blue] at (3,0.9) {$A = [1, 5)$};
|
||||||
|
|
||||||
|
% Highlight the set C in red
|
||||||
|
\foreach \x in {-1,0,1,2}
|
||||||
|
\filldraw[red] (\x,0) circle (2pt); % Solid circles for C
|
||||||
|
\node[above, red] at (0.5,0) {$C = \{ z \in \mathbb{Z} \mid -1 \leq z < 3 \}$};
|
||||||
|
|
||||||
|
% Highlight the intersection A ∩ C in green
|
||||||
|
\foreach \x in {1,2}
|
||||||
|
\filldraw[green] (\x,-0.7) circle (2pt); % Solid circles for A ∩ C
|
||||||
|
\node[below, green] at (1.5,-0.8) {$A \cap C = \{1, 2\}$};
|
||||||
|
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
|
||||||
|
\item $C \backslash B$ steht für den Ausdruck " $C$ ohne $B$ ". Bevor wir überlegen, welche Elemente hier wegfallen - kannst du $C$ noch vereinfacht darstellen?
|
||||||
|
|
||||||
|
Einfach geschrieben ist $C$ nur $\{-1,0,1,2\}$. Für $C$ ohne $B$ müssen wir also aus $C$ alle Elemente entfernen, die auch in $B$ liegen was bleibt über?
|
||||||
|
|
||||||
|
Da man die Elemente direkt vergleichen kann, sieht man, dass nur die 2 in beiden Mengen enthalten ist. Also müssen wir die aus $C$ rausnehmen und sind fertig:
|
||||||
|
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
C \backslash B & =\{-1,0,1,2\}\} \backslash\{2,3,4\} \\
|
||||||
|
& =\{-1,0,1\}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
|
||||||
|
\end{enumerate}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{tikzpicture}
|
||||||
|
% Draw the number line
|
||||||
|
\draw[->] (-2,0) -- (5,0) node[anchor=north] {x}; % arrow at the end
|
||||||
|
|
||||||
|
% Draw ticks and labels
|
||||||
|
\foreach \x in {-1,0,1,2,3}
|
||||||
|
\draw (\x,0.1) -- (\x,-0.1) node[below] {\x};
|
||||||
|
|
||||||
|
% Highlight the elements -1, 0, 1, 2
|
||||||
|
\filldraw (-1,0) circle (2pt); % Solid circle at -1
|
||||||
|
\filldraw (0,0) circle (2pt); % Solid circle at 0
|
||||||
|
\filldraw (1,0) circle (2pt); % Solid circle at 1
|
||||||
|
\filldraw (2,0) circle (2pt); % Solid circle at 2
|
||||||
|
|
||||||
|
% Open circle at 3 (exclusive)
|
||||||
|
\draw[thick] (3,0) circle (2pt); % Open circle at 3
|
||||||
|
|
||||||
|
% Label the set C
|
||||||
|
\node[above] at (1,0.3) {$C = \{ z \in \mathbb{Z} \mid -1 \leq z < 3 \}$};
|
||||||
|
|
||||||
|
\end{tikzpicture}
|
||||||
|
|
||||||
|
|
||||||
|
11111111111111111111111111111111111111111111111111
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
11111111111111111111111111111111111111111111111111
|
||||||
|
|
||||||
|
|
||||||
|
\paragraph{1.}\mbox{}\\
|
||||||
|
Gegeben sind die Mengen $M_1=\{26 ; 13 ; 5 ; 32\}, M_2=\{4 ; 5 ; 32 ; 35\}$ und $M_3=\{4 ; 1 ; 8 ; 19 ; 23 ; 5 ; 26\}$.
|
||||||
|
|
||||||
|
Bestimmen Sie die Menge $\left(M_1 \cap M_2\right) \cup M_3$.
|
||||||
|
|
||||||
|
|
||||||
|
Man liest die Formel $\left(M_1 \cap M_2\right) \cup M_3$ von links nach rechts und bestimmt zuerst die Schnittmenge der Mengen $M_1$ und $M_2$:
|
||||||
|
|
||||||
|
$$
|
||||||
|
M_1 \cap M_2=\{26 ; 13 ; \mathbf{5} ; \mathbf{32}\} \cap\{4 ; \mathbf{5} ; \mathbf{32} ; 35\}=\{5 ; 32\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Nun vereinigt man die eben bestimmte Menge $M_1 \cap M_2$ mit der Menge $M_3=\{4 ; 1 ; 8 ; 19 ; 23 ; 5 ; 26\}$, um die gesuchte Menge zu erhalten:
|
||||||
|
|
||||||
|
$$
|
||||||
|
\left(M_1 \cap M_2\right) \cup M_3=\{\mathbf{5} ; \mathbf{32}\} \cup\{\mathbf{4} ; \mathbf{1} ; \mathbf{8} ; \mathbf{19} ; \mathbf{23} ; 5 ; \mathbf{26}\}=\{1 ; 4 ; 5 ; 8 ; 19 ; 23 ; 26 ; 32\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
\paragraph{2.}\mbox{}\\
|
||||||
|
Bestimmen Sie die Schnittmenge der Menge $M=\{8 ; 11 ; 14 ; 17 ; 20 ; \ldots\}$ und der Menge der ungeraden natürlichen Zahlen.
|
||||||
|
|
||||||
|
|
||||||
|
Man betrachtet zuerst die Menge $M$ . Sie enthält unendlich viele Zahlen, von denen die ersten fünf in aufzählender Mengenschreibweise angegeben sind. Da sich zwei aufeinanderfolgende Elemente immer um 3 unterscheiden, liegt ein Bildungsgesetz vor, mit dessen Hilfe man weitere Elemente der Menge M angeben kann.
|
||||||
|
$$
|
||||||
|
M=\{8 ; \textcolor{red}{11} ; 14 ; \textcolor{red}{17} ; 20 ; \textcolor{red}{23} ; 26 ; \textcolor{red}{29} ; 32 ; \textcolor{red}{35}; \ldots\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Die Menge $\mathbb{N}=\{1 ; 2 ; 3 ; 4 ; \ldots\}$ der natürlichen Zahlen ist bekannt.
|
||||||
|
Die Menge der ungeraden natürlichen Zahlen besteht aus allen natürlichen Zahlen, die sich nicht durch 2 teilen lassen.
|
||||||
|
$$
|
||||||
|
\text { \{ungerade natürliche Zahlen }\}=\{1 ; 3 ; 5 ; 7 ; 9 ; \textcolor{red}{11} ; 13 ; 15 ; \textcolor{red}{17} ; 19 ; \ldots\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Schließlich schneidet man die Menge M mit den ungeraden natürlichen Zahlen, d.h. man bestimmt alle Zahlen in M, die ungerade sind. Da diese Schnittmenge wieder unendlich groß ist, genügt es, die ersten Elemente in aufzählender Mengenschreibweise anzugeben.
|
||||||
|
$$
|
||||||
|
M \cap\{\text { ungerade natürliche Zahlen }\}=\{11 ; 17 ; 23 ; 29 ; 35 ; \ldots\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
\paragraph{3.}\mbox{}\\
|
||||||
|
Schreiben Sie den folgenden Ausdruck als ein einziges Intervall.
|
||||||
|
$$
|
||||||
|
(-\infty,-2) \cap[-6,7]
|
||||||
|
$$
|
||||||
|
|
||||||
|
Dieses Intervall stellt den Durchschnitt von zwei Intervallen dar.
|
||||||
|
Anstelle des Symbols $\cap$ kann das Verbindungswort 'und' benutzt werden.
|
||||||
|
Damit besteht der Durchschnitt aus allen Punkten, die in $(-\infty,-2)$ und auch in $[-6,7]$ liegen.
|
||||||
|
Es gibt keine Lücke zwischen diesen zwei Intervallen.
|
||||||
|
Das Intervall, das den Durchschnitt der gegebenen Intervalle bildet, hat zwei Endpunkte.
|
||||||
|
Sie sind -2 und -6 .
|
||||||
|
Das einzige Intervall, das $(-\infty,-2) \cap[-6,7]$ darstellt, ist $[-6,-2)$.
|
||||||
|
|
||||||
|
\paragraph{4.}\mbox{}\\
|
||||||
|
Schreiben Sie den folgenden Ausdruck als ein einziges Intervall.
|
||||||
|
$$
|
||||||
|
(-\infty, 6) \cup[6,10)
|
||||||
|
$$
|
||||||
|
|
||||||
|
Dieses Intervall stellt die Vereinigung von zwei Intervallen dar.
|
||||||
|
Anstelle des Symbols $\cup$ kann das Verbindungswort 'oder' verwendet werden.
|
||||||
|
Damit besteht die Vereinigung aus allen Punkten, die in $(-\infty, 6)$ oder in $[6,10)$ liegen.
|
||||||
|
Es gibt keine Lücke zwischen diesen zwei Intervallen.
|
||||||
|
|
||||||
|
Das Intervall, dass die Vereinigung der gegebenen Intervalle beschreibt, hat einen Endpunkt. Es ist $10$.
|
||||||
|
|
||||||
|
Das einzelne Intervall, das $(-\infty, 6) \cup[6,10)$ darstellt, ist $(-\infty, 10)$.
|
||||||
|
|
||||||
|
|
||||||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||||
|
\newpage
|
||||||
|
|
||||||
|
\textbf{Schreiben} Sie den folgenden Ausdruck als ein einziges Intervall.
|
||||||
|
$$
|
||||||
|
(-\infty, 6) \cup[6,10)
|
||||||
|
$$
|
||||||
|
|
||||||
|
Dieses Intervall stellt die Vereinigung von zwei Intervallen dar.
|
||||||
|
Anstelle des Symbols $\cup$ kann das Verbindungswort 'oder' verwendet werden.
|
||||||
|
|
||||||
|
Damit besteht die Vereinigung aus allen Punkten, die in $(-\infty, 6)$ oder in $[6,10)$ liegen.
|
||||||
|
|
||||||
|
Es gibt keine Lücke zwischen diesen zwei Intervallen. Das Intervall, dass die Vereinigung der gegebenen Intervalle beschreibt, hat einen Endpunkt. Es ist $10$ .
|
||||||
|
|
||||||
|
Das einzelne Intervall, das $(-\infty, 6) \cup[6,10)$ darstellt, ist $(-\infty, 10)$.
|
||||||
|
|
||||||
|
%\begin{figure}[h]
|
||||||
|
% \centering
|
||||||
|
%\begin{tikzpicture}
|
||||||
|
% \begin{axis}[
|
||||||
|
% hide axis,
|
||||||
|
% xmin=-10, xmax=12,
|
||||||
|
% ymin=-1, ymax=1,
|
||||||
|
% width=15cm,
|
||||||
|
% height=3cm,
|
||||||
|
% axis x line=middle,
|
||||||
|
% xtick=\empty,
|
||||||
|
% ytick=\empty,
|
||||||
|
% enlargelimits
|
||||||
|
% ]
|
||||||
|
% % Das Intervall (-∞, 6)
|
||||||
|
% \addplot[domain=-10:5.9, samples=2, thick, blue] {0};
|
||||||
|
%
|
||||||
|
% % Das Intervall [6, 10)
|
||||||
|
% \addplot[domain=6:9.9, samples=2, thick, red] {0};
|
||||||
|
%
|
||||||
|
% % Offene Kreise bei x=6 und x=10
|
||||||
|
% \node[draw, fill=white, circle, inner sep=1.5pt] at (axis cs: 6, 0) {};
|
||||||
|
% \node[draw, circle, inner sep=1.5pt] at (axis cs: 10, 0) {};
|
||||||
|
%
|
||||||
|
% % Punkt x=6 als geschlossener Kreis
|
||||||
|
% \node[fill, circle, inner sep=1.5pt] at (axis cs: 6, 0) {};
|
||||||
|
%
|
||||||
|
% % Beschriftungen
|
||||||
|
% \node[below] at (axis cs: 0, 0) {0}
|
||||||
|
% \node[below] at (axis cs: 6, 0) {6};
|
||||||
|
% \node[below] at (axis cs: 10, 0) {10};
|
||||||
|
%
|
||||||
|
% % Pfeile am Ende des Intervalls
|
||||||
|
% \draw[thick,-latex] (axis cs: -10,0) -- (axis cs: -10.5,0);
|
||||||
|
% \draw[thick,-latex] (axis cs: 10,0) -- (axis cs: 10.5,0);
|
||||||
|
% \end{axis}
|
||||||
|
%\end{tikzpicture}
|
||||||
|
% %\caption{Caption}
|
||||||
|
% %\label{fig:enter-label}
|
||||||
|
%\end{figure}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\newpage
|
||||||
|
|
||||||
|
3333
|
||||||
|
|
||||||
|
Schreiben Sie den folgenden Ausdruck als ein einziges Intervall.
|
||||||
|
$$
|
||||||
|
(-\infty,-2) \cap[-6,7]
|
||||||
|
$$
|
||||||
|
|
||||||
|
Dieses Intervall stellt den Durchschnitt von zwei Intervallen dar.
|
||||||
|
Anstelle des Symbols $\cap$ kann das Verbindungswort 'und' benutzt werden.
|
||||||
|
Damit besteht der Durchschnitt aus allen Punkten, die in $(-\infty,-2)$ und auch in $[-6,7]$ liegen. Es gibt keine Luecke zwischen diesen zwei Intervallen.
|
||||||
|
|
||||||
|
Das Intervall, das den Durchschnitt der gegebenen Intervalle bildet, hat zwei Endpunkte.
|
||||||
|
Sie sind -2 und -6 .
|
||||||
|
Das einzige Intervall, das $(-\infty,-2) \cap[-6,7]$ darstellt, ist $[-6,-2)$.
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
44444
|
||||||
|
|
||||||
|
Schreiben Sie den folgenden Ausdruck als ein einziges Intervall.
|
||||||
|
$$
|
||||||
|
(-\infty,-2) \cap[-6,7]
|
||||||
|
$$
|
||||||
|
|
||||||
|
Dieses Intervall stellt den Durchschnitt von zwei Intervallen dar.
|
||||||
|
Anstelle des Symbols $\cap$ kann das Verbindungswort 'und' benutzt werden.
|
||||||
|
Damit besteht der Durchschnitt aus allen Punkten, die in $(-\infty,-2)$ und auch in $[-6,7]$ liegen.
|
||||||
|
Es gibt keine Luecke zwischen diesen zwei Intervallen.
|
||||||
|
Das Intervall, das den Durchschnitt der gegebenen Intervalle bildet, hat zwei Endpunkte.
|
||||||
|
Sie sind -2 und -6 .
|
||||||
|
Das einzige Intervall, das $(-\infty,-2) \cap[-6,7]$ darstellt, ist $[-6,-2)$.
|
||||||
|
|
||||||
|
|
||||||
|
55555
|
||||||
|
|
||||||
|
|
||||||
|
Schreiben Sie den folgenden Ausdruck als ein einziges Intervall.
|
||||||
|
$$
|
||||||
|
(-\infty, 6) \cup[6,10)
|
||||||
|
$$
|
||||||
|
|
||||||
|
Dieses Intervall stellt die Vereinigung von zwei Intervallen dar.
|
||||||
|
Anstelle des Symbols U kann das Verbindungswort 'oder' verwendet werden.
|
||||||
|
Damit besteht die Vereinigung aus allen Punkten, die in $(-\infty, 6)$ oder in $[6,10)$ liegen. Es gibt keine Luecke zwischen diesen zwei Intervallen.
|
||||||
|
|
||||||
|
Das Intervall, dass die Vereinigung der gegebenen Intervalle beschreibt, hat einen Endpunkt. Es ist 10 .
|
||||||
|
|
||||||
|
Das einzelne Intervall, das $(-\infty, 6) \cup[6,10)$ darstellt, ist $(-\infty, 10)$.
|
||||||
|
|
||||||
|
|
||||||
|
6666666
|
||||||
|
|
||||||
|
|
||||||
|
Sei $\Omega=\{a, b, c, d, e, f, g, h, i, j, k\}$
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
& X=\{a, d, e, h, i, k\} \\
|
||||||
|
& Y=\{a, c, f, h, i\} \\
|
||||||
|
& Z=\{a, d, e, i, k\}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Bestimmen Sie $\mathrm{X} \cup(\mathrm{Z} \cap \mathrm{Y})^{\prime}$.
|
||||||
|
|
||||||
|
Um $\mathrm{X} \cup(\mathrm{Z} \cap \mathrm{Y})^{\prime}$ zu bestimmen, bestimmen Sie zuerst den Ausdruck innerhalb der Klammern.
|
||||||
|
Bestimmen Sie $Z \cap Y$.
|
||||||
|
Der Durchschnitt besteht aus allen Elementen, die in beiden Mengen $Z$ und $Y$ sind.
|
||||||
|
Die Elemente 'a' und 'i' sind in der Menge $\mathrm{Z}$ und auch in der Menge $\mathrm{Y}$.
|
||||||
|
Somit sind sie in $\mathrm{Z} \cap \mathrm{Y}$.
|
||||||
|
$$
|
||||||
|
Z \cap Y=\{a, i\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Bestimmen Sie jetzt das Komplement von $\mathrm{Z} \cap \mathrm{Y}$, symbolisiert durch $(\mathrm{Z} \cap \mathrm{Y})^{\prime}$.
|
||||||
|
Diese Menge enthaelt alle Elemente aus $\boldsymbol{\Omega}$ die nicht $\mathbf{z u}\{\mathrm{a}, \mathrm{i}\}$ gehoeren.
|
||||||
|
Die Menge $\{k, j, h, g, f, e, b, c, d\}$ repraesentiert $(Z \cap Y)$ '.
|
||||||
|
Finden Sie jetzt die Vereinigung der Mengen $X$ und $\{k, j, h, g, f, e, b, c, d\}$.
|
||||||
|
Mit anderen Worten: Fuegen Sie zu der Liste der Elemente in $\{k, j, h, g, f, e, b, c, d\}$ diejenigen Elemente hinzu, die Sie noch nicht benannt haben und die in $X$ enthalten sind. Zum Beispiel ist, 'i' in der Menge $\mathrm{X}$, aber nicht in $(\mathrm{Z} \cap \mathrm{Y})$ '.
|
||||||
|
$$
|
||||||
|
X \cup(Z \cap Y)^{\prime}=\{e, f, h, g, i, k, j, b, c, d, a\} \text {. }
|
||||||
|
$$
|
||||||
|
|
||||||
|
Dies ist die Menge $\Omega$.
|
||||||
|
|
||||||
|
|
||||||
|
8888888888
|
||||||
|
Bestimmen Sie die Menge $\left(A^{\prime} \cup B^{\prime}\right) \cap C$, wenn folgendes gegeben ist.
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
& \Omega=\{x \mid x \in \mathbb{N} \text { und } x<13\} \\
|
||||||
|
& A=\{x \mid x \in \mathbb{N} \text { und } x \text { ist ungerade und } x<13\} \\
|
||||||
|
& B=\{x \mid x \in \mathbb{N} \text { und } x \text { ist gerade und } x<13\} \\
|
||||||
|
& C=\{x \mid x \in \mathbb{N} \text { und } x<8\}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Um $\left(A^{\prime} \cup B^{\prime}\right) \cap C$ zu bestimmen, beginnen Sie damit die Menge innerhalb der Klammern, $\left(A^{\prime} \cup B^{\prime}\right)$, zu finden.
|
||||||
|
Zuerst muessen Sie A' bestimmen, das Komplement von A, das alle Elemente aus $\Omega$ enthaelt, die nicht in $\mathrm{A}$ sind.
|
||||||
|
$$
|
||||||
|
A^{\prime}=\{2,4,6,8,10,12\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Als Naechstes muessen Sie $\mathrm{B}^{\prime}$ bestimmen, das Komplement von $\mathrm{B}$, das alle Elemente aus $\Omega$ enthaelt, die nicht in $\mathrm{B}$ sind.
|
||||||
|
$$
|
||||||
|
B^{\prime}=\{1,3,5,7,9,11\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Identifizieren Sie jetzt die Elemente der Menge $\left(A^{\prime} \cup B^{\prime}\right.$ ) indem Sie alle Elemente in der Menge $A^{\prime}$ auflisten und dann die Elemente in der Menge $B^{\prime}$ hinzufuegen.
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
\left(A^{\prime} \cup B^{\prime}\right) & =\{2,4,6,8,10,12\} \cup\{1,3,5,7,9,11\} \\
|
||||||
|
& =\{1,2,3,4,5,6,7,8,9,10,11,12\}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Beenden Sie jetzt das Problem, indem Sie $\left(A^{\prime} \cup B^{\prime}\right) \cap C$ bestimmen.
|
||||||
|
Der Durchschnitt ist diejenige Menge, die alle Elemente enthaelt, die den beiden Mengen $A^{\prime} \cup B^{\prime}$ und $C^{\prime}$ gemeinsam sind.
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
\left(A^{\prime} \cup B^{\prime}\right) \cap C= & \{1,2,3,4,5,6,7,8,9,10,11,12\} \cap\{1,2,3,4,5,6,7\} \\
|
||||||
|
& =\{1,2,3,4,5,6,7\}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
|
||||||
|
999999999999
|
||||||
|
|
||||||
|
Bestimmen Sie die Menge $\left(A^{\prime} \cup B^{\prime}\right) \cap C$, wenn folgendes gegeben ist.
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
& \Omega=\{x \mid x \in \mathbb{N} \text { und } x<13\} \\
|
||||||
|
& A=\{x \mid x \in \mathbb{N} \text { und } x \text { ist ungerade und } x<13\} \\
|
||||||
|
& B=\{x \mid x \in \mathbb{N} \text { und } x \text { ist gerade und } x<13\} \\
|
||||||
|
& C=\{x \mid x \in \mathbb{N} \text { und } x<8\}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Um $\left(A^{\prime} \cup B^{\prime}\right) \cap C$ zu bestimmen, beginnen Sie damit die Menge innerhalb der Klammern, ( $\left.A^{\prime} \cup B^{\prime}\right)$, zu finden.
|
||||||
|
Zuerst muessen Sie A' bestimmen, das Komplement von A, das alle Elemente aus $\Omega$ enthaelt, die nicht in A sind.
|
||||||
|
$$
|
||||||
|
A^{\prime}=\{2,4,6,8,10,12\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Als Naechstes muessen Sie B' bestimmen, das Komplement von B, das alle Elemente aus $\Omega$ enthaelt, die nicht in B sind.
|
||||||
|
$$
|
||||||
|
B^{\prime}=\{1,3,5,7,9,11\}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Identifizieren Sie jetzt die Elemente der Menge ( $\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}$ ) indem Sie alle Elemente in der Menge $\mathrm{A}^{\prime}$ auflisten und dann die Elemente in der Menge $\mathrm{B}^{\prime}$ hinzufuegen.
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
\left(A^{\prime} \cup B^{\prime}\right) & =\{2,4,6,8,10,12\} \cup\{1,3,5,7,9,11\} \\
|
||||||
|
& =\{1,2,3,4,5,6,7,8,9,10,11,12\}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
Beenden Sie jetzt das Problem, indem Sie ( $\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}$ ) $\cap \mathrm{C}$ bestimmen.
|
||||||
|
Der Durchschnitt ist diejenige Menge, die alle Elemente enthaelt, die den beiden Mengen $\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}$ und $\mathrm{C}^{\prime}$ gemeinsam sind.
|
||||||
|
$$
|
||||||
|
\begin{aligned}
|
||||||
|
\left(A^{\prime} \cup B^{\prime}\right) \cap C & =\{1,2,3,4,5,6,7,8,9,10,11,12\} \cap\{1,2,3,4,5,6,7\} \\
|
||||||
|
& =\{1,2,3,4,5,6,7\}
|
||||||
|
\end{aligned}
|
||||||
|
$$
|
||||||
|
|
||||||
|
1010101010101010
|
||||||
|
|
||||||
|
Von den Achtklaesslern der Paxton-Schule spielten 16 Basketball, 9 Volleyball, 10 Fussball, 1 spielte nur Basketball und Fussball, 2 spielten nur Volleyball und Fussball, 1 spielte nur Basketball und Volleyball und 2 spielten Volleyball, Basketball und Fussball.
|
||||||
|
|
||||||
|
Wieviele spielten eine oder mehrere der drei Sportarten?
|
||||||
|
|
||||||
|
Zeichnen Sie ein leeres Diagramm, markieren Sie einen Kreis mit $B$ für Basketball, einen mit $F$ für Fussball und einen mit $V$ für Volleyball.
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{10_1.png}
|
||||||
|
%\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
\newpage
|
||||||
|
|
||||||
|
Tragen Sie als erstes die Anzahl der Schueler in den Durchschnitt aller drei Kreise ein, die alle drei Sportarten betreiben. $2$ spielten alle drei Sportarten.
|
||||||
|
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{10_2.png}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
\newpage
|
||||||
|
|
||||||
|
Tragen Sie jetzt ein, wie viele Spieler zwei Sportarten betreiben. In der Aufgabe war gegeben, dass 2 nur Volleyball und Fussball spielen.
|
||||||
|
Diese Zahl muss in den Bereich geschrieben werden, der der Durchschnitt von V und F ist, aber nicht die 2 schon frueher Erwaehnten enthaelt.
|
||||||
|
Der Grund ist: Die Aufgabe spricht hier nur von Volleyball und Fussball, d.h. Basketball ist ausgeschlossen
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{10_3.png}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
Machen Sie dasselbe fuer die anderen Sportarten. 1 spielt nur Basketball und Fussball.
|
||||||
|
1 spielt nur Basketball und Volleyball.
|
||||||
|
Diese Werte muessen im Diagramm platziert werden
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{10_4.png}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
\newpage
|
||||||
|
|
||||||
|
Die Aufgabe sagt, dass zehn Fussball spielen.
|
||||||
|
Es gibt bereits 5 in dem Fussball-Kreis.
|
||||||
|
Deshalb muss eine 5 dort in den Kreis mit F geschrieben werden, wo keine anderen Werte sind.
|
||||||
|
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{10_5.png}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
Neun spielen Volleyball. 5 sind bereits im Diagramm.
|
||||||
|
Es bleiben 4 uebrig.
|
||||||
|
Eine 4 ist in dem Bereich platziert worden, der Nur-Volleyball repraesentiert.
|
||||||
|
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{10_6.png}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
|
||||||
|
Die Aufgabe besagt, dass 16 Basketball spielen.
|
||||||
|
Es sind bereits 4 Basketballspieler eingetragen.
|
||||||
|
Somit bleiben 16 - 4 oder 12 uebrig, die nur Basketball spielen.
|
||||||
|
|
||||||
|
\newpage
|
||||||
|
|
||||||
|
\begin{figure}[h]
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{10_7..png}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
Um zu bestimmen, wie viele der Achtklaessler eine oder mehrere der drei Sportarten spielen, addieren Sie alle Eintraege aus dem Diagramm.
|
||||||
|
Es sind insgesamt 27 Spieler.
|
||||||
|
|
||||||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||||
|
|
||||||
|
\textbf{Frage 10}
|
||||||
|
|
||||||
|
Sei $\Omega=\{0,1,2,3,4,5, \ldots\}, A=\{1,2,3,4, \ldots\}$ und $B=\{6,12,18,24, \ldots\}$.
|
||||||
|
Bestimmen Sie die Menge $\mathbf{A} \cup \mathbf{B}$.
|
||||||
|
|
||||||
|
Die Vereinigung der Menge A und der Menge B, symbolisiert durch $\mathbf{A} \cup \mathbf{B}$, ist die Menge, die alle Elemente enthaelt, die Mitglieder von der Menge $\mathbf{A}$ oder der Menge $B$ (oder von beiden Mengen) sind.
|
||||||
|
|
||||||
|
Beachten Sie , dass alle Mengen unendliche Mengen sind.
|
||||||
|
Die Menge A ist die Menge aller positiven ganzen Zahlen und die Menge B ist die Menge alle positiven ganzen Zahlen, die Vielfaches von 6 sind.
|
||||||
|
Da jedes positive ganzzahlige Vielfache von 6 auch eine positive ganze Zahl ist, ist jedes Element aus B auch ein Element aus der Menge A.
|
||||||
|
Daher ist die Vereinigung $\mathbf{A} \cup \mathbf{B}$ dasselbe wie die Menge $\mathbf{A}$, da alle Elemente aus $\mathbf{B}$ auch in der Menge $\mathbf{A}$ enthalten sind.
|
||||||
|
Daher ist $A \cup B=\{1,2,3,4, \ldots\}$.
|
||||||
|
|
||||||
|
\begin{figure}
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{Frage10_Vorlage.pdf}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
|
||||||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||||
|
\textbf{Frage 11}
|
||||||
|
|
||||||
|
\begin{figure}
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{Frage11.png}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
|
||||||
|
Nutzen Sie Venn-Diagramme, um zu bestimmen, ob die folgenden Mengen gleich sind fuer alle Mengen $A, B$ und $C$.
|
||||||
|
$A \cap(B \cup C)$
|
||||||
|
$(A \cap B) \cup C^{\prime}$
|
||||||
|
|
||||||
|
Wenn Sie das Venn-Diagramm auf der rechten Seite als Referenz verwenden, so besteht die Menge $A \cap(B \cup C)$ aus den Regionen II, IV, V.
|
||||||
|
Wenn Sie das Venn-Diagramm auf der rechten Seite als Referenz verwenden, so besteht die Menge (A $\cap$ B) $\cup \mathrm{C}^{\prime}$ aus den Regionen I, II, III, V, VIII.
|
||||||
|
Daher sind die folgenden Mengen nicht fuer alle Mengen $\mathrm{A}, \mathrm{B}$ und C gleich.
|
||||||
|
$A \cap(B \cup C)$
|
||||||
|
$(A \cap B) \cup C^{\prime}$
|
||||||
|
|
||||||
|
|
||||||
|
\begin{figure}
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{Frage12_Vorlage.pdf}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\textbf{Frage 13}
|
||||||
|
|
||||||
|
|
||||||
|
Verwenden Sie die Symbole A, B, $\cup, \cap$ und ', wenn noetig, um die schattierte Region zu beschreiben.
|
||||||
|
|
||||||
|
Dieses Venn-Diagramm hat 4 verschiedene Regionen, die wie rechts bezeichnet sind.
|
||||||
|
Um eine Mengenbeschreibung fuer das gegebene Diagramm zu entwickeln, bestimmen Sie zunaechst, welche Regionen schattiert sind
|
||||||
|
|
||||||
|
In dem gegebenen Venn-Diagramm ist Region 1 schattiert.
|
||||||
|
|
||||||
|
Die Menge $A \cap B^{\prime}$ fuehrt zur Schattierung der Region 1, was die gegebene Abbildung produziert.
|
||||||
|
\begin{figure}
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{Frage13_Vorlage.pdf}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||||
|
|
||||||
|
\textbf{Frage 14}
|
||||||
|
|
||||||
|
|
||||||
|
Verwenden Sie die Symbole A, B, U, $\cap$ und ', wenn noetig, um die schattierte Region zu beschreiben.
|
||||||
|
|
||||||
|
verschiedene Regionen, die wie rechts bezeichnet sind.
|
||||||
|
Um eine Mengenbeschreibung fuer das gegebene Diagramm zu entwickeln, bestimmen Sie zunaechst, welche Regionen schattiert sind.
|
||||||
|
|
||||||
|
In dem gegebenen Venn-Diagramm sind die Regionen 2,3,4,5,6 schattiert.
|
||||||
|
|
||||||
|
Daher fuehrt die Menge $(A \cap C) \cup B$ zur Schattierung der Regionen 2,3,4,5,6, die das gegebene Bild produziert.
|
||||||
|
\begin{figure}
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=0.5\linewidth]{Frage14_Vorlage.pdf}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
\begin{figure}
|
||||||
|
\centering
|
||||||
|
\includegraphics[width=1\linewidth]{Aufgabe15_Vorlage.pdf}
|
||||||
|
\caption{Enter Caption}
|
||||||
|
\label{fig:enter-label}
|
||||||
|
\end{figure}
|
||||||
|
|
||||||
|
\textbf{Frage 15}
|
||||||
|
Ein Personalbuero moechte die Beschaeftigten nach Geschlecht, Alter und Ausbildung klassifizieren.
|
||||||
|
Menge $A=$ maennlich
|
||||||
|
Menge $B=$ sind aelter als 45 Jahre
|
||||||
|
Menge $C=$ haben ein MBA
|
||||||
|
|
||||||
|
|
||||||
|
Bestimmen Sie die Region des Diagramms, die die maennlichen Beschaeftigten enthaelt, die aelter sind als 45 Jahre und ein MBA haben. Beschreiben Sie die Region, indem Sie die Mengen $A$, $B$ und $C$ mit den Operationen Vereinigung, Durchschnitt und Komplement verwenden. Lokalisieren Sie auch die Region im Diagramm.
|
||||||
|
|
||||||
|
Entscheiden Sie, wo jede Beschreibung der Beschaeftigten im \textbf{Venn-Diagramm} zu finden ist.
|
||||||
|
Die maennlichen Beschaeftigten gehoeren zur Menge $A$.
|
||||||
|
Beschaeftigte, die aelter sind als 45 Jahre, gehoeren zur Menge $B$.
|
||||||
|
Beschaeftigte, die ein MBA haben, gehoeren zur Menge $C$.
|
||||||
|
Daher gehoeren die maennlichen Beschaeftigten, die aelter sind als 45 Jahre und die ein MBA haben, $A \cap B \cap C$.
|
||||||
|
Die Regionen I, II, IV und V sind in $A$ enthalten.
|
||||||
|
|
||||||
|
Region V repraesentiert den Durchschnitt aller oben aufgelisteten Regionen. $A\cap B \cap C$ ist in Region V platziert.
|
||||||
60
plot06a.py
Normal file
60
plot06a.py
Normal file
@@ -0,0 +1,60 @@
|
|||||||
|
# -*- coding: utf-8 -*-
|
||||||
|
#from __future__ import unicode_literals
|
||||||
|
#import matplotlib.pyplot as plt
|
||||||
|
#from sympy import symbols, sqrt
|
||||||
|
#from sympy.plotting import plot
|
||||||
|
|
||||||
|
#http://www.scipy-lectures.org/intro/matplotlib/matplotlib.html
|
||||||
|
|
||||||
|
import numpy as np
|
||||||
|
import matplotlib.pyplot as plt
|
||||||
|
#x,y = symbols('x,y')
|
||||||
|
#y = sqrt(64-x**2)
|
||||||
|
|
||||||
|
#title="$y=\sqrt{64-x^2}$"
|
||||||
|
#graph = plot((y),(x,-10,10),(y,-1,12),ylabel='$y$',xlabel="$y=\sqrt{64-x^2}$", show=True)
|
||||||
|
|
||||||
|
plt.rc('text', usetex=True)
|
||||||
|
plt.rc('font', family='serif')
|
||||||
|
|
||||||
|
X = np.linspace(-8, 8, 512, endpoint=True)
|
||||||
|
|
||||||
|
C, S = np.cos(X), np.sin(X)
|
||||||
|
|
||||||
|
Y=np.sqrt(64-X**2)
|
||||||
|
|
||||||
|
#plt.plot(X, C)
|
||||||
|
|
||||||
|
#plt.figure(figsize=(5, 5), dpi=600)
|
||||||
|
plt.ylim(Y.min() * 1.1, Y.max() * 1.1)
|
||||||
|
plt.xlim(X.min() * 1.1, X.max() * 1.1)
|
||||||
|
|
||||||
|
ax = plt.gca() # gca stands for 'get current axis'
|
||||||
|
ax.spines['right'].set_color('none')
|
||||||
|
ax.spines['top'].set_color('none')
|
||||||
|
ax.xaxis.set_ticks_position('bottom')
|
||||||
|
ax.spines['bottom'].set_position(('data',0))
|
||||||
|
ax.yaxis.set_ticks_position('left')
|
||||||
|
ax.spines['left'].set_position(('data',0))
|
||||||
|
|
||||||
|
#x
|
||||||
|
plt.arrow(8.5, -0.003, 0.1, 0, head_width=0.1, head_length=0.25,clip_on=False) #width=0.015, color="k", clip_on=False, head_width=0.12, head_length=0.22)
|
||||||
|
|
||||||
|
#y
|
||||||
|
plt.arrow(0.0, 8.7, 0, 0.1, clip_on=False, head_width=0.1, head_length=0.25)
|
||||||
|
|
||||||
|
|
||||||
|
plt.yticks([2,4,6,8])
|
||||||
|
plt.xticks([-8,-6,-4,-2,0,2,4,6,8])#),rotation='vertical')
|
||||||
|
|
||||||
|
plt.margins(0.2)
|
||||||
|
|
||||||
|
|
||||||
|
plt.plot(X, Y, color="slateblue")
|
||||||
|
|
||||||
|
plt.title(r"$\displaystyle y=\sqrt{64-x^2}$" ,fontsize=12, color='slateblue',y=0.15,x=0.3)
|
||||||
|
|
||||||
|
#plt.show()
|
||||||
|
#graph.save('plot06a.png')
|
||||||
|
|
||||||
|
plt.savefig('plot06a.pdf', dpi=600)
|
||||||
59
plot07a.py
Normal file
59
plot07a.py
Normal file
@@ -0,0 +1,59 @@
|
|||||||
|
# -*- coding: utf-8 -*-
|
||||||
|
#from __future__ import unicode_literals
|
||||||
|
#import matplotlib.pyplot as plt
|
||||||
|
#from sympy import symbols, sqrt
|
||||||
|
#from sympy.plotting import plot
|
||||||
|
|
||||||
|
#http://www.scipy-lectures.org/intro/matplotlib/matplotlib.html
|
||||||
|
|
||||||
|
import numpy as np
|
||||||
|
import matplotlib.pyplot as plt
|
||||||
|
|
||||||
|
plt.rc('text', usetex=True)
|
||||||
|
plt.rc('font', family='serif')
|
||||||
|
|
||||||
|
x = np.linspace(-10, 30, 400, endpoint=True)
|
||||||
|
|
||||||
|
|
||||||
|
def f(x):
|
||||||
|
with np.errstate(divide='ignore', invalid='ignore'):
|
||||||
|
return np.divide(7,(9-x))
|
||||||
|
|
||||||
|
y=f(x)
|
||||||
|
|
||||||
|
#plt.plot(X, C)
|
||||||
|
|
||||||
|
plt.figure(figsize=(9, 6), dpi=600)
|
||||||
|
plt.ylim(-25, 25)
|
||||||
|
# plt.xlim(x.min() * 1.1, x.max() * 1.1)
|
||||||
|
#
|
||||||
|
ax = plt.gca() # gca stands for 'get current axis'
|
||||||
|
ax.spines['right'].set_color('none')
|
||||||
|
ax.spines['top'].set_color('none')
|
||||||
|
ax.xaxis.set_ticks_position('bottom')
|
||||||
|
ax.spines['bottom'].set_position(('data',0))
|
||||||
|
ax.yaxis.set_ticks_position('left')
|
||||||
|
ax.spines['left'].set_position(('data',0))
|
||||||
|
#
|
||||||
|
# #x
|
||||||
|
# plt.arrow(8.5, -0.003, 0.1, 0, head_width=0.1, head_length=0.25,clip_on=False) #width=0.015, color="k", clip_on=False, head_width=0.12, head_length=0.22)
|
||||||
|
#
|
||||||
|
# #y
|
||||||
|
# plt.arrow(0.0, 8.7, 0, 0.1, clip_on=False, head_width=0.1, head_length=0.25)
|
||||||
|
#
|
||||||
|
#
|
||||||
|
plt.yticks([2,4,6,8])
|
||||||
|
plt.xticks([-10,-8,-6,-4,-2,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30])#),rotation='vertical')
|
||||||
|
#
|
||||||
|
# plt.margins(0.2)
|
||||||
|
#
|
||||||
|
#
|
||||||
|
plt.plot(x, y, color="slateblue")
|
||||||
|
#
|
||||||
|
# plt.title(r"$\displaystyle y=f\left(t\right)=\frac{7}{9-t}$" ,fontsize=12, color='slateblue',y=0.15,x=0.3)
|
||||||
|
|
||||||
|
plt.savefig('plot07a.pdf', dpi=600)
|
||||||
|
#plt.show()
|
||||||
|
#graph.save('plot06a.png')
|
||||||
|
|
||||||
|
|
||||||
53
uebungenlimit01.tex
Normal file
53
uebungenlimit01.tex
Normal file
@@ -0,0 +1,53 @@
|
|||||||
|
%!TEX root=main.tex
|
||||||
|
\newpage
|
||||||
|
|
||||||
|
\section{Übungsaufgaben Grenzwerte}
|
||||||
|
|
||||||
|
\subsection{Bestimme, wie sich die Funktion $f$ im Unendlichen verhält}
|
||||||
|
|
||||||
|
|
||||||
|
\subsubsection{Aufgabe 1}
|
||||||
|
|
||||||
|
\begin{description}
|
||||||
|
|
||||||
|
\item[Verhalten gegen $+{\infty}$ ] \marginpar{Es wird nur die höchste Potenz betrachtet}$\mathop {\lim }\limits_{x \to +\infty }x^4-x^3=\mathop {\lim }\limits_{x \to +\infty }x^4=\left(+\infty\right)^4={\infty}$
|
||||||
|
|
||||||
|
\item[Verhalten gegen $-{\infty}$] $\mathop {\lim }\limits_{x \to -\infty }x^4-x^3=\mathop {\lim }\limits_{x \to -\infty }x^4=\left(-\infty\right)^4={\infty}$
|
||||||
|
|
||||||
|
\end{description}
|
||||||
|
|
||||||
|
|
||||||
|
\subsubsection{Aufgabe 2}
|
||||||
|
%http://mathenexus.zum.de/html/analysis/grenzwerte/Grenzwertplusminusunend_Ueb.htm
|
||||||
|
\begin{description}
|
||||||
|
\item[Verhalten gegen $+{\infty}$ ] $\rightarrow $ für $x>0$ \newline\newline
|
||||||
|
$f ( x ) = \frac { 1 } { x } \cdot \sqrt { x ^ { 2 } + 1 } = \mathop {\lim }\limits_{x \to \infty}\frac { 1 } { x } \cdot \sqrt { x ^ { 2 } + 1 }=\mathop {\lim }\limits_{x \to \infty}$
|
||||||
|
|
||||||
|
\item[Verhalten gegen $-{\infty}$ ]
|
||||||
|
$f ( x ) = \frac { 1 } { x } \cdot \sqrt { x ^ { 2 } + 1 }$
|
||||||
|
|
||||||
|
\end{description}
|
||||||
|
|
||||||
|
\paragraph{Grenzwerte in der Unendlichkeit mit Quadratwurzeln}
|
||||||
|
|
||||||
|
Da sich obige Aufgabe im negativ Unendlichen nicht so recht erklärt hat hier noch ein paar Ausführungen dazu:
|
||||||
|
|
||||||
|
\begin{itemize}
|
||||||
|
\item Wenn $ x $ positiv ist $ x=\sqrt{x^2} $ zum Beispiel, wenn $ x=3 $, dann $ x=3=\sqrt{9} $
|
||||||
|
|
||||||
|
\item Wenn $ x $ negativ ist $ x=-\sqrt{x^2} $ dem gegenüber ist also wenn $ x=-3 $ dann $ x=-3=-\sqrt{9} $
|
||||||
|
\end{itemize}
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
|
||||||
|
\textbf{Wichtig}: Man muss sich merken, das wenn $x=-\sqrt{x^2}$ wenn $x \rightarrow -\infty$ ist, muss man automatisch auf die negativen Werte von $x$ schauen muss.
|
||||||
|
|
||||||
|
\textbf{Beispiel 1}: $\mathop {\lim }\limits_{x \to \infty} \frac { \sqrt { 5 x ^ { 2 } + 2 x } } { x }$
|
||||||
|
Da man hier $x \rightarrow \infty$ untersucht, sind nur positive Werte von $x$ interessant und man nutzt $x=\sqrt{x^2}$.
|
||||||
|
|
||||||
|
$\lim _ { x \rightarrow \infty } \frac { \sqrt { 5 x ^ { 2 } + 2 } x } { x } = \lim _ { x \rightarrow \infty } \frac { \frac { \sqrt { 5 x ^ { 2 } + 2 } x } { \sqrt { x ^ { 2 } } } } { \frac { x } { x } } = \lim _ { x \rightarrow \infty } \frac { \sqrt { \frac { 5 x ^ { 2 } + 2 } { x ^ { 2 } } } } { 1 } = \lim _ { x \rightarrow \infty } \sqrt { 5 + \frac { 2 } { x } } = \lim _ { x \rightarrow \infty } \sqrt { 5 + 0 }=5$
|
||||||
|
|
||||||
|
\textbf{Beispiel 2}: $\mathop {\lim }\limits_{x \to \infty} \frac { \sqrt { 5 x ^ { 2 } + 2 x } } { x }$
|
||||||
Reference in New Issue
Block a user