3412 lines
129 KiB
TeX
3412 lines
129 KiB
TeX
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%% ********************************************************************************
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%% AUTHOR: Raj Dahya
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%% CREATED: November 2020
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%% EDITED: —
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%% TYPE: Notizen
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%% TITLE: Lösungen zu diversen Aufgaben im Kurs
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%% DOI: —
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%% DEPARTMENT: Fakultät for Mathematik und Informatik
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%% INSTITUTE: Universität Leipzig
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%% ********************************************************************************
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%% ********************************************************************************
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%% DOCUMENT STRUCTURE:
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%% ~~~~~~~~~~~~~~~~~~~
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%%
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%% — root.tex;
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%% |
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%% — parameters.tex;
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%% |
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%% — src/index.tex;
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%% |
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%% — ########;
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%% |
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%% — src/setup-type.tex;
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%% |
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%% — src/setup-packages.tex;
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%% |
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%% — src/setup-parameters.tex;
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%% |
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%% — src/setup-macros.tex;
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%% |
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%% — src/setup-environments.tex;
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%% |
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%% — src/setup-layout.tex;
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%% |
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%% — src/setup-localmacros.tex;
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%% |
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%% — front/index.tex;
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%% |
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%% — front/title.tex;
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%% |
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%% — front/foreword.tex;
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%% |
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%% — front/contents.tex;
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%% |
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%% — body/index.tex;
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%% |
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%% — body/uebung/ueb1.tex;
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%% |
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%% — body/uebung/ueb2.tex;
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%% |
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%% — body/uebung/ueb3.tex;
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%% |
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%% — body/ska/ska4.tex;
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%% |
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%% — body/quizzes/quiz1.tex;
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%% |
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%% — body/quizzes/quiz2.tex;
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%% |
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%% — body/quizzes/quiz3.tex;
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%% |
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%% — back/index.tex;
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%% |
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%% — ./back/quelle.bib;
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%%
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%% DOCUMENT-RANDOM-SEED: 5637845
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%% ********************************************************************************
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%% ********************************************************************************
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%% FILE: root.tex
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%% ********************************************************************************
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%% ********************************************************************************
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%% FILE: parameters.tex
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%% ********************************************************************************
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%% ********************************************************************************
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%% FILE: src/index.tex
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%% ********************************************************************************
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%% ********************************************************************************
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%% FILE: ########
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%% ********************************************************************************
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\makeatletter
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%% ********************************************************************************
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%% FILE: src/setup-type.tex
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%% ********************************************************************************
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\documentclass[
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12pt,
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a4paper,
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oneside,
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openright,
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center,
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chapterbib,
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crosshair,
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fleqn,
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headcount,
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headline,
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indent,
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indentfirst=false,
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portrait,
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phonetic,
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oldernstyle,
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onecolumn,
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sfbold,
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upper,
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]{scrbook}
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%% ********************************************************************************
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%% FILE: src/setup-packages.tex
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%% ********************************************************************************
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\PassOptionsToPackage{T2A,OT1}{fontenc} % T1,OT1,T2A,OT2
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\PassOptionsToPackage{utf8}{inputenc} % utf8
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\PassOptionsToPackage{british,english,ngerman,russian}{babel}
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\PassOptionsToPackage{
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english,
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ngerman,
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russian,
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capitalise,
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}{cleveref}
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\PassOptionsToPackage{
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bookmarks=true,
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bookmarksopen=false,
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bookmarksopenlevel=0,
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bookmarkstype=toc,
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colorlinks=false,
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raiselinks=true,
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hyperfigures=true,
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}{hyperref}
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\PassOptionsToPackage{
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reset,
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left=1in,
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right=1in,
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top=20mm,
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bottom=20mm,
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heightrounded,
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}{geometry}
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\PassOptionsToPackage{
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framemethod=TikZ,
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}{mdframed}
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\PassOptionsToPackage{normalem}{ulem}
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\PassOptionsToPackage{
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amsmath,
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thmmarks,
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}{ntheorem}
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\PassOptionsToPackage{table}{xcolor}
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\PassOptionsToPackage{
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all,
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color,
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curve,
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frame,
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import,
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knot,
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line,
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movie,
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rotate,
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textures,
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tile,
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tips,
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web,
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xdvi,
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}{xy}
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\usepackage{amsfonts}
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\usepackage{amsmath}
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\usepackage{amssymb}
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\usepackage{ntheorem} % <— muss nach den ams* Packages vorkommen!!
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\usepackage{array}
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\usepackage{babel}
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\usepackage{bbding}
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\usepackage{bbm}
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\usepackage{calc}
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\usepackage{sectsty}
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\usepackage{titlesec}
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\usepackage{fancyhdr}
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\usepackage{footmisc}
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\usepackage{geometry}
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\usepackage{graphicx}
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\usepackage{ifpdf}
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\usepackage{ifthen}
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\usepackage{ifnextok}
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\usepackage{longtable}
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\usepackage{nameref}
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\usepackage{nowtoaux}
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\usepackage{paralist}
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\usepackage{enumerate} %% nach [paralist]
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\usepackage{pgf}
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\usepackage{pgfplots}
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\usepackage{proof}
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\usepackage{refcount}
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\usepackage{relsize}
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\usepackage{savesym}
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\usepackage{stmaryrd}
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\usepackage{yfonts} %% <— Altgotische Fonts
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\usepackage{tikz}
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\usepackage{xy}
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\usepackage{undertilde}
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\usepackage{ulem} %% <– f\"ur besseren \underline-Befehl (\ul)
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\usepackage{xcolor}
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\usepackage{xspace}
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\usepackage{xstring}
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\usepackage{hyperref}
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\usepackage{cleveref} % must vor hyperref geladen werden.
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\pgfplotsset{compat=newest}
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\usetikzlibrary{math}
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\usetikzlibrary{
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angles,
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arrows,
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automata,
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calc,
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decorations,
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decorations.pathmorphing,
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decorations.pathreplacing,
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positioning,
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patterns,
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quotes,
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}
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%% \var ≈ alter Befehl
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%% \xvar ≈ wie das neue Package \var interpretieren soll.
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\savesymbol{Diamond}
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\savesymbol{emptyset}
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\savesymbol{ggg}
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\savesymbol{int}
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\savesymbol{lll}
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\savesymbol{RectangleBold}
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\savesymbol{langle}
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\savesymbol{rangle}
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\savesymbol{hookrightarrow}
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\savesymbol{hookleftarrow}
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\savesymbol{Asterisk}
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\usepackage{mathabx}
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\usepackage{wasysym}
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\let\varemptyset=\emptyset
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\restoresymbol{x}{Diamond}
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\restoresymbol{x}{emptyset}
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\restoresymbol{x}{ggg}
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\restoresymbol{x}{int}
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\restoresymbol{x}{lll}
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\restoresymbol{x}{RectangleBold}
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\restoresymbol{x}{langle}
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\restoresymbol{x}{rangle}
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\restoresymbol{x}{hookrightarrow}
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\restoresymbol{x}{hookleftarrow}
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\restoresymbol{x}{Asterisk}
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\ifpdf
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\usepackage{pdfcolmk}
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\fi
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\usepackage{mdframed}
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%% Force-Import aus MnSymbol
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\DeclareFontFamily{U}{MnSymbolA}{}
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\DeclareFontShape{U}{MnSymbolA}{m}{n}{
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<-6> MnSymbolA5
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<6-7> MnSymbolA6
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<7-8> MnSymbolA7
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<8-9> MnSymbolA8
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<9-10> MnSymbolA9
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<10-12> MnSymbolA10
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<12-> MnSymbolA12
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}{}
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\DeclareFontShape{U}{MnSymbolA}{b}{n}{
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<-6> MnSymbolA-Bold5
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<6-7> MnSymbolA-Bold6
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<7-8> MnSymbolA-Bold7
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<8-9> MnSymbolA-Bold8
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<9-10> MnSymbolA-Bold9
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<10-12> MnSymbolA-Bold10
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<12-> MnSymbolA-Bold12
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}{}
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\DeclareSymbolFont{MnSyA}{U}{MnSymbolA}{m}{n}
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\DeclareMathSymbol{\lcirclearrowright}{\mathrel}{MnSyA}{252}
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\DeclareMathSymbol{\lcirclearrowdown}{\mathrel}{MnSyA}{255}
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\DeclareMathSymbol{\rcirclearrowleft}{\mathrel}{MnSyA}{250}
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\DeclareMathSymbol{\rcirclearrowdown}{\mathrel}{MnSyA}{251}
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\DeclareFontFamily{U}{MnSymbolC}{}
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\DeclareSymbolFont{MnSyC}{U}{MnSymbolC}{m}{n}
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\DeclareFontShape{U}{MnSymbolC}{m}{n}{
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<-6> MnSymbolC5
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<6-7> MnSymbolC6
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<7-8> MnSymbolC7
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<8-9> MnSymbolC8
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<9-10> MnSymbolC9
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<10-12> MnSymbolC10
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<12-> MnSymbolC12%
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}{}
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\DeclareMathSymbol{\powerset}{\mathord}{MnSyC}{180}
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%% ********************************************************************************
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%% FILE: src/setup-parameters.tex
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%% ********************************************************************************
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\def\boolwahr{true}
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\def\boolfalsch{false}
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\def\boolleer{}
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\let\documenttwosided\boolfalsch
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\let\boolinappendix\boolfalsch
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\let\boolinmdframed\boolfalsch
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\let\eqtagset\boolfalsch
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\let\eqtaglabel\boolleer
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\let\eqtagsymb\boolleer
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\newcount\bufferctr
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\newcount\bufferreplace
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\newlength\rtab
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\newlength\gesamtlinkerRand
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\newlength\gesamtrechterRand
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\newlength\ownspaceabovethm
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\newlength\ownspacebelowthm
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\setlength{\rtab}{0.025\textwidth}
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\setlength{\ownspaceabovethm}{0.5\baselineskip}
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\setlength{\ownspacebelowthm}{0.5\baselineskip}
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\setlength{\gesamtlinkerRand}{0pt}
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\setlength{\gesamtrechterRand}{0pt}
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\def\secnumberingpt{$\cdot$}
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\def\secnumberingseppt{.}
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\def\subsecnumberingseppt{}
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\def\thmnumberingpt{$\cdot$}
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\def\thmnumberingseppt{}
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\def\thmForceSepPt{.}
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|
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|
|||
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\definecolor{leer}{gray}{1}
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\definecolor{hellgrau}{gray}{0.85}
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\definecolor{dunkelgrau}{gray}{0.5}
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\definecolor{maroon}{rgb}{0.6901961,0.1882353,0.3764706}
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\definecolor{dunkelgruen}{rgb}{0.015625,0.363281,0.109375}
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\definecolor{dunkelrot}{rgb}{0.5450980392,0,0}
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\definecolor{dunkelblau}{rgb}{0,0,0.5450980392}
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\definecolor{blau}{rgb}{0,0,1}
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\definecolor{newresult}{rgb}{0.6,0.6,0.6}
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\definecolor{improvedresult}{rgb}{0.9,0.9,0.9}
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\definecolor{hervorheben}{rgb}{0,0.9,0.7}
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\definecolor{starkesblau}{rgb}{0.1019607843,0.3176470588,0.8156862745}
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\definecolor{achtung}{rgb}{1,0.5,0.5}
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\definecolor{frage}{rgb}{0.5,1,0.5}
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\definecolor{schreibweise}{rgb}{0,0.7,0.9}
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\definecolor{axiom}{rgb}{0,0.3,0.3}
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|
|||
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%% ********************************************************************************
|
|||
|
|
%% FILE: src/setup-macros.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
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%% ****************************************************************
|
|||
|
|
%% TEX:
|
|||
|
|
%% ****************************************************************
|
|||
|
|
|
|||
|
|
\def\let@name#1#2{\expandafter\let\csname #1\expandafter\endcsname\csname #2\endcsname\relax}
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|||
|
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\DeclareRobustCommand\crfamily{\fontfamily{ccr}\selectfont}
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|||
|
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\DeclareTextFontCommand{\textcr}{\crfamily}
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|||
|
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|
|||
|
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\def\nichtzeigen#1{\phantom{#1}}
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|||
|
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|
|||
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%% ****************************************************************
|
|||
|
|
%% SPACING:
|
|||
|
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%% ****************************************************************
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|||
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|
|||
|
|
\def\ifthenelseleer#1#2#3{\ifthenelse{\equal{#1}{}}{#2}{#1#3}}
|
|||
|
|
\def\bedingtesspaceexpand#1#2#3{\ifthenelseleer{\csname #1\endcsname}{#3}{#2#3}}
|
|||
|
|
\def\voritemise{\leavevmode\nvraum{1}}
|
|||
|
|
\def\hraum{\null\hfill\null}
|
|||
|
|
\def\vraum{\null\vfill\null}
|
|||
|
|
\def\nvraum{\@ifnextchar\bgroup{\nvraum@c}{\nvraum@bes}}
|
|||
|
|
\def\nvraum@c#1{\vspace*{-#1\baselineskip}}
|
|||
|
|
\def\nvraum@bes{\vspace*{-\baselineskip}}
|
|||
|
|
\def\erlaubeplatz{\relax\ifmmode\else\@\xspace\fi}
|
|||
|
|
\def\entferneplatz{\relax\ifmmode\else\expandafter\@gobble\fi}
|
|||
|
|
|
|||
|
|
%% ****************************************************************
|
|||
|
|
%% TAGS / BEZEICHNUNGEN / LABELLING:
|
|||
|
|
%% ****************************************************************
|
|||
|
|
|
|||
|
|
\def\send@toaux#1{\@bsphack\protected@write\@auxout{}{\string#1}\@esphack}
|
|||
|
|
|
|||
|
|
%% \rlabel{LABEL}[CTR]{CREF-SHORT}{CREF-LONG}{DISPLAYTEXT}
|
|||
|
|
\def\rlabel#1[#2]#3#4#5{#5\rlabel@aux{#1}[#2]{#3}{#4}{#5}}
|
|||
|
|
\def\rlabel@aux#1[#2]#3#4#5{%
|
|||
|
|
\send@toaux{\newlabel{#1}{{\@currentlabel}{\thepage}{{\unexpanded{#5}}}{#2.\csname the#2\endcsname}{}}}\relax%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
%% \tag@rawscheme{CREF-SHORT}{CREF-LONG}[CTR]{LEFT-BRKT}{RIGHT-BRKT} [LABEL]{DISPLAYTEXT}
|
|||
|
|
\def\tag@rawscheme#1#2[#3]#4#5{\@ifnextchar[{\tag@rawscheme@{#1}{#2}[#3]{#4}{#5}}{\tag@rawscheme@{#1}{#2}[#3]{#4}{#5}[*]}}
|
|||
|
|
\def\tag@rawscheme@#1#2[#3]#4#5[#6]{\@ifnextchar\bgroup{\tag@rawscheme@@{#1}{#2}[#3]{#4}{#5}[#6]}{\tag@rawscheme@@{#1}{#2}[#3]{#4}{#5}[#6]{}}}
|
|||
|
|
\def\tag@rawscheme@@#1#2[#3]#4#5[#6]#7{%
|
|||
|
|
\ifthenelse{\equal{#6}{*}}{%
|
|||
|
|
\ifthenelse{\equal{#7}{\boolleer}}{\refstepcounter{#3}#4\csname the#3\endcsname#5}{#4#7#5}%
|
|||
|
|
}{%
|
|||
|
|
\refstepcounter{#3}#4%
|
|||
|
|
\ifthenelse{\equal{#7}{\boolleer}}{\rlabel{#6}[#3]{#1}{#2}{\csname the#3\endcsname}}{\rlabel{#6}[#3]{#1}{#2}{#7}}%
|
|||
|
|
#5%
|
|||
|
|
}%
|
|||
|
|
}
|
|||
|
|
%% \tag@scheme{CREF-SHORT}{CREF-LONG}[CTR] [LABEL]{DISPLAYTEXT}
|
|||
|
|
\def\tag@scheme#1#2[#3]{\tag@rawscheme{#1}{#2}[#3]{\upshape(}{\upshape)}}
|
|||
|
|
|
|||
|
|
%% \eqtag[LABEL]{DISPLAYTEXT}
|
|||
|
|
\def\eqtag@post#1{\makebox[0pt][r]{#1}}
|
|||
|
|
\def\eqtag@pre{\tag@scheme{Eq}{Equation}[Xe]}
|
|||
|
|
\def\eqtag{\@ifnextchar[{\eqtag@}{\eqtag@[*]}}
|
|||
|
|
\def\eqtag@[#1]{\@ifnextchar\bgroup{\eqtag@@[#1]}{\eqtag@@[#1]{}}}
|
|||
|
|
\def\eqtag@@[#1]#2{\eqtag@post{\eqtag@pre[#1]{#2}}}
|
|||
|
|
|
|||
|
|
\def\eqcref#1{\text{(\ref{#1})}}
|
|||
|
|
\def\ptcref#1{\ref{#1}}
|
|||
|
|
\def\punktlabel#1{\label{it:#1:\beweislabel}}
|
|||
|
|
\def\punktcref#1{\eqcref{it:#1:\beweislabel}}
|
|||
|
|
\def\crefit#1#2{\cref{#1}~\eqcref{it:#2:#1}}
|
|||
|
|
\def\Crefit#1#2{\Cref{#1}~\eqcref{it:#2:#1}}
|
|||
|
|
|
|||
|
|
%% UNDER/OVERSET BEFEHLE
|
|||
|
|
\def\opfromto[#1]_#2^#3{\underset{#2}{\overset{#3}{#1}}}
|
|||
|
|
\def\textoverset#1#2{\overset{\text{#1}}{#2}}
|
|||
|
|
\def\textunderset#1#2{\underset{#2}{\text{#1}}}
|
|||
|
|
\def\crefoverset#1#2{\textoverset{\cref{#1}}{#2}}
|
|||
|
|
\def\Crefoverset#1#2{\textoverset{\Cref{#1}}{#2}}
|
|||
|
|
\def\crefunderset#1#2{\textunderset{#2}{\cref{#1}}}
|
|||
|
|
\def\Crefunderset#1#2{\textunderset{#2}{\Cref{#1}}}
|
|||
|
|
\def\eqcrefoverset#1#2{\textoverset{\eqcref{#1}}{#2}}
|
|||
|
|
\def\eqcrefunderset#1#2{\textunderset{#2}{\eqcref{#1}}}
|
|||
|
|
\def\mathclap#1{#1}
|
|||
|
|
\def\oberunterset#1{\@ifnextchar^{\oberunterset@oben{#1}}{\oberunterset@unten{#1}}}
|
|||
|
|
\def\oberunterset@oben#1^#2_#3{\underset{\mathclap{#3}}{\overset{\mathclap{#2}}{#1}}}
|
|||
|
|
\def\oberunterset@unten#1_#2^#3{\underset{\mathclap{#2}}{\overset{\mathclap{#3}}{#1}}}
|
|||
|
|
\def\breitunderbrace#1_#2{\underbrace{#1}_{\mathclap{#2}}}
|
|||
|
|
\def\breitoverbrace#1^#2{\overbrace{#1}^{\mathclap{#2}}}
|
|||
|
|
\def\breitunderbracket#1_#2{\underbracket{#1}_{\mathclap{#2}}}
|
|||
|
|
\def\breitoverbracket#1^#2{\overbracket{#1}^{\mathclap{#2}}}
|
|||
|
|
|
|||
|
|
\def\generatenestedsecnumbering#1#2#3{%
|
|||
|
|
\expandafter\gdef\csname thelong#3\endcsname{%
|
|||
|
|
\expandafter\csname the#2\endcsname%
|
|||
|
|
\secnumberingpt%
|
|||
|
|
\expandafter\csname #1\endcsname{#3}%
|
|||
|
|
}%
|
|||
|
|
\expandafter\gdef\csname theshort#3\endcsname{%
|
|||
|
|
\expandafter\csname #1\endcsname{#3}%
|
|||
|
|
}%
|
|||
|
|
}
|
|||
|
|
\def\generatenestedthmnumbering#1#2#3{%
|
|||
|
|
\expandafter\gdef\csname the#3\endcsname{%
|
|||
|
|
\expandafter\csname the#2\endcsname%
|
|||
|
|
\thmnumberingpt%
|
|||
|
|
\expandafter\csname #1\endcsname{#3}%
|
|||
|
|
}%
|
|||
|
|
\expandafter\gdef\csname theshort#3\endcsname{%
|
|||
|
|
\expandafter\csname #1\endcsname{#3}%
|
|||
|
|
}%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
%% ****************************************************************
|
|||
|
|
%% ALLG. MACROS:
|
|||
|
|
%% ****************************************************************
|
|||
|
|
|
|||
|
|
\def\+#1{\addtocounter{#1}{1}}
|
|||
|
|
\def\setcounternach#1#2{\setcounter{#1}{#2}\addtocounter{#1}{-1}}
|
|||
|
|
\def\textsubscript#1{${}_{\textup{#1}}$}
|
|||
|
|
\def\rome#1{\overline{\underline{#1}}}
|
|||
|
|
\def\textTODO{\text{[{\large\textcolor{red}{More work needed!}}]}}
|
|||
|
|
\def\hlineEIGENpt{\hdashline[0.5pt/5pt]}
|
|||
|
|
\def\clineEIGENpt#1{\cdashline{#1}[0.5pt/5pt]}
|
|||
|
|
|
|||
|
|
\def\forcepunkt#1{#1\IfEndWith{#1}{.}{}{.}}
|
|||
|
|
\def\lateinabkuerzung#1#2{%
|
|||
|
|
\expandafter\gdef\csname #1\endcsname{\emph{#2}\@ifnextchar.{\entferneplatz}{\erlaubeplatz}}
|
|||
|
|
}
|
|||
|
|
\def\deutscheabkuerzung#1#2{%
|
|||
|
|
\expandafter\gdef\csname #1\endcsname{{#2}\@ifnextchar.{\entferneplatz}{\erlaubeplatz}}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
%% ****************************************************************
|
|||
|
|
%% MATHE
|
|||
|
|
%% ****************************************************************
|
|||
|
|
|
|||
|
|
\def\matrix#1{\left(\begin{array}{#1}}
|
|||
|
|
\def\endmatrix{\end{array}\right)}
|
|||
|
|
\def\smatrix{\left(\begin{smallmatrix}}
|
|||
|
|
\def\endsmatrix{\end{smallmatrix}\right)}
|
|||
|
|
|
|||
|
|
\def\multiargrekursiverbefehl#1#2#3#4#5#6#7#8{%
|
|||
|
|
\expandafter\gdef\csname#1\endcsname #2##1#4{\csname #1@anfang\endcsname##1#3\egroup}
|
|||
|
|
\expandafter\def\csname #1@anfang\endcsname##1#3{#5##1\@ifnextchar\egroup{\csname #1@ende\endcsname}{#7\csname #1@mitte\endcsname}}
|
|||
|
|
\expandafter\def\csname #1@mitte\endcsname##1#3{#6##1\@ifnextchar\egroup{\csname #1@ende\endcsname}{#7\csname #1@mitte\endcsname}}
|
|||
|
|
\expandafter\def\csname #1@ende\endcsname##1{#8}
|
|||
|
|
}
|
|||
|
|
\multiargrekursiverbefehl{svektor}{[}{;}{]}{\begin{smatrix}}{}{\\}{\\\end{smatrix}}
|
|||
|
|
\multiargrekursiverbefehl{vektor}{[}{;}{]}{\begin{matrix}{c}}{}{\\}{\\\end{matrix}}
|
|||
|
|
\multiargrekursiverbefehl{vektorzeile}{}{,}{;}{}{&}{}{}
|
|||
|
|
\multiargrekursiverbefehl{matlabmatrix}{[}{;}{]}{\begin{smatrix}\vektorzeile}{\vektorzeile}{;\\}{;\end{smatrix}}
|
|||
|
|
|
|||
|
|
\def\cases[#1]#2{\left\{\begin{array}[#1]{#2}}
|
|||
|
|
\def\endcases{\end{array}\right.}
|
|||
|
|
|
|||
|
|
\def\BeweisRichtung[#1]{\@ifnextchar\bgroup{\@BeweisRichtung@c[#1]}{\@BeweisRichtung@bes[#1]}}
|
|||
|
|
\def\@BeweisRichtung@bes[#1]{{\bfseries(#1).~}}
|
|||
|
|
\def\@BeweisRichtung@c[#1]#2#3{{\bfseries(#2#1#3).~}}
|
|||
|
|
\def\erzeugeBeweisRichtungBefehle#1#2{
|
|||
|
|
\expandafter\gdef\csname #1text\endcsname##1##2{\BeweisRichtung[#2]{##1}{##2}}
|
|||
|
|
\expandafter\gdef\csname #1\endcsname{%
|
|||
|
|
\@ifnextchar\bgroup{\csname #1@\endcsname}{\csname #1text\endcsname{}{}}%
|
|||
|
|
}
|
|||
|
|
\expandafter\gdef\csname #1@\endcsname##1##2{%
|
|||
|
|
\csname #1text\endcsname{\punktcref{##1}}{\punktcref{##2}}%
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
\erzeugeBeweisRichtungBefehle{hinRichtung}{$\Longrightarrow$}
|
|||
|
|
\erzeugeBeweisRichtungBefehle{herRichtung}{$\Longleftarrow$}
|
|||
|
|
\erzeugeBeweisRichtungBefehle{hinherRichtung}{$\Longleftrightarrow$}
|
|||
|
|
|
|||
|
|
\def\cal#1{\mathcal{#1}}
|
|||
|
|
\def\brkt#1{\langle{}#1{}\rangle}
|
|||
|
|
\def\mathfrak#1{\mbox{\usefont{U}{euf}{m}{n}#1}}
|
|||
|
|
\def\kurs#1{\textit{#1}}
|
|||
|
|
\def\rectangleblack{\text{\RectangleBold}}
|
|||
|
|
\def\rectanglewhite{\text{\Rectangle}}
|
|||
|
|
\def\squareblack{\blacksquare}
|
|||
|
|
\def\squarewhite{\Box}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: src/setup-environments.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
%% **********************************************************************
|
|||
|
|
%% CLEVEREF: ************************************************************
|
|||
|
|
|
|||
|
|
\def\crefname@full#1#2#3{\crefname{#1}{#2}{#3}\Crefname{#1}{#2}{#3}}
|
|||
|
|
\crefname@full{chapter}{Kapitel}{Kapitel}
|
|||
|
|
\crefname@full{section}{Abschnitt}{Abschnitte}
|
|||
|
|
\crefname@full{figure}{Fig.}{Fig.}
|
|||
|
|
\crefname@full{subfigure}{Fig.}{Fig.}
|
|||
|
|
|
|||
|
|
\crefname@full{proof}{Beweis}{Beweise}
|
|||
|
|
\crefname@full{thm}{Theorem}{Theoreme}
|
|||
|
|
\crefname@full{satz}{Satz}{Sätze}
|
|||
|
|
\crefname@full{claim}{Behauptung}{Behauptungen}
|
|||
|
|
\crefname@full{lemm}{Lemma}{Lemmata}
|
|||
|
|
\crefname@full{cor}{Korollar}{Korollarien}
|
|||
|
|
\crefname@full{folg}{Folgerung}{Folgerungen}
|
|||
|
|
\crefname@full{prop}{Proposition}{Propositionen}
|
|||
|
|
\crefname@full{defn}{Definition}{Definitionen}
|
|||
|
|
\crefname@full{conv}{Konvention}{Konventionen}
|
|||
|
|
\crefname@full{fact}{Fakt}{Fakten}
|
|||
|
|
\crefname@full{rem}{Bemerkung}{Bemerkungen}
|
|||
|
|
\crefname@full{qstn}{Frage}{Fragen}
|
|||
|
|
\crefname@full{e.g.}{Beipsiel}{Beipsiele}
|
|||
|
|
|
|||
|
|
%% ****************************************************************
|
|||
|
|
%% THEOREME:
|
|||
|
|
%% ****************************************************************
|
|||
|
|
|
|||
|
|
\def\qedEIGEN#1{\@ifnextchar[{\qedEIGEN@c{#1}}{\qedEIGEN@bes{#1}}}%]
|
|||
|
|
\def\qedEIGEN@bes#1{%
|
|||
|
|
\parfillskip=0pt% % so \par doesnt push \square to left
|
|||
|
|
\widowpenalty=10000% % so we dont break the page before \square
|
|||
|
|
\displaywidowpenalty=10000% % ditto
|
|||
|
|
\finalhyphendemerits=0% % TeXbook exercise 14.32
|
|||
|
|
\leavevmode% % \nobreak means lines not pages
|
|||
|
|
\unskip% % remove previous space or glue
|
|||
|
|
\nobreak% % don’t break lines
|
|||
|
|
\hfil% % ragged right if we spill over
|
|||
|
|
\penalty50% % discouragement to do so
|
|||
|
|
\hskip.2em% % ensure some space
|
|||
|
|
\null% % anchor following \hfill
|
|||
|
|
\hfill% % push \square to right
|
|||
|
|
#1% % the end-of-proof mark
|
|||
|
|
\par%
|
|||
|
|
}
|
|||
|
|
\def\qedEIGEN@c#1[#2]{%
|
|||
|
|
\parfillskip=0pt% % so \par doesnt push \square to left
|
|||
|
|
\widowpenalty=10000% % so we dont break the page before \square
|
|||
|
|
\displaywidowpenalty=10000% % ditto
|
|||
|
|
\finalhyphendemerits=0% % TeXbook exercise 14.32
|
|||
|
|
\leavevmode% % \nobreak means lines not pages
|
|||
|
|
\unskip% % remove previous space or glue
|
|||
|
|
\nobreak% % don’t break lines
|
|||
|
|
\hfil% % ragged right if we spill over
|
|||
|
|
\penalty50% % discouragement to do so
|
|||
|
|
\hskip.2em% % ensure some space
|
|||
|
|
\null% % anchor following \hfill
|
|||
|
|
\hfill% % push \square to right
|
|||
|
|
{#1~{\smaller\bfseries\upshape (#2)}}%
|
|||
|
|
\par%
|
|||
|
|
}
|
|||
|
|
\def\qedVARIANT#1#2{
|
|||
|
|
\expandafter\def\csname ennde#1Sign\endcsname{#2}
|
|||
|
|
\expandafter\def\csname ennde#1\endcsname{\@ifnextchar[{\qedEIGEN@c{#2}}{\qedEIGEN@bes{#2}}} %]
|
|||
|
|
}
|
|||
|
|
\qedVARIANT{OfProof}{$\squareblack$}
|
|||
|
|
\qedVARIANT{OfWork}{\rectangleblack}
|
|||
|
|
\qedVARIANT{OfSomething}{$\dashv$}
|
|||
|
|
\qedVARIANT{OnNeutral}{$\lozenge$} % \lozenge \bigcirc \blacklozenge
|
|||
|
|
\def\qedsymbol{\enndeOfProofSign}
|
|||
|
|
\def\proofSymbol{\enndeOfProofSign}
|
|||
|
|
|
|||
|
|
\def\ra@pretheoremwork{
|
|||
|
|
\setlength{\theorempreskipamount}{\ownspaceabovethm}
|
|||
|
|
}
|
|||
|
|
\def\rathmtransfer#1#2{
|
|||
|
|
\expandafter\def\csname #2\endcsname{\csname #1\endcsname}
|
|||
|
|
\expandafter\def\csname end#2\endcsname{\csname end#1\endcsname}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\ranewthm#1#2#3[#4]{
|
|||
|
|
%% FOR \BEGIN{THM}
|
|||
|
|
\theoremstyle{\current@theoremstyle}
|
|||
|
|
\theoremseparator{\current@theoremseparator}
|
|||
|
|
\theoremprework{\ra@pretheoremwork}
|
|||
|
|
\@ifundefined{#1@basic}{\newtheorem{#1@basic}[#4]{#2}}{\renewtheorem{#1@basic}[#4]{#2}}
|
|||
|
|
%% FOR \BEGIN{THM}[...]
|
|||
|
|
\theoremstyle{\current@theoremstyle}
|
|||
|
|
\theoremseparator{\thmForceSepPt}
|
|||
|
|
\theoremprework{\ra@pretheoremwork}
|
|||
|
|
\@ifundefined{#1@withName}{\newtheorem{#1@withName}[#4]{#2}}{\renewtheorem{#1@withName}[#4]{#2}}
|
|||
|
|
%% FOR \BEGIN{THM*}
|
|||
|
|
\theoremstyle{nonumberplain}
|
|||
|
|
\theoremseparator{\thmForceSepPt}
|
|||
|
|
\theoremprework{\ra@pretheoremwork}
|
|||
|
|
\@ifundefined{#1@star@basic}{\newtheorem{#1@star@basic}[#4]{#2}}{\renewtheorem{#1@star@basic}[#4]{#2}}
|
|||
|
|
%% FOR \BEGIN{THM*}[...]
|
|||
|
|
\theoremstyle{nonumberplain}
|
|||
|
|
\theoremseparator{\thmForceSepPt}
|
|||
|
|
\theoremprework{\ra@pretheoremwork}
|
|||
|
|
\@ifundefined{#1@star@withName}{\newtheorem{#1@star@withName}[#4]{#2}}{\renewtheorem{#1@star@withName}[#4]{#2}}
|
|||
|
|
%% GENERATE ENVIRONMENTS:
|
|||
|
|
\umbauenenv{#1}{#3}[#4]
|
|||
|
|
\umbauenenv{#1@star}{#3}[#4]
|
|||
|
|
%% TRANSFER *-DEFINITION
|
|||
|
|
\rathmtransfer{#1@star}{#1*}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\umbauenenv#1#2[#3]{%
|
|||
|
|
%% \BEGIN{THM}...
|
|||
|
|
\expandafter\def\csname #1\endcsname{\relax%
|
|||
|
|
\@ifnextchar[{\csname #1@\endcsname}{\csname #1@\endcsname[*]}%
|
|||
|
|
}
|
|||
|
|
%% \BEGIN{THM}[ANFANG]...
|
|||
|
|
\expandafter\def\csname #1@\endcsname[##1]{\relax%
|
|||
|
|
\@ifnextchar[{\csname #1@@\endcsname[##1]}{\csname #1@@\endcsname[##1][*]}%
|
|||
|
|
}
|
|||
|
|
%% \BEGIN{THM}[ANFANG][SCHLUSS]
|
|||
|
|
\expandafter\def\csname #1@@\endcsname[##1][##2]{%
|
|||
|
|
\ifx*##1%
|
|||
|
|
\def\enndeOfBlock{\csname end#1@basic\endcsname}
|
|||
|
|
\csname #1@basic\endcsname%
|
|||
|
|
\else%
|
|||
|
|
\def\enndeOfBlock{\csname end#1@withName\endcsname}
|
|||
|
|
\csname #1@withName\endcsname[##1]%
|
|||
|
|
\fi%
|
|||
|
|
\def\makelabel####1{%
|
|||
|
|
\gdef\beweislabel{####1}%
|
|||
|
|
\label{\beweislabel}%
|
|||
|
|
}%
|
|||
|
|
\ifx*##2%
|
|||
|
|
\def\enndeSymbol{\qedEIGEN{#2}}
|
|||
|
|
\else%
|
|||
|
|
\def\enndeSymbol{\qedEIGEN{#2}[##2]}
|
|||
|
|
\fi
|
|||
|
|
}
|
|||
|
|
%% \END{THM}
|
|||
|
|
\expandafter\gdef\csname end#1\endcsname{\enndeSymbol\enndeOfBlock}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
%% NEWTHEOREM EINSTELLUNGSOPTIONEN:
|
|||
|
|
%% F\"UR \theoremstyle
|
|||
|
|
%% plain Emulates original LATEX defin, except uses param \theorem...skipamount.
|
|||
|
|
%% break Header followed by line break.
|
|||
|
|
%% change Header, Number and Text are interchanged, without a line break.
|
|||
|
|
%% changebreak =change, but with a line break after Header.
|
|||
|
|
%% margin Number in left margin, without a line break.
|
|||
|
|
%% marginbreak =margin, but with a line break after the header.
|
|||
|
|
%% nonumberplain =plain, without number.
|
|||
|
|
%% nonumberbreak =break, without number.
|
|||
|
|
%% empty No number, no name. Only the optional argument is typeset.
|
|||
|
|
%% \theoremclass \theoremnumbering
|
|||
|
|
%% \theorempreskip \theorempostkip \theoremindent
|
|||
|
|
%% \theoremprework \theorempostwork
|
|||
|
|
|
|||
|
|
\def\current@theoremstyle{plain}
|
|||
|
|
\def\current@theoremseparator{\thmnumberingseppt}
|
|||
|
|
\theoremstyle{\current@theoremstyle}
|
|||
|
|
\theoremseparator{\current@theoremseparator}
|
|||
|
|
\theoremsymbol{}
|
|||
|
|
|
|||
|
|
\newtheorem{X}{X}[chapter] % for most theorems
|
|||
|
|
\newtheorem{Xe}{Xe}[chapter] % for equations
|
|||
|
|
\newtheorem*{Xdisplaynone}{Xdisplaynone}[chapter] % a dummy counter, that will never be displayed.
|
|||
|
|
\newtheorem{Xsp}{Xsp}[chapter] % for special theorems
|
|||
|
|
\generatenestedthmnumbering{arabic}{chapter}{X}
|
|||
|
|
\generatenestedthmnumbering{arabic}{chapter}{Xe}
|
|||
|
|
\generatenestedthmnumbering{Roman}{chapter}{Xsp}
|
|||
|
|
\let\theXsp\theshortXsp
|
|||
|
|
|
|||
|
|
\theoremheaderfont{\upshape\bfseries}
|
|||
|
|
\theorembodyfont{\slshape}
|
|||
|
|
|
|||
|
|
\ranewthm{thm}{Theorem}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{satz}{Satz}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{claim}{Behauptung}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{lemm}{Lemma}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{cor}{Korollar}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{folg}{Folgerung}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{prop}{Proposition}{\enndeOnNeutralSign}[X]
|
|||
|
|
|
|||
|
|
\theorembodyfont{\upshape}
|
|||
|
|
|
|||
|
|
\ranewthm{defn}{Definition}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{conv}{Konvention}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{e.g.}{Beipsiel}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{fact}{Fakt}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{rem}{Bemerkung}{\enndeOnNeutralSign}[X]
|
|||
|
|
\ranewthm{qstn}{Frage}{\enndeOnNeutralSign}[X]
|
|||
|
|
|
|||
|
|
\theoremheaderfont{\itshape\bfseries}
|
|||
|
|
\theorembodyfont{\upshape}
|
|||
|
|
|
|||
|
|
\ranewthm{proof@tmp}{Beweis}{\enndeOfProofSign}[Xdisplaynone]
|
|||
|
|
\rathmtransfer{proof@tmp*}{proof}
|
|||
|
|
|
|||
|
|
\def\behauptungbeleg@claim{%
|
|||
|
|
\iflanguage{british}{Claim}{%
|
|||
|
|
\iflanguage{english}{Claim}{%
|
|||
|
|
\iflanguage{ngerman}{Behauptung}{%
|
|||
|
|
\iflanguage{russian}{Утверждение}{%
|
|||
|
|
Claim%
|
|||
|
|
}}}}%
|
|||
|
|
}
|
|||
|
|
\def\behauptungbeleg@pf@kurz{%
|
|||
|
|
\iflanguage{british}{Pf}{%
|
|||
|
|
\iflanguage{english}{Pf}{%
|
|||
|
|
\iflanguage{ngerman}{Bew}{%
|
|||
|
|
\iflanguage{russian}{Доказательство}{%
|
|||
|
|
Pf%
|
|||
|
|
}}}}%
|
|||
|
|
}
|
|||
|
|
\def\behauptungbeleg{\@ifnextchar\bgroup{\behauptungbeleg@c}{\behauptungbeleg@bes}}
|
|||
|
|
\def\behauptungbeleg@c#1{\item[{\bfseries \behauptungbeleg@claim\erlaubeplatz #1.}]}
|
|||
|
|
\def\behauptungbeleg@bes{\item[{\bfseries \behauptungbeleg@claim.}]}
|
|||
|
|
\def\belegbehauptung{\item[{\bfseries\itshape\behauptungbeleg@pf@kurz.}]}
|
|||
|
|
|
|||
|
|
%% ****************************************************************
|
|||
|
|
%% ALTE UMGEBUNGEN:
|
|||
|
|
%% ****************************************************************
|
|||
|
|
|
|||
|
|
\newcolumntype{\RECHTS}[1]{>{\raggedleft}p{#1}}
|
|||
|
|
\newcolumntype{\LINKS}[1]{>{\raggedright}p{#1}}
|
|||
|
|
\newcolumntype{m}{>{$}l<{$}}
|
|||
|
|
\newcolumntype{C}{>{$}c<{$}}
|
|||
|
|
\newcolumntype{L}{>{$}l<{$}}
|
|||
|
|
\newcolumntype{R}{>{$}r<{$}}
|
|||
|
|
\newcolumntype{0}{@{\hspace{0pt}}}
|
|||
|
|
\newcolumntype{\LINKSRAND}{@{\hspace{\@totalleftmargin}}}
|
|||
|
|
\newcolumntype{h}{@{\extracolsep{\fill}}}
|
|||
|
|
\newcolumntype{i}{>{\itshape}}
|
|||
|
|
\newcolumntype{t}{@{\hspace{\tabcolsep}}}
|
|||
|
|
\newcolumntype{q}{@{\hspace{1em}}}
|
|||
|
|
\newcolumntype{n}{@{\hspace{-\tabcolsep}}}
|
|||
|
|
\newcolumntype{M}[2]{%
|
|||
|
|
>{\begin{minipage}{#2}\begin{math}}%
|
|||
|
|
{#1}%
|
|||
|
|
<{\end{math}\end{minipage}}%
|
|||
|
|
}
|
|||
|
|
\newcolumntype{T}[2]{%
|
|||
|
|
>{\begin{minipage}{#2}}%
|
|||
|
|
{#1}%
|
|||
|
|
<{\end{minipage}}%
|
|||
|
|
}
|
|||
|
|
\setlength{\LTpre}{\baselineskip}
|
|||
|
|
\setlength{\LTpost}{0pt}
|
|||
|
|
\def\center{\centering}
|
|||
|
|
\def\endcenter{}
|
|||
|
|
|
|||
|
|
\def\punkteumgebung@genbefehl#1#2#3{
|
|||
|
|
\punkteumgebung@genbefehl@{#1}{#2}{#3}{}{}
|
|||
|
|
\punkteumgebung@genbefehl@{multi#1}{#2}{#3}{
|
|||
|
|
\setlength{\columnsep}{10pt}%
|
|||
|
|
\setlength{\columnseprule}{0pt}%
|
|||
|
|
\begin{multicols}{\thecolumnanzahl}%
|
|||
|
|
}{\end{multicols}\nvraum{1}}
|
|||
|
|
}
|
|||
|
|
\def\punkteumgebung@genbefehl@#1#2#3#4#5{
|
|||
|
|
\expandafter\gdef\csname #1\endcsname{
|
|||
|
|
\@ifnextchar\bgroup{\csname #1@c\endcsname}{\csname #1@bes\endcsname}
|
|||
|
|
}%]
|
|||
|
|
\expandafter\def\csname #1@c\endcsname##1{
|
|||
|
|
\@ifnextchar[{\csname #1@c@\endcsname{##1}}{\csname #1@c@\endcsname{##1}[\z@]}
|
|||
|
|
}%]
|
|||
|
|
\expandafter\def\csname #1@c@\endcsname##1[##2]{
|
|||
|
|
\@ifnextchar[{\csname #1@c@@\endcsname{##1}[##2]}{\csname #1@c@@\endcsname{##1}[##2][\z@]}
|
|||
|
|
}%]
|
|||
|
|
\expandafter\def\csname #1@c@@\endcsname##1[##2][##3]{
|
|||
|
|
\let\alterlinkerRand\gesamtlinkerRand
|
|||
|
|
\let\alterrechterRand\gesamtrechterRand
|
|||
|
|
\addtolength{\gesamtlinkerRand}{##2}
|
|||
|
|
\addtolength{\gesamtrechterRand}{##3}
|
|||
|
|
\advance\linewidth -##2%
|
|||
|
|
\advance\linewidth -##3%
|
|||
|
|
\advance\@totalleftmargin ##2%
|
|||
|
|
\parshape\@ne \@totalleftmargin\linewidth%
|
|||
|
|
#4
|
|||
|
|
\begin{#2}[\upshape ##1]%
|
|||
|
|
\setlength{\parskip}{0.5\baselineskip}\relax%
|
|||
|
|
\setlength{\topsep}{\z@}\relax%
|
|||
|
|
\setlength{\partopsep}{\z@}\relax%
|
|||
|
|
\setlength{\parsep}{\parskip}\relax%
|
|||
|
|
\setlength{\itemsep}{#3}\relax%
|
|||
|
|
\setlength{\listparindent}{\z@}\relax%
|
|||
|
|
\setlength{\itemindent}{\z@}\relax%
|
|||
|
|
}
|
|||
|
|
\expandafter\def\csname #1@bes\endcsname{
|
|||
|
|
\@ifnextchar[{\csname #1@bes@\endcsname}{\csname #1@bes@\endcsname[\z@]}
|
|||
|
|
}%]
|
|||
|
|
\expandafter\def\csname #1@bes@\endcsname[##1]{
|
|||
|
|
\@ifnextchar[{\csname #1@bes@@\endcsname[##1]}{\csname #1@bes@@\endcsname[##1][\z@]}
|
|||
|
|
}%]
|
|||
|
|
\expandafter\def\csname #1@bes@@\endcsname[##1][##2]{
|
|||
|
|
\let\alterlinkerRand\gesamtlinkerRand
|
|||
|
|
\let\alterrechterRand\gesamtrechterRand
|
|||
|
|
\addtolength{\gesamtlinkerRand}{##1}
|
|||
|
|
\addtolength{\gesamtrechterRand}{##2}
|
|||
|
|
\advance\linewidth -##1%
|
|||
|
|
\advance\linewidth -##2%
|
|||
|
|
\advance\@totalleftmargin ##1%
|
|||
|
|
\parshape\@ne \@totalleftmargin\linewidth%
|
|||
|
|
#4
|
|||
|
|
\begin{#2}%
|
|||
|
|
\setlength{\parskip}{0.5\baselineskip}\relax%
|
|||
|
|
\setlength{\topsep}{\z@}\relax%
|
|||
|
|
\setlength{\partopsep}{\z@}\relax%
|
|||
|
|
\setlength{\parsep}{\parskip}\relax%
|
|||
|
|
\setlength{\itemsep}{#3}\relax%
|
|||
|
|
\setlength{\listparindent}{\z@}\relax%
|
|||
|
|
\setlength{\itemindent}{\z@}\relax%
|
|||
|
|
}
|
|||
|
|
\expandafter\gdef\csname end#1\endcsname{%
|
|||
|
|
\end{#2}#5
|
|||
|
|
\setlength{\gesamtlinkerRand}{\alterlinkerRand}
|
|||
|
|
\setlength{\gesamtlinkerRand}{\alterrechterRand}
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\ritempunkt{{\Large\textbullet}} % \textbullet, $\sqbullet$, $\blacktriangleright$
|
|||
|
|
\setdefaultitem{\ritempunkt}{\ritempunkt}{\ritempunkt}{\ritempunkt}
|
|||
|
|
\punkteumgebung@genbefehl{itemise}{compactitem}{\parskip}{}{}
|
|||
|
|
\punkteumgebung@genbefehl{kompaktitem}{compactitem}{\z@}{}{}
|
|||
|
|
\punkteumgebung@genbefehl{enumerate}{compactenum}{\parskip}{}{}
|
|||
|
|
\punkteumgebung@genbefehl{kompaktenum}{compactenum}{\z@}{}{}
|
|||
|
|
|
|||
|
|
\let\ALTthebibliography\thebibliography
|
|||
|
|
\renewenvironment{thebibliography}[1]{%
|
|||
|
|
\begin{ALTthebibliography}{#1}
|
|||
|
|
\addcontentsline{toc}{part}{\bibname}
|
|||
|
|
}{%
|
|||
|
|
\end{ALTthebibliography}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
%% ****************************************************************
|
|||
|
|
%% NEUE UMGEBUNGEN:
|
|||
|
|
%% ****************************************************************
|
|||
|
|
|
|||
|
|
\def\matrix#1{\left(\begin{array}[mc]{#1}}
|
|||
|
|
\def\endmatrix{\end{array}\right)}
|
|||
|
|
\def\smatrix{\left(\begin{smallmatrix}}
|
|||
|
|
\def\endsmatrix{\end{smallmatrix}\right)}
|
|||
|
|
\def\vector{\begin{matrix}{c}}
|
|||
|
|
\def\endvector{\end{matrix}}
|
|||
|
|
\def\svector{\begin{smatrix}}
|
|||
|
|
\def\endsvector{\end{smatrix}}
|
|||
|
|
|
|||
|
|
\def\multiargrekursiverbefehl#1#2#3#4#5#6#7#8{%
|
|||
|
|
\expandafter\gdef\csname#1\endcsname #2##1#4{\csname #1@anfang\endcsname##1#3\egroup}
|
|||
|
|
\expandafter\def\csname #1@anfang\endcsname##1#3{#5##1\@ifnextchar\egroup{\csname #1@ende\endcsname}{#7\csname #1@mitte\endcsname}}
|
|||
|
|
\expandafter\def\csname #1@mitte\endcsname##1#3{#6##1\@ifnextchar\egroup{\csname #1@ende\endcsname}{#7\csname #1@mitte\endcsname}}
|
|||
|
|
\expandafter\def\csname #1@ende\endcsname##1{#8}
|
|||
|
|
}
|
|||
|
|
\multiargrekursiverbefehl{svektor}{[}{;}{]}{\begin{smatrix}}{}{\\}{\\\end{smatrix}}
|
|||
|
|
\multiargrekursiverbefehl{vektor}{[}{;}{]}{\begin{matrix}{c}}{}{\\}{\\\end{matrix}}
|
|||
|
|
\multiargrekursiverbefehl{vektorzeile}{}{,}{;}{}{&}{}{}
|
|||
|
|
\multiargrekursiverbefehl{matlabmatrix}{[}{;}{]}{\begin{smatrix}\vektorzeile}{\vektorzeile}{;\\}{;\end{smatrix}}
|
|||
|
|
|
|||
|
|
\def\underbracenodisplay#1{%
|
|||
|
|
\mathop{\vtop{\m@th\ialign{##\crcr
|
|||
|
|
$\hfil\displaystyle{#1}\hfil$\crcr
|
|||
|
|
\noalign{\kern3\p@\nointerlineskip}%
|
|||
|
|
\upbracefill\crcr\noalign{\kern3\p@}}}}\limits%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\mathe[#1]#2{%
|
|||
|
|
\ifthenelse{\equal{\boolinmdframed}{\boolwahr}}{}{\begin{escapeeinzug}}
|
|||
|
|
\noindent%
|
|||
|
|
\let\eqtagset\boolfalsch
|
|||
|
|
\let\eqtaglabel\boolleer
|
|||
|
|
\let\eqtagsymb\boolleer
|
|||
|
|
\let\alteqtag\eqtag
|
|||
|
|
\def\eqtag{\@ifnextchar[{\eqtag@loc@}{\eqtag@loc@[*]}}%
|
|||
|
|
\def\eqtag@loc@[##1]{\@ifnextchar\bgroup{\eqtag@loc@@[##1]}{\eqtag@loc@@[##1]{}}}%
|
|||
|
|
\def\eqtag@loc@@[##1]##2{%
|
|||
|
|
\gdef\eqtagset{\boolwahr}
|
|||
|
|
\gdef\eqtaglabel{##1}
|
|||
|
|
\gdef\eqtagsymb{##2}
|
|||
|
|
}%
|
|||
|
|
\def\verticalalign{}%
|
|||
|
|
\IfBeginWith{#1}{t}{\def\verticalalign{t}}{}%
|
|||
|
|
\IfBeginWith{#1}{m}{\def\verticalalign{c}}{}%
|
|||
|
|
\IfBeginWith{#1}{b}{\def\verticalalign{b}}{}%
|
|||
|
|
\def\horizontalalign{\null\hfill\null}%
|
|||
|
|
\IfEndWith{#1}{l}{}{\null\hfill\null}%
|
|||
|
|
\IfEndWith{#1}{r}{\def\horizontalalign{}}{}%
|
|||
|
|
\begin{math}
|
|||
|
|
\begin{array}[\verticalalign]{0#2}%
|
|||
|
|
}
|
|||
|
|
\def\endmathe{%
|
|||
|
|
\end{array}
|
|||
|
|
\end{math}\horizontalalign%
|
|||
|
|
\let\eqtag\alteqtag
|
|||
|
|
\ifthenelse{\equal{\eqtagset}{\boolwahr}}{\eqtag[\eqtaglabel]{\eqtagsymb}}{}
|
|||
|
|
\ifthenelse{\equal{\boolinmdframed}{\boolwahr}}{}{\end{escapeeinzug}}%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\longmathe[#1]#2{\relax
|
|||
|
|
\let\altarraystretch\arraystretch
|
|||
|
|
\renewcommand\arraystretch{1.2}\relax
|
|||
|
|
\begin{longtable}[#1]{\LINKSRAND #2}
|
|||
|
|
}
|
|||
|
|
\def\endlongmathe{
|
|||
|
|
\end{longtable}
|
|||
|
|
\renewcommand\arraystretch{\altarraystretch}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\einzug{\@ifnextchar[{\indents@}{\indents@[\z@]}}%]
|
|||
|
|
\def\indents@[#1]{\@ifnextchar[{\indents@@[#1]}{\indents@@[#1][\z@]}}%]
|
|||
|
|
\def\indents@@[#1][#2]{%
|
|||
|
|
\begin{list}{}{\relax
|
|||
|
|
\setlength{\topsep}{\z@}\relax
|
|||
|
|
\setlength{\partopsep}{\z@}\relax
|
|||
|
|
\setlength{\parsep}{\parskip}\relax
|
|||
|
|
\setlength{\listparindent}{\z@}\relax
|
|||
|
|
\setlength{\itemindent}{\z@}\relax
|
|||
|
|
\setlength{\leftmargin}{#1}\relax
|
|||
|
|
\setlength{\rightmargin}{#2}\relax
|
|||
|
|
\let\alterlinkerRand\gesamtlinkerRand
|
|||
|
|
\let\alterrechterRand\gesamtrechterRand
|
|||
|
|
\addtolength{\gesamtlinkerRand}{#1}
|
|||
|
|
\addtolength{\gesamtrechterRand}{#2}
|
|||
|
|
}\relax
|
|||
|
|
\item[]\relax
|
|||
|
|
}
|
|||
|
|
\def\endeinzug{%
|
|||
|
|
\setlength{\gesamtlinkerRand}{\alterlinkerRand}
|
|||
|
|
\setlength{\gesamtlinkerRand}{\alterrechterRand}
|
|||
|
|
\end{list}%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\escapeeinzug{\begin{einzug}[-\gesamtlinkerRand][-\gesamtrechterRand]}
|
|||
|
|
\def\endescapeeinzug{\end{einzug}}
|
|||
|
|
|
|||
|
|
\def\programmiercode{
|
|||
|
|
\modulolinenumbers[1]
|
|||
|
|
\begin{einzug}[\rtab][\rtab]%
|
|||
|
|
\begin{linenumbers}%
|
|||
|
|
\fontfamily{cmtt}\fontseries{m}\fontshape{u}\selectfont%
|
|||
|
|
\setlength{\parskip}{1\baselineskip}%
|
|||
|
|
\setlength{\parindent}{0pt}%
|
|||
|
|
}
|
|||
|
|
\def\endprogrammiercode{
|
|||
|
|
\end{linenumbers}
|
|||
|
|
\end{einzug}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\schattiertebox@genbefehl#1#2#3{
|
|||
|
|
\expandafter\gdef\csname #1\endcsname{%
|
|||
|
|
\@ifnextchar[{\csname #1@args\endcsname}{\csname #1@args\endcsname[#3]}%]%
|
|||
|
|
}
|
|||
|
|
\expandafter\def\csname #1@args\endcsname[##1]{%
|
|||
|
|
\@ifnextchar[{\csname #1@args@l\endcsname[##1]}{\csname #1@args@n\endcsname[##1]}%]%
|
|||
|
|
}
|
|||
|
|
\expandafter\def\csname #1@args@l\endcsname[##1][##2]{%
|
|||
|
|
\@ifnextchar[{\csname #1@args@l@r\endcsname[##1][##2]}{\csname #1@args@l@n\endcsname[##1][##2]}%]%
|
|||
|
|
}
|
|||
|
|
\expandafter\def\csname #1@args@n\endcsname[##1]{%
|
|||
|
|
\let\boolinmdframed\boolwahr
|
|||
|
|
\begin{mdframed}[#2leftmargin=0,rightmargin=0,outermargin=0,innermargin=0,##1]
|
|||
|
|
}
|
|||
|
|
\expandafter\def\csname #1@args@l@n\endcsname[##1][##2]{%
|
|||
|
|
\let\boolinmdframed\boolwahr
|
|||
|
|
\begin{mdframed}[#2leftmargin=##2/2,rightmargin=##2/2,outermargin=##2/2,innermargin=##2/2,##1]
|
|||
|
|
}
|
|||
|
|
\expandafter\def\csname #1@args@l@r\endcsname[##1][##2][##3]{%
|
|||
|
|
\let\boolinmdframed\boolwahr
|
|||
|
|
\begin{mdframed}[#2leftmargin=##2,rightmargin=##3,outermargin=##2,innermargin=##3,##1]
|
|||
|
|
}
|
|||
|
|
\expandafter\gdef\csname end#1\endcsname{%
|
|||
|
|
\end{mdframed}
|
|||
|
|
\let\boolinmdframed\boolfalsch
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
\schattiertebox@genbefehl{schattiertebox}{
|
|||
|
|
splittopskip=0,%
|
|||
|
|
splitbottomskip=0,%
|
|||
|
|
frametitleaboveskip=0,%
|
|||
|
|
frametitlebelowskip=0,%
|
|||
|
|
skipabove=1\baselineskip,%
|
|||
|
|
skipbelow=1\baselineskip,%
|
|||
|
|
linewidth=2pt,%
|
|||
|
|
linecolor=black,%
|
|||
|
|
roundcorner=4pt,%
|
|||
|
|
}{
|
|||
|
|
backgroundcolor=leer,%
|
|||
|
|
nobreak=true,%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\schattiertebox@genbefehl{schattierteboxdunn}{
|
|||
|
|
splittopskip=0,%
|
|||
|
|
splitbottomskip=0,%
|
|||
|
|
frametitleaboveskip=0,%
|
|||
|
|
frametitlebelowskip=0,%
|
|||
|
|
skipabove=1\baselineskip,%
|
|||
|
|
skipbelow=1\baselineskip,%
|
|||
|
|
linewidth=1pt,%
|
|||
|
|
linecolor=black,%
|
|||
|
|
roundcorner=2pt,%
|
|||
|
|
}{
|
|||
|
|
backgroundcolor=leer,%
|
|||
|
|
nobreak=true,%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\algorithm{\schattiertebox[backgroundcolor=hellgrau,nobreak=false]}
|
|||
|
|
\def\endalgorithm{\endschattiertebox}
|
|||
|
|
|
|||
|
|
\def\tikzsetzenode#1{%
|
|||
|
|
\tikz[remember picture,baseline,overlay]{\node #1;}%
|
|||
|
|
}
|
|||
|
|
\def\tikzsetzepfeil#1{%
|
|||
|
|
\begin{tikzpicture}[remember picture,overlay,>=latex]%
|
|||
|
|
\draw #1;%
|
|||
|
|
\end{tikzpicture}%
|
|||
|
|
}
|
|||
|
|
\def\tikzsetzeoverlay#1{%
|
|||
|
|
\begin{tikzpicture}[remember picture,overlay,>=latex]%
|
|||
|
|
#1%
|
|||
|
|
\end{tikzpicture}%
|
|||
|
|
}
|
|||
|
|
\def\tikzsetzekreise[#1]#2#3{%
|
|||
|
|
\tikzsetzepfeil{%
|
|||
|
|
[rounded corners,#1]%
|
|||
|
|
([shift={(-\tabcolsep,0.75\baselineskip)}]#2)%
|
|||
|
|
rectangle%
|
|||
|
|
([shift={(\tabcolsep,-0.5\baselineskip)}]#3)
|
|||
|
|
}%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\tikzset{
|
|||
|
|
>=stealth,
|
|||
|
|
auto,
|
|||
|
|
node distance=1cm,
|
|||
|
|
thick,
|
|||
|
|
main node/.style={
|
|||
|
|
circle,draw,font=\sffamily\Large\bfseries,minimum size=0pt
|
|||
|
|
},
|
|||
|
|
state/.style={minimum size=0pt}
|
|||
|
|
loop above right/.style={loop,out=30,in=60,distance=0.5cm},
|
|||
|
|
loop above left/.style={above left,out=150,in=120,loop},
|
|||
|
|
loop below right/.style={below right,out=330,in=300,loop},
|
|||
|
|
loop below left/.style={below left,out=240,in=210,loop},
|
|||
|
|
itria/.style={
|
|||
|
|
draw,dashed,shape border uses incircle,
|
|||
|
|
isosceles triangle,shape border rotate=90,yshift=-1.45cm
|
|||
|
|
},
|
|||
|
|
rtria/.style={
|
|||
|
|
draw,dashed,shape border uses incircle,
|
|||
|
|
isosceles triangle,isosceles triangle apex angle=90,
|
|||
|
|
shape border rotate=-45,yshift=0.2cm,xshift=0.5cm
|
|||
|
|
},
|
|||
|
|
ritria/.style={
|
|||
|
|
draw,dashed,shape border uses incircle,
|
|||
|
|
isosceles triangle,isosceles triangle apex angle=110,
|
|||
|
|
shape border rotate=-55,yshift=0.1cm
|
|||
|
|
},
|
|||
|
|
litria/.style={
|
|||
|
|
draw,dashed,shape border uses incircle,
|
|||
|
|
isosceles triangle,isosceles triangle apex angle=110,
|
|||
|
|
shape border rotate=235,yshift=0.1cm
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: src/setup-layout.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\pagestyle{fancyplain}
|
|||
|
|
|
|||
|
|
\@ifundefined{setcitestyle}{%
|
|||
|
|
%% do nothing
|
|||
|
|
}{%
|
|||
|
|
\setcitestyle{numeric-comp,open={[},close={]}}
|
|||
|
|
}
|
|||
|
|
\def\crefpairconjunction{ und }
|
|||
|
|
\def\crefmiddleconjunction{, }
|
|||
|
|
\def\creflastconjunction{, und }
|
|||
|
|
|
|||
|
|
\raggedbottom %% <- pushes footers up
|
|||
|
|
\sloppy
|
|||
|
|
\def\headrulewidth{0pt}
|
|||
|
|
\def\footrulewidth{0pt}
|
|||
|
|
\setlength{\columnsep}{20pt}
|
|||
|
|
\setlength{\columnseprule}{1pt}
|
|||
|
|
\setlength{\headheight}{11pt}
|
|||
|
|
\setlength{\partopsep}{0pt}
|
|||
|
|
\setlength{\topsep}{\baselineskip}
|
|||
|
|
\setlength{\topskip}{0.5\baselineskip}
|
|||
|
|
\setlength{\footskip}{-1\baselineskip}
|
|||
|
|
\setlength{\maxdepth}{0pt}
|
|||
|
|
\renewcommand{\baselinestretch}{1}
|
|||
|
|
\renewcommand{\arraystretch}{1}
|
|||
|
|
\setcounter{LTchunksize}{\infty}
|
|||
|
|
\setlength{\abovedisplayskip}{0pt}
|
|||
|
|
\setlength{\parskip}{1\baselineskip}
|
|||
|
|
\def\firstparagraph{\noindent}
|
|||
|
|
\def\continueparagraph{\noindent}
|
|||
|
|
|
|||
|
|
\hypersetup{
|
|||
|
|
hidelinks=true,
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\partfont{\documentfont\fontseries{bx}\Huge\selectfont}
|
|||
|
|
\def\chapterfont{\documentfont\fontseries{bx}\huge\selectfont}
|
|||
|
|
\def\sectionfont{\documentfont\fontseries{bx}\Large\selectfont}
|
|||
|
|
\def\subsectionfont{\documentfont\fontseries{bx}\large\selectfont}
|
|||
|
|
|
|||
|
|
\def\thepart{\Roman{part}}
|
|||
|
|
\generatenestedsecnumbering{arabic}{part}{chapter}
|
|||
|
|
\generatenestedsecnumbering{arabic}{chapter}{section}
|
|||
|
|
\generatenestedsecnumbering{arabic}{section}{subsection}
|
|||
|
|
\generatenestedsecnumbering{arabic}{subsection}{subsubsection}
|
|||
|
|
\def\theunitnamepart{\thepart}
|
|||
|
|
\def\theunitnamechapter{\theshortchapter}
|
|||
|
|
\def\theunitnamesection{\thelongsection}
|
|||
|
|
\def\theunitnamesubsection{\thelongsubsection}
|
|||
|
|
\def\theunitnamesubsubsection{\thelongsubsubsection}
|
|||
|
|
|
|||
|
|
\def\partname{Teil\erlaubeplatz}
|
|||
|
|
\def\chaptername{Kapitel\erlaubeplatz}
|
|||
|
|
\def\sectionname{\S\erlaubeplatz}
|
|||
|
|
\def\subsectionname{}
|
|||
|
|
\def\subsubsectionname{}
|
|||
|
|
|
|||
|
|
\let\appendix@orig\appendix
|
|||
|
|
\def\appendix{%
|
|||
|
|
\appendix@orig%
|
|||
|
|
\let\boolinappendix\boolwahr
|
|||
|
|
\addcontentsline{toc}{part}{\appendixname}%
|
|||
|
|
\addtocontents{toc}{\protect\setcounter{tocdepth}{0}}
|
|||
|
|
\def\sectionname{Appendix}%
|
|||
|
|
\def\theunitnamesection{\Alph{section}}%
|
|||
|
|
}
|
|||
|
|
\def\notappendix{%
|
|||
|
|
\let\boolinappendix\boolfalse
|
|||
|
|
\addtocontents{toc}{\protect\setcounter{tocdepth}{1 }}
|
|||
|
|
\def\sectionname{}%
|
|||
|
|
\def\theunitnamesection{\arabic{section}}%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
%% \titlespacing{<sectionclassname>}
|
|||
|
|
%% {linker einzug}{platz oberhalb}{platz unterhalb}[rechter einzug]
|
|||
|
|
|
|||
|
|
\titlespacing{\section}{0pt}{\baselineskip}{\baselineskip}
|
|||
|
|
\titlespacing{\subsection}{0pt}{\baselineskip}{\baselineskip}
|
|||
|
|
\titlespacing{\subsubsection}{0pt}{\baselineskip}{\baselineskip}
|
|||
|
|
\titlespacing{\paragraph}{0pt}{0pt}{1em}
|
|||
|
|
|
|||
|
|
\titleformat{\part}[display]
|
|||
|
|
{\normalfont\headingfont\bfseries\Huge\centering}
|
|||
|
|
{%
|
|||
|
|
\ifthenelse{\equal{\partname}{}}{%
|
|||
|
|
\theunitnamepart%
|
|||
|
|
}{%
|
|||
|
|
\MakeUppercase{\partname}~\theunitnamepart%
|
|||
|
|
}%
|
|||
|
|
}{0pt}{%
|
|||
|
|
}[\thispagestyle{empty}]
|
|||
|
|
\titleformat{\chapter}[frame]
|
|||
|
|
{\normalfont\headingfont\bfseries\Large}
|
|||
|
|
{%
|
|||
|
|
\bedingtesspaceexpand{chaptername}{~}{\theunitnamechapter}%
|
|||
|
|
}{0.5em}{%
|
|||
|
|
}[\thispagestyle{empty}]%\titlerule%[2pt]%
|
|||
|
|
\titleformat{\section}[hang]
|
|||
|
|
{\normalfont\headingfont\bfseries\flushleft\large}
|
|||
|
|
{%
|
|||
|
|
\bedingtesspaceexpand{sectionname}{~}{\theunitnamesection}%
|
|||
|
|
}{0.5em}
|
|||
|
|
{%
|
|||
|
|
}
|
|||
|
|
[%
|
|||
|
|
\nvraum{0.25}%
|
|||
|
|
]
|
|||
|
|
\titleformat{\subsection}[hang]
|
|||
|
|
{\normalfont\headingfont\bfseries\flushleft\large}
|
|||
|
|
{%
|
|||
|
|
\bedingtesspaceexpand{subsectionname}{~}{\theunitnamesubsection}%
|
|||
|
|
}{0.5em}
|
|||
|
|
{%
|
|||
|
|
}
|
|||
|
|
[%
|
|||
|
|
\nvraum{0.25}%
|
|||
|
|
]
|
|||
|
|
\titleformat{\subsubsection}[hang]
|
|||
|
|
{\normalfont\headingfont\bfseries\flushleft\large}
|
|||
|
|
{%
|
|||
|
|
\bedingtesspaceexpand{subsubsectionname}{~}{\theunitnamesubsubsection}%
|
|||
|
|
}{0.5em}
|
|||
|
|
{%
|
|||
|
|
}
|
|||
|
|
[%
|
|||
|
|
\nvraum{0.25}%
|
|||
|
|
]
|
|||
|
|
|
|||
|
|
\def\rafootnotectr{20}
|
|||
|
|
\def\incrftnotectr#1{%
|
|||
|
|
\addtocounter{#1}{1}%
|
|||
|
|
\ifnum\value{#1}>\rafootnotectr\relax
|
|||
|
|
\setcounter{#1}{0}%
|
|||
|
|
\fi%
|
|||
|
|
}
|
|||
|
|
\def\footnoteref[#1]{\protected@xdef\@thefnmark{\ref{#1}}\@footnotemark}
|
|||
|
|
\let\altfootnotetext\footnotetext
|
|||
|
|
\def\footnotetext[#1]#2{\incrftnotectr{footnote}\altfootnotetext[\value{footnote}]{\label{#1}#2}}
|
|||
|
|
\let\altfootnotemark\footnotemark
|
|||
|
|
%% Undesirable solution, as the text is not hyperlinked.
|
|||
|
|
\def\footnotemark[#1]{\text{\textsuperscript{\getrefnumber{#1}}}}
|
|||
|
|
|
|||
|
|
\DefineFNsymbols*{custom}{abcdefghijklmnopqrstuvwxyz}
|
|||
|
|
\setfnsymbol{custom}
|
|||
|
|
\def\footnotelayout{\documentfont\scriptsize}
|
|||
|
|
\def\thefootnote{\fnsymbol{footnote}}
|
|||
|
|
|
|||
|
|
\def\kopfzeileleer{
|
|||
|
|
\lhead[]{}
|
|||
|
|
\chead[]{}
|
|||
|
|
\rhead[]{}
|
|||
|
|
\lfoot[]{}
|
|||
|
|
\cfoot[]{}
|
|||
|
|
\rfoot[]{}
|
|||
|
|
}
|
|||
|
|
\def\kopfzeiledefault{
|
|||
|
|
\lhead[]{}
|
|||
|
|
\lhead[]{}
|
|||
|
|
\chead[]{}
|
|||
|
|
\rhead[]{}
|
|||
|
|
\lfoot[]{}
|
|||
|
|
\cfoot{\footnotesize\thepage}
|
|||
|
|
\rfoot[]{}
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\DeclareRobustCommand\crfamily{\fontfamily{pcr}\selectfont}
|
|||
|
|
\def\headingfont{\fontfamily{cmss}\selectfont}
|
|||
|
|
\def\documentfancyfont{%
|
|||
|
|
\gdef\headingfont{\crfamily}%
|
|||
|
|
\fontfamily{ccr}\fontseries{m}\selectfont%
|
|||
|
|
}
|
|||
|
|
\def\documentfont{%
|
|||
|
|
\gdef\headingfont{\fontfamily{cmss}\selectfont}%
|
|||
|
|
\fontfamily{cmss}\fontseries{m}\selectfont%
|
|||
|
|
\renewcommand{\sfdefault}{phv}%
|
|||
|
|
\renewcommand{\ttdefault}{pcr}%
|
|||
|
|
\renewcommand{\rmdefault}{cmr}% <— funktionieren nicht mit {ptm}
|
|||
|
|
\renewcommand{\bfdefault}{bx}%
|
|||
|
|
\renewcommand{\itdefault}{it}%
|
|||
|
|
\renewcommand{\sldefault}{sl}%
|
|||
|
|
\renewcommand{\scdefault}{sc}%
|
|||
|
|
\renewcommand{\updefault}{n}%
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\allowdisplaybreaks
|
|||
|
|
\let\altcleardoublepage\cleardoublepage
|
|||
|
|
\let\cleardoublepage\clearpage
|
|||
|
|
|
|||
|
|
\def\startdocumentlayoutoptions{
|
|||
|
|
\selectlanguage{ngerman}
|
|||
|
|
\setlength{\parskip}{0.5\baselineskip}
|
|||
|
|
\setlength{\parindent}{0pt}
|
|||
|
|
\kopfzeiledefault
|
|||
|
|
\documentfont
|
|||
|
|
\normalsize
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
\def\highlightTerm#1{\emph{#1}}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: src/setup-localmacros.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
%% ****************************************************************
|
|||
|
|
%% MATHE:
|
|||
|
|
%% ****************************************************************
|
|||
|
|
|
|||
|
|
\def\reell{\mathbb{R}}
|
|||
|
|
\def\kmplx{\mathbb{C}}
|
|||
|
|
\def\Torus{\mathbb{T}}
|
|||
|
|
\def\rtnl{\mathbb{Q}}
|
|||
|
|
\def\intgr{\mathbb{Z}}
|
|||
|
|
|
|||
|
|
\def\ntrl{\mathbb{N}}
|
|||
|
|
\def\ntrlpos{\mathbb{N}}
|
|||
|
|
\def\ntrlzero{\mathbb{N}_{0}}
|
|||
|
|
\def\reellNonNeg{\reell_{+}}
|
|||
|
|
|
|||
|
|
\def\leer{\emptyset}
|
|||
|
|
\def\restr#1{\vert_{#1}}
|
|||
|
|
\def\ohne{\setminus}
|
|||
|
|
\def\Pot{\mathop{\mathcal{P}}}
|
|||
|
|
|
|||
|
|
\def\brkt#1{\langle{}#1{}\rangle}
|
|||
|
|
\def\lsim{\mathop{\sim}}
|
|||
|
|
\def\lneg{\mathop{\neg}}
|
|||
|
|
\def\land{\mathop{\wedge}}
|
|||
|
|
\def\lor{\mathop{\vee}}
|
|||
|
|
|
|||
|
|
\def\eps{\varepsilon}
|
|||
|
|
\let\altphi\phi
|
|||
|
|
\let\altvarphi\varphi
|
|||
|
|
\def\phi{\altvarphi}
|
|||
|
|
\def\varphi{\altphi}
|
|||
|
|
|
|||
|
|
\def\span{\mathop{\text{\upshape Lin}}}
|
|||
|
|
\def\dim{\mathop{\text{\upshape dim}}}
|
|||
|
|
\def\onematrix{\text{\upshape\bfseries I}}
|
|||
|
|
\def\zeromatrix{\text{\upshape\bfseries 0}}
|
|||
|
|
\def\zerovector{\text{\upshape\bfseries 0}}
|
|||
|
|
|
|||
|
|
\def\graph{\mathop{\text{\textup Gph}}}
|
|||
|
|
\def\id{\text{\textup id}}
|
|||
|
|
\def\modfn{\mathop{\text{\textup mod}}}
|
|||
|
|
\makeatother
|
|||
|
|
|
|||
|
|
\begin{document}
|
|||
|
|
\startdocumentlayoutoptions
|
|||
|
|
|
|||
|
|
%% FRONTMATTER:
|
|||
|
|
\thispagestyle{plain}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: front/index.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: front/title.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\begin{titlepage}
|
|||
|
|
\null
|
|||
|
|
|
|||
|
|
\vraum
|
|||
|
|
|
|||
|
|
\noindent\rule{\linewidth}{2pt}
|
|||
|
|
|
|||
|
|
{\hraum\LARGE Lineare Algebra I\hraum}\\
|
|||
|
|
{\hraum\LARGE $\oast$\,\rule[0.175\baselineskip]{0.65\linewidth}{1pt}\,$\oast$ \hraum}\\
|
|||
|
|
{\hraum\Large Lösungen zu diversen Aufgaben im Kurs\hraum}
|
|||
|
|
|
|||
|
|
\noindent\rule{\linewidth}{2pt}
|
|||
|
|
|
|||
|
|
\vraum
|
|||
|
|
|
|||
|
|
\noindent
|
|||
|
|
\hraum{\footnotesize Raj Dahya}\hraum\\
|
|||
|
|
\hraum{\small \itshape Fakultät für Mathematik und Informatik/Institut für Philosophie}\hraum\\
|
|||
|
|
\hraum{\small \itshape Universität Leipzig.}\hraum\\
|
|||
|
|
\hraum{\small Wintersemester 2020/2021 }\hraum
|
|||
|
|
\end{titlepage}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: front/foreword.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\chapter*{Vorwort}
|
|||
|
|
|
|||
|
|
Dieses Dokument enthält Lösungsansätze zu den Übungsserien, Selbstkontrollenaufgaben, und Quizzes.
|
|||
|
|
Diese werden natürlich \emph{nach} Abgabefristen hochgeladen
|
|||
|
|
und dienen \emph{nicht} als Musterlösungen!
|
|||
|
|
Der Zweck dieser Lösungen ist es vielmehr, Ansätze zu präsentieren,
|
|||
|
|
mit denen man seine \emph{eigenen} Versuche vergleichen kann.
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: front/contents.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\kopfzeiledefault
|
|||
|
|
\footnotesize
|
|||
|
|
\setcounter{tocdepth}{1}
|
|||
|
|
\def\contentsname{Inhaltsverzeichnis}
|
|||
|
|
|
|||
|
|
\tableofcontents
|
|||
|
|
|
|||
|
|
%% HAUPTTEXT:
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: body/index.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\setcounternach{part}{1}
|
|||
|
|
\part{Übungsserien}
|
|||
|
|
|
|||
|
|
\def\chaptername{Übungsserie}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: body/uebung/ueb1.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\setcounternach{chapter}{1}
|
|||
|
|
\chapter[Woche 1]{Woche 1}
|
|||
|
|
\label{ueb:1}
|
|||
|
|
|
|||
|
|
\textbf{ACHTUNG.}
|
|||
|
|
Diese Lösungen dienen \emph{nicht} als Musterlösungen sondern eher als Referenz.
|
|||
|
|
Hier wird eingehender gearbeitet, als generell verlangt wird.
|
|||
|
|
Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche vergleichen kann.
|
|||
|
|
|
|||
|
|
%% AUFGABE 1-1
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 1]{}
|
|||
|
|
\label{ueb:1:ex:1}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
Zu bestimmen ist die Lösungsmenge
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
L_{\alpha,\beta} &:= &\{
|
|||
|
|
\mathbf{x}\in\reell^{n}
|
|||
|
|
\mid A_{\alpha}\mathbf{x}=\mathbf{b}_{\beta}
|
|||
|
|
\}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
für $\alpha,\beta\in\reell$,
|
|||
|
|
wobei $m=3$ und $n=4$, und
|
|||
|
|
$A_{\alpha}\in\reell^{m\times n}$ und $\mathbf{b}_{\beta}\in\reell^{m}$
|
|||
|
|
durch
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rclqrcl}
|
|||
|
|
A_{\alpha} &:= &\begin{matrix}{cccc}
|
|||
|
|
1 &7 &2 &-1\\
|
|||
|
|
1 &8 &6 &-3\\
|
|||
|
|
2 &14 &\alpha &-2\\
|
|||
|
|
\end{matrix}
|
|||
|
|
&\mathbf{b}_{\beta} &:= &\begin{vector}4\\0\\\beta\\\end{vector}
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
gegeben sind.
|
|||
|
|
Um die Lösungsmenge zu bestimmen führen wir das Gaußverfahren aus:
|
|||
|
|
|
|||
|
|
\begin{algorithm}[\rtab][\rtab]
|
|||
|
|
Ursprüngliches LGS $(A_{\alpha}|b_{\beta})$:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{cccc|c}
|
|||
|
|
1 &7 &2 &-1 &4\\
|
|||
|
|
1 &8 &6 &-3 &0\\
|
|||
|
|
2 &14 &\alpha &-2 &\beta\\
|
|||
|
|
\end{matrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Wende die Zeilentransformationen
|
|||
|
|
|
|||
|
|
{\footnotesize
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
Z_{2} &\leftsquigarrow &Z_{2}-Z_{1}\\
|
|||
|
|
Z_{3} &\leftsquigarrow &Z_{3}-2\cdot Z_{1}\\
|
|||
|
|
\end{mathe}}
|
|||
|
|
|
|||
|
|
an:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{cccc|c}
|
|||
|
|
\boxed{1} &7 &2 &-1 &4\\
|
|||
|
|
0 &\boxed{1} &4 &-2 &-4\\
|
|||
|
|
0 &0 &\boxed{\alpha - 4} &0 &\beta - 8\\
|
|||
|
|
\end{matrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
\end{algorithm}
|
|||
|
|
|
|||
|
|
Die eingezeichneten Einträge markieren die ersten Einträge der Stufen.
|
|||
|
|
Es gibt also $2$ oder $3$ Stufen, je nachdem, ob ${\alpha - 4=0}$.
|
|||
|
|
Dies führt zu einem Fallunterschied:
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries {Fall} 1.}
|
|||
|
|
%% FALL 1
|
|||
|
|
\item $\alpha-4=0$. Das heißt, $\alpha=4$.
|
|||
|
|
In diesem Falle hat das augmentierte System genau $2$ Stufen
|
|||
|
|
und sieht wie folgt aus:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{cccc|c}
|
|||
|
|
\boxed{1} &7 &2 &-1 &4\\
|
|||
|
|
0 &\boxed{1} &4 &-2 &-4\\
|
|||
|
|
0 &0 &0 &0 &\beta - 8\\
|
|||
|
|
\end{matrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Dies führt zu zwei weiteren Fällen, denn die $3$. Gleichung ist jetzt genau dann lösbar,
|
|||
|
|
wenn $\beta-8=0$.
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries {Fall 1}a.}
|
|||
|
|
%% FALL 1a
|
|||
|
|
\item $\beta-8\neq 0$. Das heißt, $\beta\neq 8$.
|
|||
|
|
Dann ist die $3$. Gleichung und damit das LGS nicht lösbar.
|
|||
|
|
Darum erhalten wir $\boxed{L_{\alpha,\beta}=\leer}$.
|
|||
|
|
|
|||
|
|
%% FALL 1b
|
|||
|
|
\item $\beta-8=0$. Das heißt, $\beta=8$.
|
|||
|
|
Dann ist die $3$. Gleichung trivialerweise erfüllt.
|
|||
|
|
Das augmentierte System sieht wird zum
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{cccc|c}
|
|||
|
|
\boxed{1} &7 &2 &-1 &4\\
|
|||
|
|
0 &\boxed{1} &4 &-2 &-4\\
|
|||
|
|
0 &0 &0 &0 &0\\
|
|||
|
|
\end{matrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
und kann jetzt aufgelöst werden.
|
|||
|
|
Wir arbeiten von unten nach oben:
|
|||
|
|
|
|||
|
|
\begin{algorithm}[2\rtab][\rtab]
|
|||
|
|
Aus der ganzen Zeilenstufenform erschließt sich
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
x_{3},\, x_{4}\,\text{sind frei}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Aus der Stufenform von Gleichungen $2$ und $1$ erschließt sich
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
x_{2} &= &-4 - 4x_{3} + 2x_{4}\\
|
|||
|
|
x_{1} &= &4 - 7x_{2} - 2x_{3} + x_{4}\\
|
|||
|
|
&= &4 - 7(-4 - 4x_{3} + 2x_{4}) - 2x_{3} + x_{4}\\
|
|||
|
|
&= &32 + 26x_{3} + -13x_{4}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Zusammengefasst erhalten wir die allgemeine Form der Lösung:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\mathbf{x} &= &\begin{svector}x_{1}\\x_{2}\\x_{3}\\x_{4}\\\end{svector}\\
|
|||
|
|
&= &\begin{svector}32 + 26x_{3} + -13x_{4}\\-4 - 4x_{3} + 2x_{4}\\x_{3}\\x_{4}\\\end{svector}\\
|
|||
|
|
&= &\begin{svector}32 + 26x_{3} + -13x_{4}\\-4 - 4x_{3} + 2x_{4}\\0 + 1x_{3} + 0x_{4}\\0 + 0x_{3} + 1x_{4}\\\end{svector}\\
|
|||
|
|
&= &\begin{svector}32\\-4\\0\\0\\\end{svector}
|
|||
|
|
+ \begin{svector}26x_{3}\\-4x_{3}\\1x_{3}\\0x_{3}\\\end{svector}
|
|||
|
|
+ \begin{svector}-13x_{4}\\2x_{4}\\1x_{4}\\1x_{4}\\\end{svector}\\
|
|||
|
|
&= &\begin{svector}32\\-4\\0\\0\\\end{svector}
|
|||
|
|
+ x_{3}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
|
|||
|
|
+ x_{4}\cdot\begin{svector}-13\\2\\1\\1\\\end{svector}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
mit $x_{3}$, $x_{4}$ frei wählbar.
|
|||
|
|
\end{algorithm}
|
|||
|
|
|
|||
|
|
Also erhalten wird in diesem Falle
|
|||
|
|
$\boxed{
|
|||
|
|
L_{\alpha,\beta}=\left\{
|
|||
|
|
\begin{svector}32\\-4\\0\\0\\\end{svector}
|
|||
|
|
+ t_{1}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
|
|||
|
|
+ t_{2}\cdot\begin{svector}-13\\2\\1\\1\\\end{svector}
|
|||
|
|
\mid t_{1}, t_{2}\in\reell
|
|||
|
|
\right\}
|
|||
|
|
}$,
|
|||
|
|
oder etwas kompakter formuliert,
|
|||
|
|
${L_{\alpha,\beta}=\begin{svector}32\\-4\\0\\0\\\end{svector} + \span\left\{\begin{svector}26\\-4\\1\\0\\\end{svector}, \begin{svector}-13\\2\\1\\1\\\end{svector}\right\}}$.
|
|||
|
|
\end{enumerate}
|
|||
|
|
|
|||
|
|
%% FALL 2
|
|||
|
|
\item $\alpha-4\neq 0$. Das heißt, $\alpha\neq 4$.
|
|||
|
|
In diesem Falle hat das augmentierte System genau $3$ Stufen und diesmal ist nur $x_{4}$ frei.
|
|||
|
|
Man beachte, dass dies im Grunde genau wie Fall 1b ist, nur dass wir zusätzlich Gleichung 3 beachten und $x_{3}$ bestimmen müssen.
|
|||
|
|
|
|||
|
|
\begin{algorithm}[2\rtab][\rtab]
|
|||
|
|
Aus der Stufenform von Gleichungen $3$ ergibt sich
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
x_{3} &= &\frac{\beta-8}{\alpha-4}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Der Rest der Lösung des Gleichungssystems verhält sich genau wie im Fall 3b,
|
|||
|
|
das heißt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\mathbf{x} &= &\begin{svector}32\\-4\\0\\0\\\end{svector}
|
|||
|
|
+ x_{3}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
|
|||
|
|
+ x_{4}\cdot\begin{svector}-13\\2\\1\\1\\\end{svector}\\
|
|||
|
|
&= &\begin{svector}32\\-4\\0\\0\\\end{svector}
|
|||
|
|
+ \frac{\beta-8}{\alpha-4}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
|
|||
|
|
+ x_{4}\cdot\begin{svector}-13\\2\\1\\1\\\end{svector},\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
wobei $x_{4}$ frei wählbar ist.
|
|||
|
|
\end{algorithm}
|
|||
|
|
|
|||
|
|
Also erhalten wird in diesem Falle
|
|||
|
|
$\boxed{
|
|||
|
|
L_{\alpha,\beta}=\left\{
|
|||
|
|
\begin{svector}32\\-4\\0\\0\\\end{svector}
|
|||
|
|
+ \frac{\beta-8}{\alpha-4}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector}
|
|||
|
|
+ t\cdot\begin{svector}-13\\2\\1\\1\\\end{svector}
|
|||
|
|
\mid t\in\reell
|
|||
|
|
\right\}
|
|||
|
|
}$,
|
|||
|
|
oder etwas kompakter formuliert,
|
|||
|
|
${L_{\alpha,\beta}=\begin{svector}32\\-4\\0\\0\\\end{svector} + \frac{\beta-8}{\alpha-4}\cdot\begin{svector}26\\-4\\1\\0\\\end{svector} + \span\left\{\begin{svector}-13\\2\\1\\1\\\end{svector}\right\}}$.
|
|||
|
|
\end{enumerate}
|
|||
|
|
|
|||
|
|
Wir fassen die Lösung für alle Fälle zusammen:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
L_{\alpha,\beta} &= &\begin{cases}[m]{lcl}
|
|||
|
|
\leer &: &\alpha=4,\,\beta\neq 8\\
|
|||
|
|
\mathbf{u} + \span\{\mathbf{v},\mathbf{w}\} &: &\alpha=4,\,\beta=8\\
|
|||
|
|
\mathbf{u} + \frac{\alpha-4}{\beta-8}\mathbf{v} + \span\{\mathbf{w}\} &: &\alpha\neq 4\\
|
|||
|
|
\end{cases}
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
für alle $\alpha,\beta\in\reell$,
|
|||
|
|
wobei
|
|||
|
|
$\mathbf{u} = \begin{svector}32\\-4\\0\\0\\\end{svector}$,
|
|||
|
|
$\mathbf{v} = \begin{svector}26\\-4\\1\\0\\\end{svector}$,
|
|||
|
|
$\mathbf{w} = \begin{svector}-13\\2\\1\\1\\\end{svector}$.
|
|||
|
|
|
|||
|
|
%% AUFGABE 1-2
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 2]{}
|
|||
|
|
\label{ueb:1:ex:2}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
\begin{schattierteboxdunn}
|
|||
|
|
\begin{satz}
|
|||
|
|
\makelabel{satz:main:ueb:1:ex:2}
|
|||
|
|
Angewandt auf die erweiterte Koeffizientenmatrix eines linearen Gleichungssystems
|
|||
|
|
verändern
|
|||
|
|
die elementaren Zeilenumformungen vom Typ (I), (II) und (III)
|
|||
|
|
die Menge der Lösungen nicht.
|
|||
|
|
\end{satz}
|
|||
|
|
\end{schattierteboxdunn}
|
|||
|
|
|
|||
|
|
Wir beweisen \Cref{satz:main:ueb:1:ex:2} mithilfe der folgenden Teilergebnisse.
|
|||
|
|
|
|||
|
|
\begin{lemm}
|
|||
|
|
\makelabel{lemm:1:ueb:1:ex:2}
|
|||
|
|
Seien $m,n\in\ntrlpos$ und $A\in\reell^{m\times n}$ und $\mathbf{b}\in\reell^{m}$.
|
|||
|
|
Für $i,j\in\{1,2,\ldots,m\}$ mit $i\neq j$ bezeichne mit
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
(A|\mathbf{b}) &\overset{I;i,j}{\rightsquigarrow} &(A'|\mathbf{b}')\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
die Anwendung von Zeilentransformation (I) auf $(A|\mathbf{b})$,
|
|||
|
|
wobei Zeile${}_{i}$ und Zeile${}_{j}$ umgetauscht werden,
|
|||
|
|
was in $(A'|\mathbf{b}')$ resultiert.
|
|||
|
|
Dann für alle ${\mathbf{x}\in\reell^{n}}$,
|
|||
|
|
falls $\mathbf{x}$ eine Lösung für $(A|\mathbf{b})$ ist,
|
|||
|
|
dann ist $\mathbf{x}$ eine Lösung für $(A'|\mathbf{b}')$.
|
|||
|
|
\end{lemm}
|
|||
|
|
|
|||
|
|
\begin{einzug}[\rtab][\rtab]
|
|||
|
|
\begin{proof}
|
|||
|
|
Betrachte den Fall $i<j$.
|
|||
|
|
Es gilt
|
|||
|
|
|
|||
|
|
\begin{longtable}[mc]{RL}
|
|||
|
|
&\text{$\mathbf{x}$ eine Lösung für $(A|\mathbf{b})$}\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{i,1}x_{1} &+ &a_{i,2}x_{2} &+ &\cdots &+ &a_{i,n}x_{n} &= &b_{i})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{j,1}x_{1} &+ &a_{j,2}x_{2} &+ &\cdots &+ &a_{j,n}x_{n} &= &b_{j})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m})
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{j,1}x_{1} &+ &a_{j,2}x_{2} &+ &\cdots &+ &a_{j,n}x_{n} &= &b_{j})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{i,1}x_{1} &+ &a_{i,2}x_{2} &+ &\cdots &+ &a_{i,n}x_{n} &= &b_{i})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m})
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
&\text{da lediglich zwei Aussagen in einer Konjunktion umgetauscht werden}\\
|
|||
|
|
\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&\text{$\mathbf{x}$ eine Lösung für $(A'|\mathbf{b})'$, da $(A|\mathbf{b})\overset{I;i,j}{\rightsquigarrow}(A'|\mathbf{b}')$.}\\
|
|||
|
|
\end{longtable}
|
|||
|
|
|
|||
|
|
Der Fall $i>j$ lässt sich analog zeigen.
|
|||
|
|
Falls $i=j$ bleibt das System unverändert, sodass die Behauptung trivialerweise gilt.
|
|||
|
|
\end{proof}
|
|||
|
|
\end{einzug}
|
|||
|
|
|
|||
|
|
\begin{lemm}
|
|||
|
|
\makelabel{lemm:2:ueb:1:ex:2}
|
|||
|
|
Seien $m,n\in\ntrlpos$ und $A\in\reell^{m\times n}$ und $\mathbf{b}\in\reell^{m}$.
|
|||
|
|
Für ${i\in\{1,2,\ldots,m\}}$ und ${\alpha\in\reell\ohne\{0\}}$ bezeichne mit
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
(A|\mathbf{b}) &\overset{II;i,\alpha}{\rightsquigarrow} &(A'|\mathbf{b}')\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
die Anwendung von Zeilentransformation (II) auf $(A|\mathbf{b})$,
|
|||
|
|
wobei Zeile${}_{i}$ durch $\alpha\cdot$Zeile${}_{i}$ ersetzt wird,
|
|||
|
|
was in $(A'|\mathbf{b}')$ resultiert.
|
|||
|
|
Dann für alle ${\mathbf{x}\in\reell^{n}}$,
|
|||
|
|
falls $\mathbf{x}$ eine Lösung für $(A|\mathbf{b})$ ist,
|
|||
|
|
dann ist $\mathbf{x}$ eine Lösung für $(A'|\mathbf{b}')$.
|
|||
|
|
\end{lemm}
|
|||
|
|
|
|||
|
|
\begin{einzug}[\rtab][\rtab]
|
|||
|
|
\begin{proof}
|
|||
|
|
Es gilt
|
|||
|
|
|
|||
|
|
\begin{longtable}[mc]{RL}
|
|||
|
|
&\text{$\mathbf{x}$ eine Lösung für $(A|\mathbf{b})$}\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{i,1}x_{1} &+ &a_{i,2}x_{2} &+ &\cdots &+ &a_{i,n}x_{n} &= &b_{i})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m})
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(\alpha\cdot (a_{i,1}x_{1} &+ &a_{i,2}x_{2} &+ &\cdots &+ &a_{i,n}x_{n}) &= &\alpha\cdot b_{i})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m})
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(\alpha\cdot a_{i,1}x_{1} &+ &\alpha\cdot a_{i,2}x_{2} &+ &\cdots &+ &\alpha\cdot a_{i,n}x_{n} &= &\alpha\cdot b_{i})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m})
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
&\text{$\mathbf{x}$ eine Lösung für $(A'|\mathbf{b})'$, da $(A|\mathbf{b})\overset{II;i,\alpha}{\rightsquigarrow}(A'|\mathbf{b}')$.}
|
|||
|
|
\end{longtable}
|
|||
|
|
|
|||
|
|
Also gilt die Behauptung.
|
|||
|
|
\end{proof}
|
|||
|
|
\end{einzug}
|
|||
|
|
|
|||
|
|
\begin{lemm}
|
|||
|
|
\makelabel{lemm:3:ueb:1:ex:2}
|
|||
|
|
Seien $m,n\in\ntrlpos$ und $A\in\reell^{m\times n}$ und $\mathbf{b}\in\reell^{m}$.
|
|||
|
|
Für ${i,j\in\{1,2,\ldots,m\}}$ mit $i\neq j$ und $\alpha\in\reell$ bezeichne mit
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
(A|\mathbf{b}) &\overset{III;i,j,\alpha}{\rightsquigarrow} &(A'|\mathbf{b}')\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
die Anwendung von Zeilentransformation (III) auf $(A|\mathbf{b})$,
|
|||
|
|
wobei Zeile${}_{i}$ durch die Addition von Zeile${}_{i}$ mit $\alpha\cdot$Zeile${}_{j}$ ersetzt wird,
|
|||
|
|
was in $(A'|\mathbf{b}')$ resultiert.
|
|||
|
|
Dann für alle ${\mathbf{x}\in\reell^{n}}$,
|
|||
|
|
falls $\mathbf{x}$ eine Lösung für $(A|\mathbf{b})$ ist,
|
|||
|
|
dann ist $\mathbf{x}$ eine Lösung für $(A'|\mathbf{b}')$.
|
|||
|
|
\end{lemm}
|
|||
|
|
|
|||
|
|
\begin{einzug}[\rtab][\rtab]
|
|||
|
|
\begin{proof}
|
|||
|
|
Es gilt
|
|||
|
|
|
|||
|
|
\begin{longtable}[mc]{RL}
|
|||
|
|
&\text{$\mathbf{x}$ eine Lösung für $(A|\mathbf{b})$}\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{i,1}x_{1} &+ &a_{i,2}x_{2} &+ &\cdots &+ &a_{i,n}x_{n} &= &b_{i})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m})
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{i,1}x_{1} &+ &a_{i,2}x_{2} &+ &\cdots &+ &a_{i,n}x_{n} + \alpha\cdot b_{j} &= &b_{i} + \alpha\cdot b_{j})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m})
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{i,1}x_{1} &+ &a_{i,2}x_{2} &+ &\cdots &+ &a_{i,n}x_{n}\\
|
|||
|
|
&+\alpha\cdot a_{j,1}x_{1} &+ &\alpha\cdot a_{j,2}x_{2} &+ &\cdots &+ &\alpha\cdot a_{j,n}x_{n} &= &b_{i} + \alpha\cdot b_{j})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m})
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
&\text{da laut der $j$-ten Gleichung gilt ${b_{j}=\sum_{k=1}^{m}a_{j,k}x_{k}}$}\\
|
|||
|
|
\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&{\scriptsize
|
|||
|
|
\left\{
|
|||
|
|
\begin{array}[m]{crccccclcl}
|
|||
|
|
&(a_{1,1}x_{1} &+ &a_{1,2}x_{2} &+ &\cdots &+ &a_{1,n}x_{n} &= &b_{1})\\
|
|||
|
|
\text{und} &(a_{2,1}x_{1} &+ &a_{2,2}x_{2} &+ &\cdots &+ &a_{2,n}x_{n} &= &b_{2})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a'_{i,1}x_{1} &+ &a'_{i,2}x_{2} &+ &\cdots &+ &a'_{i,n}x_{n} &= &b'_{i})\\
|
|||
|
|
\cdots\\
|
|||
|
|
\text{und} &(a_{m,1}x_{1} &+ &a_{m,2}x_{2} &+ &\cdots &+ &a_{m,n}x_{n} &= &b_{m}),
|
|||
|
|
\end{array}
|
|||
|
|
\right.}\\
|
|||
|
|
\\
|
|||
|
|
&\text{wobei $a'_{i,k}=a_{i,k}+\alpha\cdot a_{j,k}$ für alle $k$ und $b'_{i}=b_{i}+\alpha\cdot b_{j}$}\\
|
|||
|
|
\\
|
|||
|
|
\Longrightarrow
|
|||
|
|
&\text{$\mathbf{x}$ eine Lösung für $(A'|\mathbf{b})'$, da $(A|\mathbf{b})\overset{III;i,j,\alpha}{\rightsquigarrow}(A'|\mathbf{b}')$.}
|
|||
|
|
\end{longtable}
|
|||
|
|
|
|||
|
|
Also gilt die Behauptung.
|
|||
|
|
\end{proof}
|
|||
|
|
\end{einzug}
|
|||
|
|
|
|||
|
|
Endlich können wir \Cref{satz:main:ueb:1:ex:2} beweisen:
|
|||
|
|
|
|||
|
|
\begin{proof}[von \Cref{satz:main:ueb:1:ex:2}]
|
|||
|
|
Seien $m,n\in\ntrlpos$ und $A\in\reell^{m\times n}$ und $\mathbf{b}\in\reell^{m}$.
|
|||
|
|
Seien $A'\in\reell^{m\times n}$ und $\mathbf{b}'\in\reell^{m}$,
|
|||
|
|
so dass $(A|\mathbf{b})$ durch eine Transformation der Art (I), (II) oder (III)
|
|||
|
|
aus $(A|\mathbf{b})$ entsteht.
|
|||
|
|
Das heißt, entweder
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{lrcl}
|
|||
|
|
\eqtag[eq:0:\beweislabel]
|
|||
|
|
&(A|\mathbf{b}) &\overset{I;i,j}{\rightsquigarrow} &(A'|\mathbf{b}')\\
|
|||
|
|
\text{oder} &(A|\mathbf{b}) &\overset{I;i,\alpha}{\rightsquigarrow} &(A'|\mathbf{b}')\\
|
|||
|
|
\text{oder} &(A|\mathbf{b}) &\overset{III;i,j,\alpha}{\rightsquigarrow} &(A'|\mathbf{b}')\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
gilt, für ein $i,j\in\{1,2,\ldots,m\}$ mit $i\neq j$ und $\alpha\in\reell\ohne\{0\}$.\\
|
|||
|
|
\textbf{Zu zeigen:}
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:1:\beweislabel]
|
|||
|
|
\{\mathbf{x}\in\reell^{n}\mid\mathbf{x}\text{ eine Lösung für }(A|\mathbf{b})\}
|
|||
|
|
&= &\{\mathbf{x}\in\reell^{n}\mid\mathbf{x}\text{ eine Lösung für }(A|\mathbf{b})\}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Wir zeigen dies in zwei Teile:
|
|||
|
|
|
|||
|
|
\uline{\bfseries ($\subseteq$.)}\\
|
|||
|
|
Sei $\mathbf{x}\in\reell^{n}$ ein beliebiges Element aus der linken Menge,
|
|||
|
|
d.\,h. $\mathbf{x}$ ist eine Lösung zu $(A|\mathbf{b})$.
|
|||
|
|
Laut \Cref{lemm:1:ueb:1:ex:2} + \Cref{lemm:2:ueb:1:ex:2} + \Cref{lemm:3:ueb:1:ex:2}
|
|||
|
|
und wegen \eqcref{eq:0:\beweislabel}
|
|||
|
|
erhalten wir, dass $\mathbf{x}$ eine Lösung zu $(A'|\mathbf{b}')$ ist,
|
|||
|
|
d.\,h. $\mathbf{x}$ liegt in der rechten Menge.
|
|||
|
|
Also ist die linke Menge in der rechten enthalten.
|
|||
|
|
|
|||
|
|
\uline{\bfseries ($\supseteq$.)}\\
|
|||
|
|
Man beachte zuerst, dass sich die Transformation in \eqcref{eq:0:\beweislabel} umkehren lässt---\text{und zwar durch Elementartransformationen}.
|
|||
|
|
Es ist einfach zu sehen, dass entweder
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{lrcl}
|
|||
|
|
&(A'|\mathbf{b}') &\overset{I;i,j}{\rightsquigarrow} &(A|\mathbf{b})\\
|
|||
|
|
\text{oder} &(A'|\mathbf{b}') &\overset{I;i,\alpha^{-1}}{\rightsquigarrow} &(A|\mathbf{b})\\
|
|||
|
|
\text{oder} &(A'|\mathbf{b}') &\overset{III;i,j,-\alpha}{\rightsquigarrow} &(A|\mathbf{b}).\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Die Situation ist also analog zum $\subseteq$-Teil.
|
|||
|
|
Darum gilt die $\supseteq$-Inklusion in \eqcref{eq:1:\beweislabel}.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
\clearpage
|
|||
|
|
%% AUFGABE 1-3
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 3]{}
|
|||
|
|
\label{ueb:1:ex:3}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
Für diese Aufgabe wird das Konzept der \emph{linearen Unabhängigkeit} aus Kapitel 5 angewandt.
|
|||
|
|
|
|||
|
|
\begin{defn}
|
|||
|
|
Seien $m,n\in\ntrlpos$ mit $m>n$
|
|||
|
|
und seien $A\in\reell^{m\times n}$, $\mathbf{b}\in\reell^{m}$,
|
|||
|
|
und $I\subseteq\{1,2,\ldots,m\}$.
|
|||
|
|
Bezeichne mit $(A|\mathbf{b})_{I}$ die erweiterte Koeffizientenmatrix $(A|\mathbf{b})$,
|
|||
|
|
die auf die Zeilen mit Indexes aus $I$ (in bspw. aufsteigender Reihenfolge) reduziert ist.
|
|||
|
|
\end{defn}
|
|||
|
|
|
|||
|
|
\begin{e.g.}
|
|||
|
|
Für $(A|\mathbf{b})$ gleich
|
|||
|
|
|
|||
|
|
{\scriptsize
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{ccc|c}
|
|||
|
|
-5 &0 &0 &-7\\
|
|||
|
|
4 &-6 &-10 &6\\
|
|||
|
|
-2 &-6 &-6 &9\\
|
|||
|
|
-7 &4 &-1 &-5\\
|
|||
|
|
4 &-5 &2 &-9\\
|
|||
|
|
-5 &8 &-7 &-5\\
|
|||
|
|
\end{matrix}
|
|||
|
|
\end{mathe}}
|
|||
|
|
|
|||
|
|
und $I=\{2,5,6\}$ ist $(A|\mathbf{b})_{I}$ gleich
|
|||
|
|
|
|||
|
|
{\scriptsize
|
|||
|
|
\begin{mathe}[bc]{c}
|
|||
|
|
\begin{matrix}{ccc|c}
|
|||
|
|
4 &-6 &-10 &6\\
|
|||
|
|
4 &-5 &2 &-9\\
|
|||
|
|
-5 &8 &-7 &-5\\
|
|||
|
|
\end{matrix}.
|
|||
|
|
\end{mathe}}
|
|||
|
|
|
|||
|
|
\nvraum{1}
|
|||
|
|
|
|||
|
|
\end{e.g.}
|
|||
|
|
|
|||
|
|
Mit diesem Mittel können wir nun die Hauptaussage in der Aufgabe formulieren:
|
|||
|
|
|
|||
|
|
\begin{schattierteboxdunn}
|
|||
|
|
\begin{satz}
|
|||
|
|
\makelabel{satz:main:ueb:1:ex:3}
|
|||
|
|
Seien $m,n\in\ntrlpos$ mit $m>n$
|
|||
|
|
und seien $A\in\reell^{m\times n}$ und $\mathbf{b}\in\reell^{m}$.
|
|||
|
|
Falls $(A|\mathbf{b})$ unlösbar ist,
|
|||
|
|
dann existiert $I\subseteq\{1,2,\ldots,m\}$ mit $|I|=n+1$,
|
|||
|
|
so dass $(A|\mathbf{b})_{I}$ unlösbar ist.
|
|||
|
|
\end{satz}
|
|||
|
|
\end{schattierteboxdunn}
|
|||
|
|
|
|||
|
|
\begin{einzug}[\rtab][\rtab]
|
|||
|
|
\begin{proof}[*][\Cref{\beweislabel}]
|
|||
|
|
Es stehen nun die \emph{Zeilen} der Matrix $A$ im Fokus.
|
|||
|
|
Wir verwandeln diese in Vektoren, d.\,h. setze
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\mathbf{z}^{(i)}\in\reell^{n}\,\text{die $i$-te Zeile von $A$ als Vektor geschrieben}
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
für $i\in\{1,2,\ldots,m\}$.
|
|||
|
|
Da ${\mathbf{z}^{(1)},\mathbf{z}^{(2)},\ldots,\mathbf{z}^{(m)}\in\reell^{n}}$,
|
|||
|
|
können wir eine \emph{maximale Menge} ${I_{0}\subseteq\{1,2,\ldots,m\}}$ finden,
|
|||
|
|
so dass $(\mathbf{z}^{(i)})_{i\in I_{0}}$ linear unabhängige Vektoren sind.
|
|||
|
|
Aus der Maximalität folgt, dass für jedes ${k\in\{1,2,\ldots,m\}\ohne I_{0}}$
|
|||
|
|
$(\mathbf{z}^{(i)})_{i\in I_{0}\cup\{k\}}$ \emph{linear abhängig} sind.
|
|||
|
|
Wegen der Dimension von $\reell^{n}$ gilt ${|I|\leq\min\{m,n\}=n}$.
|
|||
|
|
Aus der linearer Unabhängigkeit von den $(\mathbf{z}^{(i)})_{i\in I_{0}}$ folgt,
|
|||
|
|
dass es (eindeutige) Koeffizienten $c_{k,i}\in\reell$ für $i\in I_{0}$ gibt,
|
|||
|
|
so dass
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:1:\beweislabel]
|
|||
|
|
\mathbf{z}^{(k)} &= &\sum_{i\in I_{0}:~}c_{k,i}\mathbf{z}^{(i)}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
gilt.
|
|||
|
|
|
|||
|
|
Um nun die Hauptaussage zu zeigen, nehmen wir an, dass $(A|\mathbf{b})$ unlösbar ist.
|
|||
|
|
\textbf{Zu zeigen:} Es gibt eine Teilmenge ${I\subseteq\{1,2,\ldots,m\}}$ mit ${|I|=n+1}$,
|
|||
|
|
so dass $(A|\mathbf{b})_{I}$ unlösbar ist.
|
|||
|
|
\fbox{Angenommen, dies sei nicht der Fall.}
|
|||
|
|
Aus dieser Annahme leiten wir folgende Behauptungen ab:
|
|||
|
|
|
|||
|
|
\begin{kompaktitem}[\rtab][\rtab]
|
|||
|
|
\behauptungbeleg{1}
|
|||
|
|
Die Verhältnisse zwischen den Zeilenvektoren in \eqcref{eq:1:\beweislabel} gelten auch für die Einträge aus $\mathbf{b}$.
|
|||
|
|
Das heißt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:2:\beweislabel]
|
|||
|
|
b_{k} &= &\sum_{i\in I_{0}:~}c_{k,i}b_{i}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
für alle ${k\in\{1,2,\ldots,m+1\}\ohne I_{0}}$.\\
|
|||
|
|
\voritemise
|
|||
|
|
\belegbehauptung
|
|||
|
|
Sei $k\in\{1,2,\ldots,m+1\}\ohne I_{0}$ beliebig.
|
|||
|
|
Da $|I_{0}|\leq n<n+1$ lässt sich eine Teilmenge $I\subseteq\{1,2,\ldots,m\}$ wählen,
|
|||
|
|
mit $I\supseteq I_{0}\cup\{k\}$ und $|I|=n+1$.
|
|||
|
|
Dann per \emph{Annahme} ist $(A|\mathbf{b})_{I}$ lösbar.
|
|||
|
|
Das heißt, $\mathbf{x}\in\reell^{n}$ existiert, so dass
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:3:\beweislabel]
|
|||
|
|
b_{i} &= &\sum_{j=1}^{n}a_{i,j}x_{j}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
für alle $i\in I$ gilt.
|
|||
|
|
Da $k\in I$ und $I_{0}\subseteq I$ und wegen \eqcref{eq:1:\beweislabel} erhalten wir
|
|||
|
|
nun das Verhältnis
|
|||
|
|
|
|||
|
|
\begin{longmathe}[mc]{RCL}
|
|||
|
|
b_{k} &= &\sum_{j=1}^{n}a_{k,j}x_{j}\\
|
|||
|
|
&= &\sum_{j=1}^{n}(\mathbf{z}^{(k)})_{j}x_{j}\\
|
|||
|
|
&&\quad\text{da die Einträge der $k$-ten Zeile den Einträgen von $\mathbf{z}^{(k)}$ entsprechen}\\
|
|||
|
|
&\eqcrefoverset{eq:1:\beweislabel}{=}
|
|||
|
|
&\sum_{j=1}^{n}(\sum_{i\in I_{0}}c_{k,i}\mathbf{z}^{(i)})_{j}x_{j}\\
|
|||
|
|
&= &\sum_{j=1}^{n}\sum_{i\in I_{0}}c_{k,i}z^{(i)}_{j}x_{j}\\
|
|||
|
|
&= &\sum_{i\in I_{0}}c_{k,i}\sum_{j=1}^{n}z^{(i)}_{j}x_{j}\\
|
|||
|
|
&= &\sum_{i\in I_{0}}c_{k,i}\sum_{j=1}^{n}a_{i,j}x_{j}\\
|
|||
|
|
&&\quad\text{da die Einträge der $i$-ten Zeile den Einträgen von $\mathbf{z}^{(i)}$ entsprechen}\\
|
|||
|
|
&\eqcrefoverset{eq:3:\beweislabel}{=} &\sum_{i\in I_{0}}c_{k,i}b_{i}.\\
|
|||
|
|
\end{longmathe}
|
|||
|
|
|
|||
|
|
Darum gilt die Behauptung.
|
|||
|
|
\enndeOfSomething[Beh. 1]
|
|||
|
|
\behauptungbeleg{2}
|
|||
|
|
Es gibt eine Lösung zu $(A|\mathbf{b})$.\\
|
|||
|
|
\voritemise
|
|||
|
|
\belegbehauptung
|
|||
|
|
Da $|I_{0}|\leq n<n+1$ lässt sich eine Teilmenge $I\subseteq\{1,2,\ldots,m\}$ wählen,
|
|||
|
|
so dass $I\supseteq I_{0}$ und $|I|=n+1$.
|
|||
|
|
Dann per \emph{Annahme} ist $(A|\mathbf{b})_{I}$ lösbar.
|
|||
|
|
Das heißt, ein $\mathbf{x}\in\reell^{n}$ existiert, so dass
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:3b:\beweislabel]
|
|||
|
|
b_{i} &= &\sum_{j=1}^{n}a_{i,j}x_{j}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
für alle $i\in I$ gilt.
|
|||
|
|
Da $I\supseteq I_{0}$ können wir \textbf{Behauptung 1} und die Verhältnisse in \eqcref{eq:1:\beweislabel} anwenden.
|
|||
|
|
Für jedes ${k\in\{1,2,\ldots,m\}\ohne I}$ gilt
|
|||
|
|
|
|||
|
|
\begin{longmathe}[mc]{RCL}
|
|||
|
|
\sum_{j=1}^{n}a_{k,j}x_{j}
|
|||
|
|
&= &\sum_{j=1}^{n}(\mathbf{z}^{(k)})_{j}x_{j}\\
|
|||
|
|
&&\quad\text{da die Einträge der $k$-ten Zeile den Einträgen von $\mathbf{z}^{(k)}$ entsprechen}\\
|
|||
|
|
&\eqcrefoverset{eq:1:\beweislabel}{=}
|
|||
|
|
&\sum_{j=1}^{n}(\sum_{i\in I_{0}}c_{k,i}\mathbf{z}^{(i)})_{j}x_{j}\\
|
|||
|
|
&= &\sum_{j=1}^{n}\sum_{i\in I_{0}}c_{k,i}z^{(i)}_{j}x_{j}\\
|
|||
|
|
&= &\sum_{i\in I_{0}}c_{k,i}\sum_{j=1}^{n}z^{(i)}_{j}x_{j}\\
|
|||
|
|
&= &\sum_{i\in I_{0}}c_{k,i}\sum_{j=1}^{n}a_{i,j}x_{j}\\
|
|||
|
|
&&\quad\text{da die Einträge der $i$-ten Zeile den Einträgen von $\mathbf{z}^{(i)}$ entsprechen}\\
|
|||
|
|
&\eqcrefoverset{eq:3b:\beweislabel}{=} &\sum_{i\in I_{0}}c_{k,i}b_{i}\\
|
|||
|
|
&\textoverset{Beh. 1}{=} &b_{k}\\
|
|||
|
|
\end{longmathe}
|
|||
|
|
|
|||
|
|
Also ist $\mathbf{x}\in\reell^{n}$ nicht nur eine Lösung zu Zeile $i$ des LGS, $(A|\mathbf{b})$, für jedes $i\in I$,
|
|||
|
|
sondern auch für jedes ${i\in\{1,2,\ldots,m\}\ohne I}$.
|
|||
|
|
Das heißt, $\mathbf{x}$ ist eine Lösung des LGS $(A|\mathbf{b})$.
|
|||
|
|
Also ist $(A|\mathbf{b})$ lösbar.
|
|||
|
|
\enndeOfSomething[Beh. 2]
|
|||
|
|
\end{kompaktitem}
|
|||
|
|
|
|||
|
|
Laut \textbf{Behauptung 2} ist also $(A|\mathbf{b})$ lösbar.
|
|||
|
|
Dies ist aber ein Widerspruch!
|
|||
|
|
Darum stimmt die \emph{Annahme} oben nicht.
|
|||
|
|
Also gibt es \emph{doch} eine Teilmenge ${I\subseteq\{1,2,\ldots,m\}}$ mit ${|I|=n+1}$, so dass $(A|\mathbf{b})_{I}$ unlösbar ist.
|
|||
|
|
Damit wurde die zu zeigende Implikation bewiesen.
|
|||
|
|
\end{proof}
|
|||
|
|
\end{einzug}
|
|||
|
|
|
|||
|
|
\begin{rem}
|
|||
|
|
Falls man sich aber auf rudimentäre Mitteln beschränken will, kann man alternativ wie folgt vorgehen.
|
|||
|
|
Man wende zuerst das Gaußverfahren an und erhalte somit eine Folge
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcccccccl}
|
|||
|
|
(A^{(0)}|\mathbf{b}^{(0)})
|
|||
|
|
&\rightsquigarrow
|
|||
|
|
&(A^{(1)}|\mathbf{b}^{(1)})
|
|||
|
|
&\rightsquigarrow
|
|||
|
|
&(A^{(2)}|\mathbf{b}^{(2)})
|
|||
|
|
&\rightsquigarrow
|
|||
|
|
&\cdots
|
|||
|
|
&\rightsquigarrow
|
|||
|
|
&(A^{(N)}|\mathbf{b}^{(N)})
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
wobei $N\in\ntrl$, ${A^{(0)}=A}$, ${\mathbf{b}^{(0)}=\mathbf{b}}$,
|
|||
|
|
$(A^{(N)}|\mathbf{b}^{(N)})$ eine erweiterte Koeffizientenmatrix in Zeilenstufenform ist,
|
|||
|
|
und jede der »$\rightsquigarrow$« Übergänge jeweils eine Transformation der Art (I), (II), oder (III) bezeichnet.
|
|||
|
|
Da $m>n$ sieht nun die Zeilenstufenform, also $(A^{(N)}|\mathbf{b}^{(N)})$, folgendermaßen aus:
|
|||
|
|
|
|||
|
|
{\scriptsize
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\begin{matrix}{cccccccc|c}
|
|||
|
|
\underbrace{0\,0\,\ldots\,0}_{\ell_{1}} &\gamma_{1} &\cdots\cdots &\ast &\cdots\cdots &\cdots\cdots &\ast &\cdots\cdots &b^{(N)}_{1}\\
|
|||
|
|
0\,0\,\ldots\,0 &0 &\underbrace{0\,0\,\ldots\,0}_{\ell_{2}} &\gamma_{2} &\cdots\cdots &\cdots\cdots &\ast &\cdots\cdots &b^{(N)}_{2}\\
|
|||
|
|
\vdots & & & & & & &\vdots\\
|
|||
|
|
0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &\underbrace{0\,0\,\ldots\,0}_{\ell_{r}} &\gamma_{r} &\cdots\cdots &b^{(N)}_{r}\\
|
|||
|
|
0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &0\,0\,\ldots\,0 &0 &\cdots\cdots &b^{(N)}_{r+1}\\
|
|||
|
|
\vdots & & & & & & &\vdots\\
|
|||
|
|
0\,0\,\ldots\,0 &0 &0\,0\,\ldots\,0 &0 &\cdots\cdots &0\,0\,\ldots\,0 &0 &\cdots\cdots &b^{(N)}_{m}\\
|
|||
|
|
\end{matrix}
|
|||
|
|
\end{mathe}}
|
|||
|
|
|
|||
|
|
wobei $r\in\ntrlzero$ die Anzahl der Stufen ist,
|
|||
|
|
${\ell_{1},\ell_{2},\ldots,\ell_{r}\in\ntrlzero}$,
|
|||
|
|
und $\gamma_{1},\gamma_{2},\ldots,\gamma_{r}\in\reell\ohne\{0\}$ die Hauptkoeffizienten der Stufen sind.
|
|||
|
|
Es muss nun $0\leq r\leq \min\{m,n\}=n$ gelten.
|
|||
|
|
|
|||
|
|
Jetzt kann man leicht dafür argumentiere, dass (1) die Zeilenstufenform, $(A^{(N)}|\mathbf{b}^{(N)})$, die Implikation erfüllt.
|
|||
|
|
Dann aufgrund der Umkehrbarkeit der Elementartransformationen, reicht es aus zu zeigen, dass (2):
|
|||
|
|
wenn ${(A',\mathbf{b}')\rightsquigarrow(A'',\mathbf{b}'')}$ und wenn $(A',\mathbf{b}')$ die Implikation erfüllt,
|
|||
|
|
dann erfüllt $(A'',\mathbf{b}'')$ die Implikation.
|
|||
|
|
Dies ist nur etwas mühseliger und die Argumentation von (2) führt letzten Endes zu ähnlichen Ideen, die im Beweis oben vorkommen.
|
|||
|
|
\end{rem}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: body/uebung/ueb2.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\setcounternach{chapter}{2}
|
|||
|
|
\chapter[Woche 2]{Woche 2}
|
|||
|
|
\label{ueb:2}
|
|||
|
|
|
|||
|
|
\textbf{ACHTUNG.}
|
|||
|
|
Diese Lösungen dienen \emph{nicht} als Musterlösungen sondern eher als Referenz.
|
|||
|
|
Hier wird eingehender gearbeitet, als generell verlangt wird.
|
|||
|
|
Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche vergleichen kann.
|
|||
|
|
|
|||
|
|
%% AUFGABE 2-1
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 1]{}
|
|||
|
|
\label{ueb:2:ex:1}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
\begin{schattierteboxdunn}
|
|||
|
|
\begin{satz}[vgl. {\cite[Korollar 1.3.3]{sinn2020}}]
|
|||
|
|
\makelabel{satz:main:ueb:2:ex:1}
|
|||
|
|
Sei $V$ ein Vektorraum über $\reell$ wie $\reell^{n}$ für ein $n\in\ntrlpos$.
|
|||
|
|
Seien $\mathbf{v},\mathbf{w}\in V$ mit $\mathbf{v}\neq \mathbf{w}$ und $\mathbf{w}\neq\zerovector$
|
|||
|
|
und sei
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
L &:= &\{s\mathbf{v} + (1-s)\mathbf{w}\mid s\in\reell\}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
die Verbindungsgerade zw. $\mathbf{v}$ und $\mathbf{w}$.
|
|||
|
|
Dann gilt $\zerovector\in L$ $\Leftrightarrow$ $\exists{c\in\reell:~}\mathbf{v}=c\mathbf{w}$.
|
|||
|
|
\end{satz}
|
|||
|
|
\end{schattierteboxdunn}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Der Beweis wird in zwei Teilen gezeigt.
|
|||
|
|
|
|||
|
|
\hinRichtung Angenommen, $\zerovector\in L$.
|
|||
|
|
\textbf{Zu zeigen:} $\exists{c\in\reell:~}\mathbf{v}=c\mathbf{w}$.\\
|
|||
|
|
Per Definition von $L$ existiert ein $s\in\reell$, so dass sich $\zerovector$
|
|||
|
|
als $\zerovector=s\mathbf{v} + (1-s)\mathbf{w}$
|
|||
|
|
darstellen lässt.
|
|||
|
|
Daraus lässt sich ableiten:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\zerovector=s\mathbf{v} + (1-s)\mathbf{w}
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&s\mathbf{v} = (s-1)\mathbf{w}\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\underbrace{%
|
|||
|
|
(s=0\,\text{und}\,\mathbf{w}=s(\mathbf{w}-\mathbf{v})=\zerovector)
|
|||
|
|
}_{%
|
|||
|
|
\text{unmöglich, da $\mathbf{w}\neq\zerovector$ per Voraussetzung}
|
|||
|
|
}
|
|||
|
|
\,\text{oder}\,(s\neq 0\,\text{und}\,\mathbf{v} = ((s-1)/s)\mathbf{w})\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&s\neq 0\,\text{und}\,\mathbf{v} = ((s-1)/s)\mathbf{w}\\
|
|||
|
|
&\Longrightarrow
|
|||
|
|
&\exists{c\in\reell:~}\mathbf{v} = c\mathbf{w}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
\herRichtung Angenommen, $\mathbf{v} = c\mathbf{w}$ für ein $c\in\reell$.
|
|||
|
|
\textbf{Zu zeigen:} $\zerovector\in L$.\\
|
|||
|
|
Per Voraussetzung gilt nun $\mathbf{v}\neq\mathbf{w}$, sodass $c=1$ direkt ausgeschlossen ist.\\
|
|||
|
|
Setze nun \fbox{$s:=\frac{1}{1-c}\in\reell$}, was wohldefiniert ist, da $c\neq 1$.\\
|
|||
|
|
Man berechnet nun
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcccccccl}
|
|||
|
|
\overbrace{s\mathbf{v}+(1-s)\mathbf{w}}^{\in L,\,\text{per Definition}}
|
|||
|
|
&= &\frac{1}{1-c}c\mathbf{w}+(1-\frac{1}{1-c})\mathbf{w}
|
|||
|
|
&= &(\underbrace{\frac{c}{1-c}+1-\frac{1}{1-c}}_{=\frac{c-1}{1-c}+1=0})\mathbf{w}
|
|||
|
|
&= &0\mathbf{w}
|
|||
|
|
&= &\zerovector.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Darum gilt $\zerovector\in L$.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
%% AUFGABE 2-2
|
|||
|
|
\clearpage
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 2]{}
|
|||
|
|
\label{ueb:2:ex:2}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries (a)}
|
|||
|
|
%% AUFGABE 2-2a
|
|||
|
|
\item
|
|||
|
|
\begin{schattierteboxdunn}
|
|||
|
|
\begin{satz}
|
|||
|
|
\makelabel{satz:main:ueb:2:ex:2a}
|
|||
|
|
Seien $\mathbf{v},\mathbf{v}^{\prime},\mathbf{w},\mathbf{w}^{\prime}\in\reell^{2}$
|
|||
|
|
mit $\mathbf{w},\mathbf{w}^{\prime}\neq\zerovector$.
|
|||
|
|
Seien
|
|||
|
|
$L:=\{\mathbf{v}+t\mathbf{w}\mid t\in\reell\}$
|
|||
|
|
und
|
|||
|
|
$L^{\prime}:=\{\mathbf{v}^{\prime}+s\mathbf{w}^{\prime}\mid s\in\reell\}$.
|
|||
|
|
Angenommen, $L\neq L^{\prime}$.
|
|||
|
|
Dann sind folgende Aussagen äquivalent:
|
|||
|
|
|
|||
|
|
\begin{kompaktenum}{(i)}
|
|||
|
|
\item\punktlabel{1}
|
|||
|
|
$L\cap L^{\prime}=\leer$;
|
|||
|
|
\item\punktlabel{2}
|
|||
|
|
$\mathbf{w},\mathbf{w}^{\prime}$ sind kolinear,
|
|||
|
|
d.\,h.
|
|||
|
|
$\exists{c\in\reell:~}\mathbf{w}=c\mathbf{w}^{\prime}$.
|
|||
|
|
\end{kompaktenum}
|
|||
|
|
|
|||
|
|
\nvraum{1}
|
|||
|
|
\end{satz}
|
|||
|
|
\end{schattierteboxdunn}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Der Beweis wird in zwei Teilen gezeigt.
|
|||
|
|
|
|||
|
|
\hinRichtung{1}{2} Angenommen, $L\cap L^{\prime}=\leer$.
|
|||
|
|
\textbf{Zu zeigen:} $\exists{c\in\reell:~}\mathbf{w}=c\mathbf{w}^{\prime}$.\\
|
|||
|
|
\fbox{Angenommen, dies sei nicht der Fall.}\\
|
|||
|
|
Da $\mathbf{w},\mathbf{w}^{\prime}\neq\zerovector$ bedeutet dies,
|
|||
|
|
dass $\mathbf{w},\mathbf{w}^{\prime}$ \emph{linear unabhängig} sind. ($\to$ Warum??)\\
|
|||
|
|
Also gilt für den Untervektorraum
|
|||
|
|
$U:=\span\{\mathbf{w},\mathbf{w}^{\prime}\}$,
|
|||
|
|
dass $\dim(U)=2$.\\
|
|||
|
|
Da $U\subseteq\reell^{2}$ Vektorräume sind und $\dim(U)=2=\dim(\reell^{2})$,
|
|||
|
|
folgt hieraus, dass $U=\reell^{2}$. ($\to$ Warum??)\\
|
|||
|
|
Betrachte bspw. den Vektor
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:1-2:1:\beweislabel]
|
|||
|
|
\mathbf{\xi} &:= &\mathbf{v}^{\prime}-\mathbf{v}\in\reell^{2}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Dann $\mathbf{\xi}\in U=\span\{\mathbf{w},\mathbf{w}^{\prime}\}$.
|
|||
|
|
Folglich existieren Skalare $\alpha,\beta\in\reell$,
|
|||
|
|
so dass $\alpha\mathbf{w}+\beta\mathbf{w}^{\prime}=\mathbf{\xi}$
|
|||
|
|
gilt.\\
|
|||
|
|
Setze nun \fbox{$t:=\alpha$} und \fbox{$s:=-\beta$}.
|
|||
|
|
Dann gilt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rclcl}
|
|||
|
|
\overbrace{%
|
|||
|
|
\mathbf{v}+t\mathbf{w}
|
|||
|
|
}^{\in L}
|
|||
|
|
&= &(\mathbf{v}+t\mathbf{w})-(\mathbf{v}^{\prime}+s\mathbf{w}^{\prime})
|
|||
|
|
+\mathbf{v}^{\prime}+s\mathbf{w}^{\prime}\\
|
|||
|
|
&= &(\mathbf{v}-\mathbf{v}^{\prime})+(t\mathbf{w}-s\mathbf{w}^{\prime})
|
|||
|
|
+\mathbf{v}^{\prime}+s\mathbf{w}^{\prime}\\
|
|||
|
|
&= &(\mathbf{v}-\mathbf{v}^{\prime})+(\alpha\mathbf{w}+\beta\mathbf{w}^{\prime})
|
|||
|
|
+\mathbf{v}^{\prime}+s\mathbf{w}^{\prime}\\
|
|||
|
|
&\eqcrefoverset{eq:1-2:1:\beweislabel}{=}
|
|||
|
|
&-\mathbf{\xi}+\mathbf{\xi}
|
|||
|
|
+\mathbf{v}^{\prime}+s\mathbf{w}^{\prime}
|
|||
|
|
&= &\underbrace{%
|
|||
|
|
\mathbf{v}^{\prime}+s\mathbf{w}^{\prime}%
|
|||
|
|
}_{\in L^{\prime}}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Darum gilt $L\cap L^{\prime}\neq\leer$,
|
|||
|
|
was ein Widerspruch ist.\\
|
|||
|
|
Darum stimmt die o.\,s. Annahme nicht.
|
|||
|
|
Also sind $\mathbf{w},\mathbf{w}^{\prime}$ kolinear.
|
|||
|
|
|
|||
|
|
\hinRichtung{2}{1} Angenommen, $\mathbf{w}=c\mathbf{w}^{\prime}$ für ein $c\in\reell$.
|
|||
|
|
\textbf{Zu zeigen:} $L\cap L^{\prime}=\leer$.\\
|
|||
|
|
\fbox{Angenommen, dies sei nicht der Fall.}
|
|||
|
|
Dann existiert ein Vektor, $\mathbf{u}\in L\cap L^{\prime}$.\\
|
|||
|
|
Per Konstruktion existieren dann $s_{0},t_{0}\in\reell$,
|
|||
|
|
so dass
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcccl}
|
|||
|
|
\mathbf{v}+t_{0}\mathbf{w} &= &\mathbf{u} &= &\mathbf{v}^{\prime}+s_{0}\mathbf{w}^{\prime}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Aus der Voraussetzung für diese Richtung folgt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:2-1:1:\beweislabel]
|
|||
|
|
\mathbf{v}^{\prime} &= &\mathbf{v}+(t_{0}-s_{0}c)\mathbf{w}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Beachte, dass \fbox{$c\neq 0$}, denn sonst würde $\mathbf{w}=c\mathbf{w}^{\prime}=\zerovector$ gelten,
|
|||
|
|
was ein Widerspruch ist. Wir berechnen
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:2-1:2:\beweislabel]
|
|||
|
|
L^{\prime} &= &\{\mathbf{v}^{\prime}+s\mathbf{w}^{\prime}\mid s\in\reell\}\\
|
|||
|
|
&\eqcrefoverset{eq:2-1:1:\beweislabel}{=}
|
|||
|
|
&\{\mathbf{v}+(t_{0}-s_{0}c)\mathbf{w}+sc\mathbf{w}\mid s\in\reell\}\\
|
|||
|
|
&= &\{\mathbf{v}+(t_{0}+(s-s_{0})c)\mathbf{w}\mid s\in\reell\}\\
|
|||
|
|
&= &\{\mathbf{v}+t\mathbf{w}\mid t\in R\},\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
wobei $R=\{t_{0}+(s-s_{0})c\mid s\in\reell\}=f(\reell)$.
|
|||
|
|
Also $R=f(\reell)$, wobei ${f:\reell\to\reell}$ eine durch ${f(s)=t_{0}+(s-s_{0})c}$ definierte Funktion ist.
|
|||
|
|
Da $c\neq 0$, ist es einfach zu sehen, dass $f$ surjektiv ist (in der Tat bijektiv).
|
|||
|
|
Darum gilt $R=f(\reell)=\reell$.\\
|
|||
|
|
Aus \eqcref{eq:2-1:2:\beweislabel} folgt also
|
|||
|
|
${L^{\prime}=\{\mathbf{v}+t\mathbf{w}\mid t\in\reell\}=L}$,
|
|||
|
|
was ein Widerspruch ist.\\
|
|||
|
|
Darum stimmt die o.\,s. Annahme nicht.
|
|||
|
|
Also gilt $L\cap L^{\prime}=\leer$.
|
|||
|
|
\end{proof}
|
|||
|
|
%% AUFGABE 2-2b
|
|||
|
|
\item
|
|||
|
|
Wir zeigen nun ein minimales Beispiel dafür, dass \Cref{satz:main:ueb:2:ex:2a}
|
|||
|
|
im allgemeinen für andere Vektorräume nicht gilt.
|
|||
|
|
Betrachte den Vektorraum $\reell^{3}$.
|
|||
|
|
Betrachte die folgenden Vektoren in $\reell^{3}$:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rclqrclqrclqrcl}
|
|||
|
|
\mathbf{v} &= &\begin{svector}0\\0\\0\\\end{svector},
|
|||
|
|
&\mathbf{v}^{\prime} &= &\begin{svector}1\\0\\0\\\end{svector},
|
|||
|
|
&\mathbf{w} &= &\begin{svector}0\\1\\0\\\end{svector},
|
|||
|
|
&\mathbf{w}^{\prime} &= &\begin{svector}0\\1\\1\\\end{svector}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Bis auf 2-Dimensionalität erfüllen diese die Voraussetzungen in \Cref{satz:main:ueb:2:ex:2a}.
|
|||
|
|
Einerseits wurden $\mathbf{w}$, $\mathbf{w}^{\prime}$ so gewählt, dass sie \emph{nicht} kolinear sind.
|
|||
|
|
Dennoch schneiden sich die beiden Geraden, $L$, $L^{\prime}$, nicht,
|
|||
|
|
da
|
|||
|
|
${L\subseteq \{\mathbf{x}\in\reell^{3}\mid x_{1}=0\}=:E}$
|
|||
|
|
und
|
|||
|
|
${L^{\prime}\subseteq \{\mathbf{x}\in\reell^{3}\mid x_{1}=1\}=:E^{\prime}}$
|
|||
|
|
und offensichtlich $E\cap E'=\leer$.
|
|||
|
|
\end{enumerate}
|
|||
|
|
|
|||
|
|
%% AUFGABE 2-3
|
|||
|
|
\clearpage
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 3]{}
|
|||
|
|
\label{ueb:2:ex:3}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries (a)}
|
|||
|
|
%% AUFGABE 2-3a
|
|||
|
|
\item
|
|||
|
|
Für jedes $\gamma\in\reell$ sei die Gerade $L_{\gamma}\subseteq\reell^{2}$ gegeben durch
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
L_{\gamma} &= &\{(x,y)\in\reell^{2}\mid 2x+y=\gamma\cdot(x-3y-7)\}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
\begin{schattierteboxdunn}
|
|||
|
|
\begin{satz}
|
|||
|
|
\makelabel{satz:main:ueb:2:ex:3a}
|
|||
|
|
Es gibt exakt einen Punkt in dem Schnitt aus den Geraden, $L_{\gamma}$, $\gamma\in\reell$.
|
|||
|
|
Es gilt nämlich ${\displaystyle\bigcap_{\gamma\in\reell}L_{\gamma}=\{\mathbf{\xi}\}}$,
|
|||
|
|
wobei $\mathbf{\xi}=(1,-2)$.
|
|||
|
|
\end{satz}
|
|||
|
|
\end{schattierteboxdunn}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Wir teilen diesen Beweis in zwei Teilen auf:
|
|||
|
|
|
|||
|
|
\BeweisRichtung[$\supseteq$]
|
|||
|
|
Es reicht aus, für alle $\gamma\in\reell$ \textbf{zu zeigen}, dass $\mathbf{\xi}\in L_{\gamma}$.\\
|
|||
|
|
Fixiere also ein beliebiges $\gamma\in\reell$. Dann
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rclclcll}
|
|||
|
|
2\xi_{1}+\xi_{2}
|
|||
|
|
&= &2\cdot 1+(-2)
|
|||
|
|
&= &0,
|
|||
|
|
&&&\text{und}\\
|
|||
|
|
\gamma\cdot(\xi_{1}-3\xi_{2}-7)
|
|||
|
|
&= &\gamma\cdot(1-3(-2)-7)
|
|||
|
|
&= &\gamma\cdot 0
|
|||
|
|
&= &0.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Also ${2\xi_{1}+\xi_{2}=\gamma\cdot(\xi_{1}-3\xi_{2}-7)}$.
|
|||
|
|
Folglich gilt $\mathbf{\xi}\in L_{\gamma}$ per Konstruktion.
|
|||
|
|
|
|||
|
|
\BeweisRichtung[$\subseteq$]
|
|||
|
|
Sei ${\mathbf{\eta}:=(x,y)\in\bigcap_{\gamma\in\reell}L_{\gamma}}$ beliebig.
|
|||
|
|
\textbf{Zu zeigen:} $\mathbf{\eta}=\mathbf{\xi}$.\\
|
|||
|
|
Zu diesem Zwecke seien $\gamma_{1},\gamma_{2}\in\reell$ irgendwelche Werte mit $\gamma_{1}\neq\gamma_{2}$.
|
|||
|
|
Per Wahl gilt $\mathbf{\eta}\in L_{\gamma_{1}}\cap L_{\gamma_{2}}$.
|
|||
|
|
Also
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
2x+y &= &\gamma_{1}\cdot(x-3x-7),\,\text{und}\\
|
|||
|
|
2x+y &= &\gamma_{2}\cdot(x-3x-7).\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Wir können ganz naiv arbeiten und die Gleichungen subtrahieren.
|
|||
|
|
Dies liefert
|
|||
|
|
$(\gamma_{1}-\gamma_{2})\cdot(x-3x-7)=0$,
|
|||
|
|
woraus sich ergibt, dass
|
|||
|
|
$x-3y-7=0$
|
|||
|
|
gelten muss, da $\gamma_{1}\neq\gamma_{2}$.
|
|||
|
|
Eingesetzt in die erste Gleichung oben liefert
|
|||
|
|
$2x+y=\gamma\cdot 0=0$.
|
|||
|
|
Darum muss $\begin{svector}x\\y\\\end{svector}$
|
|||
|
|
das LGS $(A|\mathbf{b})$ lösen, wobei
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rclqrcl}
|
|||
|
|
A &= &\begin{smatrix}
|
|||
|
|
1&-3\\
|
|||
|
|
2&1\\
|
|||
|
|
\end{smatrix},
|
|||
|
|
&\mathbf{b} &= &\begin{svector}7\\0\\\end{svector}
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
\begin{algorithm}[\rtab][\rtab]
|
|||
|
|
Gaußverfahren angewandt auf $(A|\mathbf{b})$:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{cc|c}
|
|||
|
|
1 &-3 &7\\
|
|||
|
|
2 &1 &0\\
|
|||
|
|
\end{matrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Wende die Zeilentransformation
|
|||
|
|
${Z_{2}\leftsquigarrow Z_{2}-2\cdot Z_{1}}$
|
|||
|
|
an:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{cc|c}
|
|||
|
|
1 &-3 &7\\
|
|||
|
|
0 &7 &-14\\
|
|||
|
|
\end{matrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Aus der Stufenform erschließt sich
|
|||
|
|
|
|||
|
|
\begin{mathe}[bc]{rclcl}
|
|||
|
|
y &= &\frac{-14}{7} &= &-2\\
|
|||
|
|
x &= &7 + 3\cdot y &= &1.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
\end{algorithm}
|
|||
|
|
|
|||
|
|
Also
|
|||
|
|
${\mathbf{\eta}=(x, y)=(1, -2)=\mathbf{\xi}}$
|
|||
|
|
für alle $\mathbf{\eta}\in\bigcap_{\gamma\in\reell}L_{\gamma}$.
|
|||
|
|
Das heißt $\bigcap_{\gamma\in\reell}L_{\gamma}\subseteq\{\mathbf{\xi}\}$.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
\clearpage
|
|||
|
|
%% AUFGABE 2-3b
|
|||
|
|
\item
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries (i)}
|
|||
|
|
%% AUFGABE 2-3b-i
|
|||
|
|
\item
|
|||
|
|
Sei $\gamma\in\reell$. Dann gilt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
(-3,2)\in L_{\gamma}
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&2(-3)+(2)=\gamma\cdot((-3)-3(2)-7)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\gamma=\frac{-4}{-16}=\frac{1}{4}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Also ist \fbox{$\gamma=\frac{1}{4}$} der eindeutige Parameter,
|
|||
|
|
für den $(-3,2)\in L_{\gamma}$ gilt.
|
|||
|
|
|
|||
|
|
%% AUFGABE 2-3b-ii
|
|||
|
|
\item
|
|||
|
|
Sei $\gamma\in\reell$. Man beobachte, dass
|
|||
|
|
|
|||
|
|
\begin{longmathe}[mc]{RCL}
|
|||
|
|
L_{\gamma}
|
|||
|
|
&= &\{(x,y)\in\reell^{2}\mid (2-\gamma)x+(1+3\gamma)y=-7\gamma\}\\
|
|||
|
|
&= &\begin{cases}[m]{lcl}
|
|||
|
|
\{(x,y)\in\reell^{2}\mid 0x + (1+3\cdot 2)y=-7\cdot 2\}
|
|||
|
|
&: &\gamma=2\\
|
|||
|
|
\{(x,y)\in\reell^{2}\mid (2-\frac{-1}{3})x + 0y=-7\cdot\frac{-1}{3}\}
|
|||
|
|
&: &\gamma=-\frac{1}{3}\\
|
|||
|
|
\{(x,y)\in\reell^{2}\mid (2-\gamma)x+(1+3\gamma)y=-7\gamma\}
|
|||
|
|
&: &\text{sonst}
|
|||
|
|
\end{cases}\\
|
|||
|
|
&= &\begin{cases}[m]{lcl}
|
|||
|
|
\{(x,y)\in\reell^{2}\mid y=-2\}
|
|||
|
|
&: &\gamma=2\\
|
|||
|
|
\{(x,y)\in\reell^{2}\mid x=1\}
|
|||
|
|
&: &\gamma=-\frac{1}{3}\\
|
|||
|
|
\{(x,y)\in\reell^{2}\mid y=\frac{\gamma-2}{1+3\gamma}x - \frac{7\gamma}{1+3\gamma}\}
|
|||
|
|
&: &\text{sonst}
|
|||
|
|
\end{cases}.\\
|
|||
|
|
\end{longmathe}
|
|||
|
|
|
|||
|
|
Daraus folgt, dass $L_{\gamma}$
|
|||
|
|
|
|||
|
|
\begin{kompaktitem}
|
|||
|
|
\item
|
|||
|
|
parallel zur $x$-Achse für $\gamma=2$ ist,
|
|||
|
|
\item
|
|||
|
|
parallel zur $y$-Achse für $\gamma=-\frac{1}{3}$ ist,
|
|||
|
|
\item
|
|||
|
|
und ansonsten weder zur $x$- noch $y$-Achse parallel ist,
|
|||
|
|
da in diesem Falle $L_{\gamma}$ die Gerade »${y=ax+b}$« ist, wobei $a\neq 0$.
|
|||
|
|
\end{kompaktitem}
|
|||
|
|
|
|||
|
|
Also ist der gesuchte Parameterwert eindeutig \fbox{$\gamma=-\frac{1}{3}$}.
|
|||
|
|
|
|||
|
|
%% AUFGABE 2-3b-iii
|
|||
|
|
\item
|
|||
|
|
Die Gerade »$x-2y=-1$« lässt sich äquivalent
|
|||
|
|
als »$y=\frac{1}{2}x+1$
|
|||
|
|
darstellen.
|
|||
|
|
Darum wird ein Wert $\gamma\in\reell$ gesucht,
|
|||
|
|
so dass die Gerade $L_{\gamma}$ weder zur $x$- noch $y$-Achse parallel ist,
|
|||
|
|
und die die $y$-$x$-Steigung $\frac{1}{2}$ hat.
|
|||
|
|
Nach der o.\,s. Berechnung in (ii) kommt dies nur für den 3. Fall in Frage.
|
|||
|
|
Darum gilt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
L_{\gamma}\,\text{parallel zur Gerade »$x-2y=-1$«}
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\gamma\notin\{2,-\frac{1}{3}\}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
\frac{\gamma-2}{1+3\gamma}=\frac{1}{2}\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\gamma\notin\{2,-\frac{1}{3}\}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
(\gamma-2)=\frac{1}{2}(1+3\gamma)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\gamma\notin\{2,-\frac{1}{3}\}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
\gamma=-5\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\gamma=-5.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Also ist der gesuchte Parameterwert eindeutig \fbox{$\gamma=-5$}.
|
|||
|
|
\end{enumerate}
|
|||
|
|
\end{enumerate}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: body/uebung/ueb3.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\setcounternach{chapter}{3}
|
|||
|
|
\chapter[Woche 3]{Woche 3}
|
|||
|
|
\label{ueb:2}
|
|||
|
|
|
|||
|
|
\textbf{ACHTUNG.}
|
|||
|
|
Diese Lösungen dienen \emph{nicht} als Musterlösungen sondern eher als Referenz.
|
|||
|
|
Hier wird eingehender gearbeitet, als generell verlangt wird.
|
|||
|
|
Das Hauptziel hier ist, eine Variant anzubieten, gegen die man seine Versuche vergleichen kann.
|
|||
|
|
|
|||
|
|
%% AUFGABE 3-1
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 1]{}
|
|||
|
|
\label{ueb:3:ex:1}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
Wir arbeiten im Vektorraum $\reell^{3}$ und betrachten die Vektoren
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rclqrclqrclqrcl}
|
|||
|
|
\mathbf{v}_{1} &= &\begin{svector}1\\3\\1\\\end{svector}
|
|||
|
|
&\mathbf{v}_{2} &= &\begin{svector}-2\\5\\-2\\\end{svector}
|
|||
|
|
&\mathbf{w}_{1} &= &\begin{svector}4\\-3\\-3\\\end{svector}
|
|||
|
|
&\mathbf{w}_{2} &= &\begin{svector}0\\1\\1\\\end{svector}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
\textbf{Zu berechnen:}
|
|||
|
|
$U:=\span\{\mathbf{v}_{1},\mathbf{v}_{2}\}
|
|||
|
|
\cap\span\{\mathbf{w}_{1},\mathbf{w}_{2}\}$
|
|||
|
|
als Untervektorraum von $\reell^{3}$.\\
|
|||
|
|
Zu diesem Zwecke betrachte einen beliebigen Vektor, $\mathbf{\xi}\in\reell^{3}$.
|
|||
|
|
Es gilt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:0:ueb:3:ex:1]
|
|||
|
|
\mathbf{\xi}\in U
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{t_{1},t_{2},t_{3},t_{4}\in\reell:~}
|
|||
|
|
\mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
\mathbf{\xi}=t_{3}\mathbf{w}_{1}+t_{4}\mathbf{w}_{2}\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{\mathbf{t}\in\reell^{4}:~}
|
|||
|
|
\mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
=t_{3}\mathbf{w}_{1}+t_{4}\mathbf{w}_{2}\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{\mathbf{t}\in\reell^{4}:~}
|
|||
|
|
\mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
-t_{3}\mathbf{w}_{1}-t_{4}\mathbf{w}_{2}
|
|||
|
|
=\zerovector\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{\mathbf{t}\in\reell^{4}:~}
|
|||
|
|
\mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
+t_{3}\mathbf{w}_{1}+t_{4}\mathbf{w}_{2}
|
|||
|
|
=\zerovector\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{\mathbf{t}\in\reell^{4}:~}
|
|||
|
|
\mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
A\mathbf{t}=\zerovector,\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
wobei
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcccl}
|
|||
|
|
A &:= &\left(
|
|||
|
|
\mathbf{v}_{1}~
|
|||
|
|
\mathbf{v}_{2}~
|
|||
|
|
\mathbf{w}_{1}~
|
|||
|
|
\mathbf{w}_{2}
|
|||
|
|
\right)
|
|||
|
|
&= &\begin{smatrix}
|
|||
|
|
1&-2&4&0\\
|
|||
|
|
3&5&-3&1\\
|
|||
|
|
1&-2&-3&1\\
|
|||
|
|
\end{smatrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Darum ist es notwendig und hinreichend,
|
|||
|
|
die \emph{homogenen Lösungen} für $A$ zu finden,
|
|||
|
|
und daraus die Parameter abzulesen.
|
|||
|
|
|
|||
|
|
\begin{algorithm}[\rtab][\rtab]
|
|||
|
|
Homogenes Problem für $A$:\\
|
|||
|
|
Zeilentransformationen
|
|||
|
|
${Z_{2}\leftsquigarrow Z_{2}-3\cdot Z_{1}}$,
|
|||
|
|
${Z_{3}\leftsquigarrow Z_{3}-Z_{1}}$
|
|||
|
|
anwenden:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{smatrix}
|
|||
|
|
1&-2&4&0\\
|
|||
|
|
0&11&-15&1\\
|
|||
|
|
0&0&-7&1\\
|
|||
|
|
\end{smatrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Wende die Zeilentransformation
|
|||
|
|
${Z_{2}\leftsquigarrow Z_{2}-Z_{3}}$
|
|||
|
|
an:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{smatrix}
|
|||
|
|
1&-2&4&0\\
|
|||
|
|
0&11&-8&0\\
|
|||
|
|
0&0&-7&1\\
|
|||
|
|
\end{smatrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Aus der Zeilenstufenform erschließt sich, dass $t_{4}$ frei ist.
|
|||
|
|
Also $t_{4}=\alpha$ für ein frei wählbares $\alpha\in\reell$.
|
|||
|
|
Aus der Stufenform von Gleichungen $3,2,1$ erschließt sich
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
t_{3} &= &\frac{1}{7}t_{4} = \frac{1}{7}\alpha\\
|
|||
|
|
t_{2} &= &\frac{8}{11}t_{3} = \frac{8}{77}\alpha\\
|
|||
|
|
t_{1} &= &2t_{2} - 4t_{3}
|
|||
|
|
= \frac{16}{77}\alpha - \frac{4}{7}\alpha
|
|||
|
|
= -\frac{28}{77}\alpha\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Man kann o.\,E. $\alpha$ durch $\beta:=-77\alpha$ ersetzen.
|
|||
|
|
Also ist die homogene Lösung gegeben durch
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\mathbf{t} &= &\beta\begin{svector}28\\-8\\-11\\-77\\\end{svector},
|
|||
|
|
\quad\text{mit $\beta\in\reell$ frei wählbar}.
|
|||
|
|
\end{mathe}
|
|||
|
|
\end{algorithm}
|
|||
|
|
|
|||
|
|
Wir können nun \eqcref{eq:0:ueb:3:ex:1} fortsetzen und erhalten
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\eqtag[eq:1:ueb:3:ex:1]
|
|||
|
|
\mathbf{\xi}\in U
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{\mathbf{t}\in\reell^{4}:~}
|
|||
|
|
\mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
A\mathbf{t}=\zerovector\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{\mathbf{t}\in\reell^{4}:~}
|
|||
|
|
\mathbf{\xi}=t_{1}\mathbf{v}_{1}+t_{2}\mathbf{v}_{2}
|
|||
|
|
\,\text{und}\,
|
|||
|
|
\exists{\beta\in\reell:~}
|
|||
|
|
\mathbf{t}=\beta\begin{svector}28\\-8\\-11\\-77\\\end{svector}\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{\beta\in\reell:~}
|
|||
|
|
\mathbf{\xi}=\beta\cdot(
|
|||
|
|
\underbrace{
|
|||
|
|
28\mathbf{v}_{1}+-8\mathbf{v}_{2}
|
|||
|
|
}_{=:\mathbf{u}}
|
|||
|
|
)\\
|
|||
|
|
&\Longleftrightarrow &\mathbf{\xi}\in\span\{\mathbf{u}\}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
für alle $\mathbf{\xi}\in\reell^{3}$.\\
|
|||
|
|
Es gilt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcccccl}
|
|||
|
|
\mathbf{u}
|
|||
|
|
&= &28\begin{svector}1\\3\\1\\\end{svector}
|
|||
|
|
-8\begin{svector}-2\\5\\-2\\\end{svector}
|
|||
|
|
&= &\begin{svector}44\\44\\44\\\end{svector}
|
|||
|
|
&= &44\begin{svector}1\\1\\1\\\end{svector}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Aus \eqcref{eq:1:ueb:3:ex:1} ergibt sich der zu berechnende Untervektorraum
|
|||
|
|
als
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcccccccl}
|
|||
|
|
\span\{\mathbf{v}_{1},\mathbf{v}_{2}\}
|
|||
|
|
\cap\span\{\mathbf{w}_{1},\mathbf{w}_{2}\}
|
|||
|
|
&= &U
|
|||
|
|
&= &\span\{\mathbf{u}\}
|
|||
|
|
&= &\span\{44\begin{svector}1\\1\\1\\\end{svector}\}
|
|||
|
|
&= &\span\{\begin{svector}1\\1\\1\\\end{svector}\}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
%% AUFGABE 3-2
|
|||
|
|
\clearpage
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 2]{}
|
|||
|
|
\label{ueb:3:ex:2}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
Seien $X$, $Y$ nicht leere Mengen und ${f:X\to Y}$ eine Funktion.
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries (a)}
|
|||
|
|
%% AUFGABE 3-2a
|
|||
|
|
\item
|
|||
|
|
\begin{claim*}
|
|||
|
|
Die Aussage $\forall{A,B\subseteq X:~}f(A\cap B)=f(A)\cap f(B)$
|
|||
|
|
ist \fbox{\uline{nicht} allgemein gültig}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Betrachte das Beispiel $X=\{0,1\}$, $Y=\{2\}$, und ${f:X\to Y}$ mit $f(x)=2$ für alle $x\in X$.
|
|||
|
|
Für $A=\{0\}$ und $B=\{1\}$
|
|||
|
|
gilt $f(A\cap B)=f(\leer)=\leer$,
|
|||
|
|
während $f(A)\cap f(B)=\{2\}\cap\{2\}=\{2\}$.
|
|||
|
|
Also $f(A\cap B)\neq f(A)\cap f(B)$.
|
|||
|
|
Darum ist dies ein Gegenbeispiel zur Aussage.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
\text{Bemerkung.} Die Aussage ist eigentlich genau dann wahr, wenn $f$ injektiv ist.
|
|||
|
|
|
|||
|
|
%% AUFGABE 3-2b
|
|||
|
|
\item
|
|||
|
|
\begin{claim*}
|
|||
|
|
Die Aussage $\forall{A,B\subseteq X:~}f(A\cup B)=f(A)\cup f(B)$
|
|||
|
|
ist \fbox{allgemein gültig}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
Für manche (doppelte) Implikationen hier, nämlich für den Umgang mit Existenzquantoren,
|
|||
|
|
braucht man Grundkenntnisse in Prädikatenlogik 1. Stufe.
|
|||
|
|
Hierfüg gibt es zahlreiche Einführungswerke in die mathematische Logik,
|
|||
|
|
bspw. \cite{ebbinghaus2018}.
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Seien $A,B\subseteq X$ beliebige Teilmengen.
|
|||
|
|
Es reicht aus \textbf{zu zeigen},
|
|||
|
|
dass $y\in f(A\cup B)\Leftrightarrow y\in f(A)\cup f(B)$
|
|||
|
|
für alle $y\in Y$ gilt.\\
|
|||
|
|
Sei also $y\in Y$ beliebig. Es gilt
|
|||
|
|
|
|||
|
|
\begin{longmathe}[mc]{RCL}
|
|||
|
|
y\in f(A\cup B)
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in A\cup B:~}y=f(x)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in X:~}x\in A\cup B\,\text{und}\,y=f(x)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in X:~}
|
|||
|
|
(x\in A\,\text{oder}\,x\in B)
|
|||
|
|
\,\text{und}\,y=f(x)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in X:~}
|
|||
|
|
\big(
|
|||
|
|
(x\in A\,\text{und}\,y=f(x))
|
|||
|
|
\,\text{oder}\,
|
|||
|
|
(x\in B\,\text{und}\,y=f(x))
|
|||
|
|
\big)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in X:~}(x\in A\,\text{und}\,y=f(x))
|
|||
|
|
\,\text{oder}\,
|
|||
|
|
\exists{x\in X:~}(x\in B\,\text{und}\,y=f(x))\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in A:~}y=f(x)
|
|||
|
|
\,\text{oder}\,
|
|||
|
|
\exists{x\in B:~}y=f(x)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&y\in f(A)\,\text{oder}\,y\in f(B)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&y\in f(A)\cup f(B).\\
|
|||
|
|
\end{longmathe}
|
|||
|
|
|
|||
|
|
Darum gilt $f(A\cup B)=f(A)\cup f(B)$ für alle $A,B\subseteq X$.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
%% AUFGABE 3-2c
|
|||
|
|
\item
|
|||
|
|
\begin{claim*}
|
|||
|
|
Die Aussage $\forall{A\subseteq X:~}f(X\ohne A)=Y\ohne f(A)$
|
|||
|
|
ist \fbox{\uline{nicht} allgemein gültig}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Betrachte das Beispiel $X=\{0,1\}$, $Y=\{2\}$, und ${f:X\to Y}$ mit $f(x)=2$ für alle $x\in X$.
|
|||
|
|
Für $A=\{0\}$
|
|||
|
|
gilt $f(X\ohne A)=f(\{1\})=\{2\}$,
|
|||
|
|
während $Y\cap f(A)=\{2\}\ohne\{2\}=\leer$.
|
|||
|
|
Also $f(X\ohne A)\neq Y\cap f(A)$.
|
|||
|
|
Darum ist dies ein Gegenbeispiel zur Aussage.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
\text{Bemerkung.} Die Aussage ist eigentlich genau dann wahr, wenn $f$ bijektiv ist.
|
|||
|
|
Und eine leicht modifizierte Aussage,
|
|||
|
|
$\forall{A\subseteq X:~}f(X\ohne A)=f(X)\cap f(A)$,
|
|||
|
|
ist genau dann wahr, wenn $f$ injektiv ist.
|
|||
|
|
|
|||
|
|
%% AUFGABE 3-2d
|
|||
|
|
\item
|
|||
|
|
\begin{claim*}
|
|||
|
|
Die Aussage $\forall{A,B\subseteq Y:~}f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$
|
|||
|
|
ist \fbox{allgemein gültig}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Seien $A,B\subseteq Y$ beliebige Teilmengen.
|
|||
|
|
Es reicht aus \textbf{zu zeigen},
|
|||
|
|
dass $x\in f^{-1}(A\cap B)\Leftrightarrow x\in f^{-1}(A)\cap f^{-1}(B)$
|
|||
|
|
für alle $x\in X$ gilt.\\
|
|||
|
|
Sei also $x\in X$ beliebig. Es gilt
|
|||
|
|
|
|||
|
|
\begin{longmathe}[mc]{RCL}
|
|||
|
|
x\in f^{-1}(A\cap B)
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&f(x)\in A\cap B\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&f(x)\in A\,\text{und}\,f(x)\in B\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&x\in f^{-1}(A)\,\text{und}\,x\in f^{-1}(B)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&x\in f^{-1}(A)\cap f^{-1}(B).\\
|
|||
|
|
\end{longmathe}
|
|||
|
|
|
|||
|
|
Darum gilt $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$ für alle $A,B\subseteq Y$.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
%% AUFGABE 3-2e
|
|||
|
|
\item
|
|||
|
|
\begin{claim*}
|
|||
|
|
Die Aussage $\forall{A,B\subseteq Y:~}f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$
|
|||
|
|
ist \fbox{allgemein gültig}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Seien $A,B\subseteq Y$ beliebige Teilmengen.
|
|||
|
|
Es reicht aus \textbf{zu zeigen},
|
|||
|
|
dass $x\in f^{-1}(A\cup B)\Leftrightarrow x\in f^{-1}(A)\cup f^{-1}(B)$
|
|||
|
|
für alle $x\in X$ gilt.\\
|
|||
|
|
Sei also $x\in X$ beliebig. Es gilt
|
|||
|
|
|
|||
|
|
\begin{longmathe}[mc]{RCL}
|
|||
|
|
x\in f^{-1}(A\cup B)
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&f(x)\in A\cup B\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&f(x)\in A\,\text{oder}\,f(x)\in B\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&x\in f^{-1}(A)\,\text{oder}\,x\in f^{-1}(B)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&x\in f^{-1}(A)\cup f^{-1}(B).\\
|
|||
|
|
\end{longmathe}
|
|||
|
|
|
|||
|
|
Darum gilt $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ für alle $A,B\subseteq Y$.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
\end{enumerate}
|
|||
|
|
|
|||
|
|
%% AUFGABE 3-3
|
|||
|
|
\clearpage
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{Aufgabe}
|
|||
|
|
\section[Aufgabe 3]{}
|
|||
|
|
\label{ueb:3:ex:3}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries (a)}
|
|||
|
|
%% AUFGABE 3-3a
|
|||
|
|
\item
|
|||
|
|
Seien $n\in\ntrlpos$ und $v\in\reell^{n}$.
|
|||
|
|
Sei ${f:\reell^{n}\to\reell^{n}}$ durch $f(x)=x+v$ definiert.
|
|||
|
|
|
|||
|
|
\begin{claim*}
|
|||
|
|
$f$ ist \fbox{bijektiv}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Sei ${g:\reell^{n}\to\reell^{n}}$ durch $g(x)=x-v$ definiert.
|
|||
|
|
Es ist einfach zu sehen,
|
|||
|
|
dass $f\circ g=\id_{\reell^{n}}$
|
|||
|
|
und $g\circ f=\id_{\reell^{n}}$.
|
|||
|
|
Per Definition ist als $f$ eine Bijektion mit Inversem $g$.
|
|||
|
|
\end{proof}
|
|||
|
|
%% AUFGABE 3-3b
|
|||
|
|
\item
|
|||
|
|
Seien $n\in\ntrlpos$ und $X=\reell^{n}\times(\reell^{n}\ohne\{\zerovector\}$.
|
|||
|
|
Sei $Y$ die Menge aller Geraden im $\reell^{n}$.
|
|||
|
|
Sei ${f:X\to Y}$ durch $f(v,w)=\{v+t\cdot w\mid t\in\reell\}$ definiert.
|
|||
|
|
|
|||
|
|
\begin{claim*}
|
|||
|
|
$f$ ist \fbox{surjektiv} aber \fbox{nicht injektiv}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
\uwave{{\bfseries Surjektivität}}\\
|
|||
|
|
\textbf{Idee:} Folgt aus der Definition von Geraden durch Parameter.\\
|
|||
|
|
Sei $L\subseteq\reell^{n}$ eine beliebige Gerade. \textbf{Zu zeigen:} $L\in f(X)$.\\
|
|||
|
|
Nun, \emph{per Definition} einer Geraden existieren
|
|||
|
|
$u,v\in\reell^{n}$ mit $w\neq\zerovector$
|
|||
|
|
und so dass $L=\{u+t\cdot w\mid t\in\reell\}$.
|
|||
|
|
Offensichtlicht gilt $(v,w)\in X$.
|
|||
|
|
Darum gilt $L=f((v,w))\in f(X)$.
|
|||
|
|
|
|||
|
|
\uwave{{\bfseries Nichtinjektivität}}\\
|
|||
|
|
\textbf{Idee:} Wir wissen, dass verschiedene aber parallele Vektoren dieselbe Gerade definieren.\\
|
|||
|
|
Fixiere beliebiges $v,w\in\reell^{n}$
|
|||
|
|
und wähle ein $c\in\reell\ohne\{0,1\}$.\\
|
|||
|
|
Dann sind $w,cw\neq\zerovector$ verschiedene aber parallele Vektoren.\\
|
|||
|
|
Darum gilt $f((v,w))=\{v+t\cdot w\mid t\in\reell\}=\{v+tc\cdot w\mid t\in\reell\}=f((v,cw))$.\\
|
|||
|
|
Da $(v,w)\neq(v,cw)$, ist $f$ somit nicht injektiv.
|
|||
|
|
\end{proof}
|
|||
|
|
%% AUFGABE 3-3c
|
|||
|
|
\item
|
|||
|
|
Es sei $X$ die Menge aller Bücher in einem fixierten Kontext.
|
|||
|
|
Sei $Y$ die Menge alle Autor(inn)en von Büchern.
|
|||
|
|
Sei ${f:X\to\Pot(Y)}$ definiert durch
|
|||
|
|
$f(x)=\{y\mid \text{$y$ ein(e) Autor(in) vom Buch $x$}\}$
|
|||
|
|
für alle $x\in X$.
|
|||
|
|
|
|||
|
|
\begin{claim*}
|
|||
|
|
$f$ ist \fbox{nicht im Allgemeinen injektiv} und \fbox{niemals surjektiv}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
\uwave{{\bfseries Nichtsurjektivität}}\\
|
|||
|
|
\textbf{Zu zeigen:} Es gibt konstellationen von Autor(inn)en, die kein gemeinsames Buch verfasst haben.\\
|
|||
|
|
Es gibt \emph{immer} eine(n) Autor(in) eines Buchs,
|
|||
|
|
sodass $\leer\notin f(X)$ in allen Kontexten.
|
|||
|
|
Darum ist $f$ niemals surjektiv.
|
|||
|
|
|
|||
|
|
\uwave{{\bfseries Nichtinjektivität}}\\
|
|||
|
|
\textbf{Zu zeigen:} Es gibt zwei verschiedene Bücher,
|
|||
|
|
die von der gleichen Konstellation an Autor(inn)en
|
|||
|
|
verfasst wurden.
|
|||
|
|
In unserem Kontext hat bspw. $a=\text{{\itshape JK~Rowling}}$ alleine die Bücher
|
|||
|
|
${b_{1}:=\text{{\itshape »HP~and~the~Philosopher's~Stone«}}}$
|
|||
|
|
und
|
|||
|
|
${b_{2}:=\text{{\itshape »HP~and~the~Goblet~of~Fire«}}}$
|
|||
|
|
geschrieben.
|
|||
|
|
Darum $b_{1}\neq b_{2}$ und $f(b_{1})=\{a\}=f(b_{2})$.
|
|||
|
|
Also ist $f$ in unserem Kontext nicht injektiv.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
\textbf{Anmerkung.}
|
|||
|
|
Falls wir $\leer$ von der Bildmenge $\Pot(Y)$ exludieren,
|
|||
|
|
dann können wir mindestens dafür argumentieren,
|
|||
|
|
dass $f$ \fbox{nicht im Allgemeinen surjektiv} ist:
|
|||
|
|
In unserem konkreten Kontext haben bspw. {\itshape JK~Rowling} und {\itshape Oscar~Wilde} nie am selben Buch gearbeitet,
|
|||
|
|
also gilt $\{\text{JK Rowling},\,\text{Oscar Wilde}\}\notin f(X)$.
|
|||
|
|
In der Tat ist ein Kontext kaum vorstellbar,
|
|||
|
|
in dem sich \emph{alle} Autor(inn)en an einem gemeinsamen Buch beteiligt haben,
|
|||
|
|
d.\,h. $Y\in f(X)$ sowie alle „große“ Teilmengen sind fast immer ausgeschlossen.
|
|||
|
|
|
|||
|
|
%% AUFGABE 3-3d
|
|||
|
|
\item
|
|||
|
|
Seien $X$ die Menge aller in Deutschland zugelassener Kfz und
|
|||
|
|
$Y$ die Menge aller amtlicher Kennzeichen.
|
|||
|
|
Sei ${f:X\to Y}$ die Abbildung, die jedem Kfz sein Kennzeichen zuordnet.
|
|||
|
|
|
|||
|
|
\begin{claim*}
|
|||
|
|
$f$ ist \fbox{injektiv} aber \fbox{nicht im Allgemeinen surjektiv}.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
\uwave{{\bfseries Injektivität:}}
|
|||
|
|
Jedes Kennzeichen darf per Gesetz nur einem Kfz zugehören.
|
|||
|
|
\uwave{{\bfseries Nichtsurjektivität:}}
|
|||
|
|
Es besteht zwar die Chance, dass irgendwann alle Kennzeichen aufgebraucht werden,
|
|||
|
|
aber in der Praxis ist die Menge $Y$ sehr groß,
|
|||
|
|
dass dies aktuell und für eine lange Zeit nicht vorkommt.
|
|||
|
|
\end{proof}
|
|||
|
|
\end{enumerate}
|
|||
|
|
|
|||
|
|
\setcounternach{part}{2}
|
|||
|
|
\part{Selbstkontrollenaufgaben}
|
|||
|
|
|
|||
|
|
\def\chaptername{SKA Blatt}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: body/ska/ska4.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\setcounternach{chapter}{4}
|
|||
|
|
\chapter[Woche 4]{Woche 4}
|
|||
|
|
\label{ska:4}
|
|||
|
|
|
|||
|
|
%% SKA 4-1
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 1]{}
|
|||
|
|
\label{ska:4:ex:1}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
Seien $X$, $Y$ nicht leere Mengen.
|
|||
|
|
Einer Abbildung, $f:X\to Y$,
|
|||
|
|
können wir eindeutig die Relation
|
|||
|
|
$\graph(f):=\{(x,y)\in X\times Y\mid f(x)=y\}$
|
|||
|
|
zuordnen. Dies nennt sich der \textbf{Graph von $f$}
|
|||
|
|
(siehe \cite[\S{}2.3]{sinn2020}---dort wird dies mit $\Gamma_{f}$ bezeichnet).
|
|||
|
|
Hier ist $\graph(f)$ also eine Relation auf $X\times Y$.
|
|||
|
|
In der Tat \emph{definieren} setzen manche Werke Funktionen mit ihrem Graphen gleich
|
|||
|
|
(siehe bspw. \cite[S.11]{jech1997}),
|
|||
|
|
aber dies ist streng genommen nicht die ganze Wahrheit.
|
|||
|
|
|
|||
|
|
%% SKA 4-2
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 2]{}
|
|||
|
|
\label{ska:4:ex:2}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-3
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 3]{}
|
|||
|
|
\label{ska:4:ex:3}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-4
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 4]{}
|
|||
|
|
\label{ska:4:ex:4}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-5
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 5]{}
|
|||
|
|
\label{ska:4:ex:5}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-6
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 6]{}
|
|||
|
|
\label{ska:4:ex:6}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-7
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 7]{}
|
|||
|
|
\label{ska:4:ex:7}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-8
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 8]{}
|
|||
|
|
\label{ska:4:ex:8}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-9
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 9]{}
|
|||
|
|
\label{ska:4:ex:9}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-10
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 10]{}
|
|||
|
|
\label{ska:4:ex:10}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
%% SKA 4-11
|
|||
|
|
\let\altsectionname\sectionname
|
|||
|
|
\def\sectionname{SKA}
|
|||
|
|
\section[Aufgabe 11]{}
|
|||
|
|
\label{ska:4:ex:11}
|
|||
|
|
\let\sectionname\altsectionname
|
|||
|
|
|
|||
|
|
({\itshape Unter Arbeit})
|
|||
|
|
|
|||
|
|
\setcounternach{part}{3}
|
|||
|
|
\part{Quizzes}
|
|||
|
|
|
|||
|
|
\def\chaptername{Quiz}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: body/quizzes/quiz1.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\setcounternach{chapter}{1}
|
|||
|
|
\chapter[Woche 1]{Woche 1}
|
|||
|
|
\label{quiz:1}
|
|||
|
|
|
|||
|
|
\begin{claim*}
|
|||
|
|
Das LGS
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcrcr}
|
|||
|
|
-x &+ &a\cdot y &= &3\\
|
|||
|
|
a\cdot x &- &4y &= &0\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
ist genau dann lösbar, wenn $a\in\reell\ohne\{\pm 2\}$.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Sei $a\in\reell$ beliebig. Wir führen das Gaußverfahren aus:
|
|||
|
|
|
|||
|
|
\begin{algorithm}[\rtab][\rtab]
|
|||
|
|
Ursprüngliches LGS $(A_{\alpha}|b_{\beta})$:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{cc|c}
|
|||
|
|
-1 &a &3\\
|
|||
|
|
a &-4 &0\\
|
|||
|
|
\end{matrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Wende die Zeilentransformationen
|
|||
|
|
${Z_{2}\leftsquigarrow a\cdot Z_{1}+Z_{2}}$
|
|||
|
|
an:
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{c}
|
|||
|
|
\begin{matrix}{cc|c}
|
|||
|
|
1 &a &3\\
|
|||
|
|
0 &a^{2}-4 &3a\\
|
|||
|
|
\end{matrix}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
\end{algorithm}
|
|||
|
|
|
|||
|
|
Wenn $a\in\{\pm 2\}$, ist das LGS unlösbar, da in der 2. Zeile links nur $0$ Einträge stehen und rechts $\pm 6$.\\
|
|||
|
|
Wenn $a\notin\{\pm 2\}$, gibt es zwei Stufen und damit ist das LGS lösbar.\\
|
|||
|
|
Also gilt die Behauptung.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: body/quizzes/quiz2.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\setcounternach{chapter}{2}
|
|||
|
|
\chapter[Woche 2]{Woche 2}
|
|||
|
|
\label{quiz:2}
|
|||
|
|
|
|||
|
|
Sei $L$ die Gerade $\{\mathbf{v}+t\mathbf{w}\mid t\in\reell\}\subseteq\reell^{3}$,
|
|||
|
|
wobei
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rclqrcl}
|
|||
|
|
\mathbf{v} &= &\begin{svector}-4\\2\\5\\\end{svector},
|
|||
|
|
&\mathbf{w} &= &\begin{svector}2\\-6\\12\\\end{svector}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries (1)}
|
|||
|
|
%% QUIZ 2-a
|
|||
|
|
\item
|
|||
|
|
|
|||
|
|
\begin{claim*}
|
|||
|
|
Der Punkt, $\mathbf{x}=\begin{svector}-3\\-1\\11\\\end{svector}$, liegt in der Geraden, $L$.
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Es gilt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\mathbf{x}\in L
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{t\in\reell:~}
|
|||
|
|
\mathbf{x}=\mathbf{v}+t\mathbf{w}\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{t\in\reell:~}
|
|||
|
|
\mathbf{x}-\mathbf{v}=t\mathbf{w}\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{t\in\reell:~}
|
|||
|
|
\begin{svector}1\\-3\\6\\\end{svector}=t\begin{svector}2\\-6\\12\\\end{svector}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Nun ist die letzte Aussage wahr,
|
|||
|
|
da der Ausdruck innerhalb des Existenzquantors offensichtlich unter $t=\frac{1}{2}$ wahr ist.
|
|||
|
|
Darum gilt $\mathbf{x}\in L$.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
%% QUIZ 2-b
|
|||
|
|
\item
|
|||
|
|
Fixiere einen Vektor, $\mathbf{w}_{\perp}\in\reell^{3}$,
|
|||
|
|
der zu $\mathbf{w}$ normal ist.
|
|||
|
|
Z.\,B. können wir
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
\mathbf{w}_{\perp} &= &\begin{svector}3\\-1\\0\\\end{svector}\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
wählen. Dann gilt $\brkt{\mathbf{w},\mathbf{w}_{\perp}}=0$,
|
|||
|
|
sodass die Vektoren normal zueinander stehen.
|
|||
|
|
|
|||
|
|
Nun, für $\mathbf{x}\in L$ setze
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
L_{\mathbf{x}} &:= &\{\mathbf{x}+s\cdot\mathbf{w}_{\perp}\mid s\in\reell\}.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Dann gilt offensichtlich $\mathbf{x}\in L\cap L_{\mathbf{x}}$.\\
|
|||
|
|
Andererseits, da die Richtungsvektoren in den Geraden nicht linear abhängig sind,
|
|||
|
|
(da sie normal zueinander stehen),
|
|||
|
|
gilt $|L\cap L_{\mathbf{x}}|\leq 1$.\\
|
|||
|
|
Darum gilt $L\cap L_{\mathbf{x}}=\{\mathbf{x}\}$.
|
|||
|
|
\end{enumerate}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: body/quizzes/quiz3.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\setcounternach{chapter}{3}
|
|||
|
|
\chapter[Woche 3]{Woche 3}
|
|||
|
|
\label{quiz:3}
|
|||
|
|
|
|||
|
|
\begin{enumerate}{\bfseries (a)}
|
|||
|
|
%% QUIZ 3-a
|
|||
|
|
\item
|
|||
|
|
|
|||
|
|
\begin{claim*}
|
|||
|
|
Seien $X$, $Y$ beliebige Mengen und $f:X\to Y$ eine Funktion.
|
|||
|
|
Sei $B\subseteq Y$ beliebig.
|
|||
|
|
Dann gilt $f(f^{-1}(B))=f(X)\cap B$.
|
|||
|
|
Insbesondere gilt $f(f^{-1}(B))\subseteq B$
|
|||
|
|
\end{claim*}
|
|||
|
|
|
|||
|
|
\begin{proof}
|
|||
|
|
Für $y\in Y$ gilt
|
|||
|
|
|
|||
|
|
\begin{mathe}[mc]{rcl}
|
|||
|
|
y\in f(f^{-1}(B))
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in f^{-1}(B):~}f(x)=y\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in X:~}(x\in f^{-1}(B)\,\text{und}\,f(x)=y)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in X:~}(f(x)=y\,\text{und}\,x\in f^{-1}(B))\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in X:~}(y=f(x)\,\text{und}\,f(x)\in B)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&\exists{x\in X:~}(y=f(x)\,\text{und}\,y\in B)\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&(\exists{x\in X:~}y=f(x))\,\text{und}\,y\in B\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&y\in f(X)\,\text{und}\,y\in B\\
|
|||
|
|
&\Longleftrightarrow
|
|||
|
|
&y\in f(X)\cap B.\\
|
|||
|
|
\end{mathe}
|
|||
|
|
|
|||
|
|
Darum gilt $f(f^{-1}(B))=f(X)\cap B\subseteq B$.
|
|||
|
|
\end{proof}
|
|||
|
|
|
|||
|
|
%% QUIZ 3-b
|
|||
|
|
\item
|
|||
|
|
|
|||
|
|
Aus (a) folgt:
|
|||
|
|
|
|||
|
|
\begin{kompaktitem}
|
|||
|
|
\item
|
|||
|
|
$f$ \uline{surjektiv} $\Longrightarrow$
|
|||
|
|
$f(f^{-1}(B))=f(X)\cap B=Y\cap B=B$
|
|||
|
|
für alle $B\subseteq Y$;
|
|||
|
|
\item
|
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|
|
$f$ \uline{nicht surjektiv} $\Longrightarrow$
|
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|
|
$f(f^{-1}(Y))=f(X)\cap Y=f(X)\subset Y$ (strikt).
|
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|
|
\end{kompaktitem}
|
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|
|
|
|||
|
|
Darum ist es notwendig und hinreichend, eine nicht-surjektive Funktion als Beispiel zu nehmen.
|
|||
|
|
Hier ein minimales Beispiel $X=\{0\}$ und $Y=\{1,2\}$ und $B=Y$ und $f:X\to Y$ definiert durch $f(0)=1$.
|
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|
|
Dann $f(f^{-1}(B))=f(f^{-1}(Y))=f(X)=\{1\}\subset Y$ (strikt).
|
|||
|
|
\end{enumerate}
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: back/index.tex
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\bibliographystyle{alpha}
|
|||
|
|
\def\bibname{Literaturverzeichnis}
|
|||
|
|
\nocite{*}
|
|||
|
|
|
|||
|
|
\bgroup
|
|||
|
|
\footnotesize
|
|||
|
|
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
%% FILE: ./back/quelle.bib
|
|||
|
|
%% ********************************************************************************
|
|||
|
|
|
|||
|
|
\begin{thebibliography}{EFT18}
|
|||
|
|
|
|||
|
|
\bibitem[EFT18]{ebbinghaus2018}
|
|||
|
|
Heinz-Dieter Ebbinghaus, J\"org Flum, and Wolfgang Thomas.
|
|||
|
|
\newblock {\em {Einf\"uhrung in die mathematische Logik}}.
|
|||
|
|
\newblock 2018.
|
|||
|
|
|
|||
|
|
\bibitem[Jec97]{jech1997}
|
|||
|
|
Thomas Jech.
|
|||
|
|
\newblock {\em {Set Theory}}.
|
|||
|
|
\newblock Springer-Verlag, 1997.
|
|||
|
|
|
|||
|
|
\bibitem[Sin20]{sinn2020}
|
|||
|
|
Rainer Sinn.
|
|||
|
|
\newblock {Lineare Algebra I: Skript zur Veranstaltung Universit\"at Leipzig}.
|
|||
|
|
\newblock Vorlesungsskript, 2020.
|
|||
|
|
|
|||
|
|
\bibitem[Wal16]{waldmann2016}
|
|||
|
|
Stefan Waldmann.
|
|||
|
|
\newblock {\em {Lineare Algebra 1: Die Grundlagen f\"ur Studierende der
|
|||
|
|
Mathematik und Physik}}.
|
|||
|
|
\newblock Springer Berlin Heidelberg, 2016.
|
|||
|
|
|
|||
|
|
\end{thebibliography}
|
|||
|
|
\egroup
|
|||
|
|
\end{document}
|