407 lines
10 KiB
TeX
407 lines
10 KiB
TeX
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\documentclass[11pt, a4paper, twoside]{article}
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\usepackage[
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a4paper,
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headsep=5mm,
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footskip=0mm,
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top=12mm,
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left=10mm,
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right=10mm,
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bottom=10mm
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]{geometry}
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\usepackage{amsmath}
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\usepackage{gauss}
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\usepackage{nicematrix}
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\usepackage{tikz}
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\usepackage{amsfonts}
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\usepackage{makecell}
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\usepackage{multicol}
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\usepackage[noend]{algorithm2e}
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\usepackage[utf8]{inputenc}
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\usepackage{fancyhdr}
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\usepackage{tikz}
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\usetikzlibrary{arrows,automata,positioning, graphs, graphdrawing}
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\usegdlibrary {trees}
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\usepackage{hyperref}
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\hypersetup{
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colorlinks=true,
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linkcolor=blue,
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filecolor=magenta,
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urlcolor=cyan,
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pdftitle={Overleaf Example},
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pdfpagemode=FullScreen,
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}
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\setlength{\algomargin}{0pt}
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\begin{document}
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\pagestyle{fancy}
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\fancyhead{}
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\fancyhead[L]{Numerische Mathematik für die Fachrichtungen Informatik}
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\fancyhead[R]{Gero Beckmann - \url{https://github.com/Geronymos/}}
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\fancyfoot{}
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\fancyfoot[R]{\thepage}
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\newenvironment{definition}[1]{\noindent\textbf{#1:}}{}
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\section{Computergenauigkeit}
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\[
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FL = \{ +- B^e \Sigma_{l=1}^{l_m} a_l B^{-l} : e = e_{min} +
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\Sigma_{l=0}^{L_e-1} c_l B^l, a_l, c_l \in \{0, ..., B-1 \}, a \neq 0 \} \cup
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\{ 0 \} \subset \mathbb{Q} \\
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\]
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\begin{multicols}{2}
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\section{Normen und Kondition}
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\begin{align*}
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\|A\|_1 &= \max_{n=0,...,N} \Sigma_{m=0}^{N} |a_{mn}| & \text{Spaltennorm} \\
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\|A\|_2 &= \sqrt {\max \lambda \text{ von } A^T A} & \text{Spektralnorm} \\
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\|A\|_\infty &= \max_{m=0,...,N} \Sigma_{n=0}^{N} |a_{mn}| & \text{Zeilennorm} \\
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\end{align*}
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\subsection{Kondition}
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\begin{align*}
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\kappa(A) &= \|A\|\|A^{-1}\| \\
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\kappa(A) &= \frac{\max_{\|y\|=1} \|A_y\|}{\min_{\|z\|=1} \|Az\|} \\
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\kappa_2(A^TA) &= \kappa_2(A)^2 = \sqrt{\frac{\max \lambda \text{ von } A^TA}{\min \lambda}}
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\end{align*}
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\end{multicols}
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\begin{multicols}{2}
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\section{Cholesky-Zerlegung}
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\begin{enumerate}
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\item Berechne $A=LL^T$
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\item Löse durch Vorwärtssubstitution $Ly = b$
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\item Löse durch rückwärtssubstitution $L^T = y$
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\end{enumerate}
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\begin{align*}
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Ax &= b \\
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A &= \begin{pmatrix}
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l_{11} & & \\
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l_{21} & l_{22} & \\
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l_{31} & l_{32} & l_{33}
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\end{pmatrix} \begin{pmatrix}
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l_{11} & l_{21} & l_{31} \\
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& l_{22} & l_{32} \\
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& & l_{33}
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\end{pmatrix}
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\end{align*}
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\end{multicols}
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\hspace{-.6cm}
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\begin{minipage}{.42\textwidth}
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\section{LR-Zerlegung}
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\begin{enumerate}
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\item Berechne Zerlegung $A = CR$
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\item Löse $Ly = b$ durch Vorwaärtssubstitution
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\item Löse $Rx =y$ durch Rückwärtssubstitution
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\end{enumerate}
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\end{minipage}
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\hspace{-2cm}
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\begin{minipage}{.6\textwidth}
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\hspace{-10cm}
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\begin{align*}
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\begin{gmatrix}[p]
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1 & 4 & -1 \\
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3 & 0 & 5 \\
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2 & 2 & 1
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\rowops
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\add[-3]{0}{1}
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\add[-2]{0}{2}
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\end{gmatrix} \leadsto \begin{pNiceMatrix}
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1 & 4 & -1 \\
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3 & -12 & 8 \\
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2 & -6 & 3
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\CodeAfter
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\tikz \draw (2-|1) -| (4-|2);
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\end{pNiceMatrix} \begin{gmatrix}
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\\ \\
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\rowops
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\add[\frac{1}{-2}]{1}{2}
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\end{gmatrix} \leadsto \begin{pNiceMatrix}
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1 & 4 & -1 \\
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3 & -12 & 8 \\
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2 & \frac 1 2 & -1
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\CodeAfter
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\tikz \draw (2-|1) -| (3-|2) -| (4-|3);
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\end{pNiceMatrix} \\
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\Rightarrow L = \begin{pmatrix}
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1 & 0& 0 \\
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3 & 1 & 0 \\
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2 & \frac 1 2 & 1
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\end{pmatrix}, R = \begin{pmatrix}
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1 & 4 & -1 \\
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0 & -12 & 8 \\
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0 & 0 & -1
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\end{pmatrix}
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\end{align*}
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\end{minipage}
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\subsection{Mit Pivotwahl / Permutationsmatrix $PA = LR$}
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\begin{enumerate}
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\item Berechne Zerlegung $PA = LR$ durch Gauß-Elimitation
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\item Löse $Ly = Pb$ durch Vorwärtssubstition
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\item Löse $Rx = y$ durch Rückwärtssubstitution
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\end{enumerate}
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\def\rowswapfromlabel#1{#1}
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\def\rowswaptolabel#1{#1}
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\def\colswapfromlabel#1{#1}
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\def\colswaptolabel#1{#1}
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\begin{align*}
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\begin{pmatrix}
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1 \\ 2 \\ 3
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\end{pmatrix}
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\begin{gmatrix}[p]
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1 & 2 & 2 \\
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-2 & -2 & 4 \\
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2 & 4 & 2
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\rowops
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\swap[|-2| > |1|][]01
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\end{gmatrix} \leadsto
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\begin{pmatrix}
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2 \\ 1 \\ 3
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\end{pmatrix}
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\begin{gmatrix}[p]
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-2 & -2 & 4 \\
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1 & 2 & 2 \\
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2 & 4 & 2
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\rowops
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\add[\frac 1 2 ]01
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\add[1]02
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\end{gmatrix} \leadsto
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\begin{pmatrix}
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2 \\ 1 \\ 3
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\end{pmatrix}
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\begin{pNiceMatrix}
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-2 & -2 & 4 \\
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-\frac 1 2 & 1 & 4 \\
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-1 & 2 & 6
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\CodeAfter
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\tikz \draw (2-|1) -| (4-|2);
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\end{pNiceMatrix}
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\begin{gmatrix}
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\\ \\
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\rowops
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\swap[|2| > |1|]12
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\end{gmatrix} \\ \leadsto
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\begin{pmatrix}
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2 \\ 3 \\ 1
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\end{pmatrix}
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\begin{pNiceMatrix}
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-2 & -2 & 4 \\
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-1 & 2 & 6 \\
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-\frac 1 2 & 1 & 4
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\CodeAfter
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\tikz \draw (2-|1) -| (4-|2);
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\end{pNiceMatrix}
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\begin{gmatrix}
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\\ \\
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\rowops
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\add[-\frac 1 2]12
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\end{gmatrix} \leadsto
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\begin{pmatrix}
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2 \\ 3 \\ 1
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\end{pmatrix}
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\begin{pNiceMatrix}
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-2 & -2 & 4 \\
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-1 & 2 & 6 \\
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-\frac 1 2 & \frac 1 2 & 1
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\CodeAfter
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\tikz \draw (2-|1) -| (3-|2) -| (4-|3);
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\end{pNiceMatrix} \Rightarrow
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L = \begin{pmatrix}
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1 & 0 & 0 \\
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-1 & 1 & 0 \\
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-\frac12 & \frac12 &1
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\end{pmatrix},
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R = \begin{pmatrix}
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-2 & -2 & 4 \\
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0 & 2 & 6 \\
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0 & 0 & 1
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\end{pmatrix},
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P = \begin{pmatrix}
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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1 & 0 & 0
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\end{pmatrix}
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\end{align*}
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Für Eliminierung in Spalte n werden Zeilen so getauscht, dass in der n-ten
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Spaten ab dre n-ten Zeile, sodass das Betraglich größte Element in Zeile n
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steht.
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\newpage
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\begin{multicols}{2}
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\section{QR-Zerlegung $A = QR$}
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\begin{enumerate}
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\item Bestimme Matrizen Q und R durch Householder-Transformationen
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\item Löse $Qx = b$ ($Q^{-1} = Q^T$, also $c = Q^Tb$)
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\item Löse $Rx = c$ durch Rückwärtssubstitution
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\end{enumerate}
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\begin{enumerate}
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\item Bestimme Teilmatrix $A'^{(j-1)}$
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\item Berechne $v^{(j)} = {a'}_{I}^{(j-1)} + sign({a'}_{II}^{(j-1)}) \cdot \| {a'}_I^{(j-1)} \| e_I$
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\item Berechne $H'^{(j-1)} = I - \frac {2v^{(j)}v^{(j)T}} {v^{(j)T}v^{(j)}}$
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\item Bestime $H^{(j)} = \begin{pmatrix} 1 & 0 \\ 0 & H'^{(j-1)}\end{pmatrix}$
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\item Berechne $A^{(j)} = H^{(j)}A^{(j-1)}$ bis $A^{(j)} = R$
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\end{enumerate}
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\begin{align*}
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j = 1 \rightarrow j = k = min(m-1, n) \\
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Q^T = H^{(k)} \cdot ... \cdot H^{(2)} H^{(1)}
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\end{align*}
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\subsection{Minimale Fehlerquote}
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\[
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|y_i - f(x_i)|_2^2 = \Sigma_{i=1}^{N} (y_i - f(x_i))^2
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\]
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\subsection{Ausgleichssystem}
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Der Vektor $x \in \mathbb{R}^N$ löst genau dann $\|Ax -b \|_2 = min!$, falls er
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$A^TAx = A^Tb$ (Normalgleichung) löst.
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\columnbreak
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\section{Singilärwertzerlegung}
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\begin{enumerate}
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\item Rechne $S = A^TA$
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\item Berechne EW und EV von S
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\item Bilde ONB $u_1, u_2, ..., u_N$ aus EV von S
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\item Berechne $\sigma_k = \sqrt{\lambda_k}$
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\item $U = \begin{pNiceArray}{c|c|c} U_1 & ... & U_k \end{pNiceArray} =
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diag(\sqrt{\lambda_1}, ..., \sqrt{\lambda_k}) =
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diag(\sigma_1, ..., \sigma_k) = \Sigma$
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\item $V = A U \Sigma^{-1}$
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\end{enumerate}
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\subsection{Pseudoinverse }
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$A^+ = U \Sigma^{-1} V^T$ ; ist A regulär dann gilt $A^{-1} = A^+$
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\subsection{Normalengleichung}
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$|Ax-b|_2=Min!$ durch $x = A^+b$ gelöst
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\end{multicols}
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\section{Hessenbergform (rechte-obere Dreiecksmatrix ab der unterren Nebendiagonale)}
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\subsection{Tridiagonal (Nur Haupt- und Nebendiagonale)}
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\begin{align*}
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\text{TeilmatrixA }&{A'}^{(j-1)} \\
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w^{(j)} &= {a'}_{I}^{(j-1)} + sign({a'}_{Ii}^{(j-1)}) \cdot \|{a'}_{I}^{(j-1)}\|_2 \cdot e_I \\
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{Q'}^{(j-1)} &= I - \frac {2 w^{j} w^{(j)T}} {w^{(j)T} w^{(j)}} \\
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Q^{(j)} &= \begin{pmatrix} 1 & 0 \\ 0 && {Q'}^{(j-1)} \end{pmatrix} \\
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H^{(j)} &= Q^{T(j)} A^{(j-1)} Q^{(j)}
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\end{align*}
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\subsection{Jacobi-Verfahren (Lösung von Ax =b) / Gesamtschrittverfahren}
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\begin{align*}
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x_m^{k+1} &= \frac 1 {A[m;m]} (b_m - \Sigma_{n \neq m} A[m,n] x_n^k) &\text{für $m=1, ..., M$} \\
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x^{k+1} &= x^k + D^{-1} (b - Ax^k) & A = D + (L + U) \\
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& &\text{(diagonal + (strikte linke untere / rechte obere))}
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\end{align*}
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\subsection{Gauß-Seidel-verfahren / Einzelschrittverfahren}
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\begin{align*}
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x_m^{k+1} &= \frac 1 {A[m;m]} (b_m - \Sigma_{n=1}^{m-1} A[m,n] x_n^{k+1} - \Sigma_{k=m+1}^{N} A[m,n] x_n^k) \\
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x^{k+1} &= x^k + (D + L)^{-1} (b - Ax^k)
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\end{align*}
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\subsection{CG-Verfahren}
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\begin{align*}
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a
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\end{align*}
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\subsection{GMRES}
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\begin{align*}
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a
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\end{align*}
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Energienorm $\|x\|_A = \sqrt{x^TAx}$
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SKP $<x,y> = x^TAy$
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\subsection{Krylov-Raum}
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\section{Spline Interpolation}
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\begin{align*}
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& s'(a) = v_0 \text{ und } s'(b) = v_N & \text{hermitisch} \\
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& s''(a) = s''(b) = 0 & \text{natürlich} \\
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& s'(a) = s'(b) \text{ und } s''(a) = s''(b) & \text{periodisch}
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\end{align*}
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\section{Newton-Verfahren}
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\[
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x^{n+1} = x^n - \frac {f(x^n)} {f'(x^n)}
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\]
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\section{Quadraturformel}
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Gewichte $b_k \in [0,1]$, Knoten $c_k \in [0,1]$, Stützstelle $a + c_k (b-a)$
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\[
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\int_a^b f(x)dx \approx (b - a) \Sigma_{k=1}^s b_k f(a+c_k (b-a))
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\]
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\begin{tabular}{llll}
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Rechteckregel & $s=1$ & $b_1=1$ & $c_1=0$ \\
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Mittelpunktregel & $s=1$ & $b_1=1$ & $c_1 = \frac12$ \\
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Trapezregel & $s=2$ & $b_1 = b_2 = \frac12$ & $c_1 = 0, c_2 = 1$ \\
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Simpsonregel & $s=3$ & $b_1 = b_3 = \frac16, b_2 = \frac46$ & $c_1 = 0, c_2 = \frac12, c_3 = 1$
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\end{tabular}
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Symmetrische Quadraturformel $c_k = 1 - c_{s+1-k}$, $b_k = b_{s+1-k}$
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Ordung $p$ $\frac1q = \Sigma_{k=1}^S b_k c_k^{q-1}$ für alle $q=1, .., p$ nicht für $q = p+1$!
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\section{Polynom-Interpolation}
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\subsection{Lagrange}
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\begin{align*}
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& p(x) = \Sigma_{n=0}^N f_n L_n(x) &
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L_n(x) = \Pi_{j=0, j \neq n}^N \frac{x - x_j}{x_n - x_j}
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\end{align*}
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|
Lebesque-Konstante
|
||
|
|
\[
|
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|
|
\Lambda_N := \max_{x \in [a,b]} \Sigma_{n=0}^{N} |L_n(x)|
|
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|
\]
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|
\subsection{Newton-Darstellung}
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|
\begin{tabular}{c|c|c|c|c}
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||
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|
$f_n$ & 1 & 6 & -3 & 3 \\
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||
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|
\hline
|
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|
|
$x_n$ & -1 & 0 & 1 & 3
|
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|
|
\end{tabular}
|
||
|
|
|
||
|
|
\[
|
||
|
|
\begin{NiceArray}{c|cccc}
|
||
|
|
x_0 = -1 & f_0 = 1 & & & \\
|
||
|
|
x_1 = 0 & f_1 = 6 & \frac{1-6}{-1-0} = 5 & & \\
|
||
|
|
x_2 = 1 & f_2 = -3 & \frac{6+3}{0-1} = -9 & \frac{5+9}{-1-1} = -7 & \\
|
||
|
|
x_3 = 3 & f_3 = 3 & \frac{-3-3}{1-3} = 3 & \frac{-9-3}{0-3} = 4 & \frac{-7-4}{-1-3} = \frac{11}{4}
|
||
|
|
\end{NiceArray}
|
||
|
|
\]
|
||
|
|
|
||
|
|
\begin{align*}
|
||
|
|
p(x) &= 1 + 5(x-(-1)) -7(x-(-1))(x-0) + \frac{11}4 (x-(-1))(x-0)(x-1) \\
|
||
|
|
p(x) &= f_{0,0} + f_{0,1}(x-x_0) + ... + f_{0,N}(x-x_0) \cdot ... \cdot (x-x_{N-1})
|
||
|
|
\end{align*}
|
||
|
|
|
||
|
|
\end{document}
|