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Musterlösung angefangen
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195
documents/Numerik/Klausur1/Aufgabe1.tex
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195
documents/Numerik/Klausur1/Aufgabe1.tex
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\section*{Aufgabe 1}
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\subsection*{Teilaufgabe a}
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\textbf{Gegeben:}
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\[A =
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\begin{pmatrix}
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3 & 15 & 13 \\
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6 & 6 & 6 \\
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2 & 8 & 19
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\end{pmatrix}\]
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\textbf{Aufgabe:} LR-Zerlegung von $A$ mit Spaltenpivotwahl
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\textbf{Lösung:}
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\[P =
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\begin{pmatrix}
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0 & 1 & 0 \\
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1 & 0 & 0 \\
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0 & 0 & 1
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\end{pmatrix}\]
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durch scharfes hinsehen.
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Nun $L, R$ berechnen:
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\begin{align}
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&\begin{gmatrix}[p]
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6 & 6 & 6 \\
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3 & 15 & 13 \\
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2 & 8 & 19
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\rowops
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\add[\cdot (-\frac{1}{2})]{0}{1}
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\add[\cdot (-\frac{1}{3})]{0}{2}
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\end{gmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{3} & 0 & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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6 & 6 & 6 \\
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0 & 12 & 10 \\
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0 & 6 & 17
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\rowops
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\add[\cdot (-\frac{1}{2})]{1}{2}
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\end{gmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & -\frac{1}{2} & 1
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\end{pmatrix} \cdot
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\begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{3} & 0 & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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6 & 6 & 6 \\
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0 & 12 & 10 \\
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0 & 0 & 12
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\colops
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\add[\cdot (-1)]{0}{1}
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\add[\cdot (-1)]{0}{2}
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\end{gmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{12} & - \frac{1}{2} & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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6 & 0 & 0 \\
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0 & 12 & 10 \\
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0 & 0 & 12
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\colops
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\add[\cdot (-\frac{10}{12})]{1}{2}
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\end{gmatrix}
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\cdot
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\begin{pmatrix}
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1 & -1 & -1 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{pmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{12} & - \frac{1}{2} & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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6 & 0 & 0 \\
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0 & 12 & 0 \\
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0 & 0 & 12
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\colops
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\mult{0}{\cdot \frac{1}{6}}
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\mult{1}{\cdot \frac{1}{12}}
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\mult{2}{\cdot \frac{1}{12}}
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\end{gmatrix}
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\cdot
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\begin{pmatrix}
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1 & -1 & -1 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{pmatrix}
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\cdot
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\begin{pmatrix}
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1 & 0 & 0 \\
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0 & 1 & -\frac{10}{12} \\
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0 & 0 & 1
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\end{pmatrix}
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\\
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= \begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{12} & - \frac{1}{2} & 1
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\end{pmatrix} \cdot
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&\begin{gmatrix}[p]
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{gmatrix}
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\cdot
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\begin{pmatrix}
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1 & -1 & -\frac{1}{6} \\
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0 & 1 & -\frac{5}{6} \\
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0 & 0 & 1
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\end{pmatrix}
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\cdot
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\begin{pmatrix}
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\frac{1}{6} & 0 & 0 \\
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0 & \frac{1}{12} & 0 \\
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0 & 0 & \frac{1}{12}
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\end{pmatrix}
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\\
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= \underbrace{\begin{pmatrix}
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1 & 0 & 0 \\
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-\frac{1}{2} & 1 & 0 \\
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-\frac{1}{12} & - \frac{1}{2} & 1
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\end{pmatrix}}_L \cdot
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&\begin{gmatrix}[p]
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1
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\end{gmatrix}
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\cdot \underbrace{\frac{1}{72}
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\begin{pmatrix}
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12 & -6 & -1 \\
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0 & 6 & -5 \\
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0 & 0 & 6
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\end{pmatrix}}_R
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\end{align}
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ACHTUNG: Ich habe mich irgendwo verrechnet!
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Siehe \href{http://www.wolframalpha.com/input/?i=%7B%7B1%2C0%2C0%7D%2C%7B-1%2F2%2C1%2C0%7D%2C%7B-1%2F12%2C-1%2F2%2C1%7D%7D*%7B%7B12%2C-6%2C-1%7D%2C%7B0%2C6%2C-5%7D%2C%7B0%2C0%2C6%7D%7D}{WolframAlpha}
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\subsection*{Teilaufgabe b}
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\textbf{Gegeben:}
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\[A =
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\begin{pmatrix}
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9 & 4 & 12 \\
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4 & 1 & 4 \\
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12 & 4 & 17
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\end{pmatrix}\]
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\textbf{Aufgabe:} $A$ auf positive Definitheit untersuchen, ohne Eigenwerte zu berechnen.
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\textbf{Lösung:}
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Eine Matrix $A \in \mathbb{R}^{n \times n}$ heißt positiv Definit $\dots$
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\begin{align*}
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\dots & \Leftrightarrow \forall x \in \mathbb{R}^n: x^T A x > 0\\
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& \Leftrightarrow \text{Alle Eigenwerte sind größer als 0}
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\end{align*}
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Falls $A$ symmetrisch ist, gilt:
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\begin{align*}
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\text{$A$ ist pos. Definit} & \Leftrightarrow \text{alle führenden Hauptminore von $A$ sind positiv}\\
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& \Leftrightarrow \text{es gibt eine Cholesky-Zerlegung $A=GG^T$ mit $G$ ist reguläre untere Dreiecksmatrix}\\
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\end{align*}
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Mit dem Hauptminor-Kriterium gilt:
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\begin{align}
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\det(A_1) &= 9 > 0\\
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\det(A_2) &=
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\begin{vmatrix}
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9 & 4 \\
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4 & 1 \\
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\end{vmatrix} = 9 - 16 < 0\\
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&\Rightarrow \text{$A$ ist nicht positiv definit}
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\end{align}
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49
documents/Numerik/Klausur1/Aufgabe2.tex
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49
documents/Numerik/Klausur1/Aufgabe2.tex
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\section*{Aufgabe 2}
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\subsection*{Teilaufgabe a}
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\textbf{Aufgabe}
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Formulieren Sie einen Algorithmus in Pseudocode zum Lösen des Gleichungssystems
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\[Ly = b,\]
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wobei $L$ eine invertierbare, untere Dreiecksmatrix ist.
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Geben Sie die Formel zur Berechnung von $y_i$ an.
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\textbf{Lösung:} TODO! %TODO!
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\begin{algorithm}[H]
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\begin{algorithmic}
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\Require $p \in \mathbb{P}, a \in \mathbb{Z}, p \geq 3$
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\Procedure{CalculateLegendre}{$a$, $p$}
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\If{$a \geq p$ or $a < 0$}\Comment{rule (III)}
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\State \Return $\Call{CalculateLegendre}{a \mod p, p}$ \Comment{now: $a \in [0, \dots, p-1]$}
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\ElsIf{$a == 0$ or $a == 1$}
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\State \Return $a$ \Comment{now: $a \in [2, \dots, p-1]$}
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\ElsIf{$a == 2$} \Comment{rule (VII)}
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\If{$p \equiv \pm 1 \mod 8$}
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\State \Return 1
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\Else
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\State \Return -1
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\EndIf \Comment{now: $a \in [3, \dots, p-1]$}
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\ElsIf{$a == p-1$} \Comment{rule (VI)}
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\If{$p \equiv 1 \mod 4$}
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\State \Return 1
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\Else
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\State \Return -1
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\EndIf \Comment{now: $a \in [3, \dots, p-2]$}
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\ElsIf{!$\Call{isPrime}{a}$} \Comment{rule (II)}
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\State $p_1, p_2, \dots, p_n \gets \Call{Factorize}{a}$
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\State \Return $\prod_{i=1}^n \Call{CalculateLegendre}{p_i, p}$
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\Else \Comment{now: $a \in \mathbb{P}, \sqrt{p-2} \geq a \geq 3$}
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\If{$\frac{p-1}{2} \equiv 0 \mod 2$ or $\frac{a-1}{2} \equiv 0 \mod 2$}
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\State \Return $\Call{CalculateLegendre}{p, a}$
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\Else
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\State \Return $(-1) \cdot \Call{CalculateLegendre}{p, a}$
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\EndIf
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\EndIf
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\EndProcedure
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\end{algorithmic}
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\caption{Calculate Legendre symbol}
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\label{alg:calculateLegendreSymbol}
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\end{algorithm}
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\subsection*{Teilaufgabe b}
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\subsection*{Teilaufgabe c}
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1
documents/Numerik/Klausur1/Aufgabe3.tex
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1
documents/Numerik/Klausur1/Aufgabe3.tex
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\section*{Aufgabe 3}
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1
documents/Numerik/Klausur1/Aufgabe4.tex
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1
documents/Numerik/Klausur1/Aufgabe4.tex
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\section*{Aufgabe 4}
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1
documents/Numerik/Klausur1/Aufgabe5.tex
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1
documents/Numerik/Klausur1/Aufgabe5.tex
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\section*{Aufgabe 5}
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BIN
documents/Numerik/Klausur1/Klausur1.pdf
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BIN
documents/Numerik/Klausur1/Klausur1.pdf
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36
documents/Numerik/Klausur1/Klausur1.tex
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36
documents/Numerik/Klausur1/Klausur1.tex
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\documentclass[a4paper]{scrartcl}
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\usepackage{amssymb, amsmath} % needed for math
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\usepackage[utf8]{inputenc} % this is needed for umlauts
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\usepackage[ngerman]{babel} % this is needed for umlauts
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\usepackage[T1]{fontenc} % this is needed for correct output of umlauts in pdf
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\usepackage{pdfpages} % Signatureinbingung und includepdf
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\usepackage{geometry} % [margin=2.5cm]layout
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\usepackage{hyperref} % links im text
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\usepackage{color}
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\usepackage{framed}
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\usepackage{enumerate} % for advanced numbering of lists
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\usepackage{marvosym} % checkedbox
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\usepackage{wasysym}
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\usepackage{braket} % for \Set{}
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\usepackage{pifont}% http://ctan.org/pkg/pifont
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\usepackage{gauss}
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\usepackage{algorithm,algpseudocode}
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\usepackage{parskip}
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\allowdisplaybreaks
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\newcommand{\cmark}{\ding{51}}%
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\newcommand{\xmark}{\ding{55}}%
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\hypersetup{
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pdfauthor = {Martin Thoma},
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pdfkeywords = {Numerik,KIT},
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pdftitle = {Numerik Klausur1 - Musterlösung}
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}
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\begin{document}
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\include{Aufgabe1}
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\include{Aufgabe2}
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\include{Aufgabe3}
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\include{Aufgabe4}
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\include{Aufgabe5}
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\end{document}
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7
documents/Numerik/Klausur1/Makefile
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7
documents/Numerik/Klausur1/Makefile
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SOURCE = Klausur1
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make:
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pdflatex -shell-escape $(SOURCE).tex -output-format=pdf
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make clean
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clean:
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rm -rf $(TARGET) *.class *.html *.log *.aux *.out *.pyg
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1
documents/Numerik/Klausur1/Readme.md
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1
documents/Numerik/Klausur1/Readme.md
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Dies ist eine inoffizielle Lösung für [Klausur1.pdf von Dr. Daniel Weiß](http://www.math.kit.edu/ianm3/lehre/numainfing2013s/seite/uebnuminfing).
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