\documentclass[11pt,a4paper]{article}
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\usepackage{amsmath}
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\date{ge-\LaTeX{} am \today}
\author{Prüfung von R. Jeltsch}
\title{Numerische Mathematik, 1./2. Vordiplom Herbst 2002}

%eigene definitionen
\newcommand{\abs}[1]{\lvert#1\rvert}
\newcommand{\norm}[1]{\lVert#1\rVert}
\newcommand{\serarrow}{\overset{\searrow}{\rightarrow}}

\begin{document}
\markright{Numerische Mathematik, 1./2. Vordiplom Herbst 2002}
\pagestyle{myheadings}

\maketitle

\section{}
\subsection{}
\begin{equation*}\begin{split}
det(A) - \underbrace{\lambda}_{=1} E & =
\begin{vmatrix}
0  & -1 &   2\\
-1 &  2 &  -4\\
2  & -4 & a-1\\
\end{vmatrix}\\
& = (-1)(-4)2 + 2(-1)(-4)-(-1)(-1)(a-1)-(2\,2\,2)\\
& = 9 - a \overset{!}{=} 0 \Rightarrow a = 9\\
\Rightarrow A & =
\begin{pmatrix}
1  & -1 &  2\\
-1 &  3 & -4\\
2  & -4 &  9\\
\end{pmatrix}
\end{split}\end{equation*}

\subsection{}
\begin{equation*}\begin{split}
cond_x(A) & = \norm{A^{-1}}_x \norm{A}_x = \kappa\\
\norm{A}_2 & = \sqrt{\max \abs{ EW von A^TA } }
\end{split}\end{equation*}
Da $A$ symmetrisch ist $\lambda_A^2 = \lambda_{A^TA}$ (\label{A^2}), wir
suchen also $\max\abs{A}$ und $\max\abs{A^{-1}}$
\begin{equation*}\begin{split}
\chi_A & =
(1-\lambda)(3-\lambda)(9-\lambda)+8+8-(1-\lambda)16-(9-\lambda)-2(3-\lambda)2\\
& = -\lambda^3+13\lambda^2-18\lambda+6
\end{split}\end{equation*}
Da ein $EW = 1$ Polynomdivision mit $(\lambda-1)$
\begin{equation*}\begin{split}
\Rightarrow & (\lambda-1)(-\lambda^2+12\lambda-6)\\
\Rightarrow & \lambda_{A_1} = 1\\
& \lambda_{A_2} = \frac{-12+\sqrt{120}}{-2} = 6+\sqrt{30}\\
& \lambda_{A_3} = \frac{-12-\sqrt{120}}{-2} = 6-\sqrt{30}\\
\Rightarrow & \max\abs{EW} = 6+\sqrt{30}
\end{split}\end{equation*}
Da $AA^{-1} = E$ ist auch $diag(A)diag(A^{-1}) = E$ und daher gilt
$\frac{1}{\lambda_A} = \lambda_{A^{-1}}$
\begin{equation*}\begin{split}
\Rightarrow & \lambda_{A^{-1}_1} = 1\\
& \lambda_{A^{-1}_2} = \frac{1}{6+\sqrt{30}}\\
& \lambda_{A^{-1}_3} = \frac{1}{6-\sqrt{30}}\\
\Rightarrow & \max\abs{EW} = \frac{1}{6-\sqrt{30}}
\end{split}\end{equation*}
\begin{equation*}\begin{split}
cond_2(A) & = \left( 6+\sqrt{30}\right) \frac{1}{6-\sqrt{30}} =
\frac{\left(6+\sqrt{30}\right)^2}
{\left(6-\sqrt{30}\right)\left(6+\sqrt{30}\right ) }\\
& = 11 + 2\sqrt{30} \approx 21.95
\end{split}\end{equation*}

\subsection{}
\begin{equation*}\begin{split}
R =
\begin{pmatrix}
r_{11}  & r_{12} & r_{13}\\
0       & r_{22} & r_{23}\\
0       &      0 & r_{33}\\
\end{pmatrix}
=
\begin{pmatrix}
\sqrt{a_{11}} &  \frac{a_{12}}{r_{11}} & \frac{a_{13}}{r_{11}}\\
            0 & \sqrt{a_{22}-r_{12}^2} & \frac{a_{23}-r_{12} r_{13}}{r_{22}}\\
            0 &                      0 & \sqrt{a_{33}-r_{13}^2-r_{23}^2}\\
\end{pmatrix}\\
\begin{pmatrix}
\sqrt{1} &    \frac{-1}{1} & \frac{2}{1}\\
       0 & \sqrt{3-(-1)^2} & \frac{-4-(-1)2}{\sqrt{2}}\\
       0 &               0 & \sqrt{9-2^2-(-\sqrt{2})^2}\\
\end{pmatrix}
=
\begin{pmatrix}
1 &       -1 &  2\\
0 & \sqrt{2} & -\sqrt{2}\\
0 &        0 &  \sqrt{3}\\
\end{pmatrix}
\end{split}\end{equation*}

\subsection{}
\begin{equation*}\begin{split}
cond_2(R) & = \sqrt{\max \abs{ EW von (R^{-1})^T R^{-1} }}
\sqrt{\max \abs{ EW von R^TR } }\\
& = \sqrt{\max \abs{ EW von A^{-1} }} \sqrt{\max \abs{ EW von A } }\\
& \overset{\star}{=} \sqrt{\sqrt{\max \abs{ EW von (A^{-1})^T A^{-1}
}}} \sqrt{\sqrt{\max \abs{ EW von A^TA}}}\\
& = \sqrt{\sqrt{\max \abs{ EW von (A^{-1})^T A^{-1} }}
\sqrt{\max \abs{ EW von A^TA}}}\\
& = \sqrt{cond_2(A)}
\end{split}\end{equation*}
Der Schritt bei $\star$ folgt aus \ref{A^2} da eine Cholesky-Zerlegung
nur möglich ist, wenn die Matrix symmetrisch positiv definit ist.

\section{}
\subsection{}
\subsection{}

\section{}
\subsection{}

Fehler $\epsilon$ von: $T(n)$ wenn $I = \frac{2}{\sqrt{\pi}} \int f(x)\,dx :$
mit $f(x) = e^{-x^2}$
\begin{equation*}\begin{split}
\epsilon & = \abs{I-T(n)} \leq \frac{h^2}{12}(b-a)
\frac{2}{\sqrt{\pi}} \max_{x \in[a,b]}\abs{f''(x)}\\
\mbox{Anzahl Stützstellen} & = n\\
h & = \frac{b-a}{n}\\
\mbox{gesucht: } g(x,\epsilon) & = n
\end{split}\end{equation*}

\begin{displaymath}
f'(x) = -2x e^{-x^2} \Rightarrow f''(x) = (4x^2-2)e^{-x^2}
\end{displaymath}
\begin{displaymath}
\epsilon \leq \frac{\left(\frac{x-0}{n}\right)^2}{12}(x-0)\frac{2}{\sqrt{\pi}}
\max_{x \in[a,b]}\abs{(4x^2-2)e^{-x^2}}
\end{displaymath}
\begin{equation*}\begin{split}
\max_{x \in[a,b]}\abs{(4x^2-2)e^{-x^2}}
& \Rightarrow \left((4x^2-2)e^{-x^2}\right)'
= 8x e^{-x^2}-(4 x^2 -2)2x e^{-x^2}\\
& = (-4 x^2 + 4x + 2)2x e^{-x^2} \overset{!}{=} 0\\
& \Rightarrow (-2 x^2 + 2x + 1)x e^{-x^2} \overset{!}{=} 0\\
& \Rightarrow x = 0 \Rightarrow x = 0\\
& \Rightarrow e^{-x^2} = 0 \Rightarrow x = \infty\\
& \Rightarrow -2 x^2 + 2x + 1 = 0 \Rightarrow x = \frac{-2\pm\sqrt{12}}{-4} =
\frac{1}{2}\pm\frac{\sqrt{12}}{4}\\
\end{split}\end{equation*}
Einsetzen der Resultate in $\max_{x \in[a,b]}\abs{f(x)''}$ ergibt $x = 0$ als
Lösung und $\abs{f(0)''} = 2$.
\begin{displaymath}
\epsilon \leq \frac{\left(\frac{x-0}{n}\right)^2}{12}(x-0)\frac{2}{\sqrt{\pi}}2
 =
\frac{x^3}{3 \sqrt{\pi} n^2}
\Rightarrow n \geq \sqrt{\frac{x^3}{3 \sqrt{\pi} \epsilon}}
\end{displaymath}

\subsection{}
\begin{displaymath}
n \geq \sqrt{\frac{x^3}{3 \sqrt{\pi} \epsilon}} =
\sqrt{\frac{1^3}{3 \sqrt{\pi} 0.02}} \approx 3.07 \Rightarrow n = 4
\end{displaymath}
\begin{displaymath}
I = \frac{2}{\sqrt{\pi}} \int f(x)\,dx \mbox{ mit } f(x) = e^{-x^2}
\end{displaymath}
\begin{alignat*}{3}
f(0)                      &= 1 &&\\
f\left(\frac{1}{4}\right) &= e^{-\frac{1}{16}} &&\approx 0.939\\
f\left(\frac{1}{2}\right) &= e^{-\frac{1}{4}}  &&\approx 0.779\\
f\left(\frac{3}{4}\right) &= e^{-\frac{9}{16}} &&\approx 0.570\\
f(1)                      &= e^{-1}            &&\approx 0.368\\
\end{alignat*}
\begin{equation*}\begin{split}
T_{normal}(n) & =
\frac{b-a}{n}
\left(\frac{1}{2}f(x_0) + \sum_{i=1}^{n-1}f(x_i)+\frac{1}{2}f(x_n)\right)
\mbox{ohne } \frac{2}{\sqrt{\pi}}\\
& \Rightarrow \frac{2}{\sqrt{\pi}} \frac{1}{4}
\left(\frac{1}{2}1 +
e^{-\frac{1}{16}} +
e^{-\frac{1}{4}} +
e^{-\frac{9}{16}} +
\frac{1}{2} e^{-1} \right) \approx 0.838 \pm 0.02
\end{split}\end{equation*}

\section{}
\subsection{}
Periodischer Spline $M_0 = M_n$
\begin{displaymath}
\begin{pmatrix}
        2 & \lambda_1 &      0 & \cdots &      0 &         \mu_1\\
    \mu_2 &         2 & \ddots &        &        &             0\\
        0 &    \ddots & \ddots &        &        &        \vdots\\
   \vdots &           &        & \ddots & \ddots &             0\\
        0 &           &        & \ddots &      2 & \lambda_{n-1}\\
\lambda_n &         0 & \ddots &      0 &  \mu_n &             2\\
\end{pmatrix}
\begin{pmatrix}
M_1\\
\vdots\\
\vdots\\
\vdots\\
\vdots\\
M_n\\
\end{pmatrix}
=
\begin{pmatrix}
d_1\\
\vdots\\
\vdots\\
\vdots\\
\vdots\\
d_n\\
\end{pmatrix}
\end{displaymath}
\begin{displaymath}
\mu_i = \frac{h_i}{h_i + h_{i+1}} = \frac{1}{2} \qquad
\lambda_i = 1 - \mu_i = \frac{1}{2}
\end{displaymath}
\begin{displaymath}
d_i = \frac{6}{h_i}f[x_i,x_{i+1},x_{i+2}])
\Rightarrow \vec d =
\begin{pmatrix}
-6\\
-30\\
6\\
-30
\end{pmatrix}
\end{displaymath}
Schema der dividierten Differenzen
\begin{displaymath}
\begin{array}{rrrrrr}
x_i &         f_i &           & f[x_i,x_{i+1}] & &f[x_i,x_{i+1},x_{i+2}]\\
  0 &          -2 &           &                & &\\
  1 & \frac{5}{2} & \serarrow &    \frac{5}{2} & &\\
  2 &           4 & \serarrow &    \frac{3}{2} & \serarrow & -1\\
  3 & \frac{1}{2} & \serarrow &   -\frac{7}{2} & \serarrow & -5\\
  4 &          -2 & \serarrow &   -\frac{5}{2} & \serarrow &  1\\
  5 & \frac{5}{2} & \serarrow &    \frac{5}{2} & \serarrow & -5\\
  6 &           4 & \serarrow &    \frac{3}{2} & \serarrow & -1\\
\vdots &   \vdots &           &         \vdots &           & \vdots\\
\end{array}
\end{displaymath}
Momentengleichung
\begin{displaymath}
\begin{pmatrix}
          2 & \frac{1}{2} &           0 & \frac{1}{2}\\
\frac{1}{2} &           2 & \frac{1}{2} &           0\\
          0 & \frac{1}{2} &           2 & \frac{1}{2}\\
\frac{1}{2} &           0 & \frac{1}{2} &           2\\
\end{pmatrix}
\begin{pmatrix}
M_1\\
M_2\\
M_3\\
M_4\\
\end{pmatrix}
=
\begin{pmatrix}
-6\\
-30\\
6\\
-30\\
\end{pmatrix}
\end{displaymath}
Gauss-Algorithmus
\begin{equation*}\begin{split}
\begin{pmatrix}
          2 & \frac{1}{2} &           0 & \frac{1}{2}\\
\frac{1}{2} &           2 & \frac{1}{2} &           0\\
          0 & \frac{1}{2} &           2 & \frac{1}{2}\\
\frac{1}{2} &           0 & \frac{1}{2} &           2\\
\end{pmatrix}
\Rightarrow
\begin{matrix}
\begin{array}{c}
\begin{array}{c}
2\\
\end{array}\\
\begin{array}{c|}
\hline
\frac{1}{4}\\
          0\\
\frac{1}{4}\\
\end{array}
\end{array}
&
\begin{array}{ccc}
 \frac{1}{2} &           0 & \frac{1}{2}\\
\frac{15}{8} & \frac{1}{2} & -\frac{1}{8}\\
 \frac{1}{2} &           2 &  \frac{1}{2}\\
-\frac{1}{8} & \frac{1}{2} & \frac{15}{8}\\
\end{array}
\end{matrix}
\Rightarrow\\
\begin{matrix}
\begin{array}{c}
\begin{array}{c}
2\\
\end{array}\\
\begin{array}{c|}
\hline
\frac{1}{4}\\
\end{array}\\
\begin{array}{c}
          0\\
\frac{1}{4}\\
\end{array}
\end{array}
&
\begin{array}{l}
\begin{array}{c}
\frac{1}{2}\\
\frac{15}{8}\\
\end{array}\\
\begin{array}{c|}
\hline
\frac{4}{15}\\
-\frac{1}{15}
\end{array}
\end{array}
&
\begin{array}{cc}
            0 &   \frac{1}{2}\\
  \frac{1}{2} &  -\frac{1}{8}\\
\frac{28}{15} &  \frac{8}{15}\\
 \frac{8}{15} & \frac{28}{15}\\
\end{array}
\end{matrix}
\Rightarrow
\begin{matrix}
\begin{array}{c}
\begin{array}{c}
2\\
\end{array}\\
\begin{array}{c|}
\hline
\frac{1}{4}\\
\end{array}\\
\begin{array}{c}
          0\\
\frac{1}{4}\\
\end{array}
\end{array}
&
\begin{array}{l}
\begin{array}{c}
\frac{1}{2}\\
\frac{15}{8}\\
\end{array}\\
\begin{array}{c|}
\hline
\frac{4}{15}
\end{array}\\
\begin{array}{c}
-\frac{1}{15}
\end{array}
\end{array}
&
\begin{array}{c}
\begin{array}{c}
0\\
\frac{1}{2}\\
\frac{28}{15}
\end{array}\\
\begin{array}{c|}
\hline
\frac{2}{7}\\
\end{array}
\end{array}
&
\begin{array}{c}
 \frac{1}{2}\\
-\frac{1}{8}\\
\frac{8}{15}\\
\frac{12}{7}\\
\end{array}
\end{matrix}
\end{split}\end{equation*}
Vorwärtseinsetzen
\begin{displaymath}
\begin{pmatrix}
          1 &             0 &           0 & 0\\
\frac{1}{4} &             1 &           0 & 0\\
          0 &  \frac{4}{15} &           1 & 0\\
\frac{1}{4} & -\frac{1}{15} & \frac{2}{7} & 1\\
\end{pmatrix}
\vec y =
\begin{pmatrix}
-6\\
-30\\
6\\
-30\\
\end{pmatrix}
\Rightarrow \vec y =
\begin{pmatrix}
-6\\
-\frac{57}{2}\\
\frac{68}{5}\\
-\frac{240}{7}\\
\end{pmatrix}
\end{displaymath}
Rückwärtseinsetzen
\begin{displaymath}
\begin{pmatrix}
2 &  \frac{1}{2} &             0 &  \frac{1}{2}\\
0 & \frac{15}{8} &   \frac{1}{2} & -\frac{1}{8}\\
0 &            0 & \frac{28}{15} & \frac{8}{15}\\
0 &            0 &             0 & \frac{12}{7}\\
\end{pmatrix}
\vec M =
\begin{pmatrix}
-6\\
-\frac{57}{2}\\
\frac{68}{5}\\
-\frac{240}{7}\\
\end{pmatrix}
\Rightarrow \vec M =
\begin{pmatrix}
7\\
-20\\
13\\
-20\\
\end{pmatrix}
\end{displaymath}
\subsection{}
Wendestellen
\begin{multline*}
S(x)\Bigr\rvert_{[x_{i-1},x_i]} =\\
M_{i-1} \frac{(x_i - x)^3}{6 h_i} + M_{i} \frac{(x - x_{i-1})^3}{6 h_i}
+ C_i\left( x - \frac{x_{i-1}+x_i}{2}\right) + D_i\\
\forall x \in [x_{i-1},x_i]
\end{multline*}
\begin{eqnarray*}
S'(x)\Bigr\rvert_{[x_{i-1},x_i]} & = &
M_{i-1} (-3) \frac{(x_i - x)^2}{6 h_i} + M_{i} 3 \frac{(x - x_{i-1})^2}{6 h_i}
+ C_i\\
S''(x)\Bigr\rvert_{[x_{i-1},x_i]} & = &
M_{i-1} 6 \frac{(x_i - x)}{6 h_i} + M_{i} 6 \frac{(x - x_{i-1})}{6 h_i}
\end{eqnarray*}
\begin{eqnarray*}
\Rightarrow  M_{i-1}x_i - M_{i-1}x + M_{i}x - M_{i}x_{i-1} & = & 0\\
x (M_{i} - M_{i-1}) & = & M_{i}x_{i-1} - M_{i-1}x_i\\
x & = & \frac{M_{i}x_{i-1} - M_{i-1}x_i}{M_{i} - M_{i-1}}
\end{eqnarray*}
\begin{eqnarray*}
i = 1 \Rightarrow x  & = & \frac{20}{27}\\
i = 2 \Rightarrow x  & = & \frac{34}{27}\\
i = 3 \Rightarrow x  & = & \frac{86}{33}\\
i = 4 \Rightarrow x  & = & \frac{112}{33}
\end{eqnarray*}
Spline
\begin{displaymath}
\begin{array}{ll}
C_i = \frac{f[x_i]-f[x_{i-1}]}{h_i}-\frac{h_i}{6}(M_i - M_{i-1}) &
D_i = \frac{f[x_i]-f[x_{i-1}]}{2}-\frac{h_i^2}{12}(M_i + M_{i-1})\\
i = 1 \Rightarrow  0 = \frac{\frac{5}{2}-(-2)}{1}-\frac{1}{6}(7-(-20)) &
\Rightarrow \frac{10}{3}\\
i = 2 \Rightarrow  6 & \Rightarrow  \frac{7}{3}\\
i = 3 \Rightarrow -9 & \Rightarrow -\frac{7}{6}\\
i = 4 \Rightarrow  3 & \Rightarrow -\frac{2}{3}\\
\end{array}
\end{displaymath}
\begin{displaymath}
S(x)\Bigr\rvert_{[0,1]} =
-20 \frac{(1 - x)^3}{6} + 7 \frac{(x - 0)^3}{6}
+ 0 ( x - 0) + \frac{10}{3}\\
\end{displaymath}




\section{}
\subsection{}
\subsection{}

\section{}
\subsection{}
\subsection{}
\subsection{}
\subsection{}


\vfill
\begin{flushleft}
Fehler in dieser Musterlösung bitte per e-mail an vordiplome@vmp.ethz.ch
 melden.\\
\textcopyleft{} Thomas Kuster\\
thomas@fam-kuster.ch
\end{flushleft}

\end{document}
