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\begin{center}
  {\Large Übungsblatt 4} \\
  Jan Hendrik Dithmar \\
  Übungsgruppe 1 - Mo, 14-16 Uhr \\
  Janko Böhm
\end{center}

\newpage

\section* {Aufgabe 1}
$$ A=\left(
\begin{array}{rrr}
3 & -1 & -1 \\
-1 & 4 & -2 \\
-1 & -2 & 4
\end{array} \right) $$
\subsubsection*{(a)}
$\displaystyle K_j = \{\mu \in \SetC \; | \; |\mu - a_{jj}| \le \sum_{\underset{k\ne j}{k=1}}^3 |a_{kj}| \}\, \; \; j=1,2,3,...$
\begin{eqnarray*}
K_1 & = & \{\mu \; \; | \; \; |\mu - 3| \le 2\} \Rightarrow 0 \not\in K_1 \\
K_2 & = & \{\mu \; \; | \; \; |\mu - 4| \le 3\} \Rightarrow 0 \not\in K_2 \\
K_3 & = & \{\mu \; \; | \; \; |\mu - 4| \le 3\} \Rightarrow 0 \not\in K_3
\end{eqnarray*}
$\displaystyle \Rightarrow 0 \not\in \bigcup_{j=1}^3 K_j \Rightarrow A$ ist invertierbar.

\subsubsection*{(b)}
Spektralnorm: $\displaystyle \| A \|_2 = \sqrt{\lambda_{max}(^tA\cdot A)}$ \\
Da $A$ symmetrisch ist, ist $\|A\|_2 = \lambda_{max}(A)$. \\
Hierbei ist $\displaystyle \lambda_{max}(A) = \underset{j}{max} \{|\lambda_i| \; | \; \lambda_j \text{ ist Eigenwert von }A\}$ \\
Berechne die Eigenwerte von A:
\begin{eqnarray*}
det(A-t\cdot E_3) & = & 0 \\
det\left(
\begin{array}{ccc}
3-t & -1 & -1 \\
-1 & 4-t & -2 \\
-1 & -2 & 4-t
\end{array} \right) & = & (3-t)\cdot(4-t)^2-2-2-(4-t)-4\cdot(3-t)-(4-t) \\
& = & (3-t)\cdot(16-8t+t^2)-4-4+t-12+4t-4+t \\
& = & 48-24t+3t^2-16t+8t^2-t^3-24+6t \\
& = & -t^3 + 11t^2-34t+24 \\
& = & (-1)\cdot(t^3-11t^2+34t-24) \\ \\ \\
t^3-11t^2+34t-24 & = & 0 \\
t_1 & = & 1 \text{ ist Lösung} \\ \\
t^3-11t^2+34t-24 & = & (t^2-10t+24)\cdot(t-1)
\end{eqnarray*}
Die Eigenwerte von $A$ sind demnach:
$$ \lambda_1 = 1, \; \; \lambda_2=6, \; \; \lambda_3=4 $$
$$ \Rightarrow \|A|_2 = \lambda_{max}(A) = max\{1;6;4\} = 6$$
Also: $\displaystyle \|A\|_2 = 6 \le 7$ (w) \hfill $\Box$

\section*{Aufgabe 2}
\subsubsection*{(b)}
\begin{eqnarray*}
f_1(t) & = & f(t) \\ \\
f(t) & = & (cos(t),sin(t),t) \\
f'(t) & = & (-sin(t),cos(t),1) \\
f''(t) & = & (-cos(t),-sin(t),0) \\ \\
\|f'(t)\|^2 & = & sin^2(t)+cos^2(t)+1=2 \\ \\
\sum_{j=1}^3 f_j'(t) \cdot f_j''(t) & = & f_1'(t)\cdot f_1''(t) + f_2'(t)\cdot f_2''(t) f_3'(t)\cdot f_3''(t) \\
& = & -sin(t)\cdot(-cos(t))+cos(t)\cdot(-sin(t)) +1 \cdot 0 \\
& = & 0 \\ \\
\stackrel{A2(a)}{\Rightarrow} \kappa(t) & = & \frac{1}{2} \cdot \left\|(-cos(t),-sin(t),0) - \underbrace{\frac{1}{2}\cdot(-sin(t),cos(t),1)\cdot0}_{=(0,0,0)}\right\| \\
\kappa(t) & = & \frac{1}{2}\cdot\underbrace{\sqrt{cos^2(t)+sin^2(t)+0^2}}_{=1} = \frac{1}{2}
\end{eqnarray*}

\begin{eqnarray*}
f_3(t) & = & h(t) \\ \\
h(t) & = & (t,t^2,t^3) \\
h'(t) & = & (1,2t,3t^2) \\
h''(t) & = & (0,2,6t) \\ \\
\|h'(t)\|^2 & = & 1+4t^2+9t^4 \\ \\
\sum_{j=1}^3 h_j'(t)\cdot h_j''(t) & = & h_1'(t)\cdot h_1''(t) + h_2'(t)\cdot h_2''(t) + h_3'(t)\cdot h_3''(t) \\
& = & 1 \cdot 0 + 2t\cdot2 + 3t^2\cdot 6t \\
& = & 4t + 18 t^3 \\ \\
\stackrel{A2(a)}{\Rightarrow} \kappa(t) & = & \frac{1}{1+4t^2+9t^4} \cdot \left\|(0,2,6t) - \frac{4t+18t^3}{1+4t^2+9t^4} \cdot (1,2t,3t^2)\right\| \\
\kappa(t) & = & \frac{1}{1+4t^2+9t^4} \cdot \left\|\frac{(0,2+8t^2+18t^4,6t+24t^3+54t^4)-(4t+18t^3,8t^2+36t^4)}{1+4t^2+9t^4}\right\| \\
\kappa(t) & = & \frac{1}{(1+4t^2+9t^4)^2} \cdot \left\|(-4t-18t^3, 2-18t^4, 6t+12t^3)\right\| \\
\kappa(t) & = & \frac{1}{(1+4t^2+9t^4)^2} \cdot \sqrt{(-4t-18t^3)^2 + (2-18t-4)^2 + (6t+12t^3)^2} \\
\kappa(t) & = & \frac{1}{(1+4t^2+9t^4)^2} \cdot \sqrt{16t^2+144t^4+324t^6+4-72t^4+324t^8+36t^2+144t^4+144t^6} \\
\kappa(t) & = & \frac{1}{(1+4t^2+9t^4)^2} \cdot \sqrt{52t^2+216t^4+468t^6+324t^8+4} \\ \\
\text{Zwischenrechnung:} & & \\
\text{Radikand} & = & 4\cdot(1+13t^2+54t^4+117t^6+81t^8) \\
& = & 4\cdot(1+8t^2+34t^4+72t^6+81t^8) + 4\cdot(5t^2+20t^4+45t^6) \\
& = & 4\cdot(1+4t^2+9t^4)^2 + 4\cdot5t^2\cdot(1+4t^2+9t^4) \\
& = & 4\cdot(1+4t^2+9t^4)\cdot[1+4t^2+9t^4+5t^2] \\
& = & 4\cdot(1+4t^2+9t^4)\cdot(1+9t^2+9t^4) \\ \\
\Rightarrow \kappa(t) & = & \frac{2\cdot\sqrt{1+4t^2+9t^4}}{(1+4t^2+9t^4)^2} \cdot \sqrt{1+9t^2+9t^4} \\
\Leftrightarrow \kappa(t) & = & \frac{2\cdot\sqrt{1+9t^2+9t^4}}{(1+4t^2+9t^4)^{\frac{3}{2}}}
\end{eqnarray*}
\hfill $\Box$

\section*{Aufgabe 4}
Sei $a \in \SetR_{>0}$.
$$ f:\SetR\rightarrow\SetR^2, \; f(t) = (e^{at}\cdot cos(t), e^{at}\cdot sin(t)) $$
Bemerkung: Die Kurve $f$ ist eine logarithmische Spirale.

\subsubsection*{(a)}
Bestimme die Bogenlänge von $f(]-\infty,t]), t\in\SetR$ \\
Lösung: Die Bogenlänge von $f(]-\infty,t])$ sei $L(t)$. \\
Es gilt:
\begin{eqnarray*}
L(t) & = & \int_{-\infty}^t \|f'(\tau)\|d\tau = \lim_{u\rightarrow -\infty}\int_u^t \|f'(\tau)\|d\tau, \; u<t \\ \\
f'(\tau) & = & (a\cdot e^{a\tau}cos(\tau) + e^{a\tau}(-sin(\tau)), a\cdot e^{a\tau}sin(\tau) + e^{a\tau}cos(\tau)) \\
f'(\tau) & = & (e^{a\tau}\cdot(a\cdot cos(\tau)-sin(\tau)), e^{a\tau}\cdot(a\cdot sin(\tau)+cos(\tau))) \\ \\
\Rightarrow \|f'(\tau)\|^2 & = & e^{2a\tau}\cdot(a\cdot cos(\tau)-sin(\tau))^2 + e^{2a\tau}\cdot(a\cdot sin(\tau)+cos(\tau))^2 \\
& = & e^{2a\tau} \cdot (a^2\cdot cos^2(\tau) - 2a\cdot sin(\tau)cos(\tau)+sin^2(\tau)) + e^{2a\tau}\cdot(a^2\cdot sin^2(\tau)+2a\cdot sin(\tau)cos(\tau) + cos^2(\tau)) \\
& = & e^{2a\tau}\cdot [a^2 \cdot \underbrace{(cos^2(\tau)+sin^2(\tau))}_{=1} + \underbrace{(sin^2(\tau)+cos^2(\tau))}_{=1}] \\
\|f'(\tau)\|^2 & = & e^{2a\tau}\cdot(a^2+1) \\
\Rightarrow \|f'(\tau)\| & = & \sqrt{e^{2a\tau}\cdot(a^2+1)} \\
\Leftrightarrow \|f'(\tau)\| & = & \sqrt{a^2+1} \cdot e^{a\tau} \\
\Rightarrow \int_u^t\|f'(\tau)\| d\tau & = & \int_u^t \sqrt{a^2+1}\cdot e^{a\tau} d\tau \\
& = & \sqrt{a^2+1} \cdot \left[\frac{1}{a}\cdot e^{a\tau}\right]_u^t = \frac{\sqrt{a^2+1}}{a} \cdot \left(e^{at}-e^{au}\right) \\
\Rightarrow L(t) & = & lim_{u\rightarrow -\infty}\left(\frac{\sqrt{a^2+1}}{a} \cdot \left(e^{at}-e^{au}\right)\right), \; \; a>0 \\
\Leftrightarrow L(t) & = & \frac{1}{a}\cdot\sqrt{a^2+1}\cdot e^{at}
\end{eqnarray*}

\subsubsection*{(b)}
Bestimme den Winkel zwischen der Geraden durch $0$ und $f(t)$ und den Tangentialvektor von $f$ und $f(t)$ \\
Die Gerade $g(\tau)$ durch $0$ und $f(t)=(e^{at}\cdot cos(t),e^{at}\cdot sin(t))$ ist
$$ g(\tau) = (0,0)+\tau\cdot(e^{at}\cdot cos(t), e^{at}\cdot sin(t)) = \tau\cdot f(t) $$
Es gilt:
\begin{eqnarray*}
g(1) & = & (e^{at}\cdot cos(t), e^{at}\cdot sin(t)) = f(t), \; t\in\SetR \\
g'(\tau) & = & f(t) \; \forall \tau \\
\Rightarrow g'(1) & = & (e^{at}cos(t),e^{at}sin(t))
\end{eqnarray*}
Der Tangentialvektor von $f$ in $f(t)$, d.h. zum Parameterwert t, ist
$$ f'(t) = (e^{at}\cdot(a\cdot cos(t)-sin(t)), e^{at}\cdot(a\cdot sin(t)+cos(t))) $$
Der Winkel $\vartheta$ berechnet sich aus
$$ cos(\vartheta) = \frac{<f'(t),g'(1)>}{\|f'(t)\|\cdot\|g'(1)\|}, \; \; \vartheta \in [0, \pi] $$
\begin{eqnarray*}
<f'(t),g'(1)> & = & e^{at}\cdot(a\cdot cos(t)-sin(t))\cdot e^{at}\cdot cos(t) + e^{at}\cdot(a\cdot sin(t)+cos(t)) \cdot e^{at}\cdot sin(t) \\
& = & e^{2at}\cdot[a\cdot cos^2(t)-sin(t)cos(t)+a \cdot sin^2(t)+sin(t)cos(t)] \\
& = & e^{2at} \cdot a \cdot (cos^2(t)+sin^2(t)) \\
& = & a \cdot e^{2at} \\
\|f'(t)\| & = & \sqrt{a^2+1}\cdot e^{at} \\
\|g'(1)\| & = & \sqrt{e^{2at}\cdot cos^2(t) + e^{2at}\cdot sin^2(t)} = e^{at} \\
\Rightarrow cos(\vartheta) & = & \frac{a\cdot e^{2at}}{\sqrt{a^2+1}\cdot e^{at}\cdot e^{at}} = \frac{a}{\sqrt{a^2+1}} \\
\Rightarrow \vartheta & = & arccos\left(\frac{a}{\sqrt{a^2+1}}\right)
\end{eqnarray*}
Bemerkung 1: $\displaystyle a>0 \Rightarrow 0<\frac{a}{\sqrt{a^2+1}}<1 \Rightarrow \vartheta \in \left[0; \frac{\pi}{2}\right]$ \\
Bemerkung 2: Der Schnittwinkel $\vartheta$ ist von $t$ unabhängig, d.h. jede Gerade durch $0$ schneidet die logarithmische Spirale (bei festem $a$) unter dem gleichen Winkel.

\section*{Aufgabe 5}
$a,b,c,d\in\SetR^3$ \\
Vorbetrachtung: \\
Sei $e=(e_1,e_2,e_3)$ die kanonische Basis des $\SetR^3$. Betrachte
\begin{eqnarray*}
det(a,b,e) & := & \left|
\begin{array}{ccc}
a_1 & b_1 & e_1 \\
a_2 & b_2 & e_2 \\
a_3 & b_3 & e_3
\end{array}\right| \\
& = & a_1b_2e_3 + a_3b_1e_2 + a_2b_3e_1 - a_3b_2e_1 - a_1b_3e_2 - a_2b_1e_3 \\
& = & (a_2b_3 - a_3b_2)e_1 + (a_3b_1-a_1b_3)e_2 + (a_1b_2-a_2b_1)e_3 \\
& = & \left(
\begin{array}{c}
a_2b_3 - a_3b_2 \\
a_3b_1 - a_1b_3 \\
a_1b_2 - a_2b_1
\end{array} \right) \\
& = & a \times b
\end{eqnarray*}

\subsubsection*{(a)}
\begin{eqnarray*}
a \times (b \times c) & = & \left|
\begin{array}{ccc}
a_1 & (b_2c_3-b_3c_2) & e_1 \\
a_2 & (b_3c_1-b_1c_3) & e_2 \\
a_3 & (b_1c_2-b_2c_1) & e_3
\end{array} \right| \\
& = & a_1(b_3c_1-b_1c_3)e_3 + a_3(b_2c_3-b_3c_2)e_2 + a_2(b_1c_2-b_2c_1)e_1 \\
& & -a_3(b_3c_1-b_1c_3)e_1 - a_1(b_1c_2-b_2c_1)e_2 - a_2(b_2c_3-b_3c_2)e_3 \\
& = & \left(
\begin{array}{c}
a_2(b_1c_2-b_2c_1)-a_3(b_3c_1-b_1c_3) \\
a_3(b_2c_3-b_3c_2)-a_1(b_1c_2-b_2c_1) \\
a_1(b_3c_1-b_1c_3)-a_2(b_2c_3-b_3c_2)
\end{array} \right) \\
& = & \left(
\begin{array}{c}
b_1(a_2c_2 + a_3b_3) - c_1(a_2b_2 + a_3b_3) \\
b_2(a_1c_1 + a_3b_3) - c_2(a_1b_1 + a_3b_3) \\
b_3(a_1c_1 + a_2c_2) - c_3(a_1b_1 + a_2b_2)
\end{array} \right) + \underbrace{ \left(
\begin{array}{c}
b_1(a_1c_1) - c_1(a_1b_1) \\
b_2(a_2c_2) - c_2(a_2b_2) \\
b_3(a_3c_3) - c_3(a_3b_3)
\end{array} \right)}_{=\left(\begin{array}{c}0\\0\\0\end{array}\right)} \\
& = & \left(
\begin{array}{c}
b_1(a_1c_1+a_2c_2+a_3c_3) - c_1(a_1b_1+a_2b_2+a_3b_3) \\
b_2(a_1c_1+a_2c_2+a_3c_3) - c_2(a_1b_1+a_2b_2+a_3b_3) \\
b_3(a_1c_1+a_2c_2+a_3c_3) - c_3(a_1b_1+a_2b_2+a_3b_3)
\end{array} \right) \\
& = & \left(
\begin{array}{c}
b_1 \cdot <a,c> \\
b_2 \cdot <a,c> \\
b_3 \cdot <a,c>
\end{array} \right) - \left(
\begin{array}{c}
c_1 \cdot <a,b> \\
c_2 \cdot <a,b> \\
c_3 \cdot <a,b>
\end{array} \right) \\
& = & \left(
\begin{array}{c}
b_1 \\
b_2 \\
b_3
\end{array} \right) \cdot <a,c> - \left(
\begin{array}{c}
c_1 \\
c_2 \\
c_3
\end{array} \right) \cdot <a,b> \\
& = & b<a,c> - c<a,b>
\end{eqnarray*}

\subsubsection*{(b)}
\begin{eqnarray*}
<a \times b, c \times d> & = & <\left(
\begin{array}{c}
a_2b_3 - a_3b_2 \\
a_3b_1 - a_1b_3 \\
a_1b_2 - a_2b_1
\end{array} \right), \left(
\begin{array}{c}
c_2d_3 - c_3d_2 \\
c_3d_1 - c_1d_3 \\
c_1d_2 - c_2d_1
\end{array} \right)> \\
& = & (a_2b_3-a_3b_2)(c_2d_3-c_3d_2)+(a_3b_1-a_1b_3)(c_3d_1-c_1d_3) \\
& & +(a_1b_2-a_2b_1)(c_1d_2-c_2d_1) \\
& = & a_2c_2b_3d_3+a_3c_3b_2d_2+a_3c_3b_1d_1+a_1c_1b_3d_3 \\
& & +a_1c_1b_2d_2+a_2c_2b_1d_1 \\
& & -b_3c_3a_2d_2-b_2c_2a_3d_3-b_1c_1a_3d_3-b_3c_3a_1d_1 \\
& & -b_2c_2a_1d_1-b_1c_1a_2d_2
\end{eqnarray*}
Addition von $0$ in der Form $(a_1c_1b_1d_1-b_1c_1a_1d_1)+(a_2c_2b_2d_2-b_2c_2a_2d_2)+(a_3c_3b_3d_3-b_3c_3a_3d_3)$, Umordnen von Summen und Klammern ergibt:
\begin{eqnarray*}
<a \times b, c \times d> & = & (a_1c_1b_1d_1+a_1c_1b_2d_2+a_1c_1b_3d_3 \\
& & + a_2c_2b_1d_1+a_2c_2b_2d_2+a_2c_2b_3d_3 \\
& & + a_3c_3b_1d_1+a_3c_3b_2d_2+a_3c_3b_3d_3) \\
& & -(b_1c_1a_1d_1+b_1c_1a_2d_2+b_1c_1a_3d_3 \\
& & +b_2c_2a_1d_1+b_2c_2a_2d_2+b_2c_2a_3d_3 \\
& & +b_3c_3a_1d_1+b_3c_3a_2d_2+b_3c_3a_3d_3) \\
& = & a_1c_1 \cdot <b,d> + a_2c_2 \cdot <b,d> + a_3c_3 \cdot <b,d> \\
& & -(b_1c_1 \cdot <a,d> + b_2c_2 \cdot <a,d> + b_3c_3 \cdot <a,d>) \\
& = & (a_1c_1+a_2c_2+a_3c_3) \cdot <b,d> \\
& & - (b_1c_1+b_2c_2+b_3c_3) \cdot <a,d> \\
& = & <a,c> \cdot <b,d> - <b,c> \cdot <a,d>
\end{eqnarray*}


\end{document}