\documentclass[11pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{ngerman}
\usepackage{textcomp}
\usepackage{amsmath}
\usepackage{graphics}

\date{\today{}}
\author{Thomas Kuster}
\title{UMNW, Mathematik 1, Lösung Serie 1}

\begin{document}
\markright{UMNW, Mathematik 1, Lösung Serie 1}
\pagestyle{myheadings}

\maketitle

\section{}
\begin{eqnarray*}
0.001m \cdot 2^x & = & 150 \cdot 10^{9} m\\
2^x & = & \frac{150 \cdot 10^{9}m}{0.001m}\\
x \cdot \ln{2} & = & \ln{\frac{150 \cdot 10^{9}}{0.001}}\\
x & = & {\frac{\ln{1.5 \cdot 10^{14}}}{\ln{2}}}\\
x & \approx & 47.09\\
& \Rightarrow & 48 \mbox{ mal Falten (aufrunden)}
\end{eqnarray*}

\section{}
\begin{eqnarray*}
N_0 \cdot \lambda^k & = & 2 N_0\\
\lambda^k & = & 2\\
k \ln{\lambda} & = & \ln{2}\\
k & = & \frac{\ln{2}}{ \ln{\lambda}}\
\end{eqnarray*}
Lösen mit Octave (http://www.octave.org)
\begin{verbatim}
octave:1> lambda = (1.01:0.01:1.1)
lambda =
1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10

octave:2> log(2)./log(lambda)
69.6607
35.0028
23.4498
17.6730
14.2067
11.8957
10.2448
 9.0065
 8.0432
 7.2725

octave:3> title('Verdoppelungszeit');
octave:4> xlabel('Lambda');
octave:5> ylabel('Jahre');
octave:6> plot(lambda,log(2)./log(lambda))
\end{verbatim}
\includegraphics{serie1_plot0.eps}

\section{}

\subsection{}
\begin{eqnarray*}
48503000 \cdot 1.025^{2005-1985} & = & x\\
79477813 & \approx & x\\
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
\left((48503000 \cdot 1.025^{10}) 1.02^{5}\right) 1.015^{5}  & = & x\\
73847929 & \approx & x\\
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
\left((48503000 \cdot 1.025^{10}) 1.03^{5}\right) 1.035^{5}  & = & x\\
85486025 & \approx & x\\
\end{eqnarray*}

\subsection{}
\begin{eqnarray*}
N_0 \cdot p & = & N_k\\
p & = & \frac{N_k}{N_0}\\
\end{eqnarray*}
\begin{eqnarray*}
p_1 & = & \frac{79477813}{48503000} \Rightarrow 63.9\%\\
p_2 & = & \frac{73847929}{48503000} \Rightarrow 52.3\%\\
p_3 & = & \frac{85486025}{48503000} \Rightarrow 76.2\%\\
\end{eqnarray*}

\section{}

\subsection{}
\begin{eqnarray*}
N_0\lambda^k + \sum_{i=0}^{k-1}{m \lambda^i} & = & x\\
8 \cdot 10^7 \cdot 0.99^k + \sum_{i=0}^{k-1}{3 \cdot 10^5 \cdot 0.99^i} & = & x\\
\end{eqnarray*}
\begin{verbatim}
octave:1> 8*10^7 *0.99^10 + sum(3*10^5*0.99.^(0:(10-1)))
ans =  7.5219e+07
\end{verbatim}

\subsection{}
\begin{eqnarray*}
N_0\lambda^k + \sum_{i=0}^{k-1}{m \lambda^i} & = & x\\
10^5 \cdot 1.03^k + \sum_{i=0}^{k-1}{- 4 \cdot 10^3 \cdot 1.03^i} & = & x\\
\end{eqnarray*}
\begin{verbatim}
octave:1> p = 100000
p = 100000
octave:2> for i=1:50
> p = [p 10^5*1.03^i - 4*10^3*sum(1.03.^(0:(i-1)))];
> end
octave:3> title('Fischentwicklung');
octave:4> xlabel('Jahre');
octave:5> ylabel('Anzahl');
octave:6> plot(p)
\end{verbatim}
\includegraphics{serie1_plot1.eps}

oder
\begin{eqnarray*}
N_0\lambda^k + \sum_{i=0}^{k-1}{m \lambda^i} & = & x\\
N_0\lambda^k + m\frac{\lambda^{k}-1}{\lambda-1} & = & x\\
10^5 \cdot 1.03^k - 4 \cdot 10^3 \frac{1.03^{k}-1}{1.03-1} & = & x\\
\end{eqnarray*}
\begin{verbatim}
octave:1> k = (1:50);
octave:2> N = 10^5*1.03.^k + (-4*10^3)*(1.03.^k-1)./(1.03-1)
octave:3> plot(k,N);
\end{verbatim}

Wann hat es keine Fische mehr?
\begin{eqnarray*}
N_0\lambda^k + m \frac{\lambda^{k}-1}{\lambda-1} & \overset{!}{=} & 0\\
m \frac{\lambda^{k}-1}{\lambda-1} & = & - N_0\lambda^k\\
\lambda^{k}-1 & = & \frac{(- N_0\lambda^k)(\lambda-1)}{m}\\
- 1 & = & \frac{(- N_0\lambda^k)(\lambda-1)}{m} - \lambda^{k}\\
1 & = & \left(\frac{N_0(\lambda-1)}{m} +1 \right) \lambda^{k}\\
1 & = & \frac{N_0\lambda - N_0 + m}{m} \lambda^{k}\\
\frac{m}{N_0\lambda - N_0 + m} & = & \lambda^{k}\\
\ln{\frac{m}{N_0\lambda - N_0 + m}} & = & k \ln{\lambda}\\
\frac{\ln\frac{m}{N_0\lambda - N_0 + m}}{\ln{\lambda}} & = & k\\
\frac{\ln{4}}{\ln{1.03}} & \approx & 46.9 \mbox{Jahre}\\
\end{eqnarray*}

\end{document}
