Merge branch 'master' of https://github.com/AndreasFMueller/SeminarSpezielleFunktionen

70 files changed, 12478 insertions, 568 deletions

diff --git a/buch/papers/0f1/teil1.tex b/buch/papers/0f1/teil1.tex
index 2ca9647..f697f45 100644
--- a/buch/papers/0f1/teil1.tex
+++ b/buch/papers/0f1/teil1.tex
@@ -6,12 +6,12 @@
 \section{Mathematischer Hintergrund

 \label{0f1:section:mathHintergrund}}

 \rhead{Mathematischer Hintergrund}

-Basierend auf den Herleitungen des vorhergehenden Kapitels \ref{buch:rekursion:section:hypergeometrische-funktion}, werden im nachfolgenden Abschnitt nochmals die Resultate

+Basierend auf den Herleitungen des vorhergehenden Abschnittes \ref{buch:rekursion:section:hypergeometrische-funktion}, werden im nachfolgenden Abschnitt nochmals die Resultate

 beschrieben.

 

 \subsection{Hypergeometrische Funktion

 \label{0f1:subsection:hypergeometrisch}}

-Als Grundlage der umgesetzten Algorithmen dient die Hypergeometrische Funktion $\mathstrut_0F_1$. Diese ist eine Unterfunktion der allgemein definierten Funktion $\mathstrut_pF_q$.

+Als Grundlage der umgesetzten Algorithmen dient die hypergeometrische Funktion $\mathstrut_0F_1$. Diese ist eine Anwendung der allgemein definierten Funktion $\mathstrut_pF_q$.

 

 \begin{definition}

 	\label{0f1:math:qFp:def}

@@ -42,7 +42,7 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$
     \mathstrut_0F_1

     \biggl(

     \begin{matrix}

-    \\

+    \\-

     b_1

     \end{matrix}

     ;

@@ -60,22 +60,22 @@ Angewendet auf die Funktion $\mathstrut_pF_q$ ergibt sich für $\mathstrut_0F_1$
 

 \subsection{Airy Funktion

 \label{0f1:subsection:airy}}

-Die Airy-Funktion $Ai(x)$ und die verwandte Funktion $Bi(x)$ werden als Airy-Funktion bezeichnet. Sie werden zur Lösung verschiedener physikalischer Probleme benutzt, wie zum Beispiel zur Lösung der Schrödinger-Gleichung. \cite{0f1:wiki-airyFunktion}

+Die Funktion Ai(x) und die verwandte Funktion Bi(x) werden als Airy-Funktion bezeichnet. Sie werden zur Lösung verschiedener physikalischer Probleme benutzt, wie zum Beispiel zur Lösung der Schrödinger-Gleichung \cite{0f1:wiki-airyFunktion}.

 

 \begin{definition}

     \label{0f1:airy:differentialgleichung:def}

     Die Differentialgleichung

     $y'' - xy = 0$

-    heisst die {\em Airy-Differentialgleichung}. \cite{0f1:wiki-airyFunktion} 

+    heisst die {\em Airy-Differentialgleichung}.

 \end{definition}

 

-Die Airy Funktion lässt sich auf verschiedene Arten darstellen. \cite{0f1:wiki-airyFunktion} 

-Als hypergeometrische Funktion berechnet, ergibt sich wie in Kapitel \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $A(0)=1$ und $A'(0)=0$, sowie $B(0)=0$ und $B'(0)=0$.

+Die Airy Funktion lässt sich auf verschiedene Arten darstellen. 

+Als hypergeometrische Funktion berechnet, ergibt sich wie in Abschnitt \ref{buch:differentialgleichungen:section:hypergeometrisch} hergeleitet, folgende Lösungen der Airy-Differentialgleichung zu den Anfangsbedingungen $Ai(0)=1$ und $Ai'(0)=0$, sowie $Bi(0)=0$ und $Bi'(0)=0$.

 

 \begin{align}

 \label{0f1:airy:hypergeometrisch:eq}

 Ai(x)

-=

+=&

 \sum_{k=0}^\infty

 \frac{1}{(\frac23)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k

 =

@@ -84,7 +84,7 @@ Ai(x)
 \biggr).

 \\

 Bi(x)

-=

+=&

 \sum_{k=0}^\infty

 \frac{1}{(\frac43)_k} \frac{1}{k!}\biggl(\frac{x^3}{9}\biggr)^k

 =

@@ -95,7 +95,7 @@ x\cdot\mathstrut_0F_1\biggl(
 \qedhere

 \end{align}

 

-In diesem speziellem Fall wird die Airy Funktion $Ai(x)$ \eqref{0f1:airy:hypergeometrisch:eq}

-benutzt, um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen.

+Um die Stabilität der Algorithmen zu $\mathstrut_0F_1$ zu überprüfen, wird in diesem speziellem Fall die Airy Funktion $Ai(x)$ \eqref{0f1:airy:hypergeometrisch:eq}

+benutzt.

 

 

diff --git a/buch/papers/0f1/teil2.tex b/buch/papers/0f1/teil2.tex
index 9269961..15a1c44 100644
--- a/buch/papers/0f1/teil2.tex
+++ b/buch/papers/0f1/teil2.tex
@@ -6,12 +6,12 @@
 \section{Umsetzung

 \label{0f1:section:teil2}}

 \rhead{Umsetzung}

-Zur Umsetzung wurden drei verschiedene Ansätze gewählt.\cite{0f1:code} Dabei wurde der Schwerpunkt auf die Funktionalität und eine gute Lesbarkeit des Codes gelegt.

-Die Unterprogramme wurde jeweils, wie die GNU Scientific Library, in C geschrieben. Die Zwischenresultate wurden vom Hauptprogramm in einem CSV-File gespeichert. Anschliessen wurde mit der Matplot-Libray in Python die Resultate geplottet.

+Zur Umsetzung wurden drei verschiedene Ansätze gewählt \cite{0f1:code}. Dabei wurde der Schwerpunkt auf die Funktionalität und eine gute Lesbarkeit des Codes gelegt.

+Die Unterprogramme wurde jeweils, wie die GNU Scientific Library, in C geschrieben. Die Zwischenresultate wurden vom Hauptprogramm in einem CSV-File gespeichert. Anschliessen wurde mit der Matplot-Library in Python die Resultate geplottet.

 

 \subsection{Potenzreihe

 \label{0f1:subsection:potenzreihe}}

-Die naheliegendste Lösung ist die Programmierung der Potenzreihe. Allerdings ist ein Problem dieser Umsetzung \ref{0f1:listing:potenzreihe}, dass die Fakultät im Nenner schnell grosse Werte annimmt und so der Bruch gegen Null strebt. Spätesten ab $k=167$ stösst diese Umsetzung  \eqref{0f1:umsetzung:0f1:eq} an ihre Grenzen, da die Fakultät von $168$ eine Bereichsüberschreitung des \textit{double} Bereiches darstellt. \cite{0f1:double}

+Die naheliegendste Lösung ist die Programmierung der Potenzreihe. Allerdings ist ein Problem dieser Umsetzung \ref{0f1:listing:potenzreihe}, dass die Fakultät im Nenner schnell grosse Werte annimmt und so der Bruch gegen Null strebt. Spätesten ab $k=167$ stösst diese Umsetzung  \eqref{0f1:umsetzung:0f1:eq} an ihre Grenzen, da die Fakultät von $168$ eine Bereichsüberschreitung des \textit{double} Bereiches darstellt \cite{0f1:double}.

 

 \begin{align}

     \label{0f1:umsetzung:0f1:eq}

@@ -34,23 +34,22 @@ Ein endlicher Kettenbruch ist ein Bruch der Form
 \begin{equation*}

 a_0 + \cfrac{b_1}{a_1+\cfrac{b_2}{a_2+\cfrac{b_3}{a_3+\cdots}}}

 \end{equation*}

-in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen darstellen.

+in welchem $a_0, a_1,\dots,a_n$ und $b_1,b_2,\dots,b_n$ ganze Zahlen sind.

 Die Kurzschreibweise für einen allgemeinen Kettenbruch ist 

 \begin{equation*}

 	a_0 + \frac{a_1|}{|b_1} + \frac{a_2|}{|b_2} + \frac{a_3|}{|b_3} + \cdots

 \end{equation*}

-und ist somit verknüpfbar mit der Potenzreihe.

-\cite{0f1:wiki-kettenbruch}

-Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies\cite{0f1:wiki-fraction}:

+\cite{0f1:wiki-kettenbruch}.

+Angewendet auf die Funktion $\mathstrut_0F_1$ bedeutet dies \cite{0f1:wiki-fraction}:

 \begin{equation*}

 	\mathstrut_0F_1(;c;z) = 1 + \frac{z}{c\cdot1!} + \frac{z^2}{c(c+1)\cdot2!} + \frac{z^3}{c(c+1)(c+2)\cdot3!} + \cdots

 \end{equation*}

-Nach allen Umformungen ergibt sich folgender, irregulärer Kettenbruch \eqref{0f1:math:kettenbruch:0f1:eq}

+Umgeformt ergibt sich folgender Kettenbruch

 \begin{equation}

 	\label{0f1:math:kettenbruch:0f1:eq}

 	\mathstrut_0F_1(;c;z) = 1 + \cfrac{\cfrac{z}{c}}{1+\cfrac{-\cfrac{z}{2(c+1)}}{1+\cfrac{z}{2(c+1)}+\cfrac{-\cfrac{z}{3(c+2)}}{1+\cfrac{z}{5(c+4)} + \cdots}}},

 \end{equation}

-der als Code \ref{0f1:listing:kettenbruchIterativ} umgesetzt wurde. 

+der als Code (siehe: Listing \ref{0f1:listing:kettenbruchIterativ})  umgesetzt wurde. 

 \cite{0f1:wolfram-0f1}

 

 \lstinputlisting[style=C,float,caption={Iterativ umgesetzter Kettenbruch.},label={0f1:listing:kettenbruchIterativ},  firstline=8]{papers/0f1/listings/kettenbruchIterativ.c}

@@ -138,7 +137,7 @@ Nach vollständiger Induktion ergibt sich für den Schritt $k$, die Matrix
 \end{equation}

 Und Schlussendlich kann der Näherungsbruch

 \[

-\frac{Ak}{Bk}

+\frac{A_k}{B_k}

 \] 

 berechnet werden.

 

@@ -166,7 +165,7 @@ B_{k+1} &= B_{k-1} \cdot b_k + B_k \cdot a_k
 Näherungsbruch: \qquad$\displaystyle\frac{A_k}{B_k}$

 \end{itemize}

 

-Ein grosser Vorteil dieser Umsetzung \ref{0f1:listing:kettenbruchRekursion} ist, dass im Vergleich zum Code \ref{0f1:listing:kettenbruchIterativ} eine Division gespart werden kann und somit weniger Rundungsfehler entstehen können.

+Ein grosser Vorteil dieser Umsetzung als Rekursionsformel ist \ref{0f1:listing:kettenbruchRekursion}, dass im Vergleich zum Code \ref{0f1:listing:kettenbruchIterativ} eine Division gespart werden kann und somit weniger Rundungsfehler entstehen können.

 

 %Code

 \lstinputlisting[style=C,float,caption={Rekursionsformel für Kettenbruch.},label={0f1:listing:kettenbruchRekursion},  firstline=8]{papers/0f1/listings/kettenbruchRekursion.c}
\ No newline at end of file
diff --git a/buch/papers/0f1/teil3.tex b/buch/papers/0f1/teil3.tex
index 2855e26..72b1b21 100644
--- a/buch/papers/0f1/teil3.tex
+++ b/buch/papers/0f1/teil3.tex
@@ -6,19 +6,19 @@
 \section{Auswertung

 \label{0f1:section:teil3}}

 \rhead{Resultate}

-Im Verlauf des Seminares hat sich gezeigt, 

+Im Verlauf dieser Arbeit hat sich gezeigt, 

 das ein einfacher mathematischer Algorithmus zu implementieren gar nicht so einfach ist.

 So haben alle drei umgesetzten Ansätze Probleme mit grossen negativen $z$ in der Funktion $\mathstrut_0F_1(;c;z)$.

-Ebenso kann festgestellt werden,dass je grösser der Wert $z$ in $\mathstrut_0F_1(;c;z)$ wird, desto mehr weichen die berechneten Resultate von den Erwarteten ab. \cite{0f1:wolfram-0f1}

+Ebenso kann festgestellt werden, dass je grösser der Wert $z$ in $\mathstrut_0F_1(;c;z)$ wird, desto mehr weichen die berechneten Resultate von den Erwarteten ab \cite{0f1:wolfram-0f1}.

 

 \subsection{Konvergenz

 \label{0f1:subsection:konvergenz}}

 Es zeigt sich in Abbildung \ref{0f1:ausblick:plot:airy:konvergenz}, dass schon nach drei Iterationen ($k = 3$) die Funktionen schon genaue Resultate im Bereich von -2 bis 2 liefert. Ebenso kann festgestellt werden, dass der Kettenbruch schneller konvergiert und im positiven Bereich sogar mit der Referenzfunktion $Ai(x)$ übereinstimmt. Da die Rekursionsformel \ref{0f1:listing:kettenbruchRekursion} eine Abwandlung des Kettenbruches ist, verhalten sich die Funktionen in diesem Fall gleich.

 

-Erst wenn mehrere Durchläufe gemacht werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen, bezüglich Konvergenz überlegen. 

-Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach einschwingt. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von k bis zum Abbruch kleiner.

-\ref{0f1:ausblick:plot:konvergenz:positiv}

-Dieses Phänomen ist auf die Lösung der Rekursionsformel zurück zu führen.\ref{0f1:math:loesung:eq} Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen.

+Erst wenn mehrerer Iterationen gemacht werden, um die Genauigkeit zu verbessern, ist der Kettenbruch den anderen zwei Algorithmen, bezüglich Konvergenz überlegen. 

+Interessant ist auch, dass die Rekursionsformel nahezu gleich schnell wie die Potenzreihe konvergiert, aber sich danach einschwingt \ref{0f1:ausblick:plot:konvergenz:positiv}. Dieses Verhalten ist auch bei grösseren $z$ zu beobachten, allerdings ist dann die Differenz zwischen dem ersten lokalen Minimum von k bis zum Abbruch kleiner

+\ref{0f1:ausblick:plot:konvergenz:positiv}.

+Dieses Phänomen ist auf die Lösung der Rekursionsformel zurück zu führen\eqref{0f1:math:loesung:eq}. Da im Gegensatz die ganz kleinen Werte nicht zu einer Konvergenz wie beim Kettenbruch führen, sondern sich noch eine Zeit lang durch die Multiplikation aufschwingen.

 

 Ist $z$ negativ wie im Abbild \ref{0f1:ausblick:plot:konvergenz:negativ}, führt dies zu einer Gegenseitigen Kompensation von negativen und positiven Termen so bricht die Rekursionsformel hier zusammen mit der Potenzreihe ab.

 Die ansteigende Differenz mit anschliessender, ist aufgrund der sich alternierenden Termen mit wechselnden Vorzeichens zu erklären.

@@ -27,10 +27,10 @@ Die ansteigende Differenz mit anschliessender, ist aufgrund der sich alternieren
 \label{0f1:subsection:Stabilitaet}}

 Verändert sich der Wert von z in $\mathstrut_0F_1(;c;z)$ gegen grössere positive Werte, wie zum Beispiel $c = 800$ liefert die Kettenbruch-Funktion \ref{0f1:listing:kettenbruchIterativ} \verb+inf+ zurück. Dies könnte durch ein Abbruchkriterien abgefangen werden. Allerdings würde das, bei grossen Werten zulasten der Genauigkeit gehen. Trotzdem könnte, je nach Anwendung, auf ein paar Nachkommastellen verzichtet werden.

 

-Wohingegen die Potenzreihe \ref{0f1:listing:potenzreihe} das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät und irgendwann gibt es eine Bereichsüberschreitung von \verb+double+. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück.

-Die Rekursionformel \ref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Stabilität zu gewährleisten, muss wie in \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden.

+Wohingegen die Potenzreihe \eqref{0f1:listing:potenzreihe} das Problem hat, dass je mehr Terme berechnet werden, desto schneller wächst die Fakultät und irgendwann gibt es eine Bereichsüberschreitung von \verb+double+. Schlussendlich gibt das Unterprogramm das Resultat \verb+-nan(ind)+ zurück.

+Die Rekursionformel \eqref{0f1:listing:kettenbruchRekursion} liefert für sehr grosse positive Werte die genausten Ergebnisse, verglichen mit der GNU Scientific Library. Wie schon vermutet ist die Rekursionsformel, im positivem Bereich, der stabilste Algorithmus. Um die Stabilität zu gewährleisten, muss wie in Abbild \ref{0f1:ausblick:plot:konvergenz:positiv} dargestellt, die Iterationstiefe $k$ genug gross gewählt werden.

 

-Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Fakultät im Nenner, was zum Phänomen der Auslöschung führt.\cite{0f1:SeminarNumerik} Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen.

+Im negativem Bereich sind alle gewählten und umgesetzten Ansätze instabil. Grund dafür ist die Potenz von z, was zum Phänomen der Auslöschung führt \cite{0f1:SeminarNumerik}. Schön zu beobachten ist dies in der Abbildung \ref{0f1:ausblick:plot:airy:stabilitaet} mit der Airy-Funktion als Test. So sind sowohl der Kettenbruch, als auch die Rekursionsformel bis ungefähr $\frac{-15^3}{9}$ stabil. Dies macht auch Sinn, da beide auf der gleichen mathematischen Grundlage basieren. Danach verhält sich allerdings die Instabilität unterschiedlich. Das unterschiedliche Verhalten kann damit erklärt werden, dass beim Kettenbruch jeweils eine zusätzliche Division stattfindet. Diese Unterschiede sind auch in Abbildung \ref{0f1:ausblick:plot:konvergenz:positiv} festzustellen.

 

 

 

diff --git a/buch/papers/fm/00_modulation.tex b/buch/papers/fm/00_modulation.tex
index dc99b40..e2ba39f 100644
--- a/buch/papers/fm/00_modulation.tex
+++ b/buch/papers/fm/00_modulation.tex
@@ -18,10 +18,14 @@ Mathematisch wird dann daraus
     \omega_i = \omega_c + \frac{d \varphi(t)}{dt}
 \]
 mit der Ableitung der Phase\cite{fm:NAT}.
-Mit diesen drei parameter ergeben sich auch drei modulationsarten, die Amplitudenmodulation welche \(A_c\) benutzt, 
-die Phasenmodulation \(\varphi\) und dann noch die Momentankreisfrequenz \(\omega_i\):
-\newline
-\newline
+Mit diesen drei Parameter ergeben sich auch drei Modulationsarten, die Amplitudenmodulation, welche \(A_c\) benutzt, 
+die Phasenmodulation \(\varphi\) und dann noch die Momentankreisfrequenz \(\omega_i\): 
+\begin{itemize}
+    \item AM
+    \item PM
+    \item FM
+\end{itemize}
+
 To do: Bilder jeder Modulationsart
 
 
diff --git a/buch/papers/fm/01_AM.tex b/buch/papers/fm/01_AM.tex
index 921fcf2..21927f5 100644
--- a/buch/papers/fm/01_AM.tex
+++ b/buch/papers/fm/01_AM.tex
@@ -17,8 +17,8 @@ Dies sieht man besonders in der Eulerischen Formel
 \[
     x_c(t) = \frac{A_c}{2} \cdot e^{j\omega_ct}\;+\;\frac{A_c}{2} \cdot e^{-j\omega_ct}.
 \]
-Dabei ist die negative Frequenz der zweiten komplexen Schwingung zwingend erforderlich, damit in der Summe immer ein reelwertiges Trägersignal ergibt.
-Nun wird der parameter \(A_c\) durch das  Moduierende Signal \(m(t)\) ersetzt, wobei so \(m(t) \leqslant |1|\) normiert wurde.
+Dabei ist die negative Frequenz der zweiten komplexen Schwingung zwingend erforderlich, damit in der Summe immer ein reellwertiges Trägersignal ergibt.
+Nun wird der Parameter \(A_c\) durch das  Modulierende Signal \(m(t)\) ersetzt, wobei so \(m(t) \leqslant |1|\) normiert wurde.
 \newline
 \newline
 TODO:
diff --git a/buch/papers/fm/03_bessel.tex b/buch/papers/fm/03_bessel.tex
index 760cdc4..5f85dc6 100644
--- a/buch/papers/fm/03_bessel.tex
+++ b/buch/papers/fm/03_bessel.tex
@@ -3,11 +3,11 @@
 %
 % (c) 2020 Prof Dr Andreas Müller, Hochschule Rapperswil
 %
-\section{FM und Besselfunktion 
+\section{FM und Bessel-Funktion 
 \label{fm:section:proof}}
 \rhead{Herleitung}
-Die momentane Trägerkreisfrequenz \(\omega_i\) wie schon in (ref) beschrieben ist, bringt die Vorigen Kapittel beschreiben. (Ableitung \(\frac{d \varphi(t)}{dt}\) mit sich).
-Diese wiederum kann durch \(\beta\sin(\omega_mt)\) ausgedrückt werden, wobei es das Modulierende Signal \(m(t)\) ist.
+Die momentane Trägerkreisfrequenz \(\omega_i\), wie schon in (ref) beschrieben ist, bringt die Ableitung \(\frac{d \varphi(t)}{dt}\) mit sich.
+Diese wiederum kann durch \(\beta\sin(\omega_mt)\) ausgedrückt werden, wobei es das modulierende Signal \(m(t)\) ist.
 Somit haben wir unser \(x_c\) welches 
 \[
 \cos(\omega_c t+\beta\sin(\omega_mt))
@@ -15,7 +15,7 @@ Somit haben wir unser \(x_c\) welches
 ist.
 
 \subsection{Herleitung}
-Das Ziel ist es unser moduliertes Signal mit der Besselfunktion so auszudrücken:
+Das Ziel ist, unser moduliertes Signal mit der Bessel-Funktion so auszudrücken:
 \begin{align}
     x_c(t)
     = 
@@ -24,6 +24,7 @@ Das Ziel ist es unser moduliertes Signal mit der Besselfunktion so auszudrücken
     \sum_{k= -\infty}^\infty J_{k}(\beta) \cos((\omega_c+k\omega_m)t)
     \label{fm:eq:proof}
 \end{align}
+
 \subsubsection{Hilfsmittel}
 Doch dazu brauchen wir die Hilfe der Additionsthoerme 
 \begin{align}
@@ -42,7 +43,7 @@ Doch dazu brauchen wir die Hilfe der Additionsthoerme
     \cos(A-B)-\cos(A+B)
     \label{fm:eq:addth3}
 \end{align}
-und die drei Besselfunktions indentitäten,
+und die drei Bessel-Funktionsindentitäten,
 \begin{align}
     \cos(\beta\sin\phi)
     &=
@@ -51,13 +52,13 @@ und die drei Besselfunktions indentitäten,
     \\
     \sin(\beta\sin\phi)
     &=
-    J_0(\beta) + 2\sum_{k=1}^\infty J_{2k+1}(\beta) \cos((2k+1)\phi)
+    2\sum_{k=0}^\infty J_{2k+1}(\beta) \cos((2k+1)\phi)
     \label{fm:eq:besselid2}
     \\
     J_{-n}(\beta) &= (-1)^n J_n(\beta)
     \label{fm:eq:besselid3}
 \end{align}
-welche man im Kapitel (ref), ref, ref findet.
+welche man im Kapitel \eqref{buch:fourier:eqn:expinphireal}, \eqref{buch:fourier:eqn:expinphiimaginary}, \eqref{buch:fourier:eqn:symetrie} findet.
 
 \subsubsection{Anwenden des Additionstheorem}
 Mit dem \eqref{fm:eq:addth1} wird aus dem modulierten Signal
@@ -66,63 +67,111 @@ Mit dem \eqref{fm:eq:addth1} wird aus dem modulierten Signal
     =
     \cos(\omega_c t + \beta\sin(\omega_mt))
     =
-    \cos(\omega_c t)\cos(\beta\sin(\omega_m t))-\sin(\omega_c)\sin(\beta\sin(\omega_m t)).
+    \cos(\omega_c t)\cos(\beta\sin(\omega_m t))-\sin(\omega_ct)\sin(\beta\sin(\omega_m t)).
     \label{fm:eq:start}
 \]
+%-----------------------------------------------------------------------------------------------------------
 \subsubsection{Cos-Teil}
 Zu beginn wird der Cos-Teil
-\[
-    \cos(\omega_c)\cos(\beta\sin(\omega_mt))  
-\]
+\begin{align*}
+    c(t) 
+    &=
+    \cos(\omega_c t)\cdot\cos(\beta\sin(\omega_mt))  
+\end{align*}
 mit hilfe der Besselindentität \eqref{fm:eq:besselid1} zum
 \begin{align*}
-    \cos(\omega_c t) \cdot \bigg[\, J_0(\beta) + 2\sum_{k=1}^\infty J_{2k}(\beta) \cos( 2k \omega_m t)\, \bigg]
-    &=\\
-    J_0(\beta)\cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) 
-    \underbrace{2\cos(\omega_c t)\cos(2k\omega_m t)}_{\text{Additionstheorem}}
+    c(t)
+    &=
+    \cos(\omega_c t) \cdot \bigg[ J_0(\beta) + 2\sum_{k=1}^\infty J_{2k}(\beta) \cos( 2k \omega_m t)\, \bigg]
+    \\
+    &=
+    J_0(\beta) \cdot \cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) \underbrace{2\cos(\omega_c t)\cos(2k\omega_m t)}_{\text{Additionstheorem \eqref{fm:eq:addth2}}}
+\end{align*}
+%intertext{}   Funktioniert nicht.
+wobei mit dem Additionstheorem \eqref{fm:eq:addth2} \(A = \omega_c t\) und \(B = 2k\omega_m t \) ersetzt wurden.
+\begin{align*}
+    c(t)
+    &=
+    J_0(\beta) \cdot \cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) \{ \underbrace{\cos((\omega_c - 2k \omega_m) t)} \,+\, \cos((\omega_c + 2k \omega_m) t) \}
+    \\
+    &=
+    \sum_{k=-\infty}^{-1} J_{2k}(\beta)  \overbrace{\cos((\omega_c +2k \omega_m) t)} 
+    \,+\,J_0(\beta)\cdot \cos(\omega_c t+ 2\cdot0 \omega_m) 
+    \,+\, \sum_{k=1}^\infty J_{2k}(\beta)\cos((\omega_c + 2k \omega_m) t) 
 \end{align*}
-wobei mit dem Additionstheorem \eqref{fm:eq:addth2} \(A = \omega_c t\) und \(B = 2k\omega_m t \) zum
-\[
-    J_0(\beta)\cdot \cos(\omega_c t) + \sum_{k=1}^\infty J_{2k}(\beta) \{ \cos((\omega_c - 2k \omega_m) t)+\cos((\omega_c + 2k \omega_m) t) \}
-\]
 wird.
-Wenn dabei \(2k\) durch alle geraden Zahlen von \(-\infty \to \infty\) mit \(n\) substituiert erhält man den vereinfachten Term
-\[
-    \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t),
+Das Minus im Ersten Term wird  zur negativen Summe \(\sum_{-\infty}^{-1}\) ersetzt.
+Da \(2k\) immer gerade ist, wird es durch alle negativen und positiven Ganzzahlen \(n\) ersetzt:
+\begin{align*}
+    \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n \omega_m) t),
     \label{fm:eq:gerade}
-\]
-dabei gehen nun die Terme von \(-\infty \to \infty\), dabei bleibt n Ganzzahlig.
-
+\end{align*}
+%----------------------------------------------------------------------------------------------------------------
 \subsubsection{Sin-Teil}
 Nun zum zweiten Teil des Term \eqref{fm:eq:start}, den Sin-Teil
-\[
-    \sin(\omega_c)\sin(\beta\sin(\omega_m t)).
-\]
+\begin{align*}
+    s(t) 
+    &=
+    -\sin(\omega_c t)\cdot\sin(\beta\sin(\omega_m t)).
+\end{align*}
 Dieser wird mit der \eqref{fm:eq:besselid2} Besselindentität zu
 \begin{align*}
-    \sin(\omega_c t) \cdot \bigg[ J_0(\beta) + 2 \sum_{k=1}^\infty J_{ 2k + 1}(\beta) \cos(( 2k + 1) \omega_m t) \bigg]
-    &=\\
-    J_0(\beta) \cdot \sin(\omega_c t) + \sum_{k=1}^\infty J_{2k+1}(\beta) \underbrace{2\sin(\omega_c t)\cos((2k+1)\omega_m t)}_{\text{Additionstheorem}}.
+    s(t) 
+    &=
+    -\sin(\omega_c t) \cdot \bigg[ 2 \sum_{k=0}^\infty J_{ 2k + 1}(\beta) \cos(( 2k + 1) \omega_m t) \bigg]
+    \\
+    &=
+    \sum_{k=0}^\infty -1 \cdot J_{2k+1}(\beta) 2\sin(\omega_c t)\cos((2k+1)\omega_m t).
 \end{align*}
-Auch hier wird ein Additionstheorem \eqref{fm:eq:addth3} gebraucht, dabei ist \(A = \omega_c t\) und \(B = (2k+1)\omega_m t \), 
-somit wird daraus
-\[
-    J_0(\beta) \cdot \sin(\omega_c) + \sum_{k=1}^\infty J_{2k+1}(\beta) \{ \underbrace{\cos((\omega_c-(2k+1)\omega_m) t)}_{\text{neg.Teil}} - \cos((\omega_c+(2k+1)\omega_m) t) \}
-\]dieser Term.
-Wenn dabei \(2k +1\) durch alle ungeraden Zahlen von \(-\infty \to \infty\) mit \(n\) substituiert.
-Zusätzlich dabei noch die letzte Besselindentität \eqref{fm:eq:besselid3} brauchen, ist bei allen ungeraden negativen \(n : J_{-n}(\beta) = -1\cdot J_n(\beta)\).
-Somit wird neg.Teil zum Term \(-\cos((\omega_c+(2k+1)\omega_m) t)\) und die Summe vereinfacht sich zu
+Da \(2k + 1\) alle ungeraden positiven Ganzzahlen entspricht wird es durch \(n\) ersetzt. 
+Wird die Besselindentität \eqref{fm:eq:besselid3} gebraucht, so ersetzten wird \(J_{-n}(\beta) = -1\cdot J_n(\beta)\) ersetzt:
+\begin{align*}
+    s(t)
+    &=
+    \sum_{n=0}^\infty J_{-n}(\beta) \underbrace{2\sin(\omega_c t)\cos(n \omega_m t)}_{\text{Additionstheorem \eqref{fm:eq:addth3}}}.
+\end{align*}
+Auch hier wird ein Additionstheorem \eqref{fm:eq:addth3} gebraucht, dabei ist \(A = \omega_c t\) und \(B = n \omega_m t \), 
+somit wird daraus:
+\begin{align*}
+    s(t)
+    &=
+    \sum_{n=0}^\infty J_{-n}(\beta) \{ \underbrace{\cos((\omega_c - n\omega_m) t)} \,-\, \cos((\omega_c + n\omega_m) t) \}
+    \\
+    &=
+   \sum_{n=- \infty}^{0} J_{n}(\beta)  \overbrace{\cos((\omega_c + n \omega_m) t)}
+   \,-\, \sum_{n=0}^\infty J_{-n}(\beta) \cos((\omega_c + n\omega_m) t)
+\end{align*}
+Auch hier wurde wieder eine zweite Summe \(\sum_{-\infty}^{-1}\) gebraucht um das Minus zu einem Plus zu wandeln.
+Wenn \(n = 0 \) ist der Minuend gleich dem Subtrahend und somit dieser Teil \(=0\), das bedeutet \(n\) ended bei \(-1\) und started bei \(1\).
+\begin{align*}
+    s(t)
+    &=
+    \sum_{n=- \infty}^{-1} J_{n}(\beta)  \cos((\omega_c + n \omega_m) t)
+    \underbrace{\,-\, \sum_{n=1}^\infty J_{-n}(\beta)} \cos((\omega_c + n\omega_m) t)
+\end{align*}
+Um aus diesem Subtrahend eine Addition zu kreiernen, wird die Besselindentität \eqref{fm:eq:besselid3} gebraucht, 
+jedoch so \(-1 \cdot J_{-n}(\beta) = J_n(\beta)\) und daraus wird dann:
+\begin{align*}
+    s(t)
+    &=
+    \sum_{n=- \infty}^{-1} J_{n}(\beta)  \cos((\omega_c + n \omega_m) t)
+    \,+\, \sum_{n=1}^\infty J_{n}(\beta) \cos((\omega_c + n\omega_m) t)
+\end{align*}
+Da \(n\) immer ungerade ist und \(0\) nicht zu den ungeraden zahlen zählt, kann man dies so vereinfacht 
 \[
-     \sum_{n\, \text{ungerade}} -1 \cdot J_{n}(\beta) \cos((\omega_c + n\omega_m) t).
-     \label{fm:eq:ungerade}
+    s(t)
+    =
+    \sum_{n\, \text{ungerade}} -1 \cdot J_{n}(\beta) \cos((\omega_c + n\omega_m) t).
+    \label{fm:eq:ungerade}
 \]
-Substituiert man nun noch \(n \text{mit} -n \) so fällt das \(-1\) weg.
-
+schreiben.
+%------------------------------------------------------------------------------------------
 \subsubsection{Summe Zusammenführen}
 Beide Teile \eqref{fm:eq:gerade} Gerade 
 \[
     \sum_{n\, \text{gerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t)
-\]und \eqref{fm:eq:ungerade} Ungerade 
+\]
+und \eqref{fm:eq:ungerade} Ungerade 
 \[
     \sum_{n\, \text{ungerade}} J_{n}(\beta) \cos((\omega_c + n\omega_m) t)
 \]
@@ -134,13 +183,12 @@ ergeben zusammen
 \]
 Somit ist \eqref{fm:eq:proof} bewiesen.
 \newpage
-
-%----------------------------------------------------------------------------
+%-----------------------------------------------------------------------------------------
 \subsection{Bessel und Frequenzspektrum}
-Um sich das ganze noch einwenig Bildlicher vorzustellenhier einmal die Besselfunktion \(J_{k}(\beta)\) in geplottet.
+Um sich das ganze noch einwenig Bildlicher vorzustellenhier einmal die Bessel-Funktion \(J_{k}(\beta)\) in geplottet.
 \begin{figure}
 	\centering
-%	\input{./PyPython animation/bessel.pgf}
+	\input{papers/fm/Python animation/bessel.pgf}
 	\caption{Bessle Funktion \(J_{k}(\beta)\)}
 	\label{fig:bessel}
 \end{figure}
@@ -151,7 +199,7 @@ Nun einmal das Modulierte FM signal im Frequenzspektrum mit den einzelen Summen
 TODO
 Hier wird beschrieben wie die Bessel Funktion der FM im Frequenzspektrum hilft, wieso diese gebrauch wird und ihre Vorteile.
 \begin{itemize}
-    \item Zuerest einmal die Herleitung von FM zu der Besselfunktion
+    \item Zuerest einmal die Herleitung von FM zu der Bessel-Funktion
     \item Im Frequenzspektrum darstellen mit Farben, ersichtlich machen. 
     \item Parameter tuing der Trägerfrequenz, Modulierende frequenz und Beta. 
 \end{itemize}
diff --git a/buch/papers/fm/Python animation/Bessel-FM.ipynb b/buch/papers/fm/Python animation/Bessel-FM.ipynb
index 6f099a7..74f1011 100644
--- a/buch/papers/fm/Python animation/Bessel-FM.ipynb
+++ b/buch/papers/fm/Python animation/Bessel-FM.ipynb
@@ -2,7 +2,7 @@
  "cells": [
   {
    "cell_type": "code",
-   "execution_count": 1,
+   "execution_count": 4,
    "metadata": {},
    "outputs": [],
    "source": [
@@ -13,7 +13,7 @@
     "import scipy.fftpack\n",
     "import matplotlib as mpl\n",
     "# Use the pgf backend (must be set before pyplot imported)\n",
-    "#mpl.use('pgf')\n",
+    "mpl.use('pgf')\n",
     "import matplotlib.pyplot as plt\n",
     "from matplotlib.widgets import Slider\n",
     "def fm(beta):\n",
@@ -70,39 +70,26 @@
     "xf = fftfreq(N, 1 / 1000)\n",
     "plt.plot(xf, np.abs(yf_old))\n",
     "#plt.xlim(-150, 150)\n",
-    "plt.show()"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 118,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "image/png": 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",
-      "text/plain": [
-       "<Figure size 432x288 with 1 Axes>"
-      ]
-     },
-     "metadata": {
-      "needs_background": "light"
-     },
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
+    "plt.show()\n",
+    "\n",
     "fm(1)"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": 29,
+   "execution_count": 5,
    "metadata": {},
    "outputs": [
     {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.7651976865579666\n"
+     ]
+    },
+    {
      "data": {
-      "image/png": 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",
+      "image/png": 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",
       "text/plain": [
        "<Figure size 432x288 with 1 Axes>"
       ]
@@ -111,13 +98,6 @@
       "needs_background": "light"
      },
      "output_type": "display_data"
-    },
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "0.7651976865579666\n"
-     ]
     }
    ],
    "source": [
@@ -133,8 +113,8 @@
     "plt.xlabel(' $ \\\\beta $ ')\n",
     "plt.plot(x, y)\n",
     "plt.legend()\n",
-    "plt.show()\n",
-    "#plt.savefig('bessel.pgf', format='pgf')\n",
+    "#plt.show()\n",
+    "plt.savefig('bessel.pgf', format='pgf')\n",
     "print(sc.jv(0,1))"
    ]
   },
diff --git a/buch/papers/fm/Python animation/bessel.pgf b/buch/papers/fm/Python animation/bessel.pgf
new file mode 100644
index 0000000..cc7af1e
--- /dev/null
+++ b/buch/papers/fm/Python animation/bessel.pgf
@@ -0,0 +1,2057 @@
+%% Creator: Matplotlib, PGF backend
+%%
+%% To include the figure in your LaTeX document, write
+%%   \input{<filename>.pgf}
+%%
+%% Make sure the required packages are loaded in your preamble
+%%   \usepackage{pgf}
+%%
+%% Also ensure that all the required font packages are loaded; for instance,
+%% the lmodern package is sometimes necessary when using math font.
+%%   \usepackage{lmodern}
+%%
+%% Figures using additional raster images can only be included by \input if
+%% they are in the same directory as the main LaTeX file. For loading figures
+%% from other directories you can use the `import` package
+%%   \usepackage{import}
+%%
+%% and then include the figures with
+%%   \import{<path to file>}{<filename>.pgf}
+%%
+%% Matplotlib used the following preamble
+%%   \usepackage{fontspec}
+%%   \setmainfont{DejaVuSerif.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}]
+%%   \setsansfont{DejaVuSans.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}]
+%%   \setmonofont{DejaVuSansMono.ttf}[Path=\detokenize{/home/joshua/.local/lib/python3.8/site-packages/matplotlib/mpl-data/fonts/ttf/}]
+%%
+\begingroup%
+\makeatletter%
+\begin{pgfpicture}%
+\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{6.000000in}{4.000000in}}%
+\pgfusepath{use as bounding box, clip}%
+\begin{pgfscope}%
+\pgfsetbuttcap%
+\pgfsetmiterjoin%
+\pgfsetlinewidth{0.000000pt}%
+\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}%
+\pgfsetstrokecolor{currentstroke}%
+\pgfsetstrokeopacity{0.000000}%
+\pgfsetdash{}{0pt}%
+\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}%
+\pgfpathlineto{\pgfqpoint{6.000000in}{0.000000in}}%
+\pgfpathlineto{\pgfqpoint{6.000000in}{4.000000in}}%
+\pgfpathlineto{\pgfqpoint{0.000000in}{4.000000in}}%
+\pgfpathlineto{\pgfqpoint{0.000000in}{0.000000in}}%
+\pgfpathclose%
+\pgfusepath{}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\pgfsetbuttcap%
+\pgfsetmiterjoin%
+\definecolor{currentfill}{rgb}{1.000000,1.000000,1.000000}%
+\pgfsetfillcolor{currentfill}%
+\pgfsetlinewidth{0.000000pt}%
+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{currentstroke}%
+\pgfsetstrokeopacity{0.000000}%
+\pgfsetdash{}{0pt}%
+\pgfpathmoveto{\pgfqpoint{0.750000in}{0.500000in}}%
+\pgfpathlineto{\pgfqpoint{5.400000in}{0.500000in}}%
+\pgfpathlineto{\pgfqpoint{5.400000in}{3.520000in}}%
+\pgfpathlineto{\pgfqpoint{0.750000in}{3.520000in}}%
+\pgfpathlineto{\pgfqpoint{0.750000in}{0.500000in}}%
+\pgfpathclose%
+\pgfusepath{fill}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\pgfpathrectangle{\pgfqpoint{0.750000in}{0.500000in}}{\pgfqpoint{4.650000in}{3.020000in}}%
+\pgfusepath{clip}%
+\pgfsetrectcap%
+\pgfsetroundjoin%
+\pgfsetlinewidth{0.803000pt}%
+\definecolor{currentstroke}{rgb}{0.690196,0.690196,0.690196}%
+\pgfsetstrokecolor{currentstroke}%
+\pgfsetdash{}{0pt}%
+\pgfpathmoveto{\pgfqpoint{0.750000in}{0.500000in}}%
+\pgfpathlineto{\pgfqpoint{0.750000in}{3.520000in}}%
+\pgfusepath{stroke}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\pgfsetbuttcap%
+\pgfsetroundjoin%
+\definecolor{currentfill}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetfillcolor{currentfill}%
+\pgfsetlinewidth{0.803000pt}%
+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{currentstroke}%
+\pgfsetdash{}{0pt}%
+\pgfsys@defobject{currentmarker}{\pgfqpoint{0.000000in}{-0.048611in}}{\pgfqpoint{0.000000in}{0.000000in}}{%
+\pgfpathmoveto{\pgfqpoint{0.000000in}{0.000000in}}%
+\pgfpathlineto{\pgfqpoint{0.000000in}{-0.048611in}}%
+\pgfusepath{stroke,fill}%
+}%
+\begin{pgfscope}%
+\pgfsys@transformshift{0.750000in}{0.500000in}%
+\pgfsys@useobject{currentmarker}{}%
+\end{pgfscope}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\definecolor{textcolor}{rgb}{0.000000,0.000000,0.000000}%
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diff --git a/buch/papers/fm/packages.tex b/buch/papers/fm/packages.tex
index f0ca8cc..7bbbe35 100644
--- a/buch/papers/fm/packages.tex
+++ b/buch/papers/fm/packages.tex
@@ -8,3 +8,4 @@
 % following example
 %\usepackage{packagename}
 \usepackage{xcolor}
+\usepackage{pgf}
diff --git a/buch/papers/kreismembran/images/Saite.pdf b/buch/papers/kreismembran/images/Saite.pdf
new file mode 100644
index 0000000..0f87c93
--- /dev/null
+++ b/buch/papers/kreismembran/images/Saite.pdf
diff --git a/buch/papers/kreismembran/images/mask_absorber.png b/buch/papers/kreismembran/images/mask_absorber.png
new file mode 100644
index 0000000..5d0cccf
--- /dev/null
+++ b/buch/papers/kreismembran/images/mask_absorber.png
diff --git a/buch/papers/kreismembran/images/mask_disk.png b/buch/papers/kreismembran/images/mask_disk.png
new file mode 100644
index 0000000..4b38163
--- /dev/null
+++ b/buch/papers/kreismembran/images/mask_disk.png
diff --git a/buch/papers/kreismembran/images/sim_1_1.png b/buch/papers/kreismembran/images/sim_1_1.png
new file mode 100644
index 0000000..84c7c1f
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_1_1.png
diff --git a/buch/papers/kreismembran/images/sim_1_2.png b/buch/papers/kreismembran/images/sim_1_2.png
new file mode 100644
index 0000000..ac6312a
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_1_2.png
diff --git a/buch/papers/kreismembran/images/sim_1_3.png b/buch/papers/kreismembran/images/sim_1_3.png
new file mode 100644
index 0000000..9388074
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_1_3.png
diff --git a/buch/papers/kreismembran/images/sim_1_4.png b/buch/papers/kreismembran/images/sim_1_4.png
new file mode 100644
index 0000000..e25b4a0
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_1_4.png
diff --git a/buch/papers/kreismembran/images/sim_1_5.png b/buch/papers/kreismembran/images/sim_1_5.png
new file mode 100644
index 0000000..638ec92
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_1_5.png
diff --git a/buch/papers/kreismembran/images/sim_1_6.png b/buch/papers/kreismembran/images/sim_1_6.png
new file mode 100644
index 0000000..7678da5
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_1_6.png
diff --git a/buch/papers/kreismembran/images/sim_2_1.png b/buch/papers/kreismembran/images/sim_2_1.png
new file mode 100644
index 0000000..c3c7a03
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_2_1.png
diff --git a/buch/papers/kreismembran/images/sim_2_2.png b/buch/papers/kreismembran/images/sim_2_2.png
new file mode 100644
index 0000000..91f3d41
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_2_2.png
diff --git a/buch/papers/kreismembran/images/sim_2_3.png b/buch/papers/kreismembran/images/sim_2_3.png
new file mode 100644
index 0000000..e04475b
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_2_3.png
diff --git a/buch/papers/kreismembran/images/sim_2_4.png b/buch/papers/kreismembran/images/sim_2_4.png
new file mode 100644
index 0000000..5b203c6
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_2_4.png
diff --git a/buch/papers/kreismembran/images/sim_2_5.png b/buch/papers/kreismembran/images/sim_2_5.png
new file mode 100644
index 0000000..ec76085
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_2_5.png
diff --git a/buch/papers/kreismembran/images/sim_2_6.png b/buch/papers/kreismembran/images/sim_2_6.png
new file mode 100644
index 0000000..9c475eb
--- /dev/null
+++ b/buch/papers/kreismembran/images/sim_2_6.png
diff --git a/buch/papers/kreismembran/references.bib b/buch/papers/kreismembran/references.bib
index acf8f90..3d9d0c1 100644
--- a/buch/papers/kreismembran/references.bib
+++ b/buch/papers/kreismembran/references.bib
@@ -52,6 +52,14 @@
         url = {https://doi.org/10.1016/j.acha.2017.11.004}
 }
 
+@book{kreismembran:Digital_Image_processing,
+	edition = {Fourth Edition},
+	title = {Digital Image Processing},
+	publisher = {Pearson},
+	author = {Rafael C. Gozales and Richard E. Woods},
+	date = {2018},
+}
+
 @book{lokenath_debnath_integral_2015,
 	edition = {Third Edition},
 	title = {Integral Tansforms and Their Applications},
diff --git a/buch/papers/kreismembran/teil0.tex b/buch/papers/kreismembran/teil0.tex
index bb8188d..6f55358 100644
--- a/buch/papers/kreismembran/teil0.tex
+++ b/buch/papers/kreismembran/teil0.tex
@@ -5,51 +5,57 @@
 %
 \section{Einleitung\label{kreismembran:section:teil0}}
 \rhead{Membran}
-Eine Membran oder selten ein Schwingblatt ist laut Duden \cite{kreismembran:Duden:Membran} ein "dünnes Blättchen aus Metall, Papier o. Ä., das durch seine Schwingungsfähigkeit geeignet ist, Schallwellen zu übertragen ...". 
-Ein dünnes Blättchen aus Metall zeig jedoch nicht die selben dynamischen Eigenschaften auf, wie ein gespanntes Stück Papier. 
+Eine Membran oder selten ein Schwingblatt ist laut Duden \cite{kreismembran:Duden:Membran} ein ``dünnes Blättchen aus Metall, Papier o. Ä., das durch seine Schwingungsfähigkeit geeignet ist, Schallwellen zu übertragen ...''. 
+Ein dünnes Blättchen aus Metall zeig jedoch nicht die selben dynamischen Eigenschaften wie ein gespanntes Stück Papier. 
 Beschreibt man das dynamische Verhalten, muss zwischen einer dünnen Platte und einer Membrane unterschieden werden \cite{kreismembran:membrane_vs_thin_plate}. 
 Eine dünne Platte zum Beispiel aus Metall, wirkt selbst entgegen ihrer Deformation sobald sie gekrümmt wird. 
-Eine Membran auf der anderen Seite besteht aus einem Material welches sich ohne Kraftaufwand verbiegen lässt wie zum Beispiel Papier. 
-Bevor Papier als schwingende Membran betrachtet werden kann wird jedoch noch eine Spannung $ T $ benötigt welche das Material daran hindert aus der Ruhelage gebracht zu werden. 
+Eine Membran auf der anderen Seite besteht aus einem Material, welches sich ohne Kraftaufwand verbiegen lässt wie zum Beispiel Papier. 
+Bevor Papier als schwingende Membran betrachtet werden kann, wird jedoch noch eine Spannung $ T $ benötigt welche das Material daran hindert aus der Ruhelage gebracht zu werden. 
 
 Ein geläufiges Beispiel einer Kreismembran ist eine runde Trommel. 
-Sie besteht herkömmlicher weise aus einem Leder (Fell), welches auf einen offenen Zylinder (Zargen) aufgespannt wird. 
+Sie besteht herkömmlicherweise aus einem Leder (Fell), welches auf einen offenen Zylinder (Zargen) aufgespannt wird. 
 Das Leder alleine erzeugt nach einem Aufschlag keine hörbaren Schwingungen. 
-Sobald das Fell jedoch über den Zargen gespannt wird, kann das Fell auf verschiedensten weisen weiter schwingen, was für den Klang der Trommel verantwortlich ist. 
-Wie genau diese Schwingungen untersucht werden können wird in der Folgenden Arbeit Diskutiert.
-
+Sobald das Fell jedoch über den Zargen gespannt wird, kann das Fell auf verschiedensten Weisen weiter schwingen, was für den Klang der Trommel verantwortlich ist. 
+Wie genau diese Schwingungen untersucht werden können wird in der folgenden Arbeit diskutiert.
+	
 
-\paragraph{Annahmen} 
+\subsection{Annahmen} \label{kreimembran:annahmen}
 Um die Wellengleichung herzuleiten \cite{kreismembran:wellengleichung_herleitung}, muss ein Modell einer Membran definiert werden. 
-Das untersuchte Modell einer Membrane Erfüllt folgende Eigenschaften:
-\begin{enumerate}[i]
+Das untersuchte Modell erfüllt folgende Eigenschaften:
+\begin{enumerate}[i)]
 	\item Die Membran ist homogen. 
 	Dies bedeutet, dass die Membran über die ganze Fläche die selbe Dichte $ \rho $  und Elastizität hat. 
 	Durch die konstante Elastizität ist die ganze Membran unter gleichmässiger Spannung $ T $.
 	\item Die Membran ist perfekt flexibel. 
-	Daraus folgt, dass die Membran ohne Kraftaufwand verbogen werden kann. 
-	Die Membran ist dadurch nicht allein stehend schwing-fähig, hierzu muss sie gespannt werden mit einer Kraft $ T $.
-	\item Die Membran kann sich nur in Richtung ihrer Normalen in kleinem Ausmass Auslenken.
-	Auslenkungen in der ebene der Membran sind nicht möglich.
+	Damit ist gemeint, dass die Membran ohne Kraftaufwand verbogen werden kann. 
+	Die Membran ist dadurch nicht allein stehend schwingfähig, hierzu muss sie gespannt werden mit einer Kraft $ T $.
+	\item Die Membran kann sich nur in Richtung ihrer Normalen in kleinem Ausmass auslenken.
+	Auslenkungen in der Ebene der Membran sind nicht möglich.
 	\item Die Membran erfährt keine Art von Dämpfung. 
 	Die Membran wird also nicht durch ihr umliegendes Medium abgebremst noch erfährt sie Wärmeverluste durch Deformation.
-	Die resultierende Schwingung wird daher nicht gedämpft sein.
 	
 \end{enumerate}
 
 \subsection{Wellengleichung} Um die Wellengleichung einer Membran herzuleiten wird vorerst eine schwingende Saite betrachtet.
-Es lohnt sich das Verhalten einer Saite zu beschreiben da eine Saite das selbe Verhalten wie eine Membran aufweist mit dem Unterschied einer fehlenden Dimension.
+Es lohnt sich das Verhalten einer Saite zu beschreiben, da eine Saite das selbe Verhalten wie eine Membran aufweist mit dem Unterschied einer fehlenden Dimension.
 Die Verbindung zwischen Membran und Saite ist intuitiv ersichtlich, stellt man sich einen Querschnitt einer Trommel vor.
-%Wie analog zur Membran kann eine Saite erst unter Spannung schwingen. 
+\begin{figure}
+	
+	\begin{center}		
+		\includegraphics[width=5cm,angle=-90]{papers/kreismembran/images/Saite.pdf}
+		\caption{Infinitesimales Stück einer Saite}
+		\label{kreismembran:im:Saite}
+	\end{center}	
+\end{figure}
 
-Abbildung \ref{TODO} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert.
-Wie für die Membran ist die Annahme iii gültig, keine Bewegung in die Richtung $ \hat{x} $.
+Abbildung \ref{kreismembran:im:Saite} ist ein infinitesimales Stück einer Saite mit Länge $ dx $ skizziert.
+Wie für die Membran ist die Annahme iii) gültig, keine Bewegung in die Richtung $ \hat{x} $.
 Um dies zu erfüllen muss der Punkt $ P_1 $ gleich stark in Richtung $ -\hat{x} $ gezogen werden wie der Punkt $ P_2 $ in Richtung $ \hat{x} $ gezogen wird. Ist $ T_1 $ die Kraft welche mit Winkel $ \alpha $ auf Punkt $ P_1 $ wirkt sowie $ T_2 $ und $ \beta$ das analoge für Punkt $ P_2 $ ist, so können die Kräfte 
 \begin{equation}\label{kreismembran:eq:no_translation}
 	T_1 \cos \alpha = T_2 \cos \beta = T
 \end{equation}
 gleichgesetzt werden. 
-Das dynamische verhalten der senkrechten Auslenkung $ u(x,t) $ muss das newtonsche Gesetz 
+Das dynamische Verhalten der senkrechten Auslenkung $ u(x,t) $ muss das newtonsche Gesetz 
 \begin{equation*}
 	\sum F = m \cdot a
 \end{equation*} 
@@ -69,14 +75,18 @@ geschrieben werden.
 Der $ \tan \alpha $ entspricht der örtlichen Ableitung von $ u(x,t) $ an der Stelle $ x_0 $ und analog der $ \tan \beta $ für die Stelle $ x_0 + dx $.
 Die Gleichung wird dadurch zu
 \begin{equation*}
-	\frac{\partial u}{\partial x} \big\vert_{x_0 + dx} - \frac{\partial u}{\partial x} \big\vert_{x_0} = \frac{\rho dx}{T} \frac{\partial^2 u}{\partial t^2}.
+	\frac{\partial u}{\partial x} \bigg|_{x_0 + dx} - \frac{\partial u}{\partial x} \bigg|_{x_0} = \frac{\rho dx}{T} \frac{\partial^2 u}{\partial t^2}.
 \end{equation*} 
 Durch die Division mit $ dx $ entsteht 
 \begin{equation*}
-	\frac{1}{dx} \bigg[\frac{\partial u}{\partial x} \big\vert_{x_0 + dx} - \frac{\partial u}{\partial x} \big\vert_{x_0}\bigg] = \frac{\rho}{T}\frac{\partial^2 u}{\partial t^2}.
+	\frac{1}{dx} \left[\frac{\partial u}{\partial x} \bigg|_{x_0 + dx} - \frac{\partial u}{\partial x} \bigg|_{x_0}\right] = \frac{\rho}{T}\frac{\partial^2 u}{\partial t^2}.
 \end{equation*}
-Auf der Linken Seite der Gleichung wird die Differenz der Steigungen durch die Intervall-Länge geteilt, in anderen Worten die zweite Ableitung von $ u(x,t) $ nach $ x $ berechnet. Der Term $ \frac{\rho}{T} $ wird mit $ c^2 $ ersetzt, da der Bruch für eine gegebene Membran eine positive Konstante sein muss. Somit resultiert die, in der Literatur gebräuchliche Form 
+Auf der linken Seite der Gleichung wird die Differenz der Steigungen durch die Intervalllänge geteilt, in anderen Worten die zweite Ableitung von $ u(x,t) $ nach $ x $ berechnet. 
+Der Term $ \frac{\rho}{T} $ wird durch $ c^2 $ ersetzt, da der Bruch für eine gegebene Membran eine positive Konstante sein muss. 
+Somit resultiert die in der Literatur gebräuchliche Form 
 \begin{equation}
+	\label{kreismembran:Ausgang_DGL}
 	\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} = \Delta u.
 \end{equation}
-In dieser Form ist die Gleichung auch gültig für eine Membran. Für den Fall einer Membran muss lediglich die Ableitung in zwei Dimensionen gerechnet werden.
\ No newline at end of file
+In dieser Form ist die Gleichung auch gültig für eine Membran. 
+Für den Fall einer Membran muss lediglich der Laplace-Operator $\Delta$ in zwei Dimensionen gerechnet werden.
\ No newline at end of file
diff --git a/buch/papers/kreismembran/teil1.tex b/buch/papers/kreismembran/teil1.tex
index 39ca598..f0d478f 100644
--- a/buch/papers/kreismembran/teil1.tex
+++ b/buch/papers/kreismembran/teil1.tex
@@ -7,7 +7,7 @@
 \section{Lösungsmethode 1: Separationsmethode 
 	\label{kreismembran:section:teil1}}
 \rhead{Lösungsmethode 1: Separationsmethode}
-An diesem Punkt bleibt also nur noch die Lösung der partiellen Differentialgleichung. In diesem Kapitel wird sie mit Hilfe der Separationsmethode gelöst.
+An diesem Punkt bleibt also nur noch die Lösung der partiellen Differentialgleichung. In diesem Abschnitt wird sie mit Hilfe der Separationsmethode gelöst.
 
 \subsection{Aufgabestellung\label{sub:aufgabestellung}}
 Wie im vorherigen Abschnitt gezeigt, lautet die partielle Differentialgleichung, die die Schwingungen einer Membran beschreibt:
@@ -30,37 +30,36 @@ Da es sich um eine Kreisscheibe handelt, werden Polarkoordinaten verwendet, so d
 ergibt.
 
 Es wird eine runde elastische Membran berücksichtigt, die das Gebiet $\Omega$ abdeckt und am Rand $\Gamma$ befestigt ist.
-Es wird daher davon ausgegangen, dass die Membran aus einem homogenen Material von vernachlässigbarer Dicke gefertigt ist.
-Die Membran kann verformt werden, aber innere elastische Kräfte wirken den Verformungen entgegen. Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eingespannten homogenen schwingenden Membran.
+Es wirken keine äusseren Kräfte. Es handelt sich somit von einer kreisförmligen eingespannten homogenen schwingenden Membran nach den Annahmen von \ref{kreimembran:annahmen}.
 
 Daher ist die Membranabweichung im Punkt $(r,\varphi)$ $\in$ $\overline{\rm \Omega}$ zum Zeitpunkt $t$:
 \begin{align*}
 	u: \overline{\rm \Omega} \times \mathbb{R}_{\geq 0} &\longrightarrow \mathbb{R}\\
 	(r,\varphi,t) &\longmapsto u(r,\varphi,t)
 \end{align*}
-Da die Membran am Rand befestigt ist, kann es keine Schwingungen geben, so dass die \textit{Dirichlet-Randbedingung} \cite{prof_dr_horst_knorrer_kreisformige_2013}
-\begin{equation*}
-	u\big|_{\Gamma} = 0 \quad \text{für} \quad 0 \leq \varphi \leq 2\pi,\quad t \geq 0
-\end{equation*}
-gilt.
+Um die Vergleichbarkeit der beiden nachfolgend vorgestellten Lösungsverfahren in Abschnitt \ref{kreismembran:vergleich} zu vereinfachen, werden keine Randbedingungen vorgegeben.
 
-Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt:
+Um eine eindeutige Lösung bestimmen zu können, werden die folgenden Anfangsbedingungen festgelegt zur zeit $t = \text{0}$:
 \begin{align*}
 	u(r,\varphi, 0) &= f(r,\varphi)\\
 	u_t(r,\varphi, 0) &= g(r,\varphi).
 \end{align*}
 
 \subsection{Lösung\label{sub:lösung1}}
+Nun wird das in Abschnitt \ref{sub:aufgabestellung} vorgestellte Problem mit Hilfe der varibalen Trennungsmethode gelöst.
 \subsubsection{Ansatz der Separation der Variablen\label{subsub:ansatz_separation}}
-Daher muss an dieser Stelle von einer Separation der Variablen ausgegangen werden:
+Bezug muss an dieser Stelle von einer Separation der Variablen ausgegangen werden:
 \begin{equation*}
 	u(r,\varphi, t) = F(r)G(\varphi)T(t)
 \end{equation*}
-Dank der Randbedingungen kann also gefordert werden, dass $F(R)=0$ ist, und natürlich, dass $G(\varphi)$ $2\pi$ periodisch ist. Eingesetzt in der Differenzialgleichung ergibt sich:
+Dank der Randbedingungen kann gefordert werden, dass $F(R)=0$ ist, und natürlich, dass $G(\varphi)$ $2\pi$ periodisch ist. Eingesetzt in der Differenzialgleichung ergibt sich:
 \begin{equation*}
-	\frac{1}{c^2}\frac{T''(t)}{T(t)}=\frac{F''(r)}{F(r)}+\frac{1}{r}\frac{F'(r)}{F(r)}+\frac{1}{r^2}\frac{G''(\varphi)}{G(\varphi)}.
+	\frac{1}{c^2}\frac{T''(t)}{T(t)}=-\kappa^2=\frac{F''(r)}{F(r)}+\frac{1}{r}\frac{F'(r)}{F(r)}+\frac{1}{r^2}\frac{G''(\varphi)}{G(\varphi)}.
 \end{equation*}
-Da die linke Seite nur von $t$ und die rechte Seite nur von $r$ und $\varphi$ abhängt, müssen sie gleich einer reellen Zahl sein. Aus physikalischen Gründen suchen wir nach Lösungen, die weder exponentiell in der Zeit wachsen noch exponentiell abklingen. Dies bedeutet, dass die Konstante negativ sein muss, also schreibt man $k=-k^2$. Daraus ergeben sich die folgenden zwei Gleichungen:
+Da die linke Seite nur von $t$ und die rechte Seite nur von $r$ und $\varphi$ abhängt, müssen sie gleich einer reellen Zahl sein. 
+Laut Annahme iv) in \ref{kreimembran:annahmen} erfährt die Membran keine Dämpfung.
+Daher werden Lösungen gesucht, die weder exponentiell in der Zeit wachsen noch exponentiell abklingen. 
+Dies bedeutet, dass die Konstante negativ sein muss, also schreibt man $-\kappa^2$. Daraus ergeben sich die folgenden zwei Gleichungen:
 \begin{align*}
 	T''(t) + c^2\kappa^2T(t) &= 0\\
 	r^2\frac{F''(r)}{F(r)} + r \frac{F'(r)}{F(r)} +\kappa^2 r^2 &= - \frac{G''(\varphi)}{G(\varphi)}.
@@ -72,14 +71,14 @@ In der zweiten Gleichung hängt die linke Seite nur von $r$ ab, während die rec
 \end{align*}
 
 \subsubsection{Lösung für $G(\varphi)$\label{subsub:lösung_G}}
-Da für die Zweite Gelichung Lösungen von Schwingungen erwartet werden, für die $G''(\varphi)=-\omega^2 G(\varphi)$ gilt, schreibt die gemeinsame Konstante als $-\nu^2$, was die Formeln später vereinfacht. Also:
+Da für die zweite Gleichung Lösungen von Schwingungen erwartet werden, für die $G''(\varphi)=-\omega^2 G(\varphi)$ gilt, schreibt man die gemeinsame Konstante als $\nu=-\omega^2$, was die Formeln später vereinfacht. Also:
 \begin{equation*}
  G(\varphi) = C_n \cos(\varphi) + D_n \sin(\varphi)
  \label{eq:cos_sin_überlagerung}
 \end{equation*}
 
 \subsubsection{Lösung für $F(r)$\label{subsub:lösung_F}}
-Die Gleichung für $F$ hat die Gestalt
+Die Gleichung für $F$ hat die Gestalt (verweis auf \ref{buch:differentialgleichungen:bessel-operator})
 \begin{align}
 	r^2F''(r) + rF'(r) + (\kappa^2 r^2 - n^2)F(r) = 0 
 	\label{eq:2nd_degree_PDE}
@@ -90,19 +89,9 @@ Wir bereits in Kapitel \ref{buch:differntialgleichungen:section:bessel} gezeigt,
 \end{equation*}
 Lösungen der Besselschen Differenzialgleichung
 \begin{equation*}
-	x^2 y'' + xy' + (x^2 - \nu^2)y = 0
-\end{equation*}
-Die Funktionen $F(r) = J_n(\kappa r)$ lösen also die Differentialgleichung \eqref{eq:2nd_degree_PDE}. Die
-Randbedingung $F(R)=0$ impliziert, dass $\kappa R$ eine Nullstelle der Besselfunktion
-$J_n$ sein muss. Man kann zeigen, dass die Besselfunktionen $J_n, n \geq 0$, alle unendlich
-viele Nullstellen
-\begin{equation*}
-	\alpha_{1n} < \alpha_{2n} < ...
-\end{equation*}
-haben, und dass $\underset{\substack{m\to\infty}}{\text{lim}} \alpha_{mn}=\infty$. Somit ergibt sich, dass $\kappa = \frac{\alpha_{mn}}{R}$ für ein $m\geq 1$, und dass
-\begin{equation*}
-	F(r) = J_n (\kappa_{mn}r) \quad \text{mit} \quad \kappa_{mn}=\frac{\alpha_{mn}}{R}
+	x^2 y'' + xy' + (\kappa^2 - \nu^2)y = 0
 \end{equation*}
+Die Funktionen $F(r) = J_n(\kappa r)$ lösen die Differentialgleichung \eqref{eq:2nd_degree_PDE}.
 
 \subsubsection{Lösung für $T(t)$\label{subsub:lösung_T}}
 Die Differenzialgleichung $T''(t) + c^2\kappa^2T(t) = 0$, wird auf ähnliche Weise gelöst wie $G(\varphi)$. 
@@ -115,7 +104,21 @@ Durch Überlagerung aller Ergebnisse erhält man die Lösung
 \end{align}
 
 Dabei sind $m$ und $n$ ganze Zahlen, wobei $m$ für die Anzahl der Knotenkreise und $n$
-für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membran, in denen es keine Bewegung oder Vibration gibt. Wenn der nicht schwingende Bereich ein Kreis ist, nennt man ihn einen Knotenkreis, und wenn er eine Linie ist, nennt man ihn ebenfalls eine Knotenlinie. $Jn(\kappa_{mn}r)$ ist die Besselfunktion $n$-ter Ordnung, wobei $\kappa mn$ die Wellenzahl und $r$ der Radius ist. $a_{mn}$ und $b_{mn}$ sind die zu bestimmenden Konstanten.
+für die Anzahl der Knotenlinien steht. Es gibt bestimmte Bereiche auf der Membran, in denen es keine Bewegung oder Vibration gibt. Wenn der nicht schwingende Bereich ein Kreis ist, nennt man ihn einen Knotenkreis, und wenn er eine Linie ist, nennt man ihn ebenfalls eine Knotenlinie; siehe Abbildung \ref{buch:pde:kreis:fig:pauke}. $Jn(\kappa_{mn}r)$ ist die Besselfunktion $n$-ter Ordnung, wobei $\kappa mn$ die Wellenzahl und $r$ der Radius ist. $a_{mn}$ und $b_{mn}$ sind die zu bestimmenden Konstanten.
+
+\begin{figure}
+	\centering
+	\includegraphics[width=\textwidth]{chapters/090-pde/bessel/pauke.pdf}
+	%\includegraphics{chapters/090-pde/bessel/pauke.pdf}
+	\caption{Vorzeichen der Lösungsfunktionen und Knotenlinien
+		für verschiedene Werte von $\mu$ und $k$.
+		Die Bereiche, in denen die Lösungsfunktion positiv sind, ist 
+		rot dargestellt, die negativen Bereiche blau.
+		In jeder Darstellung gibt es genau $k+\mu$ Knotenlinien.
+		Die Radien der kreisförmigen Knotenlinien müssen aus den Nullstellen
+		der Besselfunktionen berechnet werden.
+		\label{buch:pde:kreis:fig:pauke}}
+\end{figure}
 
 
-An diesem Punkt stellte sich die Frage, ob es möglich wäre, die partielle Differentialgleichung mit einer anderen Methode als der der Trennung der Variablen zu lösen. Nach einer kurzen Recherche wurde festgestellt, dass die beste Methode die Transformationsmethode ist, genauer gesagt die Anwendung der Hankel-Transformation. Im nächsten Kapitel wird daher diese Integraltransformation vorgestellt und entwickelt, und es wird erläutert, warum sie für diese Art von Problem geeignet ist.
+An diesem Punkt stellte sich die Frage, ob es möglich wäre, die partielle Differentialgleichung mit einer anderen Methode als der der Trennung der Variablen zu lösen. Nach einer kurzen Recherche wurde festgestellt, dass eine weitere Methode die Transformationsmethode ist, genauer gesagt die Anwendung der Hankel-Transformation. Im nächsten Kapitel wird daher diese Integraltransformation vorgestellt und entwickelt, und es wird erläutert, warum sie für diese Art von Problem geeignet ist.
diff --git a/buch/papers/kreismembran/teil2.tex b/buch/papers/kreismembran/teil2.tex
index 6efda49..4fb139c 100644
--- a/buch/papers/kreismembran/teil2.tex
+++ b/buch/papers/kreismembran/teil2.tex
@@ -11,30 +11,30 @@ Er studierte auch Funktionen, jetzt Hankel- oder Bessel- Funktionen genannt, der
 Die Hankel-Transformation, die die Bessel-Funktion enthält, taucht natürlich bei achsensymmetrischen Problemen auf, die in zylindrischen Polarkoordinaten formuliert sind.
 In diesem Abschnitt werden die Theorie der Transformation und einige Eigenschaften der Grundoperationen erläutert.
 
-\subsubsection{Hankel-Transformation \label{subsub:hankel_tansformation}}
+\subsubsection{Definition der Hankel-Transformation \label{subsub:hankel_tansformation}}
 Wir führen die Definition der Hankel-Transformation \cite{lokenath_debnath_integral_2015} aus der zweidimensionalen Fourier-Transformation und ihrer Umkehrung ein, die durch:
 \begin{align}
-	\mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) \; dx dy,\label{equation:fourier_transform}\\
-	\mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r}))}F(k,l) \; dx dy \label{equation:inv_fourier_transform}
+	\mathscr{F}\{f(x,y)\} & = F(k,l)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i( \bm{\kappa}\cdot \mathbf{r})}f(x,y) \; dx \; dy,\label{equation:fourier_transform}\\
+	\mathscr{F}^{-1}\{F(x,y)\} & = f(x,y)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{i(\bm{\kappa}\cdot \mathbf{r})}F(k,l) \; dx \; dy \label{equation:inv_fourier_transform}
 \end{align}
-wo $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Polarkoordinaten sind für diese Art von Problemen am besten geeignet, mit $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$ findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach:
+wo $\mathbf{r}=(x,y)$ und $\bm{\kappa}=(k,l)$. Polarkoordinaten sind für diese Art von Problem am besten geeignet, mit $(x,y)=r(\cos\theta,\sin\theta)$ und $(k,l)=\kappa(\cos\phi,\sin\phi)$ findet man $\bm{\kappa}\cdot\mathbf{r}=\kappa r(\cos(\theta-\phi))$ und danach:
 \begin{align}
 	F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}r \; dr \int_{0}^{2\pi}e^{-ikr\cos(\theta-\phi)}f(r,\theta) \; d\phi.
 	\label{equation:F_ohne_variable_wechsel}
 \end{align}
-Dann wird angenommen dass, $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, und es wird eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \eqref{equation:F_ohne_variable_wechsel} zu reduzieren:
+Dann wird angenommen dass, $f(r,\theta)=e^{in\theta}f(r)$, was keine strenge Einschränkung ist, weil die \textit{Fourier-Theorie} besagt, dass sich jede Funktion durch Überlagerung solcher Terme darstellen lässt. Es wird auch eine Änderung der Variabeln vorgenommen $\theta-\phi=\alpha-\frac{\pi}{2}$, um \eqref{equation:F_ohne_variable_wechsel} zu reduzieren:
 \begin{align}
 	F(k,\phi)=\frac{1}{2\pi}\int_{0}^{\infty}rf(r) \; dr \int_{\phi_{0}}^{2\pi+\phi_{0}}e^{in(\phi-\frac{\pi}{2})+i(n\alpha-kr\sin\alpha)} \; d\alpha,
 	\label{equation:F_ohne_bessel}
 \end{align}
 wo $\phi_{0}=(\frac{\pi}{2}-\phi)$.
 
-Unter Verwendung der Integraldarstellung der Besselfunktion vom Ordnung $n$ \eqref{buch:fourier:eqn:bessel-integraldarstellung}
+Unter Verwendung der Integraldarstellung
 \begin{equation*}
 	J_n(\kappa r)=\frac{1}{2\pi}\int_{\phi_{0}}^{2\pi + \phi_{0}}e^{i(n\alpha-\kappa r \sin \alpha)} \; d\alpha
 	\label{equation:bessel_n_ordnung}
 \end{equation*}
-\eqref{equation:F_ohne_bessel} wird sie zu:
+ der Besselfunktion vom Ordnung $n$ \eqref{buch:fourier:eqn:bessel-integraldarstellung} wird \eqref{equation:F_ohne_bessel} zu:
 \begin{align}
 	F(k,\phi)&=e^{in(\phi-\frac{\pi}{2})}\int_{0}^{\infty}rJ_n(\kappa r) f(r) \; dr  \nonumber \\ 
 	&=e^{in(\phi-\frac{\pi}{2})}\tilde{f}_n(\kappa),
@@ -47,37 +47,28 @@ wo $\tilde{f}_n(\kappa)$ ist die \textit{Hankel-Transformation} von $f(r)$ und i
 \end{align}
 
 \subsubsection{Inverse Hankel-Transformation \label{subsub:inverse_hankel_tansformation}}
-Ähnlich verhält es sich mit der inversen Fourier Transformation in Form von polaren Koordinaten unter der Annahme $f(r,\theta)=e^{in\theta}f(r)$ mit \eqref{equation:F_mit_bessel_step_2}, wird die inverse Fourier Transformation \eqref{equation:inv_fourier_transform}:
+Wie bei der Entwicklung der Hankel-Transformation können auch für die Umkehrformel Analogien zur Fourier-Transformation hergestellt werden. Vergleicht man die beiden Transformationen, so stellt man fest, dass sie sehr ähnlich sind, wenn man den Term $J_n(\kappa r)$ der Hankel-Transformation durch $e^{-i( \bm{\kappa}\cdot \mathbf{r})}$ der Fourier-Transformation ersetzt. Diese beide Funktionen sind orthogonal, und bei orthogonalen Matrizen genügt bekanntlich die Transponierung, um sie zu invertieren. Da das Skalarprodukt der Bessel-Funktionen jedoch nicht dasselbe ist wie das der Exponentialfunktionen, muss man durch $\kappa\; d\kappa$ statt nur durch $d\kappa$ integrieren, um die Umkehrfunktion zu erhalten.
 
-\begin{align*}
-	e^{in\theta}f(r)&=\frac{1}{2\pi}\int_{0}^{\infty}\kappa \; d\kappa \int_{0}^{2\pi}e^{i\kappa r \cos (\theta - \phi)}F(\kappa,\phi) \; d\phi  \\
-	&= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) \; d\kappa \int_{0}^{2\pi}e^{in(\phi - \frac{\pi}{2})- i\kappa r \cos (\theta - \phi)} \; d\phi,
-\end{align*}
-was durch den Wechsel der Variablen $\theta-\phi=-(\alpha+\frac{\pi}{2})$ und $\theta_0=-(\theta+\frac{\pi}{2})$,
-
-\begin{align*}
-	&= \frac{1}{2\pi}\int_{0}^{\infty}\kappa \tilde{f}_n(\kappa) \; d\kappa \int_{\theta_0}^{2\pi+\theta_0}e^{in(\theta + \alpha - i\kappa r \sin\alpha)} \; d\alpha \\
-	&= e^{in\theta}\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) \; d\kappa,
-\end{align*}
-
-von \eqref{equation:bessel_n_ordnung} also ist, die inverse \textit{Hankel-Transformation} so definiert:
+Von \eqref{equation:hankel} also ist, die inverse \textit{Hankel-Transformation} so definiert:
 \begin{align}
 	\mathscr{H}^{-1}_n\{\tilde{f}_n(\kappa)\}=f(r)=\int_{0}^{\infty}\kappa J_n(\kappa r) \tilde{f}_n(\kappa) \; d\kappa.
 	\label{equation:inv_hankel}
 \end{align}
 
-Anstelle von $\tilde{f}_n(\kappa)$, wird häufig für die Hankel-Transformation verwendet, indem die Ordnung angegeben wird.
-\eqref{equation:hankel} und \eqref{equation:inv_hankel} Integralen existieren für eine grosse Klasse von Funktionen, die normalerweise in physikalischen Anwendungen benötigt werden.
-Alternativ kann auch die berühmte Hankel-Transformationsformel verwendet werden, 
+Anstelle von $\tilde{f}_n(\kappa)$, wird häufig einfach $\tilde{f}(\kappa)$ für die Hankel-Transformation verwendet, indem die Ordnung angegeben wird.
+Die Integrale \eqref{equation:hankel} und \eqref{equation:inv_hankel} existieren für bestimmte grosse Klassen von Funktionen, die normalerweise in physikalischen Anwendungen vorkommen.
+
+Alternativ dazu kann die berühmte Hankel-Integralformel 
 
 \begin{align*}
 	f(r) = \int_{0}^{\infty}\kappa J_n(\kappa r) \; d\kappa \int_{0}^{\infty} p J_n(\kappa p)f(p) \; dp,
 	\label{equation:hankel_integral_formula}
 \end{align*}
-um die Hankel-Transformation \eqref{equation:hankel} und ihre Inverse \eqref{equation:inv_hankel} zu definieren.
+verwendet werden, um die Hankel-Transformation \eqref{equation:hankel} und ihre Umkehrung \eqref{equation:inv_hankel} zu definieren.
+
 Insbesondere die Hankel-Transformation der nullten Ordnung ($n=0$) und der ersten Ordnung ($n=1$) sind häufig nützlich, um Lösungen für Probleme mit der Laplace Gleichung in einer achsensymmetrischen zylindrischen Geometrie zu finden.
 
-\subsection{Operative Eigenschaften der Hankel-Transformation\label{sub:op_properties_hankel}}
+\subsection{Operatoreigenschaften der Hankel-Transformation \label{sub:op_properties_hankel}}
 In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation aufgeführt. Der Beweis für ihre Gültigkeit wird jedoch nicht analysiert.
 
 \begin{satz}{Skalierung:}
@@ -88,7 +79,7 @@ In diesem Kapitel werden die operativen Eigenschaften der Hankel-Transformation
 	\end{equation*}
 \end{satz}
 
-\begin{satz}{Persevalsche Relation (Skalarprodukt bleibt erhalten):}
+\begin{satz}{Parsevalsche Relation:}
 Wenn $\tilde{f}(\kappa)=\mathscr{H}_n\{f(r)\}$ und $\tilde{g}(\kappa)=\mathscr{H}_n\{g(r)\}$, dann:
 
 \begin{equation*}
@@ -103,7 +94,7 @@ Wenn $\tilde{f}_n(\kappa)=\mathscr{H}_n\{f(r)\}$, dann:
 	&\mathscr{H}_n\{f'(r)\}=\frac{\kappa}{2n}\left[(n-1)\tilde{f}_{n+1}(\kappa)-(n+1)\tilde{f}_{n-1}(\kappa)\right], \quad n\geq1, \\
 	&\mathscr{H}_1\{f'(r)\}=-\kappa \tilde{f}_0(\kappa),
 \end{align*}
-bereitgestellt dass $[rf(r)]$ verschwindet als $r\to0$ und $r\to\infty$.
+vorausgesetzt dass $[rf(r)]$ verschwindet wenn $r\to0$ und $r\to\infty$.
 \end{satz}
 
 \begin{satz}
diff --git a/buch/papers/kreismembran/teil3.tex b/buch/papers/kreismembran/teil3.tex
index 7d5648a..276f911 100644
--- a/buch/papers/kreismembran/teil3.tex
+++ b/buch/papers/kreismembran/teil3.tex
@@ -40,7 +40,7 @@ bekommt man:
 	\tilde{u}(\kappa,0)=\tilde{f}(\kappa), \quad 
 	\tilde{u}_t(\kappa,0)=\tilde{g}(\kappa).
 \end{equation*}
-Die allgemeine Lösung für diese Transformation lautet, wie in Gleighung \eqref{eq:cos_sin_überlagerung} gesehen, wie folgt
+Die allgemeine Lösung für diese Gleichung lautet, wie in Abschnitt \eqref{eq:cos_sin_überlagerung} gesehen, wie folgt
 
 \begin{equation*}
 	\tilde{u}(\kappa,t)=\tilde{f}(\kappa)\cos(c\kappa t) + \frac{1}{c\kappa}\tilde{g}(\kappa)\sin(c\kappa t).
@@ -60,7 +60,7 @@ Es wird in Folgenden davon ausgegangen, dass sich die Membran verformt und zum Z
 \end{equation*}
 so dass $\tilde{g}(\kappa)\equiv 0$ und
 \begin{equation*}
-	\tilde{f}(\kappa)=Aa\int_{0}^{\infty}r(a^2 + r^2)^{-\frac{1}{2}} J_0 (\kappa r) \; dr=\frac{Aa}{\kappa}e^{-a\kappa}
+	\tilde{f}(\kappa)=Aa\int_{0}^{\infty}r(a^2 + r^2)^{-\frac{1}{2}} J_0 (\kappa r) \; dr=\frac{Aa}{\kappa}e^{-a\kappa}.
 \end{equation*}
 Die formale Lösung  \eqref{eq:formale_lösung} lautet also
 \begin{align*}
@@ -68,7 +68,7 @@ Die formale Lösung  \eqref{eq:formale_lösung} lautet also
 	&=AaRe\left\{r^2+\left(a+ict\right)^2\right\}^{-\frac{1}{2}}
 \end{align*}
 
-Nimmt man jedoch die allgemeine Lösung mit Summationen, 
+Nimmt man jedoch die allgemeine Lösung durch Überlagerung, 
 
 \begin{align}
 	u(r, t) = \displaystyle\sum_{m=1}^{\infty} J_0 (k_{m}r)[a_{m}\cos(c \kappa_{m} t)+b_{m}\sin(c \kappa_{m} t)]
@@ -78,7 +78,8 @@ kann man die Lösungsmethoden 1 und 2 vergleichen.
 
 \subsection{Vergleich der Analytischen Lösungen
 \label{kreismembran:vergleich}}
-Bei der Analyse der Gleichungen \eqref{eq:lösung_endliche_generelle} und \eqref{eq:lösung_unendliche_generelle} fällt sofort auf, dass die Gleichung \eqref{eq:lösung_unendliche_generelle} nicht mehr von $m$ und $n$ abhängt, sondern nur noch von $n$ \cite{nishanth_p_vibrations_2018}. Das macht Sinn, denn $n$ beschreibt die Anzahl der Knotenlinien, und in einer unendlichen Membran gibt es keine. Tatsächlich werden $a_{m0}$, $b_{m0}$ und $\kappa_{m0}$ in $a_m$, $b_m$ bzw. $\kappa_m$ umbenannt. Die beiden Termen $\cos(n\varphi)$ und $\sin(n\varphi)$ verschwinden ebenfalls, da für $n=0$ der $\cos(n\varphi)$ gleich 1 und der $\sin(n \varphi)$ gleich 0 ist.
+Bei der Analyse der Gleichungen \eqref{eq:lösung_endliche_generelle} und \eqref{eq:lösung_unendliche_generelle} fällt sofort auf, dass die Gleichung \eqref{eq:lösung_unendliche_generelle} nicht mehr von $m$ und $n$ abhängt, sondern nur noch von $n$ \cite{nishanth_p_vibrations_2018}. 
+Das macht Sinn, denn $n$ beschreibt die Anzahl der Knotenlinien, welche unter der Annahme einer rotationssymmetrischen Lösung nicht vorhanden sein können. Tatsächlich werden $a_{m0}$, $b_{m0}$ und $\kappa_{m0}$ in $a_m$, $b_m$ bzw. $\kappa_m$ umbenannt. Die beiden Termen $\cos(n\varphi)$ und $\sin(n\varphi)$ verschwinden ebenfalls, da für $n=0$ der $\cos(n\varphi)$ gleich 1 und der $\sin(n \varphi)$ gleich 0 ist.
 Die Funktion hängt also nicht mehr von der Besselfunktionen $n$-ter Ordnung ab, sondern nur von der $0$-ter Ordnung. 
 
 
diff --git a/buch/papers/kreismembran/teil4.tex b/buch/papers/kreismembran/teil4.tex
index c124354..74bb87d 100644
--- a/buch/papers/kreismembran/teil4.tex
+++ b/buch/papers/kreismembran/teil4.tex
@@ -5,12 +5,191 @@
 %
 \section{Lösungsmethode 3: Simulation 
 	\label{kreismembran:section:teil4}}
-\paragraph{TODO Einleitung}
 
 Um numerisch das Verhalten einer Membran zu ermitteln, muss eine numerische Darstellung definiert werden.
-Die Membran wird hier in Form der Matrix $  A $ digitalisiert.
-Jedes Element  $ A_{ij} $ steh für die Auslenkung der Membran $ u(x,y,t) $ an der Stelle $ \{x,y\}=\{i,j\} $.
-Die zeitliche Dimension wird in Form des Array $  X[] $ aus $ v \times A $ Matrizen dargestellt.
-Das Element auf Zeile $ i $, Spalte $ j $ der $ w $-ten Matrix von $ X[] $ also $ X[w]_{ij} $ entspricht der Auslenkung $ u(i,j,w) $.
+Die Membran wird hier in Form der Matrix $  U $ digitalisiert.
+Jedes Element  $ U_{ij} $ steh für die Auslenkung der Membran $ u(x,y,t) $ an der Stelle $ \{x,y\}=\{i,j\} $.
+Zwischen benachbarten Elementen in der Matrix $ U $ liegt immer der Abstand $ dh $, eine Inkrementierung von $ i $ oder $ j $ entspricht somit einem Schritt in Richtung $ x $ oder $ y $ von Länge $ dh $ auf der Membran.
+Die zeitliche Dimension wird in Form des Array $  U[] $ aus $ z \times U $ Matrizen dargestellt, wobei $ z $ der Anzahl Zeitschritten entspricht.
+Das Element auf Zeile $ i $, Spalte $ j $ der $ w $-ten Matrix von $ U[] $ also $ U[w]_{ij} $ entspricht somit der Auslenkung $ u(i,j,w) $.
+Da die DGL von Zweiter Ordnung ist, reicht eine Zustandsvariabel pro Membran-Element nicht aus. 
+Es wird neben der Auslenkung auch die Geschwindigkeit jedes Membran-Elementes benötigt um den Zustand eindeutig zu beschreiben. 
+Dazu existiert neben $ U[] $ ein analoger Array $ V[] $ welcher die Geschwindigkeiten aller Membran-Elementen repräsentiert. 
+$ V[w]_{ij} $ entspricht also $ \dot{u}(i,j,w) $. 
+Der Zustand einer Membran zum Zeitpunkt $ w $ wird mit $ X[w] $ beschrieben, was $ U[w] $ und $ V[w] $ beinhaltet.
+
+\subsection{Propagation}
+Um das Verhalten der Membran zu berechnen, muss aus einem gegebenen Zustand $ X[w] $ der Folgezustand $ X[w+1] $ gerechnet werden können, wobei dazwischen ein Zeitintervall $ dt $ vergeht. 
+Die Berechnung von Folgezuständen kann anschliessend repetiert werden über das zu untersuchende Zeitfenster.
+Die Folgeposition $ U[w+1] $ ergibt sich als 
+\begin{equation}
+	U[w+1] =  U[w] + dt \cdot V[w],
+\end{equation} 
+also die Ausgangslage $ + $ die Strecke welche während des Zeitintervall mit der Geschwindigkeit des Elementes zurückgelegt wurde. 
+Neben der Position muss auch die Geschwindigkeit aktualisiert werden.
+Analog zur Folgeposition wird
+\begin{equation*}
+	V[w+1] = V[w] + dt \cdot \frac{\partial^2u}{\partial t^2}.
+\end{equation*}    
+Die Beschleunigung $ \frac{\partial^2u}{\partial t^2} $  eines Elementes ist durch die DGL \ref{kreismembran:Ausgang_DGL} gegeben als
+\begin{equation*}
+	\frac{\partial^2u}{\partial t^2} = \Delta u \cdot c^2.
+\end{equation*}
+Die Geschwindigkeit des Folgezustandes kann somit mit
+\begin{equation}
+	V[w+1] = V[w] + dt \cdot \Delta_h U \cdot c^2
+\end{equation}
+berechnet werden.
+Während $ c^2 $ lediglich eine Material spezifische Konstante ist, muss noch erläutert werden, wie der diskrete Laplace-Operator für $ \Delta_h u $ definiert ist.
+
+\subsection{Diskreter Laplace-Operator $\Delta_h$}
+Die diskrete Ableitung zweiter Ordnung kann mit Hilfe der Taylor-Reihen-Entwicklung als
+\begin{equation*}
+	\frac{\partial^2f}{\partial x^2} \approx \frac{f(x+dx)-2f(x)+f(x-dx)}{dx^2}
+\end{equation*}
+approximiert werden \cite{kreismembran:Digital_Image_processing}.
+Dank der Linearität der Ableitung kann die Ableitung einer weiteren Dimension addiert werden.
+Daraus folgt für den zweidimensionalen Fall 
+\begin{equation*}
+	\Delta_h u= \frac{u(x+dh,y,t)+u(x,y+dh,t)-4f(x)+u(x-dh,y,t)+u(x,y-dh,t)}{dh^2}.
+\end{equation*}
+Um $ \Delta_h $ auf eine Matrix anwenden zu können wird die Gleichung in Form einer Filtermaske 
+ \begin{equation}
+ 	\Delta_h u= \frac{1}{dh^2}
+ 	\left[ {\begin{array}{ccc}
+ 			0 & 1  & 0\\
+ 			1 & -4 & 1\\
+ 			0 & 1  & 0\\
+ 	\end{array} } \right]
+ \end{equation}
+formuliert.
+Die Filtermaske kann dann auf jedes Element einzeln angewendet werden mit einer Matrizen-Faltung um $ \Delta_h U[] $ zu berechnen.
+
+\subsection{Simulation: Kreisförmige Membran}
+Als Beispiel soll nun eine schwingende kreisförmige Membran simuliert werden.
+\paragraph{Initialisierung}
+Die Anzahl der simulierten Elementen soll $ m \times n $ was dementsprechend die Dimensionen von $ U $ und $ V $ vorgibt.
+Als Anfangsbedingung wird eine Membran gewählt, welche bei $ t=0 $ mit einer Gauss-Kurve ausgelenkt wird.
+Die Membran soll sich zu Beginn nicht bewegen, also wird $ V[0] $ mit Nullen initialisiert.
+Die Auslenkung kann kompakt erreicht werden, wenn $ U[0] $ als Null-Matrix mit einer $ 1 $ in der Mitte initialisiert wird. 
+Diese Matrix wird anschliessend mit einer Filtermaske in Form einer Gauss-Glocke gefaltet.
+Die Faltung mit einer Gauss-Glocke ist in Programmen wie Matlab eine Standartfunktion, da dies einm Tiefpassfilter in der Bildverarbeitung entspricht.
+
+\paragraph{Rand}
+Bislang ist die definierte Matrix rechteckig.
+Um eine kreisförmige Membran zu simulieren muss der Rand angepasst werden.
+Da in den meisten Programme keine Möglichkeit besteht, mit runden Matrizen zu rechnen, wird der Rand in der Berechnung des Folgezustandes implementiert.
+Der Rand bedeutet, das Membran-Elemente auf dem Rand sich nicht Bewegen können.
+Die Position sowie die Geschwindigkeit aller Elemente welche nicht auf der definierten Membran sind müssen zu beliebiger Zeit $0$ entsprechen.
+Hierzu wird eine Maske $M$ erstellt. 
+Diese Maske besteht aus einer binären Matrix von identischer Dimension wie $ U $ und $ V $. 
+Ist in der Matrix $M$ eine $1$ abgebildet so ist an jener stelle ein Element der Membran, ist es eine $0$ so befindet sich dieses Element auf dem Rand oder ausserhalb der Membran.
+In dieser Anwendung ist $M$ eine Matrix mit einem Kreis voller $1$ umgeben von $0$ bis an den Rand der Matrix.
+Die Maske wird angewendet indem das Resultat des nächsten Zustandes noch mit der Maske elementweise multipliziert wird. 
+Der Folgezustand kann also mit den Gleichungen
+\begin{align}
+	\label{kreismembran:eq:folge_U} 
+	U[w+1] &= (U[w] + dt \cdot V[w])*M\\
+	\label{kreismembran:eq:folge_V}
+	V[w+1] &= (V[w] + dt \cdot \Delta_h u \cdot c^2)*M
+\end{align}
+berechnet werden.
+\paragraph{Simulation}
+Mit den gegebenen Gleichungen \ref{kreismembran:eq:folge_U} und \ref{kreismembran:eq:folge_V} das Verhalten der Membran mit einem Loop über das zu untersuchende Zeitintervall berechnet werden. 
+In der Abbildung \ref{kreismembran:im:simres_rund} sind Simulationsresultate zu sehen.
+Die Erste Figur zeigt die Ausgangslage gefolgt von den Auslenkungen nach jeweils $ 50 $ weiteren Iterationsschritten.
+Es ist zu erkennen, wie sich die Störung vom Zentrum an den Rand ausbreitet.
+Erreicht die Störung den Rand wird sie reflektiert und nähert sich dem Zentrum. 
+\begin{figure}
+	
+	\begin{center}
+		
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_1.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_2.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_3.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_4.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_5.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_1_6.png}
+		\caption{Simulations Resultate einer kreisförmigen Membran. Simuliert mit $ 200 \times 200 $ Elementen, dargestellt sind die Auslenkungen nach jeweils $ 50 $ Iterationsschritten.}
+		\label{kreismembran:im:simres_rund}
+		
+	\end{center}	
+\end{figure} 
+\subsection{Simulation: Unendliche Membran}
+
+Um eine unendlich grosse Membran zu simulieren könnte der unpraktische weg gewählt werden die Matrix unendlich gross zu definieren, dies wird jedoch spätestens bei der numerischen Berechnung seine Probleme mit sich bringen.
+Etwas geeigneter ist es die Matrix so gross wie möglich zu definieren wie es die Kapazitäten erlauben.
+Wenn anschliessend nur das Verhalten im Zentrum, bei der Störung beobachtet wird, verhaltet sich die Membran wie eine unendliche. 
+Dies aber nur bis die Störung am Rand reflektiert wird und wieder das innere zu beobachtende Zentrum beeinflusst.
+Soll erst gar keine Reflexion entstehen, muss ein Absorber modelliert werden welcher die Störung möglichst ohne Reflexion aufnimmt.
+
+\paragraph{Absorber}
+Sehr knapp formuliert entstehen Reflexionen, wenn eine Welle von einem Material in ein anderes Material mit unterschiedlichen Eigenschaften eindringen möchte.
+Je unterschiedlicher und abrupter der Übergang zwischen den Materialien umso ausgeprägter die Reflexion.
+In diesem Fall sind die Eigenschaften vorgegeben.
+Im Zentrum soll sich die Membran verhalten, wie von der DGL vorgegeben, am Rand jedoch muss sich jedes Membran-Element in der Ausgangslage befinden. 
+Der Spielraum welcher dem Absorber übrig bleibt ist die Art der Überganges.
+Bei der endlichen kreisförmigen Membran hat die Maske $M$ ein binärer Übergang von Membran zu Rand bezweckt.
+Anstelle dieses abrupten Wechsels wird nun eine Maske definiert, welche graduell von Membran $1$ zu Rand-Element $0$ wechselt.
+Die Elemente werden auf Basis ihres Abstand $r$ zum Zentrum definiert. 
+Der Abstand entspricht 
+\begin{equation*}
+	r(i,j) = \sqrt{|i-\frac{m}{2}|^2+|j-\frac{n}{2}|^2},
+\end{equation*} 
+wobei $ m $ und $n$ den Dimensionen der Matrix entsprechen.
+Für einen Stufenlosen Übergang werden die Elemente der Maske auf 
+
+\begin{align}
+	M_{ij} = \begin{cases} 1-e^{(r(i,j)-b)a} & \text{wenn $x > b$} \\
+		0 & \text{sonst} \end{cases}
+\end{align}
+gesetzt.
+Der Parameter $a > 0$ bestimmt wie Steil der Übergang sein soll, $b$ bestimmt  wie weit weg vom Zentrum sich der Übergang befindet.
+In der Abbildung \ref{kreismembran:im:masks} ist der Unterschied der beiden Masken zu sehen. 
+\begin{figure}
+	
+	\begin{center}
+		
+		\includegraphics[width=0.45\textwidth]{papers/kreismembran/images/mask_disk.png}
+		\includegraphics[width=0.45\textwidth]{papers/kreismembran/images/mask_absorber.png}
+		\caption{Vergleich von Masken: Links Binär für eine endliche Membran, rechts mit Absorber für eine unendliche Membran}
+		\label{kreismembran:im:masks}
+	\end{center}	
+\end{figure} 
+\paragraph{Simulation}
+Bis auf die Absorber-Maske kann nun identisch zur endlichen Membran simuliert werden.
+Auch hier wurde eine Gauss-Glocke als Anfangsbedingung gewählt. 
+Die Simulationsresultate von Abbildung \ref{kreismembran:im:simres_unendlich}
+
+\begin{figure}
+	
+	\begin{center}
+		
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_1.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_2.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_3.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_4.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_5.png}
+		\includegraphics[width=0.32\textwidth]{papers/kreismembran/images/sim_2_6.png}
+		\caption{Simulations Resultate einer unendlichen Membran. Simuliert mit $ 200 \times 200 $ Elementen, dargestellt sind die Auslenkungen nach jeweils $ 50 $ Iterationsschritten.}
+		\label{kreismembran:im:simres_unendlich}
+		
+	\end{center}	
+\end{figure}
+zeigen deutlich wie die Störung vom Zentrum weg verläuft.
+Nähert sich die Störung dem Rand, so wird sie immer stärker abgeschwächt.
+Die Wirkung des Absorber ist an der letzten Figur zu erkennen, in welcher kaum noch Auslenkungen zu sehen sind.
+Dieses Verhalten spricht für den Absorber-Ansatz, es soll jedoch erwähnt sein, dass der Übergangsbereich eine sanft ansteigende Dämpfung in das System bringt.
+Die DGL \ref{kreismembran:Ausgang_DGL} welche simuliert wird geht jedoch von der Annahme \ref{kreimembran:annahmen} iv) aus, dass die Membran keine Art von Dämpfung erfährt.
+
+\section{Schlusswort}
+Auch wenn ein Physikalisches Verhalten bereits durch Annahmen und Annäherungen deutlich vereinfacht wird, bestehen auch dann noch eine Vielzahl von Lösungsansätzen.
+Lösungen einer unendlich grosse Membran scheinen fern der Realität zu sein, doch dies darf es im Sinne der Mathematik.
+Und wer weis, für eine Ameise auf einem Trampolin ist eine unendliche Membran vielleicht eine ganz gute Annäherung. 
+
+
+
+
+
+
 
-\paragraph{title}
\ No newline at end of file
diff --git a/buch/papers/lambertw/Bilder/Abstand.py b/buch/papers/lambertw/Bilder/Abstand.py
new file mode 100644
index 0000000..d787c34
--- /dev/null
+++ b/buch/papers/lambertw/Bilder/Abstand.py
@@ -0,0 +1,18 @@
+# -*- coding: utf-8 -*-
+"""
+Created on Sat Jul 30 23:09:33 2022
+
+@author: yanik
+"""
+
+import numpy as np
+import matplotlib.pyplot as plt
+
+phi = np.pi/2
+t = np.linspace(0, 10, 10**5)
+x0 = 1
+
+def D(t):
+    return np.sqrt(x0**2+2*x0*t*np.cos(phi)+2*t**2-2*t**2*np.sin(phi))
+
+plt.plot(t, D(t))
diff --git a/buch/papers/lambertw/Bilder/Intuition.pdf b/buch/papers/lambertw/Bilder/Intuition.pdf
new file mode 100644
index 0000000..739b02b
--- /dev/null
+++ b/buch/papers/lambertw/Bilder/Intuition.pdf
diff --git a/buch/papers/lambertw/Bilder/Strategie.pdf b/buch/papers/lambertw/Bilder/Strategie.pdf
index 0de3001..b5428f5 100644
--- a/buch/papers/lambertw/Bilder/Strategie.pdf
+++ b/buch/papers/lambertw/Bilder/Strategie.pdf
diff --git a/buch/papers/lambertw/Bilder/Strategie.py b/buch/papers/lambertw/Bilder/Strategie.py
index b9b41bf..975e248 100644
--- a/buch/papers/lambertw/Bilder/Strategie.py
+++ b/buch/papers/lambertw/Bilder/Strategie.py
@@ -34,7 +34,8 @@ ax.quiver(X, Y, U, W, angles='xy', scale_units='xy', scale=1, headwidth=5, headl
 
 ax.plot([V[0], (VZ+V)[0]], [V[1], (VZ+V)[1]], 'k--')
 ax.plot(np.vstack([V, Z])[:, 0], np.vstack([V, Z])[:,1], 'bo', markersize=10)
-
+ax.set_xlabel("x", size=20)
+ax.set_ylabel("y", size=20)
 
 ax.text(2.5, 4.5, "Visierlinie", size=20, rotation=10)
 
@@ -44,9 +45,9 @@ plt.rcParams.update({
     "font.serif": ["New Century Schoolbook"],
 })
 
-ax.text(1.6, 4.3, r"$\vec{v}$", size=30)
-ax.text(0.6, 3.9, r"$V$", size=30, c='b')
-ax.text(5.1, 4.77, r"$Z$", size=30, c='b')
+ax.text(1.6, 4.3, r"$\dot{v}$", size=20)
+ax.text(0.65, 3.9, r"$V$", size=20, c='b')
+ax.text(5.15, 4.85, r"$Z$", size=20, c='b')
 
 
 
diff --git a/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png
index 90758cd..e6e7c1e 100644
--- a/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png
+++ b/buch/papers/lambertw/Bilder/VerfolgungskurveBsp.png
diff --git a/buch/papers/lambertw/Bilder/konvergenz.py b/buch/papers/lambertw/Bilder/konvergenz.py
new file mode 100644
index 0000000..dac99a7
--- /dev/null
+++ b/buch/papers/lambertw/Bilder/konvergenz.py
@@ -0,0 +1,20 @@
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Jul 31 14:34:13 2022
+
+@author: yanik
+"""
+
+import numpy as np
+import matplotlib.pyplot as plt
+
+t = 0
+phi = np.linspace(np.pi/2, 3*np.pi/2, 10**5)
+x0 = 1
+y0 = -2
+
+def D(t):
+    return (x0+t*np.cos(phi))*np.cos(phi)+(y0+t*(np.sin(phi)-1))*(np.sin(phi)-1)/(np.sqrt((x0+t*np.cos(phi))**2+(y0+t*(np.sin(phi)-1))**2))
+
+
+plt.plot(phi, D(t))
\ No newline at end of file
diff --git a/buch/papers/lambertw/Bilder/lambertAbstandBauchgefühl.py b/buch/papers/lambertw/Bilder/lambertAbstandBauchgefühl.py
new file mode 100644
index 0000000..3a90afa
--- /dev/null
+++ b/buch/papers/lambertw/Bilder/lambertAbstandBauchgefühl.py
@@ -0,0 +1,58 @@
+# -*- coding: utf-8 -*-
+"""
+Created on Sun Jul 31 13:32:53 2022
+
+@author: yanik
+"""
+
+import numpy as np
+import matplotlib.pyplot as plt
+import scipy.special as sci
+
+W = sci.lambertw
+
+
+t = np.linspace(0, 1.2, 1000)
+x0 = 1
+y0 = 1
+
+r0 = np.sqrt(x0**2+y0**2)
+chi = (r0+y0)/(r0-y0)
+
+x = x0*np.sqrt(1/chi*W(chi*np.exp(chi-4*t/(r0-y0))))
+eta = (x/x0)**2
+y = 1/4*((y0+r0)*eta+(y0-r0)*np.log(eta)-r0+3*y0)
+
+ymin= (min(y)).real
+xmin = (x[np.where(y == ymin)][0]).real
+
+
+#Verfolger
+plt.plot(x, y, 'r--')
+plt.plot(xmin, ymin, 'bo', markersize=10)
+
+#Ziel
+plt.plot(np.zeros_like(t), t, 'g--')
+plt.plot(0, ymin, 'bo', markersize=10)
+
+
+plt.plot([0, xmin], [ymin, ymin], 'k--')
+#plt.xlim(-0.1, 1)
+#plt.ylim(1, 2)
+plt.ylabel("y")
+plt.xlabel("x")
+plt.grid(True)
+plt.quiver(xmin, ymin, -0.2, 0, scale=1)
+
+plt.text(xmin+0.1, ymin-0.1, "Verfolgungskurve", size=20, rotation=20, color='r')
+plt.text(0.01, 0.02, "Fluchtkurve", size=20, rotation=90, color='g')
+
+plt.rcParams.update({
+    "text.usetex": True,
+    "font.family": "serif",
+    "font.serif": ["New Century Schoolbook"],
+})
+
+plt.text(xmin-0.11, ymin-0.08, r"$\dot{v}$", size=20)
+plt.text(xmin-0.02, ymin+0.05, r"$V$", size=20, c='b')
+plt.text(0.02, ymin+0.05, r"$Z$", size=20, c='b')
\ No newline at end of file
diff --git a/buch/papers/lambertw/teil0.tex b/buch/papers/lambertw/teil0.tex
index 8fa8f9b..6632eca 100644
--- a/buch/papers/lambertw/teil0.tex
+++ b/buch/papers/lambertw/teil0.tex
@@ -7,7 +7,7 @@
 \label{lambertw:section:Was_sind_Verfolgungskurven}}
 \rhead{Was sind Verfolgungskurven?}
 %
-Verfolgungskurven tauchen oft auf bei Fragen wie "Welchen Pfad begeht ein Hund während er einer Katze nachrennt?".
+Verfolgungskurven tauchen oft auf bei Fragen wie ``Welchen Pfad begeht ein Hund während er einer Katze nachrennt?''.
 Ein solches Problem hat im Kern immer ein Verfolger und sein Ziel.
 Der Verfolger verfolgt sein Ziel, das versucht zu entkommen.
 Der Pfad, den der Verfolger während der Verfolgung begeht, wird Verfolgungskurve genannt.
@@ -27,15 +27,15 @@ Daraus folgt, dass eine Strategie zwei dieser drei Parameter festlegen muss, um
 %
 \begin{table}
     \centering
-    \begin{tabular}{|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|}
+    \begin{tabular}{|>{$}l<{$}|>{$}c<{$}|>{$}c<{$}|>{$}c<{$}|}
         \hline
         \text{Strategie}&\text{Geschwindigkeit}&\text{Abstand}&\text{Richtung}\\
         \hline
         \text{Jagd}
-        & \text{konstant} & \text{-} & \text{direkt auf Ziel hinzu}\\
+        & \text{konstant} & \text{-} & \text{direkt auf Ziel zu}\\
         
         \text{Beschattung}
-        & \text{-} & \text{konstant} & \text{direkt auf Ziel hinzu}\\
+        & \text{-} & \text{konstant} & \text{direkt auf Ziel zu}\\
         
         \text{Vorhalt}
         & \text{konstant} & \text{-} & \text{etwas voraus Zielen}\\
@@ -59,7 +59,7 @@ Der Verfolger und sein Ziel werden als Punkte $V$ und $Z$ modelliert.
 In der Abbildung \ref{lambertw:grafic:pursuerDGL2} ist das Problem dargestellt,
 wobei $v$ der Ortsvektor des Verfolgers, $z$ der Ortsvektor des Ziels und $\dot{v}$ der Geschwindigkeitsvektor des Verfolgers ist.
 Der Geschwindigkeitsvektor entspricht dem Richtungsvektors des Verfolgers.
-Die konstante Geschwindigkeit kann man mit der Gleichung
+Die konstante Geschwindigkeit kann man mit
 %
 \begin{equation}
     |\dot{v}|
@@ -67,38 +67,54 @@ Die konstante Geschwindigkeit kann man mit der Gleichung
     \text{,}\quad A\in\mathbb{R}^+
 \end{equation}
 %
-darstellen. Der Geschwindigkeitsvektor kann mit der Gleichung
-%
+darstellen. Der Geschwindigkeitsvektor muss auf das Ziel zeigen, woraus folgt
+\begin{equation}
+    \dot{v}
+    \quad||\quad
+    z-v
+    \text{.}
+\end{equation}
+Um den Richtungsvektor zu konstruieren kann der Einheitsvektor parallel zu $z-v$ um $|\dot{v}|$ gestreckt werden, was zu
 \begin{equation}
-    \frac{z-v}{|z-v|}\cdot|\dot{v}|
+    \dot{v}
     =
+    |\dot{v}|\cdot e_{z-v}
+\end{equation}
+führt. Dies kann noch ausgeschrieben werden zu
+\begin{equation}
     \dot{v}
+    =
+    |\dot{v}|\cdot\frac{z-v}{|z-v|}
+    \text{.}
+    \label{lambertw:richtungsvektor}
 \end{equation}
 %
-beschrieben werden, wenn die Jagdstrategie verwendet wird.
-Die Differenz der Ortsvektoren $v$ und $z$ ist ein Vektor der vom Punkt $V$ auf $Z$ zeigt.
-Da die Länge dieses Vektors beliebig sein kann, wird durch Division durch den Betrag, ein Einheitsvektor erzeugt.
 Aus dem Verfolgungsproblem ist auch ersichtlich, dass die Punkte $V$ und $Z$ nicht am gleichen Ort starten und so eine Division durch Null ausgeschlossen ist.
 Wenn die Punkte $V$ und $Z$ trotzdem am gleichen Ort starten, ist die Lösung trivial.
-%
-Nun wird die Gleichung mit $\dot{v}$ skalar multipliziert, um das Gleichungssystem von zwei auf eine Gleichung zu reduzieren. Somit ergeben sich
+
+Nun wird die Gleichung mit $\dot{v}$ skalar multipliziert, um das Gleichungssystem von zwei auf eine Gleichung zu reduzieren. Somit ergibt sich
 \begin{align}
     \frac{z-v}{|z-v|}\cdot|\dot{v}|\cdot\dot{v}
     &=
     |\dot{v}|^2
-    \\
+\end{align}
+was algebraisch zu
+\begin{align}
     \label{lambertw:pursuerDGL}
     \frac{z-v}{|z-v|}\cdot \frac{\dot{v}}{|\dot{v}|}
     &=
-    1 \text{.}
+    1
 \end{align}
-Die Lösungen dieser Differentialgleichung sind die gesuchten Verfolgungskurven, insofern der Verfolger die Jagdstrategie verwendet.
+umgeformt werden kann.
+Die Lösungen dieser Differentialgleichung sind die gesuchten Verfolgungskurven, sofern der Verfolger die Jagdstrategie verwendet.
 %
 \subsection{Ziel
 \label{lambertw:subsection:Ziel}}
 Als nächstes gehen wir auf das Ziel ein.
 Wie der Verfolger wird auch unser Ziel sich strikt an eine Fluchtstrategie halten, welche von Anfang an bekannt ist.
-Diese Strategie kann als Parameterdarstellung der Position nach der Zeit beschrieben werden.
+Als Strategie eignet sich eine definierte Fluchtkurve oder ähnlich wie beim Verfolger ein Verhalten, das vom Verfolger abhängig ist.
+Ein vom Verfolger abhängiges Verhalten führt zu einem gekoppeltem DGL-System, das schwierig zu lösen sein wird.
+Eine definierte Fluchtkurve kann mit einer Parameterdarstellung der Position nach der Zeit beschrieben werden.
 Zum Beispiel könnte ein Ziel auf einer Geraden flüchten, welches auf einer Ebene mit der Parametrisierung
 %
 \begin{equation}
diff --git a/buch/papers/lambertw/teil1.tex b/buch/papers/lambertw/teil1.tex
index 2733759..e8eca2c 100644
--- a/buch/papers/lambertw/teil1.tex
+++ b/buch/papers/lambertw/teil1.tex
@@ -17,9 +17,10 @@ Nun gilt es zu definieren, wann das Ziel erreicht wird.
 Da sowohl Ziel und Verfolger als Punkte modelliert wurden, gilt das Ziel als erreicht, wenn die Koordinaten des Verfolgers mit denen des Ziels bei einem diskreten Zeitpunkt $t_1$ übereinstimmen.
 Somit gilt es
 %
-\begin{equation*}
+\begin{equation}
     z(t_1)=v(t_1)
-\end{equation*}
+    \label{bedingung_treffer}
+\end{equation}
 %
 zu lösen.
 Die Parametrisierung von $z(t)$ ist im Beispiel definiert als
@@ -29,18 +30,18 @@ Die Parametrisierung von $z(t)$ ist im Beispiel definiert als
     \left( \begin{array}{c} 0 \\ t \end{array} \right)\text{.}
 \end{equation}
 %
-Die Parametrisierung von $v(t)$ ist von den Startbedingungen abhängig. Deshalb wird die obige Bedingung jeweils für die unterschiedlichen Startbedingungen separat analysiert.
+Die Parametrisierung von $v(t)$ ist von den Startbedingungen abhängig. Deshalb wird die Bedingung \eqref{bedingung_treffer} jeweils für die unterschiedlichen Startbedingungen separat analysiert.
 %
-\subsection{Anfangsbedingung im \RN{1}-Quadranten}
+\subsection{Anfangsbedingung im ersten Quadranten}
 %
-Wenn der Verfolger im \RN{1}-Quadranten startet, dann kann $v(t)$ mit den Gleichungen aus \eqref{lambertw:eqFunkXNachT}, welche
-\begin{align*}
+Wenn der Verfolger im ersten Quadranten startet, dann kann $v(t)$ mit den Gleichungen aus \eqref{lambertw:eqFunkXNachT}, welche
+\begin{align}
     x\left(t\right)
     &=
     x_0\cdot\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right) \right)} \\
     y(t)
     &=
-    \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(r_0-y_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\\
+    \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\\
     \chi
     &=
     \frac{r_0+y_0}{r_0-y_0}, \quad
@@ -50,9 +51,10 @@ Wenn der Verfolger im \RN{1}-Quadranten startet, dann kann $v(t)$ mit den Gleich
     r_0
     =
     \sqrt{x_0^2+y_0^2}
-\end{align*}
+    \text{.}
+\end{align}
 %
-Der Folger ist durch
+Der Verfolger ist durch
 \begin{equation}
     v(t)
     =
@@ -61,9 +63,9 @@ Der Folger ist durch
 \end{equation}
 %
 parametrisiert, wobei $y(t)$ viel komplexer ist als $x(t)$.
-Daher wird das Problem in zwei einzelne Teilprobleme zerlegt, wodurch die Bedingung der $x$- und $y$-Koordinaten einzeln überprüft werden müssen. Es entstehen daher folgende Bedingungen
+Daher wird das Problem in zwei einzelne Teilprobleme zerlegt, wodurch die Bedingung der $x$- und $y$-Koordinaten einzeln überprüft werden müssen. Es entstehen daher die Bedingungen
 %
-\begin{align*}
+\begin{align}
     0
     &=
     x(t)
@@ -74,34 +76,40 @@ Daher wird das Problem in zwei einzelne Teilprobleme zerlegt, wodurch die Beding
     &=
     y(t)
     =
-    \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(r_0-y_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\text{,}
-\end{align*}
+    \frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right)\text{,}
+\end{align}
 %
-welche Beide gleichzeitig erfüllt sein müssen, damit das Ziel erreicht wurde.
+welche beide gleichzeitig erfüllt sein müssen, damit das Ziel erreicht wurde.
 Zuerst wird die Bedingung der $x$-Koordinate betrachtet.
-Da $x_0 \neq 0$ und $\chi \neq 0$ mit
+Da $x_0 \neq 0$ und $\chi \neq 0$ kann
 \begin{equation}
     0
     =
     x_0\sqrt{\frac{1}{\chi}W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)\right)}
 \end{equation}
-ist diese Bedingung genau dann erfüllt, wenn
+algebraisch zu
 \begin{equation}
     0
     =
     W\left(\chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)\right)
-    \text{.}
 \end{equation}
-%
+umgeformt werden.
 Es ist zu beachten, dass $W(x)$ die Lambert W-Funktion ist, welche im Kapitel  \eqref{buch:section:lambertw} behandelt wurde.
-Diese Gleichung entspricht genau den Nullstellen der Lambert W-Funktion. Da die Lambert W-Funktion genau eine Nullstelle bei
-\begin{equation}
+Diese Gleichung entspricht genau den Nullstellen der Lambert W-Funktion. Mit der einzigen Nullstelle der Lambert W-Funktion bei
+\begin{equation*}
     W(0)=0
+    \text{,}
+\end{equation*}
+kann die Bedingung weiter vereinfacht werden zu
+\begin{equation}
+    0
+    =
+    \chi\cdot \exp\left( \chi-\frac{4t}{r_0-y_0}\right)
+    \text{.}
 \end{equation}
-%
 Da $\chi\neq0$ und die Exponentialfunktion nie null sein kann, ist diese Bedingung unmöglich zu erfüllen.
 Beim Grenzwert für $t\rightarrow\infty$ geht die Exponentialfunktion gegen null.
-Dies nützt nicht viel, da unendlich viel Zeit vergehen müsste damit ein Einholen möglich wäre.
+Dies nützt nicht viel, da unendlich viel Zeit vergehen müsste, damit ein Einholen möglich wäre.
 Somit kann nach den gestellten Bedingungen das Ziel nie erreicht werden.
 %
 %
@@ -136,7 +144,7 @@ Somit kann nach den gestellten Bedingungen das Ziel nie erreicht werden.
 %Somit kann nach den gestellten Bedingungen das Ziel nie erreicht werden.
 %
 \subsection{Anfangsbedingung $y_0<0$}
-Da die Geschwindigkeit des Verfolgers und des Ziels übereinstimmen, kann der Verfolgers niemals das Ziel einholen.
+Da die Geschwindigkeit des Verfolgers und des Ziels übereinstimmen, kann der Verfolger niemals das Ziel einholen.
 Dies kann veranschaulicht werden anhand
 %
 \begin{equation}
@@ -184,7 +192,7 @@ was aufgelöst zu
 führt.
 Somit wird das Ziel immer erreicht bei $t_1$, wenn der Verfolger auf der positiven $y$-Achse startet.
 \subsection{Fazit}
-Durch die Symmetrie der Fluchtkurve an der $y$-Achse führen die Anfangsbedingungen in den Quadranten \RN{1} und \RN{2} zu den gleichen Ergebnissen. Nun ist klar, dass lediglich Anfangspunkte auf der positiven $y$-Achse oder direkt auf dem Ziel dazu führen, dass der Verfolger das Ziel bei $t_1$ einholt.
+Durch die Symmetrie der Fluchtkurve an der $y$-Achse führen die Anfangsbedingungen im ersten und zweiten Quadranten zu den gleichen Ergebnissen. Nun ist klar, dass lediglich Anfangspunkte auf der positiven $y$-Achse oder direkt auf dem Ziel dazu führen, dass der Verfolger das Ziel bei $t_1$ einholt.
 Bei allen anderen Anfangspunkten wird der Verfolger das Ziel nie erreichen.
 Dieses Resultat ist aber eher akademischer Natur, weil der Verfolger und das Ziel als Punkt betrachtet wurden.
 Wobei aber in Realität nicht von Punkten sondern von Objekten mit einer räumlichen Ausdehnung gesprochen werden kann.
@@ -193,20 +201,149 @@ Falls dies stattfinden sollte, wird dies als Treffer interpretiert.
 Mathematisch kann dies mit
 %
 \begin{equation}
-    |v-z|<a_{min} \text{,}\quad a_{min}\in\mathbb{R}^+
+    |v-z|<a_{\text{min}} \text{,}\quad a_{\text{min}}\in\mathbb{R}^+
 \end{equation}
 %
-beschrieben werden, wobei $a_{min}$ dem Trefferradius entspricht.
+beschrieben werden, wobei $a_{\text{min}}$ dem Trefferradius entspricht.
 Durch quadrieren verschwindet die Wurzel des Betrages, womit
 %
 \begin{equation}
-    |v-z|^2<a_{min}^2 \text{,}\quad a_{min}\in \mathbb{R}^+
+    |v-z|^2<a_{\text{min}}^2 \text{,}\quad a_{\text{min}}\in \mathbb{R}^+
+    \label{lambertw:minimumAbstand}
 \end{equation}
 %
 die neue Bedingung ist.
-Da sowohl der Betrag als auch $a_{min}$ grösser null sind, bleibt die Aussage unverändert.
-
-
-
+Da sowohl der Betrag als auch $a_{\text{min}}$ grösser null sind, bleibt die Aussage unverändert.
+%
+\subsection{trügerische Intuition}%verleitende/trügerische/verführerisch
+In der Grafik \ref{lambertw:grafic:intuition} ist eine Mögliche Verfolgungskurve dargestellt, wobei für die Startbedingung der erste-Quadrant verwendet wurde.
+Als erste Intuition für den Punkt bei dem $|v-z|$ minimal ist bietet sich der tiefste Punkt der Verfolgungskurve an, bei dem der y-Anteil des Richtungsvektors null entspricht.
+Es kann argumentiert werden, dass weil die Geschwindigkeiten gleich gross sind und $\dot{v}$ sich aus einem $y$- als auch einem $x$-Anteil zusammensetzt und $\dot{z}$ nur ein $y$-Anteil besitzt, der Abstand nur grösser werden kann, wenn $e_y\cdot z>e_y\cdot v$.
+Aus diesem Argument würde folgen, dass beim tiefsten Punkt der Verfolgungskurve im Beispiel den minimalen Abstand befindet.
+%
+\begin{figure}
+	\centering
+	\includegraphics[scale=0.4]{./papers/lambertw/Bilder/Intuition.pdf}
+	\caption{Intuition}
+	\label{lambertw:grafic:intuition}
+\end{figure}
+%
 
+Dieses Argument kann leicht überprüft werden, indem lokal alle relevanten benachbarten Punkte betrachtet und das Vorzeichen der Änderung des Abstandes überprüft wird.
+Dafür wird ein Ausdruck benötigt, der den Abstand und die benachbarten Punkte beschreibt.
+Der Richtungsvektor wird allgemein mit dem Winkel $\alpha \in[ 0, 2\pi)$
+Die Ortsvektoren der Punkte können wiederum mit
+\begin{align}
+	v
+	&=
+	t\cdot\left(\begin{array}{c} \cos (\alpha) \\ \sin (\alpha) \end{array}\right) +\left(\begin{array}{c} x_0 \\ y_0 \end{array}\right)
+	\\
+	z
+	&=
+	\left(\begin{array}{c} 0 \\ t \end{array}\right)
+\end{align}
+beschrieben werden. Der Verfolger wurde allgemein für jede Richtung $\alpha$ definiert, um alle unmittelbar benachbarten Punkte beschreiben zu können.
+Da der Abstand
+\begin{equation}
+	a
+	=
+	|v-z|
+	\geq
+	0
+\end{equation}
+ist, kann durch quadrieren ohne Informationsverlust die Rechnung vereinfacht werden zu
+\begin{equation}
+	a^2
+	=
+	|v-z|^2
+	=
+	(t\cdot\cos(\alpha)+x_0)^2+t^2(\sin(\alpha)-1)^2
+	\text{.}
+\end{equation}
+Der Abstand im Quadrat abgeleitet nach der Zeit ist
+\begin{equation}
+	\frac{d a^2}{d t}
+	=
+	2(t\cdot\cos (\alpha)+x_0)\cdot\cos(\alpha)(\alpha)+2t(\sin(\alpha)-1)^2
+	\text{.}
+\end{equation}
+Da nur die unmittelbar benachbarten Punkten von Interesse sind, wird die Ableitung für $t=0$ untersucht. Dabei kann die Ableitung in 
+\begin{align}
+	\frac{d a^2}{d t}
+	&=
+	2x_0\cos(\alpha)
+	\\
+	\frac{d a^2}{d t}
+	&<
+	0\Leftrightarrow\alpha\in\left( \frac{\pi}{2}, \frac{3\pi}{2}\right)
+	\\
+	\frac{d a^2}{d t}
+	&>
+	0\Leftrightarrow\alpha\in\left[0, \frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2}, 2\pi\right)
+	\\
+	\frac{d a^2}{d t}
+	&=
+	0\Leftrightarrow\alpha\in\left\{ \frac{\pi}{2}, \frac{3\pi}{2}\right\}
+\end{align}
+unterteilt werden.
+Von Interesse ist lediglich das Intervall $\alpha\in\left( \frac{\pi}{2}, \frac{3\pi}{2}\right)$, da der Verfolger sich stets in die negative $y$-Richtung bewegt.
+In diesem Intervall ist die Ableitung negativ, woraus folgt, dass jeglicher unmittelbar benachbarte Punkt, den der Verfolger als nächstes begehen könnte, stets näher am Ziel ist als zuvor.
+Dies bedeutet, dass der Scheitelpunkt der Verfolgungskurve nie ein lokales Minimum bezüglich des Abstandes sein kann.
+%
+\subsection{Wo ist der Abstand minimal?}
+Damit der Verfolger das Ziel erreicht muss die Bedingung \eqref{lambertw:minimumAbstand} erfüllt sein.
+Somit ist es ausreichend zu zeigen, dass
+\begin{equation}
+    \operatorname{min}(|z-v|)<a_\text{min}
+    \label{lambertw:Bedingung:abstandMinimal}
+\end{equation}
+erfüllt ist.
 
+Für folgende Betrachtung wurde für den Verfolger die Jagdstrategie mit $|\dot{v}|=|\dot{z}|$ gewählt.
+Das Minimum des Abstandes kann mit
+\begin{equation}
+    0=\frac{d|z-v|}{dt}
+\end{equation}
+gefunden werden.
+Mithilfe $(z-v)(z-v)=|z-v|^2$ kann die Gleichung umgeformt werden zu
+\begin{equation}
+    0=\frac{d(\sqrt{(z-v)(z-v)})}{dt}
+    \text{.}
+\end{equation}
+Jetzt kann die Ableitung leicht ausgeführt werden, womit
+\begin{equation}
+    0=(\dot{z}-\dot{v})\frac{z-v}{\sqrt{(z-v)(z-v)}}
+\end{equation}
+entsteht.
+In dieser Gleichung kann $(z-v)(z-v)=|z-v|^2$ nochmals angewendet werden, wodurch die Gleichung zu
+\begin{equation}
+    0=(\dot{z}-\dot{v})\frac{z-v}{|z-v|}
+\end{equation}
+umgeformt werden kann.
+Nun ist die Struktur der Gleichung \eqref{lambertw:richtungsvektor} erkennbar.
+Wird dies ausgenutzt folgt
+\begin{equation}
+    0=(\dot{z}-\dot{v})\frac{\dot{v}}{|\dot{v}|}
+    \text{.}
+\end{equation}
+Durch algebraische Umwandlung kann die Gleichung in die Form
+\begin{equation}
+    \dot{z}\dot{v}=|\dot{v}|^2
+\end{equation}
+gebracht werden.
+Da $|\dot{v}|=|\dot{z}|$ folgt
+\begin{equation}
+    \cos(\alpha)=1
+    \text{,}
+\end{equation}
+wobei $\alpha$ der Winkel zwischen den Richtungsvektoren ist.
+Mit $|\dot{z}|=|\dot{v}|=1$ entsteht
+\begin{equation}
+    \cos(\alpha)=1
+    \text{,}
+\end{equation}
+woraus folgt, dass nur bei $\alpha=0$, wenn $\alpha \in [0,2\pi)$, ein lokales als auch globales Minimum vorhanden sein kann.
+$\alpha=0$ bedeutet, dass $\dot{v}=\dot{z}$ sein muss.
+Da die Richtungsvektoren bei $t\rightarrow\infty$ immer in die gleiche Richtung zeigen ist dort die Bedingung immer erfüllt.
+Dies entspricht gerade dem einen Rand von $t$, der andere Rand bei $t=0$ muss auch auf lokales bzw. globales Minimum untersucht werden.
+Daraus folgt, dass die Bedingung \eqref{lambertw:Bedingung:abstandMinimal} lediglich für den Abstand bei $t=\{0, \infty\}$ überprüft werden muss.
\ No newline at end of file
diff --git a/buch/papers/lambertw/teil4.tex b/buch/papers/lambertw/teil4.tex
index c79aa0c..1053dd1 100644
--- a/buch/papers/lambertw/teil4.tex
+++ b/buch/papers/lambertw/teil4.tex
@@ -6,15 +6,15 @@
 \section{Beispiel einer Verfolgungskurve
 \label{lambertw:section:teil4}}
 \rhead{Beispiel einer Verfolgungskurve}
-In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve mit der Verfolgungsstrategie 1 beschreiben. Dafür werden zuerst Bewegungsraum, Anfangspositionen und Bewegungsverhalten definiert, in einem nächsten Schritt soll eine Differentialgleichung dafür aufgestellt und anschliessend gelöst werden.
+In diesem Abschnitt wird rechnerisch das Beispiel einer Verfolgungskurve mit der Verfolgungsstrategie ``Jagd'' beschreiben. Dafür werden zuerst Bewegungsraum, Anfangspositionen und Bewegungsverhalten definiert, in einem nächsten Schritt soll eine Differentialgleichung dafür aufgestellt und anschliessend gelöst werden.
 
 \subsection{Anfangsbedingungen definieren und einsetzen
 	\label{lambertw:subsection:Anfangsbedingungen}}
-Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(v = 1\), beginnend beim Ursprung des Kartesischen Koordinatensystems. Der Verfolger \(V\) startet auf einem beliebigen Punkt im ersten Quadranten und bewegt sich auch mit konstanter Geschwindigkeit \(|\dot{V}| = 1\) in Richtung Ziel. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden:
+Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter Geschwindigkeit \(|\dot{z}| = 1\), beginnend beim Ursprung des Kartesischen Koordinatensystems. Der Verfolger \(V\) startet auf einem beliebigen Punkt im ersten Quadranten und bewegt sich auch mit konstanter Geschwindigkeit \(|\dot{v}| = 1\) in Richtung Ziel. Diese Anfangspunkte oder Anfangsbedingungen können wie folgt formuliert werden:
 \begin{equation}
 	Z
 	=
-	\left( \begin{array}{c} 0 \\ v \cdot t \end{array} \right)
+	\left( \begin{array}{c} 0 \\ |\dot{z}| \cdot t \end{array} \right)
 	=
 	\left( \begin{array}{c} 0 \\ t \end{array} \right)
 	,\:
@@ -22,13 +22,13 @@ Das zu verfolgende Ziel \(Z\) bewegt sich entlang der \(y\)-Achse mit konstanter
 	=
 	\left( \begin{array}{c} x \\ y \end{array} \right)
 	\:\text{und}\:\:
-	\bigl| \dot{V} \bigl|
+	|\dot{v}|
 	=
 	1.
 	\label{lambertw:Anfangsbed}
 \end{equation}
 Wir haben nun die Anfangsbedingungen definiert, jetzt fehlt nur noch eine DGL, welche die fortlaufende Änderung der Position und Bewegungsrichtung des Verfolgers beschreibt. 
-Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} definiert, und zwar Gleichung \eqref{lambertw:pursuerDGL}. Wenn man die Startpunkte einfügt, ergibt sich folgender Ausdruck:
+Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} definiert, und zwar Gleichung \eqref{lambertw:pursuerDGL}. Wenn man die Startpunkte einfügt, ergibt sich der Ausdruck
 \begin{equation}
 	\frac{\left( \begin{array}{c} 0-x \\ t-y \end{array} \right)}{\sqrt{x^2 + (t-y)^2}}
 	\cdot
@@ -42,37 +42,38 @@ Diese DGL haben wir bereits in Kapitel \ref{lambertw:subsection:Verfolger} defin
 	\label{lambertw:subsection:DGLvereinfach}}
 Nun haben wir eine Gleichung, es stellt sich aber die Frage, ob es überhaupt eine geschlossene Lösung dafür gibt. Eine Funktion welche die Beziehung \(y(x)\) beschreibt oder sogar \(x(t)\) und \(y(t)\) liefert. Zum jetzigen Zeitpunkt mag es nicht trivial scheinen, aber mit den gewählten Anfangsbedingungen \eqref{lambertw:Anfangsbed} ist es möglich eine geschlossene Lösung für die Gleichung \eqref{lambertw:eqMitAnfangsbed} zu finden.
 
-Auf dem Weg dahin muss die definierte DGL zuerst wesentlich vereinfacht werden, sei es mittels algebraischer Umformungen oder mit den Tools aus der Analysis. Da die nächsten Schritte sehr algebralastig sind und sie das Lesen dieses Papers einfach nur mühsam machen würden, werden wir uns hier nur die wesentlichsten Schritte konzentrieren, welche notwendig sind, um den Lösungsweg nachvollziehen zu können.
+Auf dem Weg dahin muss die definierte DGL zuerst wesentlich vereinfacht werden, sei es mittels algebraischer Umformungen oder mit den Tools aus der Analysis. Da die nächsten Schritte sehr algebralastig sind und sie das Lesen dieses Papers träge machen würden, werden wir uns hier nur auf die wesentlichsten Schritte konzentrieren, welche notwendig sind, um den Lösungsweg nachvollziehen zu können.
 
 \subsubsection{Skalarprodukt auflösen
 	\label{lambertw:subsubsection:SkalProdAufl}}
-Zuerst müssen wir den Bruch und das Skalarprodukt in \eqref{lambertw:eqMitAnfangsbed} wegbringen, damit wir eine. Dies führt zu:
+Zuerst müssen wir den Bruch und das Skalarprodukt in \eqref{lambertw:eqMitAnfangsbed} wegbringen, damit wir eine viel handlichere Differentialgleichung erhalten. Dies führt zu
 \begin{equation}
 		-x \cdot \dot{x} + (t-y) \cdot \dot{y}
 		= \sqrt{x^2 + (t-y)^2}.
 		\label{lambertw:eqOhneSkalarprod}
 \end{equation}
-Im letzten Schritt, fällt die Nützlichkeit des Skalarproduktes in der Verfolgungsgleichung \eqref{lambertw:pursuerDGL} markant auf. Anstatt zwei gekoppelte Differentialgleichungen zu erhalten, eine für die \(x\) und die andere für die \(y\)-Komponente, erhält man einen einzigen Ausdruck, was in der Regel mit weniger Lösungsaufwand verbunden ist.
+Im letzten Schritt, fällt die Nützlichkeit des Skalarproduktes in der Verfolgungsgleichung \eqref{lambertw:pursuerDGL} markant auf. Anstatt zwei gekoppelte Differentialgleichungen zu erhalten, eine für die \(x\)- und die andere für die \(y\)-Komponente, erhält man einen einzigen Ausdruck, was in der Regel mit weniger Lösungsaufwand verbunden ist.
 
 \subsubsection{Quadrieren und Gruppieren
 	\label{lambertw:subsubsection:QuadUndGrup}}
-Mit der Quadratwurzel in \ref{lambertw:eqOhneSkalarprod} kann man nichts anfangen, sie steht nur im Weg, also muss man sie loswerden. Wenn man dies macht, kann \eqref{lambertw:eqOhneSkalarprod} auf folgende Form gebracht werden: 
+Mit der Quadratwurzel in \eqref{lambertw:eqOhneSkalarprod} kann man nichts anfangen, sie steht nur im Weg, also muss man sie loswerden. Wenn man dies macht, kann \eqref{lambertw:eqOhneSkalarprod} auf die Form  
 \begin{equation}
 	\left(\dot{x}^2-1\right) \cdot x^2 -2x \left(t-y\right) \dot{x}\dot{y} + \left(\dot{y}^2-1\right) \cdot \left(t-y\right)^2
-	=0.
+	=0
 	\label{lambertw:eqOhneWurzel}
 \end{equation}
+gebracht werden.
 Diese Form mag auf den ersten Blick nicht gerade nützlich sein, aber man kann sie mit einer Substitution weiter vereinfachen.
 
 \subsubsection{Wichtige Substitution
 	\label{lambertw:subsubsection:WichtSubst}}
-Wenn man beachtet, dass die Geschwindigkeit des Verfolgers konstant und gleich 1 ist, dann kann man folgende Gleichung aufstellen: 
+Wenn man beachtet, dass die Geschwindigkeit des Verfolgers konstant und gleich 1 ist, dann ergibt sich die Beziehung  
 \begin{equation}
 	\dot{x}^2 + \dot{y}^2 
 	= 1.
 	\label{lambertw:eqGeschwSubst}
 \end{equation}
-Umformungen der Gleichung \eqref{lambertw:eqGeschwSubst} können in \eqref{lambertw:eqOhneWurzel} erkannt werden. Ersetzt führen sie zu folgendem Ausdruck:
+Umformungen der Gleichung \eqref{lambertw:eqGeschwSubst} können in \eqref{lambertw:eqOhneWurzel} erkannt werden. Wenn man sie ersetzt, erhält man
 \begin{equation}
 		\dot{y}^2 \cdot x^2 +2x \left(t-y\right) \dot{x}\dot{y} + \dot{x}^2 \cdot \left(t-y\right)^2
 		=0.
@@ -82,27 +83,31 @@ Diese unscheinbare Substitution führt dazu, dass weitere Vereinfachungen durchg
 
 \subsubsection{Binom erkennen und vereinfachen
 	\label{lambertw:subsubsection:BinomVereinfach}}
-Versteckt im Ausdruck \eqref{lambertw:eqGeschwSubstituiert} befindet sich die erste binomische Formel, welche zu folgender Gleichung führt:
+Versteckt im Ausdruck \eqref{lambertw:eqGeschwSubstituiert} befindet sich die erste binomische Formel, wobei
 \begin{equation}
 	(x \dot{y} + (t-y) \dot{x})^2
-	= 0.
+	= 0
 	\label{lambertw:eqAlgVerinfacht}
 \end{equation}
-Da der linke Term gleich Null ist, muss auch der Inhalt des Quadrates gleich Null sein, somit folgt eine weitere Vereinfachung, welche zu einer im Vergleich zu \eqref{lambertw:eqOhneSkalarprod} wesentlich einfacheren DGL führt:
+die faktorisierte Darstellung davon ist.
+Da der linke Term gleich Null ist, muss auch der Inhalt des Quadrates gleich Null sein. Es ergibt sich eine weitere Vereinfachung, welche zu der im Vergleich zu \eqref{lambertw:eqOhneSkalarprod} wesentlich einfacheren DGL 
 \begin{equation}
 	x \dot{y} + (t-y) \dot{x}
-	= 0.
+	= 0
 	\label{lambertw:eqGanzVerinfacht}
 \end{equation}
-Kompakt, ohne Wurzelterme und Quadrate, nur elementare Operationen und Ableitungen. Nun stellt sich die Frage wie es weiter gehen soll, bei der Gleichung \eqref{lambertw:eqGanzVerinfacht} scheinen keine weiteren Vereinfachungen möglich zu sein. Wir brauchen einen neuen Ansatz, um unser Ziel einer möglichen Lösung zu verfolgen.
+führt.
+Kompakt, ohne Wurzelterme und Quadrate, nur elementare Operationen und Ableitungen. 
+
+Nun stellt sich die Frage wie es weiter gehen soll, bei der Gleichung \eqref{lambertw:eqGanzVerinfacht} scheinen keine weiteren Vereinfachungen möglich zu sein. Wir brauchen einen neuen Ansatz, um unser Ziel einer möglichen Lösung zu verfolgen.
 
 \subsection{Zeitabhängigkeit loswerden
 	\label{lambertw:subsection:ZeitabhLoswerden}}
-Der nächste logischer Schritt scheint irgendwie die Zeitabhängigkeit in der Gleichung \eqref{lambertw:eqGanzVerinfacht} loszuwerden, aber wieso? Nun, wie am Anfang von Abschnitt \ref{lambertw:subsection:DGLvereinfach} beschrieben, suchen wir eine Lösung der Art \(y(x)\), dies ist natürlich erst möglich wenn wir die Abhängigkeit nach \(t\) eliminieren können.
+Der nächste logische Schritt scheint irgendwie die Zeitabhängigkeit in der Gleichung \eqref{lambertw:eqGanzVerinfacht} loszuwerden, aber wieso? Nun, wie am Anfang von Abschnitt \ref{lambertw:subsection:DGLvereinfach} beschrieben, suchen wir eine Lösung der Art \(y(x)\), dies ist natürlich erst möglich wenn wir die Abhängigkeit nach \(t\) eliminieren können.
 
 \subsubsection{Zeitliche Ableitungen loswerden
 	\label{lambertw:subsubsection:ZeitAbleit}}
-Der erste Schritt auf dem Weg zur Funktion \(y(x)\), ist es die zeitlichen Ableitungen los zu werden, dafür wird \eqref{lambertw:eqGanzVerinfacht} beidseitig mit \(\dot{x}\) dividiert, was erlaubt ist, weil diese Änderung ungleich Null ist:
+Der erste Schritt auf dem Weg zur Funktion \(y(x)\) ist, die zeitlichen Ableitungen los zu werden, dafür wird \eqref{lambertw:eqGanzVerinfacht} beidseitig durch \(\dot{x}\) dividiert, was erlaubt ist, weil diese Änderung ungleich Null ist:
 \begin{equation}
 	x \frac{\dot{y}}{\dot{x}} + (t-y) \frac{\dot{x}}{\dot{x}}
 	= 0.
@@ -126,30 +131,31 @@ Nach dem die Eigenschaft \eqref{lambertw:eqQuotZeitAbleit} in \eqref{lambertw:eq
 
 \subsubsection{Variable \(t\) eliminieren
 	\label{lambertw:subsubsection:ZeitAbleit}}
-Hier wäre es natürlich passend, wenn man die Abhängigkeit nach \(t\) komplett wegbringen könnte. Um dies zu erreichen, muss man auf die Definition der Bogenlänge zurückgreifen. 
-Die Strecke \(s\) entspricht 
+Hier wäre es natürlich passend, wenn man die Abhängigkeit nach \(t\) komplett wegbringen könnte, aber wie?
+Wir wissen, dass sich der Verfolger mit Geschwindigkeit 1 bewegt, also legt er in der Zeit \(t\) die Strecke \(1\cdot t = t\) zurück. Längen und Strecken können auch mit der Bogenlänge repräsentiert werden, somit kann Zeit und zurückgelegte Strecke in der Gleichung  
 \begin{equation}
 	s
 	= 
-	v \cdot t
+	|\dot{v}| \cdot t
 	=
 	1 \cdot t
 	=
 	t
 	=
-	\int_{\displaystyle x_0}^{\displaystyle x_{\text{end}}}\sqrt{1+y^{\prime\, 2}} \: dx.
+	\int_{\displaystyle x_0}^{\displaystyle x_{\text{end}}}\sqrt{1+y^{\prime\, 2}} \: dx
 	\label{lambertw:eqZuBogenlaenge}
 \end{equation}
-
+verbunden werden.
+  
 Nicht gerade auffällig ist die Richtung, in welche hier integriert wird. Wenn der Verfolger sich wie vorgesehen am Anfang im ersten Quadranten befindet, dann muss sich dieser nach links bewegen, was nicht der üblichen Integrationsrichtung entspricht. Um eine Integration wie üblich von links nach rechts ausführen zu können, müssen die Integrationsgenerzen vertauscht werden, was in einem Vorzeichenwechsel resultiert. 
 
-Wenn man nun \eqref{lambertw:eqZuBogenlaenge} in die DGL \eqref{lambertw:DGLmitT} einfügt, dann ergibt sich folgender Ausdruck:
+Wenn man nun \eqref{lambertw:eqZuBogenlaenge} in die DGL \eqref{lambertw:DGLmitT} einfügt, dann ergibt sich der neue Ausdruck
 \begin{equation}
 	x y^{\prime} - \int\sqrt{1+y^{\prime\, 2}} \: dx - y
 	= 0.
 	\label{lambertw:DGLohneT}
 \end{equation}
-Um das Integral los zu werden, leitet man den vorherigen Ausdruck \eqref{lambertw:DGLohneT} nach \(x\) ab und erhaltet folgende DGL zweiter Ordnung \eqref{lambertw:DGLohneInt}:
+Um das Integral los zu werden, leitet man \eqref{lambertw:DGLohneT} nach \(x\) ab und erhält die DGL zweiter Ordnung 
 \begin{align}
 	y^{\prime}+ xy^{\prime\prime} - \sqrt{1+y^{\prime\, 2}} - y^{\prime}
 	&= 0, \\
@@ -157,16 +163,17 @@ Um das Integral los zu werden, leitet man den vorherigen Ausdruck \eqref{lambert
 	&= 0.
 	\label{lambertw:DGLohneInt}
 \end{align}
-Nun sind wir unserem Ziel einen weiteren Schritt näher. Die Gleichung \eqref{lambertw:DGLohneInt} mag auf den ersten Blick nicht gerade einfach sein, aber im Nächsten Abschnitt werden wir sehen, dass sie relativ einfach zu lösen ist.
+Nun sind wir unserem Ziel einen weiteren Schritt näher. Die Gleichung \eqref{lambertw:DGLohneInt} mag auf den ersten Blick nicht gerade einfach sein, aber im nächsten Abschnitt werden wir sehen, dass sie relativ einfach zu lösen ist.
 
 \subsection{Differentialgleichung lösen
 	\label{lambertw:subsection:DGLloes}}
-Die Gleichung \eqref{lambertw:DGLohneInt} ist eine DGL zweiter Ordnung, in der \(y\) nicht vorkommt. Sie kann mittels der Substitution \(y^{\prime} = u\) in eine DGL erster Ordnung umgewandelt werden:
+Die Gleichung \eqref{lambertw:DGLohneInt} ist eine DGL zweiter Ordnung, in der \(y\) nicht vorkommt. Sie kann mittels der Substitution \(y^{\prime} = u\) in die DGL
 \begin{equation}
 	xu^{\prime} - \sqrt{1+u^2}
-	= 0.
+	= 0
 	\label{lambertw:DGLmitU}
 \end{equation}
+erster Ordnung umgewandelt werden.
 Diese Gleichung ist separierbar, was sie viel handlicher macht. In der separierten Form
 \begin{equation}
 	\int{\frac{1}{\sqrt{1+u^2}}\:du} 
@@ -174,7 +181,7 @@ Diese Gleichung ist separierbar, was sie viel handlicher macht. In der separiert
 	\int{\frac{1}{x}\:dx},
 \end{equation}
 lässt sich die Gleichung mittels einer Integrationstabelle sehr rasch lösen. 
-Mit dem Ergebnis: 
+Das Ergebnis ist 
 \begin{align}
 	\operatorname{arsinh}(u)
 	&=
@@ -184,20 +191,23 @@ Mit dem Ergebnis:
 	\operatorname{sinh}(\operatorname{ln}(x) + C).
 	\label{lambertw:loesDGLmitU}
 \end{align}
-Wenn man in \eqref{lambertw:loesDGLmitU} die Substitution rückgängig macht, erhält man folgende DGL erster Ordnung, die bereits separiert ist:
+Wenn man in \eqref{lambertw:loesDGLmitU} die Substitution rückgängig macht, erhält man die DGL 
 \begin{equation}
 	y^{\prime}
 	=
-	\operatorname{sinh}(\operatorname{ln}(x) + C).
+	\operatorname{sinh}(\operatorname{ln}(x) + C)
 	\label{lambertw:loesDGLmitY}
 \end{equation}
-Ersetzt man den \(\operatorname{sinh}\) mit seiner exponentiellen Definition \(\operatorname{sinh}(x)=\frac{1}{2}(e^x-e^{-x})\), so resultiert auf sehr einfache Art folgende Lösung für \eqref{lambertw:loesDGLmitY}:
+erster Ordnung, die bereits separiert ist.
+Ersetzt man den \(\operatorname{sinh}\) durch seine exponentiellen Definition \(\operatorname{sinh}(x)=\frac{1}{2}(e^x-e^{-x})\), so resultiert auf sehr einfache Art die Lösung 
 \begin{equation}
 	y
 	=
-	C_1 + C_2 x^2 - \frac{\operatorname{ln}(x)}{8 \cdot C_2}.
+	C_1 + C_2 x^2 - \frac{\operatorname{ln}(x)}{8 \cdot C_2}
 \end{equation}
-Nun haben wir eine Lösung, aber wie es immer mit Lösungen ist, stellt sich die Frage, ob sie überhaupt plausibel ist. Dieser Frage werden wir im nächsten Abschnitt nachgehen.
+für \eqref{lambertw:loesDGLmitY}.
+
+Nun haben wir eine Lösung, aber wie es immer mit Lösungen ist, stellt sich die Frage, ob sie überhaupt plausibel ist.
 
 \subsection{Lösung analysieren
 	\label{lambertw:subsection:LoesAnalys}}
@@ -210,37 +220,34 @@ Nun haben wir eine Lösung, aber wie es immer mit Lösungen ist, stellt sich die
 	}
 \end{figure}
 
-Das Resultat, wie ersichtlich, ist folgende Funktion \eqref{lambertw:funkLoes} welche mittels Anfangsbedingungen parametrisiert werden kann: 
+Das Resultat, wie ersichtlich, ist die Funktion  
 \begin{equation}
 	{\color{red}{y(x)}}
 	=
-	C_1 + C_2 {\color{darkgreen}{x^2}} {\color{blue}{-}} \frac{\color{blue}{\operatorname{ln}(x)}}{8 \cdot C_2}.
+	C_1 + C_2 {\color{darkgreen}{x^2}} {\color{blue}{-}} \frac{\color{blue}{\operatorname{ln}(x)}}{8 \cdot C_2},
 	\label{lambertw:funkLoes}
 \end{equation}
-Für die Koeffizienten \(C_1\) und \(C_2\) ergibt sich ein Anfangswertproblem, welches für deren Bestimmung gelöst werden muss. Zuerst soll aber eine qualitative Intuition oder Idee für das Aussehen der Funktion \(y(x)\) geschaffen werden:
+für welche die Koeffizienten \(C_1\) und \(C_2\) aus den Anfangsbedingungen bestimmt werden können. Zuerst soll aber eine qualitative Intuition oder Idee für das Aussehen der Funktion \(y(x)\) geschaffen werden:
 \begin{itemize}
 	\item
 	Für grosse \(x\)-Werte, welche in der Regel in der Nähe von \(x_0\) sein sollten, ist der quadratisch Term in der Funktion \eqref{lambertw:funkLoes} dominant. 
 	\item
-	Für immer kleiner werdende \(x\) geht der Verfolger in Richtung \(y\)-Achse, wobei seine Steigung stetig sinkt, was Sinn macht wenn der Verfolgte entlang der \(y\)-Achse steigt. Irgendwann werden Verfolger und Ziel auf gleicher Höhe sein, also gleiche \(y\) aber verschiedene \(x\)-Koordinate besitzen.
+	Für immer kleiner werdende \(x\) geht der Verfolger in Richtung \(y\)-Achse, wobei seine Steigung stetig sinkt, was Sinn macht wenn der Verfolgte entlang der \(y\)-Achse steigt. Irgendwann werden Verfolger und Ziel auf gleicher Höhe sein, also gleiche \(y\)- aber verschiedene \(x\)-Koordinate besitzen.
+	In diesem Punkt findet ein Monotoniewechsel in der Kurve \eqref{lambertw:funkLoes} statt, was zu einem Minimum führt.
 	\item
 	Für \(x\)-Werte in der Nähe von \(0\) ist das asymptotische Verhalten des Logarithmus dominant, dies macht auch Sinn, da sich der Verfolgte auf der \(y\)-Achse bewegt und der Verfolger ihm nachgeht.
-	\item
-	Aufgrund des Monotoniewechsels in der Kurve \eqref{lambertw:funkLoes} muss diese auch ein Minimum aufweisen. Es stellt sich nun die Frage: Wo befindet sich dieser Punkt? 
-	Eine Abschätzung darüber kann getroffen werden und zwar, dass dieser dann entsteht, wenn \(A\) und \(P\) die gleiche \(y\)-Koordinaten besitzen. In diesem Moment ändert die Richtung der \(y\)-Komponente der Geschwindigkeit des Verfolgers, somit auch sein Vorzeichen und dadurch entsteht auch das Minimum.
 \end{itemize}
 Alle diese Eigenschaften stimmen mit dem überein, was man von einer Kurve dieser Art erwarten würde, welche durch die Grafik \ref{lambertw:BildFunkLoes} repräsentiert wurde.
 
 \subsection{Anfangswertproblem 
 	\label{lambertw:subsection:AllgLoes}}
-In diesem Abschnitt soll eine Parameterfunktion hergeleitet werden, bei der jeder beliebige Anfangspunkt im ersten Quadranten eingesetzt werden kann, ausser der Ursprung im Koordinatensystem. Diese Aufgabe erfordert ein Anfangswertproblem.
+In diesem Abschnitt soll eine Parameterfunktion hergeleitet werden, bei der jeder beliebige Anfangspunkt im ersten Quadranten eingesetzt werden kann, ausser der Ursprung im Koordinatensystem. Diese Aufgabe ist ein Anfangswertproblem für \(y(x)\).
 
-Das Lösen des Anfangswertproblems ist ein Problem aus der Algebra, auf welches hier nicht explizit eingegangen wird. Zur Vollständigkeit und Nachvollziehbarkeit, wird aber das Gleichungssystem präsentiert, welches notwendig ist, um das Anfangswertproblem zu lösen.
+Das Lösen des Anfangswertproblems ist ein Problem aus der Analysis, auf welches hier nicht explizit eingegangen wird. Zur Vollständigkeit und Nachvollziehbarkeit, wird aber das Gleichungssystem präsentiert, welches notwendig ist, um das Anfangswertproblem zu lösen.
 
 \subsubsection{Anfangswerte bestimmen
 	\label{lambertw:subsubsection:Anfangswerte}}
-Der erste Schritt auf dem Weg zur gesuchten Parameterfunktion ist, die Anfangswerte \eqref{lambertw:eq1Anfangswert} zu definieren.
-Die Anfangswerte sind:
+Der erste Schritt auf dem Weg zur gesuchten Parameterfunktion ist, die Anfangswerte 
 \begin{equation}
 	y(x)\big \vert_{t=0}
 	=
@@ -255,15 +262,17 @@ und
 	=
 	y^{\prime}(x_0)
 	=
-	\frac{y_0}{x_0}.
+	\frac{y_0}{x_0}
 	\label{lambertw:eq2Anfangswert}
 \end{equation}
+zu definieren.
 Der zweite Anfangswert \eqref{lambertw:eq2Anfangswert} mag nicht grade offensichtlich sein. Die Erklärung dafür ist aber simpel: Der Verfolger wird sich zum Zeitpunkt \(t=0\) in Richtung Koordinatenursprung bewegen wollen, wo sich das Ziel befindet. Somit entsteht das Steigungsdreieck mit \(\Delta x = x_0\) und \(\Delta y = y_0\).
 
 \subsubsection{Gleichungssystem aufstellen und lösen
 	\label{lambertw:subsubsection:GlSys}}
-Wenn man die Anfangswerte \eqref{lambertw:eq1Anfangswert} und \eqref{lambertw:eq2Anfangswert} in die Gleichung \eqref{lambertw:funkLoes} und deren Ableitung \(y^{\prime}(x)\) einsetzt, dann ergibt sich folgendes Gleichungssystem:
+Wenn man die Anfangswerte \eqref{lambertw:eq1Anfangswert} und \eqref{lambertw:eq2Anfangswert} in die Gleichung \eqref{lambertw:funkLoes} und deren Ableitung \(y^{\prime}(x)\) einsetzt, dann ergibt sich das Gleichungssystem
 \begin{subequations}
+	\label{lambertw:eqGleichungssystem}
 	\begin{align}
 		y_0
 		&=
@@ -272,9 +281,8 @@ Wenn man die Anfangswerte \eqref{lambertw:eq1Anfangswert} und \eqref{lambertw:eq
 		&=
 		2 \cdot  C_2 x_0 - \frac{1}{8 \cdot C_2 \cdot x_0}.
 	\end{align}
-	\label{lambertw:eqGleichungssystem}
 \end{subequations}
-Damit die gesuchte Funktion im ersten Quadranten bleibt, werden nur die positiven Lösungen des Gleichungssystems gewählt, welche wie folgt aussehen:
+Damit die gesuchte Funktion im ersten Quadranten bleibt, werden nur die positiven Lösungen  
 \begin{subequations}
 	\begin{align}
 		\label{lambertw:eqKoeff1}
@@ -284,16 +292,17 @@ Damit die gesuchte Funktion im ersten Quadranten bleibt, werden nur die positive
 		\label{lambertw:eqKoeff2}
 		C_2
 		&=
-		\frac{\sqrt{x_0^2 + y_0^2} + y_0}{4x_0^2}.
+		\frac{\sqrt{x_0^2 + y_0^2} + y_0}{4x_0^2}
 	\end{align}
 \end{subequations}
+des Gleichungssystems gewählt.
 \subsubsection{Gesuchte Parameterfunktion aufstellen
 	\label{lambertw:subsubsection:ParamFunk}}
-Wenn man die Koeffizienten \eqref{lambertw:eqKoeff1} und \eqref{lambertw:eqKoeff2} in die Funktion \eqref{lambertw:funkLoes} einsetzt, dann ergibt sich nach dem Vereinfachen die gesuchte Parameterfunktion:
+Wenn man die Koeffizienten \eqref{lambertw:eqKoeff1} und \eqref{lambertw:eqKoeff2} in die Funktion \eqref{lambertw:funkLoes} einsetzt, dann ergibt sich beim Vereinfachen die gesuchte Parameterfunktion
 \begin{equation}
 	y(x)
 	=
-	\frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right).
+	\frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(y_0-r_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right).
 	\label{lambertw:eqAllgLoes}
 \end{equation}
 Damit die Funktion \eqref{lambertw:eqAllgLoes} trotzdem übersichtlich bleibt, wurden Anfangssteigung \(\eta\) und Anfangsentfernung \(r_0\) wie folgt definiert:
@@ -316,27 +325,28 @@ In diesem Abschnitt werden algebraischen Umformungen ein wenig detaillierter als
 
 \subsubsection{Zeitabhängigkeit wiederherstellen
 	\label{lambertw:subsubsection:ZeitabhWiederherst}}
-Der erste Schritt ist es herauszufinden, wie die Zeitabhängigkeit wieder hineingebracht werden kann. Dafür greifen wir auf die letzte Gleichung zu, in welcher \(t\) noch enthalten war, und zwar DGL \eqref{lambertw:DGLmitT}, welche zur Übersichtlichkeit hier nochmals aufgeführt wird:
+Der erste Schritt ist es herauszufinden, wie die Zeitabhängigkeit wieder hineingebracht werden kann. Dafür greifen wir auf die letzte Gleichung zu, in welcher \(t\) noch enthalten war, und zwar DGL 
 \begin{equation}
 	x y^{\prime} + t - y
-	= 0.
+	= 0
 	\label{lambertw:eqDGLmitTnochmals}
 \end{equation}
+aus dem Abschnitt \eqref{lambertw:subsection:ZeitabhLoswerden}, welche zur Übersichtlichkeit hier nochmals aufgeführt wurde.
 Wie in \eqref{lambertw:eqDGLmitTnochmals} zu sehen ist, werden \(y\) und deren Ableitung \(y^{\prime}\) benötigt, diese sind:
 \begin{subequations}
+	\label{lambertw:eqFunkUndAbleit}
 	\begin{align}
+		\label{lambertw:eqFunkUndAbleit1}
 		y
 		&=
-		\frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right), \\
-		\label{lambertw:eqFunkUndAbleit1}
+		\frac{1}{4}\left(\left(y_0+r_0\right)\eta+\left(y_0-r_0\right)\operatorname{ln}\left(\eta\right)-r_0+3y_0\right), \\
 		y^\prime
 		&=
-		\frac{1}{2}\left(\left(y_0+r_0\right)\frac{x}{x_0^2}+\left(r_0-y_0\right)\frac{1}{x}\right).
+		\frac{1}{2}\left(\left(y_0+r_0\right)\frac{x}{x_0^2}+\left(y_0-r_0\right)\frac{1}{x}\right).
 	\end{align}
-	\label{lambertw:eqFunkUndAbleit}
 \end{subequations}
 
-Wenn man diese Gleichungen \ref{lambertw:eqFunkUndAbleit} in die DGL \label{lambertw:eqDGLmitTnochmals} einfügt, vereinfacht und nach \(t\) auflöst, dann ergibt sich folgenden Ausdruck:
+Wenn man diese Gleichungen \eqref{lambertw:eqFunkUndAbleit} in die DGL \eqref{lambertw:eqDGLmitTnochmals} einfügt, vereinfacht und nach \(t\) auflöst, dann ergibt sich der Ausdruck
 \begin{equation}
 	-4t
 	=
@@ -348,17 +358,20 @@ Wenn man diese Gleichungen \ref{lambertw:eqFunkUndAbleit} in die DGL \label{lamb
 	\label{lambertw:subsubsection:UmformBisZumZiel}}
 Mit dem Ausdruck \eqref{lambertw:eqFunkUndAbleitEingefuegt}, welcher Terme mit \(x\) und \(t\) verbindet, kann nun nach der gesuchten Variable \(x\) aufgelöst werden. 
 
-
-In einem nächsten Schritt wird alles mit \(x\) auf die eine Seite gebracht, der Rest auf die andere Seite und anschliessend beidseitig exponentiert, was wie folgt aussieht:
-\begin{align}
+In einem nächsten Schritt wird alles mit \(x\) auf die eine Seite gebracht, der Rest auf die andere Seite und anschliessend beidseitig exponenziert, sodass man 
+\begin{equation}
 	-4t+\left(y_0+r_0\right)
-	&=
-	\left(y_0+r_0\right)\eta+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right), \\
+	=
+	\left(y_0+r_0\right)\eta+\left(r_0-y_0\right)\operatorname{ln}\left(\eta\right)
+\end{equation}
+und anschliessend
+\begin{equation}
 	e^{\displaystyle -4t+\left(y_0+r_0\right)}
-	&=
-	e^{\displaystyle \left(y_0+r_0\right)\eta}\cdot\eta^{\displaystyle \left(r_0-y_0\right)}.
+	=
+	e^{\displaystyle \left(y_0+r_0\right)\eta}\cdot\eta^{\displaystyle \left(r_0-y_0\right)}
 	\label{lambertw:eqMitExp}
-\end{align}
+\end{equation}
+erhält.
 Auf dem rechten Term von \eqref{lambertw:eqMitExp} beginnen wir langsam eine ähnliche Struktur wie \(\eta e^\eta\) zu erkennen, dies schreit nach der Struktur die benötigt wird um \(\eta\) mittels der Lambert-\(W\)-Funktion \(W(x)\) zu erhalten. Dies macht durchaus Sinn, wenn wir die Funktion \(x(t)\) finden wollen und \(W(x)\) die Umkehrfunktion von \(x e^x\) ist. 
 
 Die erste Sache die uns in \eqref{lambertw:eqMitExp} stört ist, dass \(\eta\) als Potenz da steht. Dieses Problem können wir loswerden, indem wir beidseitig mit \(\:\displaystyle \frac{1}{r_0-y_0}\:\) potenzieren:
@@ -368,30 +381,32 @@ Die erste Sache die uns in \eqref{lambertw:eqMitExp} stört ist, dass \(\eta\) a
 	\eta\cdot \operatorname{exp}\left(\displaystyle \frac{y_0+r_0}{r_0-y_0}\eta\right).
 	\label{lambertw:eqOhnePotenz}
 \end{equation}
-Das nächste Problem auf welches wir in \eqref{lambertw:eqOhnePotenz} treffen ist, dass \(\eta\) nicht alleine im Exponent steht. Dies kann elegant mit folgender Substitution gelöst werden:
+Das nächste Problem auf welches wir in \eqref{lambertw:eqOhnePotenz} treffen ist, dass \(\eta\) nicht alleine im Exponent steht. Dies kann elegant mit der Substitution 
 \begin{equation}
 	\chi
 	=
-	\frac{y_0+r_0}{r_0-y_0}.
+	\frac{y_0+r_0}{r_0-y_0}
 	\label{lambertw:eqChiSubst}
 \end{equation}
+gelöst werden.
 Es gäbe natürlich andere Substitutionen wie z.B. 
 \[\displaystyle \chi=\frac{y_0+r_0}{r_0-y_0}\cdot\eta,\] 
-die auf dasselbe Ergebnis führen würden, aber \eqref{lambertw:eqChiSubst} liefert in einem Schritt die kompakteste Lösung. Also fahren wir mit der Substitution \eqref{lambertw:eqChiSubst} weiter, setzen diese in die Gleichung \eqref{lambertw:eqOhnePotenz} ein und multiplizieren beidseitig mit \(\chi\). Daraus erhalten wir folgende Gleichung:
+die auf dasselbe Ergebnis führen würden, aber \eqref{lambertw:eqChiSubst} liefert in einem Schritt die kompakteste Lösung. Also fahren wir mit der Substitution \eqref{lambertw:eqChiSubst} weiter, setzen diese in die Gleichung \eqref{lambertw:eqOhnePotenz} ein und multiplizieren beidseitig mit \(\chi\). Daraus erhalten wir die Gleichung
 \begin{equation}
 	\chi\cdot \operatorname{exp}\left(\displaystyle \chi-\frac{4t}{r_0-y_0}\right)
 	=
 	\chi\eta\cdot e^{\displaystyle \chi\eta}.
 	\label{lambertw:eqNachSubst}
 \end{equation}
-Nun sind wir endlich soweit, dass wir die angedeutete Lambert-\(W\)-Funktion \(W(x)\)einsetzen können. Wenn wir beidseitig \(W(x)\) anwenden, dann erhalten wir folgenden Ausdruck:
+Nun sind wir endlich soweit, dass wir die angedeutete Lambert-\(W\)-Funktion \(W(x)\) einsetzen können. Wenn wir beidseitig \(W(x)\) anwenden, dann erhalten wir den Ausdruck
 \begin{equation}
 	W\left(\chi\cdot \operatorname{exp}\left(\displaystyle \chi-\frac{4t}{r_0-y_0}\right)\right)
 	=
 	\chi\eta.
 \end{equation}
-Nach dem Auflösen nach \(x\) welches in \(\eta\) enthalten ist, erhalten wir die gesuchte \(x(t)\)-Funktion \eqref{lambertw:eqFunkXNachT}. Dieses \(x(t)\) in Kombination mit \eqref{lambertw:eqFunkUndAbleit1} liefert die Position des Verfolgers zu jedem Zeitpunkt. Das Gleichungspaar \eqref{lambertw:eqFunktionenNachT}, besteht aus folgenden Gleichungen:
+Nach dem Auflösen nach \(x\) welches in \(\eta\) enthalten ist, erhalten wir die gesuchte \(x(t)\)-Funktion \eqref{lambertw:eqFunkXNachT}. Dieses \(x(t)\) in Kombination mit \eqref{lambertw:eqFunkUndAbleit1} liefert die Position des Verfolgers zu jedem Zeitpunkt. Das Gleichungspaar besteht also aus den Gleichungen
 \begin{subequations}
+	\label{lambertw:eqFunktionenNachT}
 	\begin{align}
 		\label{lambertw:eqFunkXNachT}
 		x(t)
@@ -402,15 +417,14 @@ Nach dem Auflösen nach \(x\) welches in \(\eta\) enthalten ist, erhalten wir di
 		=
 		y(t)
 		&=
-		\frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(r_0-y_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right).
+		\frac{1}{4}\left(\left(y_0+r_0\right)\left(\frac{x(t)}{x_0}\right)^2+\left(y_0-r_0\right)\operatorname{ln}\left(\left(\frac{x(t)}{x_0}\right)^2\right)-r_0+3y_0\right).
 	\end{align}
-	\label{lambertw:eqFunktionenNachT}
 \end{subequations}
 Nun haben wir unser letztes Ziel erreicht und sind in der Lage eine Verfolgung rechnerisch sowie graphisch zu repräsentieren.
 
 \subsubsection{Hinweise zur Lambert-\(W\)-Funktion
 	\label{lambertw:subsubsection:HinwLambertW}}
-Wir sind aber noch nicht ganz fertig, eine Frage muss noch beantwortet werden. Und zwar wieso, dass man schon bei der Gleichung \eqref{lambertw:eqFunkUndAbleitEingefuegt} weiss, dass die Lambert-\(W\)-Funktion zum Einsatz kommen wird.
+Wir sind aber noch nicht ganz fertig, eine Frage muss noch beantwortet werden. Und zwar wieso, man schon bei der Gleichung \eqref{lambertw:eqFunkUndAbleitEingefuegt} weiss, dass die Lambert-\(W\)-Funktion zum Einsatz kommen wird.
 Nun, der Grund dafür ist die Struktur
 \begin{equation}
 	y
@@ -420,4 +434,4 @@ Nun, der Grund dafür ist die Struktur
 \end{equation}
 bei welcher \(p(x)\) eine beliebige Potenz von \(x\) darstellt. 
 
-Jedes Mal wenn \(x\) gesucht ist und in einer Struktur der Art \eqref{lambertw:eqEinsatzLambW} vorkommt, dann kann mit ein paar Umformungen die Struktur \(f(x)e^{f(x)}\) erzielt werden. Wie bereits in diesem Abschnitt \ref{lambertw:subsection:FunkNachT} gezeigt wurde, kann \(x\) nun mittels der \(W(x)\)-Funktion aufgelöst werden. Erstaunlicherweise ist \eqref{lambertw:eqEinsatzLambW} eine Struktur die oftmals vorkommt, was die Lambert-\(W\)-Funktion so wichtig macht.
\ No newline at end of file
+Jedes Mal wenn \(x\) gesucht ist und in einer Struktur der Art \eqref{lambertw:eqEinsatzLambW} vorkommt, dann kann mit ein paar Umformungen die Struktur \(f(x)e^{f(x)}\) erzielt werden. Wie bereits in diesem Abschnitt \ref{lambertw:subsection:FunkNachT} gezeigt wurde, kann \(x\) nun mittels der \(W(x)\)-Funktion aufgelöst werden. Erstaunlicherweise ist \eqref{lambertw:eqEinsatzLambW} eine Struktur die oft vorkommt, was die Lambert-\(W\)-Funktion so wichtig macht.   
\ No newline at end of file
diff --git a/buch/papers/zeta/analytic_continuation.tex b/buch/papers/zeta/analytic_continuation.tex
index 0ccc116..d45a6ae 100644
--- a/buch/papers/zeta/analytic_continuation.tex
+++ b/buch/papers/zeta/analytic_continuation.tex
@@ -3,16 +3,16 @@
 
 Die analytische Fortsetzung der Riemannschen Zetafunktion ist äusserst interessant.
 Sie ermöglicht die Berechnung von $\zeta(-1)$ und weiterer spannender Werte.
-So liegen zum Beispiel unendlich viele Nullstellen der Zetafunktion bei $\Re(s) = 0.5$.
-Diese sind relevant für die Primzahlverteilung und sind Gegenstand der Riemannschen Vermutung.
+So liegen zum Beispiel unendlich viele Nullstellen der Zetafunktion bei $\Re(s) = \frac{1}{2}$.
+Wie bereits erwähnt sind diese Gegenstand der Riemannschen Vermutung.
 
 Es werden zwei verschiedene Fortsetzungen benötigt.
 Die erste erweitert die Zetafunktion auf $\Re(s) > 0$.
-Die zweite verwendet eine Spiegelung an der $\Re(s) = 0.5$ Linie und erschliesst damit die ganze komplexe Ebene.
+Die zweite verwendet eine Spiegelung an der $\Re(s) = \frac{1}{2}$ Geraden und erschliesst damit die ganze komplexe Ebene.
 Eine grafische Darstellung dieses Plans ist in Abbildung \ref{zeta:fig:continuation_overview} zu sehen.
 \begin{figure}
     \centering
-    \input{papers/zeta/continuation_overview.tikz.tex}
+    \input{papers/zeta/images/continuation_overview.tikz.tex}
     \caption{
         Die verschiedenen Abschnitte der Riemannschen Zetafunktion.
         Die originale Definition von \eqref{zeta:equation1} ist im grünen Bereich gültig.
@@ -23,7 +23,7 @@ Eine grafische Darstellung dieses Plans ist in Abbildung \ref{zeta:fig:continuat
 \end{figure}
 
 \subsection{Fortsetzung auf $\Re(s) > 0$} \label{zeta:subsection:auf_bereich_ge_0}
-Zuerst definieren die Dirichletsche Etafunktion als
+Zuerst definieren wir die Dirichletsche Etafunktion als
 \begin{equation}\label{zeta:equation:eta}
     \eta(s)
     =
@@ -36,26 +36,40 @@ Diese Etafunktion konvergiert gemäss dem Leibnitz-Kriterium im Bereich $\Re(s)
 Wenn wir es nun schaffen, die sehr ähnliche Zetafunktion durch die Etafunktion auszudrücken, dann haben die gesuchte Fortsetzung.
 Zuerst wiederholen wir zweimal die Definition der Zetafunktion \eqref{zeta:equation1}, wobei wir sie einmal durch $2^{s-1}$ teilen
 \begin{align}
-    \zeta(s)
+    \color{red}
+        \zeta(s)
     &=
     \sum_{n=1}^{\infty}
-    \frac{1}{n^s} \label{zeta:align1}
+    \color{red}
+        \frac{1}{n^s} \label{zeta:align1}
     \\
-    \frac{1}{2^{s-1}}
-    \zeta(s)
+    \color{blue}
+        \frac{1}{2^{s-1}}
+        \zeta(s)
     &=
     \sum_{n=1}^{\infty}
-    \frac{2}{(2n)^s}. \label{zeta:align2}
+    \color{blue}
+        \frac{2}{(2n)^s}. \label{zeta:align2}
 \end{align}
 Durch Subtraktion der beiden Gleichungen \eqref{zeta:align1} minus \eqref{zeta:align2}, ergibt sich
 \begin{align}
-    \left(1 - \frac{1}{2^{s-1}} \right)
+    \left({\color{red}1} - {\color{blue}\frac{1}{2^{s-1}}} \right)
     \zeta(s)
     &=
-    \frac{1}{1^s}
-    \underbrace{-\frac{2}{2^s} + \frac{1}{2^s}}_{-\frac{1}{2^s}}
-    + \frac{1}{3^s}
-    \underbrace{-\frac{2}{4^s} + \frac{1}{4^s}}_{-\frac{1}{4^s}}
+    {\color{red}\frac{1}{1^s}}
+    \underbrace{
+    -
+    {\color{blue}\frac{2}{2^s}}
+    +
+    {\color{red}\frac{1}{2^s}}
+    }_{\displaystyle{-\frac{1}{2^s}}}
+    +
+    {\color{red}\frac{1}{3^s}}
+    \underbrace{-
+    {\color{blue}\frac{2}{4^s}}
+    +
+    {\color{red}\frac{1}{4^s}}
+    }_{\displaystyle{-\frac{1}{4^s}}}
     \ldots
     \\
     &= \eta(s).
@@ -75,7 +89,7 @@ Wir beginnen damit, die Gammafunktion für den halben Funktionswert zu berechnen
     =
     \int_0^{\infty} t^{\frac{s}{2}-1} e^{-t} dt.
 \end{equation}
-Nun substituieren wir $t$ mit $t = \pi n^2 x$ und $dt=\pi n^2 dx$ und erhalten
+Nun substituieren wir $t = \pi n^2 x$ und $dt=\pi n^2 dx$ und erhalten
 \begin{equation}
     \Gamma \left( \frac{s}{2} \right)
     =
@@ -87,86 +101,33 @@ Nun substituieren wir $t$ mit $t = \pi n^2 x$ und $dt=\pi n^2 dx$ und erhalten
 \end{equation}
 Analog zum Abschnitt \ref{zeta:section:zusammenhang_mit_gammafunktion} teilen wir durch $(\pi n^2)^{\frac{s}{2}}$
 \begin{equation}
-    \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}} n^s}
+    \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}}
+    \frac{1}{n^s}
      =
     \int_0^{\infty}
     x^{\frac{s}{2}-1}
     e^{-\pi n^2 x}
     \,dx,
 \end{equation}
-und finden Zeta durch die Summenbildung $\sum_{n=1}^{\infty}$
-\begin{equation}
+und finden $\zeta(s)$ durch die Summenbildung über alle $n$
+\begin{align}
     \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}}
     \zeta(s)
-    =
+    &=
     \int_0^{\infty}
     x^{\frac{s}{2}-1}
     \sum_{n=1}^{\infty}
     e^{-\pi n^2 x}
-    \,dx. \label{zeta:equation:integral1}
-\end{equation}
-Die Summe kürzen wir ab als $\psi(x) = \sum_{n=1}^{\infty} e^{-\pi n^2 x}$.
-Im Abschnitt \ref{zeta:subsec:poisson_summation} wird die poissonsche Summenformel $\sum f(n) = \sum F(n)$ bewiesen.
-In unserem Problem ist $f(n) =  e^{-\pi n^2 x}$ und die zugehörige Fouriertransformation $F(n)$ ist
-\begin{equation}
-    F(n)
-    =
-    \mathcal{F}
-    (
-    e^{-\pi n^2 x}
-    )
-    =
-    \frac{1}{\sqrt{x}}
-    e^{\frac{-n^2 \pi}{x}}.
-\end{equation}
-Dadurch ergibt sich
-\begin{equation}\label{zeta:equation:psi}
-    \sum_{n=-\infty}^{\infty}
-    e^{-\pi n^2 x}
-    =
-    \frac{1}{\sqrt{x}}
-    \sum_{n=-\infty}^{\infty}
-    e^{\frac{-n^2 \pi}{x}},
-\end{equation}
-wobei wir die Summen so verändern müssen, dass sie bei $n=1$ beginnen und wir $\psi(x)$ erhalten als
-\begin{align}
-    2
-    \sum_{n=1}^{\infty}
-    e^{-\pi n^2 x}
-    +
-    1
-    &=
-    \frac{1}{\sqrt{x}}
-    \left(
-    2
-    \sum_{n=1}^{\infty}
-    e^{\frac{-n^2 \pi}{x}}
-    +
-    1
-    \right)
+    \,dx\label{zeta:equation:integral1}
     \\
-    2
-    \psi(x)
-    +
-    1
     &=
-    \frac{1}{\sqrt{x}}
-    \left(
-    2
-    \psi\left(\frac{1}{x}\right)
-    +
-    1
-    \right)
-    \\
+    \int_0^{\infty}
+    x^{\frac{s}{2}-1}
     \psi(x)
-    &=
-    - \frac{1}{2}
-    + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}}
-    + \frac{1}{2 \sqrt{x}}.\label{zeta:equation:psi}
+    \,dx,
 \end{align}
-Diese Gleichung wird später wichtig werden.
-
-Zunächst teilen wir nun das Integral aus \eqref{zeta:equation:integral1} auf als
+wobei die Summe $\sum_{n=1}^{\infty} e^{-\pi n^2 x}$ als $\psi(x)$ abgekürzt wird.
+Zunächst teilen wir nun das Integral auf in zwei Teile
 \begin{equation}\label{zeta:equation:integral2}
     \int_0^{\infty}
     x^{\frac{s}{2}-1}
@@ -178,109 +139,20 @@ Zunächst teilen wir nun das Integral aus \eqref{zeta:equation:integral1} auf al
     x^{\frac{s}{2}-1}
     \psi(x)
     \,dx
-    }_{I_1}
+    }_{\displaystyle{I_1}}
     +
     \underbrace{
     \int_1^{\infty}
     x^{\frac{s}{2}-1}
     \psi(x)
     \,dx
-    }_{I_2}
-    =
-    I_1 + I_2,
-\end{equation}
-wobei wir uns nun auf den ersten Teil $I_1$ konzentrieren werden.
-Dabei setzen wir die Definition von $\psi(x)$ aus \eqref{zeta:equation:psi} ein und erhalten
-\begin{align}
-    I_1
-    =
-    \int_0^{1}
-    x^{\frac{s}{2}-1}
-    \psi(x)
-    \,dx
-    &=
-    \int_0^{1}
-    x^{\frac{s}{2}-1}
-    \left(
-    - \frac{1}{2}
-    + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}}
-    + \frac{1}{2 \sqrt{x}}
-    \right)
-    \,dx
-    \\
-    &=
-    \int_0^{1}
-    x^{\frac{s}{2}-\frac{3}{2}}
-    \psi \left( \frac{1}{x} \right)
-    + \frac{1}{2}
-    \biggl(
-    x^{\frac{s}{2}-\frac{3}{2}}
-    -
-    x^{\frac{s}{2}-1}
-    \biggl)
-    \,dx
-    \\
-    &=
-    \underbrace{
-    \int_0^{1}
-    x^{\frac{s}{2}-\frac{3}{2}}
-    \psi \left( \frac{1}{x} \right)
-    \,dx
-    }_{I_3}
-    +
-    \underbrace{
-    \frac{1}{2}
-    \int_0^1
-    x^{\frac{s}{2}-\frac{3}{2}}
-    -
-    x^{\frac{s}{2}-1}
-    \,dx
-    }_{I_4}. \label{zeta:equation:integral3}
-\end{align}
-Dabei kann das zweite Integral $I_4$ gelöst werden als
-\begin{equation}
-    I_4
-    =
-    \frac{1}{2}
-    \int_0^1
-    x^{\frac{s}{2}-\frac{3}{2}}
-    -
-    x^{\frac{s}{2}-1}
-    \,dx
+    }_{\displaystyle{I_2}}
     =
-    \frac{1}{s(s-1)}.
+    I_1 + I_2.
 \end{equation}
-Das erste Integral $I_3$ aus \eqref{zeta:equation:integral3} mit $\psi \left(\frac{1}{x} \right)$ ist nicht lösbar in dieser Form.
-Deshalb substituieren wir $x = \frac{1}{u}$ und $dx = -\frac{1}{u^2}du$.
-Die untere Integralgrenze wechselt ebenfalls zu $x_0 = 0 \rightarrow u_0 = \infty$.
-Dies ergibt
-\begin{align}
-    I_3
-    =
-    \int_{\infty}^{1}
-    \left(
-    \frac{1}{u}
-    \right)^{\frac{s}{2}-\frac{3}{2}}
-    \psi(u)
-    \frac{-du}{u^2}
-    &=
-    \int_{1}^{\infty}
-    \left(
-    \frac{1}{u}
-    \right)^{\frac{s}{2}-\frac{3}{2}}
-    \psi(u)
-    \frac{du}{u^2}
-    \\
-    &=
-    \int_{1}^{\infty}
-    x^{(-1) \left(\frac{s}{2}+\frac{1}{2}\right)}
-    \psi(x)
-    \,dx,
-\end{align}
-wobei wir durch Multiplikation mit $(-1)$ die Integralgrenzen tauschen dürfen.
-Es ist zu beachten das diese Grenzen nun identisch mit den Grenzen des zweiten Integrals von \eqref{zeta:equation:integral2} sind.
-Wir setzen beide Lösungen ein in Gleichung \eqref{zeta:equation:integral3} und erhalten
-\begin{equation}
+Abschnitt \ref{zeta:subsubsec:intcal} beschreibt wie das Integral $I_1$ umgestellt werden kann um ebenfalls die Integrationsgrenzen $1$ und $\infty$ zu bekommen.
+Die Lösung, beschrieben in Gleichung \eqref{zeta:equation:intcal_res}, lautet
+\begin{equation*}
     I_1
     =
     \int_0^{1}
@@ -294,8 +166,8 @@ Wir setzen beide Lösungen ein in Gleichung \eqref{zeta:equation:integral3} und
     \,dx
     +
     \frac{1}{s(s-1)}.
-\end{equation}
-Dieses Resultat setzen wir wiederum ein in \eqref{zeta:equation:integral2}, um schlussendlich
+\end{equation*}
+Dieses Resultat setzen wir nun ein in \eqref{zeta:equation:integral2}, um schlussendlich
 \begin{align}
     \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}}
     \zeta(s)
@@ -356,17 +228,21 @@ Somit haben wir die analytische Fortsetzung gefunden als
     \zeta(s)
     =
     \frac{\Gamma \left( \frac{1-s}{2} \right)}{\pi^{\frac{1-s}{2}}}
-    \zeta(1-s).
+    \zeta(1-s),
 \end{equation}
-%TODO Definitionen und Gleichungen klarer unterscheiden
+was einer Spiegelung an der $\Re(s) = \frac{1}{2}$ Geraden entspricht.
+Eine ganz ähnliche Spiegelungseigenschaft wurde bereits in Abschnitt \ref{buch:funktionentheorie:subsection:gammareflektion} für die Gammafunktion gefunden.
+
+\subsection{Berechnung des Integrals $I_1 = \int_0^{1} x^{\frac{s}{2}-1} \psi(x) \,dx$} \label{zeta:subsubsec:intcal}
 
-\subsection{Poissonsche Summenformel} \label{zeta:subsec:poisson_summation}
+Ziel dieses Abschnittes ist, zu zeigen wie das Integral $I_1$ aus Gleichung \eqref{zeta:equation:integral2} durch ein neues Integral mit den Integrationsgrenzen $1$ und $\infty$ ersetzt werden kann.
+Da dieser Schritt ziemlich aufwendig ist, wird er hier in einem eigenen Abschnitt behandelt.
+Zunächst wird die poissonsche Summenformel hergeleitet \cite{zeta:online:poisson}, da diese verwendet werden kann um $\psi(x)$ zu berechnen.
 
-Der Beweis für Gleichung \ref{zeta:equation:psi} folgt direkt durch die poissonsche Summenformel.
-Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta Funktion.
+Um die poissonsche Summenformel zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta Funktion.
 
 \begin{lemma}
-    Die Fourierreihe der periodischen Dirac Delta Funktion $\sum \delta(x - 2\pi k)$ ist
+    Die Fourierreihe der periodischen Dirac $\delta$ Funktion $\sum \delta(x - 2\pi k)$ ist
     \begin{equation} \label{zeta:equation:fourier_dirac}
         \sum_{k=-\infty}^{\infty}
         \delta(x - 2\pi k)
@@ -437,8 +313,8 @@ Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta F
         \underbrace{
         \sum_{k=-\infty}^{\infty}
         e^{-i 2\pi x k}
-        }_{\text{\eqref{zeta:equation:fourier_dirac}}}
-        \, dx,
+        }_{\displaystyle{\text{\eqref{zeta:equation:fourier_dirac}}}}
+        \, dx, \label{zeta:equation:1934}
     \end{align}
     und verwenden die Fouriertransformation der Dirac Funktion aus \eqref{zeta:equation:fourier_dirac}
     \begin{align}
@@ -454,7 +330,7 @@ Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta F
         \sum_{k=-\infty}^{\infty}
         \delta(x + k).
     \end{align}
-    Wenn wir dies einsetzen und erhalten wir den gesuchten Beweis für die poissonsche Summenformel
+    Wenn wir dies einsetzen in Gleichung \eqref{zeta:equation:1934} erhalten wir
     \begin{equation}
         \sum_{k=-\infty}^{\infty}
         F(k)
@@ -472,6 +348,190 @@ Um diese zu beweisen, berechnen wir zunächst die Fourierreihe der Dirac Delta F
         \, dx
         =
         \sum_{k=-\infty}^{\infty}
-        f(k).
+        f(k),
     \end{equation}
+    was der gesuchte Beweis für die poissonsche Summenformel ist.
 \end{proof}
+
+Erinnern wir uns nochmals an unser Integral aus Gleichung \eqref{zeta:equation:integral2}
+\begin{align*}
+    I_1
+    &=
+    \int_0^{1}
+    x^{\frac{s}{2}-1}
+    \sum_{n=1}^{\infty}
+    e^{-\pi n^2 x}
+    \,dx
+    \\
+    &=
+    \int_0^{1}
+    x^{\frac{s}{2}-1}
+    \psi(x)
+    \,dx
+    .
+\end{align*}
+
+Wir wenden nun diese poissonsche Summenformel $\sum f(n) = \sum F(n)$ an auf $\psi(x)$.
+In unserem Problem ist also $f(n) =  e^{-\pi n^2 x}$ und die zugehörige Fouriertransformation $F(n)$ ist
+\begin{equation}
+    F(n)
+    =
+    \mathcal{F}
+    (
+    e^{-\pi n^2 x}
+    )
+    =
+    \frac{1}{\sqrt{x}}
+    e^{\frac{-n^2 \pi}{x}}.
+\end{equation}
+Dadurch ergibt sich
+\begin{equation}\label{zeta:equation:psi}
+    \sum_{n=-\infty}^{\infty}
+    e^{-\pi n^2 x}
+    =
+    \frac{1}{\sqrt{x}}
+    \sum_{n=-\infty}^{\infty}
+    e^{\frac{-n^2 \pi}{x}},
+\end{equation}
+wobei wir die Summen so verändern müssen, dass sie bei $n=1$ beginnen und wir $\psi(x)$ erhalten als
+\begin{align}
+    2
+    \sum_{n=1}^{\infty}
+    e^{-\pi n^2 x}
+    +
+    1
+    &=
+    \frac{1}{\sqrt{x}}
+    \Biggl(
+    2
+    \sum_{n=1}^{\infty}
+    e^{\frac{-n^2 \pi}{x}}
+    +
+    1
+    \Biggr)
+    \\
+    2
+    \psi(x)
+    +
+    1
+    &=
+    \frac{1}{\sqrt{x}}
+    \left(
+    2
+    \psi\left(\frac{1}{x}\right)
+    +
+    1
+    \right)
+    \\
+    \psi(x)
+    &=
+    - \frac{1}{2}
+    + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}}
+    + \frac{1}{2 \sqrt{x}}.\label{zeta:equation:psi}
+\end{align}
+Diese Form von $\psi(x)$ eingesetzt in $I_1$ ergibt
+\begin{align}
+    I_1
+    =
+    \int_0^{1}
+    x^{\frac{s}{2}-1}
+    \psi(x)
+    \,dx
+    &=
+    \int_0^{1}
+    x^{\frac{s}{2}-1}
+    \Biggl(
+    - \frac{1}{2}
+    + \frac{\psi\left(\frac{1}{x} \right)}{\sqrt{x}}
+    + \frac{1}{2 \sqrt{x}}
+    \Biggr)
+    \,dx
+    \\
+    &=
+    \int_0^{1}
+    x^{\frac{s}{2}-\frac{3}{2}}
+    \psi \left( \frac{1}{x} \right)
+    + \frac{1}{2}
+    \biggl(
+    x^{\frac{s}{2}-\frac{3}{2}}
+    -
+    x^{\frac{s}{2}-1}
+    \biggl)
+    \,dx
+    \\
+    &=
+    \underbrace{
+    \int_0^{1}
+    x^{\frac{s}{2}-\frac{3}{2}}
+    \psi \left( \frac{1}{x} \right)
+    \,dx
+    }_{\displaystyle{I_3}}
+    +
+    \underbrace{
+    \frac{1}{2}
+    \int_0^1
+    x^{\frac{s}{2}-\frac{3}{2}}
+    -
+    x^{\frac{s}{2}-1}
+    \,dx
+    }_{\displaystyle{I_4}}. \label{zeta:equation:integral3}
+\end{align}
+Darin kann für das zweite Integral $I_4$ eine Lösung gefunden werden als
+\begin{equation}
+    I_4
+    =
+    \frac{1}{2}
+    \int_0^1
+    x^{\frac{s}{2}-\frac{3}{2}}
+    -
+    x^{\frac{s}{2}-1}
+    \,dx
+    =
+    \frac{1}{s(s-1)}.
+\end{equation}
+Das erste Integral $I_3$ aus \eqref{zeta:equation:integral3} mit $\psi \left(\frac{1}{x} \right)$ ist hingegen nicht lösbar in dieser Form.
+Deshalb substituieren wir $x = \frac{1}{u}$ und $dx = -\frac{1}{u^2}du$.
+Die untere Integralgrenze wechselt ebenfalls zu $x_0 = 0 \rightarrow u_0 = \infty$.
+Dies ergibt
+\begin{align}
+    I_3
+    =
+    \int_{\infty}^{1}
+    \left(
+    \frac{1}{u}
+    \right)^{\frac{s}{2}-\frac{3}{2}}
+    \psi(u)
+    \frac{-du}{u^2}
+    &=
+    \int_{1}^{\infty}
+    \left(
+    \frac{1}{u}
+    \right)^{\frac{s}{2}-\frac{3}{2}}
+    \psi(u)
+    \frac{du}{u^2}
+    \\
+    &=
+    \int_{1}^{\infty}
+    x^{(-1) \left(\frac{s}{2}+\frac{1}{2}\right)}
+    \psi(x)
+    \,dx,
+\end{align}
+wobei wir durch Multiplikation mit $(-1)$ die Integralgrenzen tauschen dürfen.
+Es ist zu beachten das diese Grenzen nun identisch mit den Grenzen des zweiten Integrals $I_2$ von \eqref{zeta:equation:integral2} sind.
+Wir setzen beide Lösungen in Gleichung \eqref{zeta:equation:integral3} ein und erhalten
+\begin{equation}
+    I_1
+    =
+    \int_0^{1}
+    x^{\frac{s}{2}-1}
+    \psi(x)
+    \,dx
+    =
+    \int_{1}^{\infty}
+    x^{(-1) \left(\frac{s}{2}+\frac{1}{2}\right)}
+    \psi(x)
+    \,dx
+    +
+    \frac{1}{s(s-1)}. \label{zeta:equation:intcal_res}
+\end{equation}
+Diese Form des Integrals $I_1$ hat die gewünschten Integrationsgrenzen und ein essentieller Bestandteil des Beweises der Funktionalgleichung in Abschnitt \ref{zeta:subsection:auf_ganz}.
diff --git a/buch/papers/zeta/einleitung.tex b/buch/papers/zeta/einleitung.tex
index 3b70531..828678d 100644
--- a/buch/papers/zeta/einleitung.tex
+++ b/buch/papers/zeta/einleitung.tex
@@ -1,11 +1,41 @@
 \section{Einleitung} \label{zeta:section:einleitung}
 \rhead{Einleitung}
 
-Die Riemannsche Zetafunktion ist für alle komplexe $s$ mit $\Re(s) > 1$ definiert als
+Die Riemannsche Zetafunktion $\zeta(s)$ ist für alle komplexe $s$ mit $\Re(s) > 1$ definiert als
 \begin{equation}\label{zeta:equation1}
     \zeta(s)
     =
     \sum_{n=1}^{\infty}
     \frac{1}{n^s}.
 \end{equation}
+Die Zetafunktion ist bekannt als Bestandteil der Riemannschen Vermutung, welche besagt das alle nichttrivialen Nullstellen der Zetafunktion einen Realteil von $\frac{1}{2}$ haben.
+Mithilfe dieser Vermutung kann eine gute Annäherung an die Primzahlfunktion gefunden werden.
+Die Primzahlfunktion steigt immer an, sobald eine Primzahl vorkommt.
+Eine Darstellung davon ist in Abbildung \ref{fig:zeta:primzahlfunktion} zu finden.
+Die Riemannsche Vermutung ist eines der ungelösten Millennium-Probleme der Mathematik, auf deren Lösung eine Belohnung von einer Million Dollar ausgesetzt ist \cite{zeta:online:millennium}.
+Auf eine genauere Beschreibung der Riemannschen Vermutung wird im Rahmen dieses Papers nicht eingegangen.
+\begin{figure}
+    \centering
+    \input{papers/zeta/images/primzahlfunktion2.tex}
+    \caption{Die Primzahlfunktion von $0$ bis $30$.}
+    \label{fig:zeta:primzahlfunktion}
+\end{figure}
 
+Der grundlegende Zusammenhang der Primzahlen und der Zetafunktion wird im ersten Abschnitt \ref{zeta:section:eulerprodukt} über das Eulerprodukt gezeigt.
+Danach folgt die Verbindung zur bereits bekannten Gammafunktion in Abschnitt \ref{zeta:section:zusammenhang_mit_gammafunktion}.
+Schlussendlich folgt die Beschreibung der analytischen Fortsetzung die gesamte komplexe Ebene in Abschnitt \ref{zeta:section:analytische_fortsetzung}.
+
+Diese analytische Fortsetzung wird für die Riemannsche Vermutung benötigt, ermöglicht aber auch andere interessante Aussagen.
+So findet sich zum Beispiel immer wieder die aberwitzige Behauptung, das die Summe aller natürlichen Zahlen
+\begin{equation*}
+    \sum_{n=1}^{\infty} n
+    =
+    \sum_{n=1}^{\infty}
+    \frac{1}{n^{-1}}
+    =
+    -\frac{1}{12}
+\end{equation*}
+sei.
+Obwohl diese Behauptung offensichtlich falsch ist, hat sie doch ihre Berechtigung, wie durch die analytische Fortsetzung gezeigt werden wird.
+
+Die folgenden mathematischen Herleitungen sind, sofern nicht anders gekennzeichnet, eigene Darstellungen basierend auf den überaus umfangreichen Wikipedia-Artikeln auf Deutsch \cite{zeta:online:wiki_de} und Englisch \cite{zeta:online:wiki_en} sowie einer Video-Playlist \cite{zeta:online:mryoumath}.
diff --git a/buch/papers/zeta/euler_product.tex b/buch/papers/zeta/euler_product.tex
index a6ed512..9c08dd2 100644
--- a/buch/papers/zeta/euler_product.tex
+++ b/buch/papers/zeta/euler_product.tex
@@ -1,9 +1,9 @@
 \section{Eulerprodukt} \label{zeta:section:eulerprodukt}
 \rhead{Eulerprodukt}
 
-Das Eulerprodukt stellt die Verbindung der Zetafunktion und der Primzahlen her.
-Diese Verbindung ist sehr wichtig, da durch sie eine Aussage zur Primzahlverteilung gemacht werden kann.
-Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche eines der grössten ungelösten Probleme der Mathematik ist.
+Das Eulerprodukt stellt die gesuchte Verbindung der Zetafunktion und der Primzahlen her.
+Wie der Name bereits sagt, wurde das Eulerprodukt bereits 1727 von Euler entdeckt.
+Um daraus die Riemannsche Vermutung herzuleiten, wäre aber noch einiges mehr nötig.
 
 \begin{satz}
     Für alle Zahlen $s$ mit $\Re(s) > 1$ ist die Zetafunktion identisch mit dem unendlichen Eulerprodukt
@@ -28,9 +28,9 @@ Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche
         =
         \prod_{p \in P}
         \sum_{k_i=0}^{\infty}
-        \left(
+        \biggl(
         \frac{1}{p_i^s}
-        \right)^{k_i}
+        \biggr)^{k_i}
         =
         \prod_{p \in P}
         \sum_{k_i=0}^{\infty}
@@ -53,33 +53,34 @@ Die Verteilung der Primzahlen ist Gegenstand der Riemannschen Vermutung, welche
         \sum_{k_1=0}^{\infty}
         \sum_{k_2=0}^{\infty}
         \ldots
-        \left(
+        \biggl(
         \frac{1}{p_1^{k_1}}
         \frac{1}{p_2^{k_2}}
         \ldots
-        \right)^s.
+        \biggr)^s.
         \label{zeta:equation:eulerprodukt2}
     \end{align}
     Der Fundamentalsatz der Arithmetik (Primfaktorzerlegung) besagt, dass jede beliebige Zahl $n \in \mathbb{N}$ durch eine eindeutige Primfaktorzerlegung beschrieben werden kann
     \begin{equation}
         n = \prod_i p_i^{k_i} \quad \forall \quad n \in \mathbb{N}.
     \end{equation}
-    Jeder Summand der Summen in \eqref{zeta:equation:eulerprodukt2} ist somit eine Zahl $n$.
-    Da die Summen alle möglichen Kombinationen von Exponenten und Primzahlen in \eqref{zeta:equation:eulerprodukt2} enthält haben wir
+    Jeder Summand der Summen in \eqref{zeta:equation:eulerprodukt2} ist somit der Kehrwert genau einer natürlichen Zahl $n \in \mathbb{N}$.
+    Da die Summen alle möglichen Kombinationen von Exponenten und Primzahlen in \eqref{zeta:equation:eulerprodukt2} enthält, haben wir
     \begin{equation}
         \sum_{k_1=0}^{\infty}
         \sum_{k_2=0}^{\infty}
         \ldots
-        \left(
+        \biggl(
         \frac{1}{p_1^{k_1}}
         \frac{1}{p_2^{k_2}}
         \ldots
-        \right)^s
+        \biggr)^s
         =
         \sum_{n=1}^\infty
         \frac{1}{n^s}
         =
-        \zeta(s)
+        \zeta(s),
     \end{equation}
+    wodurch das Eulerprodukt bewiesen ist.
 \end{proof}
 
diff --git a/buch/papers/zeta/fazit.tex b/buch/papers/zeta/fazit.tex
new file mode 100644
index 0000000..027f324
--- /dev/null
+++ b/buch/papers/zeta/fazit.tex
@@ -0,0 +1,94 @@
+\section{Der Wert $\zeta(-1)$} \label{zeta:section:fazit}
+\rhead{Der Wert $\zeta(-1)$}
+
+Ganz zu Beginn dieses Papers wurde die Behauptung erwähnt, dass die Summe aller natürlichen Zahlen $-\frac{1}{12}$ sei.
+Diese Summe ist nichts anderes als die Zetafunktion am Wert $s=-1$.
+Da wir die analytische Fortsetzung mit der Funktionalgleichung \eqref{zeta:equation:functional} gefunden haben, können wir den Wert $s=-1$ einsetzen und erhalten
+\begin{align*}
+    \zeta(s)
+    &=
+    \frac{\Gamma \left( \frac{1-s}{2} \right)}{\pi^{\frac{1-s}{2}}}
+    \zeta(1-s)
+    \frac{\pi^{\frac{s}{2}}}{\Gamma \left( \frac{s}{2} \right)}
+    \\
+    \zeta(-1)
+    &=
+    \frac{\Gamma(1)}{\pi}
+    \zeta(2)
+    \frac{\pi^{-\frac{1}{2}}}{\Gamma \left( -\frac{1}{2} \right)}.
+\end{align*}
+Also fehlen uns drei Werte, $\zeta(2)$, $\Gamma(1)$ und $\Gamma(-\frac{1}{2})$.
+
+Zunächst konzentrieren wir uns auf $\zeta(2)$, welches im konvergenten Bereich der Reihe liegt und auch bekannt ist als das Basler Problem.
+Wir lösen das Basler Problem \cite{zeta:online:basel} mithilfe der parsevalschen Gleichung \cite{zeta:online:pars}
+\begin{align}
+    \int_{-\pi}^{\pi} |f(x)|^2 dx
+    &=
+    2\pi \sum_{n=-\infty}^{\infty} |c_n|^2 \\
+    c_n
+    &=
+    \frac{1}{2\pi}
+    \int_{-\pi}^{\pi}f(x)e^{-inx} dx,
+\end{align}
+welche besagt dass die Summe der quadrierten Fourierkoeffizienten einer Funktion identisch ist mit dem Integral der quadrierten Funktion.
+Wenn wir dies für $f(x) = x$ auswerten erhalten wir
+\begin{align}
+    c_n
+    &=
+    \begin{cases}
+        \frac{(-1)^n}{n} i, & \text{for } n\neq0, \\
+        0, & \text{for } n=0
+    \end{cases}
+    \\
+    \int_{-\pi}^{\pi} x^2 dx
+    &=
+    2\pi \sum_{n=-\infty}^{\infty} |c_n|^2
+    =
+    4\pi \underbrace{\sum_{n=1}^{\infty} \frac{1}{n^2}}_{\displaystyle{\zeta(2)}}.
+\end{align}
+Durch einfaches Umstellen erhalten wir somit die Lösung des Basler Problems als
+\begin{equation}
+    \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{1}{4\pi}
+    \int_{-\pi}^{\pi} x^2 dx
+    = \frac{\pi^2}{6}.
+\end{equation}
+
+Als nächstes berechnen wir $\Gamma(1)$ und $\Gamma(-\frac{1}{2})$ mithilfe der Integraldefinition der Gammafunktion (Definition \ref{buch:rekursion:def:gamma}).
+Da das Integral für $\Gamma(-\frac{1}{2})$ nicht konvergiert, wird die Reflektionsformel aus \ref{buch:funktionentheorie:subsection:gammareflektion} verwendet, welche das konvergierende Integral von  $\Gamma(\frac{3}{2})$ verwendet.
+Es ergeben sich die Werte
+\begin{align*}
+    \Gamma(1)
+    &= 1\\
+    \Gamma\biggl(-\frac{1}{2}\biggr)
+    &= \frac{\pi}{\sin\left(-\frac{\pi}{2}\right)
+    \Gamma\left(\frac{3}{2}\right)}
+    = -\frac{\sqrt{\pi}}{2}.
+\end{align*}
+
+Wenn wir diese Werte in die Funktionalgleichung einsetzen, erhalten wir das gewünschte Ergebnis
+\begin{align*}
+    \zeta(-1)
+    &=
+    \frac{\Gamma(1)}{\pi}
+    \zeta(2)
+    \frac{\pi^{-\frac{1}{2}}}{\Gamma \left( -\frac{1}{2} \right)}
+    \\
+    &=
+    \frac{1}{\pi}
+    \frac{\pi^2}{6}
+    \frac{\pi^{-\frac{1}{2}}}{
+    -\frac{\sqrt{\pi}}{2}}
+    \\
+    &=
+    -\frac{1}{12}.
+\end{align*}
+
+Weiter wurde zu Beginn dieses Papers auf die Riemannsche Vermutung hingewiesen, wonach alle nichttrivialen Nullstellen der Zetafunktion auf der $\Re(s)=\frac{1}{2}$ Geraden liegen.
+Abbildung \ref{zeta:fig:einzweitel} zeigt die Funktionswerte dieser Geraden.
+\begin{figure}
+    \centering
+    \input{papers/zeta/images/zetaplot.tex}
+    \caption{Die komplexen Werte der Zetafunktion für die kritische Gerade $\Re(s)=\frac{1}{2}$ im Bereich $\Im(s) = 0\dots40$.
+    Klar sichtbar sind die immer wiederkehrenden Nullstellen, wie sie Gegenstand der Riemannschen Vermutung sind.}
+    \label{zeta:fig:einzweitel}
+\end{figure}
diff --git a/buch/papers/zeta/images/Makefile b/buch/papers/zeta/images/Makefile
new file mode 100644
index 0000000..611662d
--- /dev/null
+++ b/buch/papers/zeta/images/Makefile
@@ -0,0 +1,10 @@
+#
+# Makefile to build images
+#
+all:	primzahlfunktion2.pdf zetaplot.pdf
+
+primzahlfunktion2.pdf:	primzahlfunktion2.tex
+	pdflatex primzahlfunktion2.tex
+
+zetaplot.pdf:	zetaplot.tex zetapath.tex 
+	pdflatex zetaplot.tex
diff --git a/buch/papers/zeta/continuation_overview.tikz.tex b/buch/papers/zeta/images/continuation_overview.tikz.tex
index 836ab1d..836ab1d 100644
--- a/buch/papers/zeta/continuation_overview.tikz.tex
+++ b/buch/papers/zeta/images/continuation_overview.tikz.tex
diff --git a/buch/papers/zeta/images/primzahlfunktion.pgf b/buch/papers/zeta/images/primzahlfunktion.pgf
new file mode 100644
index 0000000..7d4f4fc
--- /dev/null
+++ b/buch/papers/zeta/images/primzahlfunktion.pgf
@@ -0,0 +1,505 @@
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+%%
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+%%
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diff --git a/buch/papers/zeta/images/primzahlfunktion2.pdf b/buch/papers/zeta/images/primzahlfunktion2.pdf
new file mode 100644
index 0000000..8998fb8
--- /dev/null
+++ b/buch/papers/zeta/images/primzahlfunktion2.pdf
diff --git a/buch/papers/zeta/images/primzahlfunktion2.tex b/buch/papers/zeta/images/primzahlfunktion2.tex
new file mode 100644
index 0000000..7425ce5
--- /dev/null
+++ b/buch/papers/zeta/images/primzahlfunktion2.tex
@@ -0,0 +1,63 @@
+%
+% primzahlfunktion2.tex -- Primzahlfunktion, alternativer Vorschlag
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\documentclass[tikz]{standalone}
+\usepackage{amsmath}
+\usepackage{times}
+\usepackage{txfonts}
+\usepackage{pgfplots}
+\usepackage{csvsimple}
+\usetikzlibrary{arrows,intersections,math}
+\begin{document}
+\def\skala{1}
+\begin{tikzpicture}[>=latex,thick,scale=\skala]
+
+\def\dx{0.38}
+\def\dy{0.5}
+
+\foreach \x in {1,...,30}{
+	\draw[color=gray!20] ({\x*\dx},0) -- ({\x*\dx},{10.5*\dy});
+}
+\foreach \y in {1,...,10}{
+	\draw[color=gray!20] (0,{\y*\dy}) -- ({30.5*\dx},{\y*\dy});
+}
+
+\draw[->] (-0.1,0) -- ({30.8*\dx},0) coordinate[label={$x$}];
+\draw[->] (0,-0.1) -- (0,{10.9*\dy}) coordinate[label={right:$\pi(x)$}];
+
+\def\segment#1#2#3{
+	%\draw[line width=0.1pt] ({#3*\dx},0) -- ({#3*\dx},{#2*\dy});
+	\draw[color=blue,line width=1.4pt]
+		({#1*\dx},{#2*\dy}) -- ({#3*\dx},{#2*\dy});
+	\draw[color=blue,line width=0.3pt]
+		({#3*\dx},{#2*\dy}) -- ({#3*\dx},{(#2+1)*\dy});
+	\draw ({#3*\dx},-0.1) -- ({#3*\dx},0.1);
+	\node at ({(#3)*\dx},-0.1) [below] {$#3\mathstrut$};
+}
+
+\foreach \y in {2,4,...,10}{
+	\draw (-0.1,{\y*\dy}) -- (0.1,{\y*\dy});
+	\node at (-0.1,{\y*\dy}) [left] {$\y\mathstrut$};
+}
+
+\begin{scope}
+\clip (0,-0.5) rectangle ({30*\dx},{10.1*\dy});
+
+\segment{0}{0}{2}
+\segment{2}{1}{3}
+\segment{3}{2}{5}
+\segment{5}{3}{7}
+\segment{7}{4}{11}
+\segment{11}{5}{13}
+\segment{13}{6}{17}
+\segment{17}{7}{19}
+\segment{19}{8}{23}
+\segment{23}{9}{29}
+\segment{29}{10}{31}
+\end{scope}
+
+\end{tikzpicture}
+\end{document}
+
diff --git a/buch/papers/zeta/images/primzahlfunktion_paper.pgf b/buch/papers/zeta/images/primzahlfunktion_paper.pgf
new file mode 100644
index 0000000..b9d67d3
--- /dev/null
+++ b/buch/papers/zeta/images/primzahlfunktion_paper.pgf
@@ -0,0 +1,505 @@
+%% Creator: Matplotlib, PGF backend
+%%
+%% To include the figure in your LaTeX document, write
+%%   \input{<filename>.pgf}
+%%
+%% Make sure the required packages are loaded in your preamble
+%%   \usepackage{pgf}
+%%
+%% and, on pdftex
+%%   \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-}
+%%
+%% or, on luatex and xetex
+%%   \usepackage{unicode-math}
+%%
+%% Figures using additional raster images can only be included by \input if
+%% they are in the same directory as the main LaTeX file. For loading figures
+%% from other directories you can use the `import` package
+%%   \usepackage{import}
+%%
+%% and then include the figures with
+%%   \import{<path to file>}{<filename>.pgf}
+%%
+%% Matplotlib used the following preamble
+%%
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diff --git a/buch/papers/zeta/images/youtube_screenshot.png b/buch/papers/zeta/images/youtube_screenshot.png
new file mode 100644
index 0000000..434041b
--- /dev/null
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diff --git a/buch/papers/zeta/images/zeta_re_-1_plot.pgf b/buch/papers/zeta/images/zeta_re_-1_plot.pgf
new file mode 100644
index 0000000..dd15ba1
--- /dev/null
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diff --git a/buch/papers/zeta/images/zeta_re_0.5_paper.pgf b/buch/papers/zeta/images/zeta_re_0.5_paper.pgf
new file mode 100644
index 0000000..44fffce
--- /dev/null
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+%%
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+%%
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diff --git a/buch/papers/zeta/images/zeta_re_0.5_plot.pgf b/buch/papers/zeta/images/zeta_re_0.5_plot.pgf
new file mode 100644
index 0000000..3ac7df8
--- /dev/null
+++ b/buch/papers/zeta/images/zeta_re_0.5_plot.pgf
@@ -0,0 +1,1206 @@
+%% Creator: Matplotlib, PGF backend
+%%
+%% To include the figure in your LaTeX document, write
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+%%
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+%%
+%% and, on pdftex
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diff --git a/buch/papers/zeta/images/zeta_re_0_plot.pgf b/buch/papers/zeta/images/zeta_re_0_plot.pgf
new file mode 100644
index 0000000..29a844e
--- /dev/null
+++ b/buch/papers/zeta/images/zeta_re_0_plot.pgf
@@ -0,0 +1,1242 @@
+%% Creator: Matplotlib, PGF backend
+%%
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+%%   \input{<filename>.pgf}
+%%
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+%%
+%% and, on pdftex
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+\pgfpathlineto{\pgfqpoint{3.305307in}{1.239949in}}%
+\pgfpathlineto{\pgfqpoint{3.193179in}{1.217069in}}%
+\pgfpathlineto{\pgfqpoint{3.079047in}{1.200096in}}%
+\pgfpathlineto{\pgfqpoint{2.963554in}{1.189114in}}%
+\pgfpathlineto{\pgfqpoint{2.847352in}{1.184170in}}%
+\pgfpathlineto{\pgfqpoint{2.731094in}{1.185279in}}%
+\pgfpathlineto{\pgfqpoint{2.731094in}{1.185279in}}%
+\pgfusepath{stroke}%
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+\begin{pgfscope}%
+\pgfsetrectcap%
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+\pgfsetlinewidth{0.803000pt}%
+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfpathmoveto{\pgfqpoint{0.800000in}{0.528000in}}%
+\pgfpathlineto{\pgfqpoint{0.800000in}{4.224000in}}%
+\pgfusepath{stroke}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\pgfsetrectcap%
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+\pgfsetlinewidth{0.803000pt}%
+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
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+\pgfpathlineto{\pgfqpoint{5.760000in}{4.224000in}}%
+\pgfusepath{stroke}%
+\end{pgfscope}%
+\begin{pgfscope}%
+\pgfsetrectcap%
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+\pgfsetlinewidth{0.803000pt}%
+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{currentstroke}%
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+\pgfpathmoveto{\pgfqpoint{0.800000in}{0.528000in}}%
+\pgfpathlineto{\pgfqpoint{5.760000in}{0.528000in}}%
+\pgfusepath{stroke}%
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+\begin{pgfscope}%
+\pgfsetrectcap%
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+\pgfsetlinewidth{0.803000pt}%
+\definecolor{currentstroke}{rgb}{0.000000,0.000000,0.000000}%
+\pgfsetstrokecolor{currentstroke}%
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+\pgfpathmoveto{\pgfqpoint{0.800000in}{4.224000in}}%
+\pgfpathlineto{\pgfqpoint{5.760000in}{4.224000in}}%
+\pgfusepath{stroke}%
+\end{pgfscope}%
+\end{pgfpicture}%
+\makeatother%
+\endgroup%
diff --git a/buch/papers/zeta/images/zetapath.tex b/buch/papers/zeta/images/zetapath.tex
new file mode 100644
index 0000000..75e1522
--- /dev/null
+++ b/buch/papers/zeta/images/zetapath.tex
@@ -0,0 +1,2003 @@
+\def\zetapath{
+ ({-1.4604*\dx},{0.0000*\dy})
+	-- ({-1.4572*\dx},{-0.0783*\dy})
+	-- ({-1.4476*\dx},{-0.1559*\dy})
+	-- ({-1.4319*\dx},{-0.2320*\dy})
+	-- ({-1.4104*\dx},{-0.3058*\dy})
+	-- ({-1.3834*\dx},{-0.3769*\dy})
+	-- ({-1.3514*\dx},{-0.4446*\dy})
+	-- ({-1.3149*\dx},{-0.5085*\dy})
+	-- ({-1.2745*\dx},{-0.5682*\dy})
+	-- ({-1.2308*\dx},{-0.6235*\dy})
+	-- ({-1.1843*\dx},{-0.6742*\dy})
+	-- ({-1.1358*\dx},{-0.7202*\dy})
+	-- ({-1.0856*\dx},{-0.7617*\dy})
+	-- ({-1.0344*\dx},{-0.7985*\dy})
+	-- ({-0.9826*\dx},{-0.8309*\dy})
+	-- ({-0.9306*\dx},{-0.8591*\dy})
+	-- ({-0.8788*\dx},{-0.8833*\dy})
+	-- ({-0.8275*\dx},{-0.9037*\dy})
+	-- ({-0.7770*\dx},{-0.9205*\dy})
+	-- ({-0.7275*\dx},{-0.9341*\dy})
+	-- ({-0.6792*\dx},{-0.9446*\dy})
+	-- ({-0.6322*\dx},{-0.9525*\dy})
+	-- ({-0.5867*\dx},{-0.9578*\dy})
+	-- ({-0.5427*\dx},{-0.9609*\dy})
+	-- ({-0.5002*\dx},{-0.9620*\dy})
+	-- ({-0.4593*\dx},{-0.9613*\dy})
+	-- ({-0.4200*\dx},{-0.9589*\dy})
+	-- ({-0.3823*\dx},{-0.9552*\dy})
+	-- ({-0.3462*\dx},{-0.9502*\dy})
+	-- ({-0.3116*\dx},{-0.9441*\dy})
+	-- ({-0.2785*\dx},{-0.9371*\dy})
+	-- ({-0.2469*\dx},{-0.9292*\dy})
+	-- ({-0.2167*\dx},{-0.9206*\dy})
+	-- ({-0.1878*\dx},{-0.9114*\dy})
+	-- ({-0.1603*\dx},{-0.9017*\dy})
+	-- ({-0.1340*\dx},{-0.8916*\dy})
+	-- ({-0.1089*\dx},{-0.8811*\dy})
+	-- ({-0.0849*\dx},{-0.8703*\dy})
+	-- ({-0.0621*\dx},{-0.8593*\dy})
+	-- ({-0.0402*\dx},{-0.8481*\dy})
+	-- ({-0.0194*\dx},{-0.8367*\dy})
+	-- ({0.0004*\dx},{-0.8253*\dy})
+	-- ({0.0194*\dx},{-0.8137*\dy})
+	-- ({0.0375*\dx},{-0.8022*\dy})
+	-- ({0.0549*\dx},{-0.7906*\dy})
+	-- ({0.0714*\dx},{-0.7790*\dy})
+	-- ({0.0872*\dx},{-0.7675*\dy})
+	-- ({0.1024*\dx},{-0.7560*\dy})
+	-- ({0.1168*\dx},{-0.7446*\dy})
+	-- ({0.1307*\dx},{-0.7333*\dy})
+	-- ({0.1439*\dx},{-0.7221*\dy})
+	-- ({0.1566*\dx},{-0.7110*\dy})
+	-- ({0.1688*\dx},{-0.7000*\dy})
+	-- ({0.1805*\dx},{-0.6890*\dy})
+	-- ({0.1917*\dx},{-0.6783*\dy})
+	-- ({0.2024*\dx},{-0.6676*\dy})
+	-- ({0.2127*\dx},{-0.6571*\dy})
+	-- ({0.2226*\dx},{-0.6467*\dy})
+	-- ({0.2321*\dx},{-0.6364*\dy})
+	-- ({0.2412*\dx},{-0.6263*\dy})
+	-- ({0.2500*\dx},{-0.6163*\dy})
+	-- ({0.2585*\dx},{-0.6064*\dy})
+	-- ({0.2666*\dx},{-0.5967*\dy})
+	-- ({0.2745*\dx},{-0.5871*\dy})
+	-- ({0.2820*\dx},{-0.5777*\dy})
+	-- ({0.2893*\dx},{-0.5684*\dy})
+	-- ({0.2963*\dx},{-0.5592*\dy})
+	-- ({0.3031*\dx},{-0.5501*\dy})
+	-- ({0.3096*\dx},{-0.5412*\dy})
+	-- ({0.3159*\dx},{-0.5324*\dy})
+	-- ({0.3220*\dx},{-0.5237*\dy})
+	-- ({0.3279*\dx},{-0.5152*\dy})
+	-- ({0.3336*\dx},{-0.5067*\dy})
+	-- ({0.3391*\dx},{-0.4984*\dy})
+	-- ({0.3445*\dx},{-0.4902*\dy})
+	-- ({0.3496*\dx},{-0.4821*\dy})
+	-- ({0.3546*\dx},{-0.4742*\dy})
+	-- ({0.3595*\dx},{-0.4663*\dy})
+	-- ({0.3642*\dx},{-0.4586*\dy})
+	-- ({0.3687*\dx},{-0.4510*\dy})
+	-- ({0.3732*\dx},{-0.4434*\dy})
+	-- ({0.3774*\dx},{-0.4360*\dy})
+	-- ({0.3816*\dx},{-0.4287*\dy})
+	-- ({0.3857*\dx},{-0.4214*\dy})
+	-- ({0.3896*\dx},{-0.4143*\dy})
+	-- ({0.3934*\dx},{-0.4073*\dy})
+	-- ({0.3972*\dx},{-0.4003*\dy})
+	-- ({0.4008*\dx},{-0.3935*\dy})
+	-- ({0.4043*\dx},{-0.3867*\dy})
+	-- ({0.4078*\dx},{-0.3800*\dy})
+	-- ({0.4111*\dx},{-0.3734*\dy})
+	-- ({0.4144*\dx},{-0.3669*\dy})
+	-- ({0.4176*\dx},{-0.3605*\dy})
+	-- ({0.4207*\dx},{-0.3541*\dy})
+	-- ({0.4237*\dx},{-0.3478*\dy})
+	-- ({0.4267*\dx},{-0.3416*\dy})
+	-- ({0.4296*\dx},{-0.3355*\dy})
+	-- ({0.4324*\dx},{-0.3294*\dy})
+	-- ({0.4352*\dx},{-0.3234*\dy})
+	-- ({0.4379*\dx},{-0.3175*\dy})
+	-- ({0.4405*\dx},{-0.3116*\dy})
+	-- ({0.4431*\dx},{-0.3059*\dy})
+	-- ({0.4457*\dx},{-0.3001*\dy})
+	-- ({0.4482*\dx},{-0.2945*\dy})
+	-- ({0.4506*\dx},{-0.2889*\dy})
+	-- ({0.4530*\dx},{-0.2833*\dy})
+	-- ({0.4554*\dx},{-0.2778*\dy})
+	-- ({0.4577*\dx},{-0.2724*\dy})
+	-- ({0.4599*\dx},{-0.2670*\dy})
+	-- ({0.4622*\dx},{-0.2617*\dy})
+	-- ({0.4643*\dx},{-0.2565*\dy})
+	-- ({0.4665*\dx},{-0.2513*\dy})
+	-- ({0.4686*\dx},{-0.2461*\dy})
+	-- ({0.4707*\dx},{-0.2410*\dy})
+	-- ({0.4727*\dx},{-0.2359*\dy})
+	-- ({0.4747*\dx},{-0.2309*\dy})
+	-- ({0.4767*\dx},{-0.2259*\dy})
+	-- ({0.4787*\dx},{-0.2210*\dy})
+	-- ({0.4806*\dx},{-0.2161*\dy})
+	-- ({0.4825*\dx},{-0.2113*\dy})
+	-- ({0.4844*\dx},{-0.2065*\dy})
+	-- ({0.4862*\dx},{-0.2018*\dy})
+	-- ({0.4880*\dx},{-0.1971*\dy})
+	-- ({0.4898*\dx},{-0.1924*\dy})
+	-- ({0.4916*\dx},{-0.1878*\dy})
+	-- ({0.4934*\dx},{-0.1832*\dy})
+	-- ({0.4951*\dx},{-0.1786*\dy})
+	-- ({0.4968*\dx},{-0.1741*\dy})
+	-- ({0.4985*\dx},{-0.1696*\dy})
+	-- ({0.5002*\dx},{-0.1652*\dy})
+	-- ({0.5019*\dx},{-0.1608*\dy})
+	-- ({0.5035*\dx},{-0.1564*\dy})
+	-- ({0.5052*\dx},{-0.1521*\dy})
+	-- ({0.5068*\dx},{-0.1478*\dy})
+	-- ({0.5084*\dx},{-0.1435*\dy})
+	-- ({0.5100*\dx},{-0.1392*\dy})
+	-- ({0.5116*\dx},{-0.1350*\dy})
+	-- ({0.5132*\dx},{-0.1308*\dy})
+	-- ({0.5147*\dx},{-0.1267*\dy})
+	-- ({0.5163*\dx},{-0.1226*\dy})
+	-- ({0.5178*\dx},{-0.1185*\dy})
+	-- ({0.5193*\dx},{-0.1144*\dy})
+	-- ({0.5208*\dx},{-0.1103*\dy})
+	-- ({0.5224*\dx},{-0.1063*\dy})
+	-- ({0.5239*\dx},{-0.1023*\dy})
+	-- ({0.5254*\dx},{-0.0984*\dy})
+	-- ({0.5268*\dx},{-0.0944*\dy})
+	-- ({0.5283*\dx},{-0.0905*\dy})
+	-- ({0.5298*\dx},{-0.0866*\dy})
+	-- ({0.5313*\dx},{-0.0827*\dy})
+	-- ({0.5327*\dx},{-0.0789*\dy})
+	-- ({0.5342*\dx},{-0.0751*\dy})
+	-- ({0.5357*\dx},{-0.0713*\dy})
+	-- ({0.5371*\dx},{-0.0675*\dy})
+	-- ({0.5386*\dx},{-0.0637*\dy})
+	-- ({0.5400*\dx},{-0.0600*\dy})
+	-- ({0.5414*\dx},{-0.0563*\dy})
+	-- ({0.5429*\dx},{-0.0526*\dy})
+	-- ({0.5443*\dx},{-0.0489*\dy})
+	-- ({0.5458*\dx},{-0.0453*\dy})
+	-- ({0.5472*\dx},{-0.0416*\dy})
+	-- ({0.5486*\dx},{-0.0380*\dy})
+	-- ({0.5501*\dx},{-0.0344*\dy})
+	-- ({0.5515*\dx},{-0.0308*\dy})
+	-- ({0.5529*\dx},{-0.0272*\dy})
+	-- ({0.5544*\dx},{-0.0237*\dy})
+	-- ({0.5558*\dx},{-0.0202*\dy})
+	-- ({0.5572*\dx},{-0.0167*\dy})
+	-- ({0.5587*\dx},{-0.0132*\dy})
+	-- ({0.5601*\dx},{-0.0097*\dy})
+	-- ({0.5615*\dx},{-0.0062*\dy})
+	-- ({0.5630*\dx},{-0.0028*\dy})
+	-- ({0.5644*\dx},{0.0006*\dy})
+	-- ({0.5659*\dx},{0.0041*\dy})
+	-- ({0.5673*\dx},{0.0075*\dy})
+	-- ({0.5688*\dx},{0.0108*\dy})
+	-- ({0.5702*\dx},{0.0142*\dy})
+	-- ({0.5717*\dx},{0.0176*\dy})
+	-- ({0.5731*\dx},{0.0209*\dy})
+	-- ({0.5746*\dx},{0.0242*\dy})
+	-- ({0.5761*\dx},{0.0275*\dy})
+	-- ({0.5776*\dx},{0.0308*\dy})
+	-- ({0.5790*\dx},{0.0341*\dy})
+	-- ({0.5805*\dx},{0.0374*\dy})
+	-- ({0.5820*\dx},{0.0407*\dy})
+	-- ({0.5835*\dx},{0.0439*\dy})
+	-- ({0.5850*\dx},{0.0471*\dy})
+	-- ({0.5865*\dx},{0.0504*\dy})
+	-- ({0.5880*\dx},{0.0536*\dy})
+	-- ({0.5896*\dx},{0.0568*\dy})
+	-- ({0.5911*\dx},{0.0599*\dy})
+	-- ({0.5926*\dx},{0.0631*\dy})
+	-- ({0.5942*\dx},{0.0663*\dy})
+	-- ({0.5957*\dx},{0.0694*\dy})
+	-- ({0.5973*\dx},{0.0725*\dy})
+	-- ({0.5988*\dx},{0.0757*\dy})
+	-- ({0.6004*\dx},{0.0788*\dy})
+	-- ({0.6020*\dx},{0.0819*\dy})
+	-- ({0.6036*\dx},{0.0850*\dy})
+	-- ({0.6052*\dx},{0.0880*\dy})
+	-- ({0.6068*\dx},{0.0911*\dy})
+	-- ({0.6084*\dx},{0.0942*\dy})
+	-- ({0.6100*\dx},{0.0972*\dy})
+	-- ({0.6117*\dx},{0.1002*\dy})
+	-- ({0.6133*\dx},{0.1033*\dy})
+	-- ({0.6149*\dx},{0.1063*\dy})
+	-- ({0.6166*\dx},{0.1093*\dy})
+	-- ({0.6183*\dx},{0.1123*\dy})
+	-- ({0.6200*\dx},{0.1152*\dy})
+	-- ({0.6217*\dx},{0.1182*\dy})
+	-- ({0.6234*\dx},{0.1212*\dy})
+	-- ({0.6251*\dx},{0.1241*\dy})
+	-- ({0.6268*\dx},{0.1271*\dy})
+	-- ({0.6285*\dx},{0.1300*\dy})
+	-- ({0.6303*\dx},{0.1329*\dy})
+	-- ({0.6320*\dx},{0.1358*\dy})
+	-- ({0.6338*\dx},{0.1387*\dy})
+	-- ({0.6356*\dx},{0.1416*\dy})
+	-- ({0.6374*\dx},{0.1445*\dy})
+	-- ({0.6392*\dx},{0.1473*\dy})
+	-- ({0.6410*\dx},{0.1502*\dy})
+	-- ({0.6428*\dx},{0.1530*\dy})
+	-- ({0.6446*\dx},{0.1559*\dy})
+	-- ({0.6465*\dx},{0.1587*\dy})
+	-- ({0.6484*\dx},{0.1615*\dy})
+	-- ({0.6502*\dx},{0.1643*\dy})
+	-- ({0.6521*\dx},{0.1671*\dy})
+	-- ({0.6540*\dx},{0.1699*\dy})
+	-- ({0.6560*\dx},{0.1727*\dy})
+	-- ({0.6579*\dx},{0.1754*\dy})
+	-- ({0.6598*\dx},{0.1782*\dy})
+	-- ({0.6618*\dx},{0.1809*\dy})
+	-- ({0.6638*\dx},{0.1837*\dy})
+	-- ({0.6658*\dx},{0.1864*\dy})
+	-- ({0.6678*\dx},{0.1891*\dy})
+	-- ({0.6698*\dx},{0.1918*\dy})
+	-- ({0.6718*\dx},{0.1945*\dy})
+	-- ({0.6738*\dx},{0.1972*\dy})
+	-- ({0.6759*\dx},{0.1998*\dy})
+	-- ({0.6780*\dx},{0.2025*\dy})
+	-- ({0.6801*\dx},{0.2052*\dy})
+	-- ({0.6822*\dx},{0.2078*\dy})
+	-- ({0.6843*\dx},{0.2104*\dy})
+	-- ({0.6864*\dx},{0.2130*\dy})
+	-- ({0.6886*\dx},{0.2156*\dy})
+	-- ({0.6907*\dx},{0.2182*\dy})
+	-- ({0.6929*\dx},{0.2208*\dy})
+	-- ({0.6951*\dx},{0.2234*\dy})
+	-- ({0.6973*\dx},{0.2260*\dy})
+	-- ({0.6996*\dx},{0.2285*\dy})
+	-- ({0.7018*\dx},{0.2310*\dy})
+	-- ({0.7041*\dx},{0.2336*\dy})
+	-- ({0.7064*\dx},{0.2361*\dy})
+	-- ({0.7087*\dx},{0.2386*\dy})
+	-- ({0.7110*\dx},{0.2411*\dy})
+	-- ({0.7133*\dx},{0.2436*\dy})
+	-- ({0.7156*\dx},{0.2460*\dy})
+	-- ({0.7180*\dx},{0.2485*\dy})
+	-- ({0.7204*\dx},{0.2509*\dy})
+	-- ({0.7228*\dx},{0.2533*\dy})
+	-- ({0.7252*\dx},{0.2558*\dy})
+	-- ({0.7276*\dx},{0.2582*\dy})
+	-- ({0.7301*\dx},{0.2606*\dy})
+	-- ({0.7326*\dx},{0.2629*\dy})
+	-- ({0.7350*\dx},{0.2653*\dy})
+	-- ({0.7376*\dx},{0.2677*\dy})
+	-- ({0.7401*\dx},{0.2700*\dy})
+	-- ({0.7426*\dx},{0.2723*\dy})
+	-- ({0.7452*\dx},{0.2746*\dy})
+	-- ({0.7478*\dx},{0.2769*\dy})
+	-- ({0.7504*\dx},{0.2792*\dy})
+	-- ({0.7530*\dx},{0.2815*\dy})
+	-- ({0.7556*\dx},{0.2837*\dy})
+	-- ({0.7583*\dx},{0.2860*\dy})
+	-- ({0.7609*\dx},{0.2882*\dy})
+	-- ({0.7636*\dx},{0.2904*\dy})
+	-- ({0.7663*\dx},{0.2926*\dy})
+	-- ({0.7691*\dx},{0.2948*\dy})
+	-- ({0.7718*\dx},{0.2969*\dy})
+	-- ({0.7746*\dx},{0.2991*\dy})
+	-- ({0.7774*\dx},{0.3012*\dy})
+	-- ({0.7802*\dx},{0.3033*\dy})
+	-- ({0.7830*\dx},{0.3054*\dy})
+	-- ({0.7858*\dx},{0.3075*\dy})
+	-- ({0.7887*\dx},{0.3096*\dy})
+	-- ({0.7916*\dx},{0.3116*\dy})
+	-- ({0.7945*\dx},{0.3137*\dy})
+	-- ({0.7974*\dx},{0.3157*\dy})
+	-- ({0.8004*\dx},{0.3177*\dy})
+	-- ({0.8033*\dx},{0.3197*\dy})
+	-- ({0.8063*\dx},{0.3216*\dy})
+	-- ({0.8093*\dx},{0.3236*\dy})
+	-- ({0.8123*\dx},{0.3255*\dy})
+	-- ({0.8154*\dx},{0.3274*\dy})
+	-- ({0.8184*\dx},{0.3293*\dy})
+	-- ({0.8215*\dx},{0.3312*\dy})
+	-- ({0.8246*\dx},{0.3330*\dy})
+	-- ({0.8277*\dx},{0.3348*\dy})
+	-- ({0.8309*\dx},{0.3367*\dy})
+	-- ({0.8340*\dx},{0.3384*\dy})
+	-- ({0.8372*\dx},{0.3402*\dy})
+	-- ({0.8404*\dx},{0.3420*\dy})
+	-- ({0.8437*\dx},{0.3437*\dy})
+	-- ({0.8469*\dx},{0.3454*\dy})
+	-- ({0.8502*\dx},{0.3471*\dy})
+	-- ({0.8534*\dx},{0.3487*\dy})
+	-- ({0.8567*\dx},{0.3504*\dy})
+	-- ({0.8601*\dx},{0.3520*\dy})
+	-- ({0.8634*\dx},{0.3536*\dy})
+	-- ({0.8668*\dx},{0.3552*\dy})
+	-- ({0.8702*\dx},{0.3567*\dy})
+	-- ({0.8736*\dx},{0.3583*\dy})
+	-- ({0.8770*\dx},{0.3598*\dy})
+	-- ({0.8804*\dx},{0.3612*\dy})
+	-- ({0.8839*\dx},{0.3627*\dy})
+	-- ({0.8874*\dx},{0.3641*\dy})
+	-- ({0.8909*\dx},{0.3655*\dy})
+	-- ({0.8944*\dx},{0.3669*\dy})
+	-- ({0.8980*\dx},{0.3683*\dy})
+	-- ({0.9015*\dx},{0.3696*\dy})
+	-- ({0.9051*\dx},{0.3709*\dy})
+	-- ({0.9087*\dx},{0.3722*\dy})
+	-- ({0.9124*\dx},{0.3734*\dy})
+	-- ({0.9160*\dx},{0.3746*\dy})
+	-- ({0.9197*\dx},{0.3758*\dy})
+	-- ({0.9233*\dx},{0.3770*\dy})
+	-- ({0.9271*\dx},{0.3781*\dy})
+	-- ({0.9308*\dx},{0.3793*\dy})
+	-- ({0.9345*\dx},{0.3803*\dy})
+	-- ({0.9383*\dx},{0.3814*\dy})
+	-- ({0.9421*\dx},{0.3824*\dy})
+	-- ({0.9459*\dx},{0.3834*\dy})
+	-- ({0.9497*\dx},{0.3844*\dy})
+	-- ({0.9535*\dx},{0.3853*\dy})
+	-- ({0.9574*\dx},{0.3862*\dy})
+	-- ({0.9612*\dx},{0.3871*\dy})
+	-- ({0.9651*\dx},{0.3879*\dy})
+	-- ({0.9690*\dx},{0.3887*\dy})
+	-- ({0.9730*\dx},{0.3895*\dy})
+	-- ({0.9769*\dx},{0.3902*\dy})
+	-- ({0.9809*\dx},{0.3910*\dy})
+	-- ({0.9848*\dx},{0.3916*\dy})
+	-- ({0.9888*\dx},{0.3923*\dy})
+	-- ({0.9929*\dx},{0.3929*\dy})
+	-- ({0.9969*\dx},{0.3935*\dy})
+	-- ({1.0009*\dx},{0.3940*\dy})
+	-- ({1.0050*\dx},{0.3945*\dy})
+	-- ({1.0091*\dx},{0.3950*\dy})
+	-- ({1.0132*\dx},{0.3954*\dy})
+	-- ({1.0173*\dx},{0.3958*\dy})
+	-- ({1.0214*\dx},{0.3962*\dy})
+	-- ({1.0256*\dx},{0.3965*\dy})
+	-- ({1.0297*\dx},{0.3968*\dy})
+	-- ({1.0339*\dx},{0.3971*\dy})
+	-- ({1.0381*\dx},{0.3973*\dy})
+	-- ({1.0423*\dx},{0.3975*\dy})
+	-- ({1.0465*\dx},{0.3976*\dy})
+	-- ({1.0508*\dx},{0.3977*\dy})
+	-- ({1.0550*\dx},{0.3977*\dy})
+	-- ({1.0593*\dx},{0.3978*\dy})
+	-- ({1.0636*\dx},{0.3977*\dy})
+	-- ({1.0679*\dx},{0.3977*\dy})
+	-- ({1.0722*\dx},{0.3976*\dy})
+	-- ({1.0765*\dx},{0.3974*\dy})
+	-- ({1.0808*\dx},{0.3973*\dy})
+	-- ({1.0851*\dx},{0.3970*\dy})
+	-- ({1.0895*\dx},{0.3968*\dy})
+	-- ({1.0938*\dx},{0.3965*\dy})
+	-- ({1.0982*\dx},{0.3961*\dy})
+	-- ({1.1026*\dx},{0.3957*\dy})
+	-- ({1.1070*\dx},{0.3953*\dy})
+	-- ({1.1114*\dx},{0.3948*\dy})
+	-- ({1.1158*\dx},{0.3943*\dy})
+	-- ({1.1202*\dx},{0.3937*\dy})
+	-- ({1.1247*\dx},{0.3931*\dy})
+	-- ({1.1291*\dx},{0.3924*\dy})
+	-- ({1.1336*\dx},{0.3917*\dy})
+	-- ({1.1380*\dx},{0.3909*\dy})
+	-- ({1.1425*\dx},{0.3901*\dy})
+	-- ({1.1469*\dx},{0.3893*\dy})
+	-- ({1.1514*\dx},{0.3884*\dy})
+	-- ({1.1559*\dx},{0.3875*\dy})
+	-- ({1.1604*\dx},{0.3865*\dy})
+	-- ({1.1649*\dx},{0.3854*\dy})
+	-- ({1.1694*\dx},{0.3843*\dy})
+	-- ({1.1739*\dx},{0.3832*\dy})
+	-- ({1.1784*\dx},{0.3820*\dy})
+	-- ({1.1829*\dx},{0.3808*\dy})
+	-- ({1.1874*\dx},{0.3795*\dy})
+	-- ({1.1919*\dx},{0.3782*\dy})
+	-- ({1.1965*\dx},{0.3768*\dy})
+	-- ({1.2010*\dx},{0.3753*\dy})
+	-- ({1.2055*\dx},{0.3738*\dy})
+	-- ({1.2100*\dx},{0.3723*\dy})
+	-- ({1.2145*\dx},{0.3707*\dy})
+	-- ({1.2190*\dx},{0.3691*\dy})
+	-- ({1.2236*\dx},{0.3674*\dy})
+	-- ({1.2281*\dx},{0.3656*\dy})
+	-- ({1.2326*\dx},{0.3638*\dy})
+	-- ({1.2371*\dx},{0.3620*\dy})
+	-- ({1.2416*\dx},{0.3600*\dy})
+	-- ({1.2461*\dx},{0.3581*\dy})
+	-- ({1.2506*\dx},{0.3561*\dy})
+	-- ({1.2551*\dx},{0.3540*\dy})
+	-- ({1.2596*\dx},{0.3519*\dy})
+	-- ({1.2641*\dx},{0.3497*\dy})
+	-- ({1.2686*\dx},{0.3475*\dy})
+	-- ({1.2730*\dx},{0.3452*\dy})
+	-- ({1.2775*\dx},{0.3428*\dy})
+	-- ({1.2819*\dx},{0.3404*\dy})
+	-- ({1.2864*\dx},{0.3379*\dy})
+	-- ({1.2908*\dx},{0.3354*\dy})
+	-- ({1.2952*\dx},{0.3329*\dy})
+	-- ({1.2996*\dx},{0.3302*\dy})
+	-- ({1.3040*\dx},{0.3275*\dy})
+	-- ({1.3084*\dx},{0.3248*\dy})
+	-- ({1.3128*\dx},{0.3220*\dy})
+	-- ({1.3171*\dx},{0.3191*\dy})
+	-- ({1.3215*\dx},{0.3162*\dy})
+	-- ({1.3258*\dx},{0.3132*\dy})
+	-- ({1.3301*\dx},{0.3102*\dy})
+	-- ({1.3344*\dx},{0.3071*\dy})
+	-- ({1.3386*\dx},{0.3040*\dy})
+	-- ({1.3429*\dx},{0.3008*\dy})
+	-- ({1.3471*\dx},{0.2975*\dy})
+	-- ({1.3513*\dx},{0.2942*\dy})
+	-- ({1.3555*\dx},{0.2908*\dy})
+	-- ({1.3597*\dx},{0.2874*\dy})
+	-- ({1.3638*\dx},{0.2839*\dy})
+	-- ({1.3680*\dx},{0.2803*\dy})
+	-- ({1.3721*\dx},{0.2767*\dy})
+	-- ({1.3761*\dx},{0.2730*\dy})
+	-- ({1.3802*\dx},{0.2693*\dy})
+	-- ({1.3842*\dx},{0.2655*\dy})
+	-- ({1.3882*\dx},{0.2616*\dy})
+	-- ({1.3922*\dx},{0.2577*\dy})
+	-- ({1.3961*\dx},{0.2537*\dy})
+	-- ({1.4000*\dx},{0.2497*\dy})
+	-- ({1.4039*\dx},{0.2456*\dy})
+	-- ({1.4077*\dx},{0.2414*\dy})
+	-- ({1.4115*\dx},{0.2372*\dy})
+	-- ({1.4153*\dx},{0.2329*\dy})
+	-- ({1.4191*\dx},{0.2286*\dy})
+	-- ({1.4228*\dx},{0.2242*\dy})
+	-- ({1.4264*\dx},{0.2197*\dy})
+	-- ({1.4301*\dx},{0.2152*\dy})
+	-- ({1.4337*\dx},{0.2107*\dy})
+	-- ({1.4372*\dx},{0.2060*\dy})
+	-- ({1.4407*\dx},{0.2014*\dy})
+	-- ({1.4442*\dx},{0.1966*\dy})
+	-- ({1.4476*\dx},{0.1918*\dy})
+	-- ({1.4510*\dx},{0.1869*\dy})
+	-- ({1.4544*\dx},{0.1820*\dy})
+	-- ({1.4577*\dx},{0.1770*\dy})
+	-- ({1.4609*\dx},{0.1720*\dy})
+	-- ({1.4642*\dx},{0.1669*\dy})
+	-- ({1.4673*\dx},{0.1618*\dy})
+	-- ({1.4704*\dx},{0.1566*\dy})
+	-- ({1.4735*\dx},{0.1513*\dy})
+	-- ({1.4765*\dx},{0.1460*\dy})
+	-- ({1.4795*\dx},{0.1406*\dy})
+	-- ({1.4824*\dx},{0.1352*\dy})
+	-- ({1.4853*\dx},{0.1297*\dy})
+	-- ({1.4881*\dx},{0.1241*\dy})
+	-- ({1.4908*\dx},{0.1185*\dy})
+	-- ({1.4935*\dx},{0.1129*\dy})
+	-- ({1.4961*\dx},{0.1072*\dy})
+	-- ({1.4987*\dx},{0.1014*\dy})
+	-- ({1.5012*\dx},{0.0956*\dy})
+	-- ({1.5037*\dx},{0.0897*\dy})
+	-- ({1.5061*\dx},{0.0838*\dy})
+	-- ({1.5084*\dx},{0.0778*\dy})
+	-- ({1.5107*\dx},{0.0718*\dy})
+	-- ({1.5129*\dx},{0.0657*\dy})
+	-- ({1.5151*\dx},{0.0596*\dy})
+	-- ({1.5172*\dx},{0.0534*\dy})
+	-- ({1.5192*\dx},{0.0472*\dy})
+	-- ({1.5211*\dx},{0.0409*\dy})
+	-- ({1.5230*\dx},{0.0346*\dy})
+	-- ({1.5248*\dx},{0.0282*\dy})
+	-- ({1.5265*\dx},{0.0218*\dy})
+	-- ({1.5282*\dx},{0.0153*\dy})
+	-- ({1.5298*\dx},{0.0088*\dy})
+	-- ({1.5313*\dx},{0.0023*\dy})
+	-- ({1.5328*\dx},{-0.0043*\dy})
+	-- ({1.5341*\dx},{-0.0110*\dy})
+	-- ({1.5354*\dx},{-0.0177*\dy})
+	-- ({1.5366*\dx},{-0.0244*\dy})
+	-- ({1.5378*\dx},{-0.0312*\dy})
+	-- ({1.5388*\dx},{-0.0380*\dy})
+	-- ({1.5398*\dx},{-0.0448*\dy})
+	-- ({1.5407*\dx},{-0.0517*\dy})
+	-- ({1.5415*\dx},{-0.0586*\dy})
+	-- ({1.5422*\dx},{-0.0656*\dy})
+	-- ({1.5429*\dx},{-0.0726*\dy})
+	-- ({1.5434*\dx},{-0.0796*\dy})
+	-- ({1.5439*\dx},{-0.0867*\dy})
+	-- ({1.5443*\dx},{-0.0938*\dy})
+	-- ({1.5446*\dx},{-0.1010*\dy})
+	-- ({1.5448*\dx},{-0.1081*\dy})
+	-- ({1.5449*\dx},{-0.1153*\dy})
+	-- ({1.5449*\dx},{-0.1226*\dy})
+	-- ({1.5449*\dx},{-0.1298*\dy})
+	-- ({1.5447*\dx},{-0.1371*\dy})
+	-- ({1.5444*\dx},{-0.1444*\dy})
+	-- ({1.5441*\dx},{-0.1518*\dy})
+	-- ({1.5437*\dx},{-0.1591*\dy})
+	-- ({1.5431*\dx},{-0.1665*\dy})
+	-- ({1.5425*\dx},{-0.1740*\dy})
+	-- ({1.5418*\dx},{-0.1814*\dy})
+	-- ({1.5409*\dx},{-0.1888*\dy})
+	-- ({1.5400*\dx},{-0.1963*\dy})
+	-- ({1.5390*\dx},{-0.2038*\dy})
+	-- ({1.5378*\dx},{-0.2113*\dy})
+	-- ({1.5366*\dx},{-0.2188*\dy})
+	-- ({1.5353*\dx},{-0.2264*\dy})
+	-- ({1.5339*\dx},{-0.2339*\dy})
+	-- ({1.5323*\dx},{-0.2415*\dy})
+	-- ({1.5307*\dx},{-0.2491*\dy})
+	-- ({1.5289*\dx},{-0.2566*\dy})
+	-- ({1.5271*\dx},{-0.2642*\dy})
+	-- ({1.5251*\dx},{-0.2718*\dy})
+	-- ({1.5230*\dx},{-0.2794*\dy})
+	-- ({1.5209*\dx},{-0.2870*\dy})
+	-- ({1.5186*\dx},{-0.2946*\dy})
+	-- ({1.5162*\dx},{-0.3022*\dy})
+	-- ({1.5137*\dx},{-0.3098*\dy})
+	-- ({1.5111*\dx},{-0.3174*\dy})
+	-- ({1.5084*\dx},{-0.3250*\dy})
+	-- ({1.5055*\dx},{-0.3326*\dy})
+	-- ({1.5026*\dx},{-0.3402*\dy})
+	-- ({1.4995*\dx},{-0.3477*\dy})
+	-- ({1.4963*\dx},{-0.3553*\dy})
+	-- ({1.4931*\dx},{-0.3628*\dy})
+	-- ({1.4897*\dx},{-0.3704*\dy})
+	-- ({1.4862*\dx},{-0.3779*\dy})
+	-- ({1.4825*\dx},{-0.3854*\dy})
+	-- ({1.4788*\dx},{-0.3929*\dy})
+	-- ({1.4749*\dx},{-0.4003*\dy})
+	-- ({1.4710*\dx},{-0.4078*\dy})
+	-- ({1.4669*\dx},{-0.4152*\dy})
+	-- ({1.4627*\dx},{-0.4226*\dy})
+	-- ({1.4584*\dx},{-0.4300*\dy})
+	-- ({1.4539*\dx},{-0.4373*\dy})
+	-- ({1.4494*\dx},{-0.4446*\dy})
+	-- ({1.4447*\dx},{-0.4519*\dy})
+	-- ({1.4399*\dx},{-0.4591*\dy})
+	-- ({1.4350*\dx},{-0.4663*\dy})
+	-- ({1.4300*\dx},{-0.4735*\dy})
+	-- ({1.4249*\dx},{-0.4806*\dy})
+	-- ({1.4197*\dx},{-0.4877*\dy})
+	-- ({1.4143*\dx},{-0.4947*\dy})
+	-- ({1.4088*\dx},{-0.5017*\dy})
+	-- ({1.4032*\dx},{-0.5086*\dy})
+	-- ({1.3975*\dx},{-0.5155*\dy})
+	-- ({1.3916*\dx},{-0.5224*\dy})
+	-- ({1.3857*\dx},{-0.5292*\dy})
+	-- ({1.3796*\dx},{-0.5359*\dy})
+	-- ({1.3734*\dx},{-0.5426*\dy})
+	-- ({1.3671*\dx},{-0.5492*\dy})
+	-- ({1.3607*\dx},{-0.5558*\dy})
+	-- ({1.3542*\dx},{-0.5623*\dy})
+	-- ({1.3475*\dx},{-0.5687*\dy})
+	-- ({1.3408*\dx},{-0.5751*\dy})
+	-- ({1.3339*\dx},{-0.5814*\dy})
+	-- ({1.3269*\dx},{-0.5876*\dy})
+	-- ({1.3198*\dx},{-0.5937*\dy})
+	-- ({1.3126*\dx},{-0.5998*\dy})
+	-- ({1.3052*\dx},{-0.6058*\dy})
+	-- ({1.2978*\dx},{-0.6117*\dy})
+	-- ({1.2902*\dx},{-0.6176*\dy})
+	-- ({1.2826*\dx},{-0.6233*\dy})
+	-- ({1.2748*\dx},{-0.6290*\dy})
+	-- ({1.2669*\dx},{-0.6346*\dy})
+	-- ({1.2589*\dx},{-0.6401*\dy})
+	-- ({1.2508*\dx},{-0.6455*\dy})
+	-- ({1.2426*\dx},{-0.6508*\dy})
+	-- ({1.2343*\dx},{-0.6560*\dy})
+	-- ({1.2259*\dx},{-0.6611*\dy})
+	-- ({1.2173*\dx},{-0.6662*\dy})
+	-- ({1.2087*\dx},{-0.6711*\dy})
+	-- ({1.2000*\dx},{-0.6759*\dy})
+	-- ({1.1911*\dx},{-0.6806*\dy})
+	-- ({1.1822*\dx},{-0.6852*\dy})
+	-- ({1.1731*\dx},{-0.6897*\dy})
+	-- ({1.1640*\dx},{-0.6941*\dy})
+	-- ({1.1548*\dx},{-0.6984*\dy})
+	-- ({1.1454*\dx},{-0.7025*\dy})
+	-- ({1.1360*\dx},{-0.7065*\dy})
+	-- ({1.1265*\dx},{-0.7105*\dy})
+	-- ({1.1169*\dx},{-0.7143*\dy})
+	-- ({1.1072*\dx},{-0.7179*\dy})
+	-- ({1.0974*\dx},{-0.7215*\dy})
+	-- ({1.0875*\dx},{-0.7249*\dy})
+	-- ({1.0775*\dx},{-0.7282*\dy})
+	-- ({1.0674*\dx},{-0.7313*\dy})
+	-- ({1.0573*\dx},{-0.7344*\dy})
+	-- ({1.0471*\dx},{-0.7373*\dy})
+	-- ({1.0368*\dx},{-0.7400*\dy})
+	-- ({1.0264*\dx},{-0.7426*\dy})
+	-- ({1.0159*\dx},{-0.7451*\dy})
+	-- ({1.0054*\dx},{-0.7474*\dy})
+	-- ({0.9948*\dx},{-0.7496*\dy})
+	-- ({0.9841*\dx},{-0.7517*\dy})
+	-- ({0.9734*\dx},{-0.7536*\dy})
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+	-- ({0.9516*\dx},{-0.7569*\dy})
+	-- ({0.9407*\dx},{-0.7584*\dy})
+	-- ({0.9297*\dx},{-0.7596*\dy})
+	-- ({0.9186*\dx},{-0.7608*\dy})
+	-- ({0.9075*\dx},{-0.7617*\dy})
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+	-- ({0.8850*\dx},{-0.7632*\dy})
+	-- ({0.8738*\dx},{-0.7637*\dy})
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+	-- ({0.8510*\dx},{-0.7642*\dy})
+	-- ({0.8396*\dx},{-0.7642*\dy})
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+	-- ({0.1004*\dx},{-0.3000*\dy})
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+	-- ({0.0563*\dx},{-0.2050*\dy})
+	-- ({0.0508*\dx},{-0.1908*\dy})
+	-- ({0.0455*\dx},{-0.1766*\dy})
+	-- ({0.0404*\dx},{-0.1622*\dy})
+	-- ({0.0355*\dx},{-0.1476*\dy})
+	-- ({0.0309*\dx},{-0.1330*\dy})
+	-- ({0.0264*\dx},{-0.1182*\dy})
+	-- ({0.0222*\dx},{-0.1033*\dy})
+	-- ({0.0183*\dx},{-0.0882*\dy})
+	-- ({0.0145*\dx},{-0.0731*\dy})
+	-- ({0.0110*\dx},{-0.0579*\dy})
+	-- ({0.0077*\dx},{-0.0425*\dy})
+	-- ({0.0047*\dx},{-0.0271*\dy})
+	-- ({0.0019*\dx},{-0.0115*\dy})
+	-- ({-0.0006*\dx},{0.0041*\dy})
+	-- ({-0.0030*\dx},{0.0199*\dy})
+	-- ({-0.0050*\dx},{0.0357*\dy})
+	-- ({-0.0068*\dx},{0.0516*\dy})
+	-- ({-0.0084*\dx},{0.0676*\dy})
+	-- ({-0.0097*\dx},{0.0836*\dy})
+	-- ({-0.0107*\dx},{0.0998*\dy})
+	-- ({-0.0115*\dx},{0.1160*\dy})
+	-- ({-0.0120*\dx},{0.1322*\dy})
+	-- ({-0.0122*\dx},{0.1486*\dy})
+	-- ({-0.0122*\dx},{0.1649*\dy})
+	-- ({-0.0119*\dx},{0.1813*\dy})
+	-- ({-0.0114*\dx},{0.1978*\dy})
+	-- ({-0.0105*\dx},{0.2143*\dy})
+	-- ({-0.0094*\dx},{0.2309*\dy})
+	-- ({-0.0080*\dx},{0.2475*\dy})
+	-- ({-0.0064*\dx},{0.2641*\dy})
+	-- ({-0.0044*\dx},{0.2807*\dy})
+	-- ({-0.0022*\dx},{0.2973*\dy})
+	-- ({0.0003*\dx},{0.3140*\dy})
+	-- ({0.0031*\dx},{0.3307*\dy})
+	-- ({0.0061*\dx},{0.3473*\dy})
+	-- ({0.0095*\dx},{0.3640*\dy})
+	-- ({0.0131*\dx},{0.3807*\dy})
+	-- ({0.0171*\dx},{0.3973*\dy})
+	-- ({0.0213*\dx},{0.4140*\dy})
+	-- ({0.0258*\dx},{0.4306*\dy})
+	-- ({0.0306*\dx},{0.4472*\dy})
+	-- ({0.0357*\dx},{0.4637*\dy})
+	-- ({0.0411*\dx},{0.4803*\dy})
+	-- ({0.0467*\dx},{0.4967*\dy})
+	-- ({0.0527*\dx},{0.5132*\dy})
+	-- ({0.0590*\dx},{0.5296*\dy})
+	-- ({0.0655*\dx},{0.5459*\dy})
+	-- ({0.0724*\dx},{0.5621*\dy})
+	-- ({0.0795*\dx},{0.5783*\dy})
+	-- ({0.0869*\dx},{0.5945*\dy})
+	-- ({0.0947*\dx},{0.6105*\dy})
+	-- ({0.1027*\dx},{0.6264*\dy})
+	-- ({0.1110*\dx},{0.6423*\dy})
+	-- ({0.1196*\dx},{0.6581*\dy})
+	-- ({0.1285*\dx},{0.6737*\dy})
+	-- ({0.1376*\dx},{0.6893*\dy})
+	-- ({0.1471*\dx},{0.7048*\dy})
+	-- ({0.1569*\dx},{0.7201*\dy})
+	-- ({0.1669*\dx},{0.7353*\dy})
+	-- ({0.1772*\dx},{0.7504*\dy})
+	-- ({0.1878*\dx},{0.7653*\dy})
+	-- ({0.1987*\dx},{0.7801*\dy})
+	-- ({0.2099*\dx},{0.7948*\dy})
+	-- ({0.2214*\dx},{0.8093*\dy})
+	-- ({0.2331*\dx},{0.8236*\dy})
+	-- ({0.2451*\dx},{0.8378*\dy})
+	-- ({0.2574*\dx},{0.8519*\dy})
+	-- ({0.2700*\dx},{0.8657*\dy})
+	-- ({0.2828*\dx},{0.8794*\dy})
+	-- ({0.2959*\dx},{0.8929*\dy})
+	-- ({0.3093*\dx},{0.9062*\dy})
+	-- ({0.3229*\dx},{0.9193*\dy})
+	-- ({0.3368*\dx},{0.9322*\dy})
+	-- ({0.3509*\dx},{0.9449*\dy})
+	-- ({0.3653*\dx},{0.9574*\dy})
+	-- ({0.3800*\dx},{0.9697*\dy})
+	-- ({0.3949*\dx},{0.9817*\dy})
+	-- ({0.4100*\dx},{0.9936*\dy})
+	-- ({0.4254*\dx},{1.0052*\dy})
+	-- ({0.4411*\dx},{1.0165*\dy})
+	-- ({0.4569*\dx},{1.0276*\dy})
+	-- ({0.4730*\dx},{1.0385*\dy})
+	-- ({0.4894*\dx},{1.0491*\dy})
+	-- ({0.5059*\dx},{1.0595*\dy})
+	-- ({0.5227*\dx},{1.0696*\dy})
+	-- ({0.5397*\dx},{1.0795*\dy})
+	-- ({0.5569*\dx},{1.0890*\dy})
+	-- ({0.5743*\dx},{1.0983*\dy})
+	-- ({0.5919*\dx},{1.1073*\dy})
+	-- ({0.6098*\dx},{1.1160*\dy})
+	-- ({0.6278*\dx},{1.1245*\dy})
+	-- ({0.6460*\dx},{1.1326*\dy})
+	-- ({0.6644*\dx},{1.1405*\dy})
+	-- ({0.6830*\dx},{1.1480*\dy})
+	-- ({0.7017*\dx},{1.1552*\dy})
+	-- ({0.7206*\dx},{1.1622*\dy})
+	-- ({0.7397*\dx},{1.1688*\dy})
+	-- ({0.7590*\dx},{1.1750*\dy})
+	-- ({0.7784*\dx},{1.1810*\dy})
+	-- ({0.7979*\dx},{1.1866*\dy})
+	-- ({0.8176*\dx},{1.1919*\dy})
+	-- ({0.8375*\dx},{1.1969*\dy})
+	-- ({0.8574*\dx},{1.2015*\dy})
+	-- ({0.8775*\dx},{1.2058*\dy})
+	-- ({0.8978*\dx},{1.2098*\dy})
+	-- ({0.9181*\dx},{1.2133*\dy})
+	-- ({0.9385*\dx},{1.2166*\dy})
+	-- ({0.9591*\dx},{1.2195*\dy})
+	-- ({0.9797*\dx},{1.2220*\dy})
+	-- ({1.0005*\dx},{1.2242*\dy})
+	-- ({1.0213*\dx},{1.2260*\dy})
+	-- ({1.0422*\dx},{1.2274*\dy})
+	-- ({1.0632*\dx},{1.2285*\dy})
+	-- ({1.0842*\dx},{1.2292*\dy})
+	-- ({1.1053*\dx},{1.2295*\dy})
+	-- ({1.1264*\dx},{1.2294*\dy})
+	-- ({1.1476*\dx},{1.2290*\dy})
+	-- ({1.1688*\dx},{1.2282*\dy})
+	-- ({1.1901*\dx},{1.2270*\dy})
+	-- ({1.2114*\dx},{1.2254*\dy})
+	-- ({1.2327*\dx},{1.2234*\dy})
+	-- ({1.2540*\dx},{1.2211*\dy})
+	-- ({1.2753*\dx},{1.2184*\dy})
+	-- ({1.2965*\dx},{1.2152*\dy})
+	-- ({1.3178*\dx},{1.2117*\dy})
+	-- ({1.3391*\dx},{1.2078*\dy})
+	-- ({1.3603*\dx},{1.2035*\dy})
+	-- ({1.3815*\dx},{1.1988*\dy})
+	-- ({1.4027*\dx},{1.1938*\dy})
+	-- ({1.4238*\dx},{1.1883*\dy})
+	-- ({1.4448*\dx},{1.1824*\dy})
+	-- ({1.4658*\dx},{1.1762*\dy})
+	-- ({1.4867*\dx},{1.1695*\dy})
+	-- ({1.5076*\dx},{1.1625*\dy})
+	-- ({1.5283*\dx},{1.1550*\dy})
+	-- ({1.5489*\dx},{1.1472*\dy})
+	-- ({1.5695*\dx},{1.1390*\dy})
+	-- ({1.5899*\dx},{1.1304*\dy})
+	-- ({1.6102*\dx},{1.1214*\dy})
+	-- ({1.6304*\dx},{1.1120*\dy})
+	-- ({1.6504*\dx},{1.1023*\dy})
+	-- ({1.6703*\dx},{1.0921*\dy})
+	-- ({1.6901*\dx},{1.0816*\dy})
+	-- ({1.7097*\dx},{1.0707*\dy})
+	-- ({1.7291*\dx},{1.0594*\dy})
+	-- ({1.7484*\dx},{1.0477*\dy})
+	-- ({1.7675*\dx},{1.0357*\dy})
+	-- ({1.7864*\dx},{1.0233*\dy})
+	-- ({1.8051*\dx},{1.0105*\dy})
+	-- ({1.8235*\dx},{0.9974*\dy})
+	-- ({1.8418*\dx},{0.9839*\dy})
+	-- ({1.8599*\dx},{0.9700*\dy})
+	-- ({1.8777*\dx},{0.9558*\dy})
+	-- ({1.8953*\dx},{0.9412*\dy})
+	-- ({1.9127*\dx},{0.9263*\dy})
+	-- ({1.9298*\dx},{0.9110*\dy})
+	-- ({1.9467*\dx},{0.8954*\dy})
+	-- ({1.9632*\dx},{0.8795*\dy})
+	-- ({1.9796*\dx},{0.8632*\dy})
+	-- ({1.9956*\dx},{0.8466*\dy})
+	-- ({2.0114*\dx},{0.8296*\dy})
+	-- ({2.0269*\dx},{0.8124*\dy})
+	-- ({2.0420*\dx},{0.7948*\dy})
+	-- ({2.0569*\dx},{0.7770*\dy})
+	-- ({2.0715*\dx},{0.7588*\dy})
+	-- ({2.0857*\dx},{0.7403*\dy})
+	-- ({2.0996*\dx},{0.7215*\dy})
+	-- ({2.1132*\dx},{0.7025*\dy})
+	-- ({2.1264*\dx},{0.6831*\dy})
+	-- ({2.1393*\dx},{0.6635*\dy})
+	-- ({2.1519*\dx},{0.6437*\dy})
+	-- ({2.1641*\dx},{0.6235*\dy})
+	-- ({2.1759*\dx},{0.6031*\dy})
+	-- ({2.1874*\dx},{0.5825*\dy})
+	-- ({2.1985*\dx},{0.5616*\dy})
+	-- ({2.2092*\dx},{0.5404*\dy})
+	-- ({2.2195*\dx},{0.5191*\dy})
+	-- ({2.2295*\dx},{0.4975*\dy})
+	-- ({2.2390*\dx},{0.4757*\dy})
+	-- ({2.2481*\dx},{0.4537*\dy})
+	-- ({2.2569*\dx},{0.4315*\dy})
+	-- ({2.2652*\dx},{0.4091*\dy})
+	-- ({2.2731*\dx},{0.3865*\dy})
+	-- ({2.2806*\dx},{0.3638*\dy})
+	-- ({2.2877*\dx},{0.3408*\dy})
+	-- ({2.2943*\dx},{0.3177*\dy})
+	-- ({2.3005*\dx},{0.2945*\dy})
+	-- ({2.3063*\dx},{0.2711*\dy})
+	-- ({2.3116*\dx},{0.2476*\dy})
+	-- ({2.3165*\dx},{0.2239*\dy})
+	-- ({2.3210*\dx},{0.2002*\dy})
+	-- ({2.3249*\dx},{0.1763*\dy})
+	-- ({2.3285*\dx},{0.1523*\dy})
+	-- ({2.3316*\dx},{0.1283*\dy})
+	-- ({2.3342*\dx},{0.1041*\dy})
+	-- ({2.3364*\dx},{0.0799*\dy})
+	-- ({2.3381*\dx},{0.0556*\dy})
+	-- ({2.3393*\dx},{0.0312*\dy})
+	-- ({2.3401*\dx},{0.0068*\dy})
+	-- ({2.3403*\dx},{-0.0176*\dy})
+	-- ({2.3402*\dx},{-0.0421*\dy})
+	-- ({2.3395*\dx},{-0.0665*\dy})
+	-- ({2.3384*\dx},{-0.0910*\dy})
+	-- ({2.3368*\dx},{-0.1155*\dy})
+	-- ({2.3347*\dx},{-0.1400*\dy})
+	-- ({2.3322*\dx},{-0.1644*\dy})
+	-- ({2.3292*\dx},{-0.1889*\dy})
+	-- ({2.3257*\dx},{-0.2133*\dy})
+	-- ({2.3217*\dx},{-0.2376*\dy})
+	-- ({2.3172*\dx},{-0.2619*\dy})
+	-- ({2.3123*\dx},{-0.2861*\dy})
+	-- ({2.3069*\dx},{-0.3102*\dy})
+	-- ({2.3010*\dx},{-0.3342*\dy})
+	-- ({2.2946*\dx},{-0.3582*\dy})
+	-- ({2.2878*\dx},{-0.3820*\dy})
+	-- ({2.2805*\dx},{-0.4057*\dy})
+	-- ({2.2727*\dx},{-0.4293*\dy})
+	-- ({2.2645*\dx},{-0.4528*\dy})
+	-- ({2.2558*\dx},{-0.4761*\dy})
+	-- ({2.2466*\dx},{-0.4992*\dy})
+	-- ({2.2370*\dx},{-0.5222*\dy})
+	-- ({2.2269*\dx},{-0.5450*\dy})
+	-- ({2.2164*\dx},{-0.5676*\dy})
+	-- ({2.2054*\dx},{-0.5900*\dy})
+	-- ({2.1939*\dx},{-0.6122*\dy})
+	-- ({2.1820*\dx},{-0.6342*\dy})
+	-- ({2.1697*\dx},{-0.6560*\dy})
+	-- ({2.1569*\dx},{-0.6775*\dy})
+	-- ({2.1437*\dx},{-0.6988*\dy})
+	-- ({2.1301*\dx},{-0.7198*\dy})
+	-- ({2.1161*\dx},{-0.7406*\dy})
+	-- ({2.1016*\dx},{-0.7611*\dy})
+	-- ({2.0867*\dx},{-0.7813*\dy})
+	-- ({2.0714*\dx},{-0.8012*\dy})
+	-- ({2.0558*\dx},{-0.8208*\dy})
+	-- ({2.0397*\dx},{-0.8401*\dy})
+	-- ({2.0232*\dx},{-0.8591*\dy})
+	-- ({2.0063*\dx},{-0.8778*\dy})
+	-- ({1.9891*\dx},{-0.8961*\dy})
+	-- ({1.9715*\dx},{-0.9141*\dy})
+	-- ({1.9535*\dx},{-0.9318*\dy})
+	-- ({1.9352*\dx},{-0.9491*\dy})
+	-- ({1.9165*\dx},{-0.9660*\dy})
+	-- ({1.8974*\dx},{-0.9825*\dy})
+	-- ({1.8781*\dx},{-0.9987*\dy})
+	-- ({1.8584*\dx},{-1.0144*\dy})
+	-- ({1.8384*\dx},{-1.0298*\dy})
+	-- ({1.8181*\dx},{-1.0448*\dy})
+	-- ({1.7974*\dx},{-1.0593*\dy})
+	-- ({1.7765*\dx},{-1.0735*\dy})
+	-- ({1.7553*\dx},{-1.0872*\dy})
+	-- ({1.7338*\dx},{-1.1004*\dy})
+	-- ({1.7120*\dx},{-1.1133*\dy})
+	-- ({1.6900*\dx},{-1.1257*\dy})
+	-- ({1.6677*\dx},{-1.1376*\dy})
+	-- ({1.6452*\dx},{-1.1491*\dy})
+	-- ({1.6225*\dx},{-1.1601*\dy})
+	-- ({1.5995*\dx},{-1.1707*\dy})
+	-- ({1.5764*\dx},{-1.1808*\dy})
+	-- ({1.5530*\dx},{-1.1904*\dy})
+	-- ({1.5294*\dx},{-1.1995*\dy})
+	-- ({1.5057*\dx},{-1.2081*\dy})
+	-- ({1.4817*\dx},{-1.2162*\dy})
+	-- ({1.4576*\dx},{-1.2239*\dy})
+	-- ({1.4334*\dx},{-1.2310*\dy})
+	-- ({1.4090*\dx},{-1.2377*\dy})
+	-- ({1.3845*\dx},{-1.2438*\dy})
+	-- ({1.3599*\dx},{-1.2494*\dy})
+	-- ({1.3352*\dx},{-1.2545*\dy})
+	-- ({1.3103*\dx},{-1.2591*\dy})
+	-- ({1.2854*\dx},{-1.2632*\dy})
+	-- ({1.2605*\dx},{-1.2667*\dy})
+	-- ({1.2354*\dx},{-1.2698*\dy})
+	-- ({1.2103*\dx},{-1.2723*\dy})
+	-- ({1.1852*\dx},{-1.2742*\dy})
+	-- ({1.1600*\dx},{-1.2757*\dy})
+	-- ({1.1348*\dx},{-1.2766*\dy})
+	-- ({1.1097*\dx},{-1.2770*\dy})
+	-- ({1.0845*\dx},{-1.2768*\dy})
+	-- ({1.0593*\dx},{-1.2762*\dy})
+	-- ({1.0342*\dx},{-1.2750*\dy})
+	-- ({1.0091*\dx},{-1.2732*\dy})
+	-- ({0.9841*\dx},{-1.2710*\dy})
+	-- ({0.9591*\dx},{-1.2682*\dy})
+	-- ({0.9342*\dx},{-1.2649*\dy})
+	-- ({0.9094*\dx},{-1.2610*\dy})
+	-- ({0.8848*\dx},{-1.2567*\dy})
+	-- ({0.8602*\dx},{-1.2518*\dy})
+	-- ({0.8357*\dx},{-1.2464*\dy})
+	-- ({0.8114*\dx},{-1.2404*\dy})
+	-- ({0.7872*\dx},{-1.2340*\dy})
+	-- ({0.7632*\dx},{-1.2271*\dy})
+	-- ({0.7394*\dx},{-1.2196*\dy})
+	-- ({0.7157*\dx},{-1.2116*\dy})
+	-- ({0.6923*\dx},{-1.2032*\dy})
+	-- ({0.6690*\dx},{-1.1942*\dy})
+	-- ({0.6460*\dx},{-1.1847*\dy})
+	-- ({0.6231*\dx},{-1.1748*\dy})
+	-- ({0.6006*\dx},{-1.1644*\dy})
+	-- ({0.5783*\dx},{-1.1535*\dy})
+	-- ({0.5562*\dx},{-1.1421*\dy})
+	-- ({0.5344*\dx},{-1.1303*\dy})
+	-- ({0.5129*\dx},{-1.1180*\dy})
+	-- ({0.4917*\dx},{-1.1052*\dy})
+	-- ({0.4708*\dx},{-1.0920*\dy})
+	-- ({0.4502*\dx},{-1.0784*\dy})
+	-- ({0.4299*\dx},{-1.0643*\dy})
+	-- ({0.4100*\dx},{-1.0498*\dy})
+	-- ({0.3904*\dx},{-1.0349*\dy})
+	-- ({0.3711*\dx},{-1.0195*\dy})
+	-- ({0.3523*\dx},{-1.0038*\dy})
+	-- ({0.3338*\dx},{-0.9877*\dy})
+	-- ({0.3157*\dx},{-0.9712*\dy})
+	-- ({0.2979*\dx},{-0.9543*\dy})
+	-- ({0.2806*\dx},{-0.9370*\dy})
+	-- ({0.2637*\dx},{-0.9194*\dy})
+	-- ({0.2472*\dx},{-0.9014*\dy})
+	-- ({0.2311*\dx},{-0.8831*\dy})
+	-- ({0.2154*\dx},{-0.8645*\dy})
+	-- ({0.2002*\dx},{-0.8455*\dy})
+	-- ({0.1855*\dx},{-0.8262*\dy})
+	-- ({0.1712*\dx},{-0.8066*\dy})
+	-- ({0.1573*\dx},{-0.7868*\dy})
+	-- ({0.1440*\dx},{-0.7666*\dy})
+	-- ({0.1311*\dx},{-0.7462*\dy})
+	-- ({0.1187*\dx},{-0.7256*\dy})
+	-- ({0.1068*\dx},{-0.7047*\dy})
+	-- ({0.0953*\dx},{-0.6835*\dy})
+	-- ({0.0844*\dx},{-0.6621*\dy})
+	-- ({0.0740*\dx},{-0.6406*\dy})
+	-- ({0.0641*\dx},{-0.6188*\dy})
+	-- ({0.0547*\dx},{-0.5968*\dy})
+	-- ({0.0458*\dx},{-0.5747*\dy})
+	-- ({0.0375*\dx},{-0.5524*\dy})
+	-- ({0.0296*\dx},{-0.5299*\dy})
+	-- ({0.0223*\dx},{-0.5074*\dy})
+	-- ({0.0156*\dx},{-0.4847*\dy})
+	-- ({0.0093*\dx},{-0.4618*\dy})
+	-- ({0.0037*\dx},{-0.4389*\dy})
+	-- ({-0.0015*\dx},{-0.4159*\dy})
+	-- ({-0.0061*\dx},{-0.3928*\dy})
+	-- ({-0.0101*\dx},{-0.3697*\dy})
+	-- ({-0.0136*\dx},{-0.3465*\dy})
+	-- ({-0.0166*\dx},{-0.3233*\dy})
+	-- ({-0.0190*\dx},{-0.3001*\dy})
+	-- ({-0.0208*\dx},{-0.2768*\dy})
+	-- ({-0.0221*\dx},{-0.2536*\dy})
+	-- ({-0.0229*\dx},{-0.2304*\dy})
+	-- ({-0.0231*\dx},{-0.2072*\dy})
+	-- ({-0.0228*\dx},{-0.1841*\dy})
+	-- ({-0.0219*\dx},{-0.1610*\dy})
+	-- ({-0.0204*\dx},{-0.1380*\dy})
+	-- ({-0.0185*\dx},{-0.1151*\dy})
+	-- ({-0.0159*\dx},{-0.0923*\dy})
+	-- ({-0.0129*\dx},{-0.0696*\dy})
+	-- ({-0.0093*\dx},{-0.0470*\dy})
+	-- ({-0.0052*\dx},{-0.0245*\dy})
+	-- ({-0.0005*\dx},{-0.0023*\dy})
+	-- ({0.0047*\dx},{0.0199*\dy})
+	-- ({0.0104*\dx},{0.0418*\dy})
+	-- ({0.0166*\dx},{0.0635*\dy})
+	-- ({0.0233*\dx},{0.0851*\dy})
+	-- ({0.0306*\dx},{0.1064*\dy})
+	-- ({0.0383*\dx},{0.1275*\dy})
+	-- ({0.0466*\dx},{0.1484*\dy})
+	-- ({0.0553*\dx},{0.1690*\dy})
+	-- ({0.0645*\dx},{0.1893*\dy})
+	-- ({0.0742*\dx},{0.2094*\dy})
+	-- ({0.0843*\dx},{0.2291*\dy})
+	-- ({0.0950*\dx},{0.2486*\dy})
+	-- ({0.1060*\dx},{0.2677*\dy})
+	-- ({0.1176*\dx},{0.2866*\dy})
+	-- ({0.1295*\dx},{0.3050*\dy})
+	-- ({0.1419*\dx},{0.3232*\dy})
+	-- ({0.1547*\dx},{0.3410*\dy})
+	-- ({0.1679*\dx},{0.3584*\dy})
+	-- ({0.1816*\dx},{0.3754*\dy})
+	-- ({0.1956*\dx},{0.3921*\dy})
+	-- ({0.2100*\dx},{0.4083*\dy})
+	-- ({0.2248*\dx},{0.4242*\dy})
+	-- ({0.2399*\dx},{0.4396*\dy})
+	-- ({0.2554*\dx},{0.4546*\dy})
+	-- ({0.2712*\dx},{0.4691*\dy})
+	-- ({0.2874*\dx},{0.4832*\dy})
+	-- ({0.3038*\dx},{0.4969*\dy})
+	-- ({0.3206*\dx},{0.5101*\dy})
+	-- ({0.3377*\dx},{0.5228*\dy})
+	-- ({0.3550*\dx},{0.5351*\dy})
+	-- ({0.3727*\dx},{0.5468*\dy})
+	-- ({0.3906*\dx},{0.5581*\dy})
+	-- ({0.4087*\dx},{0.5689*\dy})
+	-- ({0.4270*\dx},{0.5792*\dy})
+	-- ({0.4456*\dx},{0.5889*\dy})
+	-- ({0.4644*\dx},{0.5982*\dy})
+	-- ({0.4834*\dx},{0.6069*\dy})
+	-- ({0.5025*\dx},{0.6151*\dy})
+	-- ({0.5218*\dx},{0.6228*\dy})
+	-- ({0.5413*\dx},{0.6299*\dy})
+	-- ({0.5609*\dx},{0.6365*\dy})
+	-- ({0.5806*\dx},{0.6426*\dy})
+	-- ({0.6005*\dx},{0.6481*\dy})
+	-- ({0.6204*\dx},{0.6531*\dy})
+	-- ({0.6404*\dx},{0.6575*\dy})
+	-- ({0.6605*\dx},{0.6614*\dy})
+	-- ({0.6806*\dx},{0.6647*\dy})
+	-- ({0.7007*\dx},{0.6674*\dy})
+	-- ({0.7209*\dx},{0.6696*\dy})
+	-- ({0.7411*\dx},{0.6713*\dy})
+	-- ({0.7613*\dx},{0.6723*\dy})
+	-- ({0.7814*\dx},{0.6729*\dy})
+	-- ({0.8015*\dx},{0.6728*\dy})
+	-- ({0.8216*\dx},{0.6722*\dy})
+	-- ({0.8416*\dx},{0.6711*\dy})
+	-- ({0.8615*\dx},{0.6694*\dy})
+	-- ({0.8813*\dx},{0.6672*\dy})
+	-- ({0.9010*\dx},{0.6644*\dy})
+	-- ({0.9205*\dx},{0.6610*\dy})
+	-- ({0.9400*\dx},{0.6571*\dy})
+	-- ({0.9592*\dx},{0.6527*\dy})
+	-- ({0.9783*\dx},{0.6478*\dy})
+	-- ({0.9972*\dx},{0.6423*\dy})
+	-- ({1.0159*\dx},{0.6363*\dy})
+	-- ({1.0344*\dx},{0.6298*\dy})
+	-- ({1.0527*\dx},{0.6227*\dy})
+	-- ({1.0707*\dx},{0.6152*\dy})
+	-- ({1.0885*\dx},{0.6071*\dy})
+	-- ({1.1060*\dx},{0.5986*\dy})
+	-- ({1.1232*\dx},{0.5896*\dy})
+	-- ({1.1401*\dx},{0.5801*\dy})
+	-- ({1.1567*\dx},{0.5701*\dy})
+	-- ({1.1730*\dx},{0.5597*\dy})
+	-- ({1.1889*\dx},{0.5488*\dy})
+	-- ({1.2045*\dx},{0.5374*\dy})
+	-- ({1.2197*\dx},{0.5257*\dy})
+	-- ({1.2346*\dx},{0.5135*\dy})
+	-- ({1.2491*\dx},{0.5008*\dy})
+	-- ({1.2631*\dx},{0.4878*\dy})
+	-- ({1.2768*\dx},{0.4744*\dy})
+	-- ({1.2900*\dx},{0.4606*\dy})
+	-- ({1.3029*\dx},{0.4464*\dy})
+	-- ({1.3152*\dx},{0.4319*\dy})
+	-- ({1.3271*\dx},{0.4170*\dy})
+	-- ({1.3386*\dx},{0.4018*\dy})
+	-- ({1.3496*\dx},{0.3862*\dy})
+	-- ({1.3601*\dx},{0.3704*\dy})
+	-- ({1.3701*\dx},{0.3542*\dy})
+	-- ({1.3796*\dx},{0.3378*\dy})
+	-- ({1.3886*\dx},{0.3211*\dy})
+	-- ({1.3971*\dx},{0.3041*\dy})
+	-- ({1.4051*\dx},{0.2869*\dy})
+	-- ({1.4125*\dx},{0.2694*\dy})
+	-- ({1.4195*\dx},{0.2518*\dy})
+	-- ({1.4258*\dx},{0.2339*\dy})
+	-- ({1.4316*\dx},{0.2159*\dy})
+	-- ({1.4369*\dx},{0.1977*\dy})
+	-- ({1.4416*\dx},{0.1793*\dy})
+	-- ({1.4457*\dx},{0.1608*\dy})
+	-- ({1.4493*\dx},{0.1422*\dy})
+	-- ({1.4523*\dx},{0.1235*\dy})
+	-- ({1.4547*\dx},{0.1047*\dy})
+	-- ({1.4565*\dx},{0.0858*\dy})
+	-- ({1.4577*\dx},{0.0668*\dy})
+	-- ({1.4584*\dx},{0.0478*\dy})
+	-- ({1.4584*\dx},{0.0288*\dy})
+	-- ({1.4579*\dx},{0.0098*\dy})
+	-- ({1.4568*\dx},{-0.0092*\dy})
+	-- ({1.4551*\dx},{-0.0282*\dy})
+	-- ({1.4528*\dx},{-0.0472*\dy})
+	-- ({1.4499*\dx},{-0.0661*\dy})
+	-- ({1.4464*\dx},{-0.0849*\dy})
+	-- ({1.4424*\dx},{-0.1036*\dy})
+	-- ({1.4377*\dx},{-0.1222*\dy})
+	-- ({1.4325*\dx},{-0.1407*\dy})
+	-- ({1.4267*\dx},{-0.1591*\dy})
+	-- ({1.4202*\dx},{-0.1773*\dy})
+	-- ({1.4133*\dx},{-0.1953*\dy})
+	-- ({1.4057*\dx},{-0.2131*\dy})
+	-- ({1.3976*\dx},{-0.2307*\dy})
+	-- ({1.3889*\dx},{-0.2481*\dy})
+	-- ({1.3797*\dx},{-0.2652*\dy})
+	-- ({1.3699*\dx},{-0.2821*\dy})
+	-- ({1.3595*\dx},{-0.2987*\dy})
+	-- ({1.3487*\dx},{-0.3151*\dy})
+	-- ({1.3373*\dx},{-0.3311*\dy})
+	-- ({1.3253*\dx},{-0.3468*\dy})
+	-- ({1.3129*\dx},{-0.3621*\dy})
+	-- ({1.2999*\dx},{-0.3771*\dy})
+	-- ({1.2865*\dx},{-0.3917*\dy})
+	-- ({1.2725*\dx},{-0.4060*\dy})
+	-- ({1.2581*\dx},{-0.4198*\dy})
+	-- ({1.2432*\dx},{-0.4333*\dy})
+	-- ({1.2279*\dx},{-0.4463*\dy})
+	-- ({1.2121*\dx},{-0.4589*\dy})
+	-- ({1.1959*\dx},{-0.4710*\dy})
+	-- ({1.1792*\dx},{-0.4826*\dy})
+	-- ({1.1622*\dx},{-0.4938*\dy})
+	-- ({1.1447*\dx},{-0.5045*\dy})
+	-- ({1.1269*\dx},{-0.5147*\dy})
+	-- ({1.1087*\dx},{-0.5243*\dy})
+	-- ({1.0901*\dx},{-0.5335*\dy})
+	-- ({1.0712*\dx},{-0.5420*\dy})
+	-- ({1.0520*\dx},{-0.5501*\dy})
+	-- ({1.0324*\dx},{-0.5575*\dy})
+	-- ({1.0126*\dx},{-0.5644*\dy})
+	-- ({0.9924*\dx},{-0.5708*\dy})
+	-- ({0.9721*\dx},{-0.5765*\dy})
+	-- ({0.9514*\dx},{-0.5816*\dy})
+	-- ({0.9305*\dx},{-0.5861*\dy})
+	-- ({0.9094*\dx},{-0.5900*\dy})
+	-- ({0.8881*\dx},{-0.5933*\dy})
+	-- ({0.8666*\dx},{-0.5959*\dy})
+	-- ({0.8450*\dx},{-0.5979*\dy})
+	-- ({0.8232*\dx},{-0.5993*\dy})
+	-- ({0.8013*\dx},{-0.6000*\dy})
+	-- ({0.7792*\dx},{-0.6000*\dy})
+	-- ({0.7571*\dx},{-0.5994*\dy})
+	-- ({0.7349*\dx},{-0.5981*\dy})
+	-- ({0.7126*\dx},{-0.5961*\dy})
+	-- ({0.6903*\dx},{-0.5935*\dy})
+	-- ({0.6680*\dx},{-0.5902*\dy})
+	-- ({0.6457*\dx},{-0.5862*\dy})
+	-- ({0.6234*\dx},{-0.5815*\dy})
+	-- ({0.6012*\dx},{-0.5762*\dy})
+	-- ({0.5790*\dx},{-0.5701*\dy})
+	-- ({0.5568*\dx},{-0.5634*\dy})
+	-- ({0.5348*\dx},{-0.5560*\dy})
+	-- ({0.5129*\dx},{-0.5479*\dy})
+	-- ({0.4911*\dx},{-0.5391*\dy})
+	-- ({0.4695*\dx},{-0.5297*\dy})
+	-- ({0.4481*\dx},{-0.5196*\dy})
+	-- ({0.4269*\dx},{-0.5087*\dy})
+	-- ({0.4058*\dx},{-0.4973*\dy})
+	-- ({0.3850*\dx},{-0.4851*\dy})
+	-- ({0.3645*\dx},{-0.4723*\dy})
+	-- ({0.3442*\dx},{-0.4589*\dy})
+	-- ({0.3243*\dx},{-0.4448*\dy})
+	-- ({0.3046*\dx},{-0.4300*\dy})
+	-- ({0.2853*\dx},{-0.4146*\dy})
+	-- ({0.2663*\dx},{-0.3986*\dy})
+	-- ({0.2477*\dx},{-0.3820*\dy})
+	-- ({0.2295*\dx},{-0.3647*\dy})
+	-- ({0.2117*\dx},{-0.3468*\dy})
+	-- ({0.1943*\dx},{-0.3284*\dy})
+	-- ({0.1774*\dx},{-0.3093*\dy})
+	-- ({0.1609*\dx},{-0.2897*\dy})
+	-- ({0.1449*\dx},{-0.2695*\dy})
+	-- ({0.1294*\dx},{-0.2488*\dy})
+	-- ({0.1144*\dx},{-0.2275*\dy})
+	-- ({0.0999*\dx},{-0.2057*\dy})
+	-- ({0.0860*\dx},{-0.1834*\dy})
+	-- ({0.0726*\dx},{-0.1606*\dy})
+	-- ({0.0598*\dx},{-0.1373*\dy})
+	-- ({0.0476*\dx},{-0.1135*\dy})
+	-- ({0.0360*\dx},{-0.0893*\dy})
+	-- ({0.0250*\dx},{-0.0646*\dy})
+	-- ({0.0147*\dx},{-0.0395*\dy})
+	-- ({0.0050*\dx},{-0.0140*\dy})
+	-- ({-0.0040*\dx},{0.0119*\dy})
+	-- ({-0.0124*\dx},{0.0382*\dy})
+	-- ({-0.0201*\dx},{0.0648*\dy})
+	-- ({-0.0270*\dx},{0.0918*\dy})
+	-- ({-0.0333*\dx},{0.1191*\dy})
+	-- ({-0.0388*\dx},{0.1466*\dy})
+	-- ({-0.0436*\dx},{0.1745*\dy})
+	-- ({-0.0477*\dx},{0.2027*\dy})
+	-- ({-0.0510*\dx},{0.2311*\dy})
+	-- ({-0.0535*\dx},{0.2597*\dy})
+	-- ({-0.0553*\dx},{0.2885*\dy})
+	-- ({-0.0563*\dx},{0.3175*\dy})
+	-- ({-0.0565*\dx},{0.3466*\dy})
+	-- ({-0.0559*\dx},{0.3759*\dy})
+	-- ({-0.0545*\dx},{0.4054*\dy})
+	-- ({-0.0523*\dx},{0.4349*\dy})
+	-- ({-0.0494*\dx},{0.4645*\dy})
+	-- ({-0.0455*\dx},{0.4941*\dy})
+	-- ({-0.0409*\dx},{0.5238*\dy})
+	-- ({-0.0355*\dx},{0.5535*\dy})
+	-- ({-0.0292*\dx},{0.5832*\dy})
+	-- ({-0.0221*\dx},{0.6128*\dy})
+	-- ({-0.0142*\dx},{0.6424*\dy})
+	-- ({-0.0054*\dx},{0.6719*\dy})
+	-- ({0.0042*\dx},{0.7013*\dy})
+	-- ({0.0146*\dx},{0.7305*\dy})
+	-- ({0.0258*\dx},{0.7597*\dy})
+	-- ({0.0379*\dx},{0.7886*\dy})
+	-- ({0.0508*\dx},{0.8174*\dy})
+	-- ({0.0645*\dx},{0.8459*\dy})
+	-- ({0.0790*\dx},{0.8742*\dy})
+	-- ({0.0943*\dx},{0.9022*\dy})
+	-- ({0.1105*\dx},{0.9299*\dy})
+	-- ({0.1274*\dx},{0.9573*\dy})
+	-- ({0.1452*\dx},{0.9844*\dy})
+	-- ({0.1637*\dx},{1.0111*\dy})
+	-- ({0.1830*\dx},{1.0375*\dy})
+	-- ({0.2031*\dx},{1.0634*\dy})
+	-- ({0.2239*\dx},{1.0889*\dy})
+	-- ({0.2455*\dx},{1.1140*\dy})
+	-- ({0.2679*\dx},{1.1386*\dy})
+	-- ({0.2910*\dx},{1.1627*\dy})
+	-- ({0.3148*\dx},{1.1863*\dy})
+	-- ({0.3393*\dx},{1.2094*\dy})
+	-- ({0.3645*\dx},{1.2319*\dy})
+	-- ({0.3904*\dx},{1.2539*\dy})
+	-- ({0.4169*\dx},{1.2752*\dy})
+	-- ({0.4442*\dx},{1.2960*\dy})
+	-- ({0.4720*\dx},{1.3161*\dy})
+	-- ({0.5005*\dx},{1.3355*\dy})
+	-- ({0.5296*\dx},{1.3543*\dy})
+	-- ({0.5593*\dx},{1.3724*\dy})
+	-- ({0.5896*\dx},{1.3898*\dy})
+	-- ({0.6204*\dx},{1.4065*\dy})
+	-- ({0.6518*\dx},{1.4224*\dy})
+	-- ({0.6837*\dx},{1.4376*\dy})
+	-- ({0.7161*\dx},{1.4520*\dy})
+	-- ({0.7489*\dx},{1.4656*\dy})
+	-- ({0.7823*\dx},{1.4784*\dy})
+	-- ({0.8160*\dx},{1.4903*\dy})
+	-- ({0.8503*\dx},{1.5015*\dy})
+	-- ({0.8849*\dx},{1.5118*\dy})
+	-- ({0.9198*\dx},{1.5213*\dy})
+	-- ({0.9552*\dx},{1.5298*\dy})
+	-- ({0.9908*\dx},{1.5375*\dy})
+	-- ({1.0268*\dx},{1.5443*\dy})
+	-- ({1.0631*\dx},{1.5502*\dy})
+	-- ({1.0996*\dx},{1.5552*\dy})
+	-- ({1.1364*\dx},{1.5593*\dy})
+	-- ({1.1733*\dx},{1.5624*\dy})
+	-- ({1.2105*\dx},{1.5646*\dy})
+	-- ({1.2478*\dx},{1.5658*\dy})
+	-- ({1.2853*\dx},{1.5661*\dy})
+	-- ({1.3229*\dx},{1.5655*\dy})
+	-- ({1.3606*\dx},{1.5638*\dy})
+	-- ({1.3983*\dx},{1.5613*\dy})
+	-- ({1.4361*\dx},{1.5577*\dy})
+	-- ({1.4738*\dx},{1.5531*\dy})
+	-- ({1.5116*\dx},{1.5476*\dy})
+	-- ({1.5493*\dx},{1.5411*\dy})
+	-- ({1.5870*\dx},{1.5337*\dy})
+	-- ({1.6245*\dx},{1.5252*\dy})
+	-- ({1.6620*\dx},{1.5158*\dy})
+	-- ({1.6993*\dx},{1.5054*\dy})
+	-- ({1.7364*\dx},{1.4940*\dy})
+	-- ({1.7733*\dx},{1.4816*\dy})
+	-- ({1.8100*\dx},{1.4683*\dy})
+	-- ({1.8464*\dx},{1.4540*\dy})
+	-- ({1.8826*\dx},{1.4388*\dy})
+	-- ({1.9184*\dx},{1.4226*\dy})
+	-- ({1.9539*\dx},{1.4055*\dy})
+	-- ({1.9891*\dx},{1.3874*\dy})
+	-- ({2.0238*\dx},{1.3684*\dy})
+	-- ({2.0582*\dx},{1.3484*\dy})
+	-- ({2.0921*\dx},{1.3276*\dy})
+	-- ({2.1255*\dx},{1.3059*\dy})
+	-- ({2.1585*\dx},{1.2832*\dy})
+	-- ({2.1910*\dx},{1.2597*\dy})
+	-- ({2.2229*\dx},{1.2353*\dy})
+	-- ({2.2543*\dx},{1.2101*\dy})
+	-- ({2.2850*\dx},{1.1840*\dy})
+	-- ({2.3152*\dx},{1.1571*\dy})
+	-- ({2.3448*\dx},{1.1294*\dy})
+	-- ({2.3736*\dx},{1.1009*\dy})
+	-- ({2.4019*\dx},{1.0716*\dy})
+	-- ({2.4294*\dx},{1.0416*\dy})
+	-- ({2.4562*\dx},{1.0108*\dy})
+	-- ({2.4822*\dx},{0.9793*\dy})
+	-- ({2.5075*\dx},{0.9471*\dy})
+	-- ({2.5321*\dx},{0.9142*\dy})
+	-- ({2.5558*\dx},{0.8807*\dy})
+	-- ({2.5787*\dx},{0.8465*\dy})
+	-- ({2.6008*\dx},{0.8117*\dy})
+	-- ({2.6220*\dx},{0.7762*\dy})
+	-- ({2.6423*\dx},{0.7402*\dy})
+	-- ({2.6618*\dx},{0.7037*\dy})
+	-- ({2.6803*\dx},{0.6666*\dy})
+	-- ({2.6980*\dx},{0.6290*\dy})
+	-- ({2.7147*\dx},{0.5909*\dy})
+	-- ({2.7305*\dx},{0.5524*\dy})
+	-- ({2.7453*\dx},{0.5134*\dy})
+	-- ({2.7591*\dx},{0.4740*\dy})
+	-- ({2.7719*\dx},{0.4342*\dy})
+	-- ({2.7838*\dx},{0.3941*\dy})
+	-- ({2.7946*\dx},{0.3537*\dy})
+	-- ({2.8045*\dx},{0.3129*\dy})
+	-- ({2.8133*\dx},{0.2719*\dy})
+	-- ({2.8210*\dx},{0.2306*\dy})
+	-- ({2.8278*\dx},{0.1891*\dy})
+	-- ({2.8335*\dx},{0.1474*\dy})
+	-- ({2.8381*\dx},{0.1055*\dy})
+	-- ({2.8417*\dx},{0.0636*\dy})
+	-- ({2.8442*\dx},{0.0215*\dy})
+	-- ({2.8457*\dx},{-0.0207*\dy})
+	-- ({2.8461*\dx},{-0.0629*\dy})
+	-- ({2.8454*\dx},{-0.1052*\dy})
+	-- ({2.8436*\dx},{-0.1474*\dy})
+	-- ({2.8408*\dx},{-0.1896*\dy})
+	-- ({2.8369*\dx},{-0.2317*\dy})
+	-- ({2.8320*\dx},{-0.2738*\dy})
+	-- ({2.8260*\dx},{-0.3157*\dy})
+	-- ({2.8189*\dx},{-0.3574*\dy})
+	-- ({2.8108*\dx},{-0.3990*\dy})
+	-- ({2.8016*\dx},{-0.4404*\dy})
+	-- ({2.7914*\dx},{-0.4815*\dy})
+	-- ({2.7801*\dx},{-0.5224*\dy})
+	-- ({2.7678*\dx},{-0.5629*\dy})
+	-- ({2.7545*\dx},{-0.6032*\dy})
+	-- ({2.7401*\dx},{-0.6430*\dy})
+	-- ({2.7248*\dx},{-0.6825*\dy})
+	-- ({2.7084*\dx},{-0.7216*\dy})
+	-- ({2.6911*\dx},{-0.7603*\dy})
+	-- ({2.6728*\dx},{-0.7985*\dy})
+	-- ({2.6535*\dx},{-0.8362*\dy})
+	-- ({2.6333*\dx},{-0.8734*\dy})
+	-- ({2.6122*\dx},{-0.9100*\dy})
+	-- ({2.5901*\dx},{-0.9461*\dy})
+	-- ({2.5671*\dx},{-0.9816*\dy})
+	-- ({2.5433*\dx},{-1.0165*\dy})
+	-- ({2.5186*\dx},{-1.0507*\dy})
+	-- ({2.4930*\dx},{-1.0843*\dy})
+	-- ({2.4667*\dx},{-1.1172*\dy})
+	-- ({2.4395*\dx},{-1.1494*\dy})
+	-- ({2.4115*\dx},{-1.1808*\dy})
+	-- ({2.3827*\dx},{-1.2115*\dy})
+	-- ({2.3532*\dx},{-1.2414*\dy})
+	-- ({2.3230*\dx},{-1.2706*\dy})
+	-- ({2.2921*\dx},{-1.2989*\dy})
+	-- ({2.2605*\dx},{-1.3264*\dy})
+	-- ({2.2282*\dx},{-1.3530*\dy})
+	-- ({2.1953*\dx},{-1.3788*\dy})
+	-- ({2.1618*\dx},{-1.4037*\dy})
+	-- ({2.1277*\dx},{-1.4277*\dy})
+	-- ({2.0930*\dx},{-1.4508*\dy})
+	-- ({2.0578*\dx},{-1.4729*\dy})
+	-- ({2.0221*\dx},{-1.4941*\dy})
+	-- ({1.9859*\dx},{-1.5144*\dy})
+	-- ({1.9493*\dx},{-1.5337*\dy})
+	-- ({1.9123*\dx},{-1.5520*\dy})
+	-- ({1.8748*\dx},{-1.5693*\dy})
+	-- ({1.8370*\dx},{-1.5856*\dy})
+	-- ({1.7988*\dx},{-1.6009*\dy})
+	-- ({1.7604*\dx},{-1.6152*\dy})
+	-- ({1.7216*\dx},{-1.6285*\dy})
+	-- ({1.6826*\dx},{-1.6407*\dy})
+	-- ({1.6434*\dx},{-1.6519*\dy})
+	-- ({1.6040*\dx},{-1.6620*\dy})
+	-- ({1.5644*\dx},{-1.6711*\dy})
+	-- ({1.5247*\dx},{-1.6792*\dy})
+	-- ({1.4848*\dx},{-1.6862*\dy})
+	-- ({1.4449*\dx},{-1.6921*\dy})
+	-- ({1.4050*\dx},{-1.6970*\dy})
+	-- ({1.3650*\dx},{-1.7009*\dy})
+	-- ({1.3250*\dx},{-1.7037*\dy})
+	-- ({1.2851*\dx},{-1.7054*\dy})
+	-- ({1.2453*\dx},{-1.7061*\dy})
+	-- ({1.2056*\dx},{-1.7058*\dy})
+	-- ({1.1660*\dx},{-1.7044*\dy})
+	-- ({1.1265*\dx},{-1.7020*\dy})
+	-- ({1.0873*\dx},{-1.6986*\dy})
+	-- ({1.0482*\dx},{-1.6941*\dy})
+	-- ({1.0095*\dx},{-1.6887*\dy})
+	-- ({0.9709*\dx},{-1.6822*\dy})
+	-- ({0.9327*\dx},{-1.6747*\dy})
+	-- ({0.8949*\dx},{-1.6663*\dy})
+	-- ({0.8574*\dx},{-1.6569*\dy})
+	-- ({0.8202*\dx},{-1.6466*\dy})
+	-- ({0.7835*\dx},{-1.6353*\dy})
+	-- ({0.7473*\dx},{-1.6231*\dy})
+	-- ({0.7115*\dx},{-1.6099*\dy})
+	-- ({0.6761*\dx},{-1.5959*\dy})
+	-- ({0.6413*\dx},{-1.5810*\dy})
+	-- ({0.6071*\dx},{-1.5652*\dy})
+	-- ({0.5734*\dx},{-1.5486*\dy})
+	-- ({0.5403*\dx},{-1.5312*\dy})
+	-- ({0.5078*\dx},{-1.5129*\dy})
+	-- ({0.4759*\dx},{-1.4939*\dy})
+	-- ({0.4447*\dx},{-1.4741*\dy})
+	-- ({0.4141*\dx},{-1.4535*\dy})
+	-- ({0.3843*\dx},{-1.4322*\dy})
+	-- ({0.3552*\dx},{-1.4103*\dy})
+	-- ({0.3268*\dx},{-1.3876*\dy})
+	-- ({0.2991*\dx},{-1.3643*\dy})
+	-- ({0.2722*\dx},{-1.3403*\dy})
+	-- ({0.2461*\dx},{-1.3158*\dy})
+	-- ({0.2208*\dx},{-1.2907*\dy})
+	-- ({0.1963*\dx},{-1.2650*\dy})
+	-- ({0.1726*\dx},{-1.2387*\dy})
+	-- ({0.1498*\dx},{-1.2120*\dy})
+	-- ({0.1279*\dx},{-1.1848*\dy})
+	-- ({0.1068*\dx},{-1.1572*\dy})
+	-- ({0.0865*\dx},{-1.1291*\dy})
+	-- ({0.0672*\dx},{-1.1006*\dy})
+	-- ({0.0487*\dx},{-1.0718*\dy})
+	-- ({0.0312*\dx},{-1.0426*\dy})
+	-- ({0.0146*\dx},{-1.0131*\dy})
+	-- ({-0.0011*\dx},{-0.9833*\dy})
+	-- ({-0.0159*\dx},{-0.9533*\dy})
+	-- ({-0.0298*\dx},{-0.9230*\dy})
+	-- ({-0.0427*\dx},{-0.8926*\dy})
+	-- ({-0.0547*\dx},{-0.8620*\dy})
+	-- ({-0.0658*\dx},{-0.8312*\dy})
+	-- ({-0.0759*\dx},{-0.8003*\dy})
+	-- ({-0.0851*\dx},{-0.7694*\dy})
+	-- ({-0.0933*\dx},{-0.7384*\dy})
+	-- ({-0.1007*\dx},{-0.7074*\dy})
+	-- ({-0.1070*\dx},{-0.6764*\dy})
+	-- ({-0.1125*\dx},{-0.6454*\dy})
+	-- ({-0.1170*\dx},{-0.6145*\dy})
+	-- ({-0.1206*\dx},{-0.5837*\dy})
+	-- ({-0.1233*\dx},{-0.5530*\dy})
+	-- ({-0.1251*\dx},{-0.5225*\dy})
+	-- ({-0.1260*\dx},{-0.4921*\dy})
+	-- ({-0.1261*\dx},{-0.4619*\dy})
+	-- ({-0.1252*\dx},{-0.4320*\dy})
+	-- ({-0.1235*\dx},{-0.4024*\dy})
+	-- ({-0.1209*\dx},{-0.3730*\dy})
+	-- ({-0.1175*\dx},{-0.3439*\dy})
+	-- ({-0.1132*\dx},{-0.3152*\dy})
+	-- ({-0.1081*\dx},{-0.2868*\dy})
+	-- ({-0.1022*\dx},{-0.2589*\dy})
+	-- ({-0.0956*\dx},{-0.2313*\dy})
+	-- ({-0.0882*\dx},{-0.2042*\dy})
+	-- ({-0.0800*\dx},{-0.1775*\dy})
+	-- ({-0.0711*\dx},{-0.1513*\dy})
+	-- ({-0.0614*\dx},{-0.1256*\dy})
+	-- ({-0.0511*\dx},{-0.1005*\dy})
+	-- ({-0.0401*\dx},{-0.0759*\dy})
+	-- ({-0.0284*\dx},{-0.0518*\dy})
+	-- ({-0.0161*\dx},{-0.0283*\dy})
+	-- ({-0.0032*\dx},{-0.0055*\dy})
+	-- ({0.0103*\dx},{0.0168*\dy})
+	-- ({0.0243*\dx},{0.0384*\dy})
+	-- ({0.0389*\dx},{0.0593*\dy})
+	-- ({0.0541*\dx},{0.0796*\dy})
+	-- ({0.0697*\dx},{0.0992*\dy})
+	-- ({0.0858*\dx},{0.1181*\dy})
+	-- ({0.1023*\dx},{0.1363*\dy})
+	-- ({0.1193*\dx},{0.1537*\dy})
+	-- ({0.1367*\dx},{0.1705*\dy})
+	-- ({0.1544*\dx},{0.1864*\dy})
+	-- ({0.1725*\dx},{0.2017*\dy})
+	-- ({0.1908*\dx},{0.2161*\dy})
+	-- ({0.2095*\dx},{0.2298*\dy})
+	-- ({0.2284*\dx},{0.2427*\dy})
+	-- ({0.2476*\dx},{0.2549*\dy})
+	-- ({0.2669*\dx},{0.2662*\dy})
+	-- ({0.2865*\dx},{0.2768*\dy})
+	-- ({0.3062*\dx},{0.2865*\dy})
+	-- ({0.3260*\dx},{0.2955*\dy})
+	-- ({0.3459*\dx},{0.3036*\dy})
+	-- ({0.3658*\dx},{0.3110*\dy})
+	-- ({0.3858*\dx},{0.3175*\dy})
+	-- ({0.4058*\dx},{0.3233*\dy})
+	-- ({0.4258*\dx},{0.3283*\dy})
+	-- ({0.4457*\dx},{0.3324*\dy})
+	-- ({0.4655*\dx},{0.3358*\dy})
+	-- ({0.4853*\dx},{0.3384*\dy})
+	-- ({0.5049*\dx},{0.3403*\dy})
+	-- ({0.5243*\dx},{0.3413*\dy})
+	-- ({0.5436*\dx},{0.3416*\dy})
+	-- ({0.5626*\dx},{0.3412*\dy})
+	-- ({0.5814*\dx},{0.3400*\dy})
+	-- ({0.5999*\dx},{0.3381*\dy})
+	-- ({0.6182*\dx},{0.3355*\dy})
+	-- ({0.6361*\dx},{0.3321*\dy})
+	-- ({0.6537*\dx},{0.3281*\dy})
+	-- ({0.6709*\dx},{0.3234*\dy})
+	-- ({0.6877*\dx},{0.3181*\dy})
+	-- ({0.7041*\dx},{0.3120*\dy})
+	-- ({0.7201*\dx},{0.3054*\dy})
+	-- ({0.7356*\dx},{0.2982*\dy})
+	-- ({0.7506*\dx},{0.2903*\dy})
+	-- ({0.7651*\dx},{0.2819*\dy})
+	-- ({0.7791*\dx},{0.2730*\dy})
+	-- ({0.7926*\dx},{0.2635*\dy})
+	-- ({0.8055*\dx},{0.2535*\dy})
+	-- ({0.8178*\dx},{0.2430*\dy})
+	-- ({0.8296*\dx},{0.2320*\dy})
+	-- ({0.8407*\dx},{0.2206*\dy})
+	-- ({0.8512*\dx},{0.2088*\dy})
+	-- ({0.8610*\dx},{0.1965*\dy})
+	-- ({0.8702*\dx},{0.1839*\dy})
+	-- ({0.8788*\dx},{0.1710*\dy})
+	-- ({0.8866*\dx},{0.1577*\dy})
+	-- ({0.8937*\dx},{0.1442*\dy})
+	-- ({0.9002*\dx},{0.1304*\dy})
+	-- ({0.9059*\dx},{0.1163*\dy})
+	-- ({0.9109*\dx},{0.1020*\dy})
+	-- ({0.9152*\dx},{0.0876*\dy})
+	-- ({0.9187*\dx},{0.0730*\dy})
+	-- ({0.9215*\dx},{0.0582*\dy})
+	-- ({0.9236*\dx},{0.0434*\dy})
+	-- ({0.9249*\dx},{0.0284*\dy})
+	-- ({0.9254*\dx},{0.0135*\dy})
+	-- ({0.9252*\dx},{-0.0015*\dy})
+	-- ({0.9242*\dx},{-0.0165*\dy})
+	-- ({0.9225*\dx},{-0.0314*\dy})
+	-- ({0.9200*\dx},{-0.0463*\dy})
+	-- ({0.9168*\dx},{-0.0610*\dy})
+	-- ({0.9128*\dx},{-0.0756*\dy})
+	-- ({0.9081*\dx},{-0.0901*\dy})
+	-- ({0.9026*\dx},{-0.1044*\dy})
+	-- ({0.8964*\dx},{-0.1184*\dy})
+	-- ({0.8895*\dx},{-0.1322*\dy})
+	-- ({0.8818*\dx},{-0.1458*\dy})
+	-- ({0.8734*\dx},{-0.1590*\dy})
+	-- ({0.8644*\dx},{-0.1720*\dy})
+	-- ({0.8546*\dx},{-0.1845*\dy})
+	-- ({0.8442*\dx},{-0.1967*\dy})
+	-- ({0.8331*\dx},{-0.2085*\dy})
+	-- ({0.8214*\dx},{-0.2198*\dy})
+	-- ({0.8090*\dx},{-0.2307*\dy})
+	-- ({0.7960*\dx},{-0.2411*\dy})
+	-- ({0.7824*\dx},{-0.2510*\dy})
+	-- ({0.7682*\dx},{-0.2603*\dy})
+	-- ({0.7535*\dx},{-0.2691*\dy})
+	-- ({0.7382*\dx},{-0.2773*\dy})
+	-- ({0.7224*\dx},{-0.2849*\dy})
+	-- ({0.7061*\dx},{-0.2919*\dy})
+	-- ({0.6893*\dx},{-0.2982*\dy})
+	-- ({0.6721*\dx},{-0.3039*\dy})
+	-- ({0.6544*\dx},{-0.3089*\dy})
+	-- ({0.6363*\dx},{-0.3132*\dy})
+	-- ({0.6178*\dx},{-0.3168*\dy})
+	-- ({0.5989*\dx},{-0.3196*\dy})
+	-- ({0.5798*\dx},{-0.3217*\dy})
+	-- ({0.5603*\dx},{-0.3230*\dy})
+	-- ({0.5405*\dx},{-0.3235*\dy})
+	-- ({0.5205*\dx},{-0.3232*\dy})
+	-- ({0.5003*\dx},{-0.3222*\dy})
+	-- ({0.4798*\dx},{-0.3203*\dy})
+	-- ({0.4592*\dx},{-0.3175*\dy})
+	-- ({0.4385*\dx},{-0.3140*\dy})
+	-- ({0.4176*\dx},{-0.3095*\dy})
+	-- ({0.3967*\dx},{-0.3043*\dy})
+	-- ({0.3758*\dx},{-0.2981*\dy})
+	-- ({0.3548*\dx},{-0.2911*\dy})
+	-- ({0.3338*\dx},{-0.2832*\dy})
+	-- ({0.3129*\dx},{-0.2745*\dy})
+	-- ({0.2921*\dx},{-0.2649*\dy})
+	-- ({0.2714*\dx},{-0.2544*\dy})
+	-- ({0.2508*\dx},{-0.2430*\dy})
+	-- ({0.2305*\dx},{-0.2307*\dy})
+	-- ({0.2103*\dx},{-0.2176*\dy})
+	-- ({0.1904*\dx},{-0.2036*\dy})
+	-- ({0.1707*\dx},{-0.1887*\dy})
+	-- ({0.1514*\dx},{-0.1730*\dy})
+	-- ({0.1324*\dx},{-0.1564*\dy})
+	-- ({0.1137*\dx},{-0.1390*\dy})
+	-- ({0.0955*\dx},{-0.1207*\dy})
+	-- ({0.0777*\dx},{-0.1016*\dy})
+	-- ({0.0604*\dx},{-0.0817*\dy})
+	-- ({0.0435*\dx},{-0.0610*\dy})
+	-- ({0.0272*\dx},{-0.0395*\dy})
+	-- ({0.0115*\dx},{-0.0173*\dy})
+	-- ({-0.0037*\dx},{0.0058*\dy})
+	-- ({-0.0182*\dx},{0.0295*\dy})
+	-- ({-0.0321*\dx},{0.0540*\dy})
+	-- ({-0.0454*\dx},{0.0792*\dy})
+	-- ({-0.0579*\dx},{0.1051*\dy})
+	-- ({-0.0697*\dx},{0.1316*\dy})
+	-- ({-0.0807*\dx},{0.1588*\dy})
+	-- ({-0.0910*\dx},{0.1866*\dy})
+	-- ({-0.1005*\dx},{0.2150*\dy})
+	-- ({-0.1091*\dx},{0.2439*\dy})
+	-- ({-0.1169*\dx},{0.2734*\dy})
+	-- ({-0.1238*\dx},{0.3035*\dy})
+	-- ({-0.1298*\dx},{0.3340*\dy})
+	-- ({-0.1349*\dx},{0.3649*\dy})
+	-- ({-0.1391*\dx},{0.3964*\dy})
+	-- ({-0.1423*\dx},{0.4282*\dy})
+	-- ({-0.1445*\dx},{0.4604*\dy})
+	-- ({-0.1457*\dx},{0.4929*\dy})
+	-- ({-0.1459*\dx},{0.5257*\dy})
+	-- ({-0.1451*\dx},{0.5589*\dy})
+	-- ({-0.1433*\dx},{0.5922*\dy})
+	-- ({-0.1404*\dx},{0.6258*\dy})
+	-- ({-0.1365*\dx},{0.6595*\dy})
+	-- ({-0.1314*\dx},{0.6934*\dy})
+	-- ({-0.1253*\dx},{0.7274*\dy})
+	-- ({-0.1181*\dx},{0.7615*\dy})
+	-- ({-0.1098*\dx},{0.7956*\dy})
+	-- ({-0.1004*\dx},{0.8298*\dy})
+	-- ({-0.0899*\dx},{0.8638*\dy})
+	-- ({-0.0783*\dx},{0.8978*\dy})
+	-- ({-0.0655*\dx},{0.9317*\dy})
+	-- ({-0.0517*\dx},{0.9655*\dy})
+	-- ({-0.0367*\dx},{0.9990*\dy})
+	-- ({-0.0206*\dx},{1.0324*\dy})
+	-- ({-0.0034*\dx},{1.0654*\dy})
+	-- ({0.0149*\dx},{1.0982*\dy})
+	-- ({0.0344*\dx},{1.1307*\dy})
+	-- ({0.0549*\dx},{1.1627*\dy})
+	-- ({0.0765*\dx},{1.1944*\dy})
+	-- ({0.0992*\dx},{1.2256*\dy})
+	-- ({0.1230*\dx},{1.2563*\dy})
+	-- ({0.1478*\dx},{1.2865*\dy})
+	-- ({0.1737*\dx},{1.3162*\dy})
+	-- ({0.2006*\dx},{1.3452*\dy})
+	-- ({0.2285*\dx},{1.3737*\dy})
+	-- ({0.2574*\dx},{1.4014*\dy})
+	-- ({0.2874*\dx},{1.4285*\dy})
+	-- ({0.3182*\dx},{1.4548*\dy})
+	-- ({0.3501*\dx},{1.4804*\dy})
+	-- ({0.3828*\dx},{1.5051*\dy})
+	-- ({0.4165*\dx},{1.5291*\dy})
+	-- ({0.4510*\dx},{1.5521*\dy})
+	-- ({0.4864*\dx},{1.5743*\dy})
+	-- ({0.5226*\dx},{1.5955*\dy})
+	-- ({0.5597*\dx},{1.6158*\dy})
+	-- ({0.5975*\dx},{1.6350*\dy})
+	-- ({0.6360*\dx},{1.6533*\dy})
+	-- ({0.6753*\dx},{1.6705*\dy})
+	-- ({0.7152*\dx},{1.6867*\dy})
+	-- ({0.7558*\dx},{1.7017*\dy})
+	-- ({0.7970*\dx},{1.7157*\dy})
+	-- ({0.8388*\dx},{1.7285*\dy})
+	-- ({0.8812*\dx},{1.7401*\dy})
+	-- ({0.9241*\dx},{1.7505*\dy})
+	-- ({0.9674*\dx},{1.7598*\dy})
+	-- ({1.0112*\dx},{1.7678*\dy})
+	-- ({1.0554*\dx},{1.7746*\dy})
+	-- ({1.1000*\dx},{1.7801*\dy})
+	-- ({1.1449*\dx},{1.7843*\dy})
+	-- ({1.1901*\dx},{1.7872*\dy})
+	-- ({1.2355*\dx},{1.7889*\dy})
+	-- ({1.2812*\dx},{1.7892*\dy})
+	-- ({1.3270*\dx},{1.7882*\dy})
+	-- ({1.3730*\dx},{1.7859*\dy})
+	-- ({1.4190*\dx},{1.7822*\dy})
+	-- ({1.4651*\dx},{1.7772*\dy})
+	-- ({1.5111*\dx},{1.7708*\dy})
+	-- ({1.5572*\dx},{1.7631*\dy})
+	-- ({1.6031*\dx},{1.7540*\dy})
+	-- ({1.6490*\dx},{1.7435*\dy})
+	-- ({1.6946*\dx},{1.7317*\dy})
+	-- ({1.7400*\dx},{1.7185*\dy})
+	-- ({1.7852*\dx},{1.7040*\dy})
+	-- ({1.8301*\dx},{1.6882*\dy})
+	-- ({1.8746*\dx},{1.6709*\dy})
+	-- ({1.9188*\dx},{1.6524*\dy})
+	-- ({1.9625*\dx},{1.6325*\dy})
+	-- ({2.0058*\dx},{1.6114*\dy})
+	-- ({2.0485*\dx},{1.5889*\dy})
+	-- ({2.0907*\dx},{1.5651*\dy})
+	-- ({2.1322*\dx},{1.5401*\dy})
+	-- ({2.1732*\dx},{1.5138*\dy})
+	-- ({2.2134*\dx},{1.4862*\dy})
+	-- ({2.2530*\dx},{1.4574*\dy})
+	-- ({2.2918*\dx},{1.4275*\dy})
+	-- ({2.3297*\dx},{1.3963*\dy})
+	-- ({2.3669*\dx},{1.3640*\dy})
+	-- ({2.4031*\dx},{1.3306*\dy})
+	-- ({2.4385*\dx},{1.2961*\dy})
+	-- ({2.4729*\dx},{1.2605*\dy})
+	-- ({2.5063*\dx},{1.2238*\dy})
+	-- ({2.5387*\dx},{1.1861*\dy})
+	-- ({2.5701*\dx},{1.1474*\dy})
+	-- ({2.6003*\dx},{1.1078*\dy})
+	-- ({2.6295*\dx},{1.0672*\dy})
+	-- ({2.6575*\dx},{1.0258*\dy})
+	-- ({2.6843*\dx},{0.9835*\dy})
+	-- ({2.7099*\dx},{0.9403*\dy})
+	-- ({2.7343*\dx},{0.8964*\dy})
+	-- ({2.7574*\dx},{0.8517*\dy})
+	-- ({2.7793*\dx},{0.8063*\dy})
+	-- ({2.7998*\dx},{0.7603*\dy})
+	-- ({2.8190*\dx},{0.7136*\dy})
+	-- ({2.8369*\dx},{0.6663*\dy})
+	-- ({2.8534*\dx},{0.6185*\dy})
+	-- ({2.8684*\dx},{0.5702*\dy})
+	-- ({2.8821*\dx},{0.5214*\dy})
+	-- ({2.8944*\dx},{0.4722*\dy})
+	-- ({2.9052*\dx},{0.4227*\dy})
+	-- ({2.9146*\dx},{0.3728*\dy})
+	-- ({2.9225*\dx},{0.3226*\dy})
+	-- ({2.9289*\dx},{0.2722*\dy})
+	-- ({2.9339*\dx},{0.2216*\dy})
+	-- ({2.9374*\dx},{0.1709*\dy})
+	-- ({2.9394*\dx},{0.1200*\dy})
+	-- ({2.9399*\dx},{0.0692*\dy})
+	-- ({2.9389*\dx},{0.0183*\dy})
+	-- ({2.9363*\dx},{-0.0326*\dy})
+	-- ({2.9323*\dx},{-0.0833*\dy})
+	-- ({2.9268*\dx},{-0.1339*\dy})
+	-- ({2.9198*\dx},{-0.1843*\dy})
+	-- ({2.9114*\dx},{-0.2345*\dy})
+	-- ({2.9014*\dx},{-0.2844*\dy})
+	-- ({2.8900*\dx},{-0.3340*\dy})
+	-- ({2.8771*\dx},{-0.3832*\dy})
+	-- ({2.8627*\dx},{-0.4319*\dy})
+	-- ({2.8470*\dx},{-0.4802*\dy})
+	-- ({2.8298*\dx},{-0.5280*\dy})
+	-- ({2.8112*\dx},{-0.5753*\dy})
+	-- ({2.7912*\dx},{-0.6219*\dy})
+	-- ({2.7698*\dx},{-0.6679*\dy})
+	-- ({2.7471*\dx},{-0.7131*\dy})
+	-- ({2.7231*\dx},{-0.7577*\dy})
+	-- ({2.6978*\dx},{-0.8015*\dy})
+	-- ({2.6712*\dx},{-0.8444*\dy})
+	-- ({2.6433*\dx},{-0.8865*\dy})
+	-- ({2.6143*\dx},{-0.9277*\dy})
+	-- ({2.5840*\dx},{-0.9680*\dy})
+	-- ({2.5526*\dx},{-1.0073*\dy})
+	-- ({2.5200*\dx},{-1.0455*\dy})
+	-- ({2.4863*\dx},{-1.0828*\dy})
+	-- ({2.4516*\dx},{-1.1189*\dy})
+	-- ({2.4159*\dx},{-1.1539*\dy})
+	-- ({2.3791*\dx},{-1.1878*\dy})
+	-- ({2.3414*\dx},{-1.2205*\dy})
+	-- ({2.3028*\dx},{-1.2520*\dy})
+	-- ({2.2633*\dx},{-1.2823*\dy})
+	-- ({2.2230*\dx},{-1.3113*\dy})
+	-- ({2.1819*\dx},{-1.3390*\dy})
+	-- ({2.1400*\dx},{-1.3654*\dy})
+	-- ({2.0974*\dx},{-1.3904*\dy})
+	-- ({2.0542*\dx},{-1.4141*\dy})
+	-- ({2.0103*\dx},{-1.4364*\dy})
+	-- ({1.9659*\dx},{-1.4573*\dy})
+	-- ({1.9209*\dx},{-1.4768*\dy})
+	-- ({1.8755*\dx},{-1.4948*\dy})
+	-- ({1.8296*\dx},{-1.5114*\dy})
+	-- ({1.7833*\dx},{-1.5266*\dy})
+	-- ({1.7367*\dx},{-1.5402*\dy})
+	-- ({1.6897*\dx},{-1.5524*\dy})
+	-- ({1.6426*\dx},{-1.5631*\dy})
+	-- ({1.5952*\dx},{-1.5723*\dy})
+	-- ({1.5477*\dx},{-1.5800*\dy})
+	-- ({1.5001*\dx},{-1.5862*\dy})
+	-- ({1.4524*\dx},{-1.5909*\dy})
+	-- ({1.4048*\dx},{-1.5941*\dy})
+	-- ({1.3572*\dx},{-1.5958*\dy})
+	-- ({1.3097*\dx},{-1.5960*\dy})
+	-- ({1.2623*\dx},{-1.5947*\dy})
+	-- ({1.2152*\dx},{-1.5919*\dy})
+	-- ({1.1683*\dx},{-1.5876*\dy})
+	-- ({1.1217*\dx},{-1.5818*\dy})
+	-- ({1.0754*\dx},{-1.5747*\dy})
+	-- ({1.0295*\dx},{-1.5660*\dy})
+	-- ({0.9841*\dx},{-1.5559*\dy})
+	-- ({0.9391*\dx},{-1.5445*\dy})
+	-- ({0.8947*\dx},{-1.5316*\dy})
+	-- ({0.8508*\dx},{-1.5173*\dy})
+	-- ({0.8076*\dx},{-1.5017*\dy})
+	-- ({0.7650*\dx},{-1.4848*\dy})
+	-- ({0.7231*\dx},{-1.4666*\dy})
+	-- ({0.6820*\dx},{-1.4471*\dy})
+	-- ({0.6417*\dx},{-1.4263*\dy})
+	-- ({0.6022*\dx},{-1.4043*\dy})
+	-- ({0.5635*\dx},{-1.3811*\dy})
+	-- ({0.5258*\dx},{-1.3568*\dy})
+	-- ({0.4890*\dx},{-1.3313*\dy})
+	-- ({0.4532*\dx},{-1.3048*\dy})
+	-- ({0.4185*\dx},{-1.2771*\dy})
+	-- ({0.3847*\dx},{-1.2485*\dy})
+	-- ({0.3521*\dx},{-1.2188*\dy})
+	-- ({0.3205*\dx},{-1.1882*\dy})
+	-- ({0.2902*\dx},{-1.1567*\dy})
+	-- ({0.2609*\dx},{-1.1243*\dy})
+	-- ({0.2329*\dx},{-1.0911*\dy})
+	-- ({0.2061*\dx},{-1.0571*\dy})
+	-- ({0.1806*\dx},{-1.0224*\dy})
+	-- ({0.1563*\dx},{-0.9870*\dy})
+	-- ({0.1333*\dx},{-0.9509*\dy})
+	-- ({0.1117*\dx},{-0.9142*\dy})
+	-- ({0.0914*\dx},{-0.8769*\dy})
+	-- ({0.0724*\dx},{-0.8391*\dy})
+	-- ({0.0547*\dx},{-0.8008*\dy})
+	-- ({0.0385*\dx},{-0.7622*\dy})
+	-- ({0.0236*\dx},{-0.7231*\dy})
+	-- ({0.0102*\dx},{-0.6837*\dy})
+	-- ({-0.0019*\dx},{-0.6441*\dy})
+	-- ({-0.0125*\dx},{-0.6042*\dy})
+	-- ({-0.0217*\dx},{-0.5641*\dy})
+	-- ({-0.0295*\dx},{-0.5239*\dy})
+	-- ({-0.0359*\dx},{-0.4836*\dy})
+	-- ({-0.0409*\dx},{-0.4433*\dy})
+	-- ({-0.0444*\dx},{-0.4030*\dy})
+	-- ({-0.0466*\dx},{-0.3628*\dy})
+	-- ({-0.0473*\dx},{-0.3226*\dy})
+	-- ({-0.0466*\dx},{-0.2827*\dy})
+	-- ({-0.0445*\dx},{-0.2429*\dy})
+	-- ({-0.0410*\dx},{-0.2034*\dy})
+	-- ({-0.0362*\dx},{-0.1642*\dy})
+	-- ({-0.0300*\dx},{-0.1254*\dy})
+	-- ({-0.0224*\dx},{-0.0870*\dy})
+	-- ({-0.0136*\dx},{-0.0490*\dy})
+	-- ({-0.0034*\dx},{-0.0115*\dy})
+	-- ({0.0081*\dx},{0.0255*\dy})
+	-- ({0.0208*\dx},{0.0619*\dy})
+	-- ({0.0348*\dx},{0.0976*\dy})
+	-- ({0.0500*\dx},{0.1327*\dy})
+	-- ({0.0663*\dx},{0.1671*\dy})
+	-- ({0.0839*\dx},{0.2008*\dy})
+	-- ({0.1026*\dx},{0.2336*\dy})
+	-- ({0.1224*\dx},{0.2656*\dy})
+	-- ({0.1432*\dx},{0.2968*\dy})
+	-- ({0.1651*\dx},{0.3271*\dy})
+	-- ({0.1880*\dx},{0.3564*\dy})
+	-- ({0.2119*\dx},{0.3848*\dy})
+	-- ({0.2367*\dx},{0.4122*\dy})
+	-- ({0.2625*\dx},{0.4385*\dy})
+	-- ({0.2890*\dx},{0.4638*\dy})
+	-- ({0.3164*\dx},{0.4880*\dy})
+	-- ({0.3446*\dx},{0.5111*\dy})
+	-- ({0.3735*\dx},{0.5331*\dy})
+	-- ({0.4031*\dx},{0.5539*\dy})
+	-- ({0.4334*\dx},{0.5735*\dy})
+	-- ({0.4643*\dx},{0.5919*\dy})
+	-- ({0.4957*\dx},{0.6090*\dy})
+	-- ({0.5276*\dx},{0.6250*\dy})
+	-- ({0.5601*\dx},{0.6396*\dy})
+	-- ({0.5929*\dx},{0.6530*\dy})
+	-- ({0.6262*\dx},{0.6652*\dy})
+	-- ({0.6597*\dx},{0.6760*\dy})
+	-- ({0.6936*\dx},{0.6855*\dy})
+	-- ({0.7277*\dx},{0.6937*\dy})
+	-- ({0.7620*\dx},{0.7006*\dy})
+	-- ({0.7964*\dx},{0.7062*\dy})
+	-- ({0.8309*\dx},{0.7104*\dy})
+	-- ({0.8655*\dx},{0.7133*\dy})
+	-- ({0.9001*\dx},{0.7149*\dy})
+	-- ({0.9346*\dx},{0.7152*\dy})
+	-- ({0.9690*\dx},{0.7141*\dy})
+	-- ({1.0033*\dx},{0.7117*\dy})
+	-- ({1.0374*\dx},{0.7081*\dy})
+	-- ({1.0712*\dx},{0.7031*\dy})
+	-- ({1.1048*\dx},{0.6969*\dy})
+	-- ({1.1380*\dx},{0.6894*\dy})
+	-- ({1.1709*\dx},{0.6806*\dy})
+	-- ({1.2033*\dx},{0.6706*\dy})
+	-- ({1.2353*\dx},{0.6594*\dy})
+	-- ({1.2667*\dx},{0.6470*\dy})
+	-- ({1.2976*\dx},{0.6334*\dy})
+	-- ({1.3279*\dx},{0.6186*\dy})
+	-- ({1.3576*\dx},{0.6027*\dy})
+	-- ({1.3866*\dx},{0.5857*\dy})
+	-- ({1.4149*\dx},{0.5676*\dy})
+	-- ({1.4424*\dx},{0.5485*\dy})
+	-- ({1.4692*\dx},{0.5284*\dy})
+	-- ({1.4951*\dx},{0.5072*\dy})
+	-- ({1.5201*\dx},{0.4851*\dy})
+	-- ({1.5443*\dx},{0.4621*\dy})
+	-- ({1.5675*\dx},{0.4381*\dy})
+	-- ({1.5898*\dx},{0.4133*\dy})
+	-- ({1.6111*\dx},{0.3877*\dy})
+	-- ({1.6314*\dx},{0.3614*\dy})
+	-- ({1.6506*\dx},{0.3342*\dy})
+	-- ({1.6687*\dx},{0.3064*\dy})
+	-- ({1.6858*\dx},{0.2779*\dy})
+	-- ({1.7017*\dx},{0.2488*\dy})
+	-- ({1.7165*\dx},{0.2191*\dy})
+	-- ({1.7302*\dx},{0.1889*\dy})
+	-- ({1.7427*\dx},{0.1581*\dy})
+	-- ({1.7540*\dx},{0.1270*\dy})
+	-- ({1.7640*\dx},{0.0954*\dy})
+	-- ({1.7729*\dx},{0.0635*\dy})
+	-- ({1.7806*\dx},{0.0312*\dy})
+	-- ({1.7870*\dx},{-0.0013*\dy})
+	-- ({1.7921*\dx},{-0.0340*\dy})
+	-- ({1.7960*\dx},{-0.0669*\dy})
+	-- ({1.7987*\dx},{-0.1000*\dy})
+	-- ({1.8000*\dx},{-0.1331*\dy})
+	-- ({1.8002*\dx},{-0.1662*\dy})
+	-- ({1.7990*\dx},{-0.1993*\dy})
+	-- ({1.7967*\dx},{-0.2324*\dy})
+	-- ({1.7930*\dx},{-0.2654*\dy})
+	-- ({1.7881*\dx},{-0.2982*\dy})
+	-- ({1.7820*\dx},{-0.3308*\dy})
+	-- ({1.7747*\dx},{-0.3632*\dy})
+	-- ({1.7661*\dx},{-0.3952*\dy})
+	-- ({1.7563*\dx},{-0.4270*\dy})
+	-- ({1.7454*\dx},{-0.4584*\dy})
+	-- ({1.7332*\dx},{-0.4893*\dy})
+	-- ({1.7199*\dx},{-0.5198*\dy})
+	-- ({1.7055*\dx},{-0.5497*\dy})
+	-- ({1.6900*\dx},{-0.5792*\dy})
+	-- ({1.6733*\dx},{-0.6080*\dy})
+	-- ({1.6556*\dx},{-0.6362*\dy})
+	-- ({1.6368*\dx},{-0.6637*\dy})
+	-- ({1.6171*\dx},{-0.6906*\dy})
+	-- ({1.5963*\dx},{-0.7166*\dy})
+	-- ({1.5746*\dx},{-0.7419*\dy})
+	-- ({1.5519*\dx},{-0.7664*\dy})
+	-- ({1.5283*\dx},{-0.7901*\dy})
+	-- ({1.5039*\dx},{-0.8128*\dy})
+	-- ({1.4787*\dx},{-0.8347*\dy})
+	-- ({1.4526*\dx},{-0.8556*\dy})
+	-- ({1.4258*\dx},{-0.8756*\dy})
+	-- ({1.3983*\dx},{-0.8945*\dy})
+	-- ({1.3700*\dx},{-0.9124*\dy})
+	-- ({1.3412*\dx},{-0.9293*\dy})
+	-- ({1.3117*\dx},{-0.9452*\dy})
+	-- ({1.2817*\dx},{-0.9599*\dy})
+	-- ({1.2511*\dx},{-0.9735*\dy})
+	-- ({1.2201*\dx},{-0.9860*\dy})
+	-- ({1.1886*\dx},{-0.9974*\dy})
+	-- ({1.1567*\dx},{-1.0076*\dy})
+	-- ({1.1245*\dx},{-1.0166*\dy})
+	-- ({1.0920*\dx},{-1.0245*\dy})
+	-- ({1.0592*\dx},{-1.0312*\dy})
+	-- ({1.0262*\dx},{-1.0367*\dy})
+	-- ({0.9931*\dx},{-1.0409*\dy})
+	-- ({0.9598*\dx},{-1.0440*\dy})
+	-- ({0.9264*\dx},{-1.0459*\dy})
+	-- ({0.8930*\dx},{-1.0465*\dy})
+	-- ({0.8596*\dx},{-1.0460*\dy})
+	-- ({0.8263*\dx},{-1.0442*\dy})
+	-- ({0.7930*\dx},{-1.0413*\dy})
+}
diff --git a/buch/papers/zeta/images/zetaplot.m b/buch/papers/zeta/images/zetaplot.m
new file mode 100644
index 0000000..984b645
--- /dev/null
+++ b/buch/papers/zeta/images/zetaplot.m
@@ -0,0 +1,23 @@
+%
+% zetaplot.m
+%
+% (c) 2022 Prof Dr Andreas Müller
+%
+s = 1;
+h = 0.02;
+m = 40;
+
+fn = fopen("zetapath.tex", "w");
+fprintf(fn, "\\def\\zetapath{\n");
+counter = 0;
+for y = (0:h:m)
+	if (counter > 0)
+		fprintf(fn, "\n\t--");
+	end
+	z = zeta(0.5 + i*y);
+	fprintf(fn, " ({%.4f*\\dx},{%.4f*\\dy})", real(z), imag(z));
+	counter = counter + 1;
+end
+fprintf(fn, "\n}\n");
+fclose(fn);
+
diff --git a/buch/papers/zeta/images/zetaplot.pdf b/buch/papers/zeta/images/zetaplot.pdf
new file mode 100644
index 0000000..c6d3693
--- /dev/null
+++ b/buch/papers/zeta/images/zetaplot.pdf
diff --git a/buch/papers/zeta/images/zetaplot.tex b/buch/papers/zeta/images/zetaplot.tex
new file mode 100644
index 0000000..521bb1a
--- /dev/null
+++ b/buch/papers/zeta/images/zetaplot.tex
@@ -0,0 +1,47 @@
+%
+% zetaplot.tex -- Abbildung der kritischen Geraden
+%
+% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
+%
+\documentclass[tikz]{standalone}
+\usepackage{amsmath}
+\usepackage{times}
+\usepackage{txfonts}
+\usepackage{pgfplots}
+\usepackage{csvsimple}
+\usetikzlibrary{arrows,intersections,math}
+\begin{document}
+\def\skala{1}
+\begin{tikzpicture}[>=latex,thick,scale=\skala]
+
+\def\dx{2}
+\def\dy{2}
+
+\draw[->] ({-1.6*\dx},0) -- ({3.4*\dx},0)
+	coordinate[label={$\Re\zeta(\frac12+it)$}];
+\draw[->] (0,{-2.1*\dx}) -- (0,{2.2*\dx})
+	coordinate[label={left:$\Im\zeta(\frac12+it)$}];
+
+\foreach \x in {-1,1,2,3}{
+	\node at ({\x*\dx},-0.1) [below] {$\x$};
+}
+\node at (-0.1,{1*\dy}) [above left] {$i$};
+\node at (-0.1,{2*\dy}) [left] {$2i$};
+\node at (-0.1,{-1*\dy}) [below left] {$-i$};
+\node at (-0.1,{-2*\dy}) [left] {$-2i$};
+
+\foreach \x in {-1,1,2,3}{
+	\draw ({\x*\dx},-0.1) -- ({\x*\dx},0.1);
+}
+\foreach \y in {1,2}{
+	\draw (-0.1,{\y*\dy}) -- (0.1,{\y*\dy});
+	\draw (-0.1,{-\y*\dy}) -- (0.1,{-\y*\dy});
+}
+
+\input{papers/zeta/images/zetapath.tex}
+
+\draw[color=blue,line width=1pt] \zetapath;
+
+\end{tikzpicture}
+\end{document}
+
diff --git a/buch/papers/zeta/main.tex b/buch/papers/zeta/main.tex
index caddace..de297a0 100644
--- a/buch/papers/zeta/main.tex
+++ b/buch/papers/zeta/main.tex
@@ -8,12 +8,12 @@
 \begin{refsection}
 \chapterauthor{Raphael Unterer}
 
-%TODO Einleitung
 
 \input{papers/zeta/einleitung.tex}
 \input{papers/zeta/euler_product.tex}
 \input{papers/zeta/zeta_gamma.tex}
 \input{papers/zeta/analytic_continuation.tex}
+\input{papers/zeta/fazit}
 
 \printbibliography[heading=subbibliography]
 \end{refsection}
diff --git a/buch/papers/zeta/presentation/presentation.tex b/buch/papers/zeta/presentation/presentation.tex
new file mode 100644
index 0000000..53fd305
--- /dev/null
+++ b/buch/papers/zeta/presentation/presentation.tex
@@ -0,0 +1,368 @@
+\documentclass[ngerman, aspectratio=169]{beamer}
+
+%style
+\mode<presentation>{
+	\usetheme{Frankfurt}
+}
+%packages
+\usepackage[utf8]{inputenc}
+\usepackage[english]{babel}
+\usepackage{graphicx}
+\usepackage{array}
+
+\newcolumntype{L}[1]{>{\raggedright\let\newline\\\arraybackslash\hspace{0pt}}m{#1}}
+\usepackage{ragged2e}
+
+\usepackage{bm} % bold math
+\usepackage{amsfonts}
+\usepackage{amssymb}
+\usepackage{mathtools}
+\usepackage{amsmath}
+\usepackage{multirow} % multi row in tables
+\usepackage{scrextend}
+
+\usepackage{tikz}
+
+\usepackage{algorithmic}
+
+%\usepackage{algorithm} % http://ctan.org/pkg/algorithm
+%\usepackage{algpseudocode} % http://ctan.org/pkg/algorithmicx
+
+%\usepackage{algorithmicx}
+
+
+%citations
+\usepackage[style=verbose,backend=biber]{biblatex}
+\addbibresource{references.bib}
+
+
+
+\usefonttheme[onlymath]{serif}
+
+%Beamer Template modifications
+%\definecolor{mainColor}{HTML}{0065A3} % HSR blue
+\definecolor{mainColor}{HTML}{D72864} % OST pink
+\definecolor{invColor}{HTML}{28d79b} % OST pink
+\definecolor{dgreen}{HTML}{38ad36} % Dark green
+
+%\definecolor{mainColor}{HTML}{000000} % HSR blue
+\setbeamercolor{palette primary}{bg=white,fg=mainColor}
+\setbeamercolor{palette secondary}{bg=orange,fg=mainColor}
+\setbeamercolor{palette tertiary}{bg=yellow,fg=red}
+\setbeamercolor{palette quaternary}{bg=mainColor,fg=white} %bg = Top bar, fg = active top bar topic
+\setbeamercolor{structure}{fg=black} % itemize, enumerate, etc (bullet points)
+\setbeamercolor{section in toc}{fg=black} % TOC sections
+\setbeamertemplate{section in toc}[sections numbered]
+\setbeamertemplate{subsection in toc}{%
+	\hspace{1.2em}{$\bullet$}~\inserttocsubsection\par}
+
+\setbeamertemplate{itemize items}[circle]
+\setbeamertemplate{description item}[circle]
+\setbeamertemplate{title page}[default][colsep=-4bp,rounded=true]
+\beamertemplatenavigationsymbolsempty
+
+\setbeamercolor{footline}{fg=gray}
+\setbeamertemplate{footline}{%
+	\hfill\usebeamertemplate***{navigation symbols}
+	\hspace{0.5cm}
+	\insertframenumber{}\hspace{0.2cm}\vspace{0.2cm}
+}
+
+\usepackage{caption}
+\captionsetup{labelformat=empty}
+
+%Title Page
+\title{Riemannsche Zeta Funktion}
+\author{Raphael Unterer}
+\institute{Mathematisches Seminar 2022: Spezielle Funktionen}
+
+\newcommand*{\HL}{\textcolor{mainColor}}
+\newcommand*{\RD}{\textcolor{red}}
+\newcommand*{\BL}{\textcolor{blue}}
+\newcommand*{\GN}{\textcolor{dgreen}}
+\newcommand*{\YE}{\textcolor{violet}}
+
+
+
+
+\makeatletter
+\newcount\my@repeat@count
+\newcommand{\myrepeat}[2]{%
+	\begingroup
+	\my@repeat@count=\z@
+	\@whilenum\my@repeat@count<#1\do{#2\advance\my@repeat@count\@ne}%
+	\endgroup
+}
+\makeatother
+
+
+
+
+\usetikzlibrary{automata,arrows,positioning,calc}
+
+
+\begin{document}
+
+	%Titelseite
+	\begin{frame}
+		\titlepage
+	\end{frame}
+
+	%Inhaltsverzeichnis
+% 	\begin{frame}
+% 		\frametitle{Inhalt}
+% 		\tableofcontents
+% 	\end{frame}
+
+	\section{Motivation}
+
+    \begin{frame}
+        \frametitle{Summe aller Natürlichen Zahlen}
+        \begin{equation*}
+            \sum_{n=1}^{\infty} n
+            =
+            1 + 2 + 3 + \ldots + \infty
+            =
+            - \frac{1}{12}
+        \end{equation*}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Summe aller Natürlichen Zahlen}
+        \begin{center}
+            \includegraphics[width=0.7\textwidth]{../images/youtube_screenshot.png}
+        \end{center}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Riemannsche Zeta Funktion}
+        \begin{equation*}
+            \zeta(s)
+            =
+            \sum_{n=1}^{\infty}
+            \frac{1}{n^s}
+        \end{equation*}
+        \pause
+        \begin{equation*}
+            \zeta(-1)
+            =
+            \sum_{n=1}^{\infty}
+            \frac{1}{n^{-1}}
+            =
+            \sum_{n=1}^{\infty} n
+        \end{equation*}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Originaler Definitionsbereich}
+        Wir kennen die divergierende harmonische Reihe
+        \begin{equation*}
+            \zeta(1)
+            =
+            \sum_{n=1}^{\infty}
+            \frac{1}{n}
+            \rightarrow
+            \infty,
+        \end{equation*}
+        und somit ist $\Re(s) > 1$.
+    \end{frame}
+
+    \section{Analytische Fortsetzung}
+    \begin{frame}
+        \frametitle{Plan für die Analytische Fortsetzung von $\zeta(s)$}
+        \begin{center}
+            \input{../images/continuation_overview.tikz.tex}
+        \end{center}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Fortsetzung auf $\Re(s) > 0$}
+        Dirichletsche Etafunktion ist
+        \begin{equation*}\label{zeta:equation:eta}
+            \eta(s)
+            =
+            \sum_{n=1}^{\infty}
+            \frac{(-1)^{n-1}}{n^s},
+        \end{equation*}
+        und konvergiert im Bereich $\Re(s) > 0$.
+    \end{frame}
+    \begin{frame}
+        \frametitle{Fortsetzung auf $\Re(s) > 0$}
+        \begin{align}
+            \zeta(s)
+            &=
+            \RD{
+            \sum_{n=1}^{\infty}
+            \frac{1}{n^s} \label{zeta:align1}
+            }
+            \\
+            \frac{1}{2^{s-1}}
+            \zeta(s)
+            &=
+            \BL{
+            \sum_{n=1}^{\infty}
+            \frac{2}{(2n)^s} \label{zeta:align2}
+            }
+        \end{align}
+        \pause
+        \eqref{zeta:align1} - \eqref{zeta:align2}:
+        \begin{align*}
+            \left(1 - \frac{1}{2^{s-1}} \right)
+            \zeta(s)
+            &=
+            \RD{\frac{1}{1^s}}
+            \underbrace{-\BL{\frac{2}{2^s}} + \RD{\frac{1}{2^s}}}_{-\frac{1}{2^s}}
+            + \RD{\frac{1}{3^s}}
+            \underbrace{-\BL{\frac{2}{4^s}} + \RD{\frac{1}{4^s}}}_{-\frac{1}{4^s}}
+            \ldots
+            \\
+            &= \eta(s)
+        \end{align*}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Fortsetzung auf $\Re(s) > 0$}
+        Somit haben wir die Fortsetzung gefunden als
+        \begin{equation} \label{zeta:equation:fortsetzung1}
+            \zeta(s)
+            :=
+            \left(1 - \frac{1}{2^{s-1}} \right)^{-1} \eta(s).
+        \end{equation}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Spiegelungseigenschaft für $\Re(s) < 0$}
+        \begin{equation*}\label{zeta:equation:functional}
+            \frac{\Gamma \left( \frac{s}{2} \right)}{\pi^{\frac{s}{2}}}
+            \zeta(s)
+            =
+            \frac{\Gamma \left( \frac{1-s}{2} \right)}{\pi^{\frac{1-s}{2}}}
+            \zeta(1-s).
+        \end{equation*}
+    \end{frame}
+    %TODO maybe explain gamma-fct
+
+    \section{Euler Produkt und Primzahlen}
+    \begin{frame}
+        \frametitle{Wieso ist die Zeta Funktion so bekannt?}
+        \begin{itemize}
+            \item Interessante Funktionswerte z.B. $\zeta(2) = \frac{\pi^2}{6}$
+            \item Primzahlenverteilung (Riemannhypothese)
+            \item Forschungsgebiet der analytischen Zahlentheorie seit dem 18. Jahrhundert
+            \item ...
+        \end{itemize}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Euler Produkt: Verbindung von Zeta und Primzahlen}
+        \begin{equation*}
+            \zeta(s)
+            =
+            \sum_{n=1}^\infty
+            \frac{1}{n^s}
+            =
+            \prod_{p \in P}
+            \frac{1}{1-p^{-s}}
+        \end{equation*}
+        \pause
+        Geometrische Reihe
+        \begin{equation*}
+            \prod_{p \in P}
+            \frac{1}{1-p^{-s}}
+            =
+            \prod_{p \in P}
+            \left(
+            1
+            +
+            \frac{1}{p^s}
+            +
+            \frac{1}{p^{2s}}
+            +
+            \frac{1}{p^{3s}}
+            +
+            \ldots
+            \right)
+        \end{equation*}
+        \pause
+        Erste Terme ausmultiplizieren
+        \begin{align*}
+            \left(
+            1
+            +
+            \RD{\frac{1}{2^s}}
+            +
+            \GN{\frac{1}{2^{2s}}}
+            +
+            \frac{1}{2^{3s}}
+            +
+            \ldots
+            \right)
+            \left(
+            1
+            +
+            \BL{\frac{1}{3^s}}
+            +
+            \frac{1}{3^{2s}}
+            +
+            \frac{1}{3^{3s}}
+            +
+            \ldots
+            \right)
+            \left(
+            1
+            +
+            \YE{\frac{1}{5^s}}
+            +
+            \frac{1}{5^{2s}}
+            +
+            \frac{1}{5^{3s}}
+            +
+            \ldots
+            \right)
+            \ldots
+            \\
+            =
+            1
+            +
+            \RD{\frac{1}{2^s}}
+            +
+            \BL{\frac{1}{3^s}}
+            +
+            \GN{\frac{1}{4^s}}
+            +
+            \YE{\frac{1}{5^s}}
+            +
+            \ldots
+        \end{align*}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Primzahlfunktion}
+        \begin{center}
+           \scalebox{0.5}{\input{../images/primzahlfunktion.pgf}}
+        \end{center}
+    \end{frame}
+
+
+    \section{Darstellungen}
+
+    \begin{frame}
+        \frametitle{Farbcodierung}
+        \begin{center}
+            \scalebox{0.6}{\input{zeta_color_plot.pgf}}
+        \end{center}
+    \end{frame}
+
+    \begin{frame}
+        \frametitle{Konstanter Realteil $\Re(s)=-1$ und $\Im(s)=0\ldots40$}
+        \begin{center}
+            \scalebox{0.6}{\input{../images/zeta_re_-1_plot.pgf}}
+        \end{center}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Konstanter Realteil $\Re(s)=0$ und $\Im(s)=0\ldots40$}
+        \begin{center}
+            \scalebox{0.6}{\input{../images/zeta_re_0_plot.pgf}}
+        \end{center}
+    \end{frame}
+    \begin{frame}
+        \frametitle{Konstanter Realteil $\Re(s)=0.5$ und $\Im(s)=0\ldots40$}
+        \begin{center}
+            \scalebox{0.6}{\input{../images/zeta_re_0.5_plot.pgf}}
+        \end{center}
+    \end{frame}
+
+\end{document}
+
diff --git a/buch/papers/zeta/presentation/zeta_color_plot-img0.png b/buch/papers/zeta/presentation/zeta_color_plot-img0.png
new file mode 100644
index 0000000..b8c7298
--- /dev/null
+++ b/buch/papers/zeta/presentation/zeta_color_plot-img0.png
diff --git a/buch/papers/zeta/presentation/zeta_color_plot.pgf b/buch/papers/zeta/presentation/zeta_color_plot.pgf
new file mode 100644
index 0000000..0fd7cb8
--- /dev/null
+++ b/buch/papers/zeta/presentation/zeta_color_plot.pgf
@@ -0,0 +1,402 @@
+%% Creator: Matplotlib, PGF backend
+%%
+%% To include the figure in your LaTeX document, write
+%%   \input{<filename>.pgf}
+%%
+%% Make sure the required packages are loaded in your preamble
+%%   \usepackage{pgf}
+%%
+%% and, on pdftex
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+%%
+%% and then include the figures with
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diff --git a/buch/papers/zeta/python/plot_zeta.py b/buch/papers/zeta/python/plot_zeta.py
new file mode 100644
index 0000000..53097c5
--- /dev/null
+++ b/buch/papers/zeta/python/plot_zeta.py
@@ -0,0 +1,39 @@
+import numpy as np
+from mpmath import zeta
+import matplotlib.pyplot as plt
+from matplotlib import colors
+import matplotlib
+matplotlib.use("pgf")
+matplotlib.rcParams.update(
+    {
+        "pgf.texsystem": "pdflatex",
+        "font.family": "serif",
+        "font.size": 8,
+        "text.usetex": True,
+        "pgf.rcfonts": False,
+        "axes.unicode_minus": False,
+    }
+)
+
+print(zeta(-1))
+print(zeta(-1 + 2j))
+
+re_values = np.arange(-10, 5, 0.04)
+im_values = np.arange(-20, 20, 0.04)
+plot_matrix = np.zeros((len(im_values), len(re_values), 3))
+for im_i, im in enumerate(im_values):
+    print(im_i)
+    for re_i, re in enumerate(re_values):
+        z = complex(zeta(re + 1j*im))
+        h = (np.angle(z) + np.pi) / (2*np.pi)
+        v = np.abs(z)
+        s = 1.0
+        plot_matrix[im_i, re_i] = [h, s, v]
+
+log10_v = np.log10(plot_matrix[:, :, 2])
+log10_v += np.abs(np.min(log10_v))
+plot_matrix[:, :, 2] = (log10_v) / np.max(log10_v)
+plt.imshow(colors.hsv_to_rgb(plot_matrix), extent=[re_values.min(), re_values.max(), im_values.min(), im_values.max()])
+plt.xlabel("$\Re$")
+plt.ylabel("$\Im$")
+plt.savefig(f"zeta_color_plot.pgf")
diff --git a/buch/papers/zeta/python/plot_zeta2.py b/buch/papers/zeta/python/plot_zeta2.py
new file mode 100644
index 0000000..b730703
--- /dev/null
+++ b/buch/papers/zeta/python/plot_zeta2.py
@@ -0,0 +1,31 @@
+import numpy as np
+from mpmath import zeta
+import matplotlib.pyplot as plt
+import matplotlib
+matplotlib.use("pgf")
+matplotlib.rcParams.update(
+    {
+        "pgf.texsystem": "pdflatex",
+        "font.family": "serif",
+        "font.size": 8,
+        "text.usetex": True,
+        "pgf.rcfonts": False,
+        "axes.unicode_minus": False,
+    }
+)
+# const re plot
+re_values = [-1, 0, 0.5]
+im_values = np.arange(0, 40, 0.04)
+buf = np.zeros((len(re_values), len(im_values), 2))
+for im_i, im in enumerate(im_values):
+    print(im_i)
+    for re_i, re in enumerate(re_values):
+        z = complex(zeta(re + 1j*im))
+        buf[re_i, im_i] = [np.real(z), np.imag(z)]
+
+for i in range(len(re_values)):
+    plt.figure()
+    plt.plot(buf[i,:,0], buf[i,:,1], label=f"$\Re={re_values[i]}$")
+    plt.xlabel("$\Re$")
+    plt.ylabel("$\Im$")
+    plt.savefig(f"zeta_re_{re_values[i]}_plot.pgf")
diff --git a/buch/papers/zeta/python/primzahlfunktion.py b/buch/papers/zeta/python/primzahlfunktion.py
new file mode 100644
index 0000000..9434de9
--- /dev/null
+++ b/buch/papers/zeta/python/primzahlfunktion.py
@@ -0,0 +1,24 @@
+import matplotlib.pyplot as plt
+import numpy as np
+
+primzahlfunktion = [0, 0, 0, 0]
+x = [0, 1-1e-12, 1, 2-1e-12]
+x_last = 1
+value = 0
+for i in range(2, 30, 1):
+    new_value = value + 1
+    for j in range(2, i, 1):
+        if i % j == 0:
+            new_value = value
+    value = new_value
+    primzahlfunktion.append(new_value)
+    x_last += 1
+    x.append(x_last)
+    primzahlfunktion.append(new_value)
+    x.append(x_last + 1 - 1e-12)
+
+
+plt.rcParams.update({"pgf.texsystem": "pdflatex"})
+plt.plot(x, primzahlfunktion)
+plt.show()
+
diff --git a/buch/papers/zeta/references.bib b/buch/papers/zeta/references.bib
index a4f2521..f2a2f31 100644
--- a/buch/papers/zeta/references.bib
+++ b/buch/papers/zeta/references.bib
@@ -4,32 +4,58 @@
 % (c) 2020 Autor, Hochschule Rapperswil
 %
 
-@online{zeta:bibtex,
-	title = {BibTeX},
-	url = {https://de.wikipedia.org/wiki/BibTeX},
-	date = {2020-02-06},
-	year = {2020},
-	month = {2},
-	day = {6}
+@online{zeta:online:millennium,
+	title = {The Millennium Prize Problems},
+	url = {https://www.claymath.org/millennium-problems/millennium-prize-problems},
+	year = {2022},
+	month = {8},
+	day = {4}
 }
 
-@book{zeta:numerical-analysis,
-	title = {Numerical Analysis},
-	author = {David Kincaid and Ward Cheney},
-	publisher = {American Mathematical Society},
-	year = {2002},
-	isbn = {978-8-8218-4788-6},
-	inseries = {Pure and applied undegraduate texts},
-	volume = {2}
+@online{zeta:online:wiki_en,
+	title = {Riemann zeta function},
+	url = {https://en.wikipedia.org/wiki/Riemann_zeta_function},
+	year = {2022},
+	month = {8},
+	day = {7}
+}
+@online{zeta:online:wiki_de,
+	title = {Riemannsche Zeta-Funktion},
+	url = {https://de.wikipedia.org/wiki/Riemannsche_Zeta-Funktion},
+	year = {2022},
+	month = {8},
+	day = {7}
+}
+
+@online{zeta:online:poisson,
+    title = {Deriving the Poisson Summation Formula},
+    url = {https://www.youtube.com/watch?v=4Bex-4BFYWo},
+    author = {Physics and Math Lectures},
+	year = {2022},
+	month = {8},
+	day = {7}
 }
 
-@article{zeta:mendezmueller,
-        author = { Tabea Méndez and Andreas Müller },
-        title = { Noncommutative harmonic analysis and image registration },
-        journal = { Appl. Comput. Harmon. Anal.},
-        year = 2019,
-        volume = 47,
-        pages = {607--627},
-        url = {https://doi.org/10.1016/j.acha.2017.11.004}
+@online{zeta:online:mryoumath,
+    title = {Riemann Zeta Function Playlist},
+    url = {https://www.youtube.com/playlist?list=PL32446FDD4DA932C9},
+    author = {MrYouMath},
+	year = {2022},
+	month = {8},
+	day = {7}
 }
 
+@online{zeta:online:basel,
+    title = {Basel Problem},
+	url = {https://en.wikipedia.org/wiki/Basel_problem},
+	year = {2022},
+	month = {8},
+	day = {7}
+}
+@online{zeta:online:pars,
+    title = {Parseval's identity},
+	url = {https://en.wikipedia.org/wiki/Parseval%27s_identity},
+	year = {2022},
+	month = {8},
+	day = {7}
+}
diff --git a/buch/papers/zeta/zeta_color_plot-img0.png b/buch/papers/zeta/zeta_color_plot-img0.png
new file mode 100644
index 0000000..b8c7298
--- /dev/null
+++ b/buch/papers/zeta/zeta_color_plot-img0.png
diff --git a/buch/papers/zeta/zeta_color_plot.pgf b/buch/papers/zeta/zeta_color_plot.pgf
new file mode 100644
index 0000000..0fd7cb8
--- /dev/null
+++ b/buch/papers/zeta/zeta_color_plot.pgf
@@ -0,0 +1,402 @@
+%% Creator: Matplotlib, PGF backend
+%%
+%% To include the figure in your LaTeX document, write
+%%   \input{<filename>.pgf}
+%%
+%% Make sure the required packages are loaded in your preamble
+%%   \usepackage{pgf}
+%%
+%% and, on pdftex
+%%   \usepackage[utf8]{inputenc}\DeclareUnicodeCharacter{2212}{-}
+%%
+%% or, on luatex and xetex
+%%   \usepackage{unicode-math}
+%%
+%% Figures using additional raster images can only be included by \input if
+%% they are in the same directory as the main LaTeX file. For loading figures
+%% from other directories you can use the `import` package
+%%   \usepackage{import}
+%%
+%% and then include the figures with
+%%   \import{<path to file>}{<filename>.pgf}
+%%
+%% Matplotlib used the following preamble
+%%
+\begingroup%
+\makeatletter%
+\begin{pgfpicture}%
+\pgfpathrectangle{\pgfpointorigin}{\pgfqpoint{6.400000in}{4.800000in}}%
+\pgfusepath{use as bounding box, clip}%
+\begin{pgfscope}%
+\pgfsetbuttcap%
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+\pgfsetfillcolor{currentfill}%
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+\definecolor{currentstroke}{rgb}{1.000000,1.000000,1.000000}%
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+\pgftext[left,bottom]{\includegraphics[interpolate=true,width=1.390000in,height=3.700000in]{zeta_color_plot-img0.png}}%
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diff --git a/buch/papers/zeta/zeta_gamma.tex b/buch/papers/zeta/zeta_gamma.tex
index db41676..dd422e3 100644
--- a/buch/papers/zeta/zeta_gamma.tex
+++ b/buch/papers/zeta/zeta_gamma.tex
@@ -11,7 +11,7 @@ Wir erinnern uns an die Definition der Gammafunktion in \eqref{buch:rekursion:ga
     \int_0^{\infty} t^{s-1} e^{-t} \,dt,
 \end{equation*}
 wobei die Notation an die Zetafunktion angepasst ist.
-Durch die Substitution von $t$ mit $t = nu$ und $dt = n\,du$ wird daraus
+Durch die Substitution $t = nu$ und $dt = n\,du$ wird daraus
 \begin{align*}
     \Gamma(s)
     &=
@@ -19,7 +19,7 @@ Durch die Substitution von $t$ mit $t = nu$ und $dt = n\,du$ wird daraus
     &=
     \int_0^{\infty} n^s u^{s-1} e^{-nu} \,du.
 \end{align*}
-Durch Division mit durch $n^s$ ergibt sich die Quotienten
+Durch Division durch $n^s$ ergeben sich die Quotienten
 \begin{equation*}
     \frac{\Gamma(s)}{n^s}
     =
@@ -57,5 +57,5 @@ Wenn wir dieses Resultat einsetzen in \eqref{zeta:equation:zeta_gamma1} und durc
     \frac{1}{\Gamma(s)}
     \int_0^{\infty}
     \frac{u^{s-1}}{e^u -1}
-    du \qed
+    du.
 \end{equation}